From Thompson to Baer-Suzuki: a sharp characterization of the solvable radical
Nikolai Gordeev, Fritz Grunewald, Boris Kunyavskii, Eugene Plotkin
aa r X i v : . [ m a t h . G R ] M a r FROM THOMPSON TO BAER–SUZUKI: A SHARPCHARACTERIZATION OF THE SOLVABLE RADICAL
NIKOLAI GORDEEV, FRITZ GRUNEWALD, BORIS KUNYAVSKII,EUGENE PLOTKIN
Abstract.
We prove that an element g of prime order > R ( G ) of a finite (or, more generally,a linear) group if and only if for every x ∈ G the subgroup gen-erated by g, xgx − is solvable. This theorem implies that a finite(or a linear) group G is solvable if and only if in each conjugacyclass of G every two elements generate a solvable subgroup. Introduction
The classical Baer–Suzuki theorem [Ba], [Su2], [AL] states that
Theorem 1.1 (Baer–Suzuki) . The nilpotent radical of a finite group G coincides with the collection of g ∈ G satisfying the property: forevery x ∈ G the subgroup generated by g and xgx − is nilpotent. Within past few years a lot of efforts have been made in order todescribe the solvable radical of a finite group and to establish a sharpanalogue of the Baer–Suzuki theorem with respect to the solvabilityproperty (see [Fl2], [Fl3], [GGKP1], [GGKP2]). In particular, thefollowing problem is parallel to the Baer–Suzuki result:
Problem 1.2.
Let G be a finite group with the solvable radical R ( G ).What is the minimal number k such that g ∈ R ( G ) if and only ifthe subgroup generated by x gx − , . . . , x k gx − k is solvable for every x , . . . , x k in G ?Recently (see [GGKP3]) it was proved that Theorem 1.3.
The solvable radical of a finite group G coincideswith the collection of g ∈ G satisfying the property: for everythree elements a, b, c ∈ G the subgroup generated by the conjugates g, aga − , bgb − , cgc − is solvable. Theorem 1.3 is sharp: in the symmetric groups S n ( n ≥
5) everytriple of transpositions generates a solvable subgroup.However, as mentioned by Flavell [Fl2], one can expect a preciseanalogue of the Baer–Suzuki theorem to hold for the elements of primeorder greater than 3 in R ( G ). Our main result confirms this expecta-tion: Theorem 1.4.
Let G be a finite group. An element g of prime order ℓ > belongs to R ( G ) if and only if for every x ∈ G the subgroup H = h g, xgx − i is solvable. Theorem 1.4 together with Burnside’s p α q β -theorem implies Corollary 1.5.
A finite group G is solvable if and only if in eachconjugacy class of G every two elements generate a solvable subgroup. Remark 1.6.
A standard argument (cf. [GKPS, Theorem 4.1],[GGKP3, Theorem 1.4]) shows that Theorem 1.4 and Corollary 1.5remain true for the linear groups (not necessarily finite).
Remark 1.7.
Corollary 1.5 can be viewed as an extension of a theo-rem of J. Thompson [Th], [Fl1] which states that a finite group G issolvable if and only if every two-generated subgroup of G is solvable. Remark 1.8.
The proof of Theorem 1.4 uses the classification of fi-nite simple groups (CFSG). The proof of Corollary 1.5 can be obtainedwithout classification using the above mentioned J. Thompson’s char-acterization of the minimal non-solvable groups. Flavell managed toprove, without CFSG, an analogue of Theorem 1.3 for k = 10 [Fl2]and Theorem 1.3 under the additional assumption that g ∈ G is ofprime order ℓ > Remark 1.9.
R. Guralnick informed us that Theorems 1.4 and 1.3were independently proved in forthcoming works by Guest, Guralnick,and Flavell [Gu], [FGG]. Flavell [FGG] reduced k in Problem 1.2 to7 with a proof which does not rely on CFSG. Remark 1.10.
The problem of explicit description of the solvable rad-ical of a finite group in terms of quasi-Engel sequences (see [BBGKP],[GPS]) is still open: there is no explicit analogue of Baer’s theorem oncharacterizing the nilpotent radical as the collection of Engel elements.However, a recent result by J. S. Wilson [Wi], stating the existence ofa countable set of words in two variables (in spirit of [BW]) which can
ROM THOMPSON TO BAER–SUZUKI 3 be used to describe the solvable radical, gives much hope for such acharacterization.The results of the present paper were announced in [GGKP4].
Notational conventions . Whenever possible, we maintain the nota-tion of [GGKP2] which mainly follows [St1], [Ca2]. Let G (Φ , K ) be aChevalley group where Φ is a reduced irreducible root system and K isa field. Denote by W = W (Φ) the Weyl group corresponding to Φ. De-note by ˙ w a preimage of w ∈ W in G (Φ , K ). Twisted Chevalley groupsand Suzuki and Ree groups are denoted by d G (Φ , K ) , d = 2 ,
3. Wecall Chevalley groups (twisted, untwisted, Suzuki and Ree groups) thegroups of Lie type. Chevalley groups G (Φ , K ) are denoted throughoutthe paper mostly as groups of type Φ( K ). Correspondingly, for finitefields K = F q , q = p n , they are denoted just by Φ( q ) . We adopt thenotation of [Ca2] for twisted Chevalley groups which means that weuse the symbols Φ( q ) but not Φ( q ). For example, simple unitarygroups are denoted either as A n ( q ), or as P SU n ( q ) (and not by P SU n ( q )), or as P SU n ( F ), where F is a quadratic extension of K .We use the same notation for Suzuki and Ree groups (this means thatin these cases q is not integer because q is an odd power of 2 or 3).We use the standard notation u α ( t ), α ∈ Φ, t ∈ K , for elementaryunipotent elements of G . Correspondingly, split semisimple elementswill be denoted by h α ( t ) , t ∈ K ∗ , where K ∗ is the multiplicative groupof K .We say that a finite group G is almost simple if it has a uniquenormal simple subgroup L such that L ≤ G ≤ Aut ( L ). In the clas-sification of automorphisms we follow [GLS, p. 60], [GL, p. 78]. Thismeans that all automorphisms of an adjoint group of Lie type are sub-divided to inner-diagonal automorphisms, field automorphisms, graphautomorphisms, and graph-field ones (for non-adjoint groups see [GL,p. 79]). Recall that according to [GLS, Definition 2.5.13], any fieldautomorphism of prime order ℓ > L is conjugate in Aut ( L ) to astandard one in the sense of [St1].We use the formula [ x, y ] = xyx − y − to denote the commutator.If H is a subgroup of G , we denote by H a , a ∈ G , the subgroup aHa − . For the group of fixed points of an automorphism a of agroup H we use the centralizer notation C H ( a ) (both for inner andouter automorphisms of H ). The only exception is the symbol G F ,which is traditionally used for denoting the group of fixed points of GORDEEV, GRUNEWALD, KUNYAVSKII, PLOTKIN a simple algebraic group with respect to a Frobenius endomorphism(see [Ca2]).We use below some standard language of algebraic groups ([Sp],[Hu]). Here we consider only algebraic groups defined over a finitefield F q and therefore sometimes we identify such groups with thegroups of points over the algebraic closure F q . By a Chevalley groupwe mean here the group of points of a reductive algebraic group whichis defined and quasisplit over K . Note that all groups are quasisplitover finite fields. 2. Strategy of proof
As in [GGKP1]–[GGKP3], we reduce Theorem 1.4 to the followingstatement:
Theorem 2.1.
Let G be a finite almost simple group, and let g ∈ G be of prime order > . Then there is x ∈ G such that the subgroupgenerated by g and xgx − is not solvable. Although this reduction is fairly standard, we sketch its main stepsbelow. Let S ( G ) be the set of all elements g ∈ G of prime order biggerthan 3 such that for every x ∈ G the subgroup h g, xgx − i is solvable.Obviously, any element of R ( G ) of prime order > S ( G ),and we have to prove the opposite inclusion. We may assume that G is semisimple (i.e., R ( G ) = 1), and we shall prove that G doesnot contain elements from S ( G ). Assume the contrary and consider aminimal counterexample (i.e. a semisimple group G of smallest orderwith S ( G ) = ∅ ).It is easy to see that any g ∈ G acts as an automorphism (denotedby the same letter g ) on the CR-radical V of G (see [Ro, 3.3.16]) andthat V = H × · · · × H n where all H i , ≤ i ≤ n , are isomorphic (sayto H ) nonabelian simple groups ([GGKP2, Section 2]). Suppose that g = 1 belongs to S ( G ). Let us show that g cannot act on V as anon-identity element of the symmetric group S n .Since g ∈ S ( G ), the subgroup Γ = h g, xgx − i is solvable for any x ∈ G . Take x ∈ V . Evidently, Γ contains the elements [ g, x ] = gxg − x − = g ( x ) x − and g ( x ) x − . Denote by σ the element of S n corresponding to g . ROM THOMPSON TO BAER–SUZUKI 5
Suppose σ = 1. Since the order of σ is greater than 3, we mayassume that there exist i and j such that σ ( j ) = i and σ ( i ) = 1.Take x = ( x , . . . , x n ) ∈ V such that x j = b , x i = a , where a and b are generators of the simple group H and x k = 1 for k = i, j ,1 ≤ k ≤ n . Then the group h g ( x ) x − , g ( x ) x − i is not solvable since( g ( x ) x − ) = a and ( g ( x ) x − ) = b and these elements generate asimple group H . Contradiction with the assumption that Γ is solvable.So we may assume that an element g ∈ S ( G ) acts as an automor-phism ˜ g of the simple group H . Then we consider the extension of H by ˜ g . Denote this almost simple group by G . By Theorem 2.1, G contains no elements from S ( G ). Contradiction with the choice of ˜ g .So the rest of the paper is devoted to the proof of Theorem 2.1.We refer to the property stated in Theorem 2.1 as Property (NS) (for“non-solvable”): (NS) For every g ∈ G of prime order > there is x ∈ G suchthat the subgroup generated by g and xgx − is not solvable. We use CFSG to prove, by case-by-case analysis, that every almostsimple group satisfies (NS) . Section 3 deals with alternating and spo-radic groups. In Section 4 we consider groups of Lie type of rank 1.In Section 5 the general case is treated. Finally, the exceptional case F is treated separately in Section 6.3. Alternating, symmetric, and sporadic groups
Let G be an almost simple group, L ≤ G ≤ Aut L . Lemma 3.1.
Let L = A n , n ≥ , be the alternating group on n letters.Then G satisfies ( NS ) .Proof. Clearly it is enough to consider the alternating groups: asAut ( A n ) = S n for n = 6 and [Aut ( A ) : A ] = 4, any element ofodd order in Aut ( A n ) lies in A n . So let G = A n , n ≥
5. For n = 5the proof is straightforward, so we may proceed by induction. Wemay thus assume that g acts without fixed points, so n = kℓ , where ℓ stands for the order of g , and g is a product of k disjoint cyclesof length ℓ . If k = 1, we can conjugate g = (12 . . . ℓ ) by a 3-cycle z = (123) to see that h g, zgz − i = A ℓ . For k >
1, we conjugate g by aproduct of k (cid:3) GORDEEV, GRUNEWALD, KUNYAVSKII, PLOTKIN
Lemma 3.2.
Let L be a sporadic simple group. Then G satisfies ( NS ) .Proof. As the group of outer automorphisms of any sporadic groupis of order at most 2, it is enough to treat the case where G is asimple sporadic group. Here the proof goes, word for word, as in[GGKP1, Prop. 9.1]. Namely, case-by-case analysis shows that anyelement g ∈ G of prime order ℓ > G , or its normalizer is a maximal subgroup of G .In the latter case it is enough to conjugate g by an element x notbelonging to N G ( h g i ) to ensure that h g, xgx − i = G . (cid:3) Groups of Lie rank 1
In this case our proof combines arguments of several different types.In the cases L = P SL ( q ) and L = P SU ( q ) we use the analysis of [GS]with appropriate modifications whenever needed. The case of innerautomorphisms of Suzuki and Ree groups is treated in the same spiritas in [GGKP1] (see Proposition 4.6). The case of field automorphismsof Ree groups can be reduced to the P SL -case. Finally, in the case offield automorphisms of Suzuki groups we apply a counting argumentsimilar to [GS].Before starting the proof, let us make some preparations. The fol-lowing fact is well known. Proposition 4.1.
Let G be a finite almost simple group of Lie type,and let g ∈ G be an element of prime order ℓ > . Then g is eitheran inner-diagonal or a field automorphism of L .Proof. See [GL, p. 82, 7-3] and [LLS, Proposition 1.1]. (cid:3)
Proposition 4.2.
Let L be one of the following groups: A (9) , G (3) , A (2) , A (3) , B (2) , B (3) , G (2) , G (3) , A (9) , A (9) , D (8) , D (27) , F (2) . Then G satisfies ( NS ) .Proof. We use [GGKP2, Table 1] and straightforward MAGMA com-putations with outer automorphisms of L . (cid:3) Remark 4.3.
If a group L from the above list is not simple, thecomputations have been made for its derived subgroup L ′ which issimple. ROM THOMPSON TO BAER–SUZUKI 7
So from now on we can exclude the groups listed in Proposition 4.2from the further considerations.Recall now, for the reader’s convenience, a theorem of Gow [Gow]which is essential in our argument.
Let L be a finite simple group of Lie type, and let z = 1 be a semisim-ple element in L . Let C be a conjugacy class of L consisting of regularsemisimple elements. Then there exist g ∈ C and x ∈ L such that z = [ g, x ] . Theorem 4.4.
Suppose that the Lie rank of L is 1. Then G satisfies ( NS ) .Proof . Let g ∈ G be of prime order >
3. We shall check that there is x ∈ L such that the subgroup of G generated by g and xgx − is notsolvable. Proposition 4.5. If L = P SL ( q ) , q ≥ , or L = P SU ( q ) , q > ,then G satisfies ( NS ) .Proof. In the case L = P SL ( q ) the result follows from [GS, Lemma3.1]. If L = P SU ( q ), q >
2, the result follows from the proof of [GS,Lemma 3.3] with the single exception when g is a field automorphism.In the latter case we can take g to be standard. The order of g is aprime number bigger than 3, and we may thus assume that q = 2 , g normalizes but does not centralize a subgroup of type A generated by a self-conjugate root of A . The result follows from [GS,Lemma 3.1]. (cid:3) Let L be a Suzuki group B ( q ) or a Ree group G ( q ) where q is an odd power of 2, in the Suzuki case, or of 3, in the Ree case (see,e.g., [Su1], [Kl2], [LN]). Then L = G F where G is the correspondingsimple algebraic group (of type B or G ) defined over the field F or F , and F is the appropriate Frobenius endomorphism of G ([SS],[Hu2, 1.3, 20.1]). There exists an F -stable Borel subgroup B ≤ G .The group B F will be called below a Borel subgroup of L . We fix oneof such subgroups B . Every Borel subgroup of L is of the form B a forsome a ∈ L . We will denote by T a maximal subgroup of semisimpleelements of a Borel subgroup B a . Note that T is the subgroup of F -fixed elements of an F -stable torus of G contained in an F -stable Borelsubgroup of G . Hence we will call such a group T a quasisplit torus of L . Furthermore, we denote by T any group of F -fixed elements of GORDEEV, GRUNEWALD, KUNYAVSKII, PLOTKIN an F -stable torus of G which is not contained in any F -stable Borelsubgroup of G . We call such a group a nonsplit torus of L . Note that T ∩ B a = 1 for every a ∈ L . Recall that all maximal tori in Suzuki–Reegroups are cyclic (see [Su1], [V]).For Suzuki and Ree groups, we consider the cases of inner and outerautomorphisms separately. Since all diagonal automorphisms are innerin Suzuki–Ree groups [CCNPW], the case of inner-diagonal automor-phisms is reduced to the case of inner ones. We start with the casewhere g is an inner automorphism. Proposition 4.6. If L is a Suzuki group B ( q ) , q = 2 m +1 , m ≥ ,or a Ree group G ( q ) , q = 3 m +1 , m ≥ , and g ∈ L is of primeorder > , then there exists x ∈ L such that the group h g, xgx − i isnot solvable.Proof. As the order of g is greater than 3, it cannot be unipotent, so wemay and shall assume that g is semisimple. We argue as in [GGKP1,Section 4]. Note that all tori in the Suzuki and Ree groups are cyclic,and all semisimple elements of order greater than 3 are regular [Su1],[Kl2], [LN]. By Gow’s theorem, for every semisimple element z ′ ∈ L we can find x, y ∈ L such that z = yz ′ y − = [ g, x ]. Consider two cases: • g is a generator of some maximal quasisplit torus; • g is not a generator of any maximal quasisplit torus.In the first case, choose x so that z = [ g, x ] would be a generator ofa nonsplit torus. In the second case, choose x so that z = [ g, x ] wouldbe a generator of some quasisplit torus.Note that in both cases g / ∈ h z i .With such a choice of x , let H = h g, xgx − i . By construction, wehave T ≤ H for some quasisplit torus T .First note that H is not contained in N L ( T ). Indeed, g and z cannotboth normalize T since they are of prime order > T , whereas the order of N L ( T ) /T is 2.Furthermore, H is not contained in any Borel subgroup. Indeed, ifboth g and xgx − belong to a Borel subgroup B ′ = T ′ U ′ (where T ′ isa fixed maximal quasisplit torus and U ′ is the subgroup of unipotentelements), then we are in the second case. Consider the cyclic group B ′ /U ′ . Let ¯ g and xgx − = ¯ g be the corresponding images. Then ¯ g ROM THOMPSON TO BAER–SUZUKI 9 and ¯ g are of the same order, h ¯ g i 6 = T ′ , h ¯ g i 6 = T ′ , but h ¯ g − ¯ g i ∼ = T ′ .Contradiction.The Suzuki groups have no maximal subgroups other than N L ( T ), N L ( T ), B , and Suzuki groups over smaller fields [Su1]. The subgroup H is not contained in a subgroup of the latter type since H contains amaximal torus of L . Furthermore, H is not contained in N L ( T ) since itcontains a quasisplit torus. So we conclude that H = L . Using similararguments and the list of the maximal subgroups of Ree groups [Kl2],[LN], one can show that H lies in a maximal subgroup of a Ree grouponly in two cases: either q = 3 (which is excluded by Proposition4.2), or H < P SL ( q ). In the latter case we use Proposition 4.5 toconclude that H = P SL ( q ). (cid:3) It remains to consider the case of outer automorphisms of primeorder of Suzuki and Ree groups. Any such automorphism is a fieldautomorphism (see Proposition 4.1) which is assumed to be standard.We start with the simpler case of Ree groups.
Proposition 4.7.
Let L be a Ree group G ( q ) , q = 3 m +1 , m ≥ ,and let g be a field automorphism of L of prime order ℓ > . Thenthere exists x ∈ L such that h g, xgx − i is not solvable.Proof. The group L contains a subgroup isomorphic to P SL ( K ), K = F q . This subgroup is generated by the elementary unipotent elementsof L of type u A ( t ) , u − A ( t ), t ∈ K , where u A ( t ) = u a + b ( t ϑ ) u a + b ( t ) ,ϑ : K → K , 3 ϑ = 1, and a , b are the short and long simple roots of G , respectively (see [LN]).Hence g normalizes and does not centralize P SL ( K ). Thus theassertion of the proposition follows from Proposition 4.5. (cid:3) Proposition 4.8.
Let L be a Suzuki group B ( q ℓ ) , q = 2 m +1 , m ≥ , and let g be a field automorphism of L of prime order ℓ greaterthan 3. Then there is x ∈ L such that the subgroup h g, xgx − i ∩ L isnot solvable.Proof . Denote by Γ the set of all y = xgx − , x ∈ L , such that the groupΓ y := h g, y i∩ L is solvable. We shall prove that | Γ | < |{ aga − | a ∈ L }| .Note that Γ y is invariant under the action of g because g ∈ h g, y i , gLg − = L , and Γ y = h g, y i ∩ L . Fix a Borel subgroup
B < L , a maximal quasisplit torus
T < B , andmaximal nonsplit tori T of L which are invariant under the action of g . It is known that either T = T or T = T , where the orders of cyclicgroups T , T are q ℓ + p q ℓ + 1 and q ℓ − p q ℓ + 1 respectively (see[Su1], [SS]). Recall that every maximal subgroup of L is conjugate to B , N L ( T ), N L ( T ), or is isomorphic to a Suzuki group over a smallerfield.For every y ∈ Γ the group Γ y lies in some maximal subgroup of L .So Γ y < H a where a ∈ L and H is of one of the above types.The case when H is a Suzuki group over a smaller field can beexcluded because the essential case Γ y = H = Sz ( q ′ ), q ′ | q , cannotoccur. Indeed, Sz ( q ) is solvable if and only if q = 2. Lemma 4.9.
With the above notation, we have Γ y = Sz (2) . Proof.
Assume the contrary. We have Γ y = Sz (2) = h a i ⋊ h b i , a = 1, b = 1, bab − = a . Note that Γ y is a normal subgroup in h g, y i , thesubgroup h a i coincides with the derived subgroup of Γ y = Sz (2), andit is a characteristic subgroup in Γ y = Sz (2). Since the order ℓ of g is prime >
3, we conclude that gag − = a , yay − = a , and hence γaγ − = a for every γ ∈ h g, y i . Since bab − = a , we have b / ∈ h g, y i .Contradiction. (cid:3) Lemma 4.10.
Suppose that Γ y is contained in N L ( T a ) or in N L ( T a ) where a ∈ L . Then Γ y is contained in T a ( and thus in B a ) or in T a ,respectively. To prove Lemma 4.10, we need two more auxiliary assertions.
Sublemma 4.11.
Suppose that Γ y ≤ N L ( T a ) ( respectively, Γ y ≤ N L ( T a )) . Then T y := Γ y ∩ T a ( respectively, T y := Γ y ∩ T a ) is g -invariant.Proof. Let Γ y ≤ N L ( T a ). Denote by Γ y the subgroup of Γ y generatedby the squares of the elements of Γ y . It is clear that Γ y is invariantunder g . Since | N L ( T ) /T | = 2, all elements of Γ y belong to T , soΓ y lies in T y . However, all elements of T are of odd order, thereforeΓ y = T y . Hence T y is invariant under g . In order to get the statementfor T y , we repeat the above argument with Γ y replaced by Γ y . (cid:3) ROM THOMPSON TO BAER–SUZUKI 11
Sublemma 4.12.
Suppose that Γ y is contained in N L ( T a ) or in N L ( T a ) . Then for any integer r the element [ g − r , x ] belongs to T y or to T y , respectively.Proof. If ℓ | r , the assertion is satisfied for trivial reasons, so assumethat r is prime to ℓ . Let Γ y ≤ N L ( T a ). Then z := g − r xg r x ∈ N L ( T a ).Assume z ∈ Γ y \ T y . Then g − r ( x ) = g − r xg r = zx . Furthermore, g − r can act only trivially on Γ y /T y since for r prime to ℓ the order of g − r is ℓ > y /T y is ≤
4. Hence g − r ( z ) z − ∈ T y . Thus x = ( g − r ) ℓ ( x ) = z ℓ xt with t ∈ T y , so z ℓ ∈ T y . But z ∈ N L ( T a ) \ T a and | N L ( T a ) /T a | = 2, therefore z ℓ T a . Contradiction. Hence z ∈ T y .The same proof can be given for the case Γ y ≤ N L ( T a ). (cid:3) We are now ready to prove Lemma 4.10.Let a = g , b = xgx − . Then any word w = · · · a k b m a n · · · canbe written as · · · a k + m ( a − m b m ) a n · · · = · · · g u [ g − m , x ] g v · · · , where thecommutator in the middle belongs to T y in view of Sublemma 4.12.Therefore w = g u z g u z · · · , where z , z ∈ T y . Since w ∈ L , the sum u + u + · · · is divisible by ℓ , so w = ( g u z g − u )( g u + u z g − u − u )( g u + u + u · · · ) = Y ( g v i z i g − v i ) . By Sublemma 4.11, the latter element must belong to T y .The case Γ y ≤ N ( T a ) is treated in exactly the same way. (cid:3) We thus may and shall assume that Γ y ≤ H a where H = B or H = T .We are now able to estimate the number of elements y = xgx − , x ∈ L , such that the group Γ y = h g, y i ∩ L is solvable.Denote A H := { H a | gH a g − = H a , a ∈ L } .Denote L = h L, g i . Note that h g, y i 6 = L because the group h g, y i is solvable. Hence h g, y i is contained in a proper maximal subgroup M of L . By [Kl1], M is conjugate to a subgroup of the form N L ( H a ) = h H a , g i where H a is a maximal subgroup in L invariant under g .So we may assume that h g, y i lies in a semidirect product H a ⋊ h g i and Γ y lies in some H a such that gH a g − = H a . We have yH a y − = H a because H a is normal in M . The equality yH a y − = H a canbe rewritten as g ( x − H a x ) g − = x − H a x , or, in other words, as x − H a x ∈ A H . Hence it is enough to estimate, for each H , the numberof elements in the set S H := { s ∈ L : sH a s − = H a ′ for some H a , H a ′ ∈ A H } . Lemma 4.13. | S B | ≤ q ℓ +9 . Proof.
First recall that we denote L = B ( q ℓ ). We have C L ( g ) = B ( q ). The orders of the above groups are q ℓ ( q ℓ − q ℓ + 1) and q ( q − q + 1), respectively.We have H = B = T U where U is the maximal unipotent subgroupof B . Denote by C U ( g ) the centralizer of g in U . By the definitionof S H we have S B ⊇ B (because B ∈ A B ). Furthermore, let s / ∈ B and suppose that sBs − is g -invariant. The Bruhat decompositionof L contains only two cells, hence we can represent s in the form s = u ˙ wb with u ∈ U , b ∈ B , and w the non-identity element of theWeyl group. The condition g ( sBs − ) g − = sBs − can be rewrittenas u − guB − u − g − u = B − where B − stands for the Borel subgroupopposite to B . This subgroup is invariant under g , so applying g − to the last equality we conclude that v := g − ( u − ) u normalizes B − and hence belongs to B − . On the other hand, v is a product of twoelements from U and thus belongs to U . As B − ∩ U = 1, we concludethat v = 1, i.e., u belongs to the centralizer C U ( g ). Thus the set of s / ∈ B such that sBs − is g -invariant is in one-to-one correspondencewith the set of pairs { ( b, u ) } with b ∈ B and u ∈ C U ( g ). The numberof such pairs equals | B | · | C U ( g ) | . Hence the number of s ∈ L suchthat sBs − is g -invariant equals | B | (1 + | C U ( g ) | ) = q ℓ ( q ℓ − q + 1) . This calculation should be repeated for every B a = aBa − such that g ( aBa − ) g − = aBa − , i.e., for each B a ∈ A B . Let us write the lastcondition in the form a − gaBa − g − a = B and use the Bruhat decom-position a = uwb for a , as in the above paragraph. The same compu-tation shows that the number of such groups B a equals | C U ( g ) | = q .Thus we conclude that | S B | ≤ q ℓ ( q ℓ − q + 1) q ≤ q ℓ +9 . (cid:3) In order to treat the case of nonsplit tori in a similar way we needthe following
ROM THOMPSON TO BAER–SUZUKI 13
Remark 4.14. If T and T are nonconjugate nonsplit tori in Sz ( q ℓ ),then they are cyclic groups of orders q ℓ + p q ℓ +1 and q ℓ − p q ℓ +1,respectively. These are odd numbers whose difference is 2 p q ℓ whichis a power of 2, hence they are coprime. Lemma 4.15.
Let T be a nonsplit g -stable torus of L = Sz ( q ℓ ) . Thenthe first cohomology group H ( h g i , T ) is trivial.Proof. It is enough to prove that H ( h g i , T ℓ ) = 1 where T ℓ is a Sylow ℓ -subgroup of T .Let us first prove that if T ℓ = 1, then g does not centralize T ℓ .Assume the contrary. Then T ℓ is contained in Sz ( q ). The order of T is either N − = q ℓ − p q ℓ + 1, or N + = q ℓ + p q ℓ + 1. Notethat N + and N − cannot be both divisible by ℓ (see Remark 4.14). As T ℓ < Sz ( q ), we have q + 1 ≡ ℓ ) (because the order of T ℓ divides both | Sz ( q ) | = q ( q − q + 1) and N + N − = q ℓ + 1) and ℓ does not divide ( q ℓ + 1) / ( q + 1). On the other hand, q + 1 ≡ ℓ ) implies ( q ℓ + 1) / ( q + 1) = (( q ) ℓ − − ( q ) ℓ − + · · · + 1) ≡ ℓ ). Contradiction. Thus g acts nontrivially on T ℓ .As T is a cyclic group, we finish the proof by noting that if a cyclicgroup C of the order ℓ acts nontrivially on a cyclic ℓ -group M ( ℓ odd),we have H i ( C, M ) = 1 for all i ≥ (cid:3) We are now able to repeat for the tori T = T and T the compu-tations already performed for the case of Borel subgroups. Lemma 4.16. | S T | + | S T | ≤ q ℓ +24 . Proof.
Fix a maximal nonsplit torus T invariant under g . For s ∈ L such that s T s − ∈ A T , consider z := g − s − gs . Arguing as in the proofof Lemma 4.13, we arrive at the equality z T z − = T , i.e., z ∈ N L ( T ).Since g ∈ N L ( T ) (recall that L = h L, g i ), we have gz = s − gs ∈ N L ( T ), and therefore the group h g, s − gs i is contained in N L ( T ), so h g, s − gs i ∩ L is contained in N L ( T ). By Lemma 4.10, this group iscontained in T , so z defines a cocycle with values in T . By Lemma4.15, H ( h g i , T ) = 1, therefore z = g − t − gt with t ∈ T . Therefore g ( ts − ) = ts − whence ts − = a ∈ C L ( g ) = Sz ( q ). Thus s = a − t with a ∈ Sz ( q ), t ∈ T . Therefore the number of elements s ∈ L suchthat s T s − ∈ A T is bounded by | T | · | Sz ( q ) | = | T | q ( q − q + 1) . This estimate should be repeated for each T γ ∈ A T , i.e., for every T γ such that g T γ g − = T γ . We have seen above that T γ = γ T γ − ∈ A T ifand only if γ ∈ Sz ( q ). Hence the the number of groups T γ is boundedby | Sz ( q ) | .Thus | S T | ≤ | T | · q ( q − q + 1) · ( q ( q − q + 1))= | T | q ( q − ( q + 1) ≤ | T | q . Since there are two nonconjugate nonsplit tori T and T , we get | S T | + | S T | ≤ q ( | T | + | T | ). Recall that the orders of the tori are q ℓ − p q ℓ + 1 and q ℓ + p q ℓ + 1, so each order is less than q ℓ .Thus we get the needed estimate | S T | + | S T | ≤ q ℓ +22 ≤ q ℓ +24 . (cid:3) We can now finish the proof of Proposition 4.8.By Lemmas 4.13 and 4.16, the total number of all elements s ∈ L such that sH a s − ∈ A H for some H a ∈ A H is bounded by q ℓ +24 + q ℓ +9 . If ℓ ≥ ℓ + 9 ≥ ℓ + 24, and we have q ℓ +24 + q ℓ +9 ≤ q ℓ +9 ≤ q ℓ +11 . We have proved above that if the group h g, xgx − i is solvable then h g, xgx − i ≤ h H a , g i for some g -stable maximal subgroup H a ≤ L .Moreover, in this case xH a x − is also g -stable. Hence if h g, xgx − i issolvable then there exists a g -stable maximal subgroup H a such that xH a x − is also g -stable. We have just estimated the number of those x such that there is a g -stable maximal subgroup H a for which thegroup xH a x − is also g -stable. This number is not more than q ℓ +11 .But q ℓ +11 < q ℓ < | L | = q ℓ ( q ℓ − q ℓ + 1) . Thus we can find x ∈ L such that the group h g, xgx − i is nonsolvable.Proposition 4.8 is proved. (cid:3) Theorem 4.4 now follows from Propositions 4.5, 4.6, 4.7, and 4.8. (cid:3) General case
In this section we prove the main part of Theorem 1.4 consideringalmost simple groups of Lie rank > F . The case F will be treated separately in the last section. ROM THOMPSON TO BAER–SUZUKI 15
Theorem 5.1.
Let L be a simple group of Lie type of Lie rank ≥ , L = F ( q ) , and let L ≤ G ≤ Aut L . Then G satisfies ( NS ) . Suppose that the property (NS) does not hold for some group G .We may assume for G the following property ( MC stands for “minimalcounter-example”): MC: (a) G is a finite almost simple group which does not satisfy (NS) ;(b) [ G, G ] = L is a simple group of Lie type different from F ;(c) if H is a group satisfying conditions (a) and (b), then the orderof [ H, H ] is greater than or equal to the order of L . Throughout below g ∈ G is an element of prime order ℓ > such that the group h g, xgx − i is solvable for every x ∈ L (suchan element exists according to hypothesis (a)).Suppose that g induces a field automorphism of L . Then one canfind a subgroup L = h U ± α i where U ± α are root subgroups which are g -stable but not centralized by g (this follows from the definition offield automorphism.) Since the order of g is a prime number ≥
5, thegroup L = L /Z ( L ) is a simple group of rank one, and g induceson L an automorphism of prime order ≥
5. Then the almost simplegroup G = h g, L i does not satisfy (NS) . This contradicts Theorem4.4.Thus, in view of classification of automorphisms of prime order (seeProposition 4.1), we may assume that g induces an inner-diagonalautomorphism of L , and therefore we may also assume that g ∈ G = h σ, L i where σ is a diagonal automorphism of L .5.1. Recall that the group L can be represented in the form L = [ G ( K ) , G ( K )] = G sc ( K ) /Z ( G sc ( K ))where G sc is a simple, simply connected linear algebraic group definedover a finite field K and G = G ad is the corresponding adjoint group. Lemma 5.2.
There exists a reductive algebraic group G defined overa finite field K satisfying the following conditions: (i) the centre of G is a torus and the derived group G ′ is simplyconnected; (ii) G ′ ( K ) /Z ( G ′ ( K )) ∼ = L ; (iii) there is τ ∈ G ( K ) such that h τ, G ′ ( K ) i /Z ( h τ, G ′ ( K ) i ∼ = G. Proof.
Let H be a maximal K -torus of G sc which is quasisplit over K , i.e. is contained in a K -defined Borel subgroup. Further, let˜ Z = Z ( G sc ) be the centre of G sc (here we regard ˜ Z as a finite algebraicsubgroup of G sc which is also defined over K [Sp]). We also identify˜ Z with an algebraic subgroup of H . Consider the embedding i : ˜ Z ֒ → H × G sc given by i ( z ) = ( z, z − ). The image i ( ˜ Z ) will also be denotedby ˜ Z .Define the reductive group G := ( H × G sc ) / ˜ Z . Then Z ( G ) = H , G ′ = G sc (here we identify the groups H and G sc with their images in G ). Thus we have (i), (ii).Note that there exists an automorphism ˜ σ of G sc which inducesthe given diagonal automorphism σ of L (because σ is defined by itsaction on the root subgroups). All such automorphisms are inner in G sc . Thus we may assume ˜ σ ∈ G sc . Let F denote the Frobeniusmap naturally acting on G such that G F sc = G sc ( K ), H F = H ( K ) . Then F (˜ σ ) and ˜ σ induce the same automorphism of G (check this onroot subgroups). Hence F (˜ σ )˜ σ − ∈ Z ( G ) = H . By Lang’s theorem, H ( F, H ) = 1, therefore we have F (˜ σ )˜ σ − = F ( t − ) t for some t ∈ H .Hence τ = ˜ σt ∈ G ( K ) . This gives (iii). (cid:3)
Lemma 5.3.
We have G ≤ G ( K ) . Proof.
The quotient G /Z ( G ) coincides with the adjoint group G . Let θ : G → G be the natural homomorphism of algebraic groups. Wehave θ ( G ( K )) ≤ G ( K ) . Lemma 5.2 implies G = h σ, L i ≤ θ ( G ( K )) ≤ G ( K ) . (cid:3) Let GL n , SL n be the algebraic groups such that GL n ( E ) = GL n ( E ), SL n ( E ) = SL n ( E ) for every field E . Further, let K = F q , let K be analgebraic closure of K , and let Gal( K/K ) = h τ i where τ ( α ) = α q forevery α ∈ K . Denote by σ the automorphism of GL n ( K ) , SL n ( K )given by the formula σ ( A ) = ( A − ) t . The automorphism τ of K also defines the automorphism of the matrix groups GL n ( K ), SL n ( K ) ROM THOMPSON TO BAER–SUZUKI 17 which will be denoted by the same symbol τ . For every natural number m we denote by F m the Frobenius maps: F m = ( στ ) m : GL n ( K ) → GL n ( K ) , SL n ( K ) → SL n ( K ) . Denote by U n ( q ), SU n ( q ) the quasisplit forms of GL n , SL n definedover K = F q such that U n ( q )( F q m ) = GL n ( K ) F m , SU n ( q )( F q m ) = SL n ( K ) F m . Lemma 5.4.
Suppose that L = A n − ( q ) or L = A n − ( q ) . Then inLemma 5.2 one can take G = GL n or G = U n ( q ) , respectively.Proof. For the case L = A n − ( q ) the statement is obvious. Let L = A n − ( q ). Then G sc = SU n ( q ) = U n ( q ) ′ and the centre of SU n ( q ) isa one-dimensional anisotropic torus over K = F q . Thus we have (i)and (ii). Note that the proof of Lemma 5.2 implies that (iii) holds forevery reductive group G satisfying (i) and (ii). (cid:3) G and the semisimple group G as in Section 5.1.For the Chevalley groups G ( K ) and G ( K ) denote T G = T G ( K ) and T = T ( K ) where T G and T are maximal quasisplit tori of G and G .We will assume that τ ∈ T G and σ ∈ T . Further, for the Chevalleygroup G ( K ) (or G ( K )) there exists the root system Φ correspondingto T G (or T ) which either coincides with the root system R of G oris obtained from R by twisting [Ca1]. We denote by U α and U α theroot subgroups of G ( K ) and G ( K ) corresponding to α ∈ Φ. We set U G = h U α | α ∈ Φ + i , U = h U α | α ∈ Φ + i . The groups B G = T G U G and B = T U (as well as all their conjugates) are called
Borel subgroups of G ( K ) and G ( K ). Any subgroup of G ( K ) or G ( K ) which containsa Borel subgroup is called a parabolic subgroup.Fix a simple root system Π generating Φ. For Π ′ ⊂ Π denote W Π ′ = h w α | α ∈ Π ′ i . Then P Π ′ = BW Π ′ B is a standard parabolic subgroupof G ( K ). Note that every parabolic subgroup of G ( K ) is conjugate toa standard parabolic subgroup by an element of [ G ( K ) , G ( K )]. Lemma 5.5.
The element g ∈ G = h σ, L i ≤ G ( K ) does not belong toany proper parabolic subgroup of G ( K ) .Proof. Assume to the contrary that g ∈ P where P ≤ G ( K ) is a properparabolic subgroup of G ( K ). There exists γ ∈ L = [ G ( K ) , G ( K )] such that γP Π ′ γ − = P for some Π ′ ⊂ Π. Since γ ∈ L and since wemay consider the element g ∈ G = h σ, L i up to conjugacy in G , wemay assume that P is the standard parabolic subgroup P Π ′ for someΠ ′ ⊂ Π.Let us show that g is not a unipotent element. Indeed, if g is aunipotent element, then by conjugation with some appropriate ele-ment of L we can get an element u ∈ U having a nontrivial factor u α for some α ∈ Π. Again we may assume g = u = u α v, u α = 1 , v ∈ U ,and g ∈ P = P Π ′ for Π ′ = { α } . The image g of g in the quotient L = P/Z ( P ) R u ( P ) is an element of prime order ℓ >
3. Hence L isan almost simple group of Lie type of rank one which does not satisfy (NS) . This contradicts Theorem 4.4.Let us show that δgδ − / ∈ T for every δ ∈ G ( K ). Suppose δgδ − ∈ T for some δ ∈ G ( K ). Then we may assume g ∈ T (the same argumentsas above). One can then find a group G α = h U ± α i , α ∈ Π, which isnormalized but not centralized by g . Then again we have a contradic-tion with Theorem 4.4.Let now g ∈ P = P Π ′ , g / ∈ T , g / ∈ R u ( P ). Then the image g of g in L = P/Z ( P ) R u ( P ) is not trivial. Further, there exists a simplecomponent L of L which is an almost simple group of Lie type suchthat the component g of g in L is not trivial. Obviously, L =[ L , L ] is a finite simple group of Lie type = F ( q ) and | L | > | L | .Since L is a simple component of L = P/Z ( P ) R u ( P ), the image g of g can be represented in the form g = σ ′ g where g ∈ L and σ ′ ∈ L induces a diagonal automorphism of L = [ L , L ]. Then the group G = h σ ′ , L i does not satisfy (NS) . Hence we have a contradictionwith (MC) . (cid:3) Lemma 5.6.
The element g ∈ G = h σ, L i ≤ G ( K ) ≤ G is a regularsemisimple element of G .Proof. Since the order of g is prime and g is not unipotent, g issemisimple. Let C G ( g ) be the centralizer of g in G . This is a reductivesubgroup of G [Ca2, Theorem 3.5.3]. Suppose that g is not regular.Then the identity component C G ( g ) is not a torus. Since K is a finitefield, there exists a K -defined Borel subgroup B g of C G ( g ). Again, theunipotent radical R u ( B g ) is also defined and split over K [Sp, 14.4.5].Hence R u ( B g )( K ) = 1. Since R u ( B g )( K ) ≤ C G ( g )( K ) ≤ G ( K ), onecan find a nontrivial unipotent element u ∈ C G ( K ) ( g ). However, by ROM THOMPSON TO BAER–SUZUKI 19
Lemma 5.5 the element g does not lie in any proper parabolic subgroup P ≤ G ( K ), and therefore the characteristic of K does not divide theorder of C G ( K ) ( g ) [Ca2, Proposition 6.4.5]. Contradiction. (cid:3) Lemma 5.7.
The element g does not normalize any unipotent sub-group V of G ( K ) .Proof. Assume to the contrary that gV g − = V for some unipotentsubgroup V ≤ G . Then V is a closed subgroup of G (because itis finite). The following construction is due to Borel and Tits [BT].Consider the sequence of subgroups N := N G ( V ) , V := V R u ( N ) , . . . , N i = N G ( V i − ) , V i := V i − R u ( N i )in G . All groups here are defined over K . Moreover, since V is in aBorel subgroup of G (indeed, V belongs to a p -Sylow subgroup of G which is conjugate to U ≤ B ), the last term N k = P is a parabolicsubgroup of G containing N G ( V ), and therefore g ∈ P (see [Hu1,30.3]). Since g ∈ G ( K ) and P is a parabolic subgroup defined over K , we have g ∈ P ( K ) where P ( K ) is a parabolic subgroup of G ( K ).This is a contradiction with Lemma 5.5. (cid:3) w c of the Weyl group W = W (Φ)with respect to Π is a product (taken in any order) of the reflections w α , α ∈ Π, where each reflection occurs exactly once.Let now g be a preimage of g in G ( K ), see Lemma 5.2(iii). Since g is a semisimple regular element of G (Lemma 5.6), the element g isalso semisimple and regular in G . By [St2, §
9] (see also [GoS]), forevery Coxeter element w c of G ( K ) there exists x ∈ G ( K ) such that x g x − = u ˙ w c where u ∈ U G (here ˙ w c is any preimage of w c ). We have x = hy where h ∈ T G and y ∈ G ′ ( K ). Then y g y − = u ′ ˙ w c Thus we can put the element g in the Coxeter cell B G ˙ w c B G by conjuga-tion with some element from G ′ ( K ). So we may assume g ∈ B G ˙ w c B G .Therefore we may assume g ∈ B ˙ w c B and moreover(5.1) g = u ˙ w c for some u ∈ U . In [GGKP2, Section 5], it was proved that for an element g of form(5.1) with an appropriate Coxeter element w c , there is x ∈ L suchthat [ g, x ] = u ∈ U . With this choice of x , put H = h g, xgx − i . By our assumptions, H is a solvable group. Since g, u ∈ H , there isa Hall subgroup H pℓ , where p = char( K ), such that g ∈ H pℓ . Let A be the maximal abelian normal subgroup of H pℓ , and let A p be the p -Sylow subgroup of A . Suppose that A p = 1. Then A p is normalizedby g . This contradicts Lemma 5.7. Hence A p = 1. Then | A | = ℓ s . Let A [ ℓ ] = { a ∈ A | a l = 1 } , C A [ ℓ ] ( g ) = { a ∈ A [ ℓ ] | gag − = a } . We have C A [ ℓ ] ( g ) = 1 since any operator of order ℓ acting on a vectorspace over the field F ℓ is unipotent and hence has a nontrivial fixedpoint.We have C G ( g ) ≤ N G ( ˜ T ) for some maximal torus ˜ T of G (recall that g is a regular element of G ).Consider the group C G ( K ) ( g ) [ ℓ ] generated by all elements of order ℓ in C G ( K ) ( g ). Clearly, C A [ ℓ ] ( g ) ≤ C G ( K ) ( g ) [ ℓ ] . Consider three separatecases. Case 1.
Suppose that C G ( K ) ( g ) [ ℓ ] = h g i .Then C A [ ℓ ] ( g ) = C G ( K ) ( g ) [ ℓ ] = h g i . Since C A [ ℓ ] ( g ) = h g i and A is abelian, we have A [ ℓ ] = C A [ ℓ ] ( g ). Therefore h g i = A [ ℓ ] is an H pℓ -invariant subgroup. Recall that [ g, x ] = u ∈ H is unipotent. Hencethere exists a unipotent element v ∈ H pℓ . We have vgv − = g r , < r < ℓ (indeed, g is regular and therefore r = 1, otherwise g would commutewith a unipotent element). Hence g r − = [ v, g ] ∈ [ H, H ] and therefore g ∈ [ H, H ]. On the other hand, the generators of the solvable group H = h g, xgx − i are not in [ H, H ], so g / ∈ [ H, H ]. Contradiction.
Case 2.
Suppose that h g i × h a i ≤ C G ( K ) ( g ) [ ℓ ] for some a ∈ ˜ T ( K ).Let L ( G ) and L ( ˜ T ) be the Lie algebras of G and ˜ T , respectively.Then we have a subgroup of type ℓ × ℓ in ˜ T which acts by conjugationon the linear space L = L ( G ) / L ( ˜ T ) defined over a field of characteristic p . Since q and ℓ are coprime, by Maschke’s theorem this action is ROM THOMPSON TO BAER–SUZUKI 21 diagonalizable. This implies that there exists b ∈ h a i×h g i stabilizing anonzero vector from L . Then C G ( b ) is a K -defined reductive subgroupof G of nonzero semisimple rank because the Lie algebra of C G ( b )is not equal to the Cartan subalgebra L ( ˜ T ) (see [Ca2, 1.14]). Theidentity component C G ( b ) is also defined over K [Sp, 12.1.1]. Since K is a finite field, there exists a K -defined Borel subgroup of C G ( b ).Hence the group C G ( b )( K ) is not a torus (see the proof of Lemma 5.6).Further, g ∈ ˜ T ( K ) ≤ C G ( b )( K ) (cid:8) G ( K ) . Note that g does not commute with unipotent elements of C G ( b )( K ).Then there exists a subgroup M ≤ C G ( b )( K ), which is a Chevalleygroup over some finite extension of K , such that g normalizes M butdoes not centralize it and [ M, M ] /Z ( M ) is a finite group of Lie type.There exists m ∈ M such that m ∈ M/Z ( M ) induces a diagonalautomorphism of [ M, M ] /Z ( M ) and g ∈ h m, [ M, M ] /Z ( M ) i . Thegroup h m, [ M, M ] /Z ( M ) i does not satisfy (NS) but | [ M, M ] /Z ( M ) | < | L | . This is a contradiction with (MC) . Case 3.
Suppose that h g i × h a i ≤ C G ( K ) ( g ) [ ℓ ] for some a / ∈ ˜ T ( K ).We have aga − = g in G , and thus a ∈ C G ( g ) ≤ N G ( ˜ T ). As a / ∈ ˜ T ( K ), we have a ∈ N G ( ˜ T ) \ ˜ T . Let g , a , T be preimages in G of g, a, ˜ T , respectively. Since G /Z ( G ) = G , we have(5.2) aga − = gc for some c ∈ Z ( G ). Note that C G ( g ) = T because g is regular in G and G ′ is simply connected. Since a ∈ N G ( ˜ T ) \ ˜ T , we have c = 1. Lemma 5.8.
Equality ( ) cannot hold except possibly for the cases G ′ = SL ℓ or G ′ = SU ℓ ( q ) .Proof. As a is a preimage of a and a ℓ = 1, we have a ℓ ∈ Z ( G ). Hence c ℓ = 1. Thus ℓ is the order of c (recall that c = 1). Note that ℓ dividesthe order of Z ( G ′ ) because c = [ a , g ] ∈ Z ( G ′ ). Since ℓ is a prime ≥ G ′ = SL n or SU n ( q ) for some n .Now we may assume G = GL n or G = U n ( q ) (Lemma 5.4).Choose a preimage g of g of ℓ -power order, say, ℓ s . We have g ℓ ∈ Z ( G ( K )). We have G ( K ) = GL n ( K ). Note that g is a regular elementin GL n ( K ). Therefore n ≤ ℓ because all eigenvalues of g are differentand are of the form ǫ ℓ s ǫ mℓ where ǫ ℓ s and ǫ ℓ stand for fixed roots of unity of degrees ℓ s and ℓ , respectively. Suppose that n < ℓ . Thenthe Weyl group W ( GL n ) has no elements of order ℓ . The element a is of order ℓ and, according to the hypothesis of Case 3, belongsto N G ( ˜ T ) \ ˜ T . Since every element of N G ( T ) / T coincides with someelement of W ( GL n ) [Ca2, Proposition 3.3.6], we have a / ∈ N G ( T ) \ T ,and therefore a / ∈ N G ( ˜ T ) \ ˜ T , contradiction with the choice of a . Hence G ′ = SL ℓ or G ′ = SU ℓ ( q ). (cid:3) Lemma 5.9.
The case G ′ = SL ℓ cannot occur.Proof. We have g ℓ ∈ Z ( GL ℓ ( K )). As in the previous lemma, wemay assume that g ℓ s = 1 for some s . Thus ǫ s = ℓs √ / ∈ K sinceotherwise g would be a split semisimple element of G which wouldcontradict Lemma 5.5. On the other hand, ǫ s − = ℓs − √ ∈ K since g ℓ = diag( ǫ s − , ǫ s − , . . . , ǫ s − ). Let ǫ be an ℓ th root of unity.In GL ℓ ( K ) one can represent g by a diagonal matrix of the formdiag( ǫ s ǫ, ǫ s ǫ , . . . , ǫ s ǫ ℓ ). Clearly, det ( g ) = ǫ s − and the characteristicpolynomial of g is x ℓ + ( − ℓ ǫ s − . The matrix diag( ǫ s ǫ, ǫ s ǫ , . . . , ǫ s ǫ ℓ )is conjugate over K to its companion matrix m = · · ·
00 0 1 0 0 · · · · · · · · · ǫ s − · · · ∈ GL ℓ ( K ) . Since g and m have the same characteristic polynomial and g is asemisimple matrix, we have g = y m y − for some y ∈ GL ℓ ( K ). Further, y = y d where y ∈ SL ℓ ( K ) and d is a diagonal matrix. Hence g = y − g y is a monomial matrixcorresponding to an ℓ -cycle in W ( GL ℓ ). Let now g be the imageof g in P GL n ( K ) = G ( K ). The element g is conjugate to g by anelement of P SL n ( K ) = L . Then we may assume g = g . Let M be theimage in P GL n ( K ) of all monomial matrices of GL n ( K ). Then thereexists a natural epimorphism φ : M → S ℓ . We have φ ( g ) ∈ S ℓ . Since S ℓ satisfies condition (NS) , so does M . Then there exists m ∈ M such that h g, mgm − i is not solvable which is a contradiction with thechoice of g . (cid:3) Lemma 5.10.
The case G ′ = SU ℓ ( q ) cannot occur. ROM THOMPSON TO BAER–SUZUKI 23
Proof.
The same arguments as in the previous lemma imply that theelement g ∈ U ℓ ( q )( K ) ≤ GL ℓ ( K ) is conjugate in GL ℓ ( K ) to the matrix m = · · ·
00 0 1 0 0 · · · · · · · · · ǫ s − · · · ∈ U ℓ ( q )( K ) = G ( K )for some ǫ s − ∈ ℓs − √ ∈ E = F q such that ǫ s − ǫ qs − = 1. Then theelements g and m are conjugate by an element of the group U ℓ ( q )( K )[Ca2, Proposition 3.7.3].Note that U ℓ ( q )( K ) = U ℓ ( E ) is the group of unitary matrices in GL ℓ ( E ) where E = F q , i.e., the matrices satisfying the condition( ˜ A − ) t = A where ˜ A is the matrix obtained from A by replacing allthe entries α ij with α qij .Let DU ℓ ( E ) be the set of diagonal unitary matrices over E , andlet W ℓ ≤ GL n ( K ) be the group of monomial matrices with nonzeroentries equal to 1. Then m ∈ DU ℓ ( E ) W ℓ ≤ U ℓ ( E ) . Further, it is easy to see that U ℓ ( E ) = DU ℓ ( E ) SU ℓ ( E ). Since g and m are conjugate by an element of U ℓ ( q )( K ) = U ℓ ( E ), the element g is conjugate by some element of the group SU ℓ ( E ) to some m ′ ∈ DU ℓ ( E ) W ℓ . Thus we may assume g = m ′ ∈ DU ℓ ( E ) W ℓ . Moreover,the image of g in the quotient DU ℓ ( E ) W ℓ /DU ℓ ( E ) ∼ = W ℓ ∼ = S ℓ is nottrivial. Hence, as in the previous lemma, we have a contradiction withthe choice of g . (cid:3) Theorem 5.1 is proved. (cid:3) Case F In order to complete the proof of Theorem 2.1, it remains to considerthe case of groups of type F ( q ). Theorem 6.1.
Let L be a group of type F ( q ) , q > , and L ≤ G ≤ Aut L . Then G satisfies (NS) . Proof.
Let g ∈ G. If g is a field automorphism of L , then it normalizesbut does not centralize a group of rank 1, and we can use Theorem4.4.Thus we assume that g induces an inner-diagonal automorphism of L . Note that every inner-diagonal automorphism is an inner auto-morphism in the case L = F ( q ). Hence g ∈ L . Further, one candefine Borel and parabolic subgroups in F ( q ) (see [Ca2]) because F ( q ) has a BN -pair. One can also represent L in the form G ( F ) F where G is the algebraic group of type F defined over F and F isthe Frobenius map corresponding to the group F ( q ). We can de-fine “tori” of L as groups of F -invariant elements of F -stable tori in G ( F ). Denote by T the group of F -invariant elements of an F -stablequasisplit torus. If g ∈ T , then g normalizes but does not centralize asubgroup of L which is a simple group of Lie type of rank one, and wecan use Theorem 4.4. We can also use Theorem 4.4 in the case when g belongs to a parabolic subgroup of L (see the proof of Lemma 5.5).If g does not belong to any proper parabolic subgroup P ≤ G , thenthe order of C L ( g ) is odd [Ca2, 6.4.5], and therefore (see [Gow]) wecan write every semisismple element s (up to conjugacy) in the form[ g, x ] with x ∈ L .Among maximal tori of L one can find two tori T , T satisfyingthe following conditions [Ma]:(1) T and T are cyclic groups;(2) y T y − ∩ T = 1 for every y ∈ L ;(3) N L ( T i ), i = 1 , , is the only maximal subgroup of L containing T i ;(4) the only prime divisors of | N L ( T i ) / T i | are 2 and 3.In the notation of [Ma], one can take T = T and T = T . Thesegroups are cyclic, and their orders are N − = q − √ q + q − √ q + 1and N + = q + √ q + q + √ q + 1, respectively. It is easy to checkcondition (2) by showing that N + and N − are coprime (one can seethat looking at their sum and difference).By (2), we may assume that g does not belong to a torus conjugateto one of those T , T , say, to T , but [ g, x ] is a generator of T . Sinceord g = ℓ >
3, condition (4) implies that g / ∈ N L ( T ). We have T ≤ H = h g, xgx − i (cid:2) N L ( T i ). By [Ma], we get H = L , and we haveproperty (NS) for the group L . (cid:3) ROM THOMPSON TO BAER–SUZUKI 25
Acknowledgements . Gordeev was supported in part by RFBR grant N-08-01-00756-A. Kunyavski˘ı and Plotkin were supported in part by theMinistry of Absorption (Israel) and the Minerva Foundation throughthe Emmy Noether Research Institute of Mathematics. A substantialpart of this work was done in MPIM (Bonn) during the visits of Kun-yavski˘ı and Plotkin in 2007, the visit of Plotkin in 2008, and the visit ofGordeev in 2009. The work was discussed by all the coauthors duringthe international workshops hosted by the Heinrich-Heine-Universit¨at(D¨usseldorf) in 2007 and 2008. The support of these institutions ishighly appreciated.
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Gordeev: Department of Mathematics, Herzen State PedagogicalUniversity, 48 Moika Embankment, 191186, St.Petersburg, RUSSIA
E-mail address : [email protected] Grunewald: Mathematisches Institut der Heinrich-Heine-Universit¨at D¨usseldorf, Universit¨atsstr. 1, 40225 D¨usseldorf,GERMANY
E-mail address : [email protected] Kunyavskii: Department of Mathematics, Bar-Ilan University,52900 Ramat Gan, ISRAEL
E-mail address : [email protected] Plotkin: Department of Mathematics, Bar-Ilan University, 52900Ramat Gan, ISRAEL
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