A new solvability criterion for finite groups
aa r X i v : . [ m a t h . G R ] J u l A NEW SOLVABILITY CRITERION FOR FINITEGROUPS
SILVIO DOLFI, MARCEL HERZOG, AND CHERYL E. PRAEGER
Abstract.
In 1968, John Thompson proved that a finite group G is solvable if and only if every 2-generator subgroup of G issolvable. In this paper, we prove that solvability of a finite group G is guaranteed by a seemingly weaker condition: G is solvableif for all conjugacy classes C and D of G , there exist x ∈ C and y ∈ D for which h x, y i is solvable. We also prove the followingproperty of finite nonabelian simple groups, which is the key toolfor our proof of the solvability criterion: if G is a finite nonabeliansimple group, then there exist two integers a and b which representorders of elements in G and for all elements x, y ∈ G with | x | = a and | y | = b , the subgroup h x, y i is nonsolvable. Introduction
John G. Thompson’s famous ‘N-group paper’ [T] of 1968 includedthe following important solvability criterion for finite groups:
A finite group is solvable if and only if every pair of itselements generates a solvable group.
P. Flavell [F] gave a relatively simple proof of Thompson’s result in1995. We prove that solvability of finite groups is guaranteed by aseemingly weaker condition than the solvability of all its 2-generatorsubgroups.
Theorem A.
Let G be a finite group such that, for all x, y ∈ G ,there exists an element g ∈ G for which h x, y g i is solvable. Then G issolvable. Theorem A can be rephrased as the following equivalent result.
Mathematics Subject Classification.
Key words and phrases.
Solvable groups, finite simple groups.The first author is grateful to the School of Mathematics and Statistics of theUniversity of Western Australia for its hospitality and support, while the investiga-tion was carried out. He was partially supported by the MIUR project “Teoria deigruppi e applicazioni”. The third author was supported by Federation FellowshipFF0776186 of the Australian Research Council. heorem A’. Let G be a finite group such that, for all conjugacyclasses C and D of G (possibly C = D ), there exist x ∈ C and y ∈ D for which h x, y i is solvable. Then G is solvable. Our second main result, which is the key tool for proving Theorem A,deals with the nonsolvability of certain 2-generator subgroups of finitenonabelian simple groups. For a finite group G let oe ( G ) = { m | ∃ g ∈ G with | g | = m } denote the set of element orders of G . Using the classification of finitesimple groups, we prove the following theorem. Theorem B.
Let G be a finite nonabelian simple group. Then thereexist a, b ∈ oe ( G ) , such that, for all x, y ∈ G with | x | = a , | y | = b , thesubgroup h x, y i is nonsolvable. Theorem B was proved separately for alternating groups, sporadicgroups, classical groups of Lie type and exceptional groups of Lie typein Propositions 2.1, 2.2, 4.2, 4.5, respectively. In view of Proposi-tions 2.1, 2.2, we state the following conjecture.
Conjecture. If G is a finite nonabelian simple group, then there existtwo distinct primes p, q ∈ oe ( G ) , such that, for all x, y ∈ G with | x | = p , | y | = q , the subgroup h x, y i is nonsolvable (or, maybe, even nonabeliansimple). The authors are grateful to Frank L¨ubeck, Pham Tiep and ThomasWiegel for supplying us with very important information concerningsimple groups of Lie type. We are also grateful to Bob Guralnick andGunter Malle for conveying to us results from their paper [GM] priorto its publication.1.1.
Other generalisations of Thompson’s theorem.
Several oth-er ‘Thompson-like’ results have appeared in the literature recently. Wemention here four such theorems. In the first three results, solvability ofall 2-generator subgroups is replaced by a weaker condition restrictingthe required set of solvable 2-generator subgroups, in different waysfrom our generalisation.In 2000, Guralnick and Wilson [GW] obtained a solvability criterionby restricting the proportion of 2-generator subgroups required to besolvable.
Theorem 1.1.
A finite group is solvable if and only if more than of the pairs of elements of G generate a solvable subgroup. n addition they proved similar results showing that the propertiesof nilpotency and having odd order are also guaranteed if a sufficientproportion of element pairs generate subgroups with these properties,namely more than for nilpotency, and more than for having oddorder.In contrast to this, in a paper published in 2009, Gordeev, Grune-wald, Kunyavskiˇi and Plotkin [GGKP] proved a solvability criterionwhich involved 2-generation within each conjugacy class. This resultwas also proved independently by Guest in [G, Corollary 1]. Theorem 1.2.
A finite group G is solvable if and only if, for eachconjugacy class C of G , each pair of elements of C generates a solvablesubgroup. A stronger result of this type was obtained recently by Kaplan andLevy in [KL, Theorem 4]. Their criterion involves only a limited 2-generation within the conjugacy classes of elements of odd prime-powerorder.
Theorem 1.3.
A finite group G is solvable if and only if for every x, y ∈ G , with x a p -element for some odd prime p and y a -element,the group h x, x y i is solvable. Our requirement, while ranging over all conjugacy classes, requiresonly existence of a solvable 2-generator subgroup with one generatorfrom each of two (possibly equal) classes. We know of no similar criteriain this respect.The forth result we draw attention to is in a 2006 paper of Guralnick,Kunyavski, Plotkin and Shalev [GKPS]. They proved that membershipof the solvable radical of a finite group is characterised by solvability ofcertain 2-generator subgroups. (The solvable radical R ( G ) of a finitegroup G is the largest solvable normal subgroup of G .) Theorem 1.4.
For a finite group G , the solvable radical R ( G ) coin-cides with the set of all elements x ∈ G with the property: “ for any y ∈ G , the subgroup h x, y i is solvable”. Alternating and sporadic simple groups
Theorem B for the alternating groups follows from the followingproposition.
Proposition 2.1. If m ≥ , then there exist distinct primes p and q satisfying m/ < p < q ≤ m such that, for all x, y ∈ A m with | x | = p and | y | = b , the subgroup h x, y i ∼ = A d for some d ≥ . In particular, h x, y i is nonsolvable. roof. First we remark that if n is a positive integer and π ( n ) denotesthe number of primes at most n , then the following is known (see, forexample [R, p.188]): π (2 n ) − π ( n ) > n/ (3 ln n ) for n ≥ . In particular, if n ≥
17, then π (2 n ) − π ( n ) > / (3 ln 17) > , which implies that π (2 n ) − π ( n ) ≥
3. Thus, if n ≥
17, then there areat least 3 distinct primes p , r and q satisfying n < p < r < q ≤ n .In particular, n < p < q − < q ≤ n . Hence, if m ≥
34, thenthere exist primes p and q such that m/ < p < q − < q ≤ m , andelements x, y ∈ A m of order p and q , respectively. Let x, y be any suchelements and let H = h x, y i . Let ∆ be the support of H , that is, thesubset of { , , . . . , m } consisting of all the points moved by H , andlet d := | ∆ | . Since q > p > m/
2, it follows that H is transitive on∆, and that q ≤ d ≤ m < p . By [Wi, Theorem 8.4], H is primitiveon ∆. Moreover, since d ≥ q > p + 3 > H contains only evenpermutations, it follows from a theorem of C. Jordan dating from 1873,see [Wi, Theorem 13.9], that H ∼ = A d for some d ≥
5, as claimed.It remains to deal with A m , for 5 ≤ m ≤
33. In each case we willchoose primes p and q such that m/ ≤ p < q ≤ m , and consider thesubgroup H = h x, y i generated by elements x and y of A m of order p and q , respectively. Denote by ∆ the support of H , and let d = | ∆ | .In all cases d ≥ d ≥ q > p ≥
3, and hence A d is nonsolvable.For 17 ≤ m ≤
33, and for 11 ≤ m ≤
13, let q be the largest primesuch that q ≤ m , and let p be the smallest prime such that p > m/ p ≤ q −
3, and the argument above shows that H ∼ = A d .If 14 ≤ m ≤
16, let q = 13, p = 11. If d = 13, then H ≤ A andsince by the [ATLAS, p. 104] no maximal subgroup of A has orderdivisible by 11 ·
13, it follows that H = A . If d >
13, then as before, H is a primitive group on ∆, and since d = p + k with k ≥
3, it followsby [Wi, Theorem 13.9] that H ∼ = A d .If 7 ≤ m ≤
10, let q = 7, p = 5. It follows from the lists of maximalsubgroups of A d in [ATLAS, pp. 10, 22, 37, 48] that H ∼ = A d . If m = 5 ,
6, let q = 5, p = 3. It follows from the lists of maximalsubgroups of these groups in [ATLAS, pp. 2, 4] that H = A if m = 5,and that H ∼ = A or A if m = 6. The proof of Proposition 2.1 iscomplete. (cid:3) Theorem B for the sporadic simple groups follows from the followingproposition. For compactness of notation, we write L ( q ) instead ofPSL(2 , q ) in Table 1. p q h x, y i S p q h x, y i M M , L (11) M M , M , L (11) M M , L (11) M M M M , L (23) J J J J , L (7) J J , L (19) J J HS HS, M , L (11) , M He He M cL
M cL, M , M , L (11) Suz
11 13
Suz Ly
LyRu
Ru O ′ N O ′ NCo
13 23 Co Co Co , M Co Co , M F i
11 13
F i F i F i , M , L (23) F i ′ F i ′ HN HN T h
19 31
T hB
B M
M, L (59) Table 1.
Results table for Proposition 2.2
Proposition 2.2.
Let S be a sporadic simple group as in one of therows of Table . Then for the primes p, q in the corresponding row ofTable , p, q ∈ oe ( S ) and, for all x, y ∈ S with | x | = p and | y | = q , thesubgroup h x, y i is one of the nonabelian simple groups in the row for S and column labeled h x, y i of Table .Proof. The proof uses heavily the lists of maximal subgroups of thesporadic simple groups and of other simple groups, which appear in[ATLAS] and in [ATLAS3]. For all sporadic simple groups our resultsfollow from a close examination of these lists. In particular the groupslisted in the column labeled h x, y i are the only subgroups which couldpossibly be generated by two elements, the first of order p and thesecond of order q .As an example, we describe in detail our treatment of the Higman–Sims sporadic group HS of order 2 . . . .
11. Having checked themaximal subgroups of HS , we choose the primes p = 2 and q = 11.If the subgroup X = h x, y i is not equal to HS , then X is containedin a maximal subgroup of HS of order divisible by 11. We see from[ATLAS, p. 80] that the only such maximal subgroups of HS are thesimple groups M and M . Suppose first that X is a subgroup of M . If X = M , then it is contained in a maximal subgroup of M oforder divisible by 11. By [ATLAS, p. 18], each such maximal subgroupof M is isomorphic to the simple group L (11). If X = L (11), then X is contained in a maximal subgroup of L (11) of order divisibleby 11. However, such maximal subgroups have order 5 ·
11, which is ot divisible by 2. Thus we are left with the possibilities: X = HS , X = M , X = L (11), or X is a subgroup of M . If X is a propersubgroup of M , then X is contained in a maximal subgroup of M of order divisible by 11. By [ATLAS, p. 30], the only such maximalsubgroup of M is the simple group L (11), which we have alreadyexamined. Thus, finally, X is one of the simple groups HS , M , L (11) or M , as in Table 1.Thus Proposition 2.2 is proved. (cid:3) Primitive prime divisors
In the following, q = p k is a power of a prime p . For any positiveinteger e , we say that a prime r is a primitive prime divisor of q e − r divides q e − r does not divide q i − i < e . Observe that then e is the order of q modulo the prime r ; so e divides r − r ≥ e + 1. The set of primitive primedivisors of q e − q, e ).We say that a prime r is a basic primitive prime divisor of q e −
1, if r is a primitive prime divisor of p ke −
1, that is to say, if r ∈ ppd( p, ek ).We denote by bppd( q, e ) the set of basic primitive prime divisors of q e −
1. Note that bppd( q, e ) ⊆ ppd( q, e ), and that the inclusion can bestrict; for example, ppd(2 ,
3) = { } , but bppd(2 ,
3) = ppd(2 ,
6) = ∅ .The following result of Zsigmondy [Z] will be used frequently. Theorem 3.1.
Let q ≥ and e ≥ . Then bppd( q, e ) = ∅ if and onlyif one of the following holds. (i): q is a Mersenne prime, e = 2 , and here ppd( q,
2) = ∅ ; (ii): ( q, e ) = (2 , , and here ppd(2 ,
6) = ∅ ; (iii): ( q, e ) ∈ { (4 , , (8 , } , and here ppd(4 ,
3) = { } , and ppd(8 ,
2) = { } . Next, we define the set Lpd( q, e ) of large primitive divisors of q e − r ∈ ppd( q, e ) such that r > e + 1together with the integer ( e +1) if e +1 ∈ ppd( q, e ) and ( e +1) divides q e − q, e ) of large basic primitive divisors of q e − r ∈ bppd( q, e ) such that r > e + 1, together with the integer ( e + 1) if e + 1 ∈ bppd( q, e ) and( e + 1) divides q e − Proposition 3.2.
Let q ≥ and e ≥ . Then Lbpd( q, e ) = ∅ if andonly if ( q, e ) is one of the following: (2 , , (2 , , (2 , , (2 , , (2 , , (3 , , (3 , , (4 , , (5 , . roof. Let Lpd , Lbpd , bppd , ppd denote the sets Lpd( q, e ), Lbpd( q, e ),bppd( q, e ), ppd( q, e ), respectively. We prove first that Lbpd = ∅ ifand only if both bppd = ∅ and Lpd = ∅ . If Lbpd = ∅ , then clearlybppd = ∅ and Lpd = ∅ . Conversely, suppose that bppd = ∅ andLpd = ∅ . Let r be the largest element of bppd. Then r ∈ ppd( p, ke ),so r ≥ ke + 1 ≥ e + 1. If r > e + 1, then r ∈ Lbpd, so we may assumethat r = e + 1. Then k = 1 and bppd = ppd = { e + 1 } . Let s ∈ Lpd.Since ppd = { e + 1 } , it follows that s = ( e + 1) and s divides q e − s ∈ Lbpd. Thus in both cases Lbpd is nonempty, as required.Hence Lbpd is empty if and only if either Lpd is empty or bppd isempty. Now assume that q ≥ e ≥
3. By [NP1, Theorem 2.2],the set Lpd is empty only for q = 2 and e ∈ { , , , , } , for q = 3and e ∈ { , } , and for ( q, e ) = (5 , q, e ) = { (2 , , (4 , } . So Lbpd is emptyprecisely for the values of ( q, e ) listed in the proposition. (cid:3) Remark 3.3.
Let G be a subgroup of GL( d, q ) and let m ∈ Lpd( q, e )(or m ∈ Lbpd( q, e )) with d ≥ e > d/
2. If m divides | G | , then m ∈ oe ( G ). In fact, either m = r or m = r = ( e + 1) , where r is a (basic)primitive prime divisor of q e −
1. Since e > d/
2, a Sylow r -subgroup ofGL( d, q ) is cyclic and so G has elements of order m .We say that an element g ∈ G is a bppd( q, e ) -element if the orderof g is divisible by some element of bppd( q, e ). Similarly, we say that g ∈ G is an Lbpd( q, e ) -element if the order of g is divisible by someelement of Lbpd( q, e ).We use results from [NP1] and [NP2] to deal with subgroups of lineargroups containing one or two “big” ppd-elements. We observe herethat basic primitive prime divisors will be relevant in order to excludeexamples of “subfield type” (see case (d) in the proof of Lemma 3.4),while large primitive divisors will be relevant in the proof of Lemma 4.1.Our first lemma deals with irreducible subgroups of GL( d, q ) for d ≥ Lemma 3.4.
Let G be an irreducible subgroup of GL( d, q ) , with d ≥ .Assume that x, y ∈ G are such that x is a bppd( q, e ) -element and y is a bppd( q, f ) -element, with d ≥ e > f > d/ . Then either G isnonsolvable or one of the following holds. (1): d = e = f + 1 is prime, G is conjugate to a subgroup of GL(1 , q d ) .d , and G contains no Lbpd( q, f ) -elements; (2): There exists a prime c dividing gcd( d, e, f ) such that G isconjugate to a subgroup of GL( dc , q c ) .c and the elements x c and y c lie in GL( dc , q c ) as a bppd( q c , ec ) -element and a bppd( q c , fc ) -element, respectively. roof. By assumption | x | is a multiple of an element r ∈ bppd( q, e )and | y | is a multiple of an element s ∈ bppd( q, f ). Suppose first that d = 3 and q ≤
4. Then 3 ≥ e > f > / e = 3 , f = 2 andbppd( q, = ∅ , bppd( q, = ∅ . By Theorem 3.1, q = 3 ,
4. Hence, q = 2and so r = 7 , s = 3, G ≤ GL(3 ,
2) and Lbpd( q, f ) = ∅ . By [ATLAS,p.3], either G = GL(3 ,
2) is nonsolvable or G = h x, y i ∼ = GL(1 , . d = 3 we may assume that q ≥ d, q ) = (4 ,
2) or (4 , G is nonsolvable.In these cases d = 4 ≥ e > f > e = 4 , f = 3 and bppd( q, = ∅ , bppd( q, = ∅ . If ( d, q ) = (4 , G ≤ GL(4 , ∼ = 2 . PSL(4 , . r = 5, s = 13. Since no propersubgroup of PSL(4 ,
3) is divisible by 5 ·
13, PSL(4 ,
3) is a section of G and hence G is nonsolvable, as claimed. If ( d, q ) = (4 , G ≤ GL(4 , ∼ = PSL(4 ,
2) and r = 5 , s = 7. It follows from [ATLAS,pp. 10 and 22] that G ∼ = A or A and in particular G is nonsolvable.Thus we may assume further that ( d, q ) = (4 ,
2) or (4 , q = p k for some prime p and k ≥
1. Since r ∈ ppd( p, ek ), s ∈ ppd( p, f k ) and e > f ≥
2, neither r nor s divides q −
1. Moreover,if i ≤ d and h ≤ k , then neither r nor s divides p ih − ih ≤ dk < f k < ek ).We apply [NP1, Theorem 4.7]. First we show that the cases (a),(c), (d) and (e) of [NP1, Theorem 4.7], if they occur, imply that G isnonsolvable.(a): (Classical type) G has a normal subgroup Ω, where Ω is oneof the following groups: SL( d, q ) , Sp( d, q ) ( d even) , SU( d, q ) ( q = q ) , Ω ± ( d, q ) ( d even) , and Ω ◦ ( d, q ) ( d odd) . Because of our assumptionthat ( d, q ) = (4 , , (4 ,
3) or (3 , q ) with q ≤
4, each of these classicalgroups is nonsolvable and so G is nonsolvable in case (a).(c): (Nearly Simple Groups) Here G is nearly simple, and hencenonsolvable.(d): (Subfield type) Denote by Z the subgroup of scalar matrices ofGL( d, q ) and let Z ◦ GL( d, q ) denote a group which can be defined overa subfield modulo scalars. In case (d), G is conjugate to a subgroupof Z ◦ GL( d, q ) for some proper subfield F q of F q , say q = q a and q = p h , with k = ah and a ≥
2. Since x ∈ G and r does not divide q − r divides | GL( d, q ) | . Hence r divides p ht − t ≤ d . Since h ≤ k , this contradicts ourobservation above, so case (d) does not arise.(e): (Imprimitive type) Here G preserves a direct-sum decomposition V = U ⊕ · · · ⊕ U d with dim( U i ) = 1 for i = 1 , . . . , d and G induces A d or S d on the set { U , . . . , U d } . Since A d is nonsolvable for d ≥ e may assume that d ≤
4. In particular, | G | divides ( q − d d ! andsince the primes r, s do not divide q −
1, each of them divides d !. Inparticular, r, s ≤ d ≤
4. However, e > f ≥
2, so r ≥ e + 1 ≥
4, whichimplies that r = 4, a contradiction.This leaves case (b) of [NP1, Theorem 4.7], and in that case G isconjugate to a subgroup of GL( dc , q c ) .c , for some prime c dividing d .Moreover either c = d = f + 1 = e , or c < d and c divides gcd( d, e, f ).In the former case, G ≤ GL(1 , q d ) .d and s = d = f + 1. As s doesnot divide | G | , part (1) holds. In the latter case, since each of r, s isat least f + 1 > c , it follows that x c , y c ∈ GL( dc , q c ) and have ordersdivisible by r, s , respectively. Thus part (2) holds. (cid:3) If V is a G -module and U is a G -invariant section of V (that is to say, U = V /V where V ≤ V are G -submodules of V ), we denote by G U the group of automorphisms induced by G on U . So G U ∼ = G/C G ( U ),a factor group of G , and accordingly we denote by x U the image in G U of an element x ∈ G under the relevant projection homomorphism.For a prime r and integer n , let n r denote the r -part of n , that is,the largest power of r dividing n . Remark 3.5.
Let G ≤ GL( d, q ) and let V = V ( d, q ) be the naturalmodule for G . Suppose that x is a ppd( q, e )-element of G of orderdivisible by a primitive prime divisor r of q e − x be anelement of order r in h x i . Then the following statements hold.(1) Under the action of h x i , V | h x i is a completely reducible F q h x i -module, by Maschke’s Theorem, and all the nontrivial irreducible sub-modules of V | h x i have dimension e (see for instance [H, TheoremII.3.10]). In particular, it follows that there exists at least one G -composition factor U of V on which x , and hence also x , acts non-trivially. So | x U | = r and hence x U is a ppd( q, e )-element of G U ≤ GL( d , q ) of order divisible by | x U | = r and satisfying | x U | r = | x | r ,where d = dim F q ( U ) ≥ e .(2) Suppose also that y is a ppd( q, f )-element of G of order divisibleby a primitive prime divisor s of q f −
1, and that d ≥ e > f > d/ e and f are greater than d/
2, it follows by part (1) of thisremark that there exists a unique G -composition factor U of V , ofdimension d = dim F q ( U ) ≥ e , on which both x and y act nontrivially.Moreover G U ≤ GL( d , q ), and x U is a ppd( q, e )-element of G U of ordersatisfying | x U | r = | x | r ≥ r and y U is a ppd( q, f )-element of G U of ordersatisfying | y U | s = | y | s ≥ s . e now apply Lemma 3.4 to linear groups of dimension large enoughto allow the existence of primitive prime divisors for two distinct ex-ponents, both greater than half the dimension of the linear group. Itis convenient to deal separately with certain large exponent pairs. Lemma 3.6.
Let j be a nonnegative integer, δ ∈ { , } , and let x, y ∈ GL( d, q ) be such that x is a bppd( q, e ) -element and y is a bppd( q, f ) -element. Assume that e , f , δ , and d satisfy the following conditions: e = d − j, f = d − j − δ, d ≥ j + 2 δ + 1 . If δ = 1 , we assume in addition that y is an Lbpd( q, f ) -element. Then G = h x, y i is nonsolvable.Proof. Let r ∈ bppd( q, e ) such that r divides | x | , and let s ∈ bppd( q, f )such that s divides | y | . As mentioned above, r ≡ e ) and s ≡ f ). Moreover, e > f = d − j − δ ≥ j + δ + 1 ≥
2, so in particularboth r and s are odd primes. As it is sufficient to prove that somesubgroup of G is nonsolvable, we may assume that G = h x, y i , and | x | , | y | are powers of r and s , respectively. In particular, | x | and | y | are odd. If δ = 1, then in addition we assume that y is an Lbpd( q, f )-element.By our assumptions, j ≤ d − − δ , so d ≥ e > f = d − j − δ ≥ d +12 > d . In the following, we denote by V the natural module V ( d, q ) forGL( d, q ). Case (1): G is irreducible on V . The conditions of Lemma 3.4 are satisfied. Applying that result wededuce that either G is nonsolvable, or (1) d = e = f + 1 is prime, G isconjugate to a subgroup of GL(1 , q d ) .d , and G contains no Lbpd( q, f )-elements, or (2) there is a prime c < d such that c divides gcd( d, e, f ), G is conjugate to a subgroup of GL( d/c, q c ) .c , and x c and y c lie inGL( dc , q c ) as a bppd( q c , ec )-element and a bppd( q c , fc )-element, respec-tively. If (1) holds, then j = 0 , δ = 1 and we have a contradiction, since y is an Lbpd( q, f )-element of G . Thus (2) holds. Since c is a primedividing gcd( d, e, f ), and since gcd( e, f ) = gcd( e, e − f ) ≤ e − f = δ ≤
2, it follows that c = δ = 2 and all of d, e, f, j are even. Hence G ≤ GL( d/ , q ) .
2, and since the orders of x, y are odd, we concludethat G ≤ GL( d/ , q ). Since d and j are even, our assumption that d ≥ j +2 δ +1 = 2 j +5 implies d ≥ j +3. Thus replacing ( d, q, e, f, j, δ )by ( d , q , e , f , j , δ ), all the conditions of the lemma hold, and G is ir-reducible on V ( d , q ). By the arguments above and since δ = 1, weconclude that G must be nonsolvable. Case (2): G is reducible on V . f e = d , then | x | would be a multiple of a primitive prime divisorof q d − h x i would act irreducibly on the natural module V .However G is reducible, so we must have e < d and j ≥
1. By Remark3.5(2), there exists a unique G -composition factor U of V of dimension d = dim F q ( U ), say, where d > d ≥ e = d − j , such that x U and y U are bppd( q, e )- and bppd( q, f )-elements of G U , respectively, with | x U | r = | x | r ≥ r , | y U | s = | y | s ≥ s . In particular, if y is an Lbpd( q, f )-element of G , then y U is an Lbpd( q, f )-element of G U .We claim that the irreducible group G U = h x U , y U i ≤ GL( d , q )induced by G on U satisfies the conditions of Lemma 3.6 with pa-rameters d , e, f, δ , relative to the integer j := j − d + d . Note that j > j = d − e ≥ d = j + d − j ≥ j + j +2 δ +1 > j +2 δ +1 ≥ e = d − j , f = d − j − δ hold, by the definitionof j . This proves the claim. Since G U is irreducible, it follows fromCase (1) of this proof that G U is nonsolvable. Consequently, also G isnonsolvable. (cid:3) We can now prove Theorem B for classical simple groups of largedimension.
Proposition 3.7.
Assume that S is one of the following simple groups: (1): PSL( d, q ) , with d ≥ and ( d, q ) = (6 , ; (2): PSp( d, q ) ( d even) or PSU( d, q ) ( d odd), with d ≥ and ( d, q ) / ∈ { (5 , , (6 , , (8 , } ; (3): PΩ ◦ ( d, q ) ( dq odd) or PSU( d, q ) ( d even), with d ≥ ; (4): PΩ ± ( d, q ) ( d even), with d ≥ and ( d, q ) = (10 , .Then there exist a, b ∈ oe ( S ) such that for every choice of x, y ∈ S with | x | = a and | y | = b , the group h x, y i is nonsolvable.Proof. We work with the group ˆ S defined as follows: • ˆ S = SL( d, q ) ≤ GL( d, q ), when S = PSL( d, q ); • ˆ S = Sp( d, q ) ≤ GL( d, q ), when S = PSp( d, q ); • ˆ S = SU( d, q ) ≤ GL( d, q ), when S = PSU( d, q ); • ˆ S = Ω ε ( d, q ) ≤ GL( d, q ), when S = PΩ ε ( d, q ), for ε = ± or ◦ .Let Z = Z ( ˆ S ). We prove that there exist a, b ∈ oe ( ˆ S ) with gcd( ab, | Z | )= 1 such that, for every ˆ x ∈ ˆ S of order a and every ˆ y ∈ ˆ S of order b , the group h ˆ x, ˆ y i is nonsolvable. Since a and b are coprime to | Z | , if | ˆ x | = a then also the order of ˆ x Z in S is equal to a , and similarly, if | ˆ y | = b then | ˆ y Z | = b . In addition, if h ˆ x, ˆ y i is nonsolvable then also h x, y i is nonsolvable, because it is a central factor of the nonsolvablegroup h ˆ x, ˆ y i . Thus it is sufficient to work as described with the group S . In fact, each of a and b will be either a prime or the square of aprime, and it will be easy to check that gcd( ab, | Z | ) = 1. (1): Let ˆ S = SL( d, q ), with d ≥
4. Suppose first that(1.1) ( d, q ) / ∈ { (4 , , (5 , , (5 , , (6 , , (7 , , (7 , , (7 , , (11 , , (13 , , (19 , } . Then by Theorem 3.1 there exists a ∈ bppd( q, d ), and by Proposi-tion 3.2 there exists b ∈ Lbpd( q, d − q d − q d − − S and hence a, b ∈ oe ( ˆ S ) by Remark 3.3. Also gcd( ab, | Z | ) = 1. Letnow ˆ x, ˆ y ∈ ˆ S be such that | ˆ x | = a and | ˆ y | = b . Then ˆ x is a bppd( q, d )-element of ˆ S and ˆ y is an Lbpd( q, d − S . Since d ≥ δ = 1 , j = 0, and weconclude that h ˆ x, ˆ y i is nonsolvable.We now consider the “isolated” cases left out of this proof and listedin (1.1). We deal with all these cases, except the excluded pair ( d, q ) =(6 , x, ˆ y ∈ ˆ S with | ˆ x | = a and | ˆ y | = b , where a, b are as in the following table. (Here q i denotes a prime in bppd( q, i ),and, by Theorem 3.1, for all entries in the table such a prime q i exists.) ( d, q ) (4 ,
4) (5 ,
2) (5 ,
3) (7 ,
2) (7 ,
3) (7 ,
5) (11 ,
2) (13 ,
2) (19 , a q q q q q q q q q b q q q q q q q q q Thus ˆ x, ˆ y satisfy all the conditions of Lemma 3.6 with δ = 2 , j = 0,and hence h ˆ x, ˆ y i is nonsolvable. (2): If ˆ S = Sp( d, q ) with d ≥ d even, then the order of ˆ S isdivisible by both q d − q d − −
1. By Theorem 3.1, bppd( q, d ) = ∅ and bppd( q, d − = ∅ , since by our assumptions d > d − ≥ q, d ) , ( q, d − = (2 , a ∈ bppd( q, d ) and b ∈ bppd( q, d − a, b ∈ oe ( ˆ S ).If ˆ S = SU( d, q ) ≤ GL( d, q ) with d ≥ d odd, then the order of ˆ S is divisible by both q d + 1 and q d − + 1. Here we take a ∈ bppd( q, d ) =bppd( q , d ) and b ∈ bppd( q, d −
4) = bppd( q , d − q, d ) and bppd( q, d −
4) are nonempty, since ( d, q ) = (5 , a, b ∈ oe ( ˆ S ).In both cases we apply Lemma 3.6 with δ = 2 , j = 0. It followsthat in each of these two cases, h ˆ x, ˆ y i is nonsolvable for every choice ofˆ x, ˆ y ∈ ˆ S with | ˆ x | = a and | ˆ y | = b . If ˆ S = Ω ◦ ( d, q ), with d ≥ dq odd, then both q d − − q d − − S and each has a basic primitiveprime divisor by Theorem 3.1. We take a ∈ bppd( q, d −
1) and b ∈ bppd( q, d −
3) and note that a, b ∈ oe ( ˆ S ).If ˆ S = SU( d, q ) ≤ GL( d, q ) with d ≥ d even, then the order ofˆ S is divisible by both q d − + 1 and q d − + 1. We take a ∈ bppd( q, d −
2) = bppd( q , d −
1) and b ∈ bppd( q, d −
6) = bppd( q , d − q , d −
1) and bppd( q , d −
3) are nonempty, byTheorem 3.1, since d − > d − ≥ q >
2. Thus a, b ∈ oe ( ˆ S ).In both cases we apply Lemma 3.6 with δ = 2 , j = 1. It followsthat in each of these two cases, h ˆ x, ˆ y i is nonsolvable for every choice ofˆ x, ˆ y ∈ ˆ S with | ˆ x | = a and | ˆ y | = b . (4): Assume, finally, that ˆ S = Ω ± ( d, q ) with d ≥ d even, and( d, q ) = (10 , q d − − q d − − S , and each has a basic primitive prime divisor by Theorem 3.1,since d − > d − ≥ d, q ) = (10 , a ∈ bppd( q, d − b ∈ bppd( q, d − a, b ∈ oe ( ˆ S ). By Lemma 3.6 with δ = 2 , j = 2, we conclude that h ˆ x, ˆ y i is nonsolvable for every choice ofˆ x, ˆ y ∈ ˆ S with | ˆ x | = a and | ˆ y | = b . (cid:3) Proof of Theorem B
Proposition 3.7 does not cover the classical groups in small dimen-sions, when the conditions of Lemma 3.6 do not hold. For most of thesecases we use the following result, which deals with the case where thereis at least one Lbpd( q, e )-element in G ≤ GL( d, q ), with e > d/ Lemma 4.1.
Let G be an irreducible subgroup of GL( d, q ) , with d ≥ .Assume that G contains an Lbpd( q, e ) -element for some integer e with d ≥ e > d/ . Assume also that ( d, q ) / ∈ { (3 , , (3 , , (3 , , (4 , , (4 , } . Then either G is conjugate to a subgroup of GL( d/c, q c ) .c for someprime divisor c of gcd( d, e ) , or G is nonsolvable.Proof. This lemma follows from [NP2, Theorem 3.1]. Note that thecases (b), (d) and (c)(i) of that theorem do not occur, because of theassumption that the bppd( q, e )-element is large. So either G is of clas-sical type (part (a) of [NP2, Theorem 3.1]) and hence is nonsolvablebecause of the assumptions on ( d, q ) (see the details in the proof ofLemma 3.4, case (a)), or G is of nearly simple type (part (e) of [NP2,Theorem 3.1]) and hence is nonsolvable, or G is of extension field type part (c)(ii) of [NP2, Theorem 3.1]), that is to say, G is conjugate to asubgroup of GL( d/c, q c ) .c for some prime divisor c of gcd( d, e ). (cid:3) We now complete the proof of Theorem B for classical groups of Lietype.
Proposition 4.2.
Theorem B holds for all classical finite simple groupsof Lie type.Proof. (1):
Assume first that S = PSL( d, q ) with d ≥ . Suppose first that d = 2 and q ≤
7. Theorem B holds for thegroups PSL(2 , ∼ = PSL(2 , ∼ = A by Proposition 2.1. Also, for S =PSL(2 , h x, y i = S whenever x, y ∈ S with | x | = 2 and | y | = 7.Next we consider S = PSL(2 , q ) with q ≥
8. Take a = ( q + 1) /k and b = ( q − /k , where k = gcd( q − , a ≥ b ≥ a, b ∈ oe ( S ) (see Theorems II.8.3 and II.8.4 in [H]). The classificationof the subgroups of PSL(2 , q ) (see Theorem II.8.28 in [H]) implies that h x, y i = S whenever x, y ∈ S with | x | = a and | y | = b .Suppose next that S = PSL(3 , q ). The group PSL(3 , ∼ = PSL(2 , S = PSL(3 , q ) with q = 3 or 4, thentaking x, y ∈ S with ( | x | , | y | ) = (13 ,
2) or (7 , S = h x, y i (see [ATLAS, pp. 13,23]). Hence we may assume that q ≥
5. By Proposition 3.2, there exists a ∈ Lbpd( q, q = 3 k , then k > b ∈ ppd( q,
2) which, by Theorem 3.1, is nonempty.On the other hand, if q is not a power of 3, then we choose b = p (recallthat q is a power of p ). As in the proof of Proposition 3.7, we work withˆ S = SL(3 , q ) ≤ GL(3 , q ) in order to apply Lemma 4.1. (We shall oftendo this throughout the proof without further reference.) Let ˆ x, ˆ y ∈ ˆ S ,with | ˆ x | a multiple of a and | ˆ y | a multiple of b , and let X = h ˆ x, ˆ y i . Sinceˆ x ∈ X , X is an irreducible subgroup of GL(3 , q ), and since ˆ y ∈ X , | X | does not divide | GL(1 , q ) . | . Hence, by Lemma 4.1, X is nonsolvable.It follows that h x, y i is nonsolvable for every x, y ∈ S with | x | = a and | y | = b .Thus we may assume that d ≥
4. For these groups S , Theorem B fol-lows from Proposition 3.7(1), unless ( d, q ) = (6 , S = PSL(6 , ∼ =GL(6 , a = 31 ∈ Lbpd(2 ,
5) and b = 7 ∈ bppd(2 , a, b ∈ oe ( S ). Let x, y ∈ GL(6 ,
2) with | x | = 31 and | y | = 7,and set X = h x, y i . If X is reducible on V = V (6 , x ∈ X , X acts irreducibly on some X -composition factor U of V of dimension5 and X U = h x U , y U i ≤ GL(5 , | x U | = | x | = 31and | y U | = | y | = 7, and hence X U is nonsolvable by Lemma 3.6 ap-plied with δ = 2 , j = 0. Consequently, also X is nonsolvable. If X s irreducible, then by Lemma 4.1 (applied with d = 6 , e = 5), X isnonsolvable, since gcd(6 ,
5) = 1. (2):
Next let S = PSp( d, q ) ′ with d ≥ and d even (noting that PSp(2 , q ) ∼ = PSL(2 , q ) has been dealt with above). First consider d = 4. The result for PSp(4 , ′ ∼ = A follows fromProposition 2.1. If S = PSp(4 , , ∈ oe ( S ) and no maximalsubgroup of S contains elements of both orders 5 and 9 (see [ATLAS,p. 26]). Hence S = h x, y i for all x, y ∈ S with | x | = 5 and | y | = 9.If S = PSp(4 , , ∈ oe ( S ) and (see [ATLAS, p. 44]) theonly maximal subgroups of S containing an element of order 17 areof the form PSL(2 ,
16) : 2, and every subgroup of such a group oforder divisible by both 17 and 5 contains PSL(2 , h x, y i isnonsolvable whenever x, y ∈ S with | x | = 17 and | y | = 5.So suppose that q ≥ a ∈ Lbpd( q, b = ( q − / gcd(2 , q − a, b ∈ oe ( S ),since PSp(2 , q ) ∼ = PSL(2 , q ) is isomorphic to a subgroup of S . Weconsider ˆ S = Sp(4 , q ) ≤ GL(4 , q ). Let ˆ x, ˆ y ∈ ˆ S , with | ˆ x | a multiple of a and | ˆ y | a multiple of b , and let X = h ˆ x, ˆ y i . Then ˆ x is an Lbpd( q, X , and in particular ˆ x acts irreducibly on V (4 , q ). Hence X is an irreducible subgroup of GL(4 , q ). By Lemma 4.1, either X isnonsolvable or X is (conjugate to) a subgroup of GL(2 , q ) .
2. Assumethe latter. Then h ˆ x , ˆ y i ≤ GL(2 , q ), | ˆ x | is a multiple of the largeprimitive divisor a of q − | ˆ y | is a multiple of b/ gcd( b, ≥ , q ) (see [H,Theorem II.8.27]), we conclude that h ˆ x , ˆ y i ≥ SL(2 , q ) and hence, inparticular, X = h ˆ x, ˆ y i is nonsolvable.By part (2) of Proposition 3.7, we are left with the following cases:PSp(6 ,
2) and PSp(8 , S = PSp(6 , , ∈ oe ( S ), andfor all x, y ∈ S with | x | = 15 , | y | = 7, the group X = h x, y i is non-solvable. This is seen as follows: one checks from [ATLAS, p. 46] thateach maximal subgroup of S of order divisible by 35 is isomorphic to S . Moreover, a subgroup of S generated by two elements, of orders15 and 7, contains A , and in fact is equal to A . Thus, X = S or X ∼ = A .Finally, let S = PSp(8 , ∼ = Sp(8 , < GL(8 , , ∈ oe ( S ).By [ATLAS, p. 123], each maximal subgroup of S of order divisibleby 17 · − (8 ,
2) : 2. By [ATLAS, p. 89], eachmaximal subgroup of PΩ − (8 ,
2) of order divisible by 17 is isomorphicto PSL(2 ,
16) : 2, and so contains no elements of order 7. Thus h x, y i is nonsolvable whenever x, y ∈ S with | x | = 17 and | y | = 7. (3): Assume now that S = PSU( d, q ) , with d ≥ and d odd. y Proposition 3.7, we need only consider the cases S = PSU(3 , q )with q ≥ , ∼ = 3 : Q is solvable), and S = PSU(5 , S = PSU(5 , , ∈ oe ( S ) andeach maximal subgroup of S of order divisible by 11 is isomorphic toPSL(2 ,
11) and contains no elements of order 15. Thus if x, y ∈ S with | x | = 11 and | y | = 15, then h x, y i = S .Therefore we may assume that S = PSU(3 , q ) with q ≥
3. By[ATLAS, pp. 14, 34], if S = PSU(3 , q ) with q ∈ { , } , then 7 , ∈ oe ( S )and each maximal subgroup of S of order divisible by 7 is isomorphicto PSL(2 ,
7) if q = 3, and to A if q = 5, and neither of these groupscontains an element of order 8. Thus if x, y ∈ S with | x | = 7 and | y | = 8, then h x, y i = S . So we may assume that q = 2 , ,
5. Thenby Proposition 3.1, Lbpd( q, = ∅ . Let a ∈ Lbpd( q, a ∈ oe ( S ) by Remark 3.3. Also let b = p if p = 3, and b =( q − / p = 3 (recall that q is a power of p ), and note that b ∈ oe ( S ) since PSU(2 , q ) ∼ = PSL(2 , q ) is isomorphic to a subgroup of S .Now let ˆ S = SU(3 , q ) ≤ GL(3 , q ), and note that a, b ∈ oe ( ˆ S ) andthat gcd( ab, | Z ( ˆ S ) | ) = 1. Consider ˆ x, ˆ y ∈ ˆ S with | ˆ x | a multiple of a and | ˆ y | a multiple of b , and let X = h ˆ x, ˆ y i . Since a is a primitiveprime divisor of ( q ) − X is an irreducible subgroup of GL(3 , q ).Thus by Lemma 4.1, either X is nonsolvable or X is a subgroup of X = GL(1 , q ) .
3. Suppose that X ≤ X . If p = 3 then b = p doesnot divide | X | . Hence p = 3 and b = ( q − /
2. However ˆ x ∈ X and | ˆ x | is a multiple of a ∈ Lbpd( q, | ˆ x | does not divide | X | ,a contradiction. Hence X is nonsolvable and, consequently, h x, y i isnonsolvable whenever x, y ∈ S with | x | = a and | y | = b . (4): Assume now that S = PΩ ◦ ( d, q ) , with d odd and d ≥ . Since PΩ ◦ (2 m + 1 , k ) ∼ = PSp(2 m, k ), for all m and k , we mayassume that q is odd. Also since PΩ ◦ (3 , q ) ∼ = PSL(2 , q ) and PΩ ◦ (5 , q ) ∼ =PSp(4 , q ), we may assume that d ≥
7. In this case Theorem B followsfrom Proposition 3.7. (5):
Assume now that S = PSU( d, q ) , with d even. Since PSU(2 , q ) ∼ = PSL(2 , q ), we may assume that d ≥
4, and itfollows by Proposition 3.7 that we only have to check dimensions d = 4and d = 6.Let S = PSU(4 , q ). Since PSU(4 , ∼ = PSp(4 , q ≥
3. For S = PSU(4 , , ∈ oe ( S ) and each maximal subgroup of S of order divisible by7 is isomorphic to PSL(3 , ,
3) or A , and hence contains noelements of order 9. Thus h x, y i = S whenever x, y ∈ S with | x | = 7and | y | = 9. hus we may assume that S = PSU(4 , q ) with q ≥
4. Then, byTheorem 3.1 and Proposition 3.2, both Lbpd( q ,
3) and bppd( q,
4) arenonempty. Let a ∈ Lbpd( q ,
3) and b ∈ bppd( q, a, b ∈ oe ( S ) (since ( q − q + 1) divides | S | ). Considerˆ S = SU(4 , q ) ≤ GL(4 , q ) acting on the natural module V = V (4 , q ).Observe that gcd( ab, | Z ( ˆ S ) | ) = 1 and hence a, b ∈ oe ( ˆ S ). Let ˆ x, ˆ y ∈ ˆ S with | ˆ x | a multiple of a and | ˆ y | a multiple of b , and let X = h ˆ x, ˆ y i .By Remark 3.5, there exists an X -composition factor U of V such thatboth ˆ x and ˆ y act nontrivially on U , with dim F q ( U ) ≥
3. Moreover ˆ x U isan Lbpd( q , y U is an Lbpd( q , X U (since b ≥ X acts reducibly on V . Then dim F q ( U ) =3, and hence, by Lemma 3.6 with δ = 1 , j = 0, the group X U isnonsolvable. Thus also X is nonsolvable. Therefore we may assumethat X is an irreducible subgroup of GL(4 , q ). Then, by Lemma 4.1, X is nonsolvable, as a ∈ Lbpd( q ,
3) and the extension field case cannotoccur because gcd(4 ,
3) = 1.Finally let S = PSU(6 , q ). If S = PSU(6 , , ∈ oe ( S ), each maximal subgroup of S of order di-visible by 7 ·
11 is isomorphic to M , and M has no maximal sub-group of order divisible by 7 ·
11. Hence h x, y i = S or h x, y i ∼ = M whenever x, y ∈ S with | x | = 7 and | y | = 11. Thus we may as-sume that q >
2. Then, by Theorem 3.1 and Proposition 3.2, bothLbpd( q ,
5) and bppd( q,
6) are nonempty. Let a ∈ Lbpd( q ,
5) and b ∈ bppd( q, q + 1)( q + 1) divides | S | , it follows by Re-mark 3.3 that a, b ∈ oe ( S ). Consider now ˆ S = SU(6 , q ) ≤ GL(6 , q )acting on the natural module V = V (6 , q ). Since gcd( ab, | Z ( ˆ S ) | ) = 1,it follows that a, b ∈ oe ( ˆ S ). Let ˆ x, ˆ y ∈ ˆ S with | ˆ x | a multiple of a and | ˆ y | a multiple of b , and let X = h ˆ x, ˆ y i . If X is an irreducible subgroupof GL(6 , q ), then we conclude by Lemma 4.1 that X is nonsolvable,since a ∈ Lbpd( q ,
5) and the extension field case cannot occur becausegcd(6 ,
5) = 1. So we may assume that X acts reducibly on V . By Re-mark 3.5(2), there exists an X -composition factor U of V such thatboth ˆ x and ˆ y act nontrivially on U , with dim F q ( U ) ≥
5. It followsthat dim F q ( U ) = 5. Moreover, ˆ x U is an Lbpd( q , y U isa bppd( q , X U . Hence by Lemma 3.6 with δ = 2 , j = 0,the group X U is nonsolvable and so is X . (6): Let S = PΩ − ( d, q ) , with d even and d ≥ . Since PΩ − (4 , q ) ∼ = PSL(2 , q ) and PΩ − (6 , q ) ∼ = PSU(4 , q ), we mayassume that d ≥
8. Hence we are left, by Proposition 3.7, only withthe cases PΩ − (8 , q ), for q ≥
2, and PΩ − (10 , et S = PΩ − (8 , q ). For S = PΩ − (8 , , ∈ oe ( S ) and each maximal subgroup of S of orderdivisible by 17 is isomorphic to PSL(2 ,
16) : 2, and hence contains noelements of order 7. Thus h x, y i = S whenever x, y ∈ S with | x | = 7 and | y | = 17. So we may assume that q >
2. Then, by Theorem 3.1, bothbppd( q,
8) and bppd( q,
6) are nonempty. Let a ∈ bppd( q,
8) and b ∈ bppd( q, q + 1)( q −
1) divides | S | , it follows by Remark 3.3that a, b ∈ oe ( S ). Consider now ˆ S = Ω − (8 , q ) ≤ GL(8 , q ). Sincegcd( ab, | Z ( ˆ S ) | ) = 1, it follows that a, b ∈ oe ( ˆ S ). Let ˆ x, ˆ y ∈ ˆ S with | ˆ x | a multiple of a and | ˆ y | a multiple of b , and let X = h ˆ x, ˆ y i . Then X is nonsolvable by Lemma 3.6 with δ = 2 , j = 0. Hence h x, y i isnonsolvable whenever x, y ∈ S with | x | = a and | y | = b .Finally, let S = PΩ − (10 , ≤ GL(10 , ∈ bppd(2 , ∈ bppd(2 , , ∈ oe ( S ). Then by Lemma 3.6with δ = 2 , j = 0, the subgroup h x, y i is nonsolvable whenever x, y ∈ S with | x | = 11 and | y | = 17. (7): Let S = PΩ + ( d, q ) , with d even. Since S is not nonabelian simple if d = 2 and d = 4, and PΩ + (6 , q ) ∼ =PSL(4 , q ), we may assume that d ≥
8. Then by Proposition 3.7, weonly need to consider the cases PΩ + (8 , q ), for q ≥
2, and PΩ + (10 , S = PΩ + (8 , q ). For S = PΩ + (8 , , ∈ oe ( S ) and each maximal subgroup of S of orderdivisible by 35 is isomorphic to PSp(6 , : A or A . We haveshown above that a subgroup of PSp(6 ,
2) which contains elements oforders 15 and 7 is nonsolvable. Also, in the last paragraph of the proofof Proposition 2.1, we saw that a subgroup of A containing elementsof orders 7 and 5 is A d for some d ≥
7. Thus also a subgroup of 2 : A containing such elements has a composition factor A or A . Hence h x, y i is nonsolvable whenever x, y ∈ S with | x | = 7 and | y | = 15.Thus we may assume that q > S = PΩ + (8 , , ∈ oe ( S ), each maximal subgroup of S of order divisible by7 is isomorphic to PΩ ◦ (7 , + (8 ,
2) or 2 (cid:5)
PSU(4 , (cid:5) , and thelast of these groups contains no elements of order 15. Let x, y ∈ S with | x | = 7 and | y | = 15 and let X = h x, y i . Assume that X < S and let M be a maximal subgroup of S containing X . Then M isPΩ ◦ (7 ,
3) or PΩ + (8 , M = PΩ ◦ (7 , , ∈ oe ( M ), 7 ∈ bppd(3 ,
6) and 5 ∈ bppd(3 , δ = 2 , j = 1, every subgroup of ˆ M = Ω ◦ (7 , ≤ GL(7 ,
3) con-taining elements of orders 7 and 5 is nonsolvable. Hence in this case X is nonsolvable. On the other hand, if M = PΩ + (8 , revious paragraph any subgroup of M containing elements of orders7 and 15 is nonsolvable. Thus also in this case X is nonsolvable. Sowe may assume that q ≥ S = PΩ + (8 , ∈ bppd(5 , ∈ bppd(5 ,
4) belong to oe ( S ). Let X = h ˆ x, ˆ y i ≤ Ω + (8 , ≤ GL(8 , | ˆ x | is divisible by 7 and | ˆ y | is divisible by 13. We shallprove that X is nonsolvable. Now ˆ x is a bppd(5 , X (evenif not a large one), and ˆ y is an Lbpd(5 , X . Assume firstthat X acts reducibly on V = V (8 , X -composition factor U of V of dimension d = dim F ( U ) ≥ | ˆ x U | ≥ | ˆ y U | ≥
13. Thus d ∈ { , } , and by Lemma 3.6the group X U is nonsolvable in both cases d = 6 (taking δ = 2 , j = 0)and d = 7 (taking δ = 2 , j = 1).Thus we may assume that X is an irreducible subgroup of GL(8 , ,
6) = ∅ . So we have to apply [NP2, Theorem 3.1] directly(with d = 8 , e = 6), checking each of the cases (a)–(e) of that theorem.If X is in case (a) or (e), then X is nonsolvable. Since 7 ·
13 divides | X | , neither of the cases (b) nor (d) holds for X . So we may assumethat case (c) holds for X , which implies that X is (isomorphic to) asubgroup of GL(4 , ) ·
2. Hence X := h ˆ x , ˆ y i ≤ GL(4 , ). Observethat X acts irreducibly on V (4 , ), as it acts irreducibly on V . Now,7 ∈ Lbpd(5 , X isnot a subgroup of GL(2 , ) ·
2, because | GL(2 , ) · | is not divisibleby 7. Therefore we conclude that X is nonsolvable, and hence also X is nonsolvable.Now consider S = PΩ + (8 , q ) with q = 2 , ,
5. By Theorem 3.1 andProposition 3.2, both Lbpd( q,
6) and bppd( q,
4) are nonempty. Let a ∈ Lbpd( q,
6) and b ∈ bppd( q, q − q −
1) divides | S | , itfollows by Remark 3.3 that a, b ∈ oe ( S ). Consider now ˆ S = Ω + (8 , q ) ≤ GL(8 , q ) acting on on V = V (8 , q ). Since gcd( ab, | Z ( ˆ S ) | ) = 1, it followsthat a, b ∈ oe ( ˆ S ). Let ˆ x, ˆ y ∈ ˆ S with | ˆ x | a multiple of a and | ˆ y | amultiple of b , and let X = h ˆ x, ˆ y i . Now ˆ x is an Lbpd( q, y is a bppd( q, X . By Remark 3.5(2), there is an X -composition factor U of V with d = dim F q ( U ) ∈ { , , } such that | ˆ x U | a = | ˆ x | a ≥ a and | ˆ y U | b = | ˆ y | b ≥ b . It follows by Lemma 3.6 that X U , and hence also X , is nonsolvable if d = 6 (taking δ = 2 , j =0), and also if d = 7 (taking δ = 2 , j = 1). Thus we may assumethat d = 8, that is, X is irreducible on V . Then by Lemma 4.1,either X is nonsolvable, as required, or X is (isomorphic to) a subgroupof GL(4 , q ) .
2. Suppose that X ≤ GL(4 , q ) . X = ˆ x , ˆ y i ≤ GL(4 , q ). By the above argument applied to X , we mayassume that X acts irreducibly on V = V (4 , q ). It follows then, byLemma 4.1, that either X is nonsolvable or X is (isomorphic to) asubgroup of GL(2 , q ) .
2. However, since a ∈ Lbpd( q, a is coprimeto | GL(2 , q ) . | = 2 q ( q − q − X , andhence also X , is nonsolvable.Finally let S = PΩ + (10 , , ∈ oe ( S ) andno maximal subgroup of S has order divisible by both 17 and 31. Hence h x, y i = S whenever x, y ∈ S with | x | = 17 and | y | = 31. (cid:3) We now prove Theorem B for the exceptional finite simple groupsof Lie type. We make use of five papers. The first is the paper [FS]of Feit and Seitz. They prove, in [FS, Theorem 3.1], the existenceof certain self-centralizing cyclic maximal tori in simple groups of Lietype. The second is the paper [W] of Wiegel. He gives, in [W, Table1], a list of cyclic maximal tori in exceptional groups of Lie type, withsome small cases excluded, and in [W, Section 4] he determines themaximal subgroups containing these tori, for each such group. Wewere kindly informed by Frank L¨ubeck, in a letter, that Weigel’s listof cyclic maximal tori is correct without the extra conditions on q or k , with only one exception: namely the group G (2), which haselements of order q − q + 1 = 3 in several classes of maximal tori. Weshall refer to [L] concerning this important information. The third isthe paper [GM] of Guralnick and Malle, which is still in preparation.The authors kindly informed us that their paper contains importantinformation about maximal subgroups of E (2) sc and E (3) sc . Theforth is the paper [MT] of Moret`o and Tiep. We use [MT, Lemma 2.3],in a slightly ‘extended’ form, which was kindly approved by Pham Tiep.They prove, in Lemma 2.3, that each exceptional simple group of Lietype contains elements s and s of prime orders p and p , respectively,such that their centralizers have suitable orders. The ‘extended’ versionof this lemma states not only that such elements exist, but also that thecentralizers of every element of order p or p are of the same suitableorders. Finally, the fifth paper is the paper [GK] by Guralnick andKantor, which provides in [GK, Proposition 6.2] information concerningelements of the groups excluded in [W] and of the sporadic subgroups,contained in a unique or small number of maximal subgroups.We recall that every finite simple group of Lie type occurs as a com-position factor of the group of fixed points G F , under a Frobenius map F : G → G of a connected reductive algebraic group G over the alge-braic closure F q of a field F q of order q . f we choose G to be simply connected, then every finite simple excep-tional group of Lie type is a quotient G F / Z ( G F ). Moreover, Z ( G F ) = 1unless G is of type E , E or E . The following facts are used repeat-edly. Lemma 4.3.
Let S = G F / Z ( G F ) , q be as above, and suppose that S has a cyclic maximal torus T of order divisible by a prime p , such that | S : T | is coprime to p , and C S ( y ) = T for y ∈ T of order | T | p . Thenfor each x ∈ S with | x | = | T | , the subgroup h x i is conjugate to T in S ,and in particular it is a maximal torus of order | T | .Proof. By assumption T has a unique Sylow p -subgroup, say P = h y i ,and P is a Sylow p -subgroup of S . Let x ∈ S with | x | = | T | . Then h x i contains a subgroup P of order | P | , so by Sylow’s Theorem P g = P for some g ∈ S . Then h x i g ≤ C S ( y ) which by assumption is equal to T . It follows that h x i g = T . (cid:3) The following is an immediate corollary of Lemma 4.3.
Corollary 4.4.
Let
S, T be as in Lemma , let a = | T | ∈ oe ( S ) ,and suppose that b ∈ oe ( S ) is such that each maximal subgroup of S containing T has order coprime to b . Then for each x, y ∈ S with | x | = a and | y | = b , the group h x, y i = S and, in particular, is non-solvable. We now prove
Proposition 4.5.
Theorem B holds for all exceptional finite simplegroups of Lie type.Proof.
In the following, we denote by π ( n ) the set of prime divisors ofthe positive integer n and by Φ k ( x ) the k -th cyclotomic polynomial.We consider the exceptional groups, beginning with those of smallestLie rank. Our basic proof strategy is to choose a, b ∈ oe ( S ), wherepossible, so that the hypotheses of Corollary 4.4 hold. Then we haveimmediately that Theorem B holds for a, b . We call this ‘the standardargument’. (1): Let S = B ( q ) , with q = 2 n +1 and n ≥ . Then | S | = q ( q − q + 1). Write r = 2 n +1 , so q + 1 = ( q + r +1)( q − r + 1). Since B (2) is a Frobenius group of order 20, the fieldorder q is at least 8.Let p ∈ ppd( q, q +1 = ( q + r +1)( q − r +1), the prime p divides a := q + εr + 1 where ε = ±
1. By [FS, Theorem3.1], S has a cyclic maximal torus T of order a and we note that | S : T | is coprime to p . By comparing orders, we deduce from the ‘extended’ MT, Lemma 2.3] that C S ( y ) = T for y ∈ T of order | T | p . By [S], theonly maximal subgroup of S containing T is its normaliser, of order 4 a .Then the standard argument applies for a and any b ∈ π ( q − εr + 1),since b = 2 and gcd( q + r + 1 , q − r + 1) = 1, so b does not divide 4 a . (2): Let S = G ( q ) ′ , with q = 3 n +1 and n ≥ . Then | S | = q ( q − q + 1). Write r = 3 n +1 , so q + 1 = ( q + 1)( q + r + 1)( q − r + 1). Since G (3) ′ ∼ = PSL(2 ,
8) has been already treatedin Proposition 4.2, we may assume that q ≥ p ∈ ppd( q, p divides a = q + εr + 1where ε = ±
1. By [FS, Theorem 3.1], S has a cyclic maximal torus T of order a and we note that | S : T | is coprime to p . As in (1), C S ( y ) = T for y ∈ T of order | T | p , and by [K] and [LN], the onlymaximal subgroup of S containing T is its normaliser, of order 6 a . Let b ∈ π ( q − εr + 1). Then b ∈ oe ( S ), but b does not divide 6 a , because b = 2 , q + r + 1 , q − r + 1) = 1. Thus the standard argumentapplies. (3a): Let S = F (2) ′ , the Tits group. Then | S | = 2 · · ·
13. By [ATLAS3], 13 , ∈ oe ( S ) andeach maximal subgroup of S of order divisible by 130 is isomorphic toPSL(2 , h x, y i = S whenever x, y ∈ S with | x | = 13 and | y | = 10. (3b): Let S = F ( q ) , with q = 2 n +1 and n ≥ . Then | S | = q ( q + 1)( q − q + 1)( q − r = 2 n +1 , so q + 1 q + 1 = q − q + 1 = ( q + rq + q + r + 1)( q − rq + q − r + 1)and gcd( q + rq + q + r +1 , q − rq + q − r +1) divides ( q + q +1)( q −
1) = q − p ∈ ppd( q, p divides a := q + εrq + q + εr + 1, where ε = ±
1, and | S | /a is coprime to p . By [FS, Theorem3.1], S has a cyclic maximal torus T of order a , and arguing as in (1),the hypotheses of Lemma 4.3 hold for T . By [M], the only maximalsubgroup of S containing T is N S ( T ) of order 12 a . Let b ∈ ppd( q, b divides q + 1 and hence b ∈ oe ( S ) and b does not divide a . Also b ≥ b does not divide 12. It followsthat b does not divide 12 a , and hence, the standard argument applies. (4): Let S = G ( q ) , with q > . Then | S | = q ( q − q − G (2) ′ ∼ = PSU(3 ,
3) has beenalready treated, we may assume that q ≥
3. First we deal with G (3)and G (4), which were excluded in [W].Let S = G ( q ) with q = 3 or 4. Then by [ATLAS, pp. 60–61,97], a, ∈ oe ( S ), where a = 7 if q = 3 and a = 5 if q = 4, and each aximal subgroup M of S of order divisible by 13 a is isomorphic toPSL(2 ,
13) if q = 3, and to PSU(3 ,
4) : 2 if q = 4. In either case, thederived group M ′ is generated by any pair of its elements with one oforder a and the other of order 13. Hence, if x, y ∈ S with | x | = a and | y | = 13, then h x, y i is nonsolvable.Let, now, S = G ( q ), with q ≥ p ∈ ppd( q, q + 1 = ( q − q + 1)( q + 1), p divides a := q − q + 1 =Φ ( q ) and | S | /a is coprime to p . By [W, Table I], S has a cyclicmaximal torus T of order a , and arguing as in (1), the hypotheses ofLemma 4.3 hold for T . By [W, Section 4], each maximal subgroup M of S containing T is isomorphic to SU(3 , q ) . | M | = 2 q ( q +1)( q − b ∈ ppd( q, b ∈ oe ( S ) andgcd( b, q ) = 1. Since b divides q −
1, also gcd( b, q + 1) = 1, andit follows that b does not divide | M | . Hence, the standard argumentapplies. (5): Let S = D ( q ) , with q ≥ . Then | S | = q ( q + q + 1)( q − q − q + q + 1 =( q − q + 1)( q + q + 1).Let p ∈ ppd( q, q + 1 = ( q − q +1)( q + 1), p divides a = q − q + 1 = Φ ( q ) and | S | /a is coprime to p . By [W, Table I], there exist a cyclic maximal torus T of S of order a , and arguing as in (1), the hypotheses of Lemma 4.3 hold for T . By[W, Section 4], the only maximal subgroup of S containing T is N S ( T )of order 4 a . Let b ∈ ppd( q,
6) if q = 2 and let b = 7 if q = 2. Then b = 2, b ∈ oe ( S ) and gcd( b, q − q + 1) = 1, since b divides q − q − q + 1 divides q + 1. Thus b does not divide 4 a , and hence, thestandard argument applies. (6): Let S = F ( q ) , with q ≥ . Then | S | = q ( q − q − q − q − F (2) and F (3), which were excluded in [W].Let first S = F (2). Notice that 13 , ∈ oe ( S ) and by [GK, Propo-sition 6.2], the maximal subgroups of S of order divisible by 17 areisomorphic to PSp(8 , S = F (3). Notice that both 73 ∈ ppd(3 ,
12) and 41 ∈ ppd(3 ,
8) belong to oe ( S ). By [GK, Proposition 6.2], the maximalsubgroups of S of order divisible by 73 are isomorphic to D (3) . S = F ( q ), with q ≥ p ∈ ppd( q, p divides a = q − q + 1 = Φ ( q ) and | S | /a is coprime to p . By [W, Table I], S has a cyclic maximal torus T of rder a and arguing as in (1), the hypotheses of Lemma 4.3 hold for T . By [W, Section 4], every maximal subgroup M of S containing T is isomorphic to D ( q ) . | M | = 3 q ( q + q + 1)( q − q − b ∈ ppd( q, b ∈ oe ( S ), but b does not divide | M | , since gcd( b, q ) = 1, gcd( b, ( q − q − q + q + 1 , b ) divides gcd( q − , q −
1) = q −
1, so alsogcd( q + q + 1 , b ) = 1. Hence, the standard argument applies. (7): Let S = E ( q ) , with q ≥ . Then | S | = d q ( q − q + 1)( q − q − q + 1)( q − d = gcd(3 , q + 1). Moreover, S = ˆ S/ Z ( ˆ S ), where ˆ S = G F , with G a simply connected algebraic group of exceptional type E and | Z ( ˆ S ) | = d .Let p ∈ ppd( q, q + 1 = ( q + 1)( q − q + 1), p divides a = q − q + 1. By [W, Table 1] and [L], ˆ S has acyclic maximal torus T of order a and arguing as in (1), the hypothesesof Lemma 4.3 hold for T . Note that Z ( ˆ S ) ≤ T .It follows by [W, Section 4] for q ≥ q = 2 ,
3, that every maximal subgroup M of ˆ S containing T isisomorphic to PSU(3 , q ) . | M | = d q ( q + 1)( q − b ∈ ppd( q, b ≥ b divides q + 1 and b = 3. Hence b ∈ oe ( ˆ S ) and b does not divide | M | . Thus, by the standard argument, if x ∈ ˆ S is of order a and y ∈ ˆ S is of order b , then h x, y i = ˆ S .If d = 1, then Theorem B holds for S = ˆ S . So suppose that d = 3.Then q ≡ − a = q − q + 1 ≡ , b ) = 1 andsince Z ( ˆ S ) ≤ h x i for each x ∈ ˆ S of order a , it follows that 3 divides a . Thus a/ , b ∈ oe ( S ). Let z = ˆ z Z ( ˆ S ) be an arbitrary element of S of order a/ w = ˆ w Z ( ˆ S ) be an arbitrary element of S of order b , where ˆ z, ˆ w are elements of ˆ S . Since gcd( a/ ,
3) = gcd( b,
3) = 1 and | Z ( ˆ S ) | = 3, we may always choose ˆ z of order a and ˆ w of order b . Since,as shown above, h ˆ z, ˆ w i = ˆ S , it follows that h z, w i = S . (8): Let S = E ( q ) , with q ≥ . Then | S | = d q ( q − q − q − q − q − q − d = gcd(3 , q − S = ˆ S/ Z ( ˆ S ), where ˆ S = G F ,with G a simply connected algebraic group of exceptional type E and | Z ( ˆ S ) | = d .Let p ∈ ppd( q, q − q − q + q +1), p divides a = q + q + 1. By [W, Table 1], ˆ S has a cyclic maximaltorus T of order a , and arguing as in (1), the hypotheses of Lemma 4.3hold for T . In particular, Z ( ˆ S ) ≤ T . By [W, Section 4], every maximal ubgroup M of ˆ S containing T is isomorphic to SL(3 , q ) . | M | = 3 q ( q − q − b ∈ ppd( q, b ≥ b ∈ oe ( ˆ S ), b divides q + 1 and b = 3. It follows that b does notdivide | M | . Hence, by the standard argument, if x ∈ ˆ S is of order a and y ∈ ˆ S is of order b , then h ˆ x, ˆ y i = ˆ S .If d = 1, then Theorem B holds for S = ˆ S . So suppose that d = 3.Then it follows using the same proof as for E ( q ) that if z = ˆ z Z ( ˆ S )has order a/ w = ˆ w Z ( ˆ S ) has order b , where ˆ z, ˆ w are elements ofˆ S , then h z, w i = S . (9): Let S = E ( q ) , with q ≥ . Then | S | = 1 d q Y i ∈ I ( q i − , with I = { , , , , , , } where d = gcd(2 , q − S = ˆ S/ Z ( ˆ S ), where ˆ S = G F ,with G a simply connected algebraic group of exceptional type E and | Z ( ˆ S ) | = d .Let p ∈ ppd( q, q + 1 = ( q + 1)( q − q + 1), p divides a = ( q + 1)( q − q + 1). By [W, Table 1] and [L], S has a cyclic maximal torus T of order a , and arguing as in (1), thehypotheses of Lemma 4.3 hold for T . In particular, Z ( ˆ S ) ≤ T .By [W, Section 4] for q ≥ q = 2 ,
3, every maximal subgroup M of ˆ S containing T is isomorphicto ( Z q +1 · E ( q )) . | M | = d ( q + 1) q ( q − q + 1)( q − q − q + 1)( q − d = gcd(3 , q − b ∈ ppd(3 , b ∈ oe ( ˆ S ), b divides q + 1 and b = 2. Since gcd( q + 1 , q + 1) divides q −
1, it follows that b does not divide | M | . Hence, by the standard argument, if x ∈ ˆ S is oforder a and y ∈ ˆ S is of order b , then h ˆ x, ˆ y i = ˆ S .If d = 1, then Theorem B holds for S = ˆ S . So suppose that d = 2.Then q is odd and since Z ( ˆ S ) ≤ h x i for each x ∈ ˆ S of order a , it followsthat a is even, and a/ , b ∈ oe ( S ). Let z = ˆ z Z ( ˆ S ) be an arbitraryelement of S of order a/ w = ˆ w Z ( ˆ S ) be an arbitrary elementof S of order b , where ˆ z, ˆ w are elements of ˆ S . Since gcd( b,
2) = 1 and | Z ( ˆ S ) | = 2, it follows that we may choose ˆ w of order b . Now considerˆ z . Let L = h ˆ z, Z ( ˆ S ) i . Then L is abelian and it contains an element ˆ u of order p . By [MT, Lemma 2.3], | C ˆ S (ˆ u ) | = a and as shown in (1), thiscentralizer is a cyclic maximal torus of ˆ S . Consequently, as L ≤ C ˆ S (ˆ u ), e may also choose ˆ z of order a . Since, as shown above, h ˆ z, ˆ w i = ˆ S , itfollows that h z, w i = S . (10): Let S = E ( q ) , with q ≥ . Then | S | = q Y i ∈ I ( q i − , with I = { , , , , , , , } . Let p ∈ ppd( q, q + 1 = ( q − q + 1)( q + 1)( q + q − q − q − q + q + 1)= ( q − q + 1)( q + 1)Φ ( q ) ,p divides a = Φ ( q ) and | S | /a is coprime to p . By [W, Table I], S has a cyclic maximal torus T of order a , and arguing as in (1), thehypotheses of Lemma 4.3 hold for T . By [W, Section 4], the onlymaximal subgroup of S containing T is N S ( T ) of order 30 a .Let b ∈ ppd( q, b ≡ b ≥
25 and b does not divide 30. Now b divides q + 1 and a divides q + 1, and since gcd( q + 1 , q + 1) divides q −
1, it follows that b does not divide a . Therefore, b does not divide | N S ( T ) | = 30 a , so the standard argument applies. (cid:3) We are now ready to complete the proof of Theorem B, which westate again:
Theorem B.
Let S be a nonabelian finite simple group. Then thereexist a, b ∈ oe ( S ) , such that every pair of elements of S of order a and b , respectively, generates a nonsolvable subgroup of S .Proof. This follows from Propositions 2.1, 2.2, 4.2, 4.5 and from theclassification of the finite simple groups. (cid:3) Proof of Theorem A
We finally show that Theorem A follows from Theorem B. First, werestate Theorem A.
Theorem A.
Let G be a finite group. Assume that for every x, y ∈ G there exists an element g ∈ G such that h x, y g i is solvable. Then G issolvable.Proof. Suppose that the hypothesis holds for a group G . This is clearlyequivalent to assuming that, for all pairs C, D of conjugacy classes ofa group G there exist elements x ∈ C and y ∈ D such that h x, y i issolvable.We claim that this property is inherited by factor groups: let N bea normal subgroup of G and write G = G/N . Since “overbar” is a omomorphism, it sends conjugacy classes of G onto conjugacy classesof G . Hence, given two conjugacy classes C and D of G , we may assumethat C and D are conjugacy classes of G . So, by our assumption, thereexist x ∈ C and y ∈ D such that h x, y i is solvable. Hence, x ∈ C , y ∈ D and h x, y i ≤ h x, y i is solvable. Thus the claim is proved.Theorem A holds trivially if | G | = 1. Suppose inductively that | G | > | G | .Let M be a minimal normal subgroup of G . Then by induction, G/M issolvable. If G has distinct minimal normal subgroups M , M , then G isisomorphic to a subgroup of the solvable group G/M × G/M . Thus wemay assume that G has a unique minimal normal subgroup M . If M issolvable, then G is solvable as well. We show that this must be the case:suppose to the contrary that M is nonsolvable. The characteristicallysimple group M is a direct product of isomorphic simple groups. Wehence identify M with the direct power S k of a nonabelian simplegroup S . By Theorem B, there exist a, b ∈ oe ( S ) such that for everychoice of elements x, y ∈ S with | x | = a and | y | = b , the group h x, y i is nonsolvable. In particular, h x α , y β i is nonsolvable for all α, β ∈ Aut( S ). Consider now the diagonal elements u = ( x, x, . . . , x ) and w = ( y, y, . . . , y ) of M . Recalling that G can be embedded in thewreath product of Aut( S ) by a (solvable) subgroup of the symmetricgroup S k , we see that for every g, h ∈ G we have u g = ( x α , x α , . . . , x α k )and w h = ( y β , y β , . . . , y β k ), where α i , β i ∈ Aut( S ) for i = 1 , , . . . , k .Observe also that h u g , w h i is a subdirect subgroup of the direct product k Y i =1 h x α i , y β i i . However h x α i , y β i i is a nonsolvable subgroup of S , for every i = 1 , . . . , k .It follows that h u g , w h i is nonsolvable for every choice of g and h in G ,which is the required contradiction. (cid:3) References [ATLAS] J.H. Conway, R.S. Curtis, S.P. Norton, R.A. Parker and R.A. Wilson,Atlas of Finite Groups, Clarendon Press, Oxford, 1985.[ATLAS3] R.A. Wilson et al., Atlas of Group Representations,http://brauer.maths.qmul.ac.uk/Atlas/v3/[BGK] T. Breuer, R. Guralnick and W. Kantor, Probabilistic generation of finitesimple groups, II,
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E-mail address : [email protected] Cheryl E. Praeger, School of Mathematics and Statistics,The University of Western Australia, 35 Stirling Highway, Crawley,WA 6009, Australia
E-mail address : [email protected]@maths.uwa.edu.au