SMALL MODULES FOR SMALL GROUPS

Andrew Chermak

Department of Mathematics Kansas State University Manhattan KS, 66502

December, 2001

Introduction. This paper concerns “small” modules for finite groups which are themselves “small” in a certain sense to be defined, and it is directed towards some specific applications. For the space of a few paragraphs, however, we can try to place things in a fairly broad context.

For any finite group Y and any prime p, one may consider the action induced by Y, via conjugation, on the largest normal p-subgroup O_{p}(Y) of Y. As an example, let Y be a proper parabolic subgroup in a simple group X of Lie type, in characteristic p. Then C_{Y} (O_{p}(Y)) ≤ O_{p}(Y), which is to say that Y/Z(O_{p}(Y)) is faithfully represented as a group of automorphisms of O_{p}(Y). Denote by L the subgroup of Y generated by the p-elements in Y, and assume that L properly contains O_{p}(Y). That is, assume that Y is not a Borel subgroup of X. Then O_{p}(Y) = O_{p}(L), and L/O_{p}(L) is itself a group of Lie type in characteristic p. Further, there then exists an L-invariant section V of O_{p}(L) having exponent p, and such that L/O_{p}(L) acts faithfully on V. Thus V is a module for L/O_{p}(L) over the field 𝔽_{p} of p elements, and one may then appeal to the enormous body of work concerning such representations of characteristic-p groups of Lie type, in order to obtain information about the structure of V and, indirectly, about the structure of L and even of X.

The assumption that Y be a proper parabolic subgroup of X and that Y not be a Borel subgroup implies that the Lie rank of X is at least 2. There is then a rich geometry Γ(X) associated with X, namely the building associated with the set of all parabolic subgroups of X, and which is encoded in the Dynkin diagram associated with X. From this diagram one also reads off information about the various “residues” at the proper parabolic subgroups of X, so that the geometry Γ(L) associated with the group L/O_{p}(L) is given by a sub-diagram. The point is that this geometry imposes severe restrictions on the sorts of modules V which can arise in the above context. Indeed, if we choose L so that the diagram for Γ(L) is connected then, as it happens, all such

Typeset by AMS-TEX

modules V are “small”, in the sense that |L/O_{p}(L)|^{2} > |V |. This fact can be seen as a consequence of another one, which has more to do with the particular nature of the action of L on V. Namely, in “most” cases in which Γ(L) is connected there exists an elementary abelian p-subgroup A of X/O_{p}(X), A ≠ 1, with the property that

(0.1)  |A|^{2} ≥ |V/C_{V} (A)|.

(The qualifying adjective “most” will be made more precise, further along in this introduction.) For the moment, we mention that the condition (0.1) implies the condition that |L/O_{p}(L)|^{2} > |V | as a consequence of [CD, Theorem 2.3].

Suppose now that we weaken the above hypotheses, so that the ambient group X is no longer assumed to be a simple group of Lie type. Suppose instead that we require only that O_{p}(X) = 1 and that for any non-identity p-subgroup Q of X, the group Y = N_{X}(Q) has the property that C_{Y} (O_{p}(Y)) ≤ O_{p}(Y). (That is, all p-local subgroups of X are pconstrained.) As above, let L be the group generated by the p-elements in Y. One may ask whether there exists a suitable p-local subgroup Y of X, a non-identity elementary abelian subgroup A of L/O_{p}(L), and an elementary abelian section V of L in O_{p}(L), admitting faithful action by A, such that (0.1) holds.

In raising this question, we have in mind the situation where X is a simple, or nearly simple, group in which the composition factors of the various groups Y are on the list if “known” simple groups, but in which X itself is not assumed to be on such a list. That is to say, our intention is to address issues that pertain to the Classification of the Finite Simple Groups (or the CFSG, for short).

Specifically, the aim of this paper is to provide support for a project, initiated by Ulrich Meierfrankenfeld, to classify the finite simple groups X which are of “generic characteristic p type.” We will not attempt here to outline this program, other than to say that the groups under consideration correspond roughly to the class of simple groups of Lie type, of Lie rank at least 2. The feature of Meierfrankenfeld’s project which distinguishes it most clearly from the treatment of characteristic p-type groups in the CFSG as it currently stands, and from the treatment in the ongoing revision project of Gorenstein, Lyons, and Solomon (GLS), consists in directly targeting the groups Y = N_{X}(Q) for analysis, rather than switching attention to a prime r different from p. This difference in approach can be summarized by saying that in the CFSG as it stands, the emphasis is on “semisimple” subgroups, while in the Meiefrankenfeld program the emphasis is on normalizers of unipotent subgroups. Actually, even in the old version of CFSG, and in the GLS revision, an important chapter - namely the classification of Quasithin Groups - is treated via a “unipotent” approach for the prime 2 (for which see [AS]). So it may be more correct to say that in the Meierfrankenfeld program, the unipotent approach takes center stage, rather than being relegated to the treatment of a difficult special case.

In the Meierfrankenfeld program one begins with a group X of characteristic p-type, with O_{p}(X) = 1, and in which a Sylow p-subgroup S of X is contained in at least two different maximal p-local subgroups of X (so that one has, from the outset, the rudiments of a geometry). The analysis then centers on a pair of p-local subgroups C̃ and M of X containing S, such that C̃ contains the centralizer in X of Z(S), and such that M/O_{p}(M) acts faithfully on a subgroup V of exponent p in O_{p}(M) (where V is taken to be as large as possible for this condition). The normal closure W of V in C̃ is then either an abelian p-group, a non-abelian p-group, or (of course) a group which is not a p-group. One then finds:

(0.2) If W is an abelian p-group then there exists a conjugate A of V in C̃ such that [V, A, A] = 1, and such that |A/C_{A}(V)|^{2} ≥ |V/C_{V} (A)|.

The case where W is not a p-group leads to a variant of (0.2), as follows.

(0.3) If W is not a p-group then there is a C̃-conjugate A of a subgroup of V, such that the following hold.

(i) [V, A, A, A] = 1, (ii) [C_{V} (a), A, A] = 1 for every a ∈ A − C_{A}(V), and (iii) |A/C_{A}(V)|^{2} ≥ |V/C_{V} (A)|, and if p = 2 then |A/C_{A}(V)|^{3/2} ≥ |V/C_{V} (A)|.

The case in which W is a non-abelian p-group does not lend itself to the formulation of conditions as in (0.2) or (0.3). This is the reason for the qualification concerning “most” cases in the condition (0.1).

We have, so far, explained what a small module is, for our purposes. But we have not yet explained the other half of the title of this paper. A group G is a “minimal parabolic” group (for the prime p) if a Sylow p-subgroup S of G is not normal in G and is contained in a unique maximal subgroup of G. Such groups G are the “small” groups that we will consider here, and a general description of such groups may be found in 11.1, below. Our aim in this paper is to obtain information about 𝔽_{p}G-modules V, where G is a minimal parabolic group and where V satisfies conditions as in (0.2) or (0.3). The reason for restricting attention to these groups has to do with the specific demands made by one portion of the Meierfrankenfeld project. But some of our results do in fact concern small modules for groups which are not necessarily minimal parabolics. We therefore begin by stating a series of hypotheses, of varying degrees of restrictiveness.

Hypothesis 1. G is a finite group, p is a prime, S is a Sylow p-subgroup of G, and V is a faithful 𝔽_{p}G-module. Further, we have O_{p}(G) = 1, and there exists a non-identity, elementary abelian subgroup A of S such that |A|^{2} ≥ |V/C_{V} (A)|.

Whenever Hypothesis 1 is in effect, we put:

H = O^{p}(G).

That is:

H = ⟨g ∈ G : |h| is relatively prime to p⟩.

Also, whenever Hypothesis 1 is in effect, we will use additive notation for the group operation in V (so that 0, rather than 1, denotes the identity element of V). It will be important to observe that, under Hypothesis 1, if H is quasisimple then the Fitting subgroup F (G) is contained in H, and so H = F^{∗}(G).

Hypothesis 2. In addition to Hypothesis 1, assume that V = ⟨(C_{V} (S ∩ H))^{G}⟩, and C_{V} (G) = 0. Assume also that G = ⟨S^{G}⟩.

Hypothesis 3. Let G, p, S, V, and A satisfy the conditions of Hypothesis 2. We further assume that S is contained in a unique maximal subgroup of G.

Hypothesis 4. In addition to Hypothesis 3, we assume that A can be chosen so that A acts quadratically on V. That is, we have [V, A, A] = 0.

Recall that a group X is “quasisimple” if X/Z(X) is simple and Z(X) ≤ [X, X].

Hypothesis 4′. Assume Hypothesis 2, and assume that H is quasisimple. Assume further that there is a non-identity elementary abelian p-subgroup A of G satisfiying the following conditions.

(a) [V, A, A, A] = 0.

(b) [C_{V} (a), A, A] = 0 for every non-identity element a of A.

(c) If p = 2 then |A|^{3/2} ≥ |V/C_{V} (A)|, while if p is odd then |A|^{2} ≥ |V/C_{V} (A)|.

Recall that a “component” of a group X is a quasisimple subnormal subgroup of X. We assume always that the Classification of the Finite Simple Groups applies to the components (if any) of G. Thus, to state things explicitly, we have the following:

Background Hypothesis. Whenever Hypothesis 1 holds, and K is a component of G, then K/Z(K) is an alternating group, the commutator subgroup of a group of Lie type, or one of twenty-six “sporadic” groups.

(To be scrupulously honest, it must be said that we assume also that if K is a quasisimple subgroup of SL(4, p) then K/Z(K) is given by the CFSG. But, presumably, this assumption is not essential.) We accept, as part of the above background hypothesis, the classification as it stands, of the Schur multipliers of the alternating groups, the groups of Lie type, and the sporadic groups, together with structural information of various kinds concerning these groups. The information that we need can be found in [GLS3] and, for the sporadic groups, in [A2]. Also, whenever the ATLAS of finite simple groups [CCNPW] asserts that a certain group occurs as a maximal subgroup of some group, we accept that information uncritically. This kind of information from the ATLAS will be used (and will indeed be used extensively) only in section 10 below, in dealing with groups of Lie type in characteristic different from p.

In order to state our results, we need first of all to establish some terminological conventions relating to specific groups and modules.

Let G be an alternating or symmetric group Alt(n) or Sym(n), and denote by P (n, p) the permutation module for G, of dimension n over 𝔽_{p}. The “natural module” for G in characteristic p is, by definition, the module V = [P (n, p), G]/C_{[P (n,p),G]}(G). When p = 2 we shall need also the notion of a “spin module” for G, and for this we adopt the view-point taken in [M]. Thus, set U = [P (n, 2), G]. There is then a natural G-invariant quadratic form Q on U, and if Q is degenerate then n is divisible by 4 and the radical of Q has dimension 1. In any case, set \overline{U} = U/Rad(Q), and let \overline{Q} be the (non-degenerate) form on \overline{U} induced by Q. There is then an induced action of G on the Clifford algebra C associated with \overline{Q}. An 𝔽_{2}-module V is then said to be a spin module for G if V is isomorphic to an irreducible G-submodule of C.

The class Lie(p) of “groups of Lie type in characteristic p” will have the following meaning. A group X is in Lie(p) if there is a simple, affine algebraic group \overline{X}, defined over an algebraic closure of 𝔽_{p}, and a Steinberg endomorphism σ of \overline{X}, such that X is isomorphic to the subgroup of \overline{X} generated by the elements of p-power order in C\overline{X}(σ). Having said this, we then make three exceptions. Namely, it will be convenient, for our purposes, to consider also the commutator subgroups of Sp(4, 2), ^{2}F_{4}(2), and of ^{2}G_{2}(3) to be groups of Lie type in the characteristics 2, 2, and 3, respectively.

By “a” natural module” for SL(2, p^{n}) we mean any module for SL(2, p^{n}) which is isomorphic over 𝔽_{p} to “the” natural module for SL(2, p^{n}) (and by which we mean the vector space of dimension 2 over 𝔽_{p^{n}} on which SL(2, p^{n}) acts by matrix multiplication). We also have the notion of a natural Ω_{3}(p^{n})- module and, for n even, a natural Ω_{4}^{−}(p^{n})-module for SL(2, p^{n}), given by identifying P SL(2, p^{n}) with one or the other of these orthogonal groups. In the same way, we have the notion of a natural module for SU(3, p^{n}). A natural module for a Suzuki group Sz(2^{n}) is, by definition, any irreducible module of dimension 4 over 𝔽_{2^{n}} for Sz(2^{n}) (n odd), it being known from [St2] that all such modules are isomorphic over 𝔽_{2} to a module obtained from an embedding of Sz(2^{n}) in Sp(4, 2^{n}) as the fixed points of a symplectic polarity. The notions of natural modules for SL(3, p^{n}) and Sp(4, p^{n}) should now require no further explanation. If V is a natural module for Sp(4, 2^{n}) then a “contragredient module” associated with V is, by definition, the image of V under a symplectic polarity.

We require also some terminology concerning the groups A which enter into the various hypotheses stated above. Thus, let V be a vector space over 𝔽_{p}, and let G be a subgroup of GL(V). A subgroup A of G is said to act quadratically on V (or to be a quadratic subgroup of G) if A ≠ 1 and [V, A, A] = 0. It is an entirely elementary result that such a group A is necessarily an elementary abelian p-group (for which see lemma 1.1 in [C1], for example). Let Y be a subgroup of G. We denote by Q(Y, V) the set of all quadratic subgroups A of Y. Assuming that Q(Y, V) ≠ ∅, we then define q(Y, V) to be the smallest real number q for which there exists A ∈ Q(Y, V) with |A|^{q}|C_{V} (A)| = |V |. We then set

Q^{∗}(Y, V) = {A ∈ Q(Y, V) | |A|^{q(Y,V)}|C_{V} (A)| = |V |}.

For i = 1 and 2, we say that V is an F i-module for G if there exists a non-identity elementary abelian p-subgroup A of G such that |A|^{i} ≥ |V/C_{V} (A)|, and that V is a quadratic F i-module if such an A can be chosen in Q(G, V). The group A is then said to be an F i-offender on V, and a quadratic F i-offender if also A acts quadratically on V. If |A| = p and A is an F 2-offender on V then any non-identity element of A is said to be a 2-transvection on V.

We may now state our main results.

Theorem 1. Assume Hypothesis 4. Then there exists a subgroup K of H, unique up to conjugation, such that, upon setting U = [V, K], the following conditions hold.

(a) H = K_{1} × ⋯ × K_{r} where {K_{1}, ⋯, K_{r}} = K^{S}.

(b) We have [V, K_{i}, K_{j}] = 0 whenever i ≠ j.

(c) One of the following holds:

(i) K ≅ O^{p}(SL(2, p^{n})), n ≥ 1, and U/C_{U}(K) is a natural SL(2, p^{n})-module for K, or a direct sum of two natural modules for K.

(ii) K ≅ O^{p}(O_{4}^{ϵ}(p^{n})), n ≥ 1, ϵ = ±1, and U is a natural orthogonal module for

K.

(iii) K ≅ O^{p}(SU(3, p^{n})) and U is a natural module for K.
(iv) p = 2, K ≅ O^{2}(Sz(2^{n})), and U is a natural module for K.

(v) p = 2, A ≤ C_{G}(K)K, K ≅ SL(3, 2^{n}) (resp. O^{2}(Sp(4, 2^{n})) and U is the direct sum of a natural and a dual module (resp. a natural and a contragredient module) for K. Moreover, there exists g ∈ N_{S}(K) such that g interchanges, by conjugation, the two maximal subgroups of K containing S ∩ K.

(vi) p = 2, K ≅ Alt(2^{n} + 1), n ≥ 3, and U is a natural module for K, or a

direct sum of two natural modules for K. (vii) p = 2, K ≅ Alt(9), and U is a spin module for K, of dimenson 8 over 𝔽_{2}.

Moreover, if K is not invariant under Q^{∗}(S, V) then p = 2, K ≅ ℤ_{3}, |U| = 4, and q(S, V) = 2.

Two special cases of Theorem 1 are of importance in their own right, where q(A, V) < 2, and where q(A, V) ≤ 1. Theorems 2 and 3 describe the possible outcomes in these two cases.

Theorem 2. In Theorem 1, suppose that we have q(A, V) < 2. Then one of the following holds.

(i) K ≅ O^{p}(SL(2, p^{n})), n ≥ 1, and U/C_{U}(K) is a natural SL(2, p^{n})-module for K. (ii) p = 2, KA/C_{KA}(K) ≅ O_{4}^{ϵ}(2^{n}), n ≥ 1, ϵ = ±1, and U is a natural orthogonal

module for K. Moreover, we have |A/C_{A}(K)| = 2^{n+1}, and if n = 1 then ϵ = −1. (iii) p = 2, A ≤ C_{G}(K)K, K ≅ SL(3, 2^{n}) and U is the direct sum of a natural and a

dual module for K. Moreover, there exists g ∈ N_{S}(K) such that g interchanges, by conjugation, the two maximal subgroups of K containing S ∩ K.

(iv) p = 2, K ≅ Alt(2^{n} + 1), n ≥ 3, and U is isomorphic to the natural module for K.

Theorem 3. In Theorem 1, suppose that we have q(S, V) ≤ 1. Denote by A(S, V) the set of all subgroups A of S such that |A| ≥ |V/C_{V} (A)| and such that [V, A, A] = 0. Then q(S, V) = 1 and H ≤ ⟨A(S, V)^{G}⟩ = E_{1} ⋯ E_{r} where E_{i} is a subnormal subgroup of G and where one of the following holds.

(i) E_{i} ≅ SL(2, p^{n}), n ≥ 1, and [V, E_{i}] is a natural SL(2, p^{n})-module for E_{i}.

(ii) p = 2, E_{i} ≅ Sym(2^{n} + 1), n ≥ 2, and V = C_{V} (E_{i}) ⊕ [V, E_{i}], where [V, E_{i}] is isomorphic to the natural module for E_{i}.

In Theorems 4,5, and 6 we drop the assumption that A acts quadratically.

Theorem 4. Assume Hypothesis 3, and assume that H is a quasisimple group of Lie type in characteristic p. Let A be an F 2-offender on V. Then one of the following holds.

(a) H ≅ SL(2, p^{n}) and one of the following holds. (i) V is a natural SL(2, p^{n})-module for H. (ii) V has an H- submodule U such that both U and V/U are natural SL(2, p^{n})-

modules for H. (iii) p is odd, H ≅ L_{2}(p^{n}), and V is a natural Ω_{3}(p^{n})-module for H. (iv) n is even and V is a natural Ω_{4}^{−}(p^{n/2})-module for H.

(b) H ≅ SU(3, p^{n}) and V is a natural module for H.

(c) p = 2, H ≅ Sz(2^{n}), and V is a natural module for H.

(d) p = 2, H ≅ SL(3, 2^{n}) (resp. O^{2}(Sp(4, 2^{n})) and V is the direct sum of a natural and a dual module (resp. a natural and a contragredient module) for H. Moreover, there exists g ∈ N_{S}(K) such that g interchanges, by conjugation, the two maximal subgroups of H containing S ∩ H. If A is a quadratic F 2-offender then A ≤ H, and if also H ≅ Sp(4, 2^{n}) then A is conjugate to Z(S).

Theorem 5. Assume Hypothesis 4′, and assume that H/Z(H) is isomorphic to Alt(n), n ≥ 5. Assume also that H/Z(H) is not of Lie type in characteristic p, and if p = 2 assume that n is odd. Then one of the following holds.

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

(ii) p = 2, H ≅ Alt(n), n = 5, 7, or 9, and V is a spin module for H (of dimension 4, 4, or 8, respectively). Moreover, if n = 9 then A is the direct product of two quadratic fours groups in H.

(iii) |A| = p = 3, G ≅ Alt(n), n ≠ 6, and V is a natural module for G. Moreover, A

is generated by a 3-cycle. (iv) |A| = p = 3, H ≅ SL(2, 5), and V is isomorphic to the natural SL(2, 9)-module

for H.

(v) p = 3, H ≅ Alt(9), |A| = 27, and V is a spin module for H (of dimension 8 over 𝔽_{3}). Moreover, we have |A|^{2} = |V/C_{V} (A)|.

Theorem 6. Assume Hypothesis 4′, and assume that S is contained in a unique maximal subgroup of G. Assume also that H/Z(H) is a quasisimple group of Lie type in characteristic different from p, or a sporadic group, and that there exists no isomorphism of H/Z(H) with a group of Lie type in characteristic p. Then p = 3, G/Z(H) ≅ Sp(6, 2), |A|^{2} = |V/C_{V} (A)|, and one of the following holds.

(i) |A| = 3, Z(H) = 1, and V has dimension 7 over 𝔽_{3}.

(ii) |A| = 27, |Z(H)| = 2, and V has dimension 8 over 𝔽_{3}.

We mention that the case where G is solvable is hidden in Theorem 1. Section 4, below, contains many results concerning the case where F^{∗}(G) is solvable, and which are not revealed in the statements of Theorems 1 through 6. Of these results, 4.6 is the most important, and figures strongly in the proof of 11.1. There is a certain sense in which this case forms the core of this paper, and we therefore invite the reader to glance through section 4 before undertaking a thorough reading of the whole.

We attempt to list our notational conventions, many of which stem from the ATLAS), and to list also those cases where we are torn between competing conventions. We denote by 2_{ϵ}^{1+2n} an extraspecial 2-group T whose corresponding orthogonal space T/Z(T) has “defect” ϵ, where ϵ = 1 or −1. But we also write Q_{8} and D_{8} for the quaternion and dihedral groups, respectively, of order 8, and we write X ∘ Y for the “central product” of the groups X and Y, this being the direct product, modulo the relations which identify the centers of X and Y, and which is defined only only if those centers are isomorphic. An alternative notation for the groups 2_{ϵ}^{1+2n} is given by the iterated central products D_{8}^{(n)} and Q_{8}^{(n)}.

The symbol [n] indicates a group of order n, and if p is a prime then p^{n} is often used to denote an elementary abelian group of order p^{n}. If X and Y are groups then X.Y or (X).Y or sometimes just XY is said to be the “shape” of a group G provided that G has a normal subgroup X∗ isomorphic to X, with G/X∗ isomorphic to Y, and with G not isomorphic to the direct product of X and Y. Conventional notation for the groups of Lie type will always be taken to give the corresponding adjoint version. So, the center of the group ^{2}A_{n}(q), for example, is trivial. We often write L_{n}(q) for P SL(n, q), and U_{n}(q) for P SU(n, q). In general we write [X, X] for the commutator subgroup of a group X, but in the case where X is a group of Lie type or, more specifically, a classical group, we feel free to use the “apostrophic” notation to indicate derived groups. Thus, Sp(4, 2^{n})′, or U_{3}(2)′, or ^{2}F_{4}(2)′, for example. Our hope is that other notation, which we have not had the foresight to introduce at this point, will be self- explanatory.

Section 1: Decomposability questions

1.1 Lemma. Let G be a group acting faithfully on a vector space V over 𝔽_{p}. Let T be a Sylow p-subgroup of O^{p}(G), and assume that V = ⟨C_{V} (T)^{G}⟩. Then V = [V, O^{p}(G)] + C_{V} (O^{p}(G)).

Proof. Set H = O_{p}(G), let {h_{1}, ⋯, h_{r}} be a right transversal for T in H, let u be a nonzero element of C_{V} (T), and set v = u^{h_{1}} + ⋯ + u^{h_{r}}. Each g ∈ H defines a permutation in Sym(r) by means of the formula h_{i}g ∈ T h_{ig}, and then

v^{g} = u^{h_{1}g} + ⋯ + u^{h_{r}g} = u^{h_{1g}} + ⋯ + u^{h_{rg}} = v.

If u ∉ [V, H] then v = ru + w for some w ∈ [V, H], and so ru ∈ [V, H] + C_{V} (H). As p does not divide r, we conclude that C_{V} (T) ≤ [V, H] + C_{V} (H). As V = ⟨C_{V} (T)^{G}⟩, we then have the lemma. □

1.2 Corollary. Assume Hypothesis 2, and suppose that there is a unique non-trivial irreducible constituent for G in V. Then V is irreducible.

Proof. Immediate from 1.1. □

1.3 Lemma. Let (L, U) be a pair consisting of a group L and an irreducible 𝔽_{2}[L]-module U. Assume that (L, U) is given by one of the following:

(1) L ≅ SL(2, 2^{n}), U the natural module.

(2) L ≅ Ω_{4}^{ϵ}(2^{n}), ϵ = ±1, U the natural orthogonal module.

(3) L ≅ SL(3, 2^{n}), U the natural module.

(4) L ≅ Sp(4, 2^{n})′, U the natural module for Sp(4, 2^{n}).

(5) L ≅ Alt(n), U the natural module.

Then one of the following holds.

(i) H^{1}(L, U) = 0.

(ii) L ≅ SL(2, 2^{n}) and H^{1}(L, U) ≅ 𝔽_{2^{n}}.

(iii) L ≅ SL(3, 2) and H^{1}(L, U) ≅ 𝔽_{2}.

(iv) L ≅ Sp(4, 2^{n})′ and H^{1}(L, U) ≅ 𝔽_{2^{n}}.

(v) L ≅ Alt(n), n even, and H^{1}(L, U) ≅ 𝔽_{2}.

Proof. In each of the cases to be considered, let W denote an L-module containing U, with [W, L] = U and with C_{W} (L) = 0. Suppose first that L ≅ SL(2, 2^{n}) and that U is the natural module for L. For any involution t in L we have [W, t] ≤ C_{U}(t), and hence |W/C_{W} (t)| ≤ 2^{n}. Since three involutions suffice to generate L, it follows that |W | ≤ 2^{3n}. On the other hand, one may produce an example where |W | = 2^{3n}, in the following way. Put L^{∗} = SL(2, 2^{2n}) and regard L as a subgroup of L^{∗}. Identify L^{∗} with Ω_{4}^{−}(2^{n}) and let V be the natural orthogonal module for L^{∗}. Then L centralizes a 1-dimensional singular subspace V_{0} of V, and by taking W = V/V_{0} we obtain the desired example. Thus |H^{1}(L, U)| = 2^{n}.

Next, take L = Ω_{4}^{−}(2^{n}), U the natural orthogonal module. Assume that |W/U| = 2. Put q = 2^{n} and let D be a subgroup of L of order q + 1. Then |C_{W} (D)| = 2q^{2}. For any involution t of L one then has C_{W} (⟨D, t⟩) ≠ 0. But an elementary counting argument shows that one may choose t so that ⟨D, t⟩ is not an SL(2, 2^{n})-subgroup of L, and is not a dihedral group of order 2(q + 1). Then ⟨D, t⟩ = L, by appeal to Dickson’s list of subgroups of L. That is, C_{W} (L) ≠ 0, so W is decomposable after all, and H^{1}(L, U) = 0.

Assume next that L = Ω_{4}^{+}(2^{n}), U the natural module, and form the semi-direct product H = U : L. Let K = K_{1} × K_{2} be a complement to U in H, K_{i} ≅ SL(2, 2^{n}). After a suitable conjugation, we may assume that K∩L contains a subgroup X = X_{1}×X_{2}, where X_{i} = X ∩ K_{i} is of order 2^{n} + 1. But C_{U}(X_{i}) = 0, so C_{H}(X_{i}) = X_{i}K_{3−i}. This shows that K = L, and so there is only one conjugacy class of complements to U in H. That is, H^{1}(L, U) = 0.

We next consider the case where L = SL(3, 2^{n}) or Sp(4, 2^{n})′. Let P_{1} and P_{2} be the two maximal subgroups of L containing a fixed Sylow 2-subgroup S of L. Put L_{i} = [P_{i}, P_{i}], Q_{i} = O_{2}(P_{i}), and Z_{i} = C_{Q_{i}}(L_{i}), and let the indexing be chosen so that C_{U}(L_{1}) ≠ 0.

Assume now that L ≅ SL(3, 2^{n}), n > 1. Assume that |W/U| = 2, and let K be a complement to Q_{2} in L_{2}. Then both Q_{2} and C_{U}(Q_{2}) are natural SL(2, 2^{n})- modules for K, so we have |Hom_{K}(Q_{2}, C_{U}(Q_{2}))| = 2^{n}. Then [W, L_{2}] ≤ C_{U}(Q_{2}), and so W = U + C_{W} (Q_{2}). Then [W, Q_{1} ∩ Q_{2}] ≤ [U, Q_{1}] = C_{U}(L_{1}), and then also [W, Q_{1}] = [W, ⟨(Q_{1} ∩ Q_{2})^{L_{1}}⟩] ≤ C_{U}(L_{1}). This yields |C_{W} (Q_{1})| = |C_{W} (L_{1})| = 2^{n+1}. But also, for any involution t ∈ L we have |W/C_{W} (t)| = 2^{n}. Choosing t ∈ L − L_{1}, we get L = ⟨L_{1}, t⟩, and so C_{W} (L) ≠ 0. Thus H^{1}(L, U) = 0.

One may produce an indecomposable module for L = L_{3}(2) by considering the action of Alt(7) on the natural L_{4}(2)-module, and observing that Alt(7) contains two classes of L_{3}(2)-subgroups. Thus |H^{1}(L, U)| ≥ 2 in this case. Let W be an L-module of order 32, with U = [W, L]. For any y ∈ L with |y| = 3 one has |[W, y]| = 4, and since two conjugates of y suffice to generate L we then have C_{W} (L) ≠ 0. This shows that |H^{1}(L, U)| = 2 if L ≅ L_{3}(2).

Next, suppose that L ≅ Sp(4, 2^{n}), n > 1, and assume that |W/U| > 2^{n}. We have |U/C_{U}(Z_{1})| = |[U, Z_{1}]| = 2^{n}, and [W, L_{1}] ≤ C_{U}(Z_{1}). The Three Subgroups Lemma yields [W, Z_{1}, L_{1}] = 0, and so [W, Z_{1}] = [U, Z_{1}]. Let K_{1} be a complement to Q_{1} in L_{1} such that K_{1} is generated by two conjugates of Z_{1}. Then |[W, K_{1}]|| ≤ 2^{2n}, and [W, K_{1}] is a natural SL(2, 2^{n})-module for K_{1}. By what has already been proved, we have |H^{1}(K_{1}, [W, K_{1}])| = 2^{n}, and so there exists w ∈ W − U with [w, K_{1}] = 0. Then also there exists v ∈ W − U with [v, Z_{1}] = 0. Set W_{0} = ⟨v⟩ + U. As four conjugates of Z_{1} suffice to generate L we then have C_{W_{0}}(L) ≠ 0, contrary to assumption. This shows that |H^{1}(L, U)| ≤ 2^{n}. To see that equality holds, it is enough to observe that in Sp(6, 2^{n}) there is a maximal parabolic subgroup P with O^{2′}(P/O_{2}(P)) ≅ L and with O_{2}(P) an indecomposable module for O^{2′}(P), such that C_{O_{2}(P)}(O^{2′}(P)) is of order 2^{n} and such that O_{2}(P)/C_{O_{2}(P)}(O^{2′}(P)) is a natural SL(2, 2^{n})-module for O^{2′}(P/O_{2}(P)). (This is entirely well known, but one can find this worked out in detail in lemma 5.4 of [C2].)

Suppose next that L = (Sp(4, 2))′. The natural permutation module for Alt(6) provides an example which shows that |H^{1}(L, U)| ≥ 2. Let W be an L-module of order 64 with U = [W, L], and let L_{0} be a subgroup of L, L_{0} ≅ Alt(5), such that U is the natural Ω_{4}^{−}(2)-module for L_{0}. By what has already been shown, we have H^{1}(L_{0}, U) = 0, and so W = U ⊕ C_{W} (L_{0}). Let S be a Sylow 2-subgroup of L such that |S ∩ L_{0}| = 4, and let t be an involution in S − L_{0}. Here |C_{U}(S ∩ L_{0})| = 2, and so C_{W} (L_{0}) ∩ C_{W} (t) ≠ 0. As ⟨L_{0}, t⟩ = L we then have C_{W} (L) ≠ 0, and this shows that |H^{1}(L, U)| = 2.

Finally, let L = Alt(n), with U the natural module. Suppose that n = 2m + 1 is odd. Then L is generated by m 3-cycles, each of which centralizes a subspace of codimension 2 in W. Then dim(W) ≤ 2m = dim(U). Thus W = U, and H^{1}(L, U) = 0 in this case. Now suppose that n = 2m is even. Then m − 1 3-cycles suffice to generate the subgroup K_{i} of L which fixes the point i in the natural action of L on n points. Here C_{U}(K_{i}) = 0, and dim(U) = 2m−2, so we have W = U +C_{W} (K_{i}). Set X = K_{1} ∩K_{2}, and observe that dim(C_{U}(X)) = 1. Thus C_{W} (K_{i}) is a hyperplane of C_{W} (X), for i = 1 and 2. As ⟨K_{1}, K_{2}⟩ = G we have C_{W} (K_{1}) ∩ C_{W} (K_{2}) = 0, and we conclude that |W/U| ≤ 2. On the other hand, the natural permutation module for L, modulo its fixed-points for L, provides an example of an indecomposable module W in which |W/U| = 2. Thus |H^{1}(L, U)| = 2 in this case, and the lemma is proved. □

1.4 Lemma. Let L be the group L_{2}(q), q a power of p, p odd, and let V be a module for L over 𝔽_{p}, such that [V/C_{V} (L), L] is isomorphic to the natural Ω_{3}(q)-module for L. Then V = [V, L] ⊕ C_{V} (L).

Proof. Let T be a Sylow p-subgroup of L, let D be a complement to T in N_{L}(T), let U be the natural Ω_{3}(q)-module for L, and form the semi-direct product U : L. We will show that there is a unique conjugacy class of D-invariant complements to U in UT, and the lemma will follow from a standard result on 1-cohomology. (See 17.7 in [A3].)

Let T_{1} be a D-invariant complement to U in UT, and set X = ⟨T^{U}⟩. Then X = [U, T]T, where [U, T] is a 2-dimensional 𝔽_{q}- subspace of U. Every element of order p in UT lies in X, so we have T_{1} ≤ X. Notice that D acts trivially on the 1-dimensional space [U, T]/C_{U}(T), and so the q distinct conjugates of T under [U, T] are all D-invariant. These conjugates, together with C_{U}(T), provide a partition of (C_{U}(T)T)^{#} into D-invariant sets, so if T_{1} ≤ C_{U}(T)T then T_{1} is conjugate to T. Suppose now that q > 3, so that D ≠ 1. Then [X, D] = C_{U}(T)T, and so T_{1} ≤ C_{U}(T)T in this case. Thus, we are reduced to the case p = 3. Here there are exactly 9 conjugates of T under U, and there are exactly 13 = (3^{3} − 1)/(3 − 1) subgroups of X of order 3, four of which lie in U. Thus T_{1} is conjugate to T in this case as well, and the lemma is proved. □

1.5 Lemma. Let L = L_{3}(2) and let V be an 𝔽_{2}L-module such that C_{V} (L) ≤ [V, L], and such that [V/C_{V} (L), L] is a natural L_{3}(2)-module for L. Then either V = [V, L] or C_{V} (L) = 0.

Proof. Suppose false. Then 1.1 implies that |V/[V, L]| = |C_{V} (L)| = 2, and so |V | = 32. Let t be an involution in L. Then two conjugates of t generate a Sylow 2-subgroup of L, and then three conjugates of t suffice to generate L. If t induces a transvection on [V, L] we then obtain C_{[V,L]}(L) ≠ 0, while if t induces a transvection on V/C_{V} (L) then L has a non-trivial fixed-point on V/C_{V} (L). These results are contrary to the case, so we conclude that t induces a 2- transvection on both [V, L] and V/C_{V} (L). Then C_{V} (t) ≤ [V, L], and |V/C_{V} (t)| ≥ 8. But |V/C_{V} (t)| ≤ 4 as |V | = 32. This contradiction proves the lemma. □

1.6 Lemma. Let L = SL(2, p^{n}) and let U be a module for L over 𝔽_{p}. Suppose that every irreducible constituent for L in U is a natural module for L. If p = 3 assume further that a Sylow p-subgroup of L acts quadratically on U. Then U is completely reducible.

Proof. This is an old result of Richard Niles (Theorem 3.2 in [N]). To be precise, Niles’ theorem states that if the Sylow p-subgroup S of L does not act quadratically on V, and V is indecomposable, then p = 3. But in fact, Niles’ proof makes no use of the assumption that S acts non-quadratically, and what is really proved is that if V is indecomposable then p = 3 and S acts non-quadratically. The lemma follows from this result. □

1.7 Lemma. Let L = SL(2, p^{n}), and let V = [V, L] be a non-zero module for L over 𝔽_{p}. Suppose that a Sylow p-subgroup of L acts quadratically on V, and if p = 2 assume that C_{V} (X) = 0 for some Cartan subgroup X of L. Then V is a direct sum of natural modules for L.

Proof. If p is odd then C_{U}(Z(L)) = C_{U}(L), and we may assume that C_{U}(Z(L)) = 0 in this case. Let U be an irreducible L-invariant section of V. Then U is a non-trivial L-module, since C_{U}(X) = 0 if p = 2. Let S be a Sylow p-subgroup of L, and set F = End_{L}(U). As is well known (see Corollary (a) to Theorem 46 in [St2]), F is a subfield of 𝔽_{p^{n}} and dim_{F} (C_{U}(S)) = 1. As S acts quadratically on V, by hypothesis, and as two conjugates of S generate L, it follows that dim_{F} (U) = 2, and then F = 𝔽_{p^{n}} and U is a natural module for L. Thus, we are reduced to the situation in which V has a submodule U such that both U and V/U are natural modules for L. Now apply 1.6. □

1.8 Lemma. Let L = Alt(n), n odd, and let V be an 𝔽_{2}L-module such that the only non-trivial irreducible constituents for L in V are natural modules. Then V is completely reducible.

Proof. In view of 1.3, we may assume that C_{V} (L) = 0, and then that V has a submodule U such that both U and V/U are natural modules for O^{2}(L). Let K be the stabilizer of a point for the natural action of L on n points. As n is odd, U is isomorphic, as a K-module, to the natural permutation module for K. As n − 1 is even we then have U ≠ [U, K] ≥ C_{U}(K) ≠ 0, and the same is true with V/U in place of U. As H^{1}(K, [U, K]/C_{U}(K)) ≅ 𝔽_{2}, by 1.3, it follows that C_{V/U}(K) = (U + C_{V} (K))/U ≠ 0. Let w ∈ C_{V} (K) − U and set W = ⟨w^{O_{2}(L)}⟩. Then dim(W) ≤ |O^{2}(L) : K| = n, and so W is an O^{2}(L)-invariant complement to U in V. □

Section 2: Cyclic Sylow p-subgroups

In this section we treat Hypothesis 4′, in “miniature”.

2.1 Lemma. Let V be a 4-dimensional vector space over 𝔽_{p}, p an odd prime, and set G = SL(V). Let X be a quasisimple subgroup of G, and assume that X satisfies the following conditions.

(a) Z(X) ≤ Z(G).

(b) X has cyclic Sylow p-subgroups.

(c) For any element a of X of order p, we have [V, a, a, a] = 0 and [V, a, a] ≠ 0.

Assume further that X/Z(X) is an alternating group, a group of Lie type, or one of the twenty-six sporadic groups. Then one of the following holds.

(i) p ≥ 5, X ≅ L_{2}(p), V = [V, X] ⊕ C_{V} (X), and [V, X] is a natural Ω_{3}(p)-module for X (of dimension 3).

(ii) p = 3 and X ≅ Alt(5).

Proof. The 2-rank of SL(4, p) is equal to 3, so we obtain the following information at the outset.

(1) The 2-rank of X is at most 3, and if equal to 3 then Z(X) is of even order.

Suppose first that X ∈ Lie(p). Condition (b) then implies that X ≅ L_{2}(p) or SL(2, p), and then since X is assumed to be quasisimple, we have p ≠ 3. Denote by M_{i} the space (of dimension i + 1) of homogeneous polynomials of degree i in the two variables x and y. Then M_{i} admits a natural action by SL(2, p), and {M_{i}}_{0≤i≤p−1} forms a complete set of representatives for the isomorphism classes of the irreducible modules for SL(2, p) over 𝔽_{p}. (See [St 1] or section 13 in [St 2].) Let S be a Sylow p-subgroup of SL(2, p). One checks easily that [M_{i}, S, S, S] ≠ 0 for i > 2, and so condition (c) implies that any irreducible constituent U for X in V is isomorphic to one of the modules M_{i}, 0 ≤ i ≤ 2. That is, U is either a trivial module, a natural SL(2, p)-module, or a natural Ω_{3}(p)- module for X. If there exists a constituent U such that U is a natural SL(2, p)-module then 1 ≠ Z(X) ≤ Z(G), by condition (a), so all irreducible constituents of X in V are natural SL(2, p)- modules, and then V is completely reducible for X, by 1.6. But then a acts quadratically on V, contrary to (c), so we conclude that no such constituent U exists. Thus, X has a unique non-trivial irreducible constituent U in V, and U is a natural Ω_{3}(p)-module. Now 1.2 implies that V is completely reducible, and so (i) holds.

Suppose next that X is of Lie type in characteristic different from p. We appeal to Table I in [SZ] for the list of minimal degrees of cross-characteristic projective representations of groups of Lie type, where we find that only the groups L_{2}(4), L_{2}(9), L_{3}(2), L_{3}(4), and U_{4}(2) have such representations of degree at most 4. Suppose that X/Z(X) ≅ L_{2}(4). Then p = 3 (as L_{2}(4) ∈ Lie(5)). If Z(X) ≠ 1 then there is an element f of order 4 in X such that f inverts a, with ⟨f^{2}⟩ = Z(X), and since |[V, a, a]| = 3 it follows that Z(X) centralizes [V, a, a]. But C_{V} (Z(X)) = 0, so in fact Z(X) = 1, and (ii) holds. Suppose that X/Z(X) ≅ L_{3}(2). Then p = 3 as L_{3}(2) ∈ Lie(7). As 7 does not divide the order of G, we have a contradiction in this case. We note that any central extension of L_{3}(4) has 2-rank at least 4, so (1) implies that X/Z(X) is not isomorphic to L_{3}(4). Suppose that X/Z(X) ≅ U_{4}(2). The 2-rank of U_{4}(2) is greater than 3, U_{4}(2) ≅ P Sp(4, 3), and Sp(4, 3) is the universal perfect central extension of U_{4}(2), so we have X ≅ Sp(4, 3) and p = 5. But 27 does not divide the order of SL(4, 5), so we have a contradiction. Now suppose that X/Z(X) ≅ L_{2}(9). Then X/Z(X) ∈ Lie(2) ∪ Lie(3), and so p = 5. There are then non-conjugate subgroups K_{1} and K_{2} of X, with K_{i}/Z(X) ≅ L_{2}(5), and with a ∈ K_{1} ∩ K_{2}. The argument in the preceding paragraph then shows that for each i we have K_{i} ≅ L_{2}(5), V is completely reducible for K_{i}, and [V, K_{i}] is a natural Ω_{3}(5)-module for K_{i}. Let x_{i} be an element of order 3 in K_{i}. It now follows that dim(C_{V} (x_{i})) = 2. On the other hand, a Sylow 3-subgroup of X is one of G, and evidently G contains an element x of order 3 such that C_{V} (x) = 0. As x_{1} and x_{2} represent the two conjugacy classes of elements of order 3 in X, we have a contradiction.

Suppose next that X/Z(X) is an alternating group Alt(n). The 2-rank of any central extension of Alt(8) is at least 4, so (1) implies that n ≤ 7. As Alt(5) and Alt(6) are of Lie type, we then have n = 7. By condition (b), p ≠ 3, and since Alt(7) contains L_{2}(9) we have p ≠ 5. Then p = 7, and since Alt(7) contains L_{2}(7) we have Z(X) = 1. The minimal degree of a non-trivial complex irreducible representation of Alt(6) is greater than 4, and since |Alt(6)| is not divisible by 7 the same is true of the irreducible representations of Alt(6) over 𝔽_{7}. Thus, Alt(7) is not a subgroup of G.

Suppose finally that X/Z(X) is a sporadic group. Then (1) implies that X ≅ M_{11} or 2^{·M_{12}}. Then p ≠ 3, by (b), and since M_{11} is a subgroup of 2^{·}M_{12} it will now suffice to derive a contradiction in the case that X ≅ M_{11}. In this case we have p = 5 or 11, and since M_{11} contains L_{2}(9) we conclude that p = 11. Let B be a Sylow 3-subgroup of X. Then all subgroups of order 3 in B are fused in X. But also B is a Sylow subgroup of G, and evidently there are two different classes of elements of subgroups of order 3 in G. This contradiction proves the lemma. □

2.2 Corollary. Assume Hypothesis 3, with H quasisimple, and assume that S is of order p. Suppose further that p is odd, that [V, A, A, A] = 0, and that A is not quadratic on V. Then G ≅ L_{2}(p) or Alt(5).

Proof. By Hypothesis 3 there is a unique maximal subgroup M of G containing A. As H is quasisimple, there exists g ∈ G with A^{g} ≰ M. Then ⟨A, A^{g}⟩ = G, and so G ≅ L_{2}(p) or Alt(5) by 2.1. □

2.3 Theorem. Assume Hypothesis 4′, and assume that H/Z(H) is one of the twenty-six sporadic groups. Then S is contained in more than one maximal subgroup of G.

Proof. By 2.2, |S| > p, and then a survey of the sporadic groups shows that S is noncyclic. In [A3] it is shown that if X is a group with F^{∗}(X) sporadic, and T is a non-cyclic Sylow p-subgroup of X, then either T is contained in two distinct maximal subgroups of X, or else X ≅ J_{4} and p = 11. Moreover, it is shown that if J_{4} has a faithful 112-dimensional module over 𝔽_{2} then also for p = 11 there are at least two maximal overgroups of T in J_{4}. The existence of such a module for J_{4} is now known from various sources. (?) □

Section 3: Measuring lemmas

3.1 Lemma. Let G be a group, and let V be a faithful 𝔽_{p}G-module. Suppose G has a cyclic subgroup of order p^{k} + 1. Then dim(V) ≥ 2k.

Proof. Suppose first that p^{2k} = 2^{6}. Then p^{k} + 1 = 9, and since L_{5}(2) has elementary abelian Sylow 3-subgroups we are done in this case. Also, the result evidently obtains if k = 1. Thus, we may assume that p^{2k} ≠ 2^{6} and that k > 1. The main result of [Z] then says that p^{2k} − 1 is divisible by a prime r such that r does not divide p^{i} − 1 for any i with 1 ≤ i < 2k. Then r does not divide |GL(V)| if dim(V) < 2k. □

[For sure there is a proof of 3.1 that does not use [Z].]

The next result is extracted from [CD].

3.2 Lemma. Let X be a finite group, Y a normal subgroup of X, p a prime, and V a faithful 𝔽_{p}X-module. Let r be a positive real number, and let E be an elementary abelian p-subgroup of Y, chosen so that

(*)  |E|^{r}|C_{V} (E)| is as large as possible.

Then For any x ∈ X we have |EE^{x}|^{r}|C_{V} (EE^{x})| ≥ |E|^{r}|V/C_{V} (E)|. In particular, If E is chosen to be as large as possible, subject to (*), then E is weakly closed in C_{X}(E) with respect to X.

Proof. . Set F = E^{x}, and suppose that |E|^{r}|C_{V} (E)| > |EF |^{r}|C_{V} (EF)|. We then have

\frac{|F |^{r}}{|E ∩ F |^{r}} = \frac{|EF |^{r}}{|E|^{r}} < \frac{|C_{V} (E)|}{|C_{V} (EF)|} = \frac{|C_{V} (E) + C_{V} (F)|}{|C_{V} (F)|} ≤ \frac{|C_{V} (E ∩ F)|}{|C_{V} (F)|},

and hence

(**)  |F |^{r}|C_{V} (F)| < |E ∩ F |^{r}|C_{V} (E ∩ F)|.

But |F |^{r}|C_{V} (F)| = |E|^{r}|C_{V} (E)|, and the maximality of this number, among elementary abelian subgroups of X, contradicts (**). Thus |E|^{r}|C_{V} (E)| ≤ |EF |^{r}|C_{V} (EF)|. □

3.3 Lemma. Let X be a finite group, p a prime, and V a faithful 𝔽_{p}X-module. Let A be a subgroup of X, and let r be a real number such that |A|^{r} ≥ |V/C_{V} (A)|. Then let B be a normal subgroup of A, set W = C_{V} (B), and suppose that |A/B|^{r} ≤ |W/C_{W} (A)|. Then |B|^{r} ≥ |V/C_{V} (B)|.

Proof. As |A|^{r} ≥ |V/C_{V} (A)| and |A/B|^{r} ≤ |W/C_{W} (A)|, we have

|B|^{r}|C_{V} (A)||W |/|C_{W} (A)| ≥ |V |.

But C_{W} (A) = C_{V} (A), so we obtain |B|^{r} ≥ |V/W | = |V/C_{V} (B)|, as required. □

We end this section with a result which will effectively reduce the proof of Theorem 1 to the consideration of the case where F^{∗}(G) is quasisimple or where F^{∗}(G) is a q-group for some prime q, q ≠ p.

3.4 Lemma. Assume Hypothesis 1, and let A ∈ Q(S, V) with A ∈ Q^{∗}(A, V). Let X be a non-identity subgroup of G with X = [A, X], and such that X/Φ(X) is a minimal normal subgroup of AX/Φ(X). Let A_{0} be a complement in A to C_{A}(X). Then the following hold:

(a) [V, X, C_{A}(X)] = 0.

(b) For any AX-invariant section W of [V, X] on which X acts non-trivially, we have C_{A_{0}}(W) = 1, and:

q(A_{0}, W) ≤ q(A_{0}, [V, X]) ≤ q(A, V)

(c) A_{0} ∈ Q^{∗}(A_{0}, [V, X]).

Proof. We have X ≤ ⟨A^{X}⟩, so [V, X, C_{A}(X)] = 1 since A acts quadratically on V. Put U = [V, X], and let U_{1} ≤ U_{0} be AX-invariant subspaces of U with [U_{0}, X] ≰ U_{1}. Put W = U_{0}/U_{1} and set Y = C_{X}(W). Then Y ≤ Φ(X) and so [X, C_{A_{0}}(W)] ≤ Φ(X). Thus, we get C_{A_{0}}(W) ≤ A_{0} ∩ C_{A}(X), and so C_{A_{0}}(W) = 1.

Put U^{∗} = C_{V} (C_{A}(X)) and put q = q(A, V). Suppose next that |A_{0}|^{q} < |U^{∗}/C_{U^{∗}}(A_{0})|. Since C_{U^{∗}}(A_{0}) = C_{V} (A) we then have:

|V | = |A|^{q}|C_{V} (A)| < |C_{A}(X)|^{q}|U^{∗}|

and thus q(C_{A}(X), V) < q(A, V). But this is contrary to A ∈ Q_{∗}(S, V), so we now conclude that |A_{0}|^{q} ≥ |U^{∗}/C_{U^{∗}}(A_{0})|. Then also |A_{0}|^{q} ≥ |U/C_{U}(A_{0})|, and so q(A_{0}, U) ≤ q(A, V).

Next, put r = q(A_{0}, U). We have:

|A_{0}|^{r} = |U/C_{U}(A_{0})| ≥ |U_{0}/C_{U_{0}}(A_{0})|

and also:

|C_{W} (A_{0})| ≥ |C_{U_{0}}(A_{0})U_{1}/U_{1}| = |C_{U_{0}}(A_{0})/C_{U_{1}}(A_{0})|

and thus:

|A_{0}|^{r}|C_{W} (A_{0})| ≥ |U_{0}/C_{U_{1}}(A_{0})| ≥ |W |.

This shows that q(A_{0}, W) ≤ r, and completes the proof of (b).

Next, let B ≤ A_{0} and put A_{1} = C_{A}(X)B. Since A ∈ Q_{∗}(S, V) we have q ≤ q(A_{1}, V), and so:

|A_{0}/B|^{q} = |A/A_{1}|^{q} ≥ |C_{V} (A_{1})/C_{V} (A)| ≥ |C_{U}(A_{1})/C_{U}(A)|

which shows that |A_{0}|^{q}|C_{U}(A_{0})| ≥ |B|^{q}|C_{U}(B)|. This proves (c). □

Section 4: Solvable Groups

In this section we assume Hypothesis 1. Recall that for any subgroup X of G, Q(X, V) is the set of quadratic subgroups of X, and that if Q(X, V) is non-empty then q(S, V) is the minimum, over all A ∈ Q(X, V), of the numbers q for which |A|^{q} = |V/C_{V} (A)|. Denote by Q^{∗}(X, V) the set of all A ∈ Q(X, V) such that |A|^{q(X,V)} = |V/C_{V} (A)|.

4.1 Lemma. Assume Hypothesis 1, with p odd. Let a be an element of order p in G, such that a acts quadratically on V, and let R be an a-invariant p′-subgroup of G. If p = 3 assume also that R is abelian or that |R| is odd. Then [R, a] = 1.

Proof. This is [C2, Lemma 1.2].

4.2 Lemma. Let S be a 2-group, and let a be an automorphism of S of order 3. Assume that S = [S, a], and that the following condition holds.

(*) Every a-invariant abelian subgroup of S is centralized by a.

Then Φ(S) = [S, S] = Z(S) = Ω_{1}(S) = C_{S}(a).

Proof. It will be convenient to form the semi-direct product G = S⟨a⟩. Let R denote the set of all proper subgroups R of S such that R = [R, a]. We assume that S provides a minimal counter-example to the lemma. So:

(1) For any R ∈ R we have Φ(R) = [R, R] = Z(R) = Ω_{1}(R) = C_{R}(a).

Put X = ⟨C_{S}(a)^{G}⟩. Then X ⊆ Φ(S) since S = [S, a]. Here Φ(S) = ⟨s^{2} : s ∈ S⟩ since S is a 2-group. Suppose that X ≠ Φ(S) and let g ∈ S with g^{2} ∉ X. Put T = ⟨g, g^{a}⟩Φ(S), and put R = [T, a]. Then T = RX. Now R/(R ∩ X) ≅ T/X, and so R/(R ∩ X) is not elementary abelian. If now T ≠ S, then R ∈ R and (1) then implies that R/C_{R}(a) is elementary abelian, for a contradiction. We have thus shown:

(2) Either X = Φ(S) or |S/Φ(S)| = 4.

Notice that Z(S) ⊆ X,by (*). Now let Y be a normal 2-subgroup of G properly containing Z(S), and with Y minimal for this property. Moreover, if possible, choose Y ⊆ X. If [Y, a] = 1 then [Y, S] ⊆ [Y, ⟨a^{S}⟩] = 1, contrary to the choice of Y. Thus |Y : Z(S)| = 4. Setting H = [Y, a], it then follows from (*) that H is a quaternion group, of order 8. Notice that H is invariant under C_{S}(a), and then [H, C_{S}(a)] = 1. But then 1 = [Z(S)H, C_{S}(a)] = [Y, C_{S}(a)], and so [Y, X] = 1. In particular, we have [H, X] = 1, and so H ⊈ X. By our choice of Y, it then follows that [X, a] = 1, whence [X, ⟨a^{S}⟩] = 1. Thus:

(3) We have X = C_{S}(a) = Z(S).

Suppose next that X = Φ(S). In particular, (3) then says that the nilpotence class of S is 2. Let Q denote the set of all subgroups Q of S such that Q = [Q, a] is a quaternion group, and put Z = ⟨[Q, Q] : Q ∈ Q⟩. Then Z ⊆ Ω_{1}(Z(S)). Let t be an arbitrary element of S and put D = ⟨t, t^{a}, t^{a^{2}}⟩ and R = [D, a]. Then either R ∈ Q or R = 1, and in either case we have D = C_{D}(a)R = (D ∩ X)R. Write t = xr with r ∈ R and x ∈ X. Then t^{2} = x^{2}r^{2} ∈ Φ(X)Z. But then X = Φ(S) = Φ(X)Z, and so X = Z. In particular, we now have Φ(S) = Ω_{1}(Z(S)) = [S, S]. Moreover, if t is an involution then 1 = x^{2}r^{2} = r^{2}, and so r ∈ Z(R) ⊆ X, and t ∈ X. This shows that Ω_{1}(S) = X, and thus the proposition is proved in the case that X = Φ(S). In view of (2) we then have:

(4) |S/Φ(S)| = 4, and X is a proper subgroup of Φ(S).

Here Φ(S) = [Φ(S), a]X, so it follows from (1) that Φ(S)/X is elementary abelian. Suppose that |Φ(S)/X| > 4. We may then choose a normal subgroup M of G with X ⊆ M ⊂ Φ(S), and with |Φ(S)/M| = 16. Put V = Φ(S)/M. Then G/Φ(S) operates faithfully on V, as otherwise |S/Φ(S)| > 4. Here G/Φ(S) ≅ Alt(4), and a is fixed-point-free on V. One may then verify that for every g in S − Φ(S) we have C_{V} (g) = [V, g] = C_{V} (S) = [V, S], a subgroup of V of order 4. But then g^{2} lies in [S, Φ(S)] for all g ∈ S, whereas Φ(S) = ⟨g^{2} : g ∈ S⟩. Thus we have a contradiction, proving that |Φ(S)/X| = 4. This immediately implies:

(5) S/X ≅ ℤ_{4} × ℤ_{4}.

Recall from the proof of (3) that we have a quaternion group H = [H, a] with H ⊈ X. It now follows from (5) that Φ(S) = HX. Suppose that Ω_{1}(X) ≠ Z(H), and let Z be a subgroup of X of order 2 with Z ≠ Z(H). Then observe that S/Z satisfies the condition (*), in place of S. (Indeed, if g ∈ S − C_{S}(a) then ⟨a, a^{g}⟩ = H⟨a⟩ or G.) By minimality of the counter-example, we then have [Φ(S/Z), a] = 1, which is contrary to (5). Thus, we conclude that Ω_{1}(X) = Z(H), and hence X is cyclic. Then H is the unique quaternion subgroup of HX, and so H is a normal subgroup of G. Then S = C_{S}(H)H, a central product. This violates |S/Φ(S)| = 4, and the proposition is thereby proved. □

4.3 Lemma. Assume Hypothesis 1 with G = QA, Q a normal p′- subgroup of G, and with |A| = p. Assume further that Q = [Q, A], V = [V, Q], and |V : C_{V} (A)| ≤ p^{2}. Finally, assume that A acts quadratically on V. Then one of the following holds.

(i) G ≅ SL(2, 2) and V is either a natural SL(2, 2)-module or a direct sum of two natural SL(2, 2)-modules for G.

(ii) G ≅ (3 × 3) : 2, |V | = 16, and V is the direct sum of two G-invariant subspaces

of dimension 2. (iii) G ≅ (SU(3, 2))′, and V is a natural SU(3, 2)-module for G, of order 64. (iv) G ≅ Dih(10) and |V | = 16.

(v) G ≅ SL(2, 3), V is either a natural SL(2, 3)-module or a direct sum of two natural SL(2, 3)-modules for G.

(vi) G ≅ (Q_{8}×Q_{8}) : 3, |V | = 81, and V is the direct sum of two G-invariant subspaces of dimension 2.

Proof. Let (G, V) be a counter-example with |G|+|V | as small as possible. Write A = ⟨a⟩.

Suppose first that p is odd. As Q is generated by its A-invariant Sylow subgroups, 4.1 implies that p = 3 and that C_{Q}(A) has index a power of 2 in Q. Let T be an Ainvariant Sylow 2-subgroup of Q, and put X = [T, A]. Then G = C_{G}(A)X and we have ⟨Z(X)^{G}⟩ = ⟨Z(X)^{C_{G}(A)}⟩ ≤ C_{G}(A), and so ⟨Z(X)^{G}⟩ centralizes ⟨A^{G}⟩ = G. That is, we have Z(X) ≤ Z(G).

Suppose now that X ≠ Q. The minimality of |G| + |V | then implies that outcome (v) or (vi) of the lemma holds, with (XA, [V, X]) in place of (G, V). Then [V, X] = [V, Z(X)] and C_{V} (X) = C_{V} (Z(X)). Thus [V, X] and C_{V} (X) are G-invariant. As Q = [Q, A] = ⟨X^{Q}⟩, we conclude that [C_{V} (X), Q] = 0, so C_{V} (X) = 0, and |V | = 9 or 81. Observe that in SL(4, 3) the centralizer of any element of order 3 is a {2, 3}-group. Thus C_{Q}(A) ≤ T, and thus Q = T = X. Thus, we have shown that Q = X after all.

Now 4.2 yields Φ(Q) = [Q, Q] = Z(Q) = Ω_{1}(Q) = C_{Q}(A). Suppose that |Z(Q)| = 2. Then Q is a quaternion group of order 8 and G is generated by two conjugates of A. As |V/C_{V} (A)| ≤ 9 we then have |V | ≤ 81, and then 2.3 implies that V is a direct sum of two natural SL(2, 3)-modules for G. That is, (v) holds, and (G, V) is not a counter-example. We therefore conclude that |Z(Q)| > 2. Let ⟨s, t⟩ be a fours group in Z(Q), with the generators s and t chosen so that both C_{V} (s) and C_{V} (t) are non-trivial. As G = ⟨A^{G}⟩, A acts non-trivially on both [V, s] and on C_{V} (s), and so A induces a transvection on each of these subspaces. By induction, both [V, s] and C_{V} (s) are of order 9, and then C_{V} (s) = [V, t], and V = [V, s] + [V, t]. Set Q_{1} = C_{Q}([V, s]) and Q_{2} = C_{Q}([V, t]). Then Q_{1} ∩ Q_{2} = 1, and Q/Q_{i} is a quaternion group for i = 1, 2. As Φ(Q) is non-cyclic, the 2-rank m of Q/Φ(Q) is greater than 2, and since Q = [Q, A] we have m = 4. It follows that each Q_{i} is a quaternion group, and Q = Q_{1} × Q_{2}. Thus, (vi) holds, and so the lemma holds if p is odd.

We now take p = 2. Let R be the set of all proper subgroups R of Q with R = [R, A] and with |R| > 3. For any R ∈ R, the pair (RA, [V, R]) is then given by one of the outcomes (ii) through (iv) in the statement of the lemma. We note also the following consequence of our hypothesis that V = [V, Q].

(1) No proper Q-invariant subspace of V contains [V, A].

From this it follows that:

(2) If R ∈ R and R is a normal subgroup of Q, then V = [V, R].

Let 1 ≠ x ∈ Q with x^{a} = x^{−1}. Then |V/C_{V} (x)| ≤ 16, and then ⟨a, x⟩ is dihedral of order 6 or 10. Suppose |x| = 5, and then suppose that there exists y ∈ Q − ⟨x⟩ with y^{a} = y^{−1}. Examining outcomes (i) through (iv) of the Lemma, we see that ⟨x, y⟩ ∉ R, and hence ⟨x, y⟩ = Q. Here [V, A] ≤ [V, x], so |V | ≤ |[V, x] + [V, y]| ≤ 2^{6}, and G may be identified with a subgroup of L_{6}(2). One may readily verify that no subgroup of L_{6}(2) of odd order admits faithful action by a dihedral group of order 10. Therefore x lies in Z(Q), and then (2) yields that V = [V, x] is of order 16. Then Q is cyclic, and since SL(4, 2) contains no dihedral group of order 30 it follows that G is dihedral of order 10. Thus, no element y exists as chosen above, and so (iv) holds in this case.

We now conclude that x^{3} = 1 for all x in Q with x^{a} = x^{−1}. Then Q = C_{Q}(A)T where T is an A-invariant Sylow 3-subgroup of Q. Then [Q, A] ≤ T, so since Q = [Q, A] by hypothesis we obtain [Q, A] = T, and Q is a 3-group. Put \overline{Q} = Q/Φ(Q), and suppose first that |\overline{Q}| > 9. Then any maximal subgroup of \overline{Q} is the image of some element of R, and so |\overline{Q}| = 27. Notice that for any R ∈ R we have C_{R}(A) = Φ(R) and R/C_{R}(A) = 9. It follows R ∩ Φ(Q) ≤ C_{R}(a), and then [Φ(Q), A] = 1 and Φ(Q) ≤ Z(G). Suppose that there exists R ∈ R with R non-abelian. Then (RA, [V, R]) is described by outcome (iii) of the lemma, so [V, R] = [V, Z(R)], and thus [V, R] is G-invariant. Then V = [V, R], by (2). Then |V | = 64,, and since R ≠ Q, Q is isomorphic to a Sylow 3-subgroup of L_{6}(2), of order 3^{4}. Then Q is a wreath product, contrary to |\overline{Q}| = 9. We conclude that every R ∈ R is abelian. As |\overline{Q}| = 27 it follows that Q has a generating set consisting of elements which commute pair-wise, and so Q is abelian. But with |V/C_{V} (A)| ≤ 4 we then have |Q| = 9.

We conclude that |\overline{Q} ≤ 9. If |\overline{Q}| = 3 then Q is cyclic, and (i) holds, so in fact |\overline{Q}| = 9.

Then three conjugates of a suffice to generate G, and so:

(3) |V | ≤ 64.

Suppose next that Φ(Q) ≤ Z(G). Then Φ(Q) is cyclic, of exponent at most 3, and so Q is either elementary abelian of order 9 or extraspecial of order 27. Suppose that Q is elementary abelian, and let x ∈ Q^{#} with [V, x] as large as possible. As x^{a} = x^{−1} we then have |[V, x]| = 16 and [V, a] ≤ [V, x]. Here [V, x] is G-invariant, so [V, x] = V and (ii) holds. On the other hand, suppose that Q is extraspecial. No member of R is cyclic of order 9, so the exponent of Q is 3, and then G is isomorphic to the commutator subgroup of SU(3, 2). Then |V | > 32, and (3) then yields |V | = 64. There is, up to isomorphism, a unique faithful representation of Q of degree 3 over 𝔽_{4} (by ordinary character theory), and so we obtain (iii) in this case. We therefore conclude that Φ(Q) ≰ Z(G).

Now [Φ(Q), a] ≠ 1, and we may choose y ∈ Φ(Q) with y^{a} = y^{−1}. Choose a generating set {x_{1}, x_{2}} for Q with (x_{i})^{a} = (x_{i})^{−1} for i = 1 and 2, and set R_{i} = ⟨x_{i}, y⟩. Then R_{i} ∈ R and Q = ⟨R_{1}, R_{2}⟩. Suppose y ∈ Z(Q). Then each R_{i} is abelian, and so |[V, R_{i}]| = 16, and [V, R_{1}] ≠ [V, R_{2}]. Then [V, y] ≠ [V, R_{i}] for some i, and so |[V, y]| = 4. But then Q = C_{Q}([V, y]) × ⟨y⟩, contrary to y ∈ Φ(Q). Thus y ∉ Z(Q), and we may assume that R_{1} is non-abelian. Then R_{1}A ≅ SU(3, 2)′ and |[V, R_{1}]| = |[V, y]| = 64. Then also R_{2} is non-abelian, and (3) yields [V, R_{1}] = [V, R_{2}] = V. We conclude that Q is isomorphic to a Sylow 3-subgroup of GL(V), of order 81. Then Q has an elementary abelian subgroup E of order 27, and every element of order 3 in Q is contained in E ∪ R_{1}. Then R_{2} = (R_{1} ∩ R_{2}) ∪ (E ∩ R_{2}), and so R_{2} ≤ E. As R_{2} is non-abelian, we have a contradiction at this point, proving the lemma. □

4.4 Proposition. Assume Hypothesis 1, and assume that G = QA, where Q = [Q, A] is a p′-subgroup of G and where A is a subgroup of G which satisfies the following conditions.

(a) A acts quadratically on V.

(b) |A|^{2} ≥ |V/C_{V} (A)|.

(c) For any non-identity subgroup B of A, we have |A|^{2}|C_{V} (A)| ≥ |B|^{2}|C_{V} (B)|. Denote by Y the set of subgroups Y of Q for which there exists a hyperplane B of A such that Y = [C_{Q}(B), A], and such that Y ≠ 1. Then Q is the direct product ∏{[V, Y]}_{Y ∈Y}, and [V, Q] is the direct sum ⨁{[V, Y]}_{Y ∈Y}. Further, for any Y ∈ Y and any a ∈ A − C_{A}(Y), the pair (Y ⟨a⟩, [V, Y]) is given by one of the outcomes in 4.3, with Y ⟨a⟩ in place of G and [V, Y] in place of V.

Proof. Put Q_{0} = C_{Q}(C_{V} (A)). The quadratic action of A yields [V, A, Q_{0}] = 0, while also [V, Q_{0}, A] ≤ [V, A] ≤ C_{V} (Q_{0}). The Three Subgoups Lemma then yields [V, [Q_{0}, A]] ≤ C_{V} (Q_{0}), and then [Q_{0}, A] = 1, by coprime action. Thus:

(1) C_{Q}(C_{V} (A)) ≤ C_{Q}(A).

Denote by B the set of hyperplanes B of A such that [C_{Q}(B), A] ≰ Φ(Q). For any B ∈ B put Y_{B} = [C_{Q}(B), A] and set V_{B} = C_{V} (B). Thus, Y = {Y_{B}}_{B∈B}. Fix B ∈ B. Then Y_{B} acts non-trivially on V_{B}, by (1). Choose a ∈ A − B, and set X = Y_{B}, U = V_{B}, L = X⟨a⟩, and \overline{L} = L/C_{L}(U). Condition (c) implies that |U : C_{U}(a)| ≤ p^{2}, so 4.3 applies and identifies a list of possibilities for the pair (\overline{L}, [U, X]). Further, as [V, B] is X-invariant and centralizes a, we obtain [V, B, X] = 0. Then [V, X, B] = 0, and thus V = C_{V} (X) + U. In particular, L acts faithfully on U. We have shown:

(2) For any B ∈ B we have V = C_{V} (Y_{B}) + C_{V} (B), and for any a ∈ A − B the pair (Y_{B}⟨a⟩, [V, Y_{B}]) satisfies the hypothesis (and conclusion) of 4.3.

We may also record the following result.

(3) For any B ∈ B we have [V, Y_{B}, B] = 0.

Now let X and Y be distinct elements of Y, and put D = C_{A}(X) ∩ C_{A}(Y). Then |A : D| = p^{2}, and there exist elements a, b ∈ A such that A = D⟨a, b⟩,and with C_{A}(X) = D⟨b⟩ and C_{A}(Y) = D⟨a⟩. Then also X = [X, a] and Y = [Y, b]. Set W = C_{V} (D) and set E = ⟨a, b⟩. We have C_{W} (a) = C_{W} (D⟨a⟩) admitting faithful action by Y ⟨b⟩, by (2), so C_{W} (a) is contained properly in C_{W} (E). Notice also that condition (c) yields |W/C_{W} (E)| ≤ p^{4}. Thus:

(4) |W/C_{W} (a)| ≤ p^{3} and |W/C_{W} (E)| ≤ p^{4}.

We aim now to show:

(5) [X, Y] = 1.

Suppose first that |[W, Y]| = p^{2}, so that Y ⟨b⟩ ≅ SL(2, p). Let x ∈ X, and set a′ = a^{x} and A′ = D⟨a′, b⟩. Then A′ = A^{x} acts quadratically on V, and so A′ centralizes [W, b]. Put W_{0} = [W, Y] + [W, Y]^{a′}. Thus |W_{0}| ≤ p^{3} and W_{0} is invariant under Y ⟨a′, b⟩. Here p = 2 or 3, and Y ⟨a′, b⟩ acts as a subgroup K of SL(3, p) on W_{0}, containing a copy of SL(2, p), and with K = O_{p′}(K)P where P is a p-group. However, it is easy to check that the only such subgroups K of SL(3, p) are in fact isomorphic to SL(2, p). (In L_{3}(2) we are looking for the normalizer of a non-identity 3-group, and in L_{3}(3) for the normalizer of a non-identity 2-group.) Thus Y ⟨a′, b⟩ induces an action of SL(2, p) on W_{0}, and so W_{0} = [W, Y], and [W, Y] is a^{x}-invariant. As X ≤ ⟨a^{X}⟩ we conclude that Y is X-invariant, and then [X, Y] = 1 since [X, b] = 1. Thus (5) holds in this case.

We may now assume that neither [W, X] nor [W, Y] is of order p^{2}. If |W/C_{W} (a)| = p^{3} then (4) implies that b induces a transvection on C_{W} (a) = C_{V} (D⟨a⟩, and then (2) implies that |[W, Y]| = p^{2}; which is a contradiction. We therefore conclude that |W : C_{W} (a)| = p^{2}. Further, as |[W, X]| > p^{2}, (2) shows that [W, a] ≤ [W, X]. We now set R = [C_{Q}(D), a]. Then X ≤ R, and so 4.3 applies to the pair (R⟨a⟩, [W, R]).

We have X ≤ R ⊴ C_{Q}(D)E, and Y ≤ ⟨b^{C_{Q}(D)}⟩. If [R, b] = 1 it follows that [R, Y] = 1 and we have (5). So assume that [R, b] ≠ 1. Then X is a proper subgroup of R, so 4.3 implies that R is of the form 3^{2},3^{1+2}, or Q_{8} × Q_{8}, while X ≅ (SL(2, p))′. Suppose R ≅ (SL(2, p)′ ×(SL(2, p))′. Then [W, R] = [W, X], and then [W, R, b] = 0, by (3). Then [W, R, Y] = 0, so [W, X, Y] = 0. We have [V, X] = [W, X], by (2), so [V, X, Y] = 0 in this case. On the other hand, suppose that R⟨a⟩ ≅ (SU(3, 2))′ and |[W, R]| = 64. Then X is of order 3 and |[W, X]| = 16. As b centralizes [W, X], by (3), and b leaves C_{W} (X) invariant, it follows that |[W, R]/C_{[W,R]}(b)| ≤ 2. But b either inverts Z(R) (in which case |[W, R]/C_{[W,R]}(b)| ≥ 8) or b centralizes Z(R) (in which case C_{[W,R]}(b) may be regarded as an 𝔽_{4}-space). Thus b centralizes Z(R) and [W, R, b] = 0. As above, we then obtain [V, X, Y] = 0, and thus [V, X, Y] = 0 in any case. Similarly, we have [V, Y, X] = 0, and so [X, Y] = 1 by the Three Subgroups Lemma. This proves (5).

Notice that [W, b, X] ≤ [W, b, ⟨a^{X}⟩] = 0, and so also [W, X, b] = 0. Now (5) yields [W, X, Y] = 0, and so [W, X] ∩ [W, Y] = 0. But [W, X] = [V, X] by (2), so we may conclude that X ∩ Y = 1. Thus, we have shown:

(6) For any distinct X, Y ∈ Y we have ⟨X, Y ⟩ = X × Y and [V, X] ∩ [V, Y] = 0.

Now let Y_{1}, ⋯, Y_{r} ∈ Y and put K = ⟨Y_{1}, ⋯, Y_{r}⟩. Suppose that K = Y_{1} × ⋯ × Y_{r} and that [V, K] = [V, Y_{1}] ⊕ ⋯ ⊕ [V, Y_{r}]. Let X ∈ Y with X ≠ Y_{i} for any i. Then (6) implies that [V, X] ≤ C_{V} (K), and so [V, X]∩[V, K] = 0. Then also X ∩K = 1. Induction on r then yields:

(7) ⟨⋃ Y⟩ = ∏ Y and [V, ∏ Y] = ⨁{[V, Y]}_{Y ∈Y}.

Put Q_{0} = ⟨⋃ Y⟩. It now only remains to show that Q = Q_{0}. By coprime action, for each prime divisor r of |Q| there is an A-invariant Sylow r-subgroup of Q, and then Q = C_{Q}(A)Q_{0}. Then Q_{0} is a normal subgroup of Q, and since Q = [Q, A] we obtain Q = Q_{0}, as required. □

4.5 Corollary. Let the hypotheses be as in 4.4.

(a) Assume that there is no element Y ∈ Y with |[V, Y]| = p^{2}. Then for any Y ∈ Y there is a uniquely determined 2-transvection a ∈ A with Y = [Q, a], and we have q(A, V) = 2.

(b) Assume that q(A, V) ≤ 1. Then q(A, V) = 1, and for any Y ∈ Y there is a uniquely determined transvection a ∈ A with Y = [Q, a]. Moreover, we then have Y ⟨a⟩ ≅ SL(2, p), and [V, Y ⟨a⟩] is a natural module for Y ⟨a⟩.

Proof. Suppose first that there exists no Y ∈ Y with |[V, Y]| = p^{2}. If |A| = p then (a) holds by 4.3. So assume |A| > p. Let Y = {Y_{1}, ⋯, Y_{r}}, and set U_{i} = [V, Y_{i}]. Then 4.4 yields Q = Y_{1} × ⋯ × Y_{r} and [V, Q] = U_{1} ⊕ ⋯ ⊕ U_{r}. By construction, each Y_{i} centralizes a hyperplane B_{i} of A. Since U_{i} = [U_{i}, Y_{i}] = [U_{i}, ⟨A^{Y_{i}}], quadratic action implies that [U_{i}, B_{i}] = 0. By assumption, and by 4.3, we have |U_{i}/C_{U_{i}}(A)| = p^{2} for all i. Then |B_{i}|^{2} ≥ |V/C_{V} (B_{i})| for any i, and so 4.4 applies with B_{i} in place of A. By induction, for any j with j ≠ i there is a 2-transvection b ∈ B_{i} with Y_{j} = [Q, b]. In this way we obtain [Q, A]A = Y_{1}⟨a_{1}⟩ × ⋯ × Y_{r}⟨a_{r}⟩, where a_{i} is a 2-transvection on V, centralizing U_{j} for i ≠ j. Evidently {a_{1}, ⋯, a_{r}} is the set of all 2-transvections in A, and thus (a) holds.

Suppose next that q(A, V) ≤ 1. If |A| = p then (b) follows from 4.3, so assume |A| > p. As in the preceding paragraph, define Y_{i}, B_{i}, and U_{i}, 1 ≤ i ≤ r. As |U_{i}/C_{U_{i}}(A)| ≥ p for all i, we have |B_{i}| ≥ |V/C_{V} (B_{i})| for any i, and again 4.4 applies with B_{i} in place of A. As in the preceding paragraph, induction on |A| yields (b). □

4.6 Proposition. Assume Hypothesis 1, with F^{∗}(G) = F (G), and with q(G, V) ≤ 2. Denote by A the set of elements of Q^{∗}(G, V) of minimal order, and set G_{0} = ⟨A⟩ and D = [F (G), G_{0}]. If p = 3 and q(G, V) = 2 assume further that G_{0}/Q is of odd order. There is then a G-invariant set K = {K_{1}, ⋯, K_{r}} of subgroups of D, such that the following hold.

(a) D = K_{1} × ⋯ × K_{r}.

(b) V = [V, K_{1}] ⊕ ⋯ ⊕ [V, K_{r}] ⊕ C_{V} (D).

(c) For any K ∈ K, and for U = [V, K], one of the following holds.

(i) p = 2, K ≅ ℤ_{3} and |U| = 4 or 16. (ii) p = 2, K ≅ 3_{+}^{1+2} and |U| = 64. (iii) p = 2, K ≅ ℤ_{5} and |U| = 16. (iv) p = 3, K ≅ Q_{8} and |U| = 9 or 81.

(v) p = 3, K ≅ Q_{8} ∘ Q_{8} and |U| = 81.

(d) Suppose that there exists K ∈ K such that |[V, K]| > p^{2}. Then q(G, V) = 2 and each A ∈ A is of order p

(e) Suppose that there exists K ∈ K such that K is not G_{0}-invariant. Then p = 2, |[V, K] = 4, q(G, V) = 2, and each A ∈ A is of order 2.

(f) Let A ∈ A such that A acts non-trivially on some K ∈ K. Then every member of K is A-invariant.

Proof. Set Q = F (G) and q = q(G, V). For any A ∈ A, let Y_{A} be the set of subgroups Y of Q such that 1 ≠ Y = [C_{Q}(B), A] for some hyperplane B of A. Let K_{0} be the set of subgroups K of Q such that, for some A ∈ A and some Y ∈ Y_{A}, we have K = [K, A] ≤ Y, with K ≅ (SL(2, p))′ and with |[V, K]| = p^{2}. If p = 2, let K_{1} be the set of subgroups K of Q such that K ∈ Y_{A} for some A ∈ A, and such that K is not in K_{0} and is not contained in a direct product of two elements of K_{0}. If p = 3 let Y_{1} be the set of all subgroups Y of Q such that Y ∈ Y_{A} for some A ∈ A, with Y ≅ Q_{8} and with |[V, Y]| = 81. For Y and Y′ in Y_{1}, write Y Ỹ′ if Z(Y) = Z(Y′). We then take K_{1} be the set of subgroups K of Q of the form ⟨S⟩, where S is an equivalence class for this relation. In any case, set K = K_{0} ∪ K_{1}, and observe that K is G-invariant.

Let A ∈ A, and suppose that there exists Y ∈ Y_{A} with Y ∉ K_{0}. Set B = C_{A}(Y) and let a ∈ A − B. Then |[V, Y]| ≥ p^{4} and [V, Y]/C_{[V,Y]}(a) ≥ p^{2}, by 4.4. Here [V, Y, B] = 0 by quadratic action, so |V/C_{V} (A)| ≥ p^{2}|V/C_{V} (B)|. As |A|^{q} = |V/C_{V} (A)|, we then have |B|^{q} ≥ p^{2−q}|V/C_{V} (B)|. As q ≤ 2 the minimality of q and of A then yields q = 2 and B = 1. Thus (d) is proved.

Let K_{1} and K_{2} be distinct members of K, and set X = ⟨K_{1}, K_{2}⟩, U_{i} = [V, K_{i}], and W = U_{1} + U_{2}. Suppose first that K_{i} ∈ K_{0} for both i = 1 and 2. If p = 2 then X is isomorphic to a 3- subgroup of L_{4}(2), so X = K_{1} × K_{2} and W = U_{1} ⊕ U_{2}. On the other hand, suppose that p = 3, so that each K_{i} is a quaternion group, with |[V, K_{i}]| = 9. A Sylow 2-subgroup of GL(2, 3) contains a unique quaternion group, so U_{1} ≠ U_{2}. Suppose U_{1} ∩ U_{2} ≠ 0. Then each K_{i} has a subgroup of index 2 which lies in a fixed subgroup X_{0} of GL(W) with X_{0} ≅ GL(2, 3). Then Z(K_{1}) = Z(K_{2}) and since U_{i} = [V, Z(K_{i})] we arrive back at U_{1} = U_{2} in this case. Thus, U_{1} ∩ U_{2} = 0, and X is contained in a subgroup X_{1} of GL(W) of the form 3^{4} : (GL(2, 3) × GL(2, 3). But then X is contained in a complement to O_{3}(X_{1}) in X_{1}, and since |U_{i}| = 9 we then have [K_{1}, K_{2}] = 1 and W = U_{1} ⊕ U_{2}. In particular, we have thus shown that if K = K_{0} then (a) and (b), and either (c)(i) or (c)(iv) (with |U| = p^{2}), hold.

Let K ∈ K_{0}, and suppose that there exists A ∈ A such that K is not A-invariant. Let a ∈ A − N_{A}(K) and set L = KK^{a}. As we have just seen, L is the direct product of K and K^{a}, and |[V, L]| = p^{4}. Set A_{0} = C_{A}(L). The quadratic action of A implies that L is A-invariant and that A = ⟨a⟩ × A_{0}. Then |A|^{q} = |V/C_{V} (A)| ≥ p^{2}|V/C_{V} (A_{0})|, as A_{0} centralizes [W, L]. As q ≤ 2 we then have |A_{0}|^{q} ≥ |V/C_{V} (A_{0})|. The minimality of q and of A then implies that A_{0} = 1 and q = 2. Thus:

(1) If K ≠ K_{0}, or if there exists K ∈ K_{0} which is not G_{0}-invariant, then |A| = p for all A ∈ A.

Thus, parts (a) through (e) of the lemma hold if K = K_{0}. Moreover, we may record the following result

(2) If K_{1} and K_{2} are in K_{0} then [K_{1}, K_{2}] = 1 and U_{1} ∩ U_{2} = 0.

We assume henceforth that K ≠ K_{0}. Suppose that p = 2, |K_{1}| = 5 and |U_{1}| = 16. If also |K_{2}| = 5 then, since any Sylow 5-subgroup of L_{8}(2) is abelian, we have X = K_{1}×K_{2}. In this case we have also U_{1} ∩ U_{2} = 0. On the other hand, suppose that K_{2} is a 3-group. then [K_{1}, K_{2}] = 1 since Q is nilpotent. Suppose that U_{1} ∩ U_{2} ≠ 0. As End_{K_{1}}(U_{1}) ≅ ℤ_{15} it follows that U_{1} ≤ U_{2} and that |K_{2}/C_{K_{2}}(U_{1})| = 3. Then U_{2} is not a natural SU(3, 2)- module for K_{2}, with K_{2} ≅ 3_{+}^{1+2}. By 4.4 and the definition of K, we then have |K_{2}| = 3 and U_{1} = U_{2}. By (d), which has already been proved, there exist 2-transvections a_{i} with K_{i} = [Q, a_{i}], (i = 1 and 2). But then X⟨a_{1}, a_{2}⟩ ≅ Dih(6) × Dih(10), which is not a subgroup of L_{4}(2). We therefore conclude that U_{1} ∩ U_{2} = 0 in this case.

Suppose next that p = 2, and that both K_{1} and K_{2} are 3-groups in K_{1}. By definition, we then have K_{i} ∈ Y_{A_{i}} for some A_{i} ∈ A, and then (d) implies, as above, that K_{i} = [Q, a_{i}] for some 2-transvection a_{i}. In particular, we have K_{i} ⊴ X for each i. Suppose that |K_{1}| = |K_{2}| = 3. Then X = K_{1} × K_{2}, and X⟨a_{1}, a_{2}⟩ ≅ Sym(3) × Sym(3). Suppose U_{1} = U_{2}, and set a = a_{1}a_{2} and A = ⟨a⟩. Then a is a 2-transvection, so A ∈ A, X ∈ Y_{A}, and each K_{i} is contained in the direct product of two members of K_{0}. This is contrary to the definition of K_{1}, so we conclude that U_{1} ≠ U_{2}. Suppose U_{1} ∩U_{2} ≠ 0. Then |W | = 64. Set C_{i} = C_{X}(K_{i}⟨a_{i}⟩). Then |C_{i}| = 3 and [U_{3−i}, C_{i}] = U_{3−i} for both i = 1 and 2. As X = K_{i}C_{i}, it follows that [W, C_{i}] = W. But only one cyclic subgroup of X is without fixed points on W, so C_{1} = C_{2}. As [X, ⟨a_{1}, a_{2}⟩] = X we have a contradiction at this point, and so U_{1} ∩ U_{2} = 0.

Suppose that |K_{1}| = 3 (with K_{1} ∈ K_{1}) and that K_{2} ≅ 3_{+}^{1+2}. Then K_{1} ⊴ X, so [K_{1}, K_{2}] = 1. If K_{1} ≤ K_{2} then [U_{2}, K_{1}] = U_{2} is of order 64, whereas |U_{1}| = 16, so in fact K_{1} ≰ K_{2} and X = K_{1} × K_{2}. As K_{2} acts irreducibly on U_{2} we have U_{1} ∩ U_{2} = 0. Now suppose that both K_{1} and K_{2} are isomorphic to 3_{+}^{1+2}. Then K_{i} ⊴ X and so [K_{1}, K_{2}] ≤ K_{1} ∩K_{2}. In particular, we have Z(K_{1})Z(K_{2}) ≤ Z(X). Suppose K_{1} ∩K_{2} ≠ 1. It follows that Z(K_{1}) = Z(K_{2}), and since U_{i} = [V, Z(K_{i})] we get U_{1} = U_{2}. Then X is isomorphic to a 3- subgroup of L_{6}(2), which is to say that X ≅ 3 ≀ 3. There are then exactly two maximal subgroups of X which having exponent 3, and one of these two is elementary abelian. We therefore conclude that, in fact, K_{1} ∩K_{2} = 1, and X = K_{1} ×K_{2}. Irreducible action of K_{i} on U_{i} then yields U_{1} ∩ U_{2} = 0. We have shown:

(3) If p = 2 and both K_{1} and K_{2} lie in K_{1}, or some K_{i} is of order 5, then X = K_{1} × K_{2} and W = U_{1} ⊕ U_{2}.

Continue now to assume that p = 2, let K_{1} ∈ K_{0} and let K_{2} be a 3-group in K_{1}. Let b be a 2-transvection such that K_{2} = [Q, b]. Thus K_{2} ⊴ X. Suppose |K_{2}| = 3. Then X = K_{1} × K_{2}. If U_{1} ∩ U_{2} ≠ 0 then U_{1} ≤ U_{2} and we obtain a contradiction via X = [X, b] = K_{2}. Thus, we are reduced to the case where K_{2} ≅ 3_{+}^{1+2}. Then K_{1} ≰ K_{2}, since |[V, x]| ≥ 16 for any non-identity element x of K_{2}. By the definition of K_{0}, and by (d), there exists a 2-transvection a with K_{1} = [K_{1}, a]. By (3), either K_{2} = (K_{2})^{a} or U_{2} ∩ (U_{2})^{a} = 0. As a is a 2-transvection, we conclude that K_{2} = (K_{2})^{a}, and then that [Z(K_{2}), a] = 0. As K_{1} ≤ [Q, a], we have [Q, a] ≇ 3^{1+2}, and so |[Q, a]| ≤ 9 by 4.3. It now follows that [K_{2}, a] = 1, and then also [K_{2}, K_{1}] = 1. Thus X = K_{1} × K_{2}, and we again obtain U_{1} ∩ U_{2} = 0 from the irreducible action of K_{2} on U_{2}. Thus, (a), (b) and (c) hold if p = 2.

Now let p = 3, and let K ∈ K_{1}. By definition, and by 4.5, there then exists Y ∈ Y_{1} such that K = ⟨S⟩, where S is the set of all Y′ ∈ Y_{1} with Z(Y) = Z(Y′). Suppose that |S| > 1, let Y′ ∈ S − {Y }, and set K_{0} = Y Y′. By (d), there is are 2-transvections a and a′ such that Y = [Q, a] and Y = [Q, a′], so Y and Y′ are normal subgroups of K_{0}, and K_{0} is ⟨a, a′⟩-invariant. Suppose Φ(K_{0}) ≰ Z(Y). Then [Y, Y′] ≰ Z(Y) and then Φ(K_{0}) contains both Y and As Φ(K_{0}) is a proper subgroup of K_{0}, we conclude that in fact Φ(K_{0}) = Z(Y). Then K_{0} = Z(K_{0})X where X is an extraspecial group.

Set L = K_{0}⟨a, a′⟩. Then [C_{V} (Z(Y)), L] = 0, and so L acts faithfully on the space U = [V, Z(Y)]. Suppose Z(K_{0}) ≠ Z(Y) We have [Z(K_{0}), a] = 0 by 4.1, and End_{K_{0}⟨a⟩}(U) ≅ GL(2, 3), so Z(K_{0}) is cyclic or is a fours group. If Z(K_{0}) is cyclic then Y is the unique quaternion subgroup of K_{0}, contrary to Y ≠ Y′. Thus Z(K_{0}) is a fours group ⟨z, z′⟩ where ⟨z⟩ = Z(Y), and we have K_{0} = Y × ⟨z′⟩. The set Q of quaternion subgroups of K_{0} then has cardinality 4, and it follows that ⟨a, a′⟩ induces the action of Alt(4) on Q. We now employ our hypothesis that, in the case p = 3 and q = 2, G_{0}/Q is of odd order. Thus [L ∩ Q, a] properly contains Y, which is contrary to Y = [Q, a]. This proves that Z(K_{0}) = Z(Y), so K_{0} is extraspecial of width 2. [We remark that, if the hypothesis concerning G_{0}/Q is dropped, then one has to face the possibility that L ≅ (Q_{8} × Q_{8}) : 3, with Y and Y′ equal to “diagonal” subgroups of O_{2}(L), and with C_{Q}(a) acting on O_{2}(L) by interchanging the two direct factors. This appears to lead to some serious difficulties.] If K_{0} ≅ 2_{−}^{1+4} then L/C_{L}(K_{0})K_{0} has cyclic Sylow 3-subgroups, so ⟨a⟩ and ⟨a′⟩ are congruent modulo C_{L}(K_{0})K_{0}, contrary to Y ≠ Y′. Thus X ≅ Q_{8}∘Q_{8}. Then X has just two quaternion subgroups, and [Y, Y′] = 1.

We now conclude that S is a commuting set of quaternion groups acting on [V, Z(Y)]. Then |S| ≤ 2, and either K = Y or K ≅ Q_{8} ∘ Q_{8}. Returning now to the group X = ⟨K_{1}, K_{2}⟩, suppose that both K_{1} and K_{2} lie in K_{1}. Then each K_{i} is normal in X, and |U_{i}| = 81. In particular, we have Z(K_{1})Z(K_{2}) ≤ Z(X). Here Z(K_{1}) ≠ Z(K_{2}), as K_{1} and K_{2} are defined by distinct equivalence classes on Y_{1}. Then Z(K_{i}) ≰ Z(K_{3−i}), and so X = K_{1} × K_{2}. If U_{1} = U_{2} then Z(K_{1}) and Z(K_{2}) induce identical actions on V, so in fact U_{1} ≠ U_{2}. Since there is a unique quaternion group in End_{K_{i}}(U_{i}), it follows that U_{1} ∩ U_{2} = 0.

Now suppose that K_{1} ∈ K_{0} and that K_{2} ∈ K_{1}. Then K_{2} ⊴ X, and we have [Q, a_{1}] = K_{1} or [Q, a_{1}] is a direct product of two isomorphic copies of K_{1}. So, either K_{1} ⊴ X or ⟨K_{1}^{X}⟩ ≅ K_{1} × K_{1}. Set R = N_{K_{2}}(K_{1}). Then |K_{2} : R| ≤ 2, and [K_{1}, R] ≤ K_{1} ∩ K_{2}. Set R_{0} = C_{R}(K_{1}), and suppose that R_{0} has exponent 4. As End_{K_{1}}(U_{1}) ≅ ℤ_{2}, we obtain [U_{1}, Z(K_{2})] = 1 in this case. So assume that R_{0} has exponent 2. If K_{2} ≅ Q_{8} then K_{1} ∩ K_{2} = 1, as Z(K_{1}) ≠ Z(K_{2}), and so R = R_{0} in this case. Here R has exponent 4, so we conclude that K_{2} ≅ Q_{8}∘Q_{8}, and that R ≠ R_{0}. Then K_{1}∩K_{2} ≠ 1, and so Z(K_{1}) ≤ K_{2}. Then R = C_{K_{2}}(Z(K_{1})) ≅ D_{8} × ℤ_{2}. As [R, a_{1}] ≤ [Q, a_{1}], R/C_{R}(U_{1}) acts as a group of inner automorphisms on K_{1}, and it follows that [U_{1}, Φ(R)] = 0. Thus, [U_{1}, Z(K_{2})] = 0 in any case, and so [U_{1}, K_{2}] = 0. Now U_{1} ∩ U_{2} = 0, and since [Z(K_{2}), K_{1}] = 1 we then have [U_{2}, K_{1}] = 0. Then X = K_{1} × K_{2}, and we have thereby proved (a) through (d).

Suppose that we are given K ∈ K and A ∈ A such that K is not A-invariant. Set U = [V, K] and let a ∈ A − N_{A}(K). Then U ∩ U^{a} = 0, by the foregoing. If K ∈ K_{1} then A = ⟨a⟩, a is a 2-transvection, and |U| ≥ p^{4}. But then a is not a 2-transvection, so we conclude that K ∈ K_{0}. If p = 3 we obtain [V, K⟨a⟩] = U ⊕ U^{a} ⊕ U^{a^{2}}, contrary to quadratic action. Thus p = 2. Now (e) follows from (1).

Finally, by way of contradiction to (f), suppose that we are given A ∈ A, and elements K_{1} and K_{2} of K such that 1 ≠ [K_{1}, A] ≤ K_{1} and such that K_{2} is not A-invariant. Then A = ⟨a⟩ where a is a 2-transvection, by (d). But a already acts as a 2-transvection on [V, K_{2}] ⊕ [V, K_{2}]^{a}, so [V, K_{1}, a] = 0, and so [K_{1}, a] centralizes [V, K_{1}]. Then [K_{1}, a] centralizes V, and we have the desired contradiction. □

4.7 Corollary. Assume Hypothesis 1 with G = QS, Q = O_{r}(G) for some prime q, q ≠ p, and with S a Sylow p-subgroup of G. Assume that S acts irreducibly on Q/Φ(Q), and assume that there exists A ∈ Q(S, V) with |A|^{2} ≥ |V/C_{V} (A)|. Then there is a subgroup K of Q, unique up to conjugation, such that the following hold.

(a) Q = K_{1} × ⋯ × K_{r}, where {K_{i}}_{1≤i≤r} = K^{S}.

(b) V = [V, K_{1}] ⊕ ⋯ ⊕ [V, K_{r}] ⊕ C_{V} (Q).

(c) Setting U = [V, K], one of the following holds. (i) p = 2, K ≅ ℤ_{3} and |U| = 4 or 16. (ii) p = 2, K ≅ 3_{+}^{1+2} and |U| = 64. (iii) p = 2, K ≅ ℤ_{5} and |U| = 16. (iv) p = 3, K ≅ Q_{8} and |U| = 9 or 81.

(v) p = 3, K ≅ Q_{8} ∘ Q_{8} and |U| = 81. Moreover, if K is not invariant under ⟨Q^{∗}(S, V)⟩ then p = 2, |K| = 3, |U| = 4, and

q(S, V) = 2.

Proof. Immediate from 4.6 and the irreducible action of S on Q/Φ(Q). □

The next result gives some information concerning the situation defined by Hypothesis 4′.

4.8 Lemma. Assume Hypothesis 2, with |A| of order p. Assume further that G = QA, where Q = [Q, A] is an r-group for some prime r, r ≠ p. Finally, assume that [V, A, A, A] = 0 and that [V, A, A] ≠ 0. Then p = 3 and one of the following holds.

(i) G ≅ Alt(4) and |V | = 27.

(ii) G ≅ SL(2, 3) and |V | = 81.

(iii) Q ≅ 2_{+}^{1+4} and |V | = 81.

(iv) G is a Frobenius group of order 39 and |V | = 27.

Proof. Regard G as a subgroup of SL(V). The minimal polynomial for a non-identity element a of A is then (X −1)^{3}, and the Hall-Higman Theorem ([HH, Theorem B], or [G, Theorem 11.1.1]) then yields p = 3. Also, as A is non-quadratic, we have |V/C_{V} (A)| = 9, and then since V = [V, Q] it follows that V is irreducible for G. In particular, Z(G) is cyclic. Moreover, it follows that End_{G}(V) = 𝔽_{3}, and so |Z(G)| ≤ 2.

Suppose first that Q is cyclic. Then G = ⟨A, A^{g}⟩ for some g ∈ G, and so |V | ≤ 3^{4}. Here 3 divides |Q|−1, while |Q| divides 2^{8} ·3^{6} ·5·13 = |SL(4, 3)|. It follows that |R| = 13, and that (iv) holds.

We assume henceforth that Q is non-cyclic. There then exists a non-cyclic subgroup Q_{0} = [Q_{0}, A] of Q, with Q_{0}A generated by two conjugates of A, and so |[V, Q_{0}]| ≤ 3^{4} and r = 2.

Suppose that |Q/Φ(Q)| = 4. Then we again have G = ⟨A, A^{g}⟩ for some g ∈ G, and |V | ≤ 3^{4}. For any involution t ∈ Q we then have [V, t] = V or |[V, t] = 9, and since A is not quadratic on V it follows that C_{Q}(A) ≤ Z(SL(V)). In particular, we have |C_{Q}(A)| ≤ 2. Set Y = [Φ(Q), A], and set W = [V, Y]. If Y = 1 then (i) or (ii) holds, so assume that Y ≠ 1. As A is non-quadratic we have W ≥ [V, A] and V = W + C_{V} (A). Suppose that Y A ≅ Alt(4). Then |W | = 8, and Φ(Q) is elementary abelian, of order at most 8. If Y = Φ(Q) then W = V, |Q| = 16, and Y ≤ Z(Q), whereas a Sylow 2-subgroup of L_{3}(3) is semi-dihedral of order 16, with center of order 2. Thus Y ≠ Φ(Q), and |V | = 81. There is a unique conjugacy class of elementary abelian subgroups of SL(V) of order 8, and we then find that N_{GL(V)}(Φ(Q)) ≅ 2^{4} : Sym(4) and that Q is extraspecial of width two. This is contrary to |Q/Φ(Q)| = 4, so we conclude that Y A ≇ Alt(4).

Suppose that Y A ≅ SL(2, 3). Then Z(Y) = Z(SL(V)), and Y = Φ(Q). But with Y normal in Q we have Q = C_{Q}(Y)Y, and this is contrary to Φ(Q) ≥ Y. Thus Y A ≇ SL(2, 3). As A is non-quadratic on W, induction then implies that Y is extraspecial, of width 2. As |C_{Y} (A)| ≠ 1 we have Y = Φ(Q). Let Q_{1} and Q_{2} be the two quaternion subgroups of Y. Then each Q_{i} is A-invariant, so each Q_{i} is normal in G, and we have Q = C_{Q}(Q_{i})Q_{i} = C_{Q}(Y)Y. Again, this is contrary to Y = Φ(Q), so we conclude that Q/Φ(Q)| ≥ 16.

Suppose that |Q/Φ(Q)| ≥ 64 or that Y ≠ 1. Then there are proper A-invariant subgroups R_{1} and R_{2} of Q, with R_{i} = [R_{i}, A] of order bigger than 8, with [R_{1}∩R_{2}, A] ≠ 1, and with ⟨R_{1}, R_{2}⟩ = Q. Here A acts non-quadratically on [V, R_{i}] for each i, so by induction each R_{i} is extraspecial, of width 2. Set W_{i} = [V, R_{i}] and set X = [R_{1} ∩ R_{2}, A]. Then XA is isomorphic to either Alt(4) or SL(2, 3), and |[V, X]| is either 27 or 81. As [V, X] ≤ W_{1} ∩ W_{2}, where |W_{i}| = 81, we then have |V | ≤ 3^{5}. Let T be a Sylow 2subgroup of SL(V) containing Q. As R_{i} acts irreducibly on W_{i} we have C_{T} (R_{i}) ≤ Z_{i}, and thus Z(T) = Z(R_{1}) = Z(R_{2}). Then also [V, Z(T)] = W_{1} = W_{2}, and so |V | = 81. Set D = C_{R}(A) and set D_{i} = N_{D}(R_{i}). Then |D_{i}/Z(R_{i})| ≤ 2, and if D_{i} ≠ Z(R_{i}) then D_{i} = Z(R_{i})⟨d_{i}⟩ where d_{i} interchanges the two quaternion subgroups of R_{i}. In any case, we then have R_{i} = J(R_{i}D_{i} mod Z(R_{i})), and so R_{i} is characteristic in R_{i}D_{i}. It follows that [N_{Q}(R_{i}), A] ≰ R_{i}, and so each R_{i} is normal in Q. As Aut(R_{i}) ≅ Sym(4) ≀ 2, we conclude that [R_{1}, R_{2}] = Z(R_{1}) = Z(R_{2}), and so Q = R_{1}R_{2} is extraspecial of width 3. The minimal degree of a faithful representation of Q over 𝔽_{3} is then 8, and so we have a contradiction at this point.

Suppose finally that |Q/Φ(Q)| = 16 and that Y = 1. As G = ⟨A^{G}⟩ we then have Φ(Q) ≤ Z(G). Then |Φ(Q)| = 2, as follows from the first paragraph of the proof. Suppose that Φ(Q) ≠ Z(Q), and set F = [Z(Q), A]. Then F A ≅ Alt(4) and |[V, F]| = 27. As [V, F] is G-invariant we then have |V | = 27, and R is not faithfully represented on Q. We conclude that Z(Q) = Φ(Q), and hence Q is extraspecial of width 2. Let Q_{1} and Q_{2} be the two quaternion subgroups of Q. Then Q_{i}A ≅ SL(2, 3) and [V, Q_{i}] = [V, Z(Q)] = V is of order 81. Thus (iii) holds. □

Section 5: Quadratic modules for quasisimple groups

In this section we shall be concerned mostly with the situation where F^{∗}(G) is quasisimple, and in which V is a quadratic module for G. The first main step is to establish a criterion (5.4, below) for G to contain a quadratic fours group. The proof of 5.4 requires some preliminary results, all of which, perhaps, are well known, but which we include for the sake of completeness.

Lemma 5.1. Let T be a 2-group, and let F be a subgroup of T with F ≅ ℤ_{2} × ℤ_{2}. Suppose that C_{T} (F) = F. Then T is either dihedral or semi-dihedral.

Proof. It will be enough to show that T has a cyclic subgroup X of index 2, as then we will have |C_{X}(a)| = 2 and T = X⟨a⟩ for any a ∈ F − X. If |T | ≤ 8 then T is dihedral and there is nothing to prove. So assume |T | > 8 and proceed by induction on |T |. Let R be a subgroup of T of index 2, containing F. Then R is dihedral or semi- dihedral, and we may write R = T ⟨t⟩ where Y = ⟨y⟩ is the unique maximal cyclic subgroup of R and where t ∈ F. Let x ∈ T − R. Then T = R⟨x⟩ and Y = Y^{x}. Put 2^{n} = |Y | and let z be the unique involution in Y.

Suppose first that R is semi-dihedral. Then the coset ⟨y^{2}⟩t is the set of non-central involutions of R, and forms a single conjugacy class in R, of order 2^{n−1}. Then C_{T} (t) is not contained in R, and so we may take [t, x] = 1. But also F = Z(R)⟨t⟩, and then [F, x] = 1, contrary to C_{T} (F) = F. Thus R is dihedral. Further, the same argument then shows that conjugation by x interchanges the two conjugacy classes of non-central involutions of R. In particular, it follows that C_{R}(x) contains no non-cenral involutions, and so x^{2} ∈ Y. If now ⟨x^{2}⟩ = Y then ⟨x⟩ is a cyclic subgroup of T of index 2. Setting X = ⟨x⟩ we then have C_{X}(t) = C_{X}(F) = ⟨z⟩ and then T is dihedral or semi-dihedral. Thus, we may assume:

(1) ⟨x^{2}⟩ ≠ Y.

Now assume that x has been chosen so that |C_{Y} (x)| is as large as possible, and then (subject to this condition) so that |x| is as small as possible. Let y_{0} be a generator for C_{Y} (x), and put y_{1} = x^{2}. If ⟨y_{0}⟩ ≠ ⟨y_{1}⟩ then there exists u ∈ C_{Y} (x) with u^{2} = y_{1}. But in that case we have (ux)^{2} = u^{2}x^{2} = (y_{1})^{2}, and the minimality of |x| then gives y_{1} = 1. This shows:

(2) Either ⟨x^{2}⟩ = C_{Y} (x) or |x| = 2.

Suppose |x^{2}| ≠ 2, and then use (1) to again choose u ∈ Y with u^{2} = x^{2}. Then [x, u]^{2} = [x, u^{2}] = 1, and so [x, u] = z. Then (xu)^{2} = x^{−2}u^{−2}[x, u] = u^{−4z}. Minimality of |x| then implies u^{4} = 1 and x^{2} = z. Thus, in any case we have the following result.

(3) x^{2} ∈ ⟨z⟩.

Suppose that C_{Y} (x) ≠ ⟨z⟩. Then |x| = 2 by (2) and (3). But it then also follows that x inverts no element of Y − ⟨z⟩, and so ⟨x, z⟩ = Ω_{1}(Y ⟨x⟩). Then [x, t] = z and N_{T} (F) = N_{R}(F)⟨x⟩, whence C_{T} (F) properly contains F. This is contrary to assumption, so:

(4) C_{Y} (x) = ⟨z⟩.

Now x inverts the subgroup of order 4 in Y, and then |C_{Y} (xt)| > |C_{Y} (x)|. Replacing x by xt, we then contradict the minimality of |C_{Y} (x)| in our choice of x. This contradiction proves the lemma. □

5.2 Corollary. Let X be a group, t an involution in X, and suppose that X = ⟨t^{X}⟩. Assume that Z^{∗}(X) = 1, and suppose that C_{X}(t) contains no subgroup of the form ⟨t⟩×E with |E| = 4. Then a Sylow 2-subgroup of X is dihedral or semi-dihedral.

Proof. Glauberman’s Z^{∗} Theorem implies that there is a fours group F = ⟨s, t⟩ in X, with s ∈ t^{X}. The hypothesis then yields C_{Y} (F) = F for any Sylow 2-subgroup T of X containing F. Now apply the preceding lemma. □

5.3 Proposition. Let X be a finite group acting faithfully on a vector space V over 𝔽_{2}, and assume that X = ⟨t^{X}⟩ for some involution t in X. Assume further that F^{∗}(X) is quasisimple, that O_{2}(X) = 1, and that |V/C_{V} (t)| ≤ 4. Then one of the following holds.

(i) There is a fours group E in X with t ∈ E and with [V, E, E] = 0.

(ii) A Sylow 2-subgroup of X is dihedral or semi-dihedral.

Proof. Suppose that t lies in no fours group in X which acts quadratically on V. There is then no involution s in C_{X}(t) − ⟨t⟩ with [V, t, s] = 0. As |[V, t]| ≤ 4 it then follows that C_{X}(t) contains no subgroup of the form ⟨t⟩ × F with |F | = 4. Then 5.2 applies, and yields (ii). □

Remark. The known simple groups with dihedral or semi-dihedral Sylow 2-subgroups are the groups L_{2}(q), q odd; L_{3}(q), q ≡ 3 (mod 4); U_{3}(q), q ≡ 1 (mod 4); Alt(7); and M_{11}. Using this classification, we will obtain Proposition 5.5, below, on groups which contain a 2-transvection which lies in no quadratic fours group.

Proposition 5.4. Let X, t, and V be as in 5.3, and assume that C_{V} (X) ≤ [V, X]. Assume that there does not exist a fours group E in X, containing t,and such that [V, E, E] = 0. Then one of the following holds.

(i) X ≅ Alt(5) and V is the O_{4}^{−}(2)-module for X.

(ii) X ≅ Sym(5) and [V, X]/C_{V} (X) is the natural ΓL(2, 4)-module for X.

Proof. Set X_{0} = F^{∗}(X), let S be a Sylow 2-subgroup of X containing t, and set T = S∩X_{0}. Suppose first that X_{0}/Z(X_{0}) ≅ L_{3}(q), q ≡ 3 (mod 4); or U_{3}(q), q ≡ 1 (mod 4). As S has 2-rank equal to 2, by 5.2, S contains no element which induces a field automorphism on X_{0}, and so X_{0} = X. The involutions in X form a single conjugacy class, so there is a non-abelian subgroup R of X of order q^{3} such that R = [R, t]. It then follows from 4.3 that q = 3, (so that X ≅ L_{3}(3)), and that |[V, R]| = 64. Then R⟨t⟩ is generated by three conjugates of t, and then X is generated by four conjugates of t. It follows that X is isomorphic to a subgroup of L_{8}(2), whereas 13 does not divide |L_{8}(2)|. Thus, we have a contradiction via Lagrange’s Theorem.

Suppose next that Z(X_{0}) ≠ 1. As O_{2}(X) = 1 we then have X_{0} ≅ 3^{·}Alt(6) or 3^{·}Alt(7). Set V_{0} = [V, Z(X_{0})]. Then V = V_{0} ⊕ C_{V} (Z(X)0), and then, by induction, it suffices to show that every fours group in X containing t acts quadratically on V_{0}. Here, V_{0} may be regarded as an 𝔽_{4}X_{0}-module. As 3^{·}Alt(6) is not a subgroup of SL(2, 4), we have dim_{𝔽_{4}}(V_{0}) > 2, and it follows that [Z(X_{0}), t] = 1. Then [V_{0}, t] is an 𝔽_{4}- subspace of V_{0} of dimension 1, and so [V_{0}, t, F] = 0 for any fours group F containing t, as required. Thus, we may assume henceforth that Z(X_{0}) = 1.

Suppose next that X_{0} ≅ L_{2}(q), q odd, and suppose that X_{0} ≠ X. As before, no element of S induces a field automorphism on X_{0}. Suppose that q ≡ 3 (mod 4), A Sylow 2-subgroup of X is dihedral, so all involutions in X − X_{0} are conjugate, and thus t lies in a dihedral subgroup D of X of order 2q. Then 4.3 implies that q ≤ 5, so q = 3, which is contrary to our hypothesis that q ≥ 5. Thus q ≡ 1 (mod 4).

Let λ be a generator of the Sylow 2-subgroup of 𝔽_{q}^{×}. As in the preceding paragraph, all involutions in X − X_{0} are conjugate, and so t may be identified with the involution ds, where d and s are represented in GL(2, q) by the matrices

d̃= ( \begin{matrix}λ & 0 \\ 0 & 1 \end{matrix} )  and  s̃ = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )

respectively. Set Y = ⟨s, t⟩, and let z be the central involution in Y. Then Y is dihedral, of order 2(q − 1), and since ⟨t, z⟩ is not quadratic it follows that Y acts faithfully on [V, t] + [V, t]^{s}. Any dihedral subgroup of L_{4}(2) whose order is divisible by 4 has order dividing 12 or 8, so q − 1 divides 6 or 4. The given congruence satisfied by q then yields q = 5, and thus X ≅ Sym(5).

Set F = ⟨z, t⟩, and set U = [V, F]. As zt is conjugate to t we have |U| ≤ 16, and since F is not quadratic we conclude that |U| = 8 and U ≠ [V, z]. Thus |V/C_{V} (z)| ≤ 4. As z inverts an element of order 5 in X_{0}, X_{0} has no irreducible 𝔽_{2}-modules on which z induces a transvection, and we conclude that X_{0} has a unique non-trivial irreducible constituent W in V. Suppose that W is an O_{4}^{−}(2)- module for X_{0}. Then 1.3 that W = V, and one then observes that F acts quadratically on V. Thus, W is a natural SL(2, 4)-module for X_{0}, and outcome (ii) holds in this case.

Suppose next that X = X_{0} and X ≅ L_{2}(q). Then three conjugates of t generate X, and so |V/C_{V} (X)| ≤ 64. Then q ≤ 9, by Lagrange’s Theorem applied to L_{6}(2). Suppose that q = 9, so that X ≅ Alt(6). Let t ∈ X_{1} ≤ X with X_{1} ≅ Alt(5). By induction, and by 1.3, we then have V = [V, X_{1}] ⊕ C_{V} (X_{1}), where [V, X_{1}] is a natural Sym(5)-module for X_{1}. There exists a subgroup X_{2} of X, with X_{1} ≅ X_{2}, and with X_{1} not conjugate to X_{2}, and such that X_{1} ∩ X_{2} is dihedral of order 10. Let f be an element of order 5 in X_{1} ∩ X_{2}. Then [V, X_{1}] = [V, f] = [V, X_{2}]. As X_{1}X_{2} = X we then have |[V, X]| = 16. Further, the elements of order 3 in X_{1} and in X_{2} come from different conjugacy classes in X. As [V, X] is a natural Sym(5)-module for each X_{i} it follows that no element of order 3 in X is fixed-point-free on [V, X]. This result is inconsistent with a faithful action of ℤ_{3} × ℤ_{3} on 2^{4}, and we therefore conclude that q ≠ 9.

Suppose that q = 7, so that X ≅ L_{3}(2). There are three isomorphism classes of nontrivial irreducible 𝔽_{2}-modules for X, of which one, the adjoint module for L_{3}(2), admits no 2-transvections, and of which the other two are a dual pair of natural L_{3}(2)- modules. Suppose that there is a unique non-trivial constituent for X in V. Then 1.5 implies that |V | ≤ 16, and either V/C_{V} (X) or [V, X] is a natural L_{3}(2)-module. One then observes that there is a maximal subgroup P of X containing S such that |[W, O^{2}(P)]| = 4, and then O_{2}(P) is a quadratic fours group. Thus, we have a contradiction in this case, and since |V/C_{V} (X)| ≤ 64 it follows that V has a submodule U such that both U/C_{U}(X) and V/U are irreducible X-modules of order 8. Since t induces a transvection on U, and since three conjugates of t suffice to generate X, we have |[U, X]| = 8. Similarly, we have |[V, X]/[U, X]| = 8, and since C_{V} (X) ∩ [V, X] = 0 we conclude that C_{V} (X) = 0. Let s be a conjugate of t such that [s, t] = 1, and set F = ⟨s, t⟩. As F is not quadratic we have |[V, F]| = 8, and it follows that U and V/U are not isomorphic as X-modules. We may assume that F has been chosen so that |[U, F]| = 4, and we then have [V, F] = [V, t] + [U, F] ≤ C_{V} (t). That is, we have [V, F, t] = 0, and so F is quadratic, after all.

Suppose that q = 5, so that X ≅ SL(2, 4). No involution in X induces a transvection on any irreducible 𝔽_{2}X-module, so there is a unique non-trivial irreducible constituent U for X in V. If U is an O_{4}^{−}(2)-module for X, then 1.3 yields (i). So assume that U is a natural SL(2, 4)-module. Set \overline{V} = V/C_{V} (X) and identify U with [\overline{V} , X]. We then have |\overline{V} /C\overline{V} (t)| = 4, and [\overline{V} , t] = C_{[\overline{V} ,X]}(t) = C_{[\overline{V} ,X]}(S). Set V_{0} = [V, X]. Then the inverse image in V_{0} of C_{[\overline{V} ,X]}(S) is C_{V_{0}}(S), and so S acts quadratically on V.

We are now reduced to the cases where F^{∗}(X) is isomorphic to Alt(7) or M_{11}. A Sylow 2-subgroup of Sym(7) has 2-rank equal to 3, while M_{11} has no outer automorphisms, so we have X ≅ Alt(7), or M_{11}. Then X has a subgroup X_{0} with F^{∗}(X_{0})/ ≅ Alt(6), and with S ≤ X_{0}. But it follows from what we have already shown that such an X_{0} contains a quadratic fours group on V. This completes the proof of 5.4. □

5.5 Lemma. Assume Hypothesis 3, with H quasisimple. Let A be a quadratic subgroup of H, with |A|^{2} ≥ |V/C_{V} (A)|. Then one of the following holds.

(i) H/Z(H) is a group of Lie type in characteristic p.

(ii) We have p = 2, and H is an alternating group Alt(2^{n} + 1), n ≥ 2.

(iii) We have |A| = p = 3, H ≅ SL(2, 5), and V is a natural SL(2, 9)-module for H.

Proof. If p is odd then Corollary 1.4 in [C3] yields (iii). So assume that p = 2. By 3.6 we either have (i) or G contains a fours group E, acting quadratically on V. Thus, we may assume the existence of such a quadratic fours group. If H/Z(H) is a sporadic group then the main result of [A4] shows that S is not contained in a unique maximal subgroup of G, and so Hypothesis 3 is violated in this case. Assume that H/Z(H) is of Lie type in odd characteristic (and that there is no isomorphism of H/Z(H) with a group of Lie type in characteristic 2). The main result of [MS1] then implies that H/Z(H) ≅ U_{4}(3). The 2-local subgroups of H containing S ∩ H then form a diagram geometry of type B̃_{2} (for which see [RS], for example) and it follows that S is not contained in a unique maximal subgroup of G. Thus, we are reduced to the case where H/Z(H) is an alternating group Alt(n), n ≥ 7. As S is contained in a unique maximal subgroup of G, the main theorem of [LPR] implies that n = 2^{m} + 1 for some m, m ≥ 3, □

5.6 Proposition. Let G be a symmetric group Sym(n), n ≥ 2, and let V be faithful 𝔽_{2}-module for G, such that V is isomorphic to a constituent of the natural permutation module for G. Let A be a non-identity, elementary abelian 2-subgroup of G such that |A| ≥ |V/C_{V} (A)|. Then one of the following holds.

(i) A is generated by a set of transpositions, and |A| = |V/C_{V} (A)|.

(ii) n is even, A is generated by a set of n/2 pairwise distinct transpositions, C_{V} (G) =

0, V = [V, G], and |A| = 2|V/C_{V} (A)|. (iii) n is even, A is a subgroup of index 2 in a group generated by n/2 pairwise distinct

commuting transpositions, C_{V} (G) = 0, V = [V, G], and |A| = |V/C_{V} (A)|. (iv) n = 4, A = O_{2}(G), |V | = 8, and either V = [V, G] and |A| = |V/C_{V} (A)| or

C_{V} (G) = 0 and |A| = 2|V/C_{V} (A)|.

(v) n = 6, |A| = 8, V = [V, G], and |A| = |V/C_{V} (G)|.

(vi) n = 8, |A| = 8, every non-identity element of A is the product of three pairwise disjoint transpositions, C_{V} (G) = 0, V = [V, G], and |A| = |V/C_{V} (G)|.

Proof. For any m ≥ 2, denote by P (m) the permutation module of dimension m for Sym(m) over 𝔽_{2}, P_{0}(m) the submodule of P (m) of index 2, \overline{P} (m) the quotient of P (m) by C_{P (m)}(Sym(m)), and \overline{P}_{0}(m) the image of P_{0}(m) in \overline{P} (m).

We proceed by induction on |V |. The cases where n ≤ 4 may be verified by inspection, so we assume that n ≥ 5. (As G acts faithfully on V, by assumption, we have V = P (2) if n = 2, and V ≠ \overline{P}_{0}(4) if n = 4.) Suppose that V ≠ \overline{P}_{0}(n). If n is odd then P (n) ≅ \overline{P}_{0}(n) ⊕ C_{V} (G), so we may assume that n is even. Set V_{0} = [V, G] and set \overline{V} = V/C_{V} (G). Induction implies that one of the outcomes (i) through (vi) applies with \overline{V}_{0} in place of V. Suppose that A is generated by a set A of commuting transpositions. Then |A| ≤ n/2, and

(1)  |A| ≥ |V/C_{V} (A)| ≥ |\overline{V}_{0}/C_{\overline{V}_{0}}(A)| ≥ |A|/2,

where the final inequality in (1) is strict unless |A| = n/2. Thus, (i) holds if |A| < n/2. If |A| = n/2 and V ≠ V_{0}, then V ≠ V_{0} +C_{V} (A) and |A| = |V/C_{V} (A)|, and we again have (i). Otherwise, if V = V_{0} then C_{V} (A) = C_{V} (⟨A_{0}⟩) for any subset A_{0} of A of cardinality n/2 − 1, and we have |A| = 2|V/C_{V} (A)|. Thus (ii) holds in this case. Next, take A to be a set of n/2 pairwise commuting transpositions, suppose that A is a subgroup of index 2 in ⟨A⟩, and suppose that A is not generated by a subset of A. Then C_{V} (A) = C_{V} (⟨A⟩), and so (iii) holds in this case. Next, suppose that n = 6, |A| = 8, and A is not generated by transpositions. Then A is conjugate in G to the group ⟨(1 2), (3 4)(5 6), (3 5)(4 6)⟩, and direct calculation then shows that C_{V} (A) ≤ V_{0}. As |A| = |\overline{V}_{0}/C_{\overline{V}_{0}}(A)|, we then have (v). Finally, suppose that n = 8 = |A| and that every element of A^{#} is the product of three pairwise disjoint transpositions. Here N_{G}(A) ≅ 2^{3} : L_{3}(2), and both the commutator series and the central series of A with P (8) have the shape (1, 3, 3, 1). Then |A| ≥ |V/C_{V} (A)| if and only if V = \overline{V}_{0}, and so (vi) holds in this case. Thus, we may assume henceforth that V = \overline{V}_{0} ≅ \overline{P}_{0}(n), n ≥ 5.

Let \{x_{i} : 1 ≤ i ≤ n\} be the standard basis for P (n), and identify V with \overline{P}_{0}(n). Write x_{i,j} for the image in V of x_{i} + x_{j}. Fix a ∈ A^{#} so that the number k of non-trivial orbits of a on {1, 2, ⋯, n} is as small as possible. Identify a with (1 2)(3 4) ⋯ (2k −1 k), and denote by H_{1} the pointwise stabilizer in G of {2k+1, ⋯, n} and by H_{2} the pointwise stabilizer in G of {1, ⋯, 2k}. Set K = C_{H_{1}}(A). Then C_{G}(a) = K × H_{2}, and we have K ≅ 2^{k} : Sym(k) and H_{2} ≅ Sym(m). Denote by E the subgroup of K generated by the set of pairwise disjoint transpositions whose product is a. (Thus, E = O_{2}(K) unless k = 2 or 4.) Then A ∩ E = 1, by the minimality of k. Choose a complement B to ⟨a⟩ in A. Also, set V_{i} = [V, H_{i}], and set U = C_{V_{1}}(E).

Suppose first that n = 2k, so that K = C_{G}(a) and V = V_{1}, and k ≥ 3. If k is odd then U = C_{V} (a), while if k is even we have C_{V} (a) = U + ⟨y⟩ where y = x_{1,3} + x_{5,7} + ⋯ x_{2k−3,2k−1}, and where [y, E] = U. In any case we have U ≅ \overline{P} (k) as modules for K/E, and |V/U| = 2^{k−1}. Define ℓ by ℓ = 1 if k is odd, and ℓ = 2 if k is even. By induction, we have |B| ≤ 2|U/C_{U}(B)|, and so

|A| ≥ |V/C_{V} (A)| ≥ 2^{k−ℓ}|U/C_{U}(B)| ≥ 2^{k−ℓ−1}|B| = 2^{k−ℓ−2}|A|.

Thus k = 3 or 4. Suppose k = 3. Every fours group in Sym(6) contains an element which is the product of two disjoint transpositions, so the minimality of k yields |A| = 2, and a acts as a transvection on V. As only the transpositions in Sym(6) act as transvections, we have a contradiction in this case. So assume that k = 4 and n = 8. Then |V/C_{V} (a)| = 8, and so |A| ≥ 8. There exists no elementary abelian subgroup of G of order 16, in which every non-identity element is the product of four pairwise disjoint transpositions, so we have |A| = 8. Thus, (vi) holds in this case.

We assume henceforth that n ≠ 2k. Then U = C_{V_{1}}(a), and U ≅ P (k) as modules for K/E. Also, we have V_{2} ≅ P_{0}(m) as H_{2}-modules, where m = n − 2k. Set W = V_{1} + V_{2}. Then |V/W | = 2, and C_{V} (a) = U + V_{2} ≤ W. Also, |V_{1} ∩ V_{2}| = 1 if n is odd, and 2 if n is even, and we have |V_{1}/U| = 2^{k−1}.

Set B_{1} = C_{B}(H_{2}) and let B_{1} be a complement in B to B_{2}. As U ≅ P (k) we have |U/C_{U}(B_{1})| ≥ |B_{1}|. Then

|A| ≥ |V/C_{V} (A)| = 2|W/C_{W} (A)| ≥ 2^{k}|U/C_{U}(B_{1})||V_{2}/C_{V_{2}}(B_{2})|

≥ 2^{k}|B_{1}| · 1/2|B_{2}| = 2^{k−2}|A|. (2)

Thus k ≤ 2.

Suppose k = 2. As V_{2} ≅ P_{0}(m), we may apply induction in (2) and find that |A| = |V/C_{V} (A)|, m is even, and B_{2} induces on H_{2} a group generated by m/2 pairwise disjoint transpositions. But B_{2} contains no bona fide transpositions, by the minimality of k, so we have [V_{1}, B_{2}] ≠ 0. Let b ∈ B_{2} such that b induces a transposition on {2k + 1, ⋯, n}, and such that b is non-trivial on {1, ⋯, 2k}. Then b = xt where 1 ≠ x ∈ C_{K}(A) and where t is a transposition in H_{2}. If B_{1} ≠ 1 then C_{K}(A) = B_{1}⟨a⟩, so that x ∈ A and t ∈ A, contrary to the minimality of k. Thus B_{1} = 1, and now (iii) holds.

Finally, suppose that k = 1. Then B_{1} = 1 and B = B_{2}. We may therefore choose B so that B ≤ K_{2}. Suppose that |A| > |V/C_{V} (A)|. Then (2) implies that n is even and that B is generated by a set of m/2 transpositions, and so (ii) holds in this case. So assume that |A| = |V/C_{V} (A)|. If B is generated by a set of transpositions then we have (i). So assume that B is not so generated. By induction, m is even, and either (iii) or (iv) applies with B, H_{2}, and V_{2} in the roles of A, G, and V. If (iii), then (iii) holds for G, while if (iv) then (v) holds for G. This completes the proof of the 5.6. □

5.7 Lemma. Assume Hypothesis 2 with p = 2, and with H = Alt(n), n ≥ 5. Assume that there exists a non-identity subgroup A of G such that A acts quadratically on V, with |A|^{2} ≥ |V/C_{V} (A)|. If n is even, assume that G ≅ Sym(n) and that |A| ≥ |V/C_{V} (A)|. Then one of the following holds.

(i) n = 5 or 7, each non-trivial constituent for H in V is a spin module (of dimension 4 over 𝔽_{2}), and there are at most two such constituents.

(ii) n = 6 and V is a natural Sp(4, 2)-module for G.

(iii) n = 8 and V is a spin module for G, of dimension 8. Moreover, as a module for H we have V = U ⊕ U′, where U is a natural L_{4}(2)-module for H and where U′ is dual to U.

(iv) n = 9, G = H, dim(V) = 8, and V is a spin module for G. Further, A is conjugate in G to the fours group F = ⟨(1 2)(3 4), (1 3)(2 4)⟩, and |V/C_{V} (A)| = |V/C_{V} (a)| = 16 for any a ∈ A^{#}.

(v) V is a natural Sym(n)-module for G.

(vi) n is odd and V is a direct sum of two natural Sym(n)- modules for G.

Suppose further that n is odd and that V is a natural Sym(n)-module or a direct sum of two natural Sym(n)-modules for G, and let T_{A} be the set of all transpositions t in Sym(n) such that t is a factor of some a ∈ A, in a representation of a as a product of pairwise disjoint transpositions. Set A^{∗} = ⟨T_{A}⟩. Then A^{∗} is elementary abelian, [V, A] = [V, A^{∗}], and C_{V} (A) = C_{V} (A^{∗}).

Proof. Let a ∈ A^{#}, and suppose first that a is not a member of any quadratic fours group in G. Then A = ⟨a⟩, a is a 2-transvection on V, and 5.4 implies that n = 5 and that (i) or (v) holds. If (v) holds then A = A^{∗}, so we may assume henceforth that a lies in a quadratic fours group B, and we take B ≤ A if |A| ≥ 4.

Let U be a non-trivial irreducible HAB-invariant section of V. Then Theorem 4 of [MS2] implies that U is a spin module or a natural module for HAB. Suppose first that U is a spin module, and that n ≠ 8. Let F be the fours group given in (vi). Lemmas 4.1 through 4.3 in [MS2] then establish that every quadratic subgroup of HAB of order bigger than 2 is conjugate to F, and that C_{U}(x) = 0 for any 3-cycle x in H. Notice that there exists a 3-cycle x inverted by a. If a is a 2-transvection it follows that |U| ≤ 16, and that U is the unique non-trivial irreducible constituent for G in V. In that case, 1.2 yields V = U, and n ≤ 7 (since G = Sym(8) if n = 8). If now n = 6 then the spin module V is isomorphic to the natural Sp(4, 2)-module, and so (ii) holds. Suppose n = 5 or 7. We observe that A does not act quadratically on the natural Sym(n)-module for G, and so every non- trivial irreducible constituent for HA in V is a spin module. Moreover, there are at most two such constituents in V, and we obtain (i). Thus, we may assume that a is not a 2-transvection, and hence also n ≥ 7.

Now |A| = 4 and we may take A = B = F. If n is even then |A| ≥ |V/C_{V} (A)|, by assumption, and then a is a 2-transvection. Thus, n is odd, and we have |A|^{2} ≥ |V/C_{V} (A)|. Set X = O^{2}(C_{H}(A)), and set t = 1 if n = 5, and otherwise set t = (3 4)(5 6). Then X⟨t⟩ ≅ Sym(n−4), and since all quadratic fours groups are conjugate to F it follows that X acts faithfully on [V, A]. As |[V, A]| ≤ 16, we then have n ≤ 9, and then n = 9.

Let x be a 3-cycle inverted by an element of A. As C_{U}(x) = 0 we obtain dim(U) ≤ 8 Then since |[U, A]| = |U/C_{U}(A)| we obtain dim(V) = 8, and C_{U}(a) = C_{U}(A). Thus |A|^{2} = |U/C_{U}(A)| = |V/C_{V} (A)|, and so U = [V, H]. Then U = V by 1.2, and so (iv) holds.

Now suppose that U is a spin module for HA and that n = 8. Then G ≅ Sym(8), and so also G ≅ O_{6}^{+}(2). The Clifford algebra C associated with this orthogonal group is isomorphic to the algebra of 8 × 8 matrices over 𝔽_{2}, by [Chev, II.2.1], and the spin module W for G is isomorphic to a minimal left ideal of C. Thus dim(W) = 8. Moreover, as the characteristic is even, G is isomorphic to the spinor group associated with O_{6}^{+}(2), and [Chev, II.4.2] implies that W is the direct sum of two inequivalent irreducible representations for Ω_{6}^{+}(2), which are then conjugate via O_{6}^{+}(2). As O_{6}^{+}(2) ≅ Aut(L_{4}(2)), we thereby obtain (iii). Thus, we have exhausted the possibilities where U is a spin module, and we may assume henceforth that U is a natural Sym(n)-module for HA, for every non-trivial irreducible constituent U for HA in G.

Let Ω = {1, 2, ⋯, n} be the set on which G has its natural permutation representation, and identify U with the unique non-trivial irreducible constituent in the permutation module 𝔽_{2}Ω. As A is quadratic, all orbits of A on Ω have length at most 2, and hence A^{∗} is generated by a set of pairwise disjoint transpositions. One computes that [U, A] = [U, A^{∗}] and C_{U}(A) = C_{U}(A^{∗}). Suppose n is even. Then |A^{∗}| = 2|U/C_{U}(A^{∗})|, and since |A| ≥ |V/C_{V} (A)| it follows that U is the unique non-trivial constituent for G in V. Thus,

(v) holds in this case, and we may assume that n is odd. Then |A|^{∗} = |U/C_{U}(A^{∗})|. As |A|^{2} ≥ |V/C_{V} (A)|, it follows that there are at most two non-trivial irreducible sections for G in V, and if there are two, then A = A^{∗}. If there is only one such section then U = V, by 1.2. Suppose there are two. As C_{V} (H) = 0, and as H^{1}(H, U) = 0 by 1.3, it follows that we may take U to be a submodule of V, with V/U irreducible. Then 1.8 implies that there is an H-invariant complement W to U in V. As A = A^{∗} there is a transposition a ∈ A, and we have |V/C_{V} (a)| = 4. Then W ∩ W^{a} ≠ 0, and so W = W^{a}, and W is G-invariant. This yields (vi). □

The following result is immediate from 5.6 and 5.7.

5.8 Corollary. Assume Hypothesis 2, with p = 2, and let H and A be as in 5.7. Then the following hold.

(a) If |A| > |V/C_{V} (A)| then n is even, V is a natural module for G, A is generated by a set of n/2 pairwise disjoint transpositions, and |A| = 2|V/C_{V} (A)|.

(a) If |A|^{2} < |V/C_{V} (A)|, n is odd, and n ≥ 9, then V is a natural module for G.

(c) If |A| = |V/C_{V} (A)|, n is odd, and n ≥ 9, then G ≅ Sym(n) and A is generated by a set of pairwise disjoint transpositions.

□

5.9 Lemma. Assume Hypothesis 3, and assume that H is a quasisimple group of Lie type in characteristic p. Then one of the following holds:

(i) The Lie rank of H/Z(H) is equal to 1. That is, H/Z(H) ≅ L_{2}(p^{n}), U_{3}(p^{n}),

Sz(2^{n}) (with p = 2) or (^{2}G_{2}(3^{n}))′ (with p = 3).

(ii) p = 2, and H/Z(H) ≅ L_{3}(2^{n}) or Sp(4, 2^{n})′. Moreover, there exists an element t of S such that t induces a non- trivial symmetry on the Coxeter diagram for H/Z(H).

Proof. Assume that the Lie rank of H/Z(H) is at least 2. Recall that the vertices of the Coxeter diagram associated with H/Z(H) are in one-to-one correspondence with the maximal subgroups of H which contain S ∩ H. Since G ∈ M^{∗}(S), a Sylow p-subgroup of Aut(H) then acts transitively on the vertices of this diagram. It follows (from the classification of Coxeter diagrams) that the diagram has just two vertices, so p = 2, and the diagram is of type A_{2} or B_{2}. Thus, (ii) holds. □

Section 6: Lie rank 1 in characteristic p.

In this section we will be concerned with Hypothesis 2, with the additional assumption that H is a group of Lie type, in characteristic p, and of Lie rank 1. We begin by reestablishing some well known properties of these groups.

6.1 Lemma. Let G be a group containing a normal subgroup H ≅ ^{2}G_{2}(q), q = 3^{2n+1}, and with C_{G}(H) = 1. Let S be a Sylow 3-subgroup of G and set T = S ∩ H. Then the following hold.

(a) We have |T | = q^{3}, Z(T) is elementary abelian of order q, and Ω_{1}(T) is elementary abelian of order q^{2}. Moreover, T is a trivial intersection set in H, and we have N_{H}(T) = T D where D is cyclic of order q − 1, and where D acts regularly on T/Ω_{1}(T) and on Z(T). Further, D acts irreducibly on Ω_{1}(T)/Z(T), with |C_{D}(Ω_{1}(T)/Z(T))| = 2.

(b) There is a single conjugacy class of involutions in G, and for any involution t in G we have C_{G}(t) = ⟨t⟩ × L where L is isomorphic to L_{2}(q). Further, C_{G}(t) is the unique maximal subgroup of G containing L, and L ∩ Z(T) = 1.

(c) Any two conjugates of Z(T) generate O^{3}(H).

(d) Suppose that S − H contains an element a of order 3. Set X = [Ω_{1}(T), a] and set A = ⟨a^{X}⟩. Then A = Ω_{1}(C_{S}(a)), and A is elementary abelian.

Proof. Part (a) follows from the Chevalley relations (see Tables 2.4 and 2.4.7 in [GLS3]). If q = 3 then it is known that H is isomorphic to ΓL(2, 8), and one observes in this case that (b) through (d) hold. Thus we may assume that q > 3 in proving (b) through (d).

Let t be an involution in H and set M = C_{H}(t). It is then well known (see for example [GLS3, Theorem 4.5.1]) that M = ⟨t⟩ × L, where L ≅ L_{2}(q). In particular, a Sylow 2subgroup R of H is elementary abelian, of order 8, and for any fours group F contained in R we have C_{H}(F) ≅ F × D where D is dihedral of order (q + 1)/2. It follows that ⟨t⟩ = C_{H}(L).

Let M be a maximal subgroup of H containing L, with M ≠ C_{H}(t). Then Z(M) = 1. Also, we have O_{3}(M) = 1, since every 3-local subgroup of H is contained in a Borel subgroup. For primes other than 2 or 3 the Sylow subgroups of H are cyclic, so F (M) = 1 and F^{∗}(M) is simple. A Cartan subgroup of H is cyclic of order q − 1, so F^{∗}(L) is not isomorphic to P SL(2, q′), q′ a power of 3. If F^{∗}(M) ≅ ^{2}G_{2}(q_{0}) for some q_{0} then a Cartan subgroup of L is contained in a Cartan subgroup of M, and so q_{0} = q and M = G. Thus F^{∗}(M) is not a Ree group. As q > 3, M is not isomorphic to J_{1} and M is not isomorphic to a linear group L_{2}(r), r odd, r not a power of 3. The classification ([B] or [W]) of groups with abelian Sylow 2-subgroups now yields a contradiction to the presumed existence of M, and so C_{G}(t) is the unique maximal subgroup of G containing L. Since C_{G}(z) = T for any non-identity element z of Z(T), as follows from (a), we have L ∩ Z(T) = 1, and (b) holds.

Next, let K be a subgroup of H generated by a pair of conjugates of Z(T). As C_{H}(z) = T for any non-identity element z of Z(T), we have Z(K) = 1, while O_{3}(K) = 1 as T is a T.I. set in H. As q > 3 and O_{2}(K) is elementary abelian of order at most 8, we have also O_{2}(K) = 1, and then F^{∗}(K) is simple. As in the preceding paragraph, we find that F^{∗}(K) is not isomorphic to L_{2}(r) for r odd, r not a power of 3, or to J_{1}. Hence F^{∗}(K) is a Ree group ^{2}G_{2}(q_{0}) or a linear group L_{2}(q_{0}), q_{0} a power of 3. Comparison of a Cartan subgroup of F^{∗}(K) with a Cartan subgroup of H yields q_{0} = q, and K ≅ L_{2}(q). Then K contains a subgroup isomorphic to L_{2}(3), and so a non-identity element z of Z(T) normalizes a fours group F in H. As C_{H}(F) is of order prime to 3, z then normalizes a Sylow 2-subgroup of C_{H}(F), of order 8, and then |C_{H})z)| is even. This contradiction proves (c).

Now let X, a and A be as in (d). Then the coset T a contains an element a_{0} such that a_{0} induces a field automorphism on H. A Cartan subgroup of N_{H}(T) acts regularly on T/Ω_{1}(T), so we may assume that a ∈ Xa_{0}. Thus, C_{Ω_{1}(T)}(a_{0}) = C_{Omega_{1}(T)}(a) = [X, a_{o}] = [X, a]. If g is an element of order 3 in C_{S}(a) then g ∈ T ⟨a⟩, and g = g_{1}a^{i} for some element g_{1} of C_{T} (a) and some power a^{i} of a. Then g_{1} ∈ Ω_{1}(T), and so g_{1} ∈ [X, a]. This yields (d). □

6.2 Lemma. Let G be a group containing a normal subgroup H ≅ Sz(q), q = 2^{2n+1}, and with C_{G}(H) = 1. Let S be a Sylow 2-subgroup of G and set T = S ∩ H. Then the following hold.

(a) We have |T | = q^{2}, and Z(T) = Ω_{1}(T) is elementary abelian of order q. Moreover, T is a trivial intersection set in H, and we have N_{H}(T) = T D where D is cyclic of order q − 1, and where D acts regularly on T/Z(T) and on Z(T).

(b) Any two distinct conjugates of Z(T) generate O^{2}(H)Z(T).

(c) We have S = T.

Proof. Part (a) is well known and may be justified in the same way as part (a) of the preceding lemma. Part (c) also is well known, and follows from the fact that automorphisms of Sz(q) are induced by automorphisms of Sp(4, q) (where q is an odd power of 2). If q = 2 then G = H is a Frobenius group of order 20, where (b) is obvious. So assume henceforth that we have q > 2.

Let L be a subgroup of H such that L is generated by two distinct conjugates of Z(T). For any element x of H of odd order, C_{H}(x) is cyclic of odd order, and it follows that O_{r}(L) = 1 for every odd prime r. Also O_{2}(L) = 1 as T is a trivial intersection set, so we conclude that F^{∗}(L) is a non-abelian simple group. As |L| is relatively prime to 3, L is a Suzuki group, and then since Z(T) ≤ L we have L = H. Thus (b) holds. □

6.3 Lemma. Let G be a group containing a normal subgroup H ≅ SU(3, q), q = p^{n}, and with C_{G}(H) = 1. Let S be a Sylow p-subgroup of G and set T = S ∩ H. Then the following hold.

(a) We have |T | = q^{3}, and Z(T) is elementary abelian of order q. Moreover, T is a trivial intersection set in H, and we have N_{H}(T) = T D where D is cyclic of order q^{2} − 1, and where D acts regularly on T/Z(T) and |C_{D}(Z(T)| = q − 1.

(b) Any two distinct conjugates of Z(T) generate a subgroup of H isomorphic to SL(2, q). If L = ⟨Z(T), Z(T)^{g}⟩ is such a subgroup then N_{H}(L) ≅ GU(2, q), and if q > 2 then N_{H}(L) is the unique maximal subgroup of H containing L.

(c) Suppose that S − H contains an element a of order p. If p is odd then q^{1/p} is an integer, and C_{H}(a) ≅ SU(3, q^{1/p}). If p = 2 then Ω_{1}(C_{S}(a)) = C_{Z(T)}(a)⟨a⟩ = ⟨a^{Z(T)}.

Proof. That T is a trivial intersection set ffollows from H having Lie rank 1. The rest of part (a) is standard fare and may be obtained by direct computation.

Let L be a subgroup of H generated by two conjugates of Z(T), and let V be the natural 3-dimensional module for H over F = 𝔽_{q^{2}}, endowed with the usual hermitian form. Then [V, Z(T)] is a 1-dimensional singular subspace of V whose stabilizer in H is the Borel group N_{H}(T). It follows that [V, L] is a hyperbolic plane, and hence the stabilizer in H of [V, L] is isomorphic to GU(2, q). Then L ≅ SL(2, q), and this is the first part of (b).

Let M be a subgroup of H containing L, with M ≰ N_{H}(L). We observe that L acts transitively (in fact regularly) on the set of singular points not contained in [V, L], and acts transitively on the set of singular points contained in [V, L]. The stabilizer in H of [V, L] is equal to N_{H}(L), so we conclude that M has just one orbit on the set of singular points of V. Then H = N_{H}(Z(T))M, and we have ⟨Z(T)^{M}⟩ = ⟨Z(T)^{H}⟩ ≤ M. As H is quasisimple if q ≠ 2, we then have M = G or q = 2, and this completes the proof of (b).

Suppose next that we are given an element a of order p in S − H. There is an element a_{0} of the coset T a such that a_{0} induces a field automorphism on H, and one observes that a_{0} acts freely on T/Z(T). Denote by T_{0} the inverse image in T of C_{T/Z(T)}(a_{0}). If p = 2 then a_{0} acts by inversion on T_{0}, and otherwise C_{T} (a_{0}) is isomorphic to a Sylow p-subgroup of SU(3, q^{1/p}). In either case, all elements of order p in T_{0}a_{0} are fused by T to elements of C_{Z(T)}(a_{0})a_{0}. Let D be an a_{0}-invariant complement to T in N_{H}(T). Then all elements of order p in C_{Z(T)}(a_{0})a_{0} are fused by C_{T} (a_{0}), and so ⟨a⟩ is conjugate to ⟨a_{0}⟩ in N_{H}(T). This proves (c). □

We require some of the representation theory of groups of Lie type, as developed in [St1]. Thus, let X be the universal version of a group of Lie type in characteristic p, let U be a Sylow p-subgroup of X,and let H be a complement to U in N_{X}(U). Set q′ = |H_{0}| + 1 where H_{0} is a cyclic subgroup of H of maximal order. (That is, q′ = e + 1 where e is the exponent of H.) Then q′ is a power of p. Set F = 𝔽_{q′}.

If X is a Chevalley group or a Suzuki-Ree group, set q = q′ and set d = 1.

If X is a Steinberg variation (and not a Suzuki-Ree group) obtained in the standard way from a Chevalley group Y via a graph automorphism of order d (d = 2 or 3), set q = (q′)^{1/d}.

With q defined in the above way, we write X = X(q). Then 𝔽_{q} is the usual “field of definition” of X, as expressed, for example, in the formula for the order of X as given in Table 16.1 in [A3]. (We mention, however, that for the Suzuki-Ree groups there is an alternative formalism, whereby d = 2, and where q is taken to be the irrational number (q′)^{1/2}.)

6.4 Lemma. Let X = X(q) be a group of Lie type in characteristic p, and let F = 𝔽_{q^{d}} be the field defined as above. Further, let V be an irreducible 𝔽_{p}X-module, and set F_{0} = End_{X}(V). Then the following hold.

(a) F is a splitting field for X over 𝔽_{p}.

(b) We have dim_{F_{0}}(C_{V} (U)) = 1 for any Sylow p-subgroup U of X.

Proof. Part (a) is Corollary (a) to Theorem 46 in [St2]. Theorems 44(b) and 46(b) in [St2] yield (b).

Let X = X(q) and F = 𝔽_{q^{d}} be as above, and regard X as the set of fixed points of a Steinberg endomorphism of a simple algebraic group \overline{X} over an algebraic closure of F. Let \overline{T} be a σ-invariant maximal torus of \overline{X}, let Σ be the root system of \overline{X} with respect to \overline{T}, let Π be a fundamental system for Σ, and let Λ be the weight lattice of \overline{T}. A weight λ ∈ Λ is basic if either

(1) X is not a Suzuki-Ree group and 0 ≤ ⟨λ, α⟩ for all α ∈ Π, or

(2) X is a Suzuki-Ree group and we have 0 ≤ ⟨λ, α⟩ for all short α ∈ Π, and ⟨λ, α⟩ = 0 for all long α ∈ Π. Let U be an irreducible F X-module, with high weight λ. Then U is a basic module

for X if λ is a basic weight.

6.5 Lemma. Let X = X(q) and F = 𝔽_{q^{d}} have the meanings given above, and assume that X is a Steinberg variation or a Suzuki-Ree group. Let X^{∗} = X^{∗}(q^{d}) be a Chevalley group such that X is obtained from X^{∗}, in the standard way, via a graph automorphism or, in the Suzuki-Ree cases, via the composition of a graph and a field automorphism. Then the irreducible (resp., basic irreducible) modules for X over F are the restrictions to X of the irreducible (resp., basic irreducible) modules for X^{∗}.

Proof. Let \overline{X} be a simple algebraic group such that X^{∗} = O^{p′}(C\overline{X}(σ^{∗})) where σ^{∗} is a Steinberg endomorphism of \overline{X}, and let τ be an automorphism of X^{∗} such that X = O^{p′}(C_{X^{∗}})(τ). Then σ ∘ τ is a Steinberg endomorphism of \overline{X}, and X = O^{p′}(C\overline{X}(σ ∘ τ)). Let \overline{F} be an algebraic closure of F. The theory in [St2] establishes that any irreducible \overline{F} X^{∗}-module \overline{U} is the restriction to X^{∗} of an irreducible \overline{F} \overline{X}-module, and the same is true of any irreducible \overline{F} X-module. Thus any irreducible \overline{F} X-module is the restriction to X of an irreducible \overline{F} X^{∗}-module. As F is a splitting field for both X and X^{∗}, by 6.4(a), it follows that any irreducible F X-module is the restriction to X of an irreducible F X^{∗}-module. The corresponding statement concerning basic modules is then immediate from the definition of the basic modules. □

6.6 Lemma. Let X = X(q), F = 𝔽_{q^{d}}, V, and F_{0} be as in 6.4, and set Ṽ = V ⊗_{F_{0}} F. Denote by Γ the full automorphism group of the field F, and by Γ_{0} the Galois group of F over F_{0}. Set t = |Γ|, m = |Γ_{0}|, and set r = t/m. Let {ϕ_{1}, ⋯, ϕ_{r}} be a complete set of coset representatives for Γ_{0} in Γ, and write Γ_{0} = {ψ_{1}, ⋯, ψ_{m}}. Then the following hold.

(a) There exist basic irreducible modules M_{1}, ⋯, M_{r} for X such that, upon setting

N_{i} = M_{i}^{ϕ_{i}ψ_{1}} ⊗ ⋯ ⊗ M_{i}^{ϕ_{i}ψ_{m}}

we have an isomorphism

Ṽ ≅ N_{1} ⊗ ⋯ ⊗ N_{r}

of F X-modules. (The modules M_{i} are not necessarily pairwise distinct.)

(b) With the modules M_{i} as in (a), set d_{i} = dim_{F} (M_{i}). Then

dim_{F_{0}}(V) = dim_{F} (Ṽ) = (d_{1} ⋯ d_{r})^{m}.

(c) If k is an integer such |V | ≤ |F |^{k}, then (d_{1} ⋯ d_{r})^{m} ≤ km.

Proof. Let (α_{1}, ⋯, α_{t}) be a fixed ordering of Γ, and let B be a set of representatives of the isomorphism classes of basic irreducible F X-modules. As Ṽ is an irreducible F X-module, Steinberg’s tensor product theorem states that there exists a uniquely determined sequence (M_{i_{1}}, ⋯, M_{i_{t}}) of elements of elements of B such that

(*)  Ṽ ≅ M_{i_{1}}^{α_{1}} ⊗ ⋯ ⊗ M_{i_{t}}^{α_{t}}.

Here Ṽ is a module for G × Γ_{0} over F_{0}, in a unique way compatible with the action of G. The uniqueness of the tensor decomposition in (*) then yields (a). Parts (b) and (c) are then immediate. □

6.7 Lemma. The following hold.

(a) Let X = SL(2, q), q = p^{n}, and for each i, 1 ≤ i ≤ p, let M_{i} be the X-module of homogeneous polynomials of degree i − 1 in two variables. Then {M_{i}}_{1≤i≤p} contains exactly one representative from each isomorphism class of basic irreducible F_{q}X-modules.

(b) Let X = SL(3, q), q = 2^{n}. Let V be the natural module for L, U the adjoint module, and 1 the trivial module. Then {1, V, V^{∗}, U} contains exactly one representative from each isomorphism class of basic irreducible F_{q}X-modules.

(c) Let X = Sp(4, q), q = 2^{n}. Let V be the natural module for L, V′ the contragredient module to V, and 1 the trivial module. Then {1, V, V′, V ⊗ V′} contains exactly one representative from each isomorphism class of basic irreducible F_{q}Xmodules.

Proof. The number of isomorphism classes of basic irreducible modules for a (nontwisted) Chevalley group X in characteristic p, and having a root system of rank r, is p^{r}. So, we have the correct number of modules listed in parts (a) through (c) of the lemma, and it remains to check that the given modules for the given groups are in fact irreducible and basic. In the case of SL(2, q), this result is fundamental and well known (see [St2, page 219]), and requires no further discussion here.

Set F = 𝔽_{q}, q = 2^{n}, and let \overline{F} be an algebraic closure of F. For the groups \overline{L} = SL(3, \overline{F}) or Sp(4, \overline{F}), let \overline{T} be the maximal torus given by the diagonal matrices in \overline{L}, let \overline{B} be the Borel subgroup given by the upper triangular matrices in \overline{L}, let Σ be the root system determined by \overline{T}, and let Π = \{α_{1}, α_{2}\} be the fundamental system in Σ determined by \overline{B}. Then let λ_{1} and λ_{2} be the characters of \overline{T} given in the usual way by ⟨λ_{i}, α_{j}⟩ = δ_{i,j}. We may assume that the ordering of Π has been chosen so that λ_{1} is the highest weight for the natural module V for L. Then λ_{2} is the highest weight for the dual of V if \overline{L} = SL(3, \overline{F}), or for the contragredient of V if \overline{L} = Sp(4, \overline{F}).

Take L = SL(3, q), q = 2^{n}, and let L_{0} be the natural SL(3, 2) subgroup of L (i.e. the subgroup in which all the matrix entries are in 𝔽_{2}). Then all irreducible modules for L_{0} are basic. In addition to the trivial module, the natural module, and the dual of the natural module (with high weights 0, λ_{1}, and λ_{2}, respectively), there is an irreducible module U_{0} for L_{0} of dimension 8; namely the Steinberg module, and its highest weight is λ_{1} + λ_{2}. By direct calculation, the adjoint module \overline{U} for \overline{L} contains a high weight vector v with corresponding weight λ_{1} + λ_{2}. As dim_{𝔽_{2}}(U_{0}) = dim\overline{F} (\overline{U}), it follows that \overline{U} ≅ U_{0} ⊗\overline{F} as modules for L_{0}. In particular, \overline{U} is an irreducible \overline{F} L_{0}-module, and hence also an irreducible \overline{F} L-module. The adjoint module U for L is an F -form of \overline{U}, and so U is irreducible for L with highest weight λ_{1} + λ_{2}. This weight is basic, so we have (b).

Now take L = Sp(4, q). The module V ⊗ V′ is irreducible by [St2, pages 218-19], and its highest weight is λ_{1} + λ_{2}. As this weight is basic, we then have (c). □

6.8 Lemma. Assume Hypothesis 2, with H/Z(H) isomorphic to L_{2}(q), q = p^{n}. Let A be an elementary abelian p-subgroup of S such that |A|^{2} ≥ |V/C_{V} (A)|. Then |A| ≤ |V/C_{V} (A)|, and one of the following holds.

(i) V is a natural SL(2, q)-module for H.

(ii) For any irreducible H-submodule W of V, both V and V/W are natural SL(2, q)- modules for H. Moreover, we have |A|^{2} = |V/C_{V} (A)|, and A is a Sylow psubgroup of H.

(iii) q is odd, V is a natural Ω_{3}(q)-module for H, |A|^{2} = |V/C_{V} (A)|, and A is a p-Sylow subgroup of H.

(iv) q is a perfect square, H ≅ Ω_{4}^{−}(q^{1/2}), and V is a natural orthogonal module for

H.

(v) q = 4, G ≅ O_{4}^{−}(2), and for any irreducible G- submodule W of V, both W and V/W are natural orthogonal modules for G. Moreover, we have |A|^{2} = |V/C_{V} (A)|, and A ≰ H.

Proof. Let E be an elementary abelian subgroup of S such that |E|^{2}|C_{V} (E)| is as large as possible and, subject to this condition, so that E is maximal with respect to inclusion. Then E is weakly closed in C_{G}(T), by 2.2. Let U be a non-trivial, irreducible H- invariant section of V, and set F_{0} = End_{H}(U). Set F = 𝔽_{q}, and define the integers r, m, and t, the modules M_{i} and N_{i}, and the dimensions d_{i}, (1 ≤ i ≤ r) as in 6.6. We observe that since q + 1 divides the order of H, 2.1 implies that |V | ≥ q^{2}.

Set T = S ∩ H, and suppose first that E ≰ T. Let a ∈ E − T. Then all elements of order p in the coset T a are fused by T, and so a induces a (non-trivial) field automorphism on H. Denote by T_{0} the largest subgroup of T such that [T_{0}, a, a] = 1. Then [T_{0}, a] = C_{T} (a) = C_{T} (E), and since E is weakly closed in C_{G}(E) we then have E = C_{T} (a)⟨a⟩.

We claim that two conjugates of E suffice to generate HE. Indeed, if p is odd then the claim follows from a theorem of L.E. Dickson (Theorem 2.8.4 in [Gor]) on subgroups of L_{2}(q). On the other hand, suppose that p = 2. Then an element b of C_{T} (a) lies in a dihedral subgroup X of H of order 2 · (q + 1), and X is maximal in H. Thus, our claim holds if q ≠ 4. If q = 4 then HE ≅ Sym(5), and the image of E under such an isomorphism is generated by two commuting transpositions. One confirms that Sym(5) is generated by the conjugate fours groups ⟨(12), (34)⟩ and ⟨(14)(35)⟩, so our claim holds for all q. As |E| = p · q^{1/p}, we then have |V | ≤ p^{4}q^{4/p}. As p^{2n} ≤ |V | we obtain p(1 − 2/n) ≤ 2, and then either p = 2 or p = n = 3. Moreover, we have |V | ≤ 16 · q^{2} if p = 2, and V ≤ 3^{8} if p = n = 3. We have proved the following result.

(1) If E ≰ T then E = C_{T} (a)⟨a⟩ for any a ∈ E − T, and either p = n = 3 and |V | ≤ 3^{8}, or p = 2 and |V | ≤ 16 · q^{2}.

On the other hand, suppose that E ≤ T. As N_{H}(T) acts irreducibly on T, and as E is weakly closed in C_{H}(E), we then have E = T and |E| = q. Then two conjugates of E suffice to generate H, and so |V | ≤ q^{4}. Also, in this case we have |C_{U}(E)| = |F_{0}|, 6.4(b). Thus:

(2) If E ≤ T then E = T, |V | ≤ q^{4}, and |U| ≤ q^{2}q_{0}, where q_{0} = |F_{0}|.

With (1) and (2) we conclude that |V | ≤ q^{4} in any case. Then 6.6 yields the following information.

(3) We have (d_{1} ⋯ d_{r})^{m} ≤ 4m.

Set d = d_{1} ⋯ d_{r}, and suppose that m > 1. Then d = 2 and m ≤ 4. If m = 4 then |U| = q^{4} and q ≥ p^{4}, and so (1) shows that E ≤ T. Then E = T, and we have |U| < q^{3}, by (2), so we conclude that m ≤ 3.

Suppose that m = 3. Then |U| = q^{8/3}. If E = T then (2) yields |U| ≤ q^{2}q^{1/3}, so in fact we have E ≠ T. Suppose that p = 2. Then (1) gives 16 · q^{2} ≥ q^{8/3}. Here n is divisible by 2 (as E ≠ T) and by 3 (as m = 3), and we conclude that q = 64. Moreover, we then have |U/C_{U}(E)| = |E|^{2}, so U is the unique non-trivial constituent for H in V. Then U = V, by 1.2. Now C_{V} (E ∩ H) is a subspace of V over F_{0}, while a acts F_{0}- semilinearly on V. It follows that |C_{V} (E ∩ T)| = |C_{V} (E)|^{2}, and then |E ∩ T |^{2}|C_{V} (E ∩ T)| > |E|^{2}|C_{V} (E)|. This is contrary to the definition of E, so we conclude that p ≠ 2.

With p odd, (1) shows that G ≅ ΓL(2, 27) and that |U| = 3^{8}. Here |V/C_{V} (E)| ≤ |E|^{2} = 81, while |U/C_{U}(E)| = 81 since G is generated by two conjugates of E. Thus V = U, by 1.2, and we have an F_{0}-linear action of G on U. Then G acts F -linearly on Ũ. Identify Ũ with N ⊗N^{ϕ} ⊗N^{ϕ^{2}}, where ϕ is an automorphism of F of order 3, and identify a with a standard field automorphism of H. Then a acts on Ũ by permuting the three tensor factors. Explicitly, let x = (1, 0) and y = (0, 1) be the standard basis elements for N. One may then identify a standard basis for Ũ with the set of monomials of degree 3 in the non-commuting variables x and y, and then CŨ(a) is the F -linear span of

{x^{3}, x^{2}y + xyx + yx^{2}, xy^{2} + yxy + y^{2}x, y^{3}}.

Thus dim_{F} (CŨ(a) = 4. One observes that the given basis for CŨ(a) is not invariant under E ∩ T, so dim_{F} (CŨ(E)) ≤ 3. Then also dim_{F_{0}}(C_{U}(E)) ≤ 3, and so |U/C_{U}(E)| ≥ 3^{5} > |E|^{2}. With this contradiction, we have shown that m ≠ 3, and so m ≤ 2.

If m = 2 then U is an Ω_{4}^{−}(q_{0})-module, where q_{0} = q^{1/2}. On the other hand, suppose that m = 1. Then (3) yields d ≤ 4. If d = 4 then |U| = q^{4}, so (1) and (2) imply that E ≰ T and q = 4. Then U ≅ N ⊗ N^{ϕ}, where N is the natural SL(2, 4)-module for H and where ϕ is the non-identity field automorphism. But then U is reducible as an 𝔽_{2}H-module, and contrary to the definition of U. Thus, we have d ≤ 3. This shows the following.

(4) U is a natural SL(2, q)-module, a natural Ω_{3}(q)- module, or a natural Ω_{4}^{−}(q_{0})-module (q_{0} = q^{1/2}), for H. Moreover, if E ≠ T then p = 2.

We observe that if U is not E-invariant then E ≰ T and |C_{U}(E ∩ T)| ≥ p^{2}. Then |U/C_{U}(E ∩ T)| ≥ |E ∩ T |. For any minimal HE-invariant section W of V involving U we then have

|E ∩ T |^{2}|C_{W} (E ∩ T)| > p^{2}|E ∩ T |^{2}|C_{W} (E)| ≥ |E|^{2}|C_{W} (E)|,

and

|W/C_{W} (E ∩ T)| ≥ |E ∩ T |^{2}.

Then |W | ≥ |E|^{2}|C_{W} (E)|, so |W | = |V | = |E|^{2}|C_{V} (E)|, and

|E ∩ T |^{2}|C_{V} (E ∩ T)| > |E|^{2}|C_{V} (E)|.

This is contrary to the definition of E, so we conclude that U is E-invariant. Moreover, the preceding estimates show that HE has at most two non-trivial irreducible constituents in V, and if there are two then for any choice of U we have |E| = |U/C_{U}(E)|.

If H has only one non-trivial constituent in V then 1.2 gives V = U. So assume that H has two non-trivial constituents in V. If E = T then |U/C_{U}(E)| = |U|/|F_{0}| ≤ q, so |U| ≤ q · q_{0}, and then q = q_{0} and U is a natural SL(2, q)-module for H. On the other hand, suppose that E ≠ T. Then (1) and (4) yield p = 2 and |U/C_{U}(E)| ≤ |E| = 2·q^{1/2}. As two conjugates of E generate G we then have |U| ≤ 4 · q, and since |U| ≥ q^{2} we then have q = 4 and |U| = 16. If U is a natural ΓL(2, 4)-module for G then |U/C_{U}(E)| = 8, so in fact U is a natural O_{4}^{−}(2)-module for G. Thus:

(5) Suppose that U ≠ V. Then H has exactly two non-trivial irreducible constituents in V. Moreover, these constituents are E- invariant, and for any such irreducible constituent U we have |E| = |U/C_{U}(E)|. Either both irreducible constituents are natural SL(2, q)-modules for H, or else q = 4 and both are natural O_{4}^{−}(2)- modules for G.

Suppose that U ≠ V, and suppose that there is a trivial constituent for H in V. As C_{V} (H) = 0 and V = [V, H], by hypothesis, there is then an indecomposable Hsubmodule W of V such that [W, H] is irreducible for H and such that |W/[W, H]| = p. Take U = [W, H]. Then (5) says that either U is a natural SL(2, q)-module for H, or q = 4 and U is a natural O_{4}^{−}(2)-module for G. In either case, we have W = U + C_{W} (E). Suppose that U is a natural SL(2, q)- module. As |E| = |U/C_{U}(E)| we then have E = T and W = U + C_{W} (T). As two conjugates of T generate H we then have C_{W} (H) ≠ 0, and W is decomposable. Thus q = 4 and U is an O_{4}^{−}(2)-module, but in that case we contradict 1.4. Thus, H has no trivial constituents in V.

Now let A be any elementary abelian subgroup of S such that |A|^{2} ≥ |V/C_{V} (A)|. Suppose first that |E|^{2} = |V/C_{V} (A)|. Then we may take A ≤ E (by definition of E). In particular, if V is a natural Ω_{3}(q)-module for H we have A ≤ E = T, and then A = E. Thus, (iii) holds in this case. In the same way, if H has two non-trivial irreducible constituents in V then A ≤ E = T. In the case that these constituents are natural SL(2, q)- modules we find that A = E, and (ii) holds. In the case that these constituents are natural O_{4}^{−}(2)-modules we observe that E ∩ H induces a 2-transvection on each irreducible constituent, and so |E ∩ H|^{2} < |V/C_{V} (E ∩ H|. Thus A ≠ E ∩ H, and so (v) holds in this case. If V is a natural Ω_{4}^{−}(q)-module or a natural SL(2, q)-module for H then we make no assertion about A, and we have (iv) or (i).

In order to complete the proof of the lemma, it remains to show that |A| ≤ |V/C_{V} (A)|. Suppose false. Then (i) or (iv) holds, by what has already been proved, and we then observe that A ≠ T. By [REF] we may assume that A is weakly closed in C_{G}(A), and therefore A ≰ T. Setting A_{0} = A ∩ T we then have T ≠ A_{0} ≠ 1 and |A_{0}| ≥ |V/C_{V} (A_{0})|. This condition excludes (i), and so (iv) holds. But ⟨A_{0}^{N_{H}(T)}⟩ = T, and [REF] then implies that |T | ≥ |V/C_{V} (T)|, which is not the case in (iv). Thus, we have a contradiction, and the lemma is proved. □

6.9 Lemma. Assume Hypothesis 2, with H isomorphic to the commutator subgroup of a group of Lie type and of Lie rank equal to 1, in characteristic p. Assume that H/Z(H) ≇ P SL(2, q). If G ≅ ^{2}G_{2}(3) set T = S, and otherwise set T = S ∩ H. Then |A|^{2} = |V/C_{V} (A)|, A ≤ H, and one of the following holds.

(1) (i) H ≅ SU(3, q) and V is a natural module for H. Moreover, either A = Z(T) and [V, A, A] = 0, or else p is odd, |A| = q^{2}, and [V, A, A] ≠ 0.

(2) (ii) H ≅ Sz(q) and V is a natural module for H. Moreover, we have A = Z(T) and [V, A, A] = 0.

Proof. As in the proof of the preceding lemma, we fix an elementary abelian subgroup E of S such that |E|^{2}|C_{V} (E)| is as large as possible and, subject to this condition, so that E maximal with respect to inclusion. Then 2.2 implies that E is weakly closed in C_{G}(E) with respect to G.

Suppose first that HT/Z(H) is isomorphic to ^{2}G_{2}(q), q = 3^{2n+1}. Surveying the Schur multipliers of these groups, we find that Z(H) is a 3-group, and so Z(H) = 1. We observe right away that q^{3} + 1 divides |G|, and hence |V | ≥ q^{6} by 2.1. Let D be a subgroup of N_{H}(T) of order q − 1, let d be an involution in D, and set L = O^{2}(C_{H}(t)). Then L ≅ L_{2}(q), and L ∩ T is a complement in Ω_{1}(T) to Z(T), by 6.1. Suppose that E ≰ T. Then 6.1(d) implies that E = Ω_{1}(C_{S}(a)) for any a ∈ E − T. We may choose a ∈ E − T so that a centralizes an involution in N_{H}(T), and so we may assume that [a, d] = 1. Set X = C_{E∩T} (d). Then |X| = q^{1/3}. Here q is an odd power of 3, so q ≠ 9, and a theorem of L.E. Dickson (2.8.4 in [Gor]) implies that two conjugates of X suffice to generate L. As E ∩ T ≰ X we then conclude from 6.1(b) that two conjugates of E suffice to generate HE. Then |V | = |V/C_{V} (HE)| ≤ 81 · q^{8/3}. As q ≥ 3^{3} in this case, we then have |V | ≤ q^{4/3}q^{8/3} = q^{4}, contrary to |V | ≥ q^{6}. We therefore conclude that E ≤ T. As Ω_{1}(T) is abelian, E is normal in N_{H}(T), and then E = Z(T) or E = Ω_{1}(T).

If E = Z(T) then two conjugates of E suffice to generate H, by 6.1(c), and so |V | ≤ q^{4}. We therefore conclude that E = Ω_{1}(T). Now two conjugates of E generate HE, and so |V | ≤ q^{8}. Let U be a non- trivial, irreducible L-invariant section of V. Then |U/C_{U}(E ∩ L)| ≤ q^{4} and since E ∩ L is a Sylow subgroup of L it follows from 6.4(b) that |U| ≤ q^{5}. Set F_{0} = End_{L}(U), F = 𝔽_{q}, and denote by Γ_{0} the Galois group of F over F_{0}. Define the integers t and m by |F : 𝔽_{p}| = t and |F : F_{0}| = m, and set r = t/m. Set Ũ = U ⊗_{F_{0}} F, and let modules M_{i} and N_{j}, (1 ≤ i, j ≤ m) be given as in 6.6. Write d_{i} for the dimension of M_{i} over F. As p = 3 we have d_{i} ≤ 3, and there are exactly three basic irreducible modules for SL(2, q). Thus, any M_{i} is either a trivial module, a natural SL(2, q)-module, or a natural Ω_{3}(q)-module for the universal cover SL(2, q) of L. As |U| ≤ q^{5}, 4.6(c) yields

(d_{1} ⋯ d_{r})^{m} ≤ 5m.

As Ũ affords a representation of SL(2, q) with the scalar matrix −I acting trivially, there are an even number of occurances of the natural SL(2, q)-module in the tensor decomposition of Ũ given by 6.6. As |Γ_{0}| is odd, it follows that there are an even number (possibly 0) of the non-trivial modules N_{i} for which the associated basic irreducible modules M_{i} are natural SL(2, q)-modules. Taking d_{1} to be maximal among the numbers d_{i}, we then have either d_{1} = 3 and 3^{m} ≤ 5m, or else d_{1} = d_{i} = 2 for some i ≠ 1 and 4^{m} ≤ 5m. As also m is odd, we conclude that d_{1} = 3, m = 1, F_{0} = F, and U is a natural Ω_{3}(q)-module for L. As |V | ≤ q^{8} there are then at most two non-trivial irreducible constituents for L in V.

Suppose that there L has two non-trivial constituents in V. As |U/C_{U}(E ∩ L)| = q^{2} we then have |V/C_{V} (E ∩ L)| = |V/C_{V} (E)| = q^{4}. Then C_{V} (L) ≤ C_{V} (E), and since ⟨L, E⟩ = H, by 6.1(b), we conclude that C_{V} (L) = 0. Then 1.5 implies that there are no trivial constituents for L in V, and so |V | = q^{6}. As all involutions in H are conjugate, all have non-trivial fixed points on V, and so the involution d centralizes a subspace of V of order q^{3}. But one observes that for any involution x ∈ L we have |C_{U}(x)| = q, and so |C_{V} (x)| = q^{2}. With this contradiction we conclude that L has a unique non-trivial constituent in V. Then V = U ⊕ C_{V} (L), by 1.3, and so |E ∩ L|^{2} = |V/C_{V} (L)| = q^{2}. Then |E|^{2} = |V/C_{V} (E)| = q^{4}, and so |C_{V} (E)| ≥ q^{2}. As C_{V} (E) ≤ C_{V} (E ∩ L) = C_{U}(E ∩ L) ⊕ C_{V} (L), it follows that C_{V} (L) ∩ C_{V} (E) ≠ 0, and so C_{V} (H) ≠ 0. We therefore conclude that H is not isomorphic to ^{2}G_{2}(q).

It remains to treat the cases where H/Z(H) is a Suzuki group or a 3-dimensional unitary group. Again, let U be a non-trivial, irreducible H-invariant section of V and set F_{0} = End_{H}(U). If H ≅ Sz(q) set F = 𝔽_{q}, while if H/Z(H) ≅ U_{3}(q) set F = 𝔽_{q^{2}}. Set Ũ = U ⊗_{F_{0}} F, and let Γ, Γ_{0}, M_{i}, N_{i}, d_{i}, m, r, and t be given as in 6.6.

Suppose that p = 2 and that H ≅ Sz(q). Here we have S = T, by 6.2(c). As Ω_{1}(T) = Z(T), and since N_{H}(T) acts irreducibly on Z(T), we then have E = Z(T). Then two conjugates of E suffice to generate H, by 6.2(b), and since C_{V} (H) = 0, it follows that |V | ≤ q^{4}. Then 6.6 yields (d_{1} ⋯ d_{r})^{m} ≤ 4m. But also, as q^{2} + 1 divides |H|, it follows from 2.1 that d_{i} ≥ 4 for any i for which d_{i} ≠ 1. As m is odd, we then have m = 1 and d_{i} = 1 for all but one index i. That is, U is an algebraic conjugate of a basic irreducible module for H, of dimension 4 over F. The basic irreducible modules for H are obtained by restriction from those for Sp(4, q), by 4.5, so 4.7 implies that U is a natural Sp(4, q)-module for H. One then observes that |U/C_{U}(Z(T))| = |E|^{2}, so A = E, and V = U + C_{V} (E). As H = O^{2}(H) we then have U = [V, H], and then 1.2 implies that U = V. Thus, (ii) holds in this case.

Suppose next that H/Z(H) ≅ U_{3}(q). Then q^{3} + 1 divides |H|, and so |V | ≥ q^{6}, by 3.1. Consider first the case where E ≰ T. Then 6.3(c) shows that E = (E ∩ T)⟨a⟩ where a induces a field automorphism on H. Suppose further that p is odd, so that C_{T} (a) is isomorphic to a Sylow p- subgroup of U_{3}(q^{1/p}). In particular, we note that q ≠ 9. Let X be the largest subgroup of T for which [X, a, a] = 1. Then [X ∩ Z(T), a] = C_{Z(T)}(a) = C_{Z(T)}(E), and since E is weakly closed in C_{H}(E) we then have C_{Z(T)}(a) ≤ E. We claim that E ∩ T ≰ Z(T). Suppose false, and let x ∈ X − Z(T). Then [x, a] = [x, E] ≤ C_{T} (E), and so ⟨E, E^{x}⟩ is elementary abelian and properly contains E. This is contrary to E being weakly closed in its centralizer, and proves the claim.

We next claim that HE is generated by two conjugates of E. Indeed, we have E ∩ Z(T) ≤ L where L is a subgroup of H isomorphic to SL(2, q). As q is odd and q ≠ 9, it follows from [Gor, 2.8.4] that L is generated by two conjugates of E ∩ Z(T), and then 6.3(b) yields our claim. As |E| ≤ p·q^{2/p} it follows that |V | ≤ p^{4} ·q^{8/p}. As q ≥ p^{3} we then have |V | ≤ q^{4}. But we have seen already that |V | ≥ q^{6}, so we have a contradiction at this point. On the other hand, suppose that E ≰ T and that p = 2. Then Ω_{1}(T) = Z(T), and we conclude from 6.3(c) that E ∩ T = Z(T). Let L be a subgroup of H, generated by Z(T) and a conjugate of Z(T), with L ≅ SL(2, q). If L is E-invariant then E induces inner automorphisms on L, and there is then an element a_{1} of E − Z(T) such that [L, a_{1}] = 1, and we have L_{1} = C_{H}(a_{1}). As |E − Z(T)| = q, and since there are at least q^{2} conjugates of L under T, we may therefore choose L so that L is not E-invariant. It then follows from 6.3(b) that two conjugates of E suffice to generate HE. We then obtain a contradiction to |V | ≥ q^{6}, just as in the case where p is odd.

We now have E ≤ T. Suppose that E ≤ Z(T). As N_{H}(T) acts irreducibly on Z(T) we then have E = Z(T). Now 6.3(b) implies that three conjugates of E suffice to generate H, and so |V | ≤ q^{6}. Here F = 𝔽_{q^{2}}, so 6.6 yields (d_{1} ⋯ d_{r})^{m} ≤ 3m. But also d_{i} ≥ 3 if d_{i} ≠ 1, so we obtain m = 1, and U is a natural SU(3, q)-module for H. Here |E|^{2} = |V/C_{V} (E)|, so U = [V, H], and then U = V, by 1.2. Further, E acts quadratically on V, and since |E_{0}|^{2} < |V/C_{V} (E_{0})| for every proper non- identity subgroup E_{0} of E, we conclude that A = Z(T) if A ≤ Z(T).

On the other hand, suppose that E ≰ Z(T). As already mentioned, Z(T) = Ω_{1}(T) if p = 2, so we must have p odd in this case. Now q < |E| ≤ q^{2}, and two conjugates of E suffice to generate H. Then |V | ≤ q^{8}, and so (d_{1} ⋯ d_{r})^{m} ≤ 4m. Again, as d_{i} ≥ 3 for all i for which d_{i} ≠ 1, we obtain m = 1 and U = V is an irreducible F H-module with dim_{F} (U) = 3 or 4. By 4.5, V is the restriction to H of an irreducible module for the group H^{∗} = SL(3, F). Regard H as a subgroup of H^{∗}. Then T is contained in two nonconjugate parabolic subgroups P_{1} and P_{2} of H^{∗}, and we have Z(T) ≤ O_{p}(P_{1}) ∩ O_{p}(P_{2}). Also, we have P_{i} = ⟨E^{P_{i}}⟩ for both i = 1 and 2, and not both P_{1} and P_{2} centralize C_{V} (Z(T)). It follows that C_{V} (Z(T)) ≠ C_{V} (E). But dim_{F} (C_{V} (E) ≥ 2 since |E|^{2} ≥ |V/C_{V} (E)|, so we conclude that dim_{F} (C_{V} (Z(T)) ≥ 3. Then |V/C_{V} (Z(T))| = q^{2}, and since three conjugates of Z(T) generate H we obtain |V | ≤ q^{6}. This contradiction shows that dim_{F} (V) = 3. That is, V is the restriction to H of a natural SL(3, F)-module, which is to say that V is a natural SU(3, q)-module for H. Then |E| = q^{2}, in order to achieve |E|^{2} ≥ |V/C_{V} (E)|.

Since E was chosen so that |E|^{2}|C_{V} (E)| is as large as possible, we now conclude that |A|^{2}|C_{V} (A)| = |V |. The maximality of E then yields A ≤ E for some choice of E. Then A ≤ H, and since |E_{0}|^{2} < |V/C_{V} (E_{0})| for any non-identity subgroup E_{0} of E other than E and Z(T), we conclude that A = Z(T) or A = E. Thus (i) holds, and the lemma is proved. □

We now consider the case in which A acts quadratically on V.

6.10 Lemma. Assume Hypothesis 4, and suppose that H/Z(H) is a simple group of Lie type in characteristic p and of Lie rank 1. Let A be a non-identity subgroup of S such that |A|^{2} ≥ |V/C_{V} (A)| and such that [V, A, A] = 0. Then one of the following holds.

(i) H ≅ SL(2, q), V is a natural SL(2, q)-module for H, and A ≤ H. If |A| ≥ |V/C_{V} (A)| then equality holds, and A is a Sylow subgroup of H.

(ii) H ≅ SL(2, q), V is a direct sum of two natural SL(2, q)-modules for H, A is a

Sylow subgroup of H, and |A|^{2} = |V/C_{V} (A)|. (iii) H ≅ L_{2}(q^{2}) and V is a natural Ω_{4}^{−}(q)-module for H. If A ≤ H then |A|^{2} =

|V/C_{V} (A)| and A is conjugate to a Sylow p-subgroup of a subgroup K of H with K ≅ L_{2}(q). If |A|^{2} > |V/C_{V} (A)| then p = 2, |A|^{2} = 2 · |V/C_{V} (A)| , and either:

(a) A = (A ∩ H)⟨a⟩, where a induces a field automorphism on H and where A ∩ H is a Sylow subgroup of C_{H}(a), or

(b) q = 4, |A| = 2, and A induces a transvection on V. (iv) G ≅ Sym(5), V is a direct sum of two natural O_{4}^{−}(2)-modules for G, |A|^{2} =

|V/C_{V} (A)|, and A ≰ H.

(v) H ≅ SU(3, q), V is a natural module for H, |A|^{2} = |V/C_{V} (A)|, and A is the center of a Sylow subgroup of H.

(vi) H ≅ Sz(q), V is a natural module for H, |A|^{2} = |V/C_{V} (A)|, and A is the center of a Sylow subgroup of H.

Proof. Suppose that H ≅ SL(2, q), that H has more than one non- trivial irreducible constituent in V. Then 6.8 shows that there are just two such constituents, and no trivial constituents, for H in V. Suppose that the constituents are natural SL(2, q)- modules for H. Then 6.8 says that A is a Sylow subgroup of H, and then V is a completely reducible H-module, by 1.6. Thus (ii) holds in this case. On the other hand, suppose that not both of the irreducible constituents for H in V are natural SL(2, q)-modules for H. Then 6.8 says that q = 4, that A ≰ H, and that both irreducible constituents are natural O_{4}^{−}(2)-modules for G. Let a be an element of A − H. If |A| = 2 then A = ⟨a⟩ and a induces a 2-transvection on V. Suppose instead that |A| = 4. Then |C_{V} (A)| = 16 and for any irreducible constituent U for G in V we observe that |C_{U}(A)| = 4. Since a commutes with an element g of H of order 3, and A is not g-invariant, it follows that |C_{V} (a)| = 64, and that a is a 2-transvection in this case as well. Let K be a subgroup of G generated by three conjugates of a, with K ≅ Sym(4). Then, in any case, we have |V/C_{V} (K)| = 4, and if V_{0} is a fixed irreducible submodule of V we may choose v ∈ C_{V} (K) so that v ∉ V_{0}. Now |v^{G}| = 5, and so ⟨v^{G}⟩ is a proper subspace of V. This shows that V is completely reducible, and thus (iv) holds in this case.

We are now reduced to the case where V is irreducible. If H ≅ L_{2}(q) and V is a natural Ω_{3}(q)-module for H then A ≤ H, by 6.8. But here p is odd and no non-identity element of H acts quadratically on V, so Hypothesis (4) is violated in this case. If H ≅ SL(2, q) and V is a natural SL(2, q)-module for H then (i) follows. Assume that H ≅ L_{2}(q^{2}) and that V is a natural Ω_{4}^{−}(q)-module for H. Suppose further that A ≤ H. No element of H centralizes a 3-dimensional 𝔽_{q}-subspace of V, so we have |A| ≥ q. Let K be a subgroup of H with K ≅ L_{2}(q) and with K ∩ S a Sylow p-subgroup of K. Set B = K ∩S. The conjugates of B^{#} under N_{H}(T) then form a partition of (H ∩S)^{#}. Here B acts quadratically on V, and for any b ∈ B^{#} we have C_{V} (b) = [V, b] = [V, B] = C_{V} (B) is a 2-dimensional subspace of V. Let C be a conjugate of C under N_{H}(H ∩ S), C ≠ B. Then H ∩ S = BC, and so C_{V} (BC) is of dimension 1. Then C_{V} (B) ≠ C_{V} (C), and so [V, b, c] ≠ 0 for any non-identity elements b ∈ B and c ∈ C. We conclude that A is conjugate to B and that |A|^{2} = |V/C_{V} (A)|. On the other hand, suppose that A ≰ H and that |A|^{2} > |V/C_{V} (A)|. Then the foregoing shows that either A ∩ H = 1 , or |A ∩ H|^{2} = |V/C_{V} (A ∩ H)| = |V/C_{V} (A)| and that A ∩ H is conjugate to the group B given as above. If A∩H ≠ 1 we then have (iii)(a), while if A∩H = 1 then |V/C_{V} (A)| = 2 and we have (iii)(b). Thus (iii) holds if V is a natural Ω_{4}^{−}(q)-module for H.

It remains to consider the cases in which H/Z(H) is not a linear group. We then have (v) or (vi), as follows from 6.9 and from elementary properties of the natural modules for SU(3, q) and for Sz(q). □

Section 7: Lie rank 2 in characteristic p

7.1 Lemma. Let G = L_{3}(2) and let W be an indecomposable G-module of dimension 4 over 𝔽_{2}. Then there exists a fours group A ≤ G with |C_{W} (A)| = 2.

Proof. Since G is generated by three involutions, and all involutions in G are conjugate, no involution induces a transvection on W. Put Z = C_{W} (G), and suppose first that Z ≠ 0. Then |Z| = 2 and W/Z is a natural G-module. Let P be a rank-1 parabolic subgroup of G such that C_{W/Z}(P) = 0, and put A = O_{2}(P). Then C_{W/Z}(A) is a fours group on which P acts irreducibly, and since A contains no transvections we conclude that C_{W} (A) = Z. On the other hand, suppose that Z = 0. Then W has a submodule U of order 8, on which every involution in G act as a transvection. Now let P be the rank-1 parabolic subgroup with |C_{U}(P)| = 2, and put A = O_{2}(P). Then C_{U}(A) = C_{U}(P), and since A contains no transvections on W we have C_{W} (A) ≤ U. Thus |C_{W} (A)| = 2. □

7.2 Lemma. There are exactly four isomorphism classes of irreducible modules for Alt(6) over 𝔽_{2}. These classes are represented by the principal module 1, a natural Sp(4, 2)-module U, the module U′ contragredient to V, and a module V of dimension 8 which is a direct summand of the Steinberg module for Sp(4, 2)

Proof. Set H = Alt(6) and G = Sp(4, 2), and identify H with [G, G]. There are four 2-regular conjugacy classes in H, and so, by a well known result of Brauer, there are also four isomorphism classes of irreducible modules for H. The modules 1, U, and U′ represent three of these classes (with U and U′ distinguishable by their restrictions to subgroups of order 3 in H). Let W be the Steinberg module for G. Then W is the reduction mod 2 of an irreducible complex representation of G, written over ℤ, and since H has no irreducible characters of degree 16 we conclude that W is reducible for H.

Let S be a Sylow 2-subgroup of G, and set T = S ∩ H. Then S acts freely on W, and so dim(C_{W} (T)) = 2. But W is a direct sum of G-conjugate, irreducible H-submodules, so we conclude that W has an H-submodule V of dimension 8. Now {1, U, U′, V } is a complete set of representatives of the isomorphism classes of 𝔽_{2}H-modules. □

7.3 Lemma. Assume Hypothesis 2 with p = 2, and assume that H/Z(H) ≅ L_{3}(2^{n}) or Sp(4, 2^{n})′. Suppose further that Z(H) ≠ 1. Then H ≅ SL(3, 2^{n}).

Proof. Suppose false. As O_{2}(H) = 1, a survey of the Schur multipliers of the groups in question yields H/Z(H) ≅ Alt(6) and |Z(H)| = 3. Without loss, V is irreducible, and then V may be regarded as a vector space over the field F = 𝔽_{4}. Evidently dim_{F} (V) ≥ 3. Also, we may assume that A has been taken to be as small as possible, subject to the condition that |A|^{2} ≥ |V/C_{V} (A)|. Therefore:

(1) Let B be a subgroup of index 2 in A. Then |C_{V} (B)/C_{V} (A)| ≤ 2, or B = 1.

Let X_{1} and X_{2} be the two maximal subgroups of H containing S ∩ H, and set T = N_{S}(X_{1}). Then also T = N_{S}(X_{2}), and since S is contained in a unique maximal subgroup of G we have |S : T | = 2. Set Z = C_{V} (S ∩ H), and set U_{i} = ⟨Z^{X_{i}}⟩, (i = 1, 2). As V is irreducible we have C_{V} (H) = 0, and then since X_{1} and X_{2} are fused by S, we conclude that dim_{F} (U_{i}/Z) ≥ 1, (i = 1, 2). As [U_{i}, O_{2}(X_{i})] = 0, we have [U_{1} + U_{2}, Z(S ∩ H)] = 0, and so V ≠ U_{1} + U_{2}, and dim_{F} (V) ≥ 4.

Let R be a Sylow 3-subgroup of H. Then N_{H}(R) contains a representative from each conjugacy of involutions in HT, and one observes that all involutions in HT − H act non-trivially on Z(H). That is, there does not exist a central extension of the form 3^{·}Sym(6).

Suppose that A = ⟨a⟩ is of order 2. Then a is a 2-transvection, and since dim_{F} (V) ≥ 3 it follows from the preceding paragraph that a commutes with Z(H). Suppose a ∈ T. Then a ∈ H, and three conjugates of a generate H, so |V | = |V/C_{V} (H)| ≤ 4^{3}. This is contrary to dim_{F} (V) ≥ 4, so we conclude that a ∉ T. Then a interchanges the two fours groups in S ∩ H, and so (S ∩ H)A is dihedral of order 16. Then Z(H)(S ∩ H)A is a maximal subgroup of HA, and so again, three conjugates of A suffice to generate HA, with a contradiction as before. We therefore conclude that |A| > 2.

Suppose that A ≰ T, and let a ∈ A−T. Then C_{S}(a) ≤ Z(S ∩H), and so [U_{1} +U_{2}, A∩ T] = 0. But a interchanges U_{1} and U_{2}, so |(U_{1} +U_{2})/C_{U_{1}+U_{2}}(a)| ≥ 4. This is contrary to (1), so we conclude that A ≤ T. Now suppose that A ≰ H, and let a ∈ A − H. Then a centralizes an 𝔽_{2}-hyperplane of C_{V} (A ∩ H), by (1), and so dim_{F} (C_{V} (A ∩ H) = 1. Thus |C_{V} (A)| = 2, and since |A| ≤ 8 we obtain |V | ≤ 2^{7}. but then dim_{F} (V) = 3, and we have the same contradiction as before. Thus A ≤ H, and |A| = 4. Now A ≤ O_{2}(X_{i}) for some i, and we may take i = 1. By conjugating A within X_{1} we may assume also that A ≰ O_{2}(X_{2}). Set B = A ∩ O_{2}(X_{2}). Then |B| = 2 and C_{V} (B) ≥ U_{1} + U_{2}. As O^{2′}(X_{2}) acts non-trivially on U_{2}, we have [U_{2}, A] ≠ 1, and then |U_{2}/C_{U_{2}}(A)| ≥ 4. This contradicts (1), and the lemma is thereby proved. □

7.4 Lemma. Let L = L_{3}(q) or Sp(4, q), q = 2^{n}, and identify L with the group of inner automorphisms of L. Let t be a non-inner automorphism of L of order 2. Then all involutions in Lt are conjugate via L. Moreover, the following hold.

(a) If L = L_{3}(q) and n is odd, and t is conjugate to a standard graph automorphism of L, and C_{L}(t) ≅ L_{2}(q).

(a) If L = L_{3}(q) and n is even, then t is conjugate to a standard graph automorphism (with C_{L}(t) ≅ L_{2}(q)), or a field automorphism (with C_{L}(t) ≅ L_{3}(q^{1/2})), or a graph-field automorphism (with C_{L}(t) ≅ U_{3}(q^{1/2}).

(c) If L = Sp(4, q) and n is odd, then t is conjugate to an automorphism of L which induces a non-trivial polarity on the Dynkin diagram of L, and C_{L}(t) ≅ Sz(q).

(d) If L = Sp(4, q) and n is even, then t is conjugate to a field automorphism of L, and C_{L}(t) ≅ Sp(4, q^{1/2}).

Proof. Let \overline{F} be an algebraic closure of F = 𝔽_{q}, set \overline{L} = L_{3}(\overline{F}) (resp. Sp(4, \overline{F})), and view L as the set of fixed-points for a Steinberg endomorphism σ of \overline{L}. Fix a σ- invariant Borel subgroup \overline{B} of \overline{L}, a σ-invariant maximal torus \overline{T} of \overline{B}, let Σ be the root system determined by \overline{T}, and let Π be the fundamental system of roots determined by \overline{T} and \overline{B}. Let \overline{S} be the unipotent radical of \overline{B}, and set S = C\overline{S}(σ), so that S is a Sylow 2-subgroup of L. If n = 2m is even, set r = 2^{m}, and let ϕ be the standard field automorphism of L of order 2, given by (x_{α}(v))^{ϕ} = x_{α}(v^{r}) for α ∈ Σ and v ∈ F. Let ρ be the unique automorphism (resp. angle-preserving, length-changing bijection) of Σ of order 2 which preserves Π, and let γ be the automorphism of L given, in the case of L = L_{3}(q), by

x_{α}(v)^{γ} = x_{α^{ρ}}(v),

and in the case of L = Sp(4, q) by

x_{α}(v)^{γ} = \{ \begin{matrix}x_{α^{ρ}}(v)  if α is long, \\ x_{α^{ρ}}(v^{2})  if α is short. \end{matrix}

If L = L_{3}(q) set τ = γ, and if L = Sp(4, q) set τ = γ^{n}. Notice that if L = Sp(4, q) we have Aut(L) = Inn(L)⟨γ⟩, τ = 1 if n is even, and τ is an involution if n is odd. Further, if L = Sp(4, q) and n is odd then C_{L}(τ) ≅ Sz(q). (See [GLS3, 2.5.1].) It therefore only remains to show that all involutions in St are conjugate via L.

Identify L with Inn(L). Let E and F be the two maximal (under inclusion) elementary abelian subgroups of S, and set Z = E ∩ F. Then Z = Z(T). Suppose first that F = E^{t}. Then either t ∈ Sτ or L ≅ L_{3}(q) and t ∈ Sτϕ. Set J = {zxx^{t} | z ∈ C_{Z}(t), x ∈ E}. Then J is the set of elements of S inverted by t, and so Jt is the set of involutions in St. Let μ : E × C_{Z}(t) ⟶ J be the mapping given by (x, z) ↦ zxx^{t}. Then for any (x, z) ∈ E×C_{Z}(t) we have μ^{−1}(zxx^{t}) = {(xu, zuu^{t}) | u ∈ Z}. Thus |J| = |E×C_{Z}(t)|/|Z|. If t ∈ Sτ we then have |J| = q^{2}, while if L = L_{3}(q) and t ∈ Sτϕ we have |J| = q^{3/2}. Thus, |Jt| = |J| = |S : C_{S}(τ)| if t ∈ Sτ, and again |Jt| = |J| = |S : C_{S}(τϕ)| if t ∈ Sτϕ. It follows that t ∈ τ^{S} or (τϕ)^{S}, and so we are done if t interchanges E and F.

We are now reduced to the case where t ∈ Sϕ. Let P be a maximal parabolic subgroup of L containing S, set X = O^{2′}(P), and take E = O_{2}(P). Then P is ϕ-invariant, and |C_{S/E}(ϕ)| = |S/E|^{1/2}. It follows that the number of involutions in (S/E)ϕ is equal to the number of S/E-conjugates of ϕ, and so we may take t ∈ Eϕ. But also |C_{E}(ϕ)| = |E|^{1/2}, and so the number of involutions in Eϕ is equal to |ϕ^{E}|. Thus t and ϕ are conjugate via S, and the lemma is proved. □

7.5 Proposition. Assume Hypothesis 3 with p = 2, and assume that H is isomorphic to the commutator subgroup of a group of Lie type in characteristic p, with Lie rank greater than 1. Then p = 2, and HA ≅ SL(3, q) or Sp(4, q), with q = 2^{n}. Moreover, there exists an element s of S which interchanges the two maximal parabolic subgroups of HA containing S ∩ HA, and one of the the following holds.

(i) H ≅ SL(3, q) and V is the direct sum of H-submodules U and U^{s}, where U is a natural SL(3, q)-module for H. If A ≰ H then q = 2, |A| = 4, [V, A, A] ≠ 0, and |A|^{2} = |V/C_{V} (A)|.

(ii) HA ≅ Sp(4, q) and V is the direct sum of HA-submodules U and U^{s}, where U is a natural Sp(4, 2^{n})-module for H. Further, we have |A|^{3/2} < |V/C_{V} (A)|, and if [V, A, A] = 0 then |A|^{2} = |V/C_{V} (A)| and A is conjugate in G to Z(S ∩ HA).

Proof. By 5.9 we have p = 2 and H/Z(H) ≅ L_{3}(q) or Sp(4, q) (q = 2^{n}), or Alt(6). If H is not isomorphic to Alt(6) set T = S ∩ H, and if H ≅ Alt(6) set T = (S ∩ H)C_{S}(S ∩ H). Thus, either T = S ∩ H or HT ≅ Sp(4, 2). Let P_{1} and P_{2} be the two maximal subgroups of HT containing T. We use the following notation:

H_{i} = O^{2′}(P_{i}), Z_{i} = C_{S}(H_{i}), Q_{i} = O_{2}(H_{i}), and V_{i} = ⟨C_{V} (T)^{H_{i}}⟩.

For the sake of easy reference, we record also the following basic facts.

(1) Either Z_{i} ≤ H_{i} or HT ≅ Sp(4, 2).

(2) H_{i}/Q_{i} ≅ SL(2, q), and Q_{i}/Z_{i} is a natural SL(2, q)-module for H_{i}/Q_{i}.

We list a few well known (and easily checked) facts about the Sylow 2-subgroups of SL(3, q), Sp(4, q), and Alt(6), as follows.

(3) We have T = Q_{1}Q_{2}, Z(T) = Q_{1} ∩ Q_{2}, and every elementary abelian subgroup of T is contained in Q_{1} or in Q_{2}.

As S is contained in a unique maximal subgroup of G, we have the following result.

(4) There exists s ∈ S with (H_{1})^{s} = H_{2}, and with (H_{2})^{s} = H_{1}. In particular, the number c of non-trivial irreducible constituents for H_{i} in V_{i} is independent of i.

Let X be a non-central chief factor for some H_{i} in V, and let B be a non-identity subgroup of T. Then for any i, i = 1 or 2, B is conjugate in G to a subgroup B_{1} of T with B_{1} ≰ Q_{1}. We have |X| ≥ q^{2} and |X/C_{X}(B_{1})| ≥ q, by 6.8. Thus:

(5) For any non-identity subgroup B of T, we have |V/C_{V} (B)| ≥ q^{c}.

Notice that [V_{1} + V_{2}, Z(T)] = 0. Then V ≠ V_{1} + V_{2}, and so [V, O^{2}(H_{i})] ≰ V_{i}. Since C_{V} (H) = 0, we also have [V_{i}, O^{2}(H_{i})] ≠ 0. Then (5) yields the following result.

(6) c ≥ 2, and |V/C_{V} (B)| ≥ q^{2} for any non-identity subgroup B of T.

We now choose an elementary abelian subgroup E of S in the following way. If there exists a non-identity elementary abelian subgroup F of T, with |F |^{2} ≥ |V/C_{V} (F)|, then we take E ≤ T so that |E|^{2}|C_{V} (E)| is as large as possible and, subject to these conditions, so that |E| is as large as possible. Otherwise, if no such subgroup F of T exists, we take E so that |E|^{2}|C_{V} (E)| and then |E| are as large as possible. Then E is weakly closed in C_{G}(E), by 3.2. In particular, E is invariant under N_{T} (C_{S}(E)). As C_{S}(T) ≤ T it follows that E ∩T ≠ 1. Notice that our way of choosing E then yields the following information.

(7) If E ≰ T then

|V | > |F |^{2}|C_{V} (F)| < |E|^{2}|C_{V} (E)|

for any non-identity subgroup F of E ∩ T.

Let W be a non-trivial, irreducible G-invariant section of V and let U be an irreducible HT -submodule of W. Set F_{0} = End_{HT} (U), F = 𝔽_{q}, and note that F_{0} ≤ F, by 6.4(a). Denote by Γ_{0} the Galois group of F over F_{0}, and by Γ the full automorphism group of F. Write |F_{0}| = q_{0} = 2^{m}, and set r = |Γ_{0}|. Thus n = mr. Set Ũ = Y ⊗_{F_{0}} F. Let the modules M_{i} and N_{i}, 1 ≤ r, be given as in 6.6, and set d_{i} = dim_{F} (M_{i}) and d = d_{1} ⋯ d_{r}. Further, let the indexing be given so that d_{1} ≥ ⋯ ≥ d_{r}, and denote by c_{0} the number of irreducible constituents for HT in W.

Suppose first that HT = HE and that HT is not isomorphic to Alt(6). That is, suppose that HE is a group of Lie type. With E ≤ T, it follows from (3) that E ≤ Q_{i} for some i, (say i = 1), and with HT = HE we then have ⟨E^{H_{1}}⟩ = Q_{1}, and thus E = Q_{1}. Let s ∈ S be as in (4). Then EE^{s} = T and E ∩ E^{s} = Z(T). Then 2.2 implies that |T |^{2}|C_{V} (T)| ≥ |E|^{2}|C_{V} (E)|. In particular, we have |T |^{2} ≥ |V/C_{V} (T)| and |T |^{2} ≥ |W/C_{W} (T)|. As |C_{U}(T)| = q_{0}, by 6.4(b), we then have |W | ≤ (q_{0})^{c_{0}}q^{6} (resp. |W | ≤ q_{0}^{c_{0}}q^{8}) if H/Z(H) ≅ L_{3}(q) (resp. HT ≅ Sp(4, q)). By 6.6(c) we have d^{m} ≤ 6m + 1 (resp. 8m + 1). Here d_{1} = 3 or d_{1} ≥ 8 (resp. d_{1} = 4 or d_{1} ≥ 16) by 6.7, so in fact d_{1} = 3 (resp. d_{1} = 4), and if m > 1 then m = 2 and d_{i} = 1 for 1 < i ≤ r. Further, if m = 1 and r > 1 it follows that r = 2. Thus, we have proved the following result.

(8) Suppose that HT = HE ≇ Alt(6). Then HT/Z(HT) ≅ L_{3}(q) (resp Sp(4, q)) and one of the following holds.

(i) U is a natural module of dimension 3 (resp. 4) for HA. (ii) q = q_{0}^{2}, and Ũ ≅ M ⊗_{F} M^{ϕ} where M is a natural module for HT, and where

⟨ϕ⟩ = Γ_{0}.

With U as in (8i) or (8ii) one observes that C_{U}(Q_{1}) ≇ C_{U}(Q_{2}), so U is not invariant under s. Then W contains U ⊕ U^{s}, and we obtain

6m + 2 ≥ 2 · 3^{m} (resp. 8m + 2 ≥ 2 · 4^{m}).

It follows that m = 1 and that (8i) holds. Moreover, we have shown that any non-trivial, irreducible, HE⟨s⟩-invariant section of V is isomorphic to U ⊕ U^{s}, and from this, and from the preceding estimates, it follows that there is a unique such section. Then 1.2 yields V ≅ U ⊕ U^{s}.

Suppose now that HE ≅ Sp(4, q), and then suppose that A ≤ T. Then A ≤ Q_{i} for some i, and we set A∗ = ⟨A^{H_{i}}⟩. Thus A∗ = Z(H_{i}T), or O_{2}(H_{i}T), or O_{2}(H_{i}), and it is now a straightforward matter to check that |A ∗ |^{3/2}|C_{V} (A∗)| < |V |. Then also |A|^{3/2}|C_{V} (A)| < |V |, by 2.2. Assume that |A|^{3/2}|C_{V} (A)| ≥ |V |, and let A be minimal for this property. Then A ≰ T and we may take A so that |A_{0}|^{3/2}|C_{V} (A_{0})| < |A|^{3/2}|C_{V} (A)| for any proper subgroup A_{0} of A. Then N_{A}(U) = N_{A}(U^{s}) = A. Then any element a of A − T induces a non-identity field automorphism on L, by 7.4. Then [C_{V} (T), a] ≠ 0, and so A ∩ T = A, contrary to what has just been shown. Thus |A|^{3/2}|C_{V} (A)| < |V |.

Continuing to assume that HE ≅ Sp(4, q), suppose now that A ≤ HE and that [V, A, A] = 0. Inspection of the structure of U and of U^{s} allows us to fix the indices so that C_{U}(H_{1}) ≠ 0 and C_{U^{s}}(H_{1}) = 0. By (3), and possibly after conjugation by s, we may assume that A ≤ Q_{1}. Let R_{i}, 1 ≤ i ≤ 3, be root subgroups of Q_{1}, with respect to a standard set-up of HT as a group of Lie type, with Q_{1} = R_{1}R_{2}R_{3}, and taking Z(H_{1}) = R_{1} and Z(T) = R_{1}R_{2}. Set U_{1} = [U, H_{1}]. Then C_{U}(x) ≤ U_{1} for any non-identity element x of Q_{1}. Here both U_{1}/C_{U}(H_{1}) and Q_{1}/R_{1} are natural SL(2, q)- modules for H_{1}/Q_{1}, and the commutator map defines a pairing of U_{1}/C_{U}(H_{1}) × Q_{1}/R_{1} into C_{U}(H_{1}), defined over F. Suppose that AR_{1} ∩ R_{i} ≠ 1 for both i = 2 and 3. Then C_{U}(R_{i}) = C_{U}(ER_{1} ∩ R_{i}) for any i, and thus C_{U}(AR_{1}) = C_{U}(Q_{1}) = C_{U}(H_{1}) is of order q. But we also have R_{1} = C_{Q_{1}}(U/C_{U}(H_{1})) so [U, A] ≰ C_{U}(H_{1}), and thus A fails to act quadratically on U. We therefore conclude that AR_{1} ∩ R_{i} = 1 for some i. In particular, we have |AR_{1}| ≤ q^{2}.

If |C_{U}(A)| > q^{2} then A ≤ R_{1} and C_{V} (A) = C_{U}(R_{1}) + C_{U^{∗}}(R_{1}) is of index q^{3} in V. In that case we have |V/C_{V} (A)| > |A|^{2}, so we conclude that |C_{U}(A)| ≤ q^{2}. On the other hand, we have |C_{U^{∗}}(x)| = q^{2} for any non-identity element x of Q_{1}, so we have |V/C_{V} (A)| ≥ q^{4}, and hence |A| = q^{2}. It follows that R_{1}‘A. Now view Q_{1}/R_{1} as a 2-dimensional vector space over F, for the action of H_{1}/Q_{1}. This action is doubly transitive on 1-spaces, and the 1-spaces partition the space. As R_{i} ∩ A = 1 for some i, and since the same is true for any H_{1}-conjugate of A, it follows that A is conjugate in H_{1} to R_{1}R_{2}. Thus, (ii) holds if HE ≅ Sp(4, q).

Now assume that HE ≅ SL(3, q), and suppose that A ≰ H. Then 7.4 implies that |A| ≤ 2q, and so |V/C_{V} (A)| ≤ 4q^{2}. Suppose q > 2. Then |V/C_{V} (A)| ≤ q^{3}, while 7.4 shows that |C_{V} (a)| = q^{3} for any a ∈ A − H. Then |A| > 4, so A ∩ T ≠ 1, and C_{V} (a) ≤ C_{V} (A ∩ T). But for a ∈ A − H, either V = U + U^{a} or a induces a field automorphism on L (and on V), and one checks in either case that [C_{V} (a), x] ≠ 0 for any x ∈ T. Thus, we have a contradiction in this case, and so q = 2. Then A = Z(T)⟨a⟩, where a induces a graph automorphism on H, and (i) holds. We are thus reduced to the following cases.

(9) Either E ≰ T or HE ≅ Alt(6).

Suppose that HE ≅ Alt(6), and set T_{0} = S ∩ H. By (3) and 3.2 we have E = T_{0} ∩ Q_{i} for some i, and then |V/C_{V} (E)| ≤ 16. Suppose that U is a natural Sp(4, 2)-module for H. Then U ≠ U^{s}, and one checks that |(U + U^{s})/C_{U+U^{s})}(E)| ≥ 32. Thus |V/C_{V} (E)| > 16, and we conclude that U is not a natural Sp(4, 2)-module. Then 7.2 implies that dim(U) = 8 and that U is a direct summand of the Steinberg module for Sp(4, 2). It follows that E acts freely on U, and thus 64 = |U/C_{U}(A)| > |E|^{2}. This contradiction shows that HE ≇ Alt(6), and thus E ≰ T.

Set E_{1} = E ∩ T, and suppose next that E normalizes H_{1} and H_{2}. Then |E/E_{1}| = 2, and by 7.4 there is an element b ∈ E − E_{1} such that b induces a non-identity field automorphism on H. In particular, we then have q ≥ 4. By (3) we may choose indexing so that E_{1} ≤ Q_{1}, and then [Q_{1}, b] = C_{Q_{1}}(b). We then have E = ⟨b^{Q_{1}}⟩, by 3.2. If |V_{1}/C_{V_{1}}(b)| ≥ 4 then |V/C_{V} (E_{1})| ≤ |E_{1}|^{2}, in violation of (7), so in fact |V_{1}/C_{V_{1}}(b)| = 2. Then 6.8 implies that q = 4 and that V_{1}/C_{V_{1}}(H_{1}) is a natural O_{4}^{−}(2)-module for H_{1}⟨b⟩. Now |E| = 8 (resp. 16) if H/Z(H) ≅ L_{3}(q) (resp. (Sp(4, q))′), and so |V/C_{V} (E)| ≤ 64 (resp. 2^{8}). Then also |V/C_{V} (E_{1})| ≤ 2^{5} (resp 2^{7}). Set E_{2} = A ∩ Z(T). Then |E_{1}/E_{2}| = 2, and we observe that [V_{2}, E_{2}] = [V_{2}, Z(T)] = 0, while |V_{2}/C_{V_{2}}(E_{1})| ≥ 4. Thus |V/C_{V} (E_{2})| ≤ 8 (resp 32). If H/Z(H) ≅ L_{3}(4) we then contradict (6), so in fact we have H ≅ Sp(4, 4) and |E_{2}| = 4. Let x ∈ C_{H_{1}}(b) − T, and set E_{3} = (E_{2})^{x}. Then |E_{3}/(E_{3} ∩ Z(T))| = 2, and it follows, by a repetition of the preceding argument, that |V/C_{V} (E_{3} ∩ Z(T)| ≤ 8. Thus, we violate (6) in this case as well. We have shown:

(10) There exists a ∈ E such that (H_{1})^{a} = H_{2}.

Fix a as in (10). Set E_{0} = N_{E}(H_{1}) and, as above, set E_{1} = E ∩ T. Then |E/E_{0}| = 2, |E/E_{1}| ≤ 4, and E_{1} = E ∩Z(T). Moreover, by 7.4 we have |E/E_{1}| = 2 if HT ≅ Sp(4, q). Also, we have E_{1} ≤ C_{T} (a) = C_{Z(T)}(a), and so 3.2 implies that E_{1} = [C_{Z(T)}(E_{0}), a]. Notice that |V_{1}/(V_{1} ∩ V_{2})| = |V_{1}/C_{V} (T)| ≥ q, by 6.8. If |E/E_{1}|^{2} ≤ q then |V/C_{V} (E_{1})| ≤ |E_{1}|^{2}, which is contrary to (7). Thus |E/E_{1}|^{2} > q. Then q ≤ 4.

Suppose q = 4. Then |E_{1}| = 2 and |E| = 8. Moreover, we have |V/C_{V} (E)| ≤ 64, while |V/C_{V} (E_{1})| ≥ 16 by (6). It follows that |V_{1}/C_{V} (T)| = 4, and so V_{1}/C_{V_{1}}(H_{1}) is a natural SL(2, 4)-module for H_{1}/Q_{1}. Then an element of E_{0} − E_{1} induces a field automorphism on V_{1}/C_{V_{1}}(H_{1}), and so |V_{1}/C_{V_{1}}(E)| ≥ 8. But then |V/C_{V} (E_{1})| ≤ 8, and so we have a contradiction in this case. We therefore conclude that q = 2, and H ≅ Alt(6) or L_{3}(2).

Suppose that H ≅ Alt(6). Then |A| = 4, and C_{H}(a) is a dihedral group of order 10. Let X be the subgroup of C_{H}(a) of order 5, and let c be the involution in A∩H. Suppose that X has more than one non- trivial constituent in V. Then |[V, c]| ≥ 16, and since |V/C_{V} (A)| ≤ 16 we obtain |[V, c]| = 16 and C_{V} (c) = C_{V} (A). Then [V, A, A] = 0, and [V, X, a] = 0. But also C_{V} (X) ≤ C_{V} (c) ≤ C_{V} (a) in this case, and so a centralizes V. So, we conclude that X has a single non-trivial irreducible constituent in V. We again obtain [V, X, a] = 0. As two conjugates of X suffice to generate H, we have |C_{V} (X)| ≤ 16, and so |V | ≤ 2^{8}. Let D be a dihedral group of order 8 generated by two conjugates of a. Then Z(D) = ⟨c⟩, and since c is not a 2-transvection, by (7), it follows that a is not a transvection. Then |C_{V} (X)| = 16 and |V | = 2^{8}. In particular, we have |[V, A]| ≥ 16. Moreover, we now have |V/C_{V} (a)| = 4, and |V/C_{V} (A)| = 16. Now also C_{V} (A) = C_{V} (D) and, again since D is generated by two conjugates of a, [V, D] = [V, A] is of order 16. Then [V, A, A] = 0, and then also [V, D, D] = 0, which implies that D is elementary abelian. With this, we conclude that H is not isomorphic to Alt(6).

Suppose finally that H ≅ L_{3}(2). Let U be an irreducible H-submodule of V, and suppose first that U is a natural L_{3}(2)-module. Then U ≇ U^{a} as H-modules, and we find that |(U + U^{a})/C_{U+U^{a}}(A)| = 16. In view of 1.1, it follows that V = U + U^{a}, and so outcome (i) of the lemma holds in this case. On the other hand, suppose that U is of dimension 8. Then U is the Steinberg module for H, so |U : C_{U}(A∩H)| = 16. As |A| = 4 we then have U = [V, H], C_{U}(A) = C_{U}(A ∩ H), and A acts quadratically on U. Here A∩H = Z(T), so we have also [C_{U}(Q_{1})+C_{U}(Q_{2}), A] = 0. As a interchanges H_{1} and H_{2}, it follow that [C_{U}(Q_{1}), Q_{2}] = 0, and then [C_{U}(Q_{1}), H_{1}] = 0. Then [C_{U}(T), ⟨H_{1}, H_{2}⟩] = 0, contrary to C_{V} (H) = 0. With this contradiction, the proof of the 7.5 is complete. □

Section 8: Alternating groups, p odd

Theorem. Let G = Alt(n), n ≥ 5, and let V be an irreducible G-module over 𝔽_{3}. Suppose that there exists a 3-cycle a in G such that |V/C_{V} (a)| = 9 and such that 0 = [V, a, a, a] ≠ [V, a, a]. Then V is isomorphic to the unique non-trivial constituent in the natural permutation module for G over F.

Proof. For any m, denote by P (m) the permutation module for Alt(m) over 𝔽_{3}, with basis vectors x_{1}, ⋯ x_{m} permuted in the natural way by Alt(m). Denote by P_{0}(m) the codimension-1 submodule of P (m) consisting of the vectors a_{1}x_{1} + ⋯ + a_{m}x_{m} for which a_{1} + ⋯ + a_{m} = 0. Set F (m) = C_{P (m)}(Alt(m)), and set \overline{P} (m) = P (m)/F (m). Take a to be the 3-cycle (1 2 3), and let H be the subgroup of G fixing the point 1 in the natural permutation representation for G. Similarly, let K be the subgroup of G fixing both 1 and 2, and let L be the subgroup fixing each of 1, 2, and3. Thus [a, L] = 1.

(1) We have dim F (m) = 1. Further, P (m) = P_{0}(m) ⊕ F (m) if n is not divisible by 3, and otherwise F (m) ≤ P_{0}(m).

The irreducibility of \overline{P}_{0}(m) will be assumed for m ≤ n. We shall see that the irreducibility of \overline{P}_{0}(n) will follow by induction.

(2) Let ℓ be the least integer such that 3 + 2ℓ ≥ n. Then ℓ + 1 conjugates of a generate G.

We will need the following result.

(3) Let W be an 𝔽_{3}G-submodule of V, such that P_{0}(n) is a submodule of W, and such that with dim(W/P_{0}(n)) = 1. Suppose that there exists a 3- cycle a in G such that |V/C_{V} (a)| = 9 and such that 1 = [V, a, a, a] ≠ [V, a, a]. Then C_{W} (G) ≠ 0.

The proof of (3) is as follows. If 3 divides n then P_{0}(n) ≥ F (n) and there is nothing to prove. So assume that 3 does not divide n. Suppose first that 3 + 2ℓ = n. Then dim(W/C_{W} (G)) ≤ 2ℓ + 2 = n − 1, by (2). As dim(P_{0}(n)) = n − 1 we then have W = P_{0}(n) ⊕ C_{W} (G), and C_{W} (G) has dimension 1. Suppose finally that 3 + 2ℓ = n + 1. Then 3 + 2(ℓ − 1) = n − 1 and so H is generated by ℓ 3-cycles. Then dim(W/C_{W} (H)) ≤ 2ℓ = n − 2. As dim(P_{0}(n)/C_{P_{0}(n)}(H)) = n − 2 we conclude that C_{W} (H) ≰ P_{0}(n). Let x ∈ C_{W} (H) − P_{0}(n). Assuming that [x, G] ≠ 0 we obtain |x^{G}| = n and so ⟨x^{G}⟩ is a homomorphic image of P (n). Then W ≅ P (n), and thus C_{W} (G) ≠ 0, proving (3).

Suppose now that 8.1 is false, and let n be minimal for this property. Suppose C_{V} (H) ≠ 0 and let 0 ≠ x ∈ C_{V} (H). Then |x^{G}| = n and V = ⟨x^{G}⟩, and so V is a homomorphic image of P (n). As V is irreducible we are done in this case. We may therefore assume henceforth that C_{V} (H) = 0.

Suppose that 3 divides n. As C_{V} (H) = 0 it follows from (3), and induction on n, that V ≅ P_{0}(n − 1) as a module for H. Then also V ≅ P_{0}(K) ⊕ C_{V} (K) as a module for K, where dim C_{V} (K) = 1. Further, we then have V ≅ P (n − 3) + C_{V} (L), where C_{V} (L) is of dimension 2. Here [V, L] ≅ P_{0}(n − 3), and [V, L, a] = 0. As C_{V} (L) is a-invariant, we then have C_{V} (L) = [V, a].

(4) If 3 divides n then [V, L, a] = 0, C_{V} (L) = [V, a], and C_{[V,L]}(L) = [V, a, a].

We continue to suppose that 3 divides n. Identify G with the image of G in GL(V), and let t be the element of order 2 in GL(V) given by C_{V} (t) = [V, K] and [V, t] = C_{V} (K). Then [N_{G}(K), t] = 1, where N_{G}(K) ≅ Sym(n − 2).

Set G^{∗} = ⟨t, G⟩. We claim that G^{∗} ≅ Sym(n). To prove this, it now suffices to show that a^{t} = a^{−1}, by the standard presentation of Sym(n) as a Coxeter group. This calculation can be made as follows. Take b = (3 4 5), set J = ⟨a, b⟩, and set U = [V, J]. Then J ≅ Alt(5) and U ≅ P_{0}(5) as a module for J. Identify J with Alt(5), let {x_{i} | 1 ≤ i ≤ 5} be the standard basis of P (5), and identify U with P_{0}(5). As b ∈ K, t centralizes [U, b]. Also, we have [V, a, a] = C_{[V,L]}(L) ≤ [V, K] ≤ C_{V} (t), and so C_{U}(t) ≥ [U, b] + [U, a, a] = ⟨x_{3} − x_{4}, x_{4} − x_{5}, x_{1} + x_{2} + x_{3}⟩. Now observe that [U, a] = C_{V} (L), by (4). As K = ⟨b, L⟩ we then have C_{[U,a]}(b) = C_{V} (K), and thus C_{V} (K) = ⟨x_{1} − x_{2}⟩. As t acts as −1 on C_{V} (K), the action of t on U has now been completely determined, and one may compute directly that t inverts a. In detail: on the 3-dimensional subspace of U with ordered basis (x_{3} + x_{4} + x_{5}, x_{2} − x_{1}, x_{1} + x_{2} + x_{3}) the matrices of a and of t are given respectively by

( \begin{matrix}1 & 1 & −1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} )  and  ( \begin{matrix}1 & 0 & 0 \\ 0 & −1 & 0 \\ 0 & 0 & 1 \end{matrix} ) ,

and one then checks that the matrix for t inverts the matrix for a. This shows:

(5) Suppose that 3 divides n, let t be the element of GL(V) defined above, and identify G with the image of G in GL(V). Set G^{∗} = ⟨t, G⟩. Then G^{∗} ≅ Sym(n).

We continue to suppose that 3 divides n, and we now form the semi-direct product V G^{∗}. The usefulness of (5) lies in its providing us with a means to produce complements to V in V G^{∗}, by means of generators and relations. (This would be a more difficult task if we were forced to work with the smaller group V G.)

The key to the proof of 8.1 is the following result.

(6) If 3 divides n then H^{1}(G, V) ≅ 𝔽_{3}.

The proof of (6) is as follows. Assume that 3 divides n, and denote by K^{∗} the pointwise stabilizer in G^{∗} of {1, 2}. Then K^{∗} is generated by conjugates of t, i.e. by transvections, and it follows that C_{V} (K) = C_{V} (K^{∗}). Using multiplicative notation in V G^{∗}, let w be a non-identity element of [V, t] = C_{V} (K^{∗}) and set t_{i} = tw^{i}, 0 ≤ i ≤ 3. Then [t_{i}, K^{∗}] = 1. Identify t with the transposition (1 2) in G^{∗}, and set s = (2 3). Then each t_{i} is conjugate to t via an element of C_{V} (s), and so |t_{i}s| = 3 for all i. Setting G_{i}^{∗} = ⟨t_{i}, s, K^{∗}⟩, it now follows that G_{i}^{∗} ≅ Sym(n), and that G_{i}^{∗} is a complement to V in V G^{∗}.

Suppose next that there exists x ∈ V G and indices i and j such that (G_{i}^{∗})^{x} = G_{j}. We may take x ∈ V. We have G_{i}^{∗} ∩ (G_{i}^{∗})^{x} ≥ H, so that H^{x^{−1}} ≤ G_{i}^{∗} ∩ V H = H. Thus x ∈ N_{V} (H) = C_{V} (H) = 1 and so G_{i}^{∗} = G_{j}^{∗}. Thus there are at least three conjugacy classes of complements to V in V G^{∗}, and so |H^{1}(G^{∗}, V)| ≥ 3. Now let Ĝ be a complement to V in V G^{∗}. We have H^{1}(H, V) = 0, as follows from (3), so we may assume (after conjugation) that H ≤ Ĝ, and then Ĝ = ⟨tx, H⟩ for some x ∈ V. Here [tx, K^{∗}] = 1, so x ∈ C_{V} (K) = [V, t], and so Ĝ = G_{i} for some i. This shows that |H^{1}(G^{∗}, V)| = 3. But then also |H^{1}(G, V)| = 3, as the reader may verify. This completes the proof of (6).

We may now complete the proof of 8.1. Suppose first that n ≡ 2 (mod 3). As C_{V} (H) = 0 it follows from (3), and from induction, that V ≅ P_{0}(n − 1) as a module for H. In particular, we have dim V = n − 2. As 5 does not divide the order of SL(3, 3) we then have n > 5. Now V ≅ P (n−2) as a module for K, and then V ≅ P_{0}(n−3)⊕C_{V} (L) as a module for L. As dim P_{0}(n − 3) = n − 4 we then have dim C_{V} (L) = 2. As P_{0}(n − 3) is irreducible for L we then have [V, L, a] = 0 and a induces a transvection on V, contrary to hypothesis.

Suppose next that n ≡ 1 (mod 3). Then V has an H-submodule V_{0} with V_{0} ≅ \overline{P}_{0}(n − 1), and (6) implies that either V = V_{0} or V ≅ \overline{P} (n − 1) as a module for H. In particular, we have dim V = n − 3 or n − 2. As an L- module we then have V ≅ P_{0}(n − 3) ⊕ C_{V} (L), where dim C_{V} (L) ≤ 2. We therefore obtain a contradiction as in the preceding paragraph.

Suppose that n ≡ 0 (mod 3). Then V ≅ P_{0}(n − 1) as an H-module. By (6) V is a submodule of an indecomposable G-module W with dim W/V = 1, and by (3) we have C_{W} (H) ≠ 0. Then W is spanned by x^{G} for any non-zero x ∈ C_{W} (H), and W is then a homomorphic image of P (n). Thus, W ≅ \overline{P} (n) and V ≅ \overline{P}_{0}(n), as required. □

Lemma 8.2. Let H be the group 2^{·}Alt(9) and let V be a non-trivial, irreducible Hmodule over 𝔽_{3}, of dimension at most 8. Then either dim(V) = 8 and V is a spin module for H, or dim(V) = 7 and V is isomorphic to the non-trivial constituent in the natural permutation module for H.

Proof. Let x be an element of order 3 in H such that the image of x in H/Z(H) is a 3-cycle. If x acts quadratically on V then V is a spin module of dimension 8, by [M]. So we may assume that x does not act quadratically. Let J be a matrix in Jordan canonical form which represents x on V, let m be the number of 3 × 3 blocks of J, and let n be the number of 2 × 2 blocks. As x is not quadratic, we have m ≥ 1. Set L = E(C_{H}(x)). Then L ≅ SL(2, 9) or L_{2}(9), and so any non-trivial module for L over 𝔽_{3} is of dimension at least 4. Suppose that m ≥ 2 or that n ≥ 1. As dim(V) ≤ 8, it follows that L centralizes the chain [V, x] > [V, x, x] > 0, and the chain C_{V} (x) ≥ C_{V} (x) ∩ [V, x] ≥ C_{V} (x) ∩ [V, x, x] ≥ 0. But also, with m ≥ 2 or n ≥ 1 we have dim(V/([V, x] + C_{V} (x)) ≤ 3. As [V, L] ≠ 0, we conclude that m = 1 and n = 0. Then V is isomorphic to the non-trivial constituent in the natural permutation module, by 8.1. □

Theorem 8.3. Assume Hypothesis 4′, with p odd, and assume that G/Z(G) ≅ Alt(n), n ≥ 5. If p = 5 assume also that n ≠ 5, and if p = 3 assume n ≠ 6. Then p = 3, |A|^{2} = |V/C_{V} (A)|, and one of the following holds.

(i) |A| = 3, A acts non-quadratically on V, G ≅ Alt(n), and V is isomorphic to the unique non-trivial irreducible constituent in the natural permutation module for G over 𝔽_{3}.

(ii) G ≅ SL(2, 5) and V is isomorphic to the natural SL(2, 9)-module for H.

(iii) G ≅ 2^{·}Alt(9), |A| = 27, and V is a spin module for G, of dimension 8 over 𝔽_{3}.

Proof. If A acts quadratically on V then (ii) holds, by 5.6(iii). Thus, we may assume that A is not quadratic. If V is reducible, then 2.2 shows that G has more than one non-trivial constituent in V, and then, by induction, we may assume that a proper submodule U satisfies the conclusion of the lemma. But then V = C_{V} (A) + U and [V, G] = U, so we may in fact assume that V is irreducible.

Suppose first that |A| = p, and then suppose further that p ≥ 5. Then A is not quadratic on V, as follows from [Sa] or from [C3]. Let a ∈ A^{#}, let a_{1} be a p-cycle of a, and put b = a(a_{1})^{−1}. Then two conjugates of a generate a group L = K × ⟨b⟩, where K/Z(K) ≅ Alt(p). As |V/C_{V} (a)| ≤ p^{2} there is then a homomorphism of L into SL(4, p) which is non-trivial on K. If p ≥ 13 then K contains an elementary abelian subgroup of order 81, whereas SL(4, p) contains no such subgroup. Thus 5 ≤ p ≤ 11, and one observes in these three cases that K has a subgroup K_{0} with K_{0}/Z(K_{0}) ≅ L_{2}(p). As A acts non-quadratically, and as [V, A, A, A] = 0, it follows from 6.8 that K_{0} has a unique non-trivial constituent U in V, and that U is a natural O_{3}(p)-module for K_{0}. In particular, we have Z(K_{0}) = 1, and so K ≅ Alt(p) or 3^{·}Alt(7). As Alt(6) has no faithful complex representation of degree 4, we conclude that p = 5 or K ≅ 3^{·}Alt(7). It is easy to check for any element d of order 3 in X = SL(4, 7), that any component of C_{X}(d) is isomorphic SL(2, 7) or SL(3, 7), and then since |SL(3, 7)| is not divisible by 5, we conclude that p = 5 and K ≅ Alt(5). Here K = K_{0}, and with U as above we have H^{1}(K, U) = 0 by 1.4. Thus V = U ⊕ C_{V} (K). But |U/C_{U}(A)| = 25, so [C_{V} (K), A] = 0, and so A ≤ K. That is, a is a 5-cycle. Since n ≠ 5, by hypothesis, it follows that there is a subgroup J of H with J ≅ Alt(6) or 3^{·}Alt(6), and with J generated by two conjugates of A.

Set W = [V, J]. As |SL(3, 5)| is not divisible by 9 we have W ≠ U, and so |W | = 5^{4}. As |SL(4, 5)| is not divisible by 27 we have J ≅ Alt(6). There is a subgroup K_{1} of J with N_{J}(A) ≤ K_{1} ≅ Alt(5) and with ⟨K, K_{1}⟩ = J. Just as with K, we find that [V, K_{1}] is an O_{3}(5)-module for K_{1}, and W = [V, K_{1}] ⊕ C_{W} (K_{1}). As C_{W} (K) ≠ C_{W} (K_{1}) it follows that C_{W} (N_{J}(A)) is of dimension 2. Then C_{U}(N_{J}(A)) ≠ 0. But this is contrary to the case, as one may check that an involution in K inverting A acts non-trivially on C_{U}(A). We conclude that |A| = 3.

Fix an identification of G/Z(G) with Alt(n), and suppose that aZ(G) corresponds to a 3-cycle. Then two conjugates of a generate a subgroup X of G containing Z(G) and with X/Z(G) ≅ Alt(5). Then dim([V, X]) = 4, and it follows from 3.1 that X ≅ Alt(5) and Z(G) = 1. Then 8.1 yields (i). So assume that a does not correspond to a 3-cycle. Then n ≥ 7, and one may observe that two conjugates of a suffice to generate a subgroup of G whose order is divisible by 7. (Indeed, if a corresponds to a product of two disjoint 3-cycles then a is contained in a Frobenius subgroup of G of order 21, and if a corresponds to a product of three or more pairwise disjoint 3-cycles then one has only to observe that (1 2 3)(4 5 6)(7 8 9)(1 3 5)(2 6 9)(4 8 7) is a seven-cycle.) As 7 does not divide the order of SL(4, 3), we have a contradiction at this point. Thus |A| > p.

We continue to fix an identification of G/Z(G) with Alt(n). Fix a ∈ A^{#} so as to minimize the number k of p-cycles in writing the permutation aZ(G) as a product of pairwise disjoint p-cycles. Set C_{a} = O^{2}(C_{G}(a)). Then C_{a} = EK_{1} × K_{2}, where E is a normal elementary abelian p-subgroup of C_{a} of order p^{k}, K_{1} ≅ Alt(k), and K_{2}/Z(K_{2}) ≅ Alt(n − pk). The minimality of k implies that E ∩ A = ⟨a⟩, and so p divides |C_{a}/E|.

Suppose that A acts non-trivially on some K ∈ {K_{1}E/E, K_{2}}, and if possible choose K so that K = K_{1}E/E. (Notice that A acts non-trivially on some such K if p ≥ 5.) As [V, A, A, A] = 0 we have [V, A, A] ≤ C_{V} (a), and so A acts quadratically on V/C_{V} (a).

Also, A acts quadratically on C_{V} (a), by Hypothesis 4′. There is then a non-trivial irreducible constituent U for K in V, admitting A, and on which A acts quadratically. As K/Z(K) is an alternating group it follows from [C3] that p = 3, or that p = 5 and K ≅ SL(2, 5). Suppose p = 5. Then K may be identified with a component of C_{a}, and we have 1 ≠ Z(K) ≤ Z(H) Then every non-trivial constituent for K_{0} in V is a natural SL(2, 5)-module. A Sylow 5-subgroup B of K acts quadratically on every irreducible constituent for K in V, and it then follows from 1.6 that B acts quadratically on V. This is contrary to what has aleady been shown, so we conclude that p = 3. Then [M] yields K ≅ 2^{·}Alt(m) for some m, m ≥ 4, so K = K_{2}, Z(K) = Z(G) ≠ 1, and either |A/C_{A}(K)| = 3 or m = 6. As V is irreducible for G, Z(K) acts as −1 on V, and so every irreducible constituent for K in V is non-trivial.

Set W = C_{V} (C_{A}(K)). Suppose that |A/C_{A}(K)|^{2} ≤ |W/C_{W} (A)|. Then 3.3 implies that C_{A}(K) is an F 2-offender on V. Induction on |A| then yields |C_{A}(K)| = 3 and that (i) holds with C_{A}(K) in place of A. This is contrary to 1 ≠ Z(H), so we conclude that, in fact, we have |A/C_{A}(K)|^{2} > |W/C_{W} (A)|. Then W is irreducible, m = 4 or 6, and W is a natural SL(2, 3)-module (if m = 4) or a natural SL(2, 9)-module (if m = 6). Moreover, if m = 6 then |A/C_{A}(K)| = 9. In either of our two remaining cases, the choice of K implies that k ≤ 3, and so |C_{A}(K)| ≤ 9.

Suppose m = 6. Then |A| ≤ 81 and C_{V} (A) = C_{W} (A) is of order 9, so |V | ≤ 3^{1}0. Since all irreducible constituents for K in V are natural SL(2, 9)-modules we then have |V | = 3^{8} and there are just two such constituents. Then C_{a} ≅ ⟨a⟩ × K, so k = 1 and n = 9, and |C_{V} (a)| = 3^{4}. Here [V, a] = C_{V} (a), so a acts quadratically on V. Now [M] implies (iii).

Suppose m = 4. If k = 3 then n = 13, and the minimality of k implies that A∩K = 1, and so |A| = 9. If k < 3 then A ∩ EK_{1} = ⟨a⟩, and so |A| = 9 in any case. As C_{V} (A) = C_{W} (A) is of order 3, we then have |V | ≤ 3^{5}. Then |V | = 3^{4} (and V/W is a natural SL(2, 3)-module for K). As 7 doesn’t divide the ordr of SL(4, 3) we then have n ≤ 6, and with |A| = 9 we get n = 6, contrary to hypothesis.

We have now reduced to the case where no K as above exists. Thus p = 3, and A acts trivially on C_{a}/E, whence also [K_{2}, A] = 1. Suppose k > 3. Then C_{C_{a}}(EK_{1}/E) = EK_{2}, so A ≤ EK_{2}, and then n − 3k = 3. Then |A| = 9 and, by the minimality of k, any b ∈ A − ⟨a⟩ is a product of at least k disjoint 3-cycles, k − 1 of which represent orbits of ⟨a⟩. But with k > 3 we can then choose b so that two at least of the 3-cycles in b are 3- cycles of a^{−1}, and then ab has fewer than k 3-cycles. With this contradiction we conclude that k ≤ 3.

Suppose k = 1. Then C_{a}/Z(H) ≅ 3 × Alt(n − 3), and since |A| > 3 and [A, K_{2}] = 1 we have n = 6, contrary to hypothesis.

Suppose that k = 2. Then C_{a} ≅ 3^{2} × Alt(n − 6), and EK_{1} = E. As A ∩ E = ⟨a⟩ and [A, K_{2}] = 1, we then have n = 9 and |A| = 9. Here we may identify Z(H)A/Z(H) with

⟨(1 2 3)(4 5 6), (4 5 6)(7 8 9)⟩.

There is then a unique element c of A such that c can be represented as aproduct of three disjoint 3-cycles. By [M], c is not quadratic on V, and so c acts non-trivially on V/C_{V} (c). Here ⟨c⟩ = Z(S) for some Sylow 3-subgroup S of H, and so S acts faithfully on V/C_{V} (c). Then |V/C_{V} (c)| ≥ 3^{4}, and we conclude that C_{V} (c) = C_{V} (A) has index 3^{4} in V. Now observe that since a is represented as a product of two disjoint 3-cycles, there exists a subgroup J of H with a ∈ J ≅ Alt(4), and such that every involution t in J is represented as a product of four disjoint transpositions. Suppose C_{V} (a) ≠ C_{V} (A). Then |V/C_{V} (a)| ≤ 27, from which it follows that J has a unique non-trivial constituent in V. That is, we have |[V, O_{2}(J)]| = 27, and |V/C_{V} (t)| = 9. On the other hand, t inverts an element of order 7 in H, whereas 7 does not divide |GL(4, 3)|. This shows that, in fact, we have C_{V} (a) = C_{V} (A), and thus C_{V} (A) = C_{V} (d) for all d ∈ A^{#}.

Set X = N_{H}(EA). Here EA = J(S) for a Sylow 3-subgroup S of H (so EA is elementary abelian of order 27), and we have X/EA ≅ Sym(4), with X acting irreducibly on EA. Now

X ≤ ⟨g ∈ H : A ∩ A^{g} ≠ 1⟩

and so C_{V} (A) is invariant under X. Then C_{V} (A) = C_{V} (EA). Let x be an element of EA such that x is represented by a 3-cycle, and let L be the component in C_{H}(x) (L/Z(H) ≅ Alt(6)). No 3-element of L induces a transvection on C_{V} (x), by [M] so we have C_{V} (x) = C_{V} (EA), and indeed we have shown that C_{V} (EA) = C_{V} (y) for all y ∈ (EA)^{#}. Then C_{V} (EA) is invariant under the group

X^{∗} = ⟨g ∈ H : EA ∩ (EA)^{g} ≠ 1⟩.

We now ask the reader to perform the fairly straightforward exercise of showing that X^{∗} = H. With this we have H acting on the 4-space V/C_{V} (EA), and an evident contradiction.

Finally, suppose that k = 3. Then EK_{1} ≅ ℤ_{3} ≀ ℤ_{3}, and the minimality of k implies that |A ∩ EK_{1}| ≤ 9. Suppose that |A| > 9. Then A ≰ EK_{1}, and since [A, K_{2}] = 1 we conclude that |K_{2}| = 3 and n = 12. Let S be a Sylow 3-subgroup of H containing A. Then S ≅ (3 ≀ 3) × 3, and the minimality of k = 3 implies that A ≰ J(S). Then A = C_{S}(A) ≥ Z(S), so A contains a 3- cycle, and so k = 1. We conclude from this contradiction that |A| = 9.

For any b ∈ A such that b is represented as a product of three disjoint 3-cycles, we have |V/C_{V} (b)| ≥ 3^{4}, by the same reasoning as in the treatment of the case k = 2. Then, for such an element b, we have C_{V} (b) = C_{V} (A), of index 3^{4} in V. Suppose next that A ≤ EK_{1}. In this case we may identify Z(H)A/Z(H) with the group

(1)  ⟨(1 2 3)(4 5 6)(7 8 9), (1 4 7)(2 5 8)(3 6 9)⟩.

Here every non-identity element of A has a representation as a product of three disjoint 3-cycles. We may assume that n = 9, by restricting to an appropriate subgroup of H. Set X = ⟨C_{H}(b) : b ∈ A^{#}⟩. Then C_{V} (A) is X-invariant. On the other hand, Z(H)X/Z(H) contains all the 3-cycles in the two generators given in (1), and one may check that these six 3-cycles generate Alt(9). Then C_{V} (A) = C_{V} (H), and indeed [V, H] = 0. We conclude that A ≰ EK_{1}.

Now n = 12 (as [A, K_{2}] = 1) and there are, up to conjugation, two possibilities for Z(H)A/Z(H). These are as follows:

(2)  ⟨(1 2 3)(4 5 6)(7 8 9), (4 6 5)(7 8 9)(10 11 12)⟩,  or

(3)  ⟨(1 2 3)(4 5 6)(7 8 9), (1 4 7)(2 5 8)(3 6 9)(10 11 12)⟩.

Suppose that A is as in (2). Then every element of A^{#} is represented as a product of three disjoint 3-cycles. Now set X = ⟨C_{H}(b) : b ∈ A^{#}⟩, as before. Then C_{V} (A) is X-invarian, and X contains a subgroup X_{0} ≅ 3^{4} : Alt(4), and O_{3}(X_{0}) = ⟨A^{X_{0}}⟩. Then C_{V} (A) = C_{V} (O_{3}(X_{0})), and so O_{3}(X_{0}) is an F 1- offender on V. By Thompson Replacement there then exists a quadratic F 1- offender B on V. This is contrary to [M], and we thereby conclude that A is as in (3). Take Z(H)a = (1 2 3)(4 5 6)(7 8 9) and let b ∈ A with Z(H)b = (1 4 7)(2 5 8)(3 6 9)(10 11 12). Then ⟨b⟩ is contained in a subgroup J of H with J/Z(J) ≅ Alt(4). If C_{V} (b) ≠ C_{V} (A) then, as in the case k = 2, we find that an involution t in J satisfies |[V, t] = 9 and that t inverts an element of H of order 7, contrary to 7 not being a divisor of |GL(4, 3)|. Thus C_{V} (b) = C_{V} (A). Set Y = O^{2}(C_{H}(b)). Then Y ≅ 3^{4} : Alt(4), and O_{3}(Y)Z(H)/Z(H) is generated by the 3-cycles of Z(H)b. Then a ∉ O_{3}(Y), and since we now have C_{V} (b) = C_{V} (a) we conclude that [C_{V} (b), Y] = 0. Then O_{3}(Y) is an F 1-offender on V, and again, Thompson Replacement yields a quadratic F 1-offender and a contradiction to [M]. This completes the proof of the theorem. □

Section 9: Alternating groups, p = 2

This section contains a single result, which is as follows.

9.1 Proposition. Assume Hypothesis 4′, with p = 2. Suppose that F^{∗}(G) = H ≅ Alt(n), n odd, n ≥ 7. Then there exists a ∈ A^{#} and an involution t ∈ G − ⟨a⟩ such that [V, a, t] = 0.

Proof. Fix A as in Hypothesis 4′, and assume that 9.1 is false. The following is then obvious.

(1) For any involution a ∈ A and for any subgroup D of C_{G}(a) of even order with a ∉ D, we have [V, a, D] ≠ 0.

Suppose first that V is reducible for G. Then 1.2 implies that G has at least two non-trivial constituents in V. As |A|^{3/2} ≥ |V/C_{V} (A)|, by Hypothesis 4′, there then exists a non-trivial irreducible constituent W for G in V such that |A|^{3/4} ≥ |W/C_{W} (A)|.

By Timmesfeld Replacement, there is then a quadratic subgroup B of A with |B|^{3/4} ≥ |W/C_{W} (B)|, and then 9.4 yields n even, contrary to hypothesis. Thus:

(2) V is irreducible for G.

Let a ∈ A^{#}. As n ≥ 7 it follows from (1) and from 5.5 that |V/C_{V} (a)| ≥ 8. Suppose that we have |V/C_{V} (a)| = 8. Then |[V, a]| = 8, and since L_{3}(2) has 2-rank equal to 2 it follows from (1) that the 2-rank of C_{G}(a) is at most 3. Then n = 7, and we may record this result as follows.

(3) We have |V/C_{V} (a)| ≥ 8 for any involution a in A, with equality only if n = 7.

In particular, (3) implies that |A| > 2. As A contains no quadratic fours group, we have C_{V} (a) ≠ C_{V} (A) for a ∈ A^{#}. If |A| = 4 then |V/C_{V} (A)| ≤ 8 and |V/C_{V} (a)| ≤ 4, contrary to (3). Suppose that |A| = 8. Then |V/C_{V} (A)| ≤ 16 and |V/C_{V} (a)| ≤ 8. Then (3) yields n = 7, and since the 2-rank of Alt(7) is just 2 we obtain G ≅ Sym(7). Then a may be chosen to be a transposition, whence C_{G}(a) contains a subgroup X isomorphic to Alt(5), and we have [V, a, X] = 0. This is contrary to (1), so:

(4) We have |A| ≥ 16.

Let Ω = {1, 2, ⋯, n} be the set supporting the natural permutation representation of G. We now fix a ∈ A^{#} so that the number of orbits of length 2 for ⟨a⟩ on Ω is as small as possible. Denote this number by k, and identify a with the permutation (1 2) ⋯ (2k − 1 2k). Set m = n − 2k.

Set C_{a} = O^{2}(C_{G}(a)). We then have

C_{a} ≅ 2^{k−1} : Alt(k) × Alt(m).

Denote by K_{1}^{∗} the pointwise stabilizer in C_{G}(a) of {2k + 1, ⋯, n}, and by K_{2}^{∗} the pointwise stabilizer in C_{G}(a) of {1, 2, ⋯, 2k}. Denote by E^{∗} the subgroup of C_{G}(a) generated by the set of pairwise disjoint 2-cycles whose product is a. Set K_{i} = K_{i}^{∗} ∩ C_{a}, and E = E^{∗} ∩ C_{a}. Then E is elementary abelian of order 2^{k−1}, K_{1}/E ≅ Alt(k), E/⟨a⟩ is a natural Sym(k)-module for K_{1}/E, and K_{2} ≅ Alt(m). The minimality of k yields:

(5) A ∩ E^{∗} = ⟨a⟩.

Set L = O^{2}(⟨A^{C_{G}(a)}⟩. Suppose that L is solvable, and then suppose that K_{1} is nonsolvable. Then [K_{1}, A] ≤ E, and so A ≤ E^{∗}K_{2}^{∗}. Then (4) and (5) imply that the 2-rank of K_{2}^{∗} is at least 3, so K_{2} is non-solvable, K_{2} ≤ L, and L is non-solvable, contrary to assumption. On the other hand, suppose that K_{2} is non-solvable. Then A ≤ K_{1}^{∗}, and then (4) and (5) imply that K_{1} is non-solvable and K_{1} ≤ L. Thus, under the assumption that L is solvable we conclude that C_{a} is solvable. That is, we have k ≤ 4 and, as n is odd, m ≤ 3. Suppose that k = 3 or 4. No involution in K_{2}^{∗} is the product of 3 or 4 pairwise disjoint transpositions, so the minimality of k yields A ∩ K_{2}^{∗} = 1. Let b be an involution in E^{∗}K_{2}^{∗}. As m ≤ 3, either b or ab is a product of fewer than k disjoint transpositions, and so A ∩ E^{∗}K_{2}^{∗} = ⟨a⟩. Then C_{A}(K_{1}/E) = ⟨a⟩, and |A| ≤ 8, contrary to (4). We conclude that k ≤ 2. Then n ≤ 7, again contrary to (4). This proves:

(6) L is non-solvable.

Set L_{i} = L ∩ K_{i}, i = 1, 2. Then L = L_{1} × L_{2}. Set X = C_{V} (a) and Y = [V, a]. By Hypothesis 4′, A acts quadratically on X, and hence also on Y.

Set A_{0} = C_{A}(L/O_{2}(L)). Then ⟨a⟩ ≤ A_{0} = A∩O_{2}(LA). Suppose that A_{0} ≠ ⟨a⟩. Then O_{2}(LA) ≠ E^{∗}, and it follows that either L_{1} is solvable (which includes the possibility that L_{1} = 1) or that L_{2} = 1. Suppose that L_{2} = 1. Then A ≤ K_{1}^{∗}, and L_{1} is non-solvable, by (6), and then A_{0} = A ∩ E^{∗} = ⟨a⟩. So in fact L_{2} ≠ 1, and L_{1} is solvable. Then A_{0} ≤ K_{1}^{∗}. If k = 1 we have K_{1}^{∗} = ⟨a⟩, while if k = 3 we have C_{K_{1}^{∗}}(K_{1}) = ⟨a⟩. So k = 2 or 4. Suppose k = 2. Then L_{1} = 1, K_{1}^{∗} is dihedral of order 8, and |A_{0}| ≤ 4. On the other hand, suppose that k = 4. If L_{1} ≠ 1 then A ≰ O_{2}(K_{1}^{∗})K_{2}^{∗}, and |C_{O_{2}(K_{1}^{∗})}(A)E^{∗}/E^{∗}| = 2. Therefore |A_{0}| ≤ 4 in this case. If instead we have L_{1} = 1 then |A_{0}| ≤ 8 since the 2-rank of O_{2}(K_{1}^{∗})/E^{∗} is 2. Thus:

(7) Suppose A_{0} ≠ ⟨a⟩. Then A_{0} ≤ O_{2}(K_{1}^{∗}), k = 2 or 4, and |A_{0}| ≤ 8. Moreover, we have |A_{0}| = 4 unless k = 4 and L_{1} = 1.

Fix a complement B to A_{0} in A. Set B_{1} = C_{B}(K_{2}) and let B_{2} be a complement to B_{1} in B. Set Y_{i} = [Y, L_{i}], and set U_{i} = Y_{i}/C_{Y_{i}}(L_{i}), (i = 1, 2). We will obtain the following result.

(8) Suppose that C_{B}(L_{2}) ≠ 1. Then [Y_{1}, L_{2}] = [Y_{2}, L_{1}] = 0, and

|Y/C_{Y} (B)| ≥ |U_{1}/C_{U_{1}}(B_{1})||U_{2}/C_{U_{2}}(B_{2})|.

Indeed, set D = C_{B}(L_{2}). If L_{2} = 1 then B_{1} = B and (8) follows trivially. So we may assume that L_{2} ≠ 1. As L_{2} = [A, L_{2}] = [B, L_{2}], we have

0 = [Y, B, D] = [Y, ⟨B^{L_{2}}⟩, D] ≥ [Y, L_{2}, D] ≥ [Y, L_{2}, ⟨D^{L_{1}}⟩].

As D ≠ 1, by assumption, we have L_{1} ≤ ⟨D^{L_{1}}⟩, and thus [Y, L_{2}, L_{1}] = 0. The Three Subgroups Lemma yields also [Y, L_{1}, L_{2}] = 0, so it remains only to prove the second assertion in (8). Set U = U_{1} ⊕ U_{2}. Then U is a homomorphic image of the submodule Y_{1} + Y_{2} of Y, so

|Y/C_{Y} (B)| ≥ |U/C_{U}(B)| ≥ |U_{1}/C_{U_{1}}(B_{1})||U_{2}/C_{U_{2}}(B_{2})|,

and we have (8).

The next step is to show:

(9) |B| ≤ |Y/C_{Y} (B)|.

Suppose that (9) is false. If C_{B}(L_{2}) = 1 then, since m is odd, 5.8 yields |Y_{2}/C_{Y_{2}}(B)| ≥ |B|, and then (9) follows. Thus, we may assume that C_{B}(L_{2}) ≠ 1, and then |Y/C_{Y} (B)| ≥ |U_{1}/C_{U_{1}}(B_{1})||U_{2}/C_{U_{2}}(B_{2})|, by (8). Again, as m is odd, 5.8 implies that |U_{2}/C_{U_{2}}(B_{2})| ≥ |B_{2}|, and hence |U_{1}/C_{U_{1}}(B_{1})| < |B_{1}|. In particular, we have L_{1} ≠ 1, and L_{1} = K_{1}. Notice that there exists t ∈ C_{G}(a) such that K_{1}⟨t⟩/E ≅ Sym(k). It then follows from 5.8 that k is even and that if k > 4 then U_{1} is a natural Sym(k)-module for K_{1}/E and B_{1} is generated by elements which correspond to transpositions in Sym(k). That is, either k = 4 or B contains elements which are the product of two disjoint transpositions. The minimality of k then yields k = 4. Then |B_{1}| = 2, and so |U_{1}/C_{U_{1}}(B_{1})| ≥ |B_{1}|. This completes the proof of (9).

We have |A| = |A_{0}||B|, so

(*)  |A_{0}|^{3/2}|B|^{3/2} ≥ |V/C_{V} (A)|.

Notice that V/X ≅ Y as modules for C_{G}(a). Then

(**)  |V/C_{V} (A)| = |V/X||X/C_{X}(B)| ≥ |Y ||Y/C_{Y} (B)| = |Y/C_{Y} (B)|^{2}|C_{Y} (B)|.

As |Y/C_{Y} (B)| ≤ |B|, by (9), we then have

(10) |A_{0}|^{3} ≥ |B||C_{Y} (B)|^{2}, and if |Y/C_{Y} (B)| > |B| then |A_{0}|^{3} ≥ 16|B||C_{Y} (B)|^{2}.

Suppose that A_{0} = ⟨a⟩. Then (10) yields |B| = 2, |C_{Y} (B)| = 2, and |Y/C_{Y} (B)| = 2. But L/⟨a⟩ has a non- solvable direct factor which, by (1), acts faithfully on Y, and so we have a contradiction in this case. Now (6) and (7) yield:

(11) |A_{0}| > 2, k = 2 or 4, and m ≥ 5.

Suppose next that k = 4 and that |Y | ≤ 32. Then (1) implies that L_{5}(2) has a subgroup of the form 2^{4} : 3 × Alt(m). As m ≥ 5 we may choose x ∈ L_{2} of order 5, and find that [Y, x] is a hyperplane of Y, and that [Y, x, O_{2}(K_{1})] = 0. Here O_{2}(K_{1})/⟨a⟩ is elementary abelian of order 16, and contains its centralizer in L_{5}(2), so we have a contradiction. Thus:

(12) If k = 4 then |Y | ≥ 64.

Suppose that |Y/C_{Y} (B)| > |B|. If also |A_{0}| = 4 then (10) yields |B| = 1, which is contrary to (4). Thus |A_{0}| = 8, and (7) implies that k = 4, L_{1} = 1, and A_{0}E^{∗} = O_{2}(K_{1}^{∗}).

Now (10) yields |B| ≤ 8 so |A| ≤ 64 and |V/C_{V} (A)| ≤ 2^{9}. Note also that if |B| = 4 then |V/C_{V} (A)| ≤ 2^{7}, and if |B| = 2 then |V/C_{V} (A)| ≤ 2^{6}. But |V/C_{V} (A)| ≥ |Y ||Y/C_{Y} (B)|, by (**), and so |Y | ≤ 32, contrary to (12). Thus:

(13) We have |Y/C_{Y} (B)| = |B|.

Suppose that L_{1} = 1. Then L = L_{2}, and B acts faithfully on L_{2}. Set Y_{0} = U_{2}/C_{U_{2}}(L_{2}). Notice that there exists an involution t in C_{H}(a) such that L_{2}⟨t⟩ ≅ Sym(m). As m is odd it follows from (13) and 5.7 that either Y_{0} is a natural module for LB, or m = 5 and Y_{0} is a spin module for LB, of dimension 4. Suppose that m = 5 and Y_{0} is a spin module (i.e. a natural SL(2, 4)-module). Then |B| = 4, and since |A_{0}| ≤ 8, by (7), we get |A| ≤ 2^{5}. Moreover, if |A| = 2^{5} then |A ∩ L_{2}| = 4 and so k = 2. But with k = 2 we have |A_{0}| ≤ 4, so we conclude that, in fact, |A| ≤ 16, and thus |V/C_{V} (A)| ≤ 64. As C_{V} (A) ≠ C_{V} (a), by (1), it follows that |Y | ≤ 32. By (11), we may choose a_{0} ∈ A_{0} − ⟨a⟩, and then (1) implies that ⟨a_{0}⟩ × L_{2} acts faithfully on Y. But 2 × L_{2}(4) has no faithful action on a 5-dimensional space over 𝔽_{2}, so we conclude that Y_{0} is not a spin module.

Suppose that k = 2. Then L_{1} = 1, and the preceding paragraph applies. As m is odd,

(13) and 9.4 imply that B contains an element b which acts as a transvection on U_{2}. As k ≠ 1, b ∉ K_{2}^{∗}. Notice that either K_{1}^{∗} = A_{0} or K_{1}^{∗} is dihedral of order 8. In either case we have C_{G}(A_{0}) = A_{0}K_{2}^{∗}. Thus, there exists c ∈ A_{0} so that bc ∈ K_{2}^{∗}. But [U_{2}, c] = 0, so bc is a transvection on U_{2}, and then bc is a transposition in K_{2}^{∗}. This is contrary to the minimality of k, so we now conclude that k = 4. If also |A_{0}| = 4 then (10) yields |B| ≤ 16, |A| ≤ 64, and |V/C_{V} (A)| ≤ 2^{9}. Then also |Y | ≤ 32, and we contradict (12). Thus |A_{0}| = 8. As A_{0} ≤ O_{2}(K_{1}^{∗}), by (7), it follows that E^{∗}A_{0} = O_{2}(K_{1}^{∗}), and since [B, A_{0}] = 1 we get [L_{1}, B] ≤ E. Then L_{1} = 1. As in the case k = 2, there then exists b ∈ B such that b induces a transvection on U_{2}, and we have b = xt where x ∈ O_{2}(K_{1}^{∗}) and t is a transposition in K_{2}^{∗}. Then a_{0}x ∈ E^{∗} for some a_{0} ∈ A_{0}. Then either a_{0}x or aa_{0}x is the product of fewer than three pairwise disjoint transpositions. Replacing a_{0} by aa_{0} if necessary, it follows that a_{0}b = a_{0}xt is the product of fewer than four pairwise disjoint transpositions, and we again contradict the minimality of k. This completes the proof of 9.1. □

Section 10: Lie type groups in cross characteristic

10.1 Hypothesis. Hypothesis 4′ holds, and S is contained in a unique maximal subgroup of G. Further, H/Z(H) is a quasisimple group of Lie type, whose defining characteristic r is different from p. Indeed, it is assumed that there exists no isomorphism between H/Z(H) and a group of Lie type in characteristic p

Our aim in this section is to prove the following result.

10.2 Theorem. Assume Hypothesis 10.1. Then |A| = 3, and one of the following holds.

(a) G/Z(G) ≅ L_{2}(5).

(b) G ≅ Sp(6, 2), C_{G}(A) ≅ 3 × Sp(4, 2), and |V | = 3^{7}.

We remark that there is a natural embedding of Sp(6, 2) in Ω_{7}(3), which one obtains by identifying 2 × Sp(6, 2) with the Weyl group of E_{7}. Further, it is the case that in Sp(6, 2) a Sylow 3-subgroup is contained in a unique maximal subgroup (isomorphic to Aut(U_{4}(2)).

Whenever Hypothesis 10.1 is in effect we set \overline{G} = G/Z(H). We begin the proof of 10.2 by considering the case where \overline{H} ≅ Sp(6, 2).

10.3 Lemma. Assume Hypothesis 10.1, and assume that \overline{H} ≅ Sp(6, 2). Then p = 3 and one of the following holds.

(i) |A| = 3, G ≅ Sp(6, 2), C_{G}(A) ≅ 3 × Sp(4, 2), and |V | = 3^{7}.

(ii) |A| = 27, |Z(G)| = 2, and |V | = 3^{8}.

Proof. We have G = H and |Z(G)| ≤ 2. Let U be the natural 𝔽_{2}-module for \overline{G}. There are three conjugacy classes of subgroups of order 3 in G, which are distinguished by the dimensions of the commutators [U, X], X a representive of the class. We will say that a subgroup (or element) X of order 3 is in the class 3A (resp. 3B, resp. 3C) if dim([U, X]) = 1 (resp. 2, resp. 3). There is a subgroup M of G with \overline{M} ≅ L_{2}(8) : 3 ≅ ^{2}G_{2}(3), and M contains representatives of the classes 3C (in [M, M]) and 3B (in M − [M, M]).

Suppose first that |A| = 3, so that |V/C_{V} (A)| ≤ 9. By 5.6, A is not quadratic on V, and so |V/C_{V} (A)| = 9. Then A ≰ M, as follows from 6.9. Thus, A is of type 3A. Then there is a maximal subgroup K^{∗} of G containing A, with \overline{K}^{∗} ≅ Sym(8), and such that \overline{A} is generated by a 3-cycle in \overline{K}^{∗}. Let K be a subgroup of K^{∗} containing A, with \overline{K} ≅ Alt(7). Then 8.3 implies that K ≅ Alt(7) and that [V, K] is a natural module for K, of dimension 6 over 𝔽_{3}. In particular, we now have Z(G) = 1. Let K_{0} be a subgroup of K containing A, with K_{0} ≅ Alt(6), and let b be an element of of order 3 in C_{G}(K_{0}). By [C2, Theorem 4.2], [K^{∗}, K^{∗}] is the unique proper subgroup of G which properly contains K and which is generated by conjugates of A. Since ⟨b⟩ is conjugate to A we then have ⟨K, b⟩ = G. We may then assume that K^{b} ≰ K^{∗}, and so also G = ⟨K, K^{b}⟩. Here b centralizes [V, K_{0}], which has codimension 1 in [V, K], so we get V = [V, K] + [V, K^{b}] of dimension at most 7. But Sp(6, 2) has no faithful representation on degree 6 over 𝔽_{3}, by [SZ], so dim(V) = 7. Thus, (i) holds in this case.

We next consider the various conjugacy classes of subgroups of \overline{G} of order 9. Any such class is characterized by the conjugacy classes of its four cyclic subgroups, and in this way we find that there are exactly four such classes, which we list as follows, and where we list also a subgroup of \overline{G} containing a representative of the given class.

Y_{1} = (3A, 3A, 3B, 3B) ≤ Alt(7),

Y_{2} = (3C, 3B, 3B, 3B) ≤ L_{2}(8) : 3,

Y_{3} = (3A, 3B, 3C, 3C) ≤ Sym(3) × Sym(6),

Y_{4} = (3C, 3C, 3C, 3C) ≤ 3^{1+2} : GL(2, 3).

Suppose that |A| = 9, so that |V/C_{V} (A)| ≤ 3^{4}. Then A is not of type Y_{1}, by 8.3, and A is not of type Y_{2} by 6.9. Thus A is of type Y_{3} or Y_{4}, and so A contains a subgroup ⟨a⟩ of type 3C. Set K = C_{G}(a), and observe that \overline{K} ≅ 3^{1+2} : SL(2, 3). Theorem 1.3 of [C3] implies that any quadratic element of order 3 in G is of type 3A, so a is not quadratic. Set W = [V, a]. Then dim(W) ≤ 4, and [W, a] ≠ 0, so K acts faithfully on W. It follows that dim(W) = 4, and then C_{V} (a) = C_{V} (A). As ⟨A^{K}⟩ ≥ O_{3}(K) we then have C_{W} (a) = C_{W} (O_{3}(K)). Also, by Hypothesis 4′ we have [W, A, A] = 0, so [W, A, O_{3}(K)] = 0, and then O_{3}(K) acts quadratically on W, contrary to Φ(O_{3}(K)) ≠ 1. We therefore conclude that |A| ≠ 9.

Suppose finally that |A| = 27. Let Y be a subgroup of A of type Y_{2}. Then two conjugates of Y suffice to generate a maximal subgroup M of G with M ≅ L_{2}(8) : 3, and since A ≰ M it follows that two conjugates of A suffice to generate G. As dim(V/C_{V} (A)) ≤ 6 we then have dim(V) ≤ 12. Let A ≤ L ≤ G with \overline{L} ≅ Sym(3) × Sym(6), set L_{0} = E(L), set A_{0} = C_{A}(L_{0}), and set V_{0} = [V, A_{0}] + C_{V} (A_{0}). Then A ∩ L_{0} acts quadratically on V_{0}, by Hypothesis 4′, and so 1.7 implies that [V_{0}, L_{0}] is a direct sum of natural SL(2, 9)-modules for L_{0}. As V/C_{V} (A_{0}) ≅ [V, A_{0}] as L_{0}-modules, we have [V_{0}, L_{0}] ≠ 0, so we conclude that Z(L_{0}) = Z(G) ≠ 1. We may assume that V is irreducible for G, so V = [V, Z(G)], and since dim(V) ≤ 12, V_{0} is a direct sum of one or two natural modules for L_{0}. Then also V/V_{0} is a natural module for L_{0}. If A_{0} is not quadratic on V we then have dim(V) = 12 and |C_{V} (A)| = |C_{V_{0}}(A)| = 3^{4}, so that |V/C_{V} (A)| > |A|^{2}. Thus, A_{0} is quadratic on V, and dim(V) = 8. Thus, (ii) holds in this case.

□

The following lemma is given by Table I in [SZ].

10.4 Lemma. Let X be a simple group of Lie type, in characteristic different from p, and let V be a non-trivial projective 𝔽_{p}X-module. Assume that X is not isomorphic to any of the groups in the following list L_{0}.

L_{0} = {L_{2}(4), L_{2}(9), L_{3}(2), L_{3}(4), L_{4}(2) L_{4}(3), P Sp(4, 2), U_{4}(3),
P Ω_{8}^{+}(2), P Ω_{7}(3), F_{4}(2), G_{2}(3), G_{2}(4), Sz(8)}

Then the dimension of V is at least ℓ, where ℓ is given as follows.

(1) L_{2}(q): ℓ = (1/d)(q − 1), d = (2, q − 1).

(2) L_{n}(q), n ≥ 3: ℓ = (q^{n} − 1)/(q − 1) − n.

(3) P Sp(2n, q), n ≥ 2:

ℓ = \{ \begin{matrix}(q^{n} − 1)/2  if q is odd, and \\ q(q^{n} − 1)(q^{n−1} − 1)/2(q + 1)  if q is even. \end{matrix}

(4) U_{n}(q), n ≥ 3:

ℓ = \{ \begin{matrix}q(q^{n−1} − 1)/(q + 1)  if n is odd, and \\ (q^{n} − 1)/(q + 1)  if n is even. \end{matrix}

(5) P Ω_{2n+1}(q), q odd, n ≥ 3:

ℓ = \{ \begin{matrix}(q^{2n} − 1)/(q^{2} − 1) − n  if q ≠ 3, and \\ (3^{2n} − 1)/8 − (3^{n} − 1)/2  if q = 3. \end{matrix}

(6) P Ω_{2n}^{+}(q), n ≥ 4:

ℓ = \{ \begin{matrix}q(q^{2n−2} − 1)/(q^{2} − 1) + q^{n−1} − n  if q ≠ 2, 3, and \\ q(q^{2n−2} − 1)/(q^{2} − 1) − (q^{n−1} − 1)/(q − 1) − 7δ_{2,p}  if q is even. \end{matrix}

(7) P Ω_{2n}^{−}(q), n ≥ 4: ℓ = q(q^{2n−2} − 1)/(q^{2} − 1) − q^{n−1} − n + 2.

(8) E_{6}(q): ℓ = q^{9}(q^{2} − 1).

(9) E_{7}(q): ℓ = q^{15}(q^{2} − 1).

(10) E_{8}(q): ℓ = q^{27}(q^{2} − 1).

(11) F_{4}(q): ℓ = q^{6}(q^{2} − 1).

(12) ^{2}E_{6}(q): ℓ = q^{9}(q^{2} − 1).

(13) G_{2}(q): ℓ = q(q^{2} − 1).

(14) ^{3}D_{4}(q): ℓ = q^{3}(q^{2} − 1).

(15) ^{2}F_{4}(q): ℓ = (\sqrt{q/2})q^{4}(q − 1).

(16) Sz(q): ℓ = (\sqrt{q/2})(q − 1).

(17) ^{2}G_{2}(q): ℓ = q(q − 1). □ We insert the following lemma, as support for the case where H ≅ L_{2}(q).

10.5 Lemma. Set L = L_{2}(q), q odd, and identify L with the group Inn(L) of inner automorphisms of L. Let A be an elementary abelian 2-subgroup of Aut(L). Then one of the following holds.

(i) q is a perfect square, A ≰ L, and there exists an element t of A − L such that t is conjugate via Aut(L) to a field automorphism of L. Further, we have A = (A ∩ L)⟨t⟩ and |A| ≤ 8.

(ii) A ≰ L, and LA = L⟨d⟩, where d is a diagonal automorphism of L. Here |A/(A ∩ L)| = 2, |A| ≤ 4, and all involutions in LA − L are conjugate. If |A| = 4 then LA = ⟨A, A^{g}⟩ for some g ∈ L.

(iii) A ≤ L, |A| ≤ 4, and if |A| = 4 then LA = ⟨A, A^{g}⟩ for some g ∈ L.

Proof. Set G = LA, and let T be a Sylow 2-subgroup of G containing A. Also, set G_{1} = P GL(2, q), and set G^{∗} = G_{1}T. Further let T^{∗} be a Sylow 2-subgroup of G^{∗} containing T and set T_{1} = T^{∗} ∩ G_{1}.

We first note the following.

(1) T_{1} is dihedral, and G_{1} has exactly two classes of involutions.

Suppose that q = r^{2} is a perfect square, and let α be the standard field automorphism of L of order 2, induced by the automorphism of GL(2, q) which raises matrix entries to the power r (and which we denote also by α). Our first step will be to provide a proof of the following (well known) fact.

(2) We have C_{G_{1}}(α) = C_{L}(α) ≅ P GL(2, r).

The proof is as follows. Set G̃_{1} = GL(2, q), let Z be the group of scalar matrices in G̃_{1}, and set Z_{0} = {λ^{q−1}I ∥ λ ∈ 𝔽_{q}}. Denote by C̃_{1} the set of matrices x̃ ∈ G̃_{1} such that [x̃, α] ∈ Z. As Z ≤ Z(C̃_{1}), there is a homomorphism ϕ : C̃_{1} ⟶ Z given by ϕ(x̃) = [x̃, α]. Then the image of phi is Z_{0}, and the kernel of ϕ is CG̃_{1}(α) = GL(2, r). One observes that ϕ|_{Z} maps Z onto Z_{0}, so C̃_{1} = Ker(ϕ)Z. We then have

C_{G_{1}}(α) = Ker(ϕ)Z/Z ≅ Ker(ϕ)/(Ker(ϕ) ∩ Z) ≅ P GL(2, r).

On the other hand, let ω ∈ 𝔽_{q} − 𝔽_{r} so that ⟨ω⟩ = 𝔽_{r}^{×}, and let d be the linear fractional transformation x ↦ ω^{2}x. Then ω^{r} = −ω, and so [d, α] = 1. Also, d ∈ L, and d induces an outer automorphism on the subgroup P SL(2, r) of L. Thus C_{L}(α) contains a subgroup isomorphic to P GL(2, r), and this yields (2).

Next, suppose that there exists an element t ∈ A − G_{1}. As |t| = 2 we have g^{α} = g^{−1}, and then t is conjugate either to α or to zα, where z is an involution in C_{G_{1}}(α). Suppose that t is not conjugate to α. Then z ∈ L, by (2). We now apply (1) to C_{L}(t) in place of G_{1}, and find that, up to conjugation in L, there are two choices for z. One (in L_{2}(r)) is given by the linear fractional transformation z_{1} : x ↦ −1/x^{−1}, and the other (in C_{L}(α) − L_{2}(r)) is given by z_{2} : x ↦ −ω^{2}x, where ω is as above. To show that α and z_{i}α are conjugate, we suffices to display elements g_{i} of G_{1} such that g_{i}^{α} = g_{i}z_{i}. Let ρ ∈ 𝔽_{q} with ρ^{1+r} = −1. The transformations g_{1} : x ↦ (x + ρ^{r})/(ρx + ρ^{2r}) and g_{2} : x ↦ (x − ω)/(ωx + ω^{2}) perform this trick, as the careful reader may check. Thus (i) holds if A ≰ G_{1}.

Suppose next that A ≤ G_{1}, suppose that A ≰ L, and let t ∈ A − L. As T_{1} is dihedral, we have |C_{T} (t)| = 2, and thus |A| ≤ 4. Let s be the involution in C_{T} (t), and set D = C_{L}(s). Then D is dihedral of order q − ϵ, ϵ = 1 or −1, and where q ≅ ϵ(mod 4). Further, D⟨t⟩ is dihedral, and so C_{D}(t) = ⟨s⟩. It follows that C_{L}(t) is dihedral, of twiceodd order. Now let D_{1} be a dihedral subgroup of L, containing s, of order q +ϵ. Suppose that D_{1}^{x} is t-invariant, for every x ∈ C_{L}(s). Then [D, t] ≤ N_{L}(D_{1}). As D_{1} is maximal in L, and |D ∩ D_{1}| = 2, we conclude that [D, t] = ⟨s⟩, |D| = 4, and q = 5. In this case we have G_{1} ≅ Sym(5), and one checks that two conjugates of ⟨(1 2), (3 4)⟩ suffice to generate Sym(5). On the other hand, suppose that there exists x ∈ C_{L}(s) such that D_{1}^{x} is not t-invariant. We may then take x = 1, and we conclude that G = ⟨s, t, s^{g}, t^{g}⟩, where g is a generator for the maximal cyclic subgroup of D_{1}. Thus (ii) holds in this case.

Finally, suppose that A ≤ L, and that |A| > 2. Then A is a fours group. Let s be an involution in A, and let D be a dihedral subgroup of L containing s, of twice-odd order q − 1 or q + 1. Then D is maximal, so D is not A-invariant, and so G = L = ⟨A, A^{g}⟩ for any generator g of [D, D]. Thus, (iii) holds in this case, and the lemma is proved. □

10.6 Lemma. Assume Hypothesis 10.1, and suppose that H/Z(H) ≅ L_{2}(q), q a power of r. Then q = 5, r = 3, and |V | = 81.

Proof. First of all, if |Z(H)| > 2 then |Z(H)| is divisible by 3, and H/Z(H) ≅ L_{2}(9). Then p = 2 or 5, and since L_{2}(9) ≅ (Sp(4, 2))′ we have p = 5. Then |A| = 5 and H is generated by two conjugates of A, so that |V | ≤ 5^{4}. As 3 divides |Z(H)|, V may then be regarded as a 2-dimensional space over 𝔽_{25}. But L_{2}(9) is not a subgroup of L_{2}(25), so we conclude that |Z(H)| ≤ 2.

Suppose that G contains a quadratic subgroup of order at least 3. If p is odd then 3.1 yields H ≅ SL(, 2, 5), p = 3, and V is a natural SL(2, 9)-module for H. Thus, the lemma holds in this case. If p = 2 then the main result of [MS1] implies that H/Z(H) ≅ U_{4}(3), contrary to H/Z(H) ≅ L_{2}(q). Thus, we may assume henceforth that any quadratic subgroup of G has order at most 2.

Suppose first that p = 2. Then q is odd, and since H/Z(H) is not isomorphic to a group of Lie type in characteristic 2 we have q ≥ 11. By 5.4 we have |V/C_{V} (a)| ≥ 8 for any a ∈ A^{#}. Also, |A|^{3/2} ≥ |V/C_{V} (A)| by Hypothesis 4′, and so |A| ≠ 2. If |A| = 4 we have |V/C_{V} (A)| ≤ 8, and then C_{V} (a) = C_{V} (A) for all a ∈ A^{#}, and A is a quadratic fours group. Thus |A| ≠ 4. Then 10.5 implies that |A| = 8, |A ∩ H| = 4, and there exists a ∈ A − H such that a induces a field automorphism on H/Z(H). Further, if we apply 10.5 to A ∩ H then we find that H is generated by two conjugates of A ∩ H. Here |V/C_{V} (A)| ≤ 16, so dim(V) ≤ 8. But dim(V) ≥ (q − 1)/2, by 10.4(1), and so q ≤ 17. As q is odd and a perfect square we then have q = 9, and thus H/Z(H) ≅ (Sp(4, 2))′. We therefore conclude that p is odd.

Suppose that |A| = p, so that |V/C_{V} (A)| ≤ p^{2}. Suppose also that A ≤ H. The group L_{2}(r) is a homomorphic image of L_{2}(ℤ), and L_{2}(ℤ) contains the free product ℤ_{3} ∗ ℤ_{3} as a subgroup of index 2, so L_{2}(r) is generated by two elements of order 3. Thus, if p = 3 then two conjugates of A generate a subgroup Y of H with Y/Z(Y) ≅ L_{2}(r), and 2.2 then yields r = 5. Then in any case, we may choose a conjugate B of A so that ⟨A, B⟩ is nonsolvable, and then ⟨A, B⟩ is quasisimple. By 2.2 we then have p ≥ 5 (resp. p = 3), ⟨A, B⟩ ≅ L_{2}(p) (resp. L_{2}(5)), and |[V, ⟨A, B⟩]| = p^{3} (resp. 81). Any quasisimple subgroup of H/Z(H) is isomorphic to L_{2}(5) or L_{2}(q_{1}) for some power q_{1} of r, so we have p = 3 or 5 at this point. If p = 3 and q = 5 we have the conclusion of the lemma. Assuming that the lemma is false, choose a conjugate C of A, not contained in ⟨A, B⟩, so that ⟨A, C⟩ is quasisimple, and set X = ⟨A, B, C⟩. Then |[V, X]| ≤ p^{4} (resp. 3^{6}), and X/Z(X) ≅ L_{2}(q_{1}) for some power q_{1} of r (with q_{1} ≥ 25 if p = 3). Then q_{1} > 5. If q_{1} = 7 then p = 3 and X contains a Frobenius group of order 21, which is contrary to 4.2. If q_{1} = 9 then p = 5 and we may choose B so that ⟨A, B⟩ = X, which is contrary to 2.2. Thus q ≥ 11. If p = 5 then dim([V, X]) ≤ 4, and 10.4(1) yields q ≤ 9. So in fact p = 3, dim([V, X]) ≤ 6, and 10.4(1) yields q ≤ 13. But we have seen that q_{1} ≥ 25 if p = 3, so we have a contradiction at this point. Thus, A ≰ H.

Let a ∈ A − H. By Theorem 4.9.1 of [GLS3], a induces a field automorphism on H, and thus A normalizes a Sylow r-subgroup R of H. Then also [R, A] is non-cyclic. Suppose that |A| = p. Then 2.1 implies that p = 3 and that |[R, A]| = 4. It follows that q = 8 and that HA ≅ ^{2}G_{2}(3), contrary to H ∉ Lie(p). We therefore conclude that |A| = p^{2}. Set X = C_{H}(a). Then X/Z(H) ≅ L_{2}(q_{0}) where (q_{0})^{p} = q. Hypothesis 4′ implies that A ∩ X acts quadratically on the subspace W = [V, a] + C_{V} (a) of V, and since V/C_{V} (a) is X-isomorphic to [V, a] it follows that X acts non-trivially on W. Then 5.6 yields p = 3 and X ≅ SL(2, 5), so that q = 5^{3}. Evidently, three conjugates of A suffice to generate HA, so dim(V) ≤ 12. But 10.4(1) yields dim(V) ≥ 62, so we have a final contradiction, and the lemma is proved. □

In the following lemma we shall eliminate from consideration a further number of “small” groups, including groups having exceptional Schur multipliers, and the groups in the list L_{0} from 10.4. Conspicuously missing from this expanded list are any of the groups P Sp(4, q). These will be addressed later, in 10.10.

10.7 Lemma. Assume Hypothesis 10.1, and let L be the set of groups whose members are as follows:

L_{3}(q), 3 ≤ q ≤ 5,

L_{n}(q), with n = 4 or 5, and with n + q ≤ 7,

U_{3}(q), q ≤ 7,

U_{4}(q), q ≤ 3,

U_{n}(2), n ≤ 7,

P Sp(2n, q), with n = 3 or 4, and with q ≤ 3,

Ω_{7}(3), Ω_{8}^{+}(2), Ω_{8}^{−}(2), and ^{3}D_{4}(2)

2E_{6}(2), F_{4}(2), ^{2}F_{4}(2), G_{2}(3), G_{2}(4), and Sz(8).

Assume that \overline{H} is isomorphic to a group in the list L. Then \overline{H} ≅ Sp(6, 2). Proof. Assume that \overline{H} is not isomorphic to Sp(6, 2). We shall consider the various groups individually, and we shall make free use of the ATLAS, in deciding whether a Sylow psubgroup of a particular group G is contained in a unique maximal subgroup of G. This task will be simplified by the following reduction.

(a) Set H^{∗} = HS. Then there exists at most one maximal subgroup of \overline{H}^{∗} containing \overline{S}, and at most one \overline{H}-invariant maximal subgroup of \overline{S}. Indeed, if (a) is false then the unique maximal subgroup M of G containing S contains

H. But M contains also N_{G}(S ∩ H), and the Frattini argument then yields M = G.

We may also employ the following consequence of 2.2.

(b) A Sylow p-subgroup of G has order greater than p. Most groups in L, and most primes, may be eliminated at once by recourse to (a)

and (b). The only exceptions will be given by the groups L_{3}(4) and ^{3}D_{4}(2) with p = 3, and the groups L_{3}(3) and L_{4}(3) with p = 2. Apart from the treatment of these four exceptions, the argument will consist simply of a list, in which we give the isomorphism type of \overline{H}, the order of \overline{H} and of Out(\overline{H}), and then, for each prime p different from r, and for which p^{2} divides the order of Aut(\overline{H}), a pair of distinct maximal subgroups of \overline{H}^{∗} = \overline{H}\overline{S} containing \overline{S}.

(1) \overline{H} ≅ L_{3}(3), |\overline{H}| = 2^{4} · 3^{3} · 13, |Out(\overline{H})| = 2, (p = 2). There is a unique maximal subgroup containing S if S ≰ H. This case will be treated below.

(2) \overline{H} ≅ L_{3}(4), |\overline{H}| = 2^{6} · 3^{2} · 5 · 7, |Out(\overline{H})| = 12, (p = 3).

There is a unique maximal subgroup of \overline{H}^{∗} containing \overline{S} if S ≰ H, so this case will be considered separately below. Maximal subgroups when S ≤ H:

M_{10}, U_{3}(2).

(3) \overline{H} ≅ L_{3}(5), |\overline{H}| = 2^{5} · 3 · 5^{3} · 31, |Out(H)| = 2, (p = 2). Maximal subgroups when S ≤ H:

5^{2} : GL(2, 5), 4^{2} : Sym(3).

Maximal subgroups when S ≰ H:

GL(2, 5).2, 4^{2}.(Sym(3) × 2).

(4) \overline{H} ≅ L_{4}(2), |\overline{H}| = 2^{6} · 3^{2} · 5 · 7, |Out(H)| = 2, (p = 3). Maximal subgroups:

Alt(7), 2^{4} : (Sym(3) × Sym(3)).

(5) \overline{H} ≅ L_{4}(3), |\overline{H}| = 2^{7} · 3^{6} · 5 · 13, |Out(\overline{H})| = 4, (p = 2). There may be a unique maximal subgroup of \overline{H}^{∗} containing \overline{S}, if S ≰ H. This will be treated below.

(6) \overline{H} ≅ L_{5}(2), |\overline{H}| = 2^{10} · 3^{2} · 5 · 7 · 31, |Out(\overline{H})| = 2, (p = 3). Maximal subgroups:

2^{4} : L_{4}(2), 2^{6} : (L_{3}(2) × L_{2}(2)).

(7) \overline{H} ≅ U_{3}(3) ≅ (G_{2}(2))′, |\overline{H}| = 2^{5} · 3^{3} · 7, |Out(\overline{H})| = 2, cyclic Sylow subgroups in cross characteristic.

(8) \overline{H} ≅ U_{3}(4), |\overline{H}| = 2^{6} · 3 · 5^{2} · 13, |Out(\overline{H})| = 2, (p = 5). Two maximal subgroups GU(2, 4).

(9) \overline{H} ≅ U_{3}(5), |\overline{H}| = 2^{4} · 3^{2} · 5^{4} · 7, |Out(\overline{H})| = 6, (p = 2 or 3). Maximal subgroups for p = 3, with S ≤ H:

Alt(7), M_{10}.

Maximal subgroups of \overline{H}^{∗}, for p = 3, with S ≰ H:

6^{2} : Sym(3), 3^{2} : SL(2, 3).

\overline{S}-invariant maximal subgroups of \overline{H}, for p = 2:

M_{10}, 2.Sym(5).

(10) \overline{H} ≅ U_{4}(2) ≅ P Sp(4, 3), |\overline{H}| = 2^{6} · 3^{4} · 5, |Out(\overline{H})| = 2, cyclic Sylow subgroups in cross characteristic.

(11) \overline{H} ≅ U_{4}(3), |\overline{H}| = 2^{7} · 3^{6} · 5 · 7, |Out(\overline{H})| = 8, (p = 2).

In this case, the existence of a pair of proper subgoups of G, containing S, and generating G, is evident from the B̃_{2} geometry of 2-local subgroups of \overline{G}. See [K].

(12) \overline{H} ≅ U_{5}(2), |\overline{H}| = 2^{10} · 3^{5} · 5 · 11, |Out(\overline{H})| = 2, (p = 3).

Maximal subgroups of \overline{H} containing \overline{S}:

Sym(3) × GU(3, 2), GU(4, 2).

(13) \overline{H} ≅ U_{6}(2), |\overline{H}| = 2^{15} · 3^{6} · 5 · 7 · 11, |Out(\overline{H})| = 6, (p = 3).

Maximal subgroups of \overline{H} containing \overline{S}, if \overline{S} ≤ \overline{H}:

3^{1+4}.[2^{7}.3], U_{4}(3) : 2.

Maximal subgroups of \overline{H}^{∗} containing \overline{S}, if \overline{S} ≰ \overline{H}.

3^{1+4}.[2^{7}.3], 3^{5} : Sym(6).

(14) \overline{H} ≅ U_{7}(2), |\overline{H}| = 2^{21} · 3^{8} · 5 · 7 · 11 · 43, |Out(\overline{H})| = 2, (p = 3).

Subgroups of \overline{H} containing \overline{S} and which generate \overline{H}:

GU(6, 2), U_{4}(2) × GU(3, 2).

(16) \overline{H} ≅ P Sp(6, 3), |\overline{H}| = 2^{9} · 3^{9} · 5 · 7 · 13, |Out(\overline{H})| = 2, (p = 2).

Maximal subgroups of \overline{H} invariant under \overline{S}:

2(Alt(4) × U_{4}(2)), 2^{2+6} : 3^{3} : Sym(3).

(19) \overline{H} ≅ Sp(8, 2), |\overline{H}| = 2^{16} · 3^{5} · 5^{2} · 7 · 17, |Out(\overline{H})| = 1, (p = 3 or 5).

Maximal subgroups of \overline{H} containing \overline{S} if p = 5:

D_{4}(2), Sp(4, 4) : 2.

Maximal subgroups of \overline{H} containing \overline{S} if p = 3:

D_{4}(2), Sym(3) × Sp(6, 2).

(20) \overline{H} ≅ P Sp(8, 3), |\overline{H}| = 2^{14} · 3^{16} · 5^{2} · 7 · 13 · 41, |Out(\overline{H})| = 2, (p = 2 or 5). Maximal subgroups of \overline{H} containing \overline{S} if p = 5:

(Sp(4, 3) ∘ Sp(4, 3))2, P Sp(4, 9) : 2.

Maximal subgroups of \overline{H} invariant under \overline{S} if p = 2:

(Sp(4, 3) ∘ Sp(4, 3))2, \overline{M},

where \overline{M} is the image in \overline{H} of a subgroup of Sp(8, 3) of the form SL(2, 3) ≀ Sym(4).

(21) \overline{H} ≅ Ω_{7}(3), |\overline{H}| = 2^{9} · 3^{9} · 5 · 7 · 13, |Out(\overline{H})| = 2, (p = 2).

Maximal subgroups of \overline{H} invariant under \overline{S}:

2U_{4}(3) : 2, 2^{6} : Alt(7).

(22) \overline{H} ≅ Ω_{8}^{+}(2), |\overline{H}| = 2^{12} · 3^{5} · 5^{2} · 7, |Out(\overline{H})| = 6, (p = 3 or 5).

Maximal subgroups of \overline{H} containing \overline{S} if p = 5:

(Alt(5) × Alt(5)) : 2^{2} in three different ways.

Maximal subgroups of \overline{H} containing \overline{S} if p = 3 and S ≤ H:

(3 × U_{4}(2)) : 2, 3^{4} : 2^{3}.Sym(4).

Maximal subgroups of \overline{H}^{∗} containing \overline{S} if p = 3 and S ≰ H:

3_{+}^{1+4} : GL(2, 3), (3^{4} : 2^{3}.Sym(4)).3.

(23) \overline{H} ≅ Ω_{8}^{−}(2), |\overline{H}| = 2^{12} · 3^{4} · 5 · 7 · 17, |Out(\overline{H})| = 2, (p = 3).

Maximal subgroups of \overline{H} containing \overline{S}:

2^{6} : U_{4}(2), Sp(6, 2).

(24) \overline{H} ≅ ^{3}D_{4}(2), |\overline{H}| = 2^{12} · 3^{4} · 7^{2} · 13, |Out(\overline{H})| = 3, (p = 3 or 7).

Maximal subgroups of \overline{H} containing \overline{S} if p = 7:

(7 × L_{3}(2)) : 2, 7^{2} : SL(2, 3).

There is a unique maximal subgroup of \overline{H}^{∗} containing \overline{S} if p = 3. This case will be treated below.

(25) \overline{H} ≅ G_{2}(3), |\overline{H}| = 2^{6} · 3^{6} · 7 · 13, |Out(\overline{H})| = 2, (p = 2).

Maximal subgroups of \overline{H} invariant under \overline{S}:

(2^{3})L_{3}(2), 2_{+}^{1+4} : 3^{2}.2.

(26) \overline{H} ≅ G_{2}(4), |\overline{H}| = 2^{12} · 3^{3} · 5^{2} · 7 · 13, |Out(\overline{H})| = 2, (p = 3 or 5).

Maximal subgroups of \overline{H} containing \overline{S} if p = 5:

U_{3}(4) : 2, Alt(5) × Alt(5).

Maximal subgroups of \overline{H} containing \overline{S} if p = 3:

SL(3, 4) : 2, J_{2}.

(27) \overline{H} ≅ F_{4}(2), |\overline{H}| = 2^{24} · 3^{6} · 5^{2} · 7^{2} · 13 · 17, |Out(\overline{H})| = 2, (p = 3, 5, or 7).

Maximal subgroups of \overline{H} containing \overline{S} if p = 7:

3D_{4}(2), in two different ways.

Maximal subgroups of \overline{H} containing \overline{S} if p = 5:

2F_{4}(2), Sp(8, 2).

Maximal subgroups of \overline{H} containing \overline{S} if p = 3:

Aut(D_{4}(2)), L_{4}(3) : 2.

(28) \overline{H} ≅ (^{2}F_{4}(2))′, |\overline{H}| = 2^{11} · 3^{3} · 5^{2} · 13, |Out(\overline{H})| = 2, (p = 3 or 5).

Maximal subgroups containing \overline{S} if p = 5:

L_{2}(25), 5^{2} : 4Alt(4).

Maximal subgroups containing \overline{S} if p = 3:

L_{3}(3) : 2 in two different ways.

(29) \overline{H} ≅ Sz(8), |\overline{H}| = 2^{6} · 5 · 7 · 13, |Out(\overline{H})| = 3. This case contradicts (2).

(30) \overline{H} ≅ ^{2}E_{6}(2), |\overline{H}| = 2^{36} · 3^{9} · 5^{2} · 7^{2} · 11 · 13 · 17 · 19, |Out(\overline{H})| = 6, (p = 3, 5, or 7). Maximal subgroups of \overline{H} containing \overline{S} if p = 7:

F_{4}(2), in three different ways.

Maximal subgroups of \overline{H} containing \overline{S} if p = 5:

F_{4}(2), in three different ways.

Maximal subgroups of \overline{H} containing \overline{S} if p = 3 and S ≤ H:

F i_{22} in three different ways.

Maximal subgroups of \overline{H}^{∗} containing \overline{S} if p = 3 and S ≰ H.

3^{6} : U_{4}(2) : 2, 3^{1+6}.2^{3+6}.(Sym(3) × 3).3.

It now remains to take up the four exceptional cases: \overline{H} ≅ L_{3}(4), ^{3}D_{4}(2), L_{3}(3), and L_{3}(4).

Suppose first that \overline{H} ≅ L_{3}(4), of order 2^{6} · 3^{2} · 5 · 7. Then p = 3, by (2). Then Z(H) is a 2-group, the 3-rank of G is 2, and so |A| ≤ 9. We claim that four conjugates of A suffice to generate HA if |A| = 3, and that three conjugates suffice if |A| = 9. Indeed, suppose that A ∩ H ≠ 1, and let 1 ≠ a ∈ A ∩ H. Then two conjugates of a generate a subgroup K of H with Z(H)K/Z(H) ≅ L_{3}(2), and A ∩ K is a Sylow 3-subgroup of N_{G}(K). Here Z(H)K is maximal in H, so the claim holds in this case. On the other hand, suppose that A ∩ H = 1, and set X = C_{H}(A). Referring to the ATLAS, we find that Z(H)X/Z(H) ≅ SL(2, 4) or F r(21). Let B be a Sylow 3-subgroup of C_{G}(A). Then A is conjugate to every cyclic subgroup of B not contained in X, and so there are conjugates A_{1} and A_{2} of A such that Z(H)⟨A_{1}, A_{2}⟩ = Z(H)XA. It is left to the reader to demonstrate that two conjugates of XA suffice to generate HA (or to find a different argument) and thus to prove the claim. We conclude that |V | ≤ 3^{8}.

If S ≤ H then there are two different maximal subgroups M_{1} and M_{2} of H containing S (with \overline{M}_{1} ≅ U_{3}(2) and with \overline{M}_{2} ≅ M_{10}), contrary to (1). Thus S ≰ H. There then exists a subgroup Y of G with Y ≅ F r(21) × 3. Let b be a non-identity element of O_{3}(Y) and x a non-identity element of O_{7}(Y). Then dim([V, x]) = 6, and [V, x, b] = 0, so |V/C_{V} (b)| ≤ 9. But, as we have seen, b is conjugate to an element of Y not contained in ⟨b⟩, and in this way we violate 2.2. Thus, H/Z(H) ≇ L_{3}(4).

Suppose next that \overline{H} ≅ ^{3}D_{4}(2) and that p = 3. It so happens that Z(H) = 1, so \overline{G} = G. There are exactly two classes of elements of order 3 in H. To be consistent with ATLAS notation, we label a pair of fixed representatives of these two classes by 3A and 3B, where N_{H}(3A) ≅ Sym(3) × L_{2}(8), and where N_{H}(3B) is of the form 3_{+}^{1+2})GL(2, 3). Set L = E(C_{H}(3A)). We may assume that [3A, 3B] = 1, and then 3B ∈ E(C_{H}(3A)) since 3A is not contained in a cyclic group of order 9. Let t be an involution in N_{H}(3A) with [t, L] = 1. Then C_{H}(t) is a maximal parabolic subgroup of H, of the form 2_{+}^{1+8} : L_{2}(8). As two conjugates of 3B suffice to generate L, it follows that three conjugates of 3B suffice to generate C_{H}(t), and then four conjugates suffice to generate H. On the other hand, we observe that 3A is not central in a Sylow 3-subgroup of H, and hence 3A is not weakly closed in a Sylow 3- subgroup of N_{H}(3A). Thus, there is a conjugate of 3A which is “diagonal” in ⟨3A⟩ × L, and so C_{H}(3A) is generated by two conjugates of 3A. Then four conjugates of 3A suffice to generate a subgroup of H containing C_{H}(t). As C_{H}(t) contains no conjugate of 3A, it follows that four conjugates of 3A suffice to generate H. Thus, we have shown that for any element x ∈ H of order 3, four conjugates of x suffice to generate H.

No involution-centralizer in H contains a copy of ℤ_{3} × ℤ_{3}, and then it follows from the given structure of N_{H}(3B) that the 3-rank of H is two, and the 3-rank of Aut(H) is then at most three. In particular, we have |A| ≤ 27, and so |V/C_{V} (A)| ≤ 3^{6}. We now appeal to 10.4, where we find that dim(V) ≥ 24. It follows that four conjugates of A do not suffice to generate HA, and therefore A ∩ H = 1. Thus |A| = 3, |V/C_{V} (A)| ≤ 9, and twelve conjugates of A do not suffice to generate HA. But A is not weakly closed in S, and therefore eight conjugates of A suffice to generate HA. Thus, we have a contradiction, and H ≇ ^{3}D_{4}(2).

Suppose next that \overline{H} ≅ L_{3}(3). Then Z(G) = 1, |G/H| ≤ 2, and by (2) we have p = 2. A Sylow 2-subgroup of Aut(H) is a direct product of a semidihedral with ℤ_{2}, so the 2-rank of A is at most 3. By [MS1], G contains no quadratic fours groups, and then no element of A is a 2-transvection on V, by 5.6. Also, as A contains no quadratic fours group we have C_{V} (a) ≠ C_{V} (A) for any a ∈ A^{#}. As |A|^{3/2} ≥ |V/C_{V} (A)|, by Hypothesis 4′, we conclude that |A| = 8, that |V/C_{V} (A)| = 16, and that |V/C_{V} (a)| = 8 for all a ∈ A^{#}. Let a ∈ A − H. Then C_{H}(a) ≅ GL(2, 3), and since |[V, a]| = 8 we have [V, a, Z(C_{H}(a))] = 0. Thus A contains a quadratic fours group, and we have a contradiction in this case.

Finally, suppose that \overline{H} ≅ L_{4}(3). Then p = 2, as seen in (5). As Z(H) is a 2-group, we then have Z(H) = 1 and H ≅ L_{4}(4). A Sylow 2 subgroup of Aut(H) is contained in a unique maximal subgroup of Aut(G), of the form 2 × (GL(2, 3) ∘ GL(2, 3))2, so we have |A| ≤ 32. As in the preceding paragraph, G contains no quadratic fours groups, so |V/C_{V} (a)| ≥ 8 and C_{V} (a) ≠ C_{V} (A) for any a ∈ A^{#}, and we have |A| ≥ 8. In particular, we then have A ∩ H ≠ 1.

There are exactly two conjugacy classes of involutions in H, with representatives 2A and 2B, and where C_{H}(2A) and C_{H}(2B) have the form (4 × L_{2}(9)) : 2 and (SL(2, 3) ∘ SL(2, 3)) : 2^{2}, respectively. Moreover, an element of C_{H}(2B) interchanges the two subnormal SL(2, 3)- subgroups of C_{H}(2B). (The involution 2B lifts to an involution in SL(4, 3), while 2A lifts to an element of order 4.) Let 1 ≠ a ∈ A∩H, and set K = C_{H}(a). As H contains no quadratic fours groups, it follows that a is the unique involution in C_{K}([V, a]), and hence dim([V, a]) ≥ 4. As |A|^{3/2} ≥ |V/C_{V} (A)| > |V/C_{V} (a)|, we then have |A| = 16 or 32.

Suppose |A| = 16. Then |V/C_{V} (A)| ≤ 2^{6}, and so |V/C_{V} (a)| ≤ 2^{5}. We have dim(V) ≥ 26, by 10.4(2), so it follows that H cannot be generated by five conjugates of a. Suppose that a is of type 2B, and let L_{1} and L_{2} be two copies of L_{3}(3) in H such that ⟨L_{1}, L_{2}⟩ = H, with L_{1} ∩ L_{2} ≅ GL(2, 3). Then L_{1} ∩ L_{2} is generated by three conjugates of a, and since L_{1} ∩ L_{2} is maximal in each L_{i} it follows that five conujugates of a suffice to generate H. On the other hand, suppose that a is of type 2A. Then K lifts in SL(4, 4) to a group of the form (8 ∘ SL(2, 9)) : 2, so O^{2}(K) contains a conjugate of a, and then three conjugates of a suffice to generate O^{2}(K). We have O^{2}(K) ≤ L, where L ≅ P Sp(4, 3), and N_{L}(O^{2}(K)) is the unique maximal subgroup of L containing O^{2}(K). It follows that four conjugates of a suffice to generate L. As N_{H}(L) is the unique maximal subgroup of H containing L, it follows that five conjugates of a generate H.

We conclude that |A| = 32. Here |V/C_{V} (A)| ≤ 2^{7}, so |V/C_{V} (a)| ≤ 2^{6}. Then HA is not generated by A together with three conjugates of a. Further, we now have HA = G ≅ Aut(H), so we may choose a ∈ G − H so that H⟨a⟩ ≅ P GL(4, 3), and so that a inverts an element x of order 13 in H. Then three conjugates of a generate a subgroup N of G of the form 3^{3} : 13 : 2, while a conjugate of A contains an element b which induces a transpose-inverse automorphism of H, and such that ⟨O_{3}(N), O_{3}(N)^{b}⟩ = H. Thus, we have a contradiction in this case, and the lemma is proved. □

Having disposed of so many individual groups in 10.7, we may now start in on the general case. To this end, the following elementary lemma will be helpful.

10.8 Lemma. Let 1 ≤ n_{1} ≤ n_{2} < ⋯ ≤ n_{k} be a non-decreasing sequence of natural numbers, and let q > 1. Set N = ∑_{i=1}^{k} n_{i}. Then ∏_{i=1}^{k}(q^{n_{i}} + (−1)^{i}) < q^{N}.

Proof. Let k be the least integer for which the lemma fails. Let ℓ be the largest even integer smaller than k. Then ℓ ≥ 0 and one has ∏_{i=1}^{ℓ}(q^{n_{i}} + (−1)^{i}) < q^{M} where N = ∑_{i=1}^{ℓ} n_{i}. The reader is invited to supply the last step in the induction. □

10.9 Lemma. Let G_{0} be a group such that F^{∗}(G_{0}) is a quasisimple group of Lie type in characteristic r different from p. Assume also that O_{p}(G_{0}) = 1 and that G_{0} = F^{∗}(G_{0})A where A is an elementary abelian p-group. Set X = F^{∗}(G_{0})/Z(F^{∗}(G_{0})). Finally, assume that r does not divide |Z(F^{∗}(G_{0}))|. Then one of the following holds.

(1) X ≅ P SL(n, q) or P SU(n, q), and |G_{0}| ≤ 2p · q^{n^{2}−1}.

(2) X ≅ P Sp(2n, q), n ≥ 2, and |G_{0}| ≤ 2p · q^{n(2n+1)}.

(3) X ≅ P Ω(2n + 1, q), n ≥ 3, and |G_{0}| ≤ 2p · q^{n(2n+1)}.

(4) X ≅ P Ω_{2n}^{ϵ}(q), n ≥ 4, and |G_{0}| ≤ 3p · q^{n(2n−1)}.

(5) X ≅ ^{3}D_{4}(q) and |G_{0}| ≤ pq^{28}.

(6) X ≅ E_{6}(q) or ^{2}E_{6}(q), and |G_{0}| ≤ 2pq^{78}.

(7) X ≅ E_{7}(q) and |G_{0}| ≤ pq^{133}.

(8) X ≅ E_{8}(q) and |G_{0}| ≤ pq^{248}.

(9) X ≅ F_{4}(q) and |G_{0}| ≤ 2pq^{52}.

(10) X ≅ ^{2}F_{4}(q) and |G_{0}| ≤ pq^{26}.

(11) X ≅ G_{2}(q) and |G_{0}| ≤ pq^{14}.

(12) X ≅ ^{2}G_{2}(q) and |G_{0}| ≤ q^{7}.

(13) X ≅ Sz(q) and |G_{0}| ≤ pq^{5}. Moreover, in (1) through (13), the factor p appearing in the estimate for |G| is present only if q is a p^{th} power.

Proof. Let L = L(q) be a Chevalley group (i.e. a non-twisted group of Lie type) of universal type. Then |L| is a polynomial in q, of the form q^{M} ∏_{i=1}^{k}(q^{n_{i}}−1), and so we have |L| ≤ q^{M+N}, where N = ∑ i = 1^{k}n_{i}, and where, as it happens, M + N is the dimension of the adjoint module for L. We may of course take the sequence of natural numbers n_{1}, n_{2}, ⋯ n_{k} to be non-decreasing. Now assume that L is not of type D_{n}, n ≥ 4, and let 2L(q) be a Steinberg Variation of L. Then |^{2}L(q)| = q^{M} ∏_{i=1}^{k}(q^{n_{i}} + (−1)^{i}), and so also |^{2}L(q)| < q^{M+N} by 10.5. Suppose that L =^{2} D_{4}(q), of order q^{n(n−1)}(q^{n}+1) ∏_{i=1}^{n−1}(q^{2i}−1). Then |L| < q^{n(2n−1)}, as follows from an obvious variation on 10.5. Also, we have

|^{3}D_{4}(q)| = q^{12}(q^{8} + q^{4} + 1)(q^{6} − 1)(q^{2} − 1) = q^{12}(q^{12} − 1)(q^{6} − 1)/(q^{2} + 1),

and so |^{3}D_{4}(q)| < q^{30}/(q^{2} + 1) < q^{28}. So, in all these cases, |L| < |U| where U is the adjoint module for the untwisted version of L. Here we refer to Table 16.1 in [A1] for the group orders, and to [J] for the dimensions of the adjoint modules. From these references, the reader may check that if L is a Suzuki-Ree group then |L| < \sqrt{|U|} where U is the adjoint module for the untwisted antecedent of L.

We are given G_{0} = F^{∗}(G_{0})A where A is an elementary abelian p-group. Without loss of generality, we have A ∩ F^{∗}(G) = 1. Let A_{1} be the largest subgroup of A which induces inner-diagonal automorphisms on X, and A_{2} the largest subgroup of A which induces inner-diagonal-graph automorphisms of X. Then |A/A_{2}| ≤ p, since the group of field automorphisms of X is cyclic. Also, |A_{2}/A_{1}| = 1 if X is a twisted group or if the Dynkin diagram for X admits no symmetries. Otherwise we have |A_{2}/A_{1}| ≤ 3, with equality only if p = 3 and X is of type D_{4}. We have |A_{1}| ≤ p, and if |A_{1}| = p then p divides the order of Z(L) where L is the universal version of X. Since O_{p}(G) = 1, |A_{1}| is then “absorbed” by the estimate, given in the preceding paragraph, for |L|. In this way, we obtain (1) through (13). □

10.10 Lemma. Assume Hypothesis 10.1. Then \overline{H} is not isomorphic to P Sp(4, q) for any q.

Proof. Set G_{0} = HA. Then |G_{0}| ≤ q^{11}, as follows from 10.9. Suppose first that q is even, so that p is odd. Then 10.4 yields dim(V) ≥ q(q − 1)^{2}/2. As |A|^{2} ≥ |V/C_{V} (A)| we have also |\overline{G}_{0}| ≥ |V |, by Theorem 2.3 of [CD1], and thus

22log_{p}(q) > q(q − 1)^{2}/2.

This inequality evidently fails if q ≥ 8, so we have q = 2 or 4. The case q = 2 has already been eliminated in 10.5, so \overline{H} ≅ Sp(4, 4). Then |\overline{H}| = 2^{8} · 3^{2} · 5^{2} · 17, and Out(\overline{H}) is a 2-group. The criteria (a) and (b) in the proof of 10.7 then reduce the problem to finding a pair of distinct maximal subgroups of \overline{H} containing \overline{S}, in the case that p = 3 or that p = 5. According to the ATLAS, there are in fact two different maximal subgroups of \overline{H} of the form (SL(2, 4) × SL(2, 4)) : 2, and so we conclude that q is odd.

Now 10.4 yields

(1)  22log_{p}(q) > (q − 1)^{2}/2.

Suppose that p = 2. Then Hypothesis 4′ yields |A|^{3/2} ≥ |V/C_{V} (A)|, and so (1) improves to

(2)  33log_{2}(q) > (q − 1)^{2}.

This yields q ≤ 11. Here Z(H) is a 2-group, so Z(H) = 1 and H = \overline{H}. There is an S-invariant maximal subgroup L of H of the form (SL(2, q) ∘ SL(2, q)) : 2, where the components of L are interchanged by an involution in L. If q is congruent to 3 or 5 mod 8 then there is also a maximal S-invariant subgroup M of H of the form 2^{4} : Alt(5) (and which may be described as the set-wise stabilizer of an orthonormal basis in the Ω_{5}(q)-module for H). Thus, we are reduced to the cases where q = 7 or 9.

Denote by L^{∗} the normalizer in Aut(H) of L. If q = 7 then L^{∗} = L⟨d⟩ where d is an involution which induces a diagonal automorphism on each of the components of L. If q = 9 then L^{∗} = L⟨d, α⟩, where d is as just described and where α induces a field automorphism on each component of L. It follows that the 2-rank of S is 4 if q = 7, and at most 5 if q = 9. In particular, we have |A| ≤ 32, and |A| ≤ 16 if q = 7.

By 5.6, G contains no quadratic fours groups, and so 5.5 implies that |A| ≥ 8 and that C_{V} (A) ≠ C_{V} (a) for any a ∈ A^{#}. As |A|^{3/2} ≥ |V/C_{V} (A)| we have |V/C_{V} (A)| ≤ 2^{7}, and |V/C_{V} (A)| ≤ 2^{6} if q = 5. Then |[V, a]| ≤ 2^{6} (and |[V, a]| ≤ 2^{5} if q = 5) for any a ∈ A^{#}. Suppose that A ∩ H contains an involution a with C_{H}(a) = L. As A contains no quadratic fours groups, L/⟨a⟩ acts faithfully on [V, a], whereas [V, a] is too small to support such an action. Thus A ∩ Z(L) = 1. There are only two conjugacy classes of subgroups of order 2 in H, and one of these lifts to a class of cyclic subgroups of order 4 in SP (4, q). It follows from this that |A ∩ H| ≤ 4. Thus |A| ≤ 16, and |A| = 8 if q = 7. Fix an involution a ∈ A ∩ H. Then |V/C_{V} (a)| ≤ 32, and |V/C_{V} (a)| ≤ 8 if q = 7. On the other hand, 8.4 yields dim(V) ≥ 24 if q = 7, and dim(V) ≥ 40 if q = 9. As C_{V} (H) = 0 it follows that H cannot be generated by eight conjugates of a. We leave to the reader the task of verifying the absurdity of this outcome. □

We may now complete the proof of Theorem 10.2. Thus, assume Hypothesis 10.1. In view of 10.3, 10.5, 10.6, and 10.9, we may assume that \overline{H} is not isomorphic to L_{2}(q), or to any of the groups in the list L given by 10.6. Among the groups in L one finds all the groups of Lie type (other than those which are isomorphic to L_{2}(q)) which have an exceptional Schur multiplier. Having excluded these groups, it follows that H is a homomorphic image of the universal version, in the Lie-theoretic sense, of \overline{H}.

Set G_{0} = HA. By Hypothesis 10.1 we have |A|^{2} ≥ |V/C_{V} (A)|, and then Theorem 2.3 of [CD1] implies that |G_{0}|^{2} ≥ |V |. An upper bound for G_{0} is given by 10.8, while a lower bound for dim(V) is given by 10.4. The proof of 10.2 now reduces to a systematic comparison of log_{p}(|G_{0}|^{2}) with dim(V).

Suppose first that \overline{H} ≅ L_{n}(q), n ≥ 3. Then |G_{0}| ≤ 2pq^{n^{2}−1}, and the factor p is required only if q is a p^{th} power. Thus |G_{0}| ≤ q^{n^{2}}. By 10.4 we then have

(1)  2n^{2}log_{p}(q) > (q^{n} − 1)/(q − 1) − n.

Suppose that q = 2. Then p ≥ 3, and so 2log_{p}(q) < 1.5. Then (1) yields 1.5n^{2} > 2^{n}−n−1, and then n ≤ 5. Thus \overline{H} ∈ L in this case, and thus q ≥ 3. From (1) we have

2n^{2}q + n > q^{n−1},

and then

(2)  2n^{2} + n > q^{n−2}.

We observe that (2) is false if q ≥ 3 and n ≥ 6. As q > 2, it follows that (2) holds only if n ≤ 5. For n = 3 one checks that (1) is false for q ≥ 7, and for n = 4 we find that (1) is false if q ≥ 5. Also, (1) fails to hold if n = 5 and q ≥ 3. Thus, \overline{H} ∈ L.

Suppose next that \overline{H} ≅ U_{n}(q), n ≥ 3. As in the preceding case, we find |G_{0}| ≤ q^{n^{2}}. Referring to 10.8, we then have

(3)  2n^{2}log_{p}(q) > \{ \begin{matrix}(q^{n} − q)/(q + 1)  if n is odd, and \\ (q^{n} − 1)/(q + 1)  if n is even. \end{matrix}

Suppose q = 2. Then we replace (3) by the estimate

(4)  2n^{2}log_{3}(2) > (2^{n} − 2)/3.

If n = 8 then the left side of (4) is less than 81 and the right side is bigger than 84, so (4) fails in this case. Taking the derivative of each side of (4) with respect to n, we observe that

4nlog_{3}(2) < (ln(2))2^{n}

for all n ≥ 8, and so (4) holds only if n ≤ 7. Thus \overline{H} ∈ L if q = 2, and so q ≥ 3.

Suppose n = 3. We may take p = 2 in (3), and obtain 18log_{2}(q) > (q^{3} − q)/4, which is false if q = 7. Taking derivatives with respect to q we find that 18/(ln(2)q) < (3q^{2} −1)/4 for n ≥ 7, so q ≤ 5 if n = 3. Now suppose that n = 4. Then (3) yields 32log_{2}(q) > (q^{4} − 1)/4, which is false if q = 5. We leave it to the reader to perform the derivative test which establishes that (3) is false if n = 4 and q ≥ 5. We also leave to the reader the task of verifying that if n = 5 then q = 2. Thus, in all these cases we have \overline{H} ∈ L.

We now have n ≥ 6, and we obtain

2n^{2}log_{p}(q) > q(q^{n−1} − 1)/(q + 1)

from (3). Then also

2n^{2} > (q^{n−1} − 1)/(q + 1) > (q^{n−1} − 1)/2(q − 1) > q^{n−2}/2,

and so 4n^{2} > q^{n−2}. This result fails to hold if q = 4 and n = 6, and it is a trivial matter to check that q^{n−2} grows more rapidly than 4n^{2} as a function of q and n, independently, for q ≥ 4 and for n ≥ 6. One also checks directly that (3) fails to hold if n = 6 and q = 3. This completes the proof that \overline{H} ∈ L if \overline{H} ≅ U_{n}(q).

Suppose next that \overline{H} ≅ P Sp(2n, q), n ≥ 3 (the case n = 2 having been dealt with in 10.9). Then 10.8 yields |G_{0}| < q^{2n^{2}+n+1}, and then with 10.4 we have

(5)  2(2n^{2} + n + 1)log_{p}(q) > \{ \begin{matrix}(q^{n} − 1)/2  if q is odd, and \\ q(q^{n} − 1)(q^{n−1} − 1)/2(q + 1)  if q is even. \end{matrix}

If p = 2 then there is more than one maximal subgroup of G containing S, by Theorem A in [A1]. Thus p ≥ 3.

Suppose n = 3, and then suppose that q is even. Then (5) yields 44 > (q^{3}−1)(q−1)/2, which is false if q ≥ 4. On the other hand, suppose that q is odd. Then (5) yields 44log_{p}(q) > (q^{3} − 1)/2, which is false if q ≥ 7. If q = 5 then q is not a p^{th} power, and then |G_{0}| < 5^{10}. One then checks that 40log_{3}(5) < 60 < (5^{3} − 1)/2. We conclude that q = 2 if n = 3.

Suppose n = 4. Then q ≥ 3, by 10.6, and (5) then reduces us to the case q = 3. The only prime greater than 3 whose square divides |P Sp(8, 3)| is 5, so 2.2 yields p = 5. Then S is contained in a pair of subgroups L_{1} and L_{2} of H, with L_{1} ≅ P Sp(4, 3) ∘ P Sp(4, 3) and with L_{2} ≅ P Sp(4, 9), such that H = ⟨L_{1}, L_{2}⟩. Thus n ≥ 5.

Suppose that n = 5. If q = 2 then (5) yields 228log_{p}(2) > 155, which is false for p = 3 (as one checks on one’s pocket calculator) and which is therefore false for all p since p ≠ 2. Then also (5) fails to hold for all even q, when n = 5. If instead q is odd then (5) yields 228log_{p}(q) > q^{5} − 2, and one quickly verifies that this is false for q = 3 (where p ≥ 5) and then for all odd q and p.

Suppose q = 2. Then (5) yields

2(2n^{2} + n + 1)log_{3}(2) > (2^{n} − 1)(2^{n−1} − 1)/3,

which fails to hold for n ≥ 6. Thus, q > 2. In general, (5) yields

2(2n^{2} + n + 1) > (q^{n} − 1)/2log_{p}(q) > (q^{n} − 1)/q > q^{n−1} − 1.

This fails for n = 6 and q = 3, and since q^{n−1} − 1 grows more rapidly than 2(2n^{2} + n + 1) as a function of n, we have thereby eliminated all of the groups P Sp(2n, q) as possibilities for \overline{H}, with the exception of Sp(6, 2).

Suppose next that \overline{H} ≅ P Ω_{2n+1}(q), q odd, n ≥ 3. Then 10.8 and 10.4 yield

(6)  2(2n^{2} + n + 1)log_{p}(q) > \{ \begin{matrix}(q^{2n} − 1)/(q^{2} − 1) − n  if q ≠ 3, and \\ (3^{2n} − 1)/8 − (3^{n} − 1)/2  if q = 3. \end{matrix}

Supppose q ≠ 3. If n = 3 then (6) yields 44log_{2}(q) > q^{4} + q^{2} − 2, which fails to hold for any odd q. Also, in the case n = 4 we have 78log_{2}(q) > q^{6} + q^{4} + q^{2} − 3, which also fails for odd q. In general, (6) yields

2(2n^{2} + n + 1) > (q^{2n−2} − 1)/(q^{2} − 1)log_{p}(q) − n/log_{p}(q) > q^{2n−4}/log_{p}(q) − n,

and so 4n^{2} +3n+2 > q^{2n−5}. This fails for n ≥ 5 and all odd q, so we have a contradiction in the case q ≠ 3. On the other hand, if q = 3 then log_{p}(q) < 2, and so (6) implies that 4(2n^{2} + n + 1) > 3^{2n−2} − 3^{n−2} > 3^{2n−3}. This fails for n ≥ 4, and then since Ω_{7}(3) ∈ L we have a contradiction.

Suppose next that \overline{H} ≅ P Ω_{2n}^{+}(q), n ≥ 4. Then 10.8 and 10.4 yield (7)

2(2n^{2} − n + 2)log_{p}(q) > \{ \begin{matrix}q(q^{2n−2} − 1)/(q^{2} − 1) + q^{n−1} − n  q ≠ 2, 3, and \\ q(q^{2n−2} − 1)/(q^{2} − 1) − (q^{n−1} − 1)/(q − 1) − 7δ_{2,q}  q even. \end{matrix}

Suppose that q ≠ 2 or 3. As log_{p}(q) < q we then have

2(2n^{2} − n + 2) + n > q^{2n−4} + q^{n−2},

This fails to hold if n = 4 and q = 4, and then also if n ≥ 4 and q ≥ 4. Thus, q = 2 or 3. As log_{p}(q) < q, (7) yields

2(2n^{2} − n + 2) > q^{2n−4} − q^{n−3} − 7δ2, q/q > q^{2n−5} − 4.

These inequalities fail if n = 6 and q = 2, and then also if n ≥ 6 and q ≥ 2. They also fail if n = 5 and q = 3. As Ω_{8}^{+}(2) ∈ L, we are then reduced to the case n = 4 and q = 3, and to the case n = 5 and q = 2. One may then check directly from (7) that neither of these cases may occur.

Suppose next that \overline{H} ≅ P Ω_{2n}^{−}(q), n ≥ 4. Then 10.8 and 10.4 yield

(8)  2(2n^{2} − n + 2)log_{p}(q) > q(q^{2n−2} − 1)/(q^{2} − 1) − q^{n−1} + 2.

Then 2(2n^{2} − n + 2) > q^{2n−4} − q^{n−2}, and this fails if n ≥ 6, or if q ≥ 3. As P Ω_{8}^{−}(2) ∈ L, we are reduced to the case where n = 5 and q = 2, and in this case one checks directly that (8) fails to hold.

It remains to consider the exceptional groups and the groups ^{3}D_{4}(q). The estimates that we obtain from 10.4 and 10.8 are as follows.

3D_{4}(q): 58log_{p}(q) > q^{3}(q^{2} − 1).

E_{6}(q) or ^{2}E_{6}(q): 158log_{p}(q) > q^{9}(q^{2} − 1).

E_{7}(q): 268log_{p}(q) > q^{15}(q^{2} − 1).

E_{8}(q): 498log_{p}(q) > q^{27}(q^{2} − 1).

F_{4}(q): 106log_{p}(q) > q^{6}(q^{2} − 1) if q is odd, or q^{7}(q^{3} − 1)(q − 1)/2 if q is even.

2F_{4}(q): 54log_{p}(q) > \sqrt{q/2}(q^{4})(q − 1).

G_{2}(q) (q ≥ 3): 30log_{p}(q) > q(q^{2} − 1).

2G_{2}(q) (q ≥ 27 an odd power of 3): 14log_{p}(q) > q(q − 1).

Sz(q) (q ≥ 8 an odd power of 2): 12log_{p}(q) > \sqrt{q/2}(q − 1).

In the case of Sz(q) we may take q ≥ 32, as Sz(8) ∈ L. None of the above inequalities holds for q = 5, and then none holds for q ≥ 5. For p = 3, only the inequality for G_{2}(3) is valid, and for q = 2, only the inequalities for ^{3}D_{4}(2) and ^{2}F_{4}(2) are valid. All three of these groups are in L, so we have a final contradiction at this point, proving Theorem 10.2.

Section 11: Theorems 1 through 6

We ask the reader to recall the definitions of Q(Y, V), q(Y, V) and Q^{∗}(Y, V), from the Introduction. Recall also that, for any group G, a component of G is a subnormal quasisimple subgroup of G, and that E(G) is the product of all the componentsof G.

11.1 Lemma. Let p a prime, G a group whose order is divisible by p, and let S be a Sylow p-subgroup of G. Assume that O_{p}(G) = 1 and that S is contained in a unique maximal subgroup of G. Set H = O^{p}(G) and set \overline{G} = G/Φ(G). Then one of the following holds.

(i) H is an r-group for some prime r ≠ p, and S acts irreducibly on H/Φ(H).

(ii) \overline{H} is a direct product of non-abelian simple groups, permuted transitively by \overline{S}. Moreover, for any component \overline{L} of \overline{G}, N\overline{S}(\overline{L}) is contained in a unique maximal subgroup of N\overline{S}(\overline{L})\overline{L}.

Moreover, if there exists a faithful 𝔽_{p}G-module V with q(S, V) ≤ 2 then, in case (ii), we have H = E(G).

Proof. Denote by M be the unique maximal subgroup of G containing S, and set F = ∩{M^{g}}_{g∈G}. Then G = N_{G}(S ∩ F)F. As O_{p}(G) = 1 and G ≠ M it follows that S ∩ F = 1, and so F is a p′-group. Let M^{∗} be any maximal subgroup of G. If F ≰ M^{∗} then G = F M^{∗}, so S^{g} ≤ M^{∗} for some g ∈ G, and this implies that M^{∗} is conjugate to M. Thus F ≤ M^{∗}, and F = Φ(G). Now let N be the inverse image in G of a minimal normal subgroup of \overline{G}. Then NS = G, and so N = H. Suppose that N is a p′-group. For any prime divisor r of |N| there then exists an S-invariant Sylow r-subgroup of N, and it follows that N is an r-group. Uniqueness of M then implies that S acts irreducibly on N/Φ(N), and so (i) holds in this case. On the other hand, suppose that N is not a p′-group. Here \overline{N} is not a p-group, as N ≠ F, so \overline{N} is a direct product of simple groups.

Suppose that (ii) is false. We may then assume without loss of generality that F = 1. The uniqueness of M implies that S acts transitively on the set of components of N. Fix a component L of N, and suppose that there exist distinct maximal subgroups X_{1} and X_{2} of N_{S}(L)L containing N_{S}(L). Set X_{i}^{∗} = ⟨S^{X_{i}}⟩. Then X_{i}^{∗} is a proper subgroup of G, and ⟨X_{1}^{∗}, X_{2}^{∗}⟩ = G. As this contradicts the uniqueness of M, (ii) is proved.

Suppose now that G is non-solvable, and that [Φ(G), H] ≠ 1. Setting Q = Φ(G), we then have Q = F (G) = F^{∗}(G). Suppose also that there is a faithful 𝔽_{p}G-module V with q(G, V) ≤ 2, and set q = q(S, V). Without loss of generality, we may assume that V is irreducible for G. Denote by A the set of elements A of Q^{∗}(S, V) of minimal order, and set G_{0} = ⟨A⟩. Then H ≤ G_{0}. Let K the set of subgroups of Q given by 4.6, and set D = [Q, G_{0}]. Every member of K has a solvable automorphism group, so if H fixes every member of K we obtain [D, H] = 1. But in that case we have [Q, H, H] = 1, and then [Q, O^{p′}(H)] = 1, and so H = 1. We conclude that H acts non-trivially on K, and then 4.6(d) implies that |A| = p for any A ∈ A.

We have p ≤ 3, as follows from 4.1. Suppose first that p = 2, and let K ∈ K. By the definition of K (at the start of the proof of 4.6) there then exists A ∈ A with K = [K, A]. Then 4.6(f) says that every element of K is A-invariant. But H ≤ ⟨A^{G}⟩, so it follows that, in fact, H acts trivially on K. Thus, we have a contradiction if p = 2.

Suppose that p = 3. Fix A ∈ A, and set L = ⟨A^{H}⟩. Then L = ⟨A^{L}⟩ and F^{∗}(L) = F (L). There then exists an irreducible L-submodule U of V such that [F (L), L] acts non-trivially on U. Setting \overline{L} = L/C_{L}(U), we then have F^{∗}(\overline{L}) = F (\overline{L}), and we have \overline{A} ≅ A as |A| = 3. Set \overline{Y} = [F (\overline{L}), \overline{A}]. Then [C2, Theorem A] implies that F (\overline{L}) is a 2-group of symplectic type, that \overline{Y} is a quaternion group, and that U is an irreducible module for F (\overline{L}). Further, if the width of a largest extraspecial subgroup \overline{X} of F (\overline{L}) is n, then U is a direct sum of 2^{n−1} natural SL(2, 3)-modules for \overline{Y} \overline{A} if \overline{X} = F (\overline{L}), and a direct sum of 2^{n} such modules if \overline{X} ≠ F (\overline{L}). As q(G, V) ≤ 2, \overline{A} is generated by a 2-transvection, so we conclude that n ≤ 2, and that \overline{X} = F (\overline{L}) if n = 2. As \overline{L} is non-solvable, it follows that n = 2, F (\overline{L}) ≅ Q_{8} ∘ D_{8}, and \overline{L}/F (\overline{L}) ≅ Alt(5). Here \overline{L}/F (\overline{L}) is incident with a component of G/Q, so outcome (ii), above, implies that \overline{A} is contained in a unique maximal subgroup of \overline{L}/F (\overline{L}). But such is not the case, and we have a contradiction at this point. Thus, the final statement in the lemma holds for all p. □

11.2 Proposition. Let G be a group with O_{p}(G) = 1 and let V be a faithful 𝔽_{p}G-module. Let A ∈ Q(S, V), let K be a component of G, put E = ⟨K^{A}⟩, and assume that K ≠ E. Then p = 2, and one of the following holds:

(i) |A/C_{A}(E)| = 2, or

(ii) E ≅ Ω_{4}^{+}(2^{n}) for some n, n ≥ 2, and [V, E] is a direct sum of natural O_{4}^{+}(2^{n})- modules for EA.

Proof. Without loss, G = ⟨K, A⟩ and C_{A}(E) = 1. Let Ω = {K_{1}, ⋯, K_{t}} be the set of components of G, with K = K_{1}. Suppose first that |A| = p, and let 1 ≠ a ∈ A. Then t = p, and for each prime factor q of |K| there is an A-invariant Sylow q-subgroup Q of E with [Q, A] ≠ 1. If p is odd then Theorem 3.8.1 in [G] says that QA involves SL(2, p), and so p = 3 and q = 2. But then K is a {2, 3}-group, and hence K is solvable by a theorem of Burnside. Thus p = 2, and (i) holds. On the other hand, if |A| > p, then Theorem 2 of [C1] yields (ii). □

11.3 Lemma. Assume Hypothesis 4, and let A ∈ Q_{∗}(S, V). Suppose further that there exists a component X of G not normalized by A. Set K = ⟨X^{A}⟩, and let K^{S} = {K_{1}, ⋯, K_{r}}. Set V_{i} = [V, K_{i}], 1 ≤ i ≤ r. Then the following hold:

(i) p = 2, H = K_{1} × ⋯ × K_{r}, and V = C_{V} (H) ⊕ V_{1} ⊕ ⋯ ⊕ V_{r}, where K_{i} ≅ Ω_{4}^{+}(2^{n}), and V_{i} is a natural orthogonal module for K_{i}.

(ii) Each K_{i} is invariant under ⟨Q(S, V)⟩, and K_{i}A/C_{K_{i}A}(K_{i}) ≅ O_{4}^{+}(2^{n}).

Proof. Set U = [V, K], and let A_{0} be a complement to C_{A}(K) in A. As q(A, V) ≤ 2, 3.4 implies that q(A_{0}, U) ≤ 2. Suppose that |A_{0}| > 2. Then 11.2 yields p = 2, K ≅ Ω_{4}^{+}(2^{n}), and V = C_{V} (K) ⊕ U, where U is a direct sum of natural orthogonal modules for KA_{0}. Set q = 2^{n}, and let U_{0} be an irreducible KA_{0}-submodule of U. Then |U_{0}| = q^{4}, and |A_{0}| ≤ 2q. Further, no element of A_{0} ∩ K induces a transvection on U_{0} over 𝔽_{q}, and so |U_{0}/C_{U_{0}}(A_{0})| ≥ q^{2}. Thus |A_{0}|^{2} ≤ 4|U_{0}/C_{U_{0}}(A_{0})|, and if U ≠ U_{0} it follows that |A_{0}|^{2} ≤ 4|U/C_{U}(A_{0})|^{1/2}. But if U ≠ U_{0} we have 4|U/C_{U}(A_{0})|^{1/2} < |U/C_{U}(A_{0})|, and so q(A, V) > 2. We therefore conclude that U = U_{0}. We have H = ⟨X^{S}⟩, by 11.2, so H = ⟨K^{S}⟩. Let {K_{1}, ⋯, K_{r}} be the set of conjugates of K in G, and put V_{i} = [V, K_{i}]. Then [U_{i}, K_{j}] = 0 for i ≠ j, and so (i) holds. Part (ii) follows from 11.2.

Suppose next that |A_{0}| = 2, and let a be the involution in A_{0}. Then |U/C_{U}(a)| ≤ 4. As O_{2}(G) = 1, there is a Sylow 2-subgroup T of K of the form R × R^{a}, where R is a Sylow 2-subgroup of X, and R has 2-rank at least 2. Then C_{R}(a) has 2-rank at least 2, and so C_{R}(a) contains an involution b such that [U, a, b] = 0. Setting A_{1} = ⟨a, b⟩, we then have [U, A_{1}, A_{1}] = 0, and so 11.2 implies that K ≅ Ω_{4}^{+}(2^{n}) and that U is a direct sum of natural orthogonal modules for KA_{0}. As in the preceding paragraph, let U_{0} be an irreducible KA_{0}-submodule of U. As 2^{n} ≥ 4, we then have |A_{0}|^{2} ≤ |U_{0}/C_{U_{0}}(A_{0})|, and hence U = U_{0}. Now (i) and (ii) follow, as above. □

We may now assemble the proofs of Theorems 1 through 6.

Theorem 1. Assume Hypothesis 4. Then there exists a subgroup K of H, unique up to conjugation, such that, upon setting U = [V, K], the following conditions hold.

(a) H = K_{1} × ⋯ × K_{r} where {K_{1}, ⋯, K_{r}} = K^{S}.

(b) We have [V, K_{i}, K_{j}] = 0 whenever i ≠ j.

(c) One of the following holds:

(i) K ≅ O^{p}(SL(2, p^{n})), n ≥ 1, and U/C_{U}(K) is a natural SL(2, p^{n})-module for K, or a direct sum of two natural modules for K.

(ii) K ≅ O^{p}(O_{4}^{ϵ}(p^{n})), n ≥ 1, ϵ = ±1, and U is a natural orthogonal module for

K.

(iii) K ≅ O^{p}(SU(3, p^{n})) and U is a natural module for K.
(iv) p = 2, K ≅ O^{2}(Sz(2^{n})), and U is a natural module for K.

(v) p = 2, A ≤ C_{G}(K)K, K ≅ SL(3, 2^{n}) (resp. O^{2}(Sp(4, 2^{n})) and U is the direct sum of a natural and a dual module (resp. a natural and a contragredient module) for K. Moreover, there exists g ∈ N_{S}(K) such that g interchanges, by conjugation, the two maximal subgroups of K containing S ∩ K, and if K ≅ O^{2}(Sp(4, 2^{n})) then A is conjugate to Z(S).

(vi) p = 2, K ≅ Alt(2^{n} + 1), n ≥ 3, and U is a natural module for K, or a

direct sum of two natural modules for K. (vii) p = 2, K ≅ Alt(9), and U is a spin module for K, of dimenson 8 over 𝔽_{2}. viii p = 3, K ≅ SL(2, 5), and U is a natural SL(2, 9)-module for K.

Moreover, if K is not invariant under Q^{∗}(S, V) then p = 2, K ≅ ℤ_{3}, |U| = 4, and q(S, V) = 2.

Proof. If G is solvable then the theorem follows from 11.1 and 4.6. So assume that G is non-solvable. Then 11.1 yields H = E(G). Let A ∈ Q(S, V). If there is a component X of G such that X is not A-invariant, then the theorem follows from 11.3. So assume that X is A-invariant. If X ≅ SL(2, p^{n}), [V, X] is a direct sum of two natural modules for X, and there exists a unique conjugate Y of X such that [X, Y] = 1 and [V, X, Y] ≠ 0, set K = XY, and otherwise set K = X. Let {K_{1}, ⋯, K_{r}} be the set of conjugates of K under S. By 11.1 we then have H = K_{1} ⋯ K_{r}. Set V_{i} = [V, K_{i}], and set U = [V, K].

As O_{p}(G) = 1 we may assume that A and X have been chosen so that [X, A] ≠ 1. Choose a complement A_{0} to C_{A}(K) in A. Then q(A_{0}, U) ≤ 2, by 3.4. If K = X set A_{1} = A_{0}, and otherwise take A_{1} to be a complement in A_{0} to C_{A_{0}}(X). Then also q(A_{1}, [V, X]) ≤ 2. If p = 3, X ≅ SL(2, 5), and [V, X] is a natural SL(2, 9)-module for X then K = X and the theorem holds (with (c)(viii)), so we may assume that this special case does not obtain. Then 5.5 implies that X/Z(X) is a group of Lie type in characteristic p, or that p = 2 and X is an alternating group Alt(2^{n} + 1). If X/Z(X) is of Lie type in characteristic p then 5.9 says that the pair (X, [V, X]) is described by 6.10 or by 7.5, while if X ≅ Alt(2^{n} + 1) then 5.7 provides a description of (X, [V, X]). In particular, if [V, X] is reducible, and is not a direct sum of two non-isomorphic irreducible submodules, then X ≅ SL(2, p^{n}) or Alt(2^{n} + 1) and X is a direct sum of two natural modules for X. It follows that if End_{X}([V, X]) contains a subgroup isomorphic to X, then [V, X] is a direct sum of two isomorphic irreducible modules, and if W is an irreducible submodule of [V, X] then the dimension of W over End_{X}(W) is equal to 2. This is the case only if X ≅ SL(2, p^{n}) and [V, X] is a direct sum of two natural modules.

Suppose that [V, X, Y] ≠ 0 for some conjugate Y of X with Y ≠ X. Then the preceding discussion yields XY ≅ Ω_{4}^{+}(p^{n}), [V, X] is a natural orthogonal module for XY, and X and Y are the only conjugates of X which act non-trivially on [V, X]. By definition, we then have K = XY and U = [V, X]. The theorem follows in this case (with (c)(ii)), so we may assume henceforth that [V, X, Y] = 0 for any conjugate of X distinct from X. Then also K = X.

If K/Z(K) is a group of Lie type in characteristic p, the theorem now follows from 6.10 and 7.5. Thus, we are reduced to the case where p = 2 and K ≅ Alt(2^{n} + 1). Since Alt(5) is a group of Lie type in characteristic 2, we may assume that n ≥ 3. The theorem then follows from 5.7 (with either part (vi) or part (vii) of (c) obtaining). □

In order to prove Theorems 2 and 3, let the hypotheses be as in Theorem 1, and fix A ∈ Q^{∗}(S, V). Suppose that q^{∗}(A, V) < 2, and let K and U be as in Theorem 1, with [K, A] ≠ 1. Let A_{0} be a complement in A to C_{A}(K). Then q(A_{0}, U) < 2 by 3.4, and so we do not have the case given by (c)(viii) in Theorem 1. Thus, K is of Lie type in characteristic p, or p = 2 and K ≅ Alt(2^{n} + 1) with n ≥ 3.

Suppose that K is quasisimple. If the Lie rank of K/Z(K) is 1 then 6.10 implies that H ≅ SL(2, p^{n}), or that p = 2 and H ≅ Ω_{4}^{−}(2^{n}) or Alt(5), with U a natural module for K in each of these cases. Moreover, if H ≅ Omega_{4}^{−}(2^{n}) then 6.10 says that KA/C_{KA}(K) ≅ O_{4}^{−}(2^{n}). Thus, Theorem 2 holds in these cases. If the Lie rank of K/Z(K) is greater than 1 then 5.9 and 7.5 yield p = 2 and K ≅ SL(3, 2^{n}), and Theorem 2 holds in this case. Now suppose that K ≅ Alt(2^{n} + 1), n ≥ 3. Then 5.7 applies. In particular, the condition q(A_{0}, U) < 2 excludes the case where n = 3 and U is a spin module for K, and excludes also the case where U is a direct sum of two natural modules for K. Thus, U is a natural module for K, and we have Theorem 2 in this case.

Now assume that K is not quasisimple, and that K is non-solvable. Then Theorem 1 yields K ≅ Ω_{4}^{+}(p^{n}). Suppose that A fixes each of the components of K. As q(A_{0}, U) < 2 it is then easy to see that |A| > p, and so A_{0} ∩ C_{G}(K)K ≠ 1. Quadratic action then forces A_{0} ≤ C_{G}(K)K, and we again appeal to the quadratic action of A_{0}, to conclude that |A_{0}| ≤ p^{n}. But dim_{𝔽_{p^{n}}} (C_{U}(a)) ≤ 2 for any a ∈ A_{0}^{#}, so q(A_{0}, U) ≥ 2 in this case. We therefore conclude that p = 2 and that A_{0} interchanges the two components of K. Then 11.2 yields KA/C_{KA}(K) ≅ O_{4}^{+}(2^{n}). This is outcome (ii) of Theorem 2.

Suppose finally that K is solvable. If K is not A-invariant then Theorem 1 yields K ≅ ℤ_{3} and |U| = 4, and this result is contained in outcome (i) of Theorem 2. So assume that K is A-invariant. As q(A, V) < 2, 4.5(a) implies that |U| = p^{2}, and that K ≅ O^{p}(SL(2, p)) (p = 2 or 3). We have thus eliminated all the cases in Theorem 1 which do not remain in the statement of Theorem 2, and thus Theorem 2 is proved.

Now suppose that q(S, V) ≤ 1. If K is solvable then Theorem 3 follows from 4.5(b), so assume that K is non-solvable. We have q(A_{0}, U) ≤ 1, so 7.5 implies that K is not isomorphic to SL(3, 2^{n}) with p = 2. If p = 2 and K ≅ Ω_{4}^{+}(2^{n}), n ≥ 2, then Theorem 2 yields |A_{0}| = 2^{n+1} and |U/C_{U}(A_{0})| = 2^{2n}, and so q(A_{0}, U) > 1 in this case. If p = 2 and K ≅ Alt(2^{n} + 1), n ≥ 3, then 5.7 implies that A_{0} is generated by a set of commuting transvections on the natural module U, so that KA/C_{KA}(K) ≅ Sym(2^{n} +1) and q(A_{0}, U) = 1. Then 1 = q(A, V) = q(S, V), and ⟨A^{H}⟩ is isomorphic to a direct product of copies of Sym(2^{n} + 1). Then also ⟨A^{G}⟩ is the point-wise stabilizer in G of {K_{1}, ⋯, K_{r}}, and is a direct product of copies of Sym(2^{n} + 1). Thus, Theorem 3 holds in this case. Finally, the last case given by Theorem 2 is that in which K ≅ SL(2, p^{n}) and U is a natural module for K. Again, we have q(A_{0}, U) ≤ 1, and it follows that A_{0} is incident with a Sylow p-subgroup of KA/C_{KA}(K) and that q(A_{0}, U) = 1. We then obtain q(S, V) = 1, and ⟨A^{G}⟩ = H. This completes the proof of Theorem 3.

Theorem 4. Assume Hypothesis 3, and assume that H is a quasisimple group of Lie type in characteristic p. Let A be an F 2-offender on V. Then one of the following holds.

(a) H ≅ SL(2, p^{n}) and one of the following holds. (i) V is a natural SL(2, p^{n})-module for H. (ii) V has an H- submodule U such that both U and V/U are natural SL(2, p^{n})-

modules for H. (iii) n is even and V is a natural Ω_{4}^{−}(p^{n/2})-module for H.

(b) H ≅ SU(3, p^{n}) and V is a natural module for H.

(c) p = 2, H ≅ Sz(2^{n}), and V is a natural module for H.

(d) p = 2, H ≅ SL(3, 2^{n}) (resp. O^{2}(Sp(4, 2^{n})) and V is the direct sum of a natural and a dual module (resp. a natural and a contragredient module) for H. Moreover, there exists g ∈ N_{S}(K) such that g interchanges, by conjugation, the two maximal subgroups of H containing S ∩ H. If A is a quadratic F 2-offender then A ≤ H, and if also H ≅ Sp(4, 2^{n}) then A is conjugate to Z(S).

Proof. Immediate from 5.9, 6.8, 6.9, and 7.5. □

Theorem 5. Assume Hypothesis 4′, and assume that H/Z(H) is isomorphic to Alt(n), n ≥ 5. If p = 2 assume that n is odd. Then one of the following holds.

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

(ii) p = 2, H ≅ Alt(n), n = 5, 7, or 9, and V is a spin module for H (of dimension 4, 4, or 8, respectively). Moreover, if n = 9 then A is the direct product of two quadratic fours groups in H.

(iii) |A| = p = 3, G ≅ Alt(n), n ≠ 6, and V is a natural module for G. Moreover, A

is generated by a 3-cycle. (iv) |A| = p = 3, H ≅ SL(2, 5), and V is isomorphic to the natural SL(2, 9)-module

for H.

(v) p = 3, H ≅ Alt(9), |A| = 27, and V is a spin module for H (of dimension 8 over 𝔽_{3}). Moreover, we have |A|^{2} = |V/C_{V} (A)|.

Proof. If p is odd then 8.3 yields one of the outcomes (iii) through (v) above. So assume that p = 2 (and hence also that n is odd). Suppose that there is more than one nontrivial irreducible constituent for HA in V. As q(A, V) ≤ 3/2, there there then exists a non-trivial irreducible constituent U for HA in V such that q(A, U) ≤ 3/4. By the Timmesfeld Replacement Theorem there then exists B ∈ Q(S, U) with q(B, U) ≤ 3/4, and this is contrary to 5.7. Thus, there is a unique non-trivial constituent in V for HA, and then V is irreducible for HA, and hence also for G, by 1.2.

If n = 5 then H ≅ L_{2}(4), and we obtain (i) or (ii) from Theorem 4. So we may assume that n ≥ 7. By 9.1 there then exists a fours group F in G with A ∩ F ≠ 1 and with [V, F, F] = 0. Assuming that V is not a natural module for G, it then follows from [MS2, Theorem 4] that V is a spin module for HF. Let a be a non-identity element of A ∩ F, and identify H with Alt(n). After conjugation, we have F = ⟨(1 2)(3 4), (1 3)(2 4)⟩, and F is the unique quadratic fours group in G containing a. Denote by k the greatest integer in n/3. Then H contains a subgroup E generated by k pairwise disjoint 3-cycles. For any 3-cycle x in H we have C_{V} (x) = 0, so the dimension of V over 𝔽_{2} is a multiple of 2^{k}. In particular, if dim(V) < 8 then n = 7, and we have (ii). So we may assume that dim(V) ≥ 8 (and that n ≥ 9).

As a inverts a 3-cycle, we have also dim(V) = 2dim(C_{V} (a)). Then |V/C_{V} (a)| ≥ 16, and then since |A|^{3/2} ≥ |V/C_{V} (A)| ≥ |V/C_{V} (a)| we have |A| ≥ 8. If C_{V} (a) = C_{V} (A), and then every fours group in A which contains a is quadratic, contrary to the uniqueness of F. We therefore conclude that C_{V} (a) ≠ C_{V} (A), and this yields |A| ≥ 16.

Let K be the component in C_{H}(a), and let A_{0} be a complement to C_{A}(K) in A. Then |A_{0}| ≥ 4. For any 3-cycle y in K we have C_{[V,a]}(y) = 0, so [V, a] = [V, a, K], and Theorem 4 in [MS2] implies that every irreducible constituent for K in [V, a] is a spin module. As A_{0} acts quadratically on [V, a], by hypothesis, we conclude that |A_{0}| = 4. Then F ≤ A, we may take A_{0} to be a quadratic fours group contained in K, and we have A = F × A_{0}. Now |A|^{3/2} = 64, and since C_{V} (a) ≠ C_{V} (A) we have dim(C_{V} (a)) ≤ 5. Then dim(C_{V} (a)) = 4, and the faithful action of K on C_{V} (a) yields n = 9 or 11. Now dim(V) = 8, and since 11 is not a divisor of |L_{8}(2)| we have n = 9, and (ii) holds. □

Theorem 6. Assume Hypothesis 4′, and assume that S is contained in a unique maximal subgroup of G. Assume also that H/Z(H) is a quasisimple group of Lie type in characteristic different from p, or a sporadic group, and that there exists no isomorphism of H/Z(H) with a group of Lie type in characteristic p. Then |A| = p = 3, G ≅ Sp(6, 2), V has dimension 7 over 𝔽_{3}, and |A|^{2} = |V/C_{V} (A)|.

Proof. Immediate from 2.2 and 10.2. □

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