The General FF-module Theorem

U. Meierfrankenfeld B. Stellmacher

Department of Mathematics, Michigan State University, East Lansing MI 48824 meier@math.msu.edu Mathematisches Seminar, Christian-Albrechts-Universität, D24098 Kiel stellmacher@math.uni-kiel.de

Abstract Let p be a prime, M a finite group with O_{p}(M) = 1, V a faithful 𝔽_{p}M-module and J the subgroup of M generated by the best offenders on V. In this paper we determine structure of J and the action of J on V.

Introduction

Let p be a prime, M a finite group and V a finite dimensional 𝔽_{p}M-module, where 𝔽_{p} is the prime field in characteristic p. A subgroup A ≤ M is an offender on V if

1. A/C_{A}(V) is an elementary abelian p-group, and

2. |V/C_{V} (A)| ≤ |A/C_{A}(V)|;

and A is a non-trivial offender on V, if in addition [V, A] ≠ 0. Moreover, V is called an F F -module for M if some subgroup of M is a non-trivial offender on V. Faithful simple F F -modules for groups of Lie type in equicharacteristic have been classified by Cooperstein [Co] (the case p = 2) and Meixner [M] (the case p ≠ 2) and for arbitrary nearly simple groups by Guralnick, R. Lawther and G. Malle [GM1], [GM2], [GLM].

These results have been of great importance for the local theory of finite groups since such F F - modules are closely related to the failure of the Thompson-factorization in groups of characteristic p. In fact, for a finite group G and a normal elementary abelian p-subgroup X the elementary abelian p-subgroups of maximal order in G provide examples for offenders on X; and so G possesses non-trivial offenders on X if [X, J(S)] ≠ 1, where S ∈ Syl_{p}(G). The action of such elementary abelian subgroups have an additional property that is reflected in the following definition.

A subgroup A ≤ M is a best offender on V if

(i) A/C_{A}(V) is an elementary abelian p-group, and

(ii) |B||C_{V} (B)| ≤ |A||C_{V} (A)| for every subgroup B ≤ A.

It is easy to see (using B := C_{A}(V)) that every best offender is an offender. Indeed, a best offender A on V is an offender on every A-submodule of V; and this property characterizes best offenders (see 1.2).

In this paper we use this slightly stronger definition to derive a result about F F -modules that is free from the restriction to simple modules. It includes the above mentioned F F -module theorems, but also in these cases it gives more information about the size and action of offenders on V.

Most of the time we will treat groups like Alt(6) ≅ Sp_{4}(2)′, SU_{3}(3) ≅ G_{2}(2)′ and ^{2}F_{4}(2)′ together with the groups of Lie-Type. We therefore use the following definition.

Definition. A genuine group of Lie-type in characteristic p is a group isomorphic to O^{p′}(C\overline{K}(σ)), where \overline{K} is a semisimple \overline{𝔽_{p}}-algebraic group, \overline{𝔽_{p}} is the algebraic closure of 𝔽_{p}, and σ is Steinberg endomorphism of \overline{K}, see [GLS3, Definition 2.2.2] for details. A simple group of Lie-type in characteristic p is a non-abelian composition factor of a genuine group of Lie-type in characteristic p.

Before stating our main result we give some further definitions.

Definition. The normal subgroup of M generated by the best offenders of M on V is denoted by J_{M}(V). A non-trivial subgroup K of J_{M}(V) is a J_{M}(V)-component if K is minimal with respect to K = [K, J_{M}(V)]. The set of these components we denote by J_{M}(V).

A finite group H is a called a CK-group provided that each composition factor of H is one of the known finite simple groups.

Let S be a set of subgroups of M. We often write [V, S] and C_{V} (S) rather than [V, ⟨S⟩] and C_{V} (⟨S⟩). Similarly, we write × S rather than ×_{A∈S} A.

The 𝔽_{p}M-module V is perfect if V = [V, M], simple if V ≠ 0 and 0 is the only proper 𝔽_{p}Msubmodule of V, and quasisimple if V is perfect, O_{p}(M/C_{M}(V)) = 1 and V/C_{V} (M) is simple. Moreover, M acts simply on V if V is a simple M-module; and M acts nilpotently on V if there exists a finite series 0 = V_{0} ≤ V_{1} ≤ V_{k−1} ≤ V_{k} = V of 𝔽_{p}M-submodules of V with [V_{i}, M] ≤ V_{i−1} for all 1 ≤ i ≤ k.

Let A be a subgroup of M. Then

– A is a strong dual offender on V if A acts nilpotently on V and [V, A] = [v, A] for every v ∈ V \ C_{V} (A);

– A is a strong offender on V if A is an offender on V and C_{V} (A) = C_{V} (a) for every a ∈ A\C_{A}(V) (note that the last condition is equivalent to C_{A}(V) = C_{A}(v) for all v ∈ V \ C_{V} (A));

– A is an over-offender on V if A is an offender and |A/C_{A}(V)| > |V/C_{V} (A)|.

Finally we call V a natural 𝔽_{p}K-module for M if M/C_{M}(V) ≅ K, and there exists a quadratic, bilinear or sesquilinear form f on V left invariant by M such that for K, 𝕂 := End_{M}(V), dim_{𝕂} V and f one of the following cases holds:

K  dim_{𝕂} V  𝕂  f
SL_{n}(p^{k})  n  𝔽_{p^{k}}  zero-form
Sp_{2n}(p^{k})  2n  𝔽_{p^{k}} non-deg. symplectic
O_{n}^{ϵ}(p^{k})  n  𝔽_{p^{k}} non-deg. quadratic
Ω_{n}^{ϵ}(p^{k})  n  𝔽_{p^{k}} non-deg. quadratic
SU_{n}(p^{k})  n  𝔽_{p^{2k}}  non-deg. unitary
G_{2}(2^{k})  6  𝔽_{2^{k}} non-deg. symplectic
Sym(2n)  2n − 2  𝔽_{2}  zero-form
Alt(2n)  2n − 2  𝔽_{2}  − ^{||} −
Sym(2n + 1)  2n  𝔽_{2}  − ^{||} −
Alt(2n + 1)  2n  𝔽_{2}  − ^{||} −

In the last four cases V is meant to be the simple composition factor of the 𝔽_{2}-permutation module for Sym(2n) and Sym(2n + 1), respectively.

Note that in the above definition a non-degenerate quadratic form is a quadratic form that is nonzero on every non-zero element in the radical of the associated symmetric form. Also observe that O_{2n+1}(2^{k}) ≅ Sp_{2n}(2^{k}) and V is a central extension of a natural Sp_{2n}(2^{k})-module. This extension does not split if n > 1 or k > 1.

In general, M can have more than one natural module. For example, for n = 5, Alt(5) ≅ SL_{2}(4) ≅ Ω_{4}^{−}(2), so M has three natural modules, the natural SL_{2}(4)-module, the natural Ω_{4}^{−}(2)-module, and the natural Alt(5)-module, the latter two being isomorphic.

In addition, M ≅ SL_{n}(q), n > 2, has two natural SL_{n}(q)-modules that are not isomorphic due to the graph automorphism of SL_{n}(q). Similarly, M ≅ Spin_{8}^{+}(q) has three natural Ω_{8}^{+}(q)-modules. In the literature two of these are called half-spin modules depending which epimorphism from M to Ω_{8}^{+}(q) one chooses.

Theorem 1 (General FF-Module Theorem). Let M be a finite CK-group with O_{p}(M) = 1 and V be a faithful finite dimensional 𝔽_{p}M-module. Suppose that J := J_{M}(V) ≠ 1. Then for J := J_{M}(V), W := [V, J] + C_{V} (J)/C_{V} (J), K ∈ J and \overline{J} := J/C_{J}([W, K]) the following hold:

(a) K is either quasisimple, or p = 2 or 3 and K ≅ SL_{2}(p)′.

(b) [V, K, L] = 0 for all K ≠ L ∈ J, and W = ⨁_{K∈J} [W, K].

(c) J^{p}J′ = O^{p}(J) = F^{∗}(J) = × J.

(d) W is a faithful semisimple 𝔽_{p}J-module.

(e) If A ≤ M is a best offender on V, then A is a best offender on W.

(f) \overline{K} = \overline{F^{∗}(J)} = O^{p}(\overline{J}) and C_{J}([W, K]) = C_{J}([V, K]).

(g) Either [W, K] is a simple 𝔽_{p}K-module, or one of the following holds, where q is a power of p:

1. \overline{J} ≅ SL_{n}(q), n ≥ 3, and [W, K] ≅ N^{r} ⊕ N^{∗s}, where N is a natural SL_{n}(q)-module, N^{∗} its dual, and r, s are integers with 0 ≤ r, s < n and \sqrt{r} + \sqrt{s} ≤ \sqrt{n}.

2. J ≅ Sp_{2m}(q), m ≥ 3, and [W, K] ≅ N^{r}, where N is a natural Sp_{2m}(q)-module and r is a positive integer with 2r ≤ m + 1.

3. \overline{J} ≅ SU_{n}(q), n ≥ 8, and [W, K] ≅ N^{r}, where N is a natural SU_{n}(q)-module and r is a positive integer with 4r ≤ n.

4. \overline{J} ≅ Ω_{n}^{ϵ}(q) with p odd if n is odd, or \overline{J} ≅ O_{n}^{ϵ}(q) with p = 2 and n even.^{1} Moreover, n ≥ 10 and [W, K] ≅ N^{r}, where N is a natural Ω_{n}^{ϵ}(q)-module and r is a positive integer with 4r ≤ n − 2.

(h) If [W, K] is not a homogeneous 𝔽_{p}K module, then (g:1) holds with r ≠ 0 ≠ s and n ≥ 4.

Theorem 2 (FF-Module Theorem). Let M ≠ 1 be a finite CK-group and V be a faithful 𝔽_{p}Mmodule. Put

D := {A ≤ M | there exists 1 ≠ B ≤ A such that [V, B, A] = 0 and A and B are offenders on V }.^{2}

Suppose that V is a simple 𝔽_{p} J_{M}(V)-module and M = ⟨D⟩. Then one of the following holds, where q is a power of p:

1The odd-dimensional orthogonal groups in characteristic 2 are covered in case (g:2).

2 Note here that D contains all quadratic offenders and by the Timmesfeld Replacement Theorem [KS, 9.2.3], also all best offenders in M on V.

1. M ≅ SL_{n}(q), n ≥ 2, and V is a natural SL_{n}(q)-module.

2. M ≅ Sp_{2n}(q), n ≥ 1, and V is a natural Sp_{2n}(q)-module.

3. M ≅ SU_{n}(q), n ≥ 4, and V is a natural SU_{n}(q)-module.

4. M ≅ Ω_{2n}^{+}(q) for 2n ≥ 6, M ≅ Ω_{2n}^{−}(q) for p = 2 and 2n ≥ 6, M ≅ Ω_{2n}^{−}(q) for p odd and 2n ≥ 8, M ≅ Ω_{2n+1}(q) for p odd and 2n + 1 ≥ 7, M ≅ O_{4}^{−}(2), or M ≅ O_{2n}^{ϵ}(q) for p = 2 and 2n ≥ 6, and V is a corresponding natural module.

5. M ≅ G_{2}(q), p = 2, and V is a natural G_{2}(q)-module (of order q^{6}).

6. M ≅ SL_{n}(q)/⟨−id^{n−1}⟩, n ≥ 5, and V is the exterior square of a natural SL_{n}(q)-module.

7. M ≅ Spin_{7}(q), and V is a spin module of order q^{8}.

8. M ≅ Spin_{10}^{+}(q), and V is a half-spin module of order q^{16}.

9. M ≅ 3.Alt(6), p = 2 and |V | = 2^{6}.

10. M ≅ Alt(7), p = 2, and |V | = 2^{4}.

11. M ≅ Sym(n), p = 2, n odd, n ≥ 3, and V is a natural Sym(n)-module.

12. M ≅ Alt(n) or Sym(n), p = 2, n is even, n ≥ 6, and V is a corresponding natural module. Theorem 3 (Best Offender Theorem). Let M ≠ 1 be a finite group, T ∈ Syl_{p}(M), and V be a faithful 𝔽_{p}M-module, and let A ≤ T be an non-trivial offender on V.

(a) Suppose that M ≅ G_{2}(q), p = 2, and V is a natural G_{2}(q)-module. Then N_{M}(A) is a maximal Lie-parabolic subgroup, |A| = |V/C_{V} (A)| = q^{3}, [V, A] = C_{V} (A), and C_{T} (A) = A.

(b) Suppose that M ≅ SL_{n}(q)/⟨−id^{n−1}⟩, n ≥ 5, and V is the exterior square of the natural SL_{n}(q)- module W. Let U be the (unique) T -invariant 𝔽_{q}-hyperplane of W. Then A = C_{M}(U). In particular, A is uniquely determined in T, C_{T} (A) = A, [V, A] = C_{V} (A) and |V/C_{V} (A)| = |A| = q^{n−1}.

(c) Suppose that M ≅ Spin_{7}(q), and V is a spin module of order q^{8}. Then C_{V} (A) = [V, A], |V/C_{V} (A)| = q^{4} ≤ |A| ≤ q^{5}, and if A is maximal, then |A| = q^{5}, C_{T} (A) = A, O^{p′}(N_{M}(A))/A ≅ Sp_{4}(q), and A is uniquely determined in T.

(d) Suppose that M ≅ Spin_{10}^{+}(q), and V is a half-spin module of order q^{16}. Then [V, A] = C_{V} (A), q^{8} = |A| = |V/C_{V} (A)|, O^{p′}(N_{M}(A)/A) ≅ Spin_{8}^{+}(q), and A is uniquely determined in T.

(e) Suppose that M ≅ 3.Alt(6), p = 2 and |V | = 2^{6}. Then [V, A] = C_{V} (A), |[V, A]| = |C_{V} (A)| = 16, |V/C_{V} (A)| = |A| = 4, and A is uniquely determined in T.

(f) Suppose that M ≅ Alt(7), p = 2 and |V | = 2^{4}. Then [V, A] = C_{V} (A), |[V, A]| = |C_{V} (A)| = 4, |V/C_{V} (A)| = |A| = 4, and A is uniquely determined in T.

(g) Suppose that M ≅ Sym(n), p = 2, n odd, and V is a natural Sym(n)-module. Then every offender on V is a quadratic best offender, A is generated by commuting transpositions and |V/C_{V} (A)| = |[V, A]| = |A|.

(h) Suppose that M ≅ Alt(n) or Sym(n), p = 2, n is even, n ≥ 6, and V is a corresponding natural module. Then every offender on V is a best offender, and there exists a set of pairwise commuting transpositions t_{1}, …, t_{k} such that one of the following holds:

1. A = ⟨t_{1}, …, t_{k}⟩, and either n ≠ 2k, [V, A] ≤ C_{V} (A) and |[V, A]| = |V/C_{V} (A)| = |A| or n = 2k, [V, A] = C_{V} (A) and 2|V/C_{V} (A)| = |A|.

2. n = 2k and A = ⟨t_{1}t_{2}, t_{2}t_{3} …, t_{l−1}t_{l}, t_{l+1}, t_{l+2}, …, t_{k}⟩ for some 2 ≤ l ≤ k, [V, A] = C_{V} (A) and |V/C_{V} (A)| = |A|.

3. n = 2k and A = ⟨t_{1}t_{2}, s_{1}s_{2}, t_{3}, t_{4} …, t_{k}⟩, where s_{1}, s_{2} are transpositions distinct from t_{1} and t_{2} and s_{1}s_{2} moves the same four symbols as t_{1}t_{2}, A is not quadratic and |[V, A]| = |V/C_{V} (A)| = |A|.

4. n = 8 = |A|, A acts regularly on {1, 2, …, 8}, [V, A] = C_{V} (A) and |V/C_{V} (A)| = |A|.

In particular, if A ≤ Alt(n) and n ≠ 8, then n = 2k and A = ⟨t_{1}t_{2}, t_{2}t_{3}, …, t_{k−1}t_{k}⟩.

Note that in all cases of the FF-Module Theorem M is generated by quadratic best offenders. In the following list we give the module structure of A, V/C_{V} (A) and [V, A] considered as a

N_{M}(A)-modules in the cases (a) – (d) of the Offender Theorem, as it can be deduced from the action of M on V. Put P := O^{p′}(N_{M}(A)).

Case P/O_{p}(P)  A  [V, A]  V/C_{V} (A)  Remarks
(a)  SL_{2}(q)  U  U^{∗}  U  [U, P] a nat. SL_{2}(q)-module
(b) SL_{n−1}(q)  U  ⋀^{2}(U)  U  U a nat. SL_{n−1}(q)-module
(c)  Sp_{4}(q)  nat. Ω_{5}(q) nat. Sp_{4}(q) nat. Sp_{4}(q)  V/C_{V} (A) ≅ [V, A]
A/C_{A}(P) ≅ V̸/C_{V} (A)
(d) Spin_{8}^{+}(q) nat. Ω_{8}^{+}(q) nat. Ω_{8}^{+}(q) nat. Ω_{8}^{+}(q)  pairwise non-isom.

Acknowledgment: We would like to thank the referee for all his helpful comments and suggestions that improved the readability of our manuscript considerably.

1 Linear Algebra and Offenders

In this section p is a prime, M a finite group and V a finite dimensional 𝔽_{p}M-module.

Lemma 1.1. Let A ≤ M and W a set of A-submodules of V with V = ⨁ W. Suppose that A is a faithful offender on V but not an over-offender on W for any W ∈ W. Let W ∈ W and put A_{W} = ⋂_{W ≠U∈W} C_{A}(U). Then

(a) |A| = |V/C_{V} (A)|.

(b) A = ×_{W ∈W} A_{W} = A_{W} × C_{A}(W).

(c) |A/C_{A}(W)| = |W/C_{W} (A)| = |W/C_{W} (A_{W})| = |A_{W} |.

Proof. Since A is not an over-offender on W, |A/C_{A}(W)| ≤ |W/C_{W} (A)|, and since V = ⨁ W, |V/C_{V} (A)| = ∏_{W ∈W} |W/C_{W} (A)|. Since A is an offender on V this gives

(∗)  |A| ≥ |V/C_{V} (A)| = ∏_{W ∈W} |W/C_{W} (A)| ≥ ∏_{W ∈W} |A/C_{A}(W)|.

Put B = ×_{W ∈W} A/C_{A}(W) and let B_{W} = A/C_{A}(W) be viewed as a subgroup of B. So B is the internal direct product of the B_{W} , W ∈ W. Consider the homomorphism

ϕ : A → B, a → (aC_{A}(W))_{W ∈W}.

Since V is a faithful A-module and V = ⨁ W, ker ϕ = ⋂_{W ∈W} C_{A}(W) = C_{A}(V) = 1 and ϕ is injective. By (∗) |A| ≥ |B|. Thus ϕ is surjective and so an isomorphism. Moreover, equality holds everywhere in (∗). In particular, (a) and the first equality in (c) hold.

Let a ∈ A. Then aϕ ∈ B_{W} if and only if a ∈ C_{A}(U) for all W ≠ U ∈ W and so if and only if a ∈ A_{W}. Thus A_{W} ϕ = B_{W}. Also a ∈ C_{A}(W) if and only if the W -coordinate of aϕ is 1 and so if and only if aϕ ∈ ×_{W ≠U∈W} B_{W}. Thus C_{A}(W)ϕ = ×_{W ≠U∈W} B_{W}. Since B = ×_{W ∈W} B_{W} and ϕ is an isomorphism, (b) holds.

From (b) we get that C_{W} (A) = C_{W} (A_{W}) and |A_{W} | = |A/C_{A}(W)|. Hence the (already proved) first equality in (c) gives also the second and third equality in (c).

Lemma 1.2. Let A ≤ M. Then A is a best offender on V if and only if A is an offender on every A-submodule of V.

Proof. If A is a best offender, then by [MS1, 2.5] A is an offender on every A-submodule of V.

Conversely, suppose A is an offender on every A-submodule of V. Then A is an offender on V and so elementary abelian. Let B ≤ A and put W := C_{V} (B). Clearly

(∗)  B ≤ C_{A}(W) and C_{W} (A) = C_{V} (A).

As A is an offender on W, |W/C_{W} (A)| ≤ |A/C_{A}(W)|, and (∗) implies that

|B||W | ≤ |B||A/C_{A}(W)||C_{W} (A)| ≤ |A||C_{V} (A)|.

This shows that A is a best offender on V.

Lemma 1.3. Suppose that B is a minimal offender on V and W is a B-submodule of V. Then B is a quadratic best offender on W, and one of the following holds:

1. B is an over-offender on W.

2. [W, B] = 0.

3. C_{B}(W) = C_{B}(V) and V = W + C_{V} (B).

Proof. Let D ≤ B. Since B is a minimal offender,

|D||C_{V} (D)| ≤ |V ||C_{D}(V)| ≤ |V ||C_{B}(V)| ≤ |B||C_{V} (B)|

and so B is a best offender. By the Timmesfeld Replacement Theorem [KS, 9.2.3], C_{B}([V, B]) is a non-trivial offender on V and so by minimality B = C_{B}([V, B]). Thus B is quadratic.

Assume that B is not an over-offender on W. Then |B/C_{B}(W)| = |W/C_{W} (B)| and

|V/C_{V} (B) + W | = |V/C_{V} (B)||W/C_{W} (B)|^{−1} ≤ |B||B/C_{B}(W)|^{−1} = |C_{B}(W)|.

Hence C_{B}(W) is an offender on V, and the minimality of B gives either B = C_{B}(W) or C_{B}(W) = C_{B}(V). In the first case (2) holds. In the second case

V = C_{V} (B) + W

and (3) follows.

Lemma 1.4. Suppose that A ≤ M acts nilpotently on V. Then the following are equivalent:

(a) A is a strong dual offender on V.

(b) Let 0 ≤ U ≤ Y ≤ V be any chain of A-submodules with [Y/U, A] = 0. Then [V, A] ≤ U or Y ≤ C_{V} (A).

(c) A is a strong dual offender on V^{∗}.

Proof. Suppose (a) holds. Let U and Y be as in (b) and suppose that Y ≰ C_{V} (A). Pick v ∈ Y \ C_{V} (A). Then

[V, A] = [v, A] ≤ [Y, A] ≤ U.

Thus (a) implies (b).

Suppose next that (b) holds. To show that (a) holds, let v ∈ V \ C_{V} (A) and put Y := ⟨v^{A}⟩ and U := [v, A]. Since [v^{k}, a] = [v, a]^{k} for all k ∈ ℤ, a ∈ A, U = [⟨v⟩, A]. So Y and U are A-submodules, U ≤ Y and A centralizes Y/U. Since v ∈ Y, Y ≰ C_{V} (A) and so (b) implies that [V, A] ≤ U. Hence [v, A] = U = [V, A] and (a) holds.

By 1.8(c), (b) holds for V if and only if it holds for V^{∗} in place of V. Thus the above argument with V^{∗} in place of V shows that (b) and (c) are equivalent.

Lemma 1.5. Let A be a strong dual offender on V. Then the following hold:

(a) A is quadratic on V.

(b) A is a strong dual offender on every A-submodule of V and V^{∗}.

(c) A is best offender on V and on V^{∗}.

(d) If |[V, A]| = |A|, then A is a strong offender on V.

Proof. Since by 1.4 A is also a strong dual offender on V^{∗} it suffices to prove the statements for V.

(a): Since A acts nilpotently on V there exists v ∈ V \C_{V} (A) with [v, A] ≤ C_{V} (A). By definition of a strong dual offender we conclude that [V, A] = [v, A] ≤ C_{V} (A) and so A is quadratic.

(b): This follows immediately from the definition of a strong dual offender.

(c): Let v ∈ V \ C_{V} (A). Since A is quadratic on V, [v, A] = {[v, a] | a ∈ A} and so

(∗)  |[V, A]| = |[v, A]| = |A/C_{A}(v)| ≤ |A|.

Thus by 1.8 |V^{∗}/C_{V^{∗}}(A)| ≤ |A|. So A is an offender on V^{∗}. By (b) this is also true for any Asubmodule of V^{∗}. Thus by 1.2 A is a best offender on V^{∗}. By symmetry, A is also a best offender on V.

(d): Suppose |[V, A]| = |A|. Then by (∗)

|A| ≤ |A/C_{A}(v)| ≤ |A| for every v ∈ V \ C_{V} (A).

Hence C_{A}(v) = 1 and so C_{V} (a) = C_{V} (A) for all a ∈ A^{♯}.

Lemma 1.6. Let A be a strong offender on V. Then A is a quadratic best offender on V.

Proof. Let W be an A-submodule of V with [W, A] ≠ 0. Then C_{A}(W) = 1 and so

|W/C_{W} (A)| ≤ |V/C_{V} (A)| ≤ |A| = |A/C_{A}(W)|.

Hence A is an offender on W and so by 1.2, A is a best offender on V.

To show that A is quadratic we may assume that [V, A] ≠ 0. Put B = C_{A}([V, A]). By the Timmesfeld Replacement Theorem [KS, 9.2.3], [V, B] ≠ 0 and since A is a strong offender, C_{V} (B) = C_{V} (A). Since [V, A, B] = 0 we conclude that [V, A, A] = 0 and so A is quadratic.

Lemma 1.7. Let A be a subgroup of M. Suppose V is self-dual as an 𝔽_{p}A-module. Then A is a strong offender iff |V/C_{V} (A)| = |A| and A is a strong dual offender.

Proof. Suppose first that A is strong offender and let 1 ≠ a ∈ A. Then C_{V} (a) = C_{V} (A) and since V is self-dual, [V, a] = [V, A] by 1.8(c). Let v ∈ V \ C_{V} (A). Then C_{A}(v) = 1 and so |[v, A]| ≥ |A|. Hence

|A| ≤ |[v, A]| ≤ |[V, A]| = |[V, a]| = |V/C_{V} (a)| = |V/C_{V} (A)| ≤ |A|,

and equality holds everywhere. Thus [v, A] = [V, A] and so A is a strong dual offender.

Suppose now that |V/C_{V} (A)| = |A| and A is a strong dual offender. Since V is self-dual we get |[V, A]| = |A|. Thus by 1.5(d), A is a strong offender.

Lemma 1.8. Suppose that 𝕂 is a field and V is a 𝕂-space. The following hold for A ≤ GL_{𝕂}(V) and U a 𝕂-subspace of V:

(a) dim_{𝕂} V = dim_{𝕂} V^{∗}.

(b) dim_{𝕂} U + dim_{𝕂} U^{⊥} = dim_{𝕂} V.

(c) [V, A]^{⊥} = C_{V^{∗}}(A) and C_{V} (A)^{⊥} = [V^{∗}, A].

(d) [V, A, A] = 0 ⟺ [V^{∗}, A, A] = 0.

(e) C_{M}(C_{V} (A)) ∩ C_{M}(C_{V^{∗}}(A)) is the largest subgroup Y ≤ M with C_{V} (Y) = C_{V} (A) and [V, Y] = [V, A].

(f) If A is quadratic on V, then dim_{𝕂}[V, A] + dim_{𝕂} V/C_{V} (A) ≤ dim_{𝕂} V.

Proof. (a), (b) and (c) are well-known and easy to prove statements from linear algebra; and (e) follows from (c).

(d): [V, A, A] = 0 iff [V, A] ≤ C_{V} (A) iff C_{V} (A)^{⊥} ≤ [V, A]^{⊥} iff [V^{∗}, A] ≤ C_{V^{∗}}(A) iff [V^{∗}, A, A] = 0. (f): Since A is quadratic, [V, A] ≤ C_{V} (A). Thus

dim_{𝕂} V = dim_{𝕂}[V, A] + dim_{𝕂} C_{V} (A)/[V, A] + dim_{𝕂} V/C_{V} (A).

Lemma 1.9. Let 𝔽 be a finite field of characteristic p, V a finite dimensional 𝔽H-module, and N ⊴ H. Put 𝕂 := End_{𝔽N}(V) and suppose that V is a self-dual simple 𝔽N-module. Then the following hold:

(a) There exists an N-invariant non-degenerate symmetric, symplectic or unitary 𝕂-form s on V.

(b) There exists a homomorphism ρ : H → Aut_{𝔽}(𝕂) with h ↦ ρ_{h} such that h ∈ H acts ρ_{h}-semi-linearly on the right 𝕂-vector space V; i.e., (v + w)h = vh + wh and (vk)h = (vh)(kρ_{h}) for v, w ∈ V and k ∈ 𝕂.

(c) There exists a map λ : H → 𝕂^{♯} with h ↦ λ_{h} such that the map H → 𝕂^{♯} ⋊ Aut_{𝔽}(K), h → λ_{h}ρ_{h}

is a homomorphism and

(vh, wh)s = (v, w)sλ_{h}ρ_{h}

for all v, w ∈ V, h ∈ H.

(d) Let U be a 𝕂-subspace of V and put U^{⊥} = {v ∈ V | (u, v)s = 0 for all u ∈ U}. Then U^{⊥} is N_{H}(U)-invariant.

(e) Let U be a non-zero 𝕂-subspace of V such that C_{H}(U) acts simply on V/U^{⊥}. Then U is 1dimensional over 𝕂.

(f) Put H_{0} = ker ρ. Then s is O^{p′}(H_{0})N-invariant.

Proof. Recall that 𝕂 is a finite field of characteristic p since V is finite and simple. It is convenient to write V in the following as a right 𝕂-vector space since we write the action of 𝕂 on V from the right.

Put V^{∗} := Hom_{𝕂}(V, 𝕂) and W := Hom_{𝔽}(V, 𝔽). Let μ : 𝕂 → 𝔽 be any non-zero 𝔽-linear map and define

τ : V^{∗} → W by u → u ∘ μ.

(Recall that our mappings act from the right, so v(u ∘ μ) = (vu)μ.)

Let 0 ≠ u ∈ V^{∗}. Then V u = 𝕂 and so there exists v ∈ V with vu ∉ ker μ. Thus v.uτ = vuμ ≠ 0. In particular uτ ≠ 0 and ker τ = 0. Since τ is 𝔽-linear and

dim_{𝔽} V^{∗} = dim_{𝔽} 𝕂 dim_{𝕂} V^{∗} = dim_{𝔽} 𝕂 dim_{𝕂} V = dim_{𝔽} V = dim_{𝔽} W

we conclude that τ is an 𝔽-isomorphism. For n ∈ N, v ∈ V and u ∈ V^{∗} we have

v.unτ = v.un.μ = vn^{−1}uμ = vn^{−1}.uτ = v.uτn

and so unτ = uτn. Thus τ is an 𝔽N-isomorphism. Since V is self-dual as an 𝔽N-module, this shows that V and V^{∗} are isomorphic 𝔽N-modules. Hence the set H of 𝔽N-isomorphisms from V to V^{∗} is non-empty.

For k ∈ 𝕂 let

\overline{k} : V^{∗} → V^{∗} defined by x\overline{k} : v ↦ vk.x (x ∈ V^{∗}, v ∈ V).

Then \overline{k} ∈ End_{𝔽N}(V^{∗}) =: \overline{𝕂} and k ↦ \overline{k} induces an isomorphism of fields from 𝕂 to \overline{𝕂}. Let β ∈ H. Then β ∘ \overline{k} ∘ β^{−1} is 𝔽-linear and so

σ_{β} : 𝕂 → 𝕂 with k ↦ β ∘ \overline{k} ∘ β^{−1}

is an 𝔽-linear automorphism of 𝕂. Since β ∘ \overline{k} = kσ_{β} ∘ β we get 1^{∘}. β is σ_{β}^{−1}-semi-linear.

Let δ ∈ H and put l = δ ∘ β^{−1}. Then l is 𝔽N-linear and so l ∈ 𝕂. Thus:

2^{∘}. For all β, δ ∈ H there exists l ∈ 𝕂 with δ = l ∘ β.

It follows that

kσ_{δ} = δ ∘ \overline{k} ∘ δ^{−1} = l ∘ β ∘ \overline{k} ∘ β^{−1} ∘ l^{−1} = l ∘ kσ_{β} ∘ l^{−1}.

Since 𝕂 is commutative, this implies kσ_{δ} = kσ_{β}. Thus σ_{δ} = σ_{β} is independent from β ∈ H. So we just write σ for σ_{β}.

Let F be the set of all N-invariant non-zero functions s : V × V → 𝕂 which are 𝕂-linear in the first coordinate and 𝔽-linear in the second, where N-invariant means that (vn, wn)s = (v, w)s for all v, w ∈ V and n ∈ N. Clearly, all these forms are non-degenerate since V is a simple 𝔽N-module.

For β ∈ H define s_{β} : V × V → 𝕂, (v, w) → v.wβ. Then s_{β} ∈ F and so also F ≠ ∅. Conversely, for s ∈ F define β_{s} : V → V^{∗} by v.wβ_{s} = (v, w)s. Then β_{s} ∈ H, and (1^{∘}) applied to β_{s} gives:

3^{∘}. Each s ∈ F is a σ^{−1}-sesquilinear 𝕂-form.

Define s^{∗} : V ×V → 𝕂, (v, w) → (w, v)sσ. Then s^{∗} is N-invariant, 𝕂-linear in the first coordinate and σ-semi-linear in the second coordinate. In particular, s^{∗} ∈ F and so (3^{∘}) implies. Hence

4^{∘}. σ = σ^{−1}, and either σ = id_{𝕂} or σ has order 2.

We now will prove (a) – (f).

(a): Put t = s + s^{∗}. Then t = t^{∗}. Suppose first that t ≠ 0. If σ = id_{𝕂}, then t is an N-invariant symmetric 𝕂-form; and if |σ| = 2, then t is an N-invariant unitary 𝕂-form. So (a) holds in this case.

Suppose next that t = 0. Then s = −s^{∗}. Assume char 𝕂 = 2, then s = s^{∗} and so s is a symmetric or unitary 𝕂-form. Assume char 𝕂 ≠ 2. If σ = id_{𝕂} then s is a symplectic 𝕂-form. If |σ| = 2 pick x ∈ 𝕂 with x ≠ xσ and put y := x − xσ. Then yσ = −y. Hence (sy)^{∗} = s^{∗}.yσ = sy and so sy is a N-invariant unitary 𝕂-form on V. Again (a) hold.

(b): Since N ⊴ H, it is readily verified that for k ∈ 𝕂 and h ∈ H the map V → V, v ↦ vh^{−1}kh is in 𝕂. Thus ρ_{h} ∈ Aut_{𝔽}(𝕂) where

v.kρ_{h} = vh^{−1}kh for all k ∈ 𝕂, h ∈ H.

A simple calculation shows that ρ : H → Aut_{𝔽}(𝕂) with h ↦ ρ_{h} is a homomorphism and h acts ρ_{h}-semi-linearly on V.

(c): Fix h ∈ H and define

s_{h} : V × V → 𝕂, (v, w) ↦ (vh, wh)sρ_{h}^{−1}.

Using that Aut(𝕂) is abelian, it is straight forward to verify that s_{h} ∈ F. By (2^{∘}), β_{s_{h}} = k_{h} ∘ β_{s} for some k_{h} ∈ 𝕂. Thus for all v, w ∈ V

(vh, wh)sρ_{h}^{−1} = (v, w)s_{h} = v.wβ_{s_{h}} = v.wk_{h}β_{s} = (v, wk_{h})s = (v, w)s.k_{h}σ

Define λ_{h} = k_{h}σ, then

(vh, wh)s = (v, w)sλ_{h}ρ_{h}.

It is readily verified that the map H → 𝕂^{♯} ⋊ Aut_{𝔽}(𝕂), h → λ_{h}ρ_{h} is a homomorphism.

(d): Let v ∈ U^{⊥}, h ∈ N_{H}(U) and u ∈ U. Then

(u, vh)s = (uh^{−1}, v)sλ_{h}ρ_{h} = 0.

(e): Let D be a 1-dimensional 𝕂-subspace of U. Then by (d), D^{⊥} is C_{H}(U)-invariant. Since U^{⊥} ≤ D^{⊥} and C_{H}(U) is simple on V/U^{⊥} we get U^{⊥} = D^{⊥} and U = D.

(f) For a, b ∈ H_{0} the homomorphism given in (c) yields

λ_{ab}ρ_{ab} = λ_{ab} = λ_{a}ρ_{a}λ_{b}ρ_{b} = λ_{a}λ_{b}.

Hence λ |_{H_{0}} is a homomorphism from H_{0} in 𝕂^{♯}. Since 𝕂^{♯} is a p′-group, (f) follows.

2 J-Components

In this section p is a prime, M is a finite group with O_{p}(M) = 1, and V is a finite dimensional faithful 𝔽_{p}M-module such that J_{M}(V) ≠ 1.

Notation 2.1. Put J := J_{M}(V) and J := J_{M}(V). Let I be the set of solvable J-components, K be the set of perfect J-components, E := ⟨K⟩, and I := ⟨I⟩.

Lemma 2.2. The following hold:

(a) C_{M}(J/Z(J)) = C_{M}(J).

(b) Let N be a J-invariant subgroup of M with [N, J] ≠ 1. Then there exists K ∈ J with K ≤ N.

(c) J ≠ ∅, J = I ∪ K, and K is the set of components of J.

(d) Let K ∈ I. Then either p = 2, K ≅ C_{3} ≅ SL_{2}(2)′, and [V, K] ≅ 𝔽_{2}^{2}, or p = 3, K ≅ Q_{8} ≅ SL_{2}(3)′, and [V, K] ≅ 𝔽_{3}^{2}.

(e) [W, K] = [W, K, K] for every K ∈ J and every K-submodule W of V.

(f) [K, F] = 1 and [V, K, F] = 0 for every K, F ∈ J with K ≠ F.

(g) C_{J}(IE) = Z(J), or p = 2 and C_{J}(IE) = Z(J)I. So in both cases C_{J}(IE) is an abelian p′-group.

(h) Let U ≤ M and K ∈ J. Then either [K, U] = 1 or [W, K] ≤ [W, [K, U]] for every K-submodule W ≤ V.

Proof. (a) Put R = C_{M}(J/Z(J)) and let T be a p-subgroup of J. Since O_{p}(M) = 1, O_{p}(Z(J)) = 1 and so Z(J) is a p′-group, Since [Z(J), T] = 1, we conclude that T = O_{p}(Z(J)T). So, as [R, T] ≤ Z(J), R normalizes T and [R, T] ≤ T ∩ Z(J) = 1. Since J is generated by p-groups, this means [R, J] = 1 and so R = C_{M}(J).

(b): By (a), [N, J] ≰ Z(J). So by [MS1, 3.1] there exists K ∈ J with K ≤ [N, J].

(c) and (d) follow from [MS1, 3.2], and [MS1, 3.4], and (f) is The Other P (G, V)-Theorem in [MS1].

(e): By (c) and (d) K is generated by p′-elements. Hence (e) follows from elementary properties of coprime action.

(g): Put C := C_{J}(IE). Clearly Z(J) ≤ C. Hence, by (b) either C = Z(J), or there exists a J-component in C. Assume the latter case. Then by (c) and (d), p = 2 and I ≤ C. The action of C on [V, I] shows that C = IC_{C}([V, I]). But now again (b), this time applied to C_{C}([V, I]), gives C_{C}([V, I]) ≤ Z(J) and thus C = Z(J)I.

(h): Note that K[K, U] = K^{u}[K, U] for every u ∈ U. Assume first that U ≰ N_{M}(K). Then there exists u ∈ U \ N_{U}(K), and by (f) [W, K] ≤ C_{W} (K^{u}). Now (e) yields

[W, K] = [W, K, K] ≤ [W, K, K^{u}[K, U]] = [W, K, [K, U]] ≤ [W, [K, U]].

Assume now that U ≤ N_{M}(K), [K, U] ≠ 1 and [W, K] ≠ 0. Then 1 ≠ [K, U] ⊴ K. By (c) and (d) K is a component, or K ≅ C_{3}, or K ≅ Q_{8}. In the first case K ≤ [K, U], and (h) follows. In the other two cases by (d) [W, K] = [V, K] is a faithful simple K-module, so [V, K] = [V, [K, U]].

Lemma 2.3. Let A be a best offender of M on V and K ∈ J. Then the following hold:

(a) [K, A] = K or [K, A] = 1.

(b) If [K, A] ≠ 1, then there exists a best offender A_{0} ≤ A such that K = [K, A_{0}], [[V, K], A_{0}, A] = 0, and A_{0} is quadratic on [V, K].

Proof. (a) is obvious since K ⊴ J and by 2.2 either K is quasisimple or isomorphic to C_{3} or Q_{8}.

(b): This is essentially [MS1, 3.3], but since our assumption is slightly weaker we repeat the proof: By (a) [K, A] = K and by 2.2(e) [V, K] = [V, K, K], so [V, K, A] ≠ 0. The Timmesfeld Replacement Theorem [MS1, 2.7] with W := [V, K] gives a best offender A_{0} ≤ A satisfying [W, A_{0}, A] = 0 and [W, A_{0}] ≠ 0. The first property shows that A_{0} is quadratic on W. Suppose that [K, A_{0}] = 1. Then by [MS1, 2.9], [W, A_{0}] = 0, a contradiction. Thus [K, A_{0}] ≠ 1 and by (a), K = [K, A_{0}].

Lemma 2.4. Let K ∈ J and A be a subgroup of M such that [V, A, A] = 0 and [K, A] ≠ 1. Suppose that X is a perfect K-submodule of V and \overline{X} is a non-zero K-factor module of X. Then

C_{A}(X) = C_{A}(K) = C_{A}(\overline{X}).

Proof. Put L := [K, A]. The quadratic and faithful action of A shows that A is an elementary abelian p-subgroup. Hence A_{0} := C_{A}(K) centralizes ⟨K, A⟩ and so also L. The quadratic action of A gives

[V, L] ≤ [V, ⟨A^{K}⟩] = ⟨[V, A]^{K}⟩ ≤ C_{V} (A_{0}).

As [K, A] ≠ 1, 2.2(h) yields X = [X, K] ≤ [X, L] ≤ C_{V} (A_{0}) and A_{0} ≤ C_{A}(X) ≤ C_{A}(\overline{X}). Conversely, [X, [K, C_{A}(\overline{X})]] ≠ X since \overline{X} ≠ 0. Hence again 2.2(h) implies that C_{A}(\overline{X}) ≤ C_{A}(K).

Lemma 2.5. Let K ∈ J and 𝕂 := End_{K}(V). Suppose that V is a simple K-module and M is generated by quadratic offenders on V. Then the following hold:

(a) 𝕂 is a finite field.

(b) M acts 𝕂-linearly on V, or |V | = 4 and M ≅ SL_{2}(2).

(c) F^{∗}(M) = Z(M)K, and C_{M}(K) = Z(M) if |V | > 4.

Proof. (a): By Schur’s Lemma 𝕂 is a finite division ring, so by Wedderburn’s Theorem 𝕂 is a field.

(b): Let A ≤ M be a quadratic offender and suppose A does not act 𝕂-linearly on V. Then by [MS3, 2.14], |A| = 2. Since |A| is an offender we get |V/C_{V} (A)| = 2. Since A does not act 𝕂-linearly, there exists 0 ≠ k ∈ 𝕂 which is inverted by a ∈ A^{♯}; and since k acts fixed-point-freely on V, |C_{V} (a)|^{2} = |V |. This implies |𝕂| = 4 = |V |. Hence M ≅ SL_{2}(2) and (b) is proved.

(c): Suppose K is solvable. Then by 2.2 |V | = 4 or |V | = 9 and (c) is obvious. So we may assume that K is not solvable and so by 2.2 K is a component of M; in particular F^{∗}(M) = KC_{F^{∗}(M)}(K). By (b) M acts 𝕂-linearly on V, so C_{M}(K) ≤ Z(M), and F^{∗}(M) = KC_{F^{∗}(M)}(K) = KZ(M).

Lemma 2.6. Let K ∈ J and X be a perfect K-submodule of V, and let A be a best offender of M on V such that [K, A] ≠ 1. Then A normalizes X.

Proof. By 2.3(b) there exists a best offender A_{0} ≤ A such that [K, A_{0}] = K, [[V, K], A_{0}, A] = 0 and A_{0} is quadratic on [V, K]. Clearly A normalizes K since K ⊴ J.

We will first show that A_{0} normalizes X. Note that by 1.2 A_{0} is a best offender on W := ⟨X^{A_{0}}⟩. Let R := rad_{K}(W), that is, the intersection of the maximal K-submodules of W, and put \overline{W} := W/R. Note that W = [W, K] and so by 2.4 C_{A_{0}}(W) = C_{A_{0}}(\overline{W}) = C_{A_{0}}(K). Since A_{0} is a quadratic offender on W, we conclude that A_{0} is also a quadratic offender on \overline{W}. Thus there exists a quadratic best offender A_{1} ≤ A_{0} on \overline{W} such that [\overline{W} , A_{1}] ≠ 0 and so by 2.4 [K, A_{1}] ≠ 1.

Note that \overline{X} is a semisimple K-module. Let \overline{Y} be any simple K-submodule of \overline{X}. By [MS1, 2.10] A_{1} normalizes \overline{Y}. Moreover, since \overline{X} is a perfect K-module and [K, A_{1}] ≠ 1, 2.4 gives [\overline{Y} , A_{1}] ≠ 0. Now 0 ≠ [\overline{Y} , A_{1}] ≤ C\overline{Y} (A_{0}) shows that also A_{0} normalizes \overline{Y}. Hence, A_{0} normalizes \overline{X} and W = X + R, so W = X.

Thus A_{0} normalizes X. Let a ∈ A. Then [X, A_{0}] ≤ X ∩ X^{a} =: D. Since D is a KA_{0}-module and [X, A_{0}] ≤ D, we get from 2.2(h) X = [X, K] ≤ [X, [K, A_{0}]] ≤ D and thus X^{a} = X. So A normalizes X.

Lemma 2.7. Let K ∈ J and X be a perfect K-submodule of V, and let B be a best offender of M on V such that [K, B] = 0. Then [X, B] = 0.

Proof. Let X be a counterexample such that dim_{𝔽_{p}} X is minimal, and let W be a maximal Ksubmodule of X. We use the following notation:

Y := ⟨X^{B}⟩, U := [W, K], B_{0} := C_{B}(Y), \overline{Y} := Y/C_{Y} (K).

Note that [Y, K] = Y. Since [Y, C_{B}(\overline{Y}), K] = 0 and [C_{B}(\overline{Y}), K] ≤ [B, K] = 1, the Three Subgroups Lemma gives [Y, C_{B}(\overline{Y})] = [K, Y, C_{B}(\overline{Y})] = 0. It follows that

C_{B}(X) = B_{0} = C_{B}(\overline{Y}) = C_{B}(\overline{X}).

As B is a best offender on Y by 1.2, B is an offender on \overline{Y}.

Since U is a perfect K-module, the minimality of X gives [U, B] = 0. Thus [W, K, B] = 0 and [K, B] = 0, and the Three Subgroups Lemma yields [W, B, K] = 0. Thus [\overline{W} , B] = 0 and so C\overline{X}(b) = \overline{W} for every b ∈ B \ B_{0} since \overline{X}/\overline{W} is simple. Hence [\overline{X}, b] ≅ \overline{X}/C\overline{X}(b) = \overline{X}/\overline{W} ≅ X/W := I. This shows that [\overline{X}, B] is the direct sum of, say n, copies of I.

Put 𝔽 := End_{K}(I). Let

κ_{b} : \overline{X} → [\overline{X}, B] with \overline{x} + \overline{W} ↦ [\overline{x}, b]. (b ∈ B)

Then b ↦ κ_{b}, b ∈ B, defines to a homomorphism from B to Hom_{𝔽}(\overline{X}/\overline{W} , [\overline{X}, B]) ≅ 𝔽^{n} whose kernel is C_{B}(\overline{X}) = C_{B}(X). It follows that |B/C_{B}(X)| ≤ |𝔽|^{n}. Since B is an offender on \overline{Y} with B_{0} = C_{B}(\overline{Y}) and C\overline{X}(B) = \overline{W},

|𝔽|^{n} ≥ |B/B_{0}| ≥ |\overline{Y} /C\overline{Y} (B)| ≥ |\overline{X}C\overline{Y} (B)/C\overline{Y} (B)| = |\overline{X}/\overline{W} | = |I|,

so

(+)  dim_{𝔽} I ≤ n.

According to 1.2 and (b) there exists a best offender A on V such that [K, A] = K and A is quadratic on V. By 2.6 A normalizes X, Y and U and thus also W and X/W since W/U = C_{X/U}(K). Let b ∈ B \ C_{B}(\overline{X}). Then [X, b] is a perfect K-submodule of Y, and so again by 2.6 A normalizes [X, b] and thus also [\overline{X}, b]. Since I = X/W ≅ [\overline{X}, b] as K-module, D := Hom_{K}(I, [\overline{X}, b]) is a nontrivial p-group. Since A acts on D we get C_{D}(A) ≠ 0 and so Hom_{KA}(I, [\overline{X}, b]) ≠ 0. Thus [\overline{X}, b] is isomorphic to I as an KA-module.

By 2.4

(∗)  C_{A}(I) = C_{A}(K) = C_{A}(Y),

so 1.2 shows that A is a non-trivial quadratic offender on I. Hence by 2.5(b) A acts 𝔽-linearly on I or |I| = 4. In the latter case (∗) implies |A/C_{A}(I)| = 2 = |Y/C_{Y} (A)|, |K| = 3 and |Y | = 4. In particular [Y, B] = 0.

Assume now that A acts 𝔽-linearly on I. Let m = dim_{𝔽} I and c = dim_{𝔽} C_{I}(A). Recall that \overline{Y} = \overline{X} + [\overline{X}, B] and [\overline{X}, B] is the direct sum of n copies of KA-modules isomorphic to I. Hence

dim_{𝔽} Y/C_{Y} (A) ≥ dim_{𝔽} \overline{Y} /C\overline{Y} (A) ≥ n · dim_{𝔽} I/C_{I}(A) = n(m − c).

Since A acts quadratically on I, |A/C_{A}(I)| ≤ |Hom_{𝔽}(I/C_{I}(A), C_{I}(A))|, so |A/C_{A}(I)| ≤ |𝔽|^{c(m−c)}. On the other hand, by (∗) C_{A}(I) = C_{A}(Y) and so by (+)

|A/C_{A}(Y)| = |A/C_{A}(I)| ≤ |𝔽|^{c(m−c)} < |𝔽|^{n(m−c)} ≤ |Y/C_{Y} (A)|,

a contradiction since A is an offender.

Proposition 2.8. Let K ∈ J and X be a perfect K-submodule of V. Then J normalizes X.

Proof. This follows from 2.6 and 2.7.

Lemma 2.9. Let K ∈ J and let

X_{0} ≤ Y_{1} ≤ X_{1} ≤ Y_{2} ≤ X_{2} … ≤ Y_{n} ≤ X_{n} ≤ V

be a series of K-submodules such that X_{i} = [X_{i}, K], X_{i}/Y_{i} is a simple K-module, and [Y_{i}, K] ≤ X_{i−1} for i = 1, …, n. Then the following hold for S := ⊕_{i=1}^{n}X_{i}/Y_{i}:

(a) J acts on S and O_{p}(J̃) = 1, where J̃:= J/C_{J}(S).

(b) Every best offender on V is an offender on S; in particular J̃ is generated by offenders on S.

(c) K̃ is the unique JJ̃(S)-component of J̃.

Proof. (a): By 2.8 J normalizes every X_{i} and Y_{i} since Y_{i}/X_{i−1} = C_{X_{i}/X_{i−1}}(K), so J acts on S. Since X_{i}/Y_{i}, i ≥ 1, is a simple K-module, we also get O_{p}(J̃) = 1.

(b): Let A be a best offender on V. By 2.7 [S, A] = 0 if [K, A] = 1. In the other case 2.4 shows that

(∗)  C_{A}(K) = C_{A}(X_{i}) = C_{A}(X_{i}/Y_{i}), i = 1, …, n.

Hence in both cases C_{A}(S) = C_{A}(K).

By 1.2 A is a best offender on X_{n}. Hence

|X_{n}/C_{X_{n}}(A)| ≤ |A/C_{A}(X_{n})| = |A/C_{A}(K)| = |A/C_{A}(S)|.

On the other hand,

|X_{n}| = |X_{n}/Y_{n}||Y_{n}/X_{n−1}||X_{n−1}/Y_{n−1}| ⋯ |X_{1}/Y_{1}||Y_{1}|

and

|C_{X_{n}}(A)| ≤ |C_{X_{n}/Y_{n}}(A)||Y_{n}/X_{n−1}||C_{X_{n−1}/Y_{n−1}}(A)| ⋯ |C_{X_{1}/Y_{1}}(A)||Y_{1}|,

so

|A/C_{A}(S)| ≥ |X_{n}/C_{X_{n}}(A)| ≥ |X_{n}/Y_{n}/C_{X_{n}/Y_{n}}(A)| ⋯ |X_{1}/Y_{1}/C_{X_{1}/Y_{1}}(A)| ≥ |S/C_{S}(A)|.

This shows that A is an offender on S.

(c): There exists a best offender A on V such that [K, A] ≠ 1 and thus by (∗) also [S, A] ≠ 0. By (b) A is an offender on S, so A contains a non-trivial best offender B on S. Again (∗) yields [K, B] ≠ 1. Hence by 2.3(a), K̃ ≤ JJ̃(S) and so K̃ ⊴ JJ̃(S). Now 2.2(c) and (d) show that K̃ is a JJ̃(S)-component of J̃. Moreover, since [S, K̃] = S, 2.2(f) implies that K̃ is the unique JJ̃(S)-component of J̃.

Lemma 2.10. Let K ∈ J and L be a normal subgroup of M with L = O^{p′}(L). Then either K ≤ [K, L] ≤ L or [K, L] = 1.

Proof. If K is a component of M, this is [KS, 6.5.2]. So suppose K is solvable. Then either p = 2 and K ≅ C_{3}, or p = 3 and K ≅ Q_{8}.

We may assume that [K, L] ≠ 1. Since L = O^{p′}(L), there exists a p-subgroup T of L with [K, T] ≠ 1. If If T normalizes K, the structure of Aut(K) shows that K = [K, T] ≤ [K, L] ≤ L. So we may assume there exists t ∈ T with K ≠ K^{t}. Put L_{0} := KK^{t} ∩ L. Then L_{0} ⊴ J, and KK^{t} = KL_{0} = K^{t}L_{0} since [K, t] ≤ L. In particular [L_{0}, J] ≠ 1 since K = [K, J] ≠ K^{t}. Hence, by 2.2(b) there exists a J-component K̃ ≤ L_{0}, so K̃ ≤ KK^{t}. If K̃ = K or K^{t}, then K ≤ KK^{t} = K̃L_{0} ≤ L_{0} ≤ L. Suppose that K̃ is different from K and K^{t}. Then by 2.2(e),(f)

[V, K̃] = [V, K̃, K̃] ≤ [V, KK^{t}, K̃] = 0,

a contradiction.

Lemma 2.11. Let K ∈ J, W a K-submodule of V, \overline{V} := V/W and U a K-submodule of \overline{V}. Then the following are equivalent:

(a) U is a perfect K-module and U/C_{U}(K) is a simple K-module.

(b) U is a quasisimple K-module.

(c) U is a minimal non-trivial K-submodule of \overline{V}.

Proof. (a) ⟹ (b): Let N be the inverse image of O_{p}(K/C_{K}(U)) in K. Then U ≠ [U, N] and since U is a perfect K-module, N ≠ K. By 2.2 K is quasisimple or K is p′-group. In the first case N ≤ Z(K) and since O_{p}(K) ≤ O_{p}(M) = 1, N is a p′-group. So in any case N is a p′-group. Thus N/C_{K}(U) = 1 and so U is a quasisimple K-module.

(b) ⟹ (c): Let Y be non-zero K-submodule of U. By 2.2, K = O^{p}(K) and so C_{U}(K) = C_{U}(O^{p}(K)). Thus U/C_{U}(K) is a simple K-module. If Y ≰ C_{U}(K) we get U = Y + C_{V} (K) and so U = [U, K] = [Y, K] ≤ Y and Y = U. Thus, either Y = U or Y ≤ C_{U}(K), so Y is a minimal non-trivial K-submodule of \overline{V}.

(c) ⟹ (a): Since U is non-trivial, U ≠ C_{U}(K). Let Y be a proper K-submodule of U with C_{U}(K) ≤ Y. Then [Y, K] = 0 by minimality of U. Thus Y = C_{U}(K) and so U/C_{U}(K) is a simple K-module. Since K = O^{p}(K), [U, K, K] ≠ 1 and so U = [U, K] by minimality of U. Thus U is a perfect K-module and (a) holds.

3 Maximal Quadratic Offenders in Classical Groups

In this section 𝕂 is a field and V is an n-dimensional vector space over 𝕂. We assume that there exists a sesquilinear form f on V such that one of the following holds: (Recall here that f is non-degenerate if for each 0 ≠ v ∈ V there exists w ∈ V with f(v, w) ≠ 0.)

(i) f = 0.

(ii) f is a non-degenerate symplectic form on V; so f is bilinear and f(v, v) = 0 for v ∈ V.

(iii) f is a non-degenerate unitary form; so there exists α ∈ Aut(𝕂) such that α^{2} = id_{𝕂} ≠ α, f is linear in the first component, and f(v, w) = f(w, v)α for v, w ∈ V.

(iv) f is a symmetric bilinear form and there exists an associated non-degenerate quadratic form h on V, that is a function h : V → 𝕂 with

h(k_{1}v + k_{2}w) = k_{1}^{2}h(v) + k_{2}^{2}h(w) + k_{1}k_{2}f(v, w) for k_{1}, k_{2} ∈ 𝕂, v, w ∈ V.

(Recall here that h is non-degenerate if for each 0 ≠ v ∈ V with h(v) = 0 there exists w ∈ V with f(v, w) ≠ 0.) Also if char 𝕂 = 2, we assume that 𝕂 is perfect and so for each k ∈ 𝕂 there exists a unique element \sqrt{k} ∈ 𝕂 with \sqrt{k}^{2} = k.

By GL(V), Sp(V), GU(V), and O(V), respectively, we denote the group of automorphisms of V leaving invariant f (in the first three cases) and h in the fourth case. In the last three cases V is called a non-degenerate symplectic, unitary and orthogonal space, respectively.

We also use the notation GL_{n}(𝔽), Sp_{n}(𝔽), GU_{n}(𝔽), and O_{n}(𝔽), where n := dim V and either 𝔽 = 𝕂 or, in the unitary case, 𝔽 = 𝕂_{α}, the subfield centralized by α. In the first three cases put α = id_{𝕂}, so 𝔽 = 𝕂_{α}. If 𝔽 is finite, say |𝔽| = q, we also write GL_{n}(q), Sp_{n}(q), etc.

An element v ∈ V is called isotropic if f(v, v) = 0. A subspace U of V is called isotropic if f |_{U×U}= 0. An element v ∈ V is called singular if v isotropic and (in the fourth case) h(v) = 0. A subspace is called singular if it is isotropic and all its elements are singular.

By V^{∗} we denote the vector space dual to V, so V^{∗} := Hom_{𝕂}(V, 𝕂) and an element g ∈ GL(V) acts on V^{∗} via

xg : v ↦ (vg^{−1})x (x ∈ V^{∗}, v ∈ V).

We will use the notion of perpendicularity (and the symbol ⊥) with respect to f.

An α-sesquilinear form on V is a function g : V × V → 𝕂 such that g is 𝕂-linear in the first coordinate and α-semilinear in the second coordinate. We denote the set of α-sesquilinear forms on V be F_{α}(V). Observe that F_{α}(V) is vector space over 𝕂. Moreover, an element t ∈ GL_{𝕂}(V) acts on F_{α}(V) via

gt : (u, v) ↦ g(ut^{−1}, vt^{−1}) u, v ∈ V.

Let η ∈ {±}. An (α, η)-sesquilinear form on V is an α-sesquilinear form g with g(v, w) = ηg(w, v)α for all v, w ∈ V. F_{α,η}(V) denotes the set all (α, η)-sesquilinear forms. Note that F_{α,η}(V) is an 𝔽-subspace of F_{α}(V). ⋀_{2}(V) denotes the set of symplectic forms on V and S_{2}(V) denotes the set symmetric bilinear forms on V. So S_{2}(V) = F_{id,+}(V). Also ⋀_{2}(V) ≤ F_{id,−}(V) with equality if char 𝕂 ≠ 2.

Note that, if f ≠ 0, then f is an (α, ϵ)-sesquilinear form, where ϵ = + for M = O(V) or M = GU(V) and ϵ = − for M = Sp(V).

In the following M = GL(V), Sp(V), GU(V) and O(V), respectively. In this section we will write the action of M on V as right multiplication.

Lemma 3.1. Let U be an isotropic but not singular 𝕂-subspace of V. Let U_{0} be the set of singular vectors in U. Then G = O(V), p = 2, U_{0} is 𝕂-subspace of U and dim_{𝕂} U/U_{0} = 1. In particular, dim_{𝕂} V^{⊥} ≤ 1.

Proof. Since U is isotropic, f |_{U×U}= 0, so all elements in U are isotropic. Since U is not singular, there exists a non-singular element u in U. Since u is isotropic, we conclude that G = O(V) and h(u) ≠ 0. Then 4h(u) = h(2u) = h(u + u) = h(u) + f(u, u) + h(u) = 2h(u) and so p = 2. In particular, K is perfect and for every k ∈ 𝕂 there exists a unique \sqrt{k} such that \sqrt{k}^{2} = k. Consider the map

τ : U → 𝕂 with u → \sqrt{h(u)}.

Observe that U_{0} = ker τ. Since U is isotropic,

τ(u + v) = \sqrt{h(u + v)} = \sqrt{h(u) + f(u, v) + h(v)} = \sqrt{h(u)} + \sqrt{h(v)} = τ(u) + τ(v).

for all u, v ∈ U_{0}. Also

τ(ku) = \sqrt{h(ku)} = \sqrt{k^{2}h(u)} = kτ(u),

and so τ is 𝕂-linear. Thus U_{0} = ker τ is 𝕂-subspace and dim_{𝕂} U/U_{0} = dim_{𝕂} 𝕂 = 1.

Lemma 3.2. Suppose f ≠ 0. Let A ≤ M and U be subspace of V.

(a) V/U^{⊥} and U/U ∩ V^{⊥} are isomorphic 𝔽N_{M}(U)-modules. In particular, if f is non-degenerate, then V and V^{∗} are isomorphic 𝔽M-modules.

(b) C_{V/V^{⊥}}(A) = C_{V} (A)/V^{⊥}.

(c) C_{V} (A) = [V, A]^{⊥}.

(d) C_{M}(V/U) ≤ C_{M}(U^{⊥}); in particular C_{M}(V/U) ≤ C_{M}(U) if U is isotropic.

(e) If A acts quadratically on V/V^{⊥}, then A acts quadratically on V and [V, A] is an isotropic subspace of V.

Proof. (a): Replacing V by V/V^{⊥} and U by U + V^{⊥}/V^{⊥} we may assume that V^{⊥} = 0. For w ∈ V define w^{∗} : U → 𝕂, u ↦ f(u, w). Since f is 𝕂-linear in the first coordinate, w^{∗} ∈ U^{∗}. Define

ϕ : V → U^{∗}, v ↦ v^{∗}.

Since f is α-linear in the second coordinate, ϕ is α-linear and so 𝔽-linear. Moreover, ker ϕ = U^{⊥}. Hence dim V/U^{⊥} = dim V ϕ ≤ dim U^{∗} = dim U. This result applied to U^{⊥} gives dim V/U^{⊥⊥} ≤ dim U^{⊥} and since U ≤ U^{⊥⊥},

dim U ≤ dim U^{⊥⊥} ≤ dim V/U^{⊥} ≤ dim U.

So equality holds in the preceding inequalities. Therefore dim V ϕ = dim U^{∗} and ϕ is surjective. For g ∈ N_{M}(U) and u ∈ U:

u((wϕ)g) = (ug^{−1})(wϕ) = f(ug^{−1}, w) = f(u, wg) = u((wg)ϕ),

so (wϕ)g = (wg)ϕ. Thus (a) holds.

Put \overline{V} := V/V^{⊥} and define \overline{f} : \overline{V} → \overline{V} → 𝕂, (v + V^{⊥}, w + V^{⊥}) → f(v, w). Then \overline{f} is a non-degenerate form on \overline{V}.

(b): If V^{⊥} = 0, there is nothing to prove. So suppose V^{⊥} ≠ 0, that is G = O(V), char 𝕂 = 2, and n is odd. Let v ∈ V with \overline{v} ∈ C\overline{V} (A) and g ∈ A. Then vg = v + u for some u ∈ V^{⊥}, so

h(v) = h(vg) = h(v + u) = h(v) + f(u, v) + h(u) = h(v) + h(u).

Hence h(u) = 0. Since u ∈ V^{⊥} and h is non-degenerate this gives u = 0 and so v ∈ C_{V} (g). Thus (b) holds.

(c): By 1.8(c) and (a) we have C\overline{V} (A) = [\overline{V} , A]^{⊥}. Observe that [V, A]^{⊥} is the preimage of [\overline{V} , A]^{⊥} in V. By (b), C_{V} (A) is the preimage of C\overline{V} (A) in V. Thus (c) holds.

(d): Put C := C_{M}(V/U). Note that [V, C] ≤ U and so by (c), C_{V} (C) = [V, C]^{⊥} ≥ U^{⊥}. Hence C ≤ C_{M}(U^{⊥}). If U is, in addition, isotropic, U ≤ U^{⊥} and so C ≤ C_{M}(U).

(e): Suppose that A is quadratic on \overline{V}. Then [\overline{V} , A] ≤ C\overline{V} (A) = \overline{C_{V} (A)}. Thus [V, A, A] = 0 and [V, A] ≤ C_{V} (A) = [V, A]^{⊥} by (c). Hence [V, A] is isotropic.

Lemma 3.3. Suppose that f ≠ 0 and U is an isotropic subspace of V with U ∩ V^{⊥} = 0. Put \overline{V} := V/U^{⊥}, D := C_{GL(V)}(U^{⊥}) ∩ C_{GL(V)}(V/U) and

f_{d}(\overline{x}, \overline{y}) := f(x, [y, d]) for all d ∈ D, x, y ∈ V.

Let d ∈ D. Then

(a)

λ : D → F_{α}(\overline{V}), d ↦ f_{d}

is a ℤN_{M}(U)-module isomorphism.

(b) f(xd, yd) = f(x, y) for all x, y ∈ V if and only if f_{d} ∈ F_{α,−ϵ}(\overline{V}).

(c) Suppose M = Sp(V) then d ∈ M if and only if f_{d} ∈ S_{2}(\overline{V}).

(d) Suppose M = GU(V), then d ∈ M if and only if f_{d} ∈ F_{α,−}(\overline{V}).

(e) Suppose M = O(V) and U is singular, then d ∈ M if and only if f_{d} ∈ ⋀_{2}(\overline{V}).

(f) Suppose that M = O(V) and U is not singular. Then there exists a unique \overline{w} ∈ \overline{V} such that

h(u) = f(w, u)^{2} for all u ∈ U.

Moreover, d ∈ M if and only if d ∈ S_{2}(\overline{V}) and

f_{d}(\overline{x}, \overline{x}) = f_{d}(\overline{w}, \overline{x})^{2} for all \overline{x} ∈ \overline{V} .

Proof. Observe that f_{d} is well-defined and α-sesquilinear, so f_{d} ∈ F_{α}(\overline{V}). Note that [V, D] ≤ U ≤ U^{⊥} and so [\overline{V} , D] = 0. Thus λ is a homomorphism, and for d ∈ D, g ∈ N_{M}(U) and h ∈ F_{α}(\overline{V})

(f_{d}g)(\overline{x}, \overline{y}) = f_{d}(\overline{x}g^{−1}, \overline{y}g^{−1}) = f(xg^{−1}, [yg^{−1}, d]) = f(xg^{−1}, −yg^{−1} + yg^{−1}d)
= f(xg^{−1}, (−y + y(g^{−1}dg))g^{−1}) = f(x, −y + y(g^{−1}dg))
= f_{d^{g}}(\overline{x}, \overline{y}).

To see that λ is a ℤN_{M}(U)-module isomorphism it remains to show that λ is bijective. The injectivity follows from the fact that [V, D] ≤ U and U ∩ V^{⊥} = 0.

Let g ∈ F_{α}(\overline{V}). For u ∈ U define ϕ_{u} ∈ \overline{V}^{∗} by \overline{x}ϕ_{u} := f(x, u) for all x ∈ V. Since U ∩ V^{⊥} = 0, the map U → \overline{V}^{∗}, u ↦ ϕ_{u}, is an α-semilinear isomorphism. For w ∈ \overline{V}, the map t ↦ g(t, w) is in \overline{V}^{∗} and so there exists a unique u_{w} ∈ U with \overline{x}ϕ_{u_{w}} = f(x, u_{w}) = g(\overline{x}, w) for all x ∈ V. Define d_{g} ∈ GL(V) by d_{g}(v) := v + u\overline{v}. Clearly d_{g} ∈ D, and for all x, y ∈ V:

f_{d_{g}}(\overline{x}, \overline{y}) = f(x, [y, d_{g}]) = f(x, u\overline{y}) = g(\overline{x}, \overline{y}),

so f_{d_{g}} = g, and λ is surjective. Thus (a) holds.

To prove (b) let d ∈ D. We will determine necessary and sufficient conditions for d to be in M. Since f is an (α, ϵ)-sesquilinear form and U is isotropic,

f(xd, yd) − f(x, y) = f(x + [x, d], y + [y, d]) − f(x, y) = f(x, [y, d]) + f([x, d], y) =
f(x, [y, d]) + ϵf(y, [x, d])α = f_{d}(\overline{x}, \overline{y}) + ϵf_{d}(\overline{y}, \overline{x})α.

Thus d preserves f if and only if

(1)  f_{d}(\overline{x}, \overline{y}) = −ϵf_{d}(\overline{y}, \overline{x})α for all \overline{x}, \overline{y} ∈ \overline{V} .

That is, if and only if f_{d} ∈ F_{α,−ϵ}(\overline{V}). So (b) follows.

(c) and (d): These statements follow immediately from (b).

(d) and (e): So suppose that G = O(V) and let d ∈ D such that (1) holds. Since ϵ = 1 and α = id_{𝕂}, f_{d} is a skew-symmetric form. Then

(2)  h(xd) − h(x) = h(x + [x, d]) − h(x) = f(x, [x, d]) + h([x, d]) = f_{d}(\overline{x}, \overline{x}) + h([x, d]).

So

(3)  d ∈ O(V) if and only if d ∈ F_{id,−}(\overline{V}) and f_{d}(\overline{x}, \overline{x}) = −h([x, d]) for all x ∈ V.

If U is singular, then h([x, d] = 0 and we conclude that (d) holds. So suppose U is not singular. Then p = 2. Define δ : U → 𝕂, u ↦ \sqrt{h(u)}, and observe that δ is 𝕂-linear, so δ ∈ U^{∗}. On the other hand the map

ϕ^{∗} : \overline{V} → U^{∗}, ϕ^{∗}(\overline{v}) : u ↦ f(v, u)

is an isomorphism. Thus there exists a unique \overline{w} ∈ \overline{V} with ϕ^{∗}(\overline{w}) = δ. This gives

h(u) = δ(u)^{2} = f(w, u)^{2} for all u ∈ U.

Together with (3) we conclude that (e) holds.

Lemma 3.4. Let U be an k-dimensional isotropic subspace of V and E := C_{M}(U) ∩ C_{M}(V/U).

(a) Suppose M = GL(V). Then E ≅ U ⊗_{𝕂} (V/U)^{∗}, |E| = |𝕂|^{k(n−k)} and |V/C_{V} (E)| = |𝕂|^{n−k}.

(b) Suppose M = Sp(V). Then E ≅ S_{2}(U^{∗}), |E| = |𝕂|^{\frac{k(k+1)}{2}} and |V/C_{V} (E)| = |𝕂|^{k}.

(c) Suppose M = GU(V) Then E ≅ F_{α,−}(U^{∗}), |E| = |𝔽|^{k^{2}} and |V/C_{V} (E)| = |𝔽|^{2k}.

(d) Suppose M = O(V) and U is singular. Then E ≅ ⋀_{2}(U^{∗}), |E| = |𝕂|^{\frac{k(k−1)}{2}} , |V/C_{V} (E)| = |𝕂|^{k},

(e) Suppose M = O(V) and U is not singular. Put U_{0} := {u ∈ U | h(u) = 0}, E_{0} := C_{E}(V/U_{0}), and E_{1} := E ∩Ω_{n}(V). Then p = 2, E_{0} ≤ E_{1} ≤ E, E_{1}/E_{0} ≅ U_{0}, E_{0} ≅ ⋀_{2}(U_{0}^{∗}), and |E_{1}| = |𝕂|^{\frac{k(k−1)}{2}} . If V^{⊥} ∩ U ≠ 0 then |V/C_{V} (E)| = |𝕂|^{k−1} and E = E_{1}. If V^{⊥} ∩ U = 0 then |V/C_{V} (E)| = |𝕂|^{k} and |E/E_{1}| = 2.

Here all the isomorphisms are ℤN_{M}(U)-module isomorphisms.

Proof. Suppose first that f = 0, so M = GL(V). Then clearly E ≅ Hom_{𝕂}(V/U, U) ≅ U ⊗_{𝕂} (V/U)^{∗} and (a) holds.

Suppose next that f ≠ 0 and U ∩ V^{⊥} = 0. We apply 3.3 with the notation introduced there. Since [V, E] ≤ U, 3.2(c) gives C_{V} (E) = [V, E]^{⊥} ≥ U^{⊥} and so E ≤ D. Thus E = D ∩ M. So 3.3(c),

(d) and (e) imply (b), (c) and (d).

Suppose that G = O(V) and U is not singular. Let d ∈ D. By 3.3(f) there exists w ∈ V with

(2)  h(u) = f(w, u)^{2} for all u ∈ U.

and

(3)  d ∈ O(V) if and only if d ∈ S_{2}(\overline{V}) and f_{d}(\overline{x}, \overline{x}) = f_{d}(\overline{w}, \overline{x})^{2} for all x ∈ V

Recall from the proof of 3.3 that the map ϕ^{∗} : \overline{V} → U^{∗} with \overline{v}ϕ^{∗} : u ↦ f(v, u) is an isomorphism. For δ := \overline{w}ϕ^{∗} we get from (3) that ker δ = U_{0} = w^{⊥} ∩ U. Note that ϕ^{∗} also induces an isomorphism \overline{V} /𝕂\overline{w} → (ker δ)^{∗} = (U_{0})^{∗}.

Consider the map τ : E → \overline{V}^{∗} defined by \overline{x}τ(d) := f_{d}(\overline{w}, \overline{x}). By (3) ker τ consists of all d ∈ D such that f_{d} is a symplectic form on \overline{V} with \overline{w} ∈ radf_{d}. Also f_{d}(\overline{w}, \overline{x}) = 0 iff f(w, [x, d]) = 0 and (by (2)) iff h([x, d]) = 0. Thus d ∈ ker τ iff [V, d] ≤ U_{0}. Hence ker τ = E_{0}. As \overline{V} /𝕂\overline{w} ≅ U_{0}^{∗} we get

(5)  E_{0} = ker τ ≅ ⋀_{2}(\overline{V} /𝕂\overline{w}) ≅ ⋀_{2}(U_{0}^{∗}).

We claim that Im τ = X_{1} := \{ϕ ∈ \overline{V}^{∗} | ϕ(\overline{w}) ∈ \{0, 1\}\}.

If d ∈ E then (3) applied with \overline{x} = \overline{w} gives f_{d}(\overline{w}, \overline{w}) = f_{d}(\overline{w}, \overline{w})^{2} and so f_{w}(\overline{w}, \overline{w})^{2} ∈ \{0, 1\}. Hence Im τ ≤ X_{1}.

Conversely let ϕ ∈ \overline{V}^{∗} with ϕ(\overline{w}) = 1. Define g : \overline{V} ×\overline{V} , (\overline{x}, \overline{y}) ↦ ϕ(\overline{x})ϕ(\overline{y}). Then g is a symmetric bilinear form on \overline{V}, so by 3.3 with d_{g} := gλ^{−1}

f_{d_{g}}(\overline{w}, \overline{x}) = g(\overline{w}, \overline{x}) = ϕ(\overline{x})ϕ(\overline{w}) = ϕ(\overline{x})

and

f_{d_{g}}(\overline{x}, \overline{x}) = g(\overline{x}, \overline{x}) = ϕ(\overline{x})^{2} = g(\overline{w}, \overline{x})^{2} = f_{d_{g}}(\overline{w}, \overline{x}).

Thus by (3), d_{g} ∈ E and τ(d_{g}) = ϕ. Any ϕ ∈ \overline{V}^{∗} with ϕ(\overline{w}) = 0 can be written as a sum ϕ_{1} + ϕ_{2} where ϕ_{i} ∈ \overline{V}^{∗} and ϕ_{i}(\overline{w}) = 1. It follow that τ(E) = X_{1}.

Put X_{0} := \{ϕ ∈ \overline{V}^{∗} | ϕ(\overline{w}) = 0\}. Then X_{0} ≅ (\overline{V} /𝕂\overline{w})^{∗} ≅ U_{0}. Also |X_{1}/X_{0}| = 2 and so (e) holds. Thus we have proved all claims in the case V^{⊥} ∩ U = 0.

Suppose now that V^{⊥} ∩U ≠ 0. Then V is an orthogonal space and dim V^{⊥} = 1, so V^{⊥} ≤ U. Let Ṽ be an orthogonal space of dimension n + 1 with V ≤ Ṽ and Ṽ^{⊥} = 0; in particular, Ṽ^{⊥} ∩ U = 0. Put M̃ = O(Ṽ) and Ẽ := CM̃(U) ∩ CM̃(Ṽ /U). Then (e) holds for Ṽ, M̃ and Ẽ.

Note that in Ṽ, V^{⊥⊥} = V. Since V^{⊥} ≤ U, this gives Ẽ ≤ CM̃(V^{⊥}) ≤ NM̃(V) and we obtain a homomorphism β : Ẽ → E, e ↦ eCM̃(V). Note that ker β has order two, indeed the only non-trivial element in ker β is the transvection associated to the 1-space V^{⊥}. By Witt’s theorem, β is onto. Also ker β is not contained in Ẽ ∩ Ω(Ṽ). Thus (e) applied to M̃ shows that E ≅ Ẽ_{0}, and (e) also holds in this case.

Lemma 3.5. Let U be a isotropic subspace of V, let U_{0} be the subspace of all singular elements of U and put k = dim_{𝕂} U_{0}. Suppose that 𝕂 is finite and k ≥ 2. Put E := C_{M}(U) ∩ C_{M}(V/U), and P := O^{p′}(N_{M′}(U)), where p = char 𝕂.

(a) If M = GL(V) or GU(V) then E is a simple 𝔽_{p}P -module.

(b) If M = Sp(V) and p is odd, then E is a simple 𝔽_{p}P module.

(c) If M = O(V) and U is singular, then one of the following holds:

1. k ≥ 3 and E is a simple 𝔽_{p}P -module.

2. k = 2, P centralizes E and E is a simple 𝔽_{p}N_{M′}(U)-module.

(d) Suppose M = Sp(V) and p = 2 or M = O(V) and U is not singular. Then p = 2. Let E_{0} be the sum of the simple 𝔽_{2}P -submodules of E. Then one of the following holds:

1. k ≥ 3, E_{0} is a simple 𝔽_{2}P -module, and E_{0} ≅ ⋀_{2} U_{0}^{∗}.

2. k = 2, |𝕂| > 2 or V^{⊥} ≰ U, E_{0} = C_{E}(P). |E_{0}| = |𝕂| and N_{M′}(U) acts simply on E_{0}.

3. k = 2, |𝕂| = 2, M = Sp(V) or V^{⊥} ≤ U, and E is the direct sum of simple 𝔽_{2}P -modules of order 2 and 4.

Proof. Let S be a Sylow p-subgroup of P and D be a simple 𝔽_{p}P -submodule of E.

Assume first that M = GL(V) and put S_{0} := C_{S}(V/U)). Then S_{0} induces a Sylow p-subgroup of GL_{𝕂}(U) on U. Hence 3.4 implies that C_{E}(S_{0}) ≅ x ⊗ (V/U)^{∗} for some 0 ≠ x ∈ U. Thus C_{P} (U) acts simply on C_{E}(S_{0}) and so C_{E}(S_{0}) ≤ D. Since C_{P} (V/U) acts simply on U, we conclude that E = ⟨C_{E}(S_{0})^{C_{P} (V/U)}⟩ ≤ D. Thus E is a simple 𝔽_{p}P -module.

Assume next that f ≠ 0 and U ∩ V^{⊥} = 0. Put W := V/U^{⊥} and note that dim W = dim U. By Witt’s Theorem S induces a Sylow p-subgroup of GL_{𝕂}(U) on U and thus also on W. Thus C_{W} (S) is 1-dimensional. By 3.4 E is embedded into F_{α,−ϵ}(W). Let 1 ≠ x ∈ C_{D}(S), and let f_{x} ∈ F_{α,−ϵ}(W), f_{x} as in 3.3. Then f_{x} is invariant under S, so W/rad f_{x} possesses a non-degenerate (α, −ϵ) sesquilinear form invariant under a Sylow p-subgroup of GL(W/rad f_{x}). If follows that either W/rad f_{x} is 1-dimensional or α = id_{𝕂}, −ϵ = −1 and dim W/rad f_{x} = 2.

Suppose that M = Sp(V) and p is odd or that M = GU(V), so dim_{𝕂} U = k. Then P induces SL_{𝕂}(U) on U. Moreover dim W/rad f_{x} = 1 and N_{P} (S) acts simply on the subspace 𝔽f_{x} of F_{α,−ϵ}(W). Also for any ψ ∈ F_{α,−ϵ}(W) there exists a basis (x_{i})_{1≤i≤k} of W which is orthogonal with respect to ψ, that is, ψ(x_{i}, x_{j}) = 0 for i ≠ j. It follows that ψ is a 𝔽-linear combination of conjugates of f_{x} under P and so D = E.

Suppose that M = O(V) and U is singular. Then P induces SL_{𝕂}(U) on U. By 3.4(d) E ≅ ⋀_{2} W and f_{x} is a symplectic form. Thus dim W/rad f_{x} = 2. Let ψ ∈ ⋀_{2}(W). Then W has basis x_{i}, y_{i}, z_{s}, 1 ≤ i ≤ r and 1 ≤ s ≤ t, where ψ(x_{i}, y_{i}) = 1, ψ(y_{i}, x_{i}) = −1, and ψ(c, d) = 0 for any other pair of basis elements.

Assume that k ≥ 3. Then P acts transitively on the set of symplectic forms on W with radical of codimension 2. Hence ψ is a sum of P -conjugates of f_{x}. Thus D = E and (c:1) holds in this case. Assume that k = 2. Then P centralizes ⋀^{2} W. Also any scalar multiplication on W is induced by an element of N_{M′}(U) and so N_{M′}(U) acts simply on ⋀^{2} W. Thus (c:2) holds.

Suppose that M = O(V) and U is not singular. Put F = C_{M}(V/U_{0}). Note that F ≤ C_{M}(U_{0}^{⊥}) by 3.2(d), and so F ≤ E since U ≤ U_{0}^{⊥}. By the preceding case F ≅ ⋀_{2}(U_{0}^{∗}) and either k = 3 and F is a simple 𝔽_{p}P -module or k = 2, [F, P] = 1 and F is a simple N_{M′}(U)-module. Thus F ≤ E_{0} and it suffices to show that E_{0} ≤ F. Let \overline{w} be as in 3.3(f). The uniqueness of \overline{w} show that \overline{w} ∈ C_{W} (S). Since dim W = dim U > dim U_{0} ≥ 2 and dim W/radf_{x} ≤ 2 we have radf_{x} ≠ 0. Hence C_{radf_{x}}(S) ≠ 0 and since C_{W} (S) is 1-dimensional, \overline{w} ∈ radf_{x}. So 3.3(f) shows that f_{x} is symplectic and thus f_{x} ∈ F. Since D is simple, D ≤ F and E_{0} ≤ F.

Suppose M = Sp(V) and p = 2. Then by 3.4(b) E ≅ S_{2}(U^{∗}), and by 3.2(a) W ≅ U^{∗}, so S_{2}(U^{∗}) ≅ S_{2}(W). Since p = 2, ⋀_{2}(W) ≤ S_{2}(W). Let F be the inverse image of ⋀_{2}(W) in E. Then F ≅ ⋀_{2}(W) ≅ ⋀_{2}(U^{∗}). As seen in the case where U is singular either k ≥ 3 and E_{0} is a simple 𝔽_{p}P -module, or k = 2, [F, P] = 1 and N_{M′}(U) acts simply on F. If |𝕂| = 2 and k = 2, then |U| = 4 and |E| = 8 and it is easy to see that (d:3) holds. So suppose that |𝕂| > 2 or k > 2. We will show that D ≤ F. For this we just need to show that there exists 1 ≠ u ∈ D such that f_{u} is a symplectic form. Fix a basis (v_{i}) for W and for e ∈ E let M_{e} be the matrix (f_{e}(v_{i}, v_{j})). Then M_{e} is symmetric and e ∈ F if and only if all diagonal elements of M_{e} are zero. Moreover, dim W/radf_{e} = rankM_{e}. We may assume that f_{x} is not symplectic and so there exists v ∈ V with f_{x}(v, v) ≠ 0. Since 𝕂 is perfect we can choose v such that f_{x}(v, v) = 1. Put s = dim W/radf_{x}. Then either s = 1 and V = 𝕂v + radf_{x}, or s = 2, there exists w ∈ W with f_{x}(v, w) = 0 and f_{x}(w, w) = 1 and V = 𝕂v + 𝕂w + radf_{x}. So we can choose our basis such that f_{x}(v_{i}, v_{j}) = 1 for 1 ≤ i = j ≤ s and f_{x}(v_{i}, v_{j}) = 0 for all other i, j.

Suppose s = 1. Note that

(\begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}) + (\begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}) + (\begin{matrix}1 & 1 \\ 1 & 1 \end{matrix}) = (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix})

The three matrices on the left side of the equation all are symmetric of rank 1 and so conjugate under SL_{2}(𝕂) on it actions on S_{2}(𝕂^{2}). The matrix on the right is symplectic. Thus ⟨d^{P} ⟩ ∩ F ≠ 1 and so D ≤ F.

Suppose that s = 2 and |𝕂| > 2. Pick a, b ∈ 𝕂 \ {0, 1} with a + b = 1. Note that

(\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) + (\begin{matrix}a & b \\ b & a \end{matrix}) + (\begin{matrix}b & a \\ a & b \end{matrix}) = (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix})

The three matrices on the left side of the equation are symmetric, not symplectic and have determinant 1. So they are conjugate under SL_{2}(𝕂) on it actions on S_{2}(𝕂^{2}). The matrix on the right is symplectic and so again D ≤ F.

Suppose that s = 2, |𝕂| = 2 and k ≥ 3. We have

(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}) + (\begin{matrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{matrix}) = (\begin{matrix}0 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \end{matrix})

The two matrices on the left side of the equation are symmetric, not symplectic and have rank 2. So they are conjugate under SL_{3}(𝕂) on it actions on S_{2}(𝕂^{3}). The matrix on the right is symplectic and so again D ≤ F.

We have proved that D ≤ F. So E_{0} = F and (d:1) or (d:2) holds.

Assume finally that M = O(V), U is not singular and U ∩ V^{⊥} ≠ 0. Then p = 2 and M ≅ Sp(V/V^{⊥}). Hence the case where M = Sp(V) applied to V/V^{⊥} and U/V^{⊥} shows that (d) holds.

4 Smith’s Lemma and Ronan-Smith’s Lemma

In this section we provide a few pieces from the theory of equicharacteristic representations of groups of Lie-type. The material presented here essentially comes from [GLS3, Section 2.8] except that we are looking at representations over 𝔽_{p} rather than its algebraic closure \overline{𝔽_{p}}.

Lemma 4.1 (Steinberg’s Lemma). Let M be a genuine group of Lie-type defined over a finite field of characteristic p. Let V be a simple 𝔽_{p}M-module, S ∈ Syl_{p}(M), and B := N_{M}(S). Put 𝕂 := End_{M}(V). Then C_{V} (S) is 1-dimensional over 𝕂, 𝕂 is isomorphic to the subring of End_{𝔽_{p}}(C_{V} (S)) generated by the image of B, and C_{V} (S) is a simple 𝔽_{p}B-module.

Proof. Choose an embedding σ : 𝕂 → \overline{𝔽_{p}} and put \overline{V} = \overline{𝔽_{p}} ⊗_{𝕂} V. Then \overline{V} is a simple \overline{𝔽_{p}}M-module. Thus by [St, Theorem 46] C\overline{V} (S) is 1-dimensional over \overline{𝔽_{p}} and so C_{V} (S) is 1-dimensional over 𝕂. Define λ : B → 𝕂 by v^{b} = λ(b)v for all b ∈ B, v ∈ C_{V} (S), and let 𝔼 be the subfield of 𝕂 generated by λ(B). Let ρ ∈ Aut_{𝔼}(\overline{𝔽_{p}}). Then [St, Theorem 46] shows that \overline{V} ≅ \overline{V}^{ρ} as a 𝕂M-module. Thus ρ centralizes 𝕂 and so 𝕂 = 𝔼. Since C_{V} (S) is 1-dimensional over 𝕂 this implies that C_{V} (S) is a simple 𝔽_{p}B-module.

Let 𝔽 be a finite field of characteristic p, M a finite group, V a simple 𝔽M-module and W a simple 𝔽_{p}M-submodule. Recall that the field 𝕂 := End_{M}(W) is called the field of definition of the 𝔽M-module W.

Theorem 4.2 (Smith’s Lemma). Let M be a genuine group of Lie-type defined over a finite field of characteristic p. Let V be a simple 𝔽_{p}M-module, 𝕂 := End_{M}(V), E a parabolic subgroup of M, L := O^{p′}(E) and P = N_{M}(L). Then L = O^{p′}(P), O_{p}(E) = O_{p}(P) = O_{p}(L), and P is a Lie-parabolic subgroup of M. Moreover, C_{V} (O_{p}(P)) is a simple 𝔽_{p}P -module, an absolutely simple 𝕂L-module, and an absolutely simple 𝕂E-module.

Proof. Let S ∈ Syl_{p}(E) and B = N_{M}(S). Then P = BL = BE and so P is a Lie-parabolic subgroup of M. Since B/S is a p′-group we conclude that E = O^{p′}(P) and O_{p}(E) = O_{p}(L) = O_{p}(P).

Choose an embedding σ : 𝕂 → \overline{𝔽_{p}} and put \overline{V} = \overline{𝔽_{p}} ⊗_{𝕂} V. Then \overline{V} is a simple \overline{𝔽_{p}}M-module. Put U = C_{V} (O_{p}(P)) and \overline{U} = C\overline{V} (O_{p}(P)) = \overline{𝔽_{p}} ⊗_{𝕂} U. By [Ti] \overline{U} is a simple \overline{𝔽_{p}}P -module.

Let Y be a simple \overline{𝔽_{p}}L-submodule of \overline{U}. Then C_{Y} (S) ≠ 0, and since by [St, Theorem 46] C\overline{V} (S) is 1-dimensional over \overline{𝔽_{p}}, C\overline{V} (S) ≤ Y. Thus

\overline{U} = ⟨C\overline{U}(S)^{P} ⟩ = ⟨C\overline{U}(S)^{BL}⟩ = ⟨C\overline{U}(S)^{L}⟩ ≤ Y,

so \overline{U} is simple \overline{𝔽_{p}}L. Thus, U is an absolutely simple 𝕂L-module, and since L ≤ E, U is also an absolutely simple 𝕂E-module.

Let X be a simple 𝔽_{p}P -submodule of U. Then again 0 ≠ C_{X}(S) is B-invariant and since C_{V} (S) is a simple 𝔽_{p}B-module by 4.1, C_{V} (S) ≤ X. Since ⟨C_{V} (S)^{P} ⟩ is a 𝕂-submodule of U we conclude that X = U.

Theorem 4.3 (Ronan-Smith’s Lemma). Let M be a universal group of Lie-type defined over a finite field of characteristic p, S a Sylow p-subgroup of M, P_{1}, P_{2}, …, P_{n} the minimal Lie-parabolic subgroups of M containing S, and L_{i} = O^{p′}(P_{i}). Let V be the class of all tuples (𝕂, V_{1}, V_{2}, … V_{n}) such that

(i) 𝕂 is a finite field of characteristic p.

(ii) Each V_{i} is an absolutely simple 𝕂L_{i}-module.

(iii) 𝕂 = ⟨𝕂_{i} | 1 ≤ i ≤ n⟩, where 𝕂_{i} is the field of definition of the 𝕂L_{i}-module V_{i}.

Define two elements (𝕂, V_{1}, V_{2}, … V_{n}) and (𝕂̃, Ṽ_{1}, Ṽ_{2}, … Ṽ_{n}) of V to be isomorphic if there exists a field isomorphism σ : 𝕂̃ → 𝕂 such that V_{i} ≅ Ṽ_{i}^{σ} as an 𝕂L_{i}-module for all 1 ≤ i ≤ n. Then the map

V → (End_{M}(V), C_{V} (O_{p}(L_{i})), … C_{V} (O_{p}(L_{n}))) (V a simple 𝔽_{p}M-module)

induces a bijection between the isomorphism classes of simple 𝔽_{p}M-modules and the isomorphism classes of V.

Proof. Let V be a simple 𝔽_{p}M-module and put 𝕂 := End_{M}(V) and V_{i} := C_{V} (O_{p}(L_{i})). By Smith’s Lemma 4.2, V_{i} is an absolutely simple 𝕂L_{i}-module. Let 𝕂_{i} be the field of definition of the 𝕂L_{i}- module V_{i}. Put B := N_{M}(S). By 4.1 𝕂 is generated by the image of B in End_{𝔽_{p}}(C_{V} (S)). Moreover, each 𝕂_{i} is generated by the image of B ∩ L_{i} in C_{V} (S). Since B = ⟨B ∩ L_{i}, 1 ≤ i ≤ n⟩ we conclude that 𝕂 = ⟨𝕂_{i} | 1 ≤ i ≤ n⟩.

Clearly, if Ṽ is an 𝔽_{p}M-module isomorphic to V, then the corresponding elements of V are isomorphic.

Now let (𝕂, V_{1}, V_{2}, … V_{n}) ∈ V. Pick 0 ≠ v_{i} ∈ C_{V_{i}}(S) and define λ_{i}, n_{i} and μ_{i} as in [St, Theorem 46] applied to the \overline{𝔽_{p}}L_{i}/O_{p}(L_{i})-module \overline{V}_{i} = \overline{𝔽_{p}} ⊗_{𝕂} V_{i}. Since B/S = ×_{i=1}^{n}(B ∩ L_{i})/S, there exists a unique homomorphism λ : B → \overline{𝔽_{p}} with λ |_{B∩L_{o}}= λ_{i}. Let \overline{V} be the simple \overline{𝔽_{p}}M-module obtained from [St, Theorem 46]. Since C\overline{V} (O_{p}(V_{i})) is simple we conclude from [St, Theorem 46] applied to L_{i} that C\overline{V} (O_{p}(V_{i})) ≅ \overline{V_{i}}. Let V be a simple 𝔽_{p}M-submodule of \overline{V} and put 𝔼 = End_{M}(V). Then \overline{V} ≅ \overline{𝔽_{p}} ⊗_{𝔼} V as an \overline{𝔽_{p}}M-module. It is now easy to see that 𝔼 ≅ 𝕂, that V is send to (𝕂, V_{1}, V_{2}, … V_{n}) ∈ V and that V is unique up to isomorphism with this property.

5 Generating Genuine Groups of Lie-type

Lemma 5.1. Let G be a simple genuine group of Lie Type over a field of characteristic p, P^{+} a Lie-parabolic subgroup of G and P^{−} an opposite Lie-parabolic. Then G = ⟨O_{p}(P^{+}), O_{p}(P^{−})⟩.

Proof. Put L = ⟨O_{p}(P^{+}), O_{p}(P^{−})⟩. Since P^{+} is opposite to P^{−}, G = ⟨P^{+}, P^{−}⟩ and P^{ϵ} = O_{p}(P^{ϵ})(P^{+} ∩ P^{−}). It follows that L ⊴ L(P^{+} ∩ P^{−}) = ⟨P^{+}, P^{−}⟩ = G, and since G is simple, G = L.

Lemma 5.2. Let G ≅ G_{2}(q), p = q^{k}, P a Lie-parabolic subgroup of G with Z(O^{p′}(P)) = 1 and A ⊴ P with |A| = q^{3}. Then G = ⟨A, A^{t}⟩ for some t ∈ G.

Proof. Choose a root system Φ for G such that P is a Lie-parabolic with respect to Φ and let N/H be the corresponding Weyl-group. Let R_{l} ( R_{s}) be set root subgroups in G corresponding to the long (short) roots in Φ. Put L = ⟨R_{l}⟩. Then L is a genuine group of Lie-type of type A_{2} and P ∩ L is a Lie-parabolic subgroup of L with L ∩ A = O_{p}(P ∩ L). Since N/H ≅ D_{12} we can choose t ∈ N \ H with [t, N] ≤ H. Put K = ⟨A, A^{t}⟩. Since (P ∩ L)^{t} is opposite to P ∩ L in L, 5.1 implies that L = ⟨L ∩ A, (L ∩ A)^{t}⟩. Thus L ≤ K . Since (N ∩ L)H/H ≅ D_{6} we have N = (L ∩ H)⟨t⟩H and so N normalizes K. Since N acts transitive R_{s} and there exists R ∈ R_{s} with R ≤ A, ⟨R_{s}⟩ ≤ K. Hence G = ⟨R_{l}, R_{s}⟩ ≤ K and G = K.

Lemma 5.3. Let G ≅ SL_{n}(𝕂). Then G is generated by n root subgroups.

Proof. Let I = {1, …, n} and Φ = {e_{i} − e_{j} | i, j ∈ I, i ≠ j} by the root system for G and for ϕ ∈ Φ let Z_{ϕ} be the corresponding root subgroup. Then

(∗)  [Z_{e_{i}−e_{j}}, Z_{e_{j}−e_{k}}] = Z_{e_{i}−e_{k}} for all distinct i, j, k in I.

Put U := ⟨Z_{e_{i}−e_{i+1}} | n ≠ i ∈ I}⟩ and L := ⟨U, Z_{e_{n}−e_{1}}⟩. Let i, j ∈ I with i < j.

We will first show by induction on j − i that Z_{e_{i}−e_{j}} ∈ U. If j − i = 1, this holds by definition of U. So suppose j − i > 1 and by induction that Z_{e_{i}−e_{j−1}} ≤ U. Thus using (∗),

Z_{e_{i}−e_{j}} = [Z_{e_{i}−e_{j−1}}, Z_{e_{j−1}−e_{j}}] ≤ U.

Next we will show by downwards induction on j − i, then Z_{e_{j}−e_{i}} ≤ L. If j − i = n − 1, then j = n and i = 1 and so this holds by definition on L. So suppose j − i < n − 1.

Assume that i > 1 and by induction that Z_{e_{j}−e_{i−1}} ≤ L. Then by (∗)

Z_{e_{j}−e_{i}} = [Z_{e_{j}−e_{i−1}}, Z_{e_{i−1}−e_{i}}] ≤ U.

Assume that i = 1. Then j < n and by induction Z_{e_{j+1}−e_{i}} ≤ U. So by (∗)

Z_{e_{j}−e_{i}} = [Z_{e_{j}−e_{j+1}}, Z_{e_{j+1}−e_{i}}] ≤ U.

Thus L contains all Z_{ϕ}, ϕ ∈ Φ and so L = M.

Lemma 5.4. Let H be quasisimple with H/Z(H) ≅ Alt(6) and |Z(H)||3. Let S ∈ Syl_{2}(H), B = N_{H}(S), and M_{1} and M_{2} be the two maximal subgroups of H containing B. Let 𝕂 be a field of characteristic 2, V be a faithful 𝕂H-module, U a simple 𝕂B-submodule of V and put U_{i} := ⟨U^{M_{i}}⟩. Suppose that

(i) V = ⟨U^{M}⟩,

(ii) U = U_{1}, and

(iii) dim_{𝕂} U_{2} = 2 dim_{𝕂} U.

Then the following hold:

(a) Suppose H ≅ Alt(6), then V is a quotient of the natural even permutation module for H over 𝕂. In particular, V/C_{V} (H) is a natural 𝕂Alt(6)-module for H, dim_{𝕂} C_{V} (H) ≤ 1 and C_{V} (H) ≤ ⟨U_{2}^{M_{1}}⟩.

(b) Suppose H ∼ 3.Alt(6). Let 𝔼 be subring of End_{𝕂H}(V) generated by the images of 𝕂 and Z(H). Then 𝔼 is a field, 𝔼 = 𝕂(ξ) for ξ ∈ 𝔼 with |ξ| = 3, dim_{𝔼} U = 1 and dim_{𝔼} V = 3.

Proof. Since S ⊴ B and U is a simple 𝔽_{2}B-module, [U, S] = 0. As the Sylow 2-subgroups of Alt(6) are self-normalizing, B = SZ(H), and so U is a simple 𝕂Z(H)-module.

Since V = ⟨U^{M}⟩, Z(H) acts homogeneously on V and so the subring 𝔼 of End_{𝕂H}(V) generated by the images of 𝕂 and Z(H) is a field. Moreover, 𝔼 = 𝕂 if Z(H) = 1 or 𝕂 contains a non-trivial third root of unity; in the other case 𝔼 = 𝕂(ξ) where ξ ∈ 𝔼 \ 𝕂 with ξ^{3} = 1. Also dim_{𝔼} U = 1 and since dim_{𝕂} U_{2} = 2 dim_{𝕂} U, dim_{𝔼} U_{2} = 2.

Let A be the natural 𝔽_{2}Alt(6)-module for H with C_{A}(M_{1}) ≠ 0. Then there exists an Mequivariant bijection A^{♯} → U_{1}^{M}, a → U_{a}. We now use the fact that Alt(6) ≅ Sp_{4}(2)′ and A is also a natural Sp_{4}(2)′-module for H, so there exists an H-invariant non-degenerate symplectic form on A.

For B ⊆ A define U_{B} := ⟨U_{b} | b ∈ B^{♯}⟩ and W_{B} := U_{B^{⊥}}, where B^{⊥} is the 𝔽_{2}-subspace of A perpendicular to B with respect to the above mentioned symplectic form on A.

Let B be a singular 2-subspace of A. By Witt’s Theorem H acts transitively on the singular 2-subspaces of A and so U_{B} is a conjugate of U_{2}. In particular,

(∗)  U_{B} = U_{b} + U_{c} and U_{a+c} ≤ U_{a} + U_{c} for B = ⟨a, c⟩.

Now let a ∈ A^{♯}. Since dim_{𝔽_{2}} A = 4, a^{⊥} = ⟨a⟩ ⊕ B, where B is a non-singular 2-subspace. Then ⟨a, b⟩ is singular for every b ∈ B. Thus by (∗)

(∗∗)  W_{a} = Σ_{b∈B^{♯}}U_{⟨a,b⟩} = U_{a} + U_{B}.

Since |B^{♯}| = 3, dim_{𝔼} U_{B} ≤ 3 and so dim_{𝔼} W_{a} ≤ 4.

Now let d ∈ A \ a^{⊥} and put B := a^{⊥} ∩ d^{⊥}. Then B is a non-singular 2-space, and by (∗∗) applied to a and d, W_{a} + W_{d} = U_{a} + U_{B} + U_{d}. Thus dim_{𝔼} W_{a} + W_{d} ≤ 5.

Put W := W_{a} + W_{d}. We will show that V = W, that is U_{b} ≤ W for all b ∈ A^{♯}. Certainly U_{b} ≤ W if b ∈ a^{⊥} ∪ d^{⊥}. So suppose b ∉ a^{⊥} and b ∉ d^{⊥}.

Assume first that b ≠ a + d. Then ⟨b, d⟩ ≠ ⟨a, d⟩ and so also b^{⊥} ∩ a^{⊥} ≠ b^{⊥} ∩ d^{⊥}. Choose e ∈ b^{⊥} ∩ a^{⊥} \ d^{⊥}; in particular U_{e} ≤ W_{a}. Then e + b ≤ b^{⊥} ∩ d^{⊥}, so U_{e+b} ≤ W_{d}, and by (∗) U_{b} ≤ U_{e} + U_{e+b} ≤ W_{a} + W_{d} = W.

Assume next that b = a + d. Pick b̃ ∈ A \ (a^{⊥} ∪ d^{⊥}) with b̃ ≠ b. Put c = b + b̃. By the previous case Ub̃ ≤ W. Note that b̃ ∈ b^{⊥} and c ∈ a^{⊥}. Thus U_{c} ≤ W and by (∗) U_{b} ≤ Ub̃ +U_{c}. Hence U_{b} ≤ W.

We have shown that U_{b} ≤ W for all b ∈ A^{♯} and so W = V; in particular dim_{𝔼} V ≤ 5.

Suppose now that H ≅ Alt(6). Then Z(H) = 1 and 𝔼 = 𝕂. Let V̌ be the 𝕂H-module induced from the trivial 𝕂M_{1}-module U_{1}, and let Ǔ_{1} be the image of U_{1} in V̌. Put Ǔ_{2} := ⟨Ǔ_{1}^{M_{2}}⟩. Then Ǔ_{2}/CǓ_{2}(M_{2}) has dimension 2 over 𝕂. It follows that V̂ := V̌ /⟨CǓ_{2}(M_{2})^{H}⟩ fulfills the assumptions of (a).

Choose a faithful action of H on I := {1, 2, 3, 4, 5, 6} with

M_{1} = N_{H}({1, 2}) and M_{2} = N_{H}({{{1, 2}, {3, 4}, {5, 6}}.

Let Ṽ be the corresponding permutation module for H over 𝕂 with 𝕂 basis {b_{i} | i ∈ I}, and let Ṽ_{0} := {∑_{i∈I} k_{i}b_{i} | k_{i} ∈ 𝕂, ∑_{i∈I} k_{i} = 0} be the even permutation module. For J ⊆ I put b_{J} = ∑_{j∈J} b_{j}. Then M_{1} centralizes 𝕂b_{3456}, ⟨𝕂b_{3456}^{M_{2}}⟩ = 𝕂⟨b_{3456}, b_{1234}⟩ and Ṽ_{0} = 𝕂⟨b_{3456}^{H}⟩. Thus Ṽ_{0} and V are 𝕂H-quotients of V̂. Since dim_{𝕂} Ṽ_{0} = 5 and dim_{𝕂} V̂ ≤ 5 we conclude that V̂ is isomorphic to Ṽ_{0}. Thus V is isomorphic to a quotient of Ṽ_{0}. Observe that CṼ_{0}(H) = 𝕂⟨b_{123456}⟩ and b_{123456} = b_{1234} + b_{1235} + b_{1245} + b_{3456} ∈ 𝕂⟨b_{3456}^{M_{1}}, b_{1234}^{M_{1}}⟩. So (a) holds.

Suppose next that H ∼ 3.Alt(6). Let R be a Sylow 3-subgroup of H. The R is extraspecial of order 27. Let Y be any R-chief-factor of V. Then Z(H) = Z(R) acts non-trivially on Y and so dim_{𝔼} Y = 3. Thus dim_{𝔼} V is a multiple of three and since dim_{𝔼} V ≤ 5, dim_{𝔼} V = 3. So (b) holds.

6 Module Decompositions

Lemma 6.1. Let H be a finite group, V an 𝔽_{p}H-module, and 𝕂 := End_{H}(V). The following table lists the dimension d := dim_{𝕂}(H^{1}(H, V)) for various pairs (H, V).

H  p  V  Conditions  d
Ω_{n}^{ϵ}(p^{k}), n ≥ 3  p  V_{nat}^{∗}  n = 3, p^{k} = 2  1
”  ”  ”  n = 3, p^{k} = 5  1
”  ”  ”  n = 4, ϵ = −, p^{k} = 3 2
”  ”  ”  n = 5, p^{k} = 3  1
”  ”  ”  n = 6, ϵ = +, p^{k} = 2 1
”  ”  ”  all others  0
Sp_{2n}(p^{k})  p  V_{nat}  p = 2, (2n, p^{k}) ≠ (2, 2) 1
”  ”  ”  all others  0
SL_{n}(p^{k})  p  V_{nat}  n = 2, p = 2, k > 1 1
”  ”  ”  n = 3, p = 2, k = 1 1
”  ”  ”  all others  0
SU_{n}(p^{k}), n ≥ 3  p  V_{nat}  n = 4, p^{k} = 2  1
”  ”  ”  all others  0
G_{2}(2^{k})′  2  𝕂^{6}  −  1
G_{2}(p^{k})′  p ≠ 2  𝕂^{7}  −  0
3D_{4}(p^{k})  p  𝕂^{8}  −  0
Spin_{n}^{ϵ}(p^{k})  p (Half)-Spin  n ≥ 7  0
3.Alt(6)  2  𝕂^{3}  −  0
Alt(n), n ≥ 5  2  V_{nat}  n even  1
”  ”  ”  n odd  0
SL_{n}(p^{k}), n ≥ 5  p  ⋀^{2}(V_{nat})  −  0
SL_{n}(p^{k}), n ≥ 3  odd Sym^{2}(V_{nat})  −  0
SL_{n}(p^{2k}), n ≥ 3  p  V_{nat} ⊗ V_{nat}^{p^{k}}  n = 3, p^{2k} = 4  2
”  ”  ”  all others  0
E_{6}(p^{k})  p  𝕂^{27}  −  0
Mat_{n}, 22 ≤ n ≤ 24 2  Todd  n = 24  1
”  ”  ”  n = 22, 23  0
Mat_{n}, 22 ≤ n ≤ 24 2  Golay  n = 22  1
Mat_{n}, 22 ≤ n ≤ 24 2  Golay  n = 23, 24  0
3.Mat_{22}  2  𝔽_{4}^{6}  −  0
Mat_{11}  3  Todd  −  0
Mat_{11}  3  Golay  −  1
2.Mat_{12}  3  Todd  −  0
2.Mat_{12}  3  Golay  −  0

Proof. Let T ∈ Syl_{p}(H) and W be an 𝔽_{p}H-module with [W, H] ≤ V and C_{W} (H) ≤ V. Note that by Gaschütz’s Theorem, C_{W} (T) ≤ V.

1^{∘}. Let C ≤ H and A and B be normal p-subgroups of C with A ≤ B, and let X, Y, Z be C-submodules of W with X ≤ Y ≤ Z. Suppose that

(i) B centralizes Z/Y and Y/X.

(ii) A centralizes Z/X. (iii) Φ(B) ≤ A. Put U/X := C_{Z/X}(B). Then Z/U is isomorphic to a C-submodule of Hom_{𝔽_{p}}(B/A, Y/X). If in addition C centralizes Z/U, then Z/U embeds into Hom_{𝔽_{p}C}(B/A, Y/X).

For z ∈ Z define

z̃: B/A → Y/X with bA → [b, z] + X.

Since B/A and Y/X are 𝔽_{p}C-modules, for c ∈ C the element z̃^{c} := c^{−1}z̃c ∈ Hom_{𝔽_{p}}(B/A, Y/X) is defined, and

(bA)z̃^{c} = bA(c^{−1}z̃c) = (b^{c^{−1}}Az̃)c = ([b^{c^{−1}}, z] + X)^{c} = [b, z^{c}] + X = bAz̃^{c}.

Thus, the map

Z → Hom_{𝔽_{p}}(B/A, Y/X) with z → z̃

is C-equivariant with kernel U. So the first statement holds. The second follows from the first. Case 1. V is the dual of a natural module for H ≅ Ω_{n}^{ϵ}(q), n > 2 and q = p^{k}.

This case is covered by [Po] and [JP]. Case 2. V is a natural module for H = Sp_{2n}(q).

See [JP]. Case 3. V is a natural module for H = SL_{n}(q), q = p^{k}.

See [JP]. Case 4. V is a natural module for H = SU_{n}(q), q = p^{k}, and n ≥ 3.

If q > 3 see [JP]. So assume that q ≤ 3. If H is solvable, then H = SU_{3}(2), and Maschke’s Theorem shows that the lemma holds. Thus, assume in addition that H ≠ SU_{3}(2). Let V_{1} be a 1-dimensional singular 𝕂-subspace of V, V_{2} = V_{1}^{⊥} ≤ V, L = C_{H}(V_{1}), and L^{∗} = N_{H}(V_{1}).

Suppose for a contradiction that [V, O_{p}(L)] ≰ V_{2}. Since L centralizes W/V and V/V_{2} we conclude that O_{p}(L) ≰ O^{p}(L) and so n = 3 and q = 3. In particular, L = O_{3}(L) is extraspecial of exponent 3 and [W, Φ(L)] ≤ V_{2}. Hence, there exists g ∈ L \ Φ(L) with [W, g] ≰ V_{2}. Note that [v, g, g] ≠ 0 for every v ∈ V \ V_{2}. On the other hand |g| = 3, so g acts cubically on W. This shows that [W, g] ≤ V_{2}, which contradicts the choice of g. Thus 2^{∘}. [W, O_{p}(L)] ≤ V_{2}.

Since [V_{2}, O_{p}(L)] ≤ V_{1} we conclude that [W, O_{p}(L)′] ≤ V_{1}. Let W_{2} be maximal in W with [W_{2}, O_{p}(L)] ≤ V_{1}. In addition we use the following notation:

K^{∗} := C_{L^{∗}}(L/O_{p}(L)), K := C_{L^{∗}}(V_{2}/V_{1}), X/V_{2} := C_{W/V_{2}}(K^{∗}).

Then K ≤ K^{∗}, K^{∗}/O_{p}(L) has order q^{2} − 1 and K/O_{p}(L) has order q − 1. We will prove next: 3^{∘}. [W, L] ≤ V_{2}.

By Maschke’s Theorem and (2^{∘}), W/V_{2} = X/V_{2} ⊕V/V_{2}. Since [X, L^{∗}] ≤ X ∩V = V_{2} we conclude that [W, L] ≤ V_{2}. 4^{∘}. Either W = W_{2} + V or q = 2, n = 4 and |W/W_{2} + V | ≤ 4.

Suppose that q ≠ 2. Then O_{p}(L) = [O_{p}(L), K] and so K = O^{p}(K). Since [X, K] ≤ V_{2} and [V_{2}, K] ≤ V_{1} we have [X, K] = [X, O^{p}(K)] ≤ V_{1}. Thus X ≤ W_{2}. Since W = X + V, (4^{∘}) holds in this case.

So we may assume that q = 2. Then n > 3 since we are assuming that H ≠ SU_{3}(2). Put Z := O_{2}(L)′. Then [Z, L] = 1 and by (2^{∘}), [W, O_{2}(L), Z] ≤ [V_{2}, Z] = 0. Since by (3^{∘}) [W, L] ≤ V_{2}, we conclude from (1^{∘}) that W/V_{2} embeds into Hom_{L}(O_{2}(L)/Z, V_{2}/V_{1}).

Suppose that n > 4. Then L acts simply on O_{p}(L)/Z and on V_{2}/V_{1} and thus

q^{2} = |V/V_{2}| ≤ |W/V_{2}| ≤ |Hom_{L}(O_{p}(L)/Z, V_{2}/V_{1})| = q^{2}.

We conclude that V = W, so (4^{∘}) holds in this case.

Suppose that n = 4. Since V_{2} ≤ W_{2} and L^{∗} centralizes X/V_{2}, L^{∗} centralizes X + W_{2}/W_{2}. So by (1^{∘}) X + W_{2}/W_{2} embeds into Hom_{L^{∗}}(O_{p}(L)/Z, V_{2}/V_{1}). Since L^{∗} acts simply on O_{p}(L)/Z and on V_{2}/V_{1} we conclude as above that |X/X ∩ W_{2}| = |X + W_{2}/W_{2}| ≤ q^{2} = 4. Now W/V_{2} = X/V_{2} ⊕ V/V_{2} and V_{2} ≤ W_{2} imply

|W/(X ∩ W_{2}) + V | = |X + V/(X ∩ W_{2}) + V | = |(X/V_{2})/(X ∩ W_{2}/V_{2})| = |X/X ∩ W_{2}| =≤ 4,

so (4^{∘}) also holds in this case.

5^{∘}. Put W_{1} := C_{W_{2}}(O_{p}(L)). Then W_{2} = W_{1} + V_{2} and W_{2} + V = W_{1} + V.

Since [W_{2}, O_{p}(L)] ≤ V_{1} ≤ C_{V} (O_{p}(L)) the Three Subgroups Lemma gives that [W_{2}, Z] = 0. So by (1^{∘}) W_{2}/W_{1} embeds into Hom_{𝔽_{p}}(O_{p}(L)/Z, V_{1}). As an L-module Hom_{𝔽_{p}}(O_{p}(L)/Z, V_{1}) is a direct sum of copies of the dual of O_{p}(L)/Z. If n > 3 we conclude that W_{2}/W_{1} = [W_{2}/W_{1}, L] and so by (3^{∘}) W_{2} = W_{1} + V_{2}. Thus (5^{∘}) holds in this case. So suppose n = 3. Let Y/V_{1} = C_{W_{2}/V_{1}}(L^{∗}). Then by Maschke’s Theorem, W_{2} = Y + V_{2}.

Suppose that Y ≰ W_{1}. Then O_{p}(L)/Z ≅ V_{1} as an L^{∗}-module. Since n = 3 we have q > 2, and so L^{∗} acts simply on O_{p}(L)/Z and on V_{1}. It follows that there exists 0 ≤ l < 2k with λ^{2−p^{k}} = λ^{p^{l}}, for all 0 ≠ λ ∈ 𝔽_{p^{2k}}. Thus p^{2k} −1 divides p^{l} +p^{k} −2. Hence either p^{l} +p^{k} −2 ≤ 0 or p^{l} +p^{k} −2 ≥ p^{2k} −1. Since p^{k} = q > 2 we have p^{l} + p^{k} − 2 > 0. Moreover,

p^{l} + p^{k} − 2 ≤ p^{2k−1} + p^{2k−1} − 2 ≤ p^{2k} − 2 < p^{2k} − 1,

a contradiction. Thus Y ≤ W_{1}, and (5^{∘}) also holds for n = 3.

6^{∘}. W_{1} = V_{1} and W_{2} + V = V.

Let g ∈ H such that V_{1} is not perpendicular to V_{1}^{g} in V, so V_{1} ≰ V_{2}^{g}. Then by (3^{∘}), [W_{1}, L∩L^{g}] ≤ W_{1} ∩ V_{2}^{g} ≤ (W_{1} ∩ V) ∩ V_{2}^{g} = V_{1} ∩ V_{2}^{g} = 0. Thus W_{1} is centralizes by O_{p}(L)(L ∩ L^{g}) = L and so W_{1} ≤ C_{W} (T) ≤ V. Thus W_{1} = V_{1}, and (5^{∘}) implies (6^{∘}).

From (4^{∘}) and (6^{∘}) we see that the lemma holds in (Case 4).

Case 5. H = G_{2}(q)′, q = p^{k}, and either p = 2 and V = 𝕂^{6} or p ≠ 2 and V = 𝕂^{7}.

See [JP].

Case 6. V is a natural module for H = ^{3}D_{4}(q), q = p^{k}.

Fix a root system Φ. With respect to Φ, let C be the Cartan subgroup, N/C the Weyl-group, and L be the subgroup of H generated by the long root subgroups. Then L ≅ SL_{3}(q) and C normalizes L.

Let K ≤ H be the centralizer of a field automorphism of order 3 in H such that K ≅ G_{2}(q), each root subgroup with respect to Φ intersects K in a root subgroup of K, and N = (N ∩ K)C. Then L ≤ K and ⟨K, C⟩ contains all the root subgroups from Φ. So ⟨K, C⟩ = H. In the case q = 2, the action of C on the Lie-parabolic subgroups of H shows that also ⟨O^{2}(K), C⟩ = H.

Note that V/C_{V} (K) is a 7-dimensional K-module (over 𝕂), which is a natural module for p odd and a non-split central extension of a natural module for p = 2. By (Case 5), W = C_{W} (O^{p}(K))+V. Moreover, the action of K on V shows that C_{V} (O^{p}(K)) = C_{V} (L(N∩O^{p}(K))). So also C_{W} (O^{p}(K)) = C_{W} (L(N ∩ O^{p}(K)). Note that C acts fixed-point freely on C_{V} (L). Since C is a p′-group we get C_{W} (L) = C_{V} (L) ⊕ C_{W} (LC). Thus also W = V ⊕ C_{W} (LC). Since N normalizes C_{W} (LC) we have

C_{W} (LC) = C_{W} (LN) ≤ C_{W} (L(N ∩ O^{p}(K))) ≤ C_{W} (O^{p}(K)).

Thus C_{W} (LC) ≤ C_{W} (⟨C, O^{p}(K)⟩) = C_{W} (H) = 0 and so V = W.

Case 7. V is the (half)-spin-module for H = Spin_{n}^{ϵ}(q), q = p^{k}, n ≥ 7.

See [JP].

Case 8. H = 3.Alt(6) and V = 𝕂^{3}.

Since [V, Z(H)] ≠ 0, Maschke’s Theorem implies that V = W.

Case 9. V is a natural module for H ≅ Alt(n), n ≥ 5, p = 2.

See [As, page 74].

Case 10. V is the symmetric square of a natural module for H ≅ SL_{n}(q), q = p^{k}, p odd, n ≥ 3.

Let V_{2} := [V, T], L^{∗} := N_{H}(V_{2}), L_{1} := C_{L^{∗}}(V/V_{2}) and L := O^{p′}(L^{∗}). Then L/O_{p}(L) ≅ SL_{n−1}(q) and |L_{1}/L| = 2. Note that L = O^{p}(L) unless n = 3 = q, in which case L_{1}/O_{p}(L_{1}) ≅ GL_{2}(3). So in any case L_{1} = O^{p}(L_{1}) and thus

7^{∘}. [W, L_{1}] = V_{2} = [W, L].

Let V_{1} := C_{V} (O_{p}(L)) = [V_{2}, O_{p}(L)]. Then V_{2}/V_{1} is a natural SL_{n−1}(q)-module for L/O_{p}(L) isomorphic to O_{p}(L). Hence |Hom_{L}(O_{p}(L), V_{2}/V_{1})| = q. Let W_{2}/V_{1} := C_{W/V_{1}}(O_{p}(L)). Then by (1^{∘}) W/W_{2} embeds into Hom_{L}(O_{p}(L), V_{2}/V_{1}). Since |V/V_{2}| = q we conclude that

8^{∘}. W = W_{2} + V.

Let W_{1}/V_{1} := C_{W_{2}/V_{1}}(L). By (Case 3) H^{1}(L/O_{p}(L), V_{2}/V_{1}) = 0 and so by (8^{∘})

9^{∘}. W_{2} = W_{1} + V_{2} and W = W_{1} + V.

Note that V_{1} is the symmetric square of a natural module for L/O_{p}(L). In particular, V_{1} and O_{p}(L)) are non-isomorphic simple L/O_{p}(L)-modules and so [W_{1}, O_{p}(L)] = 1. Let W_{0} = C_{W_{1}}(L). Suppose that W_{1} ≠ W_{0} ⊕ V_{1}. By induction on n and with (Case 1) we conclude that n = 3 and q = 5. (Note here that for n = 3 V_{1} is an orthogonal Ω_{3}(q)-module for L/O_{p}(L).)

Since T/O_{5}(L) is cyclic, the Jordan Form for T on V shows that T does not act cubically on W_{1}. Pick g ∈ H with T = O_{5}(L)(O_{5}(L)^{g} ∩ T). By (9^{∘}), O_{5}(L) acts cubically on V and so T acts cubically in W_{1}, a contradiction.

Thus W_{1} = W_{0} + V_{1}. As W_{0} ≤ C_{W} (T) ≤ V we have W_{1} ≤ V, and by (9^{∘}) V = W.

Case 11. V is the alternating square of a natural module for H ≅ SL_{n}(q), q = p^{k}, n ≥ 5.

See [JP].

Case 12. H ≅ E_{6}(q), q = p^{k}, and V = 𝕂^{27}.

See [JP].

Case 13. H ≅ SL_{n}(q^{2}), q = p^{k}, and V is a simple 𝔽_{q}H-submodule of N ⊗_{𝔽_{q^{2}}} N^{σ}, where N is the natural 𝔽_{q^{2}}H-module and σ is the field automorphism of order 2 of 𝔽_{q^{2}}.

Let N_{1} := C_{N}(T), L^{∗} := N_{H}(N_{1}), and L := C_{H}(N_{1}), and let J ≤ L^{∗} with L^{∗} = O_{p}(L)J and N = N_{1} ⊕ [N, J ∩ L]. Then J ∩ L ≅ SL_{n−1}(q^{2}) and J ≅ GL_{n−1}(q^{2}). Let V_{1} = C_{V} (L) and V_{2} = [V, O_{p}(L)]. Then V_{2}/V_{1} is a natural SL_{n−1}(q^{2})-module for L/O_{p}(L) isomorphic to N/N_{1} and dual to O_{p}(L). Also V/V_{2} is isomorphic to a simple 𝔽_{q}L/O_{p}(L) submodule of N/N_{1} ⊗_{𝔽_{q^{2}}} N^{σ}/N_{1}^{σ}. We first show:

10^{∘}. Suppose n = 3 and q ≠ 2. Then Z(J) acts fix-point freely on V/V_{2}, and O_{p}(L) and V_{2}/V_{1} are not isomorphic as 𝔽_{p}Z(J)-modules.

Let j ∈ Z(J), then j acts as an 𝔽_{q^{2}}-scalar λ on N/N_{1}. It follows that j acts as λ^{−2} on N_{1}, as λ^{−3} on O_{p}(L), as λ^{q−2} on V_{2}/V_{1} and as λ^{q+1} on V/V_{2}. Since q > 2 we conclude that Z(J) is fixed-point free on V/V_{2}. Suppose that V_{2}/V_{1} and O_{p}(L) are isomorphic as 𝔽_{p}Z(J)-modules. Then there exists 0 ≤ l < 2k with λ^{−3p^{l}} = λ^{q−2} for all 0 ≠ λ ∈ 𝔽_{q^{2}} and so

p^{2k} − 1|3p^{l} + p^{k} − 2.

Since p^{k} = q > 2, the right side is positive and so

p^{2k} − 1 ≤ 3p^{l} + p^{k} − 2 ≤ 3p^{2k−1} + p^{k} − 2 ≤ 4p^{2k−1} − 2.

Thus p ≤ 3. If p = 3 we have

3^{2k} ≤ 3^{l+1} + 3^{k} − 1 ≤ 2 · 3^{m} − 1,

where m = max{l + 1, k}. Hence m = l + 1 = 2k. and so

3^{2k} − 1|3 · 3^{2k−1} + 3^{k} − 2 = (3^{2k} − 1) + 3^{k} − 1.

Therefore 3^{2k} − 1|3^{k} − 1, a contradiction.

Thus p = 2. If l = 0 we get 2^{2k} − 1 ≤ 2^{k} + 1 and q = 2^{k} = 2, contradiction. Hence l > 0 and since 2^{2k} − 1 is odd,

2^{2k} − 1|3 · 2^{l−1} + 2^{k−1} − 1.

So

2^{2k} ≤ 3 · 2^{l−1} + 2^{k−1} = 2^{l} + 2^{l−1} + 2^{k−1}.

It follows that k = 1 = l and q = 2, a contradiction.

11^{∘}. Suppose n = 3 and V ≠ W. Then q = 2 and |W/V | ≤ 4.

Since O_{p}(L) and V/V_{2} are non-isomorphic simple L-modules, [W, O_{p}(L)] ≤ V_{2}. Let W_{2}/V_{2} = C_{W/V_{2}}(L). If q ≠ 2, then by (10^{∘}) Z(J) acts fixed-point-freely on V/V_{2}, and if q = 2, then by (Case 1), H^{1}(L/O_{p}(L), V/V_{2}) = 0. So in any case W = W_{2} + V.

Let W_{1}/V_{1} = C_{W_{2}/V_{1}}(O_{p}(L)). Then W_{2}/W_{1} embeds into Hom_{L^{∗}}(O_{p}(L), V_{2}/V_{1}). By (10^{∘}) this group is trivial for q ≠ 2. For q = 2 it has order 4. So W_{2} = W_{1} if q ≠ 2 and |W_{2}/W_{1}| ≤ 4 if q = 2. It remains to show that W_{1} ≤ V.

Let W_{0} = C_{W_{1}}(O_{p}(L)). Then W_{1}/W_{0} embeds into Hom_{𝔽_{p}}(O_{p}(L), V_{1}). The latter group is as an L-module isomorphic to a direct sum of copies of the dual of O_{p}(L). Hence [W_{1}/W_{0}, L] = W_{1}/W_{0} and so W_{1} = W_{0} + V_{2}. Since W_{0} ∩ V = V_{1} and L = O^{p}(L) we have [W_{0}, L] = 0 and so W_{0} ≤ C_{V} (T) ≤ V. Thus also W_{1} ≤ V, and (11^{∘}) is proved.

12^{∘}. Suppose n = 3 and q = 2. Then | H^{1}(H, V)| = 4, and GL_{3}(4) acts fixed-point freely on H^{1}(H, V).

By (11^{∘}) | H^{1}(H, V)| ≤ 4. Let I be the simple 11-dimensional Golay code-module for M = Mat_{24} over 𝔽_{2}. Let H̃ = Mat_{21} ≅ PSL_{3}(4). Then [I, H̃] is simple of 𝔽_{2}-dimension 9 and C_{I}(H̃) = 0. Moreover, N_{M}(H̃) ≅ PGL_{3}(4) acts fixed-point freely on I/[I, H̃], so (12^{∘}) holds.

13^{∘}. Suppose n > 3. Then V = W.

Note that W/V_{2} and O^{p′}(L^{∗}/O_{p}(L)) satisfy (Case 13) for n−1, and note further that L^{∗}/O_{p}(L) ≅ GL_{n−1}(q^{2}). Moreover, for n − 1 = 3 the case described in (12^{∘}) does not occur since [W, L^{∗}] = V. Hence induction shows that H^{1}(L^{∗}/O_{p}(L), V/V_{2}) = 0. By (Case 3), also H^{1}(L^{∗}/O_{p}(L), V/V_{2}) = 0. Since n > 3, V/V_{2} and V_{2}/V_{1} are simple L^{∗}-modules not isomorphic to O_{p}(L). Also since L = O^{p}(L), H^{1}(L, V_{1}) = 0. Thus H^{1}(L^{∗}, V) = 0 and V = W.

By (11^{∘}), (12^{∘}) and (13^{∘}) the Lemma holds in case (Case 13).

Case 14. p = 2, and V is the simple Todd- or Golay code-module for H = Mat_{n}, n = 22, 23, or 24.

Let P := Mat_{n−1} ≤ H. Suppose first that H = Mat_{22} and V is the Todd-module. Put V_{1} := C_{V} (T) and L := C_{H}(V_{1}). Then L/O_{2}(L) ≅ Sym(5), and O_{2}(L) is a natural ΓSL_{2}(4)-module for L. Put V_{2} := [V, O_{2}(L)]. Then O_{2}(L) centralizes V_{2}/V_{1}, and V_{2}/V_{1} is an non-split extension of a 1-dimensional module by a natural ΓSL_{2}(4)-module for L/O_{2}(L). Moreover, V/V_{2} is a natural O_{4}^{−}(2)-module for L. Since V/V_{2} is not isomorphic to O_{2}(L) as an L-module, [W, O_{2}(L)] ≤ V_{2}. Put W_{2}/V_{2} := C_{W/V_{2}}(L). By (Case 1) W = W_{2} + V. Since V_{2}/V_{1} is indecomposable, Hom_{L}(O_{2}(L), V_{2}/V_{1}) = 0 and so [W_{2}, O_{2}(L)] ≤ V_{1}. Let W_{1} = C_{W_{2}}(O_{2}(L)). Then W_{2}/W_{1} embeds into Hom_{𝔽_{2}}(O_{2}(L), V_{1}). The latter is isomorphic to the dual of O_{2}(L) and so W_{2} = W_{1} + V_{2}. Note that [W_{1}, O^{2}(L)] = 1 and W_{1} ∩ V has order 4 with L/O^{2}(L) acting non-trivial on W_{1} ∩ V. It follows that W_{1} = C_{W_{1}}(L) + (W_{1} ∩ V) and so W_{1} ≤ C_{W} (T) + V ≤ V. Hence also W_{2} ≤ V and W = V.

Suppose next that H = Mat_{22} and V is the Golay code -module. Then |[V, P]| = 2^{9} and C_{V} (P) = 0, so V is a non-split extension for P as in case (Case 13). Thus (Case 13) shows that |W/V + C_{W} (P)| ≤ 2. Let L_{0} = Mat_{20} ≤ P and L = N_{H}(L_{0}) ∼ 2^{4}Sym(5). Then C_{V} (L_{0}) = 0 and so C_{W} (P) ≤ C_{W} (L_{0}) ≤ C_{W} (L). Since L contains a Sylow 2-subgroup of H, C_{W} (L) ≤ V and so C_{W} (P) = 0 and |W/V | ≤ 2.

Suppose next that H = Mat_{23}. Then P contains a Sylow 2-subgroup of H and so C_{W} (P) ≤ V. If V is the Todd-module, then V = [V, P] and V/C_{V} (P) is the Todd-module for P = Mat_{22}. Since P = O^{2}(P), the Mat_{22}-case implies that W = C_{W} (P) + V = V.

If V is the Golay code-module, then C_{V} (P) = 0 and [V, P] is the 10 dimensional Golay code module for P. Thus by the Mat_{22}-case, W = C_{W} (L) + V = V.

Suppose that H = Mat_{24}. Then V is simple as a P -module, so by the Mat_{23}-case, W = C_{W} (P) + V. Let w ∈ C_{W} (P). Then ⟨w^{H}⟩ is a quotient of the natural permutation module of Mat_{24}. If V is the Golay code-module, we conclude that [w, H] = 0 and so V = W. If V is the Todd module and w ≠ 0, we conclude that ⟨w^{H}⟩ = ⟨w⟩ + V is uniquely determined as an 𝔽_{2}H-module. Since |𝕂| = 2 this implies |W/V | ≤ 2.

Case 15. V = 𝔽_{4}^{6} and H = 3.Mat_{22}.

Since Z(H) ≠ 1, we have V = W.

Case 16. p = 3, V is the simple Todd- or Golay code-module for H = Mat_{11} or 2.Mat_{12}.

If H = 2.Mat_{12}, we have W = C_{W} (Z(H)) ⊕ V and so V = W. Suppose H = Mat_{11}.

Assume first that V is the Golay code-module. Let L_{0} = Mat_{10} and L = L′_{0} ≅ L_{2}(9). Then [V, L] is the natural Ω_{4}^{−}(3)-module for L and C_{V} (L) = 0. Thus by (Case 1), |W/V + C_{W} (L)| ≤ 3. Since L contains a Sylow 3-subgroup of H, C_{W} (L) ≤ V and so |W/V | ≤ 3.

Suppose next that V is the Todd-module. Let L = N_{H}(T). Then L/T is semidihedral of order 16. Let K ∈ Syl_{2}(L) and put V_{2} = [V, T] and V_{1} = C_{V} (T). Then |V/V_{1}| = 3 with D := C_{K}(V/V_{1}) dihedral of order 8. Moreover, V_{2}/V_{1} has order 9 with K acting faithfully on V_{2}/V_{1}, and V_{1} has order 9 with |C_{K}(V_{1})| = 2. Since T = [T, D], we have [W, T] ≤ V_{2}. Let W_{2}/V_{1} = C_{W/V_{1}}(T). Then W/W_{2} embeds into Hom_{D}(T, V_{2}/V_{1}). Since D acts simply on T and V_{2}/V_{1}, we conclude that Hom_{D}(T, V_{2}/V_{1}) has order 3. Thus W = W_{2} + V. Let W_{1}/V_{1} = C_{W_{2}/V_{1}}(L). By Mascke’s Theorem, W_{2} = W_{1} + V_{2}. Since V_{1} is not isomorphic to T as an L-module, [W_{1}, T] = 0 and so W_{1} ≤ V and V = W.

Definition 6.2. Let H be a finite group, V an 𝔽_{p}H-module and Q a p-subgroup of H. Then V is called a Q!-module for H if Q is not normal in H and

(Q!)  Q ⊴ N_{H}(A) for all 1 ≠ A ≤ C_{V} (Q).

Lemma 6.3. Let M ≅ SL_{n}(q), q a power of p, n ≥ 2, and let V be an 𝔽_{p}M-module. Suppose that there exists an M-submodule I in V such that the following hold:

(i) W := V/I is a natural SL_{n}(q)-module for M.

(ii) I ≅ Λ_{𝕂}^{2}W as an 𝔽_{p}M-module, where 𝕂 := End_{M}(W).

(iii) If H is a 𝕂-hyperplane in W and A := C_{M}(H) ∩ C_{M}(W/H), then C_{V} (A) ≰ I.

Then there exists x ∈ V \ W with C_{M}(x) = C_{M}(x + I/I). Moreover, V is not a Q!-module for any p-subgroup Q of M.

Proof. Put U := C_{V} (A), L = N_{M}(H) ∩ C_{M}(W/H) and T ∈ Syl_{p}(L). Note T ∈ Syl_{p}(M). We will first show:

1^{∘}. C_{V} (T) ≰ I.

The proof is by induction on n. If n = 2 then A = T and (1^{∘}) follows from (iii). Suppose that n ≥ 3. Note that L/A ≅ SL_{n−1}(q), H ≅ U/U ∩ I is a natural module for L/A and U ∩ I ≅ Λ_{𝕂}^{2}H. Let g ∈ M with H^{g} ≠ H and put R_{0} := L ∩ A^{g} and R := A(L ∩ A^{g}).

Assume that n = 3. Then T = R and I ≅ W^{∗}. In particular

[U ∩ (U^{g} + I), R] = [I, R_{0}] ∩ I ∩ U = 0.

Since |U ∩ (U^{g} + I)| = q^{2} while |U ∩ I| = q, we conclude that C_{U}(R) = C_{U}(T) ≰ I, and (1^{∘}) holds.

Suppose now that n > 3. Then C_{I}(R) = C_{U∩I}(R_{0}) and so C_{I}(R) has order q^{\binom{n−2}{2}}. On the other hand, C_{V} (A) has index q^{n} in V. Hence C_{V} (⟨A, A^{g}⟩) has index at most q^{2n} in V. Thus also |V/C_{V} (R)| ≤ q^{2n}. Note that

|V/C_{I}(R)| = q^{n+\binom{n}{2}−\binom{n−2}{2}} = q^{3n−3} > q^{2n},

where the last inequality holds since n > 3.

Thus C_{V} (R) ≰ C_{I}(R) and since C_{V} (R) ≤ U, C_{U}(R) ≰ U ∩I. Thus (U, U ∩I, L/A, H ∩H^{g}, R/A) in place of (V, I, M, H, A) fulfills the assumptions (i)-(iii) and so by induction C_{U}(T/A) ≰ U ∩ I. Thus (1^{∘}) holds.

Put Y := I + C_{V} (T) and F_{1} := C_{M}(Y/I). Then dim_{𝕂} Y/I = 1, so F_{1} = C_{M}(x + I/I) for x ∈ C_{V} (T)\I. Since T ∈ Syl_{p}(F_{1}), Gaschütz’ Theorem implies that Y = I ⊕X for some F_{1}-invariant subspace X of Y. Then [X, F_{1}] ≤ X∩I = 0. Let 0 ≠ x ∈ X. Then F_{1} ≤ C_{F_{1}}(x) ≤ C_{M}(x+I/I) = F_{1}, and so the first statement in 6.3 is proved.

Suppose V is a Q!-module. If n = 2, then [I, M] = 0 and so Q ⊴ C_{M}(I) = M, a contradiction. Thus n ≥ 3. Without loss Q ≤ T. Thus X ≤ C_{V} (Q) and so by Q! we get that Q ⊴ F_{1}. Similar Q ⊴ F_{2} := N_{M}(C_{I}(T)). Since F_{2} is the normalizer of a 2-dimensional subspace of W, we have M = ⟨F_{1}, F_{2}⟩ and so Q ⊴ M, a contradiction to the definition of a Q!-module.

Lemma 6.4. Let M = SL_{2}(𝔽), 𝔽 a field, and let Z be a maximal unipotent subgroup of M and B := N_{M}(Z). Suppose that X is an ℤM-module with [X, Z, Z] = 0 and Y is a B-submodule of C_{X}(Z) with X = ⟨Y^{M}⟩. Then for every h ∈ M \ B

X = Y + Y^{h} + C_{X}(M) = Y + Y^{h} + [Y^{h}, Z] and C_{X}(Z) = Y + [Y^{h}, Z] = Y + C_{X}(M);

in particular C_{X}(M) ≤ Y + [Y^{h}, Z].

Proof. Note that Z acts transitively on Z^{M} \ {Z} and so Z^{M} = {Z} ∪ Z^{hZ} and Y^{M} = {Y } ∪ Y^{hZ} for all h ∈ M \ B. Thus

(∗)  X = ⟨Y^{M}⟩ = Y + ⟨Y^{hZ}⟩ = Y + Y^{h} + [Y^{h}, Z].

By the quadratic action of Z, [Y^{h}, Z] ≤ C_{X}(Z). By assumption also Y ≤ C_{X}(Z) and so C_{X}(Z) = Y +[Y^{h}, Z]+C_{Y^{h}}(Z). Note that M = ⟨Z^{M}⟩ = ⟨Z, Z^{hZ}⟩ = ⟨Z, Z^{h}⟩ and so C_{Y^{h}}(Z) ≤ C_{X}(⟨Z^{h}, Z⟩) ≤ C_{X}(M). Hence C_{Y^{h}}(Z) ≤ C_{Y^{h}}(M) ≤ Y and so C_{X}(Z) = Y + [Y^{h}, Z].

Now by (∗) X = Y^{h} + C_{X}(Z) and thus C_{X}(Z^{h}) = Y^{h} + C_{X}(Z) ∩ C_{X}(Z^{h}) = Y^{h} + C_{X}(M). Hence C_{X}(Z) = Y + C_{X}(M) and X = Y^{h} + Y + C_{X}(M).

Notation 6.5. Let

CL(p) := {SL_{n}(q), SU_{n}(q), Sp_{2n}(q) (q odd), Ω_{n}^{ϵ}(q), O_{n}^{ϵ}(q) (q even)},

where q is a power of p. Let H ∈ CL(p) and Ã be the corresponding natural 𝔽_{p}H-module. Put A := Ã/CÃ(H). Note that A is a simple 𝔽_{p}H-module. Also CÃ(H) = 0 unless H = Ω_{2m+1}(2^{k})), in which case CÃ(H) is 1-dimensional, H ≅ Sp_{2m}(2^{k}), and A is the natural Sp_{2m}(2^{k})-module for H.

Furthermore set K := O^{p}(H) and 𝕂 := End_{H}(A). Then A is also a 𝕂H-module, and A is equipped with a natural sesquilinear form f if A is not the natural SL_{n}(q)-module.

The groups Sp_{2n}(2^{k}) have been excluded from the list in 6.5, since it will be more convenient for us to treat Sp_{2n}(2^{k}) as Ω_{2n+1}(2^{k}).

Lemma 6.6. Let H ∈ CL(p), V be a faithful 𝔽_{p}H-module with H-submodules A_{0} ≤ B ≤ V, and let D ≤ H. Suppose that

(i) [B, K] ≤ A_{0}, A ≅ A_{0} and V/B ≅ A or A^{∗},

(ii) D is a non-trivial quadratic best offender on V.

Then there exists a KD-submodule C in V such that A_{0} ≰ C and V = B + C.

Proof. Let D^{∗} be any non-trivial quadratic best offender on V such that KD^{∗} < H. Then we may assume by induction on H that V = B + C for a KD^{∗}-submodule C with A_{0} ≰ C. Since V/B is a perfect K-module and K = O^{p}(K), also V = B + [C, K] and [C, K] = [C, K, K]. Hence 2.6 shows that C is D-invariant, and we are done. Thus, we may assume

1^{∘}. H = KD^{∗} for every non-trivial quadratic best offender D^{∗} on V; in particular H = KD.

Note that by 1.2 D is a best offender on [V, K] and that D is a quadratic offender on V/C_{V} (K), so D contains a best offender on V/C_{V} (K). Hence we may assume that

2^{∘}. V = [V, K] and C_{V} (K) = 0.

We will now compare the action of H on V with that on the natural module Ã. According to (1^{∘}) we can choose D such that U := [Ã, D] is minimal with respect to (ii). Observe that U is a 𝕂-subspace. Put P := N_{H}(U) and E = C_{H}(U) ∩ C_{H}(Ã/U). Note that D acts quadratically on A_{0} and so also on A. By 3.2(e), D acts quadratically on Ã and U is isotropic. Thus D ≤ E. Since E acts quadratically on Ã, E is an elementary abelian p-group.

Pick D_{1} ≤ E such that first |D_{1}||C_{V} (D_{1})| is maximal among all subgroups of E and then that |D_{1}| is maximal with that property. Since D ≤ E, |D_{1}||C_{V} (D_{1})| ≥ |D||C_{V} (D)| ≥ |V | and so D_{1} is a non-trivial best offender on V. By [MS1, 2.6] D_{1} is uniquely determined in E and so D_{1} ⊴ P. By the Timmesfeld Replacement Theorem, D_{2} := C_{D_{1}}([V, D_{1}]) is a non-trivial quadratic best offender on V. Since [Ã, D_{2}] ≤ [Ã, E] ≤ U, the minimal choice of U and (1^{∘}) imply [Ã, D_{2}] = U, and so we may assume

3^{∘}. D ⊴ P.

By our hypothesis

|D| ≥ |A/C_{A}(D)||V/B/C_{V/B}(D)|.

Since A is self-dual if A is not the natural SL_{n}(q)-module, we get:

4^{∘}. |D| ≥ |A/C_{A}(D)||A^{∗}/C_{A^{∗}}(D)| and A is the natural SL_{n}(q)-module, or |D| ≥ |A/C_{A}(D)|^{2}.

Let CL be the type of H, so CL ∈ {SL, Sp, SU, Ω^{ϵ}, O^{ϵ}} and H = CL_{n}(𝕂).

Case 1. Suppose CL = SL, SU or Sp.

Recall that in these cases A = Ã and U = [A, D]. If dim_{𝕂} U = 1 we get |A/C_{A}(D)| ≥ |D|, a contradiction to (4^{∘}). Thus dim_{𝕂} U ≥ 2. By 3.5 and since by assumption p is odd in the symplectic case, P acts simply on E and so D = E. Let U_{1} be a 1-dimensional subspace of U. If H = SL_{n}(𝕂) let U_{n−1} be a hyperplane of A containing U, Z := C_{H}(A/U_{1}) ∩ C_{H}(U_{n−1}) and L := C_{H}(U_{1}) ∩ C_{H}(U/U_{n−1}). In the other cases let U_{n−1} := U_{1}^{⊥}, Z := C_{H}(U_{1}^{⊥}) and L := C_{H}(U_{1}). In either case put \overline{W} := U_{n−1}/U_{1}. Then Z is a transvection group, Z ≤ Z(L) ∩ D, O_{p}(L) = C_{L}(\overline{W}) and L/O_{p}(L) induces CL_{n−2}(\overline{W}) on \overline{W}. Moreover, if CL = SL, O_{p}(L)/Z is as an L/O_{p}(L)-module isomorphic to the direct sum of \overline{W} and its dual. And if CL = Sp or SU, then O_{p}(L)/Z ≅ \overline{W} as an L-module. Let S ∈ Syl_{p}(L) and note that S ∈ Syl_{p}(H).

5^{∘}. [V, Z, L] = 0.

Note that D = E induces C_{CL_{n−2}(\overline{W})}(\overline{U}) ∩ C_{CL_{n−2}(\overline{W})}(\overline{W} /\overline{U}) on \overline{W}. Since dim U ≥ 2 we have \overline{U} ≠ 0. It follows that either L = O_{p}(L)⟨D^{L}⟩ or D ≤ O_{p}(L), CL = SL and U = U_{n−1}.

In the first case O_{p}(L)/Z is a perfect L-module and Z ≤ Φ(O_{p}(L)), so L = ⟨D^{L}⟩. Since D is quadratic on V and Z ≤ D we have [V, Z, D] = 0, and since Z ≤ Z(L), this implies [V, Z, ⟨D^{L}⟩] = 0 and so [V, Z, L] = 0.

Now suppose CL = SL and U = U_{n−1}, so |D| = q^{n−1}. Since dim U ≥ 2, n ≥ 3. If V/B ≅ A^{∗}, then |V/B/C_{V/B}(D)| = q^{n−1} = |D|, a contradiction to (4^{∘}). Thus V/B ≅ A. Suppose for a contradiction that A_{0} ≠ B. Then by 6.1 n = 3 and q = 2. So |D| = 4. From

|V/B/C_{V/B}(D)||B/C_{B}(D)| ≤ |V/C_{V} (D)| ≤ |D| = 4

we conclude that |B/C_{B}(D)| = 2. Since H (≅ GL_{3}(2)) is generated by three conjugates of D, this gives |B/C_{B}(H)| ≤ 2^{3} = |A_{0}|. Hence |A_{0}| < |B| implies C_{B}(H) ≠ 0, which contradicts (2^{∘}).

Hence A_{0} = B and thus |V/C_{V} (D)| = q^{2}. In particular |[V, z]| = q^{2} for 1 ≠ z ∈ Z. Let h ∈ H with Z^{h} ≤ L, but Z^{h} ≰ D. Note that C_{V} (D)+B/B = C_{V/B}(Z) and |[C_{A}(D), z^{h}]| = q. Since B and V/B are isomorphic to A we conclude that |[C_{V} (D), z^{h}]| = q^{2}. Since |[V, z]| = q^{2} we get [V, z^{h}] = [C_{V} (D), z^{h}] ≤ C_{V} (D), so ⟨D^{L^{h}}⟩ ≤ C_{H}([V, Z^{h}]). In C_{H}([A, Z^{h}]) = C_{H}(U_{1}^{h}) ∼ q^{n−1}SL_{n−1}(q) we see that ⟨D^{L^{h}}⟩ = C_{H}(U_{1}^{h}). Since L^{h} ≤ C_{H}(U_{1}^{h}), also L^{h} ≤ ⟨D^{L^{h}}⟩ ≤ C_{H}([V, Z^{h}]), and so [V, Z^{h}, L^{h}] = 0 and again (5^{∘}) holds.

Put L̃ := C_{H}([V/B, Z]). Observe that [V/B, Z] is a 1-dimensional 𝕂-subspace of V/B and S ≤ L ≤ L̃. Thus by (5^{∘}), [V, Z] + B = C_{V} (S) + B = Y^{∗} ⊕ B for some Y^{∗} ≤ C_{V} (S). By Gaschütz’ Theorem there also exists a L̃-invariant complement Y to B in B + C_{V} (S), in particular [Y, L̃] ≤ Y ∩ B = 0. Let W := ⟨Y^{H}⟩ and h ∈ H.

6^{∘}. [Y^{h}, Z] ≤ Y.

If Z ≤ L̃^{h}, then [Y^{h}, Z] = 0. So assume that Z ≰ L̃^{h}. Note that there exists h^{∗} ∈ H with Y^{h} = Y^{h^{∗}} and T := ⟨Z^{h^{∗}}, Z⟩ ≅ SL_{2}(q). Without loss h = h^{∗}. Put X := ⟨Y^{T} ⟩. Then 6.4 and (5^{∘}) give

Y + C_{X}(T) = Y + [Y^{h}, Z] ≤ C_{V} (L).

Note that T normalizes neither U_{1} nor U_{n−1}, so T and L are not contained in a proper parabolic subgroup. Hence H = ⟨L, T ⟩ and C_{V} (H) = 0. Since C_{X}(T) ≤ C_{V} (L), this gives C_{X}(T) = 0, and we conclude that Y = [Y^{h}, Z].

From (6^{∘}) we get [W, Z] = Y. In particular A ≰ W, and the lemma holds in (Case 1).

Case 2. Suppose CL = Ω^{ϵ} or O^{ϵ}.

7^{∘}. If 0 ≠ Ã^{⊥} ≤ U, then dim U ≥ 4 and n ≥ 7. In the other cases dim U ≥ 5 and n ≥ 10.

Put k := dim U. Suppose first that 0 ≠ Ã^{⊥} ≤ U. By 3.4, |D| ≤ |E| ≤ q^{\frac{k(k−1)}{2}} and |A/C_{A}(D)|^{2} = |Ã/U^{⊥}|^{2} ≥ q^{2(k−1)}. Thus by (4^{∘}) \frac{k}{2} ≥ 2 and so k ≥ 4.

Suppose next that Ã^{⊥} = 0 or Ã^{⊥} ≰ U. By 3.4, |D| ≤ |E| ≤ 2q^{\frac{k(k−1)}{2}} ≤ q^{\frac{k(k−1)}{2} +1} and |A/C_{A}(D)|^{2} = |Ã/U^{⊥}|^{2} ≥ q^{2k}. Thus by (4^{∘}) \frac{k(k−1)}{2} + 1 ≥ 2k, k(k − 5) ≥ −2 and k ≥ 5.

By (7^{∘}), U contains a singular 2-space U_{2}. Put

Z := C_{H}(Ã/U_{2}), L := C_{H′}(U_{2}), and \overline{W} := U_{2}^{⊥}/U_{2}.

Then |Z| = q, Z is a long root subgroup of H in Z(L), and L induces Ω_{n−4}^{ϵ}(\overline{W}) on \overline{W}. Moreover, C_{L}(\overline{W}) = O_{p}(L), and O_{p}(L)/Z is as an L-module the direct sum of two copies of \overline{W}. Let U_{0} be the singular radical of U and E_{0} := C_{H}(Ã/U_{0}). Then Z ≤ E_{0} and by 3.5, E_{0} ≤ D. In particular, Z ≤ D. If E ≠ E_{0}, we have [Ã, E_{0}] = U_{0} ≠ U and so E_{0} < D.

8^{∘}. L = ⟨D^{L}⟩.

From 3.5 and (7^{∘}) we see that D acts non-trivially on \overline{W}. Suppose n ≥ 9. Then n − 4 ≥ 5 and so L/O_{p}(L) is simple and \overline{W} = [\overline{W} , L]. It follows that L = ⟨D^{L}⟩O_{p}(L) and then L = ⟨D^{L}⟩.

So suppose n < 9. Then (7^{∘}) implies that n = 7, 0 ≠ Ã^{⊥} ≤ U, dim U = 4. By 3.4(e), E/E_{0} ≅ U_{0}, and since E_{0} < D ⊴ P, 3.5 implies that D = E. Thus C_{H}(U_{2}^{⊥}) ≤ D. Also L/O_{p}(L) ≅ SL_{2}(q) and so L = ⟨D^{L}⟩O_{p}(L). Since O_{p}(L)/C_{H}(U^{⊥}) is a direct sum of two copies of the natural SL_{2}(q)-module \overline{W} /\overline{W}^{⊥} we again get that L = ⟨D^{L}⟩.

9^{∘}. [V, Z, L] = 0.

This follows immediately from [V, Z, D] = 0 and (8^{∘}).

Note that we can can embed [Ã, Z] in a non-degenerate subspace U_{4} of Ã of dimension 4. Put K := O^{p′}(N_{H′}(U_{4}) ∩ C_{H′}(U_{4}^{⊥})), L̂ := O^{p′}(N_{H}(Z)), and let U_{1} be a 1-subspace of U_{2}.

Then Z ≤ K and K ≅ O^{p′}(Ω_{4}^{+}(q)) ≅ SL_{2}(q) ∗ SL_{2}(q). Moreover T^{∗} := ⟨Z^{K}⟩ ≅ SL_{2}(q). Since dim Ã ≥ 7, N_{H}(U_{4}) induces O_{4}^{+}(U_{4}) on U_{4} and there exists h ∈ N_{H}(U_{4}) ∩ N_{H}(U_{1}) with T := T^{∗}^{h} ≠ T^{∗}. Then

K = T T^{∗}, T ≅ SL_{2}(q), L̂ = T L, and [T, T^{∗}] = 1.

Note that U_{1} = U_{2} ∩U_{2}^{h} = [Ã, Z, Z^{h}] ≠ 0. Put P̃ := N_{H}(U_{1}), so P̃ is the stabilizer of a 1-dimensional singular subspace of Ã.

Since U_{1} ≠ 0 also V_{1} := [V, Z, Z^{h}] ≠ 0. Note that V_{1} is centralizes by LZ^{h} and thus by a Sylow p-subgroup of P̃. Again Gaschütz’ Theorem gives a P̃-invariant complement Y to B in B + V_{1}.

Let s ∈ T^{∗} \ N_{T^{∗}}(Z). Then U_{1} + U_{1}^{s} is a singular 2-space normalized by T^{∗} and U_{1}^{s} ≰ U_{2}^{⊥}. Since O_{p}(L) is transitive on the singular 1-spaces of U_{2}^{⊥} + U_{1}^{s} not contained in U_{2}^{⊥}, and T is transitive on Ã/U_{2}^{⊥}, we get that T L is transitive on the conjugates of P̃ that do not contain Z. As in the previous case, this gives

[⟨Y^{H}⟩, Z] = [⟨Y^{sT L}⟩, Z] = ⟨[Y^{s}, Z]^{T} ⟩.

Observe that ⟨L, T^{∗}⟩ = H. Hence, 6.4 implies ⟨Y^{T^{∗}}⟩ = Y +Y^{s}. Since U_{1}^{h} = U_{1} we have Y^{h} = Y. Hence also ⟨Y^{T} ⟩ = Y + Y^{sh} since T^{h} = T^{∗}, and so [⟨Y^{H}⟩, Z] = Y + Y^{sh}. Then as in the previous case [A_{0}, Z] ≰ [⟨Y^{H}⟩, Z], so A ≰ ⟨Y^{H}⟩, and the lemma also follows in (Case 2).

7 Quadratic Modules

In this section M is a finite group, and V is a finite dimensional 𝔽_{p}M-module.

Lemma 7.1. Let V be faithful. Suppose that p is odd, A ≤ M with [V, A, A] = 0, and R is an Ainvariant p′-subgroup of M satisfying R = [R, A] ≠ 1. Then p = 3 and R is a non-abelian 2-group. If in addition |Φ(R)| = 2 and |A| = 3, then RA ≅ SL_{2}(3).

Proof. Observe that by coprime action for every prime divisor r of R there exists an A-invariant Sylow r-subgroup S_{r} in R. If [S_{r}, A] ≠ 1 then [KS, 9.1.3] implies that p = 3, r = 2 and S_{r} is not abelian. It follows that R = C_{R}(A)S_{2} and so R = [R, A] = [S_{2}, A] ≤ S_{2}.

Suppose now that |Φ(R)| = 2 and |A| = 3. Then A acts fixed-point freely on \overline{R} := R/Φ(R). Since A centralizes Z(R), this gives Z(R) = Φ(R) and R is an extraspecial 2-group. Assume that there exists an involution t ∈ R \ Φ(R). Then F := ⟨t^{A}⟩ has order at most 8. Since |\overline{F} | = 4 and F contains an involution, we conclude that F is abelian. But, as we have already seen, [F, A] has to be non-abelian.

This contradiction shows that there are no involutions in R \ Φ(R), and so R ≅ Q_{8} and RA ≅ SL_{2}(3).

Lemma 7.2. Let p = 2 and V be a faithful indecomposable M-module with C_{V} (M) = 0 and [V, M] = V. Suppose that M = Alt(n), n ≥ 5, and that A = ⟨(12)(34), (13)(24)⟩ acts quadratically on V. Then ⟨(123)⟩ acts fixed-point freely on V. Moreover, one of the following holds:

1. V is the (simple) spin module for M.

2. 4 divides n and there exists an 𝔽_{2}M-submodule in W such that W and V/W spin modules for M and V/W ≅ W^{h}, where h ∈ Sym(n) \ Alt(n).

Proof. Let E = ⟨123⟩ and B = AE ≅ Alt(4) and for 5 ≤ i ≤ n let D_{i} = C_{M}({1, 2, 3, 4, i}). Then B ≤ D_{i}, D_{i} ≅ Alt(5) and

(∗)  M = ⟨D_{5}, D_{6}, …, D_{n}⟩.

Suppose there exists 0 ≠ w ∈ V with [w, B] = 0. Then ⟨w^{D_{i}}⟩ is a quotient of the natural permutation module for D_{i} ≅ Alt(5) over 𝔽_{2}, and the quadratic action of A forces [w, D_{i}] = 0. So by (*) [w, M] = 0, which contradicts C_{V} (M) = 0.

Thus C_{V} (B) = 0. Since B/A ≅ E is a 2′-group,

C_{V} (A) = C_{V} (B) ⊕ [C_{V} (A), B] = [C_{V} (A), B] = [C_{V} (A), E],

and so E acts fixed-point freely on C_{V} (A). This result applied to the dual of V shows that E acts fixed-point freely on V/[V, A]. Since A is quadratic, [V, A] ≤ C_{V} (A) and so E acts fixed-point freely on V. Now [Me, Theorem 2] shows that (1) or (2) holds.

Corollary 7.3. Let p = 2 and M ≅ Alt(6). Suppose that all fours groups in M act quadratically on V. Then [V, M] = 0.

Proof. Since M = O^{2}(M) we may assume for a contradiction that V is a non-trivial simple module. By 7.2, (123) acts fix-point freely on V. Since there exists an automorphism of Alt(6) sending (123) to (123)(456), the same results shows that (123)(456) acts fixpoint freely. So all non-trivial elements of order three in the non-cyclic abelian 3-group ⟨(123), (456)⟩ act fixed-point freely on V, a contradiction to coprime action.

Lemma 7.4. Let p = 2 and V be faithful and simple, and let A ≤ M with [V, A, A] = 0 and |A| > 2. Put L := F^{*}(M). Suppose that M = ⟨A^{M}⟩, L is quasisimple, Z(L) ≠ 1, and L/Z(L) ≅ Alt(n), n ≥ 5. Then one of the following holds:

1. M ∼ 3.Alt(6) and |V | = 2^{6}.

2. M ∼ 3.Alt(7), |V | = 2^{12}, and AZ(L)/Z(L) is conjugate to ⟨(12)(34), (13)(24)⟩.

Proof. Since V is a faithful simple M-module, O_{2}(M) = O_{2}(L) = 1. From [Gr] we get that n = 6 or 7 and |Z(L)| = 3. Put Z := Z(L) and let 𝔽 be the subring of End(V) generated by the image of Z in End(V). Then 𝔽 is a field of order four and M acts semilinear on the 𝔽-module V. Now [V, A, A] = 0 and |A| > 2 imply that A acts 𝔽-linearly on V, see for example [MS3, 2.15]. Thus [Z, A] = 1 and Z = Z(M). Hence M = L or M/Z ≅ Mat_{10}. But M = ⟨A^{M}⟩ is generated by involutions while Mat_{10} is not, so M = L. Since A is elementary abelian and |A| > 2 we have |A| = 4.

Note that there are two conjugacy classes of fours groups in L. In any case we can choose a series of subgroups A ≤ B ≤ D ≤ H ≤ L with B ≅ Alt(4), D ≅ Alt(5) and H ∼ 3.Alt(6). Let E ∈ Syl_{3}(B). Then E ≅ C_{3} and B = AE. By Gaschütz’ Theorem, the Sylow 3-subgroups of L are not abelian and so the subgroups E = E_{1}, E_{2}, E_{3} of order three in EZ other than Z are conjugate. Since Z acts fixed-point freely on V we have V = [V, Z] = ⨁_{i=1}^{3} C_{V} (E_{i}) and so |V | = |C_{V} (E)|^{3}. In particular, C_{V} (E) ≠ 0.

We claim that C_{V} (B) ≠ 0 or [V, B] ≠ V. If C_{V} (E) ≤ C_{V} (A), then 0 ≠ C_{V} (E) ≤ C_{V} (B). So suppose C_{V} (E) ≰ C_{V} (A) and put \overline{V} = V/C_{V} (A). Then 0 ≠ \overline{C_{V} (E)} ≤ C\overline{V} (E). By coprime actions, \overline{V} = C\overline{V} (E) ⊕ [\overline{V} , E] and so \overline{V} ≠ [\overline{V} , E]. Since A centralizes \overline{V}, this give \overline{V} ≠ [\overline{V} , B] and so V ≠ [V, B], proving the claim. Note further that by 1.8(d) A is also quadratic on the dual module V^{∗}. So replacing V by its dual, if necessary, we may assume that C_{V} (B) ≠ 0.

Let W be 1-dimensional 𝔽-subspace of C_{V} (B). Then ⟨W^{D}⟩ is a quotient of the natural permutation module for D ≅ Alt(5) over 𝔽. The quadratic action of A forces [W, D] = 0. Put U = ⟨W^{H}⟩. Then U ≅ V̂ /X̂, where V̂ is the 𝔽H-module induced from the 𝔽ZD-module W and X̂ is a 𝔽Hsubmodule of V̂. Note that dim_{𝔽} V̂ = 6. Since A has a regular orbit on H/ZD, A does not act quadratically on V̂. Thus U ≠ V̂. Since H acts faithfully on V̂ /X̂ and on X̂ and since H has no faithful module of dimension less than 3, we conclude that dim_{𝔽} V̂ /X̂ = 3 = dim_{𝔽} X̂.

If n = 6, then H = L, V = U and (1) holds. So suppose that n = 7. Choose a transitive action of L on I := {1, …, 7}. Suppose first that A has an orbit J on I with |J| = 2. Put K := C_{L}(J)′. Then K ≅ Alt(5) and AK ≅ Sym(5). Note that K is contained in a conjugate of H and that all composition factors for 𝔽H on V are 3-dimensional. It follows that all non-trivial composition factor for 𝔽K on V are 2-dimensional. Since A ∩ K ≠ 1, the quadratic action of A in V shows that also the non-trivial composition factors for 𝔽KA on V are 2-dimensional, a contradiction since |KA| > |K| = |SL_{2}(4)|.

Thus A has no orbits of length 2 and so A has three fixed-points on I. Then D has two fixedpoints, say i and j. Put D^{∗} := O^{2′}(N_{L}({i, j}). Then D^{∗} ≅ Sym(5) and D ⊴ D^{∗}. Recall from above that W is a 1-dimensional subspace of C_{V} (D), so C_{V} (D) ≠ 0 and thus also C_{V} (D^{∗}) ≠ 0. Hence we may and do choose W such that [W, D^{∗}] = 0. For k ≠ l ∈ I and g ∈ G with {k, l} = {i, j}^{g} put W_{kl} = W_{lk} = W^{g}. Since N_{L}({i, j}) = ZD^{∗} ≤ N_{L}(W), W_{kl} is well-defined. Let i be the fixed-point of H. Since ⟨W^{H}⟩ is 3-dimensional and H acts triple transitively on {W_{ij} | j ∈ I \ i} we conclude that for any distinct a, b, c, d ∈ I, ⟨W^{H}⟩ = W_{ab} + W_{ac} + W_{ad}. Since V = ⟨W^{L}⟩ is now easy to see that V = ⟨W_{kl} | 1 ≤ k < l ≤ 4⟩. Thus V is at most 6-dimensional. By the action of H on V, dim_{𝔽} V is a multiple of 3, so dim_{F} V = 3 or 6. Since \frac{| L_{3}(4)|}{|Alt(7)|} = 8 and L_{3}(4) ≇ Alt(8), Alt(7) is not involved in L_{3}(4). We conclude that dim_{𝔽} V > 3 and so dim_{𝔽} V = 6, and (2) holds.

We remark that 3.Alt(7) has indeed a 6-dimensional quadratic module over 𝔽_{4}. One way to see this is to use the embedding 3.Alt(7) ≤ 3.Mat_{22} ≤ SU_{6}(2) (thanks to J. Hall for pointing out this embedding to us): Consider the block normalizer P ∼ 3.2^{4}.Alt(6) in 3.Mat_{22}. Then P has a unique proper submodule on 𝔽_{4}^{6}, namely a 3-dimensional one. In particular, O_{2}(P) acts quadratically. Alt(7) has orbits of length 7 and 15 on the 22 points. Any three points from the 7 lie in a unique block and so we can choose P to intersect 3.Alt(7) in B ∼ 3.(Alt(4) × Alt(3)).2. It follows that O_{2}(B) ≤ O_{2}(P) and so O_{2}(B) is a quadratic fours group.

Lemma 7.5. Let M = Alt(n) or Sym(n), n ≥ 5, n ≠ 6, 8, and V be a simple spin module for 𝔽_{2}M. Suppose that A is a maximal quadratic subgroup of M on V with |A| > 2. Then |V | = |C_{V} (A)|^{2} and [V, a] = [V, A] = C_{V} (A) = C_{V} (a) for all 1 ≠ a ∈ A. Moreover, one of the following holds:

1. A is conjugate to ⟨(12)(34), (13)(24)⟩.

2. M ≅ Alt(9), |A| = 8, |A| has a regular orbit of length 8 on {1, 2, …, 9} and, up to conjugation, A is unique in M, with the conjugacy class depending on the isomorphism type of V.

Proof. Let I = {1, 2, …, n} with M acting transitively on I. Let K ≤ M with K ≅ Alt(5) and K fixing n − 5 points of I. Then V is a direct sum of natural SL_{2}(4)-modules. From this we get for B ∈ Syl_{2}(K): B is a quadratic fours group, and

|V | = |C_{V} (B)|^{2} and [V, b] = [V, B] = C_{V} (B) = C_{V} (b) for all 1 ≠ b ∈ B.

Moreover, the non-trivial elements of odd order in K act fixed-point-freely on V.

Let 1 ≠ z ∈ B and let D be a quadratic subgroup with z ∈ D. Then C_{V} (B) = C_{V} (z) = C_{V} (D) and so DB is quadratic. In particular, DB is elementary abelian.

Let W be a simple 𝔽_{2}M′-submodule of V. Since A ∩ M′ ≠ 1, then 0 ≠ [W, A ∩ M′] ≤ C_{W} (A). Thus A normalizes W.

If n = 5 or 7 then all involutions in M′ are conjugate. Thus we may assume that z ∈ A. If n = 5, then A ≤ C_{M}(B) = B. If n = 7, then Sym(7) does not act on W and so A ≤ M′. Also B is a Sylow 2-subgroup of C_{M′}(B) and again A ≤ B. So the lemma holds for n = 5 and 7.

Suppose next that n ≥ 9. As in Section 4 of [MeSt2] define L_{z} := O^{2}(C_{M}(z)) and A_{z} := O_{2}(C_{L}(z)). Moreover, for t ∈ M with |t| = 2 let K_{t} be the subgroup generated by the quadratic subgroups of M containing t. Observe that [V, t, K_{t}] = 0, so every fours group of K_{t} containing t is quadratic on V. Note further that A_{z} = B and L_{z} ≅ Alt(n − 4).

According to [MeSt2, Lemma (4.3)] we have that L_{z} ≰ K_{z}. Since K_{z} ⊴ C_{M}(z) and L_{z} is simple this implies [L_{z}, K_{z}] = 1. Since B = C_{M}(L_{z}) we conclude that K_{z} ≤ B.

If z ∈ A we conclude that A = B, and case (1) of the lemma holds. So suppose z^{M} ∩ A = ∅. Let 1 ≠ a ∈ A. Then A ≤ K_{a}. If z ∈ K_{a}, then by the above observation, a ∈ K_{z} = B and so a ∈ z^{M}, contrary to the assumption. Thus z^{M} ∩ K_{a} = ∅.

Let k := |C_{I}(a)|, J = I \ C_{I}(a) and m := \frac{|J|}{2}. We now choose 1 ≠ a ∈ M′ ∩ A and so m is even and m ≥ 4. Let D be the largest subgroup of M′ which has the same orbits as a on I. Put X = C_{M}(I \ J) and Y = C_{M}(J). Then D is elementary abelian of order 2^{m−1} and Y ≤ C_{M}(a). Suppose that Y ∩ A ≠ 1 and let 1 ≠ b ∈ A ∩ Y. Then Alt(J) ≅ ⟨a^{C_{M} (b)}⟩ ≤ K_{b} and z^{M} ∩ K_{b} ≠ 1, a contradiction. Thus A ∩ Y = 1 and A ≰ ⟨a⟩Y. In particular, K_{a} ≰ ⟨a⟩Y. Since D ∩ z^{M} ≠ ∅ we have D ≰ K_{a}. Also D = [D, X] = [DY, X] and so D ≰ K_{a}Y and DY ∩ K_{a}Y = ⟨a⟩Y.

Hence DY/⟨a⟩Y is not the only minimal normal subgroup of C_{M}(a)/⟨a⟩Y. Since

C_{M}(a)/⟨a⟩Y ∼ 2^{m−1}Sym(m) or 2^{m−2}Sym(m)

(with k ≤ 1 and M = Alt(n) in the latter case) we conclude that m = 4, C_{M}(a)/⟨a⟩Y ∼ 2^{2}Sym(4) and M ≅ Alt(9). Moreover, |K_{a}/⟨a⟩| = 4 and C_{M}(a) acts transitively on (K_{a}/⟨a⟩)^{♯}. Thus K_{a} is elementary abelian of order 8 and since K_{a} ∩ z^{M} = ∅, K_{a} acts regularly on J. It follows that N_{M}(K_{a}) acts transitively on K_{a}^{♯}. Since [V, a, K_{a}] = 0 we conclude that K_{a} acts quadratically on V. Thus A = K_{a} by the maximality of A. In particular, A is unique up to conjugacy. Also if t ∈ C_{Sym(9)}(a) \ Alt(8), then A^{t} ≠ A = K_{a}. So A^{t} will not act quadratically on V, and A^{M} depends on the isomorphism type of V. Let F ∈ Syl_{5}(K). As seen above F acts fixed-point freely on V, and F is inverted by a conjugate of a. Thus C_{V} (a) = [V, a] and the quadratic action of A forces C_{V} (a) = [V, A] = C_{V} (A); in particular |V | = |C_{V} (a)|^{2}.

Lemma 7.6. Let M = G_{2}(2) or G_{2}(2)′, and let V be a non-trivial simple 𝔽_{2}M-module. Suppose there exists A ≤ M with |A| > 2 and [V, A, A] = 0. Then V is a natural G_{2}(2)- and G_{2}(2)′-module, respectively.

Proof. Since |A| > 2, there exists 1 ≠ z ∈ A ∩ M′, and since M′ has a unique class of involutions, z is 2-central. Put P_{1} := C_{M}(z), let S ∈ Syl_{2}(P_{1}), and let P_{2} be the other minimal parabolic subgroup containing S. Suppose for a contradiction that C_{V} (P_{2}) = 0.

Let Γ = P_{1}^{G} ∪ P_{2}^{G} be the generalized hexagon associated to M. Let (P_{1}, P_{2}, P_{3}, P_{4}) be a path of length 4 in Γ. Put Z := ⟨z⟩. Then

Z ≤ P_{4}, Z ≰ O_{2}(P_{4}), T := ZO_{2}(P_{4}) ∈ Syl_{2}(P_{4}), and P_{4} = ⟨Z^{P_{4}}⟩O_{2}(P_{4}).

Since C_{V} (P_{2}) = 0 and P_{2} and P_{4} are conjugate, we also have C_{V} (P_{4}) = 0, so

X := [C_{V} (O_{2}(P_{4})), Z] ≠ 0.

Note that T centralizes X, and since T is a maximal subgroup of P_{4}, C_{P_{4}}(X) = T. Since P_{4} and P_{3} are the only maximal subgroups of M containing T, it follows that C_{M}(X) ≤ P_{3}. From Z ≤ A and [V, A, A] = 0 we get A ≤ C_{M}(X) = P_{3}. So A fixes all vertices of distance two from P_{1}. But the stabilizer in P_{1} of these vertices is cyclic, a contradiction since |A| > 2 and A is elementary abelian.

Thus C_{V} (P_{2}) ≠ 0. Let M ≤ M^{∗} with M^{∗} ≅ G_{2}(2), and let V^{∗} be a simple quotient of the induced 𝔽_{2}M^{∗}-module V^{M^{∗}} and identify V with its image in V^{∗}. Let S^{∗} ∈ Syl_{2}(M^{∗}) with S ≤ S^{∗}. Put P_{i}^{∗} = P_{i}S^{∗}. Since |P_{2}^{∗}/P_{2}| ≤ 2 we get that C_{V^{∗}}(P_{2}^{∗}) ≠ 0. By Smith’s lemma 4.2 V_{i} := C_{V^{∗}}(O_{2}(P_{i}^{∗})) is a simple P_{i}^{∗}-module. It follows that V_{2} = C_{V} (P_{2}^{∗}) = C_{V} (S^{∗}) has order two, C_{V^{∗}}(P_{1}^{∗}) = 0, and V_{1} is the unique non-trivial simple P_{1}^{∗}/O_{2}(P_{1}^{∗})-module, namely the natural SL_{2}(2)-module. Thus by Ronan-Smith’s Lemma 4.3 V^{∗} is uniquely determined, and so V^{∗} is the natural G_{2}(2)-module for M^{∗}. Hence V = V^{∗} and the lemma is proved.

Remark 7.7. Let L := F^{∗}(M) and suppose that O_{2}(M) = 1, L is quasisimple and L/Z(L) ≅ U_{4}(3). Let \overline{M} = M/Z(L), S ∈ Syl_{2}(M), and Z = Ω_{1}Z(S). In the following we use some information about the structure of M which can be found for example in [ATLAS]. More precisely we use the following facts:

There exists exactly two elementary abelian subgroups Q_{1} and Q_{2} of order 2^{4} in S, and for

P_{1} = C_{L}(Z), Q_{1} := O_{2}(P_{1}), P_{2} := N_{L}(Q_{2}), and P_{3} := N_{L}(Q_{3})

the following hold:

(a) For i = 1, 2, 3, \overline{P}_{i} is a maximal subgroup of \overline{M} and has characteristic 2.

(b) \overline{P}_{1}/\overline{Q}_{1} ≅ Sym(3) × Sym(3), Q_{1} is extraspecial of order 2^{5}, and Q_{1}/Z is a simple P_{1}-module.

(c) For i = 1, 2, \overline{P}_{i}/\overline{Q}_{i} ≅ Alt(6), and Q_{i} is a natural Alt(6)-module for P_{i}.

(d) All involutions in L are conjugate.

(e) Suppose in addition that |Z(L)| = 3, M ≠ L, [Z(L), M] = 1, M = N_{M}(Q_{2})L, and that N_{M}(Q_{2}) induces inner automorphisms on \overline{P_{2}}/\overline{Q_{2}}. Put P_{i}^{∗} = N_{M}(Q_{i}) and Q_{i}^{∗} = O_{2}(P_{i}^{∗}). Then

(a) M is unique up to isomorphism and |M/L| = 2.

(b) M has two classes of involutions in M \ L with representatives a and b in Q_{2} such that C\overline{L}(a) ≅ U_{4}(2) and C\overline{L}(b) ∼ 2^{4}.3^{2}.2^{2}.

(c) P_{2}^{∗}/Q_{2}^{∗} ≅ 3.Alt(6), and Q_{2}^{∗} is the dual of the natural Ω_{5}(2)-module for P_{2}^{∗}.

(d) Q_{3}^{∗} = Q_{2} and P_{2}^{∗}/Q_{2} ≅ C_{3} × Sym(6). Lemma 7.8. Let p = 2 and V be faithful 𝔽_{2}M-module, and let Z ≤ M with |Z| = 2. Suppose that (i) M is quasisimple, O_{2}(M) = 1 and M/Z(M) ≅ U_{4}(3). (ii) C_{M}([V, Z]) ≰ Z. (iii) C_{V} (M) = 0, V = [V, M] and V is indecomposable, that is, V is not the sum of two proper

(non-zero) 𝔽_{2}M-submodules. Put P_{1} := N_{M}(Z) and Q_{1} := O_{2}(P_{1}), and let S ∈ Syl_{2}(P_{1}) and Q_{i}, i = 2, 3, be the two elementary

abelian subgroup of order 16 in S. Put P_{i} := N_{M}(Q_{i}), L_{i} := O^{2′}(P_{i}), L_{12} := ⟨Q_{3}^{P_{1}}⟩, L_{13} := ⟨Q_{2}^{P_{1}}⟩, and 𝔽 := End_{M}(V). Then we can choose {i, j} = {2, 3} such that the following hold :

(a) V is a simple M-module, |𝔽| = 4 and dim_{𝔽} V = 6.

(b) C_{V} (L_{i}) = 0 and C_{V} (L_{j}) ≠ 0.

(c) V is uniquely determined as a 𝔽_{2}M-module.^{3}

(d) There exists a non-degenerate M-invariant unitary 𝔽-form on V.

(e) Q_{1} ≤ L_{1k}, L_{1k}/Q_{1} ≅ Sym(3), k = 2, 3, and L_{1}/Q_{1} = L_{12}/Q_{1} × L_{13}/Q_{1} ≅ Sym(3) × Sym(3).

(f) L_{1j} = C_{M}([V, Z]), C_{V} (Z) = [V, Q_{1}] = [V, L_{1j}] and [V, Z] = C_{V} (Q_{1}) = C_{V} (L_{1j}).

(g) 1 ≤ [V, Z] ≤ C_{V} (Z_{1}) ≤ V is the unique chiefseries for P_{1} on V, each of the factors is 2dimensional over 𝔽, L_{1i} centralizes C_{V} (Z)/[V, Z] and L_{1j} centralizes [V, Z] and V/[V, Z].

(h) P_{i} = L_{i} and L_{i}/Q_{i} is quasisimple of shape 3.Alt(6).

(i) Q_{i} acts quadratically on V and C_{V} (Q_{i}) = [V, Q_{i}].

(j) 1 ≤ [V, Q_{i}] ≤ V is the unique chiefseries for P_{i} on V, each of the factors is 3-dimensional over 𝔽 and faithful for P_{i}/Q_{i}. Moreover, V/[V, Q_{i}] is as an 𝔽_{2}P_{i}-module isomorphic to the dual of [V, Q_{i}].

(k) L_{j}/Q_{j} is isomorphic to Alt(6).

3Note that 3^{2}. U_{4}(3) has two quotients isomorphic to M and so has two modules which fulfill the hypothesis of this lemma, except that the modules are not faithful.

(l) C_{V} (S) = C_{V} (Q_{j}) = C_{V} (L_{j}) and [V, S] = [V, Q_{j}] = [V, L_{j}].

(m) 1 ≤ C_{V} (Q_{j}) ≤ [V, Q_{j}] ≤ V is the unique chiefseries for P_{j} on V, where C_{V} (Q_{j}) and V/[V, Q_{j}] are 1-dimensional over 𝔽 and centralized by L_{j} while [V, Q_{j}]/C_{V} (Q_{j}) is a 4-dimensional natural 𝔽Alt(6)-module for L_{j}.

Proof. Let \overline{M} := M/Z(M), \{k, l\} = \{2, 3\} and P_{1k} := P_{1} ∩ P_{k}.

1^{∘}. V is an homogeneous 𝔽_{2}Z(M)-module and Z(M) is cyclic.

Since O_{2}(M) = 1, Z(M) is an abelian 2′-group. Thus V is a semisimple 𝔽_{2}Z(M)-module. Since V is indecomposable, we conclude that V is an homogeneous 𝔽_{2}Z(M) module and so Z(M) is cyclic. Thus (1^{∘}) holds.

In the following we will only use (1^{∘}) but no longer that V is indecomposable. Moreover, we make use of the properties listed in 7.7.

2^{∘}. [V, Z, Q_{1}] = 0.

By (1^{∘}) Z(M) ∩ C_{M}([V, Z]) = 1 and so by (ii) \overline{C_{M}([V, Z])} ≰ \overline{Z}. Note that \overline{P_{1}}/\overline{Q_{1}} ≅ Sym(3) × Sym(3), \overline{Q_{1}} is extra special of order 2^{5} and \overline{P_{1}} acts simply on \overline{Q_{1}}/\overline{Z}. Hence \overline{Q_{1}}/\overline{Z} is the unique minimal normal subgroup of \overline{P_{1}} and we conclude that \overline{Q_{1}} ≤ \overline{C_{M}([V, Z])}. Thus Q_{1} ≤ C_{M}([V, Z]) and (2^{∘}) holds.

3^{∘}. [V, Q_{k}, Q_{k}, L_{k}] = 1.

Observe that \overline{P_{k}}/\overline{Q_{k}} ≅ Alt(6), C\overline{M}(\overline{Q_{k}}) = \overline{Q_{k}} and \overline{Q_{k}} is a natural Alt(6)-module for \overline{P_{k}}. Since P_{1k} = N_{P_{k}}(Z) we conclude that \overline{P_{1k}}/O_{2}(\overline{P_{1k}}) ≅ Sym(3) and [Q_{k}, P_{1k}] is a hyperplane of Q_{k}. The structure of P_{1} shows that [O_{2}(P_{1k}), P_{1k}] ≤ Q_{1} and so [Q_{k}, P_{1k}] ≤ Q_{1} and |Q_{k}/Q_{k} ∩ Q_{1}| ≤ 2. In particular, P_{1k} normalizes [V, Z, Q_{1}Q_{k}], and by (2^{∘}) [V, Z, Q_{1}Q_{k}] = [V, Z, Q_{k}].

Note that Q_{1} does not contain an elementary abelian subgroup of order 2^{4}. So Q_{k} ≰ Q_{1} and Q_{1} ∩ Q_{k} = [Q_{k}, P_{1k}]. Pick g ∈ P_{k} with Q_{k} = (Q_{1} ∩ Q_{k})Z^{g}. Then by (2^{∘})

[V, Z, Q_{k}] = [V, Z, (Q_{1} ∩ Q_{k})Z^{g}] = [V, Z, Z^{g}] ≤ [V, Z^{g}] ≤ C_{V} (Q_{1}^{g}).

It follows that [V, Z, Q_{k}] is normalized by ⟨P_{1k}, Q_{k}^{g}⟩ = P_{k}. Thus [V, Z, Q_{k}] = [V, < Z^{P_{k}}⟩, Q_{k}] = [V, Q_{k}, Q_{k}] and [V, Q_{k}, Q_{k}] is centralized by ⟨Q_{1}^{gP_{k}}⟩ = L_{k}.

4^{∘}. [C_{V} (Q_{k}), Q_{1}, Q_{1}] = 0.

Let h ∈ P_{1} \ P_{1} ∩ P_{k}. Then Q_{1} = (Q_{1} ∩ Q_{k})(Q_{1} ∩ Q_{k}^{h}). Since Q_{1} normalizes C_{V} (Q_{k}), (3^{∘}) implies

[C_{V} (Q_{k}), Q_{1}, Q_{1}] = [C_{V} (Q_{k}), (Q_{1} ∩ Q_{k}^{h}), (Q_{1} ∩ Q_{k}^{h})] ≤ C_{V} (Q_{k}) ∩ [V, Q_{k}, Q_{k}]^{h} ≤ C_{V} (Q_{k}) ∩ C_{V} (L_{k}^{h}).

Since \overline{L_{k}} is a maximal subgroup of \overline{M} and Q_{k} ≰ L_{k}^{h} we have M = ⟨Q_{k}, L_{k}^{h}⟩. So

C_{V} (Q_{1}) ∩ C_{V} (Q_{k}) ≤ C_{V} (M) = 0,

and (4^{∘}) is proved.

In the next step we regard Q_{k} is a 4-dimensional symplectic space for \overline{L}_{k}/\overline{Q}_{K} ≅ Sp_{4}(2)′.

5^{∘}. |Q_{k}Q_{l}/Q_{k}| = 4 and Q_{k}Q_{l} ≠ Q_{k}Q_{1}. Moreover, Q_{k} ∩ Q_{l} is a singular subgroup of order 4 in Q_{k} (and Q_{l}), and Q_{k} ∩ Q_{l} acts quadratically on V.

Since Q_{l} is elementary abelian of order 2^{4} and no element in L_{k} acts as a transvection on Q_{k},

|Q_{k}Q_{l}/Q_{k}| = |Q_{l} ∩ Q_{k}| = 4, Q_{k} ∩ Q_{l} = [Q_{k}, Q_{l}] = C_{Q_{k}}(Q_{1}).

Hence 3.2(c) shows that Q_{l} ∩ Q_{k} is a singular subspace of Q_{k}. Moreover, Z ≤ Q_{k} ∩ Q_{l} ≤ Q_{k} ∩ Q_{1} and so by (2^{∘}), [V, Z, Q_{k} ∩ Q_{l}] = 1. Since |Q_{k} ∩ Q_{l}| = 4 and Z ≤ Q_{k} ∩ Q_{l}, this shows that Q_{k} ∩ Q_{l} is quadratic on V, and (5^{∘}) holds. 6^{∘}. [C_{V} (Q_{k}), Q_{l}, Q_{l}] = 1

By (5^{∘}) Q_{l} = (Q_{l} ∩ Q_{k})(Q_{l} ∩ Q_{k})^{g} for a suitable g ∈ P_{l} and (Q_{l} ∩ Q_{k})^{g} acts quadratically on V. Thus

[C_{V} (Q_{k}), Q_{l}, Q_{l}] = [C_{V} (Q_{k}), (Q_{l} ∩ Q_{k})^{g}, (Q_{l} ∩ Q_{k})^{g}] = 0,

and (6^{∘}) holds.

Since C_{V} (M) = 0, M = ⟨L_{2}, L_{3}⟩ and C_{V} (S) ≤ C_{V} (Q_{2}) ∩ C_{V} (Q_{3}) we can choose i ∈ {2, 3} such that [C_{V} (Q_{i}), L_{i}] ≠ 0. Let {2, 3} = {i, j}. 7^{∘}. P_{i} = L_{i}, Z(M) = Z(L_{i}) ≅ C_{3}. L_{i}/Q_{i} is quasisimple of shape 3.Alt(6) and C_{V} (L_{i}) = 0.

By (4^{∘}),(5^{∘}), (6^{∘}) all the fours groups in L_{i}/Q_{i} act quadratically on C_{V} (Q_{i}). Since [C_{V} (Q_{i}), L_{i}] ≠ 0, 7.3 shows that L_{i}/Q_{i} ≅ A̸lt(6). Hence Z(M) ∩ L_{i} ≠ 1. By [Gr] and since Z(M) is a cyclic 2′- group, Z(M) ≅ C_{3} and so Z(M) ≤ L_{i}. So P_{i} = L_{i}, and C_{V} (L_{i}) ≤ C_{V} (Z(M)) = 0. Thus L_{i}/Q_{i} is quasisimple of shape 3.Alt(6), and (7^{∘}) is proved.

In particular, (h) holds. 8^{∘}. Q_{i} acts quadratically on V.

By (3^{∘}) and (7^{∘}), [V, Q_{k}, Q_{k}] ≤ C_{V} (L_{k}) = 0. 9^{∘}. [C_{V} (Q_{i}), Q_{j}] ≤ C_{V} (L_{j}) = C_{V} (Q_{j}) and L_{j}/Q_{j} ≅ Alt(6); in particular C_{V} (L_{j}) ≠ 0.

Let g ∈ L_{j} with Z^{g} ≰ Q_{i} ∩ Q_{j}. Then Z^{g} ≤ L_{i} and Z^{g} ≰ Q_{i}. Since L_{i}/Q_{i} is quasisimple, L_{i} = ⟨Z^{gL_{i}}⟩Q_{i} and so [C_{V} (Q_{i}), Z^{g}] ≠ 0. On the other hand [C_{V} (Q_{i}), Z^{g}] is centralized by ⟨Q_{i}, Q_{1}^{g}⟩ = L_{j} and we conclude that 0 ≠ [C_{V} (Q_{i}), Q_{j}] ≤ C_{V} (L_{j}). In particular, Z(M) ≰ L_{j} and so L_{j}/Q_{j} ≅ Alt(6).

Thus C_{V} (L_{j}) ≠ 0. If [C_{V} (Q_{j}), L_{j}] ≠ 0 we could apply (7^{∘}) to j in place of i and conclude that C_{V} (L_{j}) = 0, a contradiction. Thus [C_{V} (Q_{j}), L_{j}] = 0 and (9^{∘}) holds.

In particular, (k) holds. Since C_{V} (L_{j}) ≠ 0, (b) is proved. 10^{∘}. V = ⟨C_{V} (L_{j})^{M}⟩.

By (9^{∘}) [C_{V} (Q_{i}), Q_{j}] ≤ C_{V} (L_{j}). It follows that

[C_{V} (Q_{i}), L_{i}] = [C_{V} (Q_{i}), ⟨Q_{j}^{L_{j}}⟩] ≤ ⟨C_{V} (L_{j})^{L_{i}}⟩.

On the other hand, by (7^{∘}) Z(M) ≤ Z(L_{i}), so by (1^{∘}) L_{i} does not have any central chieffactor in C_{V} (Q_{i}). Hence C_{V} (Q_{i}) = ⟨C_{V} (L_{j})^{L_{i}}⟩.

Since V = [V, M] and M = ⟨Q_{i}^{M}⟩, V = ⟨[V, Q_{i}]^{M}⟩. As Q_{i} acts quadratically we conclude that V = ⟨C_{V} (Q_{i})^{M}⟩, and as C_{V} (Q_{i}) = ⟨C_{V} (L_{j})^{L_{i}}⟩, this gives (10^{∘}). 11^{∘}. C_{V} (L_{1}) = 0.

By (9^{∘}) C_{V} (L_{1}) ≤ C_{V} (L_{j}). Since C_{V} (M) = 0 and M = ⟨L_{1}, L_{j}⟩, (11^{∘}) follows.

12^{∘}. [V, Z, L_{1j}] = 0, L_{1k}Q_{k} = O^{2′}(P_{1} ∩ P_{k}), and (e) holds.

Put P^{∗} := C_{P_{1}}([V, Z]). Since P_{1} normalizes [V, Z], P^{∗} ⊴ P_{1}. Moreover, by (11^{∘}) L_{j} ≤ C_{M}([V, Z] ∩ C_{V} (S)) and so C_{M}([V, Z] ∩ C_{V} (S)) ≤ P_{j}, since \overline{L}_{j} is a maximal subgroup of \overline{M}. It follows that P^{∗} ≤ P_{1} ∩ P_{j}.

Since Q_{i} acts quadratically on V and Z ≤ Q_{i}, [V, Z, Q_{i}] = 0. Hence L_{1j} = ⟨Q_{i}^{P_{1}}⟩ ≤ P^{∗}, so [V, Z, L_{1j}] = 0. Moreover, since L_{1j} ⊴ P_{1}, and P_{1} acts simply on Q_{1}/Z, also Q_{1} ≤ L_{1j}. Since L_{j} ∩ P_{1}/Q_{j} ≅ Sym(4) and L_{1j} = ⟨Q_{i}^{L_{1j}}⟩, we conclude that L_{1j}/Q_{1} ≅ Sym(3) and L_{1j}Q_{j} = O^{2′}(P_{1} ∩ P_{j}). In particular [L_{1j}, Q_{j}] ≤ Q_{1} and so [L_{1j}, L_{1i}] ≤ Q_{1}. Hence also L_{1i}/O_{2}(L_{ij}) ≅ Sym(3) and again by the simple action of P_{1} on Q_{1}/Z, O_{2}(L_{1i}) = Q_{1}. In addition, P_{1i} ≤ N_{P_{1}}(Q_{i}) and so L_{1i} = O^{2′}(P_{1} ∩ P_{i}) since by (7^{∘}) P_{1} ∩ P_{i}/Q_{i} ≅ C_{3} × Sym(4). Hence (12^{∘}) and (e) has been proved.

13^{∘}. Let 𝔼 be the subring of 𝔽 generated by the image of Z(M). Then 𝔼 ≅ 𝔽_{4} and [V, Z] is a direct sum of 2-dimensional simple 𝔼L_{1}-modules.

Since Z(M) ≅ C_{3}, 𝔼 ≅ 𝔽_{4}. The second statement follows from (12^{∘}) (and (e)) since L_{1j} = C_{L_{1}}([V, Z)), C_{V} (L_{1}) = 0 and L_{1}/L_{1j} ≅ Sym(3).

Let U_{j} be a 1-dimensional 𝔼-subspace of C_{V} (L_{j}). In the following we use the fact that (e) has already been proved, so we know that L_{1j} = C_{L_{1}}([V, Z]) ⊴ P_{1} and

L_{1}/Q_{1} = L_{12}/Q_{1} × L_{13}/Q_{1} ≅ Sym(3) × Sym(3);

in particular L_{1}/C_{L_{1}}([V, Z]) ≅ Sym(3).

Put U_{1} := ⟨U_{j}^{P_{1}}⟩ and U_{i} := ⟨U_{j}^{P_{i}}⟩, so [U_{j}, L_{1j}] = 0 since L_{1j} ≤ L_{j}, and

U_{1} = ⟨U_{j}^{L_{1i}}⟩ = ⟨U_{j}^{P_{1}∩P_{i}}⟩

since U_{j} is an 𝔼-space. As L_{1}/C_{L_{1}}([V, Z]) ≅ Sym(3) and C_{V} (L_{1}) = 0 we conclude that dim_{𝔼} U_{1} = 2. Since P_{i} ∩ L_{j} centralizes U_{j} and U_{1} = ⟨U_{j}^{P_{i}∩P_{1}}⟩, (7^{∘}) and 5.4 imply that dim_{𝔼} U_{i} = 3. In particular,

U_{i} = ⟨U_{1}^{P_{i}∩P_{j}}⟩.

Put W_{1} := ⟨U_{i}^{L_{1}}⟩ and W_{j} := ⟨U_{1}^{L_{j}}⟩. Since [U_{i}, L_{1i}] ≤ U_{1} and L_{1i} ⊴ L_{1} we have

[W_{1}, L_{1i}] ≤ U_{1} and W_{1} = ⟨U_{i}^{L_{1j}}⟩ = ⟨⟨U_{1}^{P_{i}∩P_{j}}⟩^{L_{1j}}⟩ ≤ W_{j}.

Put Y_{j} := C_{W_{j}}(L_{j}) and \overline{W}_{j} := W_{j}/U_{j}. Then dim_{𝔼} \overline{U}_{1} = 1, dim_{𝔼} \overline{U}_{i} = 2, and \overline{U}_{i} = ⟨\overline{U}_{1}^{P_{i}∩L_{j}}⟩. Thus, we can apply 5.4 (and (9^{∘})) with U = \overline{U}_{1}. This shows that \overline{W}_{j}/C_{\overline{W}_{j}}(L_{j}) is a natural 𝔼Alt(6)-

module and C_{\overline{W}_{j}}(L_{j}) ≤ ⟨\overline{U}_{i}^{L_{1j}}⟩ = \overline{W}_{1}; in particular dim_{𝔼} \overline{W}_{j}/C_{\overline{W}_{j}}(L_{j}) = 4. Since L_{j} = O^{2}(L_{j}) and [U_{j}, L_{j}] = 0, we also have C_{\overline{W}_{j}}(L_{j}) = \overline{Y_{j}}.

Since Y_{j} ≤ W_{1} [Y_{j}, L_{1i}] ≤ [W_{1}, L_{1i}] ≤ U_{1}. From L_{1i}L_{1j} = L_{1} we conclude that [Y_{j}U_{1}, L_{1}] ≤ U_{1}. Note that [Y_{j}U_{1}, Q_{1}] = 0 and O^{2}(L_{1})/Q_{1} is a 2′-group. So coprime action implies

Y_{j}U_{1} = C_{Y_{j}U_{1}}(O^{2}(L_{1}))[Y_{j}U_{1}, O^{2}(L_{1})].

Since C_{V} (L_{1}) = 0 also C_{V} (O^{2}(L_{1})) = 0 and so Y_{j}U_{1} = U_{1}. Thus Y_{j} ≤ C_{U_{1}}(Q_{j}) = U_{j}. Hence dim_{𝔼} W_{j}/U_{j} = 4 and since W_{1} ≤ W_{j}, dim_{𝔼} W_{1}/U_{1} = 2. It follows that dim_{𝔼} W_{j}/W_{1} = 1 and W_{j} = ⟨W_{1}^{P_{i}∩P_{j}}⟩.

Put W := ⟨W_{1}^{L_{i}}⟩ and W̌ = W/U_{i}. Then W_{j} ≤ W, dim_{𝔼} W̌_{1} = 1 and dim_{𝔼} W̌_{j} = 2. Hence (7^{∘}) and 5.4 give dim_{𝔼} W̌ = 3; in particular dim_{𝔼} W/W_{j} = 1. Since P_{1i} does not normalize W_{j}, W = ⟨W_{j}^{P_{i}∩P_{1}}⟩. Since dim_{𝔼} W_{j}/W_{1} = 1, [W_{j}, L_{1j}] ≤ W_{1} and so [W, L_{1j}] ≤ W_{1} ≤ W. Thus W is normalized by L_{i} and L_{1j}L_{1i} = L_{1}. Hence W is an 𝔼M submodule of V, dim_{𝔼} W = 6 and W = ⟨U_{j}^{M}⟩.

Note that [U_{j}, L_{j}] = 0 and U_{j} is the (up to isomorphism) unique non-trivial simple 𝔽_{2}Z(M) module. So U_{j} is uniquely determined as an 𝔽_{2}P_{j}-module. Let Ŵ be the 𝔽_{2}M-module induced from the 𝔽_{2}P_{j} module U_{j}. Put W̃ := Ŵ /⟨[Ŵ , Z, Q_{1}]^{M}⟩ and let Û_{j} be the image of U_{j} in Ŵ. Note that Z(M) acts fixed-point freely on Ŵ and so also on W̃. In particular, CW̃(M) = 0, W̃ = [W̃, M] and [W̃, Z, Q_{1}] = 0. Thus W̃ fulfills the assumption on W in this proof. Since W̃ = ⟨Ũ_{j}^{M}⟩ we conclude that dim_{𝔼} W̃ = 6. On the other hand W is as an 𝔽_{2}M-module an homomorphic image of Ŵ and so also of W̃. It follows that W ≅ W̃ as an 𝔽_{2}M-module and so W is unique up to isomorphism.

Up to now we only used (1^{∘}) to determine W. Suppose now that V is indecomposable. Then by (10^{∘}) we can choose U_{j} such that V = ⟨U_{j}^{M}⟩. Thus V = W and dim_{𝔼} V = 6. Any non-trivial 𝔽_{2}M quotient of V fulfills the same assumption and so is 6-dimensional over 𝔼. Thus V is a simple 𝔽_{2}M-module.

Let V^{∗} be the 𝔽-dual of V. Then V^{∗} = [V^{∗}, Z(M)] = [V^{∗}, M] and 0 = C_{V^{∗}}(Z(M)) = C_{V^{∗}}(L_{i}^{∗}) = C_{V^{∗}}(M) = 0. By 1.8(c) Q acts quadratically on V^{∗} and so C_{M}([V^{∗}, Z]) ≰ Z. Thus V^{∗} and i fulfill the same assumption as V and i, and V and V^{∗} are isomorphic 𝔽_{2}M-modules. Hence by 1.9(a) there exists a M-invariant non-degenerate symmetric, symplectic or unitary 𝔽-form on V^{∗}. In the symmetric or symplectic case, V would be selfdual as an 𝔽M-module and so also an 𝔼Z(M)-module, a contradiction. Thus (d) holds.

Since L_{i} acts simply on U_{i} and V/U_{i}, C_{V} (Q_{i}) = U_{i} = [V, Q_{i}] and (i) and (j) hold. Note that Z = Q′_{1} centralizes V/[V, Q, Q]. Since Q_{1} centralizes V/W_{1} and W_{1}/U_{1} we conclude that [V, Q, Q] = W_{1} = [V, Z] and [V, Q] = W_{1}. By a dual argument, C_{V} (Z) = W_{1} and C_{V} (Q_{1}) = U_{1}. Also [U_{1}, L_{1j}] = 1 and dually [V, L_{1j}] ≤ W_{1}. Thus (f) and (g) are proved.

C_{V} (Q_{j}) ≤ C_{V} (Z) = W_{1} < W_{j} and since W_{j}/U_{j} is a simple 𝔼L_{j}-module, C_{V} (Q_{j}) = U_{j}. Dually [V, Q_{j}] = W_{j} and so (l) and (m) hold. Since |U_{j}| = 4 and C_{V} (Q_{j}) is an 𝔽-subspace, |𝔽| ≤ 4 and so 𝔽 = 𝔼. Since W is unique up to isomorphism we conclude that (a) and (c) hold.

Lemma 7.9. Put L := F^{∗}(M) and suppose that

(i) V is faithful and indecomposable 𝔽_{2}M-module, C_{V} (L) = 0 and V = [V, L].

(ii) M = ⟨D ≤ M | [V, D, D] = 0, |D| > 2⟩; and

(iii) L is quasi-simple and Z(L) ≅ U_{4}(3).

Put 𝔽 := End_{M}(V) and let A be a maximal quadratic subgroup of M on V. Then

(a) V is a simple 𝔽_{2}L-module and (L, V) fulfills the assumptions on (M, V) and so also the conclusions in 7.8.

(b) M = LA.

(c) |A/A ∩ L| ≤ 2, |A ∩ L| = 2^{4} and C_{M}(A) = C_{M}(A ∩ L) = AZ(M).

(d) N_{M}(A) = N_{M}(A ∩ L) and so N_{M}(A)/A is a quasisimple group of shape 3.Alt(6).

(e) C_{V} (A ∩ L) = C_{V} (A) = [V, A] = [V, A ∩ L] is a 3-dimensional. simple module for N_{M}(A).

(f) A is unique up to conjugation under L, with the conjugacy class depending on the isomorphism type of V.

(g) Let 1 ≠ B ≤ M such that B acts quadratically on V. Then B is conjugate under L to an subgroup of A and assuming that B ≤ A one of the following holds:

(a) |B| = 2, B ≤ L and dim_{𝔽}[V, B] = dim_{𝔽} V/C_{V} (B) = 2.

(b) |B| = 2, dim_{𝔽}[V, B] = dim_{𝔽} V/C_{V} (B) = 1. and C_{V} (B)/[V, B] is natural 𝔽SU_{4}(2)-module for C_{L}(B).

(c) |B| = 4, B ≰ L, dim_{𝔽}[V, B] = dim_{𝔽} V/C_{V} (B) = 2 and dim_{𝔽}[V, b] = 1 for all b ∈ B \ L.

(d) C_{V} (B) = [V, B] = C_{V} (A) and A is the unique maximal quadratic subgroup of M containing B.

Proof. Put \overline{M} = M/Z(L). Among all A ≤ M with [V, A, A] = 0 and |A| > 2 let A be maximal. Let S ∈ Syl_{2}(M)) with A ≤ S. Since Out(\overline{L}) ≅ Dih_{8}, M/L is isomorphic to a subgroup of Dih_{8}. In particular, M = LS. Let Y be non-trivial indecomposable 𝔽_{2}L-submodule of V.

By [MeSt1, 2.3] we have C_{S∩L}([V, Z]) ≰ Z and so (L, Y) fulfills the hypothesis of 7.8 in place of (M, V). It follows that Y is a simple 𝔽_{2}L-module and so V is a semisimple 𝔽_{2}L-module.

Let W be a maximal homogeneous 𝔽_{2}L-submodule of V and suppose that A does not normalizes W. Then by [MS3, 2.11]|A/C_{A}(W)| = 2 and so C_{A}(W) ≠ 1. Since L is quasisimple we conclude that L = [L, C_{A}(W)] ≤ C_{L}(W), a contradiction to C_{V} (L) = 0. Hence A normalizes W. As A was an arbitrary maximal quadratic subgroup of order larger than 2, (ii) shows that M normalizes every maximal homogeneous 𝔽_{2}L-submodule W. Since V is indecomposable as an 𝔽_{2}M-module and semisimple as an 𝔽_{2}L-module we conclude that V = W and so V is a homogeneous 𝔽_{2}L-module. In particular, C_{L}(Y) = C_{L}(V) = 1, Z(L) ≅ C_{3} and the subring 𝔼 of End_{𝔽_{2}L}(V) generated by the image of Z(L) is a field isomorphic to 𝔽_{4}.

Put 𝔽_{0} := Z(End_{𝔽_{2}L}(V)) and note that 𝔽_{0} is field isomorphic to End_{𝔽_{2}L}(Y) and so to 𝔽_{4}. Thus 𝔽_{0} = 𝔼. Since |A| ≥ 4, we conclude from [MS3, 2.15], that A and so also M acts 𝔽_{0}-linear on V. Hence Z(L) = Z(M) and 𝔽_{0} = 𝔽.

Let Z = Z(S ∩ L), P_{1} = N_{L}(Z), Q_{1} = O_{2}(P_{1}), Q_{i}, i = 2, 3, the two elementary abelian subgroups of order 16 in S ∩ L, P_{i} = N_{L_{i}}(Q_{i}) and for i ∈ {1, 2, 3}, P_{i}^{∗} = N_{M}(Q_{i}), L_{i} = O^{2′}(P_{i}), and Q_{i}^{∗} = O_{2}(P_{i}^{∗}). Choose notation such that C_{Y} (L_{2}) = 0 and so C_{Y} (L_{3}) ≠ 0. In the following we will use the properties of P_{i}, i = 1, 2, 3, given in 7.8.

Since V is a homogeneous 𝔽_{2}L-module we conclude that also C_{V} (L_{2}) = 0 and C_{V} (L_{3}) ≠ 0. Thus S normalizes L_{2} and L_{3} and so S ≤ P_{i}^{∗} for all 1 ≤ i ≤ 3. In particular, |M/L| ≤ 4. Since P_{2}/Q_{2} ∼ 3.Alt(6) and P_{2}^{∗} centralizes Z(L) we conclude that P_{2}^{∗} induces inner automorphisms on P_{2}/Q_{2}, so P_{2}^{∗} = Q_{2}^{∗}P_{2}. Thus |M/L| ≤ 2. Since |A| ≥ 4 we get A∩L ≠ 1, and since L has unique class of involutions and |Z| = 2, we may assume that Z ≤ A ∩ L. In particular, 0 ≠ [Y, A ∩ L] ≤ C_{Y} (A) and since Y is a simple 𝔽_{2}L-module, A normalizes Y. Thus Y is an 𝔽_{2}M submodule. As this holds for all simple 𝔽_{2}L-submodules on V and V is a semisimple 𝔽_{2}L-module and an indecomposable 𝔽_{2}M-module, V = Y. Thus V is a simple 𝔽_{2}L-module and (a) holds. By 7.8(d), there exists an L-invariant non-degenerate quadratic form on V and by 1.9(f), this form is invariant under M.

Let D ≤ Q_{2} with |D| ≥ 4 and let a, b ∈ D^{♯} with a ≠ b. Note that P_{2} acts simple on [V, Q_{2}] and ⟨C_{P_{2}}(a), C_{P_{2}}(b)⟩ = P_{2}. Since 0 ≠ [V, a] < [V, Q_{2}] we conclude that [V, a] ≠ [V, b]. Since dim_{𝔽}[V, a] = 2 and dim_{𝔽}[V, Q_{2}] = 3 this gives [V, D] = [V, a] + [V, b] = [V, Q_{2}] We have proved

(∗)  [V, D] = [V, Q_{2}] for all D ≤ Q_{2} with |D| > 2.

Put L_{13} := ⟨Q_{2}^{P_{1}}⟩. Then Q_{1} ≤ L_{13}, L_{13} ≤ P_{1} ∩ P_{3}, L_{13}/Q_{1} ≅ Sym(3) and L_{13} = C_{L}([V, Z]). Put L_{13}^{∗} := C_{M}([V, Z]). Then A ≤ L_{13}^{∗} and so M = L_{13}^{∗}L and P_{1}^{∗} = L_{13}^{∗}P_{1}. Since |L_{13}/L_{13}^{∗}| ≤ 2 we conclude that O_{2}(L_{13}^{∗}) = Q_{1}^{∗}, L_{13}^{∗} = L_{13}Q_{1}^{∗} and L_{1}^{∗} = L_{1}Q_{1}^{∗}.

Put Z^{∗} := Z(Q_{1}^{∗}). Since L_{1} acts simply on Q_{1}/Z, we have [Q_{1}, Q_{1}^{∗}] ≤ Z and conclude that Q_{1}^{∗} = Z^{∗}Q_{1}. Note that [Z^{∗}, L_{1}] ≤ Z and so [Z^{∗}, O^{2}(L_{1})] = 1. Since V/C_{V} (Z) and C_{V} (Z^{∗})/[V, Z] are non-isomorphic as O^{2}(L_{1})-modules, [V, Z^{∗}] = [V, Z] and similarly C_{V} (Z^{∗}) = C_{V} (Z). It follows that [V, Z^{∗}] ≤ [V, Z] ≤ [V, A] ≤ C_{V} (A) ≤ C_{V} (Z) = C_{V} (Z^{∗}) and so Z^{∗}A is quadratic on V. Thus by maximality of A, Z^{∗} ≤ A and A = Z^{∗}(A ∩ L). We will show that A is contained in a conjugate of Q_{2}^{∗} under P_{1}. Since A = Z^{∗}(A ∩ L) it suffices to show that A ∩ L is contained in a conjugate of Q_{2} under P_{1}.

Suppose A ∩ L ≤ Q_{1}. Note that P_{1} acts transitively on fours groups of Q_{1} containing Z and so we may assume |A ∩ Q_{2}| ≥ 4. Thus using (∗),

A ≤ C_{M}([V, A ∩ Q_{2}]) = C_{M}([V, Q_{2}]) ≤ Q_{2}^{∗}.

Suppose next that A ∩ L ≰ Q_{1}. Since L_{13}/Q_{1} ≅ Sym(3) we may assume that A ∩ L ≤ Q_{1}Q_{2}. Let P̃_{1} := P_{1}/Z and let q ∈ Q_{2} \Q_{1}. Then CQ̃_{1}(q) = [Q̃_{1}, q] = { \atop Q_{1} ∩ Q_{2}}. It follow that all involutions in Q̃_{1}Q̃_{2} \ Q̃_{1} are conjugate and so Q_{2} is the unique maximal elementary subgroup of Q_{1}Q_{2} not contained in Q_{1}. Thus A ∩ L ≤ Q_{2}.

We proved that A is conjugate to a subgroup of Q_{2}^{∗} and we may assume that A ≤ Q_{2}^{∗}. Since C_{V} (Q_{2}) is the unique non-zero proper 𝔽_{2}L_{2} submodule of V, C_{V} (Q_{2}^{∗}) = [V, Q_{2}^{∗}] = C_{V} (Q_{2}) and so Q_{2}^{∗} is quadratic on V. This gives A = Q_{2}^{∗} , and all maximal quadratic subgroups of M of order at least 4 are conjugate to Q_{2}^{∗}.

It remains to proof (g). So let B be any quadratic subgroup of M. Suppose first that |B| = 2. If B ≤ L then B is conjugate to |Z| and so (g:a) holds. If B ≰ L then either C\overline{L}(B) ≅ U_{4}(2) or C\overline{L}(B) ∼ 2^{4}.3^{2}.2.

Suppose that C\overline{L}(B) ∼ 2^{4}.3^{2}.2. Then O_{2}(C_{L}(B)) is conjugate to A ∩ L and we may assume that B ≤ A and C_{M}(B) ≤ P_{2}. Note that C_{M}(B) contains a Sylow 3-subgroups of P_{2}. Since the Sylow 3-subgroups of P_{2} are extraspecial of order 3^{3} they act simply on [V, A] and we conclude that [V, B] = C_{V} (B) = [V, A] = C_{V} (A) and so (g:d) holds.

Suppose C\overline{L}(B) ≅ U_{4}(2). Let y ∈ Z^{∗} \ Z. Then [V, y] ≤ [V, Z]. The preceding paragraph shows that C\overline{L}(B) ≁ 2^{4}.3^{2}.2 and thus ⟨y⟩ is conjugate to B. So we may assume that B ≤ Z^{∗}. Thus V/C_{V} (B) and [V, B] have dimension at most two over 𝔽 and so are centralized by C_{L}(B). Thus C_{L}(B) acts faithfully on C_{V} (B)/[V, B]. Since [V, B] ≤ C_{V} (B) = [V, B]^{⊥}, the L-invariant unitary form on V gives raises to an C_{L}(B)-invariant unitary form on C_{V} (B)/[V, B]. It follows that dim_{𝔽} C_{V} (B)/[V, B] = 4 and C_{V} (B)/[V, B] is a natural SU_{4}(2)-module for C_{L}(B). Thus dim_{𝔽} V/C_{V} (B) = 1 = dim_{𝔽}[V, B] and (g:b) holds.

Suppose next that |B| > 2. Then B is contained in a maximal quadratic subgroup of order at least 4 and so we may assume that B ≤ A. If [V, B] = [V, A], then C_{V} (B) = [V, B]^{⊥} = [V, A]^{⊥} = C_{V} (A) and (g:d) holds. So suppose [V, B] < [V, A]. Then (∗) implies that |B ∩ L| = 2 and so |B| = 4. If d ∈ B \ L, then dim_{𝔽}[V, d] ≤ dim_{𝔽}[V, B] ≤ 2 and so (g:b) must hold for ⟨d⟩ in place of B. Thus (g:c) holds.

Lemma 7.10. Let M = O_{2n}^{ϵ}(q), q = 2^{k}, and V be the corresponding natural module over 𝔽_{q}. Let a ∈ M with |a| = 2. Then a ∈ Ω_{2n}^{ϵ}(q) if and only if dim_{𝔽_{q}}[V, a] is even.

Proof. This is well known, but a reference seems to be hard to come by. So here is a proof: If n = 1, this is obvious. Suppose there exists an a-invariant proper subspace W of V with V = W ⊕ W^{⊥}. Then the claim follows by induction on n. So we may assume that no such W exists. In particular v ⊥ v^{a} for all v ∈ V and so [V, a] is a singular subspace. Let C_{V} (a) = [V, a]⊕W for some 𝔽_{q}-subspace W. Since C_{V} (a) = [V, a]^{⊥}, V = W ⊕ W^{⊥} and so W = 0 and [V, A] = C_{V} (a) is maximal singular subspace of V. Thus ϵ = +. Since a normalize a maximal singular subspace, a ∈ Ω_{2n}^{+}(q). Consider the map s_{a} : V/C_{V} (a) × V/C_{V} (a) → 𝔽_{q} define by s_{a}(v + C_{V} (a), w + C_{V} (a)) = s(v, [w, a])), where s is the symmetric form on V invariant under M. Then s_{a} is a non-degenerate bilinear form. From v ⊥ v^{a} we get v ⊥ [v, a] and so s_{a} is a symplectic form. Thus dim[V, a] = dim V/C_{V} (a) is even.

Lemma 7.11. Let q be a power of p and K⊴M such that K ≅ Spin_{n}^{ϵ}(q), n ≥ 3, and C_{M}(K) = Z(K). Let V_{nat} be the natural 𝔽_{q}Ω_{n}^{ϵ}(q)-module for K, S ∈ Syl_{p}(M), U := C_{V_{nat}}(S ∩ K), L := C_{K}(U) and Q := O_{p}(L). Then the following hold:

(a) Suppose that W is a non-trivial simple 𝔽_{p}K-module with [W, Q, Q] = 0. Then W is a (half-)spin module for K.

(b) Suppose that p = 2, n even, n ≥ 6, W is a simple 𝔽_{2}M-module with [W, K] ≠ 0 and that there exists A ≤ S with [W, A, A] = 0, M = ⟨A^{M}⟩, |A| > 2, and A ≰ K. Then M ≅ O_{n}^{ϵ}(q) and W is the natural O_{n}^{ϵ}(q)-module for M.

Proof. Put T := S ∩ K, so T ∈ Syl_{p}(K), and \overline{N_{M}(Q)} := N_{M}(Q)/QZ(K), and let U_{0} be the unique 1-dimensional singular subspace of U. Then [U^{⊥}, Q] = U_{0}. Moreover U = U_{0}, if n is even or p is odd, and U = U_{0} + V^{⊥} if n is odd and p = 2. Hence

1^{∘}. U^{⊥}/U_{0} and Q are natural Ω_{n−2}^{ϵ}(q)-modules for \overline{L}.

Assume that n ≥ 5. Then there exists g ∈ K such that Y := U_{0} + U_{0}^{g} is a 2-dimensional singular subspace of U^{⊥} normalized by T. Put H := ⟨Q, Q^{g}⟩ and Z := Q ∩ Q^{g}. Then H/C_{H}(Y) ≅ SL_{2}(q), and H acts transitively on the 1-dimensional subspaces of Y. Thus H = ⟨Q^{N_{K}(Y)}⟩; in particular, T normalizes H. Moreover, QO_{p}(HT) = T ∈ Syl_{p}(HT), and using (1^{∘}):

2^{∘}. If n ≥ 5, then \overline{C_{Q^{g}}(Y)} = O_{p}(C\overline{L}(Y/U_{0})), and Z is a 1-dimensional singular subspace of Q.

(a): Put 𝕂 := End_{K}(W). By Smith’s Lemma 4.2 applied to W and its dual, C_{W} (Q) and W/[W, Q] are simple 𝕂L-modules. Since [W, Q] ≤ C_{W} (Q) we conclude that [W, Q] = C_{W} (Q). Suppose that n = 3 or 4. Then Q = T and so C_{W} (Q) and W/[W, Q] are 1-dimensional over 𝕂. Thus dim_{𝕂}(W) = 2.

If n = 3 or (n, ϵ) = (4, +) then W is a natural SL_{2}(q)-module. If (n, ϵ) = (4, −), then W is a natural SL_{2}(q^{2})-module. These are the (half-)spin modules for these groups, so (a) follows in this case.

Suppose now that n ≥ 5, so we are allowed to use the subgroups Y, H and Z constructed above. Since [W, Z, H] = 0 and Z ≠ 0 we conclude that C_{W} (HT) ≠ 0. By Smith’s Lemma 4.2 C_{W} (T) is 1-dimensional over 𝕂 and so C_{W} (T) = C_{W} (T H). Since K = ⟨L, HT ⟩ and W is simple, we have [C_{W} (T), L] ≠ 0, so [C_{W} (Q), L] ≠ 0. Now again Smith’s Lemma 4.2 and (2^{∘}) show that C_{W} (Q), \overline{L} and \overline{C_{Q^{g}}(Y)} satisfy the hypothesis in place of W, K, and Q. Thus by induction C_{W} (Q) is a (half-)spin module for \overline{L}. Together with [C_{W} (T), HT] = 0, this determines W up to isomorphism (see 4.3) and so W is a (half)-spin-module.

(b): Note that K ≅ Ω_{n}^{ϵ}(q) since p = 2, that S normalizes L, and that by (1^{∘}) Q is a natural Ω_{n−2}^{ϵ}(q)-module for L. Thus there exists an L-invariant quadratic form h (over 𝔽_{q}) on Q.

3^{∘}. There exist a, b ∈ A^{♯} with C_{Q}(a) ≠ C_{Q}(b).

Assume first that A does not act 𝔽_{q}-linearly on Q. Since Aut(𝔽_{q}) is cyclic and A is elementary abelian with |A| ≥ 4, we conclude that there exists 1 ≠ a ∈ A acting 𝔽_{q}-linearly on Q and b ∈ A acting not 𝔽_{q}-linearly on Q. Hence C_{Q}(a) is an 𝔽_{q}-subspace of Q while C_{Q}(b) is not; in particular C_{Q}(a) ≠ C_{Q}(b).

Assume now that A acts 𝔽_{q}-linearly on Q. Then \overline{AL} ≅ O_{n−2}^{ϵ}(q), and there exists a ∈ A \ K and 1 ≠ b ∈ A ∩ K. By 7.10 we conclude that C_{Q}(a) is odd dimensional and C_{Q}(b) is even dimensional over 𝔽_{q}. Hence again C_{Q}(a) ≠ C_{Q}(b).

4^{∘}. There exists D ≤ LA with D ∩ A ≰ Q, [W, D, D] = 0, and D ∩ Q ≠ 1.

Clearly A ≰ Q since A ≰ K, so if A ∩ Q ≠ 1 we can choose D = A. Suppose A ∩ Q = 1. Let a, b ∈ A as in (3^{∘}) and without loss C_{Q}(a) ≰ C_{Q}(b). Then there exists 1 ≠ d ∈ [C_{Q}(a), b] ≤ ⟨b^{C_{Q}(a)}⟩, so

[W, a, d] ≤ [W, a, ⟨b^{C_{Q}(a)}⟩] = ⟨[W, a, b]^{C_{Q}(a)}⟩ = 0.

Since A is elementary abelian, d ∈ ⟨b^{C_{Q}(a)}⟩ ≤ C_{L}(a) and so [a, d, W] = 0. Hence by the Three Subgroups Lemma also [W, d, a] = 0, and D := ⟨a, d⟩ satisfies (4^{∘}).

5^{∘}. There exists B ≤ Q and 1 ≠ e ∈ B such that [W, B, B] = 0, h(e) = 0 and B ≰ 𝔽_{q} e.

Let D be as in (4^{∘}). Pick 1 ≠ b ∈ D ∩ Q, and put E := ⟨D^{C_{L}(b)}⟩ and C := 𝔽_{q}b. Then [W, b, E] = 0.

Suppose that b^{⊥} ≤ E ∩ Q. Note that there exists u ∈ E ∩ Q \ C such that h(u) = 0 if h(b) ≠ 0. Pick such an element u and put B := ⟨b, u⟩. Since [W, b, B] = 0, B acts quadratically on W. Thus (5^{∘}) holds with e = b if h(b) = 0 and e = u if h(b) ≠ 0.

Suppose now that b^{⊥} ≰ E ∩ Q. By the action of C_{L}(b) on Q, any C_{L}(b)-submodule of Q, which contains b, either contains b^{⊥} or is contained in C. In particular E ∩Q ≤ C and [Q, E] ≤ E ∩Q ≤ C. Since Q is a natural Ω_{n−2}^{ϵ}(q)-module for L, 3.4 shows h(b) ≠ 0 and |DQ/Q| = |EQ/Q| = 2. Thus [D, C_{L}(b)] ≤ C, and since C_{L}(b) centralizes C, [D, O^{2}(C_{L}(b)] = 1. The structure of O_{n−2}(q) shows that

[Q, D] = C and C_{LD}(b)/Q ≅ C_{2} × Sp_{n−4}(q)

Put D^{∗} = C_{DL}(O^{2}(C_{L}(b))). It follows that D ≤ D^{∗}, |D^{∗}Q/Q| = 2, D^{∗} ∩ Q = C, and the q elements in D^{∗} \ Q are the transvections on V_{nat} corresponding to the q non-singular 1-spaces in the isotropic 2-space [V_{nat}, b]. Pick d ∈ D ∩ A \ Q. Then F := C_{DK}(d) ≅ C_{2} × Sp_{n−2}(q). In particular F = ⟨D^{F} ⟩. From [W, d, D] = 0 we get [W, d, F] = 0 and so [W, d, C_{Q}(d)] = 0. Pick e ∈ C_{Q}(d) \ C. Then ⟨e, d⟩ is quadratic on W and satisfies (4^{∘}) in place of D. Moreover [Q, d] ≰ 𝔽_{q} e. Hence the arguments of the previous paragraph apply to ⟨e, d⟩ in place of D, and (5^{∘}) holds.

6^{∘}. [W, Z, C_{Q}(Y)] = 0.

Let B and e be as in (5^{∘}). Since L is transitive on the singular elements of Q and since by (2^{∘}) Z is a singular subspace of Q, we may assume that e ∈ Z. Put Q_{e} := e^{⊥} in Q. Note that Q_{e} = C_{Q}(Y), so we have to show that [W, Z, Q_{e}] = 0.

Since B ≰ Z = 𝔽_{q}e we get Q_{e} ≤ ⟨B^{C_{L}(e)}⟩, so [W, e, Q_{e}] = 0. As N_{L}(Q_{e}) acts transitively on Z, we conclude that [W, Z, Q_{e}] = [W, ⟨e^{N_{L}(Q_{e})}⟩, Q_{e}] = 0.

7^{∘}. Put 𝕂 := End_{K}(W). Then W is a simple 𝔽_{2}K-module, and M acts 𝕂-linearly on W.

Let X be a simple 𝔽_{2}K-submodule of W and 𝔼 := End_{K}(X), and pick D as in (4^{∘}). Then 0 ≠ [X, D ∩ Q] ≤ C_{X}(D) and so X is D-invariant. Hence 0 ≠ [X, D ∩ A] ≤ C_{X}(A) and so X is A-invariant. Since D ∩ Q acts 𝔼-linearly on X, [X, D ∩ Q] is a non-trivial 𝔼-subspace centralized by D, so D acts 𝔼-linearly on X. Hence [X, D ∩ A] is a non-trivial 𝔼-subspace centralized by A, and A acts 𝔼-linearly on X. This also holds for each conjugate of A under M. Since M = ⟨A^{M}⟩ and W is a simple 𝔽_{2}M-module, X = W, 𝕂 = 𝔼, and M acts 𝕂-linearly on W.

8^{∘}. [W, Q, Q] ≠ 0.

Suppose [W, Q, Q] = 0. Then by (7^{∘}) and (a), W is a (half)-spin module. If ϵ = −, then 𝕂 ≅ 𝔽_{q^{2}} and since A acts 𝕂-linearly on W, we conclude that A ≤ K, a contradiction. If ϵ = +, then 𝕂 = 𝔽_{q} and so A induces a graph automorphism on K. But graph automorphisms interchange the two half-spin modules and so do not act on W, again a contradiction.

9^{∘}. W is a natural Ω_{n}^{ϵ}(q)-module for K.

Put Q_{Z} = C_{Q}(Y)C_{Q^{g}}(Y), where g is as in the definition of Y. Then by (6^{∘}) [W, Z, Q_{Z}] = 0. Let l ∈ L with Z^{l} ≰ C_{Q}(Y), so Q = C_{Q}(Y)Z^{l}. Note that L = ⟨Q_{Z}, Q_{Z}^{l}⟩. Since [W, Q, Q] ≠ 0 by (8^{∘}) and ⟨Z^{L}⟩ = Q, also [W, Z, Q] ≠ 0. Now [W, Z, C_{Q}(Y)] = 0 gives

0 ≠ [W, Z, Q] = [W, Z, C_{Q}(Y)Z^{l}] = [W, Z, Z^{l}].

Since [Z, Z^{l}] = 1, we get

0 ≠ [W, Z, Z^{l}] = [W, Z^{l}, Z] ≤ [W, Z] ∩ [W, Z^{l}] ≤ C_{W} (Q_{Z}) ∩ C_{W} (Q_{Z}^{l}) = C_{W} (L).

Thus C_{W} (L) ≠ 0, and with Smith’s Lemma 4.2 [C_{W} (S ∩ K), L] = 0.

By (6^{∘}) Z and thus also Z^{l} acts quadratically on W. On the other hand

Z^{l}O_{2}(HT) = QO_{2}(HT) ∈ Syl_{2}(HT).

Hence, T acts quadratically on C_{W} (O_{2}(HT)). So by (a) C_{W} (O_{2}(HT)) is a natural SL_{2}(q)-module for HT. Thus by Ronan-Smith’s Lemma 4.3 W is unique up to isomorphism, and (9^{∘}) holds.

From (9^{∘}) we conclude that 𝕂 = 𝔽_{q}. Since A acts 𝕂-linearly on W we infer that KA ≅ O_{2n}^{ϵ}(q), W is the natural module, and M = KA.

8 The FF-Module Theorems

In this section we use the same hypothesis and notation as in Section 2; that is, M is a finite group with O_{p}(M) = 1, V is a finite, faithful 𝔽_{p}M-module such that J = J_{M}(V) ≠ 1, and J is the set of J_{M}(V)-components of M on V.

Recall that a finite group H is p-minimal if S ∈ Syl_{p}(H) is contained in a unique maximal subgroup of H and S ⋬ H.

Lemma 8.1. Suppose that M is p-minimal and T ∈ Syl_{p}(M). Then there exist subgroups E_{1}, …, E_{r} such that the following hold:

(a) J = E_{1} × ⋯ × E_{r} and J = {E′_{1}, …, E′_{r}}.

(b) V = C_{V} (J) + ∑_{i=1}^{r}[V, E_{i}] and [V, E_{i}, E_{j}] = 0 for i ≠ j.

(c) [C_{V} (T), O^{p}(M)] ≠ 0.

(d) T is transitive on E_{1}, …, E_{r}.

(e) There are no over-offenders on V in M.

(f) E_{i} ≅ SL_{2}(q), q = p^{n}, and [V, E_{i}]/C_{[V,E_{i}]}(E_{i}) is a natural SL_{2}(q)-module for E_{i}, or p = 2, E_{i} ≅ Sym(2^{n} + 1), and [V, E_{i}] is a natural Sym(2^{n} + 1)-module for E_{i}.

(g) If A ≤ M is an offender on V, then A = (A ∩ E_{1}) × … × (A ∩ E_{r}), and each A ∩ E_{i} is an offender on V.

Proof. Using [BHS, 5.6] we see that (c) holds. Hence M and V satisfy the hypothesis of [BHS, 5.5]. This result gives subgroups E_{1}, …, E_{r} satisfying (b),(d), (f) and (g). Moreover, [BHS, 2.16] shows that every best offender on V induces inner automorphisms in E_{i} and is not an over-offender on [V, E_{i}]. The first property gives (a) and the second one (e).

The proof of Theorem 2:

Let K ∈ J, 𝕂 := End_{K}(V), and A ∈ D. From 2.8 we get:

1^{∘}. V is a simple K-module, and K is the unique J-component of M.

If K is solvable, then 2.2(d) shows that Theorem 2(1) holds for q = 2 or 3 and n = 2. Thus, we assume from now on that K is not solvable, so K is a component by 2.2(d).

By the definition of D there exists 1 ≠ B ≤ A such that B is an offender on V with

(∗)  [V, B, A] = 0.

We choose such an offender B with |B| minimal. Then B is a minimal offender and thus a quadratic best offender on V, so B ≤ J.

By (1^{∘}) and 2.2(b) [K, B] ≠ 1. Hence

2^{∘}. K = [K, B] and [V, B, A] = 0.

Since K is not solvable, we get from 2.5, applied to BK, that BK acts 𝕂-linearly on V. In particular, [V, B] is a 𝕂-subspace of V. Thus (∗) shows that A centralizes a 𝕂-subspace of V, so also A acts 𝕂-linearly on V. Since this holds for every A ∈ D, we conclude:

3^{∘}. M acts 𝕂-linearly on V, and C_{M}(K) = Z(M).

We will now prove Theorem 2 by using the information given in [GM2, Theorem B]. Observe that the bounds on the dimension of V in the cases (3) and (4) of Theorem 2 follow from 3.4.

Suppose that (KB, V) or (K, V) is one of the possibilities (1) – (12) given in Theorem 2 for (M, V). Since by (3^{∘}) M ≤ N_{GL_{𝕂}(V)}(K), then also (M, V) is on the list. Moreover, if there exists a non-trivial offender on V in K, then (3^{∘}) and [GM2] show that (K, V) is on the list. Thus, we may assume:

4^{∘}. B is a minimal best offender on V, M = KB, and there is no non-trivial offender on V in K. In particular K ≠ M.

Case 1. Suppose that p is odd.

In [Ch, Corollary C] all possibilities for M are given under the hypothesis that |V/C_{V} (B)| ≤ |B|^{2} for some non-trivial quadratic subgroup B ≤ M. It turns out that p = 3 and M ≅ SL_{2}(5), or M is a genuine group of Lie type in characteristic p. In the first case |V/C_{V} (B)| > |B|, and B is not an offender contradicting (4^{∘}). In the second case (4^{∘}) shows that M ≅ ^{2}G_{2}(3) ∼ SL_{2}(8).3. But then M has abelian Sylow 2-subgroups, which contradicts [KS, 9.1.4].

Case 2. Suppose that |B| = 2.

Then B acts as a transvection on V, and [McL] shows that (M, V) is on the list.

Case 3. Suppose that p = 2, |B| > 2, and K is not a genuine group of Lie-type in characteristic p.

Then [MeSt1], [MeSt2] and 7.4 together with (4^{∘}) show that

K ≅ Alt(n), n ≥ 6, n ≠ 8, U_{3}(3), 3. U_{4}(3), ^{2}F_{4}(2)′, , Mat_{12}, or Mat_{22}.

Except in the case K ≅ Alt(n) the corresponding module V is uniquely determined.

Suppose K ≅ Alt(n). Then [MeSt2] offers two possibilities for V. If V is the natural module for Alt(n), then M ≅ Sym(n) and V is the natural module for Sym(n). Hence (M, V) are on the list.

If V is not a natural module, then V is the (half-)spin module and n > 6. So 7.5 shows that B ≤ Alt(n) contradicting (4^{∘}).

Suppose that K ≅ U_{3}(3). Then M ≅ G_{2}(2), and 7.6 shows that (M, V) is on the list.

Suppose K ≅ ^{2}F_{4}(2)′. Then M ≅ ^{2}F_{4}(2) and so M \ K does not contain any involution, a contradiction.

Suppose K ≅ 3. U_{4}(3). Then 𝕂 = 𝔽_{4} and dim_{𝕂} V = 6. Since M acts 𝕂-linearly we get |M/K| = 2, and there exists B ≤ R ≤ M such that R ∼ 2^{4+1}3.Alt(6). Observe that every non-zero R-section of V is at least 3-dimensional over 𝕂. Hence I_{R} := C_{V} (O_{2}(R)) = C_{V} (O_{2}(R) ∩ K) is 3-dimensional over 𝕂 and V = [V, R].

Clearly B is not an over-offender on I_{R} since |BO_{2}(R)/O_{2}(R)| ≤ 4 and I_{R} is an 𝔽_{4}R-module. Thus, by 1.3 either V = I_{R} +C_{V} (B) or B ≤ O_{2}(R). In the first case [V, R] ≤ I_{R}, a contradiction. In the second case [MS1, 2.6] implies that there exists an offender 1 ≠ D ≤ O_{2}(R) with D ⊴ R. Since I_{R} and V/I_{R} are simple R-modules we get C_{V} (D) = I_{R} and 2^{5} = |O_{2}(R)| ≥ |D| ≥ |V/C_{V} (D)| = |V/I_{R}| = 2^{6}, a contradiction.

Suppose next that K ≅ Mat_{12} or Mat_{22}. Then M ≅ Aut(Mat_{12}) and Aut(Mat_{22}), respectively, and [MeSt2] shows that |B| = 4. But then |V/C_{V} (B∩K)| ≤ |V/C_{V} (B)| ≤ |B| = 4, which contradicts the action of K on V.

Case 4. Suppose p = 2, |B| > 2, and K is a genuine group of Lie type defined over a field of characteristic 2.

Recall that B ≤ T ∈ Syl_{2}(M). Let V_{0} := C_{V} (T ∩ K). Note that M is generated by the 2minimal subgroups containing T. Hence there exists T ≤ P ≤ M such that P is 2-minimal and [V_{0}, O^{2}(P)] ≠ 0.

5^{∘}. B ≤ O_{2}(P).

Suppose that P = M. Then by 8.1 (KB, V) is on the list, contrary to the assumptions. Thus P ≠ M.

Put V_{P} := C_{V} (O_{2}(P)∩K). Then V_{0} ≤ V_{P}. Put P̃ = N_{K}(O^{2′}(P ∩K)). Then P̃ is a Lie-parabolic subgroup of K, O_{2}(P) ∩ K = O_{2}(P̃) and O^{2′}(P̃) = O^{2′}(P ∩ K). Thus by Smith’s Lemma 4.2 V_{P} is a simple 𝕂(P ∩ K)-module. By (4^{∘}) O^{2}(P) ≤ P ∩ K, so C_{V} (O^{2}(P)) = 0 and V_{P} = C_{V} (O_{2}(P)). Moreover, since P is 2-minimal, C_{T} (V_{P}) = O_{2}(P).

Suppose that B ≰ O_{2}(P), so [V_{P} , B] ≠ 0. By 1.2 B is a non-trivial best offender on V_{P}, and by 8.1 B is not an over-offender on V_{P}. Hence 1.3 shows that C_{B}(V_{P}) = 1 and V = V_{P} +C_{V} (B). Again by 8.1 there exists O_{2}(P)B ≤ H ≤ P such that H/O_{2}(P) ≅ SL_{2}(|B|), U := [V_{P} , H] is a natural SL_{2}(|B|)-module, and V = U + C_{V} (B).

Put D := ⟨B^{H}⟩. Then [V, D] ≤ U, so every subgroup of V containing U is D-invariant. Since K is of local characteristic 2 and P ≠ M, there exists a minimal normal subgroup N of D in O_{2}(D)∩K. Then [V, D, N] ≤ [U, N] = 0 and [V, N, O^{2}(D)] = U. Hence, the Three Subgroups Lemma shows that [O^{2}(D), N, V] ≠ 0 and so [N, O^{2}(D)] ≠ 1. As SL_{2}(|B|) has no non-trivial simple 𝔽_{2}-module of order less than |B|^{2}, we get |N| ≥ |B|^{2}.

On the other hand for every 1 ≠ x ∈ N, U ≤ C_{V} (x) and so C_{V} (x) is D-invariant. Since N = ⟨x^{D}⟩ it follows that C_{V} (N) = C_{V} (x). Now choose y ∈ N and b ∈ B with x := [y, b] ≠ 1. Then x ∈ N ∩ ⟨B, B^{y}⟩ and C_{V} (B) ∩ C_{V} (B^{y}) ≤ C_{V} (x) and so

|V/C_{V} (N)| = |V/C_{V} (x)| ≤ |V/C_{V} (B)|^{2} ≤ |B|^{2} ≤ |N|.

Hence, N is a non-trivial offender on V in K. But this contradicts (4^{∘}), and so (5^{∘}) holds.

Since by (5^{∘}) B ≤ O_{2}(P) and since P = (P ∩ K)B, also P ∩ K is 2-minimal. Thus P ∩ K is a minimal parabolic subgroup of K fixed by B.

Let Δ be the Dynkin diagram of K and i be the node corresponding to P ∩ K. Among all B-invariant proper Γ ⊂ Δ with i in Γ and Γ connected we choose Γ maximal. Let T ∩ K ≤ L̃ be the parabolic subgroup of K corresponding to Γ and put L := O^{2′}(L̃), Q := O_{2}(L), and V_{L} := C_{V} (Q). Note that B normalizes L and thus also V_{L}. So by 1.2 B is a best offender on V_{L}. By Smith’s Lemma 4.2 V_{L} is a simple 𝔽_{2}L̃-module. Let W be a simple 𝔽_{2}L-submodule of V_{L}. By 2.6 and 1.2 B normalizes W and is a best offender on W.

6^{∘}. Either B ≤ LO_{2}(LB), or the following hold:

(a) LB/C_{LB}(W) ≅ O_{2n}^{ϵ}(q), n ≥ 3, and W is the corresponding natural module.

(b) |B/C_{B}(W)| ≥ 4.

Suppose that B ≰ LO_{2}(LB). Note that [V_{0}, O^{2}(L)] ≠ 0 since O^{2}(P) ≤ L and [V_{0}, O^{2}(P)] ≠ 0. Since Γ is connected, C_{B}(W) ≤ O_{2}(LB). Thus B is a non-trivial best offender on W. If |B/C_{B}(W)| = 2, then B is not an over-offender on W, and by 1.3 |B| = 2, a contradiction to the assumptions of (Case 4).

Hence |B/C_{B}(W)| ≥ 4, and by induction LB/C_{LB}(W) ≅ O_{2n}^{ϵ}(q) and W is the corresponding natural module. Moreover (5^{∘}) shows that LB is not 2-minimal, so n ≥ 3.

7^{∘}. B acts transitively on Δ \ Γ.

There exists a node j ∈ Δ \ Γ such that j is adjacent to some node in Γ. Now the maximality of Γ shows that Δ = Γ ∪ j^{B}.

We now discuss the possibilities for K/Z(K). Suppose first that K/Z(K) is an untwisted group of Lie type defined over 𝔽_{q}. Then (5^{∘}) shows that no element of B induces a field automorphism or graph-field automorphism in Δ. Thus B induces a graph automorphism on Δ, so Δ is of type A_{m}, D_{m}, F_{4}, or E_{6}. Since M is not 2-minimal by (5^{∘}), m ≥ 3.

If Δ is of type D_{m}, then (M, V) is in the list by 7.11(b). Assume now that Δ is not of type D_{m}, so m ≥ 4 if Δ is of type A_{m}. Since B induces a graph automorphism, (7^{∘}) yields one of the following possibilities:

(i) |Γ| = m − 2, and Δ is of type A_{m}.

(ii) |Γ| = 2, and Δ is of type F_{4}.

(iii) |Γ| = 4 or 5, and Δ is of type E_{6}.

In all cases B acts non-trivially on Γ; in particular B ≰ LO_{2}(LB). Hence (6^{∘}) shows that Γ is of type D_{n}. This rules out case (ii). Moreover, in case (i) m = 5 and Γ is of type D_{3}; and in case (iii) Γ is of type D_{4}. In particular, by (6^{∘}) in each of the remaining cases P is uniquely determined, C_{V} (O_{2}(P)) is a natural SL_{2}(q)-module for P, and [V_{0}, R] = 0 for every other minimal Lie-parabolic subgroup R of K containing T ∩ K. By Ronan-Smith’s Lemma 4.3 this determines the module V uniquely.

If Δ is of type A_{5}, then V is the exterior cube of a natural SL_{6}(q)-module. But then there exists an L-composition factor of V that is a natural SL_{4}(q)-module. This contradicts 2.8 and 7.11(b).

If Δ is of type E_{6}, then V is the adjoint module for E_{6}(q). But then V has an L-composition factor isomorphic to the adjoint module for Ω_{8}^{+}(q), a similar contradiction as above.

Suppose now that K/Z(K) is a twisted group of Lie type over 𝔽_{q^{ν}}. Then |Δ \ Γ| = 1 and B induces a field automorphism of order 2 on 𝔽_{q^{ν}} with fixed field 𝔽_{q}, so ν = 2. Since M is not 2-minimal by (5^{∘}), K has Lie rank at least 2.

In all cases (5^{∘}) shows that P/O_{2}(P) ≅ SL_{2}(q), and this excludes that K is of type ^{2}F_{4}, ^{3}D_{4} or 2A_{m}, m even. So K is of type ^{2}A_{m}, m odd, ^{2}D_{m}, or ^{2}E_{6}.

If K is of type ^{2}D_{m}, we are done by 7.11(b). Suppose that K is of type ^{2}A_{m}, m odd. Since 2A_{3} = ^{2}D_{3} we may assume in addition that m ≥ 5, so by (7^{∘}) |Γ| ≥ 2. In particular L contains a minimal parabolic subgroup R with R/O_{2}(R) ≅ SL_{2}(q^{2}), so B ≰ LO_{2}(LB). Hence (6^{∘}) implies that K is of type ^{2}A_{5}. Now as in the A_{5}-case, V is the exterior cube of the natural SU_{5}(q)-module and L has a composition factor which is a natural SU_{4}(q)-module. Since SU_{4}(q) ≅ Spin_{6}^{−}(q) this contradicts 7.11(b).

Suppose that K is of type ^{2}E_{6}. Then |Γ| = 3 and with the same argument as in the previous paragraph using (6^{∘}) L is of type ^{2}D_{4}. So Γ, P and V_{P} are uniquely determined. Now as in the E_{6}-case V is the adjoint module for K, and L has a composition factor isomorphic to the adjoint module for Ω_{8}^{−}(q), which contradicts 7.11(b).

The proof of Theorem 3:

Let B be a minimal offender in A and note that B is a quadratic best offender on V.

Case 1. The case M ≅ G_{2}(q), q = 2^{n}, V a natural G_{2}(q)-module.

We will use the following facts about the action of K on V and the structure of K, where i-subspace means 𝕂-subspace of dimension i in V:

There exists an M-invariant non-degenerate symplectic form on V (since V is self-dual and p = 2). Let M_{1} and M_{2} be the pair of maximal parabolic subgroups of M with T ≤ M_{i} and such that M_{i} normalizes an i-subspace V_{i} in V. Note that V_{i} is singular and the graph with vertices V_{1}^{M} ∪ V_{2}^{M} and inclusion as incidence relation is a generalized hexagon. Since M acts transitively on V^{♯}, V_{1}^{M} consists of all the 1-dimensional subspaces of V.

Put P_{i} := O^{2′}(M_{i}), and Q_{i} := O_{2}(P_{i}). There exist exactly two classes of involutions in M with representatives z, t ∈ T such that

(i) t ∉ Z(Q_{1}), P_{1} = Q_{1}C_{M}(t), and P_{2} = C_{M}(z).

(ii) t and z and do not fix any vertex of distance larger than 3 from V_{1} and V_{2}, respectively.

(iii) t and z fix all vertices of distance at most 3 from V_{1} and V_{2}, respectively.

We will use these properties to show 3(a).

1^{∘}. |C_{V} (z)| = q^{4}. More precisely, z centralizes exactly the 1-subspaces of distance 1 and 3 from V_{2}.

There are precisely q + 1 1-spaces of distance 1 and q^{2}(q + 1) 1-spaces of distance 3 from V_{2}. Hence by (ii) and (iii) C_{V} (z) has exactly q + 1 + q^{2}(q + 1) = q^{3} + q^{2} + q + 1 1-spaces.

2^{∘}. |C_{V} (t)| = q^{3}. More precisely, t centralizes exactly the 1-dimensional subspaces of distance 0 and 2 from V_{1}.

There is one 1-space of distance 0 and q(q + 1) 1-spaces of distance 2. Thus, as in (1^{∘}), C_{V} (t) contains exactly 1 + q(q + 1) = q^{2} + q + 1 1-spaces.

3^{∘}. Suppose t ∈ B. Then |B| = |C_{V} (B)| = |[V, B]| = q^{3}, C_{T} (B) = B, and B is uniquely determined in M_{1}.

Since C_{V} (B) ≤ C_{V} (t) and by (2^{∘}) and the quadratic action of B,

q^{3} = |[V, t]| = |[V, B]| and C_{V} (B) = C_{V} (t); in particular |B| ≥ q^{3}.

By (2^{∘}) C_{V} (t) is uniquely determined by M_{1}, so also B^{∗} := O^{p′}(C_{M_{1}}(C_{V} (t))) is uniquely determined. To prove the uniqueness of B in M_{1}, it suffices to show that |B^{∗}| ≤ q^{3} since then B = B^{∗}.

Note that [V_{2}^{g}, B^{∗}] = 0 for every g ∈ M_{1}, and so B^{∗} ≤ Q_{1} ∩ Q_{2}. Let x ∈ P_{2} \ M_{1} and D := B^{∗} ∩ B^{∗x}. Then |B^{∗}/D| ≤ q^{2} and |D| ≥ q since |Q_{2}| = q^{5} and |B^{∗}| ≥ q^{3}. On the other hand, D fixes a path of length 6 with V_{2} as midpoint, and (ii) yields |D| ≤ q. This shows that |D| = q and consequently |B^{∗}| ≤ q^{3}.

It remains to show that B = C_{T} (B). Assume that B_{0} =: C_{T} (B) > B. By Smiths’ Lemma, C_{V} (Q_{1}) = V_{1} and so [C_{V} (t), Q_{1}] ≠ 1. From [V_{2}, Q_{1}] ≤ V_{1} we get C_{V} (t) = ⟨V_{1}^{P_{1}}⟩ and [C_{V} (t), Q_{1}] = V_{1}. Thus Q_{1}/B = Q_{1}/C_{Q_{1}}(C_{V} (t)) is dual to the natural SL_{2}(q)-module C_{V} (t)/V_{1}. We claim that C_{Q_{1}}(B) ≰ B. If B_{0} ≤ Q_{1} this is obvious. And if B_{0} ≰ Q_{1} we get [Q_{1}, B_{0}] ≰ B and so again C_{Q_{1}}(B) ≰ B. Since C_{Q_{1}}(B) ⊴ P_{1} we conclude that Q_{1} = C_{Q_{1}}(B) and t ∈ Z(Q_{1}), which contradicts (i).

4^{∘}. t^{M} ∩ B ≠ ∅.

Assume that t^{M} ∩ B = ∅. Then we may assume that z ∈ B, so C_{V} (B) ≤ C_{V} (z) and by (1^{∘}) q^{2} ≤ |V/C_{V} (B)| ≤ |B|. On the other hand, by (ii) and (1^{∘}) the non-trivial elements of C_{T} (C_{V} (z)) centralize every 1-subspace of distance at most 3 from V_{2} but no singular 2-space of distance 4. Hence |C_{T} (C_{V} (z))| = q. It follows that there exists z^{g} ∈ B with C_{V} (z) ≠ C_{V} (z^{g}) and so also [V, z] ≠ [V, z^{g}]. Since [V, z] + [V, z^{g}] ≤ C_{V} (B) ≤ C_{V} (z) ∩ C_{V} (z^{g}) and |[V, z]| = q^{2}, we conclude that

|C_{V} (B)| = q^{3}, |B| = q^{3} and C_{V} (B) = C_{V} (z) ∩ C_{V} (z^{g}).

But then V_{2} and V_{2}^{g} are of distance 2, and we may assume that V_{1} = V_{2} ∩ V_{2}^{g}. Now (2^{∘}) shows that t centralizes C_{V} (B) and so C_{V} (B) = C_{V} (t). Hence also B⟨t⟩ is a quadratic offender, and (3^{∘}) yields t ∈ B, a contradiction.

5^{∘}. Case (a) of Theorem 3 holds.

According to (4^{∘}) we may assume that t ∈ B, and according to (3^{∘}) C_{T} (B) = B and so A = B. So 3(a) follows from (3^{∘}).

Case 2. The case M ≅ SL_{n}(q)/⟨−id^{n−1}⟩, n ≥ 5, and V the exterior square of a natural 𝕂SL_{n}(q)- module W.

Let U be a T -invariant 𝕂-hyperplane in W. Put R := C_{M}(W/U) and I_{R} := C_{V} (O_{p}(R)). Recall that R/O_{p}(R) ≅ SL_{n−1}(q) and O_{p}(R) is an natural SL_{n−1}(q)-module for R isomorphic to U.

We will use the following properties of the exterior square:

6^{∘}. U, O_{p}(R) and V/I_{R} are isomorphic natural SL_{n−1}(q)-modules for R.

7^{∘}. I_{R} is as an 𝔽_{p}R-module isomorphic to the exterior square of U.

If n ≥ 6, then by (7^{∘}) and induction B is not an over-offender on I_{R}. If n = 5, then SL_{4}(q) ≅ Ω_{6}^{+}(q) and I_{R} is the natural orthogonal module. Again by 3.4 B is not an over-offender. Hence, in both cases 1.3 shows that either B ∩ O_{p}(R) = 1 or B ≤ O_{p}(R).

In the first case |I_{R}/C_{I_{R}}(B)| = |B| and V = I_{R} + C_{V} (B); in particular [V, B] ≤ I_{R}. But this contradicts (6^{∘}). Thus we have B ≤ O_{p}(R). Pick b ∈ B^{♯} and put C := C_{R}(b). Then C acts as a point stabilizer on O_{p}(R) and thus by (6^{∘}) also as a point stabilizer on V/I_{R}. It follows that C_{V} (b) = I_{R} or |C_{V} (b)/I_{R}| = q.

If C_{V} (B) = I_{R}, then |B| ≥ |V/I_{R}| = q^{n−1} and B = O_{p}(R). Since C_{T} (O_{p}(R)) = O_{p}(R) we get A = B, and case (b) of Theorem 3 follows.

Assume now that |C_{V} (B)/I_{R}| = q. Then C_{V} (B) = C_{V} (b) for all 1 ≠ b ∈ B. Also q^{n−2} = |V/C_{V} (B)| ≤ |B|. Since n ≥ 5 this gives |B| > q, so there exists 1 ≠ b, b̃ ∈ B with C_{R}(b) ≠ C_{R}(b̃). Hence, C_{V} (B) = C_{V} (b) = C_{V} (b̃) is normalized by R = ⟨C_{R}(b), C_{R}(b̃)⟩, a contradiction.

Case 3. The case M ≅ Spin_{7}(q) or Spin_{10}^{+}(q) and V a corresponding spin module.

We will use the following facts about the action of M on V and the structure of M. Recall that P Ω_{5}(q) ≅ PSp_{4}(q). There exists T ≤ R ≤ M such that for I_{R} := C_{V} (O_{p}(R)) the following hold:

(i) Spin_{n}^{ϵ}(q)/⟨−id_{V} ⟩ ≅ Ω_{n}^{ϵ}(q).

(ii) R/O_{p}(R) ≅ Spin_{5}(q) resp. Spin_{8}^{+}(q).

(iii) O_{p}(R) is a natural Ω_{5}(q)- resp. Ω_{8}^{+}(q)-module for R.

(iv) I_{R} = [V, O_{p}(R)].

(v) If n = 7, then V/I_{R} and I_{R} are isomorphic natural Sp_{4}(q)-modules for R, but I_{R} is not isomorphic to O_{p}(R)/O_{p}(R) ∩ Z(R); while if n = 10, O_{p}(R), V/I_{R} and I_{R} are pairwise nonisomorphic natural Ω_{8}^{+}(q)-modules for R.

(vi) O_{p}(R) acts quadratically on V.

(vii) If n = 7 and Z is a 1-dimensional singular subspace of O_{p}(R), then C_{M}(Z)/O_{p}(C_{M}(Z)) ≅ Spin_{4}^{+}(q), and V/[V, Z] is a natural Ω_{4}^{+}(q)-module for C_{M}(Z).

Put δ = 1 if n = 7 and δ = 2 if n = 10. We first show:

8^{∘}. C_{V} (x) = I_{R} for every non-singular x ∈ O_{p}(R), and |V/C_{V} (x)| = q^{2δ} for every non-trivial singular x ∈ O_{p}(R).

Let 1 ≠ x ∈ O_{p}(R). Suppose first that x is singular in O_{p}(R). Then C_{M}(x) ≰ R and so C_{V} (x) ≠ I_{R}. Moreover, C_{R}(x) normalizes a unique proper submodule of V/I_{R}. This submodule has order q^{2δ} and so (8^{∘}) holds.

Suppose next that x is not singular. Then there exists g ∈ M such that R^{g} and R^{gx} are opposite Lie-parabolics of M. So by 5.1 M = ⟨O_{p}(R^{g}), O_{p}(R^{gx}⟩ ≤ ⟨O_{p}(R^{g}), x⟩. Thus C_{V} (O_{p}(R^{g}))∩C_{V} (x) = 0 and V = [V, O_{p}(R^{g})] + [V, x]. Since [V, O_{p}(R^{g})] ≤ C_{V} (O_{p}(R^{g})) and [V, x] ≤ C_{V} (x), this implies [V, x] = C_{V} (x) and so C_{V} (x) = C_{V} (O_{p}(R)) = I_{R}.

9^{∘}. B is conjugate to a subgroup of O_{p}(R).

Suppose not. Then B ≰ O_{p}(R). Let Z = O_{p}(R) ∩ B. If Z contains a non-singular element b, then by (8^{∘}) [V, B] ≤ C_{V} (B) ≤ C_{V} (b) = I_{R}. But then ⟨B^{R}⟩ centralizes V/I_{R}, a contradiction to (v). Thus all elements in Z are singular. By 1.3 either V = I_{R} + C_{V} (B) and [V, B] ≤ I_{R}, or B is an over-offender on I_{R}. The first possibility contradicts (v), so B is an over-offender on I_{R}. Then by 3.4

C_{I_{R}}(B) = [I_{R}, B], |C_{I_{R}}(B)| = q^{2δ} and q^{2δ} < |B/Z| = |B/B ∩ O_{p}(R)| ≤ q^{3δ}.

Put \overline{V} = V/I_{R}. Then B acts quadratically on \overline{V}. From |B/Z| > q^{2δ} and 3.4 we conclude that |\overline{V} , B]| = q^{2δ} and so also |\overline{V} /C\overline{V} (B)| = q^{2δ}. Thus |V/C_{V} (B)| ≥ q^{4δ} and so |Z| ≥ q^{δ}. Let 1 ≠ x ∈ Z. Note that [V, B] + I_{R} ≤ C_{V} (x). Since x is singular in O_{p}(R) (8^{∘}) gives |V/C_{V} (x)| = q^{2δ}. Thus C_{V} (x) = [V, B] + I_{R} and C_{R}(x) normalizes [V, B] + I_{R}. But R = ⟨C_{R}(x), C_{R}(y)⟩ for any singular x, y ∈ O_{p}(R) with 𝔽_{q}x ≠ 𝔽_{q}y and since R does not normalizes [V, B]+I_{R} we conclude that Z ≤ 𝔽_{q}x. Since |Z| ≥ q^{δ}, we conclude that Z is a 1-dimensional singular subspace of O_{p}(R). Also δ = 1 and so n = 7.

Put P := C_{M}(Z). By (vii) P/O_{p}(P) ≅ Spin_{4}^{+}(q), and C_{V} (Z)/[V, Z] is the natural Ω_{4}^{+}(q)-module for P. Thus every singular 1-space of C_{V} (Z)/[V, Z] is contained in a P -conjugate of I_{R}/[V, Z], and the conjugates of I_{R}/[V, Z] are TI-subgroups in C_{V} (Z)/[V, Z].

Since B acts quadratically on V, [V, B]/[V, Z] is a 2-dimensional isotropic subspace and thus contains a 1-dimensional singular subspace. Hence there exists g ∈ P such that [V, B] ∩ I_{R}^{g} ≰ [V, Z]. The TI-property of I_{R}/[V, Z] implies that B normalizes I_{R}^{g}, so B ≤ R^{g}.

If B ≰ O_{p}(R^{g}), then the above also applies to B and R^{g} in place of B and R, so [V, B] ∩ I_{R}^{g} is 2-dimensional and so [V, B] ∩ I_{R}^{g} = [V, Z], a contradiction. Thus, we have that B ≤ O_{p}(R^{g}), and B is not a counterexample. Hence (9^{∘}) is proved.

According to (9^{∘}) we may assume that B ≤ O_{p}(R). If B does not contain a non-singular element of O_{p}(R), then |B| ≤ q^{2δ}. So also |V/C_{V} (B)| ≤ q^{2δ} and by (8^{∘}) C_{V} (B) = C_{V} (b) for every 1 ≠ b ∈ B. On the other hand, for every such b, C_{R/O_{p}(R)}(b) is contained in a unique maximal parabolic subgroup of R/O_{p}(R). It follows that B is has order at most q, a contradiction.

Hence B contains a non-singular element b. Then by (8^{∘})

(+)  I_{R} = C_{V} (b) = [V, b] = C_{V} (B) = [V, B] and |B| ≥ |V/C_{V} (B)| = q^{4δ}

If M ≅ Spin_{10}^{+}(q), then |O_{p}(R)| = |I_{R}| = q^{8} = q^{2δ} and so by (+) B = O_{p}(R). Thus A ≤ C_{T} (O_{p}(R)) = O_{p}(R) and A = B. Since O_{p}(R) is weakly closed in T, we see that case (d) of Theorem 3 follows from (+).

So suppose M ≅ Spin_{7}(q). If A ≤ O_{p}(R), then case (c) Theorem 3 follows. So assume for a contradiction that A ≰ O_{p}(R). Observe that [B, A] = 1, |B| ≥ q^{2δ} = q^{4} and O_{p}(R) is a natural Ω_{5}(q)-module for R/O_{p}(R). We conclude that p = 2, |B| = q^{4}, B = A ∩ O_{p}(B) = C_{O_{p}(R)}(A) and |A/B| ≤ q. Thus |A| ≤ q^{5}. Since O_{p}(R)/O_{p}(R) ∩ Z(R) is not isomorphic to I_{R}, we get that |I_{R}/C_{I_{R}}(A)| = q^{2} and so |V/C_{V} (A)| = q^{6} > q^{5} = |A|. This contradiction completes (Case 3).

Case 4. The case M ≅ 3.Alt(6) and |V | = 2^{6}.

Then 𝕂 = 𝔽_{4}, |A| = 4, and C_{V} (A) is a 𝕂-hyperplane, so case (e) Theorem 3 follows.

Case 5. The case K ≅ Alt(n), n ≥ 5, and V the natural Alt(n)-module for K.

Let W be the natural permutation module for Sym(n) over 𝔽_{2} with basis w_{i}, i ∈ Ω := {1, …, n}, and W_{0} := ⟨∑_{Ω} w_{i}⟩. For Ψ ⊆ Ω put W_{Ψ} = ⟨w_{i} + w_{j} | i, j ∈ Ψ⟩ and \overline{W_{Ψ}} = W_{Ψ} + W_{0}/W_{0}. Then V ≅ \overline{W_{Ω}}.

10^{∘}. If A is a best offender, then case (g) or case (h) of Theorem 3 holds.

Suppose that A acts transitively Ω. Then n = 2^{k}, and since n ≥ 5, k ≥ 3. Note that |A| = 2^{k}, C_{W_{Ω}}(A) = W_{0}, and |\overline{W_{Ω}}| = 2^{2^{k}−2}. The commutator map

C\overline{W_{Ω}}(A) × A → W_{0} with (w + W_{0}, a) ↦ [w, a]

shows that

|C\overline{W_{Ω}}(A)| = |C\overline{W_{Ω}}(A)/\overline{C_{W_{Ω}}(A)}| ≤ |A| = 2^{k},

and so

2^{k} = |A| ≥ |V/C_{V} (A)| = |\overline{W}_{Ω}/C_{\overline{W}_{Ω}}(A)| ≥ 2^{2^{k}−k−2}.

Thus 2^{k−1} ≤ k + 1, so k = 3 and |A| = |V/C_{V} (A)| = 8 = |C_{V} (A)|. Since V is self-dual, also |[V, A]| = 8 and since [V, A] ≤ C_{V} (A), [V, A] = C_{V} (A). Hence case (h:4) of Theorem 3 holds.

So we may assume from now on that A does not act transitively on Ω. Let Ψ be an orbit of A on Ω of length say 2^{k}. Since A is a best offender, A is an offender on \overline{W_{Ψ}}, and since Ψ ≠ Ω, W_{0} ≰ W_{Ψ} and so \overline{W}_{Ψ} ≅ W_{Ψ}. Thus A is an offender on W_{Ψ}. Note that |A/C_{A}(W_{Ψ})| = |A/C_{A}(Ψ)| = 2^{k}, |W_{Ψ}| = 2^{2^{k}−1}, and |C_{W_{Ψ}}(A)| = |2|. Thus 2^{2^{k}−1−1} ≤ 2^{k}, 2^{k} ≤ k + 2 and k ≤ 2.

Suppose A has two orbits Ψ_{1} and Ψ_{2} of length four and put Λ := Ψ_{1} ∪ Ψ_{2}. Assume for a contradiction that Λ = Ω and put H := N_{M}({Ψ_{1}, Ψ_{2}}). Then H ≅ Sym(4) ≀ C_{2} and A ≤ O_{2}(H). So H acts simple on O_{2}(H). [MS1, 2.6] shows that O_{2}(H) is an offender, and the Timmesfeld Replacement theorem implies that O_{2}(H) acts quadratically on V, a contradiction. Hence Λ ≠ Ω and so W_{Λ} ≅ \overline{W}_{Λ}. Note that |A/C_{A}(W_{Λ})| = |A/C_{A}(Λ)| ≤ 16, |W_{Λ}| = 2^{7} and |C_{W_{Λ}}(A)| = 4. Thus 2^{7}/4 ≤ 16, a contradiction.

Suppose Ψ is an orbit of length 4 for A on Ω and A has a fixed-point i on Ω. Put V_{Ψi} := ⟨w_{i}+w_{j} | j ∈ Ψ⟩. Then V_{Ψ,i} is isomorphic to the permutations module for A on Ψ and is also isomorphic to \overline{V_{Ψ,i}}. Thus A is a best offender on V_{Ψ,i}. But |A/C_{A}(V_{Ψ,i})| = 4 and |V_{Ψ,i}/C_{W_{Ψ}}(A)| = 8, a contradiction.

We have proved that either all orbits of A on Ω have length 1 or 2, or A has a unique orbit of length four and all other orbits have length two.

Assume for a contradiction that C\overline{W_{Ω}}(A) ≠ C_{W_{Ω}}(A)/W_{0}. Then there exists w ∈ W_{Ω} such that 0 ≠ [w, A] ∈ W_{0}; in particular A_{0} := C_{A}(w) has index 2 in A. Let X ⊆ Ω with w = ∑_{i∈X} w_{i} and |X| even. Then there exists a ∈ A such that {X, X^{a}} is a partition of Ω, and A_{0} normalizes X and X^{a}. Note that C\overline{W_{X}} (A) = ⟨\overline{w}⟩ and that |X| ≥ 4 since n ≥ 5 and |X| is even. Thus

4 ≤ |\overline{W}_{X}/C\overline{W_{X}} (A)| ≤ |V/C_{V} (A)| ≤ |A|.

Thus A_{0} ≠ 1, and since C_{A_{0}}(X) = C_{A_{0}}(X ∩ X^{a}) = 1, A_{0} acts non-trivially on X. Since A has at most one orbit of length four on Ω we conclude that |X \ C_{X}(A_{0})| = 2. Thus |A_{0}| = 2 and |A| = 4. The Timmesfeld Replacement Theorem shows that A acts quadratically on V. But [\overline{W_{X}}, A_{0}, a] ≠ 0, a contradiction.

We have proved that C_{W_{Ω}/W_{0}}(A) = C_{W_{Ω}}(A)/W_{0}, so |V/C_{V} (A)| = |W_{Ω}/C_{W_{Ω}}(A)|. If follows that A is an offender on W_{Ω}. Let k be the number of orbits of length 2. Assume that A has an orbit of length four, then A has no fixed-point, n = 2k + 4, |C_{W_{Ω}}(A)| = 2^{k+1}, |A| ≤ 2^{k} · 4 = 2^{k+2}, and

|V/C_{V} (A)| = |W_{Ω}/C_{W_{Ω}}(A)| = 2^{n−1−(k+1)} = 2^{k+2}.

Since A is an offender, this implies |A| = 2^{k+2}, and since V is self-dual, |[V, A]| = |V/C_{V} (A)| = 2^{k+2} = |A|. As A has on orbit of length 4, A is not quadratic on W_{Ω} and since C_{W_{Ω}/W_{0}}(A) = C_{W_{Ω}}(A)/W_{0} also not quadratic on V. Hence case (h:3) of Theorem 3 holds.

Assume now that A does not have any orbit of length 4. Then [V, A] ≤ C_{V} (A) and |A| ≤ 2^{k}. Suppose A has a fixed-point in Ω. Then |V/C_{V} (A)| = 2^{k} = |[V, A]| and so |A| = 2^{k} and case (g) or (h:1) of Theorem 3 holds. So suppose A has no fixed-points and so n = 2k and |V/C_{V} (A)| = 2^{k−1} = |[V, A]|. Thus 2^{k−1} ≤ |A|.

Let t_{1}, …, t_{k} be the transpositions corresponding to the non-trivial orbits of orbits of A on Ω, say t_{i} ∈ A if and only if i > l. If l = 0, then again case (h:1) of Theorem 3 holds. Suppose l > 0. Let 1 ≤ r < s < l and put A_{rs} = C_{A}(C_{Ω}(⟨t_{r}, t_{s}⟩). Then |A/A_{rs}| ≤ 2^{k−2} and so A_{rs} ≠ 1. Since A_{ts} ≤ ⟨t_{r}, t_{s}⟩ and neither t_{r} nor t_{s} are in A we conclude that A_{rs} = ⟨t_{r}t_{s}⟩. It follows that

A = ⟨t_{1}t_{2}, t_{2}t_{3}, …, t_{l−1}t_{l}, t_{l+1}, t_{l+2}, t_{k}⟩.

Thus case (h:3) of Theorem 3 holds.

11^{∘}. Every offender in M on V is a best offender.

Let X be an offender and let Y ≤ X with |C_{V} (Y)||Y | maximal and then Y minimal. By the Timmesfeld Replacement Theorem, Y is quadratic. If |Y ||C_{V} (Y)| = |V |, then |Y ||C_{V} (Y)| = |X||C_{V} (X)| and so X is a best offender. If |Y ||C_{V} (Y)| > |V |, then (10^{∘}) shows that Y is generated by a maximal set of commuting transpositions. So X ≤ C_{M}(Y) = Y, X = Y, and X is a best offender.

Observe that (11^{∘}) together with (10^{∘}) completes (Case 5).

Case 6. The case M ≅ Alt(7) and |V | = 2^{4}.

Choose T ≤ R ≤ M with R ≅ Alt(6). Then the previous case applies to R, and we are done.

Theorem 8.2. Let M be a finite CK-group and V a faithful 𝔽_{p}M-module. Suppose that there exists K ∈ J_{M}(V) such that V = [V, K] and V is a semisimple but not simple 𝔽_{p}K-module. Then one of the following holds, where q is a power of p and J := J_{M}(V):

1. J ≅ SL_{n}(q), n ≥ 3, and V ≅ N^{r} ⊕ N^{∗s}, where N is a natural SL_{n}(q)-module, N^{∗} its dual, and r, s are integers with 0 ≤ r, s < n and \sqrt{r} + \sqrt{s} ≤ \sqrt{n}.

2. J ≅ Sp_{2m}(q), m ≥ 3, and V ≅ N^{r}, where N is a natural Sp_{2m}(q)-module and r is a positive integer with 2r ≤ m + 1.

3. J ≅ SU_{n}(q), n ≥ 8, and V ≅ N^{r}, where N is a natural SU_{n}(q)-module and r is a positive integer with 4r ≤ n.

4. J ≅ Ω_{n}^{ϵ}(q) with p odd if n is odd, or M ≅ O_{n}^{ϵ}(q) with p = 2 and n even. Moreover n ≥ 10 and V ≅ N^{r}, where N is a corresponding natural module and r is a positive integer with 4r ≤ n − 2.

In particular, if V is not a homogeneous 𝔽_{p}J module, then (1) holds with r ≠ 0 ≠ s and n ≥ 4.

Proof. By 2.2(f) K is the unique J-component of M; in particular K ⊴ M. Since V is a semisimple K-module we have

1^{∘}. V = N_{1} ⊕ ⋯ ⊕ N_{m}, m ≥ 2, where N_{i} is a perfect simple 𝔽_{p}K-module.

By 2.8 J normalizes N_{i} and by 1.2 every best offender on V is also a best offender on N_{i}. Moreover, O_{p}(J/C_{J}(N_{i})) = 1 since N_{i} is simple. Hence

2^{∘}. J/C_{J}(N_{i}) and N_{i} satisfy the hypothesis of Theorem 2.

By 2.2 K is not solvable since m ≥ 2, so K is a component of M. Now 2.5 shows that J acts 𝔽_{i}-linearly on N_{i}, where 𝔽_{i} = End_{K}(N_{i}). In particular [J, C_{J}(K)] ≤ C_{J}(N_{i}). Since K is the unique J-component and K ≰ C_{J}(N_{i})C_{J}(K), we get from 2.2(b) C_{J}(N_{i})C_{J}(K) ≤ Z(J). Another application of Theorem 2 shows that J/KC_{J}(N_{i}) is a p-group. Hence J/K is nilpotent, and since J is generated by p-elements and O_{p}(Z(J)) ≤ O_{p}(M) = 1, we get that Z(J) ≤ K. It follows:

3^{∘}. C_{J}(N_{i}) ≤ C_{J}(K) = Z(J) = Z(K).

From now on we fix a non-trivial best offender A ≤ M. By 2.3(b) there exists a minimal best offender B ≤ A such that [V, B, A] = 0; in particular B is quadratic on V.

Note that by (3^{∘}) C_{A}(N_{i}) = 1, since Z(J) is a p′-group, and that B is a best offender on N_{i} by 1.2. Now (1^{∘}) implies

|V/C_{V} (B)| = ∏_{i=1}^{m} |N_{i}/C_{N_{i}}(B)| ≤ |B|.

Since m ≥ 2 there exists N ∈ {N_{1}, …, N_{r}} such that

4^{∘}. |N/C_{N}(B)| ≤ |B|^{\frac{1}{2}} .

Put 𝔽 := End_{K}(N). Then (2^{∘}) and Theorems 2 and 3 imply:

5^{∘}. J/C_{J}(N) ≅ SL_{n}(q), Sp_{n}(q), SU_{n}(q), Ω_{n}^{ϵ}(q) or O_{n}^{ϵ}(q) (and p = 2), n := dim_{𝔽} N where q := |𝔽| if J/C_{J}(N) ≇ SU_{n}(q) and q = |𝔽|^{\frac{1}{2}} if J/C_{J}(N) ≅ SU_{n}(q). Moreover, N is the corresponding natural module.

Let N^{∗} be the 𝔽K-module dual to N. We first treat the cases where each N_{i} is isomorphic to N or N^{∗}, say V ≅ N^{r} ⊕ N^{∗s}, r + s = m.

By 1.8(d) B is quadratic on N^{∗}. Put

D := C_{J}(C_{N}(B)) ∩ C_{J}(C_{N^{∗}}(B)), k := dim_{𝔽} N/C_{N}(D), l = dim_{𝔽}[N, D].

By 1.8(c) l = dim_{𝔽} N^{∗}/C_{N^{∗}}(D), and by 1.8(d) B ≤ D, C_{V} (D) = C_{V} (B), [V, D] = [V, B], and D is a quadratic offender on V. Moreover by 1.8(f) k + l ≤ n. We get

6^{∘}. |V/C_{V} (D)| = q^{rk+sl} ≤ |D|.

Recall from 3.2 that N and N^{∗} are isomorphic 𝔽J-modules, if J/C_{J}(N) is not isomorphic to SL_{n}(q). We now treat the cases given in (5^{∘}) separately.

Case 1. Suppose that M ≅ SL_{m}(q) and V ≅ N^{r} ⊕ N^{∗s} with r + s ≥ 2. Then (1) holds.

By 3.4 |D| = q^{kl}, and (6^{∘}) gives |V/C_{V} (D)| = q^{rk+sl}. Thus V is an FF-module if and only if there exists 0 < k, l < n with rk + sl ≤ kl, that is \frac{r}{l} + \frac{s}{k} ≤ 1. Increasing l decreases \frac{r}{l} + \frac{s}{k}. So we may assume that k + l = n. Put g(k) = \frac{r}{n−k} + \frac{s}{k}. We will determine the minimal value of g(k) on the open interval (0, n). If k approaches 0 or n, g(k) approaches +∞. So f obtains a minimum value at some point m in (0, n) with g′(m) = 0. We have g′(m) = \frac{r}{(n−m)^{2}} − \frac{s}{m^{2}} . Straightforward

calculations show that m = \frac{\sqrt{s}}{\sqrt{r}+\sqrt{s}}n, n − m = \frac{\sqrt{r}}{\sqrt{r}+\sqrt{s}}n and g(m) = \frac{(\sqrt{r}+\sqrt{s})^{2}}{n}. Thus g(m) ≤ 1 if and only if \sqrt{r} + \sqrt{s} ≤ \sqrt{n}. So if V is an FF-module, then \sqrt{r} + \sqrt{s} ≤ \sqrt{n}. (We remark that with a little more effort it can be shown that there even exists an integer k in (0, n) with g(k) ≤ 1, so V is an F F -module if and only if \sqrt{r} + \sqrt{s} ≤ \sqrt{n}.)

In the remaining cases M ≅ Sp_{n}(q), SU_{n}(q), Ω_{n}^{ϵ}(q) or O_{n}^{ϵ}(q) we get from 3.2(a) that N ≅ N^{∗}. Hence k = l. Recall that [N, D] is an isotropic subspace of N by 3.2(e) since D is quadratic on N.

Case 2. Suppose that M ≅ Sp_{n}(q) and V ≅ N^{r} for some r ≥ 2. Then (2) holds.

By 3.4 |D| = q^{\binom{k+1}{2}} and so as in the case (Case 1) rk ≤ \frac{k(k+1)}{2} and 2r ≤ k + 1. Since [V, D] is isotropic and the maximal dimension of an isotropic subspace is \frac{n}{2} we get 2r ≤ \frac{n}{2} + 1. Now r ≥ 2 implies n ≥ 6, and (2) holds.

Case 3. Suppose that M ≅ SU_{n}(q) and V ≅ N^{r} with r ≥ 2. Then (3) holds.

In this case |N| = q^{2n}. By 3.4 |D| = q^{k^{2}} and as in the previous cases 2rk ≤ k^{2} and 2r ≤ k. Moreover, since k + l ≤ n and k = l, also 2k ≤ n and so 4r ≤ n. Now r ≥ 2 implies n ≥ 8.

Case 4. Suppose that M ≅ Ω_{n}^{ϵ}(q) or O_{n}^{ϵ}(q) and p = 2, with n even if p = 2, and V ≅ N^{r} for some r ≥ 2. Then (4) holds.

Suppose first that [N, D] is singular. Then by 3.4 |D| = q^{\binom{k}{2}} and so rk ≤ \binom{k}{2} and 2r ≤ k − 1. Since k + l = 2k ≤ n, we get 4r ≤ 2n − 2. Now r ≥ 2 implies (4).

Suppose next that [N, D] is not singular. Then p = 2 and so n is even, and 3.4 yields |D| ≤ 2q^{\binom{k}{2}} and as in the previous cases q^{rk} ≤ 2q^{\binom{k}{2}}. In addition, r ≥ 2 implies k ≥ 2. Then

rk ≤ log_{q} 2 + \binom{k}{2} and 2r ≤ \frac{2 log_{q} 2}{k} + k − 1.

If \frac{2 log_{q} 2}{k} ≥ 1, then q = 2 = k and r = 1, a contradiction. Thus \frac{2 log_{q} 2}{k} < 1 and 2r ≤ k − 1. Now again 2k ≤ n implies that 4r ≤ 2k − 2 ≤ n − 2. Since r ≥ 2, n ≥ 10, and (4) holds.

Case 5. Suppose V is not a direct sum of copies of N and N^{∗}.

Without loss N_{2} is neither isomorphic to N nor to N^{∗}. We will show that this leads to a contradiction.

By (4^{∘}) B is an offender on N ⊕ N. Hence we can apply the previous cases to N ⊕ N in place of V and get that dim N ≥ 3, 6, 8, and 10, respectively.

Suppose that M/C_{M}(N) ≅ SL_{n}(q) and N is the corresponding natural module. Since N_{2} is not a natural module, Theorem 2 shows that N_{2} is the exterior square of a natural module. For n = 3, N_{2} ≅ N^{∗} or N, which is not the case. Hence n ≥ 4. Since B is an over-offender on N_{2}, Theorem 3(b) shows that n = 4. In this case N_{2} is a natural Ω_{6}^{+}(q)-module for J/C_{J}(N_{2}). Hence 3.4 gives

|N_{2}/C_{N_{2}}(B)| = q^{s} < |B| ≤ q^{\binom{s}{2}},

where s is the 𝔽_{q}-dimension of a maximal singular subspace of N_{2} centralized by B. On the other hand 2s ≤ 6 and so s ≤ 3. But then s does not satisfy the above inequality.

Suppose M/C_{M}(N) ≅ Sp_{2n}(q). Then by Theorem 2 n = 3 and N_{2} is a spin module. So we get |B| ≤ q^{5} and |N_{2}/C_{N_{2}}(B)| = q^{4}. It follows that |N/C_{N}(B)| ≤ q, a contradiction to |B| ≥ q^{4}.

Suppose that K/C_{K}(N) ≅ SU_{n}(q), n ≥ 8, or Ω_{n}^{ϵ}(q), n ≥ 10. Then Theorems 2 and 3 show that every FF-module for J with an over-offender is a natural module, a contradiction.

Suppose now that V is not homogeneous as an 𝔽_{2}J-module. Then (1) holds with r ≠ 0 ≠ s. Thus \sqrt{n} ≥ \sqrt{1} + \sqrt{1} = 2, n ≥ 4 and all parts of the theorem are proved.

Theorem 8.3. Let M be a finite CK-group with O_{p}(M) = 1 and V a faithful 𝔽_{p}M-module. Put J := J_{M}(V), J := J_{M}(V) and W := [V, J] + C_{V} (J)/C_{V} (J) . Then the following hold:

(a) Let K ∈ J. Then K is either quasisimple, or p = 2 or 3 and K ≅ SL_{2}(p)′.

(b) [V, K, L] = 0 for all K ≠ L ∈ J, and W = ⨁_{K∈J} [W, K].

(c) J^{p}J′ = O^{p}(J) = F^{∗}(J) = × J.

(d) W is a faithful semisimple 𝔽_{p}J-module.

(e) C_{J}([W, K]) = C_{J}([V, K]).

Proof. (a) and the first part of (b) follow from 2.2. For the proof of the second part of (b) note that C_{W} (K) = C_{[V,J]}(K) + C_{V} (J)/C_{V} (J) since K = O^{p}(K). Thus, by the first part C_{W} (K) ∩ [W, K] ≤ C_{W} (J) = 0.

(c): Put J_{0} := J′J^{p}. First we prove:

1^{∘}. Let K ∈ J. Then J_{0} induces inner automorphism on K.

Let X be a quasisimple K-submodule of V and Y = C_{X}(K). Then we can apply 2.9 to 0 ≤ Y ≤ X ≤ V and S := X/Y. By 2.9(a) J̃:= J/C_{J}(S) and S satisfy the hypothesis of Theorem 2. We conclude that |J̃/K̃| ≤ p and so J̃_{0} ≤ K̃. Since C_{J}(K̃) = C_{J}(K) by 2.2(c), (d), (1^{∘}) holds.

Let D := ⟨J ⟩, so D = × J and D ≤ J_{0} by 2.2. Moreover, Z(J) ≤ J_{0} since Z(J) is a p′-group. By (1^{∘}) J_{0} induces inner automorphisms on D. Hence J_{0} ≤ DC_{J}(D), and by 2.2(g) J_{0} = DZ(J). Since J/J_{0} is an elementary abelian p-group, J/D is nilpotent, and since J is generated by p-elements J/D is a p-group and so D = J_{0}.

(d): Since O^{p}(J) ≤ ⟨J ⟩, J acts nilpotently on V/[V, J] and C_{V} (J). Hence C_{J}(W) acts nilpotently on V and so C_{J}(W) ≤ O_{p}(M) = 1. Thus W is faithful J-module.

By 2.8 every perfect simple K-submodule is also a simple J-submodule. Hence (d) follows if [W, K] is a semisimple K-module. So suppose for a contradiction that [W, K] is not semisimple K-module. We will use the bar-convention for the images of subgroups of V in W, so \overline{X} = X + C_{V} (D)/C_{V} (D) for X ≤ V.

Let X_{2} ≤ V be a K-submodule of W that is minimal such that X_{2} = [X_{2}, K] and \overline{X}_{2} is not a semisimple K-module. The minimality of X_{2} implies that X_{2} has a unique maximal K-submodule Y_{2} such that [Y_{2}, K] ≠ 0 and X_{2}/Y_{2} is a simple K-module.

Recall that [U, K, K] = [U, K] for every K-section of W since K is a J-component and thus is generated by p′-elements. It follows that C_{Y_{2}/C_{Y_{2}} (K)}(K) = 0. Hence there exists a K-submodule Y_{1} of Y_{2} that is maximal such that Y_{1} ≠ Y_{2} and C_{Y_{2}/Y_{1}}(K) = 0. Put X_{1} := [Y_{2}, K] + Y_{1}. Let Z_{1} be a K-submodule of Y_{2} with Y_{1} < Z_{1} < Y_{2}. Then by maximality of Y_{1}, C_{Y_{2}/Z_{1}}(K) ≠ 0. Let Z_{2} be the inverse image of C_{Y_{2}/Z_{1}}(K) in Y_{2}. Then C_{Y_{2}/Z_{2}}(K) = 0 and so by maximality of Y_{1}, Z_{2} = Y_{2}. Hence X_{1} = [Y_{2}, K] + Y_{1} ≤ Z_{1}. It follows that X_{1}/Y_{1} is the unique minimal K-submodule and Y_{2}/Y_{1} is the unique maximal K-submodule of X_{2}/Y_{1}, while X_{1}/Y_{1} and X_{2}/Y_{2} are simple K-modules, and X_{2}/X_{1} is a quasisimple K-module. In particular, K and X_{0} = Y_{1} ≤ X_{1} ≤ Y_{2} ≤ X_{2} satisfy the hypothesis of 2.9. This result shows that S := X_{1}/Y_{1} ⊕ X_{2}/Y_{2} and J̃ := J/C_{J}(S) satisfies the hypothesis of 8.2 in place of V and M. We conclude that

K̃ ≅ SL_{n}(q), n ≥ 3, Sp_{2n}(q), n ≥ 3, Ω_{n}^{ϵ}(q), n ≥ 10, or SU_{n}(q), n ≥ 8,

N := X_{1}/Y_{1} is a corresponding natural module, and X_{2}/Y_{2} is either isomorphic or dual to N. In particular, C_{K}(N) = C_{K}(S) = C_{K}(X_{2}/Y_{1}). Put 𝔽 := End_{K}(N). Note that there exists a Jinvariant symplectic, orthogonal or unitary form on N, which is non-degenerate with the exception of the natural SL_{n}(q)-module, where it is the zero-form.

Let B ≤ J be a nontrivial quadratic best offender on T := X_{2}/Y_{1} with E := [N, B] minimal. Since B is quadratic on T, by 3.2 E is an isotropic subspace of N. Put P := N_{KB}(E) and Q = ⟨B^{P} ⟩. Then [N, Q] ≤ E ≤ C_{N}(Q) and so Q is quadratic on N. In particular

Q′ ≤ C_{Q}(N) ∩ (KB)′ ≤ C_{K}(N) = C_{K}(T).

Since C_{K}(T) ≤ Z(K) is a p′-group, this implies that Q is abelian, so Q/C_{Q}(T) is elementary abelian. As Q contains an offender, [MS1, 2.6] and the Timmesfeld Replacement Theorem show that there exists R ≤ Q with R ⊴ P such that R is a quadratic best offender on T. The minimality of [N, B] gives [N, R] = E.

Put \overline{J} := J/C_{J}(N) and U := C_{K}(E) ∩ C_{K}(N/E). We will show next:

2^{∘}. \overline{U} does not possess any central \overline{P} -chief factor.

Note that \overline{R} ∩ \overline{K} ≤ \overline{U} ⊴ \overline{P}. If K̃ ≅ SL_{n}(𝔽) or SU_{n}(𝔽), then [\overline{U}, \overline{P}] ≠ 1 and \overline{P} acts simply on \overline{U}, so (2^{∘}) holds.

Suppose that K̃ ≅ Sp_{2n}(𝔽) or Ω_{2n}^{ϵ}(𝔽). Let l := dim_{𝔽} E. By 3.4

|T/C_{T} (R)| = q^{2l} ≤ |\overline{R}| ≤ q^{\binom{l+1}{2}} resp. 2q^{\binom{l}{2}}.

It follows that l ≥ 3 in the first case and l ≥ 5 in the second case. Hence 3.5 shows that \overline{P} has no central chief-factors on \overline{U} and again (2^{∘}) holds.

3^{∘}. C_{KR}(N) = C_{KR}(T).

Put C := C_{KR}(N) and R_{0} := R ∩ KC. Note that R_{0} ≤ UC. It follows that

R_{0}C/C ≤ UC/C ≅_{P} \overline{U}.

On the other hand O^{p}(\overline{P}) centralizes R_{0}C/(K ∩ R)C. Hence (2^{∘}) gives R_{0} ≤ (R ∩ K)C, so R_{0} = (R ∩ K)C_{R}(N). This shows that

KC ∩ KR = KR_{0} = KC_{R}(N).

By 2.4 C_{R}(N) = C_{R}(K) = C_{R}(T) and, as seen above, C_{K}(N) ≤ C_{K}(T), so C_{KR}(N) = C_{KR}(T).

By (3^{∘}) (KR/C_{KR}(T), T) satisfies the hypothesis of 6.6. It follows that there exists a Ksubmodule U of T with T = Y_{2}/Y_{1} + U and N ≰ U, a contradiction since N is the unique minimal K-submodule of T. Thus (d) is proved.

To proof (e) put C = C_{J}([W, K]). Since K acts faithfully on [W, K], C ∩K = 1 and so [C, K] = 1. Since [V, K] = [V, K, K] we have [W, K] = [V, K] + C_{V} (J)/C_{V} (J) and [V, K, C] ≤ C_{V} (J)). In particular, C_{J}([V, K]) ≤ C. Let c ∈ C. Then [V, K, c] ≅ [V, K]/C_{[V,K]}(c) as a K-module. But any quotient of [V, K] is a perfect K module, while any submodule of C_{V} (J) is a trivial K-module. So [V, K, c] = 0 and C ≤ C_{J}([V, K]).

The proof of Theorem 1, apart from statement (e): The first four statements (a) – (d) follow from 8.3. The statements (f) and (g) follow from 8.2.

Theorem 1 (e) will be proved at the very end of the paper.

Lemma 8.4. Let M be a finite CK-group with O_{p}(M) = 1 and V a faithful 𝔽_{p}M-module. Suppose that

(i) M = J_{M}(V) and there exists a unique J_{M}(V)-component K,

(ii) C_{V} (K) ≤ [V, K] and either C_{V} (K) ≠ 0 or V ≠ [V, K].

Let A ≤ M be a best offender on V and put W := [V, K] and \overline{V} := V/C_{V} (K). Then p = 2, and one of the following holds:

(a) M = K ≅ SL_{3}(2), V = W, |C_{V} (K)| = 2, \overline{V} is a natural SL_{3}(2)-module, |A| = 4, [\overline{V} , A]| = 2 and C_{V} (A) = [V, A] has order 4.

(b) M = K ≅ SL_{3}(2), |V/W | = 2, C_{V} (K) = 0, W is a natural SL_{3}(2)-module, |A| = 4 = |C_{W} (A)| and C_{V} (A) = [V, A] = C_{W} (A).

(c) M = K ≅ SU_{4}(2), V = W, 2 ≤ |C_{V} (K)| ≤ 4, \overline{V} is a natural SU_{4}(2)-module, A is the centralizer of a singular 2-subspace of \overline{V}, and C_{V} (A) = [V, A].

(d) M ≅ G_{2}(q), q = 2^{k}, V = W, 2 ≤ |C_{V} (K)| ≤ q, \overline{V} is a natural G_{2}(q)-module, |A| = q^{3}, and C_{V} (A) = [V, A].

(e) K ≅ Alt(2m) and M ≅ Sym(2m) or Alt(2m). For Ω = {1, 2, …, 2m} let N = {n_{Σ} | Σ ⊆ Ω} be the 2m-dimensional natural permutation module and Ñ be the 𝔽_{2}M-module defined by Ñ = N as an 𝔽_{2}-space and

n_{Σ}^{g} = n_{Σ^{g}} if |Σ| is even or g ∈ Alt(Ω), and n_{Σ}^{g} = n_{Σ^{g}} + n_{Ω} if |Σ| is odd and g ∉ Alt(Ω).

Then one of the following holds, where t_{1}, t_{2}, …, t_{m} is a maximal set of commuting transpositions:

1. M = Sym(n), V is isomorphic to N or N/C_{N}(K), and A = ⟨t_{1}, t_{2}, …, t_{k}⟩ for some 1 ≤ k ≤ m.

2. M = Sym(n), V ≅ Ñ and A = ⟨t_{1}, t_{2}, …, t_{m}⟩.

3. V ≅ [N, K] and A fulfills one of the cases (h:1) – (h:3) of Theorem 3.

(f) M = K ≅ Sp_{2m}(q), m ≥ 1, q = 2^{k}, (m, q) ≠ (1, 2), (2, 2), and \overline{W} is the direct sum of r natural Sp_{2n}(q)-modules.^{4} Moreover, the following hold:

4Observe that for m = 1, Sp_{2}(q) ≅ SL_{2}(q) and a natural Sp_{2}(q)-module is also a natural SL_{2}(q)-module.

(a) 2r ≤ m + 1, and if V ≠ W then m > 1 and 2r < m + 1.

(b) Let X be the 2m + 2-dimensional 𝔽_{q}M-module obtained from the embedding Sp_{2m}(q) ≅ Ω_{2m+1}(q) ≤ Ω_{2m+2}^{±}(q). Then V is isomorphic to an 𝔽_{p}M-section of X^{r}.

Proof. Suppose K is not quasisimple. Then K is a p′-group and V = [V, K] ⊕ C_{V} (K). Since C_{V} (K) ≤ [V, K] this gives C_{V} (K) = 0 and V = [V, K], contrary to the assumptions.

Thus K is quasisimple. By 8.3, \overline{W} is a semisimple K-module and we conclude that there exists simple K-submodule of \overline{U} of \overline{W} such that H^{1}(K, \overline{U}) ≠ 0 or H^{1}(K, \overline{U}^{∗}) ≠ 0.

Let B := C_{A}([V, A]). By the Timmesfeld Replacement Theorem, B is a non-trivial quadratic best offender on V. Note that by 2.4 and 1.2 A and B are offenders on \overline{U} and \overline{W}. Comparing 6.1 with Theorem 1(g) we see that p = 2 and the following holds:

1^{∘}. M ≅ SL_{3}(2), SU_{4}(2), G_{2}(q), Alt(2m), Sym(2m) or Sp_{2m}(q), and \overline{W} is the corresponding natural module, with the exception of the Sp_{2m}(q)-case, where \overline{W} is the direct sum of r natural modules for some integer r with 2r ≤ m + 1.

We now discuss the cases given in (1^{∘}) (and 6.1) separately.

Case 1. Suppose M ≅ SL_{3}(2) and C_{W} (K) ≠ 0.

Let 1 ≠ a ∈ A. Since W = [W, K] has order 2^{4} and K is generated by three conjugates of a, |[W, a]| = |W/C_{W} (a)| = 4. Since A is an offender we conclude that

A = B, |V/C_{V} (A)| = |A| = |C_{W} (A)| = 4.

In particular C_{W} (A) = [W, A], V = C_{V} (A) + W and |[\overline{V} , A]| = 2. The latter fact shows that V = W + C_{V} (K) and thus W = V. Hence (a) holds in this case.

Case 2. Suppose M ≅ SL_{3}(2) and C_{W} (K) = 0.

Then W is a natural module and V ≠ W. As above, for 1 ≠ a ∈ A, |V/C_{V} (a)| = |A| = 4, and C_{V} (a) = C_{W} (a) = C_{V} (A). Hence (b) holds.

Case 3. Suppose M ≅ SU_{4}(2).

Then [\overline{W} , B] is a singular subspace of \overline{W}, and 3.4 shows that |B| = 2^{4} = |\overline{W} /C\overline{W} (B)|. Thus A = B and |V/C_{V} (A)| = 2^{4}. Moreover, by 5.1 M is generated by two conjugates of A and so |V/C_{V} (K)| = 2^{8} and V = W + C_{V} (K). Hence V = W. As [V, A]/[V, A] ∩ C_{V} (K) has order 2^{4} and M is generated by two conjugates of A, C_{V} (K) ≤ [V, A]. Since C\overline{V} (A) = [\overline{V} , A] this gives C_{V} (A) = [V, A], and (c) holds.

Case 4. Suppose M ≅ G_{2}(q).

Then |A| = q^{3}, C\overline{W} (A) = [\overline{W} , A] has order q^{3}, |\overline{W} | = q^{6}, and by 5.2 M is generated by two conjugates of A A similar argument as in the SU_{4}(2) case now shows that (d) holds.

Case 5. Suppose M ≅ Alt(2m) or Sym(2m).

Since K is perfect, V is as an 𝔽_{2}K-module isomorphic to a section of the 2m-dimensional permutation module N. If V = W or C_{V} (K) = 0 we have C_{GL(V)}(K) = 1 and so V is also an 𝔽_{2}M-module isomorphic to N.

If H = Sym(n) and |V | = 2^{2m}, there are two possible isomorphism types for V, namely N and Ñ as described in (e). Note that if t is a transposition, and V ≅ Ñ, then C_{V} (t) ≤ W. Since A is an offender on \overline{W} we can apply Theorem 3(h).

Suppose that C_{V} (A) ≰ W. Then there exists a proper subset Σ of Ω = {1, 2, …, 2m} such that |Σ| is odd and |A| normalizes {Σ, Ω\Σ}. If Σ is A invariant, then A has a fixed-point on Σ. It follows from Theorem 3(h) that A is generated by transpositions, V ≇ Ñ, and (e:1) holds. So suppose for a contradiction that Σ^{a} = Ω \ Σ for some a ∈ A. Then |Σ| = m is odd. So Theorem 3(h:4) does not hold. Put A_{0} := N_{A}(Σ). Note that Supp(b) = Ω for all a ∈ A \ A_{0} and so b ∈ A_{0} for all b ∈ A with with |Supp(b)| ≤ 4. In the first three cases of Theorem 3(h), A is generated by such elements, so A = A_{0}, a contradiction.

Suppose that C_{V} (A) ≤ W. If W ≠ V we conclude that A is an over-offender on W. Thus by Theorem 3(h) A is generated by a maximal set of commuting transpositions. Hence (e:1) or (e:2) holds.

Assume that W = V. Then W ≅ [N, K]. If 2m = 8 and A acts transitively on Ω, then C_{V} (A) = C_{V} (K) and |V/C_{V} (A)| = 2^{6} ≥ 2^{3} = |A|, a contradiction. This excludes case (h:4) of Theorem 3, and (e:3) holds.

Case 6. Suppose M ≅ Sp_{2m}(q).

Since K is perfect we conclude from 6.1, (1^{∘}) and 8.2(2) that it remains to prove the second statement of (f:a). Since A is an offender on \overline{V} we may assume that C_{V} (K) = 0 and so V ≠ W.

Suppose that there exists v ∈ C_{V} (A) \ W. Then C_{K}(v) is contained in a subgroup isomorphic to O_{2m}^{ϵ}(V), and 8.2(4) shows that 4r ≤ 2m − 2. Thus 2r ≤ m − 1 < m + 1.

Suppose next that C_{V} (A) ≤ W. Since V ≠ W we conclude that A is an over-offender on W. The proof of 8.2(Case 2) now shows that r < m + 1.

Corollary 8.5. Assume the hypothesis of 8.4. Then every best offender in M on V is a best offender on [V, J] + C_{V} (J)/C_{V} (J).

Proof. According to 1.2 we may assume that V = [V, J]. Put \overline{V} := V/C_{V} (J) =: W and X := C_{V} (J). Let A be a best offender in M on V. Choose 1 ≠ B ≤ A such that |B||C_{W} (B)| is maximal and then B minimal. Since A is an offender on W, B is a quadratic best offender on W.

Suppose that C_{W} (B) = \overline{C_{V} (B)}. Since A is a best offender on V, |C_{V} (B)||B| ≤ |C_{V} (A)||A| and since B ≤ A, C_{X}(B) ≥ C_{X}(A). Thus

|C_{W} (B)||B| = \frac{|C_{V} (B)||B|}{|C_{X}(B)|} ≤ \frac{|C_{V} (A)||A|}{|C_{X}(A)|} = |\overline{C_{V} (A)}||A| ≤ |C_{W} (A)||A|,

and so A is a best offender on W.

Suppose that C_{W} (B) ≠ \overline{C_{V} (B)}. Since \overline{V} is J-semisimple by 8.3, there exists a perfect Jsubmodule Y of V such that \overline{Y} is simple and C\overline{Y} (B) ≠ \overline{C_{Y} (B)}. Note that there exists a unique Jcomponent K with [Y, K] ≠ 0. Moreover, Y = [Y, K] and Y ∩ X = C_{Y} (K) ≠ 0. Put J̃:= J/C_{J}(Y). The Three Subgroups Lemma implies that O_{p}(J̃) centralizes Y and so we can apply 8.4 to (J̃, K̃, Y) in place of (H, K, V).

In Case 8.4(d),(f) we have C_{J}(v) = C_{J}(\overline{v}) for all v ∈ V, a contradiction.

In Case 8.4(c) we get Ã = B̃ and C\overline{V} (B) = [\overline{V} , A] = \overline{C_{V} (A)} = \overline{C_{V} (B)}, contradiction.

Suppose 8.4(e) holds. Then A is generated by elements of support at most 4 and so C\overline{V} (A) = \overline{C_{V} (A)}.

Suppose that 8.4(a) holds. Then |Ã| = 4 and C\overline{Y} (A) = [\overline{Y} , A] = \overline{C_{Y} (A)}. Thus B̃ ≠ Ã and |B̃| = 2 = |\overline{Y} /C\overline{Y} (B)|. Put B_{0} = C_{B}(\overline{Y}). Then |C_{W} (B)||B| = |C_{W} (B_{0})||B_{0}|. The minimal choice of B implies B_{0} = 1 and so |B| = 2. Thus |C_{W} (B)||B| = |W |. Since A is an offender on W, this gives |C_{W} (B)||B| ≤ |C_{W} (A)||A|. Thus A is a best offender on W.

Finally Case 8.4(b) does not apply, since C_{V} (K) ≠ 0.

The proof of Theorem 1(e): This is 8.5.

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