Applications of the FF-Module Theorem and Related Results

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 and V a faithful 𝔽_{p}M-module. In this paper we investigate the structure of M and V under various assumptions related to best offenders and quadratic action.

Introduction

This paper is the last part of a series of papers ([MS1], [MS2], [MS3], [MS4]) which form together with [GM1], [GM2], [Ch], [MeSt1] and [MeSt2] the module theoretic background for the classification of the finite groups of local characteristic p outlined in [MSS1] and in particular, for the Structure Theorem [MSS2].

Most of the results we present should be of independent interest since they give further insight in the action of offenders or more generally quadratic subgroups on modules. The results we prove come in two categories:

1) Consequences of the General FF-Module Theorem, the FF-Module Theorem and the Offender Theorem (see [MS4, Theorems 1,2 and 3]): Like the Strong Dual FF-Module Theorem 3.1, the Strong FF-Module Theorem 3.2, the General Point-Stabilizer Theorem 3.6, and the Q!FF-Module Theorem 4.6.

2) Statements about modules with quadratically acting subgroups: Like the Quadratic L-Lemma 2.9, and the Minimal Asymmetric Module Theorem 5.5.

Let H be a finite group. A p-parabolic subgroup of H is a subgroup that contains a Sylow psubgroup of H; and we write parabolic subgroup if it is clear which prime p is meant. In contrast, for a genuine group of Lie type the subgroups containing a Borel subgroup are called Lie parabolic. For the definition of a genuine group of Lie type see [MS4]. Also the definitions for the various types of offenders can be found in [MS4].

Let T ∈ Syl_{p}(H) and V be an 𝔽_{p}H-module. Then P_{H}(V, T)) := O^{p′}(C_{H}(C_{V} (T))) is called the point-stabilizer of H on V with respect to T.

The quadratic action on modules will be in the center of the investigation. Therefore the reader should be familiar with the basic properties of quadratic action, for example that a faithfully and quadratically acting group on an 𝔽_{p}-vector space is an elementary abelian p-group (see [KS, 9.1.1]).

Several of the proofs require some knowledge of genuine groups of Lie type. So the reader should also have some understanding of these groups; in particular of their Lie parabolic structure.

In the following GFMT stands for the General FF-Module Theorem and FMT for the FF-Module Theorem and the Offender Theorem in [MS4]. The last two theorems we regard as one reference, so a reference like FMT(7) refers to case (7) in the FF-Module Theorem, then the reader is supposed to look up the properties of the offenders for this given case in the Offender Theorem. The reader might feel the need to have the article [MS4] at hand while following the proofs.

Acknowledgment: We would like to thank the referee for his immense work, in particular for all the comments and suggestions that helped us to bring the manuscript into a more readable form.

1 The Kieler Lemma

In this section we discuss a property of finite groups of local characteristic p that is relevant for our investigations in Section 3. Recall that for p a prime, a finite group G has characteristic p if C_{G}(O_{p}(G)) ≤ O_{p}(G); and G has local characteristic p if every p-local subgroup has characteristic p. We start with some elementary properties of groups of (local) characteristic p.

Lemma 1.1. Let G be a finite group. Then G has characteristic p if and only if F^{∗}(G) = O_{p}(G).

Proof. Suppose first that G has characteristic p. Put L = O_{p}(G). Then C_{G}(L) ≤ L. By [KS, 6.5.7(c)], E(G) = E(L)C_{E(G)}(L) ≤ L. Also O_{p′}(G) ≤ C_{G}(L) ≤ L, and since F(G) is nilpotent, F(G) = O_{p′}(F(G))O_{p}(G) ≤ L. Thus F^{∗}(G) = E(G) F(G) ≤ L ≤ F(G) ≤ F^{∗}(G) and F^{∗}(G) = L.

Suppose next that F^{∗}(G) = O_{p}(G). By [KS, 6.5.8] C_{G}(F^{∗}(G)) ≤ F^{∗}(G) and so G has characteristic p.

Lemma 1.2. Let G be a finite group of characteristic p and S ∈ Syl_{p}(G). Then the following hold:

(a) Every subnormal subgroup of G has characteristic p.

(b) Every subgroup containing S has characteristic p.

(c) G has local characteristic p.

Proof. (a): Let N ⊴⊴ G. Then F^{∗}(N) ≤ F^{∗}(G) by [KS, 6.5.7]. Now (a) follows from 1.1.

(b): Let S ≤ H ≤ G. Then O_{p}(G) ≤ O_{p}(H), and (b) follows.

(c): Let P be a p-subgroup of G with C_{O_{p}(G)}(P) ≤ P, and let Q be a p′-subgroup of C_{G}(P). Then C_{O_{p}(G)}(P) ≤ C_{O_{p}(G)}(Q) and so by Thompson’s P ×Q-Lemma [KS, 8.2.8], Q ≤ C_{G}(O_{p}(G)) ≤ O_{p}(G). Thus Q = 1 and C_{G}(P) is a p-group.

Now let L be a p-local subgroup of G, so L = N_{G}(R) for some non-trivial p-subgroup R ≤ G. Set P := O_{p}(L). Since R ≤ P,

C_{O_{p}(G)}(P) ≤ C_{O_{p}(G)}(R) ≤ L ∩ O_{p}(G) ≤ P.

It follows that C_{G}(P) is a p-group. Since C_{L}(P) ≤ C_{G}(P) and C_{L}(P) ⊴ L, we get that C_{L}(P) ≤ P.

Lemma 1.3. Let G be a finite group. Then G has characteristic p if and only if O^{p}(G) has characteristic p.

Proof. If G has characteristic p, then O^{p}(G) has characteristic p by 1.2(a).

Suppose O^{p}(G) has characteristic p. Note that O^{p}(F^{∗}(G)) ≤ O^{p}(G) ∩ F^{∗}(G) and so by [KS, 6.5.7] O^{p}(F^{∗}(G)) ≤ F^{∗}(O^{p}(G)). Thus by 1.1, O^{p}(F^{∗}(G)) is a p-group. Hence F^{∗}(G) is a p-group and so G has characteristic p by 1.1.

Lemma 1.4. Let G be a finite group of local characteristic p.

(a) Let N ⊴⊴ G and L be a p-local subgroup of G. Then N ∩ L has characteristic p.

(b) Every subnormal subgroup of G has local characteristic p.

(c) Let S ∈ Syl_{p}(G) and N ⊴ G. Then SN has local characteristic p.

Proof. (a): Let 1 ≠ P ≤ G be a p-subgroup. Then N_{G}(P) has characteristic p since G is of local characteristic p. As N_{N}(P) is subnormal in N_{G}(P) we get from 1.2(a) that N_{N}(P) has characteristic p.

(b): This follows from (a).

(c): Let 1 ≠ P ≤ S. By (a), N_{N}(P) has characteristic p. Since O^{p}(NS) ≤ N, O^{p}(N_{NS}(P)) ⊴ N_{N}(P). We conclude from 1.2(a) that O^{p}(N_{NS}(P)) has characteristic p and then by 1.3 that N_{NS}(P) has characteristic p.

Lemma 1.5 (Kieler Lemma). Let G be a finite group of local characteristic p and S ∈ Syl_{p}(G), and let E be a subnormal subgroup of G such that p||E|. Then C_{E}(Ω_{1}Z(S)) = C_{E}(Ω_{1}Z(S ∩ E)).

Proof. If E = G, there is nothing to prove. Thus, we may assume that E ≠ G. Put Z := Ω_{1}Z(S) and Z_{0} := Ω_{1}Z(S ∩ E). We proceed by induction on |G/E||G|. Let M be a maximal normal subgroup of G containing E. Then M < G, S ∩ M ≤ Syl_{p}(M), and by 1.4(b) M has local characteristic p, so by induction

C_{E}(Z_{0}) = C_{E}(Ω_{1}Z(S ∩ M)).

If E ≠ M, then also by induction

C_{M}(Ω_{1}Z(S ∩ M)) = C_{M}(Z),

and the lemma follows. Thus, we may assume that E = M. By 1.4(c) ES has local characteristic p, and so by induction we may assume

1^{∘}. E is a maximal normal subgroup of G and G = ES.

Set C := C_{G}(Z ∩ E). Since p divides |E|, S ∩ E ≠ 1, and by (1^{∘}) S ∩ E is normal in S, so 1 ≠ Z ∩ E ≤ Z_{0} and C_{G}(Z_{0}) ≤ C. Note that |C/C ∩ E| ≤ |G/E| and that N_{G}(Z ∩ E) has characteristic p. Since C ⊴ N_{G}(Z ∩ E) also C has characteristic p by 1.2(a). Hence, if |C| < |G|, then by induction

C_{E}(Z) = C_{C∩E}(Z) = C_{C∩E}(Z_{0}) = C_{E}(Z_{0}),

and the lemma follows. Thus, we may assume G = C_{G}(Z ∩ E) and so

2^{∘}. Z ∩ E ≤ Z(G), and G has characteristic p.

We now treat the cases Z ≤ E and Z ≰ E separately.

Case 1: Z ≰ E.

Put V := Ω_{1}Z(O_{p}(G)). By (2^{∘}) Z ≤ C_{G}(O_{p}(G)) ≤ O_{p}(G) and so Z ≤ V. By (1^{∘}) G = EZ and so V = (V ∩ E)Z. In particular, V = (V ∩ E) × Z_{1} for some Z_{1} ≤ Z. Since Z_{1} is S-invariant, Gaschütz’ Theorem [KS, 3.3.2] also gives a G-invariant complement Z_{2} for V ∩ E in V. But then Z_{2} ≤ Ω_{1}Z(G), and G = EZ yields

Z = Z_{2} × (Z ∩ E) and S = Z_{2} × (S ∩ E).

It follows that Z_{0} = Z ∩ E and C_{E}(Z) = C_{E}(Z_{0}), and the lemma is proved in this case.

Case 2: Z ≤ E.

Then Z ≤ Z_{0} and by (2^{∘})

3^{∘}. Z ≤ Z(G).

Hence C_{E}(Z) = E and it remains to show that E centralizes Z_{0}. Put V := Ω_{1}Z(O_{p}(E)). By 1.2(a), E has characteristic p, and so Z_{0} ≤ C_{E}(O_{p}(E)) ≤ O_{p}(E). Thus Z ≤ Z_{0} ≤ V and so also ⟨Z_{0}^{G}⟩ ≤ V. We now investigate the homomorphism

π : Z_{0} ⟶ V with x ↦ ∏_{U∈E/S∩E} x^{U},

where the product runs over all cosets U = (S ∩ E)t, t ∈ E and x^{U} := x^{t}. Note that x^{U} is well defined since S ∩ E centralizes x.

The following elementary properties of π are easy to check:

– (xπ)^{s} = (x^{s})π for all x ∈ Z_{0} and s ∈ S.

– Z_{0}π ≤ C_{V} (E) ≤ Z_{0}.

– π|_{C_{V} (E)} is the multiplication by |E/S ∩ E| and thus an automorphism.

– C_{ker π}(S) ≤ Z ≤ C_{V} (E) by (3^{∘}).

Let x ∈ Z_{0}. By the second property, xπ ∈ C_{V} (E) and so by the third xπ = yπ for some y ∈ C_{V} (E). Then y^{−1}x ∈ ker π and so Z_{0} ≤ C_{V} (E) ker π. From the third property we get that ker π∩C_{V} (E) = 1. and so Z_{0} = C_{V} (E) × ker π . Moreover, the first property implies that ker π is S-invariant. Now the last property shows that ker π = 1 and Z_{0} = C_{V} (E). Hence C_{E}(Z) = E = C_{E}(Z_{0}) and the lemma also holds in Case 2.

Corollary 1.6. Let H be a finite group, E ⊴⊴ H, and V be a finite dimensional 𝔽_{p}H-module. Then

C_{E}(C_{V} (T)) = C_{E}(C_{V} (T ∩ E)) = C_{E}(C_{[V,E]}(T ∩ E)) for T ∈ Syl_{p}(H).

Proof. Observe that the corollary holds for H if and only if it holds for H/C_{H}(V). Thus, we may assume that V is a faithful H-module.

Let G := V ⋊ H be the semidirect product of V and H, Ẽ := V ⋊ E and S := V ⋊ T. Then G is of characteristic p since V is faithful, so G is also of local characteristic p by 1.2(c). Moreover,

C_{V} (T) = Ω_{1}Z(S) and C_{V} (T ∩ E) = Ω_{1}Z(S ∩ Ẽ).

Hence 1.5 implies C_{E}(C_{V} (T)) = C_{E}(C_{V} (T ∩ E)). This is the first equality of 1.6.

Pick R ≤ C_{V} (T ∩ E) such that [V, E] + C_{V} (T ∩ E) = [V, E] ⊕ R. Then Gaschütz’ Theorem shows that [V, E] + R = [V, E] + C_{V} (E) and thus

C_{V} (T ∩ E) = C_{[V,E]+C_{V} (E)}(T ∩ E) = C_{[V,E]}(T ∩ E) + C_{V} (E).

Now also the second equality of 1.6 follows.

Lemma 1.7. Let H be a finite group, E ⊴⊴ H, and V be a finite dimensional 𝔽_{p}H-module. Let A ≤ T ∈ Syl_{p}(H) such that A normalizes E and V_{0} ≤ C_{V} (O^{p}(E)). Put \overline{V} := V/V_{0}. Then the following hold:

(a) P_{E}(V, T ∩ E) = P_{E}([V, E], T ∩ E) = P_{E}([V, O^{p}(E)], T ∩ E) = O^{p′}(E ∩ P_{H}(V, T)).

(b) P_{EA}(V, T ∩ EA) = O^{p′}(EA ∩ P_{H}(V, T)).

(c) P_{E}(\overline{V} , T ∩ E) ≤ P_{E}(V, T ∩ E) ≤ P_{H}(V, T) and O_{p}(P_{H}(V, T)) ∩ E ≤ O_{p}(P_{E}(\overline{V} , T ∩ E)).

Proof. Note that for any two subgroups X and Y of a finite group, O^{p′}(X ∩ Y) = O^{p′}(X ∩ O^{p′}(Y)). In particular,

O^{p′}(C_{E}(C_{V} (T)) = O^{p′}(E ∩ P_{H}(V, T)).

(a): By 1.6 O^{p′}(C_{E}(C_{V} (T)) = P_{E}(V, T ∩ E) = P_{E}([V, E], T ∩ E) and so

P_{E}(V, T ∩ E) = P_{E}([V, E], T ∩ E) = O^{p′}(E ∩ P_{H}(V, T)).

From P_{E}(V, T ∩E) = P_{E}([V, E], T ∩E) and induction on i we have P_{E}(V, T ∩E) = P_{E}([V, E, i], T ∩E) for all i ∈ ℤ^{+}, where [V, E, 0] := V and [V, E, i] := [[V, E, i−1], E] for i > 0. Since E acts nilpotently on V/[V, O^{p}(E)] and since E/C_{E}(V/[V, E, i]) is a p-group we have [V, O^{p}(E)] = [V, E, k] for some k ∈ ℤ^{+}. Thus

P_{E}(V, T ∩ E) = P_{E}([V, E, k], T ∩ E) = P_{E}([V, O^{p}(E)], T ∩ E).

and so (a) holds.

(b): We again apply 1.6 to E ⊴ EA and E ⊴⊴ H. Then

C_{EA}(C_{V} ((T ∩ E)A)) = AC_{E}(C_{V} ((T ∩ E)A)) = AC_{E}(C_{V} (T ∩ E)) = AC_{E}(C_{V} (T)) = C_{EA}(C_{V} (T)),

and so

\begin{aligned}P_{EA}(V, (T ∩ E)A) & = & O^{p′}(C_{EA}(C_{V} ((T ∩ E)A))) & = & O^{p′}(C_{EA}(C_{V} (T))) \\ & = & O^{p′}(EA ∩ P_{H}(V, T)). \end{aligned}

(c): By (a) P_{E}(V, T ∩ E) ≤ P_{H}(V, T) and so O_{p}(P_{H}(V, T)) ∩ E is normal in every subgroup of P_{E}(V, T ∩ E) containing T ∩ E. Thus, it suffices to show P := P_{E}(\overline{V} , T ∩ E) ≤ P_{E}(V, T ∩ E).

Clearly O^{p}(P) centralizes C_{V} (T ∩E)+V_{0}/V_{0} and V_{0}. Hence O^{p}(P) centralizes C_{V} (T ∩E). From P = O^{p}(P)(T ∩ E) we conclude that P ≤ C_{E}(C_{V} (T ∩ E)) and P = O^{p′}(P) ≤ P_{E}(V, T ∩ E).

Lemma 1.8. Let G be a finite group and N an abelian normal subgroup of G with N ∩ Φ(G) = 1. Then there exists a complement to N in G.

Proof. If |N| = 1 lemma clearly holds. So suppose that N ≠ 1. Then there exists a maximal subgroup M of G with N ≰ M. Note that N ∩M is normal in G = NM. Moreover, (N ∩M)∩Φ(G) ≤ N ∩Φ(G) = 1 and so by induction there exists a complement K to N ∩M in G. Then G = (N ∩M)K and so M = (N ∩ M)(M ∩ K). Thus

G = NM = N(N ∩ M)(M ∩ K) = N(M ∩ K)

and N ∩ (M ∩ K) = (N ∩ M) ∩ K = 1. Hence M ∩ K is a complement to N in G.

Lemma 1.9. Let G be a finite group and π be a set of primes. Suppose that O_{π}(G) = 1. Then Φ(G) = Φ(O^{π}(G)).

Proof. Put L := O^{π}(G). Since L ⊴ G, Φ(L) ≤ Φ(G). Since Φ(G) is nilpotent and O_{π}(G) = 1, Φ(G) is a π′-group and so Φ(G) ≤ L. Put \overline{G} := G/Φ(L) and N := Φ(G). Then Φ(\overline{L}) = 1 and Φ(\overline{G}) = \overline{N}. Note that Φ(\overline{N}) ≤ Φ(\overline{L}) = 1 and since \overline{N} is nilpotent, we conclude that \overline{N} is abelian. Also \overline{N} ∩Φ(\overline{L}) = 1, and so by 1.8 applied to \overline{N} and \overline{L} in place of N and G, there exists a complement to \overline{N} in \overline{L}. Since \overline{N} and \overline{G}/\overline{L} are coprime and \overline{N} is abelian, Gaschütz’ theorem implies that there exist a complement to \overline{N} in \overline{G}. Since \overline{N} = Φ(\overline{G}) this gives \overline{N} = 1 and so Φ(G) = N = Φ(L).

2 Quadratic Modules

Lemma 2.1. Let H be a group and A, B, C be subgroups of H with [A, B, C] = 1.

(a) If C normalizes [B, C, A], then [C, A, B] ≤ [B, C, A].

(b) If C normalizes [B, C, A] and [A, C, B], then [C, A, B] = [B, C, A].

Proof. Recall first that [A, B] = [B, A] and [A, B] ⊴ ⟨A, B⟩.

(a) Without loss H = ⟨A, B, C⟩. Since [A, B] ⊴ ⟨A, B⟩ and C centralizes [A, B] we conclude that [A, B] ⊴ H and that ⟨C^{H}⟩ centralizes [A, B]. In particular, [[A, B], [B, C]] = 1. Since B normalizes [B, C] we get

⟨[B, C, A]^{B}⟩ = [B, C, ⟨A^{B}⟩] = [[B, C], A[A, B]]] = [B, C, A].

Thus B normalizes [B, C, A]. Note that also A and C normalize [B, C, A] and so [B, C, A] ⊴ H. Put \overline{H} = H/[B, C, A]. Then [\overline{A}, \overline{B}, \overline{C}] = 1 and [\overline{B}, \overline{C}, \overline{A}] = 1. Thus [\overline{C}, \overline{A}, \overline{B}] = 1 by the Three Subgroups Lemma. So (a) holds.

(b) By (a) [C, A, B] ≤ [B, C, A], and by (a) with the roles of A and B interchanged, [C, B, A] ≤ [A, C, B]. Since [B, C, A] = [C, B, A] and [C, B, A] = [B, C, A] this gives [A, C, B] = [C, A, B].

Suppose that [A, B, C] = 1 for some subgroups A, B, C of H. Then the assumption in (a) holds if A is a normal subgroup of H or if C is an abelian normal subgroups of H. Similarly the assumption in (b) hold if A and B are normal subgroups of H or if C is an abelian normal subgroup of H.

We remark that in (a) [C, A, B] might be a proper subgroup of [B, C, A] (and so the conclusion in (b) no longer has to hold if C does not normalize [C, A, B].) Indeed let H = Sym(4), A = ⟨(12)(34)⟩, B = ⟨(12)⟩ and C = ⟨(23)⟩. Then [A, B] = 1 and so also [A, B, C] = 1. We have [C, A] = ⟨(23)(14)⟩, [C, A, B] = ⟨(12)(34)⟩, [B, C] = ⟨(123)⟩ and [B, C, A] = ⟨(12)(34), (13)(23)⟩ = O_{2}(H). So [B, C, A] is normal in H and [C, A, B] is a proper subgroup of [B, C, A].

Lemma 2.2. Let M be a group, 𝕂 a field, V an 𝕂M-module and A ≤ M. Suppose that M is generated by n conjugates of A in M.

(a) Suppose that dim_{𝕂} V/C_{V} (A) ≤ r. Then dim_{𝕂} V/C_{V} (M) ≤ nr. If, in addition V is a non-trivial simple 𝕂M-module, then dim_{𝕂} V ≤ nr.

(b) Suppose that dim_{𝕂}[V, A] ≤ r. Then dim_{𝕂}[V, M] ≤ nr. If, in addition V is a non-trivial simple 𝕂M-module, then dim_{𝕂} V ≤ nr.

Proof. (a) Let A_{1}, … A_{n} be conjugates of A in M with M = ⟨A_{i} | 1 ≤ i ≤ n⟩. Then

⋂_{i=1}^{n} C_{V} (A_{i}) ≤ C_{V} (M)

and dim_{𝕂} V/C_{V} (A_{i}) ≤ r. Thus dim_{𝕂} V/C_{V} (M) ≤ rn. If V is a non-trivial simple 𝕂M-module, then C_{V} (M) = 0 and so (a) is proved.

(b) follows from (a) applied the 𝕂-dual of V.

Lemma 2.3. Let M be a genuine group of Lie-type in characteristic p, V a simple 𝔽_{p}M-module, 𝕂 := End_{M}(V), S ∈ Syl_{p}(M) and 1 ≠ z ∈ Ω_{1}Z(S). Suppose that one of the following holds:

(i) There exists P ≤ M with S ≤ P and [V, z, O_{p}(P)O^{p}(P)] = 0.

(ii) [V, z, S] = 0.

(iii) [V, z] is 1-dimensional over 𝕂.

Then M ≅ SL_{n}(q), Sp_{n}(q) or SU_{n}(q), and V is a corresponding natural module.

Proof. Suppose that (i) holds. Then [V, z, O_{p}(P)] = 0 and so [V, z] ≤ C_{V} (O_{p}(P)). By Smith’s Lemma [MS4, 4.2] C_{V} (O_{p}(P)) is a simple 𝕂P -module and thus [V, z] is a simple 𝕂P -module. Hence [V, z, O^{p}(P)] = 0 implies that [V, z, P] = 0 and so dim_{𝕂}[V, z] = 1.

Suppose (ii) holds. Then [V, z] ≤ C_{V} (S). By Steinberg ’s Lemma [MS4, 4.1] dim_{𝕂} C_{V} (S) is 1-dimensional. Thus dim_{𝕂}[V, z] = 1. and (ii) implies (iii). We have proved that (i) and (ii) imply (iii). So we may assume from now on that (iii) holds.

Without loss M is universal. Let \overline{𝕂} be the algebraic closure of 𝕂 and \overline{V} = \overline{𝕂} ⊗_{𝕂} V. Let (\overline{M}, σ) be a σ-setup for M, see [GLS3, Definition 2.2.1]. So M = O^{p′}(C\overline{M}(σ)) and by [GLS3, 2.8.2], \overline{V} can be extended to rational 𝕂\overline{M}-module. Then S lies in maximal unipotent subgroup \overline{S} of \overline{M} and 1 ≠ z ∈ Ω_{1}Z(\overline{S}).

Put \overline{B} = N\overline{M}(\overline{S}). Then \overline{B} is a Borel subgroup of \overline{M} and since dim[\overline{V} , z] is 1-dimensional (over \overline{𝕂}), [\overline{V} , z, \overline{S}] = 0. Since by Steinberg ’s Lemma [MS4, 4.1] C\overline{V} (\overline{S}) is 1-dimensional we conclude that [\overline{V} , z] = C\overline{V} (\overline{S}). Thus also [\overline{V} , ⟨z\overline{B}⟩] = C\overline{V} (\overline{S}). Note that ⟨z\overline{B}⟩ contains a root subgroup \overline{Z} of \overline{M}. Then \overline{Z} ≤ Ω_{1}Z(\overline{S}) and [\overline{V} , \overline{Z}] = C_{V} (\overline{S}) is 1-dimensional. Observe that one of the following holds:

(a) Z(\overline{S}) is a long root subgroup.

(b) p = 3 and \overline{M} ≅ G_{2}(\overline{𝕂}), or p = 2, \overline{M} ≅ F_{4}(\overline{𝕂}) and Z(\overline{S}) is the product of a long and a short root subgroup.

(c) p = 2, \overline{M} ≅ B_{n}(\overline{𝕂}) or \overline{M} ≅ C_{n}(\overline{𝕂}) and Z(\overline{S}) is the product of a long and a short root subgroup.

In case (b) we may apply the graph automorphism of \overline{M}, if necessary, and assume that \overline{Z} is a long root subgroup. In case (c), since there exists an automorphism from B_{n}(\overline{𝕂}) to C_{n}(\overline{𝕂}) sending long root groups to short root groups and vice versa, we also may assume that \overline{Z} is a long root subgroup. Thus in any case \overline{Z} is a long root subgroup.

Suppose \overline{M} = A_{n−1}(\overline{𝕂}) = SL_{n}(\overline{𝕂}). Then \overline{M} is generated by n-conjugates of \overline{Z} (see for example [MS4, 5.3]) and so by 2.2 m := dim\overline{𝕂} \overline{V} ≤ n. By dimension reason we conclude that m = n and the image of image of \overline{M} in GL\overline{𝕂}(\overline{V}) is equal to SL\overline{𝕂}(\overline{V}). So \overline{V} is a natural module for \overline{M}. We conclude that for M = SL_{n}(q) or M = SU_{n}(q), V is a corresponding natural module.

Suppose that \overline{M} ≅ C_{n}(\overline{𝕂}) = Sp_{2n}(\overline{𝕂}), n ≥ 2. Let \overline{P}^{∗} the minimal Lie-parabolic subgroup of \overline{M} with \overline{B} ≤ \overline{P}^{∗} and \overline{Z} ⋬ \overline{P}^{∗}. Put \overline{P} := O^{p′}(\overline{P}^{∗}). Then \overline{P} /O_{p}(\overline{P}) ≅ SL_{2}(\overline{𝕂}). Let \overline{K} be a Levi complement to O_{p}(\overline{P}) in \overline{P} and \overline{R} = \overline{S} ∩ \overline{K}.

Note that \overline{R} is a short root subgroup of \overline{M} and so not conjugate to \overline{Z}. Nevertheless we will show that \overline{R} acts quadratically on \overline{V}. For this let \overline{N} be the natural Sp_{2n}(\overline{𝕂})-module for \overline{M} and \overline{U} := [\overline{N}, \overline{R}]. Then [\overline{N}, \overline{Z}] is 1-dimensional, \overline{U} is a 2-dimensional singular subspace of \overline{N} and [\overline{N}, \overline{Z}] ≤ \overline{U}. Put \overline{D} := C\overline{M}(\overline{U}^{⊥}). Then \overline{D} is abelian and \overline{Z} \overline{R} ≤ \overline{D} ≤ \overline{S}. Thus [\overline{V} , \overline{Z}, \overline{D}] = 0 since [\overline{V} , \overline{Z}] = C\overline{V} (\overline{S}). It

follows from [MS4, 3.4] that \overline{D} = ⟨\overline{Z}\overline{P} ⟩ and so [\overline{V} , \overline{D}, \overline{D}] = ⟨[\overline{V} , \overline{Z}, \overline{D}]^{P} ⟩ = 0. Thus \overline{D} and so also \overline{R} acts quadratically on \overline{V}.

By Smith’s Lemma [MS4, 4.2], \overline{W} := C\overline{V} (O_{p}(\overline{P})) is a simple \overline{𝕂}\overline{P} -module. Also [\overline{W} , \overline{R}] ≤ C\overline{W} (\overline{R}O_{p}(\overline{P})) = C\overline{W} (S) and so [\overline{W} , R] is 1-dimensional. Hence \overline{W} is a natural SL_{2}(\overline{𝕂})-module for \overline{P}. Put \overline{L} = O^{p′}(N\overline{M}(\overline{Z})). Then \overline{L} centralizes [\overline{V} , \overline{Z}], and Smiths’ Lemma implies that C\overline{V} (O_{p}(\overline{L})) = C\overline{V} (\overline{L}). Together with the Ronan-Smith’s Lemma [MS4, 4.3] we conclude that \overline{V} is natural Sp_{2n}(\overline{𝕂})- module for \overline{M}. If M ≅ Sp_{2n}(q) we conclude that V is a natural Sp_{2n}(q)-module. Suppose M ≅ ^{2}B_{2}(q). Then z is not a root element in \overline{M} and so [dim V, z] = 2, a contradiction.

Suppose now that \overline{M} ≅ G_{2}(\overline{𝕂}) or B_{3}(\overline{𝕂}). Let r = 2 and 3, respectively. Let Φ be a root system for \overline{M}, N/H the corresponding Weyl-group and Φ_{l} the set of long root in Φ. Then Φ_{l} is a root system of type A_{r}. Also there exists t ∈ N \ H with tH ∈ Z(N/H) and t induces a graph automorphism on Φ_{l} and so also on the subgroup \overline{L} of \overline{M} generated by long root subgroups corresponding to Φ_{l}. Let U be a non-trivial composition factor for \overline{𝕂} \overline{L}⟨t⟩ on \overline{V}. Since [\overline{V} , \overline{Z}] is 1-dimensional, U is a simple \overline{𝕂} \overline{L}-module. Hence by the A_{r}(𝕂)-case treated above, U is natural SL_{r+1}(\overline{𝕂}) module for \overline{L}. But this contradicts the action of t on \overline{L}.

Suppose \overline{M} ≅ B_{n}(\overline{𝕂}) or D_{n}(\overline{𝕂}) for n ≥ 4. Then \overline{M} has subgroup isomorphic to B_{3}(\overline{𝕂}) generated by long root subgroups, a contradiction to the B_{3}(\overline{𝕂})-case.

Suppose that \overline{M} ≅ E_{r}(\overline{𝕂}), 6 ≤ r ≤ 8. Then \overline{M} has subgroups isomorphic to D_{r−1}(\overline{𝕂}) generated by long root subgroups, a contradiction to the D_{n}(\overline{𝕂})-case.

Suppose that \overline{M} ≅ F_{4}(\overline{𝕂}). Then the long roots form a root system of type D_{4}(\overline{𝕂}), a contradiction to the D_{n}(\overline{𝕂})-case.

Definition 2.4. Let L be a group and A ≤ L. Then L is called A-minimal if L = ⟨A^{L}⟩ and A is contained in a unique maximal subgroup of L.

This definition has been motivated by the L-Lemma in [PPS], which says that in a p-minimal group H every p-subgroup A ≤ H with A ≰ O_{p}(H) is contained in a subgroup L which is AO_{p}(H)- minimal.

Lemma 2.5. Let L be a finite group, A ≤ L and suppose that L is A-minimal. Let M be the maximal subgroup of L containing A.

(a) Let K ⊴⊴ L with A ≤ K. Then K = L.

(b) M is not subnormal in L and N_{L}(M) = M.

(c) L = ⟨A, A^{x}⟩ for all x ∈ L \ M.

(d) A is not subnormal in L. In particular, A ≰ F(L), A ≠ 1, A ≠ L, and L is not nilpotent.

(e) If B ⪇ L with A ≤ B, then L is B-minimal and M is the unique maximal subgroup containing B .

(f) If K ⊴ L with L ≠ KA, then K ≤ M and L/K is AK/K-minimal with M/K being the maximal subgroup containing AK/K.

(g) If K ⊴ L such that L/K is nilpotent, then L = KA and so K ≰ M.

(h) If A is a p-group, then L is p-minimal, M contains the normalizer of a Sylow p-subgroup and O_{p}(L) ≤ ⋂ M^{L}.

Proof. (a): Suppose that K ≠ L. Then there exists a proper normal subgroup H of L with K ≤ H. But then L = ⟨A^{L}⟩ ≤ H, a contradiction.

(b): By (a) M is not subnormal in L. Thus M ≤ N_{L}(M) ⪇ L and since M is a maximal subgroup of L, N_{L}(M) = M.

(c): By (b), M ≠ M^{x}. Since M (M^{x}) is the unique maximal subgroups of L containing A (A^{x}), we conclude that ⟨A, A^{x}⟩ is not contained in any maximal subgroup of L. Thus L = ⟨A, A^{x}⟩.

(d): If A ⊴⊴ L, then by (a) L = A and so A is not contained in any maximal subgroup of L, which contradicts the definition of A-minimal. Since subgroups of nilpotent groups are subnormal, A ≰ F(L) and L is not nilpotent.

(e): Since A ≤ B ⪇ L we have B ≤ M and M is the unique maximal subgroup of L containing B. Also L = ⟨A^{L}⟩ ≤ ⟨B^{L}⟩ and so L = ⟨B^{L}⟩. Thus L is B-minimal.

(f): Put \overline{L} = L/K. Since KA ≠ L, KA ≤ M and so (e) shows that L is KA-minimal with KA ≤ M. Hence \overline{M} is the unique maximal subgroup of \overline{L} containing \overline{A}. Since L = ⟨A^{L}⟩ implies

\overline{L} = ⟨\overline{A}\overline{L}⟩, \overline{L} is \overline{A}-minimal.

(g): By (d) L/K is not AK/K-minimal. Hence (f) implies L = KA.

(h) Let S be a Sylow p-subgroups of L containing A. Then A ⊴⊴ S and so by (d), S ⋬ L. Thus N_{L}(S) ≤ M and S ≠ L. By (e), L is S-minimal and so L is p-minimal. Also O_{p}(L) ≤ S ≤ M and hence O_{p}(L) ≤ ⋂ M^{L}.

Lemma 2.6. Let H be a finite group. Suppose H has a quasisimple normal subgroup K such that K/Z(K) is a sporadic simple group or U_{4}(3). Then H is not 2-minimal and so H is not A-minimal for any 2-subgroup A of H.

Proof. Replacing H by H/C_{H}(K) we may assume that C_{H}(K) = 1. Then K is simple and H is isomorphic to a subgroup of Aut(K) containing K. Let S ∈ Syl_{2}(H). Since in all cases Out(K) is a 2-group, H = KS.

Suppose that K is a sporadic simple group. Then the list of maximal subgroups of K (for example in the ATLAS) shows that there exist distinct S-invariant maximal parabolic subgroups M_{1}, M_{2} of K. Then M_{1}S, M_{2}S are distinct maximal subgroups of H and H is not 2-minimal.

Suppose K ≅ U_{4}(3). Then K ∩S is contained in exactly three maximal subgroup P_{1}, P_{2}, P_{3} of K. Moreover, we can choose notation such that P_{1}/Q_{1} ≅ Sym(3) × Sym(3), P_{2}/Q_{2} ≅ P_{3}/Q_{3} ≅ Alt(6) and P_{2} ∩ P_{3} ≰ P_{1}, where Q_{i} = O_{p}(P_{i}). If S normalizes P_{2} and P_{3}, then P_{2}S and P_{3}S are distinct maximal subgroups of H. If S interchanges P_{2} and P_{3}, then P_{1}S and (P_{2}∩P_{3})S are distinct maximal subgroups of H. So again H is not 2-minimal.

We have proved that H is not 2-minimal and thus by 2.5(h), H is also not A-minimal for any 2-subgroup A of H.

The next lemma establishes some basic properties of the group L/O_{p}(L) if L is A-minimal for a p-subgroup A. For CK-groups L we will give the precise structure of L/O_{p}(L) in 2.9 under the hypothesis that L acts faithfully and A quadratically on an 𝔽_{p}L-module V. Moreover, in 2.10 we will give the structure of such a module V under the additional hypothesis that V is simple and one of the cases 2.9(1) and (2) holds.

Lemma 2.7. Let L be a finite group with O_{p}(L) = 1, and let 1 ≠ A ≤ L be a p-subgroup. Suppose that L is A-minimal with M being the maximal subgroup of L containing A. Then the following hold, where D := ⋂ M^{L} and H := O^{p}(L):

(a) H ≰ M and so L = HA and H ≰ D.

(b) If |A| = 2, then L ≅ D_{r^{n}}, where r is an odd prime and n ∈ ℤ^{+}.

(c) Let N ⊴ L. Then either N ≤ D, or L = NA, H ≤ N and N ≰ M.

(d) D is a p′-group, and D = Φ(L) = Φ(H).

(e) C_{L}(a) ≤ M for 1 ≠ a ∈ Z(A).

(f) H/D is the unique minimal normal subgroup of L/D.

(g) Suppose A is elementary abelian. Then one of the following holds:

1. H/D is simple and p||H/D|.

2. |A| = p, H is q-group, and H/D is an elementary abelian q-group for some prime q ≠ p.

Proof. (a): L/H is a p-group and so nilpotent. Thus (a) follows from 2.5(g).

(b): By 2.5(c) L is generated by two conjugates of A and so is a dihedral group of order 2m, and m is the power of an odd prime since L is A-minimal.

(c): If N ≤ M, then N ≤ D since N ⊴ L. So suppose N ≰ M. Then L = NA, L/N is a p-group and H = O^{p}(L) ≤ N.

(d): Let Q ∈ Syl_{p}(D) with A ≤ N_{L}(Q). Then by the Frattini argument L = DN_{L}(Q). Since D ≤ M this implies N_{L}(Q) ≰ M and so L = N_{L}(Q). Hence Q ≤ O_{p}(L) = 1 and D is a p′-group.

Suppose that there exists a maximal subgroup M̃ of L not containing D. Then L = M̃D, so M̃ contains a Sylow p-subgroup of L. Thus we may assume that A ≤ M̃ and so M = M̃ and D ≤ M̃, a contradiction. This shows that D ≤ Φ(L). Conversely Φ(L) ≤ M and so D = Φ(L). Since O_{p}(L) = 1, 1.9 gives Φ(L) = Φ(O^{p}(L)) = Φ(H), and (d) is proved.

(e): Observe that A ≤ C_{L}(a) and C_{L}(a) ≠ L.

(f): By (d) D ≤ H and by (a), H/D ≠ 1. Hence (f) is a direct consequence of (c).

(g): Assume first that D = 1. By (f) H is a minimal normal subgroup of L. So either H is an elementary abelian q-group (q a prime, q ≠ p) or H is a direct product of non-abelian simple groups. In the first case |A| = p since otherwise (e) and coprime action imply that H = ⟨C_{H}(a) | 1 ≠ a ∈ A⟩ ≤ M, a contradiction to (a). Thus (g:2) holds in this case.

In the second case let K be a component of H. Note that any A-invariant Sylow subgroup of H is contained in M and so by coprime action H is not a p′-group. By (a) we have L = HA. Thus H = ⟨K^{L}⟩ = ⟨K^{A}⟩ and K is not a p′-group. In particular, if A normalizes K we are done since then (g:1) holds. Thus we may assume that there exists a ∈ A with K^{a} ≠ K. Put

E := ⟨ ∏_{t∈⟨a⟩} k^{t} | k ∈ K⟩.

Then E ≤ C_{L}(a) and by (e) E ≤ M. On the other hand, by 2.5(h) M contains a Sylow p-subgroup of L and since K is not a p′-group, M ∩ K ≠ 1. It follows that

[M ∩ K, E] = [M ∩ K, K] = K ≤ M.

Hence H = ⟨K^{A}⟩ ≤ M, which contradicts (a). So (g) holds if D = 1.

Assume next that H/D is a q-group for some prime q (not necessarily distinct from p). Then H = DQ for Q ∈ Syl_{q}(H) and since D = Φ(H), H = Q. So H is a q-group and since O_{p}(L) = 1, q ≠ p.

In the general case we conclude that H/D is not a p-group, and since by (f) H/D is the unique minimal normal subgroup of L/D, O_{p}(L/D) = 1. If |AD/D| = p, then also |A| = p since by (d), A ∩ D = 1. By 2.5(f), \overline{L} := L/D is \overline{A}-minimal with \overline{M} being the maximal subgroup containing \overline{A}.

In particular, ⋂ \overline{M}\overline{L} = 1 and we can apply (g) to \overline{L}. We conclude that (g) also holds for L.

Notation 2.8. A finite group L is called a CK-group provided that each composition factor of L is isomorphic to one of the known finite simple groups.^{1}

Lemma 2.9 (Quadratic L-Lemma). Let L be a CK-group and V be a faithful 𝔽_{p}L-module. Suppose that there exists A ≤ L such that A acts quadratically on V and that L is AO_{p}(L)-minimal. Then one of the following holds for \overline{L} = L/O_{p}(L):

1. \overline{L} ≅ SL_{2}(p^{k}).

2. p = 2 and \overline{L} ≅ Sz(2^{k}), k > 1.

3. p = 2 and \overline{L} ≅ D_{r^{k}}, r an odd prime.

Proof. Observe that the faithful quadratic action of A implies that A is an elementary abelian psubgroup of L. Since by coprime action O_{p}(L) is the intersection of the centralizers in L of the non-trivial factors of an L-composition series of V, we also may replace V by the direct sum of the non-trivial L-composition factors on V and L by \overline{L}, so

1^{∘}. O_{p}(L) = 1, and V is a direct sum of non-trivial simple 𝔽_{p}L-modules. In particular, C_{V} (L) = 0 and V = [V, L].

We use the following notation: M is the unique maximal subgroup of L containing A, D := ⋂ M^{L}, H := O^{p}(L), and A ≤ T ∈ Syl_{p}(L). Recall from 2.5(h) that N_{L}(T) ≤ M and from 2.7(d) that

2^{∘}. D = Φ(L) = Φ(H) is a p′-group.

Suppose that |A| = 2. Then 2.7(b) implies (3). Thus we may assume:

3^{∘}. |A| > 2.

1This notion is weaker than that of a K-group often used in the literature since it does not assume anything about sections.

By 2.5(e) we may assume without loss that A is a maximal quadratic subgroup of L, so 4^{∘}. A = C_{L}([V, A]) ∩ C_{L}(V/[V, A]).

Next we prove: 5^{∘}. [V, a] = [V, A] = C_{V} (A) = C_{V} (a) for all 1 ≠ a ∈ A and |V | = |[V, b]|^{2} for all quadratic elements b ∈ L^{#}.

Let 1 ≠ b be a quadratic element in L. Since D is a p′-group there exists g ∈ L with b^{g} ∉ M. Then L = ⟨A, b^{g}⟩ and so by (1^{∘}) V = [V, L] = [V, A] + [V, b^{g}] and C_{V} (A) ∩ C_{V} (b^{g}) = C_{V} (L) = 0. Since [V, A] ≤ C_{V} (A) and [V, b] ≤ C_{V} (b) we conclude that V = [V, A] ⊕ [V, b^{g}], C_{V} (A) = [V, A] and C_{V} (b) = [V, b]. In particular,

|[V, b]| = |V/C_{V} (b)| = |V/[V, b]| and |V | = |[V, b]|^{2}.

For a = b ∈ A we also get

[V, a] ≤ [V, A] ≤ C_{V} (A) ≤ C_{V} (a) = [V, a].

Now (5^{∘}) follows. 6^{∘}. A ∩ A^{h} = 1 for all h ∈ L \ N_{L}(A).

Let h ∈ L and 1 ≠ b ∈ A ∩ A^{h}. By (5^{∘}) [V, A] = [V, b] = [V, A^{h}], and so by (4^{∘}) A = A^{h}. 7^{∘}. C_{L}(a) ≤ N_{L}(A) for all 1 ≠ a ∈ A

This follows from (6^{∘}). 8^{∘}. A ∩ M^{h} = 1 for every h ∈ L \ M.

Let h ∈ L with B := A∩M^{h} ≠ 1, and let g ∈ L such that B ≤ T^{g} ≤ M^{h}. Then C_{V} (T^{g}) ≤ C_{V} (B) and so by (5^{∘}) C_{V} (T^{g}) ≤ C_{V} (A). Since C_{V} (L) = 0 by (1^{∘}) we get ⟨A, T^{g}⟩ ≠ L. Thus T^{g} ≤ M and so A^{g} ≤ T^{g} ≤ M ∩ M^{h}. Since A^{g} lies in a unique maximal subgroup of L this gives M = M^{h} and so by 2.5(b), h ∈ M, a contradiction. 9^{∘}. If V is is not a simple 𝔽_{p}L-module, then (1) or (2) hold.

Suppose that V is not simple. Let W be any simple 𝔽_{p}L-submodule of V and put E = C_{L}(W). Since C_{V} (L) = 0, also C_{V} (H) = 0 and so [W, H] ≠ 0. Hence, H ≰ E and by 2.7(c)

E ≤ D ≤ Φ(L) ≤ O_{p′}(L).

In particular, E is a p′-group and A acts faithfully on W. By induction on dim V and by (3^{∘}) we get that L/E ≅ SL_{2}(q) or Sz(q), q = p^{k} > 2. It follows that Φ(L/E) ≤ Z(L/E) and so [Φ(L), L] ≤ E = C_{L}(W). Since this holds for all simple 𝔽_{p}L-submodules of V, (1^{∘}) and the faithful action of L on V yield [L, Φ(L)] = 1. Thus E ≤ Φ(L) ≤ Z(L). By 2.5, L = L′A and since E is a p′-group, so E ≤ L′. By [Gr] the p′-part of the Schur multipliers of SL_{2}(q) and Sz(q) is trivial. Hence E = 1 and (1) or (2) holds. 10^{∘}. Suppose that p is odd. Then (1) holds.

By (5^{∘})

dim[V, A] = min{dim[V, b] | 1 ≠ b ∈ L, [V, b, b] = 0}

and by 2.5(c) L is generated by two conjugates of A. Observe that A acts quadratically on V satisfying the above minimality condition for dim[V, A]. By (9^{∘}) we may assume that V is a simple 𝔽_{p}L-module. Hence [Ho1, 4.3] (for |A| = 3), [Ho2, 4.2] (for p = 3 and |A| > 3) and [Ho3, 2.6] (for p > 3) show that L ≅ SL_{2}(p^{k}).

We may assume from now on that p = 2.

11^{∘}. D = Z(L).

Since by (2^{∘}) D is a 2′-group, (3^{∘}) and coprime action show that

D = ⟨C_{D}(a) | 1 ≠ a ∈ A⟩ ≤ N_{L}(A).

Then [D, A] ≤ D ∩ A = 1 and thus D ≤ Z(L) since L = ⟨A^{L}⟩. Conversely, Z(L) ≤ D, and (11^{∘}) follows.

12^{∘}. H is quasisimple.

By (11^{∘}) and (2^{∘}) Z(L) = D = Φ(H) ≤ H. Put K := H′. Since H/Z(L) is simple by 2.7(g) and (3^{∘}), we conclude that H = Z(L)K and K is quasisimple. Since L/Z(L)K is a 2-group, L/K is nilpotent and so by 2.5(e) L = KA and H ≤ K; in particular, H = K and H is quasisimple.

13^{∘}. Suppose that p = 2. Then (1) or (2) holds.

Put L̃ := L/Z(L). Recall that H is quasisimple by (12^{∘}). We discuss the possibilities for H using that H̃ is a known simple group.

Suppose that H̃ is a group of Lie-type over a field in characteristic 2. Put S := T ∩ H and let Δ be the set of minimal Lie-parabolic subgroups of H̃ containing S̃. Since H ≰ M, we can choose P̃ ∈ Δ with P̃ ≰ M̃. Then L̃ = ⟨P̃, Ã⟩ and so H̃ = ⟨P̃ Ã⟩. It follows that A acts transitively on Δ and |Δ| ≤ 2. In particular, N_{L}(S) is a maximal subgroup of L containing A, and so M = N_{L}(S).

Suppose that H̃ ≅ SL_{2}(2^{k}). If A ≤ H, then (1) holds. So suppose that A ≰ H and let a ∈ A \ S. Since Out(H̃) is cyclic and A is elementary abelian, |AH/H| = 2, k = 2l for some l ∈ ℤ^{+}, and S is elementary abelian with |C_{S}(a)| = |C_{S}(A)| = 2^{l}. Thus all involutions in T \ S are conjugate under S. Since S acts regularly on S^{L} \ {S}, a Frattini argument gives M = S(M ∩ M^{x}) and S ∩ M^{x} = 1, where x ∈ L \ M. Thus T = S(T ∩ M^{x}) and T ∩ M^{x} has order 2. Since all involutions in T \ S are conjugate under S, we can choose x such that A ∩ M^{x} has order 2. But this contradicts (8^{∘}).

Suppose next that H̃ ≅ Sz(2^{k}). Then H has no outer automorphism of order 2, so A ≤ H and (2) holds.

Finally suppose that H̃ ≅ U_{3}(2^{k}), L_{3}(2^{k}) or Sp_{4}(2^{k}). Assume that A ∩ Z(S) ≠ 1. Then S ≤ C_{L}(Z(S) ∩ A) ≤ N_{L}(A) by (7^{∘}), so [S, A] is elementary abelian. The action of T on S shows that A ≤ S ≤ H. Hence H = L, and since A acts transitively on Δ, |Δ| = 1 and L/Z(L) ≅ U_{3}(2^{k}). It follows that A ≤ Ω_{1}(S) = Ω_{1}Z(S), and there exists U ≤ L with A ≤ Ω_{1}Z(S) ≤ U and U ≅ SL_{2}(2^{k}). But then U ≰ M = N_{L}(S) and U = L, a contradiction.

We have shown that A ∩ Z(S) = 1. Suppose that A ∩ H ≠ 1 and let 1 ≠ a ∈ A ∩ H. Since all involutions of H are 2-central, C_{H}(a) contains a Sylow 2-subgroup S_{0} of H. Now (7^{∘}) implies S_{0} = S since M = N_{L}(S), and so a ∈ A∩Z(S) = 1, a contradiction. Hence A∩H = 1. On the other hand |A| ≥ 4 by (3^{∘}), and so, since the Sylow 2-subgroups of Out(U_{3}(2^{k})) are cyclic, H̃ ≅ L_{3}(2^{k}) or Sp_{4}(2^{k}) and A_{0} := C_{A}(Δ) ≠ 1.

Let R be a root subgroup of H contained in Z(S). Then X := C_{R}(A_{0}) ≠ 1 and again by (7^{∘}), [X, A] ≤ H ∩ A = 1. Thus A ≤ C_{L}(X) ≤ N_{L}(R). This rules out the Sp_{4}(q)-case, so H/Z(H) ≅ L_{3}(q). It follows that R = Z(S) and |A/C_{A}(R)| = 2. But then again by (7^{∘}), 1 ≠ [R, A] ≤ [C_{H}(C_{A}(R)), A] ≤ A ∩ H = 1 a contradiction.

Suppose now that H̃ is not a group of Lie-Type over a field in characteristic 2. Since L is A-minimal, 2.6 shows that H̃ is not a sporadic group. If H̃ is a group of Lie Type over a field of odd characteristic, then by [MeSt1] H̃ ≅ U_{4}(3). But then L is not 2-minimal by 2.6, again a contradiction.

It remains to consider the case H̃ ≅ Alt(n), n ≥ 7. Since L is 2-minimal, n = 2^{k} + 1, with k ≥ 3 and { \atop M ∩ H} ≅ Alt(2^{k}). Let X be a non-trivial orbit for A on \{1, …, n\}. If |X| ≠ 2^{k}, then A ≤ N_{L}(X) ≰ M, a contradiction. So |X| = 2^{k}. Since A is elementary abelian we conclude that A acts regularly on X, so |A| = 2^{k} and Ã ≤ H̃. Moreover, Ã is uniquely determined in H̃ up to conjugation under Sym(n). It follows that Ã is contained in a subgroup H̃_{0} of H̃ isomorphic to L_{2}(2^{k}). But H̃_{0} ≰ M̃. This final contradiction completes the proof of 2.9.

Lemma 2.10. Let L ≅ SL_{2}(q) or Sz(q), q = p^{k}, where p = 2 in the latter case, and let V be a non-trivial simple 𝔽_{p}L-module. Suppose that L is A-minimal for some A ≤ L with [V, A, A] = 0. Then V is a corresponding natural module.

Proof. Put 𝕂 := End_{L}(V). Then 𝕂 is (isomorphic to) a subfield of 𝔽_{q}. Put W := 𝔽_{q} ⊗_{𝕂} V. Then W is a simple 𝔽_{q}L-module and [W, A, A] = 0. Let N be the natural 𝔽_{q}L-module. Then N is the only non-trivial basic module for L and so by Steinberg’s Tensor Product Theorem [St, 1.31], W ≅ ⨂_{σ∈Σ} N^{σ}, where Σ is a subset of Aut(𝔽_{q}). Since V is non-trivial, Σ ≠ ∅.

Suppose for a contradiction that |Σ| ≥ 2. Then q > 2 and since L is A-minimal, |A| ≥ 3. Then by [MS3, 6.5] |Σ| = 2. Let Σ = {μ, σ}. Replacing W by W^{μ^{−1}} we may assume that μ = id_{𝔽_{q}}. By [MS3, 4.9] A acts λ-dependent on N^{μ} = N and on N^{σ}, where λ is a homomorphism from A to (𝔽_{q}, +). Since A acts λ-dependent on N, A acts λ ∘ σ-dependent on N^{σ}. Thus λ = λ ∘ σ. Let 𝔽_{q_{0}} be the fixed-field of σ. It follows that A is contained in a subgroup of L isomorphic to SL_{2}(q_{0}) and Sz(q_{0}), respectively, which contradicts the assumption that L is A-minimal.

Thus Σ = {σ} for some σ in Aut(𝔽). Since N is a simple 𝔽_{p}L-module we conclude that W, N^{σ}, N and V all are isomorphic as 𝔽_{p}L-modules.

Lemma 2.11. Let H be a finite group, L⊴H and M a parabolic subgroup of H with H = LM. Let V be a finite 𝔽_{p}H-module and W an 𝔽_{p}M-submodule of V. Suppose that [X/Y, L] = 0 whenever X/Y is a simple 𝔽_{p}H-section of V with X ∩ W ≰ Y. Then [W, O^{p}(L)] = 0, and W is an H-submodule of V.

Proof. Note that L = (M ∩ L)O^{p}(L) since H = LM and M is parabolic. Hence [W, O^{p}(L)] = 0 implies that W is an H-module. Thus it suffices to show that [W, O^{p}(L)] = 0. We do this using induction on dim V.

We may assume V ≠ 0 and so V has a minimal (non-zero) H-submodule U. Then W + U/U and V/U in place of W and V satisfy the hypothesis. Thus by induction W + U is a H-submodule of V and [W, O^{p}(L)] ≤ U.

Suppose first that W ∩U ≠ 0. As U is a simple 𝔽_{p}H-section of V, we conclude that [U, O^{p}(L)] = 0. Thus [W, O^{p}(L)] = [W, O^{p}(L), O^{p}(L)] ≤ [U, O^{p}(L)] = 0.

Suppose next that W ∩ U = 0. Again since M is parabolic, Gaschütz’ Theorem shows that there exists an H-submodule D in W + U with W + U = D ⊕ U. Since [W + U, O^{p}(L)] ≤ U we get that [D, O^{p}(L)] ≤ D ∩ U = 0. On the other hand W + U/D is a simple 𝔽_{p}H-section with W ≰ D, so by our hypothesis [W, O^{p}(L)] ≤ D. Now similarly as above

[W, O^{p}(L)] = [W, O^{p}(L), O^{p}(L)] ≤ [D, O^{p}(L)] = 0.

The next two results describes situations in which dual offenders arise in a natural way. Compare this, for example, with the case c = 1 in the qrc-Lemma [MS5, 4.6].

Lemma 2.12. Let H be a finite group, T ∈ Syl_{p}(H) and R := [O_{p}(H), O^{p}(H)], and let V be an 𝔽_{p}H-module and Y be a T -submodule of V with V = ⟨Y^{H}⟩ ≠ Y. Then one of the following holds:

1. [V, R] = 0 and C_{O_{p}(H)}(Y) ⊴ H.

2. R is a non-trivial strong dual offender on Y.

3. There exist O_{p}(H)O^{p}(H)-submodules Z_{1} ≤ X_{1} ≤ Z_{2} ≤ X_{2} such that for i = 1, 2, X_{i}/Z_{i} is a non-trivial simple O^{p}(H)-module and X_{i} ∩ Y ≰ Z_{i}.

Proof. Suppose first that [Y, R] = 0. Since R is normal in H and V = ⟨Y^{H}⟩ we get [V, R] = 0. Moreover, [C_{O_{p}(H)}(Y), O^{p}(H)] ≤ R ≤ C_{O_{p}(H)}(Y) and so H = T O^{p}(H) normalizes C_{O_{p}(H)}(Y). Thus (1) holds in this case. Hence, we may assume now:

1^{∘}. [Y, R] ≠ 0.

Let W be an O_{p}(H)O^{p}(H)-submodule of V. We want to apply 2.11 with H_{0} := O_{p}(H)O^{p}(H), O^{p}(H) and T ∩O_{p}(H)O^{p}(H) in place of H, L and M. If [V, O^{p}(H)] ≰ W, then also [Y, O^{p}(H)] ≰ W since V = ⟨Y^{H}⟩. Hence 2.11, applied to V/W and Y + W/W in the roles of V and Y, gives an H_{0}-chieffactor X_{2}/Z_{2} with W ≤ Z_{2} such that [X_{2}, O^{p}(H)] ≰ Z_{2} and Y ∩ X_{2} ≰ Z_{2}. If also [Y ∩ W, O^{p}(H)] ≠ 0, then 2.11, applied to W and Y ∩ W in the roles of Y, gives an H_{0}-chieffactor X_{1}/Z_{1} of W such that [X_{1}, O^{p}(H)] ≰ Z_{1} and Y ∩ X_{2} ≰ Z_{1}. Hence (3) holds. Thus we may assume:

2^{∘}. [V, O^{p}(H)] ≤ W or [Y ∩ W, O^{p}(H)] = 0 for every O_{p}(H)O^{p}(H)-submodule W.

For W_{0} := [V, O_{p}(H)] this either gives [V, O^{p}(H)] ≤ W_{0} or [W_{0} ∩ Y, O^{p}(H)] = 0. In the first case Y + W_{0} = V since V = ⟨Y^{H}⟩ = ⟨Y^{O^{p}(H)}⟩, and so V/Y = [V/Y, O_{p}(H)]. The nilpotent action of O_{p}(H) on V/Y gives V = Y, a contradiction. Hence

3^{∘}. [W_{0} ∩ Y, O^{p}(H)] = 0.

In particular, [Y, O_{p}(H)] ≤ C_{Y} (O^{p}(H)) ≤ C_{Y} (R). By (1^{∘}) we can pick y ∈ Y \C_{Y} (R). It follows that

4^{∘}. [y, O_{p}(H)] ≤ C_{Y} (O^{p}(H)) ≤ C_{Y} (R).

Put U := ⟨y^{H_{0}}⟩. Then [Y ∩ U, O^{p}(H)] ≠ 0 and thus by (2^{∘}):

5^{∘}. [V, O^{p}(H)] ≤ U. In particular, V = Y + U and [V, R] ≤ Y + [U, R].

By (4^{∘}) [y, R] = [⟨y^{O_{p}(H)}⟩, R] = ⟨[y, R]^{O_{p}(H)}⟩ and [y, R] ≤ C_{Y} (O^{p}(H)). Thus

[y, R] = ⟨[y, R]^{H_{0}}⟩ = [⟨y^{H_{0}}⟩, R] = [U, R].

Hence by (5^{∘}), [[V, O^{p}(H)], R] ≤ [U, R] = [y, R] ≤ Y, and again by (5^{∘}), [V, R] ≤ Y and so [V, R] ≤ W_{0} ∩ Y. Now (3^{∘}) implies [V, R, O^{p}(H)] = 0, and 2.1 gives [O^{p}(H), R, V] = [V, O^{p}(H), R]. Since by coprime action R = [R, O^{p}(H)],

[R, V] = [O^{p}(H), R, V] = [V, O^{p}(H), R] ≤ [y, R].

But then

[y, R] ≤ [V, R] ≤ [y, R].

We have shown that [Y, R] = [y, R] for all y ∈ Y \ C_{Y} (R), so R is a strong dual offender on Y, and (2) holds.

Corollary 2.13. Let H be a finite group, V a 𝔽_{p}H-module, M a maximal parabolic subgroup of H, Y an M-submodule of V, and R := [O_{p}(H), O^{p}(H)]. Suppose that there exists a p-subgroup A ≤ M such that

(i) A acts quadratically on V,

(ii) H is AO_{p}(H)-minimal, and

(iii) V = ⟨Y^{H}⟩ and Y ≠ V.

Then one of the following holds:

1. [V, R] = 0 and C_{O_{p}(H)}(Y) ⊴ H.

2. R is a non-trivial dual offender on Y.

3. Let g ∈ H \ M and A_{0} be an elementary abelian p-subgroup of H/O_{p}(H) of maximal order. Then |Y^{g}/C_{Y^{g}}(A)| ≥ |A_{0}|^{2} ≥ |AO_{p}(H)/O_{p}(H)|^{2}.

Proof. Since M is parabolic, M contains a Sylow p-subgroup T of H. In particular, Y is a T - submodule and O_{p}(H) ≤ M. Hence M is the unique maximal subgroup of H containing AO_{p}(H), and we can apply 2.12. If 2.12(1) or (2) holds, then (1) or (2) holds. So we may assume that 2.12(3) holds. Choose X_{i}, Z_{i}, 1 ≤ i ≤ 2, with the properties given in 2.12(3), and pick g ∈ H \ M.

Case 1. |A_{0}| = p.

Then |A/A∩O_{p}(H)| ≤ |A_{0}| = p by the maximality of A_{0}, so |A/A∩O_{p}(H)| = p since A ≰ O_{p}(H) by (ii). Put B := A^{g^{−1}} and let H_{0} be minimal in H with B ≤ H_{0} and H = O_{p}(H)H_{0}. Replacing A by a conjugate under O_{p}(H) we may assume that A ∈ B^{H_{0}}. Pick a ∈ A \ O_{p}(H). Then

⟨a, B⟩O_{p}(H) = ⟨A, A^{g^{−1}}⟩O_{p}(H) = H,

so H_{0} = ⟨a, B⟩ by the minimality of H_{0}.

If |Y/C_{Y} (B)| ≥ p^{2}, then (3) holds. So we may assume that |Y/C_{Y} (B)| ≤ p. Put D := C_{Y} (H_{0}), and let E be maximal in Y with [E, a] ≤ D. Then H_{0} = ⟨a, B⟩ normalizes C_{E}(B). Since A and B are conjugate in H_{0} we conclude that A centralizes C_{E}(B) and so C_{E}(B) = D. From |E/C_{E}(B)| ≤ |Y/C_{Y} (B)| = p we conclude |E/D| ≤ p. Since E/D = C_{Y/D}(a) and a acts quadratically on Y/D, we get [Y, a] ≤ E and

|Y/D| = |C_{Y/D}(a)||[Y/D, a]| ≤ |E/D|^{2} ≤ p^{2}.

Put L := O_{p}(H)O^{p}(H) and note that L = O_{p}(L)(L ∩ H_{0}). Thus X_{i}/Z_{i} is a non-trivial simple L ∩ H_{0}-module. It follows that X_{i} ∩ D ≤ Z_{i} for i = 1, 2. Since X_{1} ∩ Y ≰ Z_{1}, this gives

1^{∘}. X_{1} ∩ Y ≰ D.

Moreover, if X_{2} ∩ Y ≤ (Z_{2} ∩ Y) + D then X_{2} ∩ Y = (Z_{2} ∩ Y) + (X_{2} ∩ D) ≤ Z_{2}, a contradiction. Thus

2^{∘}. Y ≠ (Z_{2} ∩ Y) + D,

and D < D + (X_{1} ∩ Y) ≤ D + (Z_{2} ∩ Y) < Y. Hence |Y/D| = p^{2}, |E/D| = p, and E = [Y, A] + D = C_{Y} (A). In particular, there are p + 1 hyperplanes of Y containing D, one of which is C_{Y} (A). Hence A is transitive on the other p hyperplanes, one of which is C_{Y} (B). Therefore,

3^{∘}. Y = C_{Y} (A) ∪ ⋃ C_{Y} (B)^{A} = ⋃_{F ∈A^{H_{0}}} C_{Y} (F).

Set U := ⋂ Z_{2}^{H}. By (2^{∘}) and (3^{∘}) there exists F ∈ A^{H_{0}} with [(Z_{2} ∩ Y) + D, F] = 0. Hence

Z_{2} ∩ Y ≤ ⋂ Z_{2}^{F} = ⋂ Z_{2}^{LF} = ⋂ Z_{2}^{H} = U ≤ Z_{2}

and so

4^{∘}. Z_{2} ∩ Y = U ∩ Y; in particular Z_{2} ∩ Y is M-invariant.

Assume that [Z_{2}∩Y, B] = 0. Then by (4^{∘}), Z_{2}∩Y is invariant under H = ⟨M, B⟩. So Z_{2}∩Y ≤ D, a contradiction to X_{1} ∩ Y ≤ Z_{2} ∩ Y and (1^{∘}).

Thus [Z_{2} ∩Y, B] ≠ 0. Since Z_{2} ∩Y = U ∩Y and |Y/C_{Y} (B)| = p this gives Y = (U ∩Y)+C_{Y} (B). It follows that Y + U = C_{Y} (B) + U is invariant under ⟨M, B⟩ = H, and ⟨B^{H}⟩ centralizes Y + U/U. Thus

[X_{2} ∩ Y, O^{p}(H)] ≤ [Y, O^{p}(H)] ≤ U ≤ Z_{2},

a contradiction since X_{2} ∩Y ≰ Z_{2} and X_{2}/Z_{2} is a non-trivial simple O^{p}(H)-module. This completes the analysis of (Case 1).

Case 2. |A_{0}| > p.

Put q := |A_{0}|. Then by 2.9 H/O_{p}(H) ≅ SL_{2}(q) or Sz(q), and M is the normalizer of a Sylow p-subgroup of H. In particular H = O_{p}(H)O^{p}(H) and by 2.10 X_{i}/Z_{i} is a natural module. Set D_{i} := C_{(Y ∩X_{i})+Z_{i}/Z_{i}}(O_{p}(M)). Then D_{i} has order q. Since H = ⟨O_{p}(M), B⟩ we have C_{D_{i}}(B) = 0. Thus Y ∩ X_{i}/(Y ∩ Z_{i}) + C_{Y ∩X_{i}}(B) has order at least q and |Y/C_{Y} (B)| ≥ q^{2}, so (3) holds. This treats (Case 2) and the lemma is proved.

3 Applications of the FF-Module Theorem

The purpose of this section is to derive some useful corollaries from GFMT and FMT. One idea is to eliminate some or most of the cases of FMT assuming that the offenders or the modules have particular properties. The properties we come up with play an important role in the classification program of the finite simple groups of characteristic p, but are also of independent interest. A typical example is the Strong Dual Offender Theorem below which can be applied in the situation described in 2.12.

As already mentioned in the introduction, we assume the reader to familiar with the notation used in GFMT and FMT.

Theorem 3.1 (Strong Dual FF-Module Theorem). Let M be a finite CK-group such that K := F^{∗}(M) is quasisimple, and let V be a faithful simple 𝔽_{p}M-module. Suppose that A ≤ M is a strong dual offender on V and M = ⟨A^{M}⟩. Then one of the following holds, where q is a power of p:

1. M ≅ SL_{n}(q) or Sp_{2n}(q), and V is a corresponding natural module.

2. p = 2, M ≅ Alt(6) or Alt(7), V is a spin-module of order 2^{4} and A ≅ ⟨(12)(34), (13)(24)⟩. (Note that in the Alt(6) case, V might also be viewed as an natural Alt(6)-module with A ≅ ⟨(12)(34), (34)(56)⟩).

3. p = 2, M ≅ O_{2n}^{ϵ}(2) or Sym(n), V is the corresponding natural module, and |A| = 2.

Proof. By [MS4, 1.5(c)] A is a quadratic best offender on V. Hence M = J_{M}(V) and K is a Jcomponent of M, so M satisfies the hypothesis of FMT. Let V^{∗} be the dual of V. By [MS4, 1.5 (c)], A is also an offender on V^{∗}. By [MS4, 1.8(a) -(c)] |V^{∗}/C_{V^{∗}}(A)| = |[V, A]| and so

(∗)  |[V, A]| ≤ |A|.

We now discuss the cases given in FMT.

Case FMT (1),(2): M ≅ SL_{n}(q) or Sp_{2n}(q), and V is a natural module.

Then (1) holds.

Case FMT (3): M ≅ SU_{n}(q), and V a is natural module.

Let f be the unitary form on V (with corresponding field automorphism α) left invariant by M. Pick 0 ≠ u ∈ [V, A] and choose v ∈ V such that f(u, v) ≠ 0 and f(u, v) + f(u, v)α ≠ 0. Observe that by [MS4, 3.2(c)] C_{V} (A) = [V, A]^{⊥}, so v ∉ C_{V} (A). Since A is a strong dual offender, there exists a ∈ A such that [v, a] = u. Hence

f(v, v) = f(va, va) = f(v + u, v + u) = f(v, v) + f(u, u) + f(v, u) + f(v, u)α.

It follows that f(u, u) ≠ 0. On the other hand, by [MS4, 3.2(e)] [V, A] is isotropic and thus f(u, u) = 0, a contradiction.

Case FMT (4): M ≅ Ω_{n}(q)^{ϵ} or p = 2 and M ≅ O_{2n}^{ϵ}(q), and V is a natural module.

Let h be the quadratic form on V left invariant by M and s be the corresponding symmetric form. Suppose there exists a singular 0 ≠ u ∈ [V, A]. Choose v ∈ V such that s(u, v) ≠ 0. As in the previous case there exists a ∈ A such that [v, a] = u. Hence

h(v) = h(va) = h(v + u) = h(v) + h(u) + s(v, u) = h(v) + s(v, u),

so s(v, u) = 0, a contradiction.

We have shown that [V, A] is totally non-singular. On the other hand, by [MS4, 3.2(e)] [V, A] is isotropic. Hence [MS4, 3.1] shows that dim_{𝕂}[V, A] = 1, and by [MS4, 3.4(e)] |A| = 2. As A is an offender, also q = 2, and (3) holds.

Case FMT (5): p = 2, M ≅ G_{2}(q) and V is a natural module.

Let A ≤ T ∈ Syl_{2}(M). We have |A| = |V/C_{V} (A)| = |C_{V} (A)| = q^{3}, C_{V} (A) = [V, A], and Z(T) ≤ C_{G}(A) ≤ A. Note that N_{M}(A) and N_{M}(Z(T)) are distinct maximal Lie parabolic subgroups of M containing T. Thus N_{M}(Z(T)) does not normalize C_{V} (A) and so C_{V} (Z(T)) ≠ C_{V} (A). Hence A is not a strong offender. On the other hand, A is a strong dual offender with |[V, A]| = |C_{V} (A)| = q^{3} = |A| and so by [MS4, 1.5(d)], A is a strong offender, a contradiction.

Case FMT (6): M ≅ SL_{n}(q), n ≥ 5, and V is the exterior square of a natural module.

In this case [V, A] = C_{V} (A) and |V/C_{V} (A)| = |A| = q^{n−1}. Thus |[V, A]| = ^{\frac{q^{\binom{n}{2}}}{q^{n−1}}} = q^{\binom{n−1}{2}} and so

by (∗), q^{\binom{n−1}{2}} ≤ q^{n−1}, a contradiction to n ≥ 5.

Case FMT (7): M ≅ Spin_{7}(q), and V is a spin module.

Then |A| ≥ q^{4} and there exists an M-invariant quadratic form on V. By the orthogonal case treated above |A| = 2, a contradiction.

Case FMT (8): M ≅ Spin_{10}^{+}(q), and V is a half-spin module.

A similar argument as in the G_{2}(q)-case gives a contradiction.

Case FMT (9): M ≅ 3.Alt(6) and |V | = 2^{6}.

We have |V/C_{V} (A)| = |A| = 4 and |[V, A]| = 16, a contradiction to (∗).

Case FMT (10): M ≅ Alt(7) and |V | = 2^{4}.

In this case (2) holds.

Case FMT (11),(12): M ≅ Alt(n) or Sym(n), and V is a natural module.

Let W be the permutation module with basis (v_{1}, v_{2}, …, v_{n}), W̃ := W/C_{W} (M) and W_{0} := [W, M]. Then V ≅ W̃_{0}. So A is a strong dual offender on W̃_{0}.

Suppose first that n ≠ 2k. Then [v_{1} +v_{n}, A] = ⟨v_{1} +v_{2}⟩ and so [W̃_{0}, A] = [ṽ_{1} +ṽ_{n}, A] = ⟨ṽ_{1} +ṽ_{2}⟩. Thus k = 1 and (3) holds.

Suppose next that n = 2k. Then we are in one of the cases (12) (1) – (4) of FMT. In case (12)

(3) A does not act quadratically on V, so this case is not possible.

Assume case (12)(4). Then n = 8 = |A|. So M ≅ Ω_{6}^{+}(2) or O_{6}^{+}(2) and V is the corresponding natural module, so the Case FMT (4) treated above gives a contradiction.

Thus, we are in case (12)(1) or (12)(2). In both cases all orbits of A on {1, 2, …, n} have length 1 or 2. Say the non-trivial orbits of A are {1, 2}, {3, 4}, …, {2k −1, 2k}. Then [W̃_{0}, A] = [ṽ_{1} +ṽ_{3}, A] ≤ ⟨ṽ_{1} + ṽ_{2}, ṽ_{3} + ṽ_{4}⟩. It follows that k = 3 and n = 6. If A ≤ M′ then M ≅ Alt(6) and (2) holds; and if A ≰ M′ then M ≅ Sym(6) ≅ Sp_{4}(2) and (1) holds.

Theorem 3.2 (Strong FF-Module Theorem). Let M be a finite CK-group such that K := F^{∗}(M) is quasisimple, and let V be a faithful simple 𝔽_{p}K-module. Suppose that A ≤ M is a strong offender on V and M = ⟨A^{M}⟩. Then one of the following holds, where q is a power of p:

1. M ≅ SL_{n}(q) or Sp_{2n}(q) and V is a corresponding natural module.

2. p = 2, M ≅ Alt(6), 3.Alt(6) or Alt(7), |V | = 2^{4}, 2^{6} or 2^{4}, respectively, and |A| = 4.

3. p = 2, M ≅ O_{2n}^{ϵ}(2) or Sym(n), V is a corresponding natural module, and |A| = 2.

Proof. By [MS4, 1.6] A is a best offender on V and so we can apply FMT again.

Assume first that V and V^{∗} are isomorphic as 𝔽_{p}M-modules. By [MS4, 1.7] A is a strong dual offender on V and we are done by 3.1.

So we may assume that V is not a selfdual 𝔽_{p}M-module. In particular, (see for example [MS4, 3.2(a)]) there does not exists a non-degenerated M-invariant symplectic, symmetric or unitary form on V. This excludes all cases but FMT (1),(6),(8),(9) and (10). The first and the last two of these remaining cases appear in the theorem.

Suppose V is the exterior square of a natural SL_{n}(q)-module W for n ≥ 5. Then A = C_{M}(U) for some hyperplane U of W. Let 1 ≠ t, s ∈ A, x ∈ W \ U, y := [x, t] and z := [x, s]. Since n > 2 we can choose s such that 𝔽_{q}y ≠ 𝔽_{q}z. Note that y ∈ U ≤ C_{W} (A) and so

[x ∧ y, t] = y ∧ y = 0 and [x ∧ y, s] = z ∧ y ≠ 0.

Thus C_{V} (t) ≠ C_{V} (A), a contradiction.

Suppose that M ≅ Spin_{10}^{+}(q) and V is the half-spin module. Let T ≤ A be a long root subgroup. Then N_{M}(T) and N_{M}(A) are distinct maximal Lie-parabolic subgroups of M and so C_{V} (T) ≠ C_{V} (A), a contradiction.

Definition 3.3. Let M be a finite group and V faithful 𝔽_{p}M-module. Then AP_{M}(V) is the set of non-trivial best offenders A of M on V such that A ≤ O_{p}(P) for some point-stabilizer P of M on V.

The next lemma will be used in Theorem 3.5 to show that there are no over-offenders in the radical of a point-stabilizer. This fact can be used nicely. For example, in pushing up problems for the radical of a point stabilizer.

Lemma 3.4. Let V be an 𝔽A-module and suppose that [V, A] is 1-dimensional over 𝔽. If A is an offender on V, then |V/C_{V} (A)| = |A/C_{A}(V)| and the canonical commutator map A/C_{A}(V) → Hom_{𝔽}(V/C_{V} (A), [V, A]) is an isomorphism.

Proof. Clearly the commutator map is an injective homomorphism. Since [V, A] is 1-dimensional

|Hom_{𝔽}(V/C_{V} (A), [V, A])| = |V/C_{V} (A)|.

Hence if A is an offender on V, then |V/C_{V} (A)| ≤ |A/C_{A}(V)| and the lemma holds.

Theorem 3.5 (Point-Stabilizer Theorem). Let M be a finite CK-group with O_{p}(M) = 1 and let V be a faithful 𝔽_{p}M-module. Suppose that M = ⟨AP_{M}(V)⟩ and that there exists a J_{M}(V)- component K with V = [V, K] and C_{V} (K) = 0. Let A ∈ AP_{M}(V) and let P be a point-stabilizer for M on V with A ≤ O_{p}(P). Then the following hold:

(a) M ≅ SL_{n}(q), Sp_{2n}(q), G_{2}(q) or Sym(n), q a power of p, where p = 2 in the last two cases, and n ≡ 2, 3 (mod 4) in the last case.

(b) V is a corresponding natural module.

(c) Put 𝔽 := End_{M}(V), q := |𝔽| and Z := C_{V} (P). Then Z is 1-dimensional over 𝔽, and one of the following holds:

1. M ≅ SL_{n}(q), [V, A] = Z, and A = C_{M}(C_{V} (A)) ∩ C_{M}(V/Z).

2. M ≅ Sp_{2n}(q), Z ≤ [V, A] ≤ Z^{⊥}, and A = C_{M}(C_{V} (A)) ∩ C_{M}(Z^{⊥}/Z).

3. M ≅ G_{2}(q), [V, A] = C_{V} (A), |V/C_{V} (A)| = |A| = q^{3}, and A ⊴ P.

4. M ≅ Sym(n), n ≡ 2, 3 (mod 4), n > 6, |A| = 2, and A ⊴ P.

(d) |V/C_{V} (A)| = |A|, and V is a simple 𝔽_{p}K-module.

Proof. By [MS4, 2.2] either K ≅ C_{3} (and p = 2) or Q_{8} (and p = 3) and |V | = p^{2}, or K is quasisimple. In the first two cases M is isomorphic to SL_{2}(2) and SL_{2}(3), respectively, and the theorem follows easily. Thus, we may assume that K is quasisimple. Since V = [V, K], GFMT implies that K is the unique J-component of V and V is a semisimple 𝔽_{p}M-module. Let T ∈ Syl_{p}(P) and note that A ≤ O_{p}(P) ≤ T ∈ Syl_{p}(M).

Let V_{1} be simple M-submodule of V. Then by GFMT(h) V is either a direct sum of copies of V_{1}, or M ≅ SL_{n}(q), n ≥ 4, and V is a direct sum of copies of V_{1} and copies of V_{1}^{∗}.

Suppose for a contradiction that V is not a direct sum of copies of V_{1}. So M ≅ SL_{n}(q), n ≥ 4, V_{1} is a natural SL_{n}(q)-module, and there exists a simple 𝔽_{p}M submodule V_{2} of V with V_{2} ≅ V_{1}^{∗}. Let {1, 2} = {i, j}, U_{i} = C_{V_{i}}(T) and H_{i} = [V_{i}, T]. Note that H_{i} is the hyperplane of V_{i} corresponding the 1-space U_{j} in V_{j} ≅ V_{i}^{∗}. Then

P = P_{M}(V_{1} ⊕ V_{2}, T) = O^{p′}(C_{M}(U_{1}) ∩ C_{M}(U_{2})) = C_{M}(U_{1}) ∩ C_{M}(V_{1}/H_{1}) = C_{M}(U_{2}) ∩ C_{M}(V_{2}/H_{2}).

It follows that H_{i}/U_{i} is a simple P -module, and so [H_{i}, O_{p}(P)] = U_{i}, i = 1, 2. Recall that A ≤ O_{p}(P).

Let D ≤ A be a non-trivial minimal offender on V. Assume that [H_{i}, D] ≠ 0. Then [H_{i}, D] = U_{i}, and U_{i} is 1-dimensional over End_{K}(V_{i}). By 3.4 D is not an over-offender on H_{i}, so by [MS4, 1.3] C_{D}(H_{i}) = 1 and V = H_{i} + C_{V} (D). But this is absurd since V_{j} ≰ V_{i} + C_{V} (D).

Thus [H_{1} + H_{2}, D] = 0, so D centralizes a hyperplane in V_{1} and V_{1}^{∗}. Hence |D| ≤ q and |V/C_{V} (D)| ≥ q^{2}, and D is not an offender, a contradiction.

We have shown that V is the direct direct sum of copies of V_{1}. It follows that P_{M}(V_{1}, T) = P. So (M, V_{1}, A) satisfies the hypothesis of Theorem 3.5. If the theorem holds for (M, V_{1}, A), then |V_{1}/C_{V_{1}}(A)| = |A| and V = V_{1}C_{V} (A). Hence [V, K] ≤ [V, ⟨A^{M}⟩] ≤ V_{1} and V = V_{1}, and we are done.

Thus, it suffices to prove 3.5 for (M, V_{1}, A), in other words we are allowed to assume that V = V_{1}, so V is a simple M-module.

We now discuss the different possibilities for V and M as listed in FMT. Put U := C_{V} (T) and 𝔽 := End_{M}(V).

FMT (1): Suppose that V is a natural SL_{n}(q)-module. Then dim_{𝔽} U = 1 and U = [V, O_{p}(P)]. Hence 3.4 gives (c:1) and (d).

FMT (2)-(4): Suppose that V is a natural Sp_{2n}(q)-, Ω_{n}^{ϵ}(q)-,O_{n}^{ϵ}(q)- or SU_{n}(q)-module for M (with p = 2 in the O_{n}^{ϵ}(q)-case). Then |𝔽| = q in the first two cases and |𝔽| = q^{2} in the last. Moreover, from the structure of these modules we get the following elementary facts: U is a 1-dimensional singular 𝔽-subspace, [U^{⊥}, O_{p}(P)] ≤ U, and C_{O_{p}(P)}(U^{⊥}) has order q, 1, 1, and q, respectively. Moreover, C_{V} (a) ≤ U^{⊥} for all a ∈ O_{p}(P)^{♯}. Thus

|C_{A}(U^{⊥})| ≤ q ≤ |𝔽|,

and

|A| ≥ |V/C_{V} (A)| = |V/U^{⊥}||U^{⊥}/C_{U^{⊥}}(A)| = |𝔽||U^{⊥}/C_{U^{⊥}}(A)|.

Now by 3.4 with U^{⊥} in place of V,

|U^{⊥}/C_{U^{⊥}}(A)| ≥ |A/C_{A}(U^{⊥})| ≥ q^{−1}|A|.

We conclude that equality holds in the three preceding equations. Hence |C_{A}(U^{⊥})| = |𝔽|, which excludes the natural Ω_{n}^{ϵ}(q)-, O_{n}^{ϵ}(q)-, and SU_{n}(q)-module. Thus (c:2) and (d) hold.

FMT (5): Suppose that V is a natural G_{2}(q)-module for M. Then q^{3} = |V/C_{V} (A)| = |A|, C_{M}(A) = A and A is quadratic on V. Put D = Z(O_{p}(P)), then D ≤ C_{M}(A) ≤ A. Since |C_{V} (D)| = q^{3} we conclude that C_{V} (A) = C_{V} (D) and then A = C_{M}(C_{V} (D)) ⊴ P. So (c:3) holds.

FMT (6): Suppose V is the exterior square of a natural SL_{n}(q)-module W with n ≥ 5. Then [W, A] is a hyperplane of W. On the other hand, P/O_{p}(P) ≅ SL_{2}(q) × SL_{n−2}(q) and [W, O_{p}(P)] is 2-dimensional. Since A ≤ O_{p}(P) and n ≥ 5 we have a contradiction.

FMT (7),(8): Suppose that V is a spin module for Spin_{7}(q) or a half-spin module for Spin_{10}^{+}(q). Let W be the natural Ω_{7}(q)- and Ω_{10}^{+}(q)-module, respectively, so |W^{⊥}| = q if q is even in the Spin_{7}(q)-case, and W^{⊥} = 0 in all other cases.

We will show that |C_{W} (A)/W^{⊥}| ≤ q^{2}. For this let A^{∗} ≤ M be a maximal best offender with A ≤ A^{∗}. Then by FMT |A^{∗}/A| ≤ q and O^{p′}(N_{M}(A))/A^{∗} ≅ Sp_{4}(q) and Spin_{8}^{+}(q), respectively. Thus N_{M}(A) = N_{M}(U) for some 1-dimensional singular subspace U of W, and A ≤ A^{∗} ≤ O_{p}(N_{M}(A)). Hence [U^{⊥}, A] ≤ U and C_{W} (A) ≤ U^{⊥}. By 3.4 applied with V = U^{⊥},

|U^{⊥}/C_{U^{⊥}}(A)| ≥ |A/C_{A}(U^{⊥})|.

Suppose that M ≅ Spin_{7}(q) and p is odd. Then |C_{A}(U^{⊥})| = 1, |U^{⊥}| = q^{6}, |A| ≥ q^{4} and W^{⊥} = 0. Thus

q^{4} ≤ |A| ≤ |U^{⊥}/C_{U^{⊥}}(A)| and q^{4}|C_{U^{⊥}}(A)| ≤ |U^{⊥}| = q^{6},

so |C_{W} (A)/W^{⊥}| = |C_{W} (A)| = |C_{U^{⊥}}(A)| ≤ q^{2}.

Suppose that M ≅ Spin_{7}(q) and p = 2. Then |C_{A}(U^{⊥})| ≤ q, |U^{⊥}| = q^{6} and |A| ≥ q^{4} and |W^{⊥}| = q. Thus

q^{3} ≤ |A/C_{A}(U^{⊥})| ≤ |U^{⊥}/C_{U^{⊥}}(A)| and q^{3}|C_{U^{⊥}}(A)| ≤ |U^{⊥}| = q^{6},

so |C_{W} (A)/W^{⊥}| = |C_{U^{⊥}}(A)/W^{⊥}| ≤ q^{2}.

Suppose that M ≅ Spin_{10}^{+}(q). Then |C_{A}(U^{⊥})| = 1, |U^{⊥}| = q^{9}, |A| = |A^{∗}| = q^{8} and W^{⊥} = 0. Thus

q^{8}|C_{U^{⊥}}(A)| = |A||C_{U^{⊥}}(A)| ≤ |U^{⊥}| = q^{9},

and |C_{W} (A)/W^{⊥}| = |C_{W} (A)| = |C_{U^{⊥}}(A)| ≤ q ≤ q^{2}.

We prove that |C_{W} (A)/W^{⊥}| ≤ q^{2} in all cases. On the other hand C_{W} (O_{p}(P)) contains maximal singular subspace of W and A ≤ O_{p}(P), so |C_{W} (A)/W^{⊥}| ≥ q^{3} a contradiction.

FMT (9): Suppose that p = 2 and M ≅ 3.Alt(6). Then |[V, O_{p}(P)]| = 4 and |[V, A]| = 16, which contradicts A ≤ O_{p}(P).

FMT (10): Suppose that M ≅ Alt(7), p = 2 and |V | = 2^{4}. Then P ≅ SL_{3}(2) and O_{p}(P) = 1, which contradicts A ≤ O_{p}(P).

FMT (11),(12): Suppose that p = 2 and V is a natural Alt(n)-module for K. Assume first that n = 5 or 8. Then K ≅ Ω_{4}^{−}(2) and Ω_{8}^{+}(2), respectively, and M ≅ O_{4}^{−}(2) or O_{8}^{+}(2) if K ≠ M. Thus, these cases have been treated earlier in case FMT(2) – (4).

Hence, we may assume that n ≥ 6 and n ≠ 8. Let (Ω_{1}, …, Ω_{l}) be the orbits of T on Ω = {1, …, n}. Note that |Ω_{i}| ≠ |Ω_{j}| for i ≠ j, so we may assume that |Ω_{1}| < ⋯ < |Ω_{l}|.

Suppose first that l = 1. Then P ≅ (Sym(\frac{n}{2}) ≀ C_{2}) ∩ M, and since n ≠ 8 and n ≥ 6, O_{p}(P) = 1, which contradicts A ≤ O_{p}(P).

Suppose now that l ≠ 1. Then P ≅ (×_{i=1}^{l} Sym(Ω_{i})) ∩ M. Observe that O_{2}(Sym(Ω_{i})) = 1 unless |Ω_{i}| = 2 or 4. Thus |A| ≤ 8.

If A is generated by transpositions, then A = Sym(Ω_{j}) for some j ∈ {1, 2} and |A| = 2. Thus either j = 1 and n ≡ 2 mod 4, or |Ω_{1}| = 1, j = 2, and n ≡ 3 mod 4. Hence (c:4) holds in this case, and since |A| = 2 obviously also (d) holds.

If A is not generated by transpositions, then we are in case (12)(2) or (12)(3) of FMT. In both cases |V/C_{V} (A)| = |A|, so (d) holds. In the first case |A| = 2^{\frac{n}{2} −1} ≤ 8, and so n = 6, since n is even and n ≠ 8. Hence |A| = 4, and A is conjugate to ⟨(12), (34)(56)⟩. Then M ≅ Sym(6) ≅ Sp_{4}(2) and (c:2) holds.

In the second case we get |A| = 2^{\frac{n}{2}} ≤ 8 and so n = 6 and |A| = 8. But then A is conjugate to ⟨(12), (34)(56), (35)(46)⟩, and again (c:2) holds.

Theorem 3.6 (General Point-Stabilizer Theorem). Let M be a finite CK-group with O_{p}(M) = 1 and let V be a faithful 𝔽_{p}M-module. Put AP := AP_{M}(V) and suppose that AP ≠ ∅. Then there exists an M-invariant set N of subnormal subgroups of M such that the following hold:

(a) ⟨AP⟩ = × N, and N = ⟨A ∈ AP | A ≤ N⟩ for all N ∈ N.

(b) For all N_{1} ≠ N_{2} ∈ N, [V, N_{1}, N_{2}] = 0.

(c) Put \overline{V} = V/C_{V} (N)). Then [\overline{V} , N] = ⨁_{N∈N} [\overline{V} , N].

(d) Let N ∈ N. Then (N, [\overline{V} , N]) satisfies the hypothesis of 3.5 in place of (M, V).

(e) For all N ∈ N, C_{V} (N) = C_{V} (O^{p}(N)) and [V, O^{p}(N)] = [V, N].

(f) Let A ∈ AP. Then

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

(b) A = ×_{N∈N} A ∩ N,

(c) A ∩ N ∈ AP for all N ∈ N with A ∩ N ≠ 1.

Proof. Let A ∈ AP and let P be a point stabilizer of M on V with A ≤ O_{p}(P). Choose T ∈ Syl_{p}(P) with A ≤ T. Then P = P_{M}(V, T). Put M_{0} := ⟨AP⟩ and T_{0} := T ∩ M_{0}. Then by 1.7(a) P_{0} := P_{M_{0}}(V, T_{0}) ≤ P and so A ≤ O_{p}(P_{0}). Thus replacing M by M_{0} we may assume that M = ⟨AP⟩. In particular M = J_{M}(V).

Put

J := J_{M}(V), \overline{V} := V/C_{V} (J) and W := [V, J] + C_{V} (J)/C_{V} (J),

and for K ∈ J,

W_{K} := [W, K] and M̃:= M/C_{M}(W_{K}).

Note that A normalizes K and thus also W_{K}. An application of GFMT gives:

1^{∘}. O^{p}(M) = F^{∗}(M) = ⟨J ⟩, and W is a faithful semisimple 𝔽_{p}M-module.

2^{∘}. W = ⨁_{K∈J} W_{K} and so C_{W} (A) = ⨁_{K∈J} C_{W_{K}} (A).

3^{∘}. F^{∗}(M̃) = K̃ = O^{p}(M̃).

4^{∘}. A is a best offender on W.

By (4^{∘}) and [MS4, 1.2] A is also a best offender on W_{K}. Put P_{1} := P_{M}(\overline{V} , T). By 1.7(c) (with H = E := M) we have P_{1} ≤ P and A ≤ O_{p}(P_{1}). Put P_{2} := P_{KT} (\overline{V} , T). Then clearly P_{2} ≤ P_{1} and so A ≤ O_{p}(P_{2}). Since K = O^{p}(KT), 1.7(a) (with E := KT) gives P_{2} = P_{KT} (W_{K}, T). Since by (3^{∘}) K̃ = O^{p}(M̃) we have M̃ = K̃T̃ and so P̃_{2} = PM̃(W_{K}, T̃). So either Ã = 1 or Ã ∈ AP M̃(W_{K}). Moreover, the semisimplicity of W given in (1^{∘}) yields W_{K} = [W_{K}, K], C_{W_{K}} (K̃) = 0 and O_{p}(M̃) = 1. Thus:

5^{∘}. If Ã ≠ 1, then (M̃, W_{K}, Ã) satisfies the hypothesis of 3.5.

In particular, by 3.5(d):

6^{∘}. |W_{K}/C_{W_{K}} (A)| = |Ã| = |A/C_{A}(W_{K})|.

Note that (6^{∘}) also holds if Ã = 1.We now use the following additional notation for K ∈ J:

K^{⊥} := ∏_{E∈J \{K}} E, A_{K} := C_{A}([W, K^{⊥}]).

From (6^{∘}) and [MS4, 1.1] applied with W = {W_{K} | K ∈ J } we conclude

7^{∘}. |W/C_{A}(W)| = |A/C_{A}(W)| and A = ×_{K∈J} A_{K} = A_{K} × C_{A}(W_{K}).

Recall that A is a best offender on W_{K}. Since A = A_{K} × C_{A}(W_{K}) and W = W_{K} + C_{A}(A_{K}) we have

8^{∘}. A_{K} is a best offender on W_{K} and on W.

Since by (7^{∘}) |W/C_{W} (A)| = |A| and also |V/C_{V} (A)| ≤ |A|, we conclude that C_{V} (J) ≤ C_{V} (A) and \overline{C_{[V,J]}(A)} = C_{W} (A), and

9^{∘}. C_{V} (J) ≤ C_{V} (A) and \overline{C_{[V,J]}(A)} = C_{W} (A), and |V/C_{V} (A)| = |A|.

Let D be an offender in M on V such that D ≤ O_{p}(P). Let B ≤ D with |B||C_{V} (B)| maximal. Then B is a best offender on V. We claim that |B||C_{V} (B)| = |V |. If B = 1, this is obvious and if B ≠ 1 this follows from (9^{∘}) applied to B instead of A. Since D is an offender, |V | ≤ |D||C_{V} (D)| and so |B||C_{V} (B)| = |V | ≤ |D||C_{V} (D)|. Thus 10^{∘}. All offenders in O_{p}(P) are best offenders on V

From (9^{∘}) and M = ⟨AP⟩, we have C_{V} (J) = C_{V} (M) and |[V, K]+C_{V} (A)/C_{V} (A)| = |W/C_{W} (A)| = |A|, so V = [V, K] + C_{V} (A) since A is an offender on V.

Put

M_{K} := C_{M}([W, K^{⊥}]), and N := {M_{K} | K ∈ J },

so A_{K} ≤ M_{K} and by (2^{∘}) also K ≤ M_{K}. We will now show the statements of the conclusion for N.

Since K^{⊥} acts faithfully on [W, K^{⊥}], [M_{K}, K^{⊥}] = 1. Thus the Three Subgroups Lemma implies [V, K^{⊥}, M_{K}] = 1 and so V = [V, K] + [V, K^{⊥}] + C_{V} (A) = [V, K] + C_{V} (A_{K}). It follows that [V, K] = [V, M_{K}] and C_{V} (K) = C_{V} (M_{K}). This gives (b), (c) and (e). Moreover, (f:a) and (f:b) follow from (7^{∘}) and (9^{∘}), respectively. Note that C_{W_{K}} (A_{K}) = C_{W_{K}} (A). So (9^{∘}) implies that |[V, K]/C_{[V,K]}(A_{K})| = |W_{K}/C_{W_{K}} (A_{K})|. Since V = [V, K] + C_{V} (A_{K}) and A_{K} is an offender on W_{K} we conclude that A_{K} an offender on V. Note that A_{K} ≤ A ≤ O_{p}(P) and so by (10^{∘}) A_{K} is a best offender on V. Thus (f:c) holds.

In particular, M_{K} is generated by best offenders, so the second part of (a) holds. The first part of (a) follows from (b) and the fact that M acts faithfully on W.

Note that M_{K}C_{M}(W_{K}) = M. Hence (d) follows from (5^{∘}), and all parts of the conclusion are proved.

Corollary 3.7. Let M be a finite CK-group with O_{p}(M) = 1 and V be a faithful 𝔽_{p}M-module. Let A be an offender in M on V such that A ≤ O_{p}(P) for some point stabilizer P for M on V. Then A is a best offender.

Proof. This is Step 10^{∘} in 3.6.

Corollary 3.8. Let L ≅ SL_{n}(q), q = p^{k}, X be a natural 𝔽_{p}SL_{n}(q)-module for L, and P be a point stabilizer of L on X. Put X^{∗} = Hom_{𝔽_{p}}(X, 𝔽_{p}), the dual of X. Suppose that V is a 𝔽_{p}L-module with 0 ≠ V = [V, L] and that there exists 1 ≠ A ≤ O_{p}(P) with |V/C_{V} (A)| ≤ |A|. Then one of the following holds:

1. V = V_{1} ⊕ … ⊕ V_{m} where V_{1}, …, V_{m} are isomorphic to X^{∗} and 1 ≤ m < n.

2. V is isomorphic X.

3. A = O_{p}(P), n = 3, q = 2, |V | = 2^{4}, and V/C_{V} (L) ≅ X.

4. A = O_{p}(P), n = 2, q = 2^{k}, V/C_{V} (L) ≅ X, and V is a quotient of the natural Ω_{3}(q)-module.

5. A = O_{p}(P), n ≥ 4, and V is isomorphic to the exterior square of X^{∗} over 𝔽_{q}. Moreover, C_{V} (A) = [V, A], and as an 𝔽_{p}P -module A is a natural 𝔽_{p}SL_{n−1}(q)-module dual to X/C_{X}(A), V/C_{V} (A)) is isomorphic to A, and C_{V} (A) is isomorphic to the exterior square of A over 𝔽_{q}.

Proof. Observe that O_{p}(P) is elementary abelian, so A is an offender on V. Put \overline{V} := V/C_{V} (O^{p}(L)). Let 1 ≠ B ≤ A such that |B||C_{V} (B)| is maximal. Then B is a best offender on A. Note that either L is quasisimple or n = 2 and q = p ≤ 3. In any case we conclude that L = ⟨B^{L}⟩ and so L = J_{L}(V). Thus we can apply GFMT to L and V. In particular, since V = [V, L], \overline{V} is a semisimple 𝔽_{p}Lmodule. Let Y be a simple 𝔽_{p}L-submodule of \overline{V}. By GFMT(d) B is a best offender on \overline{V} and so by [MS4, 1.2] B is a best offender on Y. Thus we can apply FMT also to (L, Y). Only the cases FMT(1), (4) (with n = 3 and ϵ = +) and (6) apply to our situation, since Ω_{6}^{+}(q) ≅ L_{4}(q).

Suppose first that n ≥ 5 and Y is the exterior square of a natural SL_{n}(q)-module Z. Then by GFMT(g), Y = \overline{V}, and B is the centralizer of an 𝔽_{q}-hyperplane in Z. Since B ≤ A ≤ O_{p}(P) this shows that Z is dual to X and A = B = O_{p}(P). By [MS4, 6.1], C_{V} (L) = 0 and so (5) holds.

Suppose next that n = 4 and Y is a natural Ω_{6}^{+}(q)-module for L. Note that C_{Y} (O_{p}(P)) = [Y, O_{p}(P)] is a 3-dimensional singular subspace of Y. Thus also [Y, B] is singular. Since Y is an offender we conclude from [MS4, 3.4] that [Y, B] is 3-dimensional, A = B = O_{p}(P), and |Y/C_{Y} (B)| = |B| = q^{3}. Thus Y = \overline{V}. Suppose that C_{Y} (L) ≠ 0. Then by [MS4, 6.1], q = 2, and V is isomorphic to a submodule of the permutation module for SL_{4}(2) ≅ Alt(8). But B acts transitively on the eight points permuted by Alt(8), and so C_{Y} (B) = C_{Y} (L) and |Y/C_{Y} (B)| = 2^{6} > |B|, a contradiction. Thus C_{V} (L) = 0 and (5) holds.

Suppose that Y is isomorphic to X. Since A ≤ O_{p}(P) the Point Stabilizer Theorem 3.5 shows that |Y/C_{Y} (B)| = |B|. Thus \overline{V} = Y. If C_{V} (L) = 0, (2) holds. So suppose C_{V} (L) ≠ 0. Then by [MS4, 8.4] p = 2, |C_{Y} (L)| ≤ q, and either n = 2, or n = 3 and q = 2. In the first case (4) holds. So suppose n = 3 and q = 2. Then by [MS4, 8.4] |B| = 4 and A = B = O_{2}(L). Hence (3) holds.

It remains to treat the case where Y (and every other simple L-submodule of \overline{V}) is isomorphic to the dual of X, so \overline{V} is the direct sum of m natural modules dual to X, and by GFMT(g) n > 2. Then q^{m} = |\overline{V} /C\overline{V} (B)| ≤ |B| ≤ |O_{p}(P)| = q^{n−1} and so m ≤ n − 1. If C_{V} (L) = 0, then (1) holds. So suppose C_{V} (L) ≠ 0. Since V = [V, L] there exists an 𝔽_{p}L-submodule U of V with U = [U, L], C_{U}(L) ≠ 0, and \overline{U} simple. Then [MS4, 6.1] shows n = 3 and q = 2. Since B is an offender on U, [MS4, 8.4] shows that |B| = 4 and |[\overline{U}, B]| = 2. But this is impossible, since B ≤ O_{2}(P) and \overline{U} is dual to X.

Theorem 3.9 (Rank 1 Theorem). Let M be a finite CK group with O_{p}(M) = 1 and V be a faithful finite dimensional 𝔽_{p}M-module, and let K be J_{M}(V)-component. Suppose that the point stabilizers of K on V are p-groups. Then K ≅ SL_{2}(q)′, q a power of p, and [V, K]/C_{[V,K]}(K) is a natural SL_{2}(q)′-module.

Proof. Let T ∈ Syl_{p}(K) and P = P_{K}(V, T). By 1.7(a), P is also a point stabilizer of K on [V, K] and so by 1.7(c) (applied with V := [V, K] and V_{0} := C_{[V,K]}(K)), P contains a point stabilizer P^{∗} of K on X := [V, K]/C_{[V,K]}(K).

Suppose first that X is a homogeneous K-module, that is, X ≅ Y^{m} for some simple 𝔽_{p}K-module Y. Then P^{∗} is the point stabilizer of K on Y. Since P^{∗} is a p-group we conclude from GFMT that K ≅ SL_{2}(q), Y is a natural SL_{2}(q) module and X ≅ Y. So the theorem holds in this case.

Suppose next that X is not an homogeneous K-module. Then by GFMT K ≅ SL_{n}(q), n ≥ 4, and X ≅ Y^{r} ⊕ Y^{∗s}, where Y is a natural SL_{n}(q) and r, s ∈ ℤ^{+}. Then P^{∗} = O^{p′}(C_{K}(C_{Y} (T)) ∩ C_{K}(Y/[Y, T]) and so P^{∗}/O_{p}(P^{∗}) ≅ SL_{n−2}(q), a contradiction since P^{∗} is p-group and n ≥ 4.

4 Q!-Modules

The results of this section are important for the investigation of finite groups G which possess a large subgroup. Here a p-subgroup Q ≤ G is called large if

(∗)  C_{G}(Q) ≤ Q and N_{G}(A) ≤ N_{G}(Q) for all 1 ≠ A ≤ Z(Q).

Observe that most of the simple groups of Lie type in characteristic p possess a large subgroup, namely O_{p}(N_{G}(Z)), where Z is a long root subgroup. The exceptions are the groups F_{4}(2^{n}), Sp_{4}(2^{n}) and ^{2}G_{2}(3^{n}).

For the investigation of p-local subgroups containing a large subgroup by means of their action on elementary abelian p-subgroups the above property (∗) can be easily transformed into a property of modules, so-called Q!-modules. We recall the definition here from [MS4, 6.2]:

Definition 4.1. 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).

In this section H is a finite group with O_{p}(H) = 1, Q is a p-subgroup of H, and V a faithful Q!-module for 𝔽_{p}H. Put H^{∘} := ⟨Q^{H}⟩.

Recall that a p-subgroup A ≤ H is a weakly closed subgroup of H if A is the only conjugate of A in T for some T ∈ Syl_{p}(H).

Lemma 4.2. (a) Q is a weakly closed subgroup of H.

(b) H^{∘} = ⟨Q^{h} | h ∈ H^{∘}⟩.

(c) C_{H}(H^{∘}/Z(H^{∘})) = C_{H}(H^{∘}).

(d) Let H^{∘} ≤ L ≤ H and W be a non-zero L-submodule of V. Then C_{L}(W) ≤ C_{L}(H^{∘}). In particular, C_{H^{∘}}(W) is a p′-group.

(e) C_{V} (H^{∘}) = 0.

(f) Let Q ≤ L ≤ H with Q ⋬ L. Then V is Q!-module for 𝔽_{p}L.

(g) Let L ⊴⊴ H with [L, Q] ≠ 1. Then C_{V} (⟨L^{Q}⟩) = 0.

Proof. (a): Let Q ≤ T ∈ Syl_{p}(H) and A := C_{V} (T). By Q!, Q ⊴ N_{H}(A), in particular Q ⊴ N_{H}(T). Thus, if Q^{h} ≤ T then also Q^{h} ⊴ N_{H}(T). By Burnside’s Theorem [Go, 7.2.1] any two normal subgroups of T are conjugate in H if and only if they are conjugate in N_{H}(T). Hence Q = Q^{h}.

(b): Let H_{0} := ⟨Q^{h} | h ∈ H^{∘}⟩. Then (a) shows that Q^{H} = Q^{H_{0}} and so H_{0} = H^{∘}.

(c): Set D := C_{H}(H^{∘}/Z(H^{∘})). Note that Z(H^{∘}) is a p′-group since O_{p}(H) = 1 and that Z(H^{∘}) centralizes Q. Hence Q = O_{p}(QZ(H^{∘})). Since [D, QZ(H^{∘})] ≤ Z(H^{∘}) ≤ QZ(H^{∘}) we conclude that D normalizes QZ(H^{∘}) and Q. So [D, Q] ≤ Q ∩ Z(H^{∘}) = 1 and thus [D, H^{∘}] = 1.

(d): Since C_{W} (Q) ≠ 0, Q! implies that C_{H^{∘}}(W) ≤ N_{H}(C_{W} (Q)) ≤ N_{H}(Q). Thus

[C_{H^{∘}}(W), Q] ≤ Q ∩ C_{H^{∘}}(W) ≤ O_{p}(C_{H^{∘}}(W)) ≤ O_{p}(H^{∘}) ≤ O_{p}(H) = 1.

Now (b) implies C_{H^{∘}}(W) ≤ Z(H^{∘}). This shows the additional claim since Z(H^{∘}) is a p′-group. Moreover, [C_{L}(W), H^{∘}] ≤ C_{H^{∘}}(W) ≤ Z(H^{∘}) and so by (c), C_{L}(W) ≤ C_{L}(H^{∘}).

(e): Since Q ⋬ H, Q ≠ 1 and so C_{H^{∘}}(C_{V} (H^{∘})) = H^{∘} is not a p′-group. Thus (d) implies C_{V} (H^{∘}) = 0.

(f): This follows immediately from the definition of a Q!-module.

(g): Put L^{∗} = ⟨L^{Q}⟩. Since L is subnormal in H, so is L^{∗}. Thus O_{p}(L^{∗}) ≤ O_{p}(H) = 1. If Q ⊴ QL^{∗} we get 1 ≠ [Q, L] ≤ [Q, L^{∗}] ≤ Q ∩ L^{∗} ≤ O_{p}(L^{∗}) = 1, a contradiction. Thus Q ⋬ L^{∗}Q and by (f), V is a Q!-module for L^{∗}Q. By (e) applied to L^{∗}Q, C_{V} (⟨Q^{L^{∗}Q}⟩) = 0. In particular, C_{V} (L^{∗}Q) = 0 and so also C_{V} (L^{∗}) = 0.

Recall from [MS2] that an 𝔽_{p}H-module U is called quasisimple if U = [U, H], U/C_{U}(H) is a simple H-module, and O_{p}(H/C_{H}(U)) = 1.

Lemma 4.3. Let K be a J_{H}(V)-component such that [V, K] is a quasisimple K-module. Then K ≤ [K, H^{∘}] ≤ H^{∘}.

Proof. Otherwise [MS4, 2.10] implies [K, H^{∘}] = 1. Put W = ([V, K] + C_{V} (K))/C_{V} (K). Then W is a simple K-module. Since [K, Q] = 1, C_{W} (Q) is a non-trivial K-submodule and so [W, Q] = 0. By 4.2(b) H^{∘} = ⟨Q^{H^{∘}}⟩ and thus [W, H^{∘}] = 0 since H^{∘} normalizes W. Thus [V, K, H^{∘}, K] = 0. Since [H^{∘}, K] = 1 the Three Subgroups Lemma gives [V, K, K, H^{∘}] = 1. By 4.2, C_{V} (H^{∘}) = 0, thus [V, K, K] = 0, a contradiction since [V, K] is quasisimple.

Lemma 4.4. Let Y be a p-subgroup of H with C_{Y} ([V, Y]) ≠ 1 and [H^{∘}, Y] ≠ 1. Then C_{Y} (H^{∘}) = 1.

Proof. Put X := C_{Y} (H^{∘}) and Z := C_{Y} ([V, Y])∩Z(Y). Note that Z ≠ 1. Suppose that X ≠ 1. Then [V, X] is a non-zero H^{∘}Z-submodule of V. Since Z centralizes [V, X] we conclude from 4.2(d), that Z centralizes H^{∘}. Since [V, Y, Z] = 0 and [Z, Y] = 0 the Three Subgroups Lemma gives [V, Z, Y] = 0. Thus [V, Z] is a non-zero H^{∘}Y submodule centralized by Y and another application of 4.2(d) gives [H^{∘}, Y] = 1, a contradiction.

Lemma 4.5. Suppose that one the following holds.

(i) F^{∗}(H) ≅ Alt(n), n ≥ 5, and [V, H] is a natural 𝔽_{p}Alt(n)-module for F^{∗}(H), or

(ii) H ≅ Alt(7) and |[V, H]| = 2^{4}.

Then (i) holds, and either n = p or (n, p) is one of (5, 2), (6, 2), (8, 2), (6, 3).

Proof. Clearly every 𝔽_{p}H-submodule of V is also a Q!-module, so we may assume that V = [V, H]. Since Q is not normal in H, Q ≠ 1. Moreover, replacing H by H^{∘} we may assume that H = H^{∘} and so H = O^{p′}(H).

If (ii) holds, then C_{H}(v) ≅ L_{3}(2) and so O_{2}(C_{H}(v)) = 1 for every 0 ≠ v ∈ V. For 0 ≠ v ∈ C_{V} (Q) this gives Q ≤ O_{2}(C_{H}(v)) = 1, a contradiction. Thus (i) holds.

Put K := F^{∗}(H) ≅ Alt(n). Note that either Aut(K) ≅ Sym(n), or n = 6, |Aut(K)/Inn(K)| = 4. For p ≠ 2 this implies H = O^{p′}(H) ≅ Alt(n). For p = 2 and n = 6, Sym(6) is the largest subgroup of Aut(K) acting on the natural 𝔽_{2}Alt(6)-module. So for p = 2, H ≅ Alt(6) or H ≅ Sym(n). In any case we may assume that H ≤ Sym(n). Let Ω := {1, 2, …, n} and W be the permutation module with basis (w_{i} | i ∈ Ω); and for Ψ ⊆ Ω let w_{Ψ} := ∑_{i∈Ψ} w_{i}. Set W_{0} := [W, H] and \overline{W} := W/C_{W} (H). Then \overline{W_{0}} ≅ V.

Assume first that Q does not act transitively on Ω. Let Ψ be any Q-invariant subset of Ω such that Ψ ≠ Ω, Q acts non-trivially on Ψ, and p||Ψ|. Then w_{Ψ} ∈ W_{0} and \overline{w}_{Ψ} ≠ 0. Note that Alt(Ψ) centralizes \overline{w}_{Ψ} , so Q! implies that Alt(Ψ) normalizes the image of Q in Sym(Ψ). It follows that either p = 2 and |Ψ| = 2 or 4, or p = 3 and |Ψ| = 3. In all cases Ψ is a non-trivial Q-orbit. Since this holds for all possible choices of Ψ and n ≥ 5, Ω is the union of two non-trivial Q-orbits or of one non-trivial Q-orbit and at most p − 1 fixed points of Q. This gives one of the following possibilities:

n = 2 + 4, n = 4 + 4, n = 4 + 1 for p = 2,
n = 3 + 3, n = 3 + 2 for p = 3.

Suppose p = 3 and n = 5. Say Q = ⟨(1, 2, 3)⟩. Then Q centralizes w := w_{1234} − w_{5}. Observe that w ∈ W_{0}, \overline{w} ≠ 0 and C_{H}(\overline{w}) = Alt(4), a contradiction to Q!. So (n, p) ≠ (5, 3) and the lemma holds in the intransitive case.

Assume next that Q acts transitively on Ω. Let (X_{i})_{i∈𝔽_{p}} be a Q-invariant partition of Ω into p sets of size \frac{n}{p}. Pick g ∈ Q with X_{0}^{g} ≠ X_{0} and choose notation such that X_{i}^{g} = X_{i+1} for all i ∈ 𝔽_{p}. Define w_{0} = ∑_{i∈𝔽_{p}} iw_{X_{i}}. Then

w_{0}^{g} = ∑_{i∈𝔽_{p}} iw_{i+1} = ∑_{j∈𝔽_{p}}(j − 1)w_{j} = ∑_{j∈𝔽_{p}} jw_{j} − ∑_{j∈𝔽_{p}} w_{j} = w_{0} − w_{Ω}

Thus Q centralizes \overline{w_{0}} and \overline{w}_{0} ≠ 0. Note that Alt(X_{0}) centralizes \overline{w}_{0} and so by Q!, Alt(X_{0}) normalizes Q. Thus [Alt(X_{0}), Q] is a p-group. Since Alt(X_{0})^{g} = Alt(X_{1}) this implies that Alt(X_{0}) is p-group. Hence one of the following holds: |X_{0}| = 1 and n = p; |X_{0}| = 2 = p and n = 4; or |X_{0}| = 3 = p and n = 9. In the first case the lemma holds. Since n ≥ 5, the second case is impossible. So suppose |X_{0}| = 3 = p. Put D = (Sym(X_{0}) × Sym(X_{1}) × Sym(X_{2})) ∩ Alt(n) and E = Alt(X_{0}) × Alt(X_{1}) × Alt(X_{2}). Then D/E ≅ C_{2} × C_{2} and Q acts non-trivially on D/E. Thus [D, Q] is a not 3-group, a contradiction to Q! and D ≤ C_{H}(\overline{w}_{0}).

Theorem 4.6 (Q!FF-Module Theorem). Let H be a finite group with O_{p}(H) = 1 and Q be a p-subgroup of H, and let V be a faithful Q!-module for H. Put H^{∘} := ⟨Q^{H}⟩ and J := J_{H}(V). Suppose that there exists an offender Y in H such that [H^{∘}, Y] ≠ 1 and that one of the following holds:

(i) Y is quadratic on V.

(ii) Y is a best offender on V.

(iii) C_{Y} ([V, Y]) ≠ 1.

(iv) C_{Y} (H^{∘}) = 1.

Then one of the following holds, where q is a power of p:

1. There exists an H-invariant set K of subgroups of H such that:

(a) For all K ∈ K, K ≅ SL_{2}(q) and [V, K] is a natural module for K,

(b) J = ×_{K∈K} K and V = ⨁_{K∈K}[V, K],

(c) Q acts transitively on K,

(d) H^{∘} = O^{p}(J)Q.

2. Put R := F^{∗}(J). Then

(a) R is quasisimple, R ≤ H^{∘}, and either J = R or p = 2 and J ≅ O_{2n}^{±}(q), Sp_{4}(2) or G_{2}(2).

(b) C_{V} (R) = 0, [V, R] is a semisimple J-module, and H acts faithfully on [V, R].

(c) Put J^{0} := J ∩ H^{∘}. Then one of the following holds:

1. (a) R = J^{0} ≅ SL_{n}(q), n ≥ 3, Sp_{2n}(q), n ≥ 3, SU_{n}(q), n ≥ 8, or Ω_{n}^{±}(q), n ≥ 10. (b) [V, R] is the direct sum of at least two isomorphic natural modules for R.

(c) H^{∘} = RC_{H^{∘}}(R).

(d) If V ≠ [V, R] then R ≅ Sp_{2n}(q), p = 2, and n ≥ 4.

2. (a) [V, R] is a simple R-module.

(b) Either H^{∘} = R = J^{0} or H^{∘} ≅ Sp_{4}(2), 3.Sym(6), SU_{4}(q).2 (≅ O_{6}^{−}(q) and [V, R] the natural SU_{4}(q)-module), or G_{2}(2).

(c) One of the cases FMT (1) - (9), (12) applies to (J, [V, R]), where n = 6 in case (12).

3. p = 2, J = R ≅ SL_{4}(q), H^{∘}/R has order two and induces a graph automorphism on R, and V is the direct sum of two non-isomorphic natural modules.

Proof. Note that (i) implies (iii), and by Timmesfeld’s Replacement Theorem [KS, 9.2.3], also (ii) implies (iii). By 4.4, (iii) implies (iv). So in any case: 1^{∘}. C_{Y} (H^{∘}) = 1.

Let J be the set of J-components of H, and let X be a minimal offender in Y. By [MS4, 1.3] X is a quadratic best offender and so X ≤ J. Now (1^{∘}) implies 2^{∘}. [H^{∘}, X] ≠ 1. In particular, [H^{∘}, J] ≠ 1.

According to (2^{∘}) and [MS4, 2.2(b)] there exists R ∈ J with R ≤ H^{∘}. Next we prove: 3^{∘}. [R, H^{∘}] ≠ 1.

Suppose that [R, H^{∘}] = 1. Since R ≤ H^{∘} it follows that R is abelian. So [MS4, 2.2(d)] implies that p = 2, R ≅ C_{3} and |[V, R]| = 4. Thus [V, R] is a simple R-module. Hence 4.3 gives R ≤ [R, H^{∘}] = 1, a contradiction.

According to (3^{∘}) we may choose R such that [R, Q] ≠ 1. Put R^{∗} := ⟨R^{Q}⟩. 4^{∘}. C_{V} (R^{∗}) = 0.

As a J-component, R is subnormal in H. Thus (4^{∘}) follows from 4.2(g). 5^{∘}. J = R^{Q}, so R^{∗} = ⟨J ⟩.

Otherwise there exists R̃ ∈ J \ R^{Q}. Put U := [V, ⟨R̃^{Q}⟩]. Since R^{Q} ∩ R̃^{Q} = ∅, [MS4, 2.2(f)] gives [U, ⟨R^{Q}⟩] = 0, which contradicts (4^{∘}). 6^{∘}. W := [V, R^{∗}] is a semisimple, faithful J-module, and R^{∗} = ×_{K∈J} K = F^{∗}(J) ≤ H^{∘}.

By (4^{∘}) C_{V} (R^{∗}) = 0. Hence (6^{∘}) follows from (5^{∘}) and [MS4, 8.3]. 7^{∘}. H acts faithfully on W.

Set D := C_{H}([V, R^{∗}]). Since W is a faithful J-module, R^{∗} ∩ D ≤ J ∩ D = 1. Thus [R^{∗}, D] = 1, and the Three Subgroups Lemma shows that [V, D, R^{∗}] = 0. Now (7^{∘}) follows from (4^{∘}).

Let S be a Sylow p-subgroup of H with Q ≤ S. Since R and R^{∗} are subnormal in H, R ∩ S and R^{∗} ∩ S are Sylow p-subgroups of R and R^{∗}, respectively. Since Q ≤ S, also R^{∗}Q ∩ S is a Sylow p-subgroup of R^{∗}Q. 8^{∘}. Let W_{0} be an R^{∗}Q-submodule of W and Z_{0} := C_{[W_{0},R]}(S ∩ R). Then C_{R}(Z_{0}) ≤ N_{H}(Q).

From 1.6, applied to E = R, H = R^{∗}Q, T = S ∩ R^{∗} and W_{0}, we get that C_{R}(Z_{0}) = C_{R}(C_{W_{0}}(S ∩ R^{∗})). Since C_{W_{0}}(S ∩ R^{∗}) ≤ C_{V} (Q), the Q!-property gives C_{H}(C_{W_{0}}(S ∩ R^{∗})) ≤ N_{H}(Q). So (8^{∘}) holds.

Put Z := C_{[V,R]}(S ∩ R), P := C_{R}(Z) and Ω := {v | 0 ≠ v ∈ C_{V} (Q^{h}), h ∈ H}. Note that S ∩ R ≤ P and by (8^{∘}) P ≤ N_{H}(Q); in particular [P, Q] ≤ O_{p}(R^{∗}). 9^{∘}. Suppose that R^{∗} acts transitively on Ω. Then H^{∘} = R^{∗}Q.

Let v ∈ Ω such that [v, Q] = 0. By the Frattini argument H = R^{∗}C_{H}(v) and by Q!, C_{H}(v) ≤ N_{H}(Q). Thus H = R^{∗}N_{H}(Q), R^{∗}Q is normal in H, and R^{∗}Q ≤ H^{∘} ≤ R^{∗}Q.

Case 1. Suppose that J ≠ {R}. Then (1) holds.

By (5^{∘}) R^{∗} is the direct product of at least two Q-conjugates of R. On the other hand by (8^{∘}) [C_{R}(Z), Q] ≤ Q. We conclude from the direct product that P is a p-group. By (6^{∘}), C_{[V,R]}(R) = 0 and so the Rank 1 Theorem 3.9 shows that R ≅ SL_{2}(q)′ and [V, R] is a natural module for R.

Then (5^{∘}) implies (1:a) and (1:c). Moreover, GFMT gives the first part of (1:b). To show the second part of (1:b) it suffices to show that V = [V, R]+C_{V} (R). Let A be a minimal offender on V. Since Q acts transitively on J we may assume [R, A] ≠ 1. Since A is not an over-offender on [V, R] and A is minimal we conclude from [MS4, 1.2] that C_{A}([V, R]) = 1, |A| = q = |V/C_{V} (A)|, [C_{V} (R), A] = 0 and RA ≅ SL_{2}(q). Thus RA is generated by two conjugates of A and so |V/C_{V} (RA)| ≤ q^{2}. Hence V = [V, R] + C_{V} (R) and (1:b) holds.

To prove (1:d) let Ω^{∗} be the set of elements in V such that [v, R̃] ≠ 0 for all R̃ ∈ J. Since R acts transitively on [V, R] and C_{V} (R^{∗}) = 0, and since we have already proved that V = [V, R^{∗}], R^{∗} acts transitively on Ω^{∗}. Since Q acts transitively on J we conclude that Ω = Ω^{∗}. Thus by (9^{∘}), H^{∘} = R^{∗}Q and (1:d) holds.

According to (Case 1) we may assume from now on that J = {R}. In particular, W = [V, R] and C_{V} (R) = 0.

Case 2. Suppose that R is solvable. Then (1) holds.

By [MS4, 2.2(d)]) (or FMT) p = 2 or 3, R ≅ SL_{2}(p)′ and [V, R] is a natural SL_{2}(p)′-module. Since C_{V} (R) = 0, [MS4, 8.4] (or coprime action) gives V = [V, R] and so (1) holds.

According to (Case 1) and (Case 2) we may assume from now on that R is quasisimple. We will show that (2) holds.

Case 3. Suppose that W is not a homogeneous R-module. Then (2) and (2:c:3) hold.

By GFMT we have R ≅ SL_{n}(q) and [V, R] ≅ N^{r} ⊕ N^{∗s}, where N and N^{∗} are simple natural R-submodule in W dual to each other. Moreover, r, s ≥ 1 and n ≥ 4.

Set U := C_{N}(R ∩ S) and U^{∗} = C_{N^{∗}}(R ∩ S). Then U and U^{∗} are 1-dimensional over 𝔽_{q} and P = C_{R}(Z) = C_{R}(U) ∩ C_{R}(U^{∗}). From (8^{∘}) we get [P, Q] ≤ P ∩ Q ≤ O_{p}(P).

Assume first that Q acts trivially on the Dynkin diagram of R. Put W_{0} := ⟨N^{Q}⟩. Then W_{0} ≅ N^{t′} for some t′ ≤ r, and by (8^{∘}) C_{R}(U) ≤ N_{R}(Q). By symmetry also C_{R}(U^{∗}) ≤ N_{R}(Q). Thus R = ⟨C_{R}(U), C_{R}(U^{∗})⟩ ≤ N_{M}(Q). But then [R, Q] ≤ Q ∩ R ≤ O_{p}(R) = 1, which contradicts (3^{∘}).

Assume now that Q acts non-trivially on the Dynkin diagram of R. Then clearly p = 2. Also [P, Q] ≤ O_{2}(P) implies that n = 4 and that Q does not induce a non-trivial field or field graph automorphism on R. By GFMT \sqrt{r} + \sqrt{s} ≤ \sqrt{4} = 2 and so r = s = 1. Thus W ≅ N ⊕ N^{∗}. An application of [MS4, 8.4] also gives W = V.

Let f : N × N^{∗} → 𝔽_{q} be a surjective R-invariant 𝔽_{q}-bilinear map. Then R acts transitively on Ω^{∗} = {n + n^{∗} | f(n, n^{∗}) = 0} and Ω = Ω^{∗}. Thus by (9^{∘}), H^{∘} = RQ and so (Case 3) is proved.

Case 4. Suppose that W is a homogeneous but not simple R-module. Then (2) and (2:c:1) hold.

GFMT shows that R is a genuine group of Lie-type of type A_{n} (n ≥ 2), B_{n} (n ≥ 5), C_{n} (n ≥ 3), D_{n} (n ≥ 5), ^{2}A_{n} (n ≥ 4), or ^{2}D_{n} (n ≥ 5), and W ≅ N^{r} for some natural module N and some r ≥ 2. Put U = C_{N}(S ∩ R). Then P = C_{R}(U), N_{R}(U) is a maximal parabolic subgroup of R, and P/O_{p}(P) is a group of Lie-type A_{n−1}, B_{n−1}, C_{n−1}, D_{n−1}, ^{2}A_{n−2} or ^{2}D_{n−2}, respectively. Note that [P, Q] ≤ O_{p}(P) . If R is of type A_{n}, P is not invariant under any graph automorphism of R, and if R is type D_{n}, any graph automorphism also induces a graph automorphism on P/Q. Thus Q does not induce graph automorphisms on R. Also if R has roots of more than one length, so does P/Q. Thus Q does not induces any field automorphism on R. Since Q is a p-group, Q does not induces any diagonal automorphism on R and so Q ≤ RC_{H}(R). Thus H^{∘} = RC_{H^{∘}}(R).

Suppose that V ≠ [V, R]. We apply [MS4, 8.4]. Since C_{V} (R) = 0 and [V, K] is not simple, we are in case (f) of [MS4, 8.4], and R ≅ Sp_{2n}(q), p = 2, and n ≥ 4.

Case 5. Suppose that W is a simple R-module.

Clearly we can apply FMT to (J, W). In the cases (10) – (12) of FMT 4.5 shows that one of the following holds:

J ≅ Sym(5) ≅ O_{4}^{−}(2), J ≅ Alt(8) ≅ Ω_{6}^{+}(2), J ≅ Sym(8) ≅ O_{6}^{+}(2),
J ≅ Alt(6) ≅ Sp_{4}(2)′, J ≅ Sym(6) ≅ Sp_{4}(2),

and W is the corresponding natural orthogonal or symplectic module. Thus, one of the cases (1) – (9) of FMT holds or J ≅ Alt(6). In particular, (2:a) holds. It remains to determine H^{∘}. Observe first that in all cases N_{R}(P) acts transitively on C_{W} (P). Also H = RN_{H}(P) and so the Frattini argument gives H = RC_{H}(v) ≤ RN_{H}(Q), where 0 ≠ v ∈ C_{V} (P Q). Thus H^{∘} = RQ and so O_{p}(RQ) ≤ O_{p}(H) = 1.

If Q ≤ R, (2:c:2) holds. So suppose Q ≰ R. Then some element of Q induces an outer automorphism on R. If R ≅ G_{2}(2)′, then H^{∘} ≅ G_{2}(2). If R ≅ Sp_{4}(2)′ then R ≅ Sp_{4}(2) and if R ≅ 3.Alt(6), then H ≅ 3.Sym(6) since H^{∘} acts on W. Thus again (2:c:2) holds.

In all the remaining case R is genuine group of Lie-type in characteristic p. If R has Lie-rank 1, R ≅ SL_{2}(q). Hence either W is a natural SL_{2}(q)-module and (1) holds, or W is a natural Ω_{4}^{−}(2)-module. In the latter case P ≅ Alt(4), so [P, Q] ≤ O_{2}(P) implies Q ≤ R, which is not the case.

Suppose now that R has rank at least two. Note that N_{R}(P) is a maximal parabolic subgroup of R and P = O^{p′}(N_{R}(P)). Since [P, Q] ≤ O_{p}(P) and Q ≰ R we can argue as in (Case 4) and conclude that Q induces non-trivial field automorphisms on R. Moreover, R must have root groups defined over two different fields (and so FMT implies R ≅ SU_{n}(q), n ≥ 4, or R ≅ Ω_{2n}^{ϵ}(q), 2n ≥ 6), and Q can only centralize root groups defined over the smaller field. Since Q centralizes P/O_{p}(P), this shows that all roots groups in P/O_{p}(P) must have order q .

The last condition rules out the natural Ω_{2n}^{ϵ}(q)-module and shows that n = 4 if W is a natural SU_{n}(q)-module. Also Q induces a field automorphism of order 2 and so p = 2. Thus (2:c:2) follows.

The next result is inspired by a situation that arises in applications of the quadratic L-lemma 2.9. Let L be a finite group, V be an elementary abelian normal p-subgroup with [V, O^{p}(L)] = V, and A be an elementary abelian p-subgroup with A ⊴ AO_{p}(L). Then Y and V normalize each other, so V acts quadratically on Y, and vice versa.

Suppose that H := L/C_{L}(V) and A := Y C_{L}(V)/C_{L}(V) satisfy the hypothesis of 2.9 and that V/C_{V} (O^{p}(L)) is a simple L-module. Then H ≅ SL_{2}(q), Sz(q) or D_{2r}, and by 2.10 V/C_{V} (O^{p}(L)) is a natural module for H. The structure of these modules shows that V is a strong offender on Y and

(∗)  [Y, V] = [Y, X] for all X ≤ V with |XC_{V} (Y)/C_{V} (Y)| ≥ 4.

In other words, the module Y possesses a non-trivial strong offender V with the additional property (∗). Such modules are investigated in the next result, where as usual the module is called V rather than Y.

Theorem 4.7. Let H be a finite group with O_{p}(H) = 1, and let V be a faithful Q!-module. Suppose that there exists 1 ≠ W ≤ H such that

(i) W is a strong offender on V; and

(ii) [X, W] = [V, W] for all X ≤ V with |X/C_{X}(W)| > 2.

Put H^{∘} := ⟨Q^{H}⟩, K^{∗} := ⟨W^{H}⟩, K := ⟨W^{K^{∗}}⟩ and K := K^{H}. Then

K^{∗} = × K and [V, K^{∗}] = ⨁_{R∈K}[V, R].

Moreover, one of the following holds:

1. (a) K ⊴ H, H^{∘} = O^{p}(K)Q, and C_{V} (K) = 0.

(b) K ≅ SL_{n}(q), n ≥ 3, Sp_{2n}(q), Sp_{4}(2)′, O_{2n}^{ϵ}(2) or 3.Alt(6); q a power of p , p = 2 in the last three cases; and [V, K] is a corresponding natural module.

(c) Either H^{∘} ≤ K or K ≅ Sp_{4}(2)′ and H^{∘} ≅ Sp_{4}(2) or K ≅ 3.Alt(6) and H^{∘} ≅ 3.Sym(6).

(d) If K ≅ O_{2n}^{ϵ}(2), then |W | = 2.

2. (a) Q acts transitively on K, H^{∘} = O^{p}(K^{∗})Q, and V = [V, K^{∗}].

(b) K ≅ SL_{2}(q), and [V, K] is a corresponding natural module.

3. (a) p = 2, K ≅ SL_{n}(2), n ≥ 3, V = [V, K] is the direct sum of two isomorphic natural modules for K, and |V/C_{V} (W)| = 4.

(b) K ⊴ H, K ≤ H^{∘}, and H^{∘} ≅ SL_{n}(2) or SL_{n}(2) × SL_{2}(2).

4. (a) p = 2, K ≅ SL_{n}(2), n ≥ 3, V = C_{V} (K^{∗})⊕[V, K^{∗}], [V, K] is the direct sum of two isomorphic natural modules for K, and |V/C_{V} (W)| = 4.

(b) K^{∗} ⊴ H, [K^{∗}, H^{∘}] = 1 and H^{∘} ≅ SL_{2}(2).

Proof. Set \overline{V} := V/C_{V} (K). Since W is a strong offender, [MS4, 1.6] implies that W is a quadratic best offender on V and so K^{∗} = J_{K^{∗}}(V). Set J = J_{K^{∗}}(V) and J_{W} = {J ∈ J | [J, W] ≠ 1}. GFMT implies that ⟨J ⟩ = × J, J = [J, W] for all J ∈ J_{W}, and K^{∗}/⟨J ⟩ is an elementary abelian p-group. The latter fact shows that K = ⟨J_{W} ⟩W and ⟨J_{W} ⟩ = O^{p}(K) = [O^{p}(K), W]. Pick J ∈ J_{W}, and let Y be a minimal non-trivial J-submodule of V. By [MS4, 2.11] Y is a quasisimple and so perfect as an K-module. Then [Y, W] ≠ 0, and by [MS4, 1.9] Y is K^{∗}-invariant.

Case 1. Suppose [V, O^{p}(K)] ≰ Y.

Since O^{p}(K) = [O^{p}(K), W] we conclude that [V, W] ≰ Y and [Y, W] ≠ [V, W]. Now (ii) implies that p = 2 and |Y/C_{Y} (W)| = 2. Moreover |V/C_{V} (W)| ≥ 4 and since W is an offender, |W | ≥ 4. Since W is a strong offender and [Y, W] ≠ 0, C_{W} (Y) = C_{W} (V) = 1. Thus 1 = C_{W} (Y) = C_{W} (J) = C_{W} (\overline{Y}) by [MS4, 2.4], and W acts faithfully on \overline{Y}. Since |\overline{Y} /C\overline{Y} (W)| = 2, this gives |[\overline{Y} , W]| = |W | ≥ 4. In particular, \overline{Y} is not a selfdual JW -module since (for example by [MS4, 1.8]), |\overline{Y} /C\overline{Y} (W)| = |[\overline{Y}^{∗}, W]. We now apply FMT. Then the properties |\overline{Y} /C\overline{Y} (W)| = 2 < |[\overline{Y} , W]| and \overline{Y} being not self-dual eliminates all cases apart from the case where K/C_{K}(Y) ≅ SL_{n}(2), n ≥ 3, |Y/C_{Y} (W)| = 2. Now an application of [MS4, 8.4] yields C_{Y} (K) = 0 and so:

1^{∘}. K/C_{K}(Y) ≅ SL_{n}(2), n ≥ 3, |Y/C_{Y} (W)| = 2, and Y is a natural SL_{n}(2)-module.

Assume that O^{2}(K) ≠ J. Then there exist J_{1} ∈ J_{W} with J_{1} ≠ J and a simple J_{1}-submodule Y_{1} in V satisfying (1^{∘}) in place of J and Y. Then |Y + Y_{1}/C_{Y +Y_{1}}(W)| ≥ 4, and (ii) implies that [V, W] ≤ Y + Y_{1}. Hence [V, K^{∗}] = Y + Y_{1}, Y = [V, J] and Y_{1} = [V, J_{1}]. By 4.3 JJ_{1} ≤ H^{∘} and so [H^{∘}, W] ≠ 1. Thus we can apply 4.6. But the only case in 4.6 with more than one J_{H}(V)-component is 4.6(1), where there J_{H}(V)-components are isomorphic to SL_{2}(q). This contradicts (1^{∘}).

We have shown that O^{2}(K) = J, so K = JW. Moreover, by (1^{∘}) W induces inner automorphisms in J. Since O_{2}(K) ≤ O_{2}(H) = 1, we conclude that

K = J, [V, J] ≰ Y and C_{J}([V, J]) = 1,

in particular by (1^{∘}) J ≅ SL_{n}(2), n ≥ 3. Moreover, [MS4, 8.4] shows that V = C_{V} (K) ⊕ [V, K]. Now

K^{∗} = × K, [V, K^{∗}] = ⨁_{J∈K}[V, J] and V = C_{V} (K^{∗}) ⊕ [V, K^{∗}].

follow from [MS4, 2.2].

By GFMT, [\overline{V} , J] is semisimple J-module and so there exists a minimal non-trivial J-submodule Y_{1} with Y ≠ Y_{1}. Put Y_{2} := Y. Then (1^{∘}) applies to (Y_{i}, J) for i = 1, 2. In particular, |Y_{i}/C_{Y_{i}}(W)| = 2 and |[Y_{i}, W]| = |W | > 2 for i = 1, 2. Hence, Y_{1} is not isomorphic to the dual of Y_{2} and so Y_{1} and Y_{2} are isomorphic natural modules. As above, (ii) implies [V, W] ≤ Y_{1} + Y_{2} and [V, J] = Y_{1} ⊕ Y_{2}.

If [H^{∘}, W] ≠ 1, we can apply 4.6. By (1^{∘}), Case 1 of 4.6 does not hold. Hence by 4.6(2:b) F^{∗}(⟨K⟩) is quasisimple and so K = {J} and J ⊴ H. If [H^{∘}, W] = 1, we have [J, H^{∘}] = 1. In any case H^{∘} normalizes J and so also [V, J]. Since [V, J] = Y_{1} ⊕ Y_{2}, the normalizer of the image of J in Aut([V, J]) is isomorphic to SL_{n}(2) × SL_{2}(2). Since O_{2}(H^{∘}) = 1 and H^{∘} is generated by 2-elements we conclude that H^{∘}/C_{H^{∘}}([V, J]) is isomorphic to SL_{n}(2), SL_{2}(2) or SL_{n}(2) × SL_{2}(2). By 4.2(d), C_{H^{∘}}([V, J]) ≤ Z(H^{∘}), and by [Gr] the Schur multipliers of the above groups are 2-groups, so C_{H^{∘}}([V, J]) ≤ O_{2}(H^{∘}) = 1.

Suppose J ≤ H^{∘}. Then K = J ⊴ H^{∘}, [J, H^{∘}] ≠ 1 and by 4.2(g) C_{V} (K) = C_{V} (J) = 0. Hence V = [V, K] and (3) holds.

Suppose that J ≰ H^{∘}. Then [J, H^{∘}] = 1, J ≅ SL_{n}(2) and H^{∘} ≅ SL_{2}(2). It follows that also [K^{∗}, H^{∘}] = 1, and (4) holds.

Case 2. [V, O^{p}(K)] = Y.

In this case clearly O^{p}(K) = J and so K = JW. By 4.3, J ≤ H^{∘}. Thus [H^{∘}, W] ≠ 1, and we are allowed to apply 4.6. If J ⋬ H, 4.6 shows that (2) holds. So suppose J ⊴ H and so also K ⊴ H. Comparing 4.6 with 3.2 we see that (1) holds if K ≅ S̸L_{2}(q). In the latter case we are in case (2) with K = K^{∗}.

Lemma 4.8. Let 1 ≠ A ≤ H be a strong dual offender on V. Put K^{∗} := ⟨A^{H}⟩, K := ⟨A^{K^{∗}}⟩ and K := K^{H}. Then one of the following holds.

1. (a) K ⊴ H, H^{∘} = ⟨Q^{K}⟩ and C_{V} (K) = 0.

(b) K ≅ SL_{n}(q), n ≥ 3, Sp_{2n}(q), Alt(6), or O_{2n}^{ϵ}(2), q a power of p , p = 2 in the last two cases; and [V, K] is a corresponding natural module.

(c) Either H^{∘} ≤ K or K ≅ Sp_{4}(2)′ and H^{∘} ≅ Sp_{4}(2).

(d) If K ≅ O_{2n}^{ϵ}(2), then |W | = 2.

2. (a) Q acts transitively on K and H^{∘} ≤ ⟨K⟩Q

(b) V = ⨁_{R∈K}[V, R], K ≅ SL_{2}(q), and [V, K] is a natural SL_{2}(q)-module for K.

Proof. By [MS4, 1.5] A is a quadratic best offender on V. Let J be a J_{H}(V)-component with [J, A] ≠ 1 and W a quasisimple J-submodule of V. Then 1 ≠ [W, A] ≤ W by [MS4, 2.6], and since A is a strong dual offender, [V, A] = [W, A]. Thus J is the unique J(V)-component of H not centralized by A and K = JA. Moreover, W = [V, J] = [W, K] and so by 4.3, J ≤ H^{∘}. Hence [H^{∘}, A] ≠ 1, and the lemma follows from 4.6 and 3.1.

5 Minimal asymmetric modules

In this section H is a finite group and V is an 𝔽_{p}H-module.

Definition 5.1. Let A and B be p-subgroups of H with A ≤ B. Then V is a minimal asymmetric 𝔽_{p}H-module (with respect to A ≤ B) provided that

(i) A ⊴ N_{H}(B), and B is a weakly closed subgroup of H,

(ii) [V, A, B] = [V, B, A] = 0,

(iii) ⟨A^{H}⟩ does not act nilpotently on V,

(iv) ⟨A^{F} ⟩ acts quadratically on V for every proper subgroup F of H with B ≤ F.

Note that conditions (i) and (iii) imply that A is non-trivial normal subgroup of B.

As many definitions in the previous section, the above definition is motivated be the investigation of the p-local structure of finite groups of local characteristic p. Consider the following situation:

G is a finite group of characteristic p, S ∈ Syl_{p}(G), and A is an abelian normal subgroup of S with A ≰ O_{p}(G). Put B := C_{S}(A). Then A ≤ B since A is abelian. Suppose that B is weakly closed in G and that [A, A^{g}] = 1 for all g ∈ G such that ⟨A, A^{g}⟩ is p-group. Then for every A ≤ L ≤ G with A ≤ O_{p}(L), X := ⟨A^{O_{p}(G)L}⟩ is p-group and so abelian. In particular, X acts quadratically on O_{p}(G) and ⟨A^{N_{G}(B)}⟩ is abelian. So replacing A by ⟨A^{N_{G}(B)}⟩ we may assume that A ⊴ N_{G}(B). Now let H be a subgroup of G minimal with B ≤ H and A ≰ O_{p}(H), and let V be elementary abelian Hinvariant section of O_{p}(G) not centralized by ⟨O^{p}(⟨A^{H}⟩). Then V is a minimal asymmetric module for H.

Lemma 5.2. Let V be a minimal asymmetric 𝔽_{p}H-module with respect to the subgroups A ≤ B. Suppose that C_{V} (H) = 0. Then the following hold:

(a) AA^{g} acts quadratically on V for every g ∈ H with [A, A^{g}] ≤ A ∩ A^{g}.

(b) ⟨A^{P} ⟩ acts quadratically on V for every P ≤ H with A ≤ O_{p}(P) ≤ N_{H}(A).

Proof. (a): By 5.1(ii) A acts quadratically on V. Put F := ⟨B, B^{g}⟩. Suppose first that F = H. Then [V, A, A^{g}] ≤ [V, A] ∩ [V, A^{g}] ≤ C_{V} (⟨B, B^{g}⟩) = C_{V} (H) = 0. By symmetry, [V, A^{g}, A] = 0 and so AA^{g} is quadratic.

Suppose next that F ≠ H. By 5.1(iv), ⟨A^{F} ⟩ acts quadratically on V. Since B is a weakly closed subgroup of H there exists f ∈ F with B^{gf} = B. Since A ⊴ N_{G}(B), this gives A^{gf} = A. Thus A^{g} ∈ A^{F}, and AA^{g} is quadratic.

(b): Note that A^{g} ⊴ O_{p}(P) for every g ∈ P. Hence (b) follows from (a).

Lemma 5.3. Let G be a group, T ≤ G, and g ∈ G with [T, T^{g}] = 1. Then T′ ≤ [T, ⟨g⟩].

Proof. Note that T ≤ T^{g}[T, ⟨g⟩] and so

T′ = [T, T] ≤ [T, T^{g}[T, ⟨g⟩]] = [T, [T, ⟨g⟩]] ≤ [T, ⟨g⟩].

Lemma 5.4. Let V be a faithful simple minimal asymmetric 𝔽_{p}H-module with respect to A ≤ B. Put L := ⟨A^{H}⟩ and K := F^{∗}(H). Then H = KB, K = [K, A] ≤ L, L = KA, and one of the following holds:

1. |B| = 2 and H = L ≅ D_{2r}, r an odd prime.

2. |A| = 2, L ≅ SU_{3}(2)′, B ≅ C_{4} or Q_{8}, and V is a natural SU_{3}(2)′-module for L.

3. |B| = 3, H = L ≅ SL_{2}(3), and V is a natural SL_{2}(3)-module for L.

4. K is quasisimple and not a p′-group, H = KB, V is a simple 𝔽_{p}K-module, and H acts 𝕂-linearly on V, where 𝕂 = End_{K}(V).

Proof. Let B ≤ T ∈ Syl_{p}(N_{H}(B)). From 5.1(ii) and the Three Subgroups Lemma we get that [B, A, V] = 0. Hence the faithful action of H and the quadratic action of A give A ≤ Ω_{1}Z(B). Moreover, T ∈ Syl_{p}(H) since B is weakly closed in H. Thus we have:

1^{∘}. A ≤ Ω_{1}Z(B), A ⊴ T, and T ∈ Syl_{p}(H).

Next we show:

2^{∘}. Let R be a B-invariant subgroup of H with [R, A] ≰ O_{p}(R). Then H = RB and K ≤ R. In particular, H = KB, K = [K, A], and L = KA.

Note that A ≰ O_{p}(RB); in particular ⟨A^{RB}⟩ does not act quadratically on V. Thus by 5.1(iv), RB = H and therefore O^{p}(H) ≤ R. Since V is a faithful simple H-module, O_{p}(H) = 1 and so K = O^{p}(K) ≤ O^{p}(H) ≤ R.

Again since O_{p}(H) = O_{p}(K) = 1, [K, A, A] ≠ 1 = O_{p}([K, A]) and A ≰ O_{p}([K, A]B). Hence for R = [K, A] we get H = [K, A]B and K = [K, A]. Since KA ⊴ KB = H also L = KA follows.

According to (2^{∘}) the initial statements in the conclusion hold, so it remains to establish one of (1) – (4).

Suppose that |B| = 2. By Baer’s Theorem there exists g ∈ H such that ⟨B, B^{g}⟩ is not a 2group. Choose ⟨B, B^{g}⟩ minimal with that property. Then ⟨B, B^{g}⟩ ≅ D_{2r}, r a prime, and by (2^{∘}) ⟨B, B^{g}⟩ = H, so (1) holds. Hence, we may assume from now on:

3^{∘}. |B| > 2.

Let W be a simple 𝔽_{p}L-submodule of V with [W, A] ≠ 0. Then [W, A] ≤ C_{W} (B). Since B normalizes L, B also normalizes W, and since H = KB = LB, we get W = V. So:

4^{∘}. V is a simple 𝔽_{p}L-module.

Now let W be a Wedderburn component for K on V ^{2} and suppose that V ≠ W. Since L = KA, (4^{∘}) implies V = ⟨W^{A}⟩. Hence A acts transitively on the set of Wedderburn components for K on V and therefore W^{A} = W^{B}. By [MS3, 2.11] |W^{A}| = 2 and since |B| > 2 we get N_{B}(W) ≠ 1. But [W, N_{B}(W)] ≤ C_{W} (A) = 0, and by (1^{∘}) N_{B}(W) centralizes A. So N_{B}(W) ≤ C_{B}(⟨W^{A}⟩) = C_{B}(V) = 1, a contradiction.

Thus W = V, and V a is homogeneous 𝔽_{p}K-module. It follows that the number of simple 𝔽_{p}K-submodules in V is not divisible by p, see for example [Go, 3.5.6]. Hence A normalizes a simple 𝔽_{p}K-submodule. Since L = KA, (4^{∘}) implies that V is a simple 𝔽_{p}K-module.

Put 𝕂 = End_{K}(V). Observe that by Schur’s Lemma and Wedderburn’s Theorem 𝕂 is a finite field. Moreover, H acts 𝕂-semilinearly on V. Suppose that A does not act 𝕂-linearly on V. Then by [MS3, 2.14] ^{3}, |A| = 2, and since the non-trivial element in A inverts an element in 𝕂 \ {0}, C_{V} (A) = [V, A]. Thus [V, B] ≤ C_{V} (A) = [V, A] ≤ C_{V} (B), and B acts quadratically on V. So using [MS3, 2.14] one more time, |B| = 2, a contradiction. Thus A acts 𝕂-linearly on V and [V, A] is a non-trivial 𝕂-subspace of V centralized by B. So also B acts 𝕂-linearly on V. Since H = KB, H acts 𝕂-linearly on V. We have proved:

5^{∘}. V is a simple 𝔽_{p}K-module, 𝕂 is a field, and H acts 𝕂-linearly on V.

Since the image of C_{H}(K) in End(V) is contained in 𝕂, (5^{∘}) implies that C_{H}(K) is a cyclic p′-group and C_{H}(K) = Z(H). Clearly Z(H) ≤ F^{∗}(H) = K and so Z(H) ≤ Z(K). Thus

6^{∘}. Z(K) = C_{H}(K) = Z(H) is a cyclic p′-group. In particular, C_{B}(K) = 1, and K is not abelian.

Case 1. Suppose that K is a p′-group.

By (2^{∘}), A centralizes every proper B-invariant subgroup of K, and by (6^{∘}) K is not abelian. Hence [Go, 5.3.7] shows that K is special and that K/Φ(K) is a simple 𝔽_{r}B-module. By (6^{∘}) K is extraspecial. Moreover, by coprime action C_{B}(K/Z(K)) ≤ C_{B}(K) = 1, and so K/Z(K) is a faithful simple 𝔽_{r}B-module. Hence Z(B) is cyclic and since A ≤ Ω_{1}Z(B), |A| = p.

Suppose first that p is odd. Since A is quadratic and K = [K, A] we can apply [MS4, 7.1]. Hence K ≅ Q_{8}, KA ≅ SL_{2}(3) and p = 3. As V is a simple 𝔽_{3}K-module, it has dimension 2, and (3) holds.

Suppose next that p = 2. Then r is odd, and by [Go, 5.3.9(i) and 5.3.10] K has exponent r. Let Y be a maximal abelian subgroup of K and put X = {X ≤ Y | Y = X × Z(K)}. Then K acts transitively on X. Moreover C_{V} (Z(K)) = 0, and coprime action shows that

(∗)  V = ⨁_{X∈X} C_{V} (X).

Note that N_{K}(X) = Y = XZ(K) for X ∈ X. Hence N_{K}(X) acts as 𝕂-scalar multiplication on C_{W} (X). Then (∗) and the simplicity of W as an 𝔽_{p}K-module yield dim_{𝕂} C_{W} (X) = 1 for each X ∈ X.

Let 1 ≠ t ∈ A and put X_{t} := [Y, t] and V_{t} := C_{V} (X_{t}). Since t inverts K/Z(K) and centralizes Z(K), X_{t} ∈ X, so dim_{𝕂} V_{t} = 1 and V_{t} ≤ C_{V} (t). Moreover, t does not fix any other element in X, and (∗) shows that C_{V} (t) = V_{t} ⊕ [V, t]. In particular, dim_{𝕂} C_{V} (t)/[V, t] = 1 and [C_{V} (t), B] ≤ [V, t]. Since [V, A, B] = 0, B acts quadratically on V/[V, t] and C_{V} (t); so B/C_{B}(V/[V, t]) and B/C_{B}(C_{V} (t)) are elementary abelian. Thus

[V, Φ(B), B] ≤ [V, t, B] = 0 and [V, B, Φ(B)] ≤ [C_{V} (t), Φ(B)] = 0.

2i.e., a maximal homogeneous K-submodule of V

3Note the misprint in [MS3, 2.14]: Instead of ’|𝕂| is a cubic 𝔼A-module’ it should read ’V is a cubic 𝔼A-module’

Since B acts faithfully on V, the Three Subgroups Lemma yields Φ(B) ≤ Z(B), and the quadratic action of Φ(B) on V shows that Φ(B) is elementary abelian. Since Z(B) is cyclic, this gives Φ(B) = A. In particular, there exists f ∈ B with f^{2} = t.

On the other hand, f centralizes [V, t], so [V, t] ≤ C_{V} (f). Put V_{0} := {v ∈ V | [v, f] ∈ [V, t]}. Then f acts quadratically on V_{0} and so V_{0} ≤ C_{V} (t). Moreover, the map given by v ↦ [v, f] + [V, t] shows that V/V_{0} ≅ [V, f]/[V, t]. From [V, f] ≤ C_{V} (t) = V_{t} ⊕ [V, t] and dim_{𝕂} V_{t} = 1 we conclude that

1 ≤ dim_{𝕂} V/C_{V} (t) ≤ dim_{𝕂} V/V_{0} = dim_{𝕂}[V, f]/[V, t] ≤ 1.

So dim V/C_{V} (t) = 1 = dim_{𝕂}[V, t] and dim_{𝕂} V = 3. In particular, |X | = 3 and |K| = 3^{3}. Thus (2) holds in this case.

Case 2. K is not a p′-group.

By (2^{∘}) [O_{p′}(K), A] = 1 and K = [K, A], so O_{p′}(K) ≤ Z(K) and by (6^{∘}) O_{p′}(K) ≤ Z(H). Since K = F^{∗}(K) and O_{p}(H) = 1 we conclude that there exists a component E of K with p||E/Z(E)|. Since K = [K, A] we have [E, A] ≠ 1 and by (2^{∘}) K = ⟨E^{B}⟩. Put F := C_{H}(A) ∩ C_{H}([V, A]). Note that B ≤ F and by the Three Subgroups Lemma, [V, F, A] = 0. Since Z(E) ≤ Z(K), Z(E) acts as 𝕂-scalar multiplication on V, and so F ∩ Z(E) = 1.

Suppose that E ∩ F ≠ 1. Then E ∩ F ≰ Z(E) and so A ≤ C_{H}(E ∩ F) ≤ N_{H}(E). Let U be a simple 𝔽_{p}E-submodule of V. Since E ⊴ K and V is a simple K-module, C_{V} (E) = 0 and thus [U, E] ≠ 0. As E is quasisimple this gives C_{U}(E) ≤ Z(E). Then 0 ≠ [U, F ∩ E] ≤ C_{U}(A) and so A normalizes U. Hence A normalizes all simple 𝔽_{p}E-submodules of V and since L = ⟨A^{K}⟩, the same is true for L. Thus U is L-invariant, and (4^{∘}) shows that V = U. It follows that C_{H}(E) is abelian, so K = E and (4) holds.

Suppose next that E ∩ F = 1. Then E ∩ B = 1. If B ≤ N_{H}(E), (4) holds. So we may assume that B ≰ N_{H}(E). Pick b ∈ B with E ≠ E^{b}. Then by 5.3 (E ∩ T)′ ≤ [E ∩ T, ⟨b⟩] ≤ [T, B] ≤ B. Since E ∩ B = 1 we conclude that E ∩ T is abelian. By Burnside’s Transfer Theorem,

(∗∗)  E ∩ T ≰ Z(N_{E}(T ∩ E)),

and so N_{E}(T ∩ E) is not a p-group. Put D = ⟨A^{N_{H}(T ∩K)}⟩. Since B ≤ N_{H}(T ∩ K) < H, 5.1(iv) shows that D acts quadratically on V. Hence D is an elementary abelian p-group and D ≤ F. Since N_{E}(T ∩ E) ≤ N_{E}(T ∩ K) we conclude that [N_{E}(T ∩ E), A] is p-group. Since N_{E}(T ∩ E) is not a p-group this gives A ≤ N_{H}(E) and so [N_{E}(T ∩ E), A] ≤ D ∩ E ≤ F ∩ E = 1. Thus N_{E}(T ∩ E) ≤ N_{H}(A). Since B ≤ F and N_{H}(A) normalizes F, this implies [N_{E}(T ∩ E), B] ≤ F. By 5.3 (N_{E}(T ∩ E))′ ≤ [N_{E}(T ∩ E), ⟨b⟩] ≤ F, and so E ∩ F = 1 implies that N_{E}(T ∩ E) is abelian, a contradiction to (∗∗).

Theorem 5.5 (Minimal Asymmetric Module Theorem). Let H be a CK-group, A ≤ B ≤ H and V be a faithful simple 𝔽_{p}H-module. Suppose that V is a minimal asymmetric 𝔽_{p}M-module with respect to A and B and that F^{∗}(H) is quasisimple with p||F^{∗}(H)|. Then one of the following holds for L := ⟨A^{H}⟩:

1. L ≅ SL_{n}(q), Sp_{2n}(q), SU_{n}(q),^{3}D_{4}(q), Spin_{7}(q), Spin_{8}^{−}(q), G_{2}(q)′ or Sz(q), where q is a power of p, V is the corresponding natural module for L, and A is a long root subgroup of L.

2. L ≅ Sym(2^{k} + 2), k ≥ 3, |A| = 2, A is generated by a transposition, and V is the corresponding natural module.

3. L ≅ 3.Alt(6), |A| = 2 and |V | = 2^{6}.

Proof. Observe that O_{p}(H) = 1 since V is simple and faithful. Put K := F^{∗}(H) and let S ∈ Syl_{p}(H) with B ≤ S.

1^{∘}. B ⊴ S and A ⊴ S.

The first statement follows since B is a weakly closed subgroup of H. The second then follows from the fact that A ⊴ N_{H}(B).

2^{∘}. A ≤ O_{p}(U) for every B ≤ U < H.

Observe that ⟨A^{U}⟩ is quadratic on V, so ⟨A^{U}⟩ is an elementary abelian normal p-subgroup of U.

3^{∘}. K = [K, A], H = KB and L = KA = ⟨A^{L}⟩.

See 5.4.

4^{∘}. V is a simple 𝔽_{p}K-module.

See 5.4.

5^{∘}. There exists g ∈ H such that AA^{g} acts quadratically on V and |AA^{g}| > 2.

If |A| > 2 we can choose g = 1. So suppose that |A| = 2. Then p = 2 and so by assumption |K| has even order. Since K ≤ [H, A] we conclude from Glauberman’s Z^{∗}-theorem [Gl] that there exists g ∈ H with [A, A^{g}] = 1 and A ≠ A^{g}. Now 5.2(a) shows that AA^{g} is quadratic.

6^{∘}. Suppose that A is a maximal quadratic subgroup of H. Then |A| > 2, and N_{M}(A) is the unique maximal subgroup of H containing B. In particular, H is B-minimal.

Since A is a maximal quadratic subgroup, (5^{∘}) shows that |A| > 2.

Let B ≤ U < H. Then ⟨A^{U}⟩ is quadratic and so by the maximality of A, A = ⟨A^{U}⟩ and A ⊴ U. Hence N_{H}(A) is the unique maximal subgroup of H containing B. By (3^{∘}) H = KB = ⟨A^{L}⟩B = ⟨B^{H}⟩ and so H is B-minimal.

Put \overline{H} := H/Z(K).

Case 1. Suppose that p is odd and \overline{K} is not a group of Lie type defined over a field of characteristic p.

By [Ch, Theorem A] p = 3 and the maximal quadratic subgroups of H have order 3. In particular |A| = p = 3, and A is a maximal quadratic subgroup of H. Moreover, \overline{L} ≅ PGU_{n}(2), Alt(n), n ≠ 6, D_{4}(2), G_{2}(4), Sp_{6}(2), Co_{1}, Sz, or J_{2}. Observe that \overline{L} has no outer automorphism of order 3, unless \overline{L} ≅ D_{4}(2). In the D_{4}(2)-case, C\overline{L}(\overline{A}) ≅ C_{3} × U_{4}(2) and so the conjugacy class of \overline{A} under \overline{L} is not invariant under the outer automorphism of order three. Hence in any case H = L.

According to [ATLAS] we can choose a subgroup \overline{U} of \overline{H} as in the following chart:

\overline{H}  PGU_{n}(2)  Alt(n), 3 ∤ n  Alt(3m), 3m ≥ 9  D_{4}(2)  G_{2}(4)
\overline{U} C_{3} ≀Sym(n)/C_{3} Alt(n − 1) O^{3′}(Sym(3) ≀ Sym(m)) O^{3′}(Sym(3) ≀ Sym(4)) U_{3}(3)
\overline{H}  Sp_{6}(2)  Co_{1}  Sz  J_{2}
\overline{U} Sym(3) ≀ Sym(3) 3^{6}.2.Mat_{12}  3^{5}Mat_{11}  U_{3}(3)

In each case \overline{U} contains a Sylow 3-subgroup of \overline{H} and Z(O^{3′}(U)) = 1. So we may assume that \overline{B} ≤ \overline{U}. Then by (6^{∘}), \overline{A} ⊴ \overline{U} and so \overline{A} ≤ Z(O^{3′}(\overline{U})) = 1, a contradiction.

Case 2. Suppose that p = 2 and \overline{K} is not a group of Lie type defined over a field of characteristic 2.

By (5^{∘}), [MeSt1, Theorem] and [MeSt2, Theorem 1] we have \overline{K} ≅ Alt(n), n = 7 or n > 8, U_{4}(3), Mat_{12}, Mat_{22}, Mat_{24}, J_{2}, Sz, Co_{2} or Co_{1}.

Suppose first that \overline{K} ≅ Alt(n), n = 7 or n > 8. Then \overline{H} ≅ Alt(n) or Sym(n). If B acts transitively on Ω = \{1, …, n\}, then n > 8 and there exists B ≤ U ≤ H with \overline{U} ≅ (Sym(\frac{n}{2})≀C_{2})∩\overline{H}. But then O_{2}(\overline{U}) = 1, which contradicts (2^{∘}).

Hence there exists a B-invariant proper subset Ψ of Ω. Put U := N_{H}(Ψ). Then B ≤ U < H and \overline{U} ≅ (Sym(Ψ) × Sym(Ω \ Ψ)) ∩ \overline{H}. Since \overline{A} ≤ O_{2}(\overline{U}) by (2^{∘}), we conclude that one of Ψ and Ω \ Ψ, say Ψ, has size 2 or 4. Since n is none of 4, 6 and 8, O_{2}(Sym(Ω \ Ψ)) = 1 and so A ≤ C_{H}(Ω \ Ψ).

Assume |Ψ| = 2. Then \overline{A} = O_{2}(\overline{U}), and \overline{A} is generated by a transposition. In particular, A ≰ O_{2}(N_{H}(Δ)) for any Δ ⊂ Ω \ Ψ, so B acts transitively on Ω \ Ψ. Hence (2) holds in this case.

So we may assume that B has no orbit of length less than 4. Then |Ψ| = 4, n > 8, and ⟨\overline{A}^{U}⟩ = O_{2}(\overline{U}). Since n > 8 and Z(K) is a 2′-group, [Gr] shows that Z(K) = 1. Hence O_{2}(U) acts quadratically on V. This shows that V is not the natural module and so by [MeSt2, Theorem 4] V is a spinmodule.

Note that there exists g ∈ H with Ψ ∩ Ψ^{g} = ∅ since n > 8. Hence A ≤ C_{H}(Ψ^{g}) and [A, A^{g}] = 1. Now 5.2(a) shows that AA^{g} acts quadratically on V. But AA^{g} does not act quadratically on the spinmodule by [MS4, 7.5].

Suppose that \overline{K} ≅ U_{4}(3). By [MS4, 7.7-7.9] there exists an elementary abelian subgroup Q of order 2^{4} in K such that N_{H}(Q) contains a Sylow 2-subgroup of H, Q is not quadratic on V, N_{H}(Q)/QZ(K) ≅ Alt(6) or Sym(6) and Q is the corresponding natural module for N_{H}(Q). In particular, Q is the unique non-trivial normal 2-subgroup of N_{H}(Q). Since N_{H}(Q) is contains a Sylow 2-subgroup of H, we may assume that B ≤ N_{H}(Q). Hence by (2^{∘}) A ≤ O_{2}(N_{H}(Q)) and so Q = ⟨A^{N_{H}(Q)}⟩. But then Q acts quadratically on V, a contradiction.

Suppose now that \overline{K} is a sporadic group and so \overline{K} ≅ Mat_{12}, Mat_{22}, Mat_{24}, J_{2}, Sz, Co_{2} or Co_{1}. Then by 2.6 H is not B-minimal, so by (6^{∘}) A is not a maximal quadratic subgroup of H.

Assume that |A| ≥ 4. Then H possesses a quadratic subgroup of order at least eight. Thus, by [MeSt2, Theorem 2] L = K ≅ 3.Mat_{22}. Hence there exists S ≤ U ≤ H with U/O_{2}(U) ≅ Sym(5). As O_{2}(U) ∩ K is the unique minimal normal subgroup of U, we get that ⟨A^{U}⟩ = O_{2}(U) ∩ K, so O_{2}(U) ∩ K is quadratic on V by 5.2. But this contradicts [MeSt2, Theorem 3].

Thus |A| = 2 and so A ≤ Z(S). In all the seven cases for \overline{K} given above C_{S}(S ∩ K) = Z(S ∩ K) and |Z(S ∩ K)| = 2. Hence A = Z(S ∩ K) and L = K. Put U := C_{K}(A). Then in all cases there exists g ∈ K with A^{g} ≤ U but A^{g} ≰ O_{2}(U). Now 5.2 implies that AA^{g} is quadratic and so C_{U}([V, A]) ≰ O_{2}(U). But this contradicts the action of U on V, see [MeSt2, Theorem 3].

Case 3. Suppose that \overline{K} is a group of Lie-type defined over a field of characteristic p.

Let Δ be the set of minimal Lie-parabolic subgroups of \overline{K} containing \overline{S ∩ K} and 𝕂 := End_{L}(V). By (4^{∘}), Schur’s Lemma and Wedderburn’s Theorem, 𝕂 is a finite field.

7^{∘}. Either L = K or \overline{L} ≅ Sp_{4}(2), G_{2}(2) or ^{2}G_{2}(3).

Let M_{1}, M_{2}, …, M_{t} be the maximal subgroups of H containing S. Then A ≤ O_{p}(M_{i}). In particular, [N_{K}(S ∩K), A] ≤ S ∩K. Hence no element of A induces a non-trivial field automorphism on K.

Suppose for a contradiction that A acts non-trivially on Δ. Then |Δ| > 1 and there exists a minimal Lie-parabolic P of K containing S ∩ K with A ≰ N_{H}(P). Put M_{P} = ⟨P, B⟩. Suppose that A ≤ O_{p}(M_{P}). Then [P, A] ≤ O_{p}(M_{P}) ∩ K ≤ S ∩ K ≤ P, a contradiction to A ≰ N_{H}(P). Thus A ≰ O_{p}(M_{P}) and the definition of a minimal asymmetric module implies that M_{P} = H and so B acts transitively on Δ. It follows that p = 2, |Δ| = 2, and \overline{K} ≅ L_{3}(q) or Sp_{4}(q)′. Then S ∩ K has exactly two maximal elementary abelian normal subgroups Q_{1} and Q_{2}. Moreover, since A acts non-trivially on Δ, A does not normalize Q_{i}, i = 1, 2. It follows that [Q_{1}Q_{2}, A] is not elementary abelian, a contradiction to [Q_{1}Q_{2}, A] ≤ [S, A] ≤ A.

We have shown that A acts trivially on Δ and that no element of A induces a non-trivial field automorphism. Hence either L = K or \overline{L} ≅ Sp_{4}(2), G_{2}(2), ^{2}F_{4}(2) or ^{2}G_{2}(3). But in the ^{2}F_{4}(2)-case, all involutions of L are contained in K, a contradiction since L = KA.

8^{∘}. Suppose that B acts non-trivially on Δ. Then A ≤ Z(S ∩ L).

Suppose first that p = 2 and \overline{L} ≅ L_{3}(q) or Sp_{4}(q)′. As above let Q_{1} and Q_{2} be the two maximal elementary abelian normal subgroups of S ∩ L. Recall from the structure of \overline{L} that

(i) all involutions of S ∩ L are contained in Q_{1} ∪ Q_{2},

(ii) Q_{1} ∩ Q_{2} = Z(S ∩ L), and

(iii) B is transitive on {Q_{1}, Q_{2}}.

The first property shows that there exists i, say i = 1, such that A ≤ Q_{1}. Since B normalizes A, (iii) shows that A ≤ Q_{1} ∩ Q_{2} Now (ii) gives A ≤ Z(S ∩ L).

Suppose next that p = 2 and \overline{L} ≅ F_{4}(q). Set U := C_{L}(Z(S ∩ L)). Then U/O_{2}(U) ≅ Sp_{4}(q) and O_{2}(U) ≤ O^{2}(U). Since B acts non-trivially on Δ, B acts non-trivially on U/O_{2}(U). Thus O_{2}(U) ≤ O^{2}(U) = [O^{2}(U), B] ≤ ⟨B^{U}⟩. Since [V, A ∩ Z(S ∩ L), B] = 0 we conclude that

0 ≠ [V, A ∩ Z(S ∩ L)] ≤ C_{V} (O^{2}(U)) ≤ C_{V} (O_{2}(U)).

Now 2.3 gives a contradiction.

Suppose finally that p = 2 and \overline{L} ≅ L_{n}(q), n ≥ 4, D_{n}(q), n ≥ 4 or E_{6}(q), or p = 3 and \overline{L} ≅ D_{4}(q). Put Z := Z(S∩L), U := N_{H}(Z) and Q := O_{p}(U ∩L). Then Z is a long root subgroup and Φ(Q) = Z. Note also that A_{1} := ⟨A^{U}⟩ is an elementary abelian normal subgroup of U.

Assume that \overline{L} ≅ L̸_{4}(q). Then U acts simply on Q/Z. It follows that all abelian normal psubgroups of U in Q are contained in Z and so A ≤ A_{1} ≤ Z.

So assume that \overline{L} ≅ L_{4}(q). Let P be the maximal subgroup of L with S ≤ P and P ≠ U. Then A ≤ O_{2}(P) and P acts simply on O_{2}(P). Thus O_{2}(P) = ⟨A^{P} ⟩ acts quadratically on V. Since S ∩ L ≤ ⟨O_{2}(P)^{U}⟩ we conclude that [V, Z, S ∩ L] = 0. It follows from 2.3 that V is a natural SL_{n}(q)-module for L, a contradiction, since B interchanges the two isomorphism classes of natural SL_{n}(q)-modules for L.

9^{∘}. Suppose that there exists a long root subgroup R in K such that A ≤ R, [V, A] = [V, R] and C_{V} (A) = C_{V} (R). Then A = R.

Note that [V, R, B] = [V, A, B] = 0 and [V, B] ≤ C_{V} (A) = C_{V} (R). Hence the Three Subgroup Lemma gives [R, B, V] = 0, so since V is faithful, B ≤ C_{H}(R). As B is a weakly closed subgroup of H, a Frattini argument yields

N_{H}(R) = C_{H}(R)(N_{H}(R) ∩ N_{H}(B)) ≤ N_{H}(A).

Since N_{H}(R) acts simply on R, we get R = A.

Case 4. Suppose that |Δ| = 1.

Then \overline{K} ≅ L_{2}(q), U_{3}(q), Sz(q) or ^{2}G_{2}(q)′. As ^{2}G_{2}(q)′ has abelian Sylow 2-subgroups, this group does not have any non-trivial quadratic module, see for example [Go, 3.8.4]. Hence this case is excluded. Since K is not solvable we have q > 2, so L = K and A ≤ S ∩ K by (7^{∘}).

Put P := N_{H}(L ∩ S) and Z := Z(S ∩ K). Then L has the following properties:

(i) P acts simply on Z.

(ii) Either S ∩ K = Z, or Z = Φ(S ∩ K) and P acts simply on (S ∩ K)/Z.

(iii) If \overline{L} ≅ U_{3}(q), then C_{P} (Z) acts simply on (S ∩ K)/Z.

(iv) S ∩ K is a TI-subgroup and |Z| = q.

Since ⟨A^{P} ⟩ is abelian we conclude from (i) and (ii) that Z = ⟨A^{P} ⟩, so Z acts quadratically V. Let Z ≤ U ≤ K such that U is minimal with respect to Z ≰ O_{p}(U). By (iv), O_{p}(U) = 1 and

Z ⊴ N_{U}(T) for Z ≤ T ∈ Syl_{p}(U). Thus 2.9 implies U ≅ SL_{2}(q) or Sz(q) since |Z| = q > 2. If L = U then by 2.10 V is a natural module, and (9^{∘}) implies case (1) of the theorem.

So suppose L ≠ U. Since 3 ∤ |Sz(q)|, SL_{2}(q) is not involved in Sz(q) and so \overline{L} ≅ U_{3}(q). Since A ≰ O_{p}(U) and U ≠ H, we have B ≰ U. From (iii) we conclude that S ∩ K ≤ ⟨B^{C_{P} (A)}⟩. Hence [V, A, S ∩ K] = 1, and Z = ⟨A^{P} ⟩ gives [V, Z, S ∩ K] = 1. So by 2.3 V is the natural module. Now (9^{∘}) implies case (1) of the theorem.

We assume from now on that |Δ| > 1. Let L_{i}, 1 ≤ i ≤ |Δ| be the maximal subgroups of L containing L ∩ S. Put A_{i} := ⟨A^{L_{i}}⟩ and E_{i} := O^{p′}(L_{i}).

10^{∘}. A_{i} acts quadratically on V; in particular A_{i} is an elementary abelian normal p-subgroup of L_{i}.

Suppose first that B acts trivially on Δ. Then L_{i} is B-invariant. Since L_{i}B is a proper subgroup, the definition of a minimal asymmetric modules shows that A_{i} acts quadratically on V.

Suppose next that B acts non-trivially on Δ. Then by (8^{∘}) A ≤ Z(S ∩ L) ≤ O_{p}(L_{i}). Since O_{p}(L_{i}) ≤ S ∩ L ≤ N_{H}(B) ≤ N_{H}(A), 5.2(b) shows that A_{i} acts quadratically on V.

Case 5. Suppose that |Δ| = 2.

Then \overline{K} is isomorphic to one of the following groups:

L_{3}(q), PSp_{4}(q)′, U_{4}(q), U_{5}(q), G_{2}(q)′, ^{2}F_{4}(q)′, ^{3}D_{4}(q).

Suppose that \overline{K} ≅ Sp_{4}(2)′ ≅ Alt(6). If Z(K) ≠ 1, then [MS4, 7.4] shows that |V | = 2^{6} and A ≤ K. Since A ≤ O_{p}(L_{1}) ∩ O_{p}(L_{2}) we have |A| = 2 and so case (3) holds. If Z(K) = 1, then [MeSt2, Theorem 4] shows that V is a natural Sp_{4}(2)′-module for K. Choose notation such that L_{1} is a point stabilizer for L on V. Then [V, A_{1}] is a singular subspace of V invariant under L_{1} and so |[V, A_{1}]| = 2. Thus |A_{1}| = 2, L ≅ Sp_{4}(2), A = A_{1}, and A is a long root subgroup of L. So case (1) of the theorem holds. Suppose that \overline{K} ≅ G_{2}(2)′. Then [MS4, 7.6] implies that V is natural G_{2}(q)′-module for K. Choose notation such that Z := Z(L_{1}) ≠ 1. Then Z is a long root subgroup of K, |Z| = 2, Z ≤ K, and Z is the unique non-trivial elementary abelian normal subgroup of L_{1}. Thus Z = A_{1} = A, L = ⟨Z^{M}⟩ = K ≅ G_{2}(2)′, and case (1) of the theorem holds.

Therefore, we may assume from now on that q > 2 in the PSp_{4}(q)′- and G_{2}(q)′-case. Hence by (7^{∘}) L = K.

If Z(E_{i}) ≤ Z(K) for some i, we choose our notation such that Z(E_{2}) ≤ Z(K). Otherwise we choose our notation such that [A, E_{2}] ≠ 1. Then in any case A ≰ Z(E_{2}). Put

Z_{2} := [Z(O_{p}(E_{2})), E_{2}], V_{i} := C_{V} (O_{p}(E_{i})),

and let Z be the root subgroup with Z ≤ Z(S ∩ L) and [Z, E_{2}] ≠ 1.

We use the following properties of the groups given above:

(i) Z_{2} is the unique normal subgroup of E_{2} minimal with respect to [Z_{2}, E_{2}] ≠ 1.

(ii) Z ≤ Z_{2}.

(iii) Either \overline{L} ≅ L_{3}(q) or [Z, E_{1}] = 1.

(iv) If \overline{L} ≅ PSp_{4}(q), U_{4}(q) or U_{5}(q), then O_{p}(E_{1})O^{p}(E_{1}) ≤ ⟨Z_{2}^{E_{1}}⟩. Indeed E_{1} ≤ ⟨Z_{2}^{E_{1}}⟩, except for \overline{L} ≅ U_{5}(2).

All these properties can be found in [DS] by first going to the table on page 98 to get the value of the parameter b and then to look up the properties in those chapters where this value of b is treated.

By (i) and (ii) Z ≤ Z_{2} ≤ A_{2}. Suppose that \overline{L} ≅ L_{3}(q). Then A_{i} = O_{p}(E_{i}), A ≤ A_{1} ∩ A_{2} ≤ Z, and A_{1}A_{2} = S ∩ L. Thus [V, A, S ∩ L] = 0. Hence by 2.3 V is a natural SL_{3}(q)-module and by (9^{∘}) case (1) of the theorem holds.

Suppose now that \overline{L} ≇ L_{3}(q). Then Z ⊴ E_{1} by (iii). Put W_{1} := ⟨Z_{2}^{E_{1}}⟩. Since [V, Z, Z_{2}] ≤ [V, A_{2}, A_{2}] and by (10^{∘}) [V, A_{2}, A_{2}] = 0, we get

11^{∘}. [V, Z, W_{1}] = 0.

Suppose that \overline{L} ≅ PSp_{4}(q), U_{4}(q) or U_{5}(q). Then by (iv) O_{p}(E_{1})O^{p}(E_{1}) ≤ W_{1} and thus [V, Z, O_{p}(E_{1})O^{p}(E_{1})] = 0. So by 2.3 V is a natural module. The action of E_{1} on the natural module reveals that any quadratic normal subgroup of E_{1} is contained in Z. Thus A ≤ A_{1} ≤ Z and by (9^{∘}) again case (1) of the theorem holds.

For the remaining cases

\overline{L} ≅ G_{2}(q), q > 2,^{2}F_{4}(q)′,^{3}D_{4}(q)

let Γ be the coset graph of L with respect to L_{1} and L_{2}. All the properties of the action of L on Γ we use here can be found in [DS], in Section 10 for G_{2}(q) and ^{3}D_{4}(q), in Section 12 for ^{2}F_{4}(q)′. In particular the value of b in these cases is 2, 2 and 3, respectively.

Choose a path (α_{1}, α_{2}, … α_{d}) of minimal length d − 1 such that α_{1} = L_{1}, α_{2} = L_{2} and Z ≰ O_{p}(L_{d}), where L_{j} is the stabilizer of α_{j} in L. Then d − 1 = b + 1. Note that for i = 1, 2 this is compatible with our earlier notation since L_{1} = α_{1} and L_{2} = α_{2}. Let Δ(α_{i}) be the set of neighbors of α_{i}. Put

E_{i} := O^{p′}(L_{i}), V_{i} := C_{V} (O_{p}(L_{i})), J_{i} := ⟨V_{k}^{E_{i}} | 1 ≤ k ≤ d, α_{k} ∈ Δ(α_{i})⟩,

and, for i even, Z_{i} := [Z(O_{p}(E_{i})), E_{i}]. Observe that Z ≤ O_{p}(L_{d−1}) ≤ L_{d} and by (11^{∘}) [V_{d}, Z, W_{1}] = 0, so [V_{d}, Z] is centralized by R := ⟨O_{p}(L_{d}), W_{1}⟩.

Suppose that \overline{L} ≅ G_{2}(q), q > 2, or ^{3}D_{4}(q). Then d = 4, Z_{2} ≤ O_{p}(L_{1}), Z_{2} ≰ Z(O_{p}(E_{1})) and Z(O_{p}(E_{1}))/Φ(O_{p}(E_{1})) is the unique maximal L_{1}-submodule of O_{p}(E_{1}))/Φ(O_{p}(E_{1})); in fact Φ(O_{p}(E_{1})) = Z(O_{p}(E_{1})) and O_{p}(E_{1})/Φ(O_{p}(E_{1})) is a simple E_{1}-module unless \overline{L} ≅ G_{2}(q) and p = 3 or q = 4. Since L_{1} = (L_{1} ∩ L_{2})E_{2}, L_{1} normalizes W_{1}. It follows that W_{1} = O_{p}(E_{1}) and R = ⟨E_{2}, E_{3}⟩ = L. So C_{V} (R) = 0 and [V_{4}, Z] = 0. Thus [V_{4}, E_{4}] = 0 and so also [V_{2}, E_{2}] = 0.

Now Steinberg’s Lemma shows that V_{2} is 1-dimensional over 𝕂 and V_{2} = C_{V} (S ∩ K). Moreover, W_{1} = O_{p}(E_{1}) and (11^{∘}) imply that [V, Z] ≤ V_{1}.

Since Z_{2} ≤ O_{p}(E_{1}), Z_{2} centralizes V_{1} and so [J_{2}, Z_{2}] = 0. Since by definition of Γ, α_{4} and α_{2} are conjugate under L we conclude that [J_{4}, Z_{4}] = 0. Then also [Z, Z_{4}, J_{4}] = 0 since Z ≤ L_{4} and Z_{4} ⊴ L_{4}. Hence the Three Subgroup Lemma gives [J_{4}, Z, Z_{4}] = 0. Since also O_{p}(E_{1}) centralizes [Z, J_{4}], we conclude that E_{2} = ⟨O_{p}(E_{1}), Z_{4}⟩ centralizes [J_{4}, Z]. Thus [J_{4}, Z] ≤ V_{2}, so also [J_{4}, Z] is 1-dimensional.

Let 1 ≠ z ∈ Z and α_{5} be a neighbor of α_{4} in Γ different from α_{3}. Then 0 ≠ [V_{5}, z] ≤ [J_{4}, Z] ≤ V_{2}, so [V_{5}, z] and thus also V_{5}/C_{V_{5}}(z) are 1-dimensional over 𝕂. Observe that

C_{V_{5}}(z) ≤ V_{5} ∩ V_{5}^{z} ≤ C_{V} (⟨O_{p}(E_{5}), O_{p}(E_{5})^{z}⟩ = C_{V} (E_{4}) = V_{4}.

Thus V_{5}/V_{4} is 1-dimensional. So also V_{1}/V_{2} is 1-dimensional. Since V_{2} is 1-dimensional, we get that V_{1} is 2-dimensional over 𝕂. Hence V_{1} is a natural SL_{2}(q^{ϵ})-module for E_{1}, where ϵ = 1 in the G_{2}(q)-case and ϵ = 3 in ^{3}D_{4}(q)-case. Note also that E_{2} centralizes V_{2}. By Ronan-Smith’s Lemma [MS4, 4.3] this determines V up to isomorphism and it follows that V is the natural module for L. According to (9^{∘}), in order to establish that (1) holds, it remains to show in these cases that A ≤ Z.

If \overline{L} ≅ G_{2}(q), q not a power of 3 or \overline{L} ≅ ^{3}D_{4}(q), then Z is the unique maximal elementary abelian normal subgroup of E_{1} and so A ≤ A_{1} ≤ Z. Suppose \overline{K} ≅ G_{2}(3^{k}) and A ≰ Z. Then also [A, E_{1}] ≠ 1 and the set-up is symmetric in 1 and 2. As we have seen above, V_{2} = C_{V} (S ∩ K) = C_{V} (E_{2}), so by symmetry also V_{1} = C_{V} (S ∩ K) = C_{V} (E_{1}) and V_{1} = V_{2}, a contradiction.

Suppose finally that \overline{L} ≅ ^{2}F_{4}(q)′. Then W_{1} is abelian, W_{1} ≤ O_{p}(E_{2}) and d = 5. Let X_{1} = ⋂_{α∈Δ(α_{1})} O_{p}(L_{α}), so W_{1} ≤ X_{1}. Observe that X_{1} centralizes J_{1}, so X_{7} centralizes J_{7} since α_{1} and α_{7} are conjugate. Thus by (11^{∘}), E_{5} = ⟨W_{1}, X_{7}⟩ centralizes [J_{7}, Z]. It follows that [J_{7}, Z] = V_{5}, [V_{5}, E_{5}] = 1 and [J_{7}, Z] is 1-dimensional over 𝕂. Observe that O^{p}(L_{7}) does not centralize J_{7} and so there exists a composition factor W for 𝕂L_{7} on J_{7} not centralized by O^{p}(E_{7}). But E_{7}/O_{p}(E_{7}) ≅ Sz(q). Thus by 2.3 dim_{𝕂}[W, Z] > 1, a contradiction.

Case 6. Suppose that |Δ| = t > 2.

According to (7^{∘}) L = K. We divide the groups under consideration into two classes:

(I) \overline{K} ≅ PSp_{2n}(q), q odd and n ≥ 3, PSU_{n}(q), n ≥ 6,
F_{4}(q), q odd, ^{2}E_{6}(q), E_{6}(q), E_{7}(q), E_{8}(q).
(II) \overline{K} ≅ L_{n}(q), n ≥ 4, P Ω_{n}(q), n ≥ 7, F_{4}(q), q even.

We first discuss the groups in (I). They all have the following properties in common:

(i) Z := Z(S ∩ K) is a long root subgroup of L.

(ii) N_{L}(Z) is a maximal subgroup of L (and we choose L_{t} = N_{L}(Z)).

(iii) Φ(O_{p}(L_{t})) = Z.

(iv) O_{p}(L_{t})/Z is a simple E_{t}-module.

(v) E_{t}/O_{p}(E_{t}) is quasisimple.

Since N_{L}(S ∩ K) acts simply on Z we have Z ≤ A_{i} for i = 1, …, t. Observe that by (iii) and (iv), A_{t} = Z, and so A ≤ Z. Put L̃_{t} := L_{t}/O_{p}(L_{t})Z(K) and R := ⟨A_{1}^{L_{t}}⟩. If \overline{K} ≅ PSp_{2n}(q) or PSU_{n}(q) choose L_{1} to be the normalizer of a maximal singular subspace (of the natural module). Then A_{1} ≰ O_{p}(E_{t}) and by (v) R̃ = Ẽ_{t}. In the other cases of (I) except \overline{K} ≅ E_{7}(q), choose L_{1} such that E_{1}/O_{p}(E_{1}) ≅ Ω_{m}^{ϵ}(q), (E_{1}∩E_{t})/O_{p}(E_{1}∩E_{t}) ≅ Ω_{m−2}^{ϵ}(q), and Z_{1} := ⟨Z^{E_{1}}⟩ is a natural Ω_{m}^{ϵ}(q)-module. For \overline{K} ≅ E_{7}(q) choose L_{1} such that E_{1}/O_{p}(E_{1}) ≅ E_{6}(q) and so (E_{1} ∩ E_{t})/O_{p}(E_{1} ∩ E_{t}) ≅ Spin_{10}^{+}(q), and Z_{1} := ⟨Z^{E_{1}}⟩ is a simple E_{1}-module of order q^{27}. Since Z ≤ A_{1}, we get Z_{1} ≤ A_{1} and thus [A_{1}, O_{p}(E_{t})] ≰ Z since [Z_{1}, O_{p}(E_{t})] ≰ Z. Hence (iii) shows that A_{1} ≰ O_{p}(E_{t}) and so by (v) R̃ = Ẽ_{t}.

We have shown Ẽ_{t} = R̃. Hence (iii) and (iv) imply

(∗)  E_{t} = R.

As Z ≤ A_{i} and A_{i} is quadratic on V, we get that [V, Z, R] = 0. Hence (∗) implies that [V, Z, E_{t}] = 0. So by 2.3 K ≅ Sp_{2n}(q) or SU_{n}(q), and V is the natural module. Hence (9^{∘}) implies case (1) of the theorem.

We now discuss the groups in (II). Suppose \overline{K} ≅ L_{n}(q), n > 3. Let P_{1} and P_{2} be the p-minimal subgroups of H with S ∩ K ≤ P_{i} and Z := Z(S ∩ K) not normal in P_{i}. As n > 3, P_{1} and P_{2} commute. Put P = P_{1}P_{2} and D = ⟨A^{P} ⟩. Then Z ≤ D and [V, Z, D] = 0, so D ≤ N_{L}(Z) but D ≰ O_{p}(N_{L}(Z)), and ⟨D^{N_{L}(Z)}⟩ = O^{p′}(N_{L}(Z)). Hence [V, Z, O^{p′}(N_{L}(Z))] = 0 and by 2.3 V is a natural module. Let L_{1} and L_{2} be the normalizers of a 1-dimensional subspace and hyperplane, respectively. Then A ≤ O_{p}(L_{1}) ∩ O_{p}(L_{2}) = Z, and (9^{∘}) implies case (1) of the theorem holds.

Suppose \overline{K} ≅ P Ω_{n}^{ϵ}(q), n ≥ 7. Let d be the dimension of a maximal singular subspace (of the natural module) and for i = 1, 2, d choose notation such that L_{i} normalizes a singular i-subspace. Set Z_{2} := Z(E_{2}), B_{i} := ⟨Z_{2}^{L_{i}}⟩ and, if p = 2 and n is odd, Z_{1} := Z(E_{1}). We will use the following properties:

(i) If p is odd or n is even, then Z(S ∩ K) = Z_{2}, and if p = 2 and n is odd, then Z_{1} and Z_{2} are non-conjugate root subgroups with Z(S ∩ K) = Z_{1}Z_{2}. Moreover, Z_{2} = B_{d} ∩ Z(S ∩ K) = B_{d} ∩ Z(O_{p}(L_{2})).

(ii) B_{d} is the unique minimal normal subgroup of L_{d} in O_{p}(L_{d}).

(iii) B_{d} = O_{p}(L_{d}) or O_{p}(L_{d})/B_{d} is a simple L_{d}-module.

(iv) If B_{d} ≠ O_{p}(L_{d}) and O_{p}(L_{d}) is abelian, then p = 2 and n is odd.

(v) B_{1} = O_{p}(L_{1})

(vi) If d ≥ 4 and \overline{K} ≇ Ω_{8}^{+}(q) then L_{2} = ⟨(B_{1}B_{d})^{L_{2}}⟩.

(vii) If \overline{K} ≅ Ω_{8}^{+}(q), then L_{3} normalizes a singular 4-space and L_{2} = ⟨(B_{1}B_{3}B_{4})^{L_{2}}⟩.

(viii) L_{2} acts simply on O_{p}(L_{2})/Z(O_{p}(L_{2})).

Suppose first that A_{d} ≠ B_{d}. Then B_{d} ≠ O_{p}(L_{d}), A_{d} = O_{p}(L_{d}), and O_{p}(L_{d}) is elementary abelian. Thus p = 2 and n is odd. Then Z_{1} is a root subgroup and Z_{1} ≤ A_{d} ≰ O_{2}(E_{1}), E_{1} = ⟨A_{d}^{E_{1}}⟩, and [V, Z_{1}, E_{1}] = 0. Hence by 2.3 V is a natural Sp_{n−1}(q)-module. In particular, Z_{1} is the unique maximal quadratic normal subgroup of L_{1} and so A ≤ A_{1} ≤ Z_{1}. Now (9^{∘}) implies that case (1) of the theorem holds.

Suppose that A_{d} = B_{d}. Then A ≤ B_{d} and so 1 ≠ A ∩ Z(S ∩ K) ≤ B_{d} ∩ Z(S ∩ K) = Z_{2}. Since N_{K}(S ∩ K) acts simply on Z_{2}, Z_{2} = ⟨(A ∩ Z(S ∩ K))^{N_{K}(S∩K)}⟩ and therefore Z_{2} ≤ A_{1} ∩ A_{d}. Since O_{p}(L_{1}) = B_{1} = ⟨Z_{2}^{L_{1}}⟩ we get A_{1} = O_{p}(L_{1}). Thus [MS4, 7.11] implies that V is a (half-)spin-module. If d ≥ 4 and \overline{K} ≇ Ω_{8}^{+}(q) then L_{2} = ⟨(B_{1}B_{d})^{L_{2}}⟩ = ⟨(A_{1}A_{d})^{L_{2}}⟩ and so [V, Z_{2}, L_{2}] = 0, a contradiction to 2.3. If \overline{K} ≅ Ω_{8}^{+}(q), then A_{3} = B_{3}, L_{2} = ⟨(B_{1}B_{3}B_{4})^{L_{d}}⟩ = ⟨(A_{1}A_{3}A_{4})^{L_{2}}⟩, and we obtain the same contradiction. Thus d = 3. Suppose that A ≰ Z_{2}. Since A_{3} ∩ Z(O_{p}(L_{2})) = Z_{2} we conclude that A ≰ Z(O_{p}(L_{2})). Since L_{2} acts simply on O_{p}(L_{2})/Z(O_{p}(L_{2})) we get O_{p}(L_{2}) = A_{2}Z(O_{p}(L_{2})), a contradiction since A_{2} is abelian. Hence A ≤ Z_{2} and (9^{∘}) yields case (1) of the theorem.

Suppose finally that \overline{K} ≅ F_{4}(q) and p = 2. Let Z_{1} and Z_{2} be the two root subgroups with Z(S ∩K) = Z_{1}Z_{2}. For i = 1, 2 let L_{i} := N_{L}(Z_{i}). Since Z_{1} ∩Z_{2} = 1 we can choose notation such that A ∩ Z(S ∩ K) ≰ Z_{1}. Then Y_{1} := Ω_{1}Z(O_{2}(E_{1})) ≤ ⟨(A ∩ Z(S ∩ K))^{E_{1}}⟩ ≤ A_{1}, and so Y_{1} is quadratic on V. Note that Z_{2} ≤ Y_{1}, Y_{1} ≰ O_{2}(E_{2}) and E_{2}/O_{2}(E_{2}) is quasisimple. We conclude that E_{2} ≤ ⟨Y_{1}^{E_{2}}⟩ and [V, Z_{2}, E_{2}] = 0, which contradicts 2.3.

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