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On the Finite Simple Groups All of Whose 2-Local Subgroups Are Solvable
210, 365]384 Ž1998. JA987587
JOURNAL OF ALGEBRA ARTICLE NO.
On the Finite Simple Groups All of Whose 2-Local Subgroups Are Solvable Makoto Hayashi Department of Mathematics, Aichi Uni¨ ersity of Education, Kariya 448, Japan
and Yasuhiko Tanaka Faculty of Engineering, Oita Uni¨ ersity, Oita 870-11, Japan Communicated by Walter Feit Received June 2, 1996
1. INTRODUCTION A finite group G is said to be of characteristic 2 type if each 2-local subgroup L of G satisfies the condition CLŽ O 2 Ž L.. : O 2 Ž L.. The purpose of this paper is to give an alternative proof of the following theorem, which was a result of the combined work of Janko w1x, Smith w2x, and Gorenstein and Lyons w3x. MAIN THEOREM. Let G be a nonabelian simple group of characteristic 2 type, all of whose 2-local subgroups are sol¨ able. Then G is isomorphic to one of the following groups: L2 Ž q ., Sz Ž q ., U3 Ž q . Ž where q is a power of 2., L2 Ž p . Ž where p is a Fermat or Mersenne prime with p G 5., A 6 , L3 Ž3., M11 , U3 Ž3., 2 F4 Ž2.X . Our approach to the proof of the Main Theorem is called an ‘‘amalgam method.’’ Several results have already been obtained for our purpose, and we use some of them in the whole of this paper. Therefore we will clarify here what we will assume and what we should prove. Let X be a group, and let Y be a subgroup of X. By definition, X is Y-irreducible if X has a unique maximal subgroup containing Y. A finite group G is often said to be 2-irreducible if G is S-irreducible for a Sylow 365 0021-8693r98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.
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2-subgroup S of G. Furthermore, a finite group G of even order is said to be thin if m 2, p Ž G . F 1 for all odd primes p, where the 2-local p-rank m 2, p Ž G . of G is the maximum of the p-ranks m p Ž L. as L ranges over the 2-local subgroups of G. The task of this paper is stated as the following theorem. THEOREM 1. Let G be a finite group, all of whose 2-local subgroups are sol¨ able. Let S g Syl 2 Ž G ., and let H and K be subgroups of G satisfying the following conditions: Ža. Žb. Žc. Žd.
S is a common Sylow 2-subgroup of H and K. Both H and K are S-irreducible and sol¨ able. No nonidentity subgroup of S is normal both in H and in K. CH Ž O 2 Ž H .. : O 2 Ž H . and CK Ž O 2 Ž K .. : O 2 Ž K ..
Then both H and K are thin. The next two theorems are assumed as a starting point of our analysis of 2-local subgroups. THEOREM 2 ŽTheorems A, C of w4x.. Let G be a finite group of characteristic 2 type, all of whose 2-local subgroups are sol¨ able. Let S g Syl 2 Ž G ., and assume O 2 Ž G . s 1. Then one of the following holds: Ž1. G has a strongly embedded subgroup. Ž2. S is dihedral, semidihedral, or isomorphic to a Sylow 2-subgroup of the group AutŽ Sp4 Ž2... Ž3. There exist subgroups H and K of G satisfying the abo¨ e conditions Ža. ] Žd.. THEOREM 3 ŽMain Theorem of w5, 6x.. Let Ž H, S, K . be an amalgam satisfying the abo¨ e conditions Ža. ] Žd. under some embeddings of S into H and K. If both H and K are thin, then the amalgam Ž H, S, K . or Ž K, S, H . is isomorphic to a GL3 Ž2.-amalgam, an Sp4Ž2.-amalgam, a G 2 Ž2.X-amalgam, a G 2 Ž2.-amalgam, an M12-amalgam, an AutŽ M12 .-amalgam, a 2 F4Ž2.X-amalgam, or a 2 F4Ž2.-amalgam. Let G be a nonabelian simple group satisfying the assumption of the Main Theorem. Then G satisfies the assumption of Theorem 2. If Ž1. or Ž2. occurs in Theorem 2, we can appeal to some of the existing classification theorems to identify G with one of the known simple groups. Hence we can assume that Ž3. occurs in Theorem 2, and then G satisfies the assumption of Theorem 1. Once Theorem 1 is proved, we can know the structures of the subgroups S, H, K by Theorem 3, and then identify G with one of the known simple groups, appealing again to some classification theorems.
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Our approach to Theorem 1 pursues and extends the approach to Theorem B of w4x, which applies to an alternative proof of the classification of quasi-thin finite simple groups, all of whose 2-local subgroups are solvable. In the course of the proof of Theorem 1, the following two theorems are of particular importance. THEOREM 4 ŽTheorem C of w7x.. Let Ž H, S, K . be an amalgam satisfying the abo¨ e conditions Ža. ] Žd., and the following additional condition Že.: Že.
S s w O 2 Ž H ., O 2 Ž H .xŽ O 2 Ž H . l O 2 Ž K ..w O 2 Ž K ., O 2 Ž K .x.
Then both H and K are thin. THEOREM 5 ŽSection 4.8 of w4x.. Let G be a finite group, all of whose 2-local subgroups are sol¨ able. Let H, S, K, H X , SX , K X be subgroups of G so that both Ž H, S, K . and Ž H X , SX , K X . satisfy the abo¨ e conditions Ža., Žb., Žd.. Suppose that SX s w O 2 Ž H ., O 2 Ž H .xŽ O 2 Ž H . l O 2 Ž K ..w O 2 Ž K ., O 2 Ž K .x, that H s ² S, H X :, and that K s ² S, K X :. If Ž H X , SX , K X . satisfies condition Žc., and both H X and K X are thin, then Ž H, S, K . s Ž H X , SX , K X .. Following w4x, we will write Ž H X , SX , K X . U Ž H, S, K . if the assumptions of Theorem 5 are fulfilled. Let G, S, H, K be groups satisfying the assumption of Theorem 1. Then we can actually choose a sequence Ž H0 , S0 , K 0 ., Ž H1 , S1 , K 1 ., . . . , Ž Hn , Sn , K n ., . . . of amalgams satisfying conditions Ža., Žb., Žd. such that Ž H0 , S0 , K 0 . s Ž H, S, K . and Ž Hnq1 , Snq1 , K nq 1 . U Ž Hn , Sn , K n . for all nonnegative integers n. Since S is finite, we have Sm s Smq1 for some nonnegative integer m. If no nonidentity subgroup of S m is normal in both Hm and K m , then repeated use of Theorem 5 forces that both H and K are thin. Thus we will assume in our analysis that some nonidentity subgroups of Sm is normal both in Hm and in K m , in particular that ² Hm , K m : is a solvable subgroup of G. In fact, most of the arguments of this paper are devoted to the study of the solvable groups ² Z1 , . . . , Zn : of G generated by D-irreducible subgroups Z1 , . . . , Zn with D : S and D g Syl 2 Ž Zk . for k s 1, . . . , n. Looking at our proof, we can understand that it was sufficient for us to classify the amalgams satisfying not only conditions Ža. ] Žd. but also the additional condition Že.. Related to quadratic 2-subgroups of 2-local subgroups, we will define new characteristic subgroups N `Ž n; B . for each finite 2-group B and each nonnegative integer n. We will frequently use them throughout the proof of Theorem 1 in the following manner: for a 2-local subgroup L containing Ž . a Sylow 2-subgroup S, either N `Ž n; S . 1 y L or the structure of LrO2 L is highly restricted. Precise definitions and necessary properties are given in Section 2. Section 3 contains necessary information about GF Ž2.-represen-
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tation of solvable groups, and Sections 4 and 5 are devoted to the body of the proof of Theorem 1.
2. ON FINITE 2-GROUPS Throughout this section, we will assume that B is an arbitrary finite 2-group, and that n is a nonnegative integer. DEFINITION 2.1. Let X and Y be subgroups of B. Ž1. We denote by v nŽ X, Y . the set of Y-invariant subgroups V of X possessing the following property: if w V, L, L x s 1 for an abelian subgroup L of Y, then there exists a family of generators ¨l < l g L4 of V Ždepending on L. such that < L: CLŽ ¨l .< F 2 n for all l g L. Ž2. We denote by v nU Ž X, Y . the set of elements V of v nq1Ž X, Y . possessing the following property: there exist a family of subgroups Vi < i g I 4 of V and a family of subgroups Yi < i g I 4 of Y such that Ža. V s ² Vi < i g I :, Žb. Vi g v nŽ X, Yi . for all i g I, Žc. ² v nŽ Y, V .: : Yi for all i g I. Ž3. We use the following convention: if Z is a subgroup of B and B is a set of subgroups of B, then Z l B denotes the set of elements of B contained in Z. LEMMA 2.2.
Let X, X X , Y, Y X : B with X : X X and Y : Y X .
X l v nŽ X X , Y . : v nŽ X, Y . : v nŽ X X , Y . and X l v nU Ž X X , Y . : . : v nU Ž X X , Y .. Ž2. v nŽ X, Y . = v nŽ X, Y X . and v nU Ž X, Y . = v nU Ž X, Y X .. Ž3. ZŽ C X Ž Y .. g v nŽ X, Y . : v nU Ž X, Y . : v nq1Ž X, Y .. Ž1.
v nU Ž X, Y
Proof. This follows immediately from Definition 2.1. DEFINITION 2.3. Define a sequence of subgroups Nk Ž n; B .4`ksy1 of B recursively as follows: Ž1. Ž2.
Ny1Ž n; B . s B. Nk Ž n; B . s ² v nU Ž B, Nky1Ž n; B ..:Ž k G 0..
Now, define N `Ž n; B . s Fk: o d d Nk Ž n; B . and N`Ž n; B . s ² Nk Ž n; B .< k: e¨ en:. For brevity, we put N ` s N `Ž n; B . and N` s N`Ž n; B . throughout this section.
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LEMMA 2.4. The following holds: Ž1. Ž2.
ZŽ B . : N` : N ` . In particular, N ` / 1 whene¨ er B / 1.
N ` is a characteristic subgroup of B. Ž3. ² v nU Ž B, N ` .: s N` and ² v nU Ž B, N` .: s N ` . Ž4. If N ` : Q : B, then N `Ž n; Q . s N ` , and N ` is a characteristic subgroup of Q. Proof. First, Nk Ž n; B . is a characteristic subgroup of B for all k because AutŽ B . leaves the set v nU Ž B, Nky1Ž n; B .. invariant by induction on k. So Ž2. holds. We will see the inclusion relations between the subgroups Nk Ž n; B .. Let Nk s Nk Ž n; B . for all k. Since Ny1 s B = N0 , we have Ny1 s B = N1 s ² v nU Ž B, N0 .: = ² v nU Ž B, Ny1 .: s N0 by Lemma 2.2 Ž2.. Assume that either Niy1 = Niq1 = Ni or Niy1 : Niq1 : Ni holds for some i. Then either Ni : Niq2 : Niq1 or Ni = Niq2 = Niq1 holds by Lemma 2.2 Ž2. because Nkq 1 s ² v nU Ž B, Nk .: for all k. Thus we can conclude, by induction, that B s Ny1 = N1 = ??? = N2 iy1 = ??? = N2 i = ??? = N2 = N0 . Since < B < is finite, there exists a nonnegative integer m such that N` s N2 i and N ` s N2 iy1 for all i G m. Hence, together with Lemma 2.2 Ž3., ZŽ B . : N0 : N` s N2 m : N2 my1 s N ` , ² v nU Ž B, N ` .: s ² v nU Ž B, N2 my1 .: s N2 m s N` , ² v nU Ž B, N` .: s ² v nU Ž B, N2 m .: s N2 mq1 s N ` . So Ž1. and Ž3. hold. Suppose that N ` : Q : B. Put Mk s Nk Ž n; Q .. Then Ny1 = My1 = N ` by our assumption. Assume that N2 iy1 = M2 iy1 = N `. Then N2 i s ² v nU Ž B, N2 iy1 .: : ² v nU Ž B, M2 iy1 .: : ² v nU Ž B, N ` .: s N` : N ` : Q by Lemma 2.2 Ž2. and Ž1.Ž3.. Thus ² v nU Ž B, M2 iy1 .: s ² v nU Ž Q, M2 iy1 .: s M2 i by Lemma 2.2 Ž1., so N2 i : M2 i : N` . Again by Lemma 2.2 Ž2. and Ž1.Ž3., N2 iq1 s ² v nU Ž B, N2 i .: = ² v nU Ž B, M2 i .: = ² v nU Ž Q, M2 i .: s M2 iq1 and ² v nU Ž B, N` .: s N ` = Q, so M2 iq1 = ² v nU Ž Q, N` .: s ² v nU Ž B, N` .: s N ` by Lemma 2.2 Ž1.. Thus we can conclude, by induction, that N2 iy1 = M2 iy1 = N ` for all i, so N `Ž n; Q . s N ` . This proves Ž4.. LEMMA 2.5. Let V g v nU Ž B, B ., and let Q : B. If < V: C V Ž x .< G 2 nq 2 for all x g B y Q, then N ` : Q. Proof. Since V : ² v nU Ž B, B .: : ² v nU Ž B, N ` .: s N` by Lemma 2.2 Ž2. and Lemma 2.4 Ž3., we have N ` s ² v nU Ž B, N` .: : ² v nU Ž B, V .: : ² v nq 1Ž B, V .: by Lemma 2.4 Ž3. and Lemma 2.2 Ž2.Ž3.. Let L g v nq 1Ž B, V .. Since w L, V, V x : w V, V x s 1, Definition 2.1 Ž1. shows that L is generated by the elements x with < V: C V Ž x .< F 2 nq 1 , so L : Q by our assumption. Therefore N ` : Q. LEMMA 2.6. Let V g v nŽ B, B ., and let Q : B. If < V: C V Ž x .< G 2 nq 1 for all x g B y Q, then N ` : Q.
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Proof. We have N ` : ² v nU Ž B, V .: by the same argument as in the first paragraph of the proof of Lemma 2.5 because V g v nU Ž B, B . by Lemma 2.2 Ž3.. Let L g v nU Ž B, V .. Let L i < i g I 4 and Vi < i g I 4 be families of subgroups of L and V, respectively, satisfying the conditions of Definition 2.1 Ž2.. Let i g I. Since V : ² v nŽ V, B .: : ² v nŽ V, L.: by Lemma 2.2 Ž1.Ž2., we have V : Vi by Definition 2.1 Ž2c., so L i g v nŽ B, V . by Definition 2.1 Ž2b.. Since w L i , V, V x : w V, V x s 1, Definition 2.1 Ž1. shows that L i s ² x g L i < < V: C V Ž x .< F 2 n : : Q, so L : Q by Definition 2.1 Ž2a.. Therefore N ` : Q.
3. ON GF Ž2.-REPRESENTATION OF SOLVABLE GROUPS In this section, we study GF Ž2.-representation of solvable groups. First, we review the basic structure of 2-irreducible solvable groups. LEMMA 3.1. Let G be a 2-irreducible sol¨ able group with S g Syl 2 Ž G ., and let M be the unique maximal subgroup of G containing S. Put Q s O 2 Ž G .. Then the following holds: Ž1. If N 1 G / SN Ž or, equi¨ alently, O 2 Ž G . N 1 G ., then S l N y y 1 G, Q g Syl 2 Ž QN ., and GrN is 2-irreducible and sol¨ able. y Ž2. G s O 2, p, 2 Ž G . for some odd prime p. Let P g Syl p Ž G .. Ž3. SrQ acts irreducibly and faithfully on PQrFŽ P . Q. Ž4. ZŽ SrQ . is cyclic, and an element t g S in¨ erts PQrFŽ P . Q if and only if Q / tQ g V 1Ž ZŽ SrQ ... Ž5. P l M s F Ž P .. Ž6. Ž7.
Let x g S y Q. Then x f M if and only if G s ² S, x :. O 2 Ž G . s P w Q, O 2 Ž G .x and w Q, O 2 Ž G .x s S l O 2 Ž G ..
Proof. Parts Ž1. ] Ž6. are proved in Ž2.1. of w11x. Since w Q, P x : w Q, 2Ž . 2 Ž .x Ž . w O G .x, we have P w Q, O 2 Ž G .x 1 y PQ s O 2, p G , so O G s P Q, O G . 2 2 2 Thus w Q, O Ž G .x s Q l O Ž G . s S l O Ž G ., so Ž7. holds. 2Ž
Throughout the remainder of this section, we will assume that G is a finite solvable group and that V is a nontrivial irreducible faithful GF Ž2.G-module. Let S g Syl 2 Ž G ., and put Sw k x s ² x g S < < V: C V Ž x .< F 2 k : for all positive integers k.
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Sw1x : O 2X , 2 Ž G ..
Proof. We may assume that Sw1x / 1. Then S w1x : ² P Ž G, V .:, where P Ž G, V . is the set of nonidentity elementary abelian subgroups A of G satisfying < A < < C V Ž A.< G < B < < C V Ž B .< for all subgroups B of A. We can see ² P Ž G, V .: : O 2X , 2 Ž G . by Ž3.2. and Ž3.3. of w8x because G is solvable and O 2 Ž G . : CG Ž V . s 1. LEMMA 3.3. Suppose that G is 2-irreducible and that S w2x / 1. Let H be the set of the homogeneous components of VP , where P s O Ž G .. Then there exists a normal subgroup PW for each W g H such that P s ² PW < W g H :, W s w V, PW x, and one of the following holds: Ž1. Ž2. Ž3. Ž4.
PW ( Z3 , < W < s 2 2 , and Sw1x / 1. PW ( Z3 , < W < s 2 2 , and Sw1x s 1. PW ( Z5 , < W < s 2 4 , and Sw1x s 1. PW ( Esp27 , < W < s 2 6 , and Sw1x s 1.
Proof. We first note that P g Syl p Ž G . for some odd prime p by Lemma 3.1 Ž2. because O 2 Ž G . s 1. Let t g S be an involution with < V: C V Ž t .< F 2 2 , and put T s t S . Let u g T , and define Pu s w P, u x and Vu s w V, Pu x. Then Pu y 1 P for all u g T. Since < V 1Ž ZŽ S ..< s 2 by Lemma 3.1 Ž4. and ² T : y 1 S, we have V 1Ž ZŽ S .. : ² T :, and then P s ² Pu < u g T :, because V 1Ž ZŽ S .. inverts PrFŽ P . also by Lemma 3.1 Ž4.. Hence V s Ý ug T Vu . Let I be a subset of T that is maximal subject to the following condition: if u / uX , then Vu / VuX for all u, uX g I . Then P s ² Pu < u g I :, and V s Ý ug I Vu . Let u g I . Pick an irreducible submodule X of VP contained in Vu . Then X s w X, Pu x as Pu 1 y P. So, if W g H contains X, then W s w W, Pu x : w V, Pu x s Vu . This shows that Vu is a direct sum of some members of H as VP is completely reducible. Suppose that Vu is homogeneous itself. Then the set Vu < u g I 4 coincides with the set H . For W g H , choose the unique element u g I so that W s Vu , and define PW s Pu . Then P s ² PW < W g H : and W s w V, PW x for each W g H . The structures of W and PW are described in Lemma 2.8 of w11x because < W: CW Ž u.< F < V: C V Ž u.< F 2 2 . In particular, ZŽ PW . is cyclic as W is homogeneous and CPW ŽW . : CP Ž V . s 1. Since G is irreducible on V, NG ŽW . is also irreducible on W, so the case < W < s 2 4 and PW ( Z3 is eliminated. Hence one of the three cases Ž1., Ž3., Ž4. occurs. Suppose that Vu is not homogeneous. For W g H , choose u g I so that W : Vu . Note that u does not centralize any member of H contained in Vu as C V uŽ Pu . s 0. Thus there exists W X g H such that Vu s W q W X , where < W < s < W X < s 2 2 , because < Vu : C V Ž u.< F < V: C V Ž u.< F 2 2 . Hence one of the u two cases Ž1., Ž2. occurs, and, moreover, PrCP ŽW . ( PrCP ŽW X . ( Z3 , CP Ž Vu . s CP ŽW . l CP ŽW X ., and u inverts PrCP Ž Vu .. Since Vu is not
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homogeneous, Lemma 2.8 of w11x shows that Pu ( E9 , and so P s Pu CP Ž Vu .. w x Define PW s Pu l CP ŽW X .. Then P 2 y PW ( Z3 and V, PW s W. Since CP Ž V . s 1 and < W < s 2 2 , we can conclude that PW is a unique subgroup of P with the property W s w V, PW x, and so, in particular, PW is independent on the choice of u g I . Thus Pu s PW PW X , and hence P s ² PW < W g H :. LEMMA 3.4. F 2:.
Let L : S with w V, L, L x s 0. Then V s ² ¨ g V < < L: CLŽ ¨ .<
Proof. Since O 2 Ž G . s 1, there exists an odd prime p such that w Op Ž G ., L x / 1. Put P s Op Ž G .. Let W be a homogeneous component of VP . Put N s NG ŽW ., C s CG ŽW ., and N s NrC. If L N, then < L: C l L < F 2 by Ž3.2. of w7x. Suppose that L : N. If W s V, then < L: C l L < F 2 by Ž3.3. of w7x because w V, L, L x s 0 / w V, w N, L xx for a GF Ž2. PL-module V. So suppose that W ; V. Then W s ² w g W < < L: CLŽ w .< F 2: by the inductive hypothesis, because N is irreducible on W and w W, L, L x : w V, L, L x s 0. Hence the lemma holds because V is a direct sum of the homogeneous components of VP .
4. PRELIMINARIES Throughout the remainder of this paper, let G be a finite group satisfying the assumption of Theorem 1, and let S be a Sylow 2-subgroup of G. We begin with some definitions. DEFINITION 4.1. Let D : S. Ž1. We denote by DŽ D . the set of subgroups Z of G satisfying the following conditions: Ža. D g Syl 2 Ž Z .. Žb. Z is D-irreducible and solvable. Žc. CZ Ž O 2 Ž Z .. : O 2 Ž Z .. Ž2. Let D : S and X, Y g DŽ D .. We write X :2 Y and say that X is 2-embedded into Y if w O 2 Ž X ., O 2 Ž X .x : O 2 Ž Y .. Ž3. We denote by G the set of triplets Ž X, D, Y . with the conditions D : S and X, Y g DŽ D .. Let b s Ž X, D, Y . g G. Define O 2 Ž b . s O 2 Ž² X, Y :. l D. Note that O 2 Ž b . is the largest subgroup of D that is normal in both X and Y. We call b thin if both X and Y are thin. Let b X s Ž X X , DX , Y X . g G. We write b X U b if DX s w O 2 Ž X ., O 2 Ž X .xŽ O 2 Ž X . l O 2 Ž Y ..w O 2 Ž Y ., O 2 Ž Y .x, X s ² X X , D :, and Y s ² Y X , D :.
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Ž4. Let L be a subgroup of G with R s L l S g Syl 2 Ž L.. Let D : R. We denote by GLŽ D . the set of subgroups Z of L satisfying the following conditions: Ža. D g Syl 2 Ž Z .. Žb. Z is D-irreducible. Žc. ² O 2 Ž Z ., R : s L. LEMMA 4.2. Ž1.
Let D : R : S and L g DŽ R . with w O 2 Ž L., O 2 Ž L.x : D.
GLŽ D . / B.
Let Z g GLŽ D .. Ž2. w O 2 Ž Z ., O 2 Ž Z .x s w O 2 Ž L., O 2 Ž Z .x. Ž3. CZ Ž O 2 Ž Z .. : CZ Žw O 2 Ž Z ., O 2 Ž Z .x. : O 2 Ž Z .. Ž4. Z g DŽ D .. Ž5. FO 2 Ž Z . R : O 2 Ž L. = FX g G Ž D . O 2 Ž X .. L Ž6. ² O 2 Ž Z . R : s O 2 Ž L. s ² O 2 Ž X .< X g GLŽ D .:. Proof. We first note that D g Syl 2 Ž DO 2 Ž L.. as O 2 Ž L. l R s w O 2 Ž L., O 2 Ž L.x : D by Definition 4.1 Ž1a. and Lemma 3.1 Ž7., and that DO 2 Ž L. is generated by the set GX of D-irreducible subgroups X of L with D g Syl 2 Ž X .. Let M be the unique maximal subgroup of L containing R. Let X g GX . Then X g GLŽ D . if and only if X M by Lemma 3.1 Ž6.. Put LU s Lrw O 2 Ž L., O 2 Ž L.x. Then O 2 Ž L.U has odd order by Lemma 3.1 Ž7., so O 2 Ž L.U l M U s F Ž O 2 Ž L.U . by Lemma 3.1 Ž5.. Hence we have GLŽ D . s X g GX < O 2 Ž X .U F Ž O 2 Ž L.U .4 . Thus O 2 Ž L. s ² O 2 Ž X .< X g GX : s ² O 2 Ž X .< X g GLŽ D .:, so Ž1. holds. Moreover, we have L s ² O 2 Ž Z . R : R by Definition 4.1 Ž4c., so Ž6. holds. Let Q s w O 2 Ž Z ., O 2 Ž Z .x and Y s O 2 Ž CZ Ž Q ... Part Ž2. holds because Q s O 2 Ž Z . l D : O 2 Ž L. l R s w O 2 Ž L., O 2 Ž L.x by Lemma 3.1 Ž7.. Since w O 2 Ž L., Y x : w O 2 Ž L., O 2 Ž Z .x s Q by Ž2., we have Y : CLŽ O 2 Ž L.. : O 2 Ž L. by Definition 4.1 Ž1c.. Hence Y s 1, so Ž3. holds. Part Ž4. is a consequence of Ž3. and Definition 4.1 Ž4a.Ž4b.. Let L s LrO2 Ž L., and GY s Z R or GLŽ D .. Then O 2 Ž L . s O Ž L. by Definition 4.1 Ž1b. and Lemma 3.1 Ž2.. Since wO 2 Ž X . , O 2 Ž X . x : O 2 Ž X . lO Ž L . s 1 for all X g GY , we have wFX g G Y O 2 Ž X . , O Ž L .x s wFX g GY O 2 Ž X . , ²O 2 Ž X . < X g GY :x s 1 by Ž6.. Thus FX g GY O 2 Ž X . : CR Ž O Ž L.. s 1, so Ž5. holds. LEMMA 4.3. Let D : R : S and L, LX g D Ž R . with w O 2 Ž L ., O L.xw O 2 Ž LX ., O 2 Ž LX .x : D and L 2 LX . 2Ž
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Ž1.
If Y g GLŽ D ., then there exists X g GLX Ž D . such that Y 2 X.
Ž2.
If Y g GLX Ž D ., then there exists Z g GLŽ D . such that Z 2 Y.
Proof. We first note that B / GLŽ D . : DŽ D . = GLX Ž D . / B by Lemma 4.2 Ž1.Ž4.. Let Y g GLŽ D .. Suppose that Y :2 X for all X g GLX Ž D .. Then w O 2 Ž L., O 2 Ž Y .x s w O 2 Ž Y ., O 2 Ž Y .x : O 2 Ž X . for all X g GLX Ž D . by Lemma 4.2 Ž2.. Thus w O 2 Ž L., O 2 Ž Y .x : FX g GLX Ž D . O 2 Ž X . : O 2 Ž LX . by Lemma 4.2 Ž5.. Hence w O 2 Ž L., O 2 Ž L.x : w O 2 Ž L., ² O 2 Ž Y . R :x : O 2 Ž LX . by Lemma 4.2 Ž6., a contradiction. So Ž1. holds. Let Y g GLX Ž D .. Suppose that Z :2 Y for all Z g GLŽ D .. Then w O 2 Ž L., O 2 Ž Z .x s w O 2 Ž Z ., O 2 Ž Z .x : O 2 Ž Y . for all Z g GLŽ D . by Lemma 4.2 Ž2.. Thus w O 2 Ž L., O 2 Ž L.x : w O 2 Ž L., ² O 2 Ž Z .< Z g GLŽ D .:x : O 2 Ž Y . by Lemma 4.2 Ž6.. Hence w O 2 Ž L., O 2 Ž L.x : FO 2 Ž Y . R : O 2 Ž LX . by Lemma 4.2 Ž5., a contradiction. So Ž2. holds. LEMMA 4.4. Let E : D : R : S and L g DŽ R .. Suppose Z g GLŽ D . with w O 2 Ž Z ., O 2 Ž Z .x : E. Then w V 1Ž ZŽ E .., O 2 Ž Z .x s 1 if and only if V 1Ž ZŽ R .. : ZŽ L.. Proof. Put Q s O 2 Ž L., T s w O 2 Ž Z ., O 2 Ž Z .x, U s V 1Ž ZŽ Q .., W s V 1Ž ZŽ E lQ... Then w Q, O 2 Ž Z .x s T s O 2 Ž Z . l D : E l Q by Lemma 4.2 Ž2. and Lemma 3.1 Ž7., so wU, O 2 Ž Z .x : w W, O 2 Ž Z .x, and < O 2 Ž Z .Ž E l Q .: Ž E l Q .< is odd. Hence CwW , O 2 Ž Z .x Ž O 2 Ž Z .. s 1. Suppose first w V 1Ž ZŽ E .., O 2 Ž Z .x s 1. Then w W, O 2 Ž Z .x l V 1Ž ZŽ E .. s 1, so w W, O 2 Ž Z .x s 1. Hence wU, O 2 Ž Z .x s 1, and then V 1Ž ZŽ R .. : ZŽ L. by Definition 4.1 Ž4c. because V 1Ž ZŽ R .. : U by Lemma 4.2 Ž3.. Suppose next w V 1Ž ZŽ E .., O 2 Ž Z .x / 1. Since V 1Ž ZŽ E .. : CZ ŽT . : O 2 Ž Z . by Lemma 4.2 Ž3., we have w V 1Ž ZŽ E .., O 2 Ž Z .x : T : E l Q, so 1 / w V 1Ž ZŽ E .., O 2 Ž Z .x : w W, O 2 Ž Z .x. Here, let T : N 1 VN s w V 1Ž ZŽ N .., O 2 Ž Z .x / 1. y M : Q, and suppose 2 We will show that VM s w V 1Ž ZŽ M .., O Ž Z .x / 1. By Lemma 3.1 Ž2., Z is a 2, p4 -group for some odd prime p. Let P g Syl p Ž Z .. Then O 2 Ž Z . s PT by Lemma 3.1 Ž7., and M s NCM Ž P . because w M, P x : w Q, P x : T : N. Ž . w Ž . x Thus VN s w V 1Ž ZŽ N .., P x 1 y M and C V N P s 1. Hence 1 / C V N M , P : w V 1Ž ZŽ M .., P x : VM . 2 Ž .x 2 Ž .x w w Now, since T : E l Q 11 Z yy Q and W, O 2Z / 1, we have U, O 2 2 / 1, so wU, O Ž L.x / 1. We have CwU, O 2 Ž L.x Ž O Ž L.. s 1 because < O Ž L. Q: Q < is odd by Lemma 3.1 Ž2.. Therefore V 1Ž ZŽ R .. ZŽ L. because 1 / CwU, O 2 Ž L.x Ž R . : V 1Ž ZŽ R ... LEMMA 4.5. Let D : S and X g DŽ D .. Let N be a sol¨ able subgroup of G containing X, and let M be a normal subgroup of N.
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Ž1. If O 2 Ž X . M, then M l D 1 X. y Ž2. If M is maximal subset to the condition that O 2 Ž X . M 1 N, then y w O 2 Ž X ., O 2 Ž X .x : M l D. Ž3. Let Y g DŽ D . with Y : N. If O 2 Ž X . M = O 2 Ž Y ., then Y :2 X. Proof. Part Ž1. follows from Lemma 3.1 Ž1.. For the proof of Ž2., let Q be a preimage of a minimal normal subgroup of NrM. The maximality of M forces that O 2 Ž X . : Q, so < QrM < is odd by the solvability of N. Thus w O 2 Ž X ., O 2 Ž X .x s O 2 Ž X . l D : Q l D : M l D by Lemma 3.1 Ž7., so Ž2. holds. Part Ž3. holds because w O 2 Ž Y ., O 2 Ž Y .x s O 2 Ž Y . l D : M l D : O 2 Ž X . by Lemma 3.1 Ž7. and Ž1.. LEMMA 4.6. Let D : S and D : DŽ D .. Suppose that D is a finite set and that ² D : is sol¨ able. Then there exists an element Y g D such that Y :2 X for all X g D. Proof. We will first show that if Z1 2 Z2 2 ??? 2 Zk for Z1 , Z2 , . . . , Zk g D, then Zk :2 Zi for 1 F i F k. Put N s ² D :. Take a normal subgroup M of N that is maximal subject to the condition O 2 Ž Zk . M. Then O 2 Ž Zi . M for 1 F i F k by repeated use of Lemma 4.5 Ž3.. So M l D : O 2 Ž Zi . for 1 F i F k by Lemma 4.5 Ž1.. On the other hand, w O 2 Ž Zk ., O 2 Ž Zk .x : M l D by Lemma 4.5 Ž2.. Suppose the lemma is false. If Z1 2 Z2 2 ??? 2 Z jy1 for Z1 , Z2 , . . . , Z jy1 g D, then we can pick Z j g D such that Z jy1 2 Z j and Z j / Zi for 1 F i F j. Hence, by induction, all elements of D are arranged in such a way that Z1 2 Z2 2 ??? 2 Zn . But then Zn :2 Zi for 1 F i F n, a contradiction. LEMMA 4.7. Let b s Ž X, D, Y . g G. Suppose either that b is not thin or that O 2 Ž b . / 1. Put DX s w O 2 Ž X ., O 2 Ž X .xŽ O 2 Ž X . l O 2 Ž Y ..w O 2 Ž Y ., O 2 Ž Y .x, and let X X g GX Ž DX . and Y X g GY Ž DX .. Then either X X :2 Y X or Y X :2 X X . Proof. Put D 0 s D, X 0 s X, Y0 s Y, D 1 s DX , X 1 s X X , Y1 s Y X . Inductively, put Dn s O 2 Ž X ny1 . , O 2 Ž X ny1 .
Ž O2 Ž X ny1 . l O2 Ž Yny1 . .
= O 2 Ž Yny1 . , O 2 Ž Yny1 . , and choose X n g GX ny 1Ž Dn . and Yn g GY ny 1Ž Dn .. Note that this is possible by Lemma 4.2 Ž1.. Define bn s Ž X n , Dn , Yn . for all nonnegative integers n. Then we have bn g G and bnq1 U bn for all n G 0 by Lemma 4.2 Ž4. and Definition 4.1 Ž3.Ž4c.. Since D is finite, we have Dm s Dmq1 for some nonnegative integer m. Suppose that O 2 Ž bm . s 1. Then bm is thin by
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Theorem 4. Hence repeated use of Theorem 5 forces that bm s bmy1 s ??? s b 1 s b 0 . This contradiction shows that O 2 Ž bm . / 1. Thus ² X m , Ym : is solvable, and so either X m :2 Ym or Ym :2 X m by Lemma 4.6. Let 2 F j F m, and suppose that X j :2 Yj . Then O 2 Ž Yj . = w O 2 Ž X j ., O 2 Ž X j .x s O 2 Ž X j . l Dj s O 2 Ž X j . l Djy2 by Lemma 3.1 Ž7.. Since O 2 Ž X jy2 . , O 2 Ž X j . : O 2 Ž X jy2 . , O 2 Ž X jy1 . s O 2 Ž X jy1 . , O 2 Ž X jy1 . : O 2 Ž X jy1 . by Lemma 4.2 Ž2., we have w O 2 Ž X jy2 ., O 2 Ž X j .x : w O 2 Ž X jy1 ., O 2 Ž X j .x s w O 2 Ž X j ., O 2 Ž X j .x : O 2 Ž X j ., also by Lemma 4.2 Ž2.. Thus O 2 Ž X jy1 . , O 2 Ž X j . s O 2 Ž X j . , O 2 Ž X j . : F O 2 Ž Yj .
w O 2 Ž X jy 2 . , O 2 Ž X jy 2 .xŽ O 2 Ž X jy 2 .lO 2 Ž Y jy 2 ..
: F O 2 Ž Yj .
D jy 1
O 2 Ž Yjy2 . , O 2 Ž Yjy2 . : O 2 Ž Yjy1 .
by Lemma 4.2 Ž5.. Hence we conclude that O 2 Ž X jy1 . , O 2 Ž X jy1 . : O 2 Ž X jy1 . , ² O 2 Ž X j . s ² O 2 Ž X jy1 . , O 2 Ž X j . s ² O2 Ž X j . , O 2 Ž X j .
D jy 1
:
D jy 1
D jy 1
:
: : O 2 Ž Yjy1 .
by Lemma 4.2 Ž6., so X jy1 :2 Yjy1. Similarly, Yj :2 X j implies Yjy1 :2 X jy1. Therefore, by induction on j, we have either X 1 :2 Y1 or Y1 :2 X 1. LEMMA 4.8. O 2 Ž b . / 1.
Let b s Ž X, D, Y . g G. If either X :2 Y or Y :2 X, then
Proof. By the symmetry between X and Y, we may assume that X :2 Y. Put DX s O 2 Ž Y .. Then GX Ž DX . / B by Lemma 4.2 Ž1., and D acts 4 on GX Ž DX . as DX 1 on GX Ž DX .. Then y D. Let X 1 , . . . , X n be a D-orbit 2 C X kŽ O 2 Ž X k .. : O 2 Ž X k . for 1 F k F n and O Ž X . s ² O 2 Ž X 1 ., . . . , O 2 Ž X n .: by Lemma 4.2 Ž3.Ž6.. Let 1 F k F n. We may assume that no nonidentity characteristic subgroup of DX is normal in X k because X s ² X k , D : by Definition 4.1 Ž4c.. Therefore, by the Corollary of w9x, X k is described as follows: X k s DX J Ž X k ., J Ž X k . s L k = D k = Ek , L k ( S4 = ??? = S4 , D k ( D 8 = ??? = D 8 , Ek is an elementary abelian 2-group. Put Vk s w O 2 Ž X k ., O 2 Ž X k .x. Then
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Vk s O 2 Ž X k . l DX : J Ž DX . by Lemma 3.1 Ž7. and the structure of X k . Put V s w O 2 Ž X ., O 2 Ž X .x. Then V s w O 2 Ž X ., ² O 2 Ž X k .<1 F k F n :x s ²w O 2 Ž X ., O 2 Ž X k .x<1 F k F n: s ²w O 2 Ž X k ., O 2 Ž X k .x<1 F k F n: s ² Vk <1 F k F n: : J Ž DX . by Lemma 4.2 Ž2.. We may assume that V ; J Ž DX .. Let A be a Hall 2X-subgroup of DX O 2 Ž Y ., and put W s ² V A :. Then W 1 y DA s Y. If w V, W x s 1, then w O 2 Ž X ., O 2 Ž X .x s V : W : CDX Ž V . : O 2 Ž X . by ² 2Ž . : ² : Lemma 3.1 Ž1., so W 1 y O X , Y s X, Y . a w x w Suppose that V, W / 1. Then we have Vi , Vj x / 1 for some i, j, and a g A. Let g s Ž X i , DX , X ja .. Then g g G by Lemma 4.2 Ž4.. Moreover, we have X i 2 X ja because J Ž DX . = Vi C J Ž DX .Ž Vj a . s O 2 Ž J Ž X ja .. s J Ž DX . l O 2 Ž X ja . by the structure of X j . Put E s J Ž DX .. Since Vi Vj a : E, there exist Z1 g GX iŽ E . and Z2 g GX ja Ž E . such that Ž Z1 , E, Z2 . g G and Z1 2 Z2 by Lemma 4.2 Ž1.Ž4. and Lemma 4.3. Let d s Ž Z1 , E, Z2 .. Since E U y Xi s ² Z1 , DX :, we have E U Z1 , so Z1 s ² E Z 1 : by Lemma 3.1 Ž1.. Hence y w O 2 Ž Z1 ., O 2 Ž Z1 .x : w O 2 Ž Z1 ., ² E Z 1 :x s ²w O 2 Ž Z1 ., E x Z 1 : : ²w E, E x Z 1 : : ²w V 1Ž ZŽ E .. Z 1 : : V 1Ž ZŽ O 2 Ž Z1 ... by the structure of E and Lemma 4.2 Ž3., so CE Žw O 2 Ž Z1 ., O 2 Ž Z1 .x. s O 2 Ž Z1 . by Lemma 3.1 Ž1.. Similarly, CE Žw O 2 Ž Z2 ., O 2 Ž Z2 .x. s O 2 Ž Z2 .. Thus Z1 2 Z2 implies Z2 2 Z1. Next, put EX s w O 2 Ž Z1 ., O 2 Ž Z1 .xŽ O 2 Ž Z1 . l O 2 Ž Z2 ..w O 2 Ž Z2 ., O 2 Ž Z2 .x. Then there exist Z1X g GZ 1Ž EX . and Z2X g GZ 2Ž EX . such that Ž Z1X , EX , Z2X . g G and Z1X 2 Z2X by Lemma 4.2 Ž1.Ž4. and Lemma 4.3. Since Z2 2 Z1 s ² Z1X , E :, we have EX y U Z1X . Similarly, EX y U Z2X . Thus, as above, we have CEX Žw O 2 Ž ZiX ., O 2 Ž ZiX .x. s O 2 Ž ZiX . for i s 1, 2, and then Z1X 2 Z2X implies Z2X 2 Z1X . Now, Lemma 4.7 shows that d is thin and that O 2 Ž d . s 1. Since w O 2 Ž Z1 ., O 2 Ž Z1 .x : V 1Ž ZŽ O 2 Ž Z1 ..., Theorem 3 yields that d is isomorphic either to a GL3 Ž2.-amalgam or to an Sp4 Ž2.-amalgam, and then < E < F 2 4 . This forces < V < F 2 3, and so X is thin and O 2 Ž X . DX is 2-irreducible. Now, by Proposition 2.3 of w11x, some nonidentity A-invariant subgroup, say W X , of DX is normal in O 2 Ž X . DX . So ²W X D : is normal in both X and Y as X s O 2 Ž X . D and Y s DA.
5. THE PROOF OF THEOREM 1 We will begin the proof of Theorem 1. Assume that G and S continue to satisfy the assumption of Theorem 1. Let H and K be subgroups of G satisfying conditions Ža. ] Žd. stated in Theorem 1. Put a s Ž H, S, K .. LEMMA 5.1.
a g G, H 2 K, K 2 H.
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Proof. This follows from Lemma 4.8 and conditions Ža. ] Žd. of Theorem 1. By the symmetry between H and K, we will assume the following: ASSUMPTION 5.2.
V 1Ž ZŽ S .. ZŽ H ..
LEMMA 5.3. The group H has a noncentral minimal normal subgroup. Proof. Let W s ² V 1Ž ZŽ S .. H :. Then O 2 Ž H . : CH ŽW . W O 2 Ž H . by Assumption 5.2. Thus w W, O 2 Ž H .x / 1, and O 2 Ž H .CH ŽW .rCH ŽW . has odd order by Lemma 3.1 Ž1.Ž2., so CwW , O 2 Ž H .x Ž O 2 Ž H .. s 1. Therefore a minimal normal subgroup of H contained in w W, O 2 Ž H .x is noncentral. Let V be a noncentral minimal normal subgroup of H. Put T U s w O 2 Ž H ., O 2 Ž H .xŽ O 2 Ž H . l O 2 Ž K ..w O 2 Ž K ., O 2 Ž K .x, T s O 2 Ž H .w O 2 Ž K ., O 2 Ž K .x, and Dw k x s ² t g D < < V: C V Ž t .< F 2 k : for all subgroups D of S and all positive integers k. LEMMA 5.4. integers k: Ž1. Ž2. Ž3.
Let T U : D : S. Then the following holds for all positi¨ e
V g v k Ž D, D .. N `Ž k; D . : Dw k x. `Ž . Ž . Ž . If D 1 y S, then N k; D O 2 H l O 2 K .
Proof. Let k be a positive integer. We first note that V s w V, O 2 Ž H .x : w O 2 Ž H ., O 2 Ž H .x : T U : D. Let L be an abelian subgroup of D with w V, L, L x s 1. Put H s HrCH Ž V ., and regard V as a nontrivial irreducible faithful GF Ž2. H-module. Since w V, L, L x s w V, L, L x s 1, Lemma 3.4 shows that V s ² ¨ g V < < L: CLŽ ¨ .< F 2: s ² ¨ g V < < L: CLŽ ¨ .< F 2:. Hence V g v 1Ž D, D ., and then V g v k Ž D, D . by Lemma 2.2 Ž3.. Put Q s Dw k x. Then < V: C V Ž x .< G 2 kq 1 for all x g D y Q by the definition of Q, so we have N `Ž k; D . : Q by Lemma 2.6. So Ž1. and Ž2. hold. Suppose that D y 1 S. Since 1 / T : D, we have N `Ž k; D . / 1 by Lemma 2Ž . Ž . 2.4 1 . Let L s H or K. Then D l O 2 Ž L. 1 y SO L s L because w O 2 Ž L., O 2 Ž L.x : D l O 2 Ž L. 1 S. Therefore, if N `Ž k; D . : O 2 Ž L., then y Ž .Ž . Ž . N `Ž k; D . s N `Ž k; D l O 2 Ž L.. 1 y L by Lemma 2.4 4 2 . Hence 3 holds. The next lemma is a technical one that shows special relations between elements of DŽ D . where D : S. LEMMA 5.5. Let D : S and n ) 1. Take Z1 , . . . , Zn g GH Ž D . j GK Ž D ., and put N s ² Z1 , . . . , Zn :. Assume that Zn 2 Zny1 2 ??? 2 Z2 2 Z1 ,
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that N is sol¨ able, and that Z1 g GH Ž D .. Then the following hold: Ž1. There exists an abelian normal subgroup U of N such that U : D and < U: CU Ž x .< G 2 2 for all x g D y O 2 Ž Zn .. Ž2. If D 1 S, < S: D < F 2 4 , and Zn g GLŽ D . where L s H or K, then y `Ž N 1; D . : O 2 Ž L.. Proof. Take a subgroup M of N that is maximal subject to the ² : condition O 2 Ž Z1 . M y 1 N. Then M ; N, and M l D 1 y Z1 , . . . , Zn by Lemma 4.5 Ž1.Ž3.. Put P s O Ž N mod M .. Since O 2 Ž NrM . s 1 by the maximality of M, we have P l D s M l D : O 2 Ž N . : M ; P, so O 2 Ž Z1 . : P also by the maximality of M. Let I s ² O 2 Ž Z1 . N :, E s I l D, W s V 1Ž ZŽ E ... Then w O 2 Ž Z1 ., O 2 Ž Z1 .x s O 2 Ž Z1 . l D : E : D by Lemma 3.1 Ž7.. Since E s I l P l D s I l M l D and O 2 Ž Z1 . I l M, we have Ey 1 N by Lemma 4.5 Ž1.Ž3.. Hence w W, O 2 Ž Z1 .x / 1 by Lemma 4.4 because Z1 g GH Ž D . and V 1Ž ZŽ S .. ZŽ H .. Thus 1 / w W, I x y 1 N. Next, we will show that if < S: D < F 2 4 , then CwW , I xŽ I . s 1. Take a subgroup R of N that is maximal subject to the conditions R 1 yN and R l D g Syl 2 Ž R .. Since O Ž NrR. s 1 by the maximality of R, CN r R Ž O 2 Ž NrR.. : O 2 Ž NrR. by the solvability of NrR. Let Q s O 2 Ž N mod R .. Suppose that O 2 Ž Z1 . Q. Then Q l D 1 y N by Lemma ŽŽ Q l D . R .. Hence Q l 4.5 Ž1.Ž3., so Ž Q l D . R 1 N and Q l D g Syl 2 y D : R by the maximality of R. Let W0 be a chief factor of N within QrR so that O 2 Ž Z1 . C s CN ŽW0 .. Let SX be a subgroup between Ž N l S . R and N so that SXrR g Syl 2 Ž NrR.. Then < SX
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holds. Suppose that < S: D < F 2 4 . Then Lemma 3.4 implies U g v 1Ž D, D ., as in the proof of Lemma 5.4 Ž1., so N `Ž1; D . : O 2 Ž Zn . by Lemma 2.6. `Ž Ž . . Hence, if D 1 y S and Zn g GL D where L s H or K, we have N 1; D S : FO 2 Ž Zn . : O 2 Ž L. by Lemma 4.2 Ž5., so Ž2. holds. In the remainder of the proof, we will assume that a is not thin and derive a contradiction. LEMMA 5.6. Let X g GH ŽT .. Then there exists Y g GK ŽT . such that X :2 Y 2 X and ² X, Y : is sol¨ able. Proof. Let X g GH ŽT .. Then there exists Y g GK ŽT . such that Y 2 X by Lemma 4.3 Ž2.. Put X U s O 2 Ž X .T U . Then T U g Syl 2 Ž X U . because O 2 Ž X . l T s w O 2 Ž X ., O 2 Ž X .x s w O 2 Ž H ., O 2 Ž X .x : w O 2 Ž H ., O 2 Ž H .x : T U by Lemma 3.1 Ž7. and Lemma 4.2 Ž2.. Moreover, as T s O 2 Ž H .T U , there is a one-to-one correspondence between the set of subgroups of X containing T and the set of subgroups of X U containing T U . Thus X U is 2-irreducible, and then X U g GX ŽT U .. Hence there exists Y U g GY ŽT U . such that Y U 2 X U as above, because w O 2 Ž Y ., O 2 Ž Y .x s w O 2 Ž K ., O 2 Ž Y .x : w O 2 Ž K ., O 2 Ž K .x : T U . Since X U g GH ŽT U . and Y U g GK ŽT U ., we have X U :2 Y U by Lemma 4.7. Thus w O 2 Ž X ., O 2 Ž X .x s O 2 Ž X . l T s O 2 Ž X U . l T U s w O 2 Ž X U ., O 2 Ž X U .x : O 2 Ž Y U . by Lemma 3.1 Ž7., and so w O 2 Ž X ., O 2 Ž X .x : FO 2 Ž Y U .T : O 2 Ž Y . by Lemma 4.2 Ž5.. Hence we have X : 2 Y. Thus O 2 Ž² X, Y .: / 1 by Lemma 4.8, and then ² X, Y : is solvable. LEMMA 5.7. V : O 2 Ž K .. Proof. By Lemma 5.6, we can choose X g GH ŽT . and Y g GK ŽT . so that Y 2 X and ² X, Y : is solvable. Then, by Lemma 5.5 Ž1., there exists an abelian normal subgroup U of ² X, Y : such that U : T and < U: CU Ž x .< G 2 2 for all x g T y O 2 Ž Y .. Since V g v 1ŽT, T . by Lemma 5.4 Ž1., and w V, U, U x : wU, U x s 1, we have V s ² ¨ g V < < U: CU Ž ¨ .< F 2: : O 2 Ž Y ., so V : FO 2 Ž Y . S : O 2 Ž K . by Lemma 4.2 Ž5.. LEMMA 5.8.
V 1Ž ZŽ S .. : ZŽ K ..
Proof. Suppose false. Then, as in Lemma 5.3, K has a noncentral minimal normal subgroup, say, W. Furthermore, W : O 2 Ž H . by Lemma 5.7 and the symmetry between H and K. Now, we will assume that Ž Ž .. QŽ K Ž S .. U y H, where Q K S is a characteristic subgroup of S defined in w10x. Then H has a unique noncentral chief factor within O 2 Ž H . as in the proof of Proposition 2.2 of w11x. Thus we have w O 2 Ž H ., O 2 Ž H .x s V. Hence H :2 K by Lemma 5.7, a contradiction. Let Y g GK ŽT .. Define U s ² V Y :, Z s FV Y, Tqs O 2 Ž H .w O 2 Ž Y ., O Y .x, and Yqs O 2 Ž Y .Tq. 2Ž
SOLVABLE
LEMMA 5.9.
2-LOCAL
SUBGROUPS
381
If T w1x : O 2 Ž H ., then the following hold:
Ž1. U is normal abelian 2-subgroup of Y contained in Tq. Ž2. w Z, Tqx s 1. Ž3.
Yq is 2-irreducible with Tqg Syl 2 Ž Yq. .
Ž4.
Tqs CT q Ž V .w O 2 Ž Yq. , O 2 Ž Yq.x .
Proof. First, U is a normal 2-subgroup of Y by Lemma 5.7, so U : O 2 Ž Y . : T. Let y g Y. Then V y g v 1ŽT y , T y . by Lemma 5.4 Ž1.. Since w V y , V, V x : w V, V x s 1, we have V y s ² w g V y < < V: C V Ž w .< F 2: : T w1x : O 2 Ž H . : Tq. Thus U : Tq and U is abelian, so Ž1. holds. Since H 2 K s ² S, Y : by Lemma 5.1 and Definition 4.1 Ž4c., and T s O 2 Ž H .w O 2 Ž K ., O 2 Ž K .x s CT Ž V . O 2 Ž Y . s CT Ž Z . O 2 Ž Y ., we have T / O 2 Ž Y ., and then Y s ²T Y : s ² CT Ž Z . Y : O 2 Ž Y . by Lemma 3.1 Ž1.. Thus w O 2 Ž Y ., O 2 Ž Y .x : O 2 Ž Y . : ² CT Ž Z . Y :. Hence Tq: CT Ž Z ., so Ž2. holds. Since T s Tqw O 2 Ž K ., O 2 Ž K .x and ww O 2 Ž K ., O 2 Ž K .x, O 2 Ž Y .x s w O 2 Ž Y ., 2 Ž .x O Y s T l O 2 Ž Y . : Tq by Lemma 4.2 Ž2. and Lemma 3.1 Ž7., we conclude that Ž3. holds and that w O 2 Ž Y ., O 2 Ž Y .x s w O 2 Ž Yq. , O 2 Ž Yq.x as in the proof of Lemma 5.6. So Ž4. also holds. LEMMA 5.10.
If T w1x : O 2 Ž H ., then U g v 2 ŽTq, Tq ..
Proof. Suppose first that w V, Tqx Z. Let V Ž1. s C V mod Z ŽTq. , V Ž2. s q C V mod V Ž1. ŽTq. , and U Ž i. s ²Ž V Ž i. . Y : for i s 1, 2. Then Z ; V Ž1. ; V Ž2. Ž2. by our assumption, and so wU , w O 2 Ž Yq. , O 2 Ž Yq.xx Z. By Lemma 5.9 Ž4.. Suppose that wU, L, L x s 1 for an abelian subgroup L of Tq. Let LX s L l O 2 Ž Yq. . Note that Yq is a 2-irreducible solvable group acting on an abelian 2-group U Ž2. by Lemma 5.9 Ž3.Ž1.. Thus < L: LX < F 2 by Ž3.9. of w7x. Since V y g Tql v 1ŽT y , T y . : v 1ŽTq, O 2 Ž Yq.. for all y g Yq by Lemma 2.2 Ž1.Ž2., we have U g v 1ŽTq, O 2 Ž Yq.. by Ž2.2. of w7x, so U s ² u g U < < LX : CLX Ž u.< F 2: s ² u g U < < L: CLŽ u.< F 2 2 :. Thus U g v 2 ŽTq, Tq . by Lemma 5.9 Ž1.. 4 Suppose next that w V, Tqx : Z. Let GX s X g GH ŽTq. < TqU y X q, and q. Ž Ž .. Ž ² < Ž define VX s C V O 2 X for X g GH T . Then V s VX X g GH T .: s ² VX < X g GX : because FX g G ŽT q . w V U , O 2 Ž X .x s w V U , FX g G ŽT q . O 2 Ž X .x H H : w V U , O 2 Ž H .x s 1 for the dual module V U of V. q Let X g GX , and put UX s ² VXY : . Then it suffices to prove that UX g v 2 ŽTq, Tq . by Lemma 5.9 Ž1. and Ž2.2. of w7x because U s ²UX < X g GX :. Since w V, ŽTq. 2 x : w V, Tq, Tq x s 1, and CT Ž V . s O 2 Ž H . by Ž3.1.Ž1., TqrO 2 Ž H . is elementary abelian. Thus < TqrO2 Ž X .< s 2 by Lemma 3.1 Ž4. because O 2 Ž H . : O 2 Ž X . ; Tq. Choose a family of subgroups VXŽ i. < i g I 4
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of VX so that the following hold for all i g I Ž1. Ž2. Ž3.
VXŽ i. is a Tq-invariant subgroup of order 2 or 2 2 . VXŽ i. Z. < VXŽ i. : Z XŽ i. < s 2 where Z XŽ i. s w VXŽ i., Tq x. q
Let i g I, and put UXŽ i. s ²Ž VXŽ i. . Y : . Then UXŽ i.rZ XŽ i. g v 1ŽTqrZ XŽ i., Tqr Z XŽ i. . by Ž3.10. of w7x because VXŽ i.rZ XŽ i. : V 1Ž ZŽTqrZ XŽ i. ... Thus UXŽ i. g v 2 ŽTq, Tq . by Ž2.3. of w7x because < Z XŽ i. < F 2. Moreover, we have Z g v 0 ŽTq, Tq . : v 2 ŽTq, Tq . by Lemma 5.9 Ž2. and Lemma 2.2 Ž3.. Therefore UX s ²UXŽ i. < i g I : Z g v 2 ŽTq, Tq . by Ž2.2. of w7x. LEMMA 5.11. The following hold: Ž1. T w2x O 2 Ž H .. Ž2. If T w1x : O 2 Ž H ., then T s T w2x. Proof. To prove Ž1. and Ž2., we may assume that T w1x : O 2 Ž H .. Then N `Ž1; T . : T w1x : O 2 Ž H . : Tq by Lemma 5.4 Ž2., so N `Ž1; Tq. s N `Ž1; T . : O 2 Ž H . by Lemma 2.4 Ž4.. Let min x g TyO 2 ŽY . < U: CU Ž x .< s 2 d, and let x g T y O 2 Ž Y . with x 2 g O 2 Ž Y . and < U: CU Ž x .< s 2 d. Since T s O 2 Ž H .w O 2 Ž K ., O 2 Ž K .x s CT Ž V . O 2 Ž Y ., there exists an element y g CT Ž V . such that y ' x Žmod O 2 Ž Y ... Let P be a Hall 2X-subgroup of Y. Take an element g g P y F Ž P . such that k s w g, y xy1 f F Ž P . O 2 Ž Y .. Then 2 d s < U: y1 y1 y1 y1 y1 y1 y1 y1 CU Ž x .< G < V g : V g l V g x < s < V g : V g l V g y < s < V g : V g l yy V g < s < V: V l V k <. Since y g CT Ž V ., we have w V l V k , ² y, y k :x s 1 so w V l V k , O 2 Ž Y ., ² y, y k :x s 1 s w² y, y k :, V l V k , O 2 Ž Y .x as V, V k 1 y O 2 Ž Y .. Hence w O 2 Ž Y ., ² y, y k :, V l V k x s 1 by the three-subgroup lemma, so w O 2 Ž Y ., ² y, y k :x : T w d x. Since ² y, y k : F Ž P . O 2 Ž Y ., we have O 2 Ž Y . s ²² y, y k :T :, so w O 2 Ž Y ., O 2 Ž Y .x : T w d x. Therefore w O 2 Ž K ., O 2 Ž K .x s w O 2 Ž K ., ² O 2 Ž Y . S :x s ²w O 2 Ž K ., O 2 Ž Y .x S : s ²w O 2 Ž Y ., O 2 Ž Y .x S : : ²T w d x S : s T w d x by Lemma 4.2 Ž6.Ž2., and so T s CT Ž V .w O 2 Ž K ., O 2 Ž K .x s T w d x. Suppose that d G 2. Let L g v 1ŽTq, U .. Since w L, U, U x : wU, U x s 1, we have L s ² x g L < < U: CU Ž x .< F 2:, so L : O 2 Ž Y .. Thus ² v 1ŽTq, U .: : Tql O 2 Ž Y . s O 2 Ž Yq. . Hence U g v U1 ŽTq,Tq . by Definition 2.1 Ž2. because V y g Tql v 1ŽT, O 2 Ž Y .. : v 1ŽTq, O 2 Ž Yq.. for all y g Yq, and U s ² V y < y g Yq: g v 2 ŽTq, Tq .. Now, suppose that d G 3. Then we have N `Ž1; Tq. : O 2 Ž Yq. : O 2 Ž Y . by Lemma 2.5, so N `Ž1; T . : FO 2 Ž Y . S : O 2 Ž K .. This contradicts Lemma 5.4 Ž3., and hence d F 2. Therefore Ž1. and Ž2. hold. Let H s HrCH Ž V ., and regard V as a nontrivial irreducible faithful GF Ž2. H-module. Put P s O 2, 2X Ž H .. Then we have P s O Ž H . s O 2 Ž H . by Lemma 3.1 Ž1.Ž2..
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LEMMA 5.12. The group T acts intransiti¨ ely on the set H of homogenous components of VP . Proof. First, assume that T w1x O 2 Ž H .. Let X g GH ŽT .. Since T w1x 1 y S, and FO 2 Ž X . S : O 2 Ž H . by Lemma 4.2 Ž5., there exists an element t g T y O 2 Ž X . such that < V: C V Ž t .< s 2. Since O 2 Ž X . C X Ž V . by Lemma 4.2 Ž6., we have C X Ž V . : F Ž O 2, 2X Ž X .mod O 2 Ž X .. by Lemma 3.1 Ž1.Ž2.Ž5., and so w O 2 Ž X ., t x C X Ž V .. Thus there exists a unique element W g H such that 1 / O 2 Ž X . , t : w P, t x s PW ( Z3 by Lemma 3.3, so X = PW , and P is an elementary abelian 3-group by Lemma 3.3. Hence CH Ž V . s F Ž P mod O 2 Ž H .. as above. Suppose that T is transitive on H . Then X = ² P W < W g H : s P s O 2 Ž H . , so O 2 Ž H . : O 2 Ž X . C H Ž V . s O 2 Ž X . O 2 Ž H . : O 2 Ž X .T s X. Thus O 2 Ž X . s O 2 Ž H .. By Lemma 5.6, there exists an element Y g GT Ž K . such that X :2 Y. Then w O 2 Ž H ., O 2 Ž H .x : O 2 Ž H . l T s O 2 Ž X . l T s w O 2 Ž X ., O 2 Ž X .x : O 2 Ž Y ., by Lemma 3.1 Ž7., and hence w O 2 Ž H ., O 2 Ž H .x : F O 2 Ž Y . S : O 2 Ž K . by Lemma 4.2 Ž5.. This shows that H :2 K, which contradicts Lemma 5.1. Next, assume that T w1x : O 2 Ž H .. Then N `Ž1; T . : T w1x : O 2 Ž H . by Lemma 5.4 Ž2.. Suppose that < H < s 1. Since T w2x O 2 Ž H . s CT Ž V . by Lemma 5.11 Ž1. and Lemma 3.1 Ž1., P ( Z3 , Z5 , or Esp27 by Lemma 3.3. Thus < S: T < F < S: O 2 Ž H .< s < S < F 2 4 because S is isomorphic to a subgroup of AutŽ P .. Suppose that < H < G 2 and that T is transitive on H . Since T s T w2x by Lemma 5.11 Ž2., T is generated by elements inducing transpositions on the set H . So < H < s 2, and < W < s 2 2 for all W g H . Hence < S: T < F < S O 2 Ž H .< s < S < F 2 3. We conclude that < S: T < F 2 4 if T is transitive on H . Now, by Lemma 5.6, choose X g GH ŽT . and Y g GK ŽT . so that Y :2 X and ² X, Y : is solvable. Then N `Ž1; T . : O 2 Ž K . by Lemma 5.5 Ž2., which contradicts Lemma 5.4 Ž3.. Let D be a maximal subgroup of S containing T so that D has precisely two orbits H1 and H2 on the set H . LEMMA 5.13. There exists Y g GK Ž D . such that ² Y, O 2 Ž H .: is sol¨ able. Proof. Put Vi s ² Hi : for i s 1, 2. Then there exist D-invariant subgroups Pi Ž i s 1, 2. of P containing CH Ž V . such that P s P1 P2, w Vi , Pix s Vi , w Vi , P3yix s 1 Ž i s 1, 2., and S interchanges P1 and P2 by Lemma 5.11 Ž1. and Lemma 3.3. Put Zi s DPi Ž i s 1, 2., and U s C V Ž D .. 2 Let Z g GK Ž D ., and put E s O 2 Ž Z .. Since U : V 1Ž ZŽ D .. : V 1Ž ZŽ E .. and w O 2 Ž Z ., O 2 Ž Z .x : E : D, Lemma 4.4 shows that wU, O 2 Ž Z .x : w V 1Ž ZŽ E .., O 2 Ž Z .x s 1 because V 1Ž ZŽ S .. : ZŽ K .. Thus wU, O 2 Ž K .x s wU, ² O 2 Ž Z .< Z g GK Ž D .:x s 1 by Lemma 4.2 Ž6.. Therefore ² Z1 , O 2 Ž K .: is solvable as it is contained in CG ŽU ..
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Since H s ² Z1 , S : and Z1 is generated by the set of its D-irreducible subgroups, there exists an element ZX g GH Ž D . such that ZX : Z1. Let D s ZX 4 j GK Ž D .. Then, by Lemma 4.6, there exists an element Y g D such that Y :2 X for all X g D. Suppose that Y s ZX . Then w O 2 Ž ZX ., O 2 Ž ZX .x : FX g G Ž D . O 2 Ž X . : O 2 Ž K . by Lemma 4.2 Ž5., so K w O 2 Ž H . , O 2 Ž H .x s w O 2 Ž H . , ² O 2 Ž Z X . S :x s ²w O 2 Ž H . , O 2 Ž Z X .x S : s ²w O 2 Ž ZX ., O 2 Ž ZX .x S : : O 2 Ž K . by Lemma 4.2 Ž6.Ž2.. This shows that H :2 K, a contradiction. Therefore Y g GK Ž D . and Y :2 ZX . Let EX s O 2 Ž ZX .. Since V1 : V 1Ž ZŽ EX .. and w O 2 Ž Y ., O 2 Ž Y .x : EX , Lemma 4.4 shows that w V1 , O 2 Ž Y .x : w V 1Ž ZŽ EX .., O 2 Ž Y .x s 1 because V 1Ž ZŽ S .. : ZŽ K .. Thus ² O 2 Ž H ., D, O 2 Ž Y .: is solvable as it is contained in NG Ž V1 .. Now, we reach a final contradiction. Choose Y g GK Ž D . so that ² Y, O 2 Ž H .: is solvable. Then there exist elements X, Z g GH Ž D . such that Z 2 Y 2 X by Lemma 4.3. Put N s ² X, Y, Z :. Since N is solvable and < S: D < s 2, Lemma 5.5 Ž2. shows that N `Ž1; D . : O 2 Ž H . l O 2 Ž K ., which contradicts Lemma 5.4 Ž3.. This completes the proof of Theorem 1.
REFERENCES 1. Z. Janko, Nonsolvable finite groups all of whose 2-local subgroups are solvable, J. Algebra 21 Ž1972., 458]517. 2. F. Smith, Finite simple groups all of whose 2-local subgroups are solvable, J. Algebra 34 Ž1975., 481]520. 3. D. Gorenstein and R. Lyons, Nonsolvable finite groups with solvable 2-local subgroups, J. Algebra 38 Ž1976., 453]522. 4. K. Gomi and M. Hayashi, A pushing-up approach to the quasithin simple finite groups with solvable 2-local subgroups, J. Algebra 146 Ž1992., 412]426. 5. K. Gomi and Y. Tanaka, On pairs of groups having a common 2-subgroup of odd indices I, Sci. Papers College Arts Sci. Uni¨ . Tokyo 35 Ž1985., 11]30. 6. K. Gomi and Y. Tanaka, On pairs of groups having a common 2-subgroup of odd indices II, Sci. Papers College Arts Sci. Uni¨ . Tokyo 40 Ž1990., 37]47. 7. M. Hayashi and Y. Tanaka, Amalgams of solvable groups, J. Fac. Sci. Uni¨ . Tokyo Sect. IA Math. 39 Ž1992., 309]338. 8. K. Gomi, Sylow 2-intersection, 2-fusion, and 2-factorizations in finite groups of characteristic 2-type, J. Math. Soc. Japan 35 Ž1983., 571]588. 9. K. Gomi, A pushing up theorem for groups of characteristic 2-type, Sci. Papers College Arts Sci. Uni¨ . Tokyo 37 Ž1987., 73]102. 10. K. Gomi, Characteristic pairs for 2-groups, J. Algebra 94 Ž1985., 488]510. 11. Y. Tanaka, Amalgams of quasithin solvable groups, Japan J. Math. 17 Ž1991., 203]266.
JOURNAL OF ALGEBRA ARTICLE NO.
On the Finite Simple Groups All of Whose 2-Local Subgroups Are Solvable Makoto Hayashi Department of Mathematics, Aichi Uni¨ ersity of Education, Kariya 448, Japan
and Yasuhiko Tanaka Faculty of Engineering, Oita Uni¨ ersity, Oita 870-11, Japan Communicated by Walter Feit Received June 2, 1996
1. INTRODUCTION A finite group G is said to be of characteristic 2 type if each 2-local subgroup L of G satisfies the condition CLŽ O 2 Ž L.. : O 2 Ž L.. The purpose of this paper is to give an alternative proof of the following theorem, which was a result of the combined work of Janko w1x, Smith w2x, and Gorenstein and Lyons w3x. MAIN THEOREM. Let G be a nonabelian simple group of characteristic 2 type, all of whose 2-local subgroups are sol¨ able. Then G is isomorphic to one of the following groups: L2 Ž q ., Sz Ž q ., U3 Ž q . Ž where q is a power of 2., L2 Ž p . Ž where p is a Fermat or Mersenne prime with p G 5., A 6 , L3 Ž3., M11 , U3 Ž3., 2 F4 Ž2.X . Our approach to the proof of the Main Theorem is called an ‘‘amalgam method.’’ Several results have already been obtained for our purpose, and we use some of them in the whole of this paper. Therefore we will clarify here what we will assume and what we should prove. Let X be a group, and let Y be a subgroup of X. By definition, X is Y-irreducible if X has a unique maximal subgroup containing Y. A finite group G is often said to be 2-irreducible if G is S-irreducible for a Sylow 365 0021-8693r98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.
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2-subgroup S of G. Furthermore, a finite group G of even order is said to be thin if m 2, p Ž G . F 1 for all odd primes p, where the 2-local p-rank m 2, p Ž G . of G is the maximum of the p-ranks m p Ž L. as L ranges over the 2-local subgroups of G. The task of this paper is stated as the following theorem. THEOREM 1. Let G be a finite group, all of whose 2-local subgroups are sol¨ able. Let S g Syl 2 Ž G ., and let H and K be subgroups of G satisfying the following conditions: Ža. Žb. Žc. Žd.
S is a common Sylow 2-subgroup of H and K. Both H and K are S-irreducible and sol¨ able. No nonidentity subgroup of S is normal both in H and in K. CH Ž O 2 Ž H .. : O 2 Ž H . and CK Ž O 2 Ž K .. : O 2 Ž K ..
Then both H and K are thin. The next two theorems are assumed as a starting point of our analysis of 2-local subgroups. THEOREM 2 ŽTheorems A, C of w4x.. Let G be a finite group of characteristic 2 type, all of whose 2-local subgroups are sol¨ able. Let S g Syl 2 Ž G ., and assume O 2 Ž G . s 1. Then one of the following holds: Ž1. G has a strongly embedded subgroup. Ž2. S is dihedral, semidihedral, or isomorphic to a Sylow 2-subgroup of the group AutŽ Sp4 Ž2... Ž3. There exist subgroups H and K of G satisfying the abo¨ e conditions Ža. ] Žd.. THEOREM 3 ŽMain Theorem of w5, 6x.. Let Ž H, S, K . be an amalgam satisfying the abo¨ e conditions Ža. ] Žd. under some embeddings of S into H and K. If both H and K are thin, then the amalgam Ž H, S, K . or Ž K, S, H . is isomorphic to a GL3 Ž2.-amalgam, an Sp4Ž2.-amalgam, a G 2 Ž2.X-amalgam, a G 2 Ž2.-amalgam, an M12-amalgam, an AutŽ M12 .-amalgam, a 2 F4Ž2.X-amalgam, or a 2 F4Ž2.-amalgam. Let G be a nonabelian simple group satisfying the assumption of the Main Theorem. Then G satisfies the assumption of Theorem 2. If Ž1. or Ž2. occurs in Theorem 2, we can appeal to some of the existing classification theorems to identify G with one of the known simple groups. Hence we can assume that Ž3. occurs in Theorem 2, and then G satisfies the assumption of Theorem 1. Once Theorem 1 is proved, we can know the structures of the subgroups S, H, K by Theorem 3, and then identify G with one of the known simple groups, appealing again to some classification theorems.
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Our approach to Theorem 1 pursues and extends the approach to Theorem B of w4x, which applies to an alternative proof of the classification of quasi-thin finite simple groups, all of whose 2-local subgroups are solvable. In the course of the proof of Theorem 1, the following two theorems are of particular importance. THEOREM 4 ŽTheorem C of w7x.. Let Ž H, S, K . be an amalgam satisfying the abo¨ e conditions Ža. ] Žd., and the following additional condition Že.: Že.
S s w O 2 Ž H ., O 2 Ž H .xŽ O 2 Ž H . l O 2 Ž K ..w O 2 Ž K ., O 2 Ž K .x.
Then both H and K are thin. THEOREM 5 ŽSection 4.8 of w4x.. Let G be a finite group, all of whose 2-local subgroups are sol¨ able. Let H, S, K, H X , SX , K X be subgroups of G so that both Ž H, S, K . and Ž H X , SX , K X . satisfy the abo¨ e conditions Ža., Žb., Žd.. Suppose that SX s w O 2 Ž H ., O 2 Ž H .xŽ O 2 Ž H . l O 2 Ž K ..w O 2 Ž K ., O 2 Ž K .x, that H s ² S, H X :, and that K s ² S, K X :. If Ž H X , SX , K X . satisfies condition Žc., and both H X and K X are thin, then Ž H, S, K . s Ž H X , SX , K X .. Following w4x, we will write Ž H X , SX , K X . U Ž H, S, K . if the assumptions of Theorem 5 are fulfilled. Let G, S, H, K be groups satisfying the assumption of Theorem 1. Then we can actually choose a sequence Ž H0 , S0 , K 0 ., Ž H1 , S1 , K 1 ., . . . , Ž Hn , Sn , K n ., . . . of amalgams satisfying conditions Ža., Žb., Žd. such that Ž H0 , S0 , K 0 . s Ž H, S, K . and Ž Hnq1 , Snq1 , K nq 1 . U Ž Hn , Sn , K n . for all nonnegative integers n. Since S is finite, we have Sm s Smq1 for some nonnegative integer m. If no nonidentity subgroup of S m is normal in both Hm and K m , then repeated use of Theorem 5 forces that both H and K are thin. Thus we will assume in our analysis that some nonidentity subgroups of Sm is normal both in Hm and in K m , in particular that ² Hm , K m : is a solvable subgroup of G. In fact, most of the arguments of this paper are devoted to the study of the solvable groups ² Z1 , . . . , Zn : of G generated by D-irreducible subgroups Z1 , . . . , Zn with D : S and D g Syl 2 Ž Zk . for k s 1, . . . , n. Looking at our proof, we can understand that it was sufficient for us to classify the amalgams satisfying not only conditions Ža. ] Žd. but also the additional condition Že.. Related to quadratic 2-subgroups of 2-local subgroups, we will define new characteristic subgroups N `Ž n; B . for each finite 2-group B and each nonnegative integer n. We will frequently use them throughout the proof of Theorem 1 in the following manner: for a 2-local subgroup L containing Ž . a Sylow 2-subgroup S, either N `Ž n; S . 1 y L or the structure of LrO2 L is highly restricted. Precise definitions and necessary properties are given in Section 2. Section 3 contains necessary information about GF Ž2.-represen-
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tation of solvable groups, and Sections 4 and 5 are devoted to the body of the proof of Theorem 1.
2. ON FINITE 2-GROUPS Throughout this section, we will assume that B is an arbitrary finite 2-group, and that n is a nonnegative integer. DEFINITION 2.1. Let X and Y be subgroups of B. Ž1. We denote by v nŽ X, Y . the set of Y-invariant subgroups V of X possessing the following property: if w V, L, L x s 1 for an abelian subgroup L of Y, then there exists a family of generators ¨l < l g L4 of V Ždepending on L. such that < L: CLŽ ¨l .< F 2 n for all l g L. Ž2. We denote by v nU Ž X, Y . the set of elements V of v nq1Ž X, Y . possessing the following property: there exist a family of subgroups Vi < i g I 4 of V and a family of subgroups Yi < i g I 4 of Y such that Ža. V s ² Vi < i g I :, Žb. Vi g v nŽ X, Yi . for all i g I, Žc. ² v nŽ Y, V .: : Yi for all i g I. Ž3. We use the following convention: if Z is a subgroup of B and B is a set of subgroups of B, then Z l B denotes the set of elements of B contained in Z. LEMMA 2.2.
Let X, X X , Y, Y X : B with X : X X and Y : Y X .
X l v nŽ X X , Y . : v nŽ X, Y . : v nŽ X X , Y . and X l v nU Ž X X , Y . : . : v nU Ž X X , Y .. Ž2. v nŽ X, Y . = v nŽ X, Y X . and v nU Ž X, Y . = v nU Ž X, Y X .. Ž3. ZŽ C X Ž Y .. g v nŽ X, Y . : v nU Ž X, Y . : v nq1Ž X, Y .. Ž1.
v nU Ž X, Y
Proof. This follows immediately from Definition 2.1. DEFINITION 2.3. Define a sequence of subgroups Nk Ž n; B .4`ksy1 of B recursively as follows: Ž1. Ž2.
Ny1Ž n; B . s B. Nk Ž n; B . s ² v nU Ž B, Nky1Ž n; B ..:Ž k G 0..
Now, define N `Ž n; B . s Fk: o d d Nk Ž n; B . and N`Ž n; B . s ² Nk Ž n; B .< k: e¨ en:. For brevity, we put N ` s N `Ž n; B . and N` s N`Ž n; B . throughout this section.
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LEMMA 2.4. The following holds: Ž1. Ž2.
ZŽ B . : N` : N ` . In particular, N ` / 1 whene¨ er B / 1.
N ` is a characteristic subgroup of B. Ž3. ² v nU Ž B, N ` .: s N` and ² v nU Ž B, N` .: s N ` . Ž4. If N ` : Q : B, then N `Ž n; Q . s N ` , and N ` is a characteristic subgroup of Q. Proof. First, Nk Ž n; B . is a characteristic subgroup of B for all k because AutŽ B . leaves the set v nU Ž B, Nky1Ž n; B .. invariant by induction on k. So Ž2. holds. We will see the inclusion relations between the subgroups Nk Ž n; B .. Let Nk s Nk Ž n; B . for all k. Since Ny1 s B = N0 , we have Ny1 s B = N1 s ² v nU Ž B, N0 .: = ² v nU Ž B, Ny1 .: s N0 by Lemma 2.2 Ž2.. Assume that either Niy1 = Niq1 = Ni or Niy1 : Niq1 : Ni holds for some i. Then either Ni : Niq2 : Niq1 or Ni = Niq2 = Niq1 holds by Lemma 2.2 Ž2. because Nkq 1 s ² v nU Ž B, Nk .: for all k. Thus we can conclude, by induction, that B s Ny1 = N1 = ??? = N2 iy1 = ??? = N2 i = ??? = N2 = N0 . Since < B < is finite, there exists a nonnegative integer m such that N` s N2 i and N ` s N2 iy1 for all i G m. Hence, together with Lemma 2.2 Ž3., ZŽ B . : N0 : N` s N2 m : N2 my1 s N ` , ² v nU Ž B, N ` .: s ² v nU Ž B, N2 my1 .: s N2 m s N` , ² v nU Ž B, N` .: s ² v nU Ž B, N2 m .: s N2 mq1 s N ` . So Ž1. and Ž3. hold. Suppose that N ` : Q : B. Put Mk s Nk Ž n; Q .. Then Ny1 = My1 = N ` by our assumption. Assume that N2 iy1 = M2 iy1 = N `. Then N2 i s ² v nU Ž B, N2 iy1 .: : ² v nU Ž B, M2 iy1 .: : ² v nU Ž B, N ` .: s N` : N ` : Q by Lemma 2.2 Ž2. and Ž1.Ž3.. Thus ² v nU Ž B, M2 iy1 .: s ² v nU Ž Q, M2 iy1 .: s M2 i by Lemma 2.2 Ž1., so N2 i : M2 i : N` . Again by Lemma 2.2 Ž2. and Ž1.Ž3., N2 iq1 s ² v nU Ž B, N2 i .: = ² v nU Ž B, M2 i .: = ² v nU Ž Q, M2 i .: s M2 iq1 and ² v nU Ž B, N` .: s N ` = Q, so M2 iq1 = ² v nU Ž Q, N` .: s ² v nU Ž B, N` .: s N ` by Lemma 2.2 Ž1.. Thus we can conclude, by induction, that N2 iy1 = M2 iy1 = N ` for all i, so N `Ž n; Q . s N ` . This proves Ž4.. LEMMA 2.5. Let V g v nU Ž B, B ., and let Q : B. If < V: C V Ž x .< G 2 nq 2 for all x g B y Q, then N ` : Q. Proof. Since V : ² v nU Ž B, B .: : ² v nU Ž B, N ` .: s N` by Lemma 2.2 Ž2. and Lemma 2.4 Ž3., we have N ` s ² v nU Ž B, N` .: : ² v nU Ž B, V .: : ² v nq 1Ž B, V .: by Lemma 2.4 Ž3. and Lemma 2.2 Ž2.Ž3.. Let L g v nq 1Ž B, V .. Since w L, V, V x : w V, V x s 1, Definition 2.1 Ž1. shows that L is generated by the elements x with < V: C V Ž x .< F 2 nq 1 , so L : Q by our assumption. Therefore N ` : Q. LEMMA 2.6. Let V g v nŽ B, B ., and let Q : B. If < V: C V Ž x .< G 2 nq 1 for all x g B y Q, then N ` : Q.
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Proof. We have N ` : ² v nU Ž B, V .: by the same argument as in the first paragraph of the proof of Lemma 2.5 because V g v nU Ž B, B . by Lemma 2.2 Ž3.. Let L g v nU Ž B, V .. Let L i < i g I 4 and Vi < i g I 4 be families of subgroups of L and V, respectively, satisfying the conditions of Definition 2.1 Ž2.. Let i g I. Since V : ² v nŽ V, B .: : ² v nŽ V, L.: by Lemma 2.2 Ž1.Ž2., we have V : Vi by Definition 2.1 Ž2c., so L i g v nŽ B, V . by Definition 2.1 Ž2b.. Since w L i , V, V x : w V, V x s 1, Definition 2.1 Ž1. shows that L i s ² x g L i < < V: C V Ž x .< F 2 n : : Q, so L : Q by Definition 2.1 Ž2a.. Therefore N ` : Q.
3. ON GF Ž2.-REPRESENTATION OF SOLVABLE GROUPS In this section, we study GF Ž2.-representation of solvable groups. First, we review the basic structure of 2-irreducible solvable groups. LEMMA 3.1. Let G be a 2-irreducible sol¨ able group with S g Syl 2 Ž G ., and let M be the unique maximal subgroup of G containing S. Put Q s O 2 Ž G .. Then the following holds: Ž1. If N 1 G / SN Ž or, equi¨ alently, O 2 Ž G . N 1 G ., then S l N y y 1 G, Q g Syl 2 Ž QN ., and GrN is 2-irreducible and sol¨ able. y Ž2. G s O 2, p, 2 Ž G . for some odd prime p. Let P g Syl p Ž G .. Ž3. SrQ acts irreducibly and faithfully on PQrFŽ P . Q. Ž4. ZŽ SrQ . is cyclic, and an element t g S in¨ erts PQrFŽ P . Q if and only if Q / tQ g V 1Ž ZŽ SrQ ... Ž5. P l M s F Ž P .. Ž6. Ž7.
Let x g S y Q. Then x f M if and only if G s ² S, x :. O 2 Ž G . s P w Q, O 2 Ž G .x and w Q, O 2 Ž G .x s S l O 2 Ž G ..
Proof. Parts Ž1. ] Ž6. are proved in Ž2.1. of w11x. Since w Q, P x : w Q, 2Ž . 2 Ž .x Ž . w O G .x, we have P w Q, O 2 Ž G .x 1 y PQ s O 2, p G , so O G s P Q, O G . 2 2 2 Thus w Q, O Ž G .x s Q l O Ž G . s S l O Ž G ., so Ž7. holds. 2Ž
Throughout the remainder of this section, we will assume that G is a finite solvable group and that V is a nontrivial irreducible faithful GF Ž2.G-module. Let S g Syl 2 Ž G ., and put Sw k x s ² x g S < < V: C V Ž x .< F 2 k : for all positive integers k.
SOLVABLE
LEMMA 3.2.
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Sw1x : O 2X , 2 Ž G ..
Proof. We may assume that Sw1x / 1. Then S w1x : ² P Ž G, V .:, where P Ž G, V . is the set of nonidentity elementary abelian subgroups A of G satisfying < A < < C V Ž A.< G < B < < C V Ž B .< for all subgroups B of A. We can see ² P Ž G, V .: : O 2X , 2 Ž G . by Ž3.2. and Ž3.3. of w8x because G is solvable and O 2 Ž G . : CG Ž V . s 1. LEMMA 3.3. Suppose that G is 2-irreducible and that S w2x / 1. Let H be the set of the homogeneous components of VP , where P s O Ž G .. Then there exists a normal subgroup PW for each W g H such that P s ² PW < W g H :, W s w V, PW x, and one of the following holds: Ž1. Ž2. Ž3. Ž4.
PW ( Z3 , < W < s 2 2 , and Sw1x / 1. PW ( Z3 , < W < s 2 2 , and Sw1x s 1. PW ( Z5 , < W < s 2 4 , and Sw1x s 1. PW ( Esp27 , < W < s 2 6 , and Sw1x s 1.
Proof. We first note that P g Syl p Ž G . for some odd prime p by Lemma 3.1 Ž2. because O 2 Ž G . s 1. Let t g S be an involution with < V: C V Ž t .< F 2 2 , and put T s t S . Let u g T , and define Pu s w P, u x and Vu s w V, Pu x. Then Pu y 1 P for all u g T. Since < V 1Ž ZŽ S ..< s 2 by Lemma 3.1 Ž4. and ² T : y 1 S, we have V 1Ž ZŽ S .. : ² T :, and then P s ² Pu < u g T :, because V 1Ž ZŽ S .. inverts PrFŽ P . also by Lemma 3.1 Ž4.. Hence V s Ý ug T Vu . Let I be a subset of T that is maximal subject to the following condition: if u / uX , then Vu / VuX for all u, uX g I . Then P s ² Pu < u g I :, and V s Ý ug I Vu . Let u g I . Pick an irreducible submodule X of VP contained in Vu . Then X s w X, Pu x as Pu 1 y P. So, if W g H contains X, then W s w W, Pu x : w V, Pu x s Vu . This shows that Vu is a direct sum of some members of H as VP is completely reducible. Suppose that Vu is homogeneous itself. Then the set Vu < u g I 4 coincides with the set H . For W g H , choose the unique element u g I so that W s Vu , and define PW s Pu . Then P s ² PW < W g H : and W s w V, PW x for each W g H . The structures of W and PW are described in Lemma 2.8 of w11x because < W: CW Ž u.< F < V: C V Ž u.< F 2 2 . In particular, ZŽ PW . is cyclic as W is homogeneous and CPW ŽW . : CP Ž V . s 1. Since G is irreducible on V, NG ŽW . is also irreducible on W, so the case < W < s 2 4 and PW ( Z3 is eliminated. Hence one of the three cases Ž1., Ž3., Ž4. occurs. Suppose that Vu is not homogeneous. For W g H , choose u g I so that W : Vu . Note that u does not centralize any member of H contained in Vu as C V uŽ Pu . s 0. Thus there exists W X g H such that Vu s W q W X , where < W < s < W X < s 2 2 , because < Vu : C V Ž u.< F < V: C V Ž u.< F 2 2 . Hence one of the u two cases Ž1., Ž2. occurs, and, moreover, PrCP ŽW . ( PrCP ŽW X . ( Z3 , CP Ž Vu . s CP ŽW . l CP ŽW X ., and u inverts PrCP Ž Vu .. Since Vu is not
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homogeneous, Lemma 2.8 of w11x shows that Pu ( E9 , and so P s Pu CP Ž Vu .. w x Define PW s Pu l CP ŽW X .. Then P 2 y PW ( Z3 and V, PW s W. Since CP Ž V . s 1 and < W < s 2 2 , we can conclude that PW is a unique subgroup of P with the property W s w V, PW x, and so, in particular, PW is independent on the choice of u g I . Thus Pu s PW PW X , and hence P s ² PW < W g H :. LEMMA 3.4. F 2:.
Let L : S with w V, L, L x s 0. Then V s ² ¨ g V < < L: CLŽ ¨ .<
Proof. Since O 2 Ž G . s 1, there exists an odd prime p such that w Op Ž G ., L x / 1. Put P s Op Ž G .. Let W be a homogeneous component of VP . Put N s NG ŽW ., C s CG ŽW ., and N s NrC. If L N, then < L: C l L < F 2 by Ž3.2. of w7x. Suppose that L : N. If W s V, then < L: C l L < F 2 by Ž3.3. of w7x because w V, L, L x s 0 / w V, w N, L xx for a GF Ž2. PL-module V. So suppose that W ; V. Then W s ² w g W < < L: CLŽ w .< F 2: by the inductive hypothesis, because N is irreducible on W and w W, L, L x : w V, L, L x s 0. Hence the lemma holds because V is a direct sum of the homogeneous components of VP .
4. PRELIMINARIES Throughout the remainder of this paper, let G be a finite group satisfying the assumption of Theorem 1, and let S be a Sylow 2-subgroup of G. We begin with some definitions. DEFINITION 4.1. Let D : S. Ž1. We denote by DŽ D . the set of subgroups Z of G satisfying the following conditions: Ža. D g Syl 2 Ž Z .. Žb. Z is D-irreducible and solvable. Žc. CZ Ž O 2 Ž Z .. : O 2 Ž Z .. Ž2. Let D : S and X, Y g DŽ D .. We write X :2 Y and say that X is 2-embedded into Y if w O 2 Ž X ., O 2 Ž X .x : O 2 Ž Y .. Ž3. We denote by G the set of triplets Ž X, D, Y . with the conditions D : S and X, Y g DŽ D .. Let b s Ž X, D, Y . g G. Define O 2 Ž b . s O 2 Ž² X, Y :. l D. Note that O 2 Ž b . is the largest subgroup of D that is normal in both X and Y. We call b thin if both X and Y are thin. Let b X s Ž X X , DX , Y X . g G. We write b X U b if DX s w O 2 Ž X ., O 2 Ž X .xŽ O 2 Ž X . l O 2 Ž Y ..w O 2 Ž Y ., O 2 Ž Y .x, X s ² X X , D :, and Y s ² Y X , D :.
SOLVABLE
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Ž4. Let L be a subgroup of G with R s L l S g Syl 2 Ž L.. Let D : R. We denote by GLŽ D . the set of subgroups Z of L satisfying the following conditions: Ža. D g Syl 2 Ž Z .. Žb. Z is D-irreducible. Žc. ² O 2 Ž Z ., R : s L. LEMMA 4.2. Ž1.
Let D : R : S and L g DŽ R . with w O 2 Ž L., O 2 Ž L.x : D.
GLŽ D . / B.
Let Z g GLŽ D .. Ž2. w O 2 Ž Z ., O 2 Ž Z .x s w O 2 Ž L., O 2 Ž Z .x. Ž3. CZ Ž O 2 Ž Z .. : CZ Žw O 2 Ž Z ., O 2 Ž Z .x. : O 2 Ž Z .. Ž4. Z g DŽ D .. Ž5. FO 2 Ž Z . R : O 2 Ž L. = FX g G Ž D . O 2 Ž X .. L Ž6. ² O 2 Ž Z . R : s O 2 Ž L. s ² O 2 Ž X .< X g GLŽ D .:. Proof. We first note that D g Syl 2 Ž DO 2 Ž L.. as O 2 Ž L. l R s w O 2 Ž L., O 2 Ž L.x : D by Definition 4.1 Ž1a. and Lemma 3.1 Ž7., and that DO 2 Ž L. is generated by the set GX of D-irreducible subgroups X of L with D g Syl 2 Ž X .. Let M be the unique maximal subgroup of L containing R. Let X g GX . Then X g GLŽ D . if and only if X M by Lemma 3.1 Ž6.. Put LU s Lrw O 2 Ž L., O 2 Ž L.x. Then O 2 Ž L.U has odd order by Lemma 3.1 Ž7., so O 2 Ž L.U l M U s F Ž O 2 Ž L.U . by Lemma 3.1 Ž5.. Hence we have GLŽ D . s X g GX < O 2 Ž X .U F Ž O 2 Ž L.U .4 . Thus O 2 Ž L. s ² O 2 Ž X .< X g GX : s ² O 2 Ž X .< X g GLŽ D .:, so Ž1. holds. Moreover, we have L s ² O 2 Ž Z . R : R by Definition 4.1 Ž4c., so Ž6. holds. Let Q s w O 2 Ž Z ., O 2 Ž Z .x and Y s O 2 Ž CZ Ž Q ... Part Ž2. holds because Q s O 2 Ž Z . l D : O 2 Ž L. l R s w O 2 Ž L., O 2 Ž L.x by Lemma 3.1 Ž7.. Since w O 2 Ž L., Y x : w O 2 Ž L., O 2 Ž Z .x s Q by Ž2., we have Y : CLŽ O 2 Ž L.. : O 2 Ž L. by Definition 4.1 Ž1c.. Hence Y s 1, so Ž3. holds. Part Ž4. is a consequence of Ž3. and Definition 4.1 Ž4a.Ž4b.. Let L s LrO2 Ž L., and GY s Z R or GLŽ D .. Then O 2 Ž L . s O Ž L. by Definition 4.1 Ž1b. and Lemma 3.1 Ž2.. Since wO 2 Ž X . , O 2 Ž X . x : O 2 Ž X . lO Ž L . s 1 for all X g GY , we have wFX g G Y O 2 Ž X . , O Ž L .x s wFX g GY O 2 Ž X . , ²O 2 Ž X . < X g GY :x s 1 by Ž6.. Thus FX g GY O 2 Ž X . : CR Ž O Ž L.. s 1, so Ž5. holds. LEMMA 4.3. Let D : R : S and L, LX g D Ž R . with w O 2 Ž L ., O L.xw O 2 Ž LX ., O 2 Ž LX .x : D and L 2 LX . 2Ž
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Ž1.
If Y g GLŽ D ., then there exists X g GLX Ž D . such that Y 2 X.
Ž2.
If Y g GLX Ž D ., then there exists Z g GLŽ D . such that Z 2 Y.
Proof. We first note that B / GLŽ D . : DŽ D . = GLX Ž D . / B by Lemma 4.2 Ž1.Ž4.. Let Y g GLŽ D .. Suppose that Y :2 X for all X g GLX Ž D .. Then w O 2 Ž L., O 2 Ž Y .x s w O 2 Ž Y ., O 2 Ž Y .x : O 2 Ž X . for all X g GLX Ž D . by Lemma 4.2 Ž2.. Thus w O 2 Ž L., O 2 Ž Y .x : FX g GLX Ž D . O 2 Ž X . : O 2 Ž LX . by Lemma 4.2 Ž5.. Hence w O 2 Ž L., O 2 Ž L.x : w O 2 Ž L., ² O 2 Ž Y . R :x : O 2 Ž LX . by Lemma 4.2 Ž6., a contradiction. So Ž1. holds. Let Y g GLX Ž D .. Suppose that Z :2 Y for all Z g GLŽ D .. Then w O 2 Ž L., O 2 Ž Z .x s w O 2 Ž Z ., O 2 Ž Z .x : O 2 Ž Y . for all Z g GLŽ D . by Lemma 4.2 Ž2.. Thus w O 2 Ž L., O 2 Ž L.x : w O 2 Ž L., ² O 2 Ž Z .< Z g GLŽ D .:x : O 2 Ž Y . by Lemma 4.2 Ž6.. Hence w O 2 Ž L., O 2 Ž L.x : FO 2 Ž Y . R : O 2 Ž LX . by Lemma 4.2 Ž5., a contradiction. So Ž2. holds. LEMMA 4.4. Let E : D : R : S and L g DŽ R .. Suppose Z g GLŽ D . with w O 2 Ž Z ., O 2 Ž Z .x : E. Then w V 1Ž ZŽ E .., O 2 Ž Z .x s 1 if and only if V 1Ž ZŽ R .. : ZŽ L.. Proof. Put Q s O 2 Ž L., T s w O 2 Ž Z ., O 2 Ž Z .x, U s V 1Ž ZŽ Q .., W s V 1Ž ZŽ E lQ... Then w Q, O 2 Ž Z .x s T s O 2 Ž Z . l D : E l Q by Lemma 4.2 Ž2. and Lemma 3.1 Ž7., so wU, O 2 Ž Z .x : w W, O 2 Ž Z .x, and < O 2 Ž Z .Ž E l Q .: Ž E l Q .< is odd. Hence CwW , O 2 Ž Z .x Ž O 2 Ž Z .. s 1. Suppose first w V 1Ž ZŽ E .., O 2 Ž Z .x s 1. Then w W, O 2 Ž Z .x l V 1Ž ZŽ E .. s 1, so w W, O 2 Ž Z .x s 1. Hence wU, O 2 Ž Z .x s 1, and then V 1Ž ZŽ R .. : ZŽ L. by Definition 4.1 Ž4c. because V 1Ž ZŽ R .. : U by Lemma 4.2 Ž3.. Suppose next w V 1Ž ZŽ E .., O 2 Ž Z .x / 1. Since V 1Ž ZŽ E .. : CZ ŽT . : O 2 Ž Z . by Lemma 4.2 Ž3., we have w V 1Ž ZŽ E .., O 2 Ž Z .x : T : E l Q, so 1 / w V 1Ž ZŽ E .., O 2 Ž Z .x : w W, O 2 Ž Z .x. Here, let T : N 1 VN s w V 1Ž ZŽ N .., O 2 Ž Z .x / 1. y M : Q, and suppose 2 We will show that VM s w V 1Ž ZŽ M .., O Ž Z .x / 1. By Lemma 3.1 Ž2., Z is a 2, p4 -group for some odd prime p. Let P g Syl p Ž Z .. Then O 2 Ž Z . s PT by Lemma 3.1 Ž7., and M s NCM Ž P . because w M, P x : w Q, P x : T : N. Ž . w Ž . x Thus VN s w V 1Ž ZŽ N .., P x 1 y M and C V N P s 1. Hence 1 / C V N M , P : w V 1Ž ZŽ M .., P x : VM . 2 Ž .x 2 Ž .x w w Now, since T : E l Q 11 Z yy Q and W, O 2Z / 1, we have U, O 2 2 / 1, so wU, O Ž L.x / 1. We have CwU, O 2 Ž L.x Ž O Ž L.. s 1 because < O Ž L. Q: Q < is odd by Lemma 3.1 Ž2.. Therefore V 1Ž ZŽ R .. ZŽ L. because 1 / CwU, O 2 Ž L.x Ž R . : V 1Ž ZŽ R ... LEMMA 4.5. Let D : S and X g DŽ D .. Let N be a sol¨ able subgroup of G containing X, and let M be a normal subgroup of N.
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Ž1. If O 2 Ž X . M, then M l D 1 X. y Ž2. If M is maximal subset to the condition that O 2 Ž X . M 1 N, then y w O 2 Ž X ., O 2 Ž X .x : M l D. Ž3. Let Y g DŽ D . with Y : N. If O 2 Ž X . M = O 2 Ž Y ., then Y :2 X. Proof. Part Ž1. follows from Lemma 3.1 Ž1.. For the proof of Ž2., let Q be a preimage of a minimal normal subgroup of NrM. The maximality of M forces that O 2 Ž X . : Q, so < QrM < is odd by the solvability of N. Thus w O 2 Ž X ., O 2 Ž X .x s O 2 Ž X . l D : Q l D : M l D by Lemma 3.1 Ž7., so Ž2. holds. Part Ž3. holds because w O 2 Ž Y ., O 2 Ž Y .x s O 2 Ž Y . l D : M l D : O 2 Ž X . by Lemma 3.1 Ž7. and Ž1.. LEMMA 4.6. Let D : S and D : DŽ D .. Suppose that D is a finite set and that ² D : is sol¨ able. Then there exists an element Y g D such that Y :2 X for all X g D. Proof. We will first show that if Z1 2 Z2 2 ??? 2 Zk for Z1 , Z2 , . . . , Zk g D, then Zk :2 Zi for 1 F i F k. Put N s ² D :. Take a normal subgroup M of N that is maximal subject to the condition O 2 Ž Zk . M. Then O 2 Ž Zi . M for 1 F i F k by repeated use of Lemma 4.5 Ž3.. So M l D : O 2 Ž Zi . for 1 F i F k by Lemma 4.5 Ž1.. On the other hand, w O 2 Ž Zk ., O 2 Ž Zk .x : M l D by Lemma 4.5 Ž2.. Suppose the lemma is false. If Z1 2 Z2 2 ??? 2 Z jy1 for Z1 , Z2 , . . . , Z jy1 g D, then we can pick Z j g D such that Z jy1 2 Z j and Z j / Zi for 1 F i F j. Hence, by induction, all elements of D are arranged in such a way that Z1 2 Z2 2 ??? 2 Zn . But then Zn :2 Zi for 1 F i F n, a contradiction. LEMMA 4.7. Let b s Ž X, D, Y . g G. Suppose either that b is not thin or that O 2 Ž b . / 1. Put DX s w O 2 Ž X ., O 2 Ž X .xŽ O 2 Ž X . l O 2 Ž Y ..w O 2 Ž Y ., O 2 Ž Y .x, and let X X g GX Ž DX . and Y X g GY Ž DX .. Then either X X :2 Y X or Y X :2 X X . Proof. Put D 0 s D, X 0 s X, Y0 s Y, D 1 s DX , X 1 s X X , Y1 s Y X . Inductively, put Dn s O 2 Ž X ny1 . , O 2 Ž X ny1 .
Ž O2 Ž X ny1 . l O2 Ž Yny1 . .
= O 2 Ž Yny1 . , O 2 Ž Yny1 . , and choose X n g GX ny 1Ž Dn . and Yn g GY ny 1Ž Dn .. Note that this is possible by Lemma 4.2 Ž1.. Define bn s Ž X n , Dn , Yn . for all nonnegative integers n. Then we have bn g G and bnq1 U bn for all n G 0 by Lemma 4.2 Ž4. and Definition 4.1 Ž3.Ž4c.. Since D is finite, we have Dm s Dmq1 for some nonnegative integer m. Suppose that O 2 Ž bm . s 1. Then bm is thin by
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Theorem 4. Hence repeated use of Theorem 5 forces that bm s bmy1 s ??? s b 1 s b 0 . This contradiction shows that O 2 Ž bm . / 1. Thus ² X m , Ym : is solvable, and so either X m :2 Ym or Ym :2 X m by Lemma 4.6. Let 2 F j F m, and suppose that X j :2 Yj . Then O 2 Ž Yj . = w O 2 Ž X j ., O 2 Ž X j .x s O 2 Ž X j . l Dj s O 2 Ž X j . l Djy2 by Lemma 3.1 Ž7.. Since O 2 Ž X jy2 . , O 2 Ž X j . : O 2 Ž X jy2 . , O 2 Ž X jy1 . s O 2 Ž X jy1 . , O 2 Ž X jy1 . : O 2 Ž X jy1 . by Lemma 4.2 Ž2., we have w O 2 Ž X jy2 ., O 2 Ž X j .x : w O 2 Ž X jy1 ., O 2 Ž X j .x s w O 2 Ž X j ., O 2 Ž X j .x : O 2 Ž X j ., also by Lemma 4.2 Ž2.. Thus O 2 Ž X jy1 . , O 2 Ž X j . s O 2 Ž X j . , O 2 Ž X j . : F O 2 Ž Yj .
w O 2 Ž X jy 2 . , O 2 Ž X jy 2 .xŽ O 2 Ž X jy 2 .lO 2 Ž Y jy 2 ..
: F O 2 Ž Yj .
D jy 1
O 2 Ž Yjy2 . , O 2 Ž Yjy2 . : O 2 Ž Yjy1 .
by Lemma 4.2 Ž5.. Hence we conclude that O 2 Ž X jy1 . , O 2 Ž X jy1 . : O 2 Ž X jy1 . , ² O 2 Ž X j . s ² O 2 Ž X jy1 . , O 2 Ž X j . s ² O2 Ž X j . , O 2 Ž X j .
D jy 1
:
D jy 1
D jy 1
:
: : O 2 Ž Yjy1 .
by Lemma 4.2 Ž6., so X jy1 :2 Yjy1. Similarly, Yj :2 X j implies Yjy1 :2 X jy1. Therefore, by induction on j, we have either X 1 :2 Y1 or Y1 :2 X 1. LEMMA 4.8. O 2 Ž b . / 1.
Let b s Ž X, D, Y . g G. If either X :2 Y or Y :2 X, then
Proof. By the symmetry between X and Y, we may assume that X :2 Y. Put DX s O 2 Ž Y .. Then GX Ž DX . / B by Lemma 4.2 Ž1., and D acts 4 on GX Ž DX . as DX 1 on GX Ž DX .. Then y D. Let X 1 , . . . , X n be a D-orbit 2 C X kŽ O 2 Ž X k .. : O 2 Ž X k . for 1 F k F n and O Ž X . s ² O 2 Ž X 1 ., . . . , O 2 Ž X n .: by Lemma 4.2 Ž3.Ž6.. Let 1 F k F n. We may assume that no nonidentity characteristic subgroup of DX is normal in X k because X s ² X k , D : by Definition 4.1 Ž4c.. Therefore, by the Corollary of w9x, X k is described as follows: X k s DX J Ž X k ., J Ž X k . s L k = D k = Ek , L k ( S4 = ??? = S4 , D k ( D 8 = ??? = D 8 , Ek is an elementary abelian 2-group. Put Vk s w O 2 Ž X k ., O 2 Ž X k .x. Then
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Vk s O 2 Ž X k . l DX : J Ž DX . by Lemma 3.1 Ž7. and the structure of X k . Put V s w O 2 Ž X ., O 2 Ž X .x. Then V s w O 2 Ž X ., ² O 2 Ž X k .<1 F k F n :x s ²w O 2 Ž X ., O 2 Ž X k .x<1 F k F n: s ²w O 2 Ž X k ., O 2 Ž X k .x<1 F k F n: s ² Vk <1 F k F n: : J Ž DX . by Lemma 4.2 Ž2.. We may assume that V ; J Ž DX .. Let A be a Hall 2X-subgroup of DX O 2 Ž Y ., and put W s ² V A :. Then W 1 y DA s Y. If w V, W x s 1, then w O 2 Ž X ., O 2 Ž X .x s V : W : CDX Ž V . : O 2 Ž X . by ² 2Ž . : ² : Lemma 3.1 Ž1., so W 1 y O X , Y s X, Y . a w x w Suppose that V, W / 1. Then we have Vi , Vj x / 1 for some i, j, and a g A. Let g s Ž X i , DX , X ja .. Then g g G by Lemma 4.2 Ž4.. Moreover, we have X i 2 X ja because J Ž DX . = Vi C J Ž DX .Ž Vj a . s O 2 Ž J Ž X ja .. s J Ž DX . l O 2 Ž X ja . by the structure of X j . Put E s J Ž DX .. Since Vi Vj a : E, there exist Z1 g GX iŽ E . and Z2 g GX ja Ž E . such that Ž Z1 , E, Z2 . g G and Z1 2 Z2 by Lemma 4.2 Ž1.Ž4. and Lemma 4.3. Let d s Ž Z1 , E, Z2 .. Since E U y Xi s ² Z1 , DX :, we have E U Z1 , so Z1 s ² E Z 1 : by Lemma 3.1 Ž1.. Hence y w O 2 Ž Z1 ., O 2 Ž Z1 .x : w O 2 Ž Z1 ., ² E Z 1 :x s ²w O 2 Ž Z1 ., E x Z 1 : : ²w E, E x Z 1 : : ²w V 1Ž ZŽ E .. Z 1 : : V 1Ž ZŽ O 2 Ž Z1 ... by the structure of E and Lemma 4.2 Ž3., so CE Žw O 2 Ž Z1 ., O 2 Ž Z1 .x. s O 2 Ž Z1 . by Lemma 3.1 Ž1.. Similarly, CE Žw O 2 Ž Z2 ., O 2 Ž Z2 .x. s O 2 Ž Z2 .. Thus Z1 2 Z2 implies Z2 2 Z1. Next, put EX s w O 2 Ž Z1 ., O 2 Ž Z1 .xŽ O 2 Ž Z1 . l O 2 Ž Z2 ..w O 2 Ž Z2 ., O 2 Ž Z2 .x. Then there exist Z1X g GZ 1Ž EX . and Z2X g GZ 2Ž EX . such that Ž Z1X , EX , Z2X . g G and Z1X 2 Z2X by Lemma 4.2 Ž1.Ž4. and Lemma 4.3. Since Z2 2 Z1 s ² Z1X , E :, we have EX y U Z1X . Similarly, EX y U Z2X . Thus, as above, we have CEX Žw O 2 Ž ZiX ., O 2 Ž ZiX .x. s O 2 Ž ZiX . for i s 1, 2, and then Z1X 2 Z2X implies Z2X 2 Z1X . Now, Lemma 4.7 shows that d is thin and that O 2 Ž d . s 1. Since w O 2 Ž Z1 ., O 2 Ž Z1 .x : V 1Ž ZŽ O 2 Ž Z1 ..., Theorem 3 yields that d is isomorphic either to a GL3 Ž2.-amalgam or to an Sp4 Ž2.-amalgam, and then < E < F 2 4 . This forces < V < F 2 3, and so X is thin and O 2 Ž X . DX is 2-irreducible. Now, by Proposition 2.3 of w11x, some nonidentity A-invariant subgroup, say W X , of DX is normal in O 2 Ž X . DX . So ²W X D : is normal in both X and Y as X s O 2 Ž X . D and Y s DA.
5. THE PROOF OF THEOREM 1 We will begin the proof of Theorem 1. Assume that G and S continue to satisfy the assumption of Theorem 1. Let H and K be subgroups of G satisfying conditions Ža. ] Žd. stated in Theorem 1. Put a s Ž H, S, K .. LEMMA 5.1.
a g G, H 2 K, K 2 H.
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Proof. This follows from Lemma 4.8 and conditions Ža. ] Žd. of Theorem 1. By the symmetry between H and K, we will assume the following: ASSUMPTION 5.2.
V 1Ž ZŽ S .. ZŽ H ..
LEMMA 5.3. The group H has a noncentral minimal normal subgroup. Proof. Let W s ² V 1Ž ZŽ S .. H :. Then O 2 Ž H . : CH ŽW . W O 2 Ž H . by Assumption 5.2. Thus w W, O 2 Ž H .x / 1, and O 2 Ž H .CH ŽW .rCH ŽW . has odd order by Lemma 3.1 Ž1.Ž2., so CwW , O 2 Ž H .x Ž O 2 Ž H .. s 1. Therefore a minimal normal subgroup of H contained in w W, O 2 Ž H .x is noncentral. Let V be a noncentral minimal normal subgroup of H. Put T U s w O 2 Ž H ., O 2 Ž H .xŽ O 2 Ž H . l O 2 Ž K ..w O 2 Ž K ., O 2 Ž K .x, T s O 2 Ž H .w O 2 Ž K ., O 2 Ž K .x, and Dw k x s ² t g D < < V: C V Ž t .< F 2 k : for all subgroups D of S and all positive integers k. LEMMA 5.4. integers k: Ž1. Ž2. Ž3.
Let T U : D : S. Then the following holds for all positi¨ e
V g v k Ž D, D .. N `Ž k; D . : Dw k x. `Ž . Ž . Ž . If D 1 y S, then N k; D O 2 H l O 2 K .
Proof. Let k be a positive integer. We first note that V s w V, O 2 Ž H .x : w O 2 Ž H ., O 2 Ž H .x : T U : D. Let L be an abelian subgroup of D with w V, L, L x s 1. Put H s HrCH Ž V ., and regard V as a nontrivial irreducible faithful GF Ž2. H-module. Since w V, L, L x s w V, L, L x s 1, Lemma 3.4 shows that V s ² ¨ g V < < L: CLŽ ¨ .< F 2: s ² ¨ g V < < L: CLŽ ¨ .< F 2:. Hence V g v 1Ž D, D ., and then V g v k Ž D, D . by Lemma 2.2 Ž3.. Put Q s Dw k x. Then < V: C V Ž x .< G 2 kq 1 for all x g D y Q by the definition of Q, so we have N `Ž k; D . : Q by Lemma 2.6. So Ž1. and Ž2. hold. Suppose that D y 1 S. Since 1 / T : D, we have N `Ž k; D . / 1 by Lemma 2Ž . Ž . 2.4 1 . Let L s H or K. Then D l O 2 Ž L. 1 y SO L s L because w O 2 Ž L., O 2 Ž L.x : D l O 2 Ž L. 1 S. Therefore, if N `Ž k; D . : O 2 Ž L., then y Ž .Ž . Ž . N `Ž k; D . s N `Ž k; D l O 2 Ž L.. 1 y L by Lemma 2.4 4 2 . Hence 3 holds. The next lemma is a technical one that shows special relations between elements of DŽ D . where D : S. LEMMA 5.5. Let D : S and n ) 1. Take Z1 , . . . , Zn g GH Ž D . j GK Ž D ., and put N s ² Z1 , . . . , Zn :. Assume that Zn 2 Zny1 2 ??? 2 Z2 2 Z1 ,
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379
that N is sol¨ able, and that Z1 g GH Ž D .. Then the following hold: Ž1. There exists an abelian normal subgroup U of N such that U : D and < U: CU Ž x .< G 2 2 for all x g D y O 2 Ž Zn .. Ž2. If D 1 S, < S: D < F 2 4 , and Zn g GLŽ D . where L s H or K, then y `Ž N 1; D . : O 2 Ž L.. Proof. Take a subgroup M of N that is maximal subject to the ² : condition O 2 Ž Z1 . M y 1 N. Then M ; N, and M l D 1 y Z1 , . . . , Zn by Lemma 4.5 Ž1.Ž3.. Put P s O Ž N mod M .. Since O 2 Ž NrM . s 1 by the maximality of M, we have P l D s M l D : O 2 Ž N . : M ; P, so O 2 Ž Z1 . : P also by the maximality of M. Let I s ² O 2 Ž Z1 . N :, E s I l D, W s V 1Ž ZŽ E ... Then w O 2 Ž Z1 ., O 2 Ž Z1 .x s O 2 Ž Z1 . l D : E : D by Lemma 3.1 Ž7.. Since E s I l P l D s I l M l D and O 2 Ž Z1 . I l M, we have Ey 1 N by Lemma 4.5 Ž1.Ž3.. Hence w W, O 2 Ž Z1 .x / 1 by Lemma 4.4 because Z1 g GH Ž D . and V 1Ž ZŽ S .. ZŽ H .. Thus 1 / w W, I x y 1 N. Next, we will show that if < S: D < F 2 4 , then CwW , I xŽ I . s 1. Take a subgroup R of N that is maximal subject to the conditions R 1 yN and R l D g Syl 2 Ž R .. Since O Ž NrR. s 1 by the maximality of R, CN r R Ž O 2 Ž NrR.. : O 2 Ž NrR. by the solvability of NrR. Let Q s O 2 Ž N mod R .. Suppose that O 2 Ž Z1 . Q. Then Q l D 1 y N by Lemma ŽŽ Q l D . R .. Hence Q l 4.5 Ž1.Ž3., so Ž Q l D . R 1 N and Q l D g Syl 2 y D : R by the maximality of R. Let W0 be a chief factor of N within QrR so that O 2 Ž Z1 . C s CN ŽW0 .. Let SX be a subgroup between Ž N l S . R and N so that SXrR g Syl 2 Ž NrR.. Then < SX
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holds. Suppose that < S: D < F 2 4 . Then Lemma 3.4 implies U g v 1Ž D, D ., as in the proof of Lemma 5.4 Ž1., so N `Ž1; D . : O 2 Ž Zn . by Lemma 2.6. `Ž Ž . . Hence, if D 1 y S and Zn g GL D where L s H or K, we have N 1; D S : FO 2 Ž Zn . : O 2 Ž L. by Lemma 4.2 Ž5., so Ž2. holds. In the remainder of the proof, we will assume that a is not thin and derive a contradiction. LEMMA 5.6. Let X g GH ŽT .. Then there exists Y g GK ŽT . such that X :2 Y 2 X and ² X, Y : is sol¨ able. Proof. Let X g GH ŽT .. Then there exists Y g GK ŽT . such that Y 2 X by Lemma 4.3 Ž2.. Put X U s O 2 Ž X .T U . Then T U g Syl 2 Ž X U . because O 2 Ž X . l T s w O 2 Ž X ., O 2 Ž X .x s w O 2 Ž H ., O 2 Ž X .x : w O 2 Ž H ., O 2 Ž H .x : T U by Lemma 3.1 Ž7. and Lemma 4.2 Ž2.. Moreover, as T s O 2 Ž H .T U , there is a one-to-one correspondence between the set of subgroups of X containing T and the set of subgroups of X U containing T U . Thus X U is 2-irreducible, and then X U g GX ŽT U .. Hence there exists Y U g GY ŽT U . such that Y U 2 X U as above, because w O 2 Ž Y ., O 2 Ž Y .x s w O 2 Ž K ., O 2 Ž Y .x : w O 2 Ž K ., O 2 Ž K .x : T U . Since X U g GH ŽT U . and Y U g GK ŽT U ., we have X U :2 Y U by Lemma 4.7. Thus w O 2 Ž X ., O 2 Ž X .x s O 2 Ž X . l T s O 2 Ž X U . l T U s w O 2 Ž X U ., O 2 Ž X U .x : O 2 Ž Y U . by Lemma 3.1 Ž7., and so w O 2 Ž X ., O 2 Ž X .x : FO 2 Ž Y U .T : O 2 Ž Y . by Lemma 4.2 Ž5.. Hence we have X : 2 Y. Thus O 2 Ž² X, Y .: / 1 by Lemma 4.8, and then ² X, Y : is solvable. LEMMA 5.7. V : O 2 Ž K .. Proof. By Lemma 5.6, we can choose X g GH ŽT . and Y g GK ŽT . so that Y 2 X and ² X, Y : is solvable. Then, by Lemma 5.5 Ž1., there exists an abelian normal subgroup U of ² X, Y : such that U : T and < U: CU Ž x .< G 2 2 for all x g T y O 2 Ž Y .. Since V g v 1ŽT, T . by Lemma 5.4 Ž1., and w V, U, U x : wU, U x s 1, we have V s ² ¨ g V < < U: CU Ž ¨ .< F 2: : O 2 Ž Y ., so V : FO 2 Ž Y . S : O 2 Ž K . by Lemma 4.2 Ž5.. LEMMA 5.8.
V 1Ž ZŽ S .. : ZŽ K ..
Proof. Suppose false. Then, as in Lemma 5.3, K has a noncentral minimal normal subgroup, say, W. Furthermore, W : O 2 Ž H . by Lemma 5.7 and the symmetry between H and K. Now, we will assume that Ž Ž .. QŽ K Ž S .. U y H, where Q K S is a characteristic subgroup of S defined in w10x. Then H has a unique noncentral chief factor within O 2 Ž H . as in the proof of Proposition 2.2 of w11x. Thus we have w O 2 Ž H ., O 2 Ž H .x s V. Hence H :2 K by Lemma 5.7, a contradiction. Let Y g GK ŽT .. Define U s ² V Y :, Z s FV Y, Tqs O 2 Ž H .w O 2 Ž Y ., O Y .x, and Yqs O 2 Ž Y .Tq. 2Ž
SOLVABLE
LEMMA 5.9.
2-LOCAL
SUBGROUPS
381
If T w1x : O 2 Ž H ., then the following hold:
Ž1. U is normal abelian 2-subgroup of Y contained in Tq. Ž2. w Z, Tqx s 1. Ž3.
Yq is 2-irreducible with Tqg Syl 2 Ž Yq. .
Ž4.
Tqs CT q Ž V .w O 2 Ž Yq. , O 2 Ž Yq.x .
Proof. First, U is a normal 2-subgroup of Y by Lemma 5.7, so U : O 2 Ž Y . : T. Let y g Y. Then V y g v 1ŽT y , T y . by Lemma 5.4 Ž1.. Since w V y , V, V x : w V, V x s 1, we have V y s ² w g V y < < V: C V Ž w .< F 2: : T w1x : O 2 Ž H . : Tq. Thus U : Tq and U is abelian, so Ž1. holds. Since H 2 K s ² S, Y : by Lemma 5.1 and Definition 4.1 Ž4c., and T s O 2 Ž H .w O 2 Ž K ., O 2 Ž K .x s CT Ž V . O 2 Ž Y . s CT Ž Z . O 2 Ž Y ., we have T / O 2 Ž Y ., and then Y s ²T Y : s ² CT Ž Z . Y : O 2 Ž Y . by Lemma 3.1 Ž1.. Thus w O 2 Ž Y ., O 2 Ž Y .x : O 2 Ž Y . : ² CT Ž Z . Y :. Hence Tq: CT Ž Z ., so Ž2. holds. Since T s Tqw O 2 Ž K ., O 2 Ž K .x and ww O 2 Ž K ., O 2 Ž K .x, O 2 Ž Y .x s w O 2 Ž Y ., 2 Ž .x O Y s T l O 2 Ž Y . : Tq by Lemma 4.2 Ž2. and Lemma 3.1 Ž7., we conclude that Ž3. holds and that w O 2 Ž Y ., O 2 Ž Y .x s w O 2 Ž Yq. , O 2 Ž Yq.x as in the proof of Lemma 5.6. So Ž4. also holds. LEMMA 5.10.
If T w1x : O 2 Ž H ., then U g v 2 ŽTq, Tq ..
Proof. Suppose first that w V, Tqx Z. Let V Ž1. s C V mod Z ŽTq. , V Ž2. s q C V mod V Ž1. ŽTq. , and U Ž i. s ²Ž V Ž i. . Y : for i s 1, 2. Then Z ; V Ž1. ; V Ž2. Ž2. by our assumption, and so wU , w O 2 Ž Yq. , O 2 Ž Yq.xx Z. By Lemma 5.9 Ž4.. Suppose that wU, L, L x s 1 for an abelian subgroup L of Tq. Let LX s L l O 2 Ž Yq. . Note that Yq is a 2-irreducible solvable group acting on an abelian 2-group U Ž2. by Lemma 5.9 Ž3.Ž1.. Thus < L: LX < F 2 by Ž3.9. of w7x. Since V y g Tql v 1ŽT y , T y . : v 1ŽTq, O 2 Ž Yq.. for all y g Yq by Lemma 2.2 Ž1.Ž2., we have U g v 1ŽTq, O 2 Ž Yq.. by Ž2.2. of w7x, so U s ² u g U < < LX : CLX Ž u.< F 2: s ² u g U < < L: CLŽ u.< F 2 2 :. Thus U g v 2 ŽTq, Tq . by Lemma 5.9 Ž1.. 4 Suppose next that w V, Tqx : Z. Let GX s X g GH ŽTq. < TqU y X q, and q. Ž Ž .. Ž ² < Ž define VX s C V O 2 X for X g GH T . Then V s VX X g GH T .: s ² VX < X g GX : because FX g G ŽT q . w V U , O 2 Ž X .x s w V U , FX g G ŽT q . O 2 Ž X .x H H : w V U , O 2 Ž H .x s 1 for the dual module V U of V. q Let X g GX , and put UX s ² VXY : . Then it suffices to prove that UX g v 2 ŽTq, Tq . by Lemma 5.9 Ž1. and Ž2.2. of w7x because U s ²UX < X g GX :. Since w V, ŽTq. 2 x : w V, Tq, Tq x s 1, and CT Ž V . s O 2 Ž H . by Ž3.1.Ž1., TqrO 2 Ž H . is elementary abelian. Thus < TqrO2 Ž X .< s 2 by Lemma 3.1 Ž4. because O 2 Ž H . : O 2 Ž X . ; Tq. Choose a family of subgroups VXŽ i. < i g I 4
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of VX so that the following hold for all i g I Ž1. Ž2. Ž3.
VXŽ i. is a Tq-invariant subgroup of order 2 or 2 2 . VXŽ i. Z. < VXŽ i. : Z XŽ i. < s 2 where Z XŽ i. s w VXŽ i., Tq x. q
Let i g I, and put UXŽ i. s ²Ž VXŽ i. . Y : . Then UXŽ i.rZ XŽ i. g v 1ŽTqrZ XŽ i., Tqr Z XŽ i. . by Ž3.10. of w7x because VXŽ i.rZ XŽ i. : V 1Ž ZŽTqrZ XŽ i. ... Thus UXŽ i. g v 2 ŽTq, Tq . by Ž2.3. of w7x because < Z XŽ i. < F 2. Moreover, we have Z g v 0 ŽTq, Tq . : v 2 ŽTq, Tq . by Lemma 5.9 Ž2. and Lemma 2.2 Ž3.. Therefore UX s ²UXŽ i. < i g I : Z g v 2 ŽTq, Tq . by Ž2.2. of w7x. LEMMA 5.11. The following hold: Ž1. T w2x O 2 Ž H .. Ž2. If T w1x : O 2 Ž H ., then T s T w2x. Proof. To prove Ž1. and Ž2., we may assume that T w1x : O 2 Ž H .. Then N `Ž1; T . : T w1x : O 2 Ž H . : Tq by Lemma 5.4 Ž2., so N `Ž1; Tq. s N `Ž1; T . : O 2 Ž H . by Lemma 2.4 Ž4.. Let min x g TyO 2 ŽY . < U: CU Ž x .< s 2 d, and let x g T y O 2 Ž Y . with x 2 g O 2 Ž Y . and < U: CU Ž x .< s 2 d. Since T s O 2 Ž H .w O 2 Ž K ., O 2 Ž K .x s CT Ž V . O 2 Ž Y ., there exists an element y g CT Ž V . such that y ' x Žmod O 2 Ž Y ... Let P be a Hall 2X-subgroup of Y. Take an element g g P y F Ž P . such that k s w g, y xy1 f F Ž P . O 2 Ž Y .. Then 2 d s < U: y1 y1 y1 y1 y1 y1 y1 y1 CU Ž x .< G < V g : V g l V g x < s < V g : V g l V g y < s < V g : V g l yy V g < s < V: V l V k <. Since y g CT Ž V ., we have w V l V k , ² y, y k :x s 1 so w V l V k , O 2 Ž Y ., ² y, y k :x s 1 s w² y, y k :, V l V k , O 2 Ž Y .x as V, V k 1 y O 2 Ž Y .. Hence w O 2 Ž Y ., ² y, y k :, V l V k x s 1 by the three-subgroup lemma, so w O 2 Ž Y ., ² y, y k :x : T w d x. Since ² y, y k : F Ž P . O 2 Ž Y ., we have O 2 Ž Y . s ²² y, y k :T :, so w O 2 Ž Y ., O 2 Ž Y .x : T w d x. Therefore w O 2 Ž K ., O 2 Ž K .x s w O 2 Ž K ., ² O 2 Ž Y . S :x s ²w O 2 Ž K ., O 2 Ž Y .x S : s ²w O 2 Ž Y ., O 2 Ž Y .x S : : ²T w d x S : s T w d x by Lemma 4.2 Ž6.Ž2., and so T s CT Ž V .w O 2 Ž K ., O 2 Ž K .x s T w d x. Suppose that d G 2. Let L g v 1ŽTq, U .. Since w L, U, U x : wU, U x s 1, we have L s ² x g L < < U: CU Ž x .< F 2:, so L : O 2 Ž Y .. Thus ² v 1ŽTq, U .: : Tql O 2 Ž Y . s O 2 Ž Yq. . Hence U g v U1 ŽTq,Tq . by Definition 2.1 Ž2. because V y g Tql v 1ŽT, O 2 Ž Y .. : v 1ŽTq, O 2 Ž Yq.. for all y g Yq, and U s ² V y < y g Yq: g v 2 ŽTq, Tq .. Now, suppose that d G 3. Then we have N `Ž1; Tq. : O 2 Ž Yq. : O 2 Ž Y . by Lemma 2.5, so N `Ž1; T . : FO 2 Ž Y . S : O 2 Ž K .. This contradicts Lemma 5.4 Ž3., and hence d F 2. Therefore Ž1. and Ž2. hold. Let H s HrCH Ž V ., and regard V as a nontrivial irreducible faithful GF Ž2. H-module. Put P s O 2, 2X Ž H .. Then we have P s O Ž H . s O 2 Ž H . by Lemma 3.1 Ž1.Ž2..
SOLVABLE
2-LOCAL
SUBGROUPS
383
LEMMA 5.12. The group T acts intransiti¨ ely on the set H of homogenous components of VP . Proof. First, assume that T w1x O 2 Ž H .. Let X g GH ŽT .. Since T w1x 1 y S, and FO 2 Ž X . S : O 2 Ž H . by Lemma 4.2 Ž5., there exists an element t g T y O 2 Ž X . such that < V: C V Ž t .< s 2. Since O 2 Ž X . C X Ž V . by Lemma 4.2 Ž6., we have C X Ž V . : F Ž O 2, 2X Ž X .mod O 2 Ž X .. by Lemma 3.1 Ž1.Ž2.Ž5., and so w O 2 Ž X ., t x C X Ž V .. Thus there exists a unique element W g H such that 1 / O 2 Ž X . , t : w P, t x s PW ( Z3 by Lemma 3.3, so X = PW , and P is an elementary abelian 3-group by Lemma 3.3. Hence CH Ž V . s F Ž P mod O 2 Ž H .. as above. Suppose that T is transitive on H . Then X = ² P W < W g H : s P s O 2 Ž H . , so O 2 Ž H . : O 2 Ž X . C H Ž V . s O 2 Ž X . O 2 Ž H . : O 2 Ž X .T s X. Thus O 2 Ž X . s O 2 Ž H .. By Lemma 5.6, there exists an element Y g GT Ž K . such that X :2 Y. Then w O 2 Ž H ., O 2 Ž H .x : O 2 Ž H . l T s O 2 Ž X . l T s w O 2 Ž X ., O 2 Ž X .x : O 2 Ž Y ., by Lemma 3.1 Ž7., and hence w O 2 Ž H ., O 2 Ž H .x : F O 2 Ž Y . S : O 2 Ž K . by Lemma 4.2 Ž5.. This shows that H :2 K, which contradicts Lemma 5.1. Next, assume that T w1x : O 2 Ž H .. Then N `Ž1; T . : T w1x : O 2 Ž H . by Lemma 5.4 Ž2.. Suppose that < H < s 1. Since T w2x O 2 Ž H . s CT Ž V . by Lemma 5.11 Ž1. and Lemma 3.1 Ž1., P ( Z3 , Z5 , or Esp27 by Lemma 3.3. Thus < S: T < F < S: O 2 Ž H .< s < S < F 2 4 because S is isomorphic to a subgroup of AutŽ P .. Suppose that < H < G 2 and that T is transitive on H . Since T s T w2x by Lemma 5.11 Ž2., T is generated by elements inducing transpositions on the set H . So < H < s 2, and < W < s 2 2 for all W g H . Hence < S: T < F < S O 2 Ž H .< s < S < F 2 3. We conclude that < S: T < F 2 4 if T is transitive on H . Now, by Lemma 5.6, choose X g GH ŽT . and Y g GK ŽT . so that Y :2 X and ² X, Y : is solvable. Then N `Ž1; T . : O 2 Ž K . by Lemma 5.5 Ž2., which contradicts Lemma 5.4 Ž3.. Let D be a maximal subgroup of S containing T so that D has precisely two orbits H1 and H2 on the set H . LEMMA 5.13. There exists Y g GK Ž D . such that ² Y, O 2 Ž H .: is sol¨ able. Proof. Put Vi s ² Hi : for i s 1, 2. Then there exist D-invariant subgroups Pi Ž i s 1, 2. of P containing CH Ž V . such that P s P1 P2, w Vi , Pix s Vi , w Vi , P3yix s 1 Ž i s 1, 2., and S interchanges P1 and P2 by Lemma 5.11 Ž1. and Lemma 3.3. Put Zi s DPi Ž i s 1, 2., and U s C V Ž D .. 2 Let Z g GK Ž D ., and put E s O 2 Ž Z .. Since U : V 1Ž ZŽ D .. : V 1Ž ZŽ E .. and w O 2 Ž Z ., O 2 Ž Z .x : E : D, Lemma 4.4 shows that wU, O 2 Ž Z .x : w V 1Ž ZŽ E .., O 2 Ž Z .x s 1 because V 1Ž ZŽ S .. : ZŽ K .. Thus wU, O 2 Ž K .x s wU, ² O 2 Ž Z .< Z g GK Ž D .:x s 1 by Lemma 4.2 Ž6.. Therefore ² Z1 , O 2 Ž K .: is solvable as it is contained in CG ŽU ..
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HAYASHI AND TANAKA
Since H s ² Z1 , S : and Z1 is generated by the set of its D-irreducible subgroups, there exists an element ZX g GH Ž D . such that ZX : Z1. Let D s ZX 4 j GK Ž D .. Then, by Lemma 4.6, there exists an element Y g D such that Y :2 X for all X g D. Suppose that Y s ZX . Then w O 2 Ž ZX ., O 2 Ž ZX .x : FX g G Ž D . O 2 Ž X . : O 2 Ž K . by Lemma 4.2 Ž5., so K w O 2 Ž H . , O 2 Ž H .x s w O 2 Ž H . , ² O 2 Ž Z X . S :x s ²w O 2 Ž H . , O 2 Ž Z X .x S : s ²w O 2 Ž ZX ., O 2 Ž ZX .x S : : O 2 Ž K . by Lemma 4.2 Ž6.Ž2.. This shows that H :2 K, a contradiction. Therefore Y g GK Ž D . and Y :2 ZX . Let EX s O 2 Ž ZX .. Since V1 : V 1Ž ZŽ EX .. and w O 2 Ž Y ., O 2 Ž Y .x : EX , Lemma 4.4 shows that w V1 , O 2 Ž Y .x : w V 1Ž ZŽ EX .., O 2 Ž Y .x s 1 because V 1Ž ZŽ S .. : ZŽ K .. Thus ² O 2 Ž H ., D, O 2 Ž Y .: is solvable as it is contained in NG Ž V1 .. Now, we reach a final contradiction. Choose Y g GK Ž D . so that ² Y, O 2 Ž H .: is solvable. Then there exist elements X, Z g GH Ž D . such that Z 2 Y 2 X by Lemma 4.3. Put N s ² X, Y, Z :. Since N is solvable and < S: D < s 2, Lemma 5.5 Ž2. shows that N `Ž1; D . : O 2 Ž H . l O 2 Ž K ., which contradicts Lemma 5.4 Ž3.. This completes the proof of Theorem 1.
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