On medium-rank Lie primitive and maximal subgroups of exceptional groups of Lie type
aa r X i v : . [ m a t h . G R ] F e b On medium-rank Lie primitive and maximalsubgroups of exceptional groups of Lie type
David A. Craven
Author address:
School of Mathematics, University of Birmingham, Birmingham,B15 2TT, United Kingdom
Email address : [email protected] ontents Chapter 1. Introduction 1Chapter 2. Notation and preliminaries 5Chapter 3. Subgroup structure of exceptional algebraic groups 17Chapter 4. Techniques for proving the results 274.1. The strategy 274.2. An example 29Chapter 5. Modules for groups of Lie type 33Chapter 6. Rank 4 groups for E +8 − D F and F E E B G G E E F iiiv CONTENTS Chapter 12. Difficult cases 17512.1. PSL (2) ≤ E (5) ≤ E (4) ≤ E E (3) 21113.3. The case G (3) ′ (3) 212Bibliography 213 bstract We study embeddings of groups of Lie type H in characteristic p into excep-tional algebraic groups G of the same characteristic. We exclude the case where H is of type PSL . A subgroup of G is Lie primitive if it is not contained in anyproper, positive-dimensional subgroup of G .With a few possible exceptions, we prove that there are no Lie primitive sub-groups H in G , with the conditions on H and G given above. The exceptions arefor H one of PSL (3), PSU (3), PSL (4), PSU (4), PSU (8), PSU (2), PSp (2) ′ and B (8), and G of type E . No examples are known of such Lie primitiveembeddings.We prove a slightly stronger result, including stability under automorphisms of G . This has the consequence that, with the same exceptions, any almost simplegroup with socle H , that is maximal inside an almost simple exceptional group ofLie type F , E , E , E and E , is the fixed points under the Frobenius map of acorresponding maximal closed subgroup inside the algebraic group.The proof uses a combination of representation-theoretic, algebraic group-theoretic, and computational means. Received by the editor 7th December, 2019.2020
Mathematics Subject Classification.
Primary 20D06, 20E28, 20G41.
Key words and phrases.
Maximal subgroups, exceptional groups, finite simple groups.The author is a Royal Society University Research Fellow, and gratefully acknowledges thefinancial support of the Society. v HAPTER 1
Introduction
This paper deals with four interconnected questions, progress on all of whichcan be achieved with the same analysis:(1) Classify the maximal subgroups of finite simple groups of Lie type;(2) Understand the subgroup structure of the simple algebraic groups;(3) Study the morphism extension problem : given a simple algebraic group G and an algebraic group H , and a Frobenius endomorphism σ of H ,when does a homomorphism H σ → G extend to a morphism H → G ofalgebraic groups;(4) Generalize Steinberg’s restriction theorem for GL n (that every simplemodule for a simple group of Lie type—in other words, an irreduciblesubgroup of GL n —is the restriction of one for the algebraic group) toarbitrary semisimple algebraic groups.Here we analyse the first problem for G a finite exceptional group of Lie type anda potential maximal subgroup that is the normalizer of a simple group of Lie type H in the same characteristic as G , but is not of type PSL . We more or less give acomplete answer for such maximal subgroups, leaving only eight possible maximalsubgroups in E .The determination of the maximal subgroups of the exceptional groups of Lietype is one of the major outstanding problems in our understanding of the structureof the finite simple groups. This task has occupied the endeavours of mathemati-cians over several decades; this paper is the third in a series by the author, whoseeventual goal is the completion of this determination.By a result of Borovik [ Bor89 ] and Liebeck–Seitz [
LS90 ], all maximal sub-groups of almost simple exceptional groups of Lie type that are not themselvesalmost simple groups are known, so we apply the classification of finite simplegroups. Alternating maximal subgroups were nearly classified in [
Cra17 ], andthere has been more work since then (there are no examples in F , E and E , andthere is one conjugacy class of maximal Alt(7) in E (5), but Alt(6) in E ( q ) andAlt(6) , Alt(7) in E are unresolved), and cross-characteristic Lie type and sporadicgroups that embed in exceptional groups were determined in [ LS99 ], with someanalysis of these in various papers in characteristic 0, and particularly [
Lit18 ] inpositive characteristic. (These will be treated in a later paper.) The remainingpotential maximal subgroups, which form the bulk of the groups to be examined,are Lie type groups in the same characteristic as the exceptional group.Broadly speaking, all embeddings of a group of Lie type H = H ( q ) into anexceptional algebraic group G of characteristic p | q are contained in proper, σ -stable, positive-dimensional subgroups of G if q > q = 16 for H of typePSL and PSU ) or if the rank of H is more than half of the rank of G . If H is of type PSL , B or G then much larger bounds than 9 are needed to guaranteethe result, for example q > G = E . (This was proved in [ LS98 ].)This paper does not deal with the case of PSL (which for G = F , E , E , butnot E , was considered in [ Cra ]), but with all of the other groups H . Recall thata finite subgroup H of an algebraic group is called Lie primitive if it does not liein a proper positive-dimensional subgroup.We prove the following result. In fact, we prove something slightly stronger thatinvolves stability under automorphisms of G (which is needed for the application tomaximal subgroups), but this will suffice for applications to the morphism extensionproblem. Theorem . Let G be the exceptional algebraic group E in characteristic p > , over an algebraically closed field. Let H = H ( q ) be a simple group of Lietype with p | q , and suppose that H is a subgroup of G . Furthermore, suppose that H does not have the same type as G , and that H is not PSL ( q ) . If N G ( H ) isLie primitive in G , then H is one of the groups PSL ( q ) for q = 3 , , PSU ( q ) for q = 3 , , , PSp (2) ′ , B (8) , and PSU (2) .Moreover, if σ is a Frobenius endomorphism on G and H ≤ G σ , then there isa σ -stable proper positive-dimensional subgroup X such that N G σ ( H ) ≤ X σ unless H is one of the above groups. (Note that there are no known examples of such groups.)We will define the slight strengthening of this result now. Let G be a sim-ple, simply connected algebraic group and σ a Frobenius endomorphism of G . Asubgroup K of G = G σ is strongly imprimitive if there is some σ -stable positive-dimensional subgroup X of G such that N ¯ G ( K ) is contained in N ¯ G ( X σ ) for all al-most simple groups ¯ G with socle G/Z ( G ). Of course, if H = N ¯ G ( H ) then stronglyimprimitive is the same as Lie imprimitive. The theorem above is also true withstrong imprimitivity in place of Lie imprimitivity.The second statement of this theorem (with strong imprimitivity) yields anapplication to maximal subgroups of the finite groups E ( q ), namely that the max-imal subgroups of the form N ¯ G ( H ) for H a Lie type group in characteristic p | q are known except for H on the list above, and H ∼ = PSL ( q ) for p | q .The author believes that one may be able to use the techniques in this paper tosolve the outstanding cases of PSU (2) and PSU (8), but that PSp (2) and B (8)cannot be resolved using these methods. For the rest he is not sure.We also consider the other exceptional groups F , E and E , obtaining thesame theorem as above, but with a very short list of possibilities. (We do not needto consider G because, by work of Kleidman [ Kle88 ] and Cooperstein [
Coo81 ]all maximal subgroups of the finite groups G ( q ) are known, and hence it is easy todetermine all Lie primitive subgroups. Indeed, we use G as an example in Section4.2.) Using [ Cra ] we can also include PSL subgroups in the list. (In the caseof F and p ≥ Mag90 ], and for E and H = PSL (11) , PSL (3) , PSU (3) this result was obtained by Aschbacher [ Asc ].)
Theorem . Let G be the exceptional algebraic group F , E or E in char-acteristic p > , over an algebraically closed field. Let H = H ( q ) be a simple groupof Lie type with p | q , and suppose that H is a subgroup of G . Furthermore, supposethat H does not have the same type as G . If H is not strongly imprimitive in G ,then G = E and H is one of PSL (7) , PSL (8) , and PSL (25) . . INTRODUCTION 3 (The group PSL (25) is always strongly imprimitive, and this case is resolvedin future work of the author. It will form part of a partial classification of maximalsubgroups of groups of type E . No example is known of the other two possiblesubgroups.)The author believes that these theorems represent close to the limit of whatcan be achieved simply using the techniques developed in the series of papers[ Cra17, Cra ] he has produced (others on PSL in E and non-defining character-istic embeddings are planned). More or less these techniques involve studying therestriction of small-dimensional modules for G to H , the eigenvalues of semisimpleelements and Jordan block structures of unipotent elements, and proper subgroupsof H known to be in positive-dimensional subgroups, to find a subspace of a mod-ule, normally either the Lie algebra L ( G ) or the minimal module for G , that isstabilized by both H and a positive-dimensional subgroup X of G . Thus h X , H i is not all of G , so H is contained in a proper positive-dimensional subgroup of G .If one chooses the subspace judiciously, one gets the refined statement on strongimprimitivity. To push these theorems still further probably requires more detaileduse of the subgroup structure of exceptional algebraic groups, and perhaps com-puter proofs that explicitly compute inside finite groups of Lie type, as has beendone for certain other cases.The structure of this article is as follows: the next chapter sets up the notationand a few preliminary results on elements of algebraic groups, while results on thesubgroup structure of exceptional algebraic groups are in Chapter 3. Chapter 4 goesthrough the techniques used to prove the results in this article. We give informationabout modules for small-rank groups of Lie type over small fields, i.e., the possiblesubgroups H , in the following chapter, and then we move on to the proof. Chapter6 considers rank-4 subgroups of E , Chapter 7 deals with subgroups of rank 3, andChapter 8 with subgroups of rank 2. After this, Chapters 9, 10 and 11 consider thesubgroups of E , E and F respectively. Chapter 12 gives the proofs of the difficultcases that were too long to fit in the general flow of the text. For the cases for E that act irreducibly on M ( E ) (or whose normalizer does), so PSL (3), PSU (3)and G (3) ′ , we have to approach them using completely different methods, andthis takes place in Chapter 13.There are a large number of supplementary materials that come with this ar-ticle, consisting of Magma programs, and also list of traces of semisimple elementsof various orders on low-dimensional modules. For many sections, these prove theassertions about module structures, actions of unipotent elements, and so on. InChapter 13 they also provide the matrices of the groups involved and prove theassertions in that chapter.The author would like to thank the anonymous referee for their fast, detailedand thoughtful report, which has greatly improved the exposition of this paper.HAPTER 2 Notation and preliminaries
Our aim for this chapter is to collate some notation and some general resultsthat have been used in the past, particularly by the author in [
Cra17, Cra ], toprove that a given subgroup is not maximal in a finite exceptional group of Lietype.For the rest of this paper, p will be a prime and G will denote a simple,simply connected exceptional algebraic group over an algebraically closed field k ofcharacteristic p . Let σ denote a Frobenius endomorphism of G and let G = G σ denote the fixed points of G under σ , a finite group of Lie type. The precisetypes of groups G that we are interested in are F ( q ), E ( q ), E ( q ), E ( q ) and E ( q ). Note that these are the simply connected groups, so for example E ( q ) hasa non-trivial centre for q odd. Write X for the set of maximal closed, positive-dimensional subgroups of G , and let X σ denote the set of fixed points X σ for X a σ -stable member of X . The members of the set X σ , together with subgroupsof the same type as G (e.g., E ( q ) and E ( q ) inside E ( q )) form almost allof the maximal subgroups of G . If ¯ G denotes an almost simple group such that G/Z ( G ) ∼ = F ∗ ( ¯ G ) via some isomorphism φ , we extend our notation of X σ to denotethe set of subgroups N ¯ G (( Z ( G ) X σ /Z ( G )) φ ) for X ∈ X σ . For X an algebraic group,write X ◦ for the connected component of the identity. Write o ( x ) for the order ofan element x of a group.A subgroup of G is Lie primitive if it does not lie in any proper, positive-dimensional subgroup of G , it is called G -irreducible if it does not lie in any properparabolic subgroup of G , and G -completely reducible (often abbreviated to G -cr)if, whenever it lies in a parabolic subgroup, it lies in a corresponding Levi subgroup.Of course, Lie primitive subgroups are G -irreducible, and both types of subgroupare vacuously G -completely reducible.For a given (simple) subgroup H of G , (with the exceptions given in Theorem1.1) we prove that H is contained in a member of X , that if H is contained in thefixed points G = G σ then H is contained in a member of X σ , and furthermorethat N ¯ G ( H ) is contained inside a member of the collection X σ for any ¯ G . This isa strong form of Lie imprimitivity, which we will unimaginatively refer to as strongimprimitivity . Normally we will prove this using general theorems—for example,if H stabilizes a line on the non-trivial composition factor of the adjoint modulethen H is strongly imprimitive (Lemma 3.5)—but occasionally we will have to provestrong imprimitivity via the stabilization of more complicated subspaces than lines,and rarely have to work directly with automorphisms.Our notation for modules is the same as in the previous papers in this series[ Cra17, Cra ]. Write soc( V ) for the socle of a module V and soc i ( V ) for thepreimage in V of the socle soc( V / soc i − ( V )) of the quotient module V / soc i − ( V ), rad( V ) for the radical of a module V , top( V ) for the top of a module V , i.e., thequotient module V / rad( V ), V ↑ H for the induction of V to H and V ↓ H for therestriction of V to H . Write V H for the fixed points of V under the action of H ,and P ( V ) for the projective cover of V . The heart of a module V is the quotientmodule rad( V ) / soc(rad( V )). Write k or 1 for the trivial kH -module. If V is a kH -module, we often consider its 1-cohomology, i.e., the module Ext kH ( k, V ). Forbrevity we sometimes conflate the 1-cohomology itself with its dimension, as it isonly the dimension of 1-cohomology that is of interest to us.In order to save space, when we give a module’s socle layers, we distinguishbetween the layers using ‘/’, so that a module with socle A and second socle B ⊕ C would be written B, C/A . We also use the notation
B/A for the quotient of B by A , but whenever we do so we say that we mean the quotient. Write cf( V ) for themultiset of composition factors of V .If I is a set of irreducible kH -modules, write I ′ for the set of (isomorphismclass representatives of) irreducible kH -modules that are not in I . The I -radical ofa kH -module V is the largest submodule of V whose composition factors lie in I ,and the I -residual is the smallest submodule whose quotient only has compositionfactors in I . The I -heart of V is the I ′ -residual of the quotient of V by its I ′ -radical.(Thus the socle and top of the I -heart consist solely of modules in I . Note that the I -heart is only an analogue of the heart, and there is no choice of I for which the I -heart coincides with the heart.) Write Λ i ( V ) and S i ( V ) for the i th exterior andsymmetric power of V respectively. Let V ∗ denote the dual of V , and if we meaneither V or its dual, write V ± .An important exception to this is when we consider submodules of projectives.We will often be interested in the I -radical of a projective cover P ( V ) of an irre-ducible kH -module V , but with V not itself in I , so of course the I -radical is 0.What we want in this case is the preimage in P ( V ) of the I -radical of the quo-tient module P ( V ) /V , but rather than writing out that whole phrase, we simplywrite the I -radical of P ( V ), accepting that this means the slightly different moduleabove.For a given dominant weight λ , write L ( λ ) for the simple module and W ( λ ) forthe Weyl module with that highest weight. Denote by M ( G ) a non-trivial Weylmodule of smallest dimension for G , and by L ( G ) the adjoint module for G . Formost primes M ( G ) and L ( G ) are irreducible, but for p = 2 , G theyhave a single trivial composition factor and a single non-trivial composition factor.Moreover, if p = 3 and G = G , or p = 2 and G = F , then L ( G ) does not havea unique non-trivial composition factor. In these cases the adjoint module has twocomposition factors of the same dimension, one of which is M ( G ). If M ( G ) hasa unique non-trivial composition factor, then denote it by M ( G ) ◦ , and similarlyfor L ( G ) ◦ , and set L ( G ) ◦ = L ( G ) otherwise. The module M ( G ) ◦ is called the minimal module in many of our references, and here. We give the description ofthese in terms of highest weights and the dimension of their simple constituent inTable 2.1. (The labelling of highest weights is consistent with [ Bou02 ], and is usedby most of our references.) For both M ( G ) and L ( G ), the decomposition of theaction of a unipotent element u into Jordan blocks was given in [ Law95 ], althoughwhen M ( G ) or L ( G ) has a trivial composition factor the action on M ( G ) ◦ or L ( G ) ◦ was not recorded there, but was given in [ Cra , Lemmas 6.1–6.3]. Table2.2 lists the unipotent classes for F in characteristic 3 and their actions on the . NOTATION AND PRELIMINARIES 7 G M ( G ) L ( G ) dim( M ( G ) ◦ ) dim( L ( G ) ◦ ) G W (10) W (01) 7 − δ p, F W ( λ ) W ( λ ) 26 − δ p, E L ( λ ) or L ( λ ) W ( λ ) 27 78 − δ p, E L ( λ ) W ( λ ) 56 133 − δ p, E N/A L ( λ ) N/A 248 Table 2.1.
Highest weights and dimensions of simple constituentsof minimal and adjoint modules for exceptional groups.modules M ( F ) ◦ , M ( F ) and M ( E ). Table 2.3 lists some unipotent classes of E in characteristic 3, and for all other classes one obtains the action on L ( E ) ◦ byremoving a block of size 1 from the action on L ( E ). For E in characteristic 2, onealways obtains the action of a unipotent element on L ( E ) ◦ by removing a block ofsize 1 from the action on L ( E ). We use the Bala–Carter–Pommerening label forunipotent classes as stated in [ Law95 ] without any further comment throughoutthis text.We will now define the concept of encapsulation, which we need occasionally inour arguments. If ( a , . . . , a r ) and ( b , . . . , b s ) are sequences with a i ≥ a i +1 > b i ≥ b i +1 >
0, then ( a i ) encapsulates ( b i ) if r ≥ s and a i ≥ b i for all 1 ≤ i ≤ s . If u is a unipotent element of a group G and V and W are kG -modules, then the actionof u on V encapsulates that of W if, when written as non-increasing sequences, thesizes of Jordan blocks of the action of u on V encapsulates that of W . (Note thatthis is not the same as the dominance order. Indeed, no partition encapsulates anyother partition of the same size.)The reason we introduce this is the following easy, but important, lemma. Lemma . Let G be a finite group and let u be a p -element of G . Let V bea kG -module and let W be a subquotient of V . The action of u on V encapsulatesthat of W . Proof.
We may of course assume that G = h u i . Let V be a minimal coun-terexample to the statement. As encapsulation is transitive, we may assume that W is a submodule or quotient of dimension 1 less than that of V . By taking duals ifnecessary, W is a submodule of V . By choice of minimal counterexample, we mayremove any summand of V that is contained in W . Thus W is indecomposable(equivalent to having 1-dimensional socle), and in this case it is clear from Jordannormal form that there are exactly two modules of dimension n + 1 that contain ablock of size n as a submodule, namely ( n +1) and ( n, n ). (cid:3) Looking at the tables in [
Law95 ], one sees that for all but finitely many primesthe Jordan blocks of a given unipotent class on M ( G ), the ‘generic’ action, andsimilarly with L ( G ). If a unipotent class acts on one of these modules with thegeneric action, then u is said to be generic for this module. Note that if u is genericfor one of M ( G ) and L ( G ) then it need not be generic for the other. The non-generic classes of elements of order p for G = E and p = 3 , , Law95 ], given here for the reader’sconvenience. (We also include unipotent elements of orders 4 and 8 in Tables 2.5and 2.8.)
2. NOTATION AND PRELIMINARIES
Class in F Act. on M ( F ) ◦ Act. on 25 / M ( E ) = 1 / / A , , , ˜ A , , , , , , A + ˜ A , , , , , , A , , , A + ˜ A , , , , , ˜ A , ˜ A + A , , B , , , , , , C ( a ) 5 , , , , , , , , , , , F ( a ) 5 , , , , , , B , , , C , F ( a ) 9 , , , , , , , , F ( a ) 9 , , , , , F , , , , , , Table 2.2.
Actions of unipotent elements on M ( F ) ◦ and its ex-tensions for F in characteristic 3. (Horizontal lines separate ele-ments of different orders.)Class in E Action on L ( E ) ◦ Action on L ( E )2 A , , , A + A , , , A , , , , , , , , , E ( a ) 9 , , , , , , , , E ( a ) 9 , , E , , , , , , Table 2.3.
Actions of unipotent elements on L ( E ) ◦ and L ( E )for E in characteristic 3, where one does not obtain the formerfrom the latter by removing a trivial Jordan block Class Action on L ( E )3 A , , A , , A , A + A , , A + 2 A , , A + 3 A , , A , A + A , , A + 2 A , Table 2.4.
Non-generic unipotent classes of elements of order 3in E in characteristic 3In this work we will be considering embeddings of finite groups into algebraicgroups G . A significant restriction on a homomorphism H → G is to considermodules V for G , and compute the possible composition factors for the restriction . NOTATION AND PRELIMINARIES 9 Class Action on L ( E ) Class Action on L ( E ) A , , A + 2 A , , A + A , , , A + A , , , A + 2 A , , , D ( a ) 4 , , A + 3 A , , , A + A (2)2 , , , A , , D ( a ) + A , , , A + A , , , A + A + A , , , A + 2 A , , , D ( a ) + A , A , , A , A + A , , Table 2.5.
Unipotent classes of elements of order 4 in E in char-acteristic 2 Class Action on L ( E ) Class Action on L ( E ) A , , D ( a ) + A , , A + A , , , , A , , A + 2 A , , , , A , A + A , , , , A + A , , , A + 2 A , , , , A + 2 A , , , D ( a ) 5 , , A + A , , D ( a ) + A , , , , A + A + A , , , A + A , , , , A + A , A + A + A , , , , Table 2.6.
Non-generic unipotent classes of elements of order 5in E in characteristic 5 Class Action on L ( E ) Class Action on L ( E ) A , , , D ( a ) 7 , , , A + A , , , , , , D ( a ) + A , , , , A , , , , , D + A , , , A + 2 A , , , , , , D ( a ) + A , , , , , A + A , , , E ( a ) 7 , , , A + A + A , , , , , , E ( a ) + A , , , , , , A , , D ( a ) 7 , , , , A + A , , , , , , E ( a ) 7 , , , , , A + A , , , , , E ( a ) 7 , , D , A , D + A , , , A + A , Table 2.7.
Non-generic unipotent classes of elements of order 7in E in characteristic 7 V ↓ H of V to H . The traces of semisimple elements of H on V must match thoseof semisimple classes of G , i.e., to each H -conjugacy class C we must assign a G -conjugacy class ¯ C of semisimple elements such that the traces of elements of C and ¯ C on V are the same. Moreover, by requiring that the power map is consistentwith this assignment, i.e., that if x ∈ C then the class assigned to x n is the classcontaining y n for y ∈ ¯ C , we obtain a slightly stronger restriction than merely Class Action Class Action D , , , D ( a ) 8 , , , D + A , , , E ( a ) 8 , , , , A , , , , , E ( a ) + A , , , , , , A + A , , , , , , , E ( a ) 8 , , , , , A + 2 A , , , , , , E ( a ) 8 , , , A + A , , , , , A , , , , D + A , , , , A + A , , , , , A + A + A , , , , , , D , , , D ( a ) 8 , , , D + A , , , D ( a ) + A , , , , D + A , , , , D + A (2)2 , , , , D ( a ) 8 , , , , A + A , , , , E ( a ) 8 , , , , D ( a ) + A , , , D + A (2)2 , , , A , , , D ( a ) 8 , A + A , , , A , Table 2.8.
Unipotent classes of elements of order 8 in E in char-acteristic 2 Class Action on L ( λ ) Action on L ( λ ), p = 3 Action on L ( λ ), p = 5 A , , , D , A , , , , , A + D , , , , , , A , , , D , A + A , , , , , , A + D , , , A , , , , , A + D , , , , , , D ( a ) 5 , , , , D , , , A , , , , D ( a ) 7 , , , , , D , , , Table 2.9.
Unipotent classes in D , with actions on L ( λ ) and L ( λ ) for p = 3 , x and y on V coincide.)A putative set of composition factors for V ↓ H is called conspicuous if allsemisimple elements satisfy the requirements above. In practice, we will some-times only be able to check this condition on some element orders, as we only havedata on the eigenvalues for some orders, in which case we will say that the set isconspicuous for some set of semisimple elements of H , for example of all elementsof order at most m for some m . Our first restriction when proving our results willnormally be to calculate the conspicuous sets of composition factors for M ( G ) ↓ H or L ( G ) ↓ H . . NOTATION AND PRELIMINARIES 11 Class Eigenspace dimensions on L ( E ) Centralizer2A 136 , E A
2B 120 , D
3A 80 , , A
3B 86 , , E A
3C 92 , , D T
3D 134 , , E T , , , , A T , , , , A A T
5C 48 , , , , A A , , , , A D T , , , , E A T , , , , D T
5G 68 , , , , D T , , , , E T Table 2.10.
Classes of semisimple elements of small orders in E and their centralizers. (The number in brackets is the number ofconjugacy classes in the subgroup generated by an element.)The conspicuous sets of composition factors for V ↓ H may be split into twocollections. A warrant for a set of composition factors for V ↓ H is a proper positive-dimensional subgroup X < G and a homomorphism φ : H → X such that thecomposition factors of V ↓ Hφ coincide with our given set. If H < G and thecomposition factors of V ↓ H are unwarranted, then H is necessarily Lie primitive in G . Thus if we are presented with a warranted set of composition factors we needto show that all corresponding copies of H in G must be (strongly) imprimitive in G , and if we have an unwarranted set of composition factors we need to show thatthere is no copy of H in G with those factors. In practice, in the second case wewill often show that H must be imprimitive instead, and therefore indirectly showthat H does not exist.While for small groups H and G it is possible to perform the computationsneeded to determine the conspicuous sets of composition factors by hand, for larger H and G , and in practice even for small H and G , we will use a computer to avoiderrors in hand solutions to linear algebra problems.A subgroup H ≤ G is a blueprint for a module V if there exists a positive-dimensional subgroup X of G stabilizing the same subspaces of V as H . An element x ∈ G is called a blueprint for V if h x i is a blueprint for V . Of course, if L ≤ H ≤ G and L is a blueprint for V , so is H .If H acts irreducibly on V then H is always a blueprint, but this informationis often not helpful. From [ LS04b ], we know all possibilities for a finite subgroupacting irreducibly on either M ( G ) ◦ or L ( G ) ◦ , and since in this article we areinterested in H ≤ G with H a Lie type group in characteristic p other than PSL ( q ),the only possibilities for H are PSL (3), PSU (3) or G (3) ′ with p = 3 and G = E .These cannot be dealt with using blueprints, and indeed cannot be dealt with at allusing the standard methods in this paper, and we have to resort to more underhandmeans. The case H ∼ = G (3) and its derived subgroup was proved to always liein an irreducible G subgroup of E in [ Asc ], but the proof involves an explicit analysis of the trilinear form on M ( E ), and there does not appear to be a way todeal with it representation-theoretically. These three groups will be considered inChapter 13.If u is a unipotent element, then there is an easy sufficient criterion for u to bea blueprint for either M ( G ) or L ( G ). Lemma
Cra17 , Lemmas 1.2 and 1.3]) . Let u be a non-trivial unipotentelement in G and let V be either M ( G ) ⊕ M ( G ) ∗ or L ( G ) . If u is generic for V then u is a blueprint for V . Work on whether semisimple elements are blueprints has been done in thepast, and we are able to give some results that prove that a semisimple element x ∈ G is always a blueprint for M ( G ) or L ( G ) if its order is sufficiently large.The next result is a combination of results from the author in [ Cra , Theorem 5.9and Proposition 6.10] for M ( G ), and Liebeck–Seitz and Lawther from [ LS98 ] and[
Law14 ] for L ( G ). Definition . Let G be one of G , F , E , E , E . Define the set T ( G ) ofpositive integers as follows: the odd elements of T ( G ) are T ( G ) odd = { , . . . , } G = G , { , . . . , } G = F , { , . . . , } G = E , { , . . . , } G = E , { , . . . , , , , , , , , , , , , , , , , } G = E , and the even elements of T ( G ) are T ( G ) even = { , . . . , } G = G , { , . . . , } G = F , { , . . . , , } G = E , { , . . . , , , , } G = E , { , . . . , , , , , , } G = E . Theorem . Let x be a semisimple element in G . (1) If the image of x in the adjoint form of G has order not in T ( G ) , then x is a blueprint for L ( G ) . (2) If G = E and o ( x ) is odd and at least , then x is a blueprint for M ( E ) . (3) If G = E and o ( x ) > , and x centralizes a -space on M ( E ) , then x is a blueprint for M ( E ) . (4) If G = E and o ( x ) is odd and at least , then x is a blueprint for M ( E ) ⊕ M ( E ) ∗ . (5) If G = E , x is real, and o ( x ) > , then x is a blueprint for M ( E ) ⊕ M ( E ) ∗ . (6) If G = F and o ( x ) > , then x is a blueprint for M ( F ) . (7) If G = G and o ( x ) > , then x is a blueprint for M ( G ) . We also consider the concept of pressure, and we take the definition and mainresult from [
Cra , Lemma 2.2]. Let M be a collection of simple modules for a . NOTATION AND PRELIMINARIES 13 group H , such that Ext kH ( M, M ′ ) = 0 for all M, M ′ ∈ M . The M -pressure of a kH -module V is the quantity X W ∈ cf( V ) X M ∈M dim(Ext kH ( M, W )) − |{ W ∈ cf( V ) | W ∈ M}| , where as mentioned above cf( V ) denotes the multiset of composition factors of V . Proposition . Let M be a set as above, and let V be a module. (1) If the M -pressure of V is negative then V has a simple submodule that isa member of M . (2) Suppose that
Ext kH ( A, B ) = Ext kH ( B, A ) for all simple modules A, B in cf( V ) . (This is the case whenever there is an automorphism that sendsall simple modules to their duals, hence in all simple groups of Lie typein defining characteristic, where the graph automorphism or the identitydoes this.) (a) If the M -pressure of V is then V has either a submodule or quotientthat is a member of M . (b) If the M -pressure of V is equal to n > , and there are no submodulesor quotients of V that are members of M , then the M -pressure ofany subquotient of V lies in the interval [ − n, n ] . This is normally used in the case where M is simply the set { k } containingthe trivial module, in which case the M -pressure is just referred to as pressure (see[ Cra17 , Lemma 1.8], and also [
LST96 , Lemma 1.2] for what might be the originaluse of the idea). Note that if H has no quotient of order p , i.e., O p ( H ) = H , thenExt kH ( k, k ) = 0 and so { k } satisfies the required condition. In particular, if H isa simple group and V is a module for H with negative pressure, then H stabilizesa line on V . This is important because the line stabilizers of M ( G ) and L ( G ) areknown to be positive dimensional.Given H ≤ G , and a proper subgroup L of H , we might already know that L is imprimitive. Thus we want to find all possibilities for L (knowing that it liesinside some X ), and we want to use information about L to restrict the possibilitiesfor H . The next proposition shows that we normally need only consider connectedpositive-dimensional subgroups X , rather than for example the normalizer of atorus. At this strength, this result appears in [ Cra17 , Lemma 1.12], althoughweaker versions have appeared in the literature before.
Proposition . Let H be a simple subgroup of G . If H is contained insidea member of X σ , then there is some σ -stable member Y of X such that H ≤ Y ◦ ,possibly unless G = E , and either H ∼ = PSU (3) , or p = 2 and H ∼ = PSL (8) . The next result finds a subquotient in the ‘middle’ of a self-dual module.
Lemma . Let V be a self-dual module for a finite group G over a field k ,and suppose that S , . . . , S r are non-isomorphic, simple, self-dual kG -modules, eachappearing in V with odd multiplicity. Then S ⊕ S ⊕ · · · ⊕ S r is a subquotient of V . Proof.
Proceed by induction on dim( V ), the case dim( V ) = 1 being clear.We may assume without loss of generality that V is equal to its { S , . . . , S r } -heart. If V ∼ = V ⊕ V then each S i appears with odd multiplicity in exactly one of the V j ,whence induction applied to each V j yields a subquotient of each V j , whose sum isa subquotient of V and has the required form.Therefore we may assume that V is indecomposable. In particular, soc( V ) ≤ rad( V ), and soc( V ) and top( V ) are isomorphic. Thus the heart of V , which is thequotient rad( V ) / soc( V ), again has each of the S i as a composition factor with oddmultiplicity. The heart of V is a module of smaller dimension satisfying the condi-tions of the lemma, hence satisfies the conclusion, and therefore V does. (Note that V cannot have only two socle layers as it contains an odd number of compositionfactors isomorphic to each S i , and hence the heart of V is non-zero.) (cid:3) This has a nice corollary, using pressure.
Corollary . Let V be a self-dual module for a finite group G = O p ( G ) over a field k , and suppose that S , . . . , S r are non-isomorphic, simple, self-dual kG -modules, each appearing in V with odd multiplicity. If none of the S i is thetrivial module, and the pressure of the sum of the S i is greater than the pressure of V , then G stabilizes a line on V . Proof.
This is immediate upon application of Lemma 2.7, and then Proposi-tion 2.5. (cid:3)
We can also obtain a result about modules that will be useful occasionally,which is that in a self-dual module, ‘half of the composition factors are in thebottom of the module’. This incredibly vague statement can be firmed up as thefollowing.
Proposition . Let V be a self-dual kH -module, let I be a set of simple kH -modules, closed under duality, and let W denote the I -heart of V . Let V denote the I ′ -radical of V . Then W is a submodule of the quotient V /V , and if S is a self-dualsimple module not in I and W contains no copies of S , then the multiplicity of S in V is at most twice that of S in V . Proof.
It is clear that W is a submodule of V /V . For the second part, since V is self-dual there is a submodule V such that the quotient V /V is isomorphicto V ∗ , i.e., V is the I -residual of V . We have that the quotient V / ( V ∩ V ) isisomorphic to W , hence V has a filtration0 ⊆ V ∩ V ⊆ V ⊆ V. All composition factors of V isomorphic to S appear either in V ∩ V ⊂ V or in V /V ∼ = V ∗ , and so the result holds. (cid:3) We include a lemma, which is certainly well-known, but which evaded theauthor’s brief search of the literature. (The lemma is trivial for p = 2.) Lemma . Let p = 2 . The module S ( M ( D n )) has a trivial submodule.Consequently, if V is a self-dual kH -module for a finite group H and S ( V ) H = 0 ,then there is no homomorphism from H to D n via the module V . Proof.
There are many proofs of this. The quickest is to note that Λ ( M ( D n ))of course does have a trivial submodule (as p = 2), and that S ( M ( D n )) has asubmodule (the image under the Frobenius endomorphism of) M ( D n ) with quotientΛ ( M ( D n )). Since the natural module for D n has no 1-cohomology (see [ JP76 ])it must be that S ( M ( D n )) has a trivial submodule. The consequence is clear. (cid:3) . NOTATION AND PRELIMINARIES 15 We need one easy result from non-abelian cohomology, the generalization toarbitrary p -groups of the standard result from cohomology that all complements ina semidirect product are conjugate if and only if the 1-cohomology is zero. Althoughwe state it in the case where P is a finite p -group, the same proof works for algebraicgroups. Lemma . Let P be a finite p -group and let H be a finite group that acts on P , with G = P ⋊ H . Suppose that there is an H -stable series P ≤ P ≤ · · · ≤ P r = P of normal subgroups of P such that P i /P i − is elementary abelian for all ≤ i ≤ r .If, viewed as an F p H -module, we have that H ( H, P i /P i − ) = 0 for all ≤ i ≤ r ,then all complements to P in G are conjugate. Proof.
Let H and H be complements to P in G . Proceed by induction on r , the case r = 1 being the standard result from cohomology theory. Note that P is normal in G , so by induction the result holds for G/P . Since H P /P and H P /P are complements to P/P in G/P , there exists g ∈ G such that( H P ) g = H P . Thus we may assume that H P = H P . But then H and H are complements to P in H P , whence since H ( H, P ) = 0 we have that H and H are conjugate, as claimed. (cid:3) HAPTER 3
Subgroup structure of exceptional algebraic groups
Let G be a simple, simply connected, exceptional algebraic group, so one of G , F , E , E and E . The collection of maximal closed, positive-dimensionalsubgroups of G , i.e., the set X , is known by work of Liebeck and Seitz [ LS04a ] .Broadly speaking, it consists of parabolic subgroups, maximal-rank subgroups, anda small collection of G -irreducible reductive subgroups.To move from subgroups of G to subgroups of G , and more generally subgroupsof ¯ G , we need the following theorem. Theorem
Bor89, LS90 ]) . Let H be a maximalsubgroup of ¯ G . One of the following holds: (IMP) H is a member of X σ ; (EX) H is an exotic r -local subgroup of ¯ G . (BOR) H is the Borovik subgroup (Alt(5) × Sym(6)) ⋊ ; (AS) H is an almost simple group. The members of X are known from [ LS04a ], and in theory from here it ispossible to write down a complete list of the members of (IMP). This has beendone in the case of maximal-rank subgroups in [
LSS92 ], but there are some minorerrors in the tables there with regards to the ‘decorations’ that are placed on thegroup. (For example, a maximal-rank subgroup of type A in E is of the form(3 · PSL ( q ) · .
2, if 3 | ( q − CLSS92 ], and the Borovik subgroup is unique up toconjugation.A priori, there are infinitely many possibilities in (AS), as PSL ( p a ) lies inside E in characteristic p for all a . If H is not a Lie type group in characteristic p , the (finitely many) possibilities are given in [ LS99 ] (note that B (8) for p =5 is missing from E ). The work of Litterick [ Lit18 ] and the author [
Cra17 ]has eliminated some of the subgroups in the list from [
LS99 ], and various papersthat have appeared in the literature have considered specific examples, often incharacteristic 0.If H is a Lie type group in characteristic p in (AS) above, then in [ LS98 ]Liebeck and Seitz proved that H must be one of the following:(MED) H = H ( q ) for q ≤ H of untwisted rank (i.e., the rank of theuntwisted group) at most half of that of G ;(LU3) H ∼ = PSL (16) or H ∼ = PSU (16);(RK-1) H ∼ = PSL ( q ), B ( q ) or G ( q ) for q ≤ gcd(2 , p − · t ( G ). There is a maximal subgroup missing in [
LS04a ], namely a copy of F lying in E incharacteristic 3. It acts on L ( E ) as the direct sum of L ( λ ) of dimension 52 and L ( λ ) ofdimension 196. This subgroup was very recently discovered by the author, and details will appearelsewhere soon.
178 3. SUBGROUP STRUCTURE OF EXCEPTIONAL ALGEBRAIC GROUPS
In [
Cra ] the author considered H ∼ = PSL ( q ) for G = F , E and E , provingthat there are no such examples in (AS) for G = F and G = E , and for G = E we may restrict q to be one of 7, 8 and 25. (There are still examples of PSL ( q )in (IMP) for F and E .) Recent work of Alexander Ryba and the author haseliminated the case q = 25.This article eliminates possibility (LU3), eliminates G ( q ) from (RK-1), se-verely restricts the examples that may occur in (MED) to fewer than ten groups,all for E , and eliminates B ( q ) from (RK-1) unless G = E and q = 8.In order to do this, we must show that for any simple subgroup H of ¯ G , N ¯ G ( H )lies inside a member of X σ , or an exotic r -local subgroup, or in the Borovik sub-group. In fact, we will always show that H ≤ G is strongly imprimitive, so werequire some techniques to prove this.The primary method is to prove that H stabilizes a subspace of some module V for G whose stabilizer is proper and positive dimensional. This proves that H liesin a member of X , and if H = H σ and if the subspace is σ -stable then H lies insidea member of X σ , but this still says nothing about N ¯ G ( H ), the potential maximalalmost simple group in Theorem 3.1. For that we need to prove that H stabilizeseach subspace in an orbit under the action of ¯ G of subspaces of some module V , andthe intersection of the stabilizers of all subspaces in the orbit is positive dimensional.Note that in order for this to be even well defined, the module V must be stabilizedby ¯ G ; as diagonal automorphisms stabilize all highest weight modules, and fieldautomorphisms act as semilinear transformations of highest-weight modules, inpractice this means checking that the module is graph-stable for E and F ( p = 2),and there is no condition for E and E . Because semilinear transformations areallowed, a graph-stable module can be taken to mean a semisimple k G -module V whose highest weights are invariant, up to applying a power of the Frobeniusendomorphism, under the graph automorphism.Thus for E we examine L ( G ) or M ( G ) ⊕ M ( G ) ∗ , which is graph-stable, andfor F and p = 2 we choose the sum of the composition factors of L ( G ), i.e., L ( λ ) ⊕ L ( λ ), which is also graph-stable. (Compare with the modules mentioned inTheorem 2.4 above.) Notice that in both of the later cases the graph automorphismwill interchange the two submodules.The theorem which proves this in its most general form is found in [ Cra ,Proposition 4.3], but it is based entirely on work of Liebeck and Seitz [
LS98 ], andwas also featured in the work of Litterick [
Lit18 ]. Define Aut + ( G ) to be the groupgenerated by inner, diagonal, graph, and p -power field automorphisms of G . As in[ Cra , Chapter 3], when p = 2 and G = F we need to be a bit more careful, as thegraph automorphism powers to a field automorphism, so we can add a single graphautomorphism to Aut + ( G ), but not all of them simultaneously. This distinction ispurely academic, but it means that when we consider N Aut + ( G ) ( H )-stability for asubgroup H ≤ G σ , then the definition of Aut + ( G ) technically depends on σ . Thiswill never be of importance from now on, and so we will never mention it again. Theorem . Let G be a simple algebraic group over an algebraically closedfield, and let V be a graph-stable module. If φ ∈ N Aut + ( G ) ( H ) then φ permutes the H -invariant subspaces of V . If W is a φ -orbit of H -invariant subspaces and G W denotes the intersection of the stabilizers G W for W ∈ W , then H ≤ G W and G W is φ -stable. . SUBGROUP STRUCTURE OF EXCEPTIONAL ALGEBRAIC GROUPS 19 Equipped with this, we must find some subspaces for H to stabilize. If H is a blueprint for a graph-stable module V (and does not stabilize exactly thesame subspaces as G ), then certainly H is contained in a member of X , and theintersection of all subspaces it stabilizes, in particular a single orbit, is positivedimensional. In addition, if H = H σ then N ¯ G ( H ) is also contained in a member of X σ (of the almost simple group ¯ G ). Corollary . Let H be a subgroup of G , and let V be a graph-stable k G -module. If H is a blueprint for V then either H is strongly imprimitive or H and G stabilize the same subspaces of V . In particular, if V is irreducible then H isstrongly imprimitive or H acts irreducibly on V . The other possibility is that H stabilizes a line on M ( G ) ◦ or L ( G ) ◦ . If H stabilizes a line on L ( G ) ◦ then we can say something about H . Lemma
Sei91 , (1.3)]) . If H stabilizes a line on L ( G ) ◦ then H is containedin either a maximal-rank subgroup or a maximal parabolic subgroup of G . If H stabilizes a line on either L ( G ) ◦ or M ( G ) ◦ , [ Cra , Propositions 4.5 and 4.6]prove that H is strongly imprimitive, with some mild conditions, which are alwaysgoing to be satisfied by an almost simple subgroup such that N ¯ G ( H ) is maximal. Lemma . Let H be a finite subgroup of G . Suppose that C G ( H ) = Z ( H ) and H has no subgroup of index . If H stabilizes a line on L ( G ) ◦ or M ( G ) ◦ then H is strongly imprimitive. If G = F and p = 2 , and either H or its imageunder the graph automorphism stabilizes a line on M ( F ) , then H stabilizes a lineor hyperplane on L ( F ) and is strongly imprimitive. (The consequence is simply because if H stabilizes a line on M ( F ) then it alsostabilizes a hyperplane. The composition factors of L ( F ) are M ( F ) and its imageunder the graph automorphism, and so H must also stabilize a line or hyperplaneon L ( F ).)This means that if the composition factors of M ( G ) ↓ H or L ( G ) ↓ H are unwar-ranted, and also force H to stabilize a line on the module, then H cannot embedwith those factors. One example is the following. Corollary . There does not exist a finite subgroup of E over any fieldwhose composition factors on L ( E ) have dimensions , , . Proof.
Let H be such a subgroup. Clearly H stabilizes a line on L ( E ), soby Lemma 3.5 lies in a member of X . However, no proper positive-dimensionalsubgroup of E has a composition factor of dimension at least 246 (see for examplethe tables in [ Tho16 ]) so we obtain a contradiction. (cid:3)
On one occasion we will show that a subgroup is contained in a parabolicsubgroup but not inside a Levi complement of it (it is not G -completely reducible).The next result comes from [ LMS05 , Proposition 2.2 and Remark 2.4], and isstated in our language in [
Cra , Proposition 4.1].
Lemma . Let H be a finite subgroup of G . If H is not G -completely reduciblethen H is strongly imprimitive. When p = 2 and G = F , the presence of a graph automorphism means weoften have to be careful not just to find a positive-dimensional subgroup containing our given subgroup H , but to find one that is stable under the automorphismsin N ¯ G ( H ), to eliminate almost simple groups from being potential maximal sub-groups. The next lemma gives us a criterion that enables us to ignore the graphautomorphism when proving normalizer-stability. We could always prove the resultsa different way, but using this makes it significantly easier. Lemma . Let p = 2 . Let H be a subgroup of the algebraic group F . If H isstabilized by φ ∈ Aut + ( G ) such that φ induces a graph automorphism on G σ (i.e., φ does not act as a semilinear automorphism on M ( F ) = L ( λ ) ), then there is anautomorphism τ of H such that the module L ( λ ) ↓ H is isomorphic to the conjugatemodule L ( λ ) ↓ H τ .In particular, if the composition factors of H on M ( F ) and the quotient module L ( F ) /M ( F ) do not have the same dimensions then N Aut + ( G ) ( H ) cannot inducea graph automorphism on G . Proof.
Immediate. (cid:3)
In a similar vein, we need to understand a bit more carefully how semilineartransformations act on vector spaces, which are needed in the case where ¯ G inducesfield automorphisms on G . Lemma . Let H be a finite group and let V be a faithful kH -module. Supposethat σ is a semilinear transformation of V that is H -equivariant. If W and W ′ are kH -submodules of V such that W σ = W ′ , then there exists an automorphism φ of H such that W ′ ∼ = W φ , and then V = V φ , and in particular the composition factorsof V are invariant under φ . Proof.
Since σ is H -equivariant, we may form the subgroup h H, σ i of thegroup of semilinear transformations of V . This group has H as a normal subgroup,and hence σ induces an automorphism φ of H . Furthermore, V σ = V , so V φ = V and the result follows. (cid:3) With some rank-2 subgroups H of E for q ≥ x thatcertainly isn’t in N G ( H ) and stabilizes a subspace of L ( E ) also stabilized by H .Hence we can take the subgroup h H, x i , which is larger than H and smaller than G , to show that H is not maximal in a finite group. With a bit more care, this canbe used to show that H is strongly imprimitive. We do this now.The proof of this uses forward references. The reason for this is that we place H inside a finite group K , and want that K is strongly imprimitive. We willonly use this result in Chapter 8, so can use results for ranks 3 and 4 from theprevious chapters. Forward referencing is not strictly required, but proving thisresult without it takes considerably longer, as one has to determine precisely whichgroups can contain which, and then prove results for these groups. Lemma . Let H be a finite simple group. Let G = E , and let x be asemisimple element of G . Suppose that one of the following holds: (1) H ∼ = PSL (8) and x has order ; (2) H ∼ = PSL (9) , PSU (9) and x has order ; (3) H ∼ = PSU (16) and x has order ; (4) H ∼ = PSp (8) and x has order ; (5) H ∼ = B (32) and x has order ; (6) H ∼ = B (128) and x has order ; . SUBGROUP STRUCTURE OF EXCEPTIONAL ALGEBRAIC GROUPS 21 (7) H ∼ = PSL (81) and x has order ; (8) H ∼ = PSU (4) and x has order , with x ∈ H .If O is a σ -stable, N Aut + ( G ) ( H ) -orbit of subspaces of L ( E ) , each subspace stabilizedby H , and the members of O are also stabilized by x , then H is strongly imprimitive.If O is stabilized by H and x , but not necessarily by N Aut + ( G ) ( H ) , then H is Lieimprimitive, but not necessarily strongly imprimitive (i.e., H lies in a member of X ). Proof.
We first check that x cannot normalize H , i.e., H E h H, x i .By [ CCN + , pp.74, 78, 79], we see that Aut( H ) contains no element of order o ( x ) for H one of PSL (8), PSL (9), PSU (9), and since 3 ∤ | Out(PSU (16)) | wesee that the same holds for H ∼ = PSU (16). For H ∼ = PSp (8), Out( H ) is cyclic oforder 3 so | H ∩ h x i| is either 65 or 195, so 65 as 195 is not the order of an element of H . However, x induces the field automorphism on H , whose centralizer is PSp (2)and must contain x of order 65, a contradiction.The Suzuki groups famously do not have elements of order 3, and their auto-morphisms consist only of field automorphisms, which have orders 5 and 7 in thecases above, so this proves the claim for them. For H ∼ = PSL (81), the intersection H ∩ h x i must have order 41; Out( H ) is dihedral of order 8, with the field automor-phism of order 2 as the centre of it. But the centralizer in H of the field involutionis PSL (9), which of course does not contain an element of order 41, so we are donein this case as well. Finally, 3 ∤ | Out(PSU (4)) | and PSU (4) does not contain anelement of order 195.Write K = h H, x i , and therefore by the above K N G ( H ). Let ¯ K be theintersection of the stabilizers of the members of O , and note that H ≤ K ≤ ¯ K .If ¯ K is infinite then we apply Theorem 3.2 to see that H is strongly imprimitive.Thus we may assume that K and ¯ K are both finite. By Theorem 3.1, if ¯ K (andhence H ) is not strongly imprimitive, then ¯ K lies inside an exotic local subgroup,or the Borovik subgroup, or ¯ K is almost simple.The non-cyclic composition factors of the exotic local subgroups (found in[ CLSS92 ]) are PSL (2) and PSL (5), neither of which contains H , so cannot con-tain ¯ K , and similarly H is not contained in the Borovik subgroup either. Thus ¯ K isalmost simple. If ¯ K is not of Lie type in characteristic p , then the possibilities for ¯ K appear in [ LS99 ]; none of these contains H (certainly they don’t contain elementsof order o ( x )) and so ¯ K must be Lie type in characteristic p , ¯ K ( q ) for some q , andthis is not H . If q > K is greater than 4 then ¯ K is a blueprint for L ( E ) as we saw above, and we are done. If ¯ K stabilizes a line on L ( E ) then sodoes H and again H is strongly imprimitive by Lemma 3.5. If the rank of ¯ K is 3or 4 then either ¯ K is strongly imprimitive or it is PSU (2), which does not contain H , by the collection of propositions from Chapters 6 and 7, and Section 12.1, so¯ K has rank 2. The only possibility now is that q = 8 , K = G ( q ), but thisstabilizes a line on L ( E ) by Proposition 8.6 below, hence so does H and we aredone again by Lemma 3.5. (cid:3) For H ∼ = PSL (16), we want the order of x to be 255, but there is an elementof order 255 in PGL (16) (but not in PSL (16)). We have to be slightly morecareful: since Out( H ) has a normal subgroup of order 3 and index 8, all elementsof order 255 lie in PGL (16) ≤ Aut( H ). Furthermore, the centralizer in PGL (16)of a cyclic subgroup h x i of order 85 (unique up to conjugacy) is of order 255 (it is a Singer cycle) and hence there are exactly two elements of order 255 in Aut( H )cubing to a given element of order 85 in PSL (16). The analogue of Lemma 3.10is therefore the following, with an identical proof. Similar statements also hold for H ∼ = PSp (9). Lemma . Let H be one of PSL (16) and PSp (9) . Suppose that H is asubgroup of G = E , and let x , x , x be three distinct elements of G of order if H ∼ = PSL (16) , and order if H ∼ = PSp (9) . If O is a σ -stable, N Aut + ( G ) ( H ) -orbit of subspaces of L ( E ) , each subspace stabilized by H , and the members of O are also stabilized by all x i , then H is strongly imprimitive. Remark . This is how we will use Lemmas 3.10 and 3.11. Let H be asubgroup of G , and let S , . . . , S r be a set of representatives for the compositionfactors of L ( E ) ↓ H . For some of these, say S i for i ∈ I , we can find an element x i ∈ G satisfying the order hypothesis in one of these two lemmas, and stabilizingthe eigenspaces of y on L ( E ) ↓ H that comprise S i , where y is a generator of h x i i∩ H .Thus if L ( E ) ↓ H possesses a submodule S isomorphic to S i then x i stabilizes it.(We may have that different S i can be stabilized by the same x i .)By one of the two lemmas, this means that H is Lie imprimitive. If S is stableunder N Aut + ( G ) ( H ) then H is also strongly imprimitive, but this will not alwaysbe the case. For this to not be the case, by Lemma 3.9 the composition factors of L ( E ) ↓ H must be stable under an outer automorphism of H (in fact one inducedby an element of N Aut + ( G ) ( H )). Furthermore, the image of S under the elementsof N Aut + ( G ) ( H ) are other simple submodules that must, by symmetry, also appearin I . If there is an element x that stabilizes all of the eigenspaces comprisingall of these different modules S simultaneously, then H is strongly imprimitive.Otherwise we may assume that both S and its image under the potential elementof N Aut + ( G ) ( H ) are submodules of L ( E ) ↓ H when analysing further, and indeedthere is an automorphism of H that maps S to this other module in its inducedaction on L ( E ) ↓ H .In conclusion, if the composition factors of L ( E ) ↓ H are not stable underan outer automorphism of H , then we obtain that H is strongly imprimitive orsoc( L ( E ) ↓ H ) consists solely of simple submodules S i for i I . If there is suchan outer automorphism, then we still obtain this conclusion if there is an elementstabilizing all of the eigenspaces from the orbit of S simultaneously, and otherwisewe may only conclude that H is Lie imprimitive or the condition above on the socleholds. In this case though, we may assume that every simple module in the orbitof S under this outer automorphism appears in the socle of L ( E ) ↓ H .We now describe the composition factors of copies of Alt(6) in E on L ( E ) for k of characteristic 2. The simple modules for Alt(6) are 1, 4 , 4 , 8 and 8 , withthe two 8s merging into a 16 in Sym(6). If we have equal numbers of 8 and 8 , wewrite 16 to make reading the list easier. Proposition . Let H ∼ = Alt(6) be a subgroup of G = E in characteristic . The warranted sets of composition factors of L ( E ) ↓ H are as follows: , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,
32. SUBGROUP STRUCTURE OF EXCEPTIONAL ALGEBRAIC GROUPS 23 If H embeds in E then the composition factors of L ( E ) ↓ H are either warrantedor possibly , , , . (No example is known with these factors.) Proof.
By [
Cra17 , Proposition 6.3], the composition factors of L ( E ) ↓ H areeither warranted or 16 , , , , thus we may assume that H lies inside apositive-dimensional subgroup X of G .If X is D then we use a computer to determine the action of H on the half-spinmodule. Since L ( D ) has the same composition factors as Λ ( M ( D )) this is easyto compute, and L ( E ) is the sum of L ( D ) and a half-spin module. This yieldsthe following table, where representations of Alt(6) (taken up to automorphism forbrevity) that extend to Sym(6) are given first, and then those that cannot (inside D ) are given afterwards.Factors on M ( D ) Factors on L ( E )16 16 , , , , , , or 4 , , , , , , or 16 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , (There is a third set of factors, 16 , , , , that has a correct Brauer characteron L ( E ) for the second row of the table, but since the factors on Λ ( M ( D )) are4 , , , a submodule of L ( E ) ↓ D , we see that this cannot occur.)If H ≤ A then, although H need not stabilize a line on M ( A ), there is anothercopy of H with the same composition factors on M ( A )—and hence on L ( E )—that does stabilize a line on M ( A ), hence inside an A -parabolic subgroup of A ,which is inside D .If H is contained in an E -parabolic subgroup of E then as we only considercomposition factors we may assume that H is contained in the E -Levi subgroup.By [ Cra17 , Proposition 6.3], H stabilizes a line on M ( E ), hence lies inside a D -parabolic of E —contained in D —or an E -parabolic of E , thus an E -Levi by thearguments above. The same proposition shows that H stabilizes a line on M ( E ) aswell, hence lies inside a D -parabolic of E —again, inside D —or inside F . Onceagain, the same [ Cra17 , Proposition 6.3] states that up to graph automorphism H stabilizes a line on M ( F ), hence lies in a parabolic subgroup of F or C , henceinside C .There are four embeddings of H into C , and we list the actions of H on M ( C ), together with the 16- and 26-dimensional modules L ( λ ) and L ( λ ). As L ( E ) restricts to F as the sum of eight copies of M ( F ), one of M ( F ) twistedby the graph automorphism, and fourteen trivial factors, it is easy to computethe factors of H on L ( E ). (We use the fact that C acts on M ( F ) as L ( λ )and on L ( F ) with factors L ( λ ), L ( λ ), L (0) and L (2 λ ), note the twist on thislast module.) We again need only work up to automorphism. Thus we need the composition factors on M ( C ), L ( λ ) and L ( λ ), from which we can deduce thetwo possible actions on L ( E ). These are as follows. M ( C ) L ( λ ) L ( λ ) First action on L ( E ) Second action on L ( E )4 , , , , , , , , ,
16 16 , , , , , , , , , , , , , , , , , , , , , , , , , , Finally, suppose that H embeds in any other maximal positive-dimensionalsubgroup X . If X has a factor A , A or G then it may be removed, and if theresulting subgroup may be placed inside a previously considered subgroup, then weare done. This completes the proof for X one of E A , E A , F G , and all butone parabolic subgroup, leaving us with A A and the A A -parabolic of A A ,so just A A . In this case, H stabilizes a line on both M ( A )s, so H is containedin an A A -parabolic of A A , which is contained in A , completing the proof forembeddings of Alt(6).We need also consider the triple cover 3 · Alt(6), embedding in a Lie type groupwhose simply connected version has a centre of order 3 but that does not occur in G , i.e., A and E A . The faithful simple modules for 3 · Alt(6) are 3 , 3 and 9(and their duals, on which the centre acts as the inverse). If H acts irreduciblyon M ( A ), then the composition factors of L ( E ) ↓ H are 16 , , , , which wehave seen before. Otherwise H lies in a copy of ( A ) which lies inside E A .Inside E A , we obtain from the proof of Proposition 10.3 below the composi-tion factors of 3 · Alt(6) on M ( E ) and L ( E ), and together with the action of H on M ( A ) being 3 ∗ or 3 ∗ , we obtain the following.Action on M ( E ) Action on L ( E ) Action on L ( E )3 , , , , , , or 8 , , , , , , , , , , , , , or 8 , , , , These have already occurred, so we gain no new possibilities. (cid:3)
We also need the stabilizers of lines on the minimal module for E ( q ) and E ( q ).(See [ LS87 , Lemmas 5.4 and 4.3].)
Lemma . Let G = E ( q ) . There are three orbits of lines of the action of G on M ( E ) , with line stabilizers as follows: (1) F ( q ) acting on M ( E ) as L ( λ ) ⊕ L (0) ( L (0) /L ( λ ) /L (0) in characteristic ); (2) a D -parabolic subgroup; q D ( q ) . ( q − , acting uniserially as L ( λ ) /L ( λ ) /L (0);(3) a subgroup q .B ( q ) . ( q − acting indecomposably as L (0) /L ( λ ) /L (0) , L ( λ ) . Lemma . Let G = E ( q ) . There are five orbits of lines of the action of G on M ( E ) , with line stabilizers as follows: (1) E ( q ) . (the graph automorphism) acting semisimply with compositionfactors of dimensions , , ; (2) E ( q ) . (the graph automorphism) acting semisimply with compositionfactors of dimensions , , ; . SUBGROUP STRUCTURE OF EXCEPTIONAL ALGEBRAIC GROUPS 25 (3) an E -parabolic subgroup q .E ( q ) . ( q − acting uniserially as L (0) /L ( λ ) /L ( λ ) /L (0);(4) a subgroup q .B ( q ) . ( q − acting indecomposably as L ( λ ) , L (0) /L ( λ ) /L ( λ ) , L (0);(5) a subgroup q .F ( q ) . ( q − acting indecomposably as L (0) , L (0) /L ( λ ) /L ( λ ) /L (0) , L (0) . In the algebraic group, there are still three orbits for E , with stabilizers: F ;one of the two conjugacy classes of D -parabolic subgroup; the preimage of B inthe Levi subgroup of the other class of D -parabolic subgroups. There are fourorbits of lines on M ( E ), with stabilizers: an E -Levi subgroup; an E -parabolicsubgroup; the preimage of B in the Levi subgroup of a D -parabolic; the preimageof F in the Levi subgroup of an E -parabolic subgroup. This can easily be seenfrom the previous two lemmas.HAPTER 4 Techniques for proving the results
This short chapter, as with [
Cra , Chapter 9], aims to give a general overviewof the ideas used in proving Theorems 1.1 and 1.2. We will outline the strategy inthe first section, and then use the example of G to illustrate the methods in thenext section. First, we use Theorem 3.1 to define which particular subgroups we need to showare strongly imprimitive. These are: groups of semisimple rank at most half thatof G and field size at most 9, PSL (16) and PSU (16), and groups PSL ( q ), B ( q )and G ( q ) for all q such that the groups do not contain a semisimple element notin T ( G ) (see Definition 2.3) using Theorem 2.4.For those groups H that remain, we show that H always stabilizes the mem-bers of some N Aut + ( G ) ( H )-orbit of subspaces of either M ( G ) or L ( G ) that is alsostabilized by a positive-dimensional subgroup. We then apply Theorem 3.2 to seethat H is strongly imprimitive, as desired. In particular, if H is a blueprint forsome module then this is usually enough.If G is not E then we can use Theorem 2.4 to prove that some of the groupsabove are blueprints for the minimal module. However, there are two issues withthis plan. Now it is possible for our subgroup H to act irreducibly on M ( G ),whereas it cannot act irreducibly on L ( G ), and so it is obviously a blueprint.(Such groups are enumerated in [ LS04b ], and include for example PSL (3) in E in characteristic 3.) In addition, the minimal module is not always stable underAut + ( G ), in particular when there is a graph automorphism, so we cannot alwaysuse this. With these two caveats, the remaining groups that possess a semisimpleelement as described in Theorem 2.4 are strongly imprimitive.Thus we assume that we have a ‘small’ group H . Using the tables of knowncharacter values that semisimple elements of G can take on M ( G ) and L ( G ), weproduce a list of candidate possibilities for the composition factors for M ( G ) ↓ H and L ( G ) ↓ H , the conspicuous sets. (If G is E then we just have L ( G ).) Notethat these have to be consistent in two different ways. First, the traces need to beconsistent with the power map. This is equivalent to stating that the eigenvaluesof semisimple elements of H on the modules need to match those of semisimpleclasses of G , not just the traces. The second is that specifying traces of elementson M ( G ) will force the elements of H to belong to particular classes of G , possiblywith some ambiguity if more than one G -class has the same eigenvalues on M ( G ).These classes have corresponding eigenvalues on L ( G ), and those must correspondto a conspicuous set of factors as well.If we have elements x of large order, say much larger than 30 or 40, we donot tend to have tables of their eigenvalues, as there are too many to compute
278 4. TECHNIQUES FOR PROVING THE RESULTS easily. What we can do, if the order of x is composite, is use the roots trick : if x n has smaller order than x , then find an element of a maximal torus T in theclass to which x n belongs (which we know from the eigenvalues of x n ) and thenconsider all n th roots of x n in T of order that of x . This much smaller set canbe enumerated and eigenvalues found. They can then be used to compute theconspicuous sets of factors. If this is still too many, if x has order mn then we canspecify the eigenvalues of x n and x m , run through all elements of order n in T witheigenvalues that of x m and multiply them by our given element of T correspondingto x n to find all possibilities for the eigenvalues of x . (This is used exactly once inthis article, for B (512). Usually n is one of 2 , , ,
7, and so there are few enough n th roots to just use all of them. In the one case where we don’t, n = 13.)We therefore obtain a list of sets of composition factors for M ( G ) and/or L ( G ).We then must go through each in turn, so we will call them ‘Case 1’, ‘Case 2’, andso on. Many proofs subdivide into these cases, and we will write Case 1 and so onat the start of each new set of composition factors.Thus at this point we may assume that we are given a conspicuous set ofcomposition factors for M ( G ) or L ( G ). We first compute its pressure, so thesum of the 1-cohomologies of the composition factors, minus the number of trivialcomposition factors. If we have a trivial composition factor and this pressure isnon-positive, then H stabilizes a line or hyperplane on the module by Proposition2.5, and then H is strongly imprimitive by Lemma 3.5. Many conspicuous sets ofcomposition factors fail this easy test.We are then left with those cases where we need to study the potential modulestructure of M ( G ) ↓ H or L ( G ) ↓ H . For this we have a few tools: importantly, L ( G ) ◦ is always self-dual (unless G = F and p = 2), and M ( G ) ◦ is as well unless G = E .In particular, if H does not stabilize a line on L ( G ) ◦ then H does not stabilize ahyperplane either. Often we argue by contradiction that H must stabilize a line on L ( G ) ◦ , and assuming the contrary will necessarily mean that H does not stabilizea hyperplane either. We will usually not explicitly note this each time we use thisextra information, as we will use it so frequently.The possible Jordan block structures of unipotent elements of G are tabulatedin [ Law95 ], so our unipotent elements of H must act in a way in those tables. Fur-thermore, Lemma 2.2 states that if H contains an element in some of the unipotentclasses of G , the generic ones, then H is always a blueprint for that module. Thuswe may assume that H acts irreducibly on the module in question, or is stronglyimprimitive.We also have subgroups of H , particularly centralizers of semisimple elements.The centralizers of semisimple elements of low order in exceptional groups areknown, and tabulated in various places, for example [ Fre98 ] for E . If a propersubgroup L of H is known to be imprimitive, either because we will prove it inthis paper or for some other reason, one may work through the members of X totry to understand the module structures of M ( G ) ↓ L and L ( G ) ↓ L . This must becompatible with the module structures for H .In particular, if we can show that some kL -submodule of the space is stabilizedby both H and a positive-dimensional subgroup X (but not the whole of G ), then h H, X i also stabilizes that space, and must be proper in G . This proves that H isnot Lie primitive, and if the same holds for an orbit of spaces under N Aut + ( G ) ( H ),then this shows that H is strongly imprimitive by Theorem 3.2. One good way .2. AN EXAMPLE 29 to do this is in the case L = h x i : we look for elements of T that stabilize all ofthe eigenspaces of the action of x that appear in the k h x i -submodule. Then thoseelements of T must also stabilize the submodule. Find enough such elements andthe stabilizer of the subspace must be positive dimensional. The roots trick is auseful tool to look for these elements of T .To constrain the structure of M ( G ) ↓ H and L ( G ) ↓ H , we often must use thecomputer package Magma, which allows us to construct the largest possible modulewith a given socle and certain composition factors or other restrictions. The mainidea is that of a ‘pyx’. If W is a kH -module, then a pyx for W is simply any module V such that W is a subquotient of V . Normally, we will construct a pyx V for W byforcing soc( V ) = soc( W ), and then build W according to some universal property.The most obvious one is that V is the cf( W )-radical of P (soc( W )), which of coursemust be an overmodule of W . More generally, if we have a filtration of W , then wemay construct V by taking repeated radicals with composition factors accordingto the filtration of W . This will in general be smaller than the cf( W )-radical, andhopefully small enough that one can gain information about W from V .We will now see these techniques in action, as we (almost) prove that everydefining-characteristic subgroup of G is strongly imprimitive. We use the techniques and preliminary lemmas on a toy example: G . Whilethe maximal subgroups of G ( q ) are known [ Coo81, Kle88 ], this will illustratesome of the points involved. We also deliberately do not do it in the easiest way inorder to show more of the techniques we will use.Let G be of type G in characteristic p , and let H ≤ G be a simple group thatis of Lie type in characteristic p . We suppose that H ∼ = PSL ( q ) with q a power of p . Step 1 : Use Theorem 2.4 to bound q .Theorem 2.4 states that H is a blueprint for L ( G ), and hence strongly imprimi-tive, if H contains a semisimple element of order at least 13. Since H contains asemisimple element of order ( q + 1) / p = 2, in which case it possesses anelement of order q + 1), we have that H is blueprint for L ( G ) whenever q ≥ q ≤
23 (and we can exclude q = 16 as well).So our set of q is { , , , , , , , , , } . Step 2 : Switch to M ( G ) and use Theorem 2.4 to exclude more cases.We cannot use Theorem 2.4 when p = 3 as M ( G ) is not graph stable in this case.Also, if p ≥ LS04b , Table 1.2] we see that H can act irreducibly on M ( G ). If H contains a semisimple element of order at least 5 then Theorem 2.4applies, but that is true for PSL ( q ) for all q except q = 5 ,
7. Thus in all othercases we may assume that H acts irreducibly on M ( G ) (and thus this excludes q = 4 , Step 3 : Use unipotent actions to eliminate more cases.Let u be an element of order p in H . If u is generic for L ( G ) then u is a blueprintfor L ( G ) by Lemma 2.2, and hence so is H . Thus H is strongly imprimitive.Consulting [ Law95 , Table 2], we see that this is the case whenever p ≥
11. If
Module 1A 2A 4A 5A 5B1 1 1 1 1 13 − λ + 1 ¯ λ + 13 − λ + 1 λ + 14 4 0 − − −
19 9 1 1 − − Table 4.1.
Brauer character table for PSL (9) and p = 3. Here, λ is the sum of a fifth root of unity and its inverse. q = 5, we see from [ Law95 , Table 1] that u is generic for M ( G ), and H cannotact irreducibly on M ( G ), so H is a blueprint for M ( G ) and again H is stronglyimprimitive.Thus q must be either 7 or 9. Step 4 : Completing q = 9 using pressure and graph automorphism.We now look at the conspicuous sets of composition factors for M ( G ) ↓ H . The sim-ple modules for PSL (9) are of dimensions 1, 3, 3, 4 and 9. If there is a trivial factorthen the dimensions must be 3 , , ; since the 3s have zero 1-cohomology, inboth cases H has negative pressure on M ( G ) and so H stabilizes a line on M ( G )by Proposition 2.5, hence is strongly imprimitive by Lemma 3.5. (In fact, we notethat if H stabilizes a line on M ( G ) then its stabilizes a line on L ( G ), and henceis strongly imprimitive. This distinction will be used for F and p = 2 in Chapter11.) The other case is that the composition factors are of dimensions 4 and 3, andup to field automorphism of H there is a unique such set of composition factors.As M ( G ) is self-dual, the restriction to H must be semisimple.The Brauer character table of PSL (9) is given in Table 4.1. If H acts on M ( G ) as 3 ⊕
4, then we can use this table, and the traces of semisimple elementson M ( G ) and L ( G ), to deduce the composition factors of the image of H underthe graph automorphism on M ( G ). This is because L ( G ) has as compositionfactors M ( G ) and its image under the graph automorphism.The traces of elements of orders 1, 2, 4, 5 and 5 of H on M ( G ) are 7, − − λ and ¯ λ respectively. This uniquely determines the semisimple classes of G towhich these elements of H belong, and their traces on L ( G ) are given by 14, − λ + 2 and λ + 2 respectively. These traces yield composition factors 1 , , , , H on the factor of L ( G ) other than M ( G ) is 1 ⊕ ⊕ . Inparticular, either H or its image under the graph automorphism stabilizes a line on M ( G ). Hence H always stabilizes a line on L ( G ), and is strongly imprimitive byLemma 3.5. Step 5 : Completing q = 7 using L ( G ).As H acts irreducibly on M ( G ) we cannot do much, so we switch to L ( G ). Aswith Step 4, we compute the traces of elements of H on M ( G ), which are 7, −
1, 1and − L ( G )of 14, − − .2. AN EXAMPLE 31 Module 1A 2A 3A 4A1 1 1 1 13 3 − − −
17 7 − − Table 4.2.
Brauer character table for PSL (7) and p = 7.We normally use a computer to calculate the conspicuous sets of compositionfactors, as it is well suited to linear algebra problems. In this case the set ofcomposition factors is 5 , , ,
3, as we can see from Table 4.2.To understand how these composition factors assemble into the module for L ( G ) ↓ H , we use the action of the unipotent element u . This acts on the module7 with a single Jordan block, hence lies in class G by [ Law95 , Table 1], i.e., it isregular unipotent. The action of this class on L ( G ) is given in [ Law95 , Table 2],and the blocks are 7 . Thus H acts projectively on L ( G ), and thus must act asthe projective module P (3), which has structure3 / , / . At this stage the representation theory has done all that it can: it has deter-mined completely the actions of H on the minimal and adjoint modules. This typeof subgroup was referred to as a Serre embedding in [
Cra ], and is not amenable toour methods, being solved in general in [
BT19 ]. (Note that this case is mentionedin [
LS04b ] but does not seem to appear in their proof, although it is of coursehandled in [
Kle88 ].)With such a small case we cannot showcase all of our techniques, in particularthe construction of a pyx. In the final step the projective was the whole module,so the pyx is just the projective module itself.We also cannot use the roots trick to find elements of large order that stabilizethe subspace 3 above. Although they exist (the maximal A subgroup stabilizesa subspace decomposition 3 / /
3) since they do not stabilize all copies of 3 in themodule, one will normally not be able to prove that some semisimple element in T of large order stabilizes exactly the 3 in the socle. Indeed, in this case we cannot.HAPTER 5 Modules for groups of Lie type
In this chapter we will discuss facts about the subgroups H that we will con-sider. In particular, we will give the dimensions of the simple modules for H when q = 2 , , , q = 4 , , ,
16 can be deduced from Steinberg’stensor product theorem) and the dimensions of the 1-cohomologies of the simplemodules for q a prime, and for q = 8 , ,
16 for those groups that we need. For H of ranks 2 and 3, and rank 4 for q = 2 , ,
5, we give the dimensions of the modulesand the dimensions of the 1-cohomologies of simple modules of dimension at most124, and of self-dual modules of dimension at most 248. These are given in Tables5.1, 5.2, 5.3 and 5.4.For G = E , we also need the dimensions of all faithful modules (of dimensionat most 28, or self-dual and at most 56) for a central extension 2 · H for H of rankat most 3 and q odd, which we give in Table 5.6. Finally, for G = E , we needdimensions of faithful modules (of dimension at most 27) for a central extension3 · H for H of rank at most 3 and q not a power of 3, which are in Table 5.7.One may compute Ext between simple modules in Magma easily: if A and B are modules, Ext ( A , B ) computes the space of extensions. Furthermore, extensionsbetween simple modules for PSL ( q ), PSU ( q ), PSp ( q ) and B ( q ) for q even aregiven in [ Sin92a, Sin92b ], and extensions for G ( q ) and G ( q ) for q a power of 3 in[ Sin93 ]. We tabulate Ext ( M, k ) here for M a simple module, as these are the ex-tensions that we need the most. However, we will use information about extensionsbetween other simple modules, and we use Magma to compute this information,together with modules realizing the extensions, using the two commands V,rho:=Ext(A,B);E:=MaximalExtension(A,B,V,rho); (This requires that B , but not necessarily A , is simple.)If a simple module M is of dimension n , and is the only kH -module of thatdimension (e.g., the Steinberg module) then we just denote it by n .If M is not self-dual but it and its dual are the only modules of this dimensionthen we label M and its dual by n and n ∗ , and use n ± to mean both or either n and n ∗ (it is clear from context which we mean). However, when giving precise modulestructures, we will need to be specific about exactly which of the two modules is n and which is n ∗ . Of course, we can make a global choice for one module (asduality induces an automorphism, so we can never fix modules precisely, only upto duality of all modules simultaneously), but then all other modules are fixed. Wenow describe, for each group, the choices that we make in this article. (We onlyinclude the modules that actually appear in our sets of composition factors, as theseare all that is required. Thus, for example, for PSU (7) we do not need a label for33 ± .)
334 5. MODULES FOR GROUPS OF LIE TYPE
Group Modules of dimension at most 248PSL (2) 1 , ± , (2) ′ , , , , G (2) ′ , , , , PSL (2) 1 , ± , , , ± , (2) 1 , ± , , , ± , (2) 1 , , , , , , (2) 1 , ± , ± , , ± , ± , PSU (2) 1 , ± , ± , , ± , ± , PSp (2) 1 , , , , , , , PΩ +8 (2) 1 , , , , , , , , , , , PΩ − (2) 1 , , , , , , , , , , , D (2) 1 , , , , , , , , , , , F (2) 1 , , , , F (2) ′ , , Table 5.1.
Simple modules of dimension at most 248 (or dualpair up to dimension 124) for groups over a field of two elements.Bold indicates a 1-dimensional 1-cohomology, underlined indicatesa 2-dimensional.Group Modules of dimension at most 248PSL (3) 1 , ± , ± , , ± , (3) 1 , ± , ± , , ± , (3) 1 , , , , , G (3) 1 , , , , , , , , G (3) ′ , , , , PSL (3) 1 , , ± , , , , ± , , (3) 1 , , , ± , , (3) 1 , , , , , , , , PSp (3) 1 , , , , , , (3) 1 , ± , ± , ± , , ± , ± , ± , , ± , ± , PSU (3) 1 , ± , ± , ± , , ± , ± , ± , , ± , ± , Ω (3) 1 , , , , , , (3) 1 , , , PΩ +8 (3) 1 , , , , , PΩ − (3) 1 , , , , , , , , , , D (3) 1 , i , , i , i , i , , j F (3) 1 , , , Table 5.2.
Simple modules of dimension at most 248 (or dualpair up to dimension 124) for groups over a field of three elements.Bold indicates a 1-dimensional 1-cohomology, underlined indicates2-dimensional, and bold and underlined indicates 3-dimensional.Here 1 ≤ i ≤ ≤ j ≤ D (3). . MODULES FOR GROUPS OF LIE TYPE 35 Group Modules of dimension at most 248PSL (5) 1 , ± , ± , , ± , ± , ± , ± , , ± , ± , ± , , ± , (5) 1 , , ± , , ± , , (5) 1 , , , , , , , , , , , G (5) 1 , , , , , , , , , (5) 1 , , , ± , ± , , , (5) 1 , , ± , , , ± , ± , , , ± , ± , , , , (5) 1 , , , , , , , , , (5) 1 , , , , , , , (5) 1 , ± , ± , ± , , ± , ± , ± , ± , ± , ± , , ± , ± , ± , (5) 1 , ± , ± , ± , , ± , ± , ± , ± , ± , ± , , ± , ± , ± , (5) 1 , , , , , , , (5) 1 , , , +8 (5) 1 , , , , PΩ − (5) 1 , , , , , , , , , , D (5) 1 , i , , i , i , i , i , j ,F (5) 1 , , Table 5.3.
Simple modules of dimension at most 248 (or dualpair up to dimension 124) for groups over a field of five elements.Bold indicates a 1-dimensional 1-cohomology, underlined indicatesa 2-dimensional. Here 1 ≤ i ≤ ≤ j ≤ D (5).Group Modules of dimension at most 248PSL (7) 1 , , ± , , ± , ± , , ± , , (7) 1 , ± , ± , , ± , ± , ± , ± , ± , , ± , ± , ± , ± , , ± , ± , ± , ± , ± , ± , , PSp (7) 1 , , , , , , , , , , , , , , , , , G (7) 1 , , , , , , , , , (7) 1 , , ± , , , ± , ± , , , ± , , ± , , , , , (7) 1 , , , ± , , ± , , , , (7) 1 , , , , , , , , , (7) 1 , , , , , , , , Table 5.4.
Simple modules of dimension at most 248 (or dualpair up to dimension 124) for groups over a field of seven elements.Bold indicates a 1-dimensional 1-cohomology.If there are multiple modules of the same dimension up to duality then we usesubscripts to distinguish them. We place self-dual modules first, so for example20 , 20 and 20 ∗ would be three modules of dimension 20.For PSL ( q ) and PSU ( q ), we can be guided by the fact that 3 lies in a non-principal block for q = 7, and so we could like to label 6 and 3 so as to lie in thesame block. With these kind of guidelines, we make the following choices. • PSL (2): only 3 , ∗ to fix, so nothing to do. Group Modules with non-zero 1-cohomologyPSL (4) 50 ± i,j , i PSp (4) 8 , , , , , Ω ± (4), D (4) i , i,j , i PSL ( q ), q = 4 , i , ± i,i − , i,i − PSU ( q ), q = 4 , i , ± i,i − , i,i − PSp ( q ), q = 4 , i , i , i,i − PSL (4) ± PSL ( q ), q = 8 ,
16 9 ± i,i +1 PSU ( q ), q = 4 , ,
16 9 and images under field automorphismPSp ( q ), q = 4 , i B ( q ), all q i G (4) 6 i , , i,j G (8) 6 i , i,i − PSL (9), PSU (9) 7 i , ± i,j PSp (9) 64 i,j , i,j G (9), G ( q ), q > Sin93 ]) Table 5.5.
Simple modules of dimension at most 248 (or dualpair up to dimension 124) with non-zero 1-cohomology for groupsover a field of four, eight, nine and sixteen elements. Bold indicatesa 2-dimensional 1-cohomology. Only those groups for which thisinformation is needed in this article are listed.Group Modules of dimension at most 56Sp (3) 4 , , (5) 4 , , , , (7) 4 , , , , , (3) 4 ± , ± · PSL (5) 6 , ± , (7) 4 ± , ± , ± · PSU (3) 6 , ± , (5) 4 ± , ± , ± · PSU (7) 6 , ± , (3) 6 , , ( q ), q = 5 , , , ( q ), q = 3 , , (7) 8 , Table 5.6.
Simple modules of dimension at most 56 (or dual pairup to dimension 28) for 2-fold central extensions of groups of rankat most 3. • PSL (3) and PSU (3). Fix 3. Then 6 = S (3 ∗ ) and 15 = Λ (6 ∗ ). • PSL (5). Fix 3. Then 6 = S (3 ∗ ), 10 lies in 6 ⊗ ∗ , 15 lies in S (6 ∗ ),15 = Λ (6 ∗ ), 18 ∗ is the socle of 3 ⊗ , and 35 lies in Λ (10 ∗ ). Finally,39 is the socle of Λ (15 ∗ ). . MODULES FOR GROUPS OF LIE TYPE 37 Group Modules of dimension at most 273 · PSL (4) 3 , ∗ , , ∗ · PSL (7) 3 , , , , , · PSU (5) 3 , , , , · PSp (2) ′ , ∗ , Table 5.7.
Simple modules of dimension at most 27 for 3-foldcentral extensions of groups of rank at most 3. (We only includemodules from the one of the two (dual) faithful blocks.) • PSU (5). Fix signs so that 3 , , , , ,
39 all lie in the same block.Then 10 lies in 6 ⊗ ∗ , 15 lies in S (6 ∗ ), 15 = Λ (6 ∗ ) and then 35 lies inΛ (10 ∗ ). • PSL (7). Fix 10. Then 28 lies in S (10 ∗ ) and 35 lies in Λ (10 ∗ ). Themodule Λ (27) has a summand 10 ∗ / / ∗ . • PSU (7). Fix 3. Set 6 = S (3 ∗ ), 10 lies in 3 ⊗ ∗ , 15 lies in S (6 ∗ ) and15 = Λ (6 ∗ ). Both 21 and 24 lie in 3 ∗ ⊗ ∗ , and 28 lies in S (10 ∗ ). Wehave that 35 lies in Λ (10 ∗ ), 42 lies in Λ (15 ∗ ), and 71 lies in Λ (28 ∗ ). • PSL (2) and PSU (2). Fix 4. Then 20 lies in 4 ⊗ • PSL (3). In the principal block, we only need to fix 10 ± and 45 ± . Fix 10.Then 45 = Λ (10 ∗ ). • PSU (3). In the principal block, only 45 needs to be fixed. • PSL ( q ) and PSU ( q ) for q = 5 ,
7. We do not need to specify the moduleshere. • PSL (2) and PSU (2). Fix 5. Set 10 = Λ (5 ∗ ). Then 40 lies in Λ (10),and 40 lies in 5 ⊗ ∗ . • PSL (3) and PSU (3). Fix 5. Set 10 = Λ (5 ∗ ) and set 30 via 5 ⊗ ∗ =10 / /
10. We have S (10) = 5 / / (10) = 45 . • PSL ( q ) and PSU ( q ) for q = 5 ,
7. We do not need to specify the moduleshere.We also have some labelling difficulties for Sp and G .For G (5) and G (7), there are two 77-dimensional simple modules, which welabel 77 and 77 . The first of these has highest weight label 30 and the second 02.For G (3), fix 7 and let 27 i be defined by S (7 i ) ∼ = 27 i ⊕ (5) and Sp (7) there are two simple modules of dimension 35. Themodule 35 has highest weight 21 and the module 35 has highest weight 40. ForSp (2 n ) and the Suzuki groups, fix 4 and let a generator for the outer automor-phism group cycle the 4 i , so that Λ (4 i ) ∼ = 1 / i +1 /
1. This is the one case where wedeviate from the convention below that the image under the standard Frobeniusmap of 4 i should be 4 i +1 , it is 4 i +2 in this case.When p = 2 ,
3, we will also have to consider modules for groups H ( q ) for q aproper power of p . In this case we need a notation for the modules. As in [ Cra ],and above, we give the dimension of the module, and with a subscript if there ismore than one module of that dimension. The subscript is chosen as follows:If M is a p -restricted module of dimension n , we write n for this module (and n ∗ for its dual if it is not self-dual). The image of n under the field automorphismwill be denoted n , then n , and so on. Modules that are not twists of p -restricted modules are a tensor product oftwists of p -restricted modules. If M is the tensor product of n i and m j , and n ≤ m ,then M is denoted by nm i,j , and so on for products of more than two modules.The dual of this module is again denoted by nm ∗ i,j . (It looks like it is impossible torecover n and m from this, but the dimensions of simple modules are sparse and n and m are both dimensions of simple modules.)If neither n i nor m j is self-dual, then we need to distinguish between the fourmodules n i ⊗ m j , n ∗ i ⊗ m ∗ j , n i ⊗ m ∗ j , n ∗ i ⊗ m j . We denote the first two by nm i,j and nm ∗ i,j , as stated before. The other two aredenoted by nm i,j and nm ∗ i,j respectively. Thus, for PSL (9), the module 18 denotes the tensor product 3 ⊗ ∗ , and 18 denotes the tensor product 3 ⊗ ∗ .None of our subscripts exceeds 9, and therefore we may also omit the comma inthe subscript and write 18 , for example. We will usually do so for space reasons.Although this is of course not enough to determine all modules uniquely for allgroups of Lie type, it is enough to determine all modules in this paper uniquely,except for PSL (4), where we state exactly what we mean in that section.(We could of course use highest-weight notation, which is unambiguous. How-ever, highest-weight notation suffers from a difficulty, in that it does not easilyshow you the dimension of the module, which is important to us here. Since wealso occasionally deal with groups that are not of Lie type, we prefer to keep ageneral module notation that is not tied to algebraic groups. When discussingpositive-dimensional subgroups of algebraic groups G on the other hand, we willusually stick to highest-weight notation, with the exception of the minimal and Liealgebra modules M ( G ) and L ( G ).)We occasionally deviate from these guidelines when expedient, which is whenthe presence of a centre for the simply connected group means that not all modulesare faithful. For example, in PSL (4), there are only two modules of dimension 9,swapped by the graph automorphism, and we can label them simply as 9 and 9 ∗ ,rather than as 9 and 9 ∗ . One other case is for PSL ( q ) and PSU ( q ) for q = 8 , (4) and PSU (4) we state in Chapter 6 what our notation is,as there are four simple modules for PSL (2) and PSU (2) of dimension 40, so ourabove notation cannot be used.HAPTER 6 Rank 4 groups for E E are the easiest case to consider, we do these first.This chapter considers Lie type groups of untwisted rank 4, so PSL , PSU , PΩ ,PSp , PΩ +8 , PΩ − , D , F and F . In many of these cases, there are elements oforder roughly q (but note that we might need to divide by the order of the centre).Theorem 2.4 states that if H has an element of order greater than 1312 then H isa blueprint for L ( E ), and 7 = 2401. Thus for fields of size 7, 8 and 9, there is agood chance that we can find an element of order greater than 1312.In this chapter we prove that H is one of the groups above over a field of sizeat most 9, and H ≤ G , then either H is strongly imprimitive or H ∼ = PSL (2).This latter case is significantly more difficult than the other cases dealt with inthis chapter, indeed the whole article, and is postponed until Section 12.1. In thatsection we will prove that PSL (2) is also strongly imprimitive.Throughout this chapter, and the rest of the article, u will denote a unipo-tent element of H of order p coming from the smallest class, i.e., with the largestcentralizer. (This is the class most likely to belong to a generic class of G .) PSL Proposition . Let H ∼ = PSL ( q ) for some q ≤ . (1) If q = 2 then either H stabilizes a line on L ( E ) or the composition factorsof L ( E ) ↓ H are , ∗ , , ∗ , , , ∗ , (5 , ∗ ) . (2) If q = 3 , , , , then H is a blueprint for L ( E ) . (3) If q = 4 then either H stabilizes a line on L ( E ) or H is a blueprint for L ( E ) . Proof. q = 7 , , : These contain an element of order ( q − / ( q − T ( E ) (see Definition 2.3). Hence H is a blueprint for L ( E ) by Theorem2.4. q = 5 : We use the traces of elements of order up to 13 to find conspicuous sets ofcomposition factors and obtain two possibilities for H on L ( E ), namely23 , (10 , ∗ ) , (5 , ∗ ) , , , ∗ , , ∗ , , , ∗ , , ∗ , . From Table 5.3, we see that L ( E ) ↓ H has non-positive pressure, and therefore inboth cases H stabilizes a line on L ( E ) by Proposition 2.5.In fact, we can prove more. The only extension between the composition factorsabove is between 23 and 1, so most of these factors are summands. The action ofthe unipotent element u on (10 ⊕ ∗ ) ⊕ ⊕ (5 ⊕ ∗ ) ⊕ ⊕ ⊕ and 45 ⊕ ∗ ⊕ ⊕ ∗ ⊕ ⊕ ∗ ⊕ ⊕ ∗ has Jordan blocks 2 , and 3 , , respectively.
390 6. RANK 4 GROUPS FOR E This means that u must lie in class A and 2 A respectively by [ Law95 , Table 9],both of which are generic (see also Table 2.6 to see that u must be generic). Thus H is a blueprint for L ( E ) by Lemma 2.2. q = 3 : We use the traces of elements of order up to 13 to find conspicuous sets ofcomposition factors, and obtain two conspicuous sets of composition factors for H on L ( E ), namely24 , (10 , ∗ ) , (5 , ∗ ) , , , ∗ , , ∗ , , (10 , ∗ ) , , ∗ . This yields two cases.
Case 1 : None of the composition factors has an extension with any other, so L ( E ) ↓ H is semisimple. The element u acts on L ( E ) with Jordan blocks 3 , , .Hence u comes from the generic class A (see [ Law95 , Table 9]), and therefore H is a blueprint for L ( E ) by Lemma 2.2. Case 2 : The modules 24, 45 and 45 ∗ have no extensions with other compositionfactors so split off as summands. The only extensions arise from Ext kH (30 ,
5) andExt kH (30 ,
10) (and their duals), both 1-dimensional, and so we must have two othersummands, one consisting of 30 , , ,
5, and one of 30 ∗ , ∗ , ∗ , ∗ .Thus 45 ⊕ ∗ ⊕ ⊕ is a summand of L ( E ) ↓ H , with u acting on this modulewith blocks 3 , , . The two remaining summands are dual to one another, sothe remaining blocks in the action of u on L ( E ) must come in pairs. The onlypossibility in [ Law95 , Table 9] is the generic class 2 A , and hence H is a blueprintfor L ( E ) by Lemma 2.2 again. q = 2 : There are three conspicuous sets of composition factors for L ( E ) ↓ H :24 , (10 , ∗ ) , (5 , ∗ ) , , , , ∗ , (10 , ∗ ) , (5 , ∗ ) , , , ∗ , , ∗ , , , ∗ , (5 , ∗ ) . (The second of these is an artefact of the fact that there is a non-rational class ofelements of order 5 in G with trace 2 on L ( E ), so ‘looks’ rational. The second setof factors therefore cannot exist.)From Table 5.1, 40 , 40 ∗ and 74 are the only composition factors with non-zero1-cohomology. Thus the first two cases have negative pressure (and the second doesnot exist), and hence H stabilizes a line on L ( E ) in the first case by Proposition2.5. Thus we are left with the third case, as claimed. q = 4 : Up to field automorphism there are three sets of composition factors for L ( E ) ↓ H that are conspicuous for elements of order up to 11:24 , (10 , ∗ ) , (5 , ∗ ) , , , , ∗ , , , , ∗ , , , , ∗ , (5 , ∗ ) , , , ∗ , , , , ∗ , , , , where 40 ± ,i are the four 2-restricted, 40-dimensional modules— L (1010), L (0101), L (0011) and L (1100)—and up to field automorphism the four 50-dimensional mod-ules are L (2100), L (0012), L (1020) and L (0201), consistent with our notation fromChapter 5. Case 1 : None of 24 , 10 ± and 5 ± has non-zero 1-cohomology, so H has negativepressure, and thus stabilizes a line on L ( E ) by Proposition 2.5. .2. PSU Cases 2 and 3 : Let x ∈ H have order 315. In these cases, x , x , x and x ,of orders 7, 35, 105 and 315 are all uniquely determined up to conjugacy by theireigenvalues on L ( E ), by a computer check using the roots trick (see Chapter 4).We find a copy of H inside the A A maximal-rank subgroup acting with thesecomposition factors (in the second case the actions of H on each M ( A ) factor are L (0002) and L (0001), and in the third case H acts as L (0001) on both factors), andit clearly lies inside a copy of SL (16). We therefore find, in SL (16), an element oforder 4095 = 315 ×
13 that powers to our element in H and with the same numberof distinct eigenvalues—171 in the second case and 149 in the third—as the elementof order 315 on L ( E ). This proves that x , and hence H , is a blueprint for L ( E ),as claimed. (cid:3) PSU We continue with our definition of u from the start of the chapter. Proposition . Let H ∼ = PSU ( q ) for some q ≤ . (1) If q = 2 then either H stabilizes a line on L ( E ) or H stabilizes a -space of L ( E ) stabilized by a positive-dimensional subgroup of G , and inaddition H is strongly imprimitive. (2) If q = 3 , , , , , then H is a blueprint for L ( E ) . Proof. q = 7 , , : Note that PSU (7) contains elements of order 2400, thegroup PSU (8) contains elements of order 3641, and PSU (9) contains elements oforder 1181. None of these lies in T ( E ) (see Definition 2.3), so if H ∼ = PSU ( q ) for q = 7 , , H is a blueprint for L ( E ) by Theorem 2.4. q = 5 : There are two sets of composition factors for L ( E ) ↓ H conspicuous forelements of order at most 13, which are23 , (10 , ∗ ) , (5 , ∗ ) , , , ∗ , , ∗ , , , ∗ , , ∗ , . There is an extension between 23 and 1, but there are no other extensions betweencomposition factors of L ( E ) ↓ H , so in both cases H stabilizes a line on L ( E ),and moreover most of the module is semisimple. The action of u on the summandconsisting of all factors that are not 23 or 1 has blocks 2 , and 3 , , respectively. Thus u comes from class A in the first case, and 3 A in the second,as we can see from [ Law95 , Table 9]. (See Table 2.6 to see that they are generic,at least.) These two classes are generic, so H is a blueprint for L ( E ) by Lemma2.2. q = 3 : We use the traces of elements of order up to 10 to find conspicuous sets ofcomposition factors, and obtain two sets of factors:24 , (10 , ∗ ) , (5 , ∗ ) , , , ∗ , , ∗ , , (10 , ∗ ) , , ∗ . Case 1 : There are no extensions between the composition factors of L ( E ) ↓ H , so L ( E ) ↓ H is semisimple; then u acts on L ( E ) with blocks 3 , , , so comes fromclass A , which is generic. Thus H is a blueprint for L ( E ) by Lemma 2.2. Case 2 : The extensions are slightly different to the PSL (3) case: the 45- and24-dimensional factors have no extensions with other composition factors, so breakoff as summands, as before. However, this time the only extensions are between5 and 30, 10 and 30, and 30 and 30 ∗ (and their duals), so we do not necessarilyobtain two separate summands. E The action of u on 45 ⊕ ∗ ⊕ ⊕ is 3 , , . If u comes from the genericclass 2 A then H is a blueprint for L ( E ) by Lemma 2.2, so we assume not. Ex-amining [ Law95 , Table 9] (or Table 2.4), we see that u comes from class 3 A ,acting with Jordan blocks 3 , , . Thus u acts on the remaining summand V of L ( E ) ↓ H —with composition factors 30 , ∗ , (10 , ∗ ) , , ∗ —with Jordan blocks3 , , . Suppose that 30 / ∗ or its image under the graph automorphism is nota subquotient of V : then as in the case of PSL (3), V breaks up as two dual sum-mands, V ⊕ V , but then we cannot have thirteen blocks of size 1, a contradiction.Thus (without loss of generality) 30 / ∗ is a subquotient of V .We now construct a module W by first building the largest submodule of P (30 ∗ )with 30 / ∗ as a submodule and whose quotient has composition factors only from5 , ∗ , , ∗ , yielding the module 5 ∗ , ∗ , / ∗ . We then take the dual of this, so30 / , , ∗ , and perform the same action, placing as many copies of 5 , ∗ , , ∗ on top of this module as we can, while remaining a submodule of P (5 ⊕ ⊕ ∗ ).This yields the module W = 5 ∗ , ∗ , / , , ∗ . A pyx for V is a sum of W and a semisimple module consisting of copies of 10 ± and 5 ± , on which u acts with blocks 2 , and 2 , respectively. As u acts on W with Jordan blocks 3 , , , it is not possible to produce the 23 blocks of size 3needed for the action of u on V , which is a contradiction. Thus u comes from thegeneric class 2 A , as needed. q = 2 : There are two conspicuous sets of composition factors for L ( E ) ↓ H :24 , (10 , ∗ ) , (5 , ∗ ) , , , ∗ , , ∗ , , , ∗ , (5 , ∗ ) . The first has pressure −
14 and hence H stabilizes a line on L ( E ) by Proposition2.5, so we concentrate on the second case. Let L be a copy of 3 × SU (2) inside H :the trace of the central element z ∈ L is 14 on L ( E ), and so C G ( z ) has type D T (see Table 2.10). Thus L ′ ∼ = SU (2) lies inside D . The composition factors of D on L ( E ) are two copies of M ( D ), one of L ( D T ) (which in characteristic 2 hasthe form 1 / /
1, but this is not important), and one copy of each of the half-spinmodules. The 1-eigenspace of the action of z on L ( E ) must be the 1 / /
1, so wecan compute which actions of L ′ on M ( D ) yield the correct composition factorson L ( D T ) and on L ( E ). The composition factors 14 and 6 , , ∗ for M ( D ) ↓ L ′ yield the correct composition factors for L ( D T ) ↓ L ′ , namely20 , ∗ , , , , ∗ , . However, the trace of elements of order 3 in L ′ on L ( E ) does not match the casewhere M ( D ) ↓ L ′ = 14, so the composition factors are 6 , , ∗ . There is a module4 / / ∗ , but the symmetric square of this module does not have a trivial submodule,whence L ′ does not lie in D with this module by Lemma 2.10. The only otherself-dual module with those composition factors is the semisimple module 6 ⊕ ⊕ ∗ .In particular, this means that L ′ lies inside the D A -Levi subgroup of D ,and this can be conjugated to lie inside the A -Levi subgroup of E . (This is sowe can more easily understand the action of L ′ on the half-spin modules; it is verypossible to do this directly.) It’s even easier to consider L ′ inside A , which actson L ( E ) with summands Λ ( M ( A )), Λ ( M ( A ) ∗ ) and M ( A ) ⊗ M ( A ) ∗ minusa trivial summand. The action of L ′ on M ( A ) is either 4 ⊕ ⊕ ⊕ ∗ ⊕ L ( E ). Embedding L ′ into a copy X of A acting as .2. PSU L (100) ⊕ L (100) ⊕ L (000) or L (100) ⊕ L (001) ⊕ L (000), in either case the action of X on L ( E ) is(000 / / ⊕ ⊕ (010 / / ⊕ (010 / / ⊕ (100 ⊕ ⊕ ⊕ (010) ⊕ ⊕ (110 ⊕ ⊕ . (Here we have suppressed the ‘ L ( − )’ to save space. The module 000 / /
000 isjust L (100) ⊗ L (001), and the module 010 / /
010 is L (100) ⊗ .)The reason for writing this like this is that we see that L ′ acts in exactly thesame way except that the 200 and 002 in the second socle layer are isomorphic to 001and 100, so in particular any semisimple submodule of L ( E ) ↓ L ′ is stabilized by X .If there is a submodule of L ( E ) ↓ H that restricts to L ′ (equivalently L ) semisimplythen we have proved that H is not Lie primitive, and since our condition is invariantunder automorphisms, strongly imprimitive via Lemma 3.9.The restrictions to L ′ of 5, 5 ∗ , 10, 10 ∗ , 40 and 40 ∗ are all semisimple, andwe in fact show that there is a submodule 10 of L ( E ) ↓ H . The module 10 hasextensions with 5, 24, 40 and 40 ∗ , so we show that none of these extensions isallowed in L ( E ) ↓ H . The restriction of 10 /
24 to L ′ is4 ∗ / / , / , , ∗ , which obviously is not a subquotient of L ( E ) ↓ L ′ . The restrictions of 10 / and10 / ∗ to L ′ are (4 ∗ / ⊕ (6 / ⊕ , (4 ∗ / ⊕ (6 / ⊕ ∗ , and since 6 /
20 and 6 /
14 are not subquotients of L ( E ) ↓ L ′ , these extensions cannotappear in L ( E ) ↓ H either. Finally the restriction of 10 / L ′ is1 ⊕ ∗ ⊕ (6 / , which is a subquotient of L ( E ) ↓ L ′ . However, the restriction of 40 ± to L ′ alreadycontains a summand 6 / ± /
6, so there cannot be more than one subquotient 6 /
4, andtherefore 10 / L ( E ) ↓ H either. Since the only compositionfactors of L ( E ) ↓ H that 10 has an extension with are 5, 24, 40 and 40 ∗ , we haveproved that 10 is actually a summand of L ( E ) ↓ H , and this is stabilized by X , asneeded. q = 4 : The simple modules for H of dimension at most 124, or self-dual and ofdimension at most 248, are of dimension 1, 24 (two of these), 50 (four) and 74 (two).There is a single set of composition factors for L ( E ) ↓ H conspicuous for elementsof order 3 and 5, and it is the set of all the simple modules of dimension 24 or 50,i.e., 50 , ∗ , , ∗ , , , where 50 ± i are the four simple modules for PSU (4) of that dimension. (Suchan embedding exists in the A A maximal-rank subgroup, and we also saw it forPSL (4).)We first show that L ( E ) ↓ H possesses a submodule of dimension 24. Notethat 24 has a single extension with 50 ± , and none with 50 ± , for some labellingfor the modules of dimension 50. However, one cannot construct a module of theform 50 / / ∗ or 50 ∗ / / , and so H must always stabilize a 24-space on L ( E ). (To see this last step, if the socle is 50 ⊕ ± then it is clear, and if it is E simple, remove the socle and top to leave a module that must have 24 ⊕ as asubmodule.) Hence 24 ⊕ is in fact a summand of L ( E ) ↓ H .Let x be an element of order 65. By a computer check, the class of x in E isdetermined by its eigenvalues on L ( E ). Next, we check the 6560 possible elementsof order 195 = 65 × x inside the normalizer of a torus, and findthat there are four elements, ˆ x , ˆ x , ˆ x , ˆ x , of order 195, such that ˆ x i = x and withexactly 91 distinct eigenvalues on L ( E ). (This is the minimal number for a rootof x .) Two of these stabilize all eigenspaces of x on 24 , and two of them stabilizeall eigenspaces of x on 24 . Therefore we see that if either 24 i is a submoduleof L ( E ) ↓ H (and we know that both are from above) then H is contained in thestabilizer of these 24-spaces, which contains elements of order 195 and a copy ofPSU (4). Thus H is contained in a member of X by Lemma 3.10, and in a memberof X σ if k contains F , as otherwise the 24 i might not be defined.We check now that H is a blueprint for L ( E ), completing the proof. Let X bea member of X containing H . The dimensions of the composition factors of X on L ( E ) must be coarser than 50 , , and this eliminates all parabolic subgroups, E A and D (as they have trivial factors on L ( E ) in characteristic 2), A (hasfactors of dimension 84 , A E (has a factor of dimension 8) and F G (hasa factor of dimension 14), leaving the A A maximal-rank subgroup, which hasfactors of dimension 50 , . As this A A is semisimple, so is L ( E ) ↓ H , and X and H stabilize the same subspaces of L ( E ), as needed. (cid:3) Ω The following proposition is easy, assuming that we have shown that Ω ( q ) isa blueprint for L ( E ), which we do in Proposition 7.3. Proposition . Let H ∼ = Ω ( q ) for some odd q ≤ . Then H is a blueprintfor L ( E ) . Proof. As L ∼ = Ω ( q ) is contained in H , and L is a blueprint for L ( E ) byProposition 7.3, so is H . This completes the proof. (cid:3) PSp We continue with our definition of u from the start of the chapter. Proposition . Let H ∼ = PSp ( q ) for some q ≤ . (1) If q = 2 then either H stabilizes a line on L ( E ) , or H is contained in an A -parabolic subgroup of G and is strongly imprimitive. (2) If q = 3 , , , , then H is a blueprint for L ( E ) . (3) If q = 4 then H stabilizes a line on L ( E ) . Proof. q = 7 , , : The group PSp (7) contains elements of order 1201,PSp (8) contains elements of order 4097, and PSp (9) contains elements of or-der 3281. None of these lies in T ( E ) (see Definition 2.3), so H is a blueprint for L ( E ) by Theorem 2.4. q = 5 : The only set of composition factors for L ( E ) ↓ H that is conspicuous forelements of orders 2 and 3 is 42 , , , , and there are no extensions betweenany of the composition factors, so L ( E ) ↓ H must be semisimple. Furthermore, theunipotent element u acts on this module with Jordan blocks 3 , , , so lies inthe generic class A . Thus H is a blueprint for L ( E ) by Lemma 2.2. .4. PSp q = 3 : There is a unique set of composition factors for L ( E ) ↓ H that is conspicuousfor elements of orders 2 and 4, which is41 , , , . This has at least seven trivial summands as the only extension between compositionfactors is between 41 and 1.To show that H is a blueprint for L ( E ), note that 36 ⊕ ⊕ ⊕ ⊕ must bea summand of L ( E ) ↓ H and an element u from the smallest unipotent class of H acts on this with Jordan blocks 3 , , . There is no non-generic class with thismany trivial summands from Table 2.4 so u is generic, and indeed u must lie inclass A by [ Law95 , Table 9]. (In fact, using other unipotent classes, one can checkthat the only embedding must be36 ⊕ ⊕ ⊕ (1 / / ⊕ ⊕ , as the semisimple possibility has unipotent elements acting with blocks that do notoccur in [ Law95 , Table 9].)Thus H is a blueprint for L ( E ) by Lemma 2.2. q = 2 : There are five conspicuous sets of composition factors for L ( E ) ↓ H :246 , , , , , , , , , , , , , , , , , , . Cases 1, 4 and 5 : The first case does not occur by Corollary 3.6. The modules 48,26 and 8 each have 1-dimensional 1-cohomology, with 16 and 1 having none (seeTable 5.1), and so the fourth and fifth set of composition factors have pressures − −
21 respectively. Thus H stabilizes a line on L ( E ) in these cases byProposition 2.5. Case 2 : By Table 5.1, we see that L ( E ) ↓ H has pressure 1, so its socle must beeither 26 or 8, if we assume that H does not stabilize a line on L ( E ). Since 128 hasno extensions with the other composition factors, it must split off as a summand.The other summand W must have the form8 / / / / / / / / / / / / / / / / . Suppose first that the socle of W is 26, and we construct a pyx for W in theobvious way: add all copies of 1, then 8, then 1, then 48 on top of 26. Already atthis stage the result of this is 1 / , , / , , / , so a pyx cannot be built. Similarly, we reverse the roles of 8 and 26, and againattempt to build a pyx. This time we start with 8, then add all copies of 1, then26, then 1, then 48, then 1, and finally 26. This yields a module1 , / , , / , , / , / , / H stabilizes a line on L ( E ), as needed. Case 3 : Assume that H does not stabilize a line on L ( E ), and that H does notlie in an A -parabolic subgroup. We will show that neither 26 nor 48 may lie inthe socle of L ( E ) ↓ H , so that it must be 8 ⊕ i for some i .Let L denote a copy of SO +8 (2) in H , and let M denote a copy of Sym(10) in H , arranged so that L ∩ M ∼ = Sym(9). Note that each of 1, 8, 26 and 48, the simplemodules for H , restrict irreducibly to each of L , M and L ∩ M . By Proposition E Cra17 , Section 10], we know that L and M both stabilize lineson L ( E ), and hence lie inside members of X . We claim that both L and M lieinside (different) A -parabolic subgroups of G . (If they were the same A -parabolicsubgroup, then H = h L, M i also lies in it.)We prove that any copy of L , M or H with these composition factors thatlies inside a member of X lies inside an A -parabolic subgroup, so assume thatone of these does not. Notice that L , M and H all require at least rank 4 for anovergroup, and do not embed in a subgroup of A or C , so if those factors appearin X in X , they may be removed. Thus we reduce to the case where, if X is aparabolic subgroup of G , then it is of type either E or D . If H , L or M lies insidea D -parabolic subgroup then it must act on M ( D ) with factors 8 , , but M ( D )appears twice in L ( E ) ↓ D , so H cannot embed in the D -parabolic subgroup withthese factors.If H , L or M embeds in the E -parabolic subgroup, then the image in the E -Levi subgroup must embed inside a proper positive-dimensional subgroup of E by Theorem 3.1 and the remarks after for H and L , and [ Lit18 , Theorem 1]for M . This proper positive-dimensional subgroup can be the A maximal-ranksubgroup, with our subgroup acting irreducibly on M ( A ). But this A acts on L ( E ) with, up to duality, four copies of Λ ( M ( A )) and one of L ( A ). Togetherthey contribute six copies of 26, which is too many. Our subgroup can embed inthe D -parabolic, but then this lies in the D -parabolic above, so we embed in the E -parabolic subgroup. But this acts on L ( E ) with (up to duality) six copies of M ( E ), and no non-trivial composition factor appears that often.Thus H , L or M embeds in a maximal-rank subgroup, and it suffices to ruleout A and D as options. If X = A then H , L or M stabilize either a line orhyperplane on M ( A ), so we lie in an A -parabolic, and if X = D then it embedsas the sum of two 8s on M ( D ) (else we lie in a D -parabolic), but this lies in an A -parabolic again, and we are done.Therefore L and M lie in different parabolics, so that L ∩ M lies in the inter-section of two parabolics, which are described in [ Car85 , Section 2.8] using doublecosets in the Weyl group. However, in this case L ∩ M must act irreducibly on M ( A ), so L ∩ M projects irreducibly onto the Levi factor of the A -parabolicsubgroup. Hence the subgroup containing L ∩ M must have A -Levi factor, andtherefore the two parabolics involved are either the same (explicitly ruled out) oropposite (as the corresponding double coset must lie in the normalizer of W ( A )inside W ( E ), and this is 2 × Sym(8)); in this latter case, L ∩ M lies in the A -Levisubgroup. In particular, L ∩ M has a known action on L ( E ) via A , given by L ( λ ) ⊕ L ( λ ) ⊕ L ( λ ) ⊕ L ( λ ) ⊕ L ( λ ) ⊕ L ( λ ) ⊕ ( L ( λ ) ⊗ L ( λ )) , and this restricts to L ∩ M as8 ⊕ ⊕ ⊕ ⊕ (1 / / ⊕ ⊕ (1 / / / / / . The Sp (2) subgroup of the A -Levi subgroup containing L ∩ M acts as8 ⊕ ⊕ ⊕ ⊕ (1 / / ⊕ ⊕ (1 / / / / / / , and the C subgroup of the A acts similarly, as8 ⊕ ⊕ ⊕ ⊕ (1 / / ⊕ ⊕ (1 / / / σ / / / , .4. PSp where 8 σ denotes the image of the natural module under the Frobenius automor-phism, so L (2 λ ), where M ( C ) = L ( λ ).Therefore we see the following facts:(1) 26 cannot be a composition factor of soc( L ( E ) ↓ H ), as it does not appearin soc( L ( E ) ↓ L ∩ M );(2) if 48 lies in soc( L ( E ) ↓ H ) then its stabilizer in G is positive dimensional,containing C , and this subgroup stabilizes all 48-dimensional simple sub-modules of L ( E ) ↓ H ;(3) if 8 ⊕ lies in soc( L ( E ) ↓ H ) then one of the 8s in this socle must also bestabilized by the C subgroup, so its stabilizer is positive dimensional.(We cannot say anything about the orbit of 8-dimensional submodules, sowe only obtain that H is contained in a member of X . However, we havealready shown that this means that H is contained in an A -parabolicsubgroup.)Thus in order to not satisfy any of these criteria, soc( L ( E ) ↓ H ) is exactly 8.Recall that M stabilizes a line on L ( E ) by [ Cra17 , Section 10], and thereforeby Frobenius reciprocity there is a map from the permutation module P M = 1 M ↑ H to L ( E ) ↓ H . We thus consider P M , which has dimension 13056. The quotient bythe { , , , } -residual of this is much smaller, and it and its dual have soclelayers 1 / / / / , / , / , / , / , / / , / / . We cannot have a copy of 26 in the second socle layer of L ( E ) ↓ H without H stabilizing a line on L ( E ), by the structures above, so with this condition weobtain a unique quotient 1 / / / / / L ( E ) ↓ H .We therefore try to add modules on top of this to build a pyx, and look for acontradiction. Let W denote this submodule.We have Ext kH (1 , W ) = 0 and Ext kH (48 , W ) = 0, so 1 and 48 cannot lie inthe socle of the quotient module L ( E ) ↓ H /W . We also have Ext kH (26 , W ) ∼ = k ,with the extension being 1 / / / / , / , which is not allowed, as we saw above. Thus the socle of the quotient module L ( E ) ↓ H /W must consist solely of 8s. There is an extension with submodule 8 andquotient W , yielding a module 1 / , / / / / kH (48 ,
8) = 0 so it cannot have anextension with 48 either. Hence there must be an extension with 26. There is nowa 2-dimensional Ext kH -group, yielding the module1 , / , / / / , / L ∩ M results in both of the new 26s splitting off assummands, so since L ( E ) ↓ L ∩ M has no 26 submodule, neither of these can occurin L ( E ) ↓ H .Hence we cannot construct L ( E ) ↓ H with a single 8 in the socle. This provesthat H is contained in a member of X , and as we saw above, this implies that H lies in an A -parabolic subgroup, as claimed. E To obtain the statement that H is strongly imprimitive, we simply applyLemma 3.7, which states that if H is non- G -cr, then H is strongly imprimitive.(In fact, since Out( H ) = 1 we only need σ -stability to obtain strong imprimitiv-ity.) If H lies in the A -Levi subgroup though, H stabilizes a line on L ( E ) and isstrongly imprimitive again, this time by Lemma 3.5. q = 4 : The modules of dimension at most 248 with non-zero 1-cohomology havedimension 8, 26 and 246. Letting L = Sp (2) ≤ H , we see that the compositionfactors of 8 ⊗ L are 26 , , , while 8 ⊗
16 = 128 and 8 ⊗
26 = 48 ⊕ L ( E ) ↓ H have dimension 1, 8, 16, 26, 48 and 64, withonly the last of these restricting reducibly to L .Note that, for H , although 8 i and 26 i still have 1-dimensional 1-cohomology,48 i no longer does, and 64 does not either. If L ( E ) ↓ H has positive pressure, then L ( E ) ↓ L must have factors 128 , , , , or 48 , , , , and so L ( E ) ↓ H has composition factors of dimension one of the following:128 , , , , , , , , , , , , , , , . If the factors are 128 , , , L ( E ) ↓ H is semisimple, and the restriction to L is determined uniquely. The action of u on this module has Jordan blocks 2 , ,which does not appear in [ Law95 , Table 9]. Thus this case cannot occur.The only set of composition factors with either of the other two multisets ofdimensions that are conspicuous for elements of order at most 9 is48 , , , , , up to field automorphism. The only extensions between composition factors in L ( E ) ↓ H arise from Ext kH (1 , i ), Ext kH (1 , ) and Ext kH (26 , ), so L ( E ) ↓ H has pressure 1. Suppose that H does not stabilize a line on L ( E ). Thus, quoti-enting out by any 48 in the socle, we may assume that the socle is either 8 or26 .The { , , , , } -radical of P (8 ) is8 / , / , / , , / , , / , / / / . Hence the socle cannot be 8 . Similarly, the { , , , , } -radical of P (26 )is 8 / , / , / , , / , , / , / , so the socle cannot be 26 either. This completes the proof that H stabilizes a lineon L ( E ). (cid:3) Ω +8 We continue with our definition of u from the start of the chapter. Proposition . Let H ∼ = PΩ +8 ( q ) for some q ≤ . (1) If q = 2 , then H stabilizes a line on L ( E ) . (2) if q = 3 , , , , then H is a blueprint for L ( E ) . Proof. q = 3 , , , : In these cases Ω ( q ) is contained in H , and Ω ( q ) is ablueprint for L ( E ) by Proposition 7.3 below, hence H is. q = 8 : In this case H contains an element of order 8 − T ( E ),so H is a blueprint for L ( E ) by Theorem 2.4. .5. Ω +8 q = 2 : There are (up to automorphism) five conspicuous sets of composition factorsfor L ( E ) ↓ H : 26 , , , , , , , , , , , , , , , , , , , , , , , . The pressures of the five cases are −
28, 0, 2, 3 and − Cases 1, 2 and 5 : The fifth case cannot occur by Corollary 3.6, and the first twocases lead to H stabilizing a line on L ( E ) by Proposition 2.5. Case 3 : Suppose that L ( E ) ↓ H has no trivial submodule, and let W denote the { , } -heart of L ( E ) ↓ H . The socle of W must be 26 or 48 (as 26 has 2-dimensional 1-cohomology, and so the socle cannot be 26 ⊕ , and if it were 48 ⊕ then this would be the entirety of W ).Suppose first that soc( W ) = 48 . The { , , } -radical of P (48 ) is1 / / , , / / , / , which does not have eight trivial factors. Since this is a pyx for W —with the 48 at the top removed—we obtain a contradiction. Thus soc( W ) = 26.The { , , , } -radical of P (26) is1 / / , , , / , / , / , / , , / W . This pyx has eight trivial composition factors buthas a trivial quotient. Any trivial quotient of W becomes a trivial quotient of L ( E ) ↓ H , so W cannot have eight trivial composition factors in the third case,another contradiction. Thus H stabilizes a line on L ( E ). Case 4 : Suppose that H does not stabilize a line on L ( E ). There are three conju-gacy classes of subgroups of H isomorphic with Alt(9), and for exactly one of themthe restriction of 48 i to it remains irreducible. Let L i denote an Alt(9) subgroupsuch that 48 i remains irreducible on restriction to L i . Since L ( E ) ↓ L i has a trivialsubmodule by [ Cra17 , Proposition 9.2], there is a non-trivial homomorphism fromthe permutation module P L i of H on L i to L ( E ) ↓ H .The module P L i has dimension 960, but the quotient by the { , i , , i } -residual has dimension 302. The quotient modulo its second radical layer is the(unique up to isomorphism) uniserial module 1 / i , and so 1 / i must be a sub-quotient of L ( E ) ↓ H as H does not stabilize a line on L ( E ). This shows that 48 i cannot lie in the socle of L ( E ) ↓ H , since then it would be a summand.Let W denote the quotient of L ( E ) ↓ H by its { , i , } -radical. This has socle48 ⊕ ⊕ , and so since each L i must stabilize a line on L ( E ), we actuallysee that W must have the submodule(1 / ) ⊕ (1 / ) ⊕ (1 / ) . From here it is easy to obtain a contradiction, and therefore see that H stabilizes aline on L ( E ). The most obvious way is to recall that L ( E ) is self-dual, and hencethere is also a subquotient 48 i /
1, so we need at least six trivial composition factorsin L ( E ) ↓ H , which is a contradiction. q = 4 : Suppose that H does not stabilize a line on L ( E ). By restricting to L = Ω +8 (2), we see that there are no composition factors of dimension 160 or 208(this is 8 i ⊗
26, which restricts to L as 48 i ⊕ i ), or 246. Thus we may assume thatthe factors of L ( E ) ↓ H are of dimensions 1, 8, 26, 48 and 64. As H ( H, i ) has E dimension 2 and all other simple modules have zero 1-cohomology, we need morethan half as many 26-dimensional composition factors as 1s in L ( E ) ↓ H . This meanswe need 64s in L ( E ) ↓ H , otherwise the dimensions of the factors of L ( E ) ↓ H matchthose of L ( E ) ↓ L , and we are done.For L , 8 ⊗ has factors 26 , , , and 8 ⊗ has factors 48 , . Hence itis easy to see that we need to have no trivial factors or 26-dimensional factors in L ( E ) ↓ H , and so the dimensions of the factors are one of64 , , , , , , , , . However, none of these is conspicuous for elements of order at most 5, and so H must always stabilize a line on L ( E ). (cid:3) Ω − We now turn to PΩ − ( q ), where we use the same embedding and element orderas for PΩ +8 ( q ) to deal with the cases q = 3 , , , ,
9. We continue with our definitionof u from the start of the chapter. Proposition . Let H ∼ = PΩ − ( q ) for some q ≤ . (1) If q = 2 , then H stabilizes a line on L ( E ) . (2) if q = 3 , , , , then H is a blueprint for L ( E ) . Proof. q = 3 , , , : In these cases Ω ( q ) is contained in H , and Ω ( q ) is ablueprint for L ( E ) by Proposition 7.3, hence H is. q = 8 : In this case H contains an element of order 8 − T ( E ), so H is a blueprint for L ( E ) by Theorem 2.4. q = 2 : There are six conspicuous sets of composition factors up to automorphism,given by 26 , , , , , , , , , , , , , , , , , , , , , , , , , , , , . As 26 has 2-dimensional 1-cohomology, and 48 has 1-dimensional 1-cohomology(see Table 5.1), in all but the third case H must stabilize a line on L ( E ) by pressurearguments (Proposition 2.5). (The last case does not occur by Corollary 3.6, andwhile the fifth case has pressure 1, it has a composition factor with 2-dimensional1-cohomology.)In the third case, suppose that H does not stabilize a line on L ( E ). Thepressure is 2 so the socle of L ( E ) ↓ H , modulo the { } -radical, is either 26 or 48 .Let W denote the quotient by the { } -radical. The { , , , } -radical of P (26)is 1 , / , / , / , / , , / , and if soc( W ) = 26 then this must be a pyx for W . But this module does nothave eight trivial factors. Thus 26 cannot be the socle. On the other hand, the { , , } -radical of P (48 ) is1 / / , , / / , / . Again, if soc( W ) = 48 then this must be a pyx for W , and does not have enoughtrivial factors either. Hence 48 cannot be the socle. Thus H stabilizes a line on L ( E ), as needed. .7. D q = 4 : This is similar to the proof for Ω +8 (4), so assume that H does not stabilizea line on L ( E ). The simple modules of dimension at most 64 that have non-zero1-cohomology are of dimension 26, with 2-dimensional 1-cohomology. Thus, justas for Ω +8 (4), we see that the possible dimensions of the composition factors of L ( E ) ↓ H are 64 , , , , , , , , . As with Ω +8 (4), there are no sets of composition factors that are conspicuous forelements of order at most 9, so H must stabilize a line in this case as well, completingthe proof. (cid:3) D We continue with our definition of u from the start of the chapter. Proposition . Let H ∼ = D ( q ) for some q ≤ . (1) If q = 2 , then H stabilizes a line on L ( E ) . (2) if q = 3 , , , , then H is a blueprint for L ( E ) . Proof. q = 7 , , : The group D ( q ) contains elements of order Φ ( q ) = q − q + 1, which for q = 7 , , T ( E ). Thus H is a blueprint for L ( E ) by Theorem 2.4. q = 5 : The group H contains G (5), which is a blueprint for L ( E ) by Proposition8.6 below. q = 3 : The simple modules of dimension at most 248 for H are 1, 8 i , 28, 35 i , 56 i ,104 i and 224 j , where 1 ≤ i ≤ ≤ j ≤
6. There are six sets of compositionfactors that are conspicuous for elements of order at most 13, four up to fieldautomorphism, and these are56 , , , , , , , , , , , . , , , , , , , , , . Each of these is semisimple because there are no extensions between compositionfactors. The unipotent element u acts on L ( E ) with blocks 3 , , in the firstthree cases and 3 , , in the last case. This means that u lies in class A or2 A by [ Law95 , Table 9], both of which are generic. Thus H is a blueprint for L ( E ) by Lemma 2.2. q = 2 : There are up to field automorphism five conspicuous sets of compositionfactors for L ( E ) ↓ H :246 , , , , , , , , , , , , , , , , , , , , , , , , . Each of these has non-positive pressure (see Table 5.1), so H stabilizes a line on L ( E ) by Proposition 2.5, as needed. (The first case cannot occur by Corollary3.6.) q = 4 : Assume that H does not stabilize a line on L ( E ). Let L be the subgroup D (2) of H . As with Ω +8 (2), we use the restriction to L to determine the factorsof L ( E ) ↓ H .From Table 5.5 we have H ( H, M ) = 0 if dim( M ) = 1 , ,
48 and H ( H, i )is 2-dimensional. Hence if the composition factors involved in L ( E ) ↓ H restrictirreducibly to L then L ( E ) ↓ H has non-positive pressure, so has a trivial submodule. E For L , we have 8 ⊗ = 1 / / , , / /
1, and 8 ⊗ = 8 / / , and the64-dimensional modules for H have zero 1-cohomology, so in order for H to notstabilize a line on L ( E ) there must be no trivial composition factors. Thus thedimensions of the composition factors of L ( E ) ↓ H are64 , , , , , , , , . (This should look familiar from Ω ± (4).) None of the 185868 such possibilities for L ( E ) ↓ H is conspicuous for elements of order 5, so H must stabilize a line on L ( E ),as needed. (cid:3) The current version of Magma (V2.24) at the time of writing will not directlycompute inside D (4) as it is given, so we explain how to coerce it to do so.Assume that we implement D (4) using ChevalleyGroup("3D",4,4) : this isgenerated by x of order 63 and y of order 12. If one tries to compute Ext betweensimple modules using Ext it will return an error, but instead implement D (4) asfollows: G1:=ChevalleyGroup("3D",4,4);M:=CompositionFactors(ExteriorSquare(GModule(G1)))[2];G:=WriteOverSmallerField(sub
For this copy of G one may now use Ext between modules.However, one still cannot use the command
ConjugacyClasses on G . Thus wetake a specific element, namely g:=(xy)^9 , where x and y are the generators of thegroup as given in Magma.This element has order 5: it has a 4-dimensional 1-eigenspace on each 8-dimensional simple module, and a 16-dimensional 1-eigenspace on all 48- and 64-dimensional simple modules. From the dimensions above, this means that the1-eigenspace of g on L ( E ) has dimension 76 or 84. However, from the list ofsemisimple classes of E , we find that the dimension of the 1-eigenspace is one of48, 52, 54, 64, 68, 82, 92 and 134. This contradiction means that there is no suchconspicuous set of composition factors. F and F Let H ∼ = F ( q ) for some 2 ≤ q ≤
9. We also require H ∼ = F ( q ) for q = 2 , u from the start of the chapter. Proposition . Let H ∼ = F ( q ) for some q ≤ , or H ∼ = F ( q ) ′ for some q = 2 , . (1) If q = 2 , , then H stabilizes a line on L ( E ) . (2) If q = 3 , , , then H is a blueprint for L ( E ) . q = 7 , , : There exists an element of order Φ ( q ) = q + 1 in H , and so if q ≥ H a blueprint for L ( E ) for q = 7 , Proof. q = 2 , , : The simple modules for H of dimension at most 248 havedimensions 1, 26 and 246. If there is a 246 then the factors are 246 , , and H cannot occur by Corollary 3.6. .8. F AND F Thus there are at least fourteen trivial composition factors and at most nine26-dimensional factors, since 248 ≡
14 mod 26. (Using a rational element of order5, which exists in H and always has trace +1 on any simple module of dimension26, and trace − , ,
23 on L ( E ), we see that 26 , is the only possibility, but wedo not need this.)If the 1-cohomology of 26-dimensional modules is at most 1 then H has negativepressure and we are done. For H ∼ = F (2) ′ this is an easy computer calculation,for F ( q ) for q = 2 , , JP76 , Section 6], and this leaves F (8). Magmacurrently cannot compute this 1-cohomology, so we need to show directly that H is a blueprint for L ( E ).Up to field automorphism, the unique set of composition factors conspicuousfor an element of order 7 (a torus element, a generator of H in the Magma imple-mentation) is 26 , , . In the Magma implementation of H , the product g ofthe two generators has order 91, and has 47 distinct eigenvalues on L ( E ).The traces of semisimple elements of G of order 13 are known, and a conjugacyclass is determined by its eigenvalues on L ( E ). Fixing an element g ′ in the group X = ( C − ) ⋊ W for W the Weyl group of type E with the same eigenvalueson L ( E ) as g , we check the 5764801 elements of order 91 in X whose seventhpower is g ′ , find 120 that have the same eigenvalues as g , and check that theyare all conjugate in X , so let g ′′ denote one of them. Thus g is determined up toconjugacy by its eigenvalues on L ( E ). Furthermore, we can check the 6561 rootsof g ′′ of order 91 × L ( E ) as g ′′ , so each stabilizethe same subspaces of L ( E ), so clearly H will be a blueprint for L ( E ). Wemake sure though, and for each one of these we can construct an element of order1365 = 273 × X whose 15th power is g ′′ and with 47 eigenvalues on L ( E ).Since 1365 T ( E ), this proves that H is a blueprint for L ( E ) by Theorem 2.4.(There are in fact many such elements powering to each of the 26 roots.) q = 3 : We have that dim(Ext kH (25 , JP76 , Section 6], and Ext kH (52 ,
1) =0 by [
V¨ol89 , Corollary 2] (see also Table 5.2). If the composition factors of L ( E ) ↓ H are not 196 ,
52, then by simple counting we must have more trivial composition fac-tors than 25-dimensional ones. Thus L ( E ) ↓ H has negative pressure, so H stabilizesa line on L ( E ) by Proposition 2.5.To prove that H also is a blueprint for L ( E ), first note that Ext kH (25 ,
52) =0 and so any 52-dimensional composition factors must break off into their ownsummand. The largest module with composition factors 1 and 25 is 1 / /
1, andany indecomposable module with these factors is a subquotient of this. This isthe action of H on M ( E ), so we know how the unipotent elements in H act on it,namely according to Table 2.2; in particular, the element u ∈ H acts with no blocksof size 3 on the summand of L ( E ) ↓ H whose composition factors have dimension1 and 25. However, the only conspicuous set of composition factors is 52 , , ,and so u has exactly one block of size 3 on L ( E ) by [ Law95 , Tables 4 and 5]. Thisproves that u lies in class A of G , which is generic, so H is a blueprint for L ( E )by Lemma 2.2.For 196 ,
52, let L denote the subgroup Spin (3) of H . This centralizes aninvolution of H , and it has trace − L ( E ), so L ≤ D . The composition factorsof L on L ( E ) have dimensions 16, 36, 84 and 112. This determines L uniquelyup to conjugacy inside D , as L acts irreducibly on the 16-dimensional module E M ( D ). Of course, L is therefore contained in a B subgroup X , which also hascomposition factors of the dimensions on L ( E ). Thus h H, X i stabilizes the 52 ⊕ L ( E ). Hence H is a blueprint for L ( E ), as claimed. q = 5 : The proof for q = 5 is easier than for q = 3: now the simple modulesof dimension at most 248 are 1, 26 and 52, and all three of these have zero 1-cohomology by [ JP76 , Section 6] and [
V¨ol89 , Corollary 2] (see also Table 5.3).Since 248 ≡
14 mod 26 there are at least fourteen trivial composition factors in L ( E ) ↓ H , which must be trivial summands, as claimed. (cid:3) We do not prove that H is a blueprint for q even, since for the Tits group thereis only one simple module of dimension 26, yielding many diagonal subspaces thatare not stabilized by an algebraic F that acts with two different 26-dimensionalcomposition factors on L ( E ).To show that F (5) is a blueprint for L ( E ), one first shows that it must becontained in the E -parabolic, then inside the E -Levi subgroup; at this point, asummand of L ( E ) ↓ H must be 26 ⊕ ⊕ ⊕ , and u must act on L ( E ) with at leasttwelve blocks of size 2 and 42 blocks of size 1, proving that it lies in a generic classby consulting [ Law95 , Table 9]. Hence H is a blueprint for L ( E ) by Lemma 2.2.Alternatively, one takes an SL (5)-Levi subgroup L of H containing u , andnotes that the composition factors of this on L ( E ) are 3 , , . The only wayto construct a block of size 5 in the action of u is from a module 1 / / L , hence u can have at most a single block of size 5 on L ( E ). Thus u is genericby consulting [ Law95 , Table 9], and again a blueprint by Lemma 2.2.HAPTER 7
Rank 3 groups for E ( q ), PSU ( q ), Ω ( q ) and PSp ( q ) for2 ≤ q ≤ q odd for Ω ( q )).We can prove that there are no Lie primitive copies of H except for PSU (2).This probably does stabilize a line on L ( E ), but it could be difficult to prove usingthese methods. There is a single set of composition factors that remains unresolved,and it is warranted.As with Chapter 6, u will denote a unipotent element of H of order p comingfrom the smallest class, i.e., with the largest centralizer. PSL We cannot use large element orders any more, unlike rank 4, so we need towork in every case. For PSL ( q ) we have a complete answer. Proposition . Let H ∼ = PSL ( q ) for some q ≤ . (1) If q = 2 , , then H stabilizes a line on L ( E ) . (2) If q = 3 , then H is a blueprint for L ( E ) . (3) If q = 5 , then H does not embed in G . Proof. q = 2 : The result was proved in [ Cra17 ]. q = 3 : There are two conspicuous sets of composition factors for L ( E ) ↓ H , whichare 19 , , (10 , ∗ ) , , , , ∗ , , , (10 , ∗ ) . The former of these has pressure −
11 (as only 19 has non-zero 1-cohomology), so L ( E ) ↓ H has at least ten trivial summands.Furthermore, the seven 15s must break off as summands as they have no ex-tensions with the other composition factors. For 6, we have that Ext kH (6 ,
19) hasdimension 2, and there are no other extensions involving it, so L ( E ) ↓ H has atleast eight 6s as summands as well; thus we have 15 ⊕ ⊕ ⊕ ⊕ ⊕ as a summand.The unipotent element u ∈ H acts on this module with blocks 3 , , , so theunipotent class of G to which u belongs can only be 2 A or 3 A from [ Law95 ,Table 9]. The class 2 A is generic, but 3 A is not: since 3 A acts on L ( E ) withblocks 3 , , , and u acts on each of 1, 6 and 10 with a block of size 1, theremaining composition factors 19 , (10 , ∗ ) , , must form an indecomposablemodule if u is from class 3 A . All extensions between these factors involve 19,so the only indecomposable module with these composition factors must be of theform 1 , , , , ∗ / / , , , , ∗ . While it is possible to make the module 19 / , , , , ∗ , one cannot construct amodule with three socle layers and with these first two socle layers. Thus u comes
556 7. RANK 3 GROUPS FOR E from class 2 A in G , which is generic, and so H is a blueprint for L ( E ) by Lemma2.2. For the second case, we restrict to L ∼ = PSp (3) inside H , and see that it hasno trivial composition factors. In Proposition 8.3 below we see that L is a blueprintfor L ( E ), and so therefore is H . q = 9 : There are many simple modules of dimension at most 248. However, al-though there are 30245 possible sets of composition factors for the self-dual module L ( E ) ↓ H , none of these is conspicuous for traces of elements of order at most 13 in G . Thus H does not embed in G . q = 5 : There are 549 possible sets of composition factors for a module of dimension248, none of which is conspicuous for elements of orders 2, 3 and 4. q = 7 : Since there are no modules with non-zero 1-cohomology of dimension atmost 248, any trivial composition factors must be summands. Using the traces ofelements of order up to 6, there are two conspicuous sets of composition factors for H , 64 , , ∗ , , , , (10 , ∗ ) , , ;both of these are semisimple because there are no extensions between the com-position factors. The element u acts with Jordan blocks 4 , , , , thus liesin the generic class 4 A of G , and 3 , , , thus lies in the generic class 2 A ,respectively. Thus H is a blueprint for L ( E ) in both cases by Lemma 2.2. q = 4 , : If L = PSL (2), then we saw in [ Cra17 , Proposition 4.2] that the possiblecomposition factors of L ( E ) ↓ L are14 , , (4 , ∗ ) , , , , (4 , ∗ ) , , (20 , ∗ ) , , , (4 , ∗ ) , , , , ∗ , , , , ∗ , . The dimensions of the composition factors of the restrictions to L of simple kH -modules are as follows.Dimension of module for H Is self-dual? Restriction of module to L
16 No 6 , ,
24 No 20 ,
436 Yes 14 , ,
56 No 20 , ,
64 ( q = 8) No 20 , , or 20 , , ,
80 No 64 , , , , , ,
84 Yes 64 , ,
96 ( q = 8) No 20 , , , , or 64 , ,
120 No 20 , , ,
196 Yes 20 , , , , From this it is easy to see, first that modules of dimension 196, 120, 96 and 80cannot occur in L ( E ) ↓ H (as factors of dimensions 120, 96 and 80 must occur inpairs), and second that in the first two sets of composition factors for L ( E ) ↓ L ,any such module for H has negative pressure (from Table 5.5 the modules withnon-zero 1-cohomology are of dimensions 14, 24 and 84), so H stabilizes a line on L ( E ) by Proposition 2.5.For q = 4, with these restrictions on the dimensions of the composition factors of L ( E ) ↓ H one manages to fairly easily compute the conspicuous sets of composition .1. PSL factors for elements of order at most 21, finding six up to field automorphism. Theseare 14 , , (4 , ∗ ) , , , , , (4 , ∗ ) , , (20 , ∗ ) , , , (4 , ∗ ) , , ∗ , , , (16 , ∗ ) , ( ¯16 , ¯16 ∗ ) , , , , , , , , ∗ , , , , , ∗ , , ∗ , , ∗ , ¯16 , ¯16 ∗ , , , , , (4 , ∗ ) , (4 , ∗ ) , , Since the only simple modules above with non-zero 1-cohomology are 14 i and 24 ± i,j (each 1-dimensional) we see that the pressures of these modules are − − − −
6, 2 and 2 respectively. Thus in the first four cases H stabilizes a line on L ( E )by Proposition 2.5. (This tallies with our earlier remarks as the first two and fourthsets of factors for H come from the first two sets for L .)For q = 8 there are many more possible 248-dimensional modules, but imposingthe requirement that the restriction to L must be one of the two remaining casesabove brings the number down to something manageable. Up to field automor-phism, there are three sets of composition factors that are conspicuous for elementsof order at most 21:(20 , ∗ ) , , , (4 , ∗ ) , , ∗ , , , , ∗ , , , . , ∗ , , ∗ , , ∗ , ¯16 , ¯16 ∗ , , , , , (4 , ∗ ) , (4 , ∗ ) , , (Notice that these are Cases 3, 5 and 6 from the q = 4 case above.) As 24 ± nolonger has non-zero 1-cohomology when q = 8 (but 24 ± still does, see Table 5.5),we see that we need only consider the set of factors involving 64 for q = 8.Thus we assume in the remaining cases that H does not stabilize a line on L ( E ). Let W denote the { i , ± i,j } -heart of L ( E ) ↓ H , with all simple summandsremoved. Note that, if H does not stabilize a line on L ( E ) then W contains alltrivial composition factors in L ( E ) ↓ H . Note also that soc( W ) consists solely ofmodules of dimensions 14 and 24. q = 4, Case 5, and q = 8 : The same proof works in both cases. Since 64 hasno extensions with the other composition factors, the two copies break off into aseparate summand. The { , , , ± } -radical of P (14 ) is14 , / , , , / , , ∗ / , / , so 14 cannot be the socle of the module W defined above. Suppose next that14 ⊕ is the socle of W , and attempt to build a pyx for W . Since there areonly two copies of 6 in L ( E ) ↓ H , we cannot have both copies of 6 in the secondsocle layer of W without them both being in the second radical layer (as W isself-dual), whence we remove a quotient (14 / ) ⊕ from W . The { , ± } -radicalof P (14 ) is simply 1 / , so since W should have four trivial factors, we obtain acontradiction. This also shows that the second socle layer cannot have no copies of6 either, for then we cannot build a pyx for W at all.Thus we can start with the module (6 / ) ⊕ and then add as many copiesof 1, 24 and 24 ∗ as possible. Doing so yields the module(1 , / , ∗ / , / ) ⊕ (1 / ) , but this has three trivial quotients, so one of the trivial factors cannot occur forpressure reasons, using Proposition 2.5. Thus we cannot build a pyx in this caseeither. E Thus, up to graph automorphism we may assume that the socle of W is ei-ther 24 or 24 ⊕ . For q = 4 , { , , } -radical of P (24 ) is simply14 / , / , so the socle cannot be 24 , as we cannot build a pyx with four triv-ial factors. If the socle is 24 ⊕ then the { , , } -radical of P (24 ) ⊕ P (14 )is (14 / , / ) ⊕ (14 / , / ) , and again we fail to build a pyx for W . Thus W cannot have four trivial compositionfactors, and hence H stabilizes a line on L ( E ). q = 4, Case 6 : We first show that L ( E ) ↓ H possesses a submodule of dimension1, 4 or 6. Thus suppose that none of 1, 4 ± i and 6 i lies in the socle of L ( E ) ↓ H . Sincethe 14 i appear with multiplicity 1, removing from the socle all composition factorswith zero 1-cohomology, and all simple summands, we obtain a module W , whichhas up to field and graph automorphism one of 24 , 24 ⊕ and 24 ⊕ ∗ as socle and four trivial composition factors.If the socle of W is 24 then we consider the { , ± i , i , i , ± , ¯16 ± , ± } -radical of P (24 ), which is of the form24 ∗ / / ∗ / , , / , , , ∗ / , , ¯16 ∗ / :this has only two trivial composition factors. Thus there must be two 24-dimensionalfactors in the socle.The { , ± i , i , i , ± , ¯16 ± } -radical of P (24 ) is1 , / , , / , , ¯16 ∗ / . On this we can place 24 ∗ or 24 ∗ .Notice that only 4 appears here out of the 4-dimensional modules. Thus W cannot have both 4 and 4 ∗ in it (as W is a submodule of the sum of this and (upto graph automorphism) the image of this under the field automorphism), so musthave neither as W is self-dual. Removing these, we take the { , i , i , ± , ¯16 ± } -radical instead of the { , ± i , i , i , ± , ¯16 ± } -radical, and this is1 / , / , , ¯16 ∗ / . We now let V denote the { ± i , i , ± i,j } -radical of L ( E ) ↓ H . As W possesses atmost two 6-dimensional composition factors, V must have at least one compositionfactor of dimension 6. The { ± i , i , ± i,j } -radicals of P (16 ) and P ( ¯16 ) are simply4 ∗ , ∗ / , ¯16 , and so there must be a 4- or 6-dimensional submodule of V , hence of L ( E ) ↓ H , aswe originally claimed.We now check that there exist elements of order 255 = 85 × x of order 85 in PSL (4) and that stabilize the eigenspaces comprising thecomposition factors 4 ± i and 6 i . Using the roots trick, we find 80 elements of order255 powering to x and stabilizing the eigenspaces comprising a given 4-dimensionalmodule, and 2186 elements that stabilize the eigenspaces for each 6 i (728 of whichstabilize the eigenspaces for both 6 i ). Indeed, one obtains more, and finds anelement of order 1785 = 7 ×
255 that powers to an element of order 255, has thesame number of distinct eigenvalues on L ( E ) as its seventh power, and stabilizesthe eigenspaces comprising both 4 and 6 . Thus in fact the stabilizer of a 1-, .1. PSL
4- or 6-space is positive dimensional, and therefore H is certainly contained in anelement of X .In fact, we wish to show that H actually stabilizes a line on L ( E ), so assumethat this is not the case. Since we have shown that H lies in a member of X , butnot necessarily X σ (and shown nothing for N G ( H )), this extra step will allow usto extend our results to N G ( H ) and N ¯ G ( H ).Let X ∈ X and suppose that H ≤ X . By Proposition 2.6 we may assumethat X is connected. If X is a maximal parabolic subgroup, then X cannot be an E -parabolic, either because there are not enough composition factors appearingwith multiplicity 2 in L ( E ) ↓ H to appear in the two copies of the 56-dimensionalfactor of L ( E ) ↓ X , or because X stabilizes a line on L ( E ) anyway, and that is whatwe are trying to prove. This also deals with the A E , A A , A D and A A A -parabolics, since the projection of H onto the Levi subgroup must lie inside afactor contained in the E -Levi subgroup (up to conjugation). Similarly, for A A , H must lie inside the A A -parabolic, hence inside the A -parabolic subgroup.If X is the A -parabolic subgroup, then it is an easy check that the embeddingof H into the A -Levi subgroup must have factors 4 ⊕ or 4 ⊕ ∗ on the naturalmodule. There is a filtration of L ( E ) ↓ X with the layers being as follows: L ( λ ) , L ( λ ) , L ( λ ) , L ( λ ) ⊗ L ( λ ) , L ( λ ) , L ( λ ) , L ( λ ) . The layers L ( λ ) , L ( λ ) have no trivial factors and no factors with non-zero 1-cohomology, and the same for their duals at the top of the module, so whether H stabilizes a line on L ( E ) depends on the middle three layers. The middle layerrestricts to H with structure(1 / / ⊕ (1 / / ⊕ ⊕ ∗ , where 16 is one of the 16-dimensional modules (depending on the action of H on M ( A )). The layer above and below are semisimple, with 24-dimensional and 4-dimensional factors. Thus in our proof above, when we construct the module W ,the 6 cannot lie in W . This is enough to remove the possibility that those threelayers can be combined to form a module with no trivial submodule or quotient, asthere is no module of the form 24 / / / , / . Thus there is a trivial submodule (and quotient) of those middle three layers, andhence of the whole module L ( E ) ↓ H .The last parabolic is the D -parabolic, so let X be a copy of this in E . Herethe action of X on L ( E ) has factors the two half-spin modules L ( λ ) and L ( λ ),a trivial module, two natural modules L ( λ ), and the exterior square W ( λ ) of thenatural (this is L ( λ ) and L (0)).The only possibility for the action of H on L ( λ ) with the right compositionfactors even on the sum L ( λ ) ⊕ ⊕ L ( λ ) is (up to field automorphism) 4 ⊕ ∗ ⊕ .(This is the only self-dual module with these factors.) This therefore lies in aparabolic subgroup of D , since 4 must be a totally isotropic subspace of M ( D )and stabilizers of totally isotropic subspaces are parabolic subgroups of D . Hence H lies inside a different parabolic subgroup of G .If X is a (connected) maximal-rank subgroup, then X = A places H inside the A -parabolic, A A places H inside the A A -parabolic, and A E and A E place H inside the E -parabolic, whence H cannot embed in X . For G F , H must lie E in the F factor, hence inside E , again a contradiction. The last case is X = D ,which in characteristic 2 stabilizes a line on L ( E ), so H stabilizes a line on L ( E ).Thus H stabilizes a line on L ( E ) in all cases. (cid:3) PSU For H ∼ = PSU ( q ), unless q = 2 we are able to prove that H is always stronglyimprimitive in G . However, for this last case of q = 2, there is one set of composi-tion factors for L ( E ) ↓ H , which is warranted, that we cannot deal with here. Wecontinue with our definition of u from the start of the chapter. Proposition . Let H ∼ = PSU ( q ) for some q ≤ . (1) If q = 2 then either H stabilizes a line on L ( E ) or the composition factorsof L ( E ) ↓ H are (20 , ∗ ) , , , (4 , ∗ ) , . (2) If q = 3 , then H does not embed in G . (3) If q = 4 , then H stabilizes a line on L ( E ) . (4) If q = 5 , then H is a blueprint for L ( E ) . Proof. q = 3 , : There are no sets of composition factors for L ( E ) ↓ H thatare conspicuous for elements of order at most 7, so H does not embed in G . q = 5 : There are two sets of composition factors for L ( E ) ↓ H that are conspicuousfor elements of order at most 8:20 , , (10 , ∗ ) , , , , , ∗ , , . In the first case, there are no extensions between any of the composition factors,so L ( E ) ↓ H is semisimple, and u acts on L ( E ) with blocks 4 , , , . Thus u lies in the generic class 4 A of G , so H is a blueprint for L ( E ) by Lemma 2.2.In the second case, if L ( E ) ↓ H is semisimple then u acts with blocks 3 , , ,so lies in the generic class 2 A , and again we apply Lemma 2.2. Furthermore,Ext kH (6 , ∼ = Ext kH (58 , ∼ = k , and all other extensions between compositionfactors are split, so the 15s and 45s must split off. As u acts on 45 ⊕ ∗ ⊕ ⊕ with blocks 4 , , , , and there are no non-generic unipotent classes withthose blocks (see Table 2.6), we see that u must belong to a generic class. Hence H is a blueprint for L ( E ), as needed. q = 9 : There are five sets of composition factors that are conspicuous for elementsof order up to 16, three up to field automorphism. These are19 , , (10 , ∗ ) , , , , (16 , ∗ ) , (16 , ∗ ) , , , , , , , ∗ , , (10 , ∗ ) , . Using a computer program and the roots trick (see Chapter 4), we check that ineach case an element x of order 328 in H is determined by its eigenvalues up toconjugacy in E , at least up to taking powers, and then that there is an element oforder 1640 = 324 × L ( E ) as x .Since 1640 T ( E ) (see Definition 2.3), this shows that H is a blueprint for L ( E )by Theorem 2.4, as claimed. q = 2 : There are seventeen conspicuous sets of composition factors for L ( E ) ↓ H ,eleven of which have positive pressure. These are64 , , ∗ , , , , ∗ , , , (20 , ∗ ) , , , (4 , ∗ ) , , .2. PSU , (20 , ∗ ) , , , , ∗ , , , (20 , ∗ ) , , , (4 , ∗ ) , , , , ∗ , , , (4 , ∗ ) , , , , ∗ , , , (4 , ∗ ) , , (20 , ∗ ) , , , (4 , ∗ ) , , (20 , ∗ ) , , , (4 , ∗ ) , , (20 , ∗ ) , , , (4 , ∗ ) , , (20 , ∗ ) , , , (4 , ∗ ) , , (20 , ∗ ) , , , (4 , ∗ ) , . We begin by using Proposition 3.13, and restricting these sets of composition factorsto the copy of L ∼ = Alt(6) in H . The restrictions are as follows:Case Restriction Case Restriction1 16 , , , , , , , , , , , , , , , , , , , , ,
10 16 , , , , , ,
11 16 , , , , , , From this we can see that only Cases 1, 3 and 10 can exist.
Cases 1 and 3 : Assume that H does not stabilize a line on L ( E ), and let W denote the { ± , } -heart of L ( E ) ↓ H . Since 4 , ∗ ,
14 are the modules with non-zero 1-cohomology, this module also has no trivial submodule or quotient, and isself-dual. To attack the first and third cases, which have pressure 2, we constructthe following submodule W ′ of P (14):(1) Take the { , , , ± } -radical of P (14);(2) On this add as many copies of 4 as possible;(3) Again, add as many copies of 1, 6, 14, 20 and 20 ∗ as possible;(4) On this add as many copies of 4 ∗ as possible;(5) Again, add as many copies of 1, 6, 14, 20 and 20 ∗ as possible.Up to application of the graph automorphism (which swaps 4 and 4 ∗ ), if soc( W ) =14 then W is a submodule of W ′ in the first and third cases, so W ′ is a pyx for W .This module has structure14 / , / , , , , ∗ / , , , , / , ∗ , , , ∗ / , / , so has four trivial composition factors, but also has a trivial quotient. Thussoc( W ) = 14, so must be (up to graph automorphism) 4 ⊕
14 or 14 ⊕ . If it is4 ⊕
14, then we take the { , , , ± } -radical of P (4) ⊕ P (14) to obtain(14 / / , , ∗ / , / , , ∗ / , / ⊕ (1 , / , ∗ / , / , which has only three trivial composition factors. Thus the socle must be 14 ⊕ : the { , ± , , ± } -radical of P (14), then with as many 14s placed on top as possible,then with all quotients other than 14 removed, is14 , / , / , ∗ , , , ∗ / , / . Two copies of this would be a pyx for W , but all four trivial composition factors inthis pyx would need to be present in W . However, we now take the { } ′ -residualof this pyx, which must lie inside W , and this is(1 / , ∗ / / ⊕ ;this has too many composition factors of dimension 4, which is a contradiction.Thus H stabilizes a line on L ( E ) in the first and third cases. E Case 10 : This is the case remaining in the proposition. q = 4 : There are, up to field automorphism, six sets of composition factors for L ( E ) ↓ H that are conspicuous for elements of order at most 17. These are14 , , (4 , ∗ ) , , , , , (4 , ∗ ) , , (20 , ∗ ) , , , (4 , ∗ ) , , ∗ , , , (16 , ∗ ) , ( ¯16 , ¯16 ∗ ) , , , , , , , , ∗ , , , , , ∗ , , ∗ , , ∗ , ¯16 , ¯16 ∗ , , , , , (4 , ∗ ) , (4 , ∗ ) , . The only simple modules from these with non-zero 1-cohomology are 14 i and24 ± i,j . Thus the fifth and sixth sets of factors have pressure 2, and the others havenegative pressure, so H stabilizes a line in the first four cases. Case 5 : Suppose that H —which has pressure 2—does not stabilize a line on L ( E ),and let W denote the { i , ± } -heart of L ( E ) ↓ H . The copies of 64 must splitoff, as they have no extensions with the other composition factors. Suppose thatsoc( W ) is either 14 or 24 . We consider the { , , , ± } -radical of P (14 ),which is 14 , / , , , / , , ∗ / , / , and the { , , } -radical of P (24 ), which is14 / , / . As neither of these contains enough trivial composition factors, we cannot havethat soc( W ) is simple.If it is 14 ⊕ , then we simply note that the { , , } -radical of P (14 )is 14 / , / , so a pyx for W cannot have enough composition factors. Finally,if soc( W ) is 14 ⊕ , then we need the { , , ± } -radical of P (14 ), which is1 , , / , ∗ / , / . Notice that this means that we must have a submodule (1 / ) ⊕ , and a quo-tient module (14 / ⊕ , and that the kernel of the quotient module contains thesubmodule (so that one may remove them both).Let v denote an element of order 4 in H that acts with a single Jordan block(of size 4) on the 4-dimensional simple modules 4 ± i . Its actions on 64 and 24 ± have Jordan blocks 4 and 4 , and on 14 and 6 it acts with blocks 4 , , v acts on L ( E ) with at least 58 blocks of size 4. Consulting Table 2.5,we see that v belongs to class 2 A and acts on L ( E ) with blocks 4 , . However,the action of v on 1 / (and its dual) has blocks 4 ,
3, so the action of v on L ( E )has at least four blocks of size at least 3, in addition to the 58 of size 4. However,this now contradicts Lemma 2.1, so we cannot build a pyx for W .Thus H stabilizes a line on L ( E ) in this case. Case 6 : We follow the corresponding proof of the case for PSL (4), at least initially.Again, H has pressure 2 on L ( E ). Again, we will show that L ( E ) ↓ H possesses asubmodule of dimension 1, 4 or 6, so suppose the contrary.Since the 14 i appear with multiplicity 1, removing from the socle all composi-tion factors with zero 1-cohomology (which are all but 14 i and 24 ± i,j , see Table 5.5),and all simple summands, we obtain a module W , which has up to field and graphautomorphism one of 24 , 24 ⊕ and 24 ⊕ ∗ as socle. .2. PSU Suppose first that soc( W ) is simple, and therefore 24 , so top( W ) ∼ = 24 ∗ . The { , ± i , i , i , ± , ¯16 ± , ± } -radical of P (24 ) is1 , / , ∗ , , / , , ¯16 ∗ / . Since this has only two trivial factors, we cannot produce a pyx for W , so the soclecannot be simple. Thus we may assume that the socle of W has two factors. Hencewe need the { , ± i , i , i , ± , ¯16 ± } -radical of P (24 ), which of course is exactlythe same module.On this we cannot place a copy of 24 or 24 ∗ (of course), but we can place asingle copy of 24 ∗ . This copy therefore must be on top of it in W . This yields amodule 24 ∗ / , / , ∗ , , / , , ¯16 ∗ / , but it has quotients other than 24 ∗ . Indeed, removing all quotients that are notof dimension 24 (which cannot occur in W ) leaves the much smaller module24 ∗ / / / , / . Thus a pyx for W is the sum of this module and its image under either the fieldor field-graph automorphism. In fact, this must be W itself, not just a pyx for W . Hence W contains no 4-dimensional factors and at most two 6-dimensionalfactors. Thus we must find at least one 6-dimensional composition factor in the { ± i , i , ± , ¯16 ± } -radical of L ( E ) ↓ H .The { ± i , i , ± , ¯16 ± } -radicals of P (16 ) and P ( ¯16 ) are simply 4 ∗ / and4 / ¯16 , so as with PSL (4), there must be a 4- or 6-dimensional submodule of L ( E ) ↓ H , as we claimed.We check that there exist elements of order 195 = 65 × x of order 65 in PSU (4) and that stabilize eigenspaces comprising thecomposition factors 4 ± i and 6 i . We find two elements of order 195 powering to x and stabilizing the eigenspaces comprising a given 4-dimensional module, and80 elements that stabilize the eigenspaces for each 6 i . Although this is enough toshow that H is contained in a member of X , one even finds elements of order1365 = 195 × L ( E ) as the element of order 195. This element isa blueprint for L ( E ) by Theorem 2.4, and so the stabilizer of a 4- or 6-dimensionalsubmodule of L ( E ) ↓ H is positive dimensional.We now have to do exactly the same as for PSL (4), and check all membersof X to see that H stabilizes a line on L ( E ), not just a 4- or 6-space. The exactsame proof works for PSU (4) as for PSL (4), so we see that H stabilizes a line on L ( E ). q = 8 : We look for sets of composition factors that are conspicuous for elementsof order at most 19, and that have either no trivials or more modules of dimension14, 24 and 84 than of dimension 1. (All simple modules of dimension at most 128,or self-dual with dimension at most 248, with non-zero 1-cohomology have thesedimensions, as we see from Table 5.5.) Since there are still 9320 possible dimensionsets for the composition factors, and billions of possible modules of dimension 248,we have to narrow down the possibilities. The six cases with non-positive pressurefor L ∼ = PSU (2) that were omitted before are14 , , (4 , ∗ ) , , , , (4 , ∗ ) , , , (20 , ∗ ) , , , (4 , ∗ ) , , E , , ∗ , , (4 , ∗ ) , , , , ∗ , , , (4 , ∗ ) , , , , ∗ , , , (4 , ∗ ) , . The dimensions of the composition factors of the restrictions of simple modulesfor H to L are slightly different from PSL (8), and are as follows.Dimension of module for H Is self-dual? Restriction of module to L
16 No 6 , ,
24 No 20 ,
436 Yes 14 , ,
56 No 20 , ,
464 No 20 , , or 20 , , ,
80 No 64 , , , , , ,
84 Yes 64 , ,
96 No 20 , , , , or 64 , ,
120 No 20 , ,
196 Yes 20 , , , , We see from this that to remove two trivial composition factors from L ( E ) ↓ H thatappear in L ( E ) ↓ L , we need at least one 14. We can in particular note that of thesix sets of factors of non-positive pressure for L ( E ) ↓ L above, only the fifth couldyield positive pressure for L ( E ) ↓ H , and then only with the dimensions84 , , , , , . There are no such conspicuous sets of composition factors. We therefore assumethat L ( E ) ↓ L comes from one of the three possible sets of factors of positive pressurefrom earlier in this proof.From this it is easy to see first that modules of dimension 196, 120, 96 and 80cannot occur in L ( E ) ↓ H , then that the number of composition factors of dimension14 and 36 put together does not exceed 4, that there are at most eight trivial factors,and so on. These sorts of easy restrictions can bring the number of possible sets ofdimensions down to something manageable, a couple of dozen. The total numberof modules is around 4 million, and we can easily check the traces of these modules.We find, up to field automorphism, two such sets of factors, but one, namely24 , ∗ , , ∗ , , ∗ , ¯16 , ¯16 ∗ , , , , , (4 , ∗ ) , (4 , ∗ ) , , has pressure 0 because 24 ± does not have non-zero 1-cohomology.The other conspicuous set of composition factors for L ( E ) ↓ H is64 , , ∗ , , , , which appeared in the case q = 4, again having pressure 2. The proof is identicalto the q = 4 case, even down to the radicals of P (14 ) and P (24 ) being the same.Hence H stabilizes a line on L ( E ), as needed. (cid:3) Ω In this section we let H ∼ = Ω ( q ) for some q ≤
9. We continue with our definitionof u from the start of the chapter. Proposition . If H ∼ = Ω ( q ) for some odd q ≤ , then H is a blueprint for L ( E ) . .4. PSp Proof. q = 3 : There is only one set of composition factors for L ( E ) ↓ H thatis conspicuous for elements of orders 2 and 4, and that is35 , , , , . Since none of these composition factors has an extension with any other, L ( E ) ↓ H is semisimple. The unipotent element u acts on L ( E ) with blocks 3 , , ;thus u comes from the generic class 2 A , proving that H is a blueprint for L ( E )by Lemma 2.2. q = 9 : As Ω (3) is contained in Ω (9), H is a blueprint for L ( E ). q = 5 : There is a single set of composition factors that is conspicuous for elementsof orders 2 and 3, and it is the same as for q = 3. Again, this is semisimple, and u acts on L ( E ) with blocks 3 , , ; thus u comes from the generic class 2 A ,proving that H is a blueprint for L ( E ) via Lemma 2.2. q = 7 : Again, elements of orders 2 and 3 yield a single conspicuous set of compo-sition factors, 35 , , , , . The only extension this time is between 26 and 1, so we have at least 35 ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ , on which u acts with blocks 3 , , . Examining [ Law95 , Table 9],we see that u must come from the generic class 2 A again, so H is a blueprint for L ( E ) by Lemma 2.2. (cid:3) PSp In this section we consider the groups PSp ( q ) for q ≤ Proposition . Let H ∼ = PSp ( q ) for some q ≤ . (1) If q is even then H stabilizes a line on L ( E ) . (2) If q is odd then H does not embed in G . Proof. q odd : For q = 3 , ,
7, there are no sets of composition factors for L ( E ) ↓ H that are conspicuous for elements of order at most 4, so H does notembed in G for those values. Since PSp (3) does not embed in G , neither doesPSp (9). q = 2 : There are seven conspicuous sets of composition factors for L ( E ) ↓ H , whichare 14 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . The 1-cohomologies of these modules are given in Table 5.1. The pressures of thesesets of composition factors are − −
1, 1, 2, 1, 0 and − Cases 1, 2, 6 and 7 : These all have non-positive pressure, so H must stabilize aline on L ( E ) by Proposition 2.5. Case 3 : Assume that H does not stabilize a line on L ( E ). Let W denote the { , } -heart of L ( E ) ↓ H . The socle of W is either 6 or 48 since W has pressure 1,by Proposition 2.5.The { , , , } -radical of P (48) is1 , , / , / , , , / , / , , / , / , , / / / . E This has only seven trivial composition factors, thus cannot be a pyx for W (withoutthe copy of 48 on top). Thus the socle cannot be 48.The dimension of Ext kH ( M , M ) for M i simple is 0 or 1, and it is 1 only for( M , M ) (or ( M , M )) one of the following pairs: (1 , , , , , , W is either 6 or 48.The { , , } -radical of P (6) is 1 / / , /
6, but both trivials are quotients ofthis, hence there cannot be three socle layers of W of the form 6 / , / / / i th and ( i + 2)th socle layers, there must bea 14 in between. For the third case, this situation occurs more than four times inthe socle layers of W , but there are only four 14s in W , a contradiction. Thus H stabilizes a line on L ( E ). Case 5 : Suppose that H does not stabilize a line on L ( E ). This has pressure 1,so let W be as in Case 3, and note that as there is a unique 48 in the compositionfactors of W , soc( W ) = 6. Ignoring any 8s and 64s, the socle structure of L ( E ) ↓ H must be 6 / , / / , / / / / / / , / / , / , with another 14 somewhere in the module. In particular, consider the quotient of L ( E ) ↓ H by the { } ′ -radical of L ( E ) ↓ H . We see that it cannot contain 64 asboth 48 and 64 appear exactly once. This quotient must have the socle structure6 / , / / , / / /
48, ignoring 8s. This is not a submodule of P (48), as we cansee from the submodule in Case 3, hence H stabilizes a line on L ( E ). Case 4 : Suppose that H does not stabilize a line on L ( E ), and let W be as before.Let L denote a copy of Alt(8) inside H , and note that the composition factors of L ( E ) ↓ L are (20 , ∗ ) , , , , . Since L stabilizes a line on L ( E ) by [ Cra17 , Proposition 8.5], it lies inside amaximal-rank or parabolic subgroup of G by Lemma 3.4.If L lies inside an A -parabolic subgroup, it acts as either 4 ⊕ or 4 ⊕ ∗ on M ( A ), and in either case the module (1 / / ⊕ is a summand of the action ofthe projection of L inside the A -Levi subgroup on L ( E ). In particular, this meansthat no 14 may lie in the socle of L ( E ) ↓ L , and this severely restricts the possiblesubmodules of L ( E ) ↓ H . We suppose that this is the case, and prove that H muststabilize a line on L ( E ) in this situation.Let V denote the { , , } -radical of W , and V the preimage in W of the { } -radical of the quotient W/V . The corresponding submodule of P (6) has structure1 , / / , , , / , and on restriction to L there is no subquotient 14 /
1. Therefore none of these 14smay appear in W , i.e., V = V . In particular, soc( W/V ) is either 48 or 48 ⊕ .Since W has pressure 2, and the submodule 1 / / , / P (6) has 1 / / / V may have at most two trivial composition factors.Writing ¯ W for the { } -heart of L ( E ) ↓ H , we therefore see that it must possess allfour 14s and at least four of the eight trivial composition factors.Therefore we consider the { , , , } -radical of P (48) above, and note that itonly possesses four 14s, so all must occur in ¯ W . Remove any quotients that are not .4. PSp
14 from this, and we find a module14 , / , / , / , / , , / / / , which must be a submodule of ¯ W . This however cannot be a subquotient of L ( E ) ↓ H , since it has a pressure 4 subquotient, namely 6 , / , / ,
6. This yields acontradiction, and so H stabilizes a line on L ( E ) whenever L lies in an A -parabolicsubgroup of G .Thus we may assume that L does not lie inside an A -parabolic subgroup of G . Let X be an A A -Levi subgroup of G , and suppose that L lies in it. If L liesin one of the A factors then L has at least 24 composition factors on L ( E ), sothis is not correct. Thus L embeds diagonally in X . In particular, it acts on thetwo modules M ( A ) as either L (100) or L (001). Since L ( E ) ↓ X has compositionfactors (101 , , , , L ( E ) ↓ L contains four subquotients 1 / / ⊗ ∗ ). Thus we may assumethat L does not lie in any A A -parabolic subgroup of G .Since L does not lie in A , A , B or G , if L lies in a central product of oneof these with another group X , then L lies in X . By [ Cra17 , Proposition 4.2], L is not contained in an E -parabolic subgroup, and so L is not contained in E A or E A either. If L ≤ A A , then L must have a trivial summand on M ( A ), so L lies inside an A A -Levi subgroup, and we use the previous paragraph. If L liesinside A then L stabilizes a line on M ( A ), so again lies inside an A -parabolicsubgroup. This deals with all maximal-rank subgroups except for D .For the parabolic subgroups, we have eliminated E , can eliminate A A asabove, and excluding A and A factors we see that lying inside A A A -parabolicmeans L lies in the A -parabolic, lying in the A D - or A E -parabolics means L lies in the E -parabolic subgroup. Since L must stabilize a line or hyperplane on M ( A ), hence if L lies in the A A -parabolic then it lies in the A - or E -parabolicsubgroup.If L lies inside the D -parabolic subgroup, then as we have considered allother parabolics of G , we may assume that L is D -irreducible. In particular,any subspace of M ( D ) that L stabilizes and acts irreducibly on must be non-singular, as totally isotropic subspaces have parabolic stabilizers, contradicting the D -irreducibility of L . In particular, there cannot be a subspace 4 ± of M ( D ) ↓ L ,as L cannot act non-singularly on a non-self-dual module, and there cannot be atrivial submodule.Thus L acts with composition factors on M ( D ) as 14 or 6 , or 6 , . In thefirst case the composition factors of L on L ( E ) are 64 , , ∗ , , , , ∗ , , bychecking traces of elements of orders 3. (In other words, L acts irreducibly on thehalf-spin modules as well as M ( D ).) In the second and third, L stabilizes a lineon M ( D ) so is not D -irreducible.If L lies inside the D maximal-rank subgroup, then we may assume that L is D -irreducible, hence again 4 is not in the socle of M ( D ) ↓ L , and neither is 1.There are no sets of composition factors that can satisfy these conditions, and so wehave proved that L always lies in an A -parabolic subgroup, completing the proof. q = 4 , : The composition factors of tensor products of simple modules for L =Sp (2) whose product has dimension at most 248 are in the table below. E Tensor product Factors6 ⊗ , , ⊗ ⊗ , , , ⊗
14 64 , , ⊗
14 11214 ⊗
14 48 , , , ⊗ , , , , By comparing with the conspicuous sets of composition factors for L ( E ) ↓ L , wetherefore see that the dimension of any factor of L ( E ) ↓ H is at most 84. Further-more, if M is a simple kH -module then the dimension of H ( H, M ) is at most 1and it is non-zero only if M has dimension 6, 48 or 84 (not all modules of dimension48 and 84 have non-zero 1-cohomology).If L ( E ) ↓ H has no trivial composition factors, then from the possibilities for L ( E ) ↓ L and the table above, we see that the dimensions of the factors of L ( E ) ↓ H are 48 , , , , , , , , , , , . There are no sets of composition factors with these dimensions that are conspicuousfor elements of order at most 13.If L ( E ) ↓ H has trivial factors then we can consider its pressure, so we eliminateall sets of dimensions that must have non-positive pressure. For each set of factorsfor L ( E ) ↓ L one obtains a small possible set of dimensions for L ( E ) ↓ H , and it iseasy to check for conspicuous sets of factors for each of these in turn.For q = 4 , L ( E ) ↓ H that are conspicuous for elements of order up to 13 and havemore factors of dimension 6, 48 and 84 than trivial factors. The first can be writtendown the same for both q = 4 and q = 8, and it is48 , , , , , ;this has pressure 0, because the 48-dimensional modules that have non-zero 1-cohomology are 48 i = 6 i ⊗ i , not 48 i,j = 6 i ⊗ j .The second set of composition factors changes a little between q = 4 and q = 8:for q = 8 it is 64 , , , , , , , , , . We aim to show that H stabilizes a line on L ( E ), so suppose that this is not thecase. This has pressure 1, and as 64 , 48 , 48 , 14 and 14 all appear withmultiplicity 1, if they are in the socle then they are summands, so can be ignored.Thus we may assume that the socle is either 6 or 6 , by quotienting out by any8 i . The { , , , , } -radicals of P (6 ) and P (6 ) are 6 / , / and 1 / respectively. Notice that the { , , , , } -heart of L ( E ) ↓ H is a module whosesocle consists of modules with multiplicity 1, hence is semisimple. But this meansthat H must stabilize a line on L ( E ) because there are not enough trivials abovethe 6 i . This completes the proof for q = 8.When q = 4, the other set of factors is64 , , , , , , , , , . One can check that there are elements of order 255 cubing to a given element x oforder 85 in H , and that stabilize all of the constituent eigenspaces of 6 , 6 , 8 and .4. PSp . Indeed, there are 2186 roots ˆ x such that ˆ x = x and ˆ x stabilizes the eigenspacescomprising 6 i , and 80 for 8 i . Moreover, 728 roots stabilize both 6 and 6 . Indeed,to show that the stabilizer of each 6 i and each 8 i is positive dimensional, one findsan element of order 1985 = 7 ×
255 whose seventh power ˆ x has the same numberof distinct eigenvalues on L ( E ) as it, and such that ˆ x is an element stabilizing theeigenspaces comprising both 6 and 8 .Thus if L ( E ) ↓ H has 1, 6 i or 8 i as a submodule then H is contained in amember of X , but of course these are the only composition factors that appearmore than once. Hence at least one of them must be in the socle of L ( E ) ↓ H , so H lies in a member of X .We want to show that H stabilizes a line on L ( E ). To see this, we run throughthe members of X , and find that H must lie inside a D maximal-rank subgroup,which stabilizes a line on L ( E ).As with PSL (4), H cannot lie inside the E -parabolic subgroup as there arenot enough factors with multiplicity 2 in L ( E ) ↓ H . The possibilities for embedding H in the A -parabolic subgroup—acting on the natural module with factors 6 , or 8 —yield the wrong factors on L ( E ). For the D -parabolic subgroup, the onlyembedding of H acts on the natural as 8 ⊕ ⊕ , clearly yielding the wrong factors.As with PSL (4), we eliminate all other parabolics by projecting onto the Levifactor, embedding in a subgroup, and using containment of parabolics.For the reductive subgroups, if H ≤ G F then H ≤ F ≤ E , and H is notcontained in A A , or the A subgroup as it is not in the A -parabolic subgroup,and of course not in A E or A E either. The only subgroup left is D , which ofcourse stabilizes a line on L ( E ), so we are done. (In this case, H must act on thenatural module as 8 ⊕ .) (cid:3) HAPTER 8
Rank 2 groups for E ( q ), PSU ( q ), PSp ( q ) and G ( q ) for2 ≤ q ≤
9, together with the Suzuki groups B ( q ) for q = 8 , , ,
512 and G ( q )for q = 3 ,
27. In addition, we consider the derived groups when these are not simple,which are PSp (2), G (2), G (3). We do not consider PSU (2), which is soluble.As to be expected, there are more groups in this situation that we cannot proveare always strongly imprimitive. In particular, we cannot for PSL ( q ) for q = 3 , ( q ) for q = 3 , ,
8, and for PSp ( q ) for q = 2. In addition, for B ( q ) wedo not manage to prove any results for q = 8. With the exception of these sevengroups, we prove that every one of the subgroups above is strongly imprimitivewhen embedded in E . One case of PSU (4) is delayed until Section 12.3 (althoughthere is another case that we cannot do), and the proof for the most difficult caseof PSL (5) is delayed until Section 12.2.Throughout this chapter, as with the previous two, we let u denote an elementof order p in the smallest conjugacy class of H . PSL For PSL ( q ), when q = 2 , H always stabilizes a line on L ( E ), but for q = 3 , H is always Lie imprimitive, nevermind strongly imprimitive. For q = 5 , , ,
16, there are some cases where H is ablueprint for L ( E ), some where it stabilizes a line on L ( E ), and others where allwe can show is that there is an orbit of subspaces under the action of N Aut + ( G ) ( H )whose simultaneous stabilizer is positive dimensional, i.e., it is strongly imprimitivevia Theorem 3.2.This proposition proves all of the results that we want except for the singlecase for q = 5, which is delayed until Section 12.2. Proposition . Let H ∼ = PSL ( q ) for some ≤ q ≤ or q = 16 . (1) If q = 2 , then H stabilizes a line on L ( E ) . (2) If q = 3 then either H stabilizes a line on L ( E ) or the composition factorsof L ( E ) ↓ H are one of (15 , ∗ ) , , , ∗ , (3 , ∗ ) , , (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , , , (15 , ∗ ) , , , ∗ , (3 , ∗ ) , . (3) If q = 4 then either H stabilizes a line on L ( E ) or the composition factorsof L ( E ) ↓ H are one of (9 , ∗ ) , , , , , (9 , ∗ ) , , , .
712 8. RANK 2 GROUPS FOR E (4) If q = 5 then either H stabilizes a line on L ( E ) or the composition factorsof L ( E ) ↓ H are , ∗ , , ∗ , , ∗ , , ∗ , , , ∗ , , ∗ . (5) If q = 8 , , then H is strongly imprimitive. Proof. q = 7 : The conspicuous sets of composition factors for L ( E ) ↓ H are71 , ∗ , , , (10 , ∗ ) , , , (28 , ∗ ) , , , , (35 , ∗ ) , , (10 , ∗ ) , , , ∗ , , , , , (10 , ∗ ) , , , , ∗ , , . Each of these has non-positive pressure (see Table 5.4), so H stabilizes a line on L ( E ) by Proposition 2.5. q = 5 : There are fifteen conspicuous sets of composition factors for L ( E ) ↓ H ,fourteen of which have a trivial composition factor. The fourteen that have atrivial factor are:8 , (3 , ∗ ) , , , ∗ , , , , (6 , ∗ ) , (3 , ∗ ) , , , ∗ , , (6 , ∗ ) , (3 , ∗ ) , , , (15 , ∗ ) , (10 , ∗ ) , , (6 , ∗ ) , , , ∗ , , , , , (35 , ∗ ) , (10 , ∗ ) , , , ∗ , , (15 , ∗ ) , , ∗ , , (6 , ∗ ) , , , , ∗ , , ∗ , , ∗ , , ∗ , (3 , ∗ ) , , , , ∗ , (15 , ∗ ) , (10 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , , , , , (6 , ∗ ) , (3 , ∗ ) , , , , ∗ , , ∗ , , ∗ , , , (10 , ∗ ) , , , (15 , ∗ ) , , ∗ , , (6 , ∗ ) , (3 , ∗ ) , . As only 39 ± and 63 have non-zero 1-cohomology, and each is 1-dimensional (seeTable 5.3), we see that all of these sets of composition factors have non-positivepressure, hence H stabilizes a line on L ( E ) in those cases by Proposition 2.5.We thus have the remaining case, where the composition factors of L ( E ) ↓ H are 39 , ∗ , , ∗ , , ∗ , , ∗ , , , ∗ , , ∗ , the unique conspicuous set with no trivial factor. This is as stated in the proposition(and dealt with in Section 12.2). q = 3 : There are fifteen conspicuous sets of composition factors for L ( E ) ↓ H :7 , (3 , ∗ ) , , , , ∗ , , , (6 , ∗ ) , (3 , ∗ ) , , , ∗ , , (6 , ∗ ) , (3 , ∗ ) , , , (15 , ∗ ) , , , ∗ , , ∗ , , (15 , ∗ ) , , , ∗ , (3 , ∗ ) , , , (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , , , , (3 , ∗ ) , , , , ∗ , , , ∗ , (3 , ∗ ) , , (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , , , (15 , ∗ ) , , , ∗ , (3 , ∗ ) , , , (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , , , , , ∗ , (3 , ∗ ) , , , (15 , ∗ ) , , (6 , ∗ ) , , , , ∗ , , (6 , ∗ ) , (3 , ∗ ) , . The pressures of the sets of composition factors are − − − − −
7, 6, −
1, 7, −
6, 6, 7, 0, −
5, 1 and 1 respectively.
Cases 1, 2, 3, 4, 5, 7, 9, 12 and 13 : In these cases H stabilizes a line on L ( E )by Proposition 2.5. .1. PSL Cases 14 and 15 : If the pressure of L ( E ) ↓ H is 1 then, upon quotienting out bythe { , ± } ′ -radical, we obtain a submodule of (up to duality) either P (7) or P (15).There are five and four trivial composition factors of these modules respectively, sothere must be a trivial module in the { , } ′ -radical. This means there is a trivialsubmodule, so H stabilizes a line on L ( E ) in these two cases as well. Case 8 : Of course, as 27 is the Steinberg module and hence projective, it splits offas a summand. The { , ± , } -radical of P (7) is3 , ∗ / , / , , ∗ / , so we need at least twice as many 7s as 1s in order for H not to stabilize a line. Inthis case we do not have that many, so H stabilizes a line on L ( E ). Cases 6, 10 and 11 : These are left over, and are stated in the proposition. q = 9 : Up to field automorphism, there are 27 sets of composition factors for L ( E ) ↓ H that are conspicuous for elements of order at most 20. Thirteen of theseare of non-positive pressure (the simple modules for H with non-zero 1-cohomologyhave dimensions 7 and 21), so H stabilizes a line on L ( E ) by Proposition 2.5.First, we note that by a computer calculation, the eigenvalues on L ( E ) of anelement x ∈ H of order 80 determine the semisimple class of G containing x . Ifthere exists an element ˆ x in G that squares to x and shares the same eigenspaces of L ( E ) as x , then we can use Lemma 3.10 to conclude that H is strongly imprimitive.Fourteen of the 27 sets of factors have this property, but there is some overlapwith the sets of non-positive pressure, and we are left with eight sets of factorswhere there is no such element ˆ x , and L ( E ) ↓ H has positive pressure. These are45 , ∗ , , (18 , ∗ ) , , , (3 , ∗ ) , , ¯45 , ¯45 ∗ , , ( ¯18 , ¯18 ∗ ) , , , (3 , ∗ ) , , , ∗ , , ∗ , ¯18 , ¯18 ∗ , , ∗ , , ∗ , ¯9 , ¯9 ∗ , , , , ∗ , (3 , ∗ ) , (3 , ∗ ) , , , ∗ , , ∗ , , ∗ , ¯18 , ¯18 ∗ , , ∗ , ¯9 , ¯9 ∗ , , , (3 , ∗ ) , (3 , ∗ ) , , , ∗ , , ∗ , (15 , ∗ ) , (¯9 , ¯9 ∗ ) , , (3 , ∗ ) , , , , ∗ , , ∗ , , ∗ , , , (3 , ∗ ) , , , , ∗ , , ∗ , , ∗ , , , , ∗ , , , ¯45 , ¯45 ∗ , ¯18 , ¯18 ∗ , ¯18 , ¯18 ∗ , , , , ∗ , . (Recall the labelling conventions from Chapter 5.)The pressures of these modules are 1, 1, 2, 4, 2, 3, 1 and 1 respectively. Let W denote the { i , ± } -heart of L ( E ) ↓ H . Cases 1, 2, 7 and 8 : These all have pressure 1, and by Lemma 2.7 they eachcontain a subquotient 7 ⊕ , which has pressure 2. Thus H stabilizes a line on L ( E ) by Proposition 2.5. Case 3 : Suppose that H does not stabilize a line on L ( E ). If 21 ± lies in the socleof W , then we need to take the { , ± i , ± , i , ± , ¯9 ± , ± , ± , ¯18 ± } -radicals of P (7 ) and P (21 ), and these are7 , / , , ∗ / , / , / , , ∗ / . E The sum of these two must be a pyx for W , but only has two trivial compositionfactors, so H must stabilize a line on L ( E ), a contradiction. Hence soc( W ) mustconsist solely of copies of 7 , so in fact only one as we only have three copies of 7 in L ( E ) ↓ H . Thus we take the { , ± i , ± , i , ± , ¯9 ± , ± , ± , ¯18 ± , ± } -radicalof P (7 ), which is(8.1) 1 / , , / , , , , ∗ , ∗ / , , , ∗ / , , ∗ / . In fact, this is not quite true. This module has an extension with 21 ± , butwhen you place either of those modules on top and then take the { ± } ′ -residual,one finds that both copies of 21 ± remain in the residual. Since only one copy of21 ± appears in L ( E ) ↓ H , this second copy at the top cannot appear in W .Thus we may work in the smaller module in (8.1). This module does not haveenough trivial composition factors, so L ( E ) ↓ H has more trivial factors than W .Hence H stabilizes a line on L ( E ). Case 5 : Assume that H does not stabilize a line on L ( E ). This case has compo-sition factors a subset of the composition factors of the third case (twisted by thefield automorphism), so the radicals above eliminate 7 , 21 and 7 ⊕ frombeing socles of W , leaving only 7 ⊕ . In this case we take the { , ± , ± } -radical(we don’t need the other modules by the above radical) of P (7 ) to obtain themodule 1 / , ∗ / , , ∗ / . Two copies of 7 may be placed on top of this module, but they fall into the thirdsocle layer, so the top trivial remains a quotient in W , a contradiction. Thus H stabilizes a line on L ( E ).The remaining two cases have pressures 3 and 4. The easiest way to proceedhere is to find elements of order 160 that stabilize the eigenspaces comprising certaincomposition factors of L ( E ) ↓ H , as we have done before. Case 4 : There are sixteen elements of order 160 stabilizing the eigenspaces com-prising 3 ⊕ ∗ ⊕ , and four that stabilize the eigenspaces comprising 3 ⊕ ∗ ⊕ ⊕ ¯9 ⊕ ¯9 ∗ ⊕ ⊕ ∗ . Furthermore, note that 18 ± and ¯18 ± lie in differentOut( H )-orbits, so no element of N Aut + ( G ) ( H ) can swap such subspaces. In partic-ular, N Aut + ( G ) ( H ) cannot induce a field automorphism on L ( E ) so, for example,an N Aut + ( G ) ( H )-orbit of a subspace of type 3 can only contain types 3 and 3 ∗ .In particular, this means that we may use Lemma 3.10 to state that H is stronglyimprimitive if any of the modules 3 ± i , 7 i , ¯9 ± and 21 ± lie in the socle of L ( E ) ↓ H .First, 9 ± has no extension with any composition factor in L ( E ) ↓ H , so theymust split off as summands. The cf( L ( E ) ↓ H )-radicals of P (18 ), P ( ¯18 ) and P (21 ) are 18 / ¯9 ∗ , ¯18 / , ¯18 / ∗ / ∗ / ∗ , / ¯18 , / , , ∗ / , ∗ / / , , ¯18 ∗ / , , ∗ / . The sum of these modules only has three trivial factors and L ( E ) ↓ H has four,so the socle must contain a module other than one of these. Thus H is stronglyimprimitive by Lemma 3.10, as needed. Case 6 : There are four elements of order 160 stabilizing the eigenspaces comprising1 ⊕ ⊕ ∗ ⊕ ⊕ ⊕ ⊕ ∗ , so if any of these modules lie in the socle of L ( E ) ↓ H then H is strongly imprimitive by Lemma 3.10. .1. PSL Thus the socle of L ( E ) ↓ H must consist of some of 49, 42 ± and 15 ± if H isnot strongly imprimitive, which we will assume. If 49 is a submodule of the soclethen it splits off as a summand as it appears exactly once and is self-dual, so wemay ignore this. The cf( L ( E ) ↓ H ) \ { ± } -radical of P (15 ) is1 , , ∗ / , / ∗ , / , and the cf( L ( E ) ↓ H ) \ { ± } -radical of P (42 ) is1 / ∗ , , ∗ / . These clearly have far too few composition factors, so soc( L ( E ) ↓ H ) contains someother module, and we are done. q = 2 : There are seven conspicuous sets of composition factors for L ( E ) ↓ H , whichare 8 , (3 , ∗ ) , , , (3 , ∗ ) , , , (3 , ∗ ) , , , , , (3 , ∗ ) , , , (3 , ∗ ) , , , (3 , ∗ ) , . The 8s split off, and the projective cover P (3) is3 / ((3 ∗ / / ∗ ) ⊕ / . From this we can see that the only indecomposable modules that possess a trivialcomposition factor but no trivial submodule or quotient are P (3) and P (3 ∗ ); thuswe need at least five times as many 3-dimensional factors as trivial factors, elsewe stabilize a line on L ( E ). Examining the sets above, this is not the case, so H stabilizes a line on L ( E ). q = 4 : There are eight conspicuous sets of composition factors for L ( E ) ↓ H up tofield automorphism:9 , ∗ , , , , (9 , ∗ ) , , , (9 , ∗ ) , , , , , (9 , ∗ ) , , , , , (9 , ∗ ) , , , , (9 , ∗ ) , , , , , (9 , ∗ ) , , , , , (9 , ∗ ) , , , . We begin by letting M denote a copy of Alt(6) inside H , restricting these to M , and then applying Proposition 3.13.Case Restriction to M , (4 , ) , , , (4 , ) , , , (4 , ) , , , (4 , ) , , , (4 , ) , , , (4 , ) , , , (4 , ) , , , (4 , ) , Cases 2, 4, 5 and 6 : These all have restrictions to M that do not exist byProposition 3.13, so H does not embed in G with these composition factors. Case 1 : This has pressure −
26, so H stabilizes a line on L ( E ) by Proposition 2.5. Case 8 : Note that the restriction to M is not warranted (but might exist), so weshow that H does not exist. We may number the conjugacy classes L , L , L ofsubgroups PSL (2) and M , M , M of subgroups Alt(6) so that the permutation E module P L i on the cosets of L i is the direct sum of the permutation module P M i on M i and a copy of 64. Thus if we exclude the summands 64 on L ( E ) ↓ H to makea module W , the fixed-point spaces of L i and M i on W have the same dimension.Fixing i , and writing L and M for L i and M i respectively, since M must be Lieprimitive, this means that the fixed-point spaces satisfy W M = W L = 0.As the composition factors of W ↓ L are (3 , ∗ ) , , , we see that W ↓ L isexactly P (3) ⊕ P (3 ∗ ) ⊕ ⊕ ⊕ A, where A is a module with composition factors (3 , ∗ ) . Since an element of order 4in L acts on L ( E ) with at least 56 blocks of size 4, from Table 2.5 we see that itacts with Jordan blocks 4 , or 4 , . This means that A is either semisimpleor the sum of four indecomposable modules of the form either 3 / ∗ or 3 ∗ /
3. Inparticular, since 9 restricts to L as 3 / ∗ /
3, this means that soc( W ) is a submoduleof 9 ⊕ ∗ (ignoring copies of 8 i ). Indeed, we may take the { i } ′ -heart of W andthis still holds (but with a different number of copies of 8), so we replace W by its { i } ′ -heart.From this we can almost determine the exact structure of L ( E ) ↓ L : it is P (3) ⊕ ⊕ P (3 ∗ ) ⊕ ⊕ P (1) ⊕ ⊕ ⊕ ⊕ A, where A is the module above.We attempt to find subgroups of G isomorphic to L with this module structure,which has composition factors on L ( E ) given by8 , (3 , ∗ ) , . Since L stabilizes a line on L ( E ), L is contained in a maximal-rank or parabolicsubgroup by Lemma 3.4.Suppose that L is contained in a subgroup X of G ; we may remove any factors A or B from X as L does not embed in these, and if L is contained in A thenit is contained in an A -parabolic subgroup of A . In fact, we saw in the case q = 2 above that if V is an indecomposable module for L with a trivial compositionfactor, then either it has a trivial quotient or submodule or V = P (3) of dimension16 (and V is not self-dual).If X is A , then we consider the action of L on M ( A ), and we obtain thefollowing table. Factors on M ( A ) Factors on L ( E )3 , , , (3 , ∗ ) , , , ∗ or 8 , , (3 , ∗ ) , , , or 3 , ∗ , , (3 , ∗ ) , , , (3 , ∗ ) , Thus L is not contained in A , or any parabolic subgroup whose Levi subgroup iscontained in A , eliminating the A - and the A A A -parabolics, and also A A and the A A -parabolic (as L must then lie in an A A -parabolic) and the A A -parabolic (as L must stabilize a line or hyperplane on M ( A ), so lie in an A A -parabolic, whose Levi lies inside A ).If X = E A then L ≤ E . We see from Proposition 9.1 below, and the factthat the composition factors of L ( E ) ↓ E are M ( E ) , L ( E ) ◦ , , that L must haveat least twelve trivial composition factors on L ( E ), so L cannot embed in X . Thus .1. PSL L does not embed in E A or the E -parabolic subgroup, or the E A -parabolic(as then it would embed in an E -parabolic).If X = E A , then X has a summand of dimension 81 on L ( E ) (the ten-sor product of the minimal modules), but all summands of L ( E ) ↓ L are of evendimension, so L X .Suppose that X is a D -parabolic subgroup: the composition factors of L ( E ) ↓ X are two copies of M ( D ), one copy each of the two (dual to one another) half-spinmodules, two trivial modules, and a copy of the 90-dimensional simple part L ( D ) ◦ of the Lie algebra.If L lies inside a D -parabolic subgroup of X then L lies inside an E -parabolic,which is a contradiction. The composition factors of L on M ( D ) with at least onetrivial factor are 8 , , (3 , ∗ ) , and 3 , ∗ , . In all cases there is a subgroup L in D acting semisimply on M ( D ), with these factors, hence lies inside a D -parabolic subgroup. Thus no copy of L with a trivial factor on M ( D ) can have thecorrect composition factors on L ( E ). Therefore L cannot have a trivial factor on M ( D ), so the composition factors are 8 , , ∗ . (These do yield a copy of L with thecorrect factors on L ( E ).) The possible actions of L , up to graph automorphism,on M ( D ) are 8 ⊕ ⊕ ∗ , ⊕ (3 / ∗ ) . By placing L inside a D D subgroup, and noting that an irreducible copy of L in D acts irreducibly on M ( D ) and the two half-spin modules (if it didn’t then upto graph automorphism L would stabilize a line on M ( D ), hence lie in a parabolic,thus cannot act irreducibly on M ( D ) up to graph automorphism as the graph mappermutes the parabolics), we see that the action of L on the half-spin modules for D is the tensor product of 8 with a module, so is projective. Using trace data, thecomposition factors can be determined, and the action on the half-spin modules is8 ⊕ ⊕ P (3) ⊕ ⊕ P (3 ∗ ) ⊕ . Thus the remainder of L ( E ) ↓ X must, upon restriction to L , form copies of 8,together with P (3), P (3 ∗ ), P (1), and the module A . We may construct L ( D ) ◦ by taking the exterior square of M ( D ) and removing a trivial summand, and theresulting module has restriction to L given by P (3) ⊕ P (3 ∗ ) ⊕ ⊕ ⊕ (3 , ∗ / , , ∗ / , ∗ ) ⊕ B, where B is either 3 ⊕ ∗ or 3 ∗ /
3, depending on the action of L on M ( D ) givenabove. However, this clearly yields a contradiction, as there is no possible extensionof the above module by 3s and trivials that can make the module P (1) ⊕ ⊕ A , whichdoes not have a 3 above the second socle layer. Thus L cannot embed in the D -parabolic subgroup.If L embeds in a D A -parabolic subgroup, then it embeds in a D A -parabolic,and this is contained in a D -parabolic subgroup. To see this, note that the actionon M ( D ) must have at least two trivial summands, hence L lies in a D -parabolicsubgroup. Thus L cannot lie in any parabolic subgroup of G .Finally, if X = D , then L must be X -irreducible, and so the action of L on M ( D ) is 8 ⊕ . The action on L ( D ) ◦ , which is the exterior square of M ( D ) witha trivial removed from top and bottom, has two trivial submodules, so as L ( E ) ↓ X has structure ( L (0) /L ( D ) ◦ /L (0)) ⊕ L ( λ ) , we see that the fixed space L ( E ) L has dimension at least 3, a contradiction. E This proves that there is no such embedding of L into G , and therefore H cannot embed in G , as needed. Cases 3 and 7 : These are left over, and are as in the proposition. q = 8 : Write x for an element in H of order 63, and we will use Lemma 3.10to show strong imprimitivity in most cases. If there are no trivial compositionfactors of L ( E ) ↓ H , then up to field automorphism there are six sets of compositionfactors that are conspicuous for elements of order at most 21, all of which are alsoconspicuous for elements of order 63. Write27 = 3 ⊗ ⊗ , = 3 ⊗ ⊗ ∗ , = 3 ⊗ ∗ ⊗ , = 3 ∗ ⊗ ⊗ . These cases are: , ∗ , ¯9 , ¯9 ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , ∗ , ¯9 , ¯9 ∗ , , , (3 , ∗ ) , (3 , ∗ ) , , ∗ , , ∗ , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , ∗ , , , (3 , ∗ ) , , ∗ , , ∗ , , ∗ , , ∗ , , ∗ , (¯9 , ¯9 ∗ ) , , , , (3 , ∗ ) , (3 , ∗ ) , (27 , ∗ ) , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , ∗ , , , (3 , ∗ ) , , ∗ , , ∗ , , ∗ , , ∗ , ¯9 , ¯9 ∗ , (9 , ∗ ) , , , , , ∗ , , ∗ , , ∗ , , ∗ , , ∗ , (9 , ∗ ) , , , , , ∗ . Case 1 : Unlike the other five cases, in this case there is no composition fac-tor that simultaneously splits off because it has no extensions with the otherfactors of L ( E ) ↓ H , and also has elements of order 189 cubing to x that pre-serve the eigenspaces that comprise it. By Lemma 3.9, since the compositionfactors of L ( E ) ↓ H are not invariant under any field automorphisms in Out( H ),the N Aut + ( G ) ( H )-orbits of simple submodules can only include a module and itsdual.There are 27 elements of order 189 that stabilize the eigenspaces that comprise3 ⊕ ∗ and nine each that stabilize the eigenspaces that comprise 3 ⊕ ∗ and 3 ⊕ ∗ .If there is a 3-dimensional submodule in L ( E ) ↓ H then H is strongly imprimitiveby Lemma 3.10.In addition there are 27 such elements that stabilize the 8 factor, 81 thatstabilize 8 , nine that stabilize 9 ⊕ ∗ and three that stabilize ¯9 ⊕ ¯9 ∗ . Thusif H is strongly imprimitive then none of these modules can lie in the socle of L ( E ) ↓ H . We therefore may assume for a contradiction that soc( L ( E ) ↓ H ) consistssolely of a subset of ¯9 ± , 9 ± , ¯9 ± and 24 ± . The cf( L ( E ) ↓ H ) \ { V ± } -radicals of P ( V ) are as follows for V one of ¯9 , 9 , ¯9 and 24 :3 ∗ / , ¯9 , , ¯9 ∗ , / ∗ , ∗ , / , , ∗ / ¯9 , ∗ / / ∗ / / , ¯9 ∗ , ¯9 ∗ / , , ¯9 ∗ , / ∗ , ∗ , / , ∗ / ¯9 , ¯9 / ∗ , ∗ / , ¯9 / ∗ , ¯9 / / ∗ / . There are three copies of 8 in L ( E ) ↓ H , and only one in the sum of modules above.Therefore soc( L ( E ) ↓ H ) must consist of modules other than those above, and H isstrongly imprimitive. Cases 2, 3 and 4 : In each case the 8 must split off as it has no extensions withother composition factors. It is stabilized by nine, 27 and nine elements of order189 cubing to x respectively. Hence H is strongly imprimitive by Lemmas 3.9 and3.10. .1. PSL Case 5 : The modules 8 and 8 split off as summands as they have no extensionswith other composition factors in L ( E ) ↓ H . There are 729 elements of order 189cubing to x and stabilizing the eigenspaces of 8 , so we are done again. Case 6 : This time there are no extensions between 3 and the other compositionfactors, so this and its dual split off as summands. There are 81 elements of order189 cubing to x and stabilizing the eigenspaces comprising 3 ⊕ ∗ , so we are doneas before.We may therefore assume that L ( E ) ↓ H has a trivial composition factor. Up tofield automorphism, there are eleven sets of composition factors for L ( E ) ↓ H thatare conspicuous for elements of order at most 21 and have positive pressure.(9 , ∗ ) , ¯9 , ¯9 ∗ , , , , , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , , ∗ , (¯9 , ¯9 ∗ ) , (9 , ∗ ) , (9 , ∗ ) , , , , , (27 , ∗ ) , (9 , ∗ ) , (9 , ∗ ) , (9 , ∗ ) , , , , , , , ∗ , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , (3 , ∗ ) , (3 , ∗ ) , , ∗ , , , , ∗ , , ∗ , (¯9 , ¯9 ∗ ) , (9 , ∗ ) , (3 , ∗ ) , , ∗ , , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , (3 , ∗ ) , , ∗ , (3 , ∗ ) , , , ∗ , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , ∗ , , , , ∗ , , ∗ , , , ∗ , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , ∗ , , , , ∗ , , ∗ , , , ∗ , , ∗ , , ∗ , (9 , ∗ ) , , , , , ∗ , . The fourth, sixth and ninth cases are not conspicuous for elements of order 63,leaving eight cases to consider.
Cases 1, 2, 5 and 8 : There are elements of order 189 that cube to x and havethe same number of distinct eigenvalues on L ( E ), so we may apply Lemma 3.10to obtain the result. Cases 3, 10 and 11 : In these cases there are elements of order 189 with only twoor four more distinct eigenvalues on L ( E ) than x , so will stabilize many of thecomposition factors of L ( E ) ↓ H .In the third case, there are elements stabilizing the constituent eigenspaces of V ⊕ V ∗ for each V a composition factor of L ( E ) ↓ H except for 27 ± , and furthermore,8 has no extensions with other composition factors of L ( E ) ↓ H , so it splits offas a summand, and H is strongly imprimitive. (In fact, there are nine elementsstabilizing all eigenspaces of all factors simultaneously, except for 27 ± of course,and 8 .)In the tenth case, there are elements stabilizing the constituent eigenspaces of V ⊕ V ∗ for each V a composition factor of L ( E ) ↓ H except for 24 ⊕ ∗ . However,24 ± has no extensions with other composition factors, and so these must split offas summands, proving the result.The eleventh case is very similar to the tenth: there are elements stabilizingthe constituent eigenspaces of V ⊕ V ∗ for each V a composition factor of L ( E ) ↓ H except for 27 ± . However, both 8 and 8 have no extensions with other compositionfactors, proving that H is strongly imprimitive. (In fact, there are nine elementsstabilizing all eigenspaces of all factors simultaneously, except for 27 ± of course,and 8 .) E Case 7 : The 64 splits off as it has no extensions with other composition factorsin L ( E ) ↓ H . We prove that H stabilizes a line on L ( E ), so assume this is not thecase.Since ¯9 ± has zero 1-cohomology, the pressure of this module is 2. Let W denotethe { ± , ± } -heart of L ( E ) ↓ H . Since H does not stabilize a line on L ( E ), themodule W has all four trivial composition factors in it. The socle of W is at mosttwo modules from 9 ± , 9 ± and 9 ± .Suppose that 9 ± does not lie in the socle of W . The cf( L ( E ) ↓ H ) \{ ± } -radicalof P (9 ) is 27 / ¯9 / ∗ / / ∗ , ∗ , / , ∗ , ¯9 / , ¯9 ∗ , / . We immediately see that soc( W ) cannot have two factors, as then a pyx for rad( W )is a sum of two of these modules (up to graph automorphism). However, W hasfour trivial factors, and the pyx only has two.In fact, soc( W ) cannot be 9 either: one cannot place a copy of 9 ∗ on topof the radical above, and can only place 9 again. However, then the quotient of W/ soc( W ) by its { } ′ -radical is again a submodule of the radical above, so westill cannot place 9 ∗ in it. Thus 9 ∗ does not lie in W at all, a clear contradiction.Thus 9 lies in soc( W ) (up to application of a graph automorphism). However,the cf( L ( E ) ↓ H ) \ { ± } -radical of P (9 ) is1 / / ∗ , / , ¯9 / . But on here we cannot place a copy of 9 ∗ , so the { ± } -heart of W is 9 ⊕ ∗ .In particular, this means that 9 cannot lie in the socle of W , which is a finalcontradiction.Thus H stabilizes a line on L ( E ), and in particular is strongly imprimitive byLemma 3.5. q = 16 : Up to field automorphism, there are ten sets of composition factors thatare conspicuous for elements of order at most 17, namely9 , ∗ , , , (9 , ∗ ) , , , , (¯9 , ¯9 ∗ ) , , ∗ , , , , (9 , ∗ ) , ¯9 , ¯9 ∗ , , , , , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , (9 , ∗ ) , , , , , , , ∗ , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , , , ∗ , , ∗ , , , , , , ∗ , , ∗ , , , , , , ∗ , (9 , ∗ ) , , ∗ , (¯9 , ¯9 ∗ ) , , ∗ , . (Here, 27 a,b,c is the module for H that is the product of 3 a , 3 ± b and 3 ± c (the othersare faithful modules for SL (16)), so 27 is 3 ⊗ ∗ ⊗ and so on.)Let x be an element of order 85 in H . Using Lemma 3.11, if we can find morethan two elements of order 255 cubing to x in G and stabilizing the same subspacesof L ( E ) as x , then H is strongly imprimitive.We can easily check that there are such elements for the first nine cases, leavingonly the tenth. In this final case, there are at least eight elements of order 255stabilizing the eigenspaces associated to each of 9 ± , 9 ± , ¯9 ± and 9 ± . Thus ifsoc( L ( E ) ↓ H ) consists of modules other than 64 and 27 ± , then we may applyLemma 3.11 to obtain that H is strongly imprimitive. Furthermore, the module64 splits off as a summand because it has no extensions with other compositionfactors, so the socle of L ( E ) ↓ H must be a sum of 64 and (up to duality) 27 . .2. PSU Thus we take the cf( L ( E ) ↓ H ) \ { ± } -radical of P (27 ), and this is¯9 / , ∗ , ¯9 ∗ , ∗ / ∗ , ¯9 / . Since this obviously does not have enough trivial composition factors, we must havemore modules in the socle, and so H is indeed strongly imprimitive, as needed. (cid:3) When q = 3, the three remaining conspicuous sets of composition factors arewarranted. A warrant for the first set of factors is given by a diagonal PSL (3) in E A acting irreducibly on both M ( A ) and M ( E ), a warrant for the second setof factors is given by a PSL (3) subgroup of D acting with factors 7 , , ∗ , on M ( D ), and a warrant for the third set of factors is given by a copy of PSL (3)inside A acting on M ( A ) as 3 ⊕ ∗ .For q = 4, both remaining sets of composition factors are warranted, lying in A . For the first, act on M ( A ) as 1 ⊕ . For the second, H does not act on M ( A ),but rather SL (4) acts faithfully and irreducibly, which yields a representation ofPSL (4) in G . PSU The next proposition proves almost all that we know about PSU ( q ). We leaveopen cases for q = 3 , ,
8, two cases for the first two and one for the third. Thesecond open case for PSU (4), that with no trivial factors, is solved in Section12.3 below, but at the time of writing no solutions exist for the other four sets ofcomposition factors given here. We continue with our definition of u from the startof the chapter. Proposition . Let H ∼ = PSU ( q ) for some ≤ q ≤ or q = 16 . (1) If q = 3 then either H stabilizes a line on L ( E ) or the composition factorsof L ( E ) ↓ H are one of (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , , , (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , . (2) If q = 4 then either H stabilizes a line on L ( E ) or the composition factorsof L ( E ) ↓ H are one of , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , (3 , ∗ ) , (3 , ∗ ) , , , ∗ , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , (3 , ∗ ) , (3 , ∗ ) . (3) If q = 5 then H stabilizes a line on L ( E ) . (4) If q = 7 then H either stabilizes a line on L ( E ) or is a blueprint for L ( E ) . (5) If q = 8 then either H is strongly imprimitive or the composition factorsof L ( E ) ↓ H are , ∗ , (9 , ∗ ) , (9 , ∗ ) , (9 , ∗ ) , , , , . (6) If q = 9 , then H is strongly imprimitive. Proof. q = 7 : The fourteen sets of composition factors for L ( E ) ↓ H that areconspicuous for elements of order up to 24 are8 , (3 , ∗ ) , , , ∗ , , , , (6 , ∗ ) , (3 , ∗ ) , , , ∗ , , (6 , ∗ ) , (3 , ∗ ) , , , (15 , ∗ ) , (10 , ∗ ) , , (6 , ∗ ) , , , ∗ , , , , , (35 , ∗ ) , , (10 , ∗ ) , , E , , ∗ , (15 , ∗ ) , (10 , ∗ ) , , , ∗ , (3 , ∗ ) , , , (10 , ∗ ) , , , , (28 , ∗ ) , , , (15 , ∗ ) , , ∗ , , (6 , ∗ ) , (3 , ∗ ) , , , ∗ , , , , ∗ , , ∗ , , ∗ , , ∗ , , , ∗ , , , (10 , ∗ ) , , , ∗ , , ∗ , , ∗ , , ∗ , . As the only simple modules with non-zero 1-cohomology have dimension 36 (seeTable 5.4), we see that all but the last set of composition factors has negativepressure, so that H stabilizes a line on L ( E ). In the last case, there are noextensions between the factors, so that L ( E ) ↓ H is semisimple, and u acts on L ( E )with blocks 6 , , , , , . Hence u lies in the generic class 2 A + 2 A by[ Law95 , Table 9], so H is a blueprint for L ( E ) by Lemma 2.2. q = 5 : The nine conspicuous sets of composition factors for L ( E ) ↓ H are10 , ∗ , , , , , , ∗ , , , , ∗ , , , , , , , , , (35 , ∗ ) , (10 , ∗ ) , , , (10 , ∗ ) , , , , (10 , ∗ ) , , , (35 , ∗ ) , , ∗ , , , , , ∗ , , ∗ , . The pressures of these modules are −
28, 2, 4, 1, −
4, 3, 0, − Cases 1, 4, 5, 7 and 8 : In the first, fifth, seventh and eighth cases H stabilizes aline on L ( E ). It also does in the fourth case since H ( H,
19) is 2-dimensional and L ( E ) ↓ H has pressure 1, by Proposition 2.5. (See Table 5.3.) Case 3 : Suppose that H does not stabilize a line on L ( E ), and let W denote the { } -heart of L ( E ) ↓ H . As W has pressure 4, there are at most two copies of 19in soc( W ). The { , , , ± } -radical of P (19) is(8.2) 8 / / , , , , , , / , , , , , ∗ / , , , , / , and the sum of two of these must be a pyx for W . This has enough trivial factors,but we can also take the { } ′ -residual of this module, and two copies of it stillmust be a pyx. This residual is19 / / , , , , , ∗ / , , , , / , and two copies of it is clearly no longer a pyx for W . Thus H stabilizes a line on L ( E ). Case 2 : Suppose that H does not stabilize a line on L ( E ). The module 10 onlyhas an extension with 63 from the composition factors of L ( E ) ↓ H , so either 10 or10 ∗ is a submodule of L ( E ) ↓ H . Let W denote the { ± , } ′ -radical of L ( E ) ↓ H .This needs to contain at least six trivial composition factors by Proposition 2.9,and since L ( E ) ↓ H has pressure 2, we cannot have a subquotient 1 ⊕ or 19 ⊕ byProposition 2.5. Since W has at least six trivial factors and at most two in eachsocle layer, it must have trivial factors in at least three different socle layers.By quotienting out by any 8s, we may assume that soc( W ) = 19, so a pyx for W is the { , , } -radical of P (19). This is a submodule of the module in (8.2),and this does not have trivial modules in three distinct layers. Thus H stabilizes aline on L ( E ). Case 6 : Suppose that H does not stabilize a line on L ( E ). The compositionfactors 10 and 10 ∗ have no extensions with 1 , ,
19, so L ( E ) ↓ H splits into twosummands. Since the pressure of L ( E ) ↓ H is 3 and H ( H,
19) is 2-dimensional, .2. PSU we cannot have 19 ⊕ as a subquotient of L ( E ) ↓ H . Quotient out by the { , ± } -radical of L ( E ) ↓ H to leave a module W , whose socle must be copies of 19, andhence simply 19. The { , , } -radical of P (19) is1 , , , , / , , , / , , , , / , and again this is a pyx for W . However, W cannot have a trivial quotient, sothose three trivial quotients can be removed from this and it is still a pyx for W .However, W has three trivial factors and the pyx has two, which is a contradiction.Thus H stabilizes a line on L ( E ), as claimed. Case 9 : We show that this does not exist. Let L denote a copy of PSL (5) in H :the restrictions of 8, 10 ± , 35 ± and 63 to L are the sum of a projective module,and 3, the zero module, the zero module, and 3 respectively. Hence if v denotes anelement of order 5 in L then there are at least 46 blocks of size 5 for the action of v on L ( E ), and as the only modules that can be made from 3 are 3 and 3 /
3, theaction of v on L ( E ) is 5 i , − i , i for some 0 ≤ i ≤
3. There is no class actingwith these blocks in [
Law95 , Table 9] (see also Table 2.6), hence H cannot embedin G with these factors. q = 3 : There are fifteen conspicuous sets of composition factors for L ( E ) ↓ H :7 , (3 , ∗ ) , , , , ∗ , , , (6 , ∗ ) , (3 , ∗ ) , , , ∗ , , (6 , ∗ ) , (3 , ∗ ) , , , , ∗ , , (6 , ∗ ) , (3 , ∗ ) , , , , ∗ , , (6 , ∗ ) , (3 , ∗ ) , , , (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , , , , (3 , ∗ ) , , , , ∗ , , , ∗ , (3 , ∗ ) , , (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , , , (15 , ∗ ) , , (6 , ∗ ) , , , (15 , ∗ ) , , (6 , ∗ ) , , , (15 , ∗ ) , , (6 , ∗ ) , , (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , , , (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , . The pressures of the above modules are − − −
16, 0, −
3, 6, 5, 7, −
6, 6, 9, 6,3, 8 and 5 respectively.
Cases 1, 2, 3, 4, 5 and 9 : For these cases H stabilizes a line on L ( E ) byProposition 2.5.The socle layers of P (6) and P (7) are61716 ⊕ ∗ ⊕ ∗ ⊕ ⊕ ⊕ ∗ ⊕ ∗ ⊕ ⊕ ∗ ⊕ ⊕ ⊕ ⊕ ∗ ⊕ ⊕ ⊕ ⊕ ∗ ⊕ ∗ ⊕ ∗ ⊕ ⊕ ⊕ ⊕ ∗ ⊕ ⊕ ∗ W denotes the { ± , } -heart of L ( E ) ↓ H , then W is a submodule of copies of P (6), P (6 ∗ ) and P (7). Therefore, trivial composition factors can occur on thesecond, fourth, sixth and eighth socle layers of W , with the eighth layer occupiedif and only if there is a projective summand. E Cases 11, 12 and 13 : In these cases the module 3 does not lie in L ( E ) ↓ H . The { ± } ′ -radicals of P (6) and P (7) are1 / , ∗ / / , / , ∗ / , / / , ∗ / / . From this it is easy to see that either H stabilizes a line on L ( E ) or we need atleast twice as many copies of 6, 6 ∗ and 7 as trivial factors in W , i.e., the pressureneeds to be at least the number of trivial factors. This latter statement does nothold, so H stabilizes a line on L ( E ) in all three cases. Case 8 : We have no 6 or 15, so we need the { , ± , } -radical of P (7), which is3 , ∗ / , / , , ∗ / . Either H stabilizes a line on L ( E ) or we need at least twice as many 7s as 1s. Thislatter statement is not the case, so H indeed stabilizes a line on L ( E ).We are left with the sixth, seventh, tenth, fourteenth and fifteenth cases, allof which have all modules from the principal block of H as composition factors, soall cf( L ( E ) ↓ H )-radicals are the entire projective modules and are therefore not ofmuch use. To help the reader, we repeat them:27 , , ∗ , , (6 , ∗ ) , (3 , ∗ ) , , , (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , , (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , , (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , , , (15 , ∗ ) , , (6 , ∗ ) , (3 , ∗ ) , . In these cases we let M denote a copy of PSL (2) in H . All simple modules for H other than 1 and 7 restrict projectively to M , and these two restrict irreducibly.The projective covers have structures 1 / / / /
7, so it is easy to determinethe possible actions of M on L ( E ) given the unipotent class to which v ∈ M oforder 3 belongs.We also require the permutation module P M on the cosets of M , which hasstructure 1 / , ∗ / ((1 / / ⊕ / , ∗ / . By Frobenius reciprocity, the spaces L ( E ) M and Hom kH ( P M , L ( E )) have thesame dimension, hence we are interested in the quotients of P M . Assuming that H does not stabilize a line on L ( E ), we have quotients1 / ± , / , ∗ , / , ∗ / / , / , ∗ / / / / ± , P M / soc( P M ) . The space of homomorphisms from P M to these quotients has dimension 1, 1, 1, 2and 3 respectively.We claim that the dimension of L ( E ) M is at most the multiplicity of 6 as acomposition factor of L ( E ) ↓ H . To see this, first note that from the structure of kM -modules, we see that if we quotient out a kM -module by a trivial submodule,the fixed-point space of it always decreases by exactly 1.Let V be a self-dual kH -module such that V H = 0. Let U ∼ = 1 / , ∗ / / U ∼ =1 / , ∗ / / / / U be the quotient P M / soc( P M ). Note that Ext kH (1 , U i ) = 0and Ext kH (6 ± , U i ) = 0 for all i = 1 , ,
3, so if U i is a submodule of V , then everysubmodule 1 or 6 ± of the quotient V /U i comes from a 1 or 6 ± submodule of V .Consider the quotient module V /U i . We see that the dimension of ( V /U i ) M isdim( V M ) − dim( U Mi ), and the number of factors 6 in V /U i is the number of thosein V minus the number in U i . However, if the socle of U i contains 6 ∗ , then thereis a corresponding quotient 6 of V that may be removed as well. Thus we obtain .2. PSU a module V ′ /U i such that the dimension of ( V ′ /U i ) M is at most the number ofcopies of 6 in V ′ /U i .We therefore quotient out by any copies of U i (and remove any necessary 6s fromthe top of V ) to obtain a new module V , which is no longer self-dual but we maystill repeat the process. If U ∼ = 1 / ± then we can quotient out by these, and againremove 6s from the top for every 1 / ∗ in the socle, as again dim(Ext kH (1 , U )) =dim(Ext kH (6 ± , U )) = 0. The last remaining case is U ∼ = 1 / , ∗ , which of coursenow has dim(Ext kH (1 , U )) = 1, but removing a single copy of this from the socleof V both reduces dim( V M ) by 1 and removes a copy of 6 from V , without havingto worry about any new trivial submodules. (Indeed, if a new trivial submoduleappears, then there was a submodule (1 / ⊕ (1 / ∗ ) in the first place, which wouldhave been removed at the previous stage.)The conclusion is that one can continually remove submodules until dim( V M ) =0, and at each stage remove a single 6 as well. Thus the number of 6s must be atleast dim( V M ), as claimed.We therefore have that if the dimension of L ( E ) M is greater than half of thenumber of 6-dimensional factors (equivalently greater than the number of copiesof 6 that are factors) then L ( E ) H = 0, as needed. For most choices of the set ofcomposition factors and unipotent class of v , this is impossible, as we will show. Case 6 : Excluding those projective summands that arise from composition factorsof L ( E ) ↓ H , the composition factors of L ( E ) ↓ M are 7 , . The projectives thatarise from composition factors have dimension 156, so yield blocks 3 . Each copyof 7 yields another two blocks of size 3, so there are at least 52 + 22 = 74 blocks ofsize 3 in the action of v on L ( E ). Thus v lies in one of 2 A (acts on L ( E ) withblocks 3 , ), 2 A + A (acts with blocks 3 , , ) and 2 A + 2 A (acts withblocks 3 , ), and the dimension of the fixed-point space L ( E ) M in the threecases must be 11, 9 and 7 respectively.(For example, if v lies in 2 A + A then the composition factors 7 , mustform a module on which v acts with blocks 3 , , . The two blocks of size 2must arise from (7 / ⊕ (1 / / / / / / L ( E ) M is always 9, as there arefifteen trivials in total. A similar argument works for the other two classes.)Thus there are only five copies of 6 in L ( E ) ↓ H , but the dimension of L ( E ) M is at least 7. Hence by the argument above, H stabilizes a line on L ( E ). Case 7 : This is very similar; instead of 7 , we have 7 , . If v lies in class2 A or 2 A + A then Hom kH ( V, L ( E ) ↓ H ) has dimension 11 or 9 respectively.Notice that v cannot belong to class 2 A + 2 A as the 2 contribution to v is(1 / ⊕ ⊕ (7 / ⊕ , and then we must construct a projective module from 7 , ,which is not possible. There are six copies of 6 in L ( E ) ↓ H , so H must stabilize aline on L ( E ). Case 10 : The proof is again very similar. We have 7 , , and the dimension ofHom kH ( V, L ( E ) ↓ H ) is 8, 6 and 4 if v lies in class 2 A , 2 A + A and 2 A + 2 A respectively. This time there are three copies of 6 in L ( E ) ↓ H , and thus H stabilizesa line on L ( E ) again. E Cases 14 and 15 : These are left over, and are stated as in the proposition. q = 9 : There are, up to field automorphism, 28 sets of composition factors for L ( E ) ↓ H that are conspicuous for elements of order at most 20. Letting x denotean element of order 80 in H , note that the eigenvalues of x , x and x on L ( E )determine the semisimple class in G , by a computer check.We will apply Lemma 3.10 to prove that some cases are strongly imprimitive:for thirteen of the 28 cases, we actually find elements of order 160 that have thesame number of distinct eigenvalues on L ( E ) as x , so H is certainly stronglyimprimitive in these cases.The remaining fifteen cases are as follows:18 , ∗ , (9 , ∗ ) , , , , ¯18 , ¯18 ∗ , (¯9 , ¯9 ∗ ) , , , , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , (6 , ∗ ) , (3 , ∗ ) , (3 , ∗ ) , , , (21 , ∗ ) , , , (6 , ∗ ) , , ∗ , , , , , , ∗ , , ∗ , , , ∗ , , (18 , ∗ ) , , , (3 , ∗ ) , , ¯45 , ¯45 ∗ , , ( ¯18 , ¯18 ∗ ) , , , (3 , ∗ ) , , , ∗ , , ∗ , ¯18 , ¯18 ∗ , , ∗ , , ∗ , ¯9 , ¯9 ∗ , , , , ∗ , (3 , ∗ ) , (3 , ∗ ) , , , ∗ , , ∗ , , ∗ , ¯18 , ¯18 ∗ , , ∗ , ¯9 , ¯9 ∗ , , , (3 , ∗ ) , (3 , ∗ ) , , , ∗ , , ∗ , , ∗ , ¯18 , ¯18 ∗ , , ∗ , ¯9 , ¯9 ∗ , , , (3 , ∗ ) , (3 , ∗ ) , , , ∗ , ¯18 , ¯18 ∗ , (15 , ∗ ) , (9 , ∗ ) , , (3 , ∗ ) , , , ∗ , , , ∗ , (¯9 , ¯9 ∗ ) , , (6 , ∗ ) , , ∗ , , , , ∗ , , ∗ , , ∗ , , , (3 , ∗ ) , , , , ∗ , , ∗ , , ∗ , , , , ∗ , , , ¯45 , ¯45 ∗ , ¯18 , ¯18 ∗ , ¯18 , ¯18 ∗ , , , , ∗ , . The pressures of these are − − − −
1, 7, 1, 1, 2, 4, 4, 2, −
1, 3, 1 and 1respectively.
Cases 1, 2, 3, 4 and 12 : In these cases H stabilizes a line on L ( E ) by Proposition2.5. Cases 6, 7, 14 and 15 : In these cases the pressure is 1. In each case, H stabilizesa line on L ( E ) by Corollary 2.8. Cases 5, 8, 9, 10, 11 and 13 : For the other cases, the quickest way to prove theresult is to find all composition factors whose constituent eigenspaces for the actionof x are stabilized by elements of order 160 that square to x . (Since semisimpleelements in E are real, if M is stabilized so is M ∗ .) We obtain the following table. .2. PSU Case number Modules stabilized by an element5 1, 3 ± , 3 ± , 7 , 7 ± , 3 ± , 6 ± , 7 , 7 , 15 ± ± , 3 ± , 7 , 7
10 1, 3 ± , 3 ± , 7 , 7 , 9 ± , ¯9 ± , 21 ± , 21 ±
11 1, 3 ± , 7 , 9 ± , 15 ±
13 1, 3 ± , 7 , 7 , 21 ± In Case 5, there are, in fact, sixteen elements of order 160 that stabilize all of theeigenspaces comprising 3 ± , 3 ± , 7 and 7 , so we even obtain N Aut + ( G ) ( H )-stabilityif one of these is a submodule of L ( E ) ↓ H . However, in Cases 9 and 10 this is notthe case, so we will have to be more careful.In the other cases, if any of the modules above lie in the socle of L ( E ) ↓ H , then H is strongly imprimitive by Lemma 3.10. Case 5 : The socle of L ( E ) ↓ H must be 49 if H is not strongly imprimitive. Weneed the { , ± i , i } -radical of P (49), which is1 , , ∗ , , ∗ / , / . Clearly we cannot produce a module with socle 49 and six copies of 7 in it, so H is strongly imprimitive in this case. Case 8 : The only extensions that involve 9 ± or 6 ± , and other composition factorsof L ( E ) ↓ H , are Ext kH (6 , ) (and its cognates such as 6 ∗ , ∗ ). Hence the factors9 and 6 break off as a separate summand. Thus, in particular, either 6 or 6 ∗ isa submodule of L ( E ) ↓ H , and hence H is strongly imprimitive. Cases 9 and 10 : We first show that H is Lie imprimitive (we cannot show strongimprimitivity directly because we cannot apply Lemma 3.10 with an N Aut + ( G ) ( H )-orbit of spaces), and then use the collection X to show that H in fact stabilizes aline on L ( E ).In Case 9, the 9-dimensional factors must split off as summands because theyhave no extensions with the other factors. For the remaining summand, the soclemust consist of at most one copy each of 18 ± , ¯18 ± , 21 ± and 21 ± , else we mayapply Lemma 3.10 to obtain that H is Lie imprimitive.The cf( L ( E ) ↓ H ) \ { V ± } -radicals of P ( V ), for V each of 18 , ¯18 , 21 and21 , are 3 / / / ∗ / ∗ / ∗ / ¯18 , / , , ¯18 ∗ / , , ∗ / , , , , / , , ∗ / . As H acts on L ( E ) with eight 3-dimensional factors we need all four of these mod-ules (up to duality) in the socle of L ( E ) ↓ H . However, then the 21 appearing inthe third layer of the first module cannot appear in L ( E ) ↓ H , and we consequentlylose a 3-dimensional factor. Since we now have only seven in total, there must beanother module in the socle, and H is Lie imprimitive.Case 10 is easier. There are no extensions between the elements of the set { ± , ¯9 ± , ± , ¯18 ± } and the other composition factors of L ( E ) ↓ H , and thereforethere is a summand of the module that has none of these as composition factors.The others are all in the table above, and therefore H is at least Lie imprimitiveby Lemma 3.10. E We now prove that H always stabilizes a line on L ( E ), and is thus stronglyimprimitive by Lemma 3.5. We will use the fact that H lies in a member of X .If H lies in E A then H stabilizes a line on L ( E ), as E A does (as p = 3).Similarly, if H lies in an E -parabolic subgroup then H stabilizes a line on L ( E ).Since H does not embed in A , if H lies in A E then H lies in E and againstabilizes a line on L ( E ).Suppose that H embeds in A . Note that it is easy to write down all compo-sition factors of kH -modules of dimension at most 9, and therefore compute thefactors of H on L ( E ), which splits as L ( λ ), L ( λ ) = L ( λ ) ∗ and L ( λ + λ ). Thedimensions of the factors of H on M ( A ) are one of: 9; 7 , ; 6 ,
3; 6 , ; 3 ; 3 , ;3 , . In each case there are multiple possible factors, but it is easy to computetheir factors on L ( E ) and none matches Cases 9 or 10. (For example, if H actsirreducibly on M ( A ) then we obtain Case 5, if it acts as 3 ⊕ then we obtainCase 7, and if it acts as 3 ⊕ ∗ ⊕ then we obtain Case 3.)If H lies in A A then it must act as 3 ± i ⊕ ⊕ on each M ( A ), which pushes H into a subgroup of type A A that has the same composition factors on L ( E )as one lying in A . In particular H cannot embed in A A either.If H embeds in D then there are fewer possible actions on M ( D ) since itis self-dual. The dimension sets are: 7 , ; 6 , ; 3 , ; 3 , . If 1 lies in M ( D ) ↓ H then there are at least six trivial factors on Λ ( M ( D )) ↓ H , which is asummand of L ( E ) ↓ H , and so we cannot have those cases. The actions 7 ⊕ ⊕ ⊕ and 7 ⊕ ⊕ ⊕ have 27 and 49 respectively in their exterior squares, so in fact H cannot embed in D either.This is enough to eliminate all parabolic subgroups: E has already been con-sidered; for A E , H lies in an E -parabolic, hence an E -parabolic; H cannotlie in a D -parabolic as H has the same factors as a subgroup D ; similarly, H cannot lie in A A - or A A A -parabolics because H would have the same factorsas a subgroup of A A ; H cannot lie in an A - or A A -parabolic subgroup as H would have the same composition factors as a subgroup of A ; H cannot lie in an A D -parabolic subgroup because H would have the same composition factors asa subgroup of A D , which is contained in D .Suppose next that H is contained in the F maximal subgroup X . The easiestmethod here appears to be to check which subsets of the composition factors forma 52-dimensional module that is conspicuous for elements of order at most 8 for F .The only sets have dimensions 18 , , , and there are several of these. Of course,this has pressure 0, so H must stabilize a line on the 52-dimensional summand L ( F ) of L ( E ) ↓ X , as needed.The only remaining subgroup is G F . In characteristic 3, this has a structure((10 , / (10 , , (01 , / (10 , ⊕ (00 , . (See, for example, [ LS04a , Table 10.1].) Suppose that H acts as 7 on M ( G ). Wecan consider the allowed composition factors of M ( F ) ↓ H by seeing which, whentensored by 7 , appear on our list (up to field automorphism, since we have fixed7 ). Indeed, only 1 and 3 are allowed, and so H cannot act as 7 on M ( G ). Theonly other possibility is that H acts as 3 ⊕ ∗ ⊕
1, in which case H stabilizes a lineon L ( E ), as claimed. Case 11 : The only extensions between any of 9 , 15 and ¯18 ∗ and other compo-sition factors of L ( E ) ↓ H are Ext kH (9 , ) and Ext kH (9 , ¯18 ∗ ), both of which .2. PSU are 1-dimensional. Thus, as in Case 8, either 9 or its dual is a submodule of L ( E ) ↓ H , and H is strongly imprimitive. Case 13 : If H is not strongly imprimitive then the socle of L ( E ) ↓ H consists solelyof copies of 15 ± , 42 ± and 49. Note that 15 has extensions with only 42 ∗ and 49from the composition factors of L ( E ) ↓ H , so if 15 ± does lie in the socle we mayquotient out by it without altering whether L ( E ) ↓ H only has allowed modules inthe socle. Similarly, if 49 lies in the socle then it is a summand, so in constructing apyx for L ( E ) ↓ H , we may assume that it is a submodule of P (42 ). We thereforerequire the cf( L ( E ) ↓ H ) \ { ± } -radical of P (42 ), which is1 / ∗ , , ∗ / . Clearly this cannot work, as it has no 7 for example, and therefore H is stronglyimprimitive.This completes the proof that H is always strongly imprimitive, as needed. q = 4 : There are 29 conspicuous sets of composition factors for L ( E ) ↓ H , of whichfourteen have positive pressure or no trivial composition factors, eight up to au-tomorphism of H . Note that—with our labelling convention from Chapter 5—thegraph automorphism acts as an automorphism of order 4,9 → ¯9 ∗ → ∗ → ¯9 . The eight sets of composition factors are(9 , ∗ ) , , , (3 , ∗ ) , , ∗ , , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , (3 , ∗ ) , (3 , ∗ ) , , , (9 , ∗ ) , , (3 , ∗ ) , (3 , ∗ ) , , , , ∗ , (9 , ∗ ) , ¯9 , ¯9 ∗ , , (3 , ∗ ) , (3 , ∗ ) , , , ∗ , (¯9 , ¯9 ∗ ) , , (3 , ∗ ) , (3 , ∗ ) , , , ∗ , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , , ∗ , , ∗ , , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , (3 , ∗ ) , (3 , ∗ ) , , , ∗ , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , (3 , ∗ ) , (3 , ∗ ) . Let W denote the { ± i , ± i,j } ′ -heart of L ( E ) ↓ H , and W the { ± i , i , ± i,j } ′ -heartof L ( E ) ↓ H . Of course, since the only simple modules with non-zero 1-cohomologyare 9 ± and ¯9 ± (see [ Sin92a ]) these contain all trivial modules, and have a trivialsubmodule if and only if L ( E ) ↓ H does. Let L ∼ = Alt(5) × H ,and write z for a central element of order 5 in L .We assume in each case that H does not stabilize a line on L ( E ) and derive acontradiction, or the case is left over and stated in the proposition. Case 1 : The pressure is 2, but the { , ± i , i , ± } -radical of P (9 ) has only twotrivial composition factors. Thus since soc( W ) can have at most two 9s, we cannotbuild a pyx for W . Hence H stabilizes a line on L ( E ) by Proposition 2.5. Case 3 : The 64 splits off as it is projective, and L ( E ) ↓ H has pressure 4. Wecompute the { ± } ′ -residual of the { , ± i , , ± } -radical of P (9 ). This is9 / / ∗ , ∗ / , / ∗ , / , , ∗ / , , ∗ / , ∗ , / . As there are only two 1s in this module, and eight in W , we must have four 9sin the socle of W . In particular, all eight 1s in the sum of four modules of the E form above are required, so we take the { } ′ -residual of this module, as all of thatmodule is a submodule of W . This residual is1 / , ∗ / , / . In particular, this means that we need eight 9s in the third socle layer of W , but W only has pressure 4. Thus H stabilizes a line on L ( E ) by Proposition 2.5. Case 4 : Again, the 64 splits off, and this time W has pressure 2 and six trivialfactors. The { , ± } ′ -radicals of P (9 ) and P (¯9 ) have five and three trivialfactors respectively, so soc( W ) cannot be simple.If ¯9 (up to duality) appears in the socle of W , then it cannot appear inthe heart of it. In this case, we replace the { , ± } ′ -radical of P (9 ) by the { , ± i , , ± } -radical of P (9 ), which possesses only two trivial factors (see Case3), another contradiction. Thus soc( W ) is (up to duality) 9 ⊕ ± .Suppose first that the socle of W is 9 ⊕ . Thus the top of W is (9 ∗ ) ⊕ , so wemay take the { ∗ } ′ -residual of the previous radical and still produce a pyx for W (from two copies of it). This residual only contains three trivial factors, so all sixof the trivial factors from the pyx lie in W . Thus we take the { } ′ -residual, as inCase 3, and find a guaranteed submodule of W . This is1 / , / ∗ , ¯9 , ¯9 ∗ / , ∗ / , ∗ / . Of course, the subquotient 9 ⊕ ¯9 ⊕ ¯9 ∗ of this already has pressure 3, so since W has pressure 2, H must stabilize a line on L ( E ) by Proposition 2.5. Thus soc( W )must be 9 ⊕ ∗ .There are five trivial composition factors in the { , ± } ′ -radical W ′ of P (9 ),in the second, fourth, fifth, sixth and eighth socle layers. We need to prove thatthree of these cannot lie in W to obtain a contradiction.Since there are two 9 ± in both the socle and the top of W , that leaves onlyone more in the heart of it. Thus we replace our { , ± } ′ -radical of P (9 ) bythe submodule obtained by adding all copies of 1, 3 ± i , ¯9 ± and 24 ± on top of 9 ,then either 9 or 9 ∗ (but not both), then 1, 3 ± i , ¯9 ± and 24 ± , then the other from9 and 9 ∗ , then 1, 3 ± i , ¯9 ± and 24 ± . Doing so yields a module with four trivialfactors, not five. Furthermore, in both cases there is a 1 ⊕ subquotient. Since thepyx for rad( W ) is the sum of this and its image under the graph automorphism,there is a subquotient 1 ⊕ in this, so we need to remove another four trivials, sotwo from each summand. This means we are left with four trivial factors from thetwo summands combined, and that is too few for W . Thus H stabilizes a line on L ( E ). Cases 2, 5 and 6 : These cases, and Case 8, which is delayed until Section 12.3,require the subgroup L , the centralizer of an element z of order 5 in H . We split L ( E ) according to the eigenspaces of the action of z , as L ( E ) ↓ L splits into fivesummands, one for each eigenvalue. Each eigenspace is then a module for thesubgroup L ′ ∼ = Alt(5), so we can use the simple k Alt(5)-modules 1, 2 , 2 and 4 todescribe it. Obviously we have to make a choice as to which eigenvalue is labelledby the fifth root of unity ζ , and we do so in such a way as to make the restrictionsto L ′ as in Table 8.1. .2. PSU Module 1-eig. ζ -eig. ζ -eig. ζ -eig. ζ -eig.1 13 ∗ ∗ / / / / ∗ ∗ / / P (2 ) 2 ⊕ ∗ ⊕ P (2 ) 1 / / / / ⊕ P (2 )24 ∗ P (2 ) 2 ⊕ / / Table 8.1.
Actions of L ′ ∼ = Alt(5) on each ζ i -eigenspace of theaction of z The permutation module P L on the cosets of L is of the form(1 , , / , ∗ , ¯9 , ¯9 ∗ / , ∗ , , ∗ / , ∗ , , ∗ / , , , ∗ , , ∗ / , ∗ , ¯9 , ¯9 ∗ / , , ) ⊕ , but (ignoring the 64) the only quotient of this not involving either 8 i at the top issimply 1 itself. The largest quotient not involving 8 is1 , / ¯9 , ¯9 ∗ / , ∗ / , , ∗ , so the image in W is a quotient of 1 , / ¯9 , ¯9 ∗ and one may check that all suchquotients have L -fixed points of dimension 1.In fact, this is generally true. One may construct all
34 (non-zero) quotientsof P L with only 9-dimensional factors in the socle, and find that they all have 1-dimensional L -fixed points. To compute all such quotients, first quotient out bythe 1, 8 and 8 . Then choose one of the 9-dimensional modules to quotient outby, and then also quotient out by any other simple submodules not of dimension9. Repeatedly do this, and at each stage it turns out that one never has two copiesof the same 9-dimensional module in the socle of the module. Thus there is anexplicit finite list of such quotients (even over the algebraically closed field), andthey can all be restricted to L . It turns out that each of these has L -fixed points ofdimension 1. Furthermore, every such quotient has either two or six socle layers,and if it has six then the top is always 1 ⊕ ⊕ , and all 8 i lie in the fifth or sixthsocle layer.From this we can prove that, if V is a module with V H = 0, then there mustbe at least as many 8-dimensional composition factors in V as the dimension n of V L , using induction on n and noting that the case n = 1 is above. If V has an8-dimensional submodule then quotient it out: both the dimension of V L and thenumber of 8 i in V decrease by 1, so we may assume that V has no 1- or 8-dimensional E submodules. We can also remove any 3-dimensional submodules without affectingeither quantity, so we assume that the socle of V has factors all of dimension 9.We may replace V by the kH -submodule generated by V L without alteringthe truth of the statement. Thus V is a (not necessarily direct) sum of V , . . . , V m ,where each V i is an image of P L . Thus the top of V is a sum of trivial and 8-dimensional simple modules. Let V ′ denote the { i } -residual of V . Since V ′ hasno trivial submodules, or 8 i submodules or quotients, it cannot have any L -fixedpoints. Thus V L has dimension at most that of the L -fixed points ( V /V ′ ) L of thequotient V /V ′ . This dimension is equal to the number of 8-dimensional factors in V /V ′ , proving the claim.We will use this fact, that if V H = 0 then dim( V L ) is bounded above by thenumber of 8-dimensional composition factors, in the remaining cases. Case 2 : We assume that L ( E ) H = 0 and work towards a contradiction. We showthat L ( E ) L has dimension at least 3. The central element z of L has trace 3 on L ( E ), so from [ Fre98 , Table 1.16] we see that the centralizer of z is A A T , andthis is a Levi subgroup of G . Since A A ≤ A , we can understand the action of L ′ on L ( E ) via A , and we see that L ′ must act with factors 2 , , on M ( A ) inorder to have the correct factors on L ( E ). Including the action of h z i , and notingthat h z i acts trivially on the summand L ( A ) ⊕ ( L (0) /L ( A ) ◦ /L (0)) of L ( E ) ↓ A A ,we obtain that L ′ acts as 2 on M ( A ) and with factors 2 , , on M ( A ). Sincethe 1-eigenspace of the action of z on L ( E ) has the structure above, and we wantto compute the fixed-point space L ( E ) L , we need to understand the action of L on M ( A ), not just the factors.From Table 8.1, we see that for the principal block, excluding copies of theprojective 4, L acts the same on both L ( E ) and W (but not W ). We claim that L must centralize at least a 3-space on L ( E ), hence on W . To see this, notethat L centralizes a 1-space on L (0) /L ( A ) ◦ /L (0), on which it acts as 1 / /
1. As L ( A ) ⊕ L (0) = M ( A ) ⊗ M ( A ) ∗ , the fixed-point space of L on M ( A ) ⊗ M ( A ) ∗ is the same as that on L ( E ), so we count that quantity.Also, ( M ( A ) ⊗ M ( A ) ∗ ) L ∼ = Hom kL ( M ( A ) , M ( A )), and this is easier tounderstand. The composition factors of M ( A ) ↓ L are 2 , , , so it has pressure −
1, hence L stabilizes a line and a hyperplane on M ( A ). If all trivial compositionfactors lie in the socle or top then we have at least two different homomorphismsfrom the top to the socle, plus the identity, so we have a 3-space of endomorphisms.Thus there is a subquotient 2 / / (up to duality). In this case we can either havetwo trivial summands, so easily enough kL -homomorphisms, or we place a trivialon the 2 / / , to make 1 / / / , but no trivial may be added to the top or socleof this, so we have a trivial summand and again a 3-space of homomorphisms, asneeded.We now claim that 8 ⊕ ⊕ is a subquotient of W . To see this, notice thatall of them must lie in the kH -submodule generated by W L , that this is a sum ofquotients of P L , and all 8 i must lie in the top of this submodule. This proves theresult.We see therefore that W is the sum of W and some 8-dimensional simplesummands.We now compute a specific submodule of both P (9 ) and P (¯9 ). In each case,we take the sum of two submodules: the first is simply the { , ± i , ± , ¯9 ± , ± } -radical of the projective; the second is the { , ± i , ± , ¯9 ± } -radical of the projective, .2. PSU on which we place as many copies of 8 and 8 as we can. On top of this sum, weplace as many copies of 1, 3 ± i , 9 ± and ¯9 ± as we can, and then take the { ± , ¯9 ± } ′ -residual of the resulting module. A sum of these must form a pyx for W . Thetwo modules have dimension 211 and in both cases all trivial and 8-dimensionalcomposition factors lie in either the second socle or the second top. Since there areeight trivial composition factors in W and W has pressure 4, we see therefore thatthere must be exactly four 9-dimensional factors in the socle (and four in the top).In particular, this means that 8 , which occurs exactly once in W , is eithera summand of W or a summand of the heart of W . But now we use the factfrom earlier, that any quotient of P L with 9-dimensional factors in the socle haseither two or six socle layers, and if it has six then 8 lies in the fifth or sixth layer.Since this does not happen in our case, we conclude that all 8 i must in fact liein the second socle and the second radical layer of W . Thus one may constructa pyx for W as a sum of submodules of P (9 ± ) and P (¯9 ± ), formed by takingthe { , ± i , ± , ¯9 ± , ± } -radical of P (9 ± ) and P (¯9 ± ), then add on the 8 j in thesecond layer, then add on any 9-dimensional modules on top of this, then take the { ± , ¯9 ± } ′ -residual of this module.The two modules have the form9 / ∗ / / ∗ / , ∗ / ∗ , ¯9 , ¯9 ∗ / , ∗ , , ∗ / , ∗ , / , ¯9 / / , ∗ , ¯9 / , , ∗ , ∗ / , ∗ , ¯9 , ¯9 ∗ / , , / ¯9 . We see that we cannot obtain a pyx for W as no sum of at most four ofthese, at most three of which have socle ¯9 ± , can have eight trivial factors. Thiscontradiction means that H does indeed stabilize a line on L ( E ), as needed. Case 5 : We proceed as in Case 2, so again assume that L ( E ) H = 0, and againthe trace is 3, so L ≤ A A T . This time the composition factors must be 2 for M ( A ) and 2 , M ( A ). There are only two modules for M ( A ) with thosefactors, up to duality: semisimple and (1 / ) ⊕ ⊕ .In the first case, the 1-eigenspace of the action of z on L ( E ) ↓ L is(1 / / ⊕ ⊕ ⊕ . Thus L ( E ) L has dimension 10, and since there are only eight 8-dimensional com-position factors, we deduce that L ( E ) H = 0, as needed.Thus we are in the second case, where 2 / M ( A ) ↓ L , andthe 1-eigenspace of z on L ( E ) is now(1 / / ⊕ ⊕ (2 / / / ⊕ ⊕ (1 / / / ) ⊕ ⊕ (2 / / / / ) . We may also compute the other eigenspaces, at least up to duality: the module(0 , λ ) restricts to L ′ as1 ⊕ ⊕ ⊕ (2 / ⊕ ⊕ (2 / / / ⊕ ⊕ (2 / , , / ⊕ , the module (1 , λ ) restricts to L ′ as2 ⊕ ⊕ ⊕ ⊕ (2 / / / ⊕ (2 / / , / ⊕ , and (1 , λ ) and (0 , λ ) restrict to L ′ as(1 / / ⊕ ⊕ (1 / / / ) and 2 ⊕ ⊕ (2 / . E From this we see that 2 does not appear as a submodule or quotient in anyeigenspace of the action of x , and hence none of 9 ± , ¯9 ± and 3 ± lies in the so-cle of L ( E ) ↓ H . Since we have no trivial submodules by assumption, the socleconsists of copies of 8 and 3 ± .From this we may determine the possible quotients of P L that can appear in L ( E ) ↓ H . The module 1 , / ¯9 , ¯9 ∗ / , ∗ / , ∗ has a copy of 2 appearing in the ζ - and ζ − -eigenspaces of z , so this cannotoccur. Removing 3 or 3 ∗ from the socle, then any modules that may not appearin soc( L ( E ) ↓ H ) yields modules1 , / ¯9 / / , , / ¯9 ∗ / ∗ / ∗ , and these have a copy of 2 in the ζ − - and ζ -eigenspace respectively. Thus only1 and 8 may be quotients of P L lying in L ( E ) ↓ H . We see from the 1-eigenspacethat there are at most five 8 summands, but unless there is a trivial submodule,exactly seven 8 submodules, and therefore seven 8 quotients. This requires ninecopies of 8 , which is too many, and therefore H stabilizes a line in this case. Case 6 : Now we have that the trace of z is 23, and so L ≤ D T . Again, we assumethat H does not stabilize a line on L ( E ). The actions of L ′ on the eigenspaces of z are given below. Eigenspace Action1 4 , , , . ζ , , , ζ , , , The 1-eigenspace comprises L ( D ) ◦ and four trivial factors, and the only actions of L ′ on M ( D ) with those factors are 2 , ,
4, and 2 , , up to field automorphism.The former case is consistent with the other eigenspaces, which are the sum of atrivial, a copy of M ( D ) and a half-spin module, whereas the trace of an element oforder 5 in L ′ is inconsistent with the latter possibilities. Thus L ′ acts as 4 ⊕ ⊕ ⊕ ⊕ on M ( D ).We can compute the action of D on L ( E ), either by a direct calculation inMagma using L ( E ) and the subsystem subgroup, or by using the unipotent class D in [ Law95 , Table 9], which acts on M ( D ) with blocks 10 , , , ,
2. A computer calculation shows that it acts on L ( D ) ◦ with blocks 16 , , , ,
6, and so comparing these with the action on L ( E ), wesee that we must have a summand L (0) , L (0) /L ( D ) ◦ /L (0) , L (0).The action of L ′ on Λ ( M ( D )) is P (2 ) ⊕ ⊕ P (2 ) ⊕ ⊕ ⊕ ⊕ (1 / , / ⊕ (1 / / ⊕ (1 / / ⊕ ⊕ . As we saw above, we need to add two trivial factors to this to get the contribution tothe 1-eigenspace. There are three unipotent classes in D with blocks 2 on M ( D )[ Law09 , Table 7]. Using a computer, we determine the actions on the half-spinmodules, and L ( D ) ◦ and its extension with the trivials.Class L ( λ ) L ( λ ) L ( λ ) L ( λ ) L (0) , L (0) /L ( λ ) /L (0) , L (0)(3 A ) ′ , , , (3 A ) ′′ , , , A + D , , Since u (which can be chosen to lie in L ) acts on the sum of the composition factorsof L ( E ) ↓ H with blocks 2 , , we see that u lies in class 2 A + D , and thereforethe extension of L ( D ) by trivials is non-split on restriction to L . Thus the actionof L on L (0) , L (0) /L ( λ ) /L (0) , L (0) must be P (2 ) ⊕ ⊕ P (2 ) ⊕ ⊕ ⊕ ⊕ (1 / / ⊕ ⊕ (1 / / ⊕ ⊕ ⊕ . Hence L ( E ) L is 8-dimensional, and there are at most two submodules 8 and twosubmodules 8 . If W is defined as in Case 2, we therefore have (at least) a 4-spaceof homomorphisms from P L to W whose image is not 8 i . Since we only have eight9-dimensional modules, and one is needed for each such submodule (and quotient),this must be exactly what occurs.Thus W consists of four 8-dimensional simple summands, and a module withfour 9-dimensional modules in the socle, four in the top, and other modules inbetween.We now examine the ζ i -eigenspaces of z on L ( E ). As L ′ acts on M ( D ) as4 ⊕ ⊕ ⊕ ⊕ , it lies inside D D ; there are two classes of subgroups L ′ inside eachof these D i acting in this way, swapped by the graph automorphism. Their actionson the half-spin modules are2 , and 2 ⊕ ⊕ ⊕ , ⊕ ⊕ . The half-spin modules for D are sums of tensor products of half-spin modules for D and D , so we get that, up to field automorphism, L ′ acts on a half-spin moduleas 4 ⊕ ⊕ P (2 ) ⊕ (1 / / ⊕ ⊕ ⊕ . Adding on a copy of M ( D ) and a trivial, we get the action on the ζ -eigenspace,which is 4 ⊕ ⊕ P (2 ) ⊕ (1 / / ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ . Examining the table at the start of this section, we see that the P (2 ) and one1 / / ± i,j , so there are likely very few extensions between modulesin L ( E ) ↓ H . The ζ -eigenspace of z on the module 1 , / ¯9 ∗ has L ′ -action 2 /
1, andthe ζ -eigenspace of z on the module 3 ∗ / ¯9 ∗ is 2 /
1, so there can be no extensionbetween 3 ∗ and ¯9 ∗ . By choosing different eigenspaces, there can be no extensionbetween 3s and 9s. The { , i , ± , ¯9 ± , ± i,j } -radical of P (¯9 ∗ ) is1 / ¯9 , ¯9 ∗ / , / ¯9 ∗ , so we can remove the top 1 from this, and just need to decide which 9-dimensionalmodule we want in the third layer. However, the module ¯9 / , / ¯9 ∗ has a trivialquotient, so we need ¯9 ∗ / , / ¯9 ∗ .Thus W must be the sum of 8 ⊕ ⊕ ⊕ and four modules of the form9 ± / , / ± , ¯9 ± / , / ¯9 ± . The 3s must split off, and there are no extensions between 24 ± i,j and the modulesabove, so they must split off, and we obtain the action M i =1 (9 i, − i / , − i / i, − i ) ⊕ (9 ∗ i, − i / , − i / ∗ i, − i ) ⊕ i, − i ⊕ ∗ i, − i ⊕ ⊕ i ⊕ i ⊕ ∗ i . (Here, we write 9 = ¯9 to make the expression easier to understand.) This existsinside E A . We show that H is contained in a positive-dimensional subgroup E stabilizing all 3-dimensional summands, hence all members of an N Aut + ( G ) ( H )-orbit of submodules of L ( E ) ↓ H .To do this, we place L inside a group A × h z i , with the A subgroup Y lyingin D , in fact inside D D . Let Y denote an A subgroup of D acting on M ( D )as L (3), and on the two half-spin modules as L (1) and L (2). Clearly Y can bechosen to contain the projection of L ′ onto the D factor.Let Y denote an A subgroup of D acting on M ( D ) as L (1) ⊕ ⊕ L (2) ⊕ .The actions on the two half-spin modules are as L (1) ⊕ ⊕ L (2) ⊕ and L (3) ⊕ L (0) ⊕ .(To see this, note that we may conjugate Y by the graph automorphism of order3 to lie in a D -Levi subgroup, and then it becomes clear.) Again, we may choose Y to contain the projection of L ′ onto D .Finally, let Y be a diagonal A in D D ≤ D acting as Y on the one factorand Y on the other. It is easy to see that Y and L ′ stabilize all of the same simplesubmodules of the ζ i -eigenspaces of z for i = 1 , , ,
4, and so in particular muststabilize each 3 ± i in L ( E ) ↓ H . This completes the proof. Cases 7 and 8 : These remaining cases are as in the statement of the proposition.Case 8 is solved in Section 12.3, but Case 7 is unsolved. q = 8 : There are, up to field automorphism, eight conspicuous sets of compositionfactors for L ( E ) ↓ H : 9 , ∗ , , , (9 , ∗ ) , , , , (¯9 , ¯9 ∗ ) , , ∗ , , , , , (¯9 , ¯9 ∗ ) , (9 , ∗ ) , , , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , (9 , ∗ ) , , , , , (27 , ∗ ) , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , (9 , ∗ ) , , , , , , , ∗ , , ∗ , (¯9 , ¯9 ∗ ) , (9 , ∗ ) , , , , , ∗ , , ∗ , , , . Note that, with the notation from Chapter 5, Ext kH (¯9 , ∗ ) is non-zero, whereasExt kH (¯9 , ) = 0.The pressures of these are −
28, 0, 4, 8, 10, 4, 6 and 0 respectively.
Cases 1, 2 and 8 : These all have non-positive pressure, so H stabilizes a line on L ( E ) by Proposition 2.5. Case 3 : The { , , , ¯9 ± , ± } -radicals of P (¯9 ) and P (9 ) are¯9 / / ¯9 ∗ , ¯9 ∗ / , , / ¯9 , ¯9 , ¯9 ∗ / , , ∗ / ¯9 , , ¯9 / , , ∗ / , , ¯9 ∗ / . Since L ( E ) ↓ H has pressure 4, we can support at most eight trivial compositionfactors above the modules with non-zero 1-cohomology, which is not enough. Thus H stabilizes a line on L ( E ), as claimed. Cases 4, 6 and 7 : We will work in the centralizer of an element z of order 3, so wegive some information about this now. Let L = C H ( z ) ∼ = h z i × PSL (8), and notethat L ′ ∼ = PSL (8). Any module for L splits as a direct sum of three summands:the 1-eigenspace, ω -eigenspace, and ω -eigenspace of the action of z , where ω = 1.We arrange the modules for H , L and the exact power of z so that the actions of L ′ on the eigenspaces of z are as in Table 8.2. If the trace of z on L ( E ) is − C G ( z ) is the group of type A , if it is 5 then C G ( z ) is of type E A , and if it is 14then C G ( z ) is of type D T (see Table 2.10). .2. PSU Module 1-eigenspace ω -eigenspace ω -eigenspace1 18 / / / / / / ⊕ ⊕ ¯9 ⊕ ∗ ⊕ ∗ ⊕ ¯9 ∗ ⊕
27 4 ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ (1 / , / ⊕ (2 / / ) Same as ω -eigenspace / , , / , / ⊕ (2 / / / / ) Table 8.2.
Actions of L ′ ∼ = PSL (8) on each ω i -eigenspace of theaction of z We also need the permutation module P L on the cosets of L , which has di-mension 3648, so it is too large to describe completely. We give some facts aboutit: • P L has simple quotients precisely 1, 8 i , 64 i,j and 512; • there is no non-trivial quotient of P L with trivial top; • there is no non-semisimple quotient of P L with top 8 ⊕ ⊕ .Suppose that V is a kH -module with composition factors of dimension at most27, and with V H = 0. We show that dim( V L ) is bounded above by the number of8-dimensional composition factors in V , proceeding by induction on the dimensionof V . We may assume that dim( V L ) >
0, of course. This proof will look verysimilar to the proof for q = 4.If V possesses a submodule of dimension 8 then we may quotient it out: thedimension of the L -fixed points must reduce by 1, as must the number of 8-dimensional factors. Thus there are no 8-dimensional quotients. We also mayassume that V is generated by V L . Let V ′ denote the { i } -residual of V . Since the kH -submodule generated by any L -fixed point is either 8 i or has a trivial and an 8 i quotient, ( V ′ ) L = 0. Thus dim( V L ) ≤ dim(( V /V ′ ) L ), where V /V ′ is the quotient.As the result holds for V /V ′ , the result holds for V , as needed.As a consequence, if L ( E ) ↓ H has no trivial submodules then in Case 6, L ( E ) L has dimension at most 3, and in Case 7, L ( E ) L also has dimension at most 3 (sincethe 64 splits off as a summand and has a 1-dimensional L -fixed space). Case 4 : The module 64 splits off as a summand as it has no extensions withother composition factors. The trace of z on L ( E ) is −
4, so that X = C G ( z ) is A . Note that X acts on L ( E ) as the sum of M ( X ) ⊗ M ( X ) ∗ (minus a trivialsummand), Λ ( M ( X )) and Λ ( M ( X ) ∗ ), where M ( X ) is the 9-dimensional minimalmodule for X . The summand M ( X ) ⊗ M ( X ) ∗ is the 1-eigenspace for z .From the composition factors of L ( E ) ↓ L , we easily deduce that the compositionfactors of V = M ( X ) ↓ L ′ are 4 , ,
1, so up to field automorphism 4 , ,
1. There E are two modules with this structure, up to duality:4 ⊕ ⊕ (2 / , ⊕ ⊕ ⊕ . In the former possibility, there is no 4-dimensional simple summand of V ⊗ V ∗ , butthere is one in the restriction of the 1-eigenspace of z on 64 . Thus the action of L ′ on V is the second possibility, so V ↓ L ′ is semisimple. Let Y denote the algebraic A containing L ′ and acting as L (0) ⊕ L (1) ⊕ ⊕ L (3) on V . Using the formulaΛ ( A ⊕ B ) = Λ ( A ) ⊕ (Λ ( A ) ⊗ B ) ⊕ ( A ⊗ Λ ( B )) ⊕ Λ ( B ) , we can easily compute the structure of L ( E ) ↓ Y , and indeed L ′ and Y stabilize thesame subspaces of L ( E ), so that L ′ and H are blueprints for L ( E ). Case 6 : This time z lies in the class with centralizer D T (see Table 2.10), so L ′ ≤ D . The action of D on L ( E ) has factors of dimensions 14, 14, 64, 64, 90,1 and 1, with the final three organized into the shape 1 / /
1. The 1 / / z , as z has trace 14 on L ( E ), and the 90 here is a submodule ofcodimension 1 in the exterior square of the natural module for D .The possible composition factors of M ( D ) ↓ L ′ are • , , , , • , , , , and • , , , .If the factors are the first possibility, then the exterior square is P (1) ⊕ P (4 ) ⊕ P (4 ) ⊕ P (4 ) ⊕ ⊕ ⊕ ⊕ ⊕ . The restriction of 1 / / L ′ must have an extra trivial summand. If the factorsare the second possibility, then the exterior square is P (4 ) ⊕ P (4 ) ⊕ P (4 ) ⊕ (1 / , / ⊕ (1 / , / ⊕ (1 / , / ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ . The restriction of 1 / / L ′ must have an extra trivial submodule and quotient(but the trivial summand is not there). In both possibilities therefore, L ( E ) L isat least 4-dimensional, and we have previously shown that it can be at most 3-dimensional without H stabilizing a line on L ( E ). Thus H stabilizes a line on L ( E ) in both of these possibilities.In the final possibility, M ( D ) ↓ L ′ has 2 i -pressure 0, so it has a submodule (andquotient as it is self-dual) 2 i for each i . These each contribute one trivial submod-ule to Λ ( M ( D )) ↓ L ′ , so we just need one further one to achieve a 4-dimensionalfixed-point space. For this, see that the alternating form that these trivial sub-modules correspond to is singular, with kernel of codimension 2, and so any linearcombination of them has kernel of codimension at most 6. However, M ( D ) hasa non-singular symmetric, hence alternating form on it, and so Λ ( M ( D )) L ′ isat least 4-dimensional. Hence again L ( E ) L has dimension at least 4, and so H stabilizes a line on L ( E ), as needed. Case 7 : The 64 must split off as a summand. The trace of z on L ( E ) is 5, andhence C G ( z ) is an A E maximal-rank subgroup. The kL ′ -module that consists ofthe 1-eigenspace of the action of z on L ( E ) has composition factors4 , , , , , , . .2. PSU This is the sum of L ( A ) ↓ L ′ and L ( E ) ↓ L ′ , and if the action of L ′ on M ( A ) hasfactors 2 i ,
1, then we must subtract 1 , i , i +1 from these factors to get the actionon L ( E ). The only i for which this produces a conspicuous set of compositionfactors is i = 3, and L ( E ) ↓ L ′ has composition factors4 , , , , , , . Using traces of semisimple elements, the corresponding factors on M ( E ) are4 , , , , , . We claim that we may assume that L ′ is contained in the product of A andthe A -Levi subgroup of E . From [ Cra ], it is known that any copy of PSL (8)lying in E lies inside a ( σ -stable) proper positive-dimensional subgroup of E , sowe can place L ′ inside A X for some maximal X ≤ E . We eliminate all X otherthan the A -parabolic subgroup first.We start with maximal-rank subgroups. It is easy to see that X = A A A ,for example because any such L ′ in this subgroup would have to have at least ninetrivial factors on M ( E ). If L ′ is contained in X = A A (and not A itself), then L ′ must act as 2 on M ( A ) and 4 ⊕ ⊕ on M ( A ). This places L ′ inside an A A -Levi subgroup, so inside some A -parabolic subgroup that contains it.Now the parabolics: as there is no copy of L ′ inside A A A , it cannot be con-tained in the A A A -parabolic subgroup. If L ′ is contained in an A A -parabolicthen the image in the Levi subgroup must be contained in A A , as above. Thismeans that L ′ lies in (up to conjugacy) the intersection of both A A -parabolicsubgroups of A A , hence of E . Thus up to conjugation we may place L ′ in an A -parabolic subgroup of E .We are left with the D -parabolic subgroups. We must distribute the factorsamong the 1-, 10- and 16-dimensional composition factors of the restriction of M ( E ) to D . As there is no unipotent class acting with blocks 2 on M ( D )(see, for example, [ Law09 , Table 7]), the 1 must lie there. The only way todistribute the composition factors of M ( E ) ↓ L ′ among the factors of D is 2 , , for M ( D ) ↓ L ′ (and hence 4 , , for the half-spin module). Note that L ′ muststabilize a line on M ( D ), else there is certainly a subquotient 1 / /
1. In this casethe involution in L ′ acts projectively on M ( D ), but no such involution lies in D .Since L ′ stabilizes a line on M ( D ), it lies in either B or a D -parabolicsubgroup. By checking traces of elements of order 3 on the two half-spin modules,we find that L ′ cannot lie in D , so it must lie in B , whence L ′ acts as 2 ⊕ ⊕ on M ( B ) (which has dimension 8 in characteristic 2).We see that one of the submodules 2 (diagonal if necessary) must be totallyisotropic, and therefore has a parabolic subgroup as a stabilizer. This places L ′ inside a proper Levi subgroup—in particular, B A —of B , hence a proper Levisubgroup—in particular, D A —of D . Thus L ′ lies inside the D A -parabolicsubgroup of E , hence inside the A A -parabolic subgroup, which we have alreadyconsidered.Of the reductive subgroups, the G maximal subgroup acts on M ( E ) acts withfactors L (01), L (10), L (20) and L (00) (of dimensions 14, 6, 6 and 1), and this isincompatible with L ′ . If L ′ ≤ A G then the only action compatible with M ( E ) isacting on M ( A ) with factors 2 , M ( G ) with factors 2 , . This places L ′ inside a parabolic subgroup of A G , and hence inside a parabolic subgroup of E .
00 8. RANK 2 GROUPS FOR E If L ′ ≤ X = F , then again by [ Cra ] there is a proper positive-dimensionalsubgroup of X containing L ′ . If it is a parabolic then L ′ is contained in a parabolicsubgroup of E . If it is B then L ′ is contained in the D -Levi subgroup. If L ′ ≤ A A then L ′ ≤ A A A ≤ E , which we know it is not. The only remainingsubgroup is C . Note that from L ( E ) ↓ L ′ and the fact that L ( E ) ↓ X is the sum oftwo copies of M ( F ) and one of the image of M ( F ) under the graph automorphism,we may compute the factors of the image of L ′ under the graph automorphism on M ( F ), and this subgroup must be contained in B . The composition factors are4 , , , a module of pressure −
5, hence centralizes a 5-space on M ( F ). Thisplaces L ′ inside a very small subgroup of B , certainly contained in another positive-dimensional subgroup of F , and so L ′ is contained in another subgroup of E , asneeded.We now move inside the A -parabolic subgroup, and place L ′ inside an A -Levisubgroup. Inside the A -parabolic, we see from above that the composition factorson M ( A ) are 4 , , and there are (up to duality) two such modules: 4 ⊕ and2 / . As the layers of the unipotent radical of the A -parabolic subgroup are L (0)and L (00100), and the restrictions of these modules to L ′ have zero 1-cohomology,we see from Lemma 2.11 that L ′ lies in the A -Levi subgroup. In particular, wemay write down the structure of L ( E ) ↓ L ′ exactly, as the restriction of L (0) ⊕ ⊕ L (00100) ⊕ ⊕ ( L (10000) ⊗ L (00001))to L ′ . As 64 is a summand of L ( E ) ↓ H , the 1-eigenspace of z on this module hasa summand of the form 1 / , / / , / , /
1, and this does not lie in L ( E ) ↓ L ′ if M ( A ) ↓ L ′ is not semisimple, we obtain that L ( E ) ↓ L ′ is exactly(1 / , / / , / , / ⊕ (2 / / / / ) ⊕ ⊕ (2 / / ) ⊕ ⊕ (1 / / ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ . In addition, the action of L ′ on L ( A ) is either (1 / / ⊕ ⊕ or 2 / / / / .This has (at least) a 4-dimensional fixed space, and we saw earlier that if L ( E ) H = 0 then it must have dimension at most 3. Thus H stabilizes a line on L ( E ) in this case as well. Case 5 : This is as stated in the proposition. q = 16 : As there are so many possible sets of composition factors, we will onlyconsider conspicuous sets of composition factors that have positive pressure or notrivial composition factors. There are, up to outer automorphism, 28 sets of com-position factors that are conspicuous for elements of order at most 17. Let x denotean element of order 255. A computer check shows that, in each of these cases, theeigenvalues of x determine the class of x in G , and the same holds for x of order85. Thus we look for elements of G of order 765 cubing to x , aiming to applyLemma 3.10. For 24 of the 28 sets of composition factors we find elements of order765 with the same number of distinct eigenvalues on L ( E ) as x , so H is stronglyimprimitive.Thus we are left with four sets of composition factors for L ( E ) ↓ H .24 , ∗ , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , ¯9 , ¯9 ∗ , , ∗ , , , , ∗ , , ∗ , , ∗ , , ∗ , , ∗ , (9 , ∗ ) , ¯9 , ¯9 ∗ , , , , , ∗ , , ∗ , , ∗ , , ∗ , , ∗ , , , , , , ∗ , , ∗ , , ∗ , , ∗ , , , , . .3. PSp (Here, 27 i,j,k means 3 i ⊗ j ⊗ k if i < j and 3 i ⊗ j ⊗ ∗ k if j < i .) Cases 3 and 4 : By [
Sin92a ] there are no extensions between 8- and 27-dimensionalmodules, or between 8- and 8-dimensional modules, so in the last two cases the 8 i must split off as summands. Furthermore, the eigenvalues of x on 8 i , with theexception of two copies of 1, are distinct from the eigenvalues of x on all othercomposition factors of L ( E ) ↓ H . Thus every root of x in a maximal torus of G must stabilize each 8 i (the 1-eigenspace of x is 8-dimensional so also preserved byevery root), and hence H is strongly imprimitive by Lemma 3.10. Case 2 : The 3 , 3 ∗ , 8 , 8 and 8 all split off as summands, as they have noextensions with the other composition factors of L ( E ) ↓ H . As with the previouscases, the eigenvalues of x on 8 , apart from 1, do not appear in the other factors,and so every element of a maximal torus containing x that powers to x must stabilize8 . (Again, the 1-eigenspace is 8-dimensional so is automatically preserved.) Sincethe factors of L ( E ) ↓ H are not preserved by a field automorphism of H , we applyLemma 3.9 to see that 8 is in its own N Aut + ( G ) ( H )-orbit. Hence H is stronglyimprimitive by Lemma 3.10. Case 1 : There are no composition factors whose eigenspaces miss all other factors,but we find exactly 729 elements of order 765 that cube to x and stabilize theeigenspaces that comprise a given composition factor of L ( E ) ↓ H . Noting that thecomposition factors of L ( E ) ↓ H are stable under an outer automorphism α of order4 of H , in fact we find 81 elements that cube to x and stabilize the eigenspacesthat make up all four simple modules in the orbit under α for a composition factorof L ( E ) ↓ H . We may therefore apply Lemma 3.10 via Lemma 3.9 to see that H isstrongly imprimitive. (cid:3) For q = 3, both conspicuous sets of composition factors in the statement ofthe proposition are warranted. A warrant for the first set of factors is given bya diagonal PSL (3) in E A acting irreducibly on both M ( A ) and M ( E ). Awarrant for the second set of factors is given by a copy of PSU (3) inside A actingon M ( A ) as 3 ⊕ ∗ . (These are the same actions, but not the same compositionfactors, as the first and third cases for PSL (3) that were described at the end ofthe previous section.) PSp When computing with H ∼ = PSp (3), we will be able to use the subgroup L = 2 ⋊ Alt(5) (which is an extraspecial type maximal subgroup in the groupSp (3), or a parabolic subgroup in PSU (2) ∼ = PSp (3)) to great effect. Because ofthis, we will give some information about the group now.The simple modules for PSp (3) are 1, 5, 10, 14 and 25, together with theprojective module 81, which we will ignore. The simple modules for L are 1, 4, 5,10 , 10 and 15, together with two 3-dimensional modules that do not appear inrestrictions except for 81. The restrictions of the simple modules for H to L are1 , , , ⊕ , ⊕ , ⊕ P (10 ) ⊕ P (10 ) ⊕ ⊕ , and the projective indecomposable modules are1 / / , / / , / / , / / , / / , .
02 8. RANK 2 GROUPS FOR E The simple modules for H restrict to L semisimply (other than 81, which is alwaysa summand anyway as it is projective) and with few factors, and the structuresof the projectives for L are very easy to understand. These can be used to placesignificant restrictions on the structure of L ( E ) ↓ H , which is difficult just using H ,as there are lots of extensions between simple modules for H (for example, both 5and 10 have self-extensions).If v denotes an element of order 3 in L , then v acts on the simple modules for L with block structures1 , , , , , , , , , . We continue with our definition of u , that it is a element of order p in H from thesmallest conjugacy class, from the start of the chapter. Proposition . Let H ∼ = PSp ( q ) for some q ≤ . (1) If q = 2 then either H stabilizes a line on L ( E ) or the composition factorsof L ( E ) ↓ H are , , , , . (2) If q = 3 , then H either stabilizes a line on L ( E ) or is a blueprint for L ( E ) . (3) If q = 4 , , then H is strongly imprimitive. (4) If q = 7 then H is a blueprint for L ( E ) . Proof. q = 7 : There are three sets of composition factors for L ( E ) ↓ H thatare conspicuous for elements of order at most 12, namely154 , , , , , , , , , , . None of the simple modules in these sets has an extension with any other, andso L ( E ) ↓ H must be semisimple in all three cases. The action of u on L ( E ) hasblocks 7 , , , , , , in the first case, 4 , , , in the second and3 , , in the third, which are the actions of the generic classes A + 2 A , 4 A and 2 A respectively (see [ Law95 , Table 9]). Thus H is a blueprint for L ( E ) byLemma 2.2, and this completes the proof. q = 5 : There are three sets of composition factors for L ( E ) ↓ H that are conspicuousfor elements of order at most 12, namely86 , , , , , , , , , , , , , . The second and third cases have non-positive pressure, so H stabilizes a line on L ( E ) in these two cases.In the first case, we consider the subgroup L ∼ = SL (5) ◦ SL (5), the derivedsubgroup of the centralizer of an involution z in H . The trace of z on L ( E ) is − L ≤ D . The composition factors of L on the 1-eigenspace of z are(4 , , (4 , , (2 , , (2 , , (2 , , (0 , , (0 , . (We use algebraic group notation here because we will embed L into an algebraic A A and so we are trying to avoid confusion.) There is a unique possibility for M ( D ) ↓ L whose exterior square has these factors, which is (1 , ⊕ (3 , Y ∼ = A A on M ( D ) extending this action, namely(1 , ⊕ (3 , A A subgroup lies in D D , acting on L ( λ ) for the two D factors as (1 ,
3) and (3 ,
1) respectively. Using the traces of elements of order 3, .3. PSp one proves that the actions of L on the two modules L ( λ ) and L ( λ ) are (1 ,
3) and(2 , ⊕ (0 ,
4) when L ( λ ) is (1 , ,
1) and (0 , ⊕ (4 ,
0) in the other case.The ( − z on L ( E ) is a half-spin module for D , which restrictsto D as a sum of products of half-spin modules. To keep z acting as − λ , λ ) and( λ , λ ). Thus we need to cosnider the tensor product of (1 ,
3) and (0 , ⊕ (4 , , ⊕ ((1 , / (1 , / (1 , ⊕ ((3 , / (5 , / (3 , . The other tensor product is just like this but with the two copies of A swapped.The restriction of this module to A (5) A (5) ∼ = L is(1 , ⊕ ((1 , / (1 , / (1 , ⊕ ((3 , / (1 , / (3 , , (where we keep the notation from algebraic groups) so L is not a blueprint, but everysemisimple kL -submodule of the ( − z on L ( E ) is a k Y -submodule.We now consider the 1-eigenspace of z , which is Λ ( M ( D )). The exteriorsquare of (1 , ⊕ (3 ,
1) for Y is((0 , / (0 , / (0 , ⊕ ((2 , / (6 , / (2 , ⊕ (2 , ⊕ (0 , ⊕ (2 , ⊕ (2 , ⊕ ⊕ (4 , ⊕ ⊕ (4 , . The restriction of this module to A (5) A (5) ∼ = L is((0 , / (0 , , (0 , / (0 , ⊕ ((2 , / (0 , , (2 , / (2 , ⊕ (2 , ⊕ (0 , ⊕ (2 , ⊕ (2 , ⊕ ⊕ (4 , ⊕ ⊕ (4 , . Thus again all semisimple kL -submodules of the 1-eigenspace of L ( E ) are k Y -submodules, and hence any submodule of L ( E ) ↓ H that restricts semisimply to L is stabilized by Y . The restriction of 10 to L is (0 , ⊕ (2 , ⊕ (1 , L is(2 , ⊕ (0 , ⊕ (1 , ⊕ (3 , ⊕ (2 , ⊕ (4 , ⊕ (3 , , and so if either 10 or 68 lies in the socle of L ( E ) ↓ H then h H, Y i is positive dimen-sional and not equal to G . However, 10 and 68 are the only composition factors toappear more than once, so at least one of them (in fact, both), must appear in thesocle, and H is strongly imprimitive. q = 3 : There are seven conspicuous sets of composition factors for L ( E ) ↓ H ,namely14 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . We will prove that only the first and last of these seven copies may exist (and infact do exist, both inside A A , the first embedding only in one factor and the lastembedding diagonally in both factors). Cases 2, 3, 4, 5 and 6 : Case 5 has negative pressure, it is fairly easy to provedirectly that Case 3 stabilizes a line on L ( E ) anyway, and more detailed argumentsusing the subgroup 2 ⋊ Alt(5) can deal with Case 2, but the others seem difficultwithout using this approach.Let L denote a copy of 2 ⋊ Alt(5) in H , with the normal subgroup 2 denotedby L . The restrictions of the seven cases for L ( E ) ↓ H to L are as follows.
04 8. RANK 2 GROUPS FOR E Case Restriction to L , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , We will show that except for the first and last cases, there is no such embedding of L in G , hence no such embedding of H in G .We suppose first that L is contained in a member of X , so we therefore considerthe members of X , and look for all copies of L inside them. Note that L possessesno (perfect) central extensions by a subgroup of order 3 or 5 by a result of Schur(see for example [ Asc00 , (33.14)]). Therefore if L is contained in, for example, A A , then we have an action of L on each M ( A ), and not a central extension of L . Note that 1, 3 i and 4 are representations of Alt(5) and not L , so at least one of M ( A )s must not consist of these factors, so is 5.First M ( A ) Second M ( A ) Action on L ( E )5 1 , , , , , , , , , , , , , , , , , , , , , , , We see that the first and last options yield embeddings of L in E with correct setsof composition factors on L ( E ), corresponding to Cases 1 and 7 respectively.If L is contained in an E -Levi subgroup then L cannot have a factor 3 i on M ( E ), as there are two copies of M ( E ) in L ( E ) ↓ E and at most one copy of 3 i in L ( E ) ↓ H . There are only two such conspicuous sets of composition factors for M ( E ) ↓ L : 10 , , and 15 , , . Since 1 ⊕ and M ( E ) ⊕ are submodules of L ( E ) ↓ E , we see that we need at least 1 in L ( E ) ↓ L , so we must be in Case 1,which we have already seen exists.If L ≤ A then again we consider its action on M ( A ), which has dimension 9.Again, we need a 5, yielding three possible actions: 5 ,
4; 5 , ,
1; 5 , . The last onelies in A , so is as above, and the other two lie in an A A -parabolic, whose Levisubgroup lies in A A , so again they appear in the previous table.If L ≤ E A acting diagonally then L must act on M ( A ) as 3 (without loss ofgenerality), and as M ( E ) does not have dimension divisible by 5, we need a copy of1, 3 i or 4. Since M ( E ) ⊗ M ( A ) appears twice up to duality (which doesn’t matterfor our group) and each 3 i appears at most once in L ( E ) ↓ H , we may assume thatneither 3 nor 3 appears in M ( E ) ⊗ M ( A ). As 1 ⊗ = 3 , 3 ⊗ = 3 ⊕ P (1)and 4 ⊗ = 3 ⊕ P (4), none of these may occur. Thus M ( E ) contains 3 , butthen M ( E ) ⊗ M ( A ) contains 4 , so that we have 4 in L ( E ) ↓ L , which is toomany. Therefore we have no such L in this case.If L is contained in D then we first assume that L acts on M ( D ) and nota central extension of L . In this case we need a set of composition factors for M ( D ) ↓ L , but then the trace of an involution in L must be 0 or ±
8. There areonly four such sets of composition factors for M ( D ) ↓ L if one renumbers the 3 i , .3. PSp and these are 5 , , , , , , , , , . The exterior square of M ( D ) in the third case has 3 as composition factors, whichis incorrect. The fourth set of composition factors yields an exterior square P (4) ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ . The set of factors of this module is not a subset of any of the composition factorsfor L ( E ) ↓ L , so this is also incorrect. The first case lies in D , and is thereforein E . Thus we have factors 10 , , M ( D ), and the composition factors on L ( D ) are 15 , , , . This is compatible with Cases 6 and 7. However, thetrace of an element of L of order 5 on M ( D ) is 1, and this semisimple class actson half-spin modules with trace −
2, and L ( D ) with trace 0, thus on L ( E ) withtrace −
2. This is consistent with Case 7, and not with Case 6, on which x actswith trace 3. (The Brauer characters of these two cases differ only on x .)On the other hand, suppose that L embeds in HSpin , but its preimage inSpin is a central extension. We first claim that in any (non-split) central extension2 · L there is an element of order 4 that squares to the central involution. In thiscase, we need to find an element of order 4 in Spin whose image in HSpin has order 2, i.e., it has trace 128 on the half-spin module. However, there is nosuch conjugacy class, as can be seen by examining the list of the sixteen classes ofelements of order 4 for D . Thus this central extension cannot occur.To see the claim, the easiest way is to construct all three central extensions of L using the SmallGroups command (they are numbers , and )and check it manually. One can see it theoretically by noting that they are quotientsof the group 2 ⋊ (2 · Alt(5)).If L ≤ E A acting diagonally, then L acts on L ( A ) ≤ L ( E ) as 3 (withoutloss of generality). Thus we may only be in Cases 3 and 6. We consider all conspic-uous sets of composition factors for L ( E ) ↓ L with no 3 and composition factors asubset of those of Cases 3 or 6. We obtain a single possibility,15 , , , , , , which can arise as Case 3. However, the action on M ( E ) is incorrect: of thethree central extensions, only one has modules of dimension 2 (needed to projectonto A ), which is ˆ L = 2 ⋊ (2 · Alt(5)). There is up to field automorphism aunique conspicuous set of composition factors for this group (that is a faithfulmodule for ˆ L ). The tensor product of this with either 2-dimensional module (recallthat M ( E ) ⊗ M ( A ) is a factor of L ( E ) ↓ E A ) has three composition factors ofdimension 15, which means we cannot be in Case 3 (as that has only three copiesof 15 in total). Hence we can eliminate this subgroup as well.We now finish with the positive-dimensional subgroups. Every Levi subgroupis contained in a maximal-rank subgroup above, so we need only consider F , G F and any subgroup X such that L X .For X = F , there are two possibilities for the action of L on M ( F ): 10 , and 15 , . In these cases, there is no elementary abelian 2-subgroup of order 8consisting of (the identity and) elements of trace 1. By [ Gri91 ], we see that L istoral, and thus L is contained in the normalizer of a torus. But the Weyl group of X is soluble, and Alt(5) is not. Thus L does not embed in X .For X = G F , first note that M ( G ) appears twice as a composition factor in L ( E ) ↓ X (as we are in characteristic 3). Therefore 3 i cannot appear in M ( G ) ↓ L .
06 8. RANK 2 GROUPS FOR E However, L does not embed in G , so we must act as L/L ∼ = Alt(5) on this factor,and this can only embed via A , which acts as M ( A ) plus its dual (and a trivial)or M ( A ) ⊗ M ( A ) ∗ minus two trivials. Both of these restrict to L to contain 3 i ,so we obtain a contradiction in this case.Finally, we consider the normalizer of a torus. Of course, if T denotes thetorus of N G ( T ) then L ∩ T is either L or 1. We may use Magma to enumerate theconjugacy classes of subgroups L and L/L in W ( E ) and find there is one of theformer and four of the latter. All but one of them centralize at least a 2-space on thereflection representation, so they lie in parabolic subgroups of G . The remainingclass is a diagonal Alt(5) lying in Sym(5) × Sym(5) = W ( A A ), and so L ≤ A A in this case.It remains to show that L is always contained in a positive-dimensional sub-group. We will do this by showing that L is always a toral subgroup, using resultsof Griess [ Gri91 ]. In Cases 5 to 7, the trace of an involution in L is −
8, so class2B in the notation of [
Gri91 ]. Thus from [
Gri91 , Table I] we see that L is toral.For the rest, we see from [ Gri91 ] that whether L is toral depends on the subset of L that consists of 2A-elements, i.e., those with centralizer E A . This is the samesubset (as the traces of elements in L are the same) for the first four cases, and soif L is toral in one of the cases then it is toral in all cases. However, clearly the L ≤ L ≤ H ≤ A ≤ A A is toral, as L is a torus of H . This completes the proofthat only Cases 1 and 7 can occur. (Notice that this chimes with the case q = 5,where these were the two cases other than the maximal B .) Case 1 : The pressure of this module is −
3, so H stabilizes a line on L ( E ). Case 7 : We first prove that either 5 or 10 is a submodule of L ( E ) ↓ H . We willthen restrict to L and show that the previous statement implies that H is containedin a member of X , and then check that in fact H is a blueprint for L ( E ).Suppose that L ( E ) ↓ H has a socle consisting only of copies of 14 and 25. Ifany of the submodules of the socle are summands they may be removed, so we mayassume that there is at most one 14 in the socle, and at most two 25s in the socle.If 14 lies in the socle then we need to take the { , , } -radical of P (14), andthis is 10 , / , / , so obviously we need more in the socle. The { , , } -radical of P (25) is25 / / / / / / , / , , / , / . Together these only have eight copies of 10, so we need 14 ⊕ ⊕ in the socle,but then all 25s lie in the socle or top of L ( E ) ↓ H . Therefore we need to take the { , } -radical of P (25), which is 5 , / , / . This yields a contradiction, and so 14 cannot be a factor of soc( L ( E ) ↓ H ). Examin-ing the Cartan matrix of the principal block of H , we find that there are nine copiesof 10 in P (25), so soc( L ( E ) ↓ H ) cannot be 25. Thus we take the { , , } -radicalof P (25), to yield the module 5 / , , / , / . This clearly does not have enough copies of 10. Thus we must have either 5 or 10in soc( L ( E ) ↓ H ), as claimed. .3. PSp We now consider the restriction to L ; as we saw above, L lies in N G ( T ) (butthen lies in A A ), inside A A or inside D .For A A , we have that L acts as 5 on both copies of M ( A ). Thus L isdetermined up to conjugacy in G , and acts as(10 / / ) ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ . (This is because L acts on L ( A ) as 10 ⊕ ⊕
4, and on M ( A ) ⊗ Λ ( M ( A )) as(10 / / ) ⊕ ⊕ B subgroup acting irreducibly on M ( A ) as L ( λ ); since L ( E ) ↓ A A restricts to a diagonal B subgroup Y acting in this waywith two isomorphic 24-dimensional summands, and four isomorphic 50-dimensionalsummands, we just need to check the structure of these two modules. We see that L ( λ ) ⊗ ∼ = L (0) ⊕ L (2 λ ) ⊕ L (2 λ ) , and since Λ ( L ( λ )) ∼ = L (2 λ ) (as we can see from the above formula), we have L ( λ ) ⊗ Λ ( L ( λ )) = L ( λ ) ⊗ L (2 λ ) ∼ = L ( λ ) ⊕ ( L (2 λ ) /L ( λ + 2 λ ) /L (2 λ )) . In terms of dimensions, this is 5 ⊗ ⊕ ⊕
14 and 5 ⊗
10 = 5 ⊕ (10 / / L and Y , we see that every submodule 5 and10 is stabilized, so either H is contained in a member of X or it does not stabilizeany of these subspaces, i.e., neither 5 nor 10 is a submodule of L ( E ) ↓ H , but wehave already shown that this is impossible. Thus H is contained in a member of X in this case.Thus we may assume that L is contained in X = D , and above we showed that L acts on M ( D ) as 10 ⊕ ⊕
1. This clearly lies inside B , and then inside B D ,the stabilizer of the 5 ⊕
10 decomposition, and then inside a copy of B B . Thediagonal B subgroup of this is contained in A A (it is subgroup 59 in [ Tho20 ,Table 13]) and so L is conjugate to the embedding in A A .In particular, this means that H is contained in a member of X . Furthermore,since L is contained in only A A and D (and the normalizer of a torus), H canonly be embedded in A A or D . As above, we see that H is the fixed points ofthe irreducible B subgroup 59 from [ Tho20 , Table 10], and the structure of thison L ( E ) is easiest to see through A A , as we computed it above. This is also thestructure of H , and so H is a blueprint for L ( E ), as needed. q = 9 : Using the traces of elements only of order 41, we find up to field automor-phism exactly seven conspicuous sets of composition factors for L ( E ) ↓ H , namely14 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . Cases 1, 2 and 3 : In each case there is a trivial composition factor. No simple kH -module of dimension at most 25 has non-zero 1-cohomology, so in each case H stabilizes a line on L ( E ). Cases 4, 5, 6 and 7 : In each of these cases the restriction to L ∼ = PSp (3) hascomposition factors 25 , , , on L ( E ). Thus L is a blueprint for L ( E ) bythe previous proof, and so therefore is H . q = 2 : This was obtained in [ Cra17 , Proposition 6.3].
08 8. RANK 2 GROUPS FOR E q = 4 : Label the simple modules 4 to 4 so that the graph automorphism, whichhas order 4, cycles them. In particular, this means that S (4 i ) = 1 / i +1 / / i +2 forall i .We have two conjugacy classes of subgroups of H isomorphic to Alt(5) × Alt(5).Write y and z for elements of orders 3 and 5 from the first factor of one of theclasses, and y and z for elements of orders 3 and 5 from the first factor of theother class. Thus the centralizer in H of y i is h y i i × L i and the centralizer in H of z i is h z i i × L i for L i ∼ = Alt(5). In particular, C H ( y i ) ′ = C H ( z i ) ′ .We will normally focus on one element for each proof, whichever works best.Write V for the 1-eigenspace of this element, and V θ i and V ζ i for the θ i - and ζ i -eigenspace of L ( E ) with respect to this element, where θ = 1 and ζ = 1. Eachof these forms a module for the appropriate L i .We label the L i and the modules for them so that the restriction of 4 to L is 1 ⊕ ⊕ , and the restriction of 4 to L is 1 ⊕ ⊕ (hence 4 restricts to L as2 ⊕ ). We also choose the z i and ζ so that 4 has zero ζ -eigenspace for z , and 4 has zero ζ -eigenspace for z .Because there are many modules for H and two classes of L i , each witheigenspaces with respect to y i and z i , we do not give complete information aboutthe actions here, and just inform the reader as and when we use a particular pieceof information.Up to automorphism, there are eight conspicuous sets of composition factorsfor L ( E ) ↓ H with positive pressure:16 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . There are also three with non-positive pressure, which stabilize a line on L ( E ) byProposition 2.5:4 , , , , , , , , , , , , , , , , Hence they are strongly imprimitive by Lemma 3.5.
Case 1 : The eigenspaces of z on L ( E ) have dimensions 64, 56, 36, 36 and 56, so z lies in the class with centralizer A T from Table 2.10. The composition factorsof V are 2 , , so L must act on M ( A ) as 2 ⊕ . It is easy to see that the copyof PSL (4) inside A acting this way is a blueprint for L ( E ) with the A subgroupof A acting on M ( A ) as L (2) ⊕ stabilizing the same eigenspaces on Λ i ( M ( A ))for i = 1 , , L ( λ i ) for i = 1 , , , , ,
7) and on M ( A ) ⊗ (i.e., L ( λ + λ )).Thus H is a blueprint for L ( E ). Case 2 : The eigenspaces of z on L ( E ) have dimensions 54 , , , ,
46, so z hascentralizer A D T from Table 2.10. The composition factors of V are 2 , , ,and this is L ( D ) ⊕ L ( A ) ⊕ L ( T ). This means that M ( D ) ↓ L cannot have 4 ⊕ ⊕ or 2 ⊕ in it. In fact, the only solution is that L acts trivially on M ( A ) (so L is contained in D ) and with factors 2 , on M ( D ). The composition factorsof L ( λ ) ↓ L are 2 , , so the module is always semisimple. (We use this notation .3. PSp for simple modules to alert the reader as to which factor of the centralizer of z thesimple module is for.)The action of D on the ζ i -eigenspaces of z are L (0) ⊕ ⊕ L ( λ ) ⊕ for ζ and L ( λ ) ⊕ ⊕ L ( λ ) for ζ , each up to duality (but since L acts the same on L ( λ )and L ( λ ) we do not work out the exact distribution).There are five possible modules for M ( D ) ↓ L , combining various summandsof the form 1 / /
1, (1 / ) ⊕ (2 / and 1. In each case the obvious A subgroup Y is a blueprint for M ( D ) = L ( λ ) and L ( D ) = L ( λ ). Hence L and Y stabilizethe same subspaces of V and any subspace of V ζ whose composition factors areonly trivials and copies of 2 . In particular, Y stabilizes any submodule 4 or 4 in L ( E ) ↓ H , since these have only composition factors 1 and 2 on the 1, ζ - and ζ -eigenspaces, and ζ ± is not an eigenvalue of them. Furthermore, since clearly 2 is not a submodule of L ( D ) (as it only appears inside 2 ⊗ = 1 / /
1) we can haveno submodule 4 in L ( E ) ↓ H .Thus if H is not Lie imprimitive then the socle and top consist of copies of 4 ,16 and 16 . The { , i , , } -radical of P (16 ) is1 , / , / / , / , and removing all quotients that cannot be quotients of L ( E ) ↓ H leaves just 4 / .Thus we consider the { , i , , } -radicals of P (4 ) and P (16 ), and thentake the { , , } ′ -residuals, to obtain the modules4 / / / / / / , / , / / , / , / / , / , / and 16 / / / , / , / , / / / / / . But we need three copies of 4 in L ( E ) ↓ H , so we need 16 ⊕ in the socle, andanother 16 ⊕ in the top, but we only have four copies of 16 in L ( E ) ↓ H . Thustwo must be summands, and of course we obtain a contradiction.To see strong imprimitivity, note that the composition factors of L ( E ) ↓ H arenot stable under any outer automorphism of H , so we may apply Lemma 3.9. Case 3 : The trace of z on L ( E ) is −
2, so z has centralizer A A . The compo-sition factors of L ( E ) ↓ A A are (1001 , , z (we are suppressing the ‘ L ( − )’), and the other four modules are(1000 , , (0100 , , (0010 , , (0001 , . These four modules can be permuted by taking duals and swapping the two A -factors, so we may assume that the first one is the restriction of the ζ -eigenspaceof z on L ( E ) to A A . With that assignment, the actions of A A on the ζ , ζ , ζ and ζ -eigenspaces must be the modules above in the same order.The action of L on V has factors 2 , , , and the action on V ζ has factors4 , , , . This means that there is only one possibility for the action of L onthe two modules M ( A ), and that is on the first copy of M ( A ) with factors 2 , ,
1. This yields nine possibilities for the action of L on L ( E ), as we can act as (1 / i ) ⊕ i , (2 i / ⊕ i , or 2 ⊕ i ⊕ L ( E ) ↓ L is stabilized by somepositive-dimensional subgroup Y of G . Any submodule of L ( E ) ↓ H that restrictssemisimply to L is therefore also stabilized by Y . The modules 1, 4 i and 16 all restrict semisimply to L , and so if the condition holds, and any of those aresubmodules of L ( E ) ↓ H , then H is strongly imprimitive by Theorem 3.2 and Lemma
10 8. RANK 2 GROUPS FOR E and 16 : the former has ζ ± -eigenspace 1 / / ζ ± -eigenspace 1 / / M ( A ) are 2 ⊕ ⊕ / ) ⊕ or its dual.We set Y to be an A subgroup of A A acting on the first module as L (1) ⊕ ⊕ L (0)and on the second as ( L (0) /L (2)) ⊕ L (2) or its dual. In this case, it is an easy checkthat all semisimple submodules of L ( E ) ↓ L are stabilized by Y . There is no copyof 2 in the second socle layer of one of V ζ or V ζ in this case (depending on whichpossibility for L one takes) so 16 cannot be a submodule of L ( E ) ↓ H . There isno copy of 2 in the second socle layer of V ζ ± either, so 16 cannot be a submoduleof L ( E ) ↓ H either. Thus H is strongly imprimitive.On the other hand, if the two copies are (1 / ) ⊕ or its dual, and 2 ⊕ ⊕ Y to act as ( L (0) /L (4)) ⊕ L (4) and L (2) ⊕ ⊕ L (0), and again we stabilizeevery semisimple submodule of L ( E ) ↓ L . As before, there is no copy of 2 in thesecond socle layer of V ζ ± and no copy of 2 in the second socle layer of either V ζ or V ζ , so neither 16 nor 16 are submodules of L ( E ) ↓ H . Thus H is stronglyimprimitive.If both modules are semisimple then we again embed in Y acting as L (1) ⊕ ⊕ L (0) and L (2) ⊕ ⊕ L (0) and obtain the same conclusion.This eliminates five possibilities for the action of L , leaving the four whereneither of the two modules is semisimple. To eliminate these, we show that L cannot embed in this fashion, by considering the centralizer of y , which is h y i× L .The trace of y on L ( E ) is −
4, hence y has centralizer A . The composition factorsof L on the 1-eigenspace of y are 4 , , , , and these are consistent with thecomposition factors of M ( A ) ↓ L being either 4 , , ,
1. But the former hasthe wrong action on the rest of L ( E ). Since there is a single trivial compositionfactor in M ( A ) ↓ L , we have very few possible modules: it is easy to see that one2 and one 2 must split off as summands, and the rest is up to duality one of2 ⊕ ⊕ , ⊕ (1 / ) , ⊕ (1 / ) , / , , / / . This yields five actions of L on L ( E ). The first three of these appear as well inthe actions of L on L ( E ) via A A , where one of the two modules is semisimple.However, if neither module M ( A ) ↓ L is semisimple, then L ( E ) ↓ L does not matchup with any of the five possibilities arising from A . Thus these four cannot yieldembeddings of L , and the proof is complete. Case 4 : The restriction L ( E ) ↓ L has composition factors 4 , , , . Wenote that by just checking all possibilities, there is no copy of L inside A that hasthose factors on L ( E ). (We recall that the composition factors of A on L ( E ) are M ( A ) ⊗ M ( A ) ∗ (minus a trivial), Λ ( M ( A )) and Λ ( M ( A ) ∗ )). Thus L cannotembed in A , or in any parabolic whose Levi subgroup is contained in A .However, the trace of z on L ( E ) is 3, so z has centralizer A A . Hence L embeds in A A , which can be conjugated into A . (In fact, L must have atrivial composition factor on M ( A ) so there is a version with the same compositionfactors on L ( E ) that lies in A A , directly in A .) This contradiction means that H cannot embed into E with these factors. Case 5 : This time the trace of z is 3, so z has centralizer A A . The compositionfactors of V are 4 , , , , and this yields composition factors of L on M ( A ) .3. PSp Module 1-eigenspace ζ -eigenspace ζ -eigenspace1 1 - -4 - 2 -4 - 14 - - 2 - 4 416 ⊕ (1 / /
1) 2 ⊕ (1 / / ⊕ ⊕ ⊕ P (2 ) ⊕ P (2 ) ⊕ ⊕ Table 8.3.
Actions of L on z -eigenspaces of simple kH -modules.and M ( A ) of 4 , , respectively. We therefore have three possibilities forthe action of L on L ( E ) and on each eigenspace of z .The L -actions of the composition factors of L ( E ) ↓ H on the eigenspaces of z are given in Table 8.3. The dimensions of V ζ and V ζ are the same, but they arenot isomorphic modules up to duality. By considering the L -composition factorswe can tell them apart, and we know that the action of A A on the ζ -eigenspaceof L ( E ) is the sum of ( λ ,
0) or its dual and ( λ ,
1) or its dual, and the action of A A on the ζ -eigenspace of L ( E ) is the sum of ( λ ,
0) or its dual and ( λ ,
1) orits dual.Suppose that L acts semisimply on M ( A ). The actions of L on ( λ ,
0) and( λ ,
1) are P (2 ) ⊕ P (2 ) ⊕ ⊕ ⊕ (1 / , / ⊕ P (2 ) ⊕ (1 / / ⊕ . From this we see that 16 cannot be a submodule of L ( E ) ↓ H , since 1 / / V ζ .In a similar vein to our proofs for PSU (4), we embed L in Y acting on M ( A )as L (6) ⊕ L (2) ⊕ L (0) and on M ( A ) as L (2). With these factors, L and Y stabilizeall simple submodules of L ( E ) ↓ L other than those isomorphic to 4 that lie in V .This means that Y stabilizes any submodule of L ( E ) ↓ H isomorphic to 1, 4 i or16 . Thus either H is strongly imprimitive or the socle contains some of 16 , 16 and 64 , but each such submodule must be a summand. Since this means thereis another submodule of L ( E ) ↓ H , this proves that H is strongly imprimitive.Thus we may assume that L acts on M ( A ) as either (1 / ) ⊕ / ⊕ z by z − we choose the first one.We prove that there is no embedding of ¯ L , the maximal subgroup Alt(5) ≀ H containing L , with this action. We first prove that if ¯ L has these compositionfactors on L ( E ) then it can only be embedded in D , and then that any suchembedding has the previous action on L ( E ).Since ¯ L is not an almost simple group, it is contained in a member of X σ . Wewill use algebraic group notation for the modules for the group ¯ L ′ ∼ = Alt(5) × Alt(5),writing (0 ,
0) for the trivial module, (1 , , ,
0) and (0 ,
2) for the four 2-dimensional modules, and so on.
12 8. RANK 2 GROUPS FOR E The only normal subgroups of ¯ L are 1, ¯ L itself and ¯ L ′ of index 2. If ¯ L embeds in a product X X and there is no subgroup isomorphic to ¯ L in X , then¯ L ′ ≤ X . We will prove that there are a few positive-dimensional subgroups thatcannot contain ¯ L ′ , and hence eliminate certain maximal subgroups from containing¯ L . For example, there is no set of composition factors for M ( A ) ↓ ¯ L ′ that yieldsa subgroup with the correct composition factors on L ( E ), and so ¯ L ′ cannot becontained in any subgroup conjugate to a subgroup of A . Using the remark aboveas well, this eliminates the A -, A A - and A A A -parabolics. Since ¯ L muststabilize a line or hyperplane on M ( A ), this means that the A A maximal-ranksubgroup, and A A -parabolic subgroups are also eliminated.Suppose that ¯ L lies in E . There are 24776 sets of composition factors for ¯ L on M ( E ), fourteen of which are conspicuous. However, none of these has traceson elements of orders 3 and 5 that fuse to the correct classes of E , so ¯ L cannotembed in E . Hence ¯ L cannot embed in an E -parabolic either.This is enough to eliminate the E A - and D A -parabolic subgroups, the E A maximal-rank subgroup ( ¯ L ′ must lie inside the E factor, but then the compositionfactors of L ( E ) ↓ ¯ L ′ must be distributed among M ( E ) and L ( E ) just as if it wereextensible to ¯ L , so cannot exist), and the F G maximal subgroup. It can alsoeliminate the E A maximal-rank subgroup, as in characteristic 2 this is the directproduct of E and A , so any subgroup ¯ L containing ¯ L ′ lies entirely inside E .The remaining maximal subgroups are the D -parabolic, the D maximal-ranksubgroup, and any maximal-rank subgroups whose component group contains aninvolution where ¯ L ′ can embed in the connected component, or ¯ L can embed inthe component group. However, ¯ L cannot embed in the Weyl group W ( E ) by adirect computer check.There are four possible sets of composition factors for M ( D ) ↓ ¯ L ′ :(1 , , (0 , , (2 , , (0 , , (2 , , (2 , , (3 , , (0 , , (1 , , (2 , , (0 , , (2 , , (0 , , (1 , , (0 , , (0 , , (3 , . Only the first two of these can yield a copy of ¯ L rather than ¯ L ′ ; thus ¯ L cannotembed in the D -parabolic subgroup. We show that in the first two cases, ¯ L doesnot embed in such a way as the restriction to L × h z i is consistent with theremaining possibility discussed above.The L -actions on the two modules ( λ ,
0) and ( λ , / ) ⊕ M ( A ) and 2 on M ( A ), are4 ⊕ ⊕ P (2 ) ⊕ P (2 ) ⊕ (1 / , / ⊕ / / / ⊕ P (2 ) . The exterior square Λ ( M ( D )) and L ( D ) differ by the precise extensionsof the trivial modules on the non-trivial factor of the Lie algebra. Thus the ζ i -eigenspaces for i = 0 of L ( D ) and Λ ( M ( D )) are the same, so we can use thisexterior square to understand the ¯ L ′ -action on L ( E ).If ¯ L ′ acts on M ( D ) as (2 , ⊕ (3 ,
2) then (2 , ⊗ (3 ,
2) is a summand of the¯ L ′ -action on the exterior square of M ( D ). The restriction to L × h z i of the k ¯ L ′ -module (2 , ⊗ (3 ,
2) has ζ -eigenspace P (2 ) ⊕ , which is not a summand ofthe options for V ζ given above.Similarly, since (1 ,
0) and (0 ,
1) have no extensions with (2 , L ′ -action on M ( D ), (1 , ⊕ (0 ,
1) must be a summand. The tensor product .3. PSp (1 , ⊗ (0 ,
1) restricts to L × h z i with ζ -eigenspace simply 2 . This therefore mustbe a summand of V ζ , but of course it is not.This deals with ¯ L being contained in D . For maximal-rank subgroups whosecomponent group contains an involution, we have the D D , A and A maximal-rank subgroups. For each of these, their connected component lies in D , and ineach case ¯ L ′ must be contained in D , so is one of the four options above. Noticethat these maximal-rank subgroups cannot stabilize lines on M ( D ), so ¯ L ′ mustact with one of the first two actions given above; but the second was eliminatedcompletely and the first yielded a contradiction with the action of L on M ( A ).Thus these cannot occur, and the result is proved. Case 6 : The 64 only has an extension with 16 , but there is no module ofthe form 64 / / ; the { } -heart of L ( E ) ↓ H is either of this form or64 ⊕ , so it is the latter. Thus there are two summands 64 in L ( E ) ↓ H . Any16-dimensional simple submodule is a summand, since it appears with multiplicity1. This time, for ease of calculation we consider the action of y rather than z .The trace of y on L ( E ) is −
4, so y has centralizer A , and the compositionfactors on V are 4 , , , , so L acts on M ( A ) with factors 2 , ,
1. It iseasy to see that any such module is the sum of 2 ⊕ and one of the nine moduleswith composition factors 2 , ,
1, as we will see in Case 8 of PSU (4), in Section12.3. These nine modules are2 ⊕ ⊕ , (1 / ) ⊕ , (2 / ⊕ , (1 / ) ⊕ , (2 / ⊕ , / , , , / , / / , / / . However, for six of these the module ( M ( A ) ⊗ M ( A ) ∗ ) ↓ L does not have a sum-mand 2 , whereas the 1-eigenspace of y on 64 has L -action isomorphic to 2 ⊕ .The remaining three modules are 2 ⊕ plus one of 2 ⊕ ⊕
1, (1 / ) ⊕ and(2 / ⊕ .In the first case, we embed L into an algebraic A subgroup Y of A actingon M ( A ) as L (0) ⊕ L (1) ⊕ L (2) ⊕ , and note that the tensor square and exteriorcube of this module are(0 / / ⊕ (0 / / ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ and (0 / / ⊕ ⊕ (1 / / ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ respectively, where we have removed the L ( − ) for clarity. This is of course theaction of A on the kL -module V θ i , so we see that L and Y stabilize the samesemisimple submodules that do not have 4 as a composition factor of V θ i and V .Hence Y stabilizes any submodule of L ( E ) ↓ H whose restriction to L is semisimplebut does not involve 4, for example, each 4 i . But one of these must be a submodule(or H stabilizes lines on L ( E ) and so is strongly imprimitive anyway) as all otherfactors are multiplicity free, hence H is strongly imprimitive by Theorem 3.2 (andLemma 3.5).The other possibility is that L acts on M ( A ) as (1 / ) ⊕ ⊕ or its dual, inwhich case we embed L into Y acting as ( L (0) /L (4)) ⊕ L (2) ⊕ . The product ofthis and its dual, and the exterior cube of this module, are(4 / / / / ⊕ (0 / / ⊕ ⊕ ⊕ ⊕ ⊕
614 8. RANK 2 GROUPS FOR E and (4 / / / / ⊕ ⊕ (0 / / ⊕ ⊕ (0 / ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ V and V θ i is therefore stabilized by Y in this case, so again H is strongly imprimitive, as before. Case 7 : The trace of z is −
2, so has centralizer A A from Table 2.10. Thecomposition factors of V are 4 , , , , so L acts on each M ( A ) as 4 ⊕ V is isomorphic to P (1) ⊕ and each V ζ i isisomorphic to P (1) ⊕ P (2 ) ⊕ P (2 ) ⊕ (1 / , / . Furthermore, both 64-dimensional modules have no extensions with the other com-position factors, so split off as summands. The action of L on the ζ i -eigenspaceof the sum 64 ⊕ is P (2 ) ⊕ P (2 ) ⊕ ⊕ , so any composition factor ofsoc( L ( E ) ↓ H ) other than the 64-dimensionals must restrict to L × h z i with onlytrivial and 4-dimensional simple submodules. The only simple modules for H thatrestrict in this way are 1 and 16 , with the latter not a composition factor of L ( E ) ↓ H . Thus H stabilizes a line on L ( E ), as needed. Case 8 : The trace of y on L ( E ) is −
4, hence y has centralizer A . The com-position factors of V are 4 , , , , but of the (up to field automorphism)thirteen possible sets of composition factors for M ( A ) ↓ L , none has these compo-sition factors. (2 , , has the correct number of 4s, but has 2 , .) Thus thiscase cannot exist. q = 8 : There are far too many sets of composition factors for a module of dimension248, so we need to make some reductions first.Write 1 a , b , c , d for the dimensions of the composition factors of L ( E ) ↓ H .By considering a unipotent element of order 4, which acts on L ( E ) with at most60 blocks of size 4 by Table 2.5, and acts only with blocks of size 4 on non-trivialsimple modules, we see that a ≥
8. By pressure we may assume that b > a , since H ( H, M ) = 0 if and only if dim( M ) = 4 for simple modules by [ Sin92b ]. Finally,the trace of a rational element of order 5 is either − a − b + c − d . These constraints yield 23 solutions for ( a, b, c, d ):(8 , , ,
2) (12 , , ,
2) (16 , , ,
2) (20 , , ,
2) (8 , , ,
1) (12 , , , , , ,
1) (20 , , ,
1) (24 , , ,
1) (28 , , ,
1) (32 , , ,
1) (36 , , , , , ,
0) (12 , , ,
0) (16 , , ,
0) (20 , , ,
0) (24 , , ,
0) (28 , , , , , ,
0) (36 , , ,
0) (40 , , ,
0) (44 , , ,
0) (48 , , , , , , , ,
23 by taking a subgroup h x i oforder 7 stable under the graph automorphism and summing the traces of x , x and x on the simple modules. This yields an integer, which we then compare to thepossible such integers from E , finding that those cases above cannot occur.We can eliminate another one using module structure: the { , i } -radical of P (4 ) is 4 / / , / , / , , / , / , / / , where S (4 i ) = 1 / i +1 / / i +2 . Thus we again see that the 23rd and 12th cases(48 , , ,
0) and (36 , , ,
1) cannot occur, but these have already been eliminated.If L ( E ) ↓ H has pressure 1, then we split L ( E ) ↓ H into sections by taking the { , i } -radical and then any 16s and 64s, and then repeating this. By the above structure,we can obtain at most four trivials before we must hit either a 16 or the top of .3. PSp the module, so we must have a ≤ c + 1), removing the fourth case (20 , , , L ( E ) ↓ H (these are all conspicuous for elements of orders 63 and 65 as well). Theseare , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . We will use Lemma 3.10, so we look for elements of order 195, powering to anelement y ∈ H of order 65, and stabilizing certain eigenspaces of y on L ( E ). Weemploy the strategy outlined in Remark 3.12.For the eighth, tenth and eleventh cases, there are elements of order 195 stabi-lizing all eigenspaces of y , so these cases are certainly strongly imprimitive. For theothers, we give the table of which modules can have their eigenspaces stabilized byroots of y . Case Modules1 1, 4 , 4 , 4 , 4 , 4 , 4 , 4 , 4 , 4 , 4 , 16 , 16 , 4 , 4 , 4 , 4 , 16 , 4 , 4 , 4 , 4 , 16 , 4 , 4 , 4 , 4 , 4 , 4 , 4 , 4 , 4 , 16 , 16 , 16 , 16 , 4 , 4 , 4 , 4 , 4 , 16 Case 1 : We have no possible automorphisms that permute the factors, so if H is not strongly imprimitive, the socle of L ( E ) ↓ H must consist of summands of L ( E ) ↓ H and 16 , so we need the { , i } radical of P (16 ). This is1 / / / , / . We apply Proposition 2.9 with the set I = { , , , } . Since each ofthese appears in L ( E ) ↓ H with multiplicity 1, the I -heart is semisimple. Thus bythe proposition at least four trivials appear in the I ′ -radical of L ( E ) ↓ H . But thisradical us a submodule of the module above, a clear contradiction. Thus L ( E ) ↓ H contains a submodule from the list in the table, and thus H is strongly imprimitive,using Lemma 3.10. Case 2 : Suppose that H is not strongly imprimitive, and we aim to apply Lemma3.10 again. Since the composition factors are invariant under an outer automor-phism swapping 4 and 4 , we remark that there are elements of order 195 stabilizingall eigenspaces of y on 4 ⊕ , but not for 4 ⊕ . In addition, the 64-dimensionalfactors have no extension with 16 and therefore split off as summands. Ignoringthese, the socle of L ( E ) ↓ H is a sum of some of 4 ⊕ , 16 , and potentially other16-dimensional modules that must be summands. We take the { , i } -radical of
16 8. RANK 2 GROUPS FOR E P (16 ), then add any copies of 16 and 16 on top, then add any copies of 1, 4 ,4 , 4 and 4 on top of that, to obtain the module1 , / , , , / , , / , / . Since this clearly does not have enough composition factors, we must have 4 ⊕ in the socle (we need both else one on its own is an orbit under N Aut + ( G ) ( H ), so H is strongly imprimitive by Theorem 3.2). They of course then can only appearin the socle and top of the module.We take the { , , , } -radical of P (4 ), then add any copies of 16 and16 on top, then add any copies of 1, 4 , 4 and 16 on top of that, to obtain themodule 4 / , / , / , / . We need eight trivial composition factors, and we can see at most seven, so thisyields a contradiction. Thus either there must be a copy of one of 1, 4 or 4 inthe socle of L ( E ) ↓ H , or a copy of 4 but not 4 (or vice versa). In particular, H is strongly imprimitive. Cases 3, 4, 5 and 6 : In these cases H is definitely Lie imprimitive, becauseall 16-dimensional composition factors appear with multiplicity 1, there are onlyextensions between modules of dimension 4 i and 4 i ± , and L ( E ) is self-dual. Thusthere must be a 4-dimensional factor in soc( L ( E ) ↓ H ). In the third, fourth and fifthcases the composition factors are not invariant under any outer automorphism, sothese are also strongly imprimitive. In the sixth case, we find no elements thatsimultaneously stabilize the eigenspaces of all of 4 , 4 and 4 , so we must use adifferent strategy.Suppose that L ( E ) ↓ H is invariant under the field automorphism (as it needsto be for H not to be strongly imprimitive). We show that H in fact stabilizes aline on L ( E ), so suppose that this is not true. As we must have a subquotient16 ⊕ ⊕ ⊕ ⊕ ⊕ by Lemma 2.7, this subquotient must form the‘middle’ of the module L ( E ) ↓ H , and at least six trivial modules must appear inthe { , , , } -radical of L ( E ) ↓ H by Proposition 2.9. (We may ignore the 64-dimensional modules in the socle as we are interested in whether H stabilizes a line.)We therefore consider the { , , , , } -radical of P (4 i ) ( i = 1 , , / i . Thus we need at least six 4-dimensional factors in the socle to have thatmany trivial factors, but the pressure of the module is only 3. Thus H stabilizesa line on L ( E ) by Proposition 2.5, and in particular is strongly imprimitive byLemma 3.5. Case 7 : This case is very easy, since every composition factor of soc( L ( E ) ↓ H ) isin the table apart from 64 , which must be a summand if it is in the socle. Thus H is always strongly imprimitive. Case 9 : We first show that H is Lie imprimitive, and then that H is stronglyimprimitive. We need to consider the cf( L ( E ) ↓ H )-radicals of P (16 ) and P (16 ),which are1 , / , , / , / , / , , / , , / , / , / . We clearly cannot fit sixteen trivial composition factors above two of these modules,and so there is another factor in soc( L ( E ) ↓ H ). Thus H is Lie imprimitive, so liesinside a member of X . .4. B To prove strong imprimitivity, note that there are elements of order 195 stabi-lizing the eigenspaces in 4 ⊕ and 4 ⊕ as well, so we may assume that theonly extra modules in the socle are 4 and 4 , else H is strongly imprimitive.We show that there is a subquotient(8.3) 16 ⊕ ⊕ ⊕ ⊕ ⊕ in L ( E ) ↓ H . To see this, we take the { , , , } -heart V of L ( E ) ↓ H , andwe show that it is semisimple. Note that V is self-dual. If 4 or 4 is a submoduleof V then it splits off as a summand, as we desire, so we may assume that soc( V ),and top( V ), have composition factors only copies of 16 and 16 . We constructedabove a pyx for V , namely a sum of the radicals of P (16 ) and P (16 ). However,the { , } ′ -residuals of these modules still contain a pyx for V , as top( V )consists solely of copies of these modules. However, these are just 16 and 16 ,so indeed V is semisimple.We now compute a few radicals. The { , , , , , } -radicals of P (4 )and P (4 ) are simply 1 / / / , / / / . There can be at most four copies of 4 and 4 (that are not summands) in the socleof L ( E ) ↓ H , and so the { , , , , , } -radical W of L ( E ) ↓ H possesses atmost two copies of 4 . However, the quotient module L ( E ) ↓ H /W must have thesubquotient from (8.3) as a submodule, and the quotient by that must be isomorphicto a submodule of W ∗ , by self-duality of L ( E ). We see then that there are notenough copies of 4 in W , as there are eight in L ( E ) ↓ H , which is a contradiction.Hence one of 4 , 4 , 4 , 4 , and 16 is a submodule of L ( E ) ↓ H .Thus H is strongly imprimitive in this, and all other, cases. (cid:3) This last remaining case of PSp (2) is not warranted. Two possible moduleactions of H on L ( E ) can be constructed in this case. A unipotent element v oforder 4 in H acts on L ( E ) with Jordan blocks either 4 , or 4 , . Assumingthat H does not stabilize a line on L ( E ), we must have submodules of P (4 i ),together with copies of 8 and 8 . The structure of P (4 ) is uniserial, and is4 / / / / / / / / and similarly for P (4 ). Therefore, if the action of v of order 4 has a block of size 2,there must be a summand 4 i / / − i / / i , so we see that there is a unique structurefor the action of H on L ( E ) in each case, namely8 ⊕ ⊕ ⊕ ⊕ (4 / / / / / / ) ⊕ ⊕ (4 / / / / / / ) ⊕ and 8 ⊕ ⊕ ⊕ ⊕ P (4 ) ⊕ ⊕ P (4 ) ⊕ ⊕ (4 / / / / ) ⊕ ⊕ (4 / / / / ) ⊕ . B As B ( q ) contains an element of order q −
1, we need only consider q = 2 n +1 for q = 8 , , , q − T ( E ).Thus here we assume that n = 1 , , , Proposition . Let H ∼ = B (2 n +1 ) for some n ≤ .
18 8. RANK 2 GROUPS FOR E (1) If n = 1 then H stabilizes a line on L ( E ) or the composition factors, upto field automorphism, are one of , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . (2) If n = 2 , then H is strongly imprimitive. (3) If n = 4 then either H stabilizes a line on L ( E ) or H is a blueprint for L ( E ) . Proof.
We first list the 23 possible distributions of dimensions of compositionfactors for L ( E ) ↓ H , based on the trace of an element of order 5 being − , , ,
2) (12 , , ,
2) (16 , , ,
2) (20 , , ,
2) (8 , , ,
1) (12 , , , , , ,
1) (20 , , ,
1) (24 , , ,
1) (28 , , ,
1) (32 , , ,
1) (36 , , , , , ,
0) (12 , , ,
0) (16 , , ,
0) (20 , , ,
0) (24 , , ,
0) (28 , , , , , ,
0) (36 , , ,
0) (40 , , ,
0) (44 , , ,
0) (48 , , , n = 1 : We assume that the trace of an element of order 5 is not 23, as then it musthave negative pressure. There are five remaining conspicuous sets of compositionfactors for L ( E ) ↓ H up to field automorphism:16 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . We can do nothing with any of these cases. n = 2 : Note that H has an element x of order 31 and an element of order 25.There are, up to field automorphism, seven sets of composition factors for L ( E ) ↓ H that are conspicuous for elements of orders 25 and 31 and have positive pressure:64 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , As with PSp (8), we use Remark 3.12 to yield that either H is strongly imprimitiveor the socle of L ( E ) ↓ H consists solely of a subcollection of the composition factors,which do not have elements of order 93 stabilizing their eigenspaces. Since noneof the sets of composition factors is invariant under the field automorphism, we donot need to distinguish between strong and Lie imprimitivity (plus σ -stability). Cases 6 and 7 : For these sets of composition factors, there exists an element oforder 93 in G , powering to x , and stabilizing the same subspaces of L ( E ) as x ,hence H is strongly imprimitive by Lemma 3.10.For the other cases, we consider elements ˆ x of order 93 that cube to x andstabilize the constituent eigenspaces of some composition factors. We include thenumber of elements of order 93 stabilizing those eigenspaces in the table. .4. B Case Factors Number of elements1 4
22 4 , 4 , 4 , 4
8, 8, 8, 83 4 , 4 , 4 , 4
2, 8, 2, 24 4 , 4 , 4 , 4
2, 2, 8, 25 4 , 4 , 4 , 4 , 4 , 16
2, 26, 2, 2, 8, 2
Cases 2 and 3 : We may remove any 64-dimensional modules from the socle of L ( E ) ↓ H without affecting whether the socle contains 1- or 4-dimensional modules.Doing this, we note that the 16s appear with multiplicity 1, so in fact L ( E ) ↓ H must have a 4-dimensional (or trivial) submodule. This is enough to complete theproof for Case 3, using Lemma 3.10.For Case 2, we may assume that the socle consists of copies of 4 only (andsimple summands of dimension 16). The { , , , } -radical of P (4 ) is1 / / / . By Lemma 2.7, the { , , ij } -heart of L ( E ) ↓ H is semisimple, whence by Propo-sition 2.9 there must be at least six trivial composition factors (and two copies of4 ) in the { , , ij } ′ -radical of L ( E ) ↓ H . However, from the above module wesee that there are at most four trivials in this, and no copies of 4 at all.Thus soc( L ( E ) ↓ H ) must contain one of the composition factors in the tableabove (or a trivial), and so H is strongly imprimitive. Case 4 : Suppose that none of the modules in the table appears in the socle of L ( E ) ↓ H , and that there are no trivial submodules either. The composition factorsof L ( E ) ↓ H that appear with non-unity multiplicity and are not in the table aboveare 16 , 16 and 4 , and by Lemma 2.7 the { , , , } -heart of L ( E ) ↓ H is semisimple.Let I be the set { , , , , , , } . The I -radicals of P (4 ), P (16 )and P (16 ) are 4 / / / / , / , / , / , / / / , / / / , / , , / , / / , / / / / / , / , / , / , / . (In fact, one may place a copy of 16 on top of the last one, but of course the othercopy of 16 would have to lie in the top, so we can ignore this.) There must betwo copies of 4 in the I -radical of L ( E ) ↓ H by Proposition 2.9, but in each of themodules above, each copy of 4 lies directly above a copy of 16 . However, thereare four copies of 4 and only two of 16 , so we obtain a contradiction.Thus L ( E ) ↓ H has a module in the socle from the table above (or a trivial),and H is strongly imprimitive. Case 5 : This is similar to, but easier than, the previous case. Now the only possiblemodules in the socle are 16 and 16 , so set I = { , , , , , , , } .The I \ { V } -radicals of P ( V ), for V = 16 , , are4 / / , , / , , / , , / , , / , , / , / / , / / / / / / , / , / , / , / , / . Neither of these has eight trivial composition factors, which they would need to doin order for the socle to be simple. (This uses Proposition 2.9, just as in Case 4.)
20 8. RANK 2 GROUPS FOR E Thus we see that the socle cannot be simple, so in fact must be 16 ⊕ . Inthis case we take the { , , , , , } -radicals of the two projectives, whichare 4 / / / , / , , / , / / , / / , / , / , / . There are only six trivials in this submodule, so we apply Proposition 2.9, as inCase 4, to conclude that this cannot be the socle either.Thus L ( E ) ↓ H has a module in the socle from the table above, and H is stronglyimprimitive. Case 1 : The two 64-dimensional modules have no extensions with 16 , and somust split off as summands. We will show that L ( E ) ↓ H always possesses eithera 1 or a 4 in the socle, so assume that this is not the case. Note that if 16 or16 lie in the socle then they are summands, so we may ignore them. Let W beobtained from L ( E ) ↓ H by removing any simple summands, so that soc( W ) is asubmodule of 16 ⊕ ⊕ ⊕ ⊕ . As the pressure of L ( E ) ↓ H is 4, and the 16 must support a 4-dimensional module above it, there are at most four factors insoc( W ). To begin with, we suppose that 16 does not lie in the socle of W . Ateach stage we will be unable to build a pyx for W with enough trivial compositionfactors.In turn we eliminate all possible socles, starting with 4 , 4 and 4 ⊕ . Thenext step is to eliminate 4 and 4 ⊕ . After that we eliminate 4 ⊕ plus a submoduleof 4 ⊕ . This means that the socle is 4 plus a non-trivial submodule of 4 ⊕ ,and we eliminate the three possibilities there. We finally briefly consider the casewhere 16 lies in the socle, and eliminate this quickly.We construct a submodule of P (4 i ) for i = 1 , ,
5, by taking all copies of 1, 16 and 4 j for j = i , together with 4 if i = 1, on top of 4 i , then any copies of 16 and 16 , then any copies of the 1, 16 , and so on again. Finally, we add copies of4 i on top of that, and remove any quotients of the form 1, 4 , 16 and 16 . Thisforms the following two modules for i = 4 , / / , , / , , / , , , / , , / , , , / , , / , / / , , / , / , , / , , / , , / , / . Let v denote an element of order 4 in H : the action of v on the sum of thecomposition factors of L ( E ) ↓ H is 4 , , so from Table 2.5 we see that v musthave Jordan blocks 4 , . The element v acts on the first module with blocks 4 ,so four of the trivial composition factors of that module cannot appear in W if 4 lies in the socle of W . Since there are eight trivial factors in W , this means thatnone of 4 , 4 and 4 ⊕ is the socle of W .If there are only copies of 4 in soc( W ) then instead of the { , , , } -residual of this module for P (4 ), we may take the { } ′ -residual. So we take the { , , , , , } -radical of P (4 ), which is4 / / / , , / , , / , / , , / , / , then add on copies of 16 and 16 , then take all copies of 1, 4 , 4 , 4 , 4 and16 on top of this module, and finally take the { } ′ -residual of this. This yieldsthe module4 / / , , / , , , / , , , , , / , , , , / , , , / , , / . .4. B Thus the socle of W cannot be 4 as there are only six trivials in this. However, ifsoc( W ) ∼ = 4 ⊕ then we cannot have any copies of 4 in the middle of this module,yielding a smaller module:4 / / / / , / / , , / , , / . As v acts projectively on this module as well, then as before two of the trivialsin this module cannot appear in W . However, this means that 4 ⊕ cannot be thesocle of W either.Thus we need both a 4 and either a 4 or a 4 in soc( W ). Suppose first that4 ⊕ lies in the socle of W . Similarly to above, we construct submodules of P (4 ), P (4 ) and P (4 ), but now imposing that all copies of 4 lie in the socle or top.This yields modules4 / / , / , / , , / , / , , , , / , , / , / / / / , / / , / / , , / , / , / , / , , / , / . The element v acts with Jordan blocks 4 , on thesecond and 4 , , W from each of these. Thus to achieve eight trivials in W , the socle mustbe 4 ⊕ ⊕ ⊕ . However, in this case we have to remove the copies of 4 , 4 and4 from the middle of the modules, and this yields the three modules4 / / / / , , , , / , , / , , / / , / / , , / , / , / , / , , / , / . The action of v on these three modules has Jordan blocks 4 , for the first, 4 , , , v must act on a submodule of this with blocks 4 , , andthe trivial composition factors only form three blocks of size 2. Thus there is asingle 4 in the socle of W .If the socle is 4 ⊕ ⊕ , then we allow copies of 4 in the interior of themodule, but not 4 or 4 . Taking the { , , , } -radical of P (4 i ), then addingon any copies of 16 and 16 , and then again any copies of 1, 4 and 4 , and finallyall copies of 4 and 4 on that, yields three modules. The { , , } ′ -residuals ofthem are4 , / , / , , , , / , , , / , , , , / , , / , / / , , , / , , / , / / , , / , / , , / , , / , , / , / . There are ten trivial factors in the sum of these, and we need eight, so if we canprove that at least three of these factors do not lie in W then we get a contradiction.Take the { } -heart of these modules, to produce three much smaller modules. Thesecond is self-dual and has structure1 / , / , , / , / , while the first and third are isomorphic, and1 / / , / / , / / / / , A and A respectively, so that the { } -heart of W is a submodule of A ⊕ A ⊕ . This nowlooks good for a contradiction: of course, W has a single 16 , and modulo projective
22 8. RANK 2 GROUPS FOR E blocks the action of v of order 4 on A is 2 and A is 2 ,
1. We end up that W losesa trivial factor that contributes to a block of size 2 in A , but we need all four ofthese to make the action of v equal to 4 , on L ( E ). This contradiction dealswith soc( W ) ∼ = 4 ⊕ ⊕ .We next consider the socle 4 ⊕ , so we allow copies of 4 and 4 in the interiorof our modules. As we have done before, we obtain the following submodules of theprojectives:4 / / , , , / , , , , / , , , , , , / , , , , / , , , , / , , / , , / , / , / , , / , , / , / The actions of v on these two modules have blocks 4 , , , , , , , so in particular we cannotmake four blocks of size 2. Thus we again obtain a contradiction to the socle of W being this module.Finally, for a socle of W of the form 4 ⊕ , we allow copies of 4 in the interiorof our modules. Proceeding as before, we obtain submodules of P (4 ) and P (4 ),which are quite large, and contain lots of trivial composition factors, too many forour standard construction above to work. We can cut the submodules of P (4 )and P (4 ) down by noting that we can take copies of modules in the followingorder: { , , , } ; { } ; { , , , } ; { , } ; { , , , } ; { } ; { , , , } . On top of these, we place copies of 4 and 4 , then take the { , } ′ -residual of these modules, to form much smaller modules:4 / / , , / , , / , , , / , , / , / / , / , / , , , / , , / , / / . As a pyx for W , the sum of these two modules has exactly eight trivial factors,so all must be present in any pyx for W . However, the action of v on these moduleshas blocks 4 , , v on L ( E ), namely 4 , . This final contradiction dealswith all cases where the socle of W consists solely of 4-dimensional factors.We have therefore shown that 16 must lie in the socle of W , and also in thetop. Remove these: as 16 has extensions with 4 and 4 , potentially both of thesedrop into the socle of W now, so quotient out by 4 if it exists. If there is no trivialsubmodule of this new module then we obtain a contradiction, as this new modulecan take the place of W as well. However, we have already shown above that sucha module cannot exist. Hence we find a submodule 1 / / , but it is easy toshow that such a module cannot exist (there is a module 1 / , / but it needsboth 4 i in the second layer). This completes the proof that the first case yields astrongly imprimitive subgroup, and therefore all cases. n = 3 : Let x be an element of order 145 in H , so that x has order 29. The tracesof elements of order 145 are not known, but those of order 29 are. In addition, thereare 48384 elements of order 5 with trace − C ) of a maximaltorus T , so given an element y of order 29 in T , there are 48384 elements ˆ y of order145 in T such that ˆ y = y and ˆ y has trace − L ( E ). These are manageablenumbers for a fixed element of order 29, but there are far too many possible modulesfor H of dimension 248 for us to test them all to find the conspicuous ones. .4. B Suppose that L ( E ) ↓ H does not possess a composition factor of dimension 64.The 1-eigenspace of x on modules of dimension 4 and 16 is zero, and there are noelements of order 29 in T with a 1-eigenspace of dimension 44 or 8, which removesthe cases (8 , , ,
0) and (44 , , ,
0) above.For the others, we look whether, given an element y ∈ T of order 29, there isan integral solution to the simultaneous linear equations that arise when writingthe dimensions of the eigenspaces of y on L ( E ) and of x on the simple modulesfor H . There are no solutions when the 1-eigenspace of y has dimensions 28, 36, 40or 48, and so we have so far eliminated Cases 13, 18, 20, 21, 22 and 23 from the listabove. For the others, very few elements y have the required property: in all caseswhere there are no 64, we find only ten (up to taking powers of y ) possibilities forthe conjugacy class of y .We run through the 48384 possible roots of y and see if there are solutions usingthe same method as above, but now with x rather than x . This yields exactly foursolutions: 16 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . In each case, one finds eight elements of order 435 in T that cube to x and stabilizethe same subspaces of L ( E ) as x , whence H is strongly imprimitive by Lemma3.10. (In fact, it is easy to show that H is actually a blueprint for L ( E ) from this.)We now look at the cases where there is a 64 in the composition factors. Using asimilar method as above, we find exactly one set of composition factors for L ( E ) ↓ H that is conspicuous for elements of order 145, and that is64 , , , , , , , , , , . Again, there are eight elements of T of order 435 that cube to x and stabilize thesame subspaces of L ( E ) as x , so we are again done and H is strongly imprimitive.Finally, we consider those sets of dimensions that have two 64s. We find sevensets of composition factors that are conspicuous for an element of order 145, whichare 64 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , In all but the first and sixth of these cases, there are eight elements of order 435in T that cube to x and stabilize the same subspaces of L ( E ), so we are done inthose cases again, and H is strongly imprimitive.In the first case, there are elements of order 435 stabilizing the constituenteigenspaces of each simple module in L ( E ) ↓ H but 64 , and since the socle can-not consist solely of that factor (as it must be a summand if it is a submodule), H is strongly imprimitive by Lemma 3.10. In the sixth case, even more is true: thereare elements of order 435 stabilizing the constituent eigenspaces of each composi-tion factor (but of course not simultaneously) and so H is strongly imprimitive by
24 8. RANK 2 GROUPS FOR E Lemma 3.10 again, in both cases via Lemma 3.9 (as the composition factors arenot stable under any outer automorphisms).This completes the proof that H is always strongly imprimitive. n = 4 : We have an element x of order 37, and a subgroup L = B (8) of H , andthese are the only things we can use to compute conspicuous sets of compositionfactors for L ( E ) ↓ H . Our strategy is to determine which classes in G of elementsof order 37 are possibilities in conspicuous sets of composition factors for L ( E ) ↓ H ,by checking that there is a solution to the linear algebra problem of the variouseigenspaces for the action of x having the required dimensions for x to belong tothat class, as we range over all possible sets of composition factors for L ( E ) ↓ H .(This is a surprisingly stringent requirement.) Given a class for x , we then rangeover all thirteen conspicuous sets of composition factors for L ( E ) ↓ L , using elementsof orders 13, 7 and 5. This time we use a linear programming solver in Magma tocheck that there is a solution to the eigenspace dimension problem for elements oforders 37, 13, 7 and 5 simultaneously, and one with non-negative integer coefficients.This yields a few possible sets of traces, but each can yield many conspicuoussets of composition factors for L ( E ) ↓ H . For these, we now fix an element x ′ oforder 37 in T with the appropriate trace on L ( E ), and then run over all (severalmillion) elements y ′ of order 13 with the appropriate trace on L ( E ) and check ifthe eigenvalues of x ′ y ′ match any of the conspicuous sets of composition factorsabove. This process takes a long, but finite, amount of time.Very few do. With no 64-dimensional factor, there are five sets of factors up tofield automorphism:16 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . (The last three of these fit into the schema16 a,b , a,c , b,d , a , b , c , d , a +1 , b +1 , . Here, a, b, c, d are distinct integers between 1 and 9, the indices are read modulo 9,and if (say) a + 1 = b then 4 b has multiplicity 9.)There is one with a single 64-dimensional factor:64 , , , , , , , , , , . There are several with two: most fit in a general schema64 a +4 ,b +4 ,c +4 , a,b , b,c , c,a , a , b , c , a − , b − , c − , . (The same rules apply for this schema as the previous one.) This yields a set offactors conspicuous for elements of orders 7, 13, 37 and 481 for ( a, b, c ) any suchtriple. Apart from these, there are up to field automorphism three more conspicuoussets of composition factors for L ( E ) ↓ H :64 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , (These are related by being of the form64 a,b,c , b,c,d , a,d , a,b +1 , c +1 ,d , a , d , b +1 , c +1 , , .5. G as with the previous case.)In each case, we check that among all possible elements of order 481 that areproducts of a fixed element of order 37 and all elements of order 13 that lie in thesame maximal torus and have the correct eigenvalues, every element of order 481with the correct eigenvalues on L ( E ) has a root of order 481 · T ( E )with the same eigenspaces on L ( E ). In other words, the element of order 481 in H , and hence H , is a blueprint for L ( E ). (We have only checked those cases withpositive pressure to save time, and so we must include the option that H stabilizesa line on L ( E ) as well.) (cid:3) It doesn’t seem possible, using the techniques from this paper, to resolve any ofthe cases for q = 8. Even the first one, where 16 and 16 do not appear, cannoteasily be dealt with, as the { , i , } -radicals of P (4 ), P (4 ) and P (4 ) are allquite sizeable, with around fifteen socle layers. In addition the maximal subgroupsof B (8) are few: the normalizer of a Sylow 2-subgroup, and subgroups 13 ⋊ ⋊ ⋊
2. The second, third and fourth of these yield no extra informationthan the action of a unipotent element, and it seems difficult to use the normalizerof a Sylow 2-subgroup, which is just contained in some parabolic subgroup of G .It seems highly likely that different techniques will be needed here. G In this section we simply prove that G (3 n +1 ) ′ must stabilize a line on L ( E ). Proposition . If H ∼ = G (3 n +1 ) ′ for some n ≥ , then H stabilizes a lineon L ( E ) . Proof. n = 0 : There are, up to field automorphism, ten conspicuous sets ofcomposition factors for L ( E ) ↓ H . As the Sylow 3-subgroup of H is cyclic, we canwrite down explicitly all indecomposable modules: the projective covers of 1 and 7are P (1) = 1 / / , P (7) = 7 / ((7 / / ⊕ / , and from this it can be seen that the only indecomposable module with a trivialcomposition factor but no trivial submodule or quotient is P (7). Therefore, in orderfor L ( E ) ↓ H not to have a trivial submodule, there must be at least five times asmany 7s as 1s. Up to field automorphism there is a single conspicuous set of factorswith this property, namely 9 , , , , , and if H does not stabilize a line on L ( E ), then L ( E ) ↓ H is the sum of a projectivemodule and either 7 ⊕ /
7. If v denotes an element of order 9 in H , then v actson L ( E ) with Jordan blocks 9 , or 9 ,
5. Examining [
Law95 , Table 9], we seethat no unipotent class has either of those Jordan block structures, so H stabilizesa line on L ( E ) in all cases. n > : Of the 10 conspicuous sets of composition factors in the previous case (upto field automorphism), only three yield conspicuous sets of factors for G (3) =PSL (8) .
3, namely 27 , , , , , , , . The simple modules for H of dimension at most 248 are of dimension 1, 7, 27,49 = 7 × ×
27. Those of dimension 49 restrict to L ∼ = G (3)
26 8. RANK 2 GROUPS FOR E as (7 / / ⊕ ⊕
1, and those of dimension 189 restrict to L as 27 ⊕ ⊕ P (7) ⊕ .Moreover, from [ Sin93 ], we see that dim( H ( H, ≤
1, and dim( H ( H, i )) = 0for i = 1 , , , L ( E ) ↓ H is at leastthe number of 49s, so H always stabilizes a line on L ( E ). (Equality is possible,for 49 , , .) (cid:3) G In this section we will prove that there are no Lie primitive subgroups G ( q ).In all cases apart from q = 7 we prove that H stabilizes a line on L ( E ), and for q = 7 we show that H is a blueprint for L ( E ). We continue with our definition of u from the start of the chapter. Proposition . Let H ∼ = G ( q ) ′ for q ≤ . (1) If q = 2 , , , , , then H stabilizes a line on L ( E ) . (2) If q = 5 , then H is a blueprint for L ( E ) . Proof. q = 7 : There are four sets of composition factors for L ( E ) ↓ H thatare conspicuous for elements of order at most 21, namely77 , , , , , , , , , , , , , , , . Case 1 : The 77 and 14s must split off as they have no extensions with the compo-sition factors of L ( E ) ↓ H , so u must act on L ( E ) with the blocks 4 , , , ,among others. There is no non-generic unipotent class in G with these blocks inits action on L ( E ) from Table 2.7, so u , and hence H , is a blueprint for L ( E ) byLemma 2.2. Case 2 : The 14 must split off as a summand because it has no extensions with theother composition factors of L ( E ) ↓ H . The { , , } -radicals of P (1) and P (26)are the modules 1 / / / , /
26 respectively. The action of u on thesemodules has blocks 3 , , and 4 , , , , so u is a generic unipotent classas it has no blocks of size 7 (see Table 2.7). Thus H is a blueprint for L ( E ). Case 3 : The only extension is between 1 and 26, and so we have a summand14 ⊕ ⊕ ⊕ ⊕ ⊕ , on which u acts with blocks 3 , , . Examining [ Law95 ,Table 9], we see that u must come from the generic unipotent class 2 A . Thus H is a blueprint for L ( E ) by Lemma 2.2. Case 4 : This must be semisimple, and so u acts on L ( E ) with blocks 3 , , .Hence, from [ Law95 , Table 9], u comes from the generic class A , hence H is ablueprint for L ( E ) by Lemma 2.2. q = 5 : There are four sets of composition factors for L ( E ) ↓ H that are conspicuousfor elements of order at most 21, namely77 , , , , , , , , , , , , , , . None of these composition factors has an extension with any other, so L ( E ) ↓ H must be semisimple. We can therefore read off the action of u , and it acts on thesemodules with blocks4 , , , , , , , , , , , , , . This means u lies in the generic class 4 A , 3 A , 2 A and A respectively, by [ Law95 ,Table 9]. Hence H both stabilizes a line on L ( E ), and is a blueprint for L ( E ) byLemma 2.2, as claimed. .6. G q = 3 : There are (up to graph automorphism) four conspicuous sets of compositionfactors for L ( E ) ↓ H , namely7 , , , , , , , , , , , , , , , . Cases 1, 2 and 3 : These have pressures − − − H stabilizes a line on L ( E ) by Proposition 2.5. Case 4 : Suppose that H does not stabilize a line on L ( E ). The pressure of L ( E ) ↓ H is 3, so there can be no 49 ⊕ as a subquotient, and hence if W denotesthe { } -heart of L ( E ) ↓ H then W is a submodule of P (49) and has three trivialfactors.Suppose that 1 , /
49 is not a submodule of W ; then 1 /
49 is, and this haspressure 1, so ignoring the copies of 7 i in W , we see that W is of the form49 / , / / /
49, but such a module is clearly not self-dual, a contradiction. Thus(again ignoring the 7 i ) W has the form 49 / / / , /
49, with one of the trivials inthe submodule 49 / , /
49 being a quotient.Let W ′ be defined as follows:(1) Construct the { , i } -radical of P (49);(2) Take all possible copies of 49 on top of this module (there are four);(3) Take all possible copies of 1 on top of this module (there are three).The { i , } -residual ¯ W of W must lie inside W ′ . Note that ¯ W has a quotient ofthe form 1 /
49, and the kernel corresponding to this quotient is 1 , /
49 with somecopies of 7 i placed on top. This will be important later.The socle structure of W ′ is1 , / , , / , , , / , , , / , , , / , and the only quotients are trivial modules. The 49 in W that is not either asubmodule or quotient therefore occurs on either the fourth or fifth socle layer.Suppose that it occurs in the fourth, so that the trivial in the fifth layer of W ′ is also in W . We take the { i , } -residual of soc ( W ′ ), which is1 / / , / , , , / , but all trivial factors of this are quotients, a contradiction. Thus the 49 in themiddle of W lies in the fifth socle layer, and the trivial lies in the sixth.The second top W ′ / rad ( W ′ ) of W ′ has the form (1 / ⊕ (1 , / , , V for the second, indecomposable, summand. Suppose for a contradictionthat V possesses a submodule V of the form 1 /
49. This is not contained in asubmodule of the form 1 , /
49, as then the rest of the socle forms a complement tothis and V is decomposable. Obviously there is some indecomposable submodule V of the form 1 / ,
49; if V ∩ V = 0 then we have a decomposition for V , so V ∩ V = 0. Certainly V V , so V ∩ V is just the 49. Then V + V has theform 1 , / ,
49 and is complemented by a simple submodule in the socle, anothercontradiction.Thus V has no submodule of the form 1 /
49, and therefore the quotient module W ′ / rad ( W ′ ) has a unique submodule of the form 1 /
49. Letting W ′′ be the { } ′ -residual of the preimage of this submodule in W ′ , we actually see that this is the { i , } -residual of soc ( W ′ ) that we saw before. However, the 49 is in the wrongsocle layer, a contradiction. Thus H stabilizes a line on L ( E ).
28 8. RANK 2 GROUPS FOR E q = 9 : From [ Sin93 ], we see that the only simple modules with non-zero 1-cohomology of dimension at most 248 are of dimension 49, and they have 1-dimensional 1-cohomology. A simple kH -module of dimension at most 248 hasdimension one of 1, 7, 27, 49 and 189, and the restriction to L ∼ = G (3) is eithersimple or 7 i ⊗ i , or 7 i ⊗ i . The former has factors 27 , , ,
1, and the latterhas two 49-dimensional and one 27-dimensional factor. Thus from the conspicuoussets of factors for L ( E ) ↓ L , we see that there are at least three trivial factors and atmost three 49-dimensional factors in L ( E ) ↓ H . Thus H has non-positive pressure,and so H stabilizes a line on L ( E ) by Proposition 2.5. q = 2 : There are four conspicuous sets of composition factors for L ( E ) ↓ H , whichare 14 , , , , , , , ∗ , , , , (32 , ∗ ) , , , ;each of these has negative pressure, so H always stabilizes a line on L ( E ) byProposition 2.5, as needed.Note that: • ⊗ , / / , • ⊗
14 = (6 / / / / ⊕ • ⊗
14 is a module with factors 14 , , , and • ⊗ is a module with factors 64 , , , , where 64 = 32 ⊕ ∗ .These will be needed for q = 4 , q = 4 : There are, up to field automorphism, seven conspicuous sets of compositionfactors for L ( E ) ↓ H :14 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . The pressures of these modules are − −
3, 3, −
2, 2, 1 and 1 respectively.
Cases 1, 2, 4 and 7 : In Cases 1, 2 and 4, H must stabilize a line on L ( E ), byProposition 2.5. Since the last case has no modules with non-zero 1-cohomologythat occur with multiplicity greater than 1, we must have a trivial submodule hereas well. Case 6:
The copies of 64 have no extensions with other modules, so split off assummands. Suppose that H does not stabilize a line on L ( E ), and let W denotethe { i , } -heart of L ( E ) ↓ H . As the pressure is 1 we may assume that the socle of W is either 6 or 6 . As there is a single 36, if one ignores any composition factors14 , the structure of the module is uniserial, of the form6 i / / − i / / / / − i / / i . The { , i , } -radicals of P (6 ) and P (6 ) are6 / , / / , / / / / , , / . We wish to place 36 on top of this, but in both cases 6 ⊕ is a quotient, so thismust be removed to stay pressure 1; hence there is a unique module of the form1 / − i / / i . For i = 2, two copies of 36 may be placed on top of this module,but they fall into the second and third socle layers, and so cannot cover the trivialmodule. For i = 1, two copies of 36 may be placed on top, lying in the secondand fifth socle layers, and a single trivial can be placed on top of this new module, .6. G in the sixth layer. Thus we can remove a copy of 36 which appears as a quotient,leaving a unique module of the form1 / / , / / / . A single 6 j can be placed on top of this module for j = 1 ,
2, but it falls into thefifth or second socle layer, as we saw in the original module above. Hence L ( E ) ↓ H must have a trivial submodule in this case. Case 5:
Suppose that H does not stabilize a line on L ( E ). Since L ( E ) ↓ H haspressure 2, no subquotient of L ( E ) ↓ H may have pressure greater than 2 or lessthan −
2, by Proposition 2.5. Let W be the { } ′ -heart of L ( E ) ↓ H . By Lemma2.7 there is a subquotient 6 ⊕ ⊕ in W . Thus there are two submodules V and V of W such that V ≤ V and the quotient V /V is this module. We willconstruct a specific submodule V now.Let V denote the { , , } -heart of W , which is a submodule of the quo-tient of W by its { , , } ′ -radical. The socle of V cannot be a submodule of14 ⊕ as then V is semisimple, so there is at least one 6 in soc( V ). On theother hand, soc( V ) cannot have 6 ⊕ as a submodule because L ( E ) ↓ H has pressure2. If the socle of V has 6 ⊕ in it then rad( V ) ∩ soc( V ) has a single copy of 6 in it.Add all copies of 1, 6 , 14 and 36 on top of this that lie in V , and then take thepreimage of this new module in W to obtain V .By construction, the quotient W/V has socle 84 ⊕ ⊕ . Let V denotethe preimage in W of soc( W/V ), so that V /V = 84 ⊕ ⊕ . Since W isself-dual, the dual of the quotient W/V is isomorphic to a submodule of W , say V . By construction of V , it has a single 6 composition factor, as does V , and wemust have that V ≤ V . Since W has the same composition factors as V ⊕ V , wesee that V contains at least two copies of 6 , and at least one copy of 36. Since allsubquotients have pressure between − V cannot have negative pressure, V has pressure 0. We note that V must have exactly one copy of 36: otherwise V contains the entire { } -heart of W , and then the { , , } ′ -residual of V must have a quotient 36, but also 6 ⊕ as a quotient as well, hence a quotientof pressure 3. Since V has pressure 0 it has no quotients 6 i or 36.The socle of V is one of 6 , 6 , 36, 6 ⊕ , 6 ⊕ , 6 ⊕
36 and 6 ⊕
36. Ifsoc( V ) = 6 ⊕
36, then V is contained in the sum of the { , , } -radicals of P (6 ) and P (36), which are1 / / , , / , / , , / . This has exactly four trivial composition factors. In order to support the 1 inthe third layer of P (36), we need the 6 in the second layer. But then V has asubmodule 6 ⊕ (6 / V ) = 36. We take the { , , } -radical of P (36), as givenabove, then add all copies of 6 on top of this as possible, and then all copies of 1,6 and 14 as possible, to create a pyx for V . We can construct a smaller pyx byremoving all 6 i quotients. This yields a module1 , , / , , , / , , / , , , / . Notice that this has exactly two copies of 6 , and V has at least that many. Thusboth of these must lie in V , and we take the { } ′ -residual of this module to find
30 8. RANK 2 GROUPS FOR E a submodule of V . This is 6 / / , / . It has a submodule of pressure 3, which is a contradiction.A similar proof works for soc( V ) = 6 , with the roles of 6 and 36 reversed. Sotake the { , , } -radical of P (6 ), add copies of 36 on top, then again all copiesof 1, 6 and 14 . Finally, remove all quotients 6 and 36 to produce the module1 / / , , , / , , , , / , , , / . However, this has a quotient 1 ⊕ , and so we need to remove three of the trivials,and we obtain a contradiction again.For soc( V ) = 6 ⊕
36, we need the { , i , } -radicals of P (6 ) and P (36),which are6 / , / / , / , , / , , , / , , , / , , / , , , / . The second and third socle layers contribute 1 ⊕ , and the fourth and fifth another1 ⊕ . Thus, upon removing two from each of these, we obtain exactly four trivialcomposition factors, the minimum needed for V .Also, the 6 and 6 in the second socle layer of P (36) cannot lie in V , and sothe two trivials in the third layer of that module cannot exist in V . This meansthat the third layer of V must be exactly 6 , but now the submodules of P (36)and P (6 ) both rely on a copy of 6 lying in the third layer to cover the trivial.Thus the 6 in V must be diagonally embedded across the two modules, but thenit cannot support two trivials in the fourth socle layer. This means that a trivialin the fifth layer of P (36) lies in V , but we saw in the soc( V ) = 36 case that the6 in the fourth layer of P (36) is not allowed to lie in V , and we know the 6 liesin the third layer of V . This means that the trivial cannot lie in V after all, acontradiction.If soc( V ) = 6 ⊕ , all copies of 6 lie in soc( W ) or top( W ). Here the first twosocle layers of V must be 1 , / , as there is no extension between 6 i and 14 i .There is no module 36 / / i , so the third socle layer must be 6 , and by choosingsummands appropriately we have (6 / / ) ⊕ (1 / ). (The first summand is notunique up to isomorphism.) On top of this we must place a trivial, and get thewell-defined module 1 / / / . There is no uniserial module 36 / / / / (thesingle 36 we can place on top of this falls to the second socle layer), but we canplace 14 on top and then a 36, to make a (again, not well-defined) module36 / , / / / . (It is not well defined because there is a 36 in the second socle layer of P (6 ), andwe can take a diagonal 36.) On top of this we must place a single trivial, and thisdoes make the module well defined, and so V must be(1 / / , / / / ) ⊕ (1 / ) . To make the module V we must add on top of this 6 ⊕ ; one may place twocopies of 84 on top of the first summand (one on the second), but they fall intothe second and fifth socle layers, so won’t cover a trivial quotient of V . This yieldsa contradiction.The next case is soc( V ) = 6 ⊕ ; the first two socle layers are either 1 , / , or 1 , , / , . There is no module 36 / / i , but there is one 36 / , / , so if .6. G there is no 14 in the second socle layer then one cannot place a 36 in V at all.Thus we can construct the module(1 / ) ⊕ (36 / , / )so far as a submodule of V , but the second summand is not well defined. We arestill yet to find at least 6 , from V . The first summand cannot be extendedfurther, and so the remaining factors pile on the second summand. The factor 6 cannot appear in the third socle layer of the second summand as then there is asubmodule 6 ⊕ (6 , / , / ) of V , which has pressure 3, a contradiction. Thus V contains the submodule 1 / / , / , which is well defined. On this we can place two copies of 6 , but they fall into thesecond and third socle layers, so we cannot construct an appropriate module V .Thus soc( V ) = 6 . There is no module 1 / / or 1 / / (there is a module1 / , / , which has no non-trivial quotients). If 36 / is a submodule of V thenit is also a quotient of W . Removing the 6 /
36 quotient from W yields a moduleof pressure 0, and if there is still a submodule 36 / then removing that yields amodule of pressure − W , but then there is asubquotient 6 ⊕ ⊕
36 of pressure 3, another contradiction.Thus the second socle of V is a submodule of 1 , / , and since one cannotplace a 36 on top of this in the third socle layer, the third socle must be a submoduleof 6 / , / . On top of this we must place another trivial, yielding1 / / , / , and thus the well-defined submodule 1 / / / is a submodule of V . We havetherefore identified the location of the 6 in V . On top of the module 1 / / / we place all copies of 1, 6 and 14 , then all copies of 36, then all copies of 1, 6 and 14 , to yield the module1 / / , / , / , , / , / , / . This has a quotient 1 ⊕ , so one of those trivial quotients needs to be removed, andall others must be present. This means that V has exactly four trivial factors, andtherefore exactly two copies of 6 .On top of 1 / / / one may place a 14 , then either 6 or 36. Suppose that36 can be placed on top. One adds on 14 and then can add two copies of 36. Onthis we place all copies of 6 and 1, then remove all quotients 6 and 36, to obtainthe module 1 / / , , / , / , / . This has four trivial composition factors, which must therefore all be present in V .But this module has no non-trivial quotients, and V cannot have two copies of 36in it.Thus one must place the second 6 on first, and then 36. (There’s no need toremove quotients 6 and 36 this time.) Doing it this way round yields the module1 , / , / , / / , / . This only has three trivial composition factors, a final contradiction.Thus there is no option for the socle of V , and we have obtained a contradictionto H not stabilizing a line on L ( E ), as needed.
32 8. RANK 2 GROUPS FOR E Case 3:
Suppose that L ( E ) H = 0. As 14 and 14 appear exactly once, they arein the ‘middle’ of the module. Formally, we use Lemma 2.7 and Proposition 2.9with I being { , } . So we construct the { , i , } -radical of L ( E ) ↓ H first,then add on 14 and 14 , and this forms an overmodule of the { , i , } -residualof L ( E ) ↓ H . Note that L ( E ) ↓ H has pressure 3.Suppose that 36 lies in the socle of L ( E ) ↓ H . The { , i } -radicals of P (36) and P (6 i ) are 6 , / , / , / , / , , / , i / / − i / , − i / i . As at least half of the trivial composition factors need to appear in the { , i } -radical of the quotient module L ( E ) ↓ H / soc( L ( E ) ↓ H ) by Proposition 2.9, we needthe socle to be 36 ⊕ i ⊕ j ; furthermore, this means that we can afford to lose atmost one trivial factor from the sum of the three modules above, but there is asubquotient 1 ⊕ in this module (the second and third socle layers), so at least twotrivials need to be excluded, a contradiction.Thus 36 does not lie in the socle of L ( E ) ↓ H . Let W denote the { } -heartof L ( E ) ↓ H . The socle of W consists of copies of 36, and if there is more thanone then at least one is a summand. Suppose that soc( W ) = 36, so that the heartof W contains (among other factors) a single copy of 36 and at most one 14 i . Inparticular, there can be no extensions between them in the heart. We saw the { , i } -radical of P (36) before, say V . We have that Ext kH (36 , V ) = 0, but this isa contradiction, since the heart of W contains a copy of 36.Therefore W has a 36 as a direct summand, so let W ′ denote the complement.This is a submodule of P (36) again, and since Ext kH (36 , V ) = 0, we see that W ′ needs to contain 14 or 14 (or both). Let V ′ be constructed by adding as manycopies of 14 and 14 on top of V , and then again adding 1- and 6-dimensionalmodules on top. The structure of V ′ is as follows:6 , / , , , , , / , , , / , , , / , , , , / . Suppose that the 14 i that appear in W ′ lie in the second socle layer: then the heartof W ′ has those 14-dimensional modules as summands, so one may remove the top36 from W ′ , then remove the top 14 i (as W ′ is self-dual, so they must be quotients),and we are left with a submodule of V again, except note that W ′ has pressure atmost 2 (since it is in a sum with 36). Since it has pressure at most 2, we cannothave both 6 i in the second socle layer of V , but each 6 i in that layer supports onetrivial from the third and fifth layers; hence W ′ must contain at most three trivialcomposition factors.On the other hand, suppose that a 14 i appears in the fifth socle layer of W ′ ,which we remind the reader is self-dual. The dual of V ′ is36 / , / , , , , / , , , / , , , / , , , , , , and we see that this 14 i appears in the second radical layer of W ′ , hence in thesecond socle layer as W ′ is self-dual, a contradiction.Thus W contains at most three trivial composition factors. However, the { , i } -radical of P (6 i ) has the form6 i / / − i / , − i / i , whence we see that the { , i } -radical of L ( E ) ↓ H contains at most six trivialfactors, and the same holds for the quotient modulo the { , i } -residual. As this .6. G yields the construction of W , we find that L ( E ) ↓ H possesses at most 6 + 6 + 3 = 15trivial factors, but it really contains 16, a contradiction.Thus H stabilizes a line on L ( E ) in all cases. q = 8 : There are 27 sets of composition factors for L ( E ) ↓ H that are conspicuousfor elements of order up to 19. One (up to field automorphism) has no trivialfactors, but has dimensions 14 , , ,
84. If v denotes the involution from thelarger of the two classes, then v acts projectively on 36-, 64- and 84-dimensionalmodules, and as 2 , on 14-dimensional modules. Thus v would have to act withat least 122 blocks of size 2 on L ( E ), which does not appear in [ Law95 , Table 9].Thus L ( E ) ↓ H has trivial factors. The remaining 24 conspicuous sets of com-position factors, eight up to field automorphism, all have non-positive pressure (themodules with non-zero 1-cohomology have dimensions 6 and 84, see Table 5.5), so H stabilizes a line on L ( E ), as claimed. (cid:3) In the case q = 7, we do not prove that H stabilizes a line on L ( E ): thesecond, third and fourth cases all have negative pressure, so it remains to check thefirst case. The only possible structure for L ( E ) ↓ H in which H does not stabilize aline is (26 / , / ⊕ ⊕ ⊕ ⊕ ;this does occur in F G , which only for p = 7 has a diagonal G acting irreduciblyon M ( F ) and M ( G ).HAPTER 9 Subgroups of E G be the algebraic group of type E and H be a subgroup withrank at most 3, q ≤
9, together with a small Ree group, a Suzuki group, PSL (16)or PSU (16). By Theorem 2.4, the following of these are blueprints for M ( E ),since they have semisimple elements of odd order greater than 75:(1) PSL ( q ) for q = 4 , , , ( q ) for q = 8 , ( q ) for q = 4 , , , ( q ) for q = 7 , (16), PSU (16), PSL (9), G (9);(6) B ( q ) for q > G ( q ) for q > (2) and PSp (2) are also alternating groups, and so are dealt within [ Cra17 ].We also have groups considered in [
Cra , Proposition 3.8], which were provedto be blueprints for M ( E ):(1) PSL ( p ), PSU ( p ), PSp ( p ), for p = 5 , ( q ), Ω ( q ) for q odd.In addition, if the Schur multipler of H is of odd order and H is a blueprint for L ( E ), then H is a blueprint for L ( E ) and M ( E ) as well: we use this to exclude G (5) and G (7).This leaves:(1) PSL ( q ) for q = 2 , , , , , ( q ) for q = 3 , , , , , ( q ) for q = 3 , , , G ( q ) ′ for q = 2 , , , G ( q ) ′ for q = 3 , B ( q ) for q = 8 , (3);(8) PSU ( q ) for q = 2 , , (2).We therefore deal with the remaining groups in turn. Throughout this chapter, H will denote one of the groups above, embedded in the simply connected form G of an algebraic group of type E , in characteristic p dividing q . If q is odd andthe Schur multiplier of H is even (for example, PSp ( q )), let ¯ H denote the centralextension 2 · H (note that, for the groups considered, this is unique), embeddedin G so that Z ( ¯ H ) = Z ( G ) (so the image of ¯ H in the adjoint algebraic group issimple). E As with the previous three chapters, we let u denote an element of order p in H belonging to the smallest conjugacy class. Proposition . Let H ∼ = PSL ( q ) for some ≤ q ≤ . (1) If q = 2 , then H stabilizes a line on either M ( E ) or L ( E ) ◦ . (2) If q = 4 then H stabilizes a line on M ( E ) . (3) If q = 5 then either H is a blueprint for M ( E ) or H acts on L ( E ) as ⊕ and is a blueprint for L ( E ) . (4) If q = 7 , then H is a blueprint for M ( E ) . Proof. q = 7 : The four conspicuous sets of composition factors for M ( E ) ↓ H are 28 , ∗ , , , (10 , ∗ ) , , , . If H acts semisimply on M ( E ), then u acts on the four modules as7 , , , , , , , , , , , , , , , , , , , and in each case the unipotent class exists and is generic by [ Law95 , Table 7].Therefore H is a blueprint for M ( E ) by Lemma 2.2. From Ext data, all but thefirst case must be semisimple. So H is a blueprint in all but the first case. In thiscase there is a module 28 / ∗ (and its image under the graph automorphism), butthe action of u on it has blocks 7 , , which is not in [ Law95 , Table 7], so H againacts semisimply on M ( E ) and we are done. q = 5 : There are twelve sets of composition factors for M ( E ) ↓ H that are conspic-uous for elements of order at most 12 (all are in fact conspicuous):6 , ∗ , (3 , ∗ ) , , , (6 , ∗ ) , (3 , ∗ ) , , (3 , ∗ ) , , , , , (3 , ∗ ) , , , ∗ , (6 , ∗ ) , (10 , ∗ ) , , , ∗ , , ∗ , , , , , , ∗ , , ∗ . , ∗ , , ∗ , , ∗ , , ∗ , (6 , ∗ ) , . The first seven of these and the eleventh and twelfth are semisimple by Ext data, so as in characteristic 7 consider the action of u . We see from this that theaction of u on each of the composition factors in these embeddings has no blocksof size 5, whence u is never non-generic (see [ Law95 , Table 7]), so H is a blueprintfor M ( E ) by Lemma 2.2.The eighth case is either semisimple or (18 / ∗ ) ⊕ (10 / ∗ ) (up to apply-ing the graph automorphism). In these two cases u acts on M ( E ) with blocks5 , , , , and 5 , , respectively, neither of which occurs in [ Law95 , Ta-ble 7]. Thus H cannot embed in G with these factors.For the ninth case, the trivial factors split off, and the remaining compositionfactors form either (8 / ⊕ (19 /
8) or a semisimple module. In either possibility,the action of u on M ( E ) would have Jordan blocks 5 , , , , , and so lies inthe class 2 A + A (see [ Law95 , Table 7]), which is generic. Thus H is a blueprintfor M ( E ) by Lemma 2.2.For the tenth case, we switch to L ( E ), where the corresponding set of compo-sition factors for L ( E ) ↓ H is 125 ,
8, which of course must be semisimple. Letting x denote an element of order 31 in H , we note that the conjugacy class of x isdetermined by its trace on L ( E ). Assuming x lies in a maximal torus T of G , welook for elements of order 93 that power to x . None stabilizes all of the eigenspacesof x on L ( E ), but eight stabilize all of the eigenspaces comprising the 8. The sta-bilizer in G of the 8-dimensional subspace of L ( E ) must contain H and these eight . SUBGROUPS OF E elements of order 93, and stabilize a 125 ⊕ L ( E ) is self-dual).In order to show that H is a blueprint, we just need that this stabilizer is containedin a proper, positive-dimensional subgroup X of G , as X cannot be irreducible on L ( E ), hence L ( E ) ↓ X must be 125 ⊕ E instead of E .) If the stabilizer is not positivedimensional, it must be contained in an exotic r -local subgroup (but clearly that isnot the case for this group, from the list in [ CLSS92 ]) or contained in an almostsimple subgroup J of G (strictly containing N G ( H ) and containing an element oforder 93), and J is Lie primitive. The rest of the results from this chapter provethat J cannot be Lie type in defining characteristic. From the tables in [ Lit18 ] wesee that J cannot act as 125 ⊕
8, and those that act irreducibly on L ( E ) are listedin [ LS04b ], and these are PSU (8), M , Ru and HS . None of these contains anelement of order 31, so J cannot exist.This shows that H is a blueprint for L ( E ), as needed. q = 3 : There are ten conspicuous sets of composition factors for H on M ( E ):27 , , (3 , ∗ ) , , , , , (6 , ∗ ) , (3 , ∗ ) , , , (3 , ∗ ) , , , ∗ , (3 , ∗ ) , , , ∗ , (6 , ∗ ) , , , (3 , ∗ ) , , , (6 , ∗ ) , , ∗ , , ∗ , , (3 , ∗ ) . The pressures of these cases are − − − − − −
2, 0, 4, 2 and 4, so in allbut the last three cases H stabilizes a line on M ( E ) by Proposition 2.5. Case 8 : We obtain the set of composition factors 27 , , , ∗ , for L ( E ) ↓ H ,which has pressure 2. However, the 27s must split off (being projective), and the { , ± , } -radical of P (7) is 3 , ∗ / , / , , ∗ / . Thus if H does not stabilize a line on L ( E ), the socle of the { ± } ′ -heart of L ( E ) ↓ H must contain four copies of 7, but this is impossible, either by pressure using Propo-sition 2.5, or because then there must be summand 7s (as there are at least four 7sin the top as well). Case 9 : Now we obtain two possible corresponding sets of composition factors for L ( E ) ↓ H , 27 , (15 , ∗ ) , , , , (15 , ∗ ) , , , ∗ , , These have pressures − −
1, so H stabilizes a line on L ( E ) by Proposition2.5. Case 10 : There is a unique set of composition factors for L ( E ),27 , (15 , ∗ ) , , , ∗ , , which has pressure 2. Suppose that L ( E ) H = 0. Letting W denote the { , ± } -heart of L ( E ) ↓ H , we consider the { , ± , , ± } -radicals of P (7) and P (15), whichare 7 / / , ∗ / / / , ∗ / / / / . We see that we can only support two trivials above each factor in soc( W ) (as thattop trivial in P (15) cannot lie in W as there is no 7 or 15 ± above it). Thus soc( W )must have at least three composition factors, whence it has pressure at least 3,contradicting Proposition 2.5.
38 9. SUBGROUPS OF E Thus H stabilizes a line on either M ( E ) or L ( E ) in all cases, as claimed. q = 2 : The four conspicuous sets of composition factors for M ( E ) ↓ H are(3 , ∗ ) , , (3 , ∗ ) , , , (3 , ∗ ) , , , . The 8s are projective and so split off as summands. The projective cover P (3) hasstructure 3 / (1 ⊕ (3 ∗ / / ∗ )) / , so we need five 3-dimensional factors for every trivial composition factor in ordernot to stabilize a line on M ( E ). Thus in the second, third and fourth cases H must stabilize a line on M ( E ). In the first case, we switch to the non-trivialcomposition factor L ( E ) ◦ of the Lie algebra L ( E ), where we find that there aretwo corresponding Brauer characters. Hence the composition factors of L ( E ) ◦ ↓ H are one of 8 , (3 , ∗ ) , , , (3 , ∗ ) , . Both of these clearly have trivial submodules, and so we are done. q = 4 : There are, up to field automorphism, just two conspicuous sets of compo-sition factors for M ( E ) ↓ H : 8 , , (9 , ∗ ) , , . The first of these is semisimple and so H stabilizes a line on M ( E ). The rest of thisproof considers the second case. This has pressure 4, and we cannot immediatelyprove that it always stabilizes a line on M ( E ), as there is a module 9 ∗ / , / H does not stabilize a line on M ( E ), and let W denote the { , ± } -heart of M ( E ) ↓ H . We note that soc( W ) cannot be 9 (up to duality) as the { , , ± } -radical of P (9) is 1 , , / , ∗ , ∗ / , , / , and so any pyx for W must have socle either 9 ⊕ ⊕ ∗ , and must have asubmodule W , of the form (1 , / ⊕ (1 , / ± ) . Step 1 : Actions of Alt(6) subgroups and unipotent elements.There are three classes of subgroups Alt(6) in H , labelled L , L , L , containingelements v , v , v of order 4 from three different H -classes. The irreducible kL i -modules are 1 , , , , , and 8 for H restricts to 8 for L i . One may arrangethe labellings so that the restriction of 9 to each L i is 4 / / . Let L denote anyof the L i , and v the corresponding v i . Since 8 is a projective kL -module, therestriction of M ( E ) to L is isomorphic to the sum of 8 ⊕ and W ↓ L .The restriction of W to L has the form1 ⊕ (1 / / / ) ⊕ ⊕ (1 / j / / − j ) , for some j = 1 , W ) is 9 ⊕ ⊕ ∗ . Note that v acts on thesum of the composition factors of M ( E ) ↓ H with Jordan blocks 4 , , and on eachsummand of W with blocks 4 , ,
1, so we see from [
Law95 , Table 7] that v actson M ( E ) with blocks 4 , . Step 2 : The dimension of W L is equal to 2. . SUBGROUPS OF E That it is at least 2 can be seen from the restriction of W to L above. For the otherdirection, note that the permutation module P L on the cosets of L has structure1 / , ∗ / , , , / , ∗ / , and the quotient of P L by its { , , ± } -residual is 1 / , ∗ / , , . If dim( W L ) = 3then there must be a 3-dimensional Hom-space from this module to M ( E ) ↓ H andalso W by Frobenius reciprocity, and we can of course quotient out by the socleof this as we assume that W has no submodule 1 or 8 . Thus there is now a 3-dimensional Hom-space from 1 / , ∗ to W . The image of any of these maps in W must lie inside the second socle layer of the { , ± } -radical of W , so W . But W L has dimension 2, as we saw above. This proves that dim( W L ) = 2. Step 3 : L is contained in an A subgroup.Since M ( E ) L has dimension 2, L is contained in an E -parabolic subgroup or a B -type subgroup by Lemma 3.15 (which, since p = 2, is of type C ). The compositionfactors of L on M ( C ) are 8 , or 4 , , or 4 , . Unless it is the last option, M ( C ) L always has dimension at least 2, hence (since M ( C ) / M ( E ) to the B -type subgroup) M ( E ) L has dimension at least3, a contradiction. But if the factors are 4 , then the traces of elements of order 3on the 32-dimensional module for C are not consistent with it having compositionfactors 8 , , so this case cannot occur.On the other hand, suppose that L is contained in an E -parabolic subgroup.By Proposition 10.3 below L stabilizes a line or hyperplane on M ( E ) as well, andso the image of L in the Levi lies in either F or a D -parabolic subgroup by Lemma3.14.We see that L cannot lie in a D -parabolic subgroup: the factors on M ( D )would have to be 8 , or 4 , , or 4 , (up to automorphism) and so inparticular L would stabilize a line on M ( D ). This means that L lies in C or a D -parabolic, so in either case there is a version of L with the same compositionfactors in C . However, in the table in the proof of Proposition 3.13 we computedthe actions of Alt(6) on M ( C ) and the 16-dimensional module, and none matchesup with these factors.Thus we may assume that the image ¯ L of L inside the E -Levi subgroup liesin F . But then this F subgroup acts on M ( E ) as 1 , /M ( F ) /M ( F ) / ,
1, so L has no fixed points on M ( F ). By Proposition 11.3 below ¯ L or its image under thegraph automorphism stabilizes a line on M ( F ), and so in particular ¯ L is not Lieprimitive in F . Thus we consider the proper, positive-dimensional subgroups X of F . The composition factors of ¯ L on M ( F ) are 8 , , , .We note that ¯ L must be F -irreducible: if ¯ L lies in an A A -parabolic thenit must lie in an A -parabolic, but both of the corresponding Levi subgroups aresimply connected A , and ¯ L lies in adjoint A only; the B -parabolic subgroupalready has four trivial composition factors, so that is not possible; the C -parabolicsubgroup acts on M ( F ) with M ( C ) as a submodule, but of course ¯ L must stabilizea line on this if ¯ L ≤ C , so this cannot work either.As B stabilizes a line on M ( F ), ¯ L cannot lie in this. We are left with A A and C . From the table in the proof of Proposition 3.13 we mentioned above, wesee that if ¯ L is contained in C then it acts irreducibly on M ( C ), and this lies inan A subgroup (subgroup 18 in [ Tho20 , Table 10/10A]). If ¯ L lies in A A thenit must lie in a diagonal A , and we see from [ Tho20 , Table 10A] that this is the
40 9. SUBGROUPS OF E same subgroup of F . This acts on M ( F ) as L (00) ⊕ L (30) ⊕ L (03) ⊕ L (11) . (Note that L (03) has an extension with L (22), but not with L (11). The judiciouschoice of the A allows us not to worry about the 8-dimensional factors, so we donot need to remove them before examining the possible structures.)We now note that the restriction of the extension L (00) /L (30) to the sub-group Alt(6) of A is 1 / / / , so that restriction induces an isomorphism on1-cohomology. In particular, this holds for the actions of the A subgroup and ¯ L on M ( E ), so every conjugacy class of subgroup L in the E -parabolic with image¯ L in the Levi is contained in an A subgroup with the above action on M ( E ). Step 4 : Action of L on W and conclusion.We actually work with the subgroup J ∼ = PSL (16) of this A . As a finite group,it is easier to construct the largest modules with certain composition factors, andit contains an element of order 93 >
75 (see Theorem 2.4). We find that the copiesof 8 must split off, and the { , ± } -radicals of P (1) and P (9 ) are1 , / , ∗ / , / . In particular, we can obtain useful information about the action of J , and hence L , on M ( E ) from this. The most important of these is that there can be no 4 i inthe fifth socle layer of W ↓ L .We now consider a pyx for W , namely(9 , ∗ , ∗ / , , / ⊕ (9 ± , ∓ , ∓ / , , / ± ) . We restrict the module 9 / , , / L , obtaining the module1 ⊕ ⊕ (4 / / / / / / ) . Clearly this contradicts the structure above, and so this 9 cannot lie in the thirdlayer of the pyx. We then restrict 9 ∗ , ∗ / , , / L , obtaining the module8 ⊕ (1 / ) ⊕ (4 / / / ⊕ (4 / / / / ) . Again, the 4-dimensional factor in the fifth layer means that we should remove oneof the copies of 9 ∗ from the top of the module. Because it lies in the highest soclelayer, there is a unique submodule of 9 ∗ , ∗ / , , / L , has no 4 in the fifth layer.We now develop a contradiction, proving that H stabilizes a line on M ( E ): if L = L then we obtain a different submodule of codimension 9 from the case when L = L . But the intersection of these two submodules is simply 1 , , /
9, so wecannot place a copy of 9 ∗ on top of this module such that it remains consistentwith the known action of L on W .This completes the proof. q = 8 : There are 49 sets of composition factors for M ( E ) ↓ H that are conspicuousfor elements of order at most 21. Let x be an element of order 63 in H . In eachcase, we check that for every conjugacy class of elements of order 21 with the sameeigenvalues as x on M ( E ) (there are sometimes one and sometimes two), and forwhich there is an element of order 63 that powers to that class and has the sameeigenvalues on M ( E ) as x , there exists an element of order 315 powering to theelement of order 63 and having the same number of distinct eigenvalues on M ( E ) . SUBGROUPS OF E as x . Since 315 >
75, this shows that H is a blueprint for M ( E ) in every caseusing Theorem 2.4. (cid:3) Proposition . Let H ∼ = PSU ( q ) for some ≤ q ≤ . (1) If q = 3 then H is strongly imprimitive. (2) If q = 4 , then H stabilizes a line on M ( E ) . (3) If q = 5 then H either is a blueprint for M ( E ) or stabilizes a line on M ( E ) , or H stabilizes a unique -space of M ( E ) , and its stabilizer isa maximal A subgroup of G . (4) If q = 7 then H is a blueprint for M ( E ) . (5) If q = 9 then H stabilizes a line on either M ( E ) or L ( E ) . Proof. q = 7 : There are eleven conspicuous sets of composition factors for M ( E ): 8 , (6 , ∗ ) , (3 , ∗ ) , , , , , ∗ , (6 , ∗ ) , , , , , ∗ , (3 , ∗ ) , , , (3 , ∗ ) , , (3 , ∗ ) , , , ∗ , (10 , ∗ ) , , , ∗ , (6 , ∗ ) , , ∗ , , ∗ , , ∗ By considering Ext , there are no extensions between any two modules in eachdecomposition, so M ( E ) ↓ H is always semisimple.This means, as with PSL (7), we consider the element u . Only 28 and 28 ∗ inthe composition factors above have a block of size 7 in the action of u , and then weget 7 , , , , , , , as in PSL (7). As we need blocks of size 7 for the actionof u to be non-generic (see [ Law95 , Table 7]), this means u always lies in a genericconjugacy class. Hence H is a blueprint for M ( E ) by Lemma 2.2, as needed. q = 5 : There are three conspicuous sets of composition factors for M ( E ) ↓ H :8 , , (10 , ∗ ) , , , , . The first case is semisimple, and u acts on M ( E ) with blocks 3 , , , which isthe generic class (3 A ) ′ by [ Law95 , Table 7]. Hence H is a blueprint for M ( E )by Lemma 2.2.In the second case there is an extension between 10 and 10 ∗ , but the 8s mustbreak off, and hence u acts with blocks 3 , , plus some other blocks. However,from [ Law95 , Table 7] we see that there is no non-generic class of elements of order5 that is compatible with this, so u is generic, and again H is a blueprint for M ( E ).In the third case, suppose that H does not stabilize a line on M ( E ): thereis no module of the form 19 / /
19 or 19 / , /
19, and so if there is a copy of 8 insoc( M ( E ) ↓ H ) then it is a summand, and there is at most one of these. Thus theaction of H on M ( E ) has one of the structures(19 / , , / ⊕ , / , , , / . Write L , L and L for representatives of the three H -classes of subgroupsPGL (5), all Aut( H )-conjugate, and let L denote one of them. Let v be a non-trivial unipotent element from L . The simple kL -modules come in pairs, two ofeach dimension, and we write 1 + , 1 − , 3 + , 3 − , 5 + and 5 − for them. The non-simpleprojectives are P (1 + ) = 1 + / − / + , P (1 − ) = 1 − / + / − ,P (3 + ) = 3 + / − , − / + , P (3 − ) = 3 − / + , + / − .
42 9. SUBGROUPS OF E The restriction of 8 to L is 5 + ⊕ + , and the restriction of 19 to L is 5 + ⊕ − ⊕ P (1 + ) ⊕ − ⊕ + . Furthermore, the restriction of 1 , /
19 to L is5 + ⊕ − ⊕ P (1 + ) ⊕ (1 + / − ) ⊕ + ⊕ − . We see that v acts on M ( E ) with at least eight blocks of size 5, whence from[ Law95 , Table 7] v must act on M ( E ) with blocks one of5 , , , , , , . Combining this with the composition factors of M ( E ) ↓ L yields the possible actionsof L ′ = PSL (5) on M ( E ), which are(9.1) 5 ⊕ ⊕ P (1) ⊕ ⊕ (3 / ⊕ ⊕ ⊕ , (1 , / ⊕ (3 / , ⊕ ⊕ (1 , / , ⊕ or P (1) ⊕ (1 , / , ⊕ P (1) ⊕ ⊕ ⊕ . There are multiple possible ways to attach + and − signs to the modules to obtainfrom this the action of L .Since 1 + / − is a submodule of M ( E ) ↓ H , the element v cannot act as in thefirst case of (9.1). In the second case of (9.1), the presence of a submodule 1 + / − and self-duality forces the signs, and in the third case of (9.1) the presence of 1 + / − and self-duality, together with the fact that (1 + ) ⊕ is a subquotient shows that themiddle and last case cannot occur. Thus the non-projective part of the action of L on M ( E ) is(9.2) (1 + , + / − ) ⊕ (3 − / + , + ) ⊕ (1 − ) ⊕ or (1 + , + / − , − ) ⊕ (1 − , − / + , + ) . Note that 8 cannot be a summand in either case, since there is no summand 3 + in either expression. Thus the structure of M ( E ) ↓ H is definitely 19 / , , , / , , /
19, this is not thecase, and so there is a unique submodule 8 , /
19 for which the 1 − is a summand,and hence the submodule V = 1 , , , /
19 is also unique up to isomorphism.However, replacing L by a different one of the L i yields a different module V . We obtain three different modules V , V and V , and the intersection of the V i is simply 1 , /
19. However, we know that we need a copy of 8 in the heart of M ( E ) ↓ H , and this yields a contradiction.In the second case from (9.2), we see that one of the two copies of 3 + fromthe two 8s must become a submodule of M ( E ) ↓ L . We again consider the module8 , , /
19, which restricts to L as(5 + ) ⊕ ⊕ − ⊕ P (1 + ) ⊕ (3 + / − ) ⊕ (3 + / − ) ⊕ + . There is therefore a unique module 8 /
19 such that the 3 + in the restriction to L splits off.As with the previous case, there is a different extension for each L i , and thesum of these extensions is a module 8 , /
19. So we obtain a unique submodule V = 1 , , , /
19. For this module, Ext kH (19 , V ) is 1-dimensional, and it producesa unique indecomposable module 19 / , , , /
19, which must be M ( E ) ↓ H .We now switch to M = SL (5), and note that M ( E ) ↓ M is5 ⊕ ⊕ P (4) ⊕ ⊕ P (3) ⊕ ⊕ ⊕ ⊕ ⊕ . . SUBGROUPS OF E Since M stabilizes a line on M ( E ), by Lemma 3.15 it lies in one of the five linestabilizers, but it clearly cannot lie in q .F ( q ) . ( q −
1) as that has four trivialquotients and submodules. If M lies in q .B ( q ) . ( q − M ( B ) we see that the composition factors of M ( B ) ↓ M are 4 , , , M ( B ) is self-dual, we see that M has a trivial summand on M ( B ), andagain M would have four trivial quotients and submodules on M ( E ).We claim that M lies in an E -Levi subgroup of G , but certainly we have atleast shown that M lies in an E -parabolic subgroup. Notice that the projection of M onto the E -Levi subgroup cannot have a trivial summand on M ( E ). However,this projection of M lies in the centralizer of an involution z , an A A subgroup.The 1-eigenspace of z is Λ ( M ( A )), and M cannot have a trivial summand onthis space. The action of M on M ( A ) is one of 5 ⊕ ⊕ or 3 /
3, with theremaining possibilities having three trivial composition factors and therefore notbeing suitable. The first two have Λ ( M ( A )) ↓ M being projective and the thirdhas a trivial summand in the restriction. Thus M ( E ) ↓ M has projective actionon the 1-eigenspace of z . In particular, H ( M, M ( E )) = 0, so M lies in a Levisubgroup, and we have that M ( E ) ↓ M is5 ⊕ P (4) ⊕ P (3) ⊕ . This centralizes an involution with centralizer A A , which acts on M ( E ) as( L ( λ ) , L (1)) ⊕ ( L ( λ ) , L (0)). Thus L ( λ ) restricts to M as P (3) ⊕
5, and the onlypossibility for M embedding in A with this action is as 5 ⊕
1. We may embed M into a diagonal A subgroup X , acting as L (1) on the A factor and as L (4) ⊕ L (0)on the A factor. This acts on M ( E ) as L (1) ⊗ ( L (0) ⊕ L (4)) ⊕ Λ ( L (0) ⊕ L (4))= L (1) ⊕ ( L (3) /L (5) /L (3)) ⊕ L (4) ⊕ ( L (2) /L (6) /L (2)) . The embedding of X into E now stabilizes the subspace of M ( E ) ↓ M given by4 ⊕ ⊕ ⊕ , and also every diagonal subspace of 5 ⊕ . In particular, X stabilizes the19 of M ( E ) ↓ H , which restricts to M as 5 ⊕ ⊕ ⊕ ⊕ . Therefore h X , H i contains H and stabilizes a 19-space, so is positive dimensional and H is not Lie primitive.In fact, H is strongly imprimitive by Theorem 3.2.We now examine the subgroups in X to check the final claim of the proposition:if H lies in a maximal parabolic subgroup X , then the dimensions of the factors showthat X is not the D -, A - or A A -parabolic, and since it does not stabilize a lineon M ( E ) it cannot be the E -parabolic subgroup. All other maximal parabolics,upon removing an A factor, lie in one of these four, so H is not contained ina parabolic subgroup. If H is contained in a reductive subgroup X then since M ( E ) ↓ H is indecomposable, X cannot be D A , A , G C , A G or A F , and ofcourse H A A . This leaves A , where the factors do have dimensions 19 , , ,and H is the fixed points under the Frobenius map of the maximal A in G (see[ LS04a , Table 10.2] for the factors). q = 3 : There are eleven conspicuous sets of composition factors for M ( E ) ↓ H ,which are: 27 , , (3 , ∗ ) , , (6 , ∗ ) , , , , (6 , ∗ ) , (3 , ∗ ) , , , , , ∗ , (6 , ∗ ) , , , ∗ , (3 , ∗ ) , , , (3 , ∗ ) , , , (3 , ∗ ) , , , (6 , ∗ ) , , ∗ , , ∗ , , (3 , ∗ ) .
44 9. SUBGROUPS OF E The simple modules with non-zero 1-cohomology are 6 , ∗ , , , ∗ as with PSL (3), so the pressures are generally higher than before.The pressures of the first eight sets of composition factors are − −
20, 4, 0, − − Cases 1, 2, 4, 5, 7 and 8 : Proposition 2.5 shows that H always stabilizes a lineon M ( E ) in these cases. Case 3 : If the composition factors of M ( E ) ↓ H are (6 , ∗ ) , , , then the actionon L ( E ) is uniquely determined, with composition factors27 , , ∗ , , ∗ , , (3 , ∗ ) , ;as there are five composition factors with non-zero 1-cohomology and four trivialfactors, the trivials must be contained in a module of the form 7 / / / / / / ∗ / / ∗ / / / / / / / /
6, with the 15 , ∗ and four 3s connected in some way. Inparticular, such a module must exist as a submodule of P (7) or P (6), as any othersubmodules can be quotiented out and the structure not change. However, P (7)and P (6) have exactly nine socle layers, as this module does, and so this is P (7)—which has dimension 162, so not possible—or P (6), which is not self-dual. Hence H stabilizes a line on L ( E ), as claimed. Thus H is strongly imprimitive by Lemma3.5. Case 9 : For 7 , (3 , ∗ ) , , the corresponding action on L ( E ) is determineduniquely, and it has factors 27 , , , ∗ , . The 27s split off as projective sum-mands, and the { , ± , } -radical of P (7) is3 , ∗ / , / , , ∗ / . Since we have four trivials, either H stabilizes a line on L ( E ) or we need four 7sbelow them, and four 7s above them, but this is impossible as we only have six 7s.So L ( E ) ↓ H has a trivial submodule, and thus H is strongly imprimitive by Lemma3.5. Case 10 : For 7 , (6 , ∗ ) , , ∗ , we again switch to the Lie algebra, where theunique corresponding set of composition factors is 27 , (15 , ∗ ) , , , thus H clearlystabilizes a line on L ( E ). Thus H is strongly imprimitive by Lemma 3.5.The remaining cases are 15 , ∗ , (6 , ∗ ) , and 15 , ∗ , , (3 , ∗ ) . These aremuch more difficult than the other cases: let L and M be copies of GU (3) andPSL (2) respectively, containing elements u and v of order 3 respectively. Case 6 : We deal with 15 , ∗ , (6 , ∗ ) , first. The { , ± , ± } -radicals of P (6)and P (15) are 6 / , ∗ / ∗ /
15 respectively, hence if H does not stabilize aline on M ( E ) then M ( E ) ↓ H must be(6 / , ∗ / ⊕ (6 ∗ / , / ∗ ) . Note that both u and v act on this module with blocks 3 , , hence lie eitherin class 2 A or class 2 A + A , by [ Law95 , Table 7]. The corresponding set ofcomposition factors for L ( E ) ↓ H is27 , , ∗ , , (6 , ∗ ) , (3 , ∗ ) , . Assume that H does not stabilize a line on L ( E ) either. . SUBGROUPS OF E There are eight non-projective irreducible modules for L , four of dimension 1and four of dimension 2. Their projective modules are1 / / , / / , / ∗ / , ∗ / / ∗ , / / , / / , ∗ / ∗ / ∗ , ∗ / ∗ / ∗ . The non-projective parts of the restrictions of the irreducible modules for H to L are as follows: Module Restriction1 1 ⊕ ∗ ∗ ⊕ ∗ ⊕ ∗ ⊕ ∗ ⊕ ∗
15 1 ⊕ ∗ ∗ ∗ ⊕ From this we obtain that the non-projective parts of the composition factors of L ( E ) ↓ H , restricted to L , are1 , , (1 , ∗ ) , (2 , ∗ ) , (2 , ∗ ) . The two possible actions of u on L ( E ) are with blocks 3 , and 3 , . In thesecond case, the 1-dimensional factors of this restriction must assemble themselvesinto the module P (1 ) ⊕ ⊕ P (1 ) ⊕ P (1 ) ⊕ P (1 ∗ ) . In the first case, one of P (1 ) and P (1 ) must become semisimple, and the restof the module must stay the same. For one of the two (dual) blocks consisting of2-dimensional modules, in the first case the structure must be P (2 ) ⊕ ⊕ (2 / ) ⊕ , or with 2 / instead of 2 / for the last two modules. In particular, P (2 ) (and P (2 ∗ )) are not submodules of L ( E ) ↓ L .We will use this information in the rest of the proof. Step 1 : v belongs to class 2 A + A .Suppose v lies in class 2 A . The action of v on L ( E ) then has blocks 2 , by[ Law95 , Table 8], and the composition factors of L ( E ) ↓ H are27 , , ∗ , , (6 , ∗ ) , (3 , ∗ ) , . (From now on we ignore the 27, since we know that splits off as a projective sum-mand.) The modules 3 ± , 6 ± and 15 ± are all projective for M , so we focus on 7 , .The projective cover P (7) for M is 7 / /
7, and the projective cover P (1) for M is1 / /
1, so we see that the 7 , summand of L ( E ) ↓ M must arrange itself into oneof two modules: P (1) ⊕ ⊕ ⊕ ⊕ or P (7) ⊕ ⊕ ⊕ ⊕ . Either way, L ( E ) M has dimension 4.However, in the proof of Proposition 8.2, we proved that dim( L ( E ) M ) was atmost the multiplicity of 6 as a composition factor in L ( E ), which is in this case 2.This is a contradiction, and hence v cannot lie in class 2 A .
46 9. SUBGROUPS OF E Let U denote the quotient 1 / , ∗ / / / / . of the permutation module P M on the cosets of M . Step 2 : U is not a submodule of L ( E ) ↓ H .Suppose that U is a submodule of L ( E ) ↓ H , and hence 6 ∗ is a quotient of L ( E ) ↓ H .Removing this 6 ∗ from the top does not affect U . Suppose first that u belongs toclass 2 A + A . The contribution to the principal block of U ↓ L is P (1 ) ⊕ P (1 ), andif u lies in class 2 A + A then the principal block part of L ( E ) ↓ L is P (1 ) ⊕ ⊕ P (1 ).Of course there is a single 1 quotient of this, which appears inside U and alsooutside the submodule with quotient 6 ∗ , which is a contradiction.If u lies in class 2 A however, then since P (1 ) is a submodule of U ↓ L we seethat the principal block part of L ( E ) ↓ L must be P (1 ) ⊕ P (1 ) ⊕ ⊕ ⊕ . Removing the 6 ∗ from the top of L ( E ) ↓ H to yield a module W , we see that W ↓ L has principal block part P (1 ) ⊕ P (1 ) ⊕ ⊕ . Removing the 6 from the socle of W yields a summand 1 / , but the quotient W/ u cannotcome from any class, we must have that U L ( E ) ↓ H . Step 3 : u belongs to class 2 A and (1 / , ∗ / / ⊕ (1 /
6) is a submodule of L ( E ) ↓ H .If u belongs to class 2 A + A , then there is a single copy of 1 in the socle of L ( E ) ↓ L , as we saw in the previous step. Hence we cannot have both 6 and 6 ∗ , ortwo copies of 6, in the socle of L ( E ) ↓ H . We know that L ( E ) M has dimension 2,hence Hom kH ( U, L ( E )) has dimension 2.The only quotients of U that can occur as submodules of L ( E ) ↓ H are 1 / ± and 1 / , ∗ / /
7, so we must have the sum of two of these as a submodule. Therestriction of this to L has two copies of 1 in the socle, contradicting our choice ofthe class of u .If (1 / ⊕ (1 / ± ) is a submodule of L ( E ) ↓ H though, L acts with a submodule(1 / ) ⊕ , which is impossible according to the structure given above. Thus wehave, up to graph automorphism, that (1 / , ∗ / / ⊕ (1 /
6) is a submodule of L ( E ) ↓ H . The restriction of this to L has principal block part(1 / / ) ⊕ (1 / ) ⊕ , so we see that the principal block part of L ( E ) ↓ L must be P (1 ) ⊕ P (1 ) ⊕ ⊕ ⊕ . Step 4 : The permutation module P L on the cosets of L .We have already proved that (1 / , ∗ / / ⊕ (1 /
6) (up to graph automorphism) isa submodule of L ( E ) ↓ H , and we know that L ( E ) L is 3-dimensional. (Recall thatwe are ignoring the 27 summand.) The module P L has structure1 / / ((1 / , ∗ / ⊕ ⊕ ∗ ) / / . SUBGROUPS OF E plus a copy of 27, which we again ignore. Since we assume that H stabilizes nolines on L ( E ), we obtain a 3-dimensional Hom-space from X to L ( E ) ↓ H , where X is 1 / / ((1 / , ∗ / ⊕ ⊕ ∗ ) / . Suppose that X is a submodule of L ( E ) ↓ H . The principal block part of X ↓ L is P (1 ) ⊕ ⊕ , so we need another quotient of X to be a submodule of L ( E ) ↓ H .Since (up to graph automorphism of H ) 6 is a submodule and 6 ∗ a quotient, wealready know where all of the 6-dimensional modules are, so any 6 ± in this othersubmodule must be in the socle or top. This submodule cannot contain either 3or 3 ∗ , or 1, or 6 ∗ either, for the reasons that we already have 1 ⊕ ∗ in the socleof L ( E ) ↓ L , H cannot stabilize a line on L ( E ), and we have accounted for bothcopies of 6 ∗ respectively. Thus we have a quotient of 1 / / /
6, and therefore either (cid:0) / / ((1 / , ∗ / ⊕ ⊕ ∗ ) / (cid:1) ⊕ (cid:0) / / / (cid:1) or (cid:0) / / ((1 / , ∗ / ⊕ ⊕ ∗ ) / (cid:1) ⊕ (cid:0) / (cid:1) ⊕ (cid:0) / (cid:1) is a submodule of L ( E ) ↓ H .In both cases, we are missing 15 , ∗ , , ∗ , , ∗ . In order for the trivial moduleat the top of X to stay in L ( E ), it must be covered by a 7 or 6 ∗ . There is noextension with submodule X and quotient 7 or 6 ∗ that does this, so we must addmore of the missing modules on top of X . We add all copies of 15, 15 ∗ and 7 thatwe can, then 3 and 3 ∗ , and finally 6 ∗ and 7, but this produces a module1 / , / , , , ∗ , ∗ / , ∗ , , , , ∗ / , , ∗ / , which clearly still has a trivial quotient. Thus X cannot be a submodule of L ( E ) ↓ H .Hence the possible images of a map from X to L ( E ) ↓ H are1 / , / / / , / / ± , / / / , ± , / / , ∗ , / / / , , ∗ . Ignoring any 3 ± in the socle, we still need a sum of three of the modules 1 / / / /
7. This means we certainly need another copy of 1 /
7, possibly with 3 ± or3 , ∗ underneath it, and another copy of 1 /
7, either on top of the 1 / ⊕ ∗ ⊕ ⊕ ⊕ , which has pressure 5, so H stabilizes a line on L ( E ) ↓ H by Proposition 2.5.Thus we have a submodule(9.3) (1 / , ∗ / / ⊕ (1 / / / ⊕ (1 / , possibly with some copies of 3 ± under the second and third summands.Let Y denote the { ± , ± } -heart of L ( E ) ↓ H , which is of course a self-dualmodule. Step 5 : Y contains a single trivial composition factor, and soc( L ( E ) ↓ H ) containsexactly one 3-dimensional submodule.From the composition factors of L ( E ) ↓ H , we see that there are at most two com-position factors of dimension 3 in the socle. (None can be summands as there is nosummand 1 ± in the action of L .)
48 9. SUBGROUPS OF E Suppose first that there are no 3-dimensional factors in the socle. In thiscase the whole of the module from (9.3) appears as a submodule of L ( E ) ↓ H , andalso 6 ∗ ⊕ ⊕ in the top. This means that Y can only have composition factors15 , ∗ , (3 , ∗ ) .Hence we need to understand the { ± , ± } -radicals of P (3) and P (15). Theseare 3 / ∗ , ∗ / ∗ / . The former of these is self-dual, so one cannot make a module with three socle layers,six composition factors, and two factors in the third layer, which is necessary tohave the reduction to L containing P (1 ) ⊕ P (1 ∗ ). This means that we must bewrong, and there is a 3-dimensional composition factor in the socle.Suppose that there is more than one, and therefore that all 3-dimensional fac-tors lie in the socle or the top. Then one must be able to put possibly 15 , ∗ , andthen two 7s on top of the module in (9.3). However, on top of this module one mayonly place 15 ∗ , and then one 7. This means that one cannot cover two of the threetrivial quotients, and this yields another contradiction.Thus there must be exactly one 3-dimensional factor in the socle, as claimed.Since one 3 ± lies below either the second or third summand in (9.3) we see that theremust be a trivial factor in Y , and there are four trivial factors in the submodulefrom (9.3) not lying below any 3 ± , so this completes the proof. Step 6 : Contradiction.We will obtain a contradiction with the action of M .First, consider the { } -heart of Y . Since Y contains factors 7 , / / / /
7, so there are other factors in the { } -heart. Takethe { , ± , ± } -radical of P (7), place as many copies of 7 on top as possible, thentake the { } ′ -residual, and this must be a pyx for the { } -heart of Y . This pyx is7 , / , , ∗ / , so the { } -heart is 7 / , , ∗ /
7, but this module is not uniquely determined by thisstructure. However, we see that the { } -heart of L ( E ) ↓ H —note, not the { } -heartof Y —must have a subquotient 6 ⊕ ∗ ⊕ ⊕ ∗ . Since 15 only has extensions with3 ∗ and 6 ∗ , the upshot of this is that if one quotients out by the { ± , ± } -radical of L ( E ) ↓ H , there must be some 15 ± in the socle.We now consider the action of M on L ( E ), aiming to see that it cannot be thesum of a projective module and (7 / ⊕ (1 / kH -submodulewhose restriction to M is projective, and similarly to kH -quotients whose restrictionto M is projective.Since 3 ± , 6 ± and 15 ± all restrict projectively to M , we may remove all of thesefrom the top and socle, so take the { , } -heart of L ( E ) ↓ H . This is a module withcomposition factors 7 , , ∗ , , ∗ , , and we can see most of its structure in (9.3). There is a submodule 1 / /
1, on which M acts projectively and so may be removed as well. We are left with a modulewith a submodule (1 / , ∗ / / ⊕ (1 / , . SUBGROUPS OF E on which we have to place 3 ⊕ ∗ first, and then two copies of 7.Notice that the { } -heart of Y is now a quotient of the remaining moduleabove, and the restriction to M of it is again projective. Thus we may removeit, and are left with an extension with quotient 7 and submodule (1 / , ∗ / / M must be (1 / ⊕ (7 /
1) plus projective. However, the uniqueextension, 7 / / , ∗ / /
7, has restriction to M given by (7 / / ⊕ / ⊕ (7 / H stabilizes a line on L ( E ). Case 11 : Here H acts on M ( E ) with composition factors 15 , ∗ , , (3 , ∗ ) —which resemble those of the module Y in the previous case—and on L ( E ) withcomposition factors 27 , (15 , ∗ ) , , (6 , ∗ ) , , ∗ , . The element z of order 4 in Z ( L ) has trace 0 on M ( E ) and − L ( E ). Consulting [ Fre01 , Table II], forexample, we see that z has centralizer A A A , inside the centralizer A D of z .Let X be the copy of A D centralizing the involution in L ′ , and Y the copy of A A A centralizing z in X . The summand of M ( E ) ↓ L consisting of 2-dimensionalfactors is the half-spin module for D , and the summand consisting of 1- and 3-dimensional factors is the tensor product of the two minimal modules for D and A . Any element of A centralizes the half-spin module. Note that the 3-pressureof M ( E ) ↓ H is 0, and so either 3 or 3 ∗ is a submodule of M ( E ) ↓ H , say 3 up toduality. Let W denote this 3-dimensional kH -submodule of M ( E ). The action of L on W is 1 ⊕ , so we aim to find an infinite subgroup of G stabilizing W . Wedo so by finding an element of X simultaneously stabilizing any given submodule1 of the product of minimal modules, and any given submodule 2 of the half-spinmodule.Of course, not all of L can embed in A , and so the projection of L ′ onto A cannot be faithful. Thus the composition factors of L ′ on M ( A ) are 1 , and L ′ acts as 1 ⊕ (if u ∈ D ) or 1 / u D ). In particular, this means that L ′ actson M ( D ) with composition factors 3 , . We therefore see that u acts on M ( D )with blocks either 3 , or 3 (there is no class in D acting with blocks 3 , ,
1, aswe see from [
Law09 , Table 7]). Therefore u lies in class either A + D or 2 A of D (from the same table). Of course, either u lies in the D or acts diagonally, andthis yields four options for the unipotent class of E containing u , using [ Law09 ,4.10]: if u ∈ D then we obtain A + 2 A or 2 A , and if u D then we obtain A + 3 A or 2 A + A . The actions of u on M ( E ) are as follows, via [ Law95 ,Table 7]. Class Blocks on M ( E ) A + 2 A , , A + 3 A , A and 2 A + A , The action of Y on M ( D ) is the sum of the two copies of Λ ( M ( A )) (i.e., M ( D ))for the two A factors. The half-spin module for D restricts to Y as the sum oftwo dual modules, each a tensor product of (up to duality) the minimal modulesfor the two A factors. In order for this to contain only 2-dimensional factors, L ′ must act on the one copy of M ( A ) with factors 3 , ,
2. Thus L ′ acts on the one as 3 ⊕ ⊕ or 2 / u projects onto the class A + D or 2 A . Notice that L is containedin N X ( Y ), and in fact L must lie inside the subgroup of index 2 that stabilizesthe two A factors (as there is an involution that swaps the two factors). This
50 9. SUBGROUPS OF E subgroup still induces a simultaneous graph automorphism on the two A factors(since GO × GO ≤ GO ). Step 1 : Eliminating classes A + 2 A and A + 3 A for u .In both of these cases the projection of u onto D lies in class A + D , and fromthe action described above, we see that L ′ must act on the two copies of M ( A ) as1 ⊕ ⊕
2. Embed L ′ in a copy L of A acting on the two factors in the obviousway, as L (0) ⊕ L (2) and L (1) ⊕ L (1) respectively. The exterior square of the secondone (a summand of M ( D ) ↓ L ) is L (2) ⊕ L (0) ⊕ , so L centralizes the fixed-pointspace M ( E ) L ′ . On the other hand, the tensor product of the two modules is L (1) ⊕ ⊕ ( L (1) /L (3) /L (1)) ⊕ , and we see that L again stabilizes every 2-space stabilized by L ′ . Thus L definitelystabilizes all submodules 3 and 3 ∗ , and H contained in a member of X , in fact isstrongly imprimitive by Theorem 3.2. Step 2 : Eliminating classes 2 A and 2 A + A for u .In these cases u acts with Jordan blocks 3 , on M ( E ), and this determines theaction of L ′ as well. It almost completely determines the action of L , and does soif we assume that 3 is a submodule of M ( E ) ↓ H . The action must be P (1 ) ⊕ P (1 ∗ ) ⊕ P (2 ) ⊕ P (2 ∗ ) ⊕ P (2 ) ⊕ P (2 ∗ ) ⊕ (2 / ) ⊕ (2 ∗ / ∗ ) ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ∗ . Since 15 and 15 ∗ contribute P (2 ) and P (2 ∗ ) respectively to this action, and3 ± restricts to L with a submodule 2 ± , there can be at most one submodule 3 andno submodules 3 ∗ of M ( E ) ↓ H . Thus, via Theorem 3.2, if the stabilizer of W ispositive dimensional, H is strongly imprimitive.Let M denote the stabilizer in G of the 1-space 1 (which lies in W ), and notethat L ≤ M . From Lemma 3.15 we see that M is an E T -parabolic subgroup, an E -Levi subgroup extended by the graph automorphism, a subgroup U · F T ora subgroup U · B T .Suppose that M is an E T -parabolic subgroup. In this case, the stabilizerhas dimension 106. Then W is contained in the M -invariant hyperplane of M ( E ),which has dimension 55 (as all copies of 2 ± i are contained in this hyperplane).To compute the dimension of the stabilizer we choose two points at random inthe 55-space, and see if they lie in the 3-space W . The dimension of the variety of3-spaces containing the specified line is 55 + 55 − − W is always positive dimensional and H is strongly imprimitive.Since there is no M -invariant complement to the M ◦ -fixed points of M ( E ), M cannot be an E -Levi subgroup (extended by the graph automorphism).Assume that M = U · F T . This group acts on M ( E ) with structure1 ∗ a , ∗ b / ∗ / / a , b (the duality is for the T part). The composition factors of L ′ on the 26-dimensional module must be 3 , , M ( E ) the factors are3 , , ), and therefore L ′ acts as a sum of 1 ⊕ ⊕ and another module withfactors of dimension 2. In particular, the action of L ′ on M ( F ) (which is theaction on U ) has 1-dimensional 1-cohomology. This is acted upon by the T factor with one regular orbit and one fixed point (see [ Ste13 , Lemma 3.2.15], or itis a simple calculation in our case). The fixed point is clearly the copy of L ′ in the F subgroup, and this has complemented 1-spaces, like the E case for M above.Thus L ′ is (up to conjugacy) the other complement. But this cannot be normalized . SUBGROUPS OF E by the torus, so L ≤ U F . However, this group centralizes a 2-space on M ( E )and L does not, a contradiction.The final case is to suppose that M is a subgroup U · B T . The module M ( E ) ↓ M has structure 1 ∗ , ∗ / / ,
11, so the 32-dimensional factor must be self-dual. This means that the toral summand T of the B T subgroup of M actstrivially on it. Since T is the toral summand from the D T -parabolic containing M , it acts as the same scalar on 1 and 11. In particular, since U · B is theline centralizer, the B T subgroup is a direct product of the two factors.If the central involution in L has trace 11 on M ( B ) then all 2-dimensionalcomposition factors lie in the submodule 32 / ,
11 of M ( E ) ↓ M , which has dimension44. As the dimension of M is 89, we see that all such 2-spaces have a positive-dimensional stabilizer in M —in particular, the 2 inside W —and thus the W haspositive-dimensional stabilizer. Hence H is strongly imprimitive.Thus we may assume that the trace must be − B have trace 8 on M ( E )). This means that the composition factors of M ( B ) ↓ L ′ are 3 , . In particular L ′ , and hence L (as all normal subgroups of L contain the central involution) acts faithfully on the 32-dimensional factor, and thus L T / T ∼ = L . Thus the projection of L onto the B factor is L itself, and the imageof L in B T is a subdirect product of GU (3) and a group of order 4. Thereforethere is a copy of L inside the B subgroup acting on M ( B ) with compositionfactors of degrees 3 , . But a non-central involution in L obviously has trace ± B on M ( B ) is one of 11, − q = 9 : There are 34 sets of composition factors for M ( E ) ↓ H that are conspicuousfor elements of order up to order 20, so we eliminate some first. Only 7 i and 21 ± i,j have non-zero 1-cohomology, so we first eliminate all sets of composition factorswith non-positive pressure (and trivial factors) using Proposition 2.5. Up to fieldand graph automorphisms, this leaves the following sets of composition factors thatare conspicuous for elements of order at most 20:7 , (3 , ∗ ) , , , ∗ , , , ∗ , , ∗ , , (6 , ∗ ) , , ∗ , , ∗ , , , ∗ , ¯18 , ¯18 ∗ , , , ∗ . For each of these, we switch to the Lie algebra:27 , , , ∗ , , , (15 , ∗ ) , ¯9 , ¯9 ∗ , , , ∗ , , , (15 , ∗ ) , , , , ∗ , , , , , ¯45 , ¯45 ∗ , , , , . All but the first case have non-positive pressure, so H stabilizes a line on L ( E ).The first case has pressure 2. However, the { , ± , , } -radical of P (7 ) is7 , / , , ∗ / , so clearly we must stabilize a line on L ( E ) in this case as well, just as for thecorresponding case for PSU (3).Thus H stabilizes a line on either M ( E ) or L ( E ) in all cases. q = 4 : From Table 5.5, we see that the only modules with non-zero 1-cohomologyare 9-dimensional, and they always have 1-dimensional 1-cohomology, so we needeither no trivial composition factors or more 9-dimensional factors than trivialfactors, else H stabilizes a line on M ( E ) by Proposition 2.5. In fact, there are
52 9. SUBGROUPS OF E twenty conspicuous sets of composition factors for M ( E ) ↓ H , eighteen of whichhave negative or zero pressure (and all have trivial factors), so we only have toconsider two sets of factors, one up to field automorphism. Thus we reduce toconsidering 9 , ∗ , ¯9 , ¯9 ∗ , (3 , ∗ ) , , ∗ , . (Note that the field-graph automorphisms act as a cycle9 → ¯9 ∗ → ∗ → ¯9 with the labelling convention from Chapter 5.)Let W denote the { ± i } ′ -heart of M ( E ) ↓ H . Suppose first that the socle of W is simple, say 9 . The { , ± i , ¯9 ± } -radical of P (9 ) is1 / ∗ , ¯9 ∗ , ¯9 / , ∗ / , ∗ / . We cannot place a copy of 9 ∗ on top of this module and so we cannot build a pyxfor W .Thus we need two 9-dimensional modules in the socle, and it cannot be 9 ⊕ ∗ ,so (up to automorphism) it is 9 ⊕ ¯9 . (To choose this we might need to replacethe composition factors by their image under the field automorphism.) This meansthat we may remove the ¯9 from the above module, and W is uniquely determinedup to isomorphism, as(9.4) (¯9 ∗ / ∗ / , ∗ / ) ⊕ (9 ∗ / / , / ¯9 ) . We cannot place a copy of 3 or 3 ∗ on top of (9.4) but we can place a single copy of3 ∗ , and two copies of 3 . We obtain two modules by taking the direct sum of (9.4)with either 3 ⊕ ∗ or 3 ⊕ ∗ . We obtain one more module by extending above by3 ∗ and below by 3 . Finally, we obtain three more modules by extending above by3 and below by 3 ∗ .Let L ∼ = Alt(5) be a subgroup of H . We can consider the fixed-point space M ( E ) L for the actions above. For three of these it has dimension 4 and for threeit has dimension 5. Those with dimension 5 have a summand of dimension 3.Note that L centralizes an element of order 5 with centralizer A A T [ Fre01 ,Table II]. Thus L ≤ A A , and the composition factors of A A on M ( E ) aregiven in [ Tho16 , Table 21]. The best way to proceed though is to embed A A in A and take the sum of M ( A ) and M ( A ). The structure of M ( E ) ↓ A is easier,and is M ( A ) ⊕ M ( A ) ∗ ⊕ Λ ( M ( A )) ⊕ Λ ( M ( A )) ∗ . The composition factors of L on M ( E ) force the action of L on M ( A ) to be 2 ,and on M ( A ) to be 2 , , ,
1, or the action of L on M ( A ) to be 2 and on M ( A ) to be 2 ,
1. If the action on M ( A ) is 2 then there are six possible modulestructures for L on M ( A ), up to duality:4 ⊕ , ⊕ ⊕ , (2 / ⊕ , (2 / ⊕ , / / , , / . For each of these we can construct the corresponding action of L on M ( E ) andcompute the fixed-point space M ( E ) L : their dimensions are 6, 10, 7, 8, 5 and 6respectively.If the action on M ( A ) is 2 then up to duality there are two module actionsfor L on M ( A ), namely 2 ⊕ ⊕ , (2 / ⊕ , . SUBGROUPS OF E and the dimensions of M ( E ) L in these cases are 10 and 7 respectively. (Theyappeared in the previous list.)Comparing to the dimension of M ( E ) L computed above, we find a unique caseof interest, of action (up to graph automorphism) of 2 ⊕ (2 / / ). However, inthis one case the precise module structures of the two actions don’t match up. Theaction of this copy of L on M ( E ) is1 ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ (2 / / ) ⊕ (2 / / ) ⊕ (1 / , / ⊕ ⊕ (1 / / / ) ⊕ (2 / / / , and this does not have a summand of dimension 3.Thus the module action for H is not correct, and so H stabilizes a line on M ( E ), as needed. q = 8 : Up to field automorphism there are four conspicuous sets of compositionfactors for M ( E ) ↓ H :8 , , (9 , ∗ ) , , , (9 , ∗ ) , , , , ∗ , , ∗ , ¯9 , ¯9 ∗ , . The first three of these are of non-positive pressure, so H stabilizes a line on M ( E )in these cases.In the final case, we first assume that the socle of M ( E ) ↓ H is simple, say 9 .Taking the { , ± , ¯9 ± } -radical of the quotient P (9 ) then adding copies of 9 ∗ ontop, yields a pyx of M ( E ) ↓ H , and this is 56-dimensional, having structure9 ∗ / , ∗ , ¯9 ∗ / , , ¯9 / . Thus this must be exactly M ( E ) ↓ H . Let L denote a subgroup PSL (8) thatcentralizes an element of order 3 in H with trace −
7; then L lies inside the centralizerof this element, which is A T , so L ≤ A , and A acts on M ( E ) as L ( λ ) ⊕ L ( λ ) ⊕ L ( λ ) ⊕ L ( λ ) . The restriction of the pyx above to L is a sum of eight 7-dimensional modules, butfor each of them their exterior square is indecomposable, so this is not compatiblewith the structure above.An element v ∈ H of order 4 acts on 9 ± i,j and ¯9 ± i,j with Jordan blocks 4 , M ( E ) with at least twelve blocks of size 4. Thus we see from [ Law95 ,Table 7] that v acts on M ( E ) with blocks 4 , , and in particular there is nosimple summand of M ( E ) ↓ H . So if a 9-dimensional module lies in the socle of thismodule, then its dual does not.If the socle is 9 ⊕ then the corresponding pyx to that above is(9 ∗ / , ¯9 / ) ⊕ (9 ∗ / , ¯9 ∗ / ) , again it has dimension 56 so must be isomorphic to M ( E ) ↓ H , and the restrictionof it to L has two summands of dimension i for each 1 ≤ i ≤
7. (The modules ofdimensions 1, 2 and 4 are simple, the others are not and are not self-dual.) Wewill try to find which of these summands lie inside L ( λ ): it cannot restrict to L as 1 ⊕ V for some V , for then ( V ⊕ V ∗ ) ⊕ appears in M ( E ) ↓ L , which is notpossible. The exterior square of the 7-dimensional summand is indecomposable, sowe are left with summands of dimensions 5 and 2, and of dimensions 4 and 3. Theexterior square of the 3-dimensional summand is also 3-dimensional, so this is notallowed, and the exterior square of the 5-dimensional summand is a 10-dimensionalindecomposable module, so there is no compatibility. Hence H cannot embed withthis socle.
54 9. SUBGROUPS OF E If the socle is 9 ⊕ ¯9 ∗ , however, we can place 1 ⊕ ⊕ ⊕ on top of this, butnot 9 ∗ and so this cannot be the socle. Up to automorphism these are the onlyoptions with two factors in the socle.If the socle has three 9-dimensional submodules, however, then the structuremust have the form 9 , , / , / , ,
9, and in particular there is a uniserial mod-ule 9 / /
9, which we cannot construct. (For example, it does not appear in thesubmodule of P (9 ) above.)This proves that H must stabilize a line on M ( E ), as needed. (cid:3) Proposition . Let H ∼ = PSp ( q ) for q = 3 , , , , or B ( q ) for q = 8 , ,and let ¯ H ∼ = Sp ( q ) for q = 3 , . (1) If q = 4 , , then H stabilizes a line on M ( E ) . (2) If q = 3 then H stabilizes a line on M ( E ) and ¯ H does not embed in G . (3) If q = 9 then H is a blueprint for M ( E ) , and ¯ H does not embed in G . Proof. q = 3 : If H embeds in G then there are two conspicuous sets of com-position factors for M ( E ) ↓ H , and they are 10 , , and 25 , . The pressuresof these are 0 and − H stabilizes a line on M ( E ) by Proposition2.5. As seen with the case q = 5 , Cra , Proposition 3.8], ¯ H cannot embedin G with the centres coinciding, because all factors of M ( E ) ↓ H are faithful, butthe trace of a non-central involution on faithful modules is 0, and the trace of aninvolution in G is ±
8, a contradiction. q = 9 : we need only consider H ∼ = PSp (9) embedding in G and not the doublecover because Sp (3) does not embed with Z ( ¯ H ) = Z ( G ). Using traces of ele-ments of order up 8, we find up to field automorphism three conspicuous sets ofcomposition factors: 10 , , , , , , , , . For each of these, an element of order 41 in H centralizes at least an 8-space on M ( E ), so by Theorem 2.4, H is a blueprint for M ( E ). q even : Note that in all cases H contains a rational element x of order 5. Thereis a unique rational class of semisimple elements of order 5 in E , with trace 6 on M ( E ). Note that the trace of x on modules of dimension 1, 4 and 16 is 1, − M ( E ) ↓ H (as dim( M ( E )) = 56), there must be at least three more trivial factorsthan 4-dimensional ones. As the pressure is the number of 4-dimensional factorsminus the number of trivial factors (see [ Sin92b ]), the pressure must be negative.Thus H stabilizes a line on M ( E ). (cid:3) Proposition . Let H ∼ = G ( q ) ′ for q = 2 , , , or G ( q ) ′ for q = 3 , . (1) If q = 2 or H ∼ = G (3) ′ then H stabilizes a line on either M ( E ) or L ( E ) ◦ . (2) If q = 3 , , but H is not G (3) ′ , then H is a blueprint for M ( E ) . (3) If q = 4 then H stabilizes a line on either M ( E ) or L ( E ) , or is ablueprint for M ( E ) . Proof. | q : Letting H ∼ = PSL (8) = G (3) ′ , the conspicuous sets of com-position factors for M ( E ) ↓ H are7 , , , , , , , , , , , , , , , , , . . SUBGROUPS OF E As H ( H,
7) is 1-dimensional and 9 i is projective, we see that in all but the firstcase H must stabilize a line on M ( E ) by Proposition 2.5, as the pressures are 8, − − − − − L ( E ) ↓ H are 9 , , , , , and as the only indecomposable module with a trivialcomposition factor and no trivial submodule or quotient is P (7) = 7 / ((7 / / ⊕ / , we see that H must stabilize a line on L ( E ).We must also deal with H ∼ = G (3) and H ∼ = G (27). Both contain L = G (3),which has composition factors on M ( E ) one of27 , , , , . The modules for H of dimension at most 56 have dimensions 1, 7, 27 and 49: thefirst three restrict irreducibly to L , and the fourth restricts as 7 ⊗ ∼ = 27 ⊕ (7 / / ⊕ M ( E ) ↓ H must have the same dimensions as thefactors of M ( E ) ↓ L .The first case is a blueprint for M ( E ) as it lies in an E -Levi subgroup. Thesecond is a blueprint for M ( E ) as it is semisimple, so u has at least fourteen blocksof size 1 in its action on M ( E ), hence by [ Law95 , Table 7] lies in a generic class(using Lemma 2.2).For the third case, any semisimple element in H must centralize an 8-space on M ( E ), since semisimple elements of H are real. As G (27) contains an elementof order 37 >
30, it is a blueprint for M ( E ) by Theorem 2.4. For the other group H ∼ = G (3), first note that, up to graph automorphism, the only conspicuous setof composition factors for M ( E ) ↓ H consisting only of 7-dimensional modules is7 , . Then H contains an element x of order 13, and its trace on M ( E ) showsthat it lies in F . Furthermore, there exist elements of order 26 in F that squareto x and have the same number (thirteen) of distinct eigenvalues on M ( F ). Thus x ∈ F is a blueprint for M ( F ) by Theorem 2.4, and hence also for M ( E ) as well.Thus so is H . This completes the proof. q = 2 : The conspicuous sets of composition factors for M ( E ) ↓ H are6 , , , , , , . The second and third cases have pressures −
14 and 0 respectively, so both stabilizelines on M ( E ) by Proposition 2.5, as needed. For the first case, the correspondingfactors on L ( E ) ◦ are 14 , , . Clearly H has at least twelve trivial summandson L ( E ) ◦ ↓ H , as needed. q = 4 : We see from Table 5.5 that the modules with non-zero 1-cohomology ofdimension at most 56 are 6 , 6 and 36, the last of which restricts to the subgroup G (2) ′ of H with factors 14 , , . Thus if the restriction of M ( E ) to G (2) ′ iseither the first or second case above, then H must stabilize a line on L ( E ) and M ( E ) respectively, by Proposition 2.5. For the third case, it either has pressure 0or has composition factors of dimension 36 , , and has pressure 2.There is a unique (up to field automorphism) such conspicuous set of composi-tion factors for M ( E ) ↓ H , which is 36 , , , . Suppose that H does not stabilizea line on M ( E ), so that M ( E ) ↓ H is a submodule of P (6 ), perhaps plus sum-mands 6 and 36. Since there is no module 36 / / , we must have a submodule1 / / / of M ( E ) ↓ H , which is unique up to isomorphism.
56 9. SUBGROUPS OF E Let L denote a copy of SL (4) in H . It centralizes an element of order 3that acts with trace 2 on M ( E ) and − L ( E ), which means it has centralizer X = A A (see [ Fre01 , Table II]), so L ≤ X . The composition factors of M ( E ) ↓ L are 9 , ∗ , ¯9 , ¯9 ∗ , (3 , ∗ ) , , ∗ , . Since Z ( L ) acts faithfully on both M ( A ) and M ( A ), we see that the restric-tions M ( A ) ↓ L and M ( A ) ↓ L have factors only of dimension 3. If M ( A ) ↓ L = 3 then M ( A ) ↓ L has factors either 3 , or 3 , ∗ , and if M ( A ) ↓ L = 3 then M ( A ) ↓ L has factors 3 , ∗ .If the restriction M ( A ) ↓ L is semisimple then there is an obvious A subgroup Y containing L , and as X acts on M ( E ) as( λ , λ ) ⊕ ( λ , λ ) ∗ ⊕ ( λ , , one can compute immediately the actions of L and Y on these three modules. Byplacing L inside Y , acting on M ( A ) as L (10) and on M ( A ) as either L (10) ⊕ L (20)or L (10) ⊕ L (02), or on M ( A ) as L (20) and on M ( A ) as L (10) ⊕ L (01), we seethat L and Y stabilize the same subspaces of M ( A ) ⊗ M ( A ) and its dual. As forany modules M, M ′ ,Λ ( M ⊕ M ′ ) ∼ = Λ ( M ) ⊕ (Λ ( M ) ⊗ M ′ ) ⊕ ( M ⊗ Λ ( M ′ )) ⊕ Λ ( M ′ ) , we compute that Y acts on Λ ( M ( A )) as L (0) ⊕ ⊕ L (12) ⊕ L (21) , L (0) ⊕ ⊕ L (30) ⊕ L (03) , L (0) ⊕ ⊕ L (10) ⊗ ⊕ L (01) ⊗ . In all cases, L and Y stabilize the same subspaces of M ( E ), so L , and H , areblueprints for M ( E ).Thus we may assume that L does not act semisimply on M ( A ), and hencethe factors of M ( A ) ↓ L are 3 , ∗ , as there are no extensions between the otherfactors. First, note that the module 1 / / / for H restricts to L with two triv-ial submodules, hence M ( E ) ↓ L has two trivial summands, not just compositionfactors. Since dim(Ext kL (3 ∗ , )) = 2, there is a module 3 ∗ , ∗ / , so 3 ∗ / is notdefined uniquely up to isomorphism. Let v denote an element of order 4 in L , whichmust act on M ( E ) with at least two blocks of size 1, as M ( E ) ↓ L has two trivialsummands. Then v acts on 3 with a single block of size 3, hence on 3 ⊕ ∗ withblocks 3 , and on 3 ∗ / with blocks either 3 or 4 ,
2. The exterior cube of theblocks 4 , , , which is incompatible with there being twotrivial summands in M ( E ) ↓ H . Thus v acts on 3 ∗ / with blocks 3 , and there isa unique such indecomposable module. However, if v came from a different classof elements of order 4 then this yields a different extension 3 ∗ / on which v actswith blocks 3 . Thus the only extension on which two different choices of v bothact correctly is the split extension 3 ∗ ⊕ .Thus H is a blueprint for M ( E ) or stabilizes a line on M ( E ), as needed. q = 8 : Since PSL (8) is a blueprint for M ( E ) by Proposition 9.1, and G (8)contains PSL (8), we are done. (cid:3) Proposition . If H ∼ = PSL (3) then H is a blueprint for M ( E ) . If ¯ H ∼ =SL (3) then ¯ H does not embed in G . Proof.
The two conspicuous sets of composition factors for M ( E ) ↓ H are10 , ∗ , , , , . . SUBGROUPS OF E In both cases there are no extensions between the factors so M ( E ) ↓ H is semisimple,and u acts on these with blocks 3 , , . This is the generic class 2 A , and so H is a blueprint for M ( E ). Since there is no embedding of Sp (3) into G withcentres coinciding, the same holds for SL (3). (cid:3) For H ∼ = PSU ( q ), we know we need to consider q = 2 , , Proposition . Let H ∼ = PSU ( q ) for q = 2 , , , and let ¯ H ∼ = 2 · PSU (3) . (1) If q = 2 , then H stabilizes a line on either M ( E ) or L ( E ) ◦ . (2) If q = 3 then H does not embed in G and ¯ H is a blueprint for M ( E ) . Proof. q = 3 : As H does not embed in E by Proposition 7.2, H does notembed in E either. If ¯ H = 2 · H embeds in G with Z ( ¯ H ) = Z ( G ), then the onlyconspicuous set of composition factors for M ( E ) ↓ H is 10 , ∗ , , and the 6s mustsplit off as summands. This means that u acts with blocks at least 2 , . Hencethis element must come from a generic class, as we see from [ Law95 , Table 7], sothat H is a blueprint for M ( E ) by Lemma 2.2. q = 2 : There are three conspicuous sets of composition factors for M ( E ) ↓ H :6 , , ∗ , , , , , (4 , ∗ ) , . The pressures of the second and third modules are − − H stabilizes lines on M ( E ) by Proposition 2.5. In the first case, weswitch to L ( E ) ◦ , which has composition factors 14 , , , so clearly H stabilizesa line on L ( E ) ◦ . q = 4 : Up to field automorphism, there are four conspicuous sets of compositionfactors for M ( E ) ↓ H :14 , , , , ∗ , , , , , ∗ , , (4 , ∗ ) , . The pressures of the first and last sets of factors are − −
12, so H stabilizeslines in these cases by Proposition 2.5. In the second and third cases we switch to L ( E ) ◦ , where we find composition factors of36 , ( ¯16 , ¯16 ∗ ) , , , , , , respectively. The pressures of these are − − H stabilizes aline on L ( E ) ◦ , as claimed. (cid:3) For H ∼ = PSp ( q ), we know from above that we need only check q = 2. Proposition . If H ∼ = PSp (2) then H stabilizes a line on either M ( E ) or L ( E ) ◦ . Proof.
There are three conspicuous sets of composition factors for M ( E ) ↓ H :8 , , , , , , , . The first two cases have pressures −
10 and 0 respectively, so H stabilizes a line on M ( E ) in both cases by Proposition 2.5. In the third case, we switch to the Liealgebra L ( E ) ◦ , on which H must act with composition factors14 , , . Clearly H stabilizes a line on L ( E ) ◦ as it has pressure − (cid:3) HAPTER 10
Subgroups of E G be a simply connected algebraic group of type E and H bea subgroup of rank at most 3, q ≤
9, together with a small Ree group, a Suzukigroup, PSL (16) or PSU (16). By embedding G inside E , if the Schur multiplierof H is not divisible by 3 and H is a blueprint for M ( E ) or L ( E ), then it is ablueprint for M ( E ) ⊕ M ( E ) ∗ or L ( E ) respectively. We may also use Theorem2.4 to eliminate H with semisimple elements of odd order greater than 75, as withthe previous chapter. We can also eliminate any subgroup containing a real elementof order at least 19. These conditions together eliminate many possibilities for H ,leaving the following:(1) PSL ( q ) for q = 2 , , , ( q ) for q = 3 , , , , ( q ) for q = 3 , G (2) ′ ;(5) G (3) ′ ;(6) B (8);(7) PSU ( q ) for q = 2 , (2).Note that if H is PSL ( q ), PSU ( q ), G ( q ) or G ( q ) for q a power of 3, then H canact irreducibly on M ( E ), but this action is self-dual. This means that H stabilizesdiagonal submodules of M ( E ) ⊕ M ( E ) ∗ that are not stabilized by G . Thus thestatement that H is a blueprint for M ( E ) ⊕ M ( E ) ∗ actually shows that H is notLie primitive. This gives a proof that all of these groups are not Lie primitive,hence strongly imprimitive, except for PSL (3), PSU (3) and G (3) ′ . These arethe three left out of the proof in [ LS04b ], where a different proof of imprimitivityfor the other subgroups is given.In this chapter we will eliminate all of the cases above except for PSL (3),PSU (3) and G (3) ′ , the first two acting irreducibly on M ( E ), and the thirdacting as the sum of three modules that are permuted by G (3), so that thatgroup (if G (3) ′ extends to it) acts irreducibly on M ( E ). These three cases willbe dealt with in Chapter 13.As with the previous chapters, in this chapter u denotes an element of order p in H belonging to the smallest conjugacy class.Our first proposition leaves open the possibility of a Lie primitive copy ofPSL (3) acting irreducibly on M ( E ), as mentioned above. See Section 13.4 forhow to prove strong imprimitivity in this case. Proposition . Let H ∼ = PSL ( q ) for q = 2 , , , , and let ¯ H ∼ = SL ( q ) for q = 4 , . (1) If q = 2 then H stabilizes a line or hyperplane on either M ( E ) or L ( E ) . E (2) If q = 3 then H stabilizes a line on either M ( E ) or L ( E ) ◦ , or is ablueprint for M ( E ) ⊕ M ( E ) ∗ , or acts irreducibly on M ( E ) . (3) If q = 4 then H stabilizes a line or hyperplane on M ( E ) , and ¯ H stabilizesa line on L ( E ) . (4) If q = 7 then H and ¯ H are blueprints for M ( E ) ⊕ M ( E ) ∗ . Proof. q = 7 : From Proposition 9.1 we see that H is a blueprint for M ( E ),and so we only consider ¯ H . There are three sets of composition factors for M ( E ) ↓ ¯ H that are conspicuous for elements of order at most 8:6 , , , , , . The second of these is not conspicuous for elements of order 16. The other twomust be semisimple, with u acting on M ( E ) with blocks 3 , , (class 2 A ) and4 , , , (class A + A ) respectively, both of which are generic (see [ Law95 ,Table 5]). This completes the proof. q = 3 : The conspicuous sets of composition factors for M ( E ) ↓ H are27 , , , , , ∗ , , , , , ∗ , , ∗ , , , (3 , ∗ ) , , , , (3 , ∗ ) , . Cases 4, 5, 6 and 8 : The cases with a trivial composition factor all have negativepressure, so H stabilizes a line on M ( E ) in all cases by Proposition 2.5. Case 2 : It is easiest to switch to the Lie algebra L ( E ) ◦ , of dimension 77, wherethe corresponding set of composition factors is27 , , (3 , ∗ ) , . As we saw in Proposition 8.1, the { , ± , } -radical of P (7) is 3 , ∗ / , / , , ∗ / L ( E ) ◦ ,which we do not have. Case 3 : There is an extension between 7 and 3 ± but none between 3-dimensionalmodules. Let v be from a unipotent class of elements of order 3 lying in PSL (3) ≤ H ; then v acts on 3 as simply 3, and on 7 as 3 ,
1. So v acts on M ( E ) with atleast eight blocks of size 3, hence on M ( E ) as 3 by [ Law95 , Table 5]. However,the { ± , } -radicals of P (7) and P (3) are 7 / , ∗ / ∗ / /
3. Thus there mustbe a subquotient 7 ⊕
7, and this contradicts the action of v . Therefore H cannotembed with these factors. Case 7 : This case must be semisimple, and then u acts with Jordan blocks 3 , , on M ( E ), which is the generic class 2 A . Case 1 : This leaves the irreducible case, as claimed. q = 2 : There are four conspicuous sets of composition factors, namely8 , , , (3 , ∗ ) , , (3 , ∗ ) , , , ∗ . The projective cover P (3) has structure3 / (1 ⊕ (3 ∗ / / ∗ )) / , so in order to have a trivial composition factor but no trivial submodule or quotientwe need a copy of P (3) (or P (3 ∗ )). Thus the first and third cases yield stabilizedlines, and the second case must be, up to duality,8 ⊕ P (3) ⊕ .
0. SUBGROUPS OF E The action of v ∈ H of order 4 on this module has blocks 4 ,
3, but this is not in[
Law95 , Table 5], a contradiction.For the fourth case, we switch to the Lie algebra, and note that the correspond-ing set of factors on L ( E ) is 8 , , which is semisimple and therefore H stabilizesa line on L ( E ). q = 4 : There are two conspicuous sets of composition factors for M ( E ) ↓ H up tofield automorphism, namely 8 , , , ∗ , , . The first case must centralize a 3-space on M ( E ), but the second case need notstabilize a line on M ( E ), as there is a module 1 , /
9, so we consider this case.There are three H -conjugacy classes of subgroups isomorphic to PSL (2), withrepresentatives L , L and L . Note that the restriction of M ( E ) to each L i has composition factors 8 , (3 , ∗ ) , L i stabilizes a line or hyperplane on M ( E ), so we may assume that at least twoof them stabilize a line on M ( E ), say L and L . This yields maps from thepermutation modules P L i to M ( E ) ↓ H .There is no module 9 ∗ / / ∗ / , /
9, but with notrivial quotients), so our module must have the form 9 ∗ / , /
9. Hence the L i -fixedline must lie in the submodule 1 /
9. There is a unique extension 1 / L , but is not the same as the extension that splits on restriction to L . Thus the only consistent option is that H itself stabilizes a line on M ( E ) (upto duality).If we have ¯ H ∼ = SL (4), there are up to field automorphism and duality threeconspicuous sets of composition factors for M ( E ) ↓ ¯ H , which are3 , ∗ , ¯9 ∗ , (3 , ∗ ) , , . (Here, all modules come from the same faithful 2-block.)For the first case, the corresponding factors for L ( E ) ↓ ¯ H are 8 , , which issemisimple, so ¯ H stabilizes a line on L ( E ).For the second case, the corresponding factors for L ( E ) ↓ ¯ H are(9 , ∗ ) , , , . The { , ± } -radical of P (9) is 9 ∗ / , /
9, and on this we may add one copy each of8 and 8 , falling into the second socle layer. On this we add as many copies of 9,9 ∗ and 1 as we can, and we obtain a module1 , , , , , / , , ∗ , ∗ , ∗ / , , , / . Of course, all of the extra trivials are quotients, so we need as many 9s as 1s inorder not to stabilize a line on L ( E ), which we do not have. Thus ¯ H stabilizes aline on L ( E ).In the third case, an element v of order 4 acts on 24 ⊕ as 4 ,
3, and so ¯ H cannot embed with these composition factors since this action does not appear in[ Law95 , Table 5]. (cid:3) If H ∼ = PSL (3) acts irreducibly on M ( E ) then the composition factors of L ( E ) ◦ ↓ H are 15 , ∗ , , , ∗ , (3 , ∗ ) , .
62 10. SUBGROUPS OF E Since the actual copy of A lying in G does not lie in a subgroup containing amaximal torus, this A cannot stabilize a line on L ( E ) ◦ by Lemma 3.4. We provein Section 13.4 that all such H lie in copies of G (3) in E . Proposition . Let H ∼ = PSU ( q ) for q = 3 , , , , , and let ¯ H ∼ = SU ( q ) for q = 5 , . (1) If q = 3 then H stabilizes a line on M ( E ) , is a blueprint for M ( E ) ⊕ M ( E ) ∗ , stabilizes a line on L ( E ) ◦ , or acts irreducibly on M ( E ) . (2) If q = 4 then H either stabilizes a line or hyperplane on M ( E ) or L ( E ) ,or is a blueprint for M ( E ) ⊕ M ( E ) ∗ . (3) If q = 5 , then H and ¯ H are blueprints for M ( E ) ⊕ M ( E ) ∗ , and ¯ H stabilizes a line on L ( E ) . (4) If q = 9 then either H is a blueprint for M ( E ) ⊕ M ( E ) ∗ or H stabilizesa line on L ( E ) ◦ . If H acts irreducibly on M ( E ) then H is a blueprintfor L ( E ) ◦ . Proof. q = 5 : There are two conspicuous sets of composition factors for M ( E ) ↓ H : 8 , and 19 ,
8. The first of these is semisimple, so H stabilizes a lineon M ( E ), and u acts on M ( E ) with blocks 3 , , . This is the generic class3 A (see [ Law95 , table 5]), so H is a blueprint for M ( E ) ⊕ M ( E ) ∗ by Lemma2.2. The second of these will be the maximal A subgroup; note that Ext kH (8 , M ( E ) ↓ H .The corresponding factors on L ( E ) are 35 , ∗ ,
8, with an extension between 8and 35. Therefore L ( E ) ↓ H is either semisimple or (up to automorphism) 35 ∗ / / u acts on L ( E ) with blocks 5 , , , , , class 2 A + A , whichis not generic for L ( E ) but is generic for M ( E ), so that H is a blueprint for M ( E ) ⊕ M ( E ) ∗ . The second case, a module 35 ∗ / /
35, does not actually exist.Therefore H is a blueprint for M ( E ) ⊕ M ( E ) ∗ , as needed.There are two conspicuous sets of composition factors for M ( E ) ↓ ¯ H : 6 , and15 , . Both of these are semisimple, and u acts on them with blocks 3 , , and4 , , , respectively, which are the generic classes 2 A and A + A respectively.Thus ¯ H is a blueprint for M ( E ) ⊕ M ( E ) ∗ by Lemma 2.2. q = 3 : Up to duality there are eight conspicuous sets of composition factors for M ( E ) ↓ H : (3 , ∗ ) , , , , , (3 , ∗ ) , , , , ∗ , , ∗ , , , , , , ∗ , , , . Cases 1, 2 and 3 : The first three have negative pressure, so H stabilizes a lineon M ( E ) by Proposition 2.5. Case 4 : This has pressure 1, so quotienting out by any 3 ± in the socle, if H does notstabilize a line on M ( E ) then the socle is one of 7, 6 or 6 ∗ . The { , ± , ± } -radicalof P (7) is 1 / , ∗ / , , ∗ / , so it cannot be 7. There is a module 6 ∗ / / / /
6, and neither 3 nor 3 ∗ may attachto this module anywhere, so either H stabilizes a line or hyperplane on M ( E ) or H acts on M ( E ) as one of(6 ∗ / / / / ⊕ ⊕ ∗ (6 ∗ / / / / ⊕ (3 ∗ /
0. SUBGROUPS OF E up to swapping modules with their duals. The action of u on these moduleshas blocks 3 , , and 3 , , respectively, with the former not appearing in[ Law95 , Table 5], and the latter being class A + A , which acts on L ( E ) withfactors 3 , , .Thus we are in the latter situation. Since H stabilizes a 3-dimensional subspace,it lies inside a proper positive-dimensional subgroup of G by [ Cra17 , Lemma 1.4],but we require strong imprimitivity. Thus we go through the list of members of X ,first checking which have module structures that are compatible with the modulestructure of M ( E ) ↓ H .For the reductive subgroups, H cannot lie in A A A as this has three sum-mands of dimension 9. If H lies in A A then H lies in A , hence in the A -parabolicsubgroup. If H lies in F then H stabilizes a line on M ( E ), which it does not.The G and C subgroups do not intersect the unipotent class A + A by [ Law09 ,Tables 31 and 32]. This leaves A G .Of the parabolic subgroups, the D -parabolic subgroups stabilize a line or hy-perplane on M ( E ), so H is not contained in them. If H is contained in an A A -parabolic subgroup then H is contained in an A -parabolic subgroup, hence inthe D -parabolic subgroup. If H is contained in the A A A -parabolic then itlies in the A A -parabolic, hence in the A -parabolic subgroup. This leaves the A -parabolic subgroup. This acts on M ( E ) as 6 / /
6, with the top and bottom6 being isomorphic and not self-dual. This is therefore not compatible with themodule structure either, as H would have to act irreducibly on M ( A ).Thus we are left with X ∼ = A G . Here the decomposition M ( E ) ↓ X is as thesum of a 21- and 6-dimensional module, fitting the structure above, but with the6-dimensional module being the symmetric square of a 3-dimensional module for A . This can only restrict to H as 1 ⊕ or 6 ± , not 3 ∓ / ± . Thus H cannot embedwith this module structure.(There is a copy of H with these factors in the A A A , F and A G sub-groups, but in each case H stabilizes a line on M ( E ).) Case 5 : This must be semisimple and u acts on M ( E ) with blocks 3 , , , whence u lies in the generic class 2 A and H is a blueprint for M ( E ) ⊕ M ( E ) ∗ by Lemma2.2. Case 6 : There is no corresponding set of composition factors on L ( E ), so thiscase cannot exist. Case 7 : We switch to L ( E ) ◦ , the composition factors being 27 , , (3 , ∗ ) , ,which has pressure 2. Of course, the 27 splits off, and the { , ± , } -radical of P (7)is 3 , ∗ / , / , , ∗ / , so we need at least three 7s in the socle (ignoring 3 , ∗ ) to cover the three trivials.Therefore H stabilizes a line on L ( E ) ◦ . Case 8 : This is the irreducible case, as for PSL (3). q = 9 : There are 36 sets of composition factors for M ( E ) ↓ H that are conspicuousfor elements of order at most 16. If M ( E ) ↓ H has real trace of an element of order73 in H , for example if M ( E ) ↓ H has the same composition factors as its dual, thenthe element, and hence H , is a blueprint for M ( E ) ⊕ M ( E ) ∗ by Theorem 2.4.(This includes the irreducible case. The proof that in this case H is a blueprint for
64 10. SUBGROUPS OF E L ( E ) ◦ is given in [ LS04b , Lemma 4.15].) This is the case for sixteen of the setsof composition factors, leaving ten up to duality, five up to field automorphism:6 , , ∗ , , , , ¯9 , , , , , , . The first two of these must be semisimple, and u acts in both cases with blocks3 , , , which belongs to the generic class 2 A . For the third case, only 6 and 9 have a non-split extension, so M ( E ) ↓ H is either semisimple or up to automorphism(6 / ) ⊕ ¯9 ⊕ , and u acts in these two possibilities with blocks 3 , , and3 , , respectively. The second does not appear in [ Law95 , Table 5], and thefirst is class 3 A , which is generic. Thus H is a blueprint for M ( E ) ⊕ M ( E ) ∗ inthese cases by Lemma 2.2.For 15 , , the corresponding factors on L ( E ) ◦ are27 , , (3 , ∗ ) , . The 27 has no extensions with the other factors so must break off, and the { , ± , } -radical of P (7 ) is 7 , / , , ∗ / . Thus we need at least twice as many copies of 7 as 1s in order for H not to stabilizea line on L ( E ) ◦ , which we do not have.In the final case, let x denote an element of order 80 in H . There are 24 distincteigenvalues for the action of x on M ( E ), and 6 (and 21 ) are sums of eigenspaces(i.e., the eigenvalues of x on 6 and 21 are disjoint), hence any semisimple elementof G powering to x preserves 6 and 21 . Thus H is a blueprint for M ( E ), placing H inside a positive-dimensional subgroup X of G stabilizing the same subspaces of M ( E ). Since no composition factor of M ( E ) ↓ H is self-dual, the same is true for M ( E ) ↓ X , whence H and X stabilize the same subspaces of M ( E ) ⊕ M ( E ) ∗ aswell. Thus H is a blueprint for this module, as required. q = 4 : The conspicuous sets of composition factors for M ( E ) ↓ H split into twogroups: those with a trivial factor and those without. Those with a trivial factorare those of M ( F ) ↓ H with a trivial added, and of those, only 9 , ∗ , , , ∗ , , H stabilizes a line on M ( E ). In the remaining cases, however, the cf( M ( E ) ↓ H )-radicals of P (9 ) are1 / and 9 ∗ / , / , and both of these have trivial quotients. Thus if H has a trivial factor on M ( E ),it stabilizes a line or hyperplane on it.From now we assume that H has no trivial composition factors on M ( E ).There are 16 such sets of composition factors, four up to field and graph automor-phism: 3 , ∗ , ∗ , , (3 ∗ ) , , ¯9 , , ∗ , , . Since H stabilizes a 3-space in all cases H always lies inside a member of X by [ Cra17 , Lemma 1.4], but we need stability under σ and also N Aut + ( G ) ( H ).Switching to L ( E ), the first three sets have corresponding factors8 , , (¯9 , ¯9 ∗ ) , , , , , ∗ , , , , ∗ , and the fourth has no corresponding set. In the first two cases therefore, H stabilizeslines on L ( E ), so we consider the third set. The two 9-dimensional modules in
0. SUBGROUPS OF E M ( E ) ↓ H split off as summands, so we consider the summand comprising the 3-dimensionals. Since Ext kH (3 , ) = 0 and there is an extension between 3 and3 ∗ , we have a module 3 / ∗ / , so M ( E ) ↓ H has a summand one of the followingthree modules, up to duality:3 ⊕ ⊕ ∗ , ⊕ (3 ∗ / ) , / ∗ / . The action of v ∈ H of order 4 on M ( E ) is 4 , , and 4 , , , and 4 , respectively. Only the last of these appears in [ Law95 , Table 3], so H acts like thison M ( E ).Now we use the fact that H is not Lie primitive to prove that H is stronglyimprimitive. The only connected members of X acting with composition factors ofdimensions compatible with 9 , are A A A , A G and G . In the second andthird cases, the A A subgroup of A G , and the A subgroup of G , also lie in A A A by [ Tho20 , Table 11A], so we assume that H is contained in X = A A A .This group acts on M ( E ) as the sum of three 9-dimensional modules, with tensorproducts (10 , , , ,
01) and (01 , , H maps into the first andsecond factors as 3 and the third as 3 , then H acts on M ( E ) in the requiredway, and all other permissible embeddings are obtained from this by dualizing andpermutations. Thus we see that H and the corresponding A subgroup actingas (10 , ,
20) stabilize the same subspaces of M ( E ) ⊕ M ( E ) ∗ , whence H is ablueprint for M ( E ) ⊕ M ( E ) ∗ , as needed. q = 8 : There are eleven conspicuous sets of composition factors for M ( E ) ↓ H , nineof which have the same composition factors as their duals, and hence are blueprintsfor M ( E ) ⊕ M ( E ) ∗ for the same reason as for q = 9, using a semisimple elementof order 21. The remaining set of composition factors (up to automorphism) is9 , , ∗ . If L denotes a copy of PSL (8) in H , then L centralizes an element of order 3whose eigenvalues on M ( E ) have multiplicities 6, 6 and 15, i.e., L lies in the A -Levi subgroup. The action of L on M ( E ) is semisimple, with factors4 , , , , , , . Since M ( A ) appears twice in M ( E ) ↓ A , we see that L acts as 2 ⊕ ⊕ on M ( A ), whence L lies in the positive-dimensional A subgroup X contained in A acting as L (1) ⊕ L (2) ⊕ L (4) on M ( A ). The action of X on Λ ( M ( A )) is easy tocompute, and is L (0) ⊕ ⊕ L (3) ⊕ L (5) ⊕ L (6) , whence L , and therefore H , are blueprints for M ( E ) ⊕ M ( E ) ∗ .If we have ¯ H instead, then there are four conspicuous sets of composition factorsfor elements of order at most 19, up to field automorphism:3 , ∗ , , , , , , , . The third and fourth cases require that an involution in ¯ H acts with blocks 2 , M ( E ), which does not appear in [ Law95 , Table 5]. The first case has com-position factors 8 , on L ( E ), of pressure −
14, so ¯ H stabilizes a line on L ( E )by Proposition 2.5. The second case has composition factors (9 , ∗ ) , , , ,which has pressure −
2, so ¯ H stabilizes a line on L ( E ) in this case as well. (cid:3)
66 10. SUBGROUPS OF E As with PSL (3), if H acts irreducibly on M ( E ) then its composition factorson L ( E ) ◦ are 15 , ∗ , , , ∗ , (3 , ∗ ) , . Again, the actual H acting irreducibly on M ( E ) cannot stabilize a line on L ( E ) ◦ .This case is dealt with in Section 13.2, where as with PSL (3), we prove that allsuch H lie in copies of G (3) in E . Proposition . If H ∼ = PSp ( q ) ′ for q = 2 , , , or H ∼ = B ( q ) for q = 8 ,then H stabilizes a line or hyperplane on M ( E ) . If ¯ H ∼ = 3 · PSp (2) ′ then ¯ H stabilizes a line on L ( E ) . Proof. q even : The proof is exactly the same as for Proposition 9.3, but thistime the trace of a rational element of order 5 is 2 on M ( E ). If H ∼ = Alt(6) then by[ Cra17 , Proposition 6.3] we see that H stabilizes a line or hyperplane on M ( E ).(Note that the proof of the result in [ Cra17 ] does yield this statement, butthe claim in [
Cra17 , Proposition 6.3] is erroneously that H stabilizes a line on M ( E ), rather than a line or hyperplane . Of course, in uses of the result thisdistinction is usually irrelevant. Moreover, with slightly more work one obtainsthe original result: if H stabilizes a hyperplane but not a line then H lies insidea D -parabolic subgroup by Lemma 3.14, which acts on M ( E ) as 1 / /
10. Thus H cannot stabilize a line on the self-dual module 10 = M ( D ). But this is clearlyimpossible, as there is no positive-pressure 10-dimensional module for H .)In the case of ¯ H , in each faithful block there are three simple modules, 3 , 3 and 9, labelled so that 3 i ⊗ ∗ i = 1 ⊕ i , where 8 i is one of the two factors into whichthe Steinberg module splits on restriction to Sp (2) ′ . As in [ Cra17 ] we find up toautomorphism four conspicuous sets of composition factors for M ( E ) ↓ ¯ H , namely3 , , , , , , , , , . Notice that each of these stabilizes a 3-space on M ( E ), hence lies in a memberof X by [ Cra17 , Lemma 1.4], but is not necessarily strongly imprimitive. We gothrough the members of X , and prove that the second and fourth cases do notoccur.Suppose that ¯ H ≤ X for some X ∈ X . Since ¯ H is a non-split extension bythe centre, so must X be, so X must be either the A -parabolic ( ¯ H cannot mapinto A , and H cannot embed in G , so A A and A G need not be considered)or A A A . In order to embed ¯ H into A A A in the correct manner, the actionof Z ( ¯ H ) on each of the three modules M ( A ) must be different. In particular, itmust act trivially on one of them, so ¯ H ≤ A A ≤ A , and it suffices to check thissubgroup. The action of ¯ H on M ( A ) has factors either 3 or 3 , . In the firstpossibility M ( E ) ↓ ¯ H has composition factors 3 , , and in the second possibility M ( E ) ↓ ¯ H has composition factors 9 , , , so these are the only two cases thatmay occur.In the first case there is a unique possible set of composition factors for L ( E ) ↓ ¯ H ,namely 8 , , and clearly ¯ H stabilizes a line on L ( E ) the module has pressure −
14. In the third case the corresponding Brauer character for L ( E ) is not uniquelydefined, so we use the fact that H embeds in A . This has a unique correspondingset of composition factors on L ( E ), and they are 8 , , , , . This set offactors has pressure − H stabilizes a line on L ( E ), as claimed.
0. SUBGROUPS OF E q = 3 : From the composition factors for M ( E ) ↓ H , we see that the only conspic-uous sets of composition factors for M ( E ) ↓ H are 10 , , and 25 , . The lattercase must be semisimple, but an element from the largest two classes of elementsof order 3 acts on 25 with blocks 3 ,
1, hence on M ( E ) with blocks 3 , . Thisaction does not appear in [ Law95 , Table 5], hence H does not embed with thesefactors.The first case has pressure 1, so if H does not stabilize a line or hyperplaneon M ( E ), then ignoring the 10 the module structure would have to be 5 / / / / { , , } -radical of P (5) has the form5 , / , , / . Thus H stabilizes a line or hyperplane on M ( E ), as needed. (cid:3) Proposition . Let H ∼ = G ( q ) ′ for q = 2 or G ( q ) ′ for q = 3 . (1) If q = 2 then H stabilizes a line or hyperplane on M ( E ) or L ( E ) , or H stabilizes a unique -space in L ( E ) and its stabilizer is a maximal G ,and N G ( H ) = G (2) is a blueprint for L ( E ) . (2) If q = 3 then either H stabilizes a line on M ( E ) , or H acts on M ( E ) and L ( E ) ◦ as ⊕ ⊕ , and ⊕ ⊕ ⊕ P (7) ⊕ (7 / respectively. Proof. q = 2 : There are two conspicuous sets of composition factors for M ( E ) ↓ H , namely 6 , and 14 , ,
1. In the first case H must stabilize a lineon M ( E ) by Proposition 2.5 (since it has pressure − v have order 8 in H . Suppose that H does not stabilize a line or hyper-plane on M ( E ). If 14 is a summand then the structure must be (6 / / ⊕
14, andthis module is uniquely determined. The element v acts on this module with blocks8 , , ,
1, which does not appear in [
Law95 , Table 5], hence 14 is not a summandof M ( E ) ↓ H . In particular, this means that up to duality we may assume that thesocle of M ( E ) ↓ H is 6.Either there is a submodule 14 / , /
6, on which 6 must be placed. If there is not then we have a submodule6 / /
6, on which 14 must be placed. In each case we can place two copies on top,and we thus obtain two modules,6 , / , / , / / , / , one of which must be a pyx for M ( E ) ↓ H . The intersection of these two submodulesof P (6) is a module 6 / , / v ∈ H oforder 8 acts with blocks 8 , , ,
1. This does not appear in [
Law95 , Table 5], andthus this module is not M ( E ) ↓ H . (But we have not yet excluded any of the othermodules.)Let L denote a copy of PSL (2) inside H . The restrictions of 6 and 14 to L are 3 ⊕ ∗ and 3 ⊕ ∗ ⊕ / / L has structure1 ⊕ (3 ∗ / ⊕ (3 / ∗ ) , and the restriction of 14 / / , / L has structure1 ⊕ (3 / ∗ / ⊕ (3 ∗ / / ∗ ) ⊕ ⊕ ∗ ⊕ ⊕ .
68 10. SUBGROUPS OF E This means that there is a unique extension with quotient 14 and submodule 6 / / L , and it turns out that this is the module 6 / , / kH -modules 14 / / / L is always(10.1) 1 ⊕ (3 / ∗ / ⊕ (3 ∗ / / ∗ ) ⊕ . Up to changing the number of 8s involved, the exact same statements hold for1 , / , / , /
6. Thus if H does not stabilize a line or hyperplane on M ( E ) then we may assume that M ( E ) ↓ L has the structure given in (10.1).Either by Proposition 11.1 below, or by computing the composition factors on L ( F ), we see that the image of L under the graph automorphism of F stabilizesa line on M ( F ), and in fact acts with composition factors 8 , . Certainly theimage under the graph automorphism cannot lie in a maximal parabolic subgroupof F , hence neither can L . Thus L lies in one of B (which stabilizes a line on M ( F )), C (and then, as in the proof of Proposition 9.1, we find that L lies in anirreducible A subgroup lying in A A from [ Tho20 , Table 10/10A]) or A A .Thus L is contained in A A , and obviously a diagonal A subgroup of A A .It must be embedded as either (10 ,
10) or (01 , M ( F )with factors 8 , and the other with factors 8 , (3 , ∗ ) , so it is of course the latter.The action is then L (11) ⊕ ( L (10) /L (02) /L (10)) ⊕ ( L (01) /L (20) /L (01)) , and note that we must have that the stabilizer of the kL -submodule 3 ⊕ ∗ containsthis diagonal A subgroup X . In particular, this means that H is contained in apositive-dimensional subgroup of E , namely h H, X i .This proves that H is not Lie primitive, but since M ( E ) is not graph-stable,we need to work a bit harder to prove strong imprimitivity. It is easier to switch to L ( E ) at this point. Let Y denote a maximal, positive-dimensional subgroup of E containing H . The action of H on L ( E ) is one of two sets of composition factors:32 , , , , , . In the second case H has pressure −
1, so H stabilizes a line on L ( E ), so we mayassume that H acts as in the first case.Note that Y cannot act irreducibly on L ( E ), and H acts with compositionfactors of dimensions 32 , ,
14, so if Y acts with three composition factors thenthey need to be of those dimensions. As H F , we must have that Y is amaximal G , which acts with dimensions 64 ,
14, and N Y ( H ) = G (2) is a blueprintfor L ( E ). Also, H is contained in an N Aut + ( G ) ( H )-stable, positive-dimensionalsubgroup of G , namely the stabilizer Y of the 14-space in L ( E ). q = 3 : The conspicuous sets of composition factors for M ( E ) ↓ H are7 , , , , , , , , . The first and third sets of composition factors of H on M ( E ) have negative pres-sure, hence H stabilizes a line on M ( E ) by Proposition 2.5. For the second set,the action of H on L ( E ) ◦ has composition factors9 , , , , ,
0. SUBGROUPS OF E and since an element v ∈ H of order 9 acts on M ( E ) as 9 , it comes from class E ( a ), so acts on L ( E ) with blocks 9 , Law95 , Tables 5 and 6]). Hencethe action on L ( E ) ◦ is 9 ⊕ ⊕ ⊕ P (7) ⊕ (7 / , as claimed in the proposition. (cid:3) The case of G (3) ′ acting as 9 ⊕ ⊕ on M ( E ) was proved to lie in themaximal G subgroup in [ Asc , Theorem 29.3], using the trilinear form on M ( E ).It appears that pure representation-theoretic methods are not powerful enough toattack this. We will give another proof of this, still using the trilinear form, inSection 13.3. Proposition . If H ∼ = PSU ( q ) for q = 2 , , then H stabilizes a line on M ( E ) . Proof. q = 2 : As we would guess from the conspicuous sets of factors for M ( E ) ↓ H , there are two conspicuous sets of factors for M ( E ) ↓ H :14 , , , , (4 , ∗ ) , . The first set of factors has pressure 0, so H stabilizes a line or hyperplane on M ( E ) by Proposition 2.5. However, the factors do not fit with the factors ofthe D -parabolic subgroup in Lemma 3.14, so H ≤ F and so stabilizes a line on M ( E ). The second has pressure −
1, so stabilizes a line on M ( E ). q = 4 : Again, there are two conspicuous sets of composition factors for M ( E ) ↓ H up to field automorphism: 14 , , , , (4 , ∗ ) , . The same pressure statements apply this time, except now Ext kH (1 , ± ) is zero, sothe trivial factors even split off as summands in both cases. (cid:3) Proposition . If H ∼ = PSp (2) is a subgroup of E , then H stabilizes aline on M ( E ) or L ( E ) . Proof.
There are two conspicuous sets of composition factors for M ( E ) ↓ H ,namely 8 , , , of pressure −
4, hence H stabilizes a line on M ( E ) by Proposition2.5, and 14 , ,
1, which has pressure 1.For the second we switch to the Lie algebra L ( E ), where the correspondingset of composition factors is 14 , , , . From Proposition 10.4 (in particular itsproof) above we see that in this case the subgroup L ∼ = G (2) of H stabilizes aline on L ( E ), whence there is a (non-zero) homomorphism from the permutationmodule P L to L ( E ) ↓ H . Assume that H does not stabilize a line on L ( E ).Since there is no module 6 / /
6, and L ( E ) ↓ H has pressure 1, we see that thereare two possible structures for the module:6 / , / / , / / , / / , / , / , / / , / / , / / , / . The module P L has structure1 , / / / / / / / / / / , , and so we see that the second structure must be the correct one, and the quotient W = 1 , / L ( E ) ↓ H . But now Ext kH (6 , W ) is 1-dimensional,but the module 6 / , /
70 10. SUBGROUPS OF E This completes the proof. (cid:3)
HAPTER 11
Subgroups of F G be the algebraic group of type F and H be a subgroup of rank 2, q ≤
9, together with a small Ree group, a Suzuki group, PSL (16) or PSU (16). Wecan exclude all groups where p is odd and where H contains a semisimple elementof order at least 19, and where p = 2 and H contains a semisimple element of orderat least 57 by Theorem 2.4. If p is odd then any subgroup that is a blueprint for M ( E ) is a blueprint for M ( F ), so those cases eliminated in the previous chapterstay eliminated here. The remaining ones are as follows.(1) PSL ( q ) for q = 2 , , ( q ) for q = 3 , , ( q ) for q = 3 , G ( q ) ′ for q = 2 , B ( q ) for q = 8 , G (3) ′ .In characteristic 2 we might prove that either H or its image under the graphautomorphism stabilizes a line on M ( F ), and then use Lemma 3.5. Note thatwe can also apply Lemma 3.8, so if we can show that H is a blueprint for M ( F )and the composition factors of H on M ( F ) and the quotient L ( F ) /M ( F ) havedifferent dimensions, then we are done.As with all of the previous chapters, here u denotes an element of order p in H belonging to the smallest conjugacy class. Proposition . Let H ∼ = PSL ( q ) for ≤ q ≤ . (1) If q = 2 , then H stabilizes a line or hyperplane on L ( F ) . (2) If q = 3 then H either stabilizes a line on M ( F ) ◦ or is a blueprint for M ( F ) ◦ . Proof. q = 2 : There are three conspicuous sets of composition factors for M ( F ) ↓ H , namely (3 , ∗ ) , , , , , (3 , ∗ ) . The pressures of the first two are −
2, so H stabilizes a line on M ( F ) by Proposition2.5. An element of order 7 in H acts with trace 5 in the first two cases and − H acts as in the third case on M ( F ), the image of H underthe graph automorphism acts as in the first or second cases on M ( F ). Thus either H or its image under the graph automorphism stabilizes a line on M ( F ), and H stabilizes a line or hyperplane on L ( F ) by Lemma 3.5. q = 4 : There are, up to field automorphism, only two conspicuous sets of com-position factors: 8 , and 9 , ∗ , . Again the first case has pressure −
2, and so H stabilizes a line on M ( F ), and again (this time up to field automorphism) the F graph automorphism swaps the two classes. (This is because the graph automor-phism does not fix any classes for q = 2.) As for q = 2 we find that H stabilizes aline or hyperplane on L ( F ), as needed. q = 3 : There are four conspicuous sets of composition factors for the action of H on the 25-dimensional module M ( F ) ◦ , given by(3 , ∗ ) , , , , , , ∗ , , ∗ , , (3 , ∗ ) . Cases 1 and 2 : These have pressures − − H stabilizesa line on M ( F ) ◦ by Proposition 2.5. Case 3 : The only classes of elements of order 3 that are non-generic for M ( F ) ◦ are A + ˜ A , ˜ A and ˜ A + A , which on M ( F ) ◦ act as 3 , or 3 , u acts on 3 as 2 ,
1, on 6 as 3 , ,
1, and on 7 as 3 , , we see that if H is not ablueprint for M ( F ) ◦ , then there are no simple summands of M ( F ) ◦ ↓ H . Thus thesocle is, up to duality, one of: 3; 6; 3 ⊕
6; and 3 ⊕ ∗ (note that Ext kH (3 ,
6) = 0but Ext kH (3 , ∗ ) ∼ = k ). In the first case we consider the { ∗ , ± , } -radical of P (3),which is 6 / ∗ / ∗ , / , and it is clearly not self-dual. The dual of this is the { ± , , } -radical of P (6 ∗ ),so we cannot have a simple socle. If the socle of M ( F ) ◦ ↓ H is 3 ⊕
6, we take the { ∗ , ∗ , } -radical of P (3) ⊕ P (6), then take the { } -residual, to obtain(3 ∗ / ∗ , / ⊕ (3 ∗ , ∗ / . This has one too many copies of 6 ∗ , and u acts on it with blocks 3 , , , so wecannot get at least seven blocks of size 3 in the action of u on M ( F ) ◦ ↓ H this way.Similarly, we do the same with socle 3 ⊕ ∗ to obtain the module3 ∗ / , / , ∗ , and u acts on this with blocks 3 , . This contradiction proves that u is genericon M ( F ) ◦ . Case 4 : The { ± , } -radical of P (3) is 3 ∗ / /
3, so we see that at least four 3-dimensional factors must split off as summands in M ( F ) ◦ ↓ H ; since u acts on 3with blocks 2 ,
1, this means that we have blocks at least 2 , in the action of u on M ( F ) ◦ , and this is enough to assure that u is generic from [ Law95 , Table 3]. (cid:3)
Proposition . Let H ∼ = PSU ( q ) for q = 3 , , . (1) If q = 3 then H stabilizes a line on either M ( F ) ◦ or L ( F ) . (2) If q = 4 then H stabilizes a line or hyperplane on L ( F ) or either H or itsimage under a graph automorphism is a blueprint for M ( F ) and is notgraph-stable. (3) If q = 8 then either H stabilizes a line or hyperplane on L ( F ) or thestabilizer of all H -invariant -spaces on the composition factors of L ( F ) is positive dimensional, and hence H is strongly imprimitive. Proof. q = 3 : There are four conspicuous sets of composition factors for M ( F ) ◦ ↓ H : (3 , ∗ ) , , , , , (3 , ∗ ) , , , ∗ , , ∗ . The first two have negative pressure, so H stabilizes a line on M ( F ) ◦ by Proposition2.5. For the third and fourth cases we switch to L ( F ), and see that H acts with
1. SUBGROUPS OF F composition factors 7 , (6 , ∗ ) , and 15 , ∗ , , , ∗ , respectively, which havepressures − H stabilizes a line on L ( F ) in both cases. q = 4 : There are, up to field automorphism, six conspicuous sets of compositionfactors for M ( F ) ↓ H :(3 , ∗ ) , , , , , (3 , ∗ ) , , (3 , ∗ ) , , ∗ , , ∗ , , ¯9 , ¯9 ∗ , . The graph automorphism swaps (up to field automorphism) the first and thirdcases, the second and fifth, and the fourth and sixth sets of factors. As there areno extensions between the factors, the first two cases are semisimple and so H stabilizes a line or hyperplane on M ( F ), hence either H stabilizes a line on L ( F )or we may assume that we have either the fourth or sixth case.In the sixth case, let x denote an element of order 15 in H . There is no elementof order 45 in G cubing to x and having the same eigenspaces on M ( F ) as x , butthere is an element of order 75 = 5 ×
15 having the same eigenvalues as x , hence x and H are blueprints for M ( F ). Furthermore, we apply Lemma 3.8 to see thatthis is enough to force σ -stability and N Aut + ( G ) ( H )-stability, as needed. q = 8 : Up to field automorphism there are three conspicuous sets of compositionfactors for M ( F ) ↓ H , namely8 , , , ∗ , , , ∗ , . Up to field automorphism, the first and second are swapped by the graph automor-phism, so as the first case stabilizes a line on M ( F ), in both cases H stabilizesa line or hyperplane on L ( F ) by Lemma 3.5. Thus we consider the third case,which is left invariant (up to field automorphism) by the graph. Since H containselements of orders 19 and 21, H is a blueprint for M ( F ) by Theorem 2.4. Theonly positive-dimensional subgroup with compatible composition factors for both H and its image under the graph is X = A A , and of course this acts semisimplywith composition factors of dimension 8, 9 and 9 on M ( F ). Thus X stabilizes one8-dimensional submodule on each composition factor of L ( F ), as does H . Thiscompletes the proof that H is strongly imprimitive, using Theorem 3.2. (cid:3) Proposition . Let H ∼ = PSp ( q ) ′ for q = 2 , , , or B ( q ) for q = 8 , . (1) If q = 2 then H stabilizes a line or hyperplane on L ( F ) . (2) If q = 4 , , then H stabilizes a line on M ( F ) . (3) If q = 3 then H does not embed in G . Proof. q even : The proof is exactly the same as for Proposition 9.3, but thistime the trace of a rational element of order 5 is 1 on M ( F ).If q = 2 then we prove that H or its image under the graph automorphismstabilizes a line on M ( F ) in [ Cra17 , Proposition 6.3], hence H stabilizes a line orhyperplane on L ( F ) via Lemma 3.5. q = 3 : The composition factors of M ( F ) ◦ ↓ H must be 10 , or 25, from the com-position factors for M ( E ) ↓ H in Proposition 10.3, but we already proved there thatthe second case cannot occur.Note that there are no positive-dimensional subgroups of G whose compositionfactors are all multiples of 5, so H does not embed in G if we prove that H cannotbe Lie primitive. If M ( F ) ◦ ↓ H is semisimple, then u acts on M ( F ) ◦ with blocks3 , , , so lies in the generic class ˜ A . Thus it is not semisimple. If 10 is a
74 11. SUBGROUPS OF F submodule then it is a summand, and the { } -radical of P (5) is 5 /
5, so we musthave 10 ⊕ (5 / ⊕
5. Then u acts on M ( F ) ◦ with blocks 3 , , , which is not avalid unipotent class. Thus 10 is not a submodule or quotient of M ( F ) ◦ ↓ H , and weconsider the { , } -radical of P (5). This is a self-dual module 5 / , /
5, so this is M ( F ) ◦ ↓ H . In this case u acts with Jordan blocks 3 , , so lies in the generic class A . (In this case a different unipotent element actually has an invalid action, butwe do not need this.) This completes the proof that H does not embed in G . (cid:3) Proposition . Let H ∼ = G ( q ) ′ for q = 2 , , or G ( q ) ′ for q = 3 . (1) If q = 2 , then H stabilizes a line or hyperplane on L ( F ) . (2) If q = 3 then H stabilizes a line on M ( F ) ◦ or an N Aut + ( G ) ( H ) -orbit of H -invariant -spaces on M ( F ) ◦ that has a positive-dimensional stabilizer,and hence is strongly imprimitive. Proof. q = 2 : The conspicuous sets of composition factors for M ( F ) ↓ H are6 , and 14 , , and these are swapped by the graph automorphism. Since H clearly stabilizes a line on M ( F ) in the first case, H stabilizes a line or hyperplaneon L ( F ) in both cases by Lemma 3.5. q = 4 : Let L = G (2) be contained in H . We see that, up to graph automorphism,we may assume that the composition factors of M ( F ) ↓ L are 6 , . Since the onlymodules of dimension at most 26 for H have dimensions 1, 6 and 14, we see that M ( F ) ↓ H has composition factors with dimensions 6 , , and again H stabilizes aline on M ( F ) up to graph automorphism, i.e., H stabilizes a line on L ( F ). q = 3 : Here the conspicuous sets of composition factors for M ( F ) ◦ ↓ H are 7 , and 9 , , H stabilizes a line on M ( F ) ◦ as it has pressure −
1, by Proposition 2.5, so we assume we are in thesecond case. We see that M ( F ) ◦ ↓ H must be semisimple. If v denotes an elementof order 9 in H , then v acts on M ( F ) ◦ with blocks 9 ,
7, so lies in class F ( a ),which is class D in E . This class acts on M ( E ) with blocks 9 , , , and so H must embed in E acting on M ( E ) as9 ⊕ ⊕ ⊕ ⊕ . Thus H is contained in the intersection of two line stabilizers in E , one of whichis G . The intersection is G ∩ X for X equal to F or a D -parabolic subgroup, ofdimensions 52 and 62 respectively, and in either case G ∩ X is positive dimensionaland contains H , so H is contained in a member of X , the maximal positive-dimensional subgroups of F .Since H lies inside a positive-dimensional subgroup of G , and acts on M ( F ) ◦ with composition factors of dimensions 9 , ,
7, we see that H can only lie in B ,acting irreducibly on M ( B ). (Such an embedding does exist.) Since the com-position factors of M ( F ) ◦ ↓ H are not stable, each of the two 9-spaces forms an N Aut + ( G ) ( H )-orbit of subspaces by Lemma 3.9. One of these is stabilized by B ,so we may apply Theorem 3.2 to see that H is strongly imprimitive. (cid:3) HAPTER 12
Difficult cases
This chapter considers three cases that have been postponed from earlier chap-ters: PSL (2), PSL (5) and PSU (4), all in E . We split this up into three sections,one for each group. PSL (2) ≤ E This proof is probably the hardest in this article, and so deserves this specialsection all on its own. Because of the many possible paths through to a positiveresult, there may well be a much shorter and conceptually easier proof, but I havenot found it.Let H ∼ = PSL (2). Proposition 6.1 showed that either H stabilizes a line on L ( E ), or the composition factors of H on L ( E ) are40 , ∗ , , ∗ , , , ∗ , (5 , ∗ ) . We will prove that H is always strongly imprimitive. We start with some prelimi-nary information, and then proceed to the proof.The non-split extensions between these simple kH -modules are, up to dualityand graph automorphism,(5 , ∗ ) , (5 , ) , (5 , ∗ ) , (10 , ∗ ) , (10 , ) , (24 , ) , (24 , ) , and in each case Ext is 1-dimensional.Write L for a PSL (2) subgroup of H , and write M for the subgroup 3 × PSL (2)of H . Let z denote a central element of order 3 in M , let θ denote a primitive cuberoot of unity. Let u , v and w denote unipotent elements of H acting on the naturalmodule with blocks 3 , and 4 , Note that this differs fromour previous convention for u . We may conjugate u to lie in M , so that we maysplit the blocks of u among the eigenspaces of z , i.e., consider the action of uz on L ( E ). Since z has trace 14 on L ( E ), it lies in the class whose centralizer in D T .By computing traces, we determine that the composition factors of M ( D ) ↓ M are(3 , ∗ ) , , and any such module has a trivial submodule.We will write the 1- and θ -eigenspace of a module V to mean the 1- and θ -eigenspace of the action of z on V . If no module is mentioned we mean theeigenspace of L ( E ), but we will usually state it.The restrictions to L and M of the simple kH -modules that we consider are asfollows: Module Restriction to L θ -eigenspace5 4 ⊕ ∗ ∗ ⊕ ∗
110 4 ∗ ⊕ ⊕ ∗ ∗ ⊕ ⊕ ∗
324 (1 / / ⊕ ⊕ ∗ ⊕ ⊕ ⊕ ⊕ ∗ ⊕ ⊕ ∗ (3 / ∗ / ⊕ ∗ ⊕ ∗ ⊕ ∗ ⊕ ⊕
20 (3 ∗ / / ∗ ) ⊕ ⊕ ∗ ⊕ (6 / / ⊕ ⊕
20 3 ⊕ ⊕ ∗ / / ∗ ) ⊕ ⊕ ∗ (6 / / ⊕ ∗ ⊕ ∗ (3 ∗ ) ⊕ ⊕ / ∗ / ⊕ ⊕ ∗ This proof takes ten steps. The first is to understand the placement of the 14s,and to prove that there are no submodules 40 and 40 ∗ of L ( E ) ↓ H .The second analyses the actions of M and u on L ( E ). In this step we constructthe actual copy of H that does embed in E (inside A A ), and show that if u belongs to class 2 A then H is strongly imprimitive. Furthermore, we determinethe possible E -classes for u , and how their Jordan blocks distribute amongst the1- and θ -eigenspaces of z . This information is relied upon heavily in later steps.Step 3 proves that there must be a subquotient 40 ⊕ ∗ in L ( E ) ↓ H , andSteps 4 and 5 prove that there is a subquotient 40 ⊕ ∗ . Steps 6 and 7 combinethese, and prove that 40 ⊕ ∗ ⊕ ⊕ ∗ is in fact a subquotient. Step 8 provesthat the copies of 24 do not lie above and below this subquotient in a particularconfiguration, and therefore 24 ⊕ is a subquotient. Finally, Steps 9 and 10 providethe final contradiction. Step 1 : Every composition factor 14 in L ( E ) ↓ L lies in a subquotient 1 / /
1. Inparticular, every kL -submodule of L ( E ) with a composition factor 14 has a 14 / ± of L ( E ) ↓ H .The subgroup L normalizes a 2-subgroup of H , hence lies inside a parabolic sub-group of G . The composition factors of L ( E ) ↓ L are(20 , ∗ ) , , , (4 , ∗ ) , , and from [ Cra17 , Propositions 4.1 and 4.2] we see that L cannot lie in an E -parabolic subgroup. Since L cannot lie in an A or A subgroup, if L lies in anyparabolic it must lie in either an A -, A A - or D -parabolic subgroup.If L lies in an A -parabolic subgroup, the composition factors on M ( A ) areeither 4 , , ∗ . Either way, L lies in an A A -parabolic subgroup of A , hencein an A A -parabolic subgroup of G . If it lies in a D -parabolic subgroup then thecomposition factors of L on M ( D ) are 6 , , ∗ , and this lies in a D A -parabolicsubgroup, which of course lies in an A A -parabolic subgroup. Hence we mayassume that L lies in an A A -parabolic subgroup. In this case the action on M ( A ) must be 4 (up to automorphism of L ) and on M ( A ) it must be 1 ⊕ ± .Recall that the A -parabolic subgroup of E acts on L ( E ) with seven layers.We do not need the precise structure here, but the layers are Λ i ( M ( A )) for i =1 , ,
3, their duals, and M ( A ) ⊗ M ( A ) ∗ .If L acts on M ( A ) as 4 ⊕ ∗ , then the first four layers of the action are4 ⊕ ∗ , (1 / / ⊕ ⊕ , ⊕ ∗ ⊕ (4 ⊕ ∗ ) ⊕ , (6 / / ⊕ (6 / ∗ / ⊕ (1 / / ⊕ (2) ≤ E and then the duals of the third to first respectively. If L acts on M ( A ) as 4 ⊕ ⊕ , (6 / / ⊕ ⊕ , (20 ∗ ) ⊕ ⊕ (4 ∗ ) ⊕ , (1 / / ⊕ , and then the duals of the third to the first. The result is the same: 14 cannot be asubmodule of L ( E ) ↓ L .If L is contained in an A A -parabolic then a similar situation unfolds, but nowthere are more layers (see, for example, [ Tho16 , Table 22]). The factors include(101 , , / / , M ( A ) is 1 ⊕ ∗ then these two modules for A A restrict to L with a summand1 / /
1. There are also factors (100 , M ( A ) is1 ⊕ ∗ in the restriction of L (0010) to L , hence againthere is a summand 1 / / L -action on (100 , / /
1. Thus the first statement alwaysholds.From the previous table, 40 restricts to L as 6 ⊕ ⊕
20, so 40 ± cannot be asubmodule of L ( E ) ↓ H .Recall that we have chosen to lie in the centralizer of z , which is D T . Since u is unipotent, u ∈ D . Step 2 : If u lies in class 2 A of D then H is strongly imprimitive. Furthermore, u can only lie in one of the following classes of E :2 A , A + A , A + 2 A , A + A , A + A (2)2 , A , D ( a ) + A , D ( a ) + A . The composition factors of M ( D ) ↓ M are (3 , ∗ ) , , and u acts on 3 ± with a singleJordan block. Since the only indecomposable module with a trivial compositionfactor and no trivial submodule or quotient is P (3), which has dimension 16, wesee that M stabilizes a line and hyperplane on M ( D ).As u acts on 3 with a block of size 3, if M ( D ) ↓ M is semisimple then u acts on M ( D ) with blocks 3 , . Note that u acts on 3 / ∗ with blocks 4 ,
2, and on 3 / u can only act on M ( D ) with blocks one of 3 , and 4 , , , and 4 , , (as 3 / ∗ / / ∗ doesnot support a symmetric form: to see this, notice that 4 , is not an acceptableunipotent action on M ( D )). On the other hand, if the trivials are not summandsthen there must be two blocks of size 4, so u must act as either 4 , or 4 , Law09 ] (including Table 8 from there to find the classes of D ), we seethat the corresponding classes of E are exactly those described in the statement.Now suppose that u lies in class 2 A . Since M is contained in the A A -Levisubgroup of D , it is easier to compute the action of M on L ( E ), and find asuitable positive-dimensional subgroup containing M , by placing M inside the A maximal-rank subgroup, acting on L ( E ) as the sum of M ( A ) ⊗ M ( A ) ∗ (minusa trivial summand), Λ ( M ( A )) and its dual. The action of M on M ( A ) is either3 ⊕ ⊕ ⊕ or 3 ⊕ ∗ ⊕ ⊕ (both give the same action on L ( E )) and this actionis a sum of modules of the form: 3; 3 ∗ ; 1; 3 ⊗
3; 3 ⊗ ∗ ; 3 ∗ ⊗ ∗ . (Recall thatΛ ( A ⊕ B ) ∼ = Λ ( A ) ⊕ Λ ( B ) ⊕ ( A ⊗ Λ ( B )) ⊕ (Λ ( A ) ⊗ B ) , and also Λ (3) = 1 and Λ (3) = 3 ∗ .)
78 12. DIFFICULT CASES
Class in D Class in E Act. M ( D ) On 1 /L ( D ) ◦ / HS ( D ) On L ( E )2 A A , , , , , , , A + A A + A , , , , , , , , A + D A + 2 A , , , , , , A + D A + A , , , , , , , , , D + A (2)2 A + A (2)2 , , , , , , , , A + D A , , , D ( a ) + A D ( a ) + A , , , , , , , , D ( a ) + A D ( a ) + A , , , Table 12.1.
Possible unipotent classes for u and their actions onthe 1- and θ -eigenspaces of L ( E )Let Y denote the algebraic A subgroup of A acting as L (10) ⊕ L (10) ⊕ L (0) ⊕ or L (10) ⊕ L (01) ⊕ L (0) ⊕ , depending on the action of M on M ( A ). We see that Y contains M , and the action of Y on L ( E ) is again a sum of copies of L (10), L (01), L (0), L (10) ⊗ L (10), L (10) ⊗ L (01) and L (01) ⊗ L (01). It is well known that L (10) ⊗ L (10) = L (01) /L (20) /L (01) , L (10) ⊗ L (01) ∼ = L (0) ⊕ L (11) , so that Y stabilizes any semisimple kM -submodule of L ( E ).If u lies in class 2 A of D , then u acts on M ( D ) with blocks 3 , , i.e., M ( D ) ↓ M is semisimple. The unipotent class 2 A of D is contained in class 2 A of E by [ Law09 , 4.13], so u acts on L ( E ) with blocks 4 , , . This is thesame as the action of u on the sum of the composition factors of L ( E ) ↓ H , so anyextension between composition factors in that module must split on restriction to h u i . Thus the following extensions are not allowed in L ( E ) ↓ H :10 ∗ / , ∗ / , / , / ∗ , / , / . Thus 40 ± , 24 and 10 ± must all be summands of L ( E ) ↓ H , and the only possibilitycompatible with the previous step is(5 / / ⊕ (5 ∗ / ∗ / ∗ ) ⊕ ⊕ ∗ ⊕ ⊕ ⊕ ⊕ ∗ . As the restriction of 5 and 5 ∗ , and 10 and 10 ∗ , to M are semisimple, they are allstabilized by Y as well, and so therefore H is strongly imprimitive by Theorem 3.2.We also wish to understand the action of u on the eigenspaces of z . Since the1-eigenspace of z has the form 1 /L ( D ) ◦ /
1, and the θ i -eigenspaces are the sumsof M ( D ) and a half-spin module HS ( D ), we are interested in the actions ofunipotent classes of D on M ( D ), 1 /L ( D ) ◦ / HS ( D ), together with theaction on L ( E ) and which class of E contains the unipotent class of D . Theseare given in Table 12.1.From this we see that if either 24 or 10 ⊕ ∗ is a summand of L ( E ) ↓ H , then u (which acts on 24 with blocks 4 , , and on 10 with blocks 3 ,
1) lies either inclass 2 A —and then H is strongly imprimitive—or A + A , so we may assume thelatter case. We also see that 24 ⊕ cannot be a summand of L ( E ) ↓ H if H is notstrongly imprimitive. Moreover, 40 cannot be a summand since u acts on 40 ⊕ ∗ with blocks 4 , , , and so u lies in class 2 A , so H is strongly imprimitive. If40 ± is a summand of L ( E ) ↓ H then 14 is a summand of L ( E ) ↓ L , contradictingStep 1. Finally, if 5 ⊕ ∗ is a summand of L ( E ) ↓ H then the class of w needs tohave at least 20 blocks of size 8 (the projective parts of all composition factors) (2) ≤ E and two blocks of size 5 (from the two summands), together with a block of sizeat least 7 (coming from the 10). The only option for that is class A , acting withblocks 8 , , , , . Step 3 : The { ± } -heart of L ( E ) ↓ H is 40 ⊕ ∗ .We examine the { ± } -heart of L ( E ) ↓ H , aiming to show that it is simply 40 ⊕ ∗ .If it is not, then the socle of it must be (up to automorphism) 40 . We computethe { ± , ± , , ± } -radical of P (40 ), add on top any copies of 40 ∗ that we can,then take the { ∗ } ′ -residual, which yields the self-dual module M with structure40 ∗ / / / / ∗ / ∗ , / . Thus the { ± } -heart of L ( E ) ↓ H is either M (up to graph automorphism of H ) or40 ⊕ ∗ ; suppose it is M . The element w of order 8 in H acts on M with blocks8 , ; thus from Table 2.8 we see that M cannot be a summand of L ( E ) ↓ H asthere is no unipotent class that acts with blocks of size 5 and at least 26 blocks ofsize 8.Consider the quotient by the { ± } ′ -radical N of L ( E ) ↓ H : this cannot be M either, as then M is a quotient, so is a submodule, and the kernel of the maponto M must complement M in L ( E ), contradicting the fact that M is not asummand. One cannot place 10 ± or 5 on top of M , but there is a single non-splitextension with 5 ∗ , yielding a module M ′ with structure40 ∗ / / / ∗ , / ∗ / ∗ , / . If M ′ is the quotient of L ( E ) ↓ H by N then N has composition factors 5 , , ∗ . If10 or 10 ∗ is not a submodule of N then it must have structure 10 / ∗ /
5. Thereforewe need to be able to place 10 underneath M ′ , and then 10 ∗ underneath that, elseone of the 10-dimensional factors must become a quotient, a contradiction.There is a unique extension first placing 10 underneath and then 10 ∗ , yieldinga module M ′′ with structure40 ∗ / / / ∗ , / ∗ / ∗ , , / ∗ , . However, w acts on M ′′ with blocks 8 , , , ,
2. There is no unipotent class inTable 2.8 whose action encapsulates this, so this is not a subquotient of L ( E ) ↓ H by Lemma 2.1. Hence 10 or 10 ∗ is a submodule of L ( E ) ↓ H .We therefore take the { ± } ′ -heart M of L ( E ) ↓ H , which has 5 as a sub-module, and M ′ as a quotient by this submodule. There is a unique non-split suchextension, a self-dual module, and w acts on it with blocks 8 , , , so M cannotbe a summand of L ( E ) ↓ H either. Thus the quotient by the { ± } -radical mustalso have a single copy of 10 ± in it. Both 10 and 10 ∗ have an extension with M ,so we need to examine each.We obtain two modules, with socle structures40 ∗ / , / / ∗ , / ∗ / ∗ , / , and 40 ∗ / / / ∗ , / ∗ / ∗ , ∗ , / , . The former may be excluded because v acts on it with blocks 4 , ,
2, so on L ( E ) with at least 61 blocks of size 4, contrary to Table2.5. To exclude the latter, we first note that w acts on this module with blocks8 , , , ,
3. We then place as many copies of 10 on the bottom of it as we can,
80 12. DIFFICULT CASES which is two: this produces a module with structure40 ∗ / / / ∗ , / ∗ / ∗ , ∗ , / , , , , on which w acts with blocks 8 , , , , ,
2. The action of w on L ( E ) must beencapsulated by this, and encapsulate the previous action by Lemma 2.1, but noclass in Table 2.8 has this property.We have finally concluded that M is not a subquotient of L ( E ) ↓ H , so thatthe { ± } -heart of L ( E ) ↓ H is 40 ⊕ ∗ , as claimed. Step 4 : If 40 ∗ / / or 40 / / ∗ is a subquotient of L ( E ) ↓ H then H is stronglyimprimitive.By applying the graph automorphism if necessary we may assume that N ∼ =40 ∗ / / is a subquotient of L ( E ) ↓ H . There are no extensions of N involv-ing 40 ± , and the { ± } -heart of L ( E ) ↓ H is 40 ⊕ ∗ , so the { ± , ± } ′ -heart N of L ( E ) ↓ H must be isomorphic to(40 ∗ / / ) ⊕ ⊕ ∗ ⊕ . Since the 24 has no extensions with 5 ± or 10 ± , it is a summand of L ( E ) ↓ H , sofrom the end of Step 2, either H is strongly imprimitive or u belongs to class A + A , with θ -eigenspace action 4 , , , . However, the action of u on the θ -eigenspace of N above has blocks 4 , , , , which is not encapsulated bythis. This contradiction completes the proof of the step. Step 5 : The { ± } -heart of L ( E ) ↓ H is semisimple.Let M denote the { ± } -heart of L ( E ) ↓ H , which by applying the graph automor-phism of H if necessary we may assume has socle 40 (if it is not 40 ⊕ ∗ ). Weconstruct the { ± , ± , , ± } -radical of the quotient module P (40 ) / , thenplace on top any copies of 40 ∗ that we can, to obtain a module with structure(12.1) (40 ∗ / ⊕ (5 , ∗ / ∗ , ∗ , ∗ / , , with 5 and 40 ∗ being quotients. From this we see that if the { ± } -heart is notsemisimple, it is (up to graph automorphism of H ) one of40 ∗ / / , ∗ / ∗ , ∗ / , / , ∗ / ∗ , ∗ / , , / , where the last module is not well defined up to isomorphism. The first of these hasalready been eliminated in the previous step.Suppose that we are in the second possibility: by Step 1, 40 ∗ cannot be aquotient of L ( E ) ↓ H , and thus the quotient by the { } ′ -radical must contain thismodule, but be strictly larger. Adding as many copies of 5 ± , 24 and 40 ± on top ofthis module as we can, we obtain the module24 / , ∗ / ∗ , ∗ , ∗ / , , / . The 14 in the restriction of the 40 ∗ to L is still a quotient of this module, so weobtain a contradiction for this case.For the third possibility, M is a module of the form 40 ∗ / ∗ , ∗ / , , / .As before, this cannot be a quotient of L ( E ) ↓ H , but we only have a single copy of24 (which cannot have any extensions with this module, as in the previous step),40 ± and 5 ± that are not already part of M . The { ± , } -heart of L ( E ) ↓ H mustbe the sum of M and 24, and so the { ± } ′ -radical of L ( E ) ↓ H must contain the24. (2) ≤ E While M is not determined uniquely up to isomorphism by the socle structure,the quotient M ′ = M / soc( M ) is. Therefore we may attempt to add copies of 40 ± and 5 ± on top of this module to form a quotient of L ( E ) ↓ H . However, havingdone this we obtain a module5 , ∗ / ∗ , ∗ , , ∗ , ∗ / , , , which still has a 40 ∗ quotient, hence we obtain a contradiction again. Thus the { ± } -heart is 40 ⊕ ∗ , as claimed. Step 6 : The { ± , ± , } ′ -heart of L ( E ) ↓ H is not (40 ∗ / / ) ⊕ (40 ∗ / ∗ / ).Suppose to the contrary that the { ± , ± , } ′ -heart is this module, denoted M .(This is unique up to isomorphism, as we will see in the next step.) We first claimthat there must be a submodule 5 ± in L ( E ) ↓ H . Since 24 has no extensions with5 ± or 10 ± , we may ignore that. Thus assuming the claim is false, 10 or 10 ∗ is asubmodule of L ( E ) ↓ H , so we take the { ± , } ′ -heart of L ( E ) ↓ H , which is themodule M with 5 and 5 ∗ attached in some way. Since Ext kH ( M , ∼ = k andExt kH ( M , ∗ ) = 0, there is a unique way to extend M by 5 and 5 ∗ , namely(40 ∗ / / / ⊕ (5 ∗ / ∗ / ∗ / ) . One can place a copy of 10 ∗ under this module, but it does not go below the 5, sothis remains a submodule. Of course, if the { ± , } ′ -heart is M ⊕ ⊕ ∗ insteadof an indecomposable module then one must have either 5 or 5 ∗ as a submodule,and so the claim holds.Write M for the { ± , } ′ -heart of L ( E ) ↓ H . Either the 10 and 10 ∗ extend M , or M is the sum of M and a module with factors 10 and 10 ∗ .If M ∼ = M ⊕ ⊕ ∗ then we must place a copy of 5 below this to lie underthe 40 , else the 14 is a submodule on restriction to L and we obtain a contradictionby Step 1. There are two such modules, and for each there is a unique self-dualmodule M ′ that can be obtained by placing 5 ∗ on top:(40 ∗ / / / ⊕ (5 ∗ / ∗ / ∗ / ) ⊕ ⊕ ∗ and(40 ∗ / / ∗ , / ⊕ (5 ∗ / ∗ / ∗ / , ) . In both cases the action of u on the 1-eigenspace of M ′ is 4 , , , , and sinceeach 24 contributes 4 to this eigenspace ( M acts on the 1-eigenspace of 24 as 1 ⊕ ⊕ u belongs must have 1-eigenspace encapsulating 4 , , , .This eliminates 2 A , A + A , A + 2 A , 2 A and D ( a ) + A from consideration(see Table 12.1). The remaining classes all act on the θ -eigenspace with exactly 22blocks.In both cases the module M ′ has a 3-dimensional Ext with 24, and the θ -eigenspace of this extended module has u -action 4 , , , , so 29 blocks in total.Since the θ -eigenspace of u on 24 has blocks 3 ,
1, removing two copies of 24 fromthis module can remove at most six blocks, so any extension by a 24 has at least23 blocks. This proves that u cannot belong to the remaining classes. Thus M cannot have this structure.The argument is very similar if M ∼ = M ⊕ (10 ∓ / ± ). The same analysisyields two possibilities for M ′ for each of 10 / ∗ and 10 ∗ / ∗ / / / ⊕ (5 ∗ / ∗ / ∗ / ) ⊕ (10 ∓ / ± ) , and5 ∗ , ∗ / , , ∗ / ∗ , ∗ , / , or 5 ∗ , ∗ / , ∗ / ∗ , ∗ , / , , .
82 12. DIFFICULT CASES
The action on the 1-eigenspace is 4 , , , , as before. Again, there is a 3-dimensional Ext with 24, and the θ -eigenspace is slightly different, with blocks4 , , , , but still 29 blocks, so the conclusion is still the same, that M cannothave this form.Thus M is obtained from M by extending by 10 and 10 ∗ in some way, ratherthan a direct sum. We have that Ext kH ( M ,
10) = 0, so if M is extended by 10 and10 ∗ , it must be 10 ∗ on the bottom and 10 on the top. There is a unique extensioninvolving 10 ∗ , and under this we must place 5, else the 40 becomes a submoduleof L ( E ) ↓ H modulo the 24, contrary to Step 1.There is a 2-dimensional Ext with 5, yielding the module(40 ∗ / / / ⊕ (40 ∗ / ∗ , ∗ / , ) . There are, up to isomorphism, three possible quotients of this by 5, depending onwhether the 5 is in the first summand, the second, or diagonal. We cannot removethe one under the first summand as that contradicts Step 1, as we have seen before.Write M ′ for the extension by a single 5 that lies in L ( E ) ↓ H , and supposethat the 5 in M ′ lies entirely below the 40 . In particular, 10 ∗ is a submodule of L ( E ) ↓ H , so 10 is a quotient and we can add 5 ∗ on top of M ′ . There is a uniquesuch module, (40 ∗ / / / ⊕ (5 ∗ / ∗ / ∗ / ∗ , ) , but it is possible to place two copies of 10 on top of this module, one on top ofthe 40 and one on top of the 10 ∗ . Since they lie in different summands, we againobtain three (rather than infinitely many) non-isomorphic extensions by 10, oneof which is clearly not self-dual. Write M ′′ for whichever of these extensions is asubquotient of L ( E ) ↓ H .For both possibilities for M ′′ , the element u acts on the 1-eigenspace of M ′′ withblocks 4 , , , . The two 24s contribute four blocks of size 4 to this, to yield4 , , , . The only unipotent classes to encapsulate this action are A + A (2)2 and D ( a ) + A , both with action on the 1-eigenspace of L ( E ) having blocks4 , , , . In particular, any extension of M ′′ involving a 24 cannot produceany new blocks of size 4.The dimension of Ext kH ( M ′′ ,
24) is 3, but in this full extension, u acts on the1-eigenspace with 21 blocks of size 4. Therefore there is a unique 2-dimensionalsubspace of this space consisting of extensions that do not produce any new blocksof size 4. The action of u on the θ -eigenspace of this module is 4 , , , , whichis 26 blocks. Removing one copy of 24 can remove at most three blocks, so u actswith at least 23 blocks on any such quotient. But u acts with only 22 blocks on L ( E ) if u comes from one of the two remaining classes, a clear contradiction. Thuswe obtain a contradiction to either structure for M ′′ , and hence for the particularstructure of M ′ .Thus the 5 we added on to make M ′ must be diagonal. We add 5 ∗ /
10 on top of M ′ and 24 as well, and the 1-eigenspaces of these modules have five blocks of size4 in them. Together with the 1-eigenspace of M ′ , which is 4 , , , , this meansthat we must encapsulate 4 , , , , so again we are forced into u lying in theclass A + A (2)2 or D ( a ) + A . These two classes act on the 1- and θ -eigenspaceswith blocks 4 , , , and 4 , , respectively. (2) ≤ E We go back a step now, and consider the { ± , ± } ′ -heart M of L ( E ) ↓ H .This is obtained from M by adding a copy of 24 above and below M . The spaceExt kH (24 , M ) is 3-dimensional, with the full extension given by(40 ∗ / , / ) ⊕ (24 / ∗ / ∗ , / ) , but the space Ext kH (24 , M ′ ) is only 2-dimensional. The missing copy of 24 is theone in the fourth socle layer, so M cannot have that copy in it either.There are, up to isomorphism, three modules that are extensions with quotient24 and submodule M , and these are(40 ∗ / , / ) ⊕ (40 ∗ / ∗ / ) , (40 ∗ / / ) ⊕ (40 ∗ / ∗ , / ) , ∗ , ∗ / , ∗ , / , . Each of these has a 2-dimensional Ext -space with 24: in the first two cases, thereis a unique self-dual module that is an extension with submodule 24, which mustbe M , whereas in the third case all but two possible extensions (of the ‘ | k | + 1’many) are self-dual, so there are many possibilities for M . However, the action of u on the 1- and θ -eigenspaces of these are the same.Note that we obtain L ( E ) ↓ H from M by adding 10 ∗ / ∗ /
10 on the top; u acts on the 1-eigenspace of each of these with blocks 4 , , M has structure(40 ∗ / , / ) ⊕ (40 ∗ / ∗ / , ) , and there is a unique extension with submodule 10 ∗ , it lying in the socle of thesecond summand. The θ -eigenspace of this larger module has u -action 4 , , , ,and the remaining modules 5, 5 ∗ and 10 contribute 1, 1 and 3 , u cannot act on the θ -eigenspace of L ( E ) withmore than sixteen blocks of size 4, contradicting u lying in one of the classes above.In the second possibility, M has structure(40 ∗ / / , ) ⊕ (40 ∗ / ∗ , / ) , and u acts on the 1-eigenspace of this with blocks 4 , , . Adding in the twoblocks of size 4 from the rest of L ( E ) ↓ H , we obtain 4 , , , which is not en-capsulated by the action of u on the 1-eigenspace of L ( E ) ↓ H , so this case cannotoccur either.Thus the third case occurs: the action of u on the 1-eigenspace of the moduleabove has blocks 4 , , , . On this we add 5 ∗ /
10, 10 ∗ / u acts on L ( E ) with blocks encapsulating 4 , , , , but this has 21 blocks of size at least3, whereas u must act with twenty blocks of size at least 3. This contradictionproves that M ′ cannot be either possibility, and this completes the proof of thestep. Step 7 : The { ± , ± , } ′ -heart of L ( E ) ↓ H is 40 ⊕ ∗ ⊕ ⊕ ∗ .If both 40 i and 40 ∗ i appear in the socle of the { ± , ± , } ′ -heart M of L ( E ) ↓ H then both 40 ± i are summands of M , hence M is the sum of the 40 ± i and the { ± − i } -heart, which is 40 − i ⊕ ∗ − i . In particular, this means that M is semisim-ple or (up to graph automorphism) the socle of M is either 40 ⊕ or 40 ⊕ ∗ . Ifthe socle is 40 ⊕ ∗ then we construct the { ± , ± , , ∗ , } -radical of P (40 ∗ ),which is 5 , / ∗ .
84 12. DIFFICULT CASES
Hence the 40 ∗ splits as a summand of M , and therefore 40 must also lie in thesocle, which is a contradiction. Thus if our conclusion does not hold then the socleof M is 40 ⊕ , so we assume that from now on. The { ± , ± , } -radicals of P (40 i ), then with any copies of 40 ∗ and 40 ∗ placed on top, then with any modulesnot 40 ∗ or 40 ∗ removed from the top, are40 ∗ / ∗ , ∗ , ∗ , ∗ / , , / , ∗ / ∗ / . (The former of these can be seen from the radical in (12.1).) A pyx for M is thesum of these two modules. Notice that M must contain the 40 ∗ , and hence containthe 5 in the second socle layer.Suppose that 40 ∗ lies in the fourth socle layer of M : then M must have thesocle structure 40 ∗ / ∗ , ∗ , ∗ / , , (24) / , , where the 24 might or might not be present. Although this module is not uniqueup to isomorphism, the submodule M ′ whose quotient is 40 ∗ / ∗ is unique up toisomorphism, and is (10 ∗ , ∗ / , , (24) / ) ⊕ . Since M has 40 as a submodule, and this cannot be a submodule of L ( E ) ↓ H byStep 1, we must be able to place 5 or 24 below this. (The only modules in L ( E ) ↓ H but not in M are 5, 5 ∗ , and one or two copies of 24.) The 24s are irrelevant though,since upon restriction to L there is still a submodule 14. Hence we must be able toplace a copy of 5 below M ′ in such a way as to stop the 14 being a submodule ofthe restriction to L .The space Ext kH ( M ′ ,
5) is 1-dimensional, but the unique non-split extensionhas structure (10 ∗ , ∗ / , , (24) / , ) ⊕ . This proves that M cannot have this form, and in particular the 40 ∗ lies in thethird socle layer of M .Thus M is a submodule of(40 ∗ , ∗ / , / ) ⊕ (40 ∗ / ∗ / ) . The 40 ∗ cannot lie entirely on the second summand by Step 6 or on the first byself-duality, and since the two summands have no composition factor in common,the diagonal submodule is unique up to isomorphism. The action of u on the θ -eigenspace is 4 , , . Adding the copy of 24, which must be a summand of L ( E ) ↓ H and on whose θ -eigenspace u acts with blocks 3 ,
1, we obtain a u -actionwith blocks 4 , , , . Since 24 is a summand, from the end of Step 2 we see that u must belong to class 2 A or A + A , acting with blocks 4 , , or 4 , , , respectively. Neither of these actions encapsulates the block structure above. Thisis a contradiction, and so M cannot have this form either. This completes theproof of the step. Step 8 : The { } -heart of L ( E ) ↓ H is 24 ⊕ .Let M denote the { } -heart of L ( E ) ↓ H , and suppose that M is not 24 ⊕ . Thus M must be a module that has 24 in the socle and top, and some 40 ± i in the heart.There is exactly one copy of 24 that may be placed on top of 40 , ∗ , , ∗ / / , ∗ , , ∗ / . (2) ≤ E Thus we assume from now on that this module is M . The restriction of M to L has a submodule 14 ⊕ , so we need two copies of 5 ± below M in order to complywith Step 1.If 10 ± is a submodule of L ( E ) ↓ H , then we add a copy of 5 and 5 ∗ above andbelow M to form the { ± } ′ -heart M , due to the conclusion of Step 1. There isa unique such module, 5 , ∗ , / , ∗ , , ∗ / , ∗ , , and u acts on this module with blocks 4 , , . From Table 12.1, the only blockstructure that encapsulates this is 4 , , , , and this comes from A + A . Weonly have 10 and 10 ∗ to add on, and M cannot be a summand of L ( E ) ↓ H fromcomparison of the u -actions, so 10 ± must attach to M to form L ( E ) ↓ H .The action of u on the 1-eigenspaces of M and L ( E ) are 4 , , and4 , , , respectively, and u acts on 10 with blocks 3 ,
1, and 1-eigenspace3 ,
1. We see that the number of blocks of sizes 3 and 4 remain the same, so there isan extension of 1 by 3 that is encapsulated by 2 , , which is clearly impossible.Thus we were incorrect in assuming that 10 ± is a submodule, and hence itcannot be. Write M for the { ± } ′ -heart of L ( E ) ↓ H , which is obtained from M byadding a submodule 10 and a quotient 10 ∗ . (We may choose this as M is invariantunder the graph automorphism.) If the 10 and 10 ∗ form a separate summand wehave the possibilities that M ∼ = M ⊕ ⊕ ∗ and M ∼ = M ⊕ (10 ∗ / and 40 ∗ are submodules of M (ignoring the 24) we need both 5 and 5 ∗ as submodules (and also quotients) of L ( E ) ↓ H , by Step 1.In the first case the actions of u on the 1- and θ -eigenspaces of M are 4 , , and 4 , respectively, and we see that the only unipotent class that encapsulatesboth is D ( a ) + A , so u acts on the θ -eigenspace with blocks 4 , . Comparingthis and the θ -eigenspace of M , we see that no extension involving M and the 5 ± can split on restriction to the θ -eigenspace of h u i .Note that Ext kH (5 , M ) is 2-dimensional, and u acts on the θ -eigenspace of thisextension with blocks 4 , , . But this means any such extension splits uponrestriction, contrary to what is needed.In the second case the actions of u on the 1- and θ -eigenspaces of M are4 , , and 4 , , A + A , 2 A and D ( a ) + A . We have that Ext kH (5 , M )is again 2-dimensional, and the action of u on the 1-eigenspace of the extension of M by 5 ⊕ is 4 , , , , so any extension of M with submodule 5 has 25 blocksin the action of u on its 1-eigenspace. This eliminates 2 A and D ( a ) + A , leavingjust A + A . Thus u acts on the 1-eigenspace of L ( E ) as 4 , , , , whereasthe action of u on 1-eigenspace of M has blocks 4 , , .There are two ways to derive a contradiction: the first is to note that theremust be a unique extension that does not have an extra block of size 4, and that isthe one where the 5 is placed on top of M . Alternatively, recall that u acts on the1-eigenspace of 5 with a single block of size 3. There is no way to add four blocksof size 3 to the action for M to produce the action for L ( E ).Thus the 10 and 10 ∗ in M are attached to M , and we assume that 10 is asubmodule of M by applying the graph automorphism. There is a unique such
86 12. DIFFICULT CASES non-split extension M ′ , with structure24 / , ∗ , , ∗ / , . One may place two copies of 10 ∗ on top of this, one into the second socle layerforming a submodule 10 ∗ /
10, and one in the third socle layer above 40 ∗ . Themodule M is a potentially diagonal submodule of these two copies of 10 ∗ (definitelynot the one solely on top of the 10, as this would not be self-dual), and we willcompute the action of u on its 1- and θ -eigenspaces.The action of M on the 1-eigenspace of M ′ is8 ⊕ ⊕ (1 / , ∗ / ⊕ ⊕ (3 / ∗ / ⊕ (3 ∗ / / ∗ ) ⊕ (3 ∗ / ⊕ ⊕ , and on the θ -eigenspace is8 ⊕ ⊕ P (3) ⊕ P (3 ∗ ) ⊕ (1 / , ∗ / ⊕ (3 / ∗ ) ⊕ ⊕ ∗ ⊕ ∗ . The 1-eigenspace of 10 is 1 ⊕ ∗ and the θ -eigenspace is 3: the extension by 10 ∗ contributing the 10 ∗ /
10 submodule splits on restriction to M , so we obtain sum-mands 1 ⊕ ∗ in the first and 3 in the second. The other copy of 10 ∗ places the 3 ∗ on top of the summand 3 with the 1 splitting off in the 1-eigenspace, and placesthe 3 on top of a summand 3 ∗ in the θ -eigenspace.Thus any extension other than the one contributing the submodule 10 ∗ / ⊕ (3 ∗ /
3) in the 1-eigenspace and 3 / ∗ in the θ -eigenspace.The actions of u on the 1- and θ -eigenspaces of M ′ are 4 , , , and 4 , , θ -eigenspaces of M the actions must be 4 , , and 4 , , respectively.Looking at Table 12.1, the 1-eigenspace action is not encapsulated by 2 A or A + A , and the θ -eigenspace is not encapsulated by A + A or A + 2 A . Wemust eliminate the remaining four.To eliminate A + A (2)2 and D ( a ) + A , which have the same action on the1-eigenspace, namely 4 , , , , note that we must add four blocks of size 3(one from each 5 ± ) to 4 , , , and this is not possible. To see this, noticethat the socle of the extended module is the sum of the socles of the submoduleand quotient, so this may be removed without loss of generality, and now we areadding four blocks of size 2 and we need to increase the number of blocks by six,which is impossible. Thus u lies in class 2 A or D ( a ) + A , each with exactly 24blocks in the 1-eigenspace of L ( E ), the same as u has on M , and 20 blocks in the θ -eigenspace.In order for 10 not to be a submodule of L ( E ) ↓ H , there must be a submodule5 ∗ . If both copies of 5 ∗ are submodules then this module is determined uniquely, be-cause Ext kH ( M ′ , ∗ ) is 2-dimensional (and because neither of the remaining classesfor u is compatible with a summand 5 ± ). This module, which has the form24 / / , ∗ , , ∗ / ∗ , ∗ , , has θ -eigenspace 4 , , ,
1, with 21 blocks. This cannot be encapsulated by thetwo remaining classes for u , which have 20 blocks. Thus there must be a 5 ∗ in boththe socle and top.The dimension of Ext kH (5 ∗ , M ′ ) is 2, so we may place two copies of both 10 ∗ and 5 ∗ on top of M ′ . The action of u on the 1-eigenspaces of the module5 ∗ , ∗ , / ∗ , ∗ , , ∗ , , ∗ / , (2) ≤ E is 4 , , , . Removing a copy of 5 ∗ ⊕ ∗ removes 3 , u onthe 1-eigenspace, i.e., at most three blocks. Thus there are at least 25 blocks in theaction of u on the 1-eigenspace of L ( E ). But the remaining two classes have 24blocks.This is a final contradiction, so 10 ± is neither a submodule nor not a submod-ule, and thus M cannot have the structure assumed. Thus M is semisimple, asclaimed. Step 9 : There is a single 5-dimensional composition factor of soc( L ( E ) ↓ H ).Since the { ± } -pressure of L ( E ) ↓ H is 2 and there is a subquotient of { ± } -pressure4, namely the sum of the 40 ± i , the subgroup H stabilizes at least one 5-space on L ( E ) by Proposition 2.5. Thus assume that there are two or more.Let M denote the { ± , ± } ′ -heart of L ( E ) ↓ H , and let M denote the { ± } ′ -heart of L ( E ) ↓ H . We know that M has a subquotient 40 ⊕ ∗ ⊕ ⊕ ∗ , withtwo copies of 24 attached somehow. We also know from Step 2 that they do notsplit off as summands of M , for then they would be summands of L ( E ) ↓ H .Thus there is a single submodule 24 of M , and some of the 40 ± i are submodules.From Step 8, the { } -heart is 24 ⊕ , and so any 40 ± i that is not a submodule is aquotient, and therefore there is a submodule M (1)1 consisting of the submodule 24and all 40 ± i that are not submodules of M . The dual of M (1)1 must be a quotientof M , and has a 24 quotient and no other composition factors in common. We seethat M must in fact be the direct sum of M (1)1 and its dual M (2)1 , and possiblysome summands 40 ± i . This leads, up to graph automorphism, to four possibilities: M (1)1 contains 40 , 40 , 40 ⊕ , and 40 ∗ ⊕ .Suppose that M is (40 / ⊕ (24 / ∗ ) ⊕ ⊕ ∗ ;the action of u on the θ -eigenspace of this module has blocks 4 , , , . FromTable 12.1, we see that no class can encapsulate this, and so this cannot be thestructure of M .The other three possibilities for M are40 ⊕ ∗ ⊕ (40 / ⊕ (24 / ∗ ) , (40 , / ⊕ (24 / ∗ , ∗ ) , (40 ∗ , / ⊕ (24 / , ∗ );in each of these cases, u acts on the 1-eigenspace of M with blocks 4 , , andon the θ -eigenspace with blocks 4 , .In each case, w acts on M with blocks 8 , , , so 5 cannot be a summandof L ( E ) ↓ H by the remark at the end of Step 2. In particular, this means that L ( E ) ↓ H has exactly two 5-dimensional factors in the socle, so that M containsno copy of 5 ± .If M is the sum of M and 10 ⊕ ∗ then in all three possibilities above u actson the θ -eigenspace of M with blocks 4 , . In each possibility Ext kH ( M ,
5) is2-dimensional. Placing both copies of 5 on the bottom of M yields a module whose θ -eigenspace has u -action 4 , , , so any extension of M with submodule 5 hasblocks 4 , ,
1. There is no unipotent class that encapsulates this action, as seenin Table 12.1, so we can exclude this structure for M .If M is the sum of M and 10 ∗ /
10, or the sum of M and 10 / ∗ , then thereare six possibilities for M , and for each of them the proof is the same. Notice
88 12. DIFFICULT CASES that 5 and 5 ∗ must both be submodules of L ( E ) ↓ H as we need to place a copyof 5 below 40 and a copy of 5 ∗ below 40 ∗ , because of Step 1. The action of u onthe 1-eigenspace of M is 4 , , and on the θ -eigenspace is 4 , ,
2. The spaceExt kH ( M ,
5) is 2-dimensional and u acts on the θ -eigenspace of this extensionwith blocks 4 , , , , hence on any extension of M with submodule 5 withblocks 4 , , ,
1, so 21 blocks in total. Examining Table 12.1, the only class thatencapsulates these is A + A , acting on the 1-eigenspace of L ( E ) with blocks4 , , , . In particular, when we add the 5 onto M , we may make no moreblocks of size 4 in the 1-eigenspace of u . This restriction picks out a unique extensionof M with submodule 5, and then a unique extension inside of the 2-dimensionalspace of extensions obtained by placing 5 ∗ on top of the resulting module. In eachcase, this results in one of the following modules, depending on M :(40 / ⊕ (5 ∗ / ∗ ) ⊕ (24 / ) ⊕ (40 ∗ / ⊕ (10 ± / ∓ ) , (40 , / , ⊕ (5 ∗ , / ∗ , ∗ ) ⊕ (10 ± / ∓ ) , (40 ∗ , / ∗ , ⊕ (5 , / , ∗ ) ⊕ (10 ± / ∓ ) . The action of u on the θ -eigenspace of each of these modules has blocks 4 , , , .We now must place a copy of 5 ∗ below these modules; in each case there arethree potential copies, and placing all of them below we obtain a u -action on the θ -eigenspace of 4 , , , , so any extension by 5 ∗ must have action 4 , , , ,which is not encapsulated by the class A + A . This contradiction proves that M cannot have this form.Thus M is obtained from M via a non-split extension involving 10 and 10 ∗ .In all three possibilities Ext kH ( M , ± ) is 1-dimensional, as is Ext kH ( M , kH ( M , ∗ ) is 2-dimensional.First we assume that there is a submodule 10 ± in L ( E ) ↓ H , so that the { ± , ± } -radical of L ( E ) ↓ H is semisimple. It is either 5 ⊕ ∗ ⊕ ± or 5 ∗ ⊕ ∗ ⊕ ± .Since there are four possibilities for this radical, and three for M , there aretwelve possibilities in total, but the same proof works for all of these. SinceExt kH ( M , ± ), Ext kH ( M ,
5) and Ext kH ( M , ∗ ) are 1-, 1- and 2-dimensional re-spectively, we may form an extension M ′ of M with submodule 10 ± ⊕ ⊕ ∗ ⊕ ∗ ,and the { ± , ± } -residual M of L ( E ) ↓ H is the quotient of M ′ by a single 5 or5 ∗ . The twelve possibilities for M become six for M ′ .In all six possibilities, u acts on the 1-eigenspace of M ′ with blocks 4 , , , and on the θ -eigenspace with blocks 4 , , , . If we remove a single 5 or 5 ∗ fromthe socle, we remove a block of size 3 from the 1-eigenspace and a block of size1 from the θ -eigenspace. Thus M has at least 21 blocks in its θ -eigenspace, andat least twenty blocks of size at least 3 in its 1-eigenspace: the only classes thatencapsulate these actions of u are A + A , A + A (2)2 and D ( a ) + A , whence u acts on the 1-eigenspace of L ( E ) with blocks 4 , , , or 4 , , , .To obtain a conclusion, note that u acts on the 1-eigenspace of the extension of M by 10 with blocks 4 , , , (already eliminating the first possible action), on10 ⊕ , , , , and on the 1-eigenspace of the extension of M by10 ⊕ ∗ ⊕ ∗ with blocks 4 , , , , so in fact the action of u on the 1-eigenspaceof M has blocks 4 , , , . This is 23 blocks in total, five short of the number ofblocks needed for L ( E ). However, the 1-eigenspace of the quotient L ( E ) ↓ H /M (which is the sum of 10 ∓ and two copies of 5 or 5 ∗ ) has u -action 3 ,
1, which is only (2) ≤ E four blocks in total. This contradiction means that the { ± , ± } -radical cannotbe semisimple, i.e., 10 ± is not a submodule of L ( E ) ↓ H .Hence, the { ± , ± } -radical must contain a summand 10 ± / ∓ , with the re-maining summand being either 5 or 5 ∗ . Since u always acts on the 1-eigenspaceof M with blocks 4 , , and on the 1-eigenspace of 10 ± / ∓ with blocks 4 , , u must have 1-eigenspace encapsulating 4 , , . Since wehave 5 and 5 ∗ left to add, one must be able to add two more blocks of size 3 tothis, so there must be either more than eighteen blocks of size 4 or more than twoblocks of size 3. This leaves only 2 A and D ( a ) + A .We have to be careful now because the actions almost work. If the radicalis 5 ⊕ (10 ∗ / M , add the single copy of 10 ∗ below, add the twocopies of 5 below, add the two copies of 10 above, and then the three copies of 5 ∗ above that. (These are the full extensions in all cases.) This constructs a module M that is a pyx for L ( E ) ↓ H , in fact an extension of L ( E ) ↓ H with submodule5 ∗ / u acts on the 1- and θ -eigenspaces of M with blocks 4 , , , , u on L ( E ) if u lies in class D ( a ) + A , and the latter does not encapsulate the θ -eigenspace of u on L ( E ) if u lies in class 2 A , so we obtain a contradiction.Suppose that the { ± , ± } -radical of L ( E ) ↓ H is 5 ∗ ⊕ (10 / ∗ ). If M is thethird possibility then one cannot place 10 below M and 5 ∗ below the 10, so wemay exclude this case. In the other two possibilities, we place 10 below M , thenthree copies of 5 ∗ below that, then two copies of 10 ∗ above M , then four copies of5 above that, to create a pyx M for L ( E ) ↓ H . To obtain L ( E ) ↓ H from M , wemust remove a copy of 5 ∗ from the bottom of it and 5 ⊕ (5 / ∗ ) from the top ofit. The action of u on the 1-eigenspace of M has blocks 4 , , ,
1, and on the θ -eigenspace it has blocks 4 , , , which does not encapsulate the class 2 A , so u must lie in D ( a ) + A . This class acts on the 1-eigenspace of L ( E ) with blocks4 , . Note that removing the quotient 5 / ∗ from M must remove a block of size4 from the action of u on the 1-eigenspace, and that leaves u acting with exactly24 blocks of size at least 3. Thus the 5 that is removed from the top, and the 5 ∗ that is removed from the bottom of M , cannot alter the number of blocks of sizeat least 3 in this action. However, removing all three copies of 5 from the bottomof M removes three blocks of size at least 3 from the action of u , so removing onemust always remove one. Thus two copies of 5 ∗ cannot lie in the socle.The third case is where the radical is 5 ⊕ (10 / ∗ ). In this case Ext kH ( M ,
5) is1-dimensional, so we may place a single 5 below, and a single 5 ∗ above, M , to makea new module M , which is a subquotient of L ( E ) ↓ H . The presence of the 5 ∗ inthe top of M means that Ext kH ( M ,
10) has dimension 2 rather than 1, so we addboth copies of 10 to the bottom of M , then three copies of 5 ∗ . One of the copies of10 remains a submodule of this extension (for the last possibility of M both do, sowe already obtain a contradiction), so we may remove it as we can identify which10 belongs in L ( E ) ↓ H . To obtain a subquotient of L ( E ) ↓ H from this module,we must remove two copies of 5 ∗ , so two blocks of size 2 from the θ -eigenspace.The θ -eigenspace of this module has blocks 4 , , , , so the θ -eigenspace of thesubquotient has at least 21 blocks in it. However, the θ -eigenspace of u on L ( E )must have 20 blocks, since we have shown that u lies in class 2 A or D ( a ) + A ,
90 12. DIFFICULT CASES both of which act with 20 blocks. This contradiction eliminates this possibility forthe radical as well.We are left with 5 ∗ ⊕ (10 ∗ /
5) as the socle. We follow exactly the same strategy.We write out the previous paragraph with altered text in italics.In this case Ext kH ( M , ∗ ) is now two copiesof ∗ below, and two copies of M , to make a new module M , which is not quite a subquotient of L ( E ) ↓ H . The presence of the two copies of M means that Ext kH ( M , ∗ ) now has dimension 3 rather than 1, so weadd all three copies of 10 ∗ to the bottom of M , then two copies of 5. Two of thecopies of 10 ∗ remain a submodule of this extension ( including in the third case ), sowe may remove them as we can identify which 10 ∗ belongs in L ( E ) ↓ H . To obtaina subquotient of L ( E ) ↓ H from this module, we must remove one copy of ∗ andtwo copies of
5, so three blocks of size 2 from the θ -eigenspace. The θ -eigenspaceof this module now has blocks 4 , , , , so the θ -eigenspace of the subquotienthas at least 21 blocks in it. However, the θ -eigenspace of u on L ( E ) must have 20blocks, since we have shown that u lies in class 2 A or D ( a ) + A , both of whichact with 20 blocks. This contradiction eliminates this possibility for the radical aswell.Having eliminated all four possible radicals, the proof of the step is complete. Step 10 : The final contradiction.From the previous steps, inside L ( E ) ↓ H we have a submodule 10 ± , a single 5 or 5 ∗ in the socle, and two 5-dimensional factors in the socle modulo the 10 ± . We alsohave three possible modules M , the { ± , ± } ′ -heart of L ( E ) ↓ H . Since 10 ± is asubmodule, the { ± } ′ -heart of L ( E ) ↓ H must have both 5 and 5 ∗ as a submodule,as we have seen in previous steps. In all cases, Ext kH ( M ,
5) is 1-dimensional, aswe saw in Step 9, and Ext kH ( M , ∗ ) is 2-dimensional. Construct the module M ′′ in all three cases by extending by 5 ⊕ ∗ ⊕ ∗ .By Step 9, there can be only one 5 ± in the socle of L ( E ) ↓ H , but there mustbe two 5-dimensional factors below the 40 ± to comply with Step 1. Thus the 10 ± that is a submodule must cover one of the three copies of 5 ± we have placed under M . However, in all cases Ext kH ( M , ± ) = Ext kH ( M ′′ , ± ), so no copy of 10 ± can be placed underneath any of the 5 ± .This final contradiction means that u does indeed lie in class 2 A , the actionof H on L ( E ) is as described in Step 2 (this can be found diagonally in A A ) and H is strongly imprimitive, as needed. Thus we have the following. Proposition . Let H ∼ = PSL (2) . If H is a subgroup of G ∼ = E incharacteristic , then H is strongly imprimitive. PSL (5) ≤ E The case in Proposition 8.1 left over for q = 5 is where the composition factorson L ( E ) ↓ H are 39 , ∗ , , ∗ , , ∗ , , ∗ , , , ∗ , , ∗ . (5) ≤ E The proof of this will take some time (although it is easier than the last one!), sowe proceed in steps. We start with a few radicals: we take the cf( L ( E ) ↓ H ) \ { V ± } -radicals of P ( V ), for V = 3 , , , , ∗ / ∗ , , , ∗ / ∗ , , ∗ / , / ∗ , , / , ∗ / , / , ∗ / ∗ , , / , ∗ , , / , ∗ / , / , ∗ , ∗ / , ∗ , ∗ , , ∗ / . We will now make a few specific observations that will be useful in the balance ofthe proof.
Step 1 : The socle of L ( E ) ↓ H contains a module not of dimension 3, 15 or 18.Assume this is not the case, and note that if we have two factors of soc( L ( E ) ↓ H )of the same dimension then they are both summands. We remove any simplesummands from L ( E ) ↓ H to make a module W , whose socle has at most one factoreach of dimension 3, 15 and 18. From these radicals we see that the socle of W cannot be simple, as there are not enough 6-dimensional factors in the radicalsabove.If 3 ± lies in the socle then the 35 ∓ in the top of the radical in (12.2) cannotoccur in any pyx for W . Thus in order to have both 35 and 35 ∗ in W , the socleof W contains 15 ⊕ ∗ or its dual. Note that the third radical above is obtainedfrom the second radical by taking duals and then the graph automorphism.If W has socle 15 ⊕ ∗ then W , and indeed L ( E ) ↓ H , is the sum of the secondradical and its dual. However, an element u from the smallest class of elements of H of order 5 acts on this module with Jordan blocks 5 , , , which does notappear in [ Law95 , Table 9] (see also Table 2.6), so the socle of W cannot have atmost two composition factors.Thus the socle of W contains three factors, one each of dimensions 3, 15 and 18.We consider the { ± , , ± , ± } -radicals of the P ( V ) for V = 3 , ,
18, yielding8 / , ∗ / , , / , ∗ / , / , / ± , 15 ± and 18 ± (except for V itself) on top of thesemodules, yielding8 , ∗ / ∗ , , ∗ / , ∗ , , / , ∗ / , / ∗ , , / ± , 15 ± and 18 ± ,to obtain the modules 15 ∗ / ∗ , / , ∗ / ∗ / , ∗ / . As there is no 35 ± in these, we obtain a final contradiction. (For the second moduleone may also add on a copy of 18, but this will extend the 3 ∗ that we added on,and so the 3 ∗ would not be in the top.) Step 2 : The centralizer of an element of order 4 in H .Let z be an element of order 4 in H with centralizer L ∼ = GL (5). The element hastrace 0 on L ( E ), so lies in the class with centralizer X = D D ≤ D (see [ Fre98 ,Table 1.16] or Table 2.10). The action of D D on M ( D ) is as M ( D ) ⊕ M ( D ),where M ( D ) is 6-dimensional.The eigenspaces of z on L ( E ) are modules for X , and it is easy to see that the 1-eigenspace of z is Λ ( M ( D )) ⊕ Λ ( M ( D )), the ( − M ( D ) ⊗ M ( D ),
92 12. DIFFICULT CASES and the ( ± i)-eigenspaces are the tensor products of half-spin modules for D and D (up to duality, and we will see that this doesn’t matter in our case).Since L ≤ C G ( z ) = X , we see that each of these eigenspaces is a module for L . The simple modules for L that have z acting as ± + , 1 − , 3 + , 3 − , 5 + and 5 − , so that P (1 + ) = 1 + / + / + , P (3 + ) = 3 + / + , − / + . (There is such a labelling for both sets of modules.) Similarly, the simple modulesfor L that have z acting as ± i have dimensions 2 and 4, two of each dimension, andwill be labelled 2 + , 2 − , 4 + and 4 − , so that P (4 + ) = 4 + / + / + , P (2 + ) = 2 + / + , − / + . (Again, there is such a labelling.) The choice of which is + and which is − isirrelevant as long as it is consistent, and we choose our labellings so that the L -action on the simple modules for H is as follows.Module 1-eigenspace ( − − i-eigenspace3 - 1 + + -3 ∗ - 1 − - 2 + − − + -6 ∗ − + - 2 + + ⊕ − - 2 − − + + ⊕ − + ⊕ − − ∗ + − ⊕ + − + ⊕ −
18 3 + − − + ⊕ + ∗ + + + ⊕ + −
35 3 − ⊕ − − ⊕ + ⊕ − − ⊕ + P (4 + ) ⊕ + ∗ − ⊕ − + ⊕ − ⊕ + P (4 + ) ⊕ + − ⊕ +
39 3 + ⊕ − ⊕ + − ⊕ − P (4 − ) 2 − ⊕ + ⊕ − ∗ + ⊕ − ⊕ + + ⊕ + − ⊕ + ⊕ − P (4 − )We now determine the possible embeddings of L ′ into D and D : this yieldsactions on M ( D ), M ( D ) and the corresponding half-spin modules, and these canbe used to produce the actions of L ′ on each eigenspace of z .The composition factors of L ′ on L ( E ) are5 , , , , , and this yields two possible sets of composition factors for M ( D ) ↓ L ′ , which are5 , , and 3 ,
1. Splitting these among M ( D ) and M ( D ), we find two candidatesfor these that have the correct composition factors on the 1-eigenspace of z , namely5 , , . These are in the following table, including the composition factors onthe half-spin modules, which are determined uniquely by the traces of elements onthe minimal modules.Case M ( D ) M ( D ) L (010) L (00010)1 5 , , , , , , , Step 3 : Case 1 with M ( D ) not semisimple.Suppose that we are in Case 1, and M ( D ) is not semisimple. This means that L ′ acts on M ( D ) as 5 ⊕ M ( D ) as 5 ⊕ P (1), and on L (00010) as 5 ⊕ ⊕ (3 / (5) ≤ E The actions of L ′ on the eigenspaces of z are as follows: on the 1-, ( − ± i)-eigenspaces L ′ acts as P (3) ⊕ ⊕ ⊕ , P (1) ⊕ ⊕ P (3) ⊕ ⊕ ⊕ , P (4) ⊕ ⊕ P (2) ⊕ ⊕ (2 / L (thathave the correct composition factors), namely P (3 + ) ⊕ ⊕ P (3 − ) ⊕ ⊕ (5 + ⊕ − ) ⊕ , and P (1 + ) ⊕ P (1 − ) ⊕ P (3 + ) ⊕ P (3 − ) ⊕ (5 + ⊕ − ) ⊕ . The action of L on the ( ± i)-eigenspace has composition factors (4 + , − ) , (2 + , − ) ,and there are two possible extensions to L : P (4 + ) ⊕ ⊕ P (4 − ) ⊕ ⊕ P (2 + ) ⊕ P (2 − ) ⊕ (2 + / − ) ,P (4 + ) ⊕ ⊕ P (4 − ) ⊕ ⊕ P (2 + ) ⊕ P (2 − ) ⊕ (2 − / + ) . (Of course, the i-eigenspace is the dual of the ( − i)-eigenspace.)From this we first see that 6, 6 ∗ and 8 cannot lie in the socle of L ( E ) ↓ H , andsecond that there are no simple summands of L ( E ) ↓ H . By Step 1, we know thatsomething other than 3, 15 and 18 lies in the socle of L ( E ) ↓ H , so either 39 or 35(up to duality).Suppose that 39 lies in the socle of L ( E ) ↓ H . The modules 35 and 39 ∗ con-tribute P (4 + ) ⊕ P (4 − ) to the ( − i)-eigenspace, and so no other module with 4 + or 4 − as a submodule of its ( − i)-eigenspace can lie in the socle of L ( E ) ↓ H . Thismeans that none of 15 ∗ , 18, 18 ∗ and 35 ∗ can lie in the socle. Similarly, using the( − . This leaves 3, 3 ∗ and 35 as possibilitiesfor other composition factors of soc( L ( E ) ↓ H ).As we saw from the radical in (12.2), the socle cannot simply be 39. Assumingit does lie in the socle of L ( E ) ↓ H , we are interested in the cf( L ( E ) ↓ H ) \{ V ± , ± } -radical of P ( V ), for V one of 3, 3 ∗ , 35 and 39, (i.e., submodules of the radicals in(12.2)) which are18 ∗ / , / ∗ , / / , / , ∗ , ∗ / , ∗ , ∗ , , ∗ / L ( E ) ↓ H cannot bea subset of { , ∗ , , } containing 39, i.e., 39 ± does not lie in soc( L ( E ) ↓ H ).Thus soc( L ( E ) ↓ H ) must contain 35, as all other options are exhausted. Theother factors of the socle are some of 3, 15 and 18 (as we have excluded 6 ± , 8 and39 ± , the ( − ∗ and 15 ∗ , and the i-eigenspace excludes 18 ∗ ).Again, we take the radicals of P ( V ) but exclude 35 ± , and this time we obtain6 ∗ , , , ∗ / ∗ , , ∗ / , ∗ , / ∗ / , ∗ / ∗ , , / , ∗ , , / , ∗ / , / , where we repeat the case V = 35 for the reader’s convenience. Only the firstcontains 6 ∗ , so 3 must lie in soc( L ( E ) ↓ H ). This means that 15 cannot lie in thesocle, as otherwise the ( − + as submodules.It also means any composition factors of the first module that lie underneath 6 ∗ also must be present. The submodule generated by 6 ∗ is6 ∗ / ∗ , / , and so in particular 18 cannot lie in the socle either.
94 12. DIFFICULT CASES
The sum of the two modules corresponding to 3 and 35 contains all compositionfactors needed, but all 8s are quotients, i.e., will be quotients of rad( L ( E ) ↓ H ).However, they are not submodules of L ( E ) ↓ H / soc( L ( E ) ↓ H ), so neither of themcan exist in L ( E ) ↓ H , which yields a contradiction. Thus this case cannot occur. Step 4 : Case 2 where M ( D ) ↓ L ′ = 3 / , / L ′ acts on M ( D ) as 3 / , /
3. In this case W = M ( D ) ↓ L ′ is either 3 ⊕ /
3. Since the exterior square of P (3) = 3 / , / P (3) ⊕ ⊕ ⊕ , we see that the 1-eigenspace of the action of z has one of threepossible L -structures:(5 + ⊕ − ) ⊕ ⊕ P (3 + ) ⊕ P (3 − ) ⊕ P (3 + ) ⊕ + ⊕ − ⊕ − ⊕ − W = 3 ⊕ ,P (3 + ) ⊕ P (3 − ) ⊕ P (3 − ) ⊕ + ⊕ − ⊕ + ⊕ + W = 3 ⊕ ,P (3 + ) ⊕ ⊕ P (3 − ) ⊕ W = 3 / , and in both cases the action on the ( − P (1 + ) ⊕ P (1 − ) ⊕ P (3 + ) ⊕ P (3 − ) ⊕ (5 + ⊕ − ) ⊕ . We also consider the action of L ′ on the half-spin modules.For the module L (010) if L ′ acts semisimply on M ( D ) then it acts as 3 ⊕ L (010), and (up to duality) as 3 / / M ( D ). On the other hand,the action on L (00010) has composition factors 4 , and u acts with blocks 5 , P (2) ⊕ (2 / L (010) and L (00010):the tensor product of P (2) with 3 / ⊕ P (4) ⊕ ⊕ P (2) ⊕ . For the rest ofthe module, we need the tensor product of 2 / /
3, with 3 / ⊕ P (2) ⊕ P (4) ⊕ , P (4) ⊕ ⊕ (2 / , P (4) ⊕ (2 / ⊕ (2 , / , respectively.As every simple kH -module in L ( E ) ↓ H has a 2-dimensional summand on someeigenspace of z , we see that there are no simple summands of L ( E ) ↓ H , as in Step3. If 8 is a submodule then it must be a summand, since both 18 / / ± could in this case be asubmodule. By Step 1 a module other than 3 ± , 15 ± and 18 ± lies in the socle.Suppose that 39 lies in the socle of L ( E ) ↓ H . In all three options for the ( ± i)-eigenspaces, there are exactly four 4s in the socle and four in the top. As in Step3, this therefore excludes any module with a 4 in the ( − i)-eigenspace from beinga submodule of L ( E ) ↓ H , i.e., 15 ∗ , 18, 18 ∗ and 35 ∗ . The single 3 − in the socleof the ( − from being in the socle of L ( E ) ↓ H aswell. This leaves 3, 3 ∗ , 6 ∗ and 35 as possibilities for other composition factors ofsoc( L ( E ) ↓ H ).In Step 3 we dealt with all of these except for 6 ∗ , so we may assume now that6 ∗ also lies in the socle. Thus we need the cf( L ( E ) ↓ H ) \ { V ± , ± , ± } -radical of P ( V ), for V one of 3, 3 ∗ , 6 ∗ , 35 and 39. These are18 ∗ / , / ∗ , ∗ , ∗ / ∗ , / , , ∗ , , ∗ / . Again, there is a single 8, and so this cannot occur. (5) ≤ E If soc( L ( E ) ↓ H ) contains 35 (but not 39 or 39 ∗ ), then the same analysis as inStep 3 excludes 3 ∗ , 15 ∗ and 18 ∗ from the socle, and 6 ∗ can be excluded via the( − and 18. Using Step 3, we may assume that6 also lies in the socle. The appropriate radicals for this case are8 , , ∗ / ∗ , , ∗ / , ∗ , / , ∗ / , ∗ , / ∗ / , ∗ , / , / . Only the first module contains 39, so 3 must also lie in the socle. From the ( − in the socle at the same time. Thusthe 15 ∗ in the third layer of P (3) must lie in L ( E ) ↓ H . The submodule generatedby this factor is 15 ∗ / /
3, so there is also an extension 39 ∗ / in L ( E ) ↓ H . Butthis does not appear anywhere in these modules, a contradiction.Thus neither 39 ± nor 35 ± lies in the socle. We must therefore have 6 (up toduality) in the socle. This has 3 − in the ( − cannot also liein the socle. This leaves 3, 3 ∗ , 15 ∗ , 18 and 18 ∗ . The appropriate radicals for thecorresponding projectives are8 , , ∗ / ∗ , , ∗ / , , , / , , ∗ / ∗ , ∗ , , / , , ∗ / , , / ∗ , / ∗ , ∗ , / , , / ∗ . Since 35 ∗ appears only once, we must have 15 ∗ in the socle of L ( E ) ↓ H . The pres-ence of 15 ∗ in the socle means that 3 ∗ cannot be there, from the ( − ∗ can be in the socle as well.But now the only module with 18 ∗ in is the first, so the socle must be 3 ⊕ ⊕ ∗ .Thus we add any copies of 3 ∗ , 6 ∗ and 15 on top of the modules correspondingto 3, 6 and 15 ∗ . However, as we said before, 8 only has extensions with 18 ± and39 ± , so all 8s in these modules will remain quotients. Since this is not allowed, weobtain a contradiction. Step 5 : Case 2 where M ( D ) ↓ L ′ = (3 / ⊕ ⊕ ( M ( D ) ↓ L ′ ), which is a summand of the 1-eigenspace of the actionof z , is of the form P (3) ⊕ ⊕ ⊕ (3 / ⊕ (1 , / , ⊕ ⊕ . Whether M ( D ) is 3 ⊕ /
3, we don’t obtain another 1 , / , z . However, the action of L rather than L ′ now produces a contradiction: theaction of L on this summand is either 1 + , − / − , + or 1 − , + / + , − , but neitherof these is self-dual, so L ( E ) ↓ L is not self-dual, a contradiction since L ( E ) isself-dual. Step 6 : Conclusion for Case 2 in non-semisimple case.Since the only self-dual modules with composition factors 3 , P (3), (3 / ⊕ ⊕ ⊕ ⊕ M ( D ) ↓ L ′ is semisimple. Thus M ( D ) ↓ L ′ = 3 / M ( D ) ↓ L ′ = 3 ⊕ ⊕
1. Thus the action of L ′ on the 1-eigenspace of z on L ( E ) is P (3) ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ . The action of L ′ on the half-spin modules L (010) and L (00010) are (up to duality)3 / ⊕ ⊕ ⊕ , and we have(3 / ⊗ , / , (3 / ⊗ P (4) ⊕ (2 / . In particular, 2 − is not a summand of the L ′ -action on the (+i)-eigenspace of z on L ( E ), hence 8 is not a summand of L ( E ) ↓ H .
96 12. DIFFICULT CASES
The module 8 has an extension only with 18 , ∗ , , ∗ from our compositionfactors. The 1-eigenspaces of z on 39 / / L ′ -action containing thesummand 3 + / + , − , so since there is at most one such subquotient of this in the1-eigenspace of L ( E ) ↓ L ′ (and then only if the P (3) is a P (3 + )) we must have atleast one 8 summand in L ( E ) ↓ H . This is a contradiction, so this possibility cannotoccur. Step 7 : M ( D ) ↓ L ′ semisimple and conclusion.The remaining case is where M ( D ) ↓ L ′ is semisimple, which of course splits intotwo cases, depending on the actions of L ′ on M ( D ) and M ( D ). In Case 1, let Y be an algebraic A subgroup of D D acting on M ( D ) as L (4) ⊕ L (0) and on M ( D ) as L (4) ⊕ L (2) ⊕ L (0) ⊕ , so containing L ′ and centralizing z . It is easy tocompute the action of Y on the ( ± z , and these areΛ ( M ( D ) ↓ Y ) ⊕ Λ ( M ( D ) ↓ Y ) ∼ = (cid:0) (2 / / ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ (cid:1) ⊕ ((2 / / ⊕ ∼ = (2 / / ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ L ( − )’), and M ( D ) ↓ Y ⊗ M ( D ) ↓ Y = (2 / / ⊕ ⊕ (0 / / ⊕ ⊕ ⊕ ⊕ ⊕ . The corresponding actions of Y on the half-spin modules are L (3) and L (4) ⊕ ⊕ L (2) ⊕ , and the tensor product of these is(3 / / ⊕ ⊕ (1 / / ⊕ ⊕ ⊕ . If W is a simple submodule of one of the eigenspaces of z on L ( E ) ↓ L ′ , then W is sta-bilized by Y unless W has dimension 5 and we are considering the ( − L ′ lies inside the summand L (0) /L (8) /L (0).Thus we immediately see that if L ( E ) ↓ H possesses a submodule 3 ± , 6 ± , 8 or 15 ± then this is stabilized by Y , and h H, Y i 6 = G , so H is not Lie primitive. If thereis an 18 ± in the socle of L ( E ) ↓ H , then we need to be a bit more careful. As Y can be embedded in D in such a way that the automorphism inducing GL (5) on h L ′ , Z i also normalizes Y , the + / − -labelling of the 5s from L ′ can be extended tothe labelling of L (4)s, so there are (say) three with + and two with − . This meansthat if 18 ∗ is a submodule then it is stabilized by Y , and if 18 ∗ is a quotient thenthe kernel of the map L ( E ) ↓ H → ∗ is stabilized by Y . Either way H is not Lieprimitive again.We will show the same statement in Case 2, but this is easier. Now the subgroup Y acts on M ( D ) as L (2) ⊕ and on M ( D ) as L (2) ⊕ ⊕ L (0). This yields actionson the ( ± z ofΛ ( M ( D ) ↓ Y ) ⊕ Λ ( M ( D ) ↓ Y ) = L (4) ⊕ ⊕ L (2) ⊕ ⊕ L (0) ⊕ and M ( D ) ↓ Y ⊗ M ( D ) ↓ Y = L (4) ⊕ ⊕ L (2) ⊕ ⊕ L (0) ⊕ , so L ′ is in fact a blueprint for the ( ± z . For thehalf-spin modules, the action of Y is L (2) ⊕ L (0) and L (3) ⊕ ⊕ L (1) ⊕ , and thetensor product of these is( L (3) /L (5) /L (3)) ⊕ ⊕ L (3) ⊕ ⊕ L (1) ⊕ . In this case, every simple submodule of L ( E ) ↓ L ′ is stabilized by Y , and thereforeany copy of 3 ± , 6 ± , 8, 15 ± or 18 ± in the socle of L ( E ) ↓ H is stabilized by Y . (4) ≤ E Thus either H is Lie imprimitive or the socle of L ( E ) ↓ H must only containcopies of 39 ± and 35 ± , as all other modules restrict semisimply to L ′ . It cannotbe simple, as we saw with the radicals at the start of the proof, and it cannot be35 ∗ ⊕
39 as there is no 15 in the sum of the corresponding radicals. Thus the soclemust be 35 ⊕
39. But the { ± , ± , , ± , ± } -radical of P (35) is6 / / , / P (39), only possesses a single 8, a con-tradiction. Thus there is another factor in the socle of L ( E ) ↓ H , and therefore H is Lie imprimitive. It must also be strongly imprimitive, since either an automor-phism of H stabilizes this factor of the socle or it interchanges it with another thatis stabilized by the same positive-dimensional subgroup above.This completes the proof of the following proposition. Proposition . Let H ∼ = PSL (5) . If H is a subgroup of G ∼ = E incharacteristic , then H is strongly imprimitive. PSU (4) ≤ E Let H ∼ = PSU (4). We left open two possible sets of composition factors for L ( E ) ↓ H in Section 8.2. One we still cannot deal with, but the other, Case 8 fromthat section, can be solved, and we do so now. The composition factors of L ( E ) ↓ H are 24 , ∗ , , ∗ , (9 , ∗ ) , (¯9 , ¯9 ∗ ) , , , (3 , ∗ ) , (3 , ∗ ) , and we will prove that H is always strongly imprimitive.The proof of this is long, and involves 81 cases. We proceed step by step, withsome preliminaries first to set the stage. Recall that z has order 5 in H , and that L is the centralizer of z in H , which is 5 × Alt(5). The trace of z is −
2, and so L ≤ X = A A . The principal block contribution of L ( E ) ↓ L has compositionfactors 4 , , , , and this requires the composition factors of L ′ on both M ( A )to be 2 , , L ( E ) ↓ H , any fixed point in L ( E ) ↓ L must come from a submodule 8 i , as there are no other quotients of thepermutation module P L with no trivial quotient. (We saw the structure of P L inthe proof of Proposition 8.2.) Furthermore, there are nine isomorphism types ofmodule for M ( A ) ↓ L ′ , which are the following:2 ⊕ ⊕ , (1 / ) ⊕ , (2 / ⊕ , (1 / ) ⊕ , (2 / ⊕ , / , , , / , / / , / / . We call these modules 1 to 9 respectively. In each case, the summands of L ( A ) ↓ L ′ are 4, 2 i , 1 / i / P (2 i ), so the only summands with a trivial submodule orquotient are 1 / i / ζ be a primitive 5th root of unity, and write V ζ i for the ζ i -eigenspace ofthe action of z on L ( E ), viewed as a module for L ′ . If there are two summands1 / i / V then there must be two submodules 8 i , hence both are summands. Tosee this, if there is a single submodule 1 / i / V then there must be a submodule8 i , and upon removing it there is no trivial quotient of V , hence no 8 i quotient.This proves that every submodule 8 i of L ( E ) ↓ H is a summand.
98 12. DIFFICULT CASES
Module 1-eig. ζ -eig. ζ -eig. ζ -eig. ζ -eig.1 13 ∗ ∗ / / / / ∗ ∗ / / P (2 ) 2 ⊕ ∗ ⊕ P (2 ) 1 / / / / ⊕ P (2 )24 ∗ P (2 ) 2 ⊕ / / Table 12.2.
Actions of L ′ ∼ = Alt(5) on each ζ i -eigenspace of theaction of z The composition factors of L ( E ) ↓ X are (1001 , , z (we are suppressing the ‘ L ( − )’), and the other four modules are(1000 , , (0100 , , (0010 , , (0001 , . These four modules can be permuted by taking duals and swapping the two A -factors, so we may assume that the first one is the restriction of the ζ -eigenspaceof z on L ( E ) to X . With that assignment, the actions of X on the ζ , ζ , ζ and ζ -eigenspaces must be the modules above in the same order.The restrictions of the simple kH -modules to L are in Table 12.2, which is thesame as Table 8.1, but has been reproduced here for the reader’s benefit.Let M be the action of L ′ on the first M ( A ), and M be the action of L ′ on the second M ( A ). A priori we have nine options for each of M and M , so81 possible options for the action of L on L ( E ) (some might be isomorphic). Wewrite ( i, j ) to mean that M is module i from the list above, and M is module j .Replacing M by M and M by M ∗ yields a cycle of length four on the kL ′ -modules V ζ i , and this corresponds to taking the field automorphism of H that maps3 ∗ ∗ . Since this operation permutes the possible modules L ( E ) ↓ H ,we may consider a single representative from each orbit under this action. It iseasy to see that the action of this 4-cycle on the 81 pairs ( i, j ) has twenty cyclesof length 4 and a single fixed point (1 , , , (2 , , (4 , , (8 , , (8 , , (8 , , (8 , , (8 , , (1 , , (2 , , (2 , , (2 , , (2 , , (1 , , (1 , , (4 , , (4 , , (6 , , (6 , , (8 , , (6 , . (The ordering chosen here is the order in which we solve the cases.) All but thelast one are easy to exclude, and the majority of the proof will be on Case (6 , (4) ≤ E We are also interested in the { ± , ¯9 ± } -heart W of L ( E ) ↓ H . For this we takethe { } ′ -radical of P (9 ), then take its { ± , ¯9 ± } ′ -residual. This has structure9 / ∗ / , ∗ / , ∗ , / ∗ , , ∗ / , ∗ / , ∗ , ∗ / ∗ , / . This has three copies of 9 in it, and we must have exactly two in W . The { } ′ -residual of it has simple top, and so we cannot have the 9 in the highest soclelayer. Removing this and again taking the { ± , ¯9 ± } -residual yields the module(12.3) 9 ∗ / , / ∗ , / , ∗ / , ∗ / ∗ , / . Notice that this has no 8 , no ¯9 ± and no 24 ± i,j . The { ± , ¯9 ± } -heart of L ( E ) ↓ H must be a submodule of the sum of the four cognates of this module, so in particularit has no copies of 24 ± i,j in it.The equivalent module with socle ¯9 is¯9 ∗ / ∗ , / ∗ , ¯9 / , ∗ / , ¯9 ∗ / , / ¯9 . Notice that W is the sum of its { ± } ′ - and { ¯9 ± } ′ -radicals, so we may just focuson one, say 9 . If W does not have a summand the whole of the above module,then the socle must contain either 9 ⊕ ∗ or 9 ⊕ , and in neither case is there a9-dimensional factor in the heart of the module. Taking the { ± i , } -radical (andthe 9 in the socle) of the module above yields3 ∗ / / ∗ , / . One may place a 9 in the fifth socle layer, on top of the 3s but not the 8 , or a9 ∗ in the third socle layer, on top of the 8 but not any of the 3s. Moreover, since W is self-dual, if we see the uniserial module9 / ∗ / / ∗ / as a subquotient then we must also have its dual.Thus we see that the number of 3s in this module is either six or none, andtherefore the number of 3s in W is twelve, six or none. Step 1 : Case (1 , L acts semisimply on both M i . In this case V is (1 / / ⊕ ⊕ (1 / / ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ . This means that all 8-dimensional factors of L ( E ) ↓ H are summands, and with thatrestriction any 248-dimensional module with the correct composition factors yieldsthis module. The V ζ i for i = 1 , , , P (2 ) ⊕ P (2 ) ⊕ ⊕ ⊕ (1 / / ⊕ (1 / / ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ . Let Y be the A subgroup of A A acting on each M ( A ) as L (0) ⊕ L (1) ⊕ L (2).The action of Y on the 1-eigenspace is(0 / / ⊕ ⊕ (0 / / ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ . (Here, as often before, we have removed the ‘ L ( − )’.) The subgroup Y contains L ′ ,and we see that the { } ′ -residual as a kL ′ -module and the { L (0) } ′ -residual as a k Y -module are the same subspace of L ( E ). Inside this subspace, L ′ and Y stabilizethe same subspaces, and so Y stabilizes the 1-eigenspace of z on any submodule 8 i .The action of Y on each ζ i -eigenspace is also the same, and is(2 / / / / ⊕ (1 / / ⊕ (0 / / ⊕ (0 / / ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ .
00 12. DIFFICULT CASES
Comparing these two modules, we see that every semisimple kL -submodule of V ζ is also a semisimple k Y -module. We can see that Y and L ′ do not stabilize exactlythe same subspaces, since there is a composition factor L (5) for example, but Y will stabilize any submodule 8 i of L ( E ) ↓ H .Thus Y stabilizes all 8-dimensional simple submodules of L ( E ) ↓ H , so H isstrongly imprimitive by Theorem 3.2. Step 2 : Cases (2 ,
2) and (4 , ,
4) first. In the case M ∼ = M ∼ = (1 / ) ⊕ , we see that L ( E ) ↓ H has a summand 8 ⊕ , but has no submodule 8 . As in the previous case, we will showthat these are stabilized by an A -subgroup of A A , and therefore H is stronglyimprimitive.Let Y act on each M ( A ) as ( L (0) /L (2)) ⊕ L (1), so that Y contains L ′ . Theaction of Y on the 1-eigenspace of z on L ( E ) is(2 / / / / ⊕ ⊕ (0 / / ⊕ ⊕ ⊕ ⊕ ⊕ . As in the previous case, every kL ′ -submodule of the form 1 / / k Y -submodule of the form 0 / /
0, so the 1-eigenspace of z on any kH -submodule8 of L ( E ) ↓ H is stabilized by Y .The ζ -eigenspace of the actions of L ′ and Y are P (2 ) ⊕ P (2 ) ⊕ ⊕ ⊕ ⊕ (1 / / ⊕ (1 / ) ⊕ ⊕ ⊕ / / ⊕ (2 / / / / ⊕ ⊕ ⊕ ⊕ (0 / / ⊕ (0 / ⊕ ⊕ ⊕ . We see that every kL ′ -submodule of the ζ -eigenspace, such that either it or whosequotient by it is simple, is stabilized by Y . Hence, every simple kL ′ -submodule ofthe ζ - and ζ -eigenspaces is stabilized by Y , and so therefore is any submodule 8 of L ( E ) ↓ H . This proves that H is strongly imprimitive.For the case (2 , ⊕ is a summand of L ( E ) ↓ H , we apply a sin-gle Frobenius automorphism to the case (4 , M and M are both (1 / ) ⊕ , and Y acts on both M ( A ) as ( L (0) /L (4)) ⊕ L (2). (Notethat this embedding works, whereas the more obvious overgroup of L ′ , acting as( L (0) /L (1)) ⊕ L (2), has a different submodule structure.)As Y is simply a Frobenius twist of the previous case, the same argument holdsfor the 1-eigenspace, so we now consider the ζ -eigenspaces of L ′ and Y . These are P (2 ) ⊕ P (2 ) ⊕ ⊕ ⊕ (1 / / , / , ) ⊕ (1 / / ⊕ (2 / ⊕ ⊕ (the dual of 2 / , / , / / / / / ⊕ (2 / / ⊕ ⊕ ⊕ (0 / / , / , ⊕ (0 / / ⊕ (4 / ⊕ ⊕ . Clearly every simple submodule of the L ′ action is stabilized by Y , and for simplequotients, one needs only to understand the module 0 / / , / ,
4, which is the dualof 4 / , / , /
0, so every simple quotient is also stabilized. Therefore in particulareach submodule 8 of L ( E ) ↓ H is stabilized by Y , and H is strongly imprimitiveagain. Step 3 : Cases (8 , , , , , M ∼ = 2 / / . The exterior square of M is 4 ⊕ (1 / , / P (2 ) ⊕ ⊕ P (2 ) ⊕ ⊕ ⊕ ⊕ (1 / , / . (4) ≤ E In particular it has no summand 2 i . Since this is the ζ -eigenspace, we see that L ( E ) ↓ H can have no simple summand whose ζ -eigenspace is 2 , for example 8 .The tensor products M ⊗ Λ ( M ) (i.e., the ζ -eigenspace) do depend on theisomorphism type of M , but again in all cases there is no summand 2 i . Hencethere can be no summand 8 i in L ( E ) ↓ H in this case, so in particular M cannothave a summand 2 i for i = 1 ,
2. However, in the first five possibilities for M wedo indeed have a summand 1 / i / i in the 1-eigenspace.This yields a contradiction. Step 4 : Case (1 , M is semisimple, we have both 1 / i / M ⊗ M ∗ . Thus8 ⊕ is a summand of L ( E ) ↓ H .However, the tensor product of Λ ( M ) with M ∗ does not have a 2-dimensionalsummand. This contradicts the fact that 8 is a summand. Thus we obtain acontradiction in this case. Step 5 : Cases (2 , , ,
6) and (2 , , M ∼ = (1 / ) ⊕ . When M is either 2 , / / , , V ζ has no 2-dimensional summand. Since M has a summand 2 , there must be a summand8 in L ( E ) ↓ H , and hence a summand 2 in V ζ , which yields a contradiction. Asimilar statement holds when M ∼ = (1 / ) ⊕ or its dual, so (2 ,
4) and (2 , summand and an 8 summand, but in both cases theonly 2-dimensional summand of M ⊗ Λ ( M ), which is V ζ , is 2 , whereas we needa 2 coming from the restriction of the 8 . Thus these two pairs cannot occur. Wealso can exclude M semisimple and M being (1 / ) ⊕ , so pair (1 , summand in V ζ , but we should have a summand 8 . Step 6 : Case (1 , M ∼ = 1 ⊕ ⊕ and M ∼ = (1 / ) ⊕ . The module V ζ is P (2 ) ⊕ P (2 ) ⊕ ⊕ ⊕ (1 / / / ) ⊕ (1 / / ⊕ (2 / ⊕ ⊕ ⊕ , and the P (2 ) is contributed by the 24 , so can be ignored. The composition factor24 cannot be a submodule of L ( E ) ↓ H since there is no submodule 1 / / L -action of the ζ -eigenspaceof the module 3 / is 2 / / /
1, which is not a subquotient of V ζ (excludingthe P (2 )). But this is the only extension between 24 and another simple kH -module, we see that 24 is a quotient of L ( E ) ↓ H . Hence 24 ∗ is a submodule.However, 24 ∗ contributes a 2 to V ζ , and having ignored the P (2 ), there is asingle submodule 2 of V ζ . Therefore in the quotient module L ( E ) ↓ H / ∗ , the1 / / / in the ζ -eigenspace of the L -action becomes a 1 / /
1; i.e., the extension24 / is not a quotient of L ( E ) ↓ H , and therefore 24 is a summand. But thisis a contradiction as 24 ↓ L is not a summand of L ( E ) ↓ L from the description of V ζ . Step 7 : Cases (4 , , ,
6) and (6 , , V ζ has structure P (2 ) ⊕ P (2 ) ⊕ ⊕ ⊕ (2 / , / , / ⊕ (1 / , ) ⊕ (2 / / / ⊕ , and we may ignore the P (2 ) as it lies inside 24 ∗ . This time the lack of a subquo-tient 1 / / / means that 24 must be a submodule, hence 24 ∗ a quotient, and
02 12. DIFFICULT CASES this quotient must contain the only 2 in the top of V ζ . We obtain a contradictionin the same way as the previous case.For (4 , ,
6) and (6 , V ζ is P (2 ) ⊕ P (2 ) ⊕ (1 / / / ) ⊕ (2 , / ⊕ (1 / / , , / , ) ⊕ ⊕ ,P (2 ) ⊕ P (2 ) ⊕ (1 / / , , / , ) ⊕ (1 / / , , / , ) ⊕ ⊕ and P (2 ) ⊕ P (2 ) ⊕ ⊕ ⊕ ⊕ (1 / , / , / , ) ⊕ (2 / P (2 ) this time, and there is no subquotient 2 / / / is a quotient, so 24 ∗ is a submodule, and this removes the (unique) 2 in the socle of the above module. Again, a contradiction ensues. Step 8 : Case (8 , M and M are 2 / / , and all V ζ i for i = 1 , , , ⊕ ⊕ P (2 ) ⊕ ⊕ P (2 ) ⊕ ⊕ (1 / , / , with the last summand being self-dual. Clearly 24 ± i,j is not a submodule of L ( E ) ↓ H as 1 / i / { ± , ¯9 ± } -heart of L ( E ) ↓ H cannot contain all the 3-dimensionals (or the 24-dimensionals) but doescontain all 8 i , we need a module of dimension 3 in the socle. We assume that itis 3 , but the same proof works in all cases with a change of eigenspace of z . Inthis case, we remove one P (2 ) from V ζ as it lies inside 24 , and see that thereis now a unique 2 in the socle of V ζ , so the quotient by this 3 does not have asubquotient 1 / / / . Thus 24 is a submodule of the quotient, so 24 / is asubmodule of L ( E ) ↓ H . In particular, 3 ∗ / ∗ is a quotient of L ( E ) ↓ H —call thekernel of this quotient W —and the ζ -eigenspace of z on this module has L ′ -action2 ⊕
4, i.e., the unique 2 quotient in V ζ comes from this quotient module. Butnow we argue as in the case (4 , must be a quotient of thesubmodule W , hence a quotient of L ( E ) ↓ H , which is a contradiction. Step 9 : Case (6 ,
8) and conclusion.This time V ζ is as in the previous case, P (2 ) ⊕ ⊕ P (2 ) ⊕ ⊕ (1 / , / ⊕ ⊕ , but V ζ is now P (2 ) ⊕ ⊕ P (2 ) ⊕ (1 / , / , / , ) ⊕ (2 / ⊕ ⊕ . Notice that this means that each V ζ i for i = 1 , , , ± or 24 , or 8 or8 . By the same argument as the previous case, we cannot have a submodule thatis 3 ± either. However, there could be a submodule 24 ∗ .Let W denote the { ± , ¯9 ± } ′ -radical of L ( E ) ↓ H . (This is non-zero since weknow that the socle cannot consist solely of 9s from the start of this proof.) Notethat W must contain at least one 3 ± and one 3 ± , and at least one 24 ± and at leastone 24 ± .For some possible socles we cannot even build a radical that works. For otherswe will need to (try to) add W on top to produce a contradiction. (4) ≤ E We now work based on the socle of W . Since V L ′ ζ i is 1-dimensional, we see thatthe socle of W is one of24 ∗ , ∗ ⊕ , ∗ ⊕ ⊕ ∗ , ∗ ⊕ ∗ , , ∗ , ⊕ ∗ . The case where soc( W ) = 24 ∗ . The { ± i , ± i,j } -radical of P (24 ∗ ) is3 ∗ / / ∗ / ∗ . We need more modules in soc( W ), as there is only one 24-dimensional factor in thisradical. Thus 24 ∗ cannot be the socle. The cases where soc( W ) = 24 ∗ ⊕ and ∗ ⊕ ⊕ ∗ . Suppose that the socle of W contains 3 ⊕ ∗ . The ζ -eigenspace of this modulehas L -action 2 ⊕ , and therefore in the quotient module L ( E ) ↓ H / (3 ⊕ ∗ ) wefind no subquotient 1 / / / in the ζ -eigenspace. But now this means—as wehave seen several times up to this point—that the 24 quotient is a summand of L ( E ) ↓ H / (3 ⊕ ∗ ), hence a summand of L ( E ) ↓ H , which is a contradiction. The case where soc( W ) = 24 ∗ ⊕ ∗ . We construct the { ± i , ± } -radical of P (3 ∗ ), which is3 ∗ / , ∗ , / , ∗ / ∗ . We must have that 24 factor in W , and this requires the 3 ∗ in the second soclelayer of P (3 ∗ ). However, the submodule 3 ∗ / / ∗ of this radical has a ζ -eigenspace1 / /
1, and this is not a submodule of V ζ (not a quotient of V ζ = V ∗ ζ ). Hencethat 3 ∗ in the third layer, hence the 3 ∗ in the fourth layer, cannot lie in W .Thus we have a guaranteed submodule W ′ of W , (24 / ∗ / ∗ ) ⊕ ∗ , and apyx for W , (3 , / , ∗ / ∗ ) ⊕ (3 ∗ / / ∗ / ∗ ) . Note that the ζ - and ζ -eigenspaces of W ′ have a trivial submodule, hence therecan be no others in L ( E ) ↓ H . In particular, this means that 9 and ¯9 cannot besubmodules of L ( E ) ↓ H .It is not possible to place a copy of 9 ∗ on top of the pyx for W , and so anycopy of 9 ∗ in the socle of W becomes a copy in the socle of L ( E ) ↓ H . Thereforethere is at most one copy of 9 ∗ in the socle of W .If there is a copy of 9 in the socle of W , then it must lie above W . It ispossible to place two copies of 9 on top of the pyx for W . Taking the { } ′ -radicals of them, and including the 24 , we obtain the module24 / ∗ , / ∗ , / ∗ / / ∗ / ∗ . The 9 cannot lie solely in the first module as that 9 furnishes the ζ -eigenspacewith another fixed point. Thus either the 9 lies solely over the second module oris diagonally embedded.One cannot extend the second copy of 9 by 3 ∗ / / ∗ , and hence W cannotcontain a summand as in (12.3). In particular, this means that the socle of W cannot be just 9 , and must contain both 9 and 9 ∗ , or just 9 ∗ .If soc( W ) = 9 ∗ then there must be a submodule 3 ∗ / / ∗ (by applying theappropriate automorphism to the module in (12.3)). Adding as many copies of 3
104 12. DIFFICULT CASES and then 3 ∗ on top of the pyx for W as possible, we manage no copies of 3 ∗ at all.Thus L ( E ) ↓ H must have a submodule(24 / ∗ / ∗ ) ⊕ ∗ ⊕ (3 ∗ / / ∗ ) . The ζ -eigenspace of this has two L ′ -fixed points though: one from the secondsummand and one from the third.Thus soc( W ) must in fact contain 9 ⊕ ∗ . Note that, in the pyx for W , thecopy of 3 in the second socle layer of the summand with socle 3 ∗ cannot occurin W , because that contributes another L ′ -fixed point to V ζ . So we now have aprecise description of W , namely(24 / ∗ / ) ⊕ (3 ∗ / / ∗ / ∗ ) . Together with 9 ∗ , this provides the trivial submodule for V ζ , V ζ and V ζ .On W we may place one copy of ¯9 , but the trivial from its ζ -eigenspace fallsinto the socle of V ζ , so this cannot occur. We may also place one copy of ¯9 ∗ , andagain the trivial from its ζ -eigenspace falls into the socle of V ζ . In particular, thismeans that the socle of W must be exactly 9 ⊕ ∗ ⊕ ¯9 ∗ .We now show that the ¯9 ∗ must actually lie in the socle of L ( E ) ↓ H . Supposenot: we add on top of W the ¯9 ∗ , then the 3 ∗ then the 3 ∗ that is a submodule of W . Doing this with all possible copies of the modules yields(3 ∗ / ∗ / ¯9 ∗ , / ∗ / ∗ ) ⊕ (3 ∗ / / ∗ / ∗ ) . However, the first summand of this has ζ -eigenspace(1 / / ⊕ , which is not a submodule of V ζ .We also reach a contradiction if ¯9 ∗ is a submodule. Then again we take themodule W ⊕ ¯9 ∗ , and then add on top the unique copy of 3 ∗ and then the uniquecopy of 3 ∗ , and this now forms the module(24 / ∗ / ∗ ) ⊕ (3 ∗ / / ∗ / ∗ ) ⊕ (3 ∗ / ∗ / ¯9 ∗ ) . Both the first and third summand have a trivial submodule for the L ′ -action onthe ζ -eigenspace, so this cannot be a submodule of L ( E ) ↓ H either.We have finally reached a complete contradiction. The case where soc( W ) = 3 . The { ± i , ± i,j } -radical W ′ of P (3 ) is3 / , ∗ , / , , ∗ , ∗ / , ∗ , / . Since W is a submodule of this, the 24 ∗ and one of the copies of 24 in the moduleabove lie in W . Since W ′ has simple top and two copies of 24 , the 3 in the fifthsocle layer of W ′ does not lie in W .We also cannot have the submodule 3 / ∗ / of W ′ in W as it has a ζ -eigenspace 1 / /
1, which isn’t a submodule of V ζ . This module is a submoduleof the minimal submodule of W ′ supporting the 3 in the fourth socle layer, hencethat does not lie in W either, and the same holds for the 24 in the fourth soclelayer as well. We may therefore replace W ′ by the module3 ∗ / , ∗ , ∗ / , ∗ , / . (4) ≤ E One cannot place 9 on top of this module, and can place a unique copy of 9 ∗ ontop of this module, namely3 ∗ / , ∗ , ∗ / , ∗ , ∗ , / . This has a 2-dimensional L ′ -fixed space on its ζ -eigenspace: one from the 9 ∗ andone from the sum of the 3 and 24 . The 3 supports the 24 ∗ , so needs to be in W , and so we cannot have a 9 ∗ in the socle of W . Thus the socle of W mustcontain exactly 9 and modules ¯9 ± . In particular, 9 lies in the socle of L ( E ) ↓ H .Since there is a single 9 and no 9 ∗ in soc( W ), we must have that the { ± } -heart of W is the full module from (12.3)9 ∗ / , / ∗ , / , ∗ / , ∗ / ∗ , / . We build this module from the ground up, on top of the module W ⊕ . We addall copies of 3 ∗ , then 3 , then 3 ∗ , then 9 , on top of our pyx for W , but this doesnot affect the pyx at all, as there are no such extensions. Thus L ( E ) ↓ H containsa submodule (24 ∗ / , / ) ⊕ (3 ∗ / / ∗ / ) . Both of these summands have an L ′ -fixed point on the ζ -eigenspace, and so weobtain a contradiction in this case. The case where soc( W ) = 3 ∗ . Notice that in the previous proof, every statement also holds if we apply a fieldautomorphism, i.e., replace every composition factor of every module by its dual.This swaps the ζ - and ζ -eigenspaces, but neither of these has a submodule 1 / / ζ - and ζ -eigenspaces, but neither of these has 1 ⊕ as a submodule.Thus the exact proof above still works. The case where soc( W ) = 3 ⊕ ∗ . The presence of 3 ⊕ ∗ as a submodule means that neither ¯9 nor ¯9 ∗ lies insoc( L ( E ) ↓ H ), as we would again have two trivial submodules on the ζ ± -eigenspaceof z . However, (at least) one of them must lie in the socle of W , so we have to beable to place one of them on top of W . We will show that this is impossible in bothcases, providing the final contradiction. We first must restrict the structure of W .The module W is a submodule of W ′ , which is(3 ∗ / , ∗ , ∗ / , ∗ , / ) ⊕ (3 / , ∗ , / , ∗ , ∗ / ∗ ) . We will assume that 24 ∗ lies in W , as either 24 or 24 ∗ must do so, and thesame proof works with all factors replaced by their duals in the other case. Thus W must contain(24 ∗ / , / ) ⊕ ∗ or (24 ∗ / / ) ⊕ (24 ∗ / ∗ ) . Each summand of W ′ can serve as the submodule for an extension with quotienteither ¯9 or ¯9 ∗ . We deal with ¯9 first, and note that to support the ¯9 on top,each summand of W ′ needs the 3 in the second socle layer.We examine the ζ -eigenspaces, which for ¯9 is simply 1, and for the twomodules above are P (2 ) ⊕ (1 / / / ) ⊕ / / / ) ⊕ (2 / to the second summand in each case yields P (2 ) ⊕ (1 / / / ) ⊕ (2 /
1) and (1 / / / ) ⊕ (2 , / .
06 12. DIFFICULT CASES
On each of these four modules we must add one or two trivial modules, but wecannot have two trivial submodules in total, and we cannot have a submodule1 / i / i = 1 ,
2. Thus we must be in the last case, where we have a submodule1 / , /
1. However, adding both copies of ¯9 on top of this module yields the ζ -eigenspace (1 / / / ) ⊕ (1 / , / ⊕
1, but the second module has dual 1 / / , ,so this is not the right module. Therefore ¯9 does not lie in the socle of W .For ¯9 ∗ , we may place two copies of ¯9 ∗ on top of W ′ ; for ease of computationwe take for W ′ the module(3 , ∗ / , , ∗ , ∗ / , ∗ , / ) ⊕ (3 / , ∗ , ∗ / , ∗ , ∗ / ∗ ) , (i.e., the { ± i , ∗ , ± } -radical of P (3 ) ⊕ P (3 ∗ )) a module that definitely contains W . On this we may place two copies of ¯9 ∗ , one in the third socle layer of eachsummand. On this we may place two copies of 3 ∗ , one in the fourth layer of eachsummand. However, we cannot place any copies of 3 ∗ on this module, so there canbe no submodule 3 ∗ / ∗ / ¯9 ∗ of W .From our work on W at the start of this proof, this means that we must havetwo copies of ¯9 ∗ in soc( W ), hence both ¯9 ∗ that we placed on top of our modelfor W ′ . We take the { ± i } -residual of this to produce a module(¯9 ∗ , ∗ / , ∗ , / ) ⊕ (¯9 ∗ / ∗ , ∗ / ∗ ) . This is not necessarily a submodule of L ( E ) ↓ H , but if we remove either 24 or24 ∗ then it is. Doing the former yields a module whose ζ -eigenspace has twotrivial submodules as a kL ′ -module, and doing the latter yields a module whose ζ -eigenspace has three 2 submodules as a kL ′ -module, neither of which lies in V ζ from the description above.Thus we cannot find a configuration that is consistent with the V ζ i , and there-fore we obtain a contradiction, completing the proof.This final case eliminates all 21 orbit representatives, and therefore the set ofcomposition factors.HAPTER 13 The trilinear form for E G is of type E in characteristic 3,although our methods apply to other characteristics. In this chapter we completethe proof that every copy of PSL (3), PSU (3) and PSL (8) ∼ = G (3) ′ in charac-teristic 3 is strongly imprimitive. From the results of Chapter 10, we are left withPSL (3) and PSU (3) acting irreducibly on M ( E ), and PSL (8) acting as a sumof three non-isomorphic 9-dimensional modules on M ( E ). (In the third case, thenormalizer acts irreducibly on M ( E ).)This chapter uses completely different methods from the previous chapters,to prove that every such subgroup H is contained in a subgroup G (3) actingirreducibly on M ( E ). Note that S ( M ( E )) G is 1-dimensional, and this yields asymmetric trilinear form on G , as is well known. Use of the trilinear form to study G and its subgroups was the focus of Aschbacher’s series of papers [ Asc, Asc87,Asc88, Asc90a, Asc90b ], and was also used in Magaard’s thesis on F [ Mag90 ].It also appears a little in work of Cohen–Wales [
CW97 ] on Lie primitive subgroupsof E ( C ).We will not define the trilinear form explicitly, because we do not deal with afixed group E but several conjugates of it, in order to give a ‘nice’ description ofcertain subgroups. Suppose that we are given a 27-dimensional vector space V over an algebraicallyclosed field k , and a symmetric trilinear form f ( − , − , − ) whose symmetry group G is (the simply connected form of) E . An element x ∈ GL( V ) lies in G if and onlyif, for all u, v, w ∈ V , f ( ux, vx, wx ) − f ( u, v, w ) = 0 . Let H be a finite group that we are attempting to embed in G , and view H as asubgroup of Y = GL( V ). If g ∈ Y and h ∈ H , then h g lies in G if and only if, forall u, v, w ∈ V , f ( ug − hg, vg − hg, wg − hg ) − f ( u, v, w ) = 0 . By replacing u, v, w by ug, vg, wg we see that h g lies in G if and only if, for all u, v, w ∈ V ,(13.1) f ( uhg, vhg, whg ) − f ( ug, vg, wg ) = 0 . This version has the benefit of not having to invert g . In particular, given a basis { v , . . . , v } for V , g may be written as a 27 × g is not inverted,some of the entries of the matrix can be chosen to be parameters from k . If this isdone, then (13.1) becomes a polynomial equation in those parameters of degree atmost 3. E Now suppose that L is a subgroup of H contained in G , and we wish to classifythe elements of the set H of all conjugates of H via elements of Y that lie in G and still contain L . In order to proceed we will assume that all subgroups of H isomorphic to L are H -conjugate to L . This will always hold for the pairs ( H, L )that we consider here and in later work, so it is no restriction.Let g ∈ Y be such that H g ≤ G and L ≤ H g . Thus L g − ≤ H , so since L g − and L are H -conjugate by assumption, there exists y ∈ H such that L g − = L y ,i.e., L = L yg . Clearly H g = H yg , and so if we are attempting to classify elementsof H , it suffices to consider g ∈ N Y ( L ).While elements of C Y ( L ) can be written, with respect to an appropriate basis,as matrices with entries 0 or parameters from k , it is less easy to do this for N Y ( L ). One special case of our situation is where all automorphisms of L inducedby elements of Y are also induced by elements of N Y ( H ), or in other words, N Y ( L ) = C Y ( L ) · N N Y ( H ) ( L ) . If this is the case then when classifying elements of H it suffices to consider g ∈ C Y ( L ). Of the three groups we consider—PSL (3), PSU (3) and G (3) ′ —two ofthese have this extra property. Of course, if g and g are elements of N Y ( L ) suchthat g g − lies in N Y ( H ), then H g = H g as well, so when constructing H itsuffices to consider cosets via the normalizer of H .Our strategy is therefore as follows:(1) Identify a subgroup L of our subgroup H ≤ Y ;(2) Determine the number of G -conjugacy classes of subgroups L with theappropriate action on V (perhaps the action on L ( E ) is also necessaryto exclude some G -classes, but not in the cases here);(3) Construct a basis of V so that one may write elements of C Y ( L ) as (rea-sonably nice) matrices with entries parameters from the field k ;(4) if H = h L, h i , and g ∈ C Y ( L ) (with parameters), and u, v, w are basiselements of V , produce 3654 = 27 · · / (3 · ·
1) equations f ( uhg, vhg, whg ) − f ( ug, vg, wg ) = 0;(5) solve these equations (modulo C Y ( H )) to yield all elements of H .If we have to consider N Y ( L ) rather than C Y ( L ), as we do for PSL (3) in Section13.4, then we have to alter this method slightly, and perform it once for every cosetof C Y ( L ) · N N Y ( H ) ( L ) in N Y ( L ). However, if Aut G ( L ) covers these cosets then wecan avoid this extra work, as all such sets H are related by N G ( L )-conjugacy.To solve the systems of cubic equations, we can either work directly with theequations themselves, or simply use Gr¨obner bases to quickly yield all solutions.(We do both of these in the supplementary materials.)We now introduce the groups H and L that we consider in this chapter. Let H be one of the groups PSL (3), PSU (3) and G (3) ′ = PSL (8) acting as mentionedat the start of this chapter. We need the subgroup L , its action on V and itsnormalizer in Y , together with the normalizer of H in Y .If H ∼ = PSL (3), then L ∼ = 13 ⋊
3, the normalizer of a Sylow 13-subgroup of H ,unique up to H -conjugacy. The simple kL -modules are 1 , , ∗ , , ∗ , , ∗ , andthe 3 ± i are projective. The action of L on M ( E ) is, up to isomorphism,(1 / / ) ⊕ (3 ⊕ ∗ ) ⊕ ⊕ (3 ⊕ ∗ ) ⊕ . The centralizer of this in Y is easy to describe: it is k × × k + × k + × GL ( k ) × GL ( k ) × GL ( k ) × GL ( k ) . (The centralizer of 1 / / ( k ) is k × × k + × k + , and the centralizer of each3 ± i ⊕ ± i in GL ( k ) is isomorphic to GL ( k ).) The normalizer of L in Y mustnormalize the 13, and so in fact N Y ( L ) is simply C Y ( L ) × (13 ⋊ N Y ( H ) is k × × H.
2, where the subgroup k × is the scalar matrices, so | Aut Y ( L ) :Aut N Y ( H ) ( L ) | = 2.If H ∼ = PSU (3), then L ∼ = 4 ⋊ Sym(3), the normalizer of a ‘maximal Φ -torus’in H (in the language of generic groups of Lie type). This is again unique up toconjugacy in H . The simple kL -modules are 1 , , , , ± , ± ,
6. The action of L on M ( E ) is (1 / / ) ⊕ ⊕ ⊕ ⊕ ∗ ⊕ ⊕ . As with the previous group, the centralizer of this in Y is easy to describe, and is( k × × k + ) × k × × k × × k × × k × × GL ( k ) . (The contribution from 1 / / is k × × k + , each 3-dimensional summand con-tributes k × , and the contribution from 6 ⊕ is GL ( k ).)The automizer of L in Y has L as a subgroup of index 2, as can be checkedin Magma as | Aut( L ) : Inn( L ) | = 2 and of course L ∼ = Inn( L ). Thus N Y ( L ) = C Y ( L ) × L.
2. The normalizer N Y ( H ) is k × × H.
2, where the subgroup k × is thescalar matrices, so Aut Y ( L ) = Aut N Y ( H ) ( L ).If H ∼ = PSL (8), then L ∼ = 2 ⋊
7, the normalizer of a Sylow 2-subgroup of H . This is also unique up to conjugacy in H . The simple kL -modules are 1 i for i = 1 , .., L on M ( E ) is7 ⊕ ⊕ M i =2 i . The centralizer of this in GL ( k ) is easy to describe, and isGL ( k ) × k × × k × × k × × k × × k × × k × . Since Aut( L ) ∼ = (2 ⋊ ⋊
3, we see that N Y ( L ) = C Y ( L ) × L.
3. Similarly, C Y ( H ) = k × × k × × k × , N Y ( H ) = C Y ( H ) × H. . Thus again Aut Y ( L ) = Aut N Y ( H ) ( L ).Our next task is to prove that L is unique up to G -conjugacy, or in one caseconjugacy in Aut( G ), i.e., allowing the graph automorphism. Lemma . Let H be one of the subgroups PSL (3) , PSU (3) and G (3) ′ ofthe algebraic subgroup G of G = E . Let L be a subgroup ⋊ , ⋊ Sym(3) and ⋊ of H respectively. If ¯ L is a subgroup of G isomorphic to L , and the actionsof L and ¯ L on M ( E ) are isomorphic, then L and ¯ L are conjugate in G for the firsttwo possibilities for L , and conjugate in G extended by the graph automorphism inthe third case. Proof.
We start with 13 ⋊
3. If M ( E ) ↓ H is irreducible, then an element x oforder 13 in H acts with trace 1, and is regular semisimple. Hence the centralizer of x is simply a torus of G . Inside N G ( T ), one sees that all powers of x are conjugate,and hence N G ( h x i ) is contained in N G ( T ) as well. But clearly L is contained in
10 13. THE TRILINEAR FORM FOR E N G ( h x i ), so is contained in N G ( T ). The fastest way to proceed now is simply tocheck that there is a single conjugacy class of subgroups L with the 13 regular andthe 3 acting with blocks 3 in the group 13 ⋊ W ( E ).(If verifying this in Magma, the best way to compute this is to find the twoclasses of elements of order 3 in W ( E ) that act projectively on M ( E ), and thentake power-conjugate presentations of the two groups 13 ⋊ PCGroup . Thenone easily enumerates all groups 13 ⋊ · W ( E ).)Next, we consider L ∼ = 4 ⋊ Sym(3). First, the 4 subgroup L is toral as anyrank-2 abelian subgroup of a semisimple algebraic group is [ Gri91 , Proposition2.13(vi)]. From the composition factors of H on L ( E ) we see that the restrictionof L ( E ) to L has a 6-dimensional centralizer. Hence C G ( L ) ◦ is a maximal torusof G . In particular, any element of G that normalizes L normalizes C G ( L ), hencenormalizes C G ( L ) ◦ , and so L is contained in the normalizer of a torus. One mayagain use Magma to enumerate such subgroups in the normalizer of a torus andcheck that they are all conjugate. (This is done in the supplementary materials.)Suppose that L ∼ = 2 ⋊
7, so that L acts on M ( E ) with the structure1 ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ . Since L stabilizes, but does not centralize, a line on M ( E ), and L is a 3 ′ -group,we see that L lies in a D T -Levi subgroup, but not inside the D , by Lemma 3.14.Since D acts on M ( E ) with composition factors of dimensions 1, 16 and 10,we see that the projection L of L onto D acts on 10 (semisimply) with factorsof dimensions 1, 1, 1 and 7, whence it lies inside B T . Thus L lies inside a B T subgroup of the D T -Levi subgroup. Indeed, there is a unique conjugacy class ofsubgroups isomorphic to L in B , necessarily irreducible on M ( B ). (This can evenbe seen inside SO (3), although note that Ω (3) possesses two classes.)Then we have a diagonal subgroup. Note that the image of L in B in factlies in G (and there is a unique class of subgroups of order 56 in G (3)), with thesubgroup 2 being an exotic 2-local subgroup of G (with normalizer 2 · PSL (2)).The centralizer of G is an A , and this forms the A G maximal subgroup of G .Thus L is diagonally embedded in A G . The projection onto A is as 7, andthe action on S ( M ( A ) ∗ ) is as the sum of all non-trivial simple modules. This canbe seen as the action of A G on M ( E ) is(02 , ⊕ (10 , . As is easy to compute, there is a unique subgroup of order 7 in A with this property.Conjugacy classes of diagonal subgroups are determined by the automizers ofthe projections onto each side. The group L is normalized by an element of order3 in G (this is the whole of Out( L )), as is the 7 in A , but the latter is notnormalized by an element of order 2 (the graph automorphism of A induces this).Thus there are two conjugacy classes of diagonal subgroups, swapped by the graphautomorphism of G (which induces the graph automorphism of A ). (cid:3) Once we have understood the embeddings of H , i.e., the set H , we will provethat each element of H is contained inside a G (3) subgroup of G that is containedin a G subgroup. Thus H is Lie imprimitive, but we need strongly imprimitive.In all cases, N G ( H ) is irreducible on M ( E ), and thus is contained in a unique G (3) ′ G subgroup X . If φ ∈ Aut + ( G ) normalizes H then it necessarily normalizes X ,whence X is N Aut + ( G ) ( H )-stable, and N G ( H ) and H are strongly imprimitive. PSU (3)Now we set H ∼ = PSU (3) and L ∼ = 4 ⋊ Sym(3). As we saw in the previoussection, overgroups of L in H are labelled by cosets of C Y ( H ) in C Y ( L ). Since H acts irreducibly, the first centralizer is simply k × . The second centralizer has tenparameters, as we saw in the previous section, labelled a, b, c , . . . , c , m , . . . , m .The parameters m , m , m , m label the entries of a matrix in GL ( k ), and sosatisfy m m = m m . The parameters a and the c i are all non-zero, and b canbe any element of k . The precise matrices we use appear in the supplementarymaterials. We choose a copy of L to make the structure of C Y ( L ) clear, and thenconjugate L into E , in particular into the normalizer of a torus. This means thatwe can pull the trilinear form back to our copy of L , and then choose an element of H not in L to test coefficients of the symmetric trilinear form, as in Section 13.1.Using Gr¨obner bases it is very easy to find the two solutions to the resultingequations. (All variables are fixed in terms of m except for m , which satisfies aquadratic, yielding the two solutions.) Thus there are exactly two subgroups H of G containing a given subgroup L . In the supplementary materials we also providea direct proof using specific equations, although that proof is long-winded.We find exactly two subgroups H containing L . We then check that the unique G (3) subgroup of Y containing each PSU (3) is also contained in E , by showingthat it leaves the form invariant. This completes the proof of the following. Proposition . If H ∼ = PSU (3) acts irreducibly on M ( E ) , then H iscontained in a copy of G (3) inside G and H is strongly imprimitive. Strong imprimitivity was shown at the end of Section 13.1. G (3) ′ Now we set H ∼ = PSL (8) and L ∼ = 2 ⋊
7. As we saw in Section 13.1, overgroupsof (one of the two options for) L in H are labelled by cosets of C Y ( H ) in C Y ( L ).We also saw there that C Y ( H ) is simply three copies of k × , and C Y ( L ) has fifteenparameters. We label these a, b, c, d, e, f, m , . . . , m . The parameters m , . . . , m label the entries of a matrix in GL ( k ), and so satisfy the determinant condition.The parameters a, b, c, d, e, f are all non-zero. We perform the same analysis as inSection 13.2. Again, we use Gr¨obner bases for an easy proof, and a painstakingelimination using the equations directly as a fast, but difficult, second proof.The precise matrices we use appear in the supplementary materials. We choosea copy of L to make the structure of C Y ( L ) clear and easy to use, and then conjugate L into E , in particular into the subgroup G , as constructed in [ Tes89 ]. Thismeans that we can pull the trilinear form back to our copy of L , and then choosean element of H not in L to test coefficients of the symmetric trilinear form, as inSection 13.1. Of course, we may choose H to also lie in G , so that the identitymatrix conjugates H into our copy of G .As we saw in Section 13.1, L is unique up to conjugacy even in the finite group E ( q ), so choosing L to lie in the split Levi subgroup D T , any copy of L hascentralizer in E ( q ) of order ( q − . (Since p = 3, Z ( G ) = 1, so this centralizerintersects C Y ( H ) trivially.) On the other hand, of the maximal positive-dimensional
12 13. THE TRILINEAR FORM FOR E subgroups of G , H only lies in the G subgroup, whence its centralizer in E ( q ) istrivial. Therefore we see at least ( q − different, but E ( q )-conjugate, subgroups H containing L . We prove in the supplementary materials that the set H containsexactly ( q − elements in it, and therefore they are all E ( q )-conjugate.In the supplementary materials we confirm this theoretical calculation. Weapply the same method as above to the trivial element of H rather than an elementof H \ L . In other words, we use the trilinear form to determine when a genericelement of C Y ( L ) lies in G . Indeed, doing so yields exactly ( q − elements, andmodulo C Y ( H ) they are the same as the elements that conjugate H into G foundabove. This confirms that all elements of H are conjugate in the finite group E ( q ),and hence are contained in copies of G (3). Proposition . If H ∼ = PSL (8) acts as the sum of three -dimensionalsimple modules on M ( E ) , then H is contained in a copy of G (3) inside G and isstrongly imprimitive. Strong imprimitivity was shown at the end of Section 13.1.(This result also appears in [
Asc , Theorem 29.3].)
PSL (3)Now we set H ∼ = PSL (3) and L ∼ = 13 ⋊
3. As we saw in Section 13.1, overgroupsof L in H are not labelled by cosets of C Y ( H ) in C Y ( L ), because there are Y -automorphisms of L that do not come from Y -automorphisms of H . However, N G ( L ) induces the whole of Aut( L ) (as mentioned in the proof of Lemma 13.1).Thus if H ′ denotes the C Y ( L )-conjugates of H that lie in G , then all N Y ( L )-conjugates of H that lie in G are contained in subgroups G (3) if and only if thesame holds for H ′ .As we saw in Section 13.1, C Y ( H ) = k × and C Y ( L ) is 19-dimensional. It nowappears difficult to work directly with the equations to determine the set H ′ . Inthe supplementary materials we use Gr¨obner bases to easily deal with the relations.We also provide a proof that works directly with the equations, but it uses asubtly different method, that we give a sketch of because it might be of independentinterest. The space of H -invariant symmetric trilinear forms on V is 3-dimensional,with basis { f , f , f } . If g is an element of C Y ( L ) and H = h L, h i , then we wishto classify those g such that the space of H g -invariant symmetric trilinear forms on V contains the form determining G . Equivalently, we can think of g as a change-of-basis matrix, so we fix h and allow the basis to change, to see when we can pushthe form of G into the 3-space of H -invariant forms.Thus we let g be an element of Y , and compute the trilinear form on ( ug, vg, wg )for u, v, w in V . We then compare that value with the value of an arbitrary formin the 3-space, αf + βf + γf for some α, β, γ ∈ k ; if the two values coincide forall u, v, w then h g − lies in G .Both of these methods yield exactly two subgroups H containing L (under theaction of C Y ( L ), so four under the action of N Y ( L )). We then check that the uniquesubgroup G (3) of Y containing each of these conjugates of H is also contained in G , by showing that it leaves the form invariant. We have the following. Proposition . If H ∼ = PSL (3) acts irreducibly on M ( E ) , then H iscontained in a copy of G (3) inside G and H is strongly imprimitive. Strong imprimitivity was shown at the end of Section 13.1. ibliography [Asc] Michael Aschbacher,
The maximal subgroups of E , preprint, 170pp.[Asc87] , The 27-dimensional module for E , I , Invent. Math. (1987), 159–195.[Asc88] , The 27-dimensional module for E , II , J. London Math. Soc. (2) (1988),275–293.[Asc90a] , The 27-dimensional module for E , III , Trans. Amer. Math. Soc. (1990),45–84.[Asc90b] , The 27-dimensional module for E , IV , J. Algebra (1990), 23–39.[Asc00] , Finite group theory , second ed., Cambridge Studies in Advanced Mathematics,vol. 10, Cambridge University Press, Cambridge, 2000.[Bor89] Alexandre Borovik,
Structure of finite subgroups of simple algebraic groups , Algebrai Logika (1989), 249–279 (Russian), English transl., Algebra and Logic (1989),no. 3, 163–182 (1990).[Bou02] Nicolas Bourbaki, Lie groups and Lie algebras. Chapters 4–6 , Elements of Mathemat-ics, Springer-Verlag, Berlin, 2002, translated from the 1968 French original by AndrewPressley.[BT19] Timothy Burness and Donna Testerman, A -type subgroups containing regular unipo-tent elements , Forum Math. Sigma (2019), e12.[Car85] Roger Carter, Finite groups of Lie type , John Wiley & Sons, New York, 1985.[CCN +
85] John Conway, Robert Curtis, Simon Norton, Richard Parker, and Robert Wilson,
Atlasof finite groups , Oxford University Press, Eynsham, 1985.[CLSS92] Arjeh Cohen, Martin Liebeck, Jan Saxl, and Gary Seitz,
The local maximal subgroupsof exceptional groups of Lie type, finite and algebraic , Proc. London Math. Soc. (3) (1992), 21–48.[Coo81] Bruce Cooperstein, Maximal subgroups of G (2 n ), J. Algebra (1981), 23–36.[Cra] David A. Craven, Maximal
PSL subgroups of exceptional groups of Lie type , Mem.Amer. Math. Soc., to appear.[Cra17] , Alternating subgroups of exceptional groups of Lie type , Proc. Lond. Math.Soc. (2017), 449–501.[CW97] Arjeh Cohen and David Wales,
Finite subgroups of F ( C ) and E ( C ), Proc. LondonMath. Soc. (3) (1997), 105–150.[Fre98] Darrin Frey, Conjugacy of
Alt and SL (2 , subgroups of E ( C ), Mem. Amer. Math.Soc. (1998), no. 634, viii+162pp.[Fre01] , Conjugacy of
Alt and SL (2 , subgroups of E ( C ), J. Group Theory (2001), 277–323.[Gri91] Robert Griess, Elementary abelian p -subgroups of algebraic groups , Geom. Dedicata (1991), 253–305.[JP76] Wayne Jones and Brian Parshall, On the cohomology of finite groups of Lie type ,Proceedings of the Conference on Finite Groups (William Scott and Fletcher Gross,eds.), Academic Press, 1976, pp. 313–328.[Kle88] Peter Kleidman,
The maximal subgroups of the Chevalley groups G ( q ) with q odd,the Ree groups G ( q ) , and their automorphism groups , J. Algebra (1988), 30–71.[Law95] Ross Lawther, Jordan block sizes of unipotent elements in exceptional algebraic groups ,Comm. Algebra (1995), 4125–4156.[Law09] , Unipotent classes in maximal subgroups of exceptional algebraic groups , J.Algebra (2009), 270–293.[Law14] ,
Sublattices generated by root differences , J. Algebra (2014), 255–263. [Lit18] Alastair Litterick,
On non-generic finite subgroups of exceptional algebraic groups ,Mem. Amer. Math. Soc. (2018), no. 1207, vi+156.[LMS05] Martin Liebeck, Benjamin Martin, and Aner Shalev,
On conjugacy classes of maximalsubgroups of finite simple groups, and a related zeta function , Duke Math. J. (2005), 541–557.[LS87] Martin Liebeck and Jan Saxl,
On the orders of maximal subgroups of the finite excep-tional groups of Lie type , Proc. London Math. Soc. (3) (1987), 299–330.[LS90] Martin Liebeck and Gary Seitz, Maximal subgroups of exceptional groups of Lie type,finite and algebraic , Geom. Dedicata (1990), 353–387.[LS98] , On the subgroup structure of exceptional groups of Lie type , Trans. Amer.Math. Soc. (1998), 3409–3482.[LS99] ,
On finite subgroups of exceptional algebraic groups , J. reine angew. Math. (1999), 25–72.[LS04a] ,
The maximal subgroups of positive dimension in exceptional algebraic groups ,Mem. Amer. Math. Soc. (2004), no. 802, vi+227.[LS04b] ,
Subgroups of exceptional algebraic groups which are irreducible on an adjointor minimal module , J. Group Theory (2004), 347–372.[LSS92] Martin Liebeck, Jan Saxl, and Gary Seitz, Subgroups of maximal rank in finite excep-tional groups of Lie type , Proc. London Math. Soc. (3) (1992), 297–325.[LST96] Martin Liebeck, Jan Saxl, and Donna Testerman, Simple subgroups of large rank ingroups of Lie type , Proc. London Math. Soc. (3) (1996), 425–457.[Mag90] Kay Magaard, The maximal subgroups of the Chevalley groups F ( F ) where F isa finite or algebraically closed field of characteristic = , Ph.D. thesis, CaliforniaInstitute of Technology, 1990.[Sei91] Gary Seitz, Maximal subgroups of exceptional algebraic groups , Mem. Amer. Math.Soc. (1991), no. 441, iv+197.[Sin92a] Peter Sin, Extensions of simple modules for SL (2 n ) and SU (2 n ), Proc. LondonMath. Soc. (1992), 265–296.[Sin92b] , Extensions of simple modules for Sp (2 n ) and Suz(2 m ), Bull. London Math.Soc. (1992), 159–164.[Sin93] , Extensions of simple modules for G (3 n ) and G (3 m ), Proc. London Math.Soc. (1993), 327–357.[Ste13] David Stewart, The reductive subgroups of F , Mem. Amer. Math. Soc. (2013),no. 1049, vi+88.[Tes89] Donna Testerman, A construction of certain maximal subgroups of the algebraic groups E and F , J. Algebra (1989), 299–322.[Tho16] Adam Thomas, Irreducible A subgroups of exceptional algebraic groups , J. Algebra (2016), 240–296.[Tho20] , The irreducible subgroups of exceptional algebraic groups , Mem. Amer. Math.Soc. (2020), no. 1307, iv+191.[V¨ol89] Helmut V¨olklein, 1 -cohomology of Chevalley groups , J. Algebra127