Sharp upper bounds on the minimal number of elements required to generate a transitive permutation group
aa r X i v : . [ m a t h . G R ] F e b SHARP UPPER BOUNDS ON THE MINIMAL NUMBER OF ELEMENTSREQUIRED TO GENERATE A TRANSITIVE PERMUTATION GROUP
GARETH TRACEY
Abstract.
The purpose of this paper is to prove that if G is a transitive permutation groupof degree n ≥
2, then G can be generated by ⌊ cn/ √ log n ⌋ elements, where c := √ /
2. Owingto the transitive group D ◦ D of degree 8, this upper bound is best possible. Our new resultimproves a 2015 paper by the author, and makes use of the recent classification of transitivegroups of degree 48. Introduction
For an arbitrary group G , let d ( G ) be the minimal size of a generating set of G . In [13], theproblem of finding numerical upper bounds for d ( G ) for an arbitrary transitive permutationgroup G of degree n is considered. It had already been proved in [9] that d ( G ) is at most cn √ log n in this case, where c is an absolute constant. This bound is shown to be asymptotically bestpossible in [7] (that is, there exists constants c , c , and an infinite family ( G n i ) ∞ i =1 of transitivegroups of degree n i , with c ≤ n i d ( G ni ) √ log n i ≤ c for all i ).In [13] it is proved that, apart from a finite list of possible exceptions, the bound d ( G ) ≤ j cn √ log n k holds, where c := √ . This bound is best possible in the sense that d ( G ) = √ n √ log n when G = D ◦ D < S and n = 8.In this paper, we remove the possible exceptions listed in [13, Theorem 1.1]. More precisely,we prove: Theorem 1.1.
Let G be a transitive permutation group of degree n . Then d ( G ) ≤ (cid:22) cn √ log n (cid:23) where c := √ . There are a number of steps in the proof of Theorem 1.1. For this reason, we will spend thenext few paragraphs outlining our general strategy for the proof.First, by [13, Theorem 5.3], we only need to prove Theorem 1.1 when G is imprimitive withminimal block size 2, and n has the form n = 2 x y y = 0 and 17 ≤ x ≤
26; or y = 1and 15 ≤ x ≤
35. Thus, in particular, G may be viewed as a subgroup in a wreath product2 ≀ G Σ , where Σ is a set of blocks for G of size 2. It follows that d ( G ) ≤ d G Σ ( M ) + d ( G Σ ),where M is the intersection of G with the base group of the wreath product, and d G Σ ( M ) isthe minimal number of elements required to generate M as a G Σ -module. Mathematics Subject Classification.
Of course, d ( G Σ ) can be bounded using induction. The bulk of this section will therefore betaken up with finding upper bounds on d G Σ ( M ). Since d G Σ ( M ) ≤ d H ( M ) for any subgroup H of G Σ , the strategy of the third author in [13] in this case involved replacing G Σ by a convenientsubgroup H of G Σ , and then deriving upper bounds on d H ( M ), usually in terms of the lengthsof the H -orbits in Σ. This approach turns out to be particularly fruitful when H is chosen tobe a soluble transitive subgroup of G Σ , whenever such a subgroup exists. When G Σ does notcontain a soluble transitive subgroup, however, the analysis becomes much more complicated.This leads to less sharp bounds, and ultimately, the omitted cases in [13, Theorem 1.1].Our approach in this section involves a careful analysis of the orbit lengths of soluble sub-groups in a minimal transitive insoluble subgroup of G , building on the work in [13] in the case n = 2 x x y z , with 0 ≤ y, z ≤
1. Finally, this information willallow us to prove Theorem 1.1, and we do so in Section 4.
Acknowledgements:
It is a pleasure to thank Derek Holt and Gordon Royle for helpfulconversations and suggestions.2.
Subdirect products of finite groups
As mentioned above, we begin preparations toward the proof of Theorem 1.1 with somestraightforward but necessary lemmas concerning subdirect products of finite groups. First, wemake the following remark concerning our notation and terminology in direct products of finitegroups:
Remark 2.1.
For a group T we will write T e for the direct product of e copies of T . Moregenerally, let G , . . . , G r be finite groups, and consider the direct product G := G e × . . . × G e r r .Fix subgroups H i ≤ G i , for 1 ≤ i ≤ r , and set e := P ri =1 k i . Then we write H e × . . . × H e r r for the subgroup { ( x , . . . , x e ) : x j ∈ H i for f i − + 1 ≤ j ≤ f i } of G , where f := 0, and f i := P ik =1 e k for i > e × . . . × S e r of S e (in its natural intransitive action) acts via automorphismson G (by permutation of coordinates). Thus we can, and do, speak of (S e × . . . × S e r )-conjugatesof subgroups of G . This will be very useful for avoiding cumbersome notation.Finally, for a group T we call a subgroup G of T e a diagonal subgroup of T e if there existsautomorphisms α i of T such that G = { ( t α , t α , . . . , t α r ) : t ∈ T } .Suppose now that G is a subgroup in a direct product G × . . . × G e of groups G i , and write π i : G → G i for the i th coordinate projections. Then G is a subdirect product in G × . . . × G e if Gπ i = G i for all i . We also introduce the following non-standard definition: Definition 2.2.
Let G i be as above, and let n be a positive integer. We say that G is a subdirectproduct of the form G = n ( G × . . . × G e ) if each of the following holds:(a) G is a subdirect product in G × . . . × G e ;(b) G contains [ G , G ] × . . . × [ G e , G e ]; and HARP UPPER BOUNDS 3 (c) G has index n in G × . . . × G e .The following is a collection of easy facts concerning subdirect products in direct products offinite groups: Lemma 2.3.
Let G , . . . , G e be finite groups, and π i : G × . . . × G e → G i be as above, and let K i be the intersection of G i with the i th coordinate subgroup of G × . . . × G e . Suppose that G is a subdirect product in G × . . . × G e . Then each of the following holds:(i) K i π i E G i for all i .(ii) If e = 2 and G and G have no common nontrivial homomorphic images, then G = G × G .(iii) Let e = P ri =1 e i be a positive partition of e , and define f := 0 , and f i := P ik =1 e k for ≤ i ≤ r . Suppose that for each ≤ i ≤ r , there exists a finite simple group T i such that G j is isomorphic to T i for all f i − + 1 ≤ j ≤ f i . Then for each ≤ i ≤ r there exists a partition P t i k =1 d i,k of e i and diagonal subgroups D i,k of T d i,k i such that G is (S e × . . . × S e r ) -conjugate to ( D , × . . . × D ,t ) × . . . × ( D r, × . . . × D r,t r ) .(iv) Let T be a nonabelian simple group, and suppose that | G i | < | T | for all i . Then G has nohomomorphic image isomorphic to T .(v) Fix < f < e . Suppose that G i is nonabelian simple for ≤ i ≤ f , and that | G j | < | G i | for all ≤ i ≤ f and f + 1 ≤ j ≤ e . Assume also that | G i | does not divide the index of G in G × . . . × G e for any ≤ i ≤ f . Then G = G × . . . × G f × L , where L is a subdirectproduct in G f +1 × . . . × G e .Proof. Part (i) is clear, while part (ii) follows from the fact that if e = 2, then G K π ∼ = G K π . Part (iii) is well-known (for example, see [11, Theorem 4.16(iii)]), so we just need to prove (iv)and (v).We begin with (iv). So assume that T is a nonabelian finite simple group, and that | G i | < | T | for all i . We prove the claim in (iv) by induction on e , with the case e = 1 being trivial. Soassume that e >
2, and write ˆ π for the projection ˆ π : G → G × . . . × G e . Suppose that N isa normal subgroup of G with G/N ∼ = T . Then since G ˆ π/N ˆ π is both a homomorphic image of G/N ∼ = T , and the subdirect product G ˆ π ≤ G × . . . × G e , the inductive hypothesis implies that N ˆ π = G ˆ π . Then since K is the kernel of ˆ π , we have | G | = | K || G ˆ π | and | N | = | K ∩ N || G ˆ π | .Hence, | G | ≥ | K | ≥ | K /K ∩ N | = | T | – a contradiction. This proves (iv).Finally, assume that the hypotheses in (v) hold. Then G may be viewed as a subdirectproduct in R × L , where R is a subdirect product in G × . . . × G f , and L is a subdirect productin G f +1 × . . . × G e . It then follows from (ii), (iii), and (iv) that G = R × L . Moreover, since | G × . . . × G f : R | divides | G × . . . × G e : G | , we must have R = G × . . . × G f , by (iii). Thiscompletes the proof of (v), whence the lemma. (cid:3) Minimal transitive groups of degree x y z In this subsection, we investigate the structure of minimal transitive groups of degree 2 x y z ,with 0 ≤ y, z ≤
1. Our main result reads as follows:
GARETH TRACEY
Theorem 3.1.
Let G be an insoluble minimal transitive permutation group of degree n =2 x y z , with ≤ y, z ≤ . Then:(i) G has at most y + z nonabelian chief factors;(ii) each nonabelian chief factor of G is isomorphic to a direct product of copies of one of A ; A ; S (4) ; O +8 (2) ; A ; A ; or L ( q ) , where q has the form q = 2 x y z − with ≤ y , z ≤ ;(iii) if G has two nonabelian chief factors, then they are isomorphic to T e and L ( p ) e , where T ∈ { A , A } , and p is a Mersenne prime; and(iv) G has a soluble subgroup F whose orbit lengths are given in Table 1. Now, Table 1 requires some explanation: In Theorem 3.1, we prove that a minimal transitivepermutation group of degree 2 x y z with 0 ≤ y, z ≤
1, has at most two nonabelian chief factors,which have to be isomorphic to direct powers of one of A ; A ; A ; A ; or L ( q ), where q hasthe form q = 2 x y z −
1. The first column of Table 1 will be a group T e , where T is one ofthese simple groups. This means that if G is the minimal transitive group under consideration,then G/R ( G ) has a minimal normal subgroup L/R ( G ) which is isomorphic to a direct productof e copies of T (recall that R ( G ) is the soluble radical of G ). The second column gives thedegree of G as a permutation group, while the third column gives the orbit lengths of a solublesubgroup F of G . Moreover, these orbit lengths will be given in one of the following ways: Rows 23–30:
In these cases, the orbit lengths are given in the form of a list with entries P fi =1 k i × l i , with the entry P fi =1 k i × l i meaning that F has k i orbits of size l i for each1 ≤ i ≤ f . Rows 1–11 and 13–21:
In these cases, the orbit lengths are given in the form αX i , βY j .Here, { X i } i and { Y j } j are positive unordered partitions whose sum is given in the fourthcolumn of the table. For example, if F is as in row 1, then { X i } i is a positive unorderedpartition of 4 (there are five such partitions) and { Y j } j is a positive unordered partitionof 2 (there are two such partitions). Thus, there are ten possibilities for the orbit lengthsof F . For instance, the orbit sizes (40 a , 100 a , 100 a ) correspond to the partitions 4 = 4and 2 = 1 + 1 of 4 and 2, respectively. Rows 12 and 22:
The orbits lengths in these cases are given in the form P e i =0 P j,k C ( i ) j × b p i X ( i,j ) k , 0 ≤ i ≤ e . By this notation we mean that ( C ( i ) j ) j is a positive orderedpartition of a particular number C for each i ; ( X ( i,j ) k ) k is a positive ordered partition ofa particular number X for each i, j ; and F has C ( i ) j orbits of size 2 b p i X ( i,j ) k for each i , j , k . The numbers C , X , b , and p are given in the fourth column of the table (in fact, C := (cid:0) e i (cid:1) in each case).As an example, assume that G is a minimal transitive group of degree n = 2
15. If aminimal normal subgroup of G is isomorphic to a direct product of copies of A , then the firstsection of Table 1 gives us the orbit lengths of a soluble subgroup F of G . For example, if F is as in the first line of the table, then F has orbit lengths 2 X i , 2 Y j , where ( X i ) i ∈{ (4) , (3 , , (2 , , (1 , , , } , and ( Y j ) j ∈ { (2) , (1 , } . Thus, the possible F -orbit lengths in thiscase are: (2
5, 2
5, 2
25, 2
15, 2
5, 2
15, 2
5, 2
5, 2
5, 2
5, 2
5, 2
25, 2
5, 2
5, 2
5, 2
5, 2 HARP UPPER BOUNDS 5 (2
5, 2
5, 2
5, 2
5, 2
25, 2 F canbe computed in an entirely analogous way. (Although these computations seem tedious, theycan be done very quickly using a computer). Remark 3.2.
Let G be as in the statement of Theorem 3.1, and suppose that we have provedthe theorem for all degrees less than n . Let K be the kernel of the action of G on a blocksystem Σ = { ∆ g : g ∈ G } for G in O. Then G Σ ∼ = G/K is minimal transitive. Thus, we maychoose a soluble subgroup
F/K of G/K ∼ = G Σ satisfying Theorem 3.1(iv) with G replaced bythe minimal transitive group G Σ If K is soluble and acts transitively on ∆, then F is soluble and the F -orbits have lengths | ∆ || Σ i | , as Σ i runs over the F Σ -orbits. Suppose that the Σ i are as in row k in Table 1. Thensince | ∆ | divides n , it is easy to see that by replacing a by | ∆ | a , F ≤ G has the required orbitlengths in order to fulfil the conclusion of (iv). (For example, if F/K is as in row 2 of Table1, then G has degree 120 | ∆ | a , and F is soluble with orbit sizes 5 | ∆ | aX i , 50 | ∆ | aY j .) This is auseful inductive tool which we will use throughout the section.We begin preparations toward the proof of Theorem 3.1 with a reduction lemma: Lemma 3.3.
Let G be a counterexample to Theorem 3.1 of minimal degree. Then all minimalnormal subgroups of G are nonabelian. Moreover, if L is a minimal normal subgroup of G and K denotes the kernel of the action of G on the set of L -orbits, then K/L is soluble.Proof.
Let L be a minimal normal subgroup of G . Also, let ∆ be an L -orbit, and let Σ := { ∆ g : g ∈ G } be the set of L -orbits in [ n ], so that G Σ ∼ = G/K . Then n = | ∆ || Σ | , and G Σ isminimal transitive. The minimality of G as a counterexample then implies that G Σ satisfies theconclusion of Theorem 3.1.Let E/L be a subgroup of
G/L which is minimal with the property that (
E/L )( K/L ) =
G/L .Then (
E/L ) ∩ ( K/L ) ≤ Φ( E/L ), so (
E/L ) ∩ ( K/L ) is soluble. But also, EK = G . Since E contains L and E Σ = G Σ is transitive, we deduce that E is transitive, whence E = G , since G is minimal transitive. Thus, K/L is soluble. This proves the second part of the lemma.Finally, assume that L is abelian. Then K is soluble. It follows that the set of nonabelianchief factors of G is same as the set of nonabelian chief factors of G/K . Thus, G = G O satisfies(i), (ii), and (iii) in Theorem 3.1. Moreover, since the theorem holds in G Σ ∼ = G/K , we maychoose a soluble subgroup
F/K of G/K satisfying Theorem 3.1(iv). Then F is soluble, since K is soluble, and the F -orbits in [ n ] have lengths | ∆ || Σ i | , as Σ i runs over the F Σ -orbits in Σ.Then F is a subgroup of G satisfying the conclusion of Theorem 3.1(iv) (see Remark 3.2). Thus,(i)–(iv) in Theorem 3.1 hold in G , contradicting the assumption that G is a counterexample. (cid:3) By Lemma 3.3, a minimal normal subgroup L of a minimal counterexample to Theorem 3.1must be a direct product of nonabelian simple groups. Since the subnormal subgroups of atransitive permutation group of degree n have orbit lengths dividing n , this motivates us tostudy the pairs ( T , H ) where T is a nonabelian simple group, H is a proper subgroup of T , and | T : H | has the form 2 x y z , for some 0 ≤ y, z ≤ T with a maximal subgroup M satisfying π ( | M | ) = π ( | T | ) are classified. In order degree of G orbit lengths of a soluble subgroup F notesA e a aX i , 100 aY j ( P i X i , P j Y j ) = (4 , a aX i , 50 aY j ( P i X i , P j Y j ) = (4 , a aX i , 50 aY j ( P i X i , P j Y j ) = (2 , a aX i , 25 aY j ( P i X i , P j Y j ) = (2 , a aX i , 50 aY j ( P i X i , P j Y j ) = (2 , a aX i , 25 aY j ( P i X i , P j Y j ) = (1 , b a b aX i P i X i = 2; b ∈ { , } ; G/L soluble60 a a , 50 a a aX i , 10 aY j ( P i X i , P j Y j ) = (2 , a aX i , 25 aY j ( P i X i , P j Y j ) = (4 , a aX i , 25 aY j ( P i X i , P j Y j ) = (2 , b ( p + 1) e a P e i =0 P j,k C ( i ) j × b p i aX ( i,j ) k P k C ( i ) j = (cid:0) e i (cid:1) and P j X ( i,j ) k = 2 foreach i, j ; b ∈ { , } ; G/L insoluble; p aMersenne prime; 1 ≤ e ≤ e A e a aX i , 50 aY j ( P i X i , P j Y j ) = (4 , a aX i , 25 aY j ( P i X i , P j Y j ) = (4 , a aX i , 25 aY j ( P i X i , P j Y j ) = (2 , a aX i P i X i = 1215 a aX i P i X i = 340 a aX i P i X i = 4; G/L soluble2 b a aX i P i X i = 2 b ; b ∈ { , } ; G/L soluble60 a aX i , 10 aY j ( P i X i , P j Y j ) = (4 , a aX i , 10 aY j ( P i X i , P j Y j ) = (2 , b X ( p + 1) e a P e i =0 P j,k C ( i ) j × b p i aX ( i,j ) k P j C ( i ) j = (cid:0) e i (cid:1) , P k X ( i,j ) k = X for each i, j ; ( X, b ) ∈ { (4 , , (4 , , (2 , } ; G/L insoluble; p a Mersenne prime; 1 ≤ e ≤ e T e b a b a + 2 b a b ∈ { , } , T ∈ { S (4) , O +8 (2) } A e e a P j =1 P e i =0 as j (cid:0) e i (cid:1) × i l j ≤ e ≤ e ; 1 ≤ j ≤
3, where ( s , l ) =(1 , s , l ) = (11 , s , l ) =(42 , e a P j =1 P e i =0 as j (cid:0) e i (cid:1) × i l j ≤ e ≤ e ; 1 ≤ j ≤
3, where ( s , l ) =(1 , s , l ) = (5 , s , l ) =(4 , e a P j =1 P e i =0 as j (cid:0) e i (cid:1) × i l j ≤ e ≤ e ; 1 ≤ j ≤
2, where ( s , l ) =(1 ,
1) and ( s , l ) = (2 , e e b a P e i =0 a b (cid:0) e i (cid:1) × i +1 b ∈ { , } , 0 ≤ e ≤ e L ( q ) e a ( q + 1) e +2 P e i =0 a (cid:0) e i (cid:1) × q i + 2 a (cid:0) e i (cid:1) × q i +1 q a Mersenne prime, 0 ≤ e ≤ e y z a ( q +1) e P e i =0 a (cid:0) e i (cid:1) × y z q i q a Mersenne prime, 1 ≤ e ≤ e x ′ a aq − y ′ − z ′ + a − y ′ − z ′ q = 2 x ′ y ′ z ′ − { y ′ , z ′ } = { , } Table 1.
Orbit lengths of a certain soluble subgroup of a minimal transitivegroup G with L := L/R ( G ) a minimal normal subgroup of G/R ( G ). HARP UPPER BOUNDS 7 to use this result, we will first need to determine the finite simple groups T with a subgroup H of index 2 x y z as above, such that π ( | M | ) = π ( | T | ) for some maximal subgroup M < T containing H . Proposition 3.4.
Let T be a nonabelian finite simple group, and suppose that T has a subgroup H of index x y z , with ≤ y, z ≤ . Let M be a maximal subgroup of T containing H , andassume that either π ( | T | ) = π ( | M | ) , or that T is an alternating group of degree less than .Then one of the following holds:(i) T = A n is an alternating group and either(a) n = 5 and H is any subgroup of T ;(b) n = 6 and | H | = 2 i , for some i ≤ ;(c) n = 7 , and either H = GL (2) (two classes) or H = 7 : 3 < GL (2) ;(d) n = 8 and either H = 2 : GL (2) (two classes); or H = 2 : (7 : 3) < : GL (2) (twoclasses); or H ∼ = GL (2) < : GL (2) (three classes); or H ∼ = 7 : 3 < : GL (2) ; or(e) n = 9 and H = L (8) . (two classes).(ii) T = L n ( q ) and either:(a) n = 2 , q has the form q = 2 x y z − , and H is a subgroup of a maximal parabolicsubgroup M of T , with | M : H | = 3 y − y z − z ; or(b) ( n, q ) = (4 , , and H = 3 : L (3) (two classes).(iii) T = S (3) with | T : H | = 40 (two classes); or T = S (2) , and | T : H | ∈ { , } .Proof. Assume first that T = A n . We claim that in order for | T : M | to have the form | T : M | = 2 x y z with 0 ≤ y , z ≤ π ( | T | ) = π ( | M | ), we must have n ≤
9. To provethis, we first list the the possibilities for M : • M = (A k × A n − k ) .
2, for some k ≤ n . • M = A n ∩ (S r ≀ S s ), where r, s > rs = n . • M = A n ∩ X , with X ≤ S n primitive in its natural action.In particular, if | M | is odd, then M must be as in the third listed case with n an odd prime, n − odd, and M = C n ⋊ C n − . However, in this case | T : M | = ( n − | T : M | = 2 x y z with 0 ≤ y , z ≤ n ≤
7. So we may assume that | M | iseven. But then either | T | = 3 or | T | = 5, since π ( | T | ) = π ( | M | ). Thus n ≤
9, as claimed. Wecan now use direct computation to quickly determine the possibilities for H .So we may assume that T is not an alternating group. If T ∈ { L (8) , L (3) , U (3) , S (8) } then direct computation quickly shows that T has no maximal subgroups M with index of theform | T : M | = 2 x y z with 0 ≤ y , z ≤
1. So we may assume that T
6∈ { L (8) , L (3) , U (3) , S (8) } .Suppose next that T = L ( p ) with p = 2 x − M must containa Sylow p -subgroup of T . Hence, M is a maximal parabolic subgroup of T ; | T : M | = p + 1 = 2 x and | M : H | = 3 y z (since | M | is odd). So assume also that T is not of the form T = L ( p )with p a Mersenne prime. Then by [8, Corollary 6], there exists a set Π of odd prime divisorsof | T | such that Π intersects π ( | T | ) \ π ( | M | ) ⊆ { , , } non-trivially. Moreover, the proof of [8,Corollary 6] shows that either T is in [8, Table 10.1] and Π is as in the third column of [8, Table GARETH TRACEY T ∈ { S n ( q ) , O n ( q ) } and the second row for T = O − n ( q )); or T isin [8, Table 10.3] and Π is as in the third column of [8, Table 10.3]; or T is in [8, Table 10.4] andΠ is as in the second column of [8, Table 10.4] (except when T = U (2), where Π := { , , } );or T is in [8, Table 10.5] and Π is as in the second column of [8, Table 10.5] (except when T = G ( q ), where Π := { p, , } , { , , } , or { , } according to whether q ≥ q = 4, or q = 3, respectively); or T is in [8, Table 10.6] and Π is as in the second column of [8, Table10.6] (except when T = M , M , or M , where Π := { , } , { , } , or { , } , respectively).Fix r to be a prime in Π ∩ ( π ( | T | ) \ π ( | M | )), so that r ∈ { , } . This implies in particular that | T | r = r .By [8, Corollary 6] and inspection of simple group orders, the case r = 3 occurs only if T = L ( p ) with p of the form p = 2 x y −
1, with 0 ≤ y ≤
1. In this case, one can see from[2, Table 8.1] that M is a maximal parabolic subgroup of T . Since 2 x is the largest power of 2dividing | T | in this case, we deduce that x = x and that H must have index dividing 3 y − y z in M .So we may assume, for the remainder of the proof, that r = 5. In what follows, for a primepower q and a positive integer n we will write q n for an arbitrary primitive prime divisor of q n −
1. By Zsigmondy’s theorem these exists for n > q, n ) = (2 , n | q n − G is a finite simple classical group, and write T = X n ( q ), where X ∈{ L , U , S , O ± , O } , and q = p f with p prime. Then using (3.1) we can deduce from [8, Tables 10.1–10.6] that T ∈ { L n ( q ) ( n ≤ , U n ( q ) ( n ≤ , S n ( q ) ( n ≤ , O ± n ( q ) ( n ≤ , O n ( q ) ( n ≤ } .Below, we examine these remaining cases.Suppose first that T = L n ( q ), with n ≤
5. If n ∈ { , } , then (3.1) and [8, Tables 10.1 and10.3] imply that either 5 is a primitive prime divisor of q − p = 5. If n ∈ { , } , then5 is a primitive prime divisor of q −
1. Suppose first that M is parabolic. Then we see from[2, Tables 8.2, 8.3, 8.8, and 8.18] that | T : M | has the form q n − q − or ( q n − q n − − q − q − . A routineexercise then shows that only the former can take the form 2 x y z with 0 ≤ y , z ≤
1, andthis can only happen if n ∈ { , } and q has the form q = 2 a b c −
1. It is then easy to see thatthe only possibilities are either ( n, q ) = (4 ,
3) or ( n, q ) = (2 , x y z −
1) with 0 ≤ y , z ≤ n, q ) = (4 , M = H = 3 : L (3). If ( n, q ) = (2 , x y z − x is the largest power of 2 dividing | T | , so x = x and H has index dividing 3 y − y z − z in M . So we may assume that M is not parabolic. Then p divides | T : M | , so p ∈ { , , } .Also, as mentioned above, r = 5 does not divide | M | . We can then quickly see from [2, Tables8.2, 8.3, 8.8, and 8.18] that the only possibility is ( n, q ) = (3 ,
2) with H having order 7 or 21.Since L (2) ∼ = L (7), this case is already listed in (ii)(a). This completes the proof in the case T = L n ( q ).The other classical cases are entirely similar: Suppose that T = U n ( q ), with 3 ≤ n ≤
6. If M is parabolic, then | T : M | is divisible by either ( q n − ( − n )( q n − − ( − n − ) q − or Q ⌊ n ⌋ i =1 ( q i +1 + 1).It is easy to see, however, that neither q + 1 not q + 1 can ever have the form 2 x y z with 0 ≤ y , z ≤
1. So M must be non-parabolic. Then p divides | T : M | , so p ∈ { , , } . HARP UPPER BOUNDS 9
Also, 5 does not divide | M | . We can then use the tables in [2, Chapter 8] to quickly find thatthe only case of a maximal non-parabolic subgroups M of T with | T : M | = 2 x y z and0 ≤ y , z ≤ n, q ) = (4 , (2) ∼ = S (3), this case is accounted for in(iii). This completes the proof in the case T = U n ( q ).The arguments in the remaining classical cases are almost identical, so we now move onto the case where G is an exceptional group of Lie type. Then r = 5 is one of the primesoccurring in the second column of the row for T in [8, Table 10.5]. By using (3.1), we see thatthe only possibilities are T ∈ { F ( q ) ′ , B ( q ) } , and 5 is a primitive prime divisor of q −
1. Ineach of these cases, the maximal subgroups of T are known: see [10] for T = F ( q ) ′ and [12]for T = B ( q ). In each case, we can quickly check that no maximal subgroup M of T with π ( | T | ) = π ( | M | ) can have | T : M | of the form 2 x y z , for 0 ≤ y , z ≤ G is a sporadic simple group. Then T = J , by [8, Table 10.6]. However,a quick check of the Web Atlas [14] shows that no maximal subgroup M of J with π ( | T | ) = π ( | M | ) can have | T : M | of the form 2 x y z , for 0 ≤ y, z ≤
1. This completes the proof. (cid:3)
We are now ready to determine the possibilities for the pairs ( T , H ), with T simple, | T : H | dividing 2 x
15, and H contained in a maximal subgroup M of T with π ( | T | ) = π ( | M | ). Proposition 3.5.
Let T be a nonabelian finite simple group, and suppose that T has a subgroup H of index n = 2 x y z , with ≤ y, z ≤ . Suppose also that T is not an alternating groupof degree less than . Let M be a maximal subgroup of T containing H , and assume that π ( | T | ) = π ( | M | ) . Then one of the following holds:(i) T = A n , and either H = A n − , or n = 16 and A ≤ H ≤ A . .(ii) T = S m (2 f ) with ( m, f ) ∈ { (4 , , (2 , } , and O − m (2 f ) E H .(iii) T = S ( q ) , with q ∈ { , } , and S ( q ) E H .(iv) T = M and H = L (11) ;(v) T = O +8 (2) and H = A (three classes, fused by a triality automorphsim); or(vi) T = M or M , and H = M or H = M , respectively.Proof. We can quickly check using direct computation that T can have no maximal subgroup M of the required index when T is L (8), L (3), U (3) or S (8). So we may assume that T isnot one of these groups. Suppose first that T = L ( p ) for a Mersenne prime p . Then using [8,Corollary 6 and Table 10.7], the possibilities for T and M are as follows.(1) T = A n and M = A k ∩ (S k × S n − k ) for some k ≤ n − k ≥ p for all primes p ≤ n .Then | A n : A n ∩ (S k × S n − k ) | = (cid:0) nk (cid:1) divides | T : H | = 2 x y z . By a well-known theoremof Sylvester and Schur (see [4]), (cid:0) nk (cid:1) has a prime divisor exceeding min { k, n − k } . Thus k ∈ { n − , n − , n − , n − } , since k ≥
10. We then quickly see that the only possibilitiesare H = A n − ; or n = 16 and k = 14, which gives us what we need.(2) T = S m ( q ) ( m , q even) or O m +1 ( q ) ( m even, q odd), and M = N T (O − m ( q )). Now, | N T (O − m ( q )) : O − m ( q ) | ≤ | T :O − m ( q ) | divides 2 x +1 y z . Also, for each of the two choices of T we get | T : O − m ( q ) | = q m ( q m − q m ( q m −
1) has the form 2 x y z for 0 ≤ y , z ≤ q = 2 f with f m ∈ { , } , since m is even. Since S (2) ∼ = S is not simple, we deduce that( m, f ) ∈ { (4 , , (2 , } .(3) T = O +2 m ( q ) ( m even, q odd) and M = N T (O m − ( q )). By [6, Corollary 2.10.4 Part (i) andTable 2.1.D], we have | N T (O m − ( q )) : O m − ( q ) | ≤
2. It follows that q m − ( q m −
1) = | T :O m − ( q ) | divides 2 x +1 y z . However, since q is odd, m ≥
4, and 0 ≤ y, z ≤
1, this cannotbe the case.(4) T = S ( q ) and M = N T (S ( q )). Then [6, Corollary 2.10.4 Part (i) and Table 2.1.D] gives | N T (S ( q )) : S ( q ) | ≤
2. It follows that q ( q −
1) = | T : S ( q ) | divides 2 x +1 y z . Thus, q ∈ { , } . We then use direct computation to see that the only possibility is S ( q ) E H .(5) In each of the remaining cases (see [8, Table 10.7]), we are given a pair ( T , Y ), where T is L (8), L (3), L (2), U (3), U (3), U (5), U (2), U (3), U (2), U (2), S (7), S (8), S (2),O +8 (2), G (3), F (2) ′ , M , M , M , HS , M cL , Co or Co , and Y is a subgroup of T containing H . Apart from when T = O +8 (2), M , M , and M , we find that | T : Y | does not divide 2 x y z , so we get a contradiction in all other cases. When T = O +8 (2),the only possibilities are H = A of index 2
15 (three T -conjugacy classes fused under thetriality automorphism). When T = M , M , or M , the only possibilities for H are H = L (11) ≤ M (of index 12), H = M ≤ M (of index 12), or H = M ≤ M (ofindex 24).Finally, assume that T = L ( p ), for some Mersenne prime p , and let M be a maximal subgroupof T containing H . Then, since | T : M | divides | T : H | = 2 x y z with 0 ≤ y, z ≤ M must beparabolic (see [2, Table 8.1]). But then π ( | T | ) = π ( | M | ), since p + 1 is the highest power of 2dividing | T | – a contradiction. This completes the proof. (cid:3) This completes our analysis of the pairs (
T, H ) with T a nonabelian simple group and H asubgroup of T of index | T : H | = 2 x y z , with 0 ≤ y, z ≤
1. The possibilities are listed inProposition 3.4(i)–(iii) and Proposition 3.5(i)–(vi).In particular, if G is a counterexample to Theorem 3.1 of minimal degree, L = T e is a minimalnormal subgroup of G , and H is a point stabilizer in L , then we know the possibilities for thegroup T , and the intersections of H with the i th coordinate subgroups in L . Our next task isto use this information to find out more about the structure of H .One of our main tools for doing this is the Frattini argument . The proof is well-known, butwe couldn’t find a reference so we include a proof here.
Lemma 3.6.
Let G be a finite group, and let L be a normal subgroup of G . Let A ≤ Aut( L ) be the image of the induced action of G on L (by conjugation). Suppose that H is a subgroupof L with the property that H and H α are L -conjugate for each α ∈ A . Then G = N G ( H ) L .Proof. Let g ∈ G . Then H g = H l for some l ∈ L , by hypothesis. Hence, gl − ∈ N G ( H ), so g ∈ N G ( H ) L , and this completes the proof. (cid:3) With the Frattini argument in mind, the next lemma will be crucial.
Lemma 3.7.
Let T be a nonabelian finite simple group, and suppose that T has a subgroup H of index n := 2 x y z , with ≤ y, z ≤ . Assume also that: HARP UPPER BOUNDS 11 • If T = A , then H is either transitive in its natural action, or H = A ; • if T = A , then H is either transitive in its natural action, or H = A is a natural pointstabilizer in T ; • if T = L ( q ) , with q of the form q = 2 x y z − , then y = y and z = z ; • if T = S (4) , then H = S (16) and H = S (16) . ; • if T = O +8 (2) , then H = A (three classes); • if T = A , then H = A and H = A . ; and • ( T, H ) is not (A , H ) , with H of shape , : (7 : 3) , GL (2) , or : GL (2) .Denote by Σ the set of right cosets of H in T . Then there exists a proper subgroup S of T withthe following properties:(i) S and S α are conjugate in T for each α ∈ Aut( T ) ;(ii) N T ( S ) Σ is transitive;(iii) N T ( S ) acts transitively on the set of cosets of any subgroup of T of -power index.Proof. Note first that if T = A n and K is a subgroup of T of 2-power index, then K = A n − and n is a power of 2, by Proposition 3.4 and 3.5. Thus, if T = A n , with n not a power of 2,and H is transitive, then setting S to be a natural point stabilizer S := A n − suffices. Also,if | T : H | is a power of 2, then setting S to be a Sylow 2-subgroup of T gives us what we need.So we may assume that we are in neither of these cases.By Proposition 3.4 and 3.5, the remaining possibilities for the pair ( T, H ) are as follows:(1) (
T, H ) = (A n , A n − ), with n = 2 x y z and y + z = 0. Set S = h (1 , . . . , y z ) , (3 y z +1 , . . . , × y z ) , . . . , ( n − y z + 1 , . . . , n ) i . Then N T ( S ) Σ is transitive, and S and S α are T -conjugate for all α ∈ Aut( T ) (this includes the case n = 6, when Out (A ) has order 4).Thus, (i) and (ii) are satisfied. Property (iii) is vacuously satisfied since T has no propersubgroup of 2-power index in this case, as mentioned at the beginning of the proof.(2) T = S (2), and O − (2) E H . Note that property (iii) is vacuously satisfied in each case,since T has no proper subgroup of 2-power index. Moreover, T has a maximal parabolicsubgroup S of shape 2 .GL (2) satisfying (i) and (ii).(3) T = L (3), H = 3 : L (3) (two classes). In this case, take S to be the unique maximalsubgroup of T of shape L (9) . .
2. Then it is easily seen that (i) and (ii) hold, for either ofthe two non-conjugate choices for H . Property (iii) holds since T has no proper subgroupsof 2-power index.(4) T = S (3), and | T : H | = 40 (two classes). Here, the unique maximal subgroup S of T ofindex 27 has properties (i) and (ii) for each of two non-conjugate choices for H . Property(iii) again holds since T has no proper subgroups of 2-power index.(5) T = S (2), and | T : H | ∈ { , } . Taking S to be the unique maximal subgroup of T of index 36 gives us (i) and (ii) in either of the cases | T : H | = 120 and | T : H | = 960.Property (iii) holds since T has no proper subgroups of 2-power index.(6) ( T, H ) = ( M , M ) or ( M , M ): In each case, let S be a subgroup of T generated bya fixed point free element of order 3. When T = M , N T ( S ) ∼ = A × S (see [14]) is amaximal subgroup of T , and acts transitively on the cosets of H . Thus, (ii) holds. Also,outer automorphisms of M fix the set of T -conjugates of S , so S and S α are T -conjugate for all α ∈ Aut( T ). Hence, (i) holds. When T = M , N T ( S ) has order 1008, and actstransitively on the cosets of H (using Magma [1], for example). Also, Out ( T ) is trivial.Thus, (i) and (ii) are again satisfied. Property (iii) is vacuously satisfied in each case, since T has no proper subgroup of 2-power index.(7) T = L ( q ), with q of the form q = 2 x y z −
1, 0 ≤ y , z ≤
1. Since we are assuming that y = y and z = z , H must be a maximal parabolic subgroup of T (see [2, Table 8.1]). Let S be a D q +1 maximal subgroup of T . Then S and S α are T -conjugate for all α ∈ Aut( T ),so (i) holds. Clearly, (ii) also holds. The group T only has a proper subgroup of 2-powerindex if y = z = 0, and in this case the only such subgroup is H . Thus, property (iii) isalso satisfied. (cid:3) Corollary 3.8.
Let G be a minimal transitive group of degree n = 2 x y z , where ≤ y, z ≤ .Suppose that G has a nonabelian minimal normal subgroup L = T e , and let H be the intersectionof L with a point stabilizer in G . Write | L : H | = 2 x y z , where x ≤ x , y ≤ y , z ≤ z .Then:(i) Either y or z is non-zero;(ii) Suppose that H = Hπ × . . . × Hπ e , and all but one of the groups Hπ i , say Hπ e , has -power index in T . Then ( T , Hπ e ) cannot satisfy the hypothesis of Lemma 3.7.Proof. Suppose that either | L : H | is a power of 2; or H = Hπ × . . . × Hπ e where all but oneof the groups Hπ i , say Hπ e , has 2-power index in T , and ( T , Hπ e ) satisfies the hypothesis ofLemma 3.7. Then we claim that there exists a proper subgroup S of L with S acting transitivelyon the cosets of H in L , and LN G ( S ) = G ( ∗ ). In particular, N G ( S ) is a proper subgroup of G , since L is a minimal normal subgroup of G . But then N G ( S ) < G acts transitively bothon an L -orbit, and on the set of L -orbits, so N G ( S ) is transitive. This contradicts the minimaltransitivity of G . Thus, it will suffice to prove the existence of such a subgroup S .Now, if | L : H | is a power of 2, then S ∈ Syl ( L ) satisfies ( ∗ ), by Lemma 3.6 and the factthat ( | L : H | , | L : S | ) = 1. So assume that H = Hπ × . . . × Hπ e , and all but one of the groups Hπ i , say Hπ e , has 2-power index in T . Assume that the pair ( T , Hπ e ) satisfies the hypothesisof Lemma 3.7, and let S be the subgroup of T exhibited therein. Set S := S e < L . Then S actstransitively on the cosets of H in L . Moreover, since Aut( L ) ∼ = Aut( T ) ≀ Sym e , Lemma 3.7(i)implies that S and S α are L -conjugate for all α ∈ Aut( L ). Thus, LN G ( S ) = G by Lemma 3.6,so S satisfies ( ∗ ), and this completes the proof. (cid:3) In what follows, recall from Definition 2.2 that a subdirect product of the form G = n ( G × . . . × G e ) is a subdirect product of index n in G × . . . × G e containing [ G , G ] × . . . × [ G e , G e ]. Corollary 3.9.
Let G be a minimal transitive group of degree n = 2 x y z , where ≤ y, z ≤ .Suppose that G has a nonabelian minimal normal subgroup L = T e , and write π i : L → T forthe i th coordinate projections. Let H be the intersection of L with a point stabilizer in G . Write | L : H | = 2 x y z , where x ≤ x , y ≤ y , z ≤ z . Then one of the following holds:(1) T = A n with n ∈ { , } and one of the following holds: HARP UPPER BOUNDS 13 (a) H is Sym e -conjugate to T e − × H e − × H e , where H e − and H e are subgroups of T with | T : H e − | = 2 x e − , | T : H e | = 2 x e , and x = x e − + x e ;(b) H is Sym e -conjugate to T e − × H e , where H e ≤ T is not transitive in its natural action; H e is not T -conjugate to a natural point stabilizer A n − ; and | T : H e | = | L : H | ;(c) T = A and H is Sym e -conjugate to T e − × J , where J < T is a subdirect product ofthe form J = (S × D ) < T ; or(d) T = A and H is Sym e -conjugate to T e − × J , where J is a diagonal subgroup of T .(2) T = A and H is Sym e -conjugate to T e × H e +1 × . . . × H e − × H e , where H e is one of thesubgroups of shape , : (7 : 3) , GL (2) , or : GL (2) in T ; H i is T -conjugate to A for all e + 1 ≤ i ≤ e − ; and ≤ e ≤ e − .(3) T = A and H is Sym e -conjugate to T e × H e +1 × . . . × H e − × H e , where H e is one of thesubgroups of shape A or A . in T ; H i is T -conjugate to A for all e + 1 ≤ i ≤ e − ;and ≤ e ≤ e − .(4) T = L ( q ) , with q of the form q = 2 x ′ y ′ z ′ − , ≤ y ′ , z ′ ≤ , and one of the followingholds:(a) y ′ = z ′ = 0 ; | q − ; H is Sym e -conjugate to T e × P × . . . × P e × H e − × H e , whereeach P i is a maximal parabolic subgroup of T ; H e − [respectively H e ] is a subgroup ofindex [resp. ] in a maximal parabolic subgroup of T ; and e + e = e − ;(b) y ′ = z ′ = 0 and H is Sym e -conjugate to T e × L , where L is a subdirect product of theform L = t ( L × Hπ e ) ; L is a subdirect product of the form L = s ( P × . . . × P e ) ;each P i is a maximal parabolic subgroup of T ; Hπ e is a subgroup of index r in a maximalparabolic subgroup of T ; rst = 3 y z ; and e + e = e − ; or(c) y ′ + z ′ = 1 , and H is Sym e -conjugate to T e − × (cid:16) q : q − r (cid:17) , where r > divides q − ,and y ′ z ′ r = 3 y z .(5) T = S (4) and H is Sym e -conjugate to T e − × H e , where H e ∈ { S (16) ( two T -classes ) , S (16) . two T -classes ) } .(6) T = O +8 (2) and H is Sym e -conjugate to T e − × H e , where H e = A (three T -classes, fusedunder a triality automorphism).Proof. By Corollary 3.8, either y or z is non-zero. Now, set H i := ( T i ∩ H ) π i , for 1 ≤ i ≤ e .Then H i E Hπ i , and | T : H i | has the form 2 a i b i c i for each i . Note also that since H isa subgroup of Hπ × . . . × Hπ e ≤ L , we have that Q ei =1 | T : Hπ i | divides | L : H | . Thus, | T : Hπ i | has the form 2 x i y i z i for each i , and y i [respectively z i ] is 1 for at most y [resp. z ] values of i . It follows that | T : Hπ i | is a power of 2 for all but at most two values of i .Let Λ := { i : | T : Hπ i | is a power of 2 } , so that | Λ | ≥ t −
2. For a subset E of { , . . . , e } , let π E : L → Q i ∈ E T i be the natural projection homomorphism onto the ‘ E -part’ of L . Then since H is a subgroup of Hπ Λ × Q i Λ Hπ i (up to permutation of coordinates), we have that | T | Λ | : Hπ Λ | Y i Λ | T : Hπ i | divides | L : H | .(3.2)Furthermore, we can determine the subgroups of 2-power index in the simple groups T fromPropositions 3.4 and 3.5: we see that either(I) Hπ i = T for all i ∈ Λ (II) T = A xi and Hπ i is T -conjugate to either T or A xi − for all i ∈ Λ; or(III) T = L ( p ), with p = 2 x i −
1, and Hπ i is either T or a maximal parabolic subgroup of T ,for all i ∈ Λ.Assume first that | Λ | = t −
2, and without loss of generality, write [ i ] \ Λ = { e − , e } . Then wemay assume, again without of loss of generality, that | T : Hπ e − | = 2 x e − | T : Hπ e | = 2 x e | T e − : Hπ Λ | is a power of 2 . (3.3)Next, from Propositions 3.4 and 3.5, we can determine all finite simple groups T which containboth a subgroup of index 2 x e − x e
5: we see that T ∈ { L ( p ) , A ,A : p a Mersenne prime, 15 | p − } . Suppose first that T = A n , for n ∈ { , } . Then since T is the only subgroup of T of 2-power index, we have Hπ i = T for all i ∈ Λ. Thus, H isSym e -conjugate to T e − × J , where J is a subdirect subgroup of Hπ e − × Hπ e , by Lemma 2.3and (3.3). We can now quickly determine, using Magma [1] for example, all subgroups J of T where J π e − , J π e < T , and | T : J | has the form 2 A B C . This yields cases (1)(a) and (1)(c)in the statement of the corollary.Suppose next that T = L ( p ), where p is a Mersenne prime. Then the only subgroups of T of 2-power index are T itself, together with the maximal parabolic subgroup P < T , of index p + 1. It follows from (3.3) and Lemma 2.3(v) that Hπ Λ = T e × L (up to permutation ofcoordinates), where L is a subdirect product in P × . . . × P e , each P i is T -conjugate to P , and e + e = e . Also, since P has odd order, (3.3) implies that that L = P × . . . × P e . Hence, Hπ λ = T e × P × . . . × P e (up to permutation of coordinates). Thus, H is a subdirect productin T e × P × . . . × P e × J , where J := Hπ { e − ,e } ≤ Hπ e − × Hπ e . Now, the only subgroupof T (up to T -conjugacy) with index 2 x e − Y of P with | P : Y | = 3 , | T : Y | = 3( p + 1); while the only subgroup of T (up to T -conjugacy) with index 2 x e − Y of P with | P : Y | = 5 , | T : Y | = 5( p + 1). Thus, J is a subdirect productin Hπ e − × Hπ e ∼ = p : p − × p : p − . Since | T : J | has the form 2 A b C and | J | is odd, we musthave J = Hπ e − × Hπ e − ∼ = p : p − × p : p − . Moreover, since | T e − × T e : H ∩ ( T e − × T e ) | has the form 2 A B C , we must have J = Hπ e − × Hπ e = H ∩ ( T e − × T e ). It follows that H isSym e -conjugate to T e × P × . . . × P e × H e − × H e , and this yields case (4)(a) in the statementof the corollary.Suppose next that | Λ | = t −
1. Without loss of generality, we may assume that [ i ] \ Λ = { e } .Hence, | T : Hπ e | = 2 x e r , with r ∈ { , , } . It then follows from (3.2) that | T e − : Hπ Λ | x e r divides | L : H | for some r ∈ { , , } .(3.4)In particular, | T | does not divide | T e − : Hπ Λ | . Suppose that case (I) occurs (recall that (I) isdefined at the end of the first paragraph of this proof). Then H is Sym e -conjugate to T e − × H e by (3.4) and Lemma 2.3(v). Suppose now that we are in case (II). Then Hπ Λ is a subdirectproduct in T e × H e +1 × . . . × H e − , where 0 ≤ e ≤ e −
1, and each H i is T -conjugateto A xi − . The analogous argument to the above then shows that H is Sym e -conjugate to T e × H e +1 × . . . × H e , where 0 ≤ e ≤ e −
1. By Corollary, 3.8, the pair ( T , H e ) cannot satisfy HARP UPPER BOUNDS 15 the hypothesis of Lemma 3.7, in either of the cases (I), (II). Since | T : H e | has the form 2 A A
5, or 2 A
15, Propositions 3.4 and 3.5 then imply that one of the following holds: • T = A n with n ∈ { , } , and H e is both intransitive and not conjugate to A ; • T = A , and H e has shape 7 : 3, 2 : (7 : 3), GL (2), or 2 : GL (2); • T = A , and H e ∈ { A , A . } ; • T = S (4), and either H e ∼ = S (16) or H e ∼ = S (16) . • T = O +8 (2), and H ∼ = A ; or • T = L ( q ) with q = 2 x ′ y ′ z ′ − | L : H | = r ( q + 1) for some r ∈ { , } (so either y ′ or z ′ is 0), and H e ∼ = q : q − r .The first five of these are cases (1)(b), (2), (3), (5), and (6), respectively, in the statementof the corollary. The last is case (4)(b) (with s = t = 1) if y ′ = z ′ = 0, and case (4)(c)otherwise. Finally, assume that case (III) holds. Then using (3.4) and Lemma 2.3 again, wesee that H is Sym e -conjugate to T e × L , where L is a subdirect product in L × Hπ e ; L is a subdirect product in P × . . . × P e ; each P i is a maximal parabolic subgroup of T ; and e + e = e −
1. Moreover, Hπ Λ = T e × L (up to permutation of coordinates). In particular, | H | = | T | e | L || H e | . Thus, | L : H | = | T e : L || T : Hπ e || Hπ e : H e | . Let s and t be the oddparts of | T e : L | and | Hπ e : H e | , respectively. Then s is the index of L in P × . . . × P e , rst = 3 y z , and one of the following must hold: • s = t = 1 and H is Sym e -conjugate to T e × L × Hπ e , L = P × . . . × P e . • s ∈ { , } , t = 1, and H is Sym e -conjugate to T e × L × Hπ e , where L has the form L = s ( P × . . . × P e ). • t ∈ { , } , s = 1, and H is Sym e -conjugate to T e × L , where L has the form L = t ( P × . . . × P e × p : p − r ).These are the cases in (4)(b) in the statement of the corollary.Finally, assume that | Λ | = t . Suppose first that case (I) holds (see the first paragraph above),so that H is a subdirect product in L = T e . Then | T | divides | L : H | by Lemma 2.3(iii), since H = L . Thus, we must have T = A , since | L : H | divides 2 x
15. Moreover, H is Sym e -conjugate to T e − × J , where J is a diagonal subgroup of T , since | T | cannot divide | L : H | .This is case (1)(d) in the statement of the corollary. Suppose next that case (II) holds. Thensince neither | A xi | nor | A xi − | can divide | L : H | in this, case another application of Lemma2.3 yields that H is Sym e -conjugate to T e × Y × . . . × Y e , where Y i is T -conjugate to A xi − for each i , and e = e + e . But then | L : H | is a power of 2 – a contradiction. Suppose nextthat case (III) holds. Then arguing as in the | Λ | = t − H is Sym e -conjugateto T e × L , where L is a subdirect product of the form L = y z ( P × . . . × P e ), with P i a maximal parabolic subgroup of T for all 1 ≤ i ≤ e , and e + e = e . If (III) holdswith 3 y z = 1, then | L : H | is a power of 2 – a contradiction. Thus, if case (III) holds then3 y z = 1. This is case (4)(b) (with r = 1) in the statement of the corollary. The proof iscomplete. (cid:3) Corollary 3.9 gives us information about the possible minimal normal subgroups L (and theiractions) in a minimal counterexample to Theorem 3.1. We would now like to analyse the orbitlengths of certain soluble subgroups of L .Since a point stabilizer H in L in this case is a direct product of subgroups X which arethemselves subdirect products of the form X = n ( X × . . . × X f ) (see Definition 2.2), ouranalysis will have two parts: First, we will show how to determine the orbit lengths of certainsubgroups F < L acting on the cosets of a subdirect product H of the form H = n ( P × . . . × P e ),where each P i is a subgroup of T . Then, we will show how to determine the orbit lengths ofcertain subgroups F < L acting on the cosets of a subgroup
H < L of the form H = H × . . . × H r ,where H i ≤ T e i , and P ri =1 e i = e (i.e. when L has ‘product action’ type). These two bits ofinformation, together with Corollary 3.9, will then allow us to prove Theorem 3.1(iv). We beginwith the first part: Lemma 3.10.
Let L = G × . . . × G e , where each G i is a finite group, and fix proper subgroups P i , S i of G i . Suppose that H ≤ P × . . . × P e is a subdirect product of the form H = n ( P × . . . × P e ) ,and that H is core-free in L . Assume that S := S × . . . × S e has orbit lengths m , . . . , m r inits action on the cosets of P := P × . . . × P e in L , where P i m i = | L : P | . Assume also that ( S i ∩ P x i i )[ P x i i , P x i i ] = P x i i for all x i ∈ G i . Then S has r orbits in its action on the cosets of H in L , with lengths nm , . . . nm r .Proof. Note first that since H is core-free in L , L may be viewed as a transitive permutationgroup acting on the set Ω of cosets of H in L . Moreover, H < P , so the set of cosets of P in L is permutation isomorphic to a set Σ of blocks for P in Ω. Note that the blocks in Σ have size n , H E P , and P/H is abelian of order n . It follows that L is isomorphic to a subgroup of thewreath product n ≀ L Σ .Now, write Σ i , 1 ≤ i ≤ r , for the S -orbits in Σ, where | Σ i | = m i . Also, fix a block ∆ i ∈ Σ i ,for each i . If Stab S (∆ i ) ∆ i is transitive for each i , then it follows that the S O -orbits have sizes | ∆ || Σ i | = nm i , for each i . Thus, we just need to prove thatStab S (∆ i ) ∆ i is transitive for each i .(3.5)To this end, fix 1 ≤ i ≤ e . Then Stab S (∆ i ) ∆ i = S ∩ P α i for some α i ∈ L . To show thatStab S (∆ i ) ∆ i is transitive, we just need to show that H α i ( S ∩ P α i ) = P α i . But α i = ( x , . . . , x e ),for some x j ∈ G j . Thus, S ∩ P α i = ( S ∩ P x ) × . . . × ( S e ∩ P x e e ). Since H α i contains[ P x , P x ] × . . . × [ P x e e , P x e e ], we have H α i ( S ∩ P α i ) = P α i by hypothesis. This proves (3.5),whence the lemma. (cid:3) Next, we prove a general lemma concerning the orbits of subgroups in a transitive permutationgroup with ‘product action’.
Lemma 3.11.
Let L = G × . . . × G e , where each G i is a finite group, and fix proper subgroups H i , S i of G i . Set S = S × . . . × S e and H = H × . . . × H e . Then:(i) Suppose that S i acts transitively on the cosets of H i in G i for ≤ i ≤ e − , and that S e has r orbits, of lengths l , . . . , l r say, in its action on the cosets of H e in G e . Then S has r orbitsin its action on the cosets of H in L , of lengths l m, . . . , l r m , where m := Q e − i =1 | G i ; H i | . HARP UPPER BOUNDS 17 (ii) Assume that there exists ≤ e ≤ e − such that S i acts transitively on the cosets of H i in G i for i ≤ e ; that S i has orbits of lengths k and k in each of the cosets spaces H i \ G i , for e + 1 ≤ i ≤ e − (in particular, | G i : H i | = | G j : H j | = k + k for all e + 1 ≤ i, j ≤ e − in this case); and that S e has r orbits, of lengths l , . . . , l r say, inits action on the cosets of H e in G e . Then S has e − e − × r orbits in its action on thecosets of H in G , with (cid:0) e − e − i (cid:1) orbits of length l j k i k e − e − i for each ≤ i ≤ e − e − ,and each ≤ j ≤ r .Proof. The proof here is routine: if S i is a group acting on a set Ω i , then the orbits of S × . . . × S e in its product action on Ω × . . . × Ω e are precisely the sets ∆ i , × . . . × ∆ i f ,f , where the∆ ,j , . . . , ∆ k j ,j are the orbits of S Ω j j , for 1 ≤ j ≤ e . (cid:3) We will also need the following corollary, which is notationally heavy, but routine.
Corollary 3.12.
Let L , H , and S be as in Lemma 3.11(i), and assume that G i = T for ≤ i ≤ e − , and G e = T k where T is a fixed nonabelian finite simple group. Assume also that S i = S for all ≤ i ≤ e − , and S e = S k where S is a fixed subgroup of T . Write n , . . . , n u for the distinct orbit lengths of S e acting on the cosets of H e in T k , and suppose that S e has f ( n j ) orbits of length n j . Let G ≤ Sym(Ω) be a finite transitive permutation group in which L is a minimal normal subgroup. Suppose that H is the intersection of L with a point stabilizerin G , and let K be the kernel of the action of G on the set of L -orbits. Let R/K be a solublesubgroup of
G/K with orbits of lengths a , . . . , a t on the set of L -orbits. Assume also that:(i) S is T -conjugate to S α for all automorphisms α of T ; and(ii) N L ( S ) is soluble (equivalently, N T ( S ) is soluble).Then F has orbit lengths a i X ( i,j ) k n j m for some positive partitions f ( n j ) = P r i,j k =1 X ( i,j ) k of f ( n j ) ,where ≤ j ≤ u , ≤ i ≤ t .Proof. Let ω ∈ Ω such that H = L ∩ Stab G ( ω ), and let ∆ be the L -orbit containing the point ω .Let Σ := { ∆ g : g ∈ G } be the set of L -orbits. Fix a subgroup B of G , and assume that B Σ hasorbits Σ , . . . , Σ r , of lengths b , . . . , b r , respectively. Let ∆ , . . . , ∆ r ∈ Σ be representatives ofthese orbits. Then B acts on each of the sets S x ∈ B ∆ xi . Suppose that Stab B (∆ i ) ∆ i has orbitslengths x i, , . . . , x i,k i . Then B Ω has orbit lengths b i x i, , . . . , b i x i,k i .Now, as in the proof of Corollary 3.8, S being T -conjugate to S α for all automorphisms α of T implies that S = S e is L -conjugate to S α for all automorphisms α ∈ Aut( T ) ≀ S e ∼ =Aut( L ). Hence, LN G ( S ) = G , by Lemma 3.6. Thus, we may choose a subgroup F of N G ( S )containing N L ( S ) with LF/L = R/L . It follows that F Σ has orbits Σ , . . . , Σ t of lengths a , . . . , a t , respectively. Fix ∆ i ∈ Σ i . Then since F normalizes S and S acts trivially on Σ, eachStab F (∆ i ) ∆ i -orbit is a union of S -orbits of the same length. Thus, by Lemma 3.11 the orbitlengths of Stab F (∆ i ) ∆ i are X ( i,j )1 n j m , . . . , X ( i,j ) r i n j m , where P r i k =1 X ( i,j ) = f ( n j ), for each i , j . Using the first paragraph above, we deduce that F O has orbits of lengths a i X ( i,j ) k n j m , for1 ≤ k ≤ r i , 1 ≤ j ≤ u , 1 ≤ i ≤ t .Finally, since both F L/L = R/L and F ∩ L = N L ( S ) are soluble, the group F is soluble.This completes the proof. (cid:3) J Orbit lengths for S = D < A acting on the cosets of J < A A two of length 10S two of length 5A one of length 5 C two of length 1, one of length 10 D one of length 1, one of length 5 Table 2.
Orbit lengths for S = D < A acting on the cosets of a subgroup J < A We are now ready to exhibit orbit lengths of particular soluble subgroups of minimal coun-terexamples to Theorem 3.1.
Lemma 3.13.
Let G be a minimal counterexample to Theorem 3.1, and let L = T e be a minimalnormal subgroup of G , where T = A . Let H be the intersection of L with a point stabilizer for G , and let a be the number of L -orbits. Then G has a soluble subgroup F such that one of thefollowing holds:(i) If H is as in case (1)(a) of Corollary 3.9, then F has orbit lengths b aX i , b aY j ,where ( X i X i , X j Y j , b , b ) ∈ { (4 , , , , (4 , , , , (2 , , , , (2 , , , − , (2 , , , , (1 , , , − } . (ii) If H is as in case (1)(b) of Corollary 3.9, and G/L is soluble, then F either has orbitlengths b aX i , where ( X i X i , b ) ∈ { (2 , , (2 , } , or aX i , aY j , where ( P i X i , P j Y j ) = (2 , .(iii) If H is as in case (1)(c) of Corollary 3.9, then F has orbit lengths aX i , aY i , where ( X i X i , X j Y j ) = (4 , . (iv) If H is as in case (1)(d) of Corollary 3.9, then F has precisely two orbits, of lengths a and a .Proof. Set S := D , S := S e , and F := N G ( S ). We note first that N T ( S ) is soluble, and D is A -conjugate to S α for all α ∈ S . Thus, (i) and (ii) in Corollary 3.12 hold. We will also showthat G modulo the kernel K of the action of G on the set of L -orbits is soluble in each case,so we can take R = G in Corollary 3.12. Since either H is Sym e -conjugate to T e − × J with J ≤ T , or H is Sym e -conjugate to T e − × J with J ≤ T , we therefore just need to computethe orbit lengths for S [respectively S ] in the case e = 2 [resp. e = 1] and then apply Lemma3.11 and Corollary 3.12.Suppose first that we are in case (i). Then either | T i : Hπ e − | = 2 x
3, and | T i : Hπ e | = 2 x x + x = x . Thus, G/K , being a minimal transitive group of 2-powerdegree, is soluble (so we can indeed take G = R in Corollary 3.12). Moreover, H e − ∈ { C , D } and H e ∈ { A , S , A } . The orbit lengths of S acting on the cosets of H e − and H e are thengiven in Table 2. If S has orbit lengths c , . . . , c k on the cosets of H e − , and d , . . . , d l on the HARP UPPER BOUNDS 19 cosets of H e , then S has orbit lengths c i d j , 1 ≤ i ≤ k , 1 ≤ j ≤ l , on the cosets of H e − × H e .Thus, we can use the table above and Lemma 3.11 and Corollary 3.12 to deduce the orbitlengths of the soluble group F = N G ( S ). For example, if H e − = A and H e = C , then S hasfour orbits of length 10, and two orbits of length 100 in its action on the cosets of H e − × H e in T . By Lemma 3.11(i), S e then has orbits of the same lengths in its action on the cosets of H in L ( m = 1 in this case). Thus, by Corollary 3.12, F has orbit lengths 10 aX i , 100 aY j , where X i , Y j ∈ N , and P i X i = 4, P j Y j = 2.We now move on to cases (ii), (iii) and (iv). In case (ii), K ≥ L , so G/K is soluble byhypothesis. In cases (iii) and (iv), | L : H | = 2 x
15, so
G/K is a minimal transitive group of2-power degree, whence soluble. Thus, we can take again G = R (and hence t = 1, a = a ) inCorollary 3.12.For part (ii), the computations of the possible orbit lengths is completely analogous to thecomputations in case (i) above (except that the cases H e ∈ { A , C , D } cannot occur, andwe also need to compute the orbit lengths of the subgroups of A of orders 1, 2, and 4). Forpart (iii), one can use Magma [1] for example, to see that S has four orbits of length 5 andfour orbits of length 25, in its action on the cosets of J = (S × D ) in T . For part (iv), S has one orbit of length 10 and one orbit of length 50 in its action on the cosets of any diagonalsubgroup in T . We now apply Corollary 3.12 in each case to complete the proof. (cid:3) Lemma 3.14.
Let G be a minimal counterexample to Theorem 3.1, and let L = T e be a minimalnormal subgroup of G , where T = A . Let H be the intersection of L with a point stabilizer for G , and let a be the number of L -orbits. Then G has a soluble subgroup F such that one of thefollowing holds:(i) If H is as in case (1)(a) of Corollary 3.9, then F has orbit lengths b aX i , b aY j , where ( X i X i , X j Y j , b ) ∈ { (4 , , , (4 , , , (2 , , } . (ii) If H is as in case (1)(b) of Corollary 3.9, and G/L is soluble, then F either has orbitlengths b aX i , where ( X i X i , b ) ∈ { (4 , , (12 , , (4 , , (3 , , (2 , } , or aX i , aY j , where ( P i X i , P j Y j ) ∈ { (4 , , (2 , } .Proof. Set S := D < A < A , S := S e , and F := N G ( S ). Then S has soluble normalizer inA , and D is A -conjugate to S α for all α ∈ Aut(A ). Thus, (i) and (ii) in Corollary 3.12 hold.Note also that, as in the proof of Lemma 3.13, G modulo the kernel K of the action of G on theset of L -orbits is soluble in each of (i) and (ii). Thus, we can take R = G in Corollary 3.12 (andhence t = 1, a = a ). Since either H is Sym e -conjugate to T e − × J with J = H e − × H e ≤ T ,or H is Sym e -conjugate to T e − × J with J = H e ≤ T , we therefore just need to compute theorbit lengths for S [respectively S ] in the case e = 2 [resp. e = 1] and then apply Lemma 3.11and Corollary 3.12.Suppose first that we are in case (i). Then, without loss of generality, we may assume that | T i : Hπ e − | = 2 x | T i : Hπ e | = 2 x
5, where x + x = x . Hence, H e − is in one of the J Orbit lengths for S = D < A acting on the cosets of J < A A × A four of length 10(A × A ) . × A ) . . (transitive) one of length 1, one of length 5A (intransitive) one of length 1, one of length 5 Table 3.
Orbit lengths for S = D < A acting on the cosets of a subgroup J < A two conjugacy classes of A in A , while H e ∈ { A × A , (A × A ) . , (A × A ) . . } . The orbitlengths of S acting on the cosets of H e − and H e are then given in Table 3. The computationsof the possible orbit lengths are then completely analogous to the computations in the proof ofLemma 3.13.If we are in case (ii), then H e is either one of the groups J in Table 3, with J intransitive, or | H e | ∈ { , , , } . We can quickly compute the orbit lengths of S in these latter cases, andpart (ii) follows as in the proof of Lemma 3.13. (cid:3) Lemma 3.15.
Let G be a minimal counterexample to Theorem 3.1, and let L = T e be a minimalnormal subgroup of G , where T ∈ { S (4) , O +8 (2) } . Let H be the intersection of L with a pointstabilizer for G , and let a be the number of L -orbits. Then G has a soluble subgroup F suchthat either n = 120 a and F has two orbits of lengths a and a ; or n = 960 a and F has twoorbits of length a and a .Proof. In these cases, H is Sym e -conjugate to T e − × H e , where ( T, H e ) is one of the pairs(S (4) , S (16)) (2classes), (O +8 (2) , A ) (3 classes), or (O +8 (2) , O (2)) (3 classes). Thus, | T : H | is 120, 960, or 120, respectively. If T = S (4), then let S be the maximal parabolic subgroupof T of index 425. If T = O +8 (2), then let S be the maximal parabolic subgroup of T of index1575. Then in each case, N T ( S ) is soluble, and S and S α are T -conjugate for all α in Aut( T ).That is, (i) and (ii) in Corollary 3.12 hold. Moreover, S has two orbits in its action on thecosets of H e in T , of lengths 24 and 96 when | T : H | = 120, and lengths 192 and 768 when | T : H | = 960.Let Σ be the set of L -orbits. Then since 15 divides | L : H | in each case, G Σ is a 2-group.Thus, as in the proof of Lemmas 3.13 and 3.14, we see that either n = 120 a and F has twoorbits of lengths 24 a and 96 a , or n = 960 a and F has two orbits of length 192 a and 768 a . (cid:3) In the next three lemmas we deal with the cases T ∈ { A , A , L ( q ) : q = 2 x ′ y ′ z ′ − } .Apart from the case q = 2 x ′ y ′ z ′ − { y ′ , z ′ } = { , } , we will find a soluble subgroup S of L with convenient orbit lengths, and then simply set F := S (we do not need to trouble tofind a soluble subgroup with fewer orbits, as we did in the T ∈ { A , A , S (4) , O +8 (2) } cases). Lemma 3.16.
Let G be a minimal counterexample to Theorem 3.1, and let L = T e be aminimal normal subgroup of G , where T = A . Let H be the intersection of L with a pointstabilizer for G , so that H is of type (2) in Corollary 3.9. That is, H is Sym e -conjugate to T e × Y × . . . × Y e × H e , where e + e = e − ; Y i is T -conjugate to A for each i ; and HARP UPPER BOUNDS 21 H e has shape , : (7 : 3) , GL (2) , or : GL (2) . Let a be the number of L -orbits, let S := 7 : 3 < T , and set F := S e < L . Then(i) Suppose that H e = 7 : 3 , and define ( s , l ) := (1 , ; ( s , l ) := (11 , ; and ( s , l ) :=(42 , . Then F has as j (cid:0) e i (cid:1) orbits of size i l j , for each ≤ i ≤ e , ≤ j ≤ .(ii) Suppose that H e ∼ = 2 : (7 : 3) or K ∼ = GL (2) , and define ( s , l ) := (1 , ; ( s , l ) := (5 , ;and ( s , l ) := (4 , . Then F has as j (cid:0) e i (cid:1) orbits of size i l j , for each ≤ i ≤ e , ≤ j ≤ .(iii) Suppose that H e = 2 : GL (2) , and define ( s , l ) := (1 , ; and ( s , l ) := (2 , . Then F has as j (cid:0) e i (cid:1) orbits of size i l j , for each ≤ i ≤ e , ≤ j ≤ .Proof. Note that S has orbits of lengths 7 and 1 in its action on the cosets of A in T , while S has s i orbits of length l i in its action on the cosets of H in T , where s i and l i are as definedin each of the listed cases. We then apply Lemma 3.11(ii) to find the orbit lengths of F in itsaction on the cosets of H in L : we see that F has s j (cid:0) e i (cid:1) orbits of size 7 i l j in each case.Finally, since this holds for each G -conjugate of H in L , and F fixes each L -orbit, we deducethat F < G has as j (cid:0) e i (cid:1) orbits of size 7 i l j in [ n ], as required. (cid:3) Lemma 3.17.
Let G be a minimal counterexample to Theorem 3.1, and let L = T e be aminimal normal subgroup of G , where T = A . Let H be the intersection of L with a pointstabilizer for G , so that H is of type (3) in Corollary 3.9. That is, H is Sym e -conjugate to T e × Y × . . . × Y e × H e , where e + e = e − ; Y i is T -conjugate to A for each i ; and H e has shape A . b , for some b ∈ { , } . Let a be the number of L -orbits, let S := C < T , andset F := S e < L . Then F has b (cid:0) e i (cid:1) a orbits of size i +1 , for each ≤ i ≤ e .Proof. The proof is identical to the A case above: S has orbits of lengths 15 and 1 in itsaction on the cosets of A in T , while S has b orbits of length 15 in its action on the cosetsof H e in T We then apply Lemma 3.11 and argue as in the proof of Lemma 3.16 to completethe proof. (cid:3)
Lemma 3.18.
Let G be a minimal counterexample to Theorem 3.1, and let L = T e be a minimalnormal subgroup of G , where T = L ( q ) , with q an odd prime power of the form q = 2 x ′ y ′ z ′ − , ≤ y ′ , z ′ ≤ . Let H be the intersection of L with a point stabilizer for G , so that H is oftype (4) in Corollary 3.9. Let a be the number of L -orbits, and set F := S e < L , where S is amaximal parabolic subgroup of T . Then(i) Suppose that H has type (4)(a) from Corollary 3.9, so that q = p is a Mersenne primewith dividing p − , and H is S e -conjugate to T e × Y × . . . × Y e × ( p : p − ) × ( p : p − ) ,where each Y i is a maximal parabolic subgroup of T . Then F := S has a (cid:0) e i (cid:1) orbits of size p i ; a (cid:0) e i (cid:1) orbits of size p i +1 ; and a (cid:0) e i (cid:1) orbits of size p i +2 , for each ≤ i ≤ e .(ii) Suppose that H has type (4)(b) from Corollary 3.9, so that q = p is a Mersenne primeand H is Sym e -conjugate to T e × L , where L is a subdirect product of the form L = t ( L × Hπ e ) ; L is a subdirect product of the form L = s ( P × . . . × P e ) ; each P i isa maximal parabolic subgroup of T ; Hπ e is a subgroup of index r in a maximal parabolicsubgroup of T ; rst = 3 y z is the odd part of | L : H | ; and e + e = e − . Then F := S has a (cid:0) e +1 i (cid:1) orbits of size y z p i for each ≤ i ≤ e + 1 . (iii) Suppose that H has type (4)(c) from Corollary 3.9, so that H is Sym e -conjugate to T e − × ( q : q − r ) , where y ′ + z ′ = 1 ; ∈ { y ′ , z ′ } ; and r = 3 − y ′ − z ′ > . Then F := N G ( S ) issoluble, and has precisely two orbits, of sizes ar and aqr .Proof. We need to determine the orbit lengths of S in its action of the cosets of the subgroups X ∈ { q : q − , q : q − , q : q − (5 | q − , q : q − (5 | q − } for part (ii); while we need todetermine the orbit lengths of S in its action of the cosets of the subgroup ( p : p − ) × ( p : p − ) < T for part (i).We first consider the orbits of the action of S ∼ = q : q − on the cosets of itself in T . This isequivalent to the action of S on the set of one dimensional subspaces in F q . It is well knownthat S has one orbit of size 1 and one orbit of size q in this action. It follows easily that if X has shape q : q − r with r a divisor of q − , then S has one orbit of size r , and one orbit ofsize rq . Next, when p = q , S has four orbits on the cosets of ( p : p − ) × ( p : p − ), of lengths15, 15 q , 15 q , and 15 q . Lemma 3.11(ii) then implies that F has (cid:0) e i (cid:1) orbits of size 15 p i ; 2 a (cid:0) e i (cid:1) orbits of size 15 p i +1 ; and a (cid:0) e i (cid:1) orbits of size 15 p i +2 in its action on the cosest of H in L , for each0 ≤ i ≤ e . Since our arguments hold with H replaced by any G -conjugate of H , we deducethat F has a (cid:0) e i (cid:1) orbits of size 15 p i ; 2 a (cid:0) e i (cid:1) orbits of size 15 p i +1 ; and a (cid:0) e i (cid:1) orbits of size 15 p i +2 in [ n ], for each 0 ≤ i ≤ e .Suppose now that we are in case (ii). Note that( S ∩ P xi )[ P x i i , P x i i ] = P x i i for all x i in T .(3.6)Thus, by the arguments above, together with Lemma 3.10, we deduce that S e has (cid:0) e i (cid:1) orbitsof length sq i on the cosets of L in T e , for each 0 ≤ i ≤ e . Hence, since S has two orbits,of sizes r and qr , on the cosets of Hπ e in T , we deduce from Lemma 3.11 that S e +10 has (cid:0) e i (cid:1) orbits of length srq i and (cid:0) e i (cid:1) orbits of length srq i +1 in its action on the cosets of L × Hπ e in T e +1 . By (3.6), we can then use Lemma 3.10 again: we get that S e +10 has (cid:0) e i (cid:1) orbits oflength rstq i and (cid:0) e i (cid:1) orbits of length rstq i +1 in its action on the cosets of L in T e +1 . Notethat rst = 3 x x and (cid:0) e i (cid:1) + (cid:0) e i − (cid:1) = (cid:0) e +1 i (cid:1) for i >
0. We can then deduce from Lemma 3.11(i)(with m = 1) that F has (cid:0) e +1 i (cid:1) orbits of size 3 y z p i in its action on the cosets of H in L , foreach 0 ≤ i ≤ e + 1. Since our arguments are independent of the choice of G -conjugate of H in L , we see as above that F has a (cid:0) e +1 i (cid:1) orbits of size 3 y z p i in [ n ], for each 0 ≤ i ≤ e + 1.Finally, the proof in case (iii) is identical to the proof of Lemma 3.15. (cid:3) Lemma 3.19.
Let G be a minimal counterexample to Theorem 3.1, and let L = T e be a minimalnormal subgroup of G , where T = A m with m ∈ { , } , and G/L is insoluble. Let H be theintersection of L with a point stabilizer for G , and let a be the number of L -orbits. Then(i) H is as in case (1)(b) of Corollary 3.9;(ii) | T : H e | = 2 x and a = 2 x ;(iii) G/L has a unique nonabelian chief factor L ′ such that L ′ ∼ = L ( p ) f , where p is a Mersenneprime; and(iv) There exist positive integers X , b , and e ≤ e ; positive partitions (cid:0) e i (cid:1) = P j C ( i ) j and X = P k X ( i,j ) k of (cid:0) e i (cid:1) and X (for each i, j ); and a soluble subgroup F of G , such that oneof the following holds: HARP UPPER BOUNDS 23 • T = A , ( X, b ) ∈ { (2 , , (2 , } , and F has C ( i ) j orbits of length b p i X ( i,j ) k , for ≤ i ≤ e , and each j, k . • T = A , ( X, b ) ∈ { (4 , , (4 , , (2 , } , and F has C ( i ) j orbits of length b p i X ( i,j ) k ,for ≤ i ≤ e , and each j, k .Proof. Let K be the kernel of the action of G on the set Σ of L -orbits, and fix an L -orbit ∆.Then since G/K is an insoluble minimal transitive group, its degree a cannot be a power of 2.Hence, either 3 or 5 must divide a . Thus, the odd part of | ∆ | = | L : H | is either 3 or 5, andit follows from Corollary 3.9, and by inspection of the subgroups of A of A , that the onlypossibility is that H is as in case (1)(b) of Corollary 3.9. Part (i) follows.Now, by Corollary 3.9(1)(b), H e is not transitive, and not equal to a natural point stabilizerA m − < T = A m . Thus, going through the subgroups of A and A , we see that | T : H e | mustbe of the form | T : H e | = 2 x
5. Hence,
G/K is an insoluble minimal transitive group of degree2 x
3. This proves (ii). Part (iii) then follows immediately from [13, Theorem 1.8].We now prove (iv). Set S := D < T , and S := S e . Then as shown in Lemmas 3.13and 3.14, the hypotheses in Corollary 3.12 hold with this choice of S . Thus we can, and do,apply Corollary 3.12, with R/K chosen to be the soluble subgroup of
G/K exhibited in Lemma3.18(ii), with p = 2 x − rst = 3 . We can then deduce (iv) by taking the possible orbitlengths of S acting on the cosets of H e from Tables 2 and 3 (noting again that H e is intransitive,and that H e = A in the A case). (cid:3) Finally, we are ready to prove Theorem 3.1.
Proof of Theorem 3.1.
Let G be a counterexample to the theorem of minimal degree, and let L ∼ = T e be a minimal normal subgroup of G , with T simple. Then T is nonabelian, by Lemma3.3. Let H be the intersection of L with a point stabilizer in G , and write | L : H | = 2 x y z ,where x ≤ x , y ≤ y , and z ≤ z . Thus, if K denotes the kernel of the action of G on the setΣ of L -orbits, then G Σ ∼ = G/K is minimal transitive of degree 2 x − x y − y z − z . Also, either y or z is non-zero, by Corollary 3.8(i). Thus, since K/L is soluble by Lemma 3.3, the minimalityof G as a counterexample implies that G has at most 1 + y − y + z − z ≤ y + z nonabelianchief factors. That is, (i) holds.Now, T is one of the groups in part (ii) of Theorem 3.1, by Corollary 3.9, so (ii) also holds.Next, we prove (iii). So assume that G has two nonabelian chief factors, and let L ′ ∼ = T ′ e ′ be anonabelian chief factor of G/L . Then the minimal transitive group G Σ ∼ = G/K has L ′ as a chieffactor, since K/L is soluble. Since minimal transitive groups of 2-power degree are 2-groups, wededuce that G Σ ∼ = G/K is minimal transitive of degree either (1) 2 x − x x − x
3. The sizeof an L -orbit in these cases is 2 x x
5, respectively. Now, by Corollary 3.9, the only possibil-ity for T when the size of an L -orbit is 2 x T = L ( p ) with p a Mersenne prime; while the onlypossibility for T when the size of an L -orbit is 2 x T ∈ { A , A , L ( p ) : p a Mersenne prime } .This gives us what we need. Moreover, inspection of Table 1 and the inductive hypothesis implythat in case (1), we must have T ′ ∈ { A , A , L ( p ′ ) : p a Mersenne prime } ; while in case (2),we have T ′ = L ( p ′ ) with p ′ a Mersenne prime. This proves (iii). Finally, in Lemmas 3.13, 3.14, 3.15, 3.16, 3.17, 3.18, and Lemma 3.19, we exhibit a solublesubgroup F of G with particular orbit lengths. These orbit lengths are as listed in Table 1, andthis completes the proof. (cid:3) The proof of Theorem 1.1
In this section, we prove Theorem 1.1. First, we require the following definitions and lemmafrom [13].
Definition 4.1.
For a positive integer s with prime factorisation s = p r p r . . . p r t t , set ω ( s ) := P r i , ω ( s ) := P r i p i , K ( s ) := ω ( s ) − ω ( s ) = P r i ( p i −
1) and e ω ( s ) = s K ( s ) (cid:18) K ( s ) j K ( s )2 k(cid:19) . For a prime p , write s p for the p -part of s . Definition 4.2.
Let p be prime, and let s be a positive integer. We define E sol ( s, p ) := min { e ω ( s ) , s p } . Lemma 4.3. [13, Corollary 4.25(ii)(b)]
Let G be a transitive permutation group of degree n ,and suppose that G is imprimitive with a minimal block of size . Let Σ be the associated systemof blocks of size n , so that G may be viewed as a subgroup of the wreath product ≀ G Σ . Let F be a soluble subgroup of G Σ , and suppose that F has orbit lengths s , . . . , s r . Then d ( G ) ≤ r X i =1 E sol ( s i ,
2) + d ( G Σ ) . Corollary 4.4.
Let G be a transitive permutation group of degree n , and suppose that G isimprimitive with a minimal block of size . Let Σ be the associated system of blocks of size n ,so that G may be viewed as a subgroup of the wreath product ≀ G Σ . Suppose that n has theform n = 2 x y z , with ≤ y, z ≤ , and that G Σ contains no soluble transitive subgroups. Let F be a soluble subgroup of G Σ with orbit lengths as exhibited in Table 1.(1) If F is as in row 12 of Table 1, then d ( G ) ≤ e + b +1 a + d ( G Σ ) .(2) If F is as in row 22 of Table 1, then d ( G ) ≤ e + b +2 a + d ( G Σ ) .(3) If F is as in row 24, 25, or 26 of Table 1, then d ( G ) ≤ e a + d ( G Σ ) ; d ( G ) ≤ e a + d ( G Σ ) ; or d ( G ) ≤ e a + d ( G Σ ) , respectively.(4) If F is as in row 27 of Table 1, then d ( G ) ≤ e +4 − b a + d ( G Σ ) .(5) If F is as in row 28 of Table 1, then d ( G ) ≤ e a + d ( G Σ ) .(6) If F is as in row 29 of Table 1, then d ( G ) ≤ e a + d ( G Σ ) .(7) If F is as in row 30 of Table 1, then d ( G ) ≤ a + d ( G Σ ) .Proof. Write s , . . . s r for the F -orbit lengths. Then by Lemma 4.3 and the definition of E sol ,we have d ( G ) ≤ r X i =1 ( s i ) + d ( G Σ ) . (4.1) HARP UPPER BOUNDS 25
Parts (3), (4), (5), and (6) now follow immediately from (4.1) and inspection of the relevantrow in Table 1.Suppose now that F is as in row 12 or 22 of Table 1. Then (4.1) implies that d ( G ) ≤ b X i,j,k C ( i ) j ( X ( i,j ) k ) + d ( G Σ ) . Since ( X ( i,j ) k ) is an ordered positive partition of either 2 or 4, the result now follows easilyby going through each of the possibilities for ( X ( i,j ) k ) k , and noting that P i,j C ( i ) j = P i (cid:0) e i (cid:1) =2 e . (cid:3) We are now ready to prove Theorem 1.1.
Proof of Theorem 1.1.
By [13, Theorem 1.1], we may assume that n has the form n = 2 x y y ∈ { , } . Furthermore, we may assume that 17 ≤ x ≤
26 if y = 0, and 15 ≤ x ≤ y = 1; that G has a block system Σ consisting of blocks of size 2; and that G Σ contains nosoluble transitive subgroup.It then follows from Theorem 3.1 that G Σ has a soluble subgroup F whose orbit lengthsoccur in one of the rows of Table 1. We can then use this fact, together with Lemma 4.3 toprove the theorem. Indeed, suppose that we have proved the theorem for all degrees properlydividing n , and that the possible orbit lengths for the soluble subgroup F are s ,j , s ,j , . . . , s r j ,j ,for 1 ≤ j ≤ t . We then compute the maximum value of P r j i =1 E sol ( s i,j ,
2) over 1 ≤ j ≤ t , andadd it to our previously obtained bound for d ( G Σ ). By Lemma 4.3, this gives a bound for d ( G ).Let us illustrate this strategy with some examples. Suppose first that n = 2
5. From Table1, we see that G Σ ≤ S has a soluble subgroup whose orbit lengths lie in one of the rows 7,18, 19, or 29 (with y = 0, z = 1 in the last case). Computing as in the n = 2
15 examplegiven at the beginning of the section, we see that the lengths of the F -orbits are one of thefollowing: • i orbits of equal lengths, i ∈ { , } ; • •
15; or • (cid:0) e i (cid:1) a orbits of size 5 p i and (cid:0) e i (cid:1) a orbits of size 5 p i +1 , for each 0 ≤ i ≤ e , where thepossibilities for e and p are ( e , p ) ∈ { ( A, − , (1 , −
1) : 1 ≤ A ≤ } , and n = 5 a ( p + 1) e .We make some remarks about the last point above. Since ( p + 1) e must divide n , we only needto consider the Mersenne primes p = 2 u − p + 1 = 2 u < k := ( n ) , and those e with 1 ≤ e ≤ (cid:4) ku (cid:5) −
1. And since we are assuming, throughout this proof, that n ≤ , theonly Mersenne primes under consideration are 2 u −
1, for u a prime less than or equal to 31. Inthe case n = 2 p −
1. So only the cases u ∈ { , } can occur.We now proceed to compute a bound for d ( G ) (for n = 2
5) in each case, by using Table1 and Corollary 4.3, as explained above. By [13, Table A.1], we have d ( G Σ ) ≤ F has 2 orbits of length 2 d ( G ) ≤ E sol (2 ,
2) + 65538 = 123274. Similarly, if F has 4 orbits of length 2 d ( G ) ≤ F has 2 orbits of length 2 d ( G ) ≤ F has 1 orbit of length 2
15 then d ( G ) ≤ F has (cid:0) e i (cid:1) d orbits of size 5 p i and (cid:0) e i (cid:1) d orbits of size 5 p i +1 , for each 0 ≤ i ≤ e , where ( e , p ) ∈ { (1 , − , (2 , − , (1 , − } and a = n p +1) e . If ( e , p ) = (1 , − F has 2 orbits of size 5, and 2 orbits of size5 ×
31 = 155. Therefore, we get d ( G ) ≤ E sol (5 ,
2) + 2 E sol (155 ,
2) + 65538 = 69634. Theother cases are entirely similar, and we see that d ( G ) is always bounded above by 126313. Since (cid:22) c √ log 2 (cid:23) = 129117, this proves Theorem 1.1 in the case n = 2 n = 2 x
15. More precisely, we will go through the necessary computations in thecase n := 2
15. In order to do this, we will first need an upper bound on d ( G Σ ). Since G Σ is a transitive permutation group of degree 2
15, we can deduce from [13, Table A.1] that d ( G Σ ) ≤ d ( G Σ ) ≤ d ( X ) ≤
16 for transitive groups X ≤ S is used. Wecan now use the exact bound d ( X ) ≤
10 for transitive X ≤ S from [5, Section 5]. Performingthe computations in [13, proof of Lemma 5.12, case 2, page 40] again with this new bound, wequickly deduce that d ( G Σ ) ≤ G Σ ≤ S is transitive.(4.2)Using (4.2) and Lemma 4.3, we can now determine an upper bound for d ( G ) when n = 2 F -orbit lengths in the case when G hasa unique nonabelian chief factor, and this chief factor is a direct product of copies of A (thesecases correspond to rows 1–11 in Table 1). Entirely analogous calculations give us the possible F -orbit lengths when G has a unique nonabelian chief factor, and this chief factor is a directproduct of copies of A (these cases correspond to rows 13–21 in Table 1) or S (4) or O +8 (2)(row 23). When F has orbit lengths as in rows 12, 22, and 24–30 of 1, we use Corollary 4.4 tobound d ( G ). Taking the maximum of these bounds for d ( G ), we get d ( G ) ≤ (cid:22) c √ log 2 (cid:23) = 97895.The remaining cases with n of the form n = 2 x
15, and 16 ≤ x ≤
35, are entirely similar,except that we use the bound d ( G Σ ) ≤ (cid:22) c x − √ log 2 x − (cid:23) for d ( G Σ ) (which holds by the inductivehypothesis). We note in particular that we get d ( G Σ ) ≤ x = 16 (this will beimportant). The result now follows in each case, except when x = 17. So assume that x = 17.By [13, Theorem 1.1], we may assume that G Σ is imprimitive with a minimal block of size 2.Let K be the kernel of the action of G Σ on a set Σ of blocks of size 2 in Σ. Assume first that G Σ ∩ K = 1. If ( G Σ ) Σ ∼ = G Σ /K contains a soluble transitive subgroup S/K , then S is asoluble transitive subgroup of G Σ – a contradiction. Thus, ( G Σ ) Σ contains no soluble transitivesubgroups. Because of this, we can use our previously obtained bound d ( G Σ ) ≤ HARP UPPER BOUNDS 27 case. If G Σ ∩ K = 1, then d ( G Σ ) = d ( G Σ ) ≤ (cid:22) c √ log 2 (cid:23) = 98547, so in either case, we get d ( G Σ ) ≤ d ( G ) as we did in the case n = 2 d ( G ) ≤ (cid:22) c √ log 2 (cid:23) = 372380. (cid:3) References [1] W. Bosma, J. Cannon, and C. Playoust. The Magma algebra system. I. The user language.
J. SymbolicComput. (1997), 235–265.[2] J.N. Bray, D.F. Holt and C.M. Roney-Dougal. The maximal subgroups of the low-dimensional finite classicalgroups. London Math. Soc. Lecture Note Ser. , Cambridge University Press, 2013.[3] J. J. Cannon and D. F. Holt. The transitive permutation groups of degree 32. Experiment. Math. (2008), 307–314.[4] P. Erd¨os. On a Theorem of Sylvester and Schur. J. London Math. Soc. (1934), 282–288.[5] D. F. Holt, G. Royle, and G. Tracey. The transitive groups of degree 48 and some applications. Availablefrom arXiv:2102.07183.[6] P. Kleidman and M. Liebeck. The subgroup structure of the finite classical groups. London Math. Soc.Lecture Note Ser. , Cambridge University Press, 1990.[7] L.G. Kov´acs and M.F. Newman. Generating transitive permutation groups. Quart. J. Math. Oxford ,(1988) 361–372.[8] M.W. Liebeck, C.E. Praeger, and J. Saxl. Transitive subgroups of primitive permutation groups. J. Algebra , (2000) 291–361.[9] A. Lucchini. Enumerating transitive finite permutation groups. Bull. London Math. Soc. , (1998) 569–577.[10] G. Malle. The maximal subgroups of F ( q ). J. Algebra (1991), 52–69.[11] C. E. Praeger, and C. Schneider, Permutation groups and Cartesian decompositions . London Math. Soc.Lecture Note Ser. , Cambridge University Press, 2018.[12] M. Suzuki. On a class of doubly transitive groups, Ann. of Math. (1962), 105–145.[13] G. Tracey. Minimal generation of transitive permutation groups. J. Algebra (2018), 40–100.[14] R. Wilson et al. A World Wide Web Atlas of finite group representations, http://brauer.maths.qmul.ac.uk/Atlas/v3/.
Mathematical Institute, University of Oxford
Email address ::