QUADRATIC PAIRS

Andrew Chermak

Department of Mathematics Kansas State University Manhattan KS, 66502

February, 2002

Introduction

This paper revisits the work of John Thompson and Chat-Yin Ho (cf. [Th], [H1], [H2]) from the early 1970’s, on the “Quadratic Pairs”. Thus, we will be concerned with the following situation.

Hypothesis 1.0. G is a finite group, p is an odd prime, and V is a faithful, irreducible module for G over the field 𝔽_{p} of p elements. Further, there is a subgroup A of G such that G = ⟨A^{G}⟩ and such that [V, A, A] = 0.

Recall that a group H is quasisimple if H = [H, H] and H/Z(H) is simple. In [Ch] the author considered the case in which, in addition to Hypothesis 1.0, it is assumed that G has no quasisimple subnormal subgroups. (Such subgroups are the components of G.) In this paper, we take up the alternative case, in which G has at least one component. By lemma 1.4 in [Ch] this amounts to making the following stronger hypothesis.

Hypothesis 1.1. In addition to Hypothesis 1.0, we assume that there is a normal, quasisimple subgroup H of G, such that C_{G}(H) = Z(G).

As an addendum to Hypothesis 1.1 we assume that, in fact, H/Z(H) is one of the groups given by the Classification of the Finite Simple Groups. That is to say, H/Z(H) is isomorphic to an alternating group, a group of Lie type, or one of twenty-six sporadic groups. We assume also that we have complete information about the Schur multipliers of the finite simple groups, so that not only H/Z(H), but H itself, is a “known” group. We shall prove the following result.

Typeset by AMS-TEX

Theorem A. Assume Hypothesis 1.1 and the above addendum. Then Z(G) ≤ H, and either G = H is a group of Lie type in characteristic p, or |A| = p = 3, and one of the following holds.

(a) G ≅ P GU(n, 2), n ≥ 5,

(b) |Z(G)| = 2, \overline{G} ≅ Alt(n), n ≥ 5 and n ≠ 6, or

(c) |Z(G)| = 2, and \overline{G} is isomorphic to D_{4}(2), G_{2}(4), Sp(6, 2), Co_{1}, Suz, or J_{2}.

We rely largely on [GLS3] for information about “known” simple or quasisimple groups, but we have made an attempt to keep to a minimum the amount of detailed information that we draw upon, and to rely on general principles as far as is practicable. For the sporadic groups, by their very nature (or by the nature of the current state of understanding of these groups) we are forced to take an opportunistic approach, but we can get by with information on conjugacy classes, centralizers, Schur multipliers, and the fact that none of the sporadic groups have outer automorphisms of odd order. That is, we require only “basic” information, such as can be found either in the ATLAS of Finite Groups [CCNPW] or in Table 5.3 in [GLS3]. Concerning the simple groups of Lie type, the situation is turned on its head, and we have made it a point to avoid appealing to the detailed information (provided in [GLS3] and elsewhere) concerning the structure of centralizers, and to rely only on information derivable from the most basic results relating the simple groups of Lie type to simple algebraic groups, and from the Coxeter diagrams of these various groups. As has already been mentioned, we take for granted the determination of the Schur multipliers. Aside from that, we need the theorem of Borel and Tits which states that p-local subgroups of simple groups of Lie type in characteristic p are contained in parabolic subgroups, and are p-constrained, and we need some results, due to Steinberg, concerning automorphisms of the groups of Lie type. Other properties of the groups of Lie type that will be needed here will be developed in section 3, below.

Whenever Hypothesis 1.1 is in effect, we denote by H the unique normal, quasisimple subgroup of G, and we set \overline{G} = G/Z(G).

By a quadratic module for a group X, we mean a module U such that [U, A, A] = 0 for some non-identity subgroup A of X such that X = ⟨A^{X}⟩. We then say that A is a quadratic subgroup of X.

In the exceptional cases (a) through (c) of Theorem 1.2, we determine the possible conjugacy classes of quadratic subgroups of order 3. The result is as follows.

Theorem B. Assume Hypothesis 1.1, and assume that G is not a group of Lie type in characteristic p. Let A be a quadratic subgroup of order 3 in G.

(a) Suppose that G = P GU(n, 2), n ≥ 5. Let ϕ : GU(n, 2) ⟶ P GU(n, 2) be the canonical homomorphism, and let U be the natural module for GU(n, 2) over the field 𝔽_{4}. Then there is an element a^{∗} of GU(n, 2) with ⟨ϕ(a^{∗})⟩ = A, such that C_{U}(a^{∗}) has codimension 1 in U.

(b) If \overline{G} ≅ Alt(n), n ≠ 6, then \overline{A} is generated by a 3-cycle in \overline{G}.

(c) If \overline{G} ≅ D_{4}(2) then C\overline{G}(\overline{A}) ≅ 3×U_{4}(2), and A lies in a maximal subgroup M of G such that O_{2}(M) is an extraspecial 2-group of order 2^{7}, with M/O_{2}(M) ≅ L_{4}(2),

and such that [O_{2}(M), A] is a quaternion group.

(d) If \overline{G} ≅ G_{2}(4) then C\overline{G}(\overline{A}) ≅ SL(3, 4).

(e) If \overline{G} ≅ Sp(6, 2) then C\overline{G}(\overline{A}) ≅ 3 × Sp(4, 2).

(f) If \overline{G} ≅ J_{2} then C\overline{G}(\overline{A}) ≅ 3^{·}Alt(6).

(g) If \overline{G} ≅ Suz then C\overline{G}(\overline{A}) ≅ 3^{·}U_{4}(3).

(f) If \overline{G} ≅ Co_{1} then C\overline{G}(\overline{A}) ≅ 3^{·}Suz. Moreover, in every case except (c), the conjugacy class of A in G is uniquely determined by the given conditions. In case (c) the class of A is uniquely determined up to conjugacy in Aut(G).

We remark, in connection with Theorem A, that there are descending chains of groups

2^{·}Co_{1} ≥ 6^{·}Suz ≥ 2^{·}G_{2}(4) ≥ 2^{·}J_{2},

and

2^{·}Co_{1} ≥ 2^{·}D_{4}(2) ≥ 2^{·}Sp(6, 2).

Denote by Λ the (24-dimensional) Leech Lattice, with automorphism group 2^{·}Co_{1}. We will show in section 9, below, that Λ/3Λ is a quadratic module for 2^{·}Co_{1}, and a quadratic module also for each of the groups in each of the above chains of subgroups. Thus, all of the groups listed in part (c) of Theorem A possess quadratic modules in characteristic 3.

Also, the groups G listed in parts (a) and (b) of Theorem A have quadratic modules in characteristic 3. For 2^{·}Alt(n) such modules have been classified in [M]. For the unitary groups in characteristic 2, and also for the exceptional groups in (c), a complete determination of the quadratic modules appears in [GMST]. The quadratic modules for the groups of Lie type in characteristic p, p odd, were determined long ago, in [PS].

The following corollary to Theorems A and B is useful for certain applications.

Corollary C. Assume Hypothesis 1.1, and assume that G is not a group of Lie type in characteristic p. Assume also that there exists a quadratic subgroup A of G such that |A|^{2} ≥ |V/C_{V} (A)|. Then p = 3, G ≅ SL(2, 5), and V is a natural SL(2, 9)-module for G.

In proving Theorem A, we can reduce immediately to the case where G = HA, as the following lemma shows.

Lemma 1.2. Let G be a minimal counter-example to Theorem A. Then G = HA.

Proof. Set G_{0} = HA. As H is quasisimple we have G_{0} = ⟨A^{G_{0}}⟩, and evidently V is a quadratic module for G_{0}. By Clifford’s theorem, there exists an irreducible H-submodule U of V on which H acts faithfully. Then also H acts faithfully on any irreducible G_{0}- submodule V_{0} of ⟨U^{A}⟩. As V is irreducible for G, Z(G) is a 3′-group, and then since C_{G}(H) = Z(G), by 1.1, we have C_{G_{0}}(V_{0}) = 1. Thus, Hypothesis 1.1 is satisfied by G_{0} and V_{0} in place of G and V. Suppose now that G ≠ G_{0}. Then A ≰ H, and since G is a minimal counter-example to Theorem A, we may appeal to Theorem A for the structure of G_{0}. The condition that A not be contained in H then yields G_{0} ≅ P GU(n, 2), where 3 divides n. But then G_{0} ≅ Aut(H), and so G = C_{G}(H)G_{0}. Hypothesis 1.1 then yields G = Z(G)G_{0}. As G = ⟨A^{G}⟩ = [G, A]A, it follows that G = G_{0}A, and so G = G_{0}. □

The structure of this paper is as follows. We begin, in section 2, by collecting together some general results about quadratic action, including a lemma of Meierfrankenfeld (lemma 2.8, below) which gives a useful characterization of the groups SL(2, p) for p > 3.

Section 3 concerns properties of quasisimple groups of Lie type. As indicated above, we have found it convenient to draw on [GLS3] for basic background material. From this background we obtain results on automorphisms, on centralizers, and on the action of certain automorphisms on Schur multipliers.

In section 4 we use the results of section 3 in order to show that, if \overline{G} is a group of Lie type (possibly of characteristic p) then \overline{a} induces an inner-diagonal automorphism on \overline{G}. We also show that if \overline{G} is an alternating group, then \overline{A} is generated by a 3-cycle, and |Z(G)| = 2. Thus, in the succeeding sections, we need only be concerned with groups of Lie type (possibly extended above by diagonal automorphisms) in characteristic different from p, and with sporadic groups.

Section 5 provides a quick treatment of the case where p is greater than 3. (Of course, the result here is not new. See [Sa] for a treatment which is based on Aschbacher’s classification [A1] of groups of Lie type in odd characteristic. Much more recently, one has Timmesfeld’s work [Ti], where the groups are not assumed to be finite, but in which the question addressed by lemma 2.8 below is left hanging.) From then on, we assume that p = 3, and section 6 is devoted to the case where \overline{H} is of Lie type in characteristic different from 3. Section 7 treats the case where p = 3 and \overline{H} is a sporadic group. Finally, sections 8 and 9 provide proofs for Theorem B and Corollary C, and establish that all of the “exceptional” groups that arise in Theorem A do indeed have quadratic modules.

It should be emphasized that this paper should in no way be construed as somehow finessing the work of Thompson and Ho. The work of Thompson was begun before the Classification was anywhere within sight, and before there was even any strong reason to believe that only a small number of finite simple groups remained to be discovered. Thompson’s work (and that of Ho) may be understood as an attempt to continue the momentum towards the Classification that had begun with the Odd Order Paper and the N-Group Paper. Their work was dropped when a powerful program leading to the Classification began to take shape. On the other hand, a new approach to at least one aspect of the Classification (concerning groups having a “generic” prime characteristic) is currently developing, under the leadership of Meierfrankenfeld. The determination of certain kinds of quadratic groups and modules, in arbitrary prime characteristic, forms one of the tools that are needed for the Meierfrankenfeld program.

Section 2. Quadratic groups

Lemma 2.1. Let G be a finite group generated by two elements x_{1} and x_{2} of odd prime order p. Suppose that there exists a faithful, irreducible G-module V over 𝔽_{p}, with [V, x_{i}, x_{i}] = 0 for both i = 1 and 2. Then one of the following holds:

(i) G ≅ SL(2, p^{n}) for some n, and V is a natural module for G, or (ii) p = 3, G ≅ SL(2, 5), and V is a natural SL(2, 9)- module for G.

Proof. Let F be a splitting field for G over 𝔽_{p}, and put Ṽ = F ⊗ V. Put Γ = Aut(F). Then Ṽ is an irreducible module for Γ × G, by [A2, result 25.7]. Let U be an irreducible F [G]-submodule of Ṽ. There is then a finite subset Σ of Aut(F), containing the identity automorphism, such that:

Ṽ = ⊕{U^{σ} : σ ∈ Σ}

Since C_{G}(Ṽ) = 1, we then have C_{G}(U) = 1. Thus, U is a faithful, irreducible F [G]- module. Theorem 3.8.1 of [G] then says that G contains a subgroup isomorphic to SL(2, p). But the point is that a much stronger statement is actually proved. Namely, the argument of [G, Theorem 3.8.1] shows that dim_{F} (U) = 2, and that, relative to a suitable basis of U, the generators x_{1} and x_{2} of G have the matrix form:

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

for some λ ∈ F. A theorem of L.E.Dickson (Theorem 2.8.4 in [G]) then implies that either G ≅ SL(2, p^{n}), where λ is a primitive (p^{n} − 1)^{th}-root of unity; or else p = 3, λ ∈ 𝔽_{9}, and G ≅ SL(2, 5). Moreover, U is then seen to be irreducible as a G-module over 𝔽_{p}, and hence U is isomorphic to V as 𝔽_{p}[G]-modules. This yields the lemma. □

Lemma 2.2. Set L = SL(2, p) with p an odd prime, let V be the natural Ω_{3}(p)-module for L, and form the semi-direct product H = V L. Then every complement to V in H is conjugate to L.

Proof. Identify V with the space of all homogeneous polynomials of degree 2 in the variables x and y, over 𝔽_{p}. Thus, {x^{2}, xy, y^{2}} is a basis for V. Let a and t be the elements of L given as follows.

a = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} )  and  t = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )

Regard x and y as the standard basis vectors (1, 0) and (0, 1) for the natural SL(2, p)- module for L. Then the action of a and of t on V is given as follows.

(x^{2})^{a} = x^{2} + 2xy + y^{2} (xy)^{a} = xy + y^{2} (y^{2})^{a} = y^{2}
(x^{2})^{t} = y^{2} (y^{2})^{t} = x^{2} (xy)t = −xy

Suppose, by way of contradiction, that we are given a complement L_{1} to V in H with L_{1} not conjugate to L. All cyclic groups of order p in V ⟨a⟩ − V are V -conjugate to each other, so we may assume that a ∈ L_{1}. Set N = N_{H}(⟨a⟩). Then N = ⟨y^{2}, a⟩D where D is the diagonal subgroup of L, of order p − 1. Thus D is a Hall p′-subgroup of N, and since any two such are conjugate in N, we may assume that D ≤ L_{1}. If p = 3 then L is the normalizer in H of a Sylow 2-subgroup of H, and there is then nothing to show. Thus, we have p > 3. Then D acts non-trivially on V, and we find that N_{H}(D) = ⟨xy, D, t⟩. Then L_{1} = ⟨a, t_{1}⟩ for some t_{1} ∈ ⟨xy⟩t. Thus t_{1} = (cxy)t for some c ∈ 𝔽_{p}^{×}.

Notice that (ta)^{3} = I. Then also (t_{1}a)^{3} = I. We will therefore have a contradiction if we can show that, in fact, (t_{1}a)^{3} is a non-zero element of V. This is indeed the case, by the following calculation:

(t_{1}a)^{2} = ((cxy)ta)((cxy)ta) = (cxy + (cxy)^{(ta)^{2}})(ta)^{2} = (−cx^{2})(ta)^{2} and then
(t_{1}a)^{3} = ((−cx^{2})(ta)^{2})((cxy)ta) = −cx^{2} + (cxy)^{ta} = −c(x^{2} + xy + y^{2}).

This contradiction proves the lemma. □

Lemma 2.3. Let p be an odd prime, let V be a vector space of dimension 4 over 𝔽_{p}, and let U be a subspace of V of dimension 2. Let H be a subgroup of GL(V) which leaves U invariant, and assume that H has the following two properties.

(1) O_{p}(H) = C_{H}(U) = C_{H}(V/U),

(2) H/O_{p}(H) ≅ SL(2, p). Then the following hold.

(a) If there exists an element a of order p in H, with [V, a, a] = 0, then there is a complement L to O_{p}(H) in H, containing a. For any such complement L we have V = U ⊕ U_{1} for some L- submodule U_{1} of V.

(b) If p > 3 and there exists a complement L to O_{p}(H) in H containing a, then [V, b, b] = 0 for any element b of order p in L.

Proof. Set N = N_{GL(V)}(U), R = O_{p}(N), \overline{N} = N/R, and M = O^{p′}(N). Then \overline{M} ≅ SL(2, p) × SL(2, p) and R is a natural Ω_{4}^{+}(p)-module for \overline{M}. Here \overline{H} is a “diagonal” copy of SL(2, p) in \overline{M}. That is, \overline{H} ≅ SL(2, p) and \overline{H} is not a direct factor of \overline{M}. If p = 3 we observe that also O_{2}(\overline{H}) acts non-trivially on both U and V/U. For any p, the above conditions determine \overline{H} up to conjugacy in \overline{N}, and we may therefore identify RH with the subgroup of GL(4, p) consisting of all matrices of the form

(*)  ( \begin{matrix}X & A \\ 0 & X \end{matrix} )

where X ∈ SL(2, p) and where A is an arbitrary 2 × 2 matrix over 𝔽_{p}. Then R = [R, H] × C_{R}(H), with |C_{R}(H)| = p and where [R, H] is a natural Ω_{3}(p)-module for \overline{H}.

Suppose that there exists a complement L to O_{p}(H) in H. Then L is a complement to R in RH. By lemma 2.2, C_{R}(H)L is determined up to conjugacy in RH. If p > 3 then L = [C_{R}(H)L, C_{R}(H)L], and so L is uniquely determined up to conjugacy in RH in this case. Thus, if p > 3 we may take L to be the set of matrices in (*) with A = 0. This yields (b).

Suppose next that [V, a, a] = 0. Every element of order p in R⟨a⟩ − R is R-conjugate to an element of ⟨a⟩C_{R}(H) not contained in C_{R}(H). Moreover, if z is a non-identity element of C_{R}(H) then no element of ⟨a⟩z is quadratic on V. Thus, up to R-conjugacy there is a unique subgroup C of order p in R⟨a⟩ such that C ≰ R. We may therefore take a to be given by a matrix as in (*), with A = 0 and with

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

Then a lies in a complement L_{1} to R in RH such that V is completely reducible as an L_{1}-module. Suppose that also a lies in a complement L to O_{p}(H) in H. Then L is R-conjugate to a subgroup of C_{R}(H)L_{1}, by 2.2. As L_{1} is the unique subgroup of C_{R}(H)L_{1} which is isomorphic to SL(2, p) and which contains a quadratic element of order p, it follows that L and L_{1} are R-conjugate. Thus, we may assume that a lies in no complement to O_{p}(H) in H. In particular, O_{p}(H) ≠ 1, H is not isomorphic to a direct product SL(2, p) × ℤ_{p}, and H ≠ RH. It follows that R ∩ H = [R, H]. If p > 3 we then have H = [RH, RH] and H contains every complement to R in RH. Thus p = 3 and the normalizer in H of a Sylow 2-subgroup of H is a complement to O_{p}(H) in H. Here all subgroups of order 3 in H − O_{3}(H) are conjugate in H, so once again we have a in a complement. This contradiction completes the proof of (a). □

Lemma 2.4. Let G be a finite group, p an odd prime, and V a faithful 𝔽_{p}[G]-module. Suppose that we are given an element a ∈ G − O_{p}(G) with [V, a, a] = 0. Then there is a subgroup H = ⟨a^{H}⟩ of G having the following properties:

(a) H ≅ SL(2, p), or else p > 3 and H ≅ ℤ_{p} × SL(2, p).

(b) V = [V, H] ⊕ C_{V} (O^{p}(H)), and [V, H] is a direct sum of natural SL(2, p)-modules

for H. In particular, there exists an element b of order p in H such that b acts quadratically on V and such that b lies in a subgroup of G which is isomorphic to SL(2, p).

Proof. Suppose false, and let a counter-example (G, V) be chosen with |G|+|V | minimal. As a ∉ O_{p}(G) the Baer-Suzuki Theorem implies that there is a conjugate b of a in G such that ⟨a, b⟩ is not a p-group. Set H = ⟨a, b⟩. Then a ∉ O_{p}(H), and we may therefore assume that a and b are conjugate in H. By minimality of |G| we then have G = H.

Suppose first that V is irreducible for G. Then 2.1 yields G ≅ SL(2, p^{n}) or SL(2, 5), and by minimality we then have G ≅ SL(2, p). Moreover, 2.1 shows also that V is a natural module for G, so we are done in this case. As (G, V) is a counter-example to 2.4, we conclude that V is reducible. By minimality of |V |, all non-central chief factors for G in V are natural SL(2, p)-modules for G. Let W_{0} ≥ W_{1} ≥ W_{2} be a descending chain of G-submodules of V, with irreducible quotients, and set X = W_{0}/W_{2} and \overline{G} = G/C_{G}(X). Suppose that X is indecomposable, and consider first the case in which exactly one of the irreducible constituents for G in X is non-trivial. Then \overline{G}/O_{p}(\overline{G}) is isomorphic to SL(2, p), and O_{p}(\overline{G}) is a natural SL(2, p)-module for \overline{G}/O_{p}(\overline{G}). Every element of order p in \overline{G}−O_{p}(\overline{G}) is then contained in a complement to O_{p}(\overline{G}) in \overline{G}, and we thereby contradict the minimality of |G|. Thus no such indecomposable section X of V exists. It follows that V = [V, G]⊕C_{V} (O^{p}(G), and that both irreducible constituents of X are non-trivial. Let U be the irreducible submodule of X. Then O_{p}(\overline{G}) = C\overline{G}(U) = C\overline{G}(X/U) as \overline{G} is generated by two conjugate elements of order p, and \overline{G}/O_{p}(\overline{G}) ≅ SL(2, p). Now 2.3(a) and minimality of |G| together imply that O_{p}(\overline{G}) = 1 and that Xis decomposable. We have thus shown that [V, G] is a direct sum of natural SL(2, p)-modules for G. □

Lemma 2.5. Let G = K ×⟨c⟩, with K ≅ SL(2, 5) and with c an element of order 3. Let G be a faithful 𝔽_{3}G-module, and suppose that there exists an element a of order 3 in G which acts quadratically on V. Then such an element a may be chosen to lie in K ∪ ⟨c⟩.

Proof. Suppose false, and let b be an element of order 3 in K, with a ∈ ⟨b, c⟩. Also, let L be a subgroup of G, containing a, with L ≅ SL(2, 3). Then [V, K] = [V, Z(K)] = [V, Z(L)] = [V, O_{2}(L)] is a direct sum of natural SL(2, 3)-modules for L, by 2.4(b). As a ∉ K, c acts quadratically on C_{V} (K), and then since c is not quadratic on V we conclude that there exists an irreducible L-submodule U of [V, K] such that c is not quadratic on the subspace W = ⟨U^{⟨c⟩}⟩ of V. Then W = U ⊕U^{c}⊕U^{c^{2}}, and then also W = U ⊕U^{x}⊕U^{x^{2}} for any x ∈ ⟨a, c⟩ − ⟨a⟩. Thus, ⟨a⟩ is the unique quadratic subgroup of order 3 in ⟨a, c⟩. But ⟨a⟩ = ⟨bc⟩ or ⟨b^{2}c⟩, where bc and b^{2}c are conjugate via K. Thus, ⟨a, c⟩ contains at least two quadratic subgroups of order 3, and we have a contradiction at this point. □

Lemma 2.6. Let G be a finite group, p an odd prime, and V a faithful 𝔽_{p}G-module. Suppose that we are given an element a of G − O_{p}(G) with [V, a, a] = 0, and suppose further that G has cyclic Sylow p-subgroups. Let g ∈ G − N_{G}(⟨a⟩), and set H = ⟨a, a^{g}⟩. Then either H ≅ SL(2, p) or p = 3 and H ≅ SL(2, 5).

Proof. Assume false, and let (G, V) be a counter-example with |G| + |V | minimal. Then V is reducible for the action of H, by 2.1. As in the proof of 2.4, let W_{0} ≥ W_{1} ≥ W_{2} be a properly descending chain of H-submodules of V, set X = W_{0}/W_{2}, and set \overline{H} = H/C_{H}(X). Assume that H acts non-trivially on X. As H has cyclic Sylow p-subgroups we then have O_{p}(\overline{H}) = 1. If one of the constituents for H in X is trivial, it then follows that X is completely reducible, and this shows that each irreducible constituent for H in [V, H] is either a natural SL(2, p)-module or, exceptionally, a natural SL(2, 9)-module for H. Now let U be an irreducible H-submodule of V, chosen if possible so that [U, H] = 0, and set Ĥ = H/C_{H}(U). Then O^{p}(H) acts non-trivially on V/U, and by the minimality of |V | we then have Ĥ ≅ SL(2, p) or SL(2, 5). Set K = C_{H}(V/U). Then K is a normal p′-subgroup of H, and K ≠ 1 as otherwise we are done. Then K acts non-trivially on U, so U is a natural SL(2, p) or SL(2, 9)-module for H, and |K| = 2. Thus H is a central extension of SL(2, p) or SL(2, 9) by a group of order 2. But for any prime power q, the 2-part of the Schur multiplier of SL(2, q) is trivial, and so H has a direct factor of order 2. This is contrary to H being generated by two elements of order p, and the lemma is thereby proved. □

Lemma 2.7. Let G be a finite group, p an odd prime, and A a subgroup of G such that G = ⟨A^{G}⟩. Suppose that G has a faithful 𝔽_{p}-module M such that [M, A, A] = 0. Then for any element a of A and any conjugate b of a in G, either ⟨a, b⟩ is a p-group, or the following hold.

(a) ⟨a, b⟩ has a normal subgroup N such that ⟨a, b⟩/N is isomorphic to one of the groups SL(2, p^{n}), (n ≥ 1), or SL(2, 5).

(b) ⟨a, b⟩ has a subgroup K = ⟨a^{K}⟩ with K isomorphic to SL(2, p) or SL(2, p) × ℤ_{p}. Moreover, if p = 3 then K ≅ SL(2, 3).

Proof. Let a ∈ A and let b ∈ a^{G}, and put L = ⟨a, b⟩. Suppose that L is not a p-group, and let W be a non-trivial irreducible constuent in M for the action of L. Setting N = C_{L}(W), it follows from lemma 2.1 that L/N is isomorphic to SL(2, p^{n}) for some n, or to SL(2, 5). Further, lemma 2.4 implies that L has a subgroup K = ⟨a^{K}⟩ with K isomorphic to SL(2, p) or ℤ_{p} × SL(2, p). □

The following result is due to Ulrich Meierfrankenfeld.

Lemma 2.8. Let G be a finite group, p a prime, p > 3, and let V be a faithful, irreducible 𝔽_{p}G-module. Suppose that G = ⟨A^{G}⟩, where A is a non-identity subgroup of G which acts quadratically on V. Suppose further that G has cyclic Sylow p- subgroups, and that all involutions in C_{G}(A) are contained in Z(G). Then G ≅ SL(2, p).

Proof. Let g ∈ G−N_{G}(A), set H = ⟨A, A^{g}⟩, and set B = N_{H}(A). As G has cyclic Sylow p-subgroups, we have H ≅ SL(2, p), by 2.5. The involution z in H is then in Z(G), and since V is irreducible we have V = [V, z]. Then 2.4(b) implies that V = ⊕_{1≤i≤n}V_{i}, where each V_{i} is a natural SL(2, p)- module for H. As p > 3, B is non-abelian and we find that Aut_{B}(V_{i}) = Aut_{H}(V_{i}) for all i. Setting D = Aut_{H}(V), it follows that Aut_{B}(V) = D.

The centralizer in G of the chain V ≥ C_{V} (A) ≥ 0 acts quadratically on V, and is therefore an elementary abelian p-subgroup of G containing A. As G has cyclic Sylow p-subgroups we therefore conclude that

(1)  A = C_{G}(C_{V} (A)) ∩ C_{G}(V/C_{V} (A)).

Now let H̃ be a subgroup of G with A ≤ H̃ ≅ H, and set B̃ = NH̃(A). The image of B in GL(C_{V} (A)) × GL(V/C_{V} (A)) is {(λI, λ^{−1}I): 0 ≠ λ ∈ 𝔽_{p}}, and the same is true of B̃. Then (1) implies that B = B̃, and so AutH̃(V) = AutB̃(V) = D. On the other hand, we have Aut_{D}(V) ≅ GL(2, p), and ⟨H, H̃⟩ ≤ Aut_{D}(V). Thus H = H̃.

For any x ∈ G − N_{G}(A), we may now conclude that ⟨A^{x}, A^{gx}⟩ = ⟨A, A^{x} (by replacing A by A^{x} in the preceding discussion). Since also H = ⟨A, A^{g}⟩ = ⟨A, A^{x}⟩, we conclude that H is invariant under ⟨G−N_{G}(A)⟩. That is, H is G-invariant, and thus H = ⟨A^{G}⟩ = G. □

Lemma 2.9. Assume Hypothesis 1.1, and let a be a non-identity element of A. Suppose that we are given a p′-subgroup Q of G, with [Q, a] = Q. Then p = 3 and Q is a non-abelian 2-group. Moreover, if Q is extraspecial then Q is a quaternion group.

Proof. Let R be an a-invariant Sylow subgroup of Q, with [R, a] ≠ 1. Then 2.6(ib) implies that p = 3, R is a 2-group, and Q = C_{Q}(a)R. But then also Q = R, since Q = [Q, a]. Further, 2.7(b) also shows that every a-invariant abelian subgroup of Q is centralized by a, and so Q itself is non-abelian. Now suppose that Q is extraspecial. Then C_{Q}(a) = Z(Q), and it follows that for any involution t in Q − Z(Q) we have ⟨t, a⟩ containing a subgroup isomorphic to Alt(4). Therefore there is no such involution t, and so Q is a quaternion group. □

Lemma 2.10. Assume Hypothesis 1.1, let a ∈ A, and let L be an a-invariant subgroup of G. Then every component of L is a-invariant.

Proof. Suppose false, and let K be a component of L which is not a-invariant. As K is quasisimple, there is a prime divisor r of |K/Z(K)| with r ∉ {2, p}. Let R be a Sylow R-subgroup of K. Then [R, a] is a non-identity r-group, and we contradict 2.9. □

Section 3: Groups of Lie type

In this section we collect the information that we need concerning automorphisms, Schur multipliers, and centralizers of semisimple elements in groups of Lie type. In doing so, we are guided to a great extent by [GLS3]. Many of the results from [GLS3] that we quote come indirectly from [St1], [St2], [St3], [Gr], and [Ma].

Let r be a prime, let \overline{F} be an algebraic closure of the field 𝔽_{r} of r elements, and let \overline{K} be a simple (linear) algebraic group defined over \overline{F}. (If also Z(\overline{K}) = 1 then we say that \overline{K} is of adjoint type.)

Fix a maximal torus \overline{T} of \overline{K}, and let Σ be the root system associated with \overline{T}. For any α in Σ, let \overline{X}_{α} = \{x_{α}(t) : t ∈ \overline{F} \} be the one- parameter subgroup (i.e. the root subgroup of \overline{K}) associated with α, and denote by X the set of all elements x_{α}(t) of \overline{K}, α ∈ Σ and t ∈ \overline{F}. The root subgroups of \overline{K} generate \overline{K}, so any endomorphism of \overline{K} is determined by its action on X.

A surjective algebraic endomorphism σ of \overline{K} is said to be a Steinberg endomorphism if C\overline{K}(σ) is finite. A finite group K is a group of Lie type (in characteristic r) if K = O^{r′}(C\overline{K}(σ)) for some simple algebraic group \overline{K} and some Steinberg endomorphism σ of \overline{K}, and we then say that (\overline{K}, σ) is a σ-setup of K. If \overline{K} is of adjoint type (i.e. if Z(\overline{K}) = 1) then Z(K) = 1, and we say also that K is of adjoint type. The class of groups of Lie type in characteristic r is denoted Lie(r).

Let q = r^{n} be a power of r, where n is a positive integer. There is then a Steinberg endomorphism ϕ_{q} of \overline{K} given on X by

(3.1)  ϕ_{q}(x_{α}(t)) = x_{α}(t^{q}).

For any isometry ρ of Σ there is an automorphism γ_{ρ} of \overline{K} given on X by

(3.2)  γ_{ρ}(x_{α}(t)) = x_{α^{ρ}}(t).

If Σ is B_{2}, F_{4}, or G_{2}, and r is 2, 2, or 3, respectively, then there is a unique anglepreserving, length-changing bijection ρ on Σ, and and there is an automorphism ψ of \overline{K} given on X by

(3.3)  ψ(x_{α}(t)) = \{ \begin{matrix}x_{α^{ρ}}(t)  if α is long \\ x_{α^{ρ}}(t^{r})  if α is short. \end{matrix}

One observes that ϕ_{r} commutes with γ_{ρ} for any isometry ρ of Σ, and in the special cases given by 3.3 one observes that ψ^{2} = ϕ_{r}.

The following result is Theorem 2.2.3 in [GLS3].

Proposition 3.4. Let K ∈ Lie(r) and let (\overline{K}, σ) be a σ-setup of K. Then there is a maximal torus \overline{T} of \overline{K}, with associated root system Σ, such that, after conjugating σ by a suitable inner automorphism of \overline{K}, one of the following holds.

(i) σ = γ_{ρ} ∘ ϕ_{q} for some isometry ρ of Σ and some positive integral power q of r. (ii) Σ = B_{2}, F_{4}, or G_{2}, with r = 2, 2, or 3, respectively, and σ = ψ^{n} for some odd

positive integer n, where ψ is as in 3.3.

The group K in 3.4 may be denoted ^{d}Σ(q), where d = |ρ| in case (i), and where d = 2 in case (ii). If d = 1 then we may write simply K = Σ(q), and we say in this case that K is a Chevalley group. If d ≠ 1 and σ is conjugate to γ_{ρ} ∘ ϕ_{q}, where ρ is a non-trivial isometry of Σ, then K is a Steinberg variation. If d = 2 and σ is conjugate to ψ^{n}, n odd, where ψ is as in 3.3, then K is a Ree-Suzuki group.

Let K = ^{d}Σ(r^{n}) ∈ Lie(r), let (\overline{K}, σ) be a σ-setup of K, and let x be an automorphism of K. We say that x is an inner-diagonal automorphism if x is the restriction to K of an inner automorphism of C\overline{K}(σ). The group of all inner-diagonal automorphisms of K is denoted Inndiag(K). We say that x is a field automorphism if x is conjugate via Inndiag(K) to a non-identity automorphism of the form ϕ_{q}|_{K}. We say that ϕ is a graph automorphism if d = 1 and x is conjugate via Inndiag(K) to an automorphism of the form (γ_{ρ})|_{K}, ρ a non-identity isometry of Σ. We say that x is a graph-field automorphism if either d = 1 and ϕ is conjugate via Inndiag(K) to an automorphism of the form γ_{ρ} ∘ ϕ_{q}) of \overline{K}, ρ a non-trivial isometry of Σ, or if d = 2 and ϕ is conjugate via Inndiag(K) to an automorphism of the form ψ^{n}, n odd, where ψ is given as in 3.3.

If Σ is not B_{2}, F_{4}, or G_{2}, with p = 2, 2, or 3, respectively, set ψ = ϕ_{r}. In any case, set Φ\overline{K} = ⟨ψ⟩. Also, denote by Γ\overline{K} the set of all γ_{ρ}, ρ an isometry of Σ.

Proposition 3.5. Let K be a group of Lie type and let (\overline{K}, σ) be a σ-setup for K. Assume that Z(\overline{K}) = 1, and identify \overline{K} with the group of inner automorphisms of \overline{K}. Denote by Aut_{1}(\overline{K}) the group of automorphisms τ of \overline{K} as an abstract group, such that either τ or τ^{−1} is an algebraic endomorphism of \overline{K}. Then the following hold.

(a) We have Aut_{1}(\overline{K}) = (Φ\overline{K} × Γ\overline{K})\overline{K}.

(b) The restriction map from C_{Aut_{1}(\overline{K})}(σ) to Aut(K) is surjective, with kernel ⟨σ⟩.

(c) We have C_{Aut_{1}(\overline{K})}(K) = ⟨σ⟩.

Proof. Parts (b) and (c) are Theorem 2.5.4 and Lemma 2.5.7, respectively, in [GLS3]. Part (a) follows from (b) and from the theorem of Steinberg [St, Theorem 30] which states that every automorphism of K is the product of inner-diagonal, field, and graph automorphisms. □

The next result is [GLS3, Proposition 4.9.1]. The proof given below supplies a few details not given in the cited reference.

Proposition 3.6. Let K = ^{d}Σ(q) ∈ Lie(r), with Z(K) = 1. Let x be a field automorphism or a graph-field automorphism of K, of prime order p, and let y ∈ Inndiag(K)x. If K is a Steinberg variation, assume that d ≠ p. Then x and y are conjugate via Inndiag(K).

Proof. Let (\overline{K}, σ) be a σ-setup of K, with Z(\overline{K}) = 1. If K is a Chevalley group or a Ree-Suzuki group (resp. a Steinberg variation) we may take σ = ψ^{n} (resp. γ_{ρ} ∘ ϕ_{r^{n}}, for some appropriate n > 0. We claim:

(1) There exists a Steinberg endomorphism τ of \overline{K}, with x ∈ ⟨τ|_{K}⟩ and with τ^{p} = σ.

Suppose that (1) holds, and let k be the integer, 1 ≤ k < p, such that τ|_{K} = x^{k}. We are given y ∈ Aut(K) with y ∈ Inndiag(K)x. Identify Inndiag(K) with C\overline{K}(σ). Then y^{k} is the restriction to K of some automorphism τ_{1} = hτ of \overline{K}, where h ∈ C\overline{K}(σ). As |y^{k}| = p, we have (τ_{1})^{p} ∈ ⟨σ⟩, by 3.5(c). As τ_{1}^{p} ≡ τ^{p} (mod Inndiag(K)), we then have τ_{1}^{p} = σ, by (1). As τ is a Steinberg endomorphism of \overline{K} we may apply Lang’s Theorem ([GLS3, Theorem 2.1.1]), and conclude that h = gτg^{−1}τ^{−1} for some g ∈ \overline{K}. Then τ_{1} = gτg^{−1}, and by taking p^{th} powers we obtain σ = gσg^{−1}. Thus g ∈ C\overline{K}(σ) = Inndiag(K), and y = gxg^{−1}, as required. Thus, it remains to establish (1).

Set Φ = Φ\overline{K} and Γ = Γ\overline{K}. By 3.4, we may take σ = ψ^{n} ∘ γ_{ρ} for some n > 0 and some isometry ρ of Σ. We are free to replace x by any Inndiag(K)-conjugate of x, and then since x is a field or graph-field automorphism of K we may take x = τ_{0}|_{K}, for some τ_{0} ∈ ΦΓ. Moreover, we have [σ, τ_{0}] = 1 by 3.5(b), and (τ_{0})^{p} ∈ ⟨σ⟩ by 3.5(c).

Suppose that τ_{0} ∈ Γ⟨σ⟩. By assumption, x is not a graph automorphism of K, so K is not a Chevalley group. If K is a Ree-Suzuki group then Γ = 1, and since τ_{0} ∉ ⟨σ⟩ we conclude that K is a Steinberg variation. Then ρ ≠ 1, and since τ_{0} and σ commute it follows that τ_{0} ∈ ⟨ρ, σ⟩. Then d = p, contrary to assumption. Thus, we conclude that τ_{0} ∉ Γ⟨σ⟩.

We have ΦΓ/⟨σ⟩Γ ≅ ℤ_{n}, and since x^{p} = 1 it now follows that p divides n. Write n = pm and set ψ_{1} = ψ^{m}. Suppose that ρ = 1. We then have (ψ_{1})^{p} = σ, and τ_{0} ∈ ⟨ψ_{1}⟩Γ. Write τ_{0} = (ψ_{1})^{k}γ, where γ ∈ Γ. As [Φ, Γ] = 1 we conclude that |γ| = 1 or p, so there exists an integer ℓ with γ^{kℓ} = γ. We then take τ = ψ_{1}γ^{ℓ}, and obtain τ^{p} = σ and τ^{k} = τ_{0}. Thus, (1) holds in this case. On the other hand, suppose that ρ ≠ 1. Then p does not divide |ρ|, by assumption, and so there exists γ ∈ ⟨γ_{ρ}⟩ with γ^{p} = ρ. Setting τ = ψ_{1}γ, we then have τ^{p} = σ. Any homomorphic image of Φ × ⟨γ_{ρ}⟩ has at most one subgroup of order p, so x ∈ ⟨τ|K⟩, and thus (1) holds in any case. □

We next consider centralizers of semisimple elements.

Lemma 3.7. Let K be a simple group of Lie type in characteristic r, and let (\overline{K}, σ) be a σ-setup of K. Identify Inndiag(K) with C\overline{K}(σ), and let x ∈ Inndiag(K) with |x| prime to r. Then the following conditions are equivalent.

(1) C\overline{K}(x) contains a non-identity unipotent element.

(2) O^{r′}(C_{K}(x)) ≠ 1.

(3) O^{r′}(C_{K}(x)) is a product L_{1} ⋯ L_{n}, (n ≥ 1), where each L_{i} is a group of Lie type in characteristic r, and where [L_{i}, L_{j}] = 1 for all i and j with i ≠ j.

Proof. Set \overline{C} = C\overline{K}(x). As |x| is relatively prime to r, x is a semisimple element of \overline{K}, and hence \overline{C} is closed, connected and reductive. Set \overline{L} = [\overline{C}, \overline{C}]. Thus \overline{C} = Z(\overline{C})\overline{L}, where \overline{L} is closed, connected and semisimple, and where Z(\overline{C}) is a torus. Then \overline{L} contains all of the unipotent elements of \overline{C}. Denote by M the set of normal, simple algebraic subgroups of \overline{M}. Then \overline{L} is the commuting product of the members of M. If M is non-empty, we write M = \{\overline{M}_{i}\}_{1≤i≤t}. Notice that if L = 1 then C_{K}(x) = C\overline{C}(σ) consists of semisimple elements, so that O^{r′}(C_{K}(x)) = 1. Thus (2) implies (1).

Suppose that M is non-empty. That is, assume that (1) holds. As σ commutes with x, \overline{C} is σ-invariant and σ then induces a permutation action on M. Let M_{1} ⋯ M_{n} be the set of orbits for σ on M, and assume (without loss) that indices have been chosen so that M_{1} = \{\overline{M}_{i}\}_{1≤i≤k}. Since any positive power of a Steinberg endomorphism is again a Steinberg endomorphism, it follows that σ^{k} induces a Steinberg endomorphism on each \overline{L}_{i}, 1 ≤ i ≤ k. For such i, set M_{i} = O^{r′}(C_{\overline{M}_{i}}(σ^{k})). Then each M_{i} is a group of Lie type in characteristic r, by definition. Now set M = M_{1} ⋯ M_{k}. Then M/Z(M) is the direct product of the images in M/Z(M) of the groups M_{i}, 1 ≤ i ≤ k, and the action of σ on M/Z(M) is given by the transitive permuting of these factors. Set L_{1} = O^{r′}(C\overline{M}(σ)). It now follows that L_{1} = O^{r′}(C_{M}(σ)) is isomorphic to a quotient of M_{1} by a subgroup of Z(M_{1}). We repeat this procedure for the remaining σ- orbits, obtaining the groups L_{1} through L_{n}.

Now set L = O^{r′}(C\overline{L}(σ)). Since central quotients of groups in Lie(r) are also in Lie(r), we conclude that L is the pairwise commuting product of the groups L_{j}, 1 ≤ j ≤ n, where each L_{j} is a member of Lie = (r). On the other hand, we have C\overline{K}(σ) = C\overline{T} (σ)K for some σ-invariant maximal torus \overline{T}, and then L = O^{r′}(C\overline{K}(σ, x)) = O^{r′}(C_{C\overline{K}(σ)}(x)) = O^{r′}(C_{K}(x)). Thus (3) holds. Clearly, (3) implies (2), and thus the lemma is proved. □

Lemma 3.8. Let r be a prime and let \overline{K} be a simple linear algebraic group over an algebraic closure \overline{F} of the field F = 𝔽_{r} of r elements. Let σ be a Steinberg endomorphism of \overline{K}, and set K = O^{r′}(C\overline{K}(σ)) (so that K is a group of Lie type in characteristic r). Let g ∈ K, and assume that either

(i) |g| = 2 and \overline{K} is not of type A_{1}, or (ii) |g| = 3 and \overline{K} is not of type A_{1} or A_{2}.

Then C\overline{K}(g) contains a non-identity unipotent element.

Proof. We may assume that |g| ≠ r as otherwise the result holds trivially. Thus g is a semisimple element of \overline{K}, and so there is a maximal torus \overline{T} of \overline{K} containing g. Let Σ be the root system for \overline{K} given by \overline{T}, Π a fundamental system in Σ, \overline{B} the corresponding Borel subgroup, and \overline{U} the unipotent radical of \overline{B}. Recall that each α ∈ Σ is a homomorphism of \overline{T} into \overline{F}^{×}, and that there is then a \overline{T} -invariant subgroup \overline{U}_{α} of \overline{U} and a parametrization

x_{α} : \overline{F}^{×} ⟶ \overline{U}_{α}

such that gx_{α}(t)g^{−1} = x_{α}(α(g)t) for all t ∈ \overline{F}^{×}.

We aim to show that, under the conditions given in (a) and (b), there exists a root α such that \overline{U}_{α} ≤ C\overline{K}(g). Suppose false, and suppose first that |g| = 2. Here α(g)^{2} = 1, so we have α(g) = −1 for all α ∈ Σ. Assuming that Σ is not A_{1}, there exist two roots α and β whose sum is a root, and we then have (α + β)(g) = α(g)β(g) = 1, for a contradiction. Now suppose that |g| = 3, and let ω be a primitive cube root of unity in \overline{F}. Then α(g) = ω or ω^{−1} for all roots α. It follows that α(g) = β(g) ≠ 1 whenever α and β are roots whose sum is again a root. In particular, we may assume that α(g) = ω for all α ∈ Π. If Σ has more than one root length then there are fundamental roots α and β such that α + 2β is a root, and we obtain (α + 2β)(g) = 1 in that case. Also, if the rank of Σ is at least 3 then there exist fundamental roots α, β, and γ whose sum is a root, yielding (α + β + γ)(g) = 1. Thus, Σ is of rank at most two, and Σ has only one root length. That is, Σ is A_{1} or A_{2}. □

Lemma 3.9. Let K be a simple group of Lie type in characteristic r, let p be a prime different from r, and let x be an element of order p in Inndiag(K). Suppose that x is contained in a non-cyclic abelian p-subgroup of Inndiag(K). Then there exists an element y of order p in C_{K}(x) such that O^{r′}(C_{K}(y)) ≠ 1.

Proof. Let E be an elementary abelian subgroup of Inndiag(K) of order p^{2}, containing x, and let (\overline{K}, σ) be a σ-setup for K. Let \overline{T} be a maximal torus of \overline{K} containing E, let \overline{B} be a Borel subgroup of \overline{K} containing \overline{T}, let Σ be the root system defined by \overline{T} and \overline{B}, and let α ∈ Σ. Then α is a homomorphism of \overline{T} into the multiplicative group of an algrebaic closure of 𝔽_{r}. The image of α is then cyclic, and so there exists a non-identity element y ∈ E ∩ Ker(α). This means that C\overline{K}(y) contains the root subgroup of \overline{B} corresponding to α. The desired result then follows from 3.7. □

We next consider normalizers of r-groups in groups K, K ∈ Lie(r).

Lemma 3.10 (Borel-Tits). Let K ∈ Lie(r) and let R be a non-identity r-subgroup of K. Then there is a parabolic subgroup P of K such that R ≤ O_{r}(P) and N_{K}(R) ≤ P.

Proof. This result, proved first in [BT], appears as [GLS3, Theorem 3.1.3(a)].

Recall that a group G is said to be r-constrained if C_{G}(O_{r}(G)) ≤ O_{r}(G).

Lemma 3.11. Let K ∈ Lie(r), let X be a subgroup of Aut(K) containing Inn(K), and let R be a non-identity r-subgroup of K. Then the following hold.

(a) Both C_{X}(R) and N_{X}(R) are r-constrained.

(b) If R = O_{r}(N_{K}(R)) then the group P = N_{K}(R) is a parablic subgroup of K, and R = O_{r}(P).

Proof. See [GLS3, Corollaries 3.1.4 and 3.1.5]. □

We now review the Schur multipliers of the groups of Lie type.

Proposition 3.12. Let K be a simple group of Lie type, in characteristic r, and let K̂ be the universal, perfect central extension of K. Set Z = Z(K). Then Z = Z_{c} × Z_{e}, where Z_{c} (the “canonical” part of Z is isomorphic to the quotient group Outdiag(K) = Inndiag(K)/Inn(K), and where Z_{e} (the “exceptional” part of Z) is equal to O_{r}(Z). Moreover, we have Z_{e} = 1 except in the following cases.

(a) |Z_{e}| = 2, and K is isomorphic to L_{2}(4), L_{3}(2), Sp(4, 2)′, L_{4}(2), Sp(6, 2), U_{4}(2), F_{4}(2), or G_{2}(4).

(b) Z_{e} ≅ ℤ_{2} × ℤ_{2}, and K is isomorphic to U_{6}(2), D_{4}(2), Sz(8), or ^{2}E_{6}(2).

(c) Z_{e} ≅ ℤ_{4} × ℤ_{4}, and K is isomorphic to L_{3}(4).

(d) |Z_{e}| = 3, and K is isomorphic to L_{2}(9), Sp(6, 3), or G_{2}(3).

(e) Z_{e} ≅ ℤ_{3} × ℤ_{3}, and K is isomorphic to U_{4}(3).

Proof. The relevant references are [St3], and [Gr]. See also chapter 6 of [GLS3]. □

Lemma 3.13. Let K be one of the groups Sz(8), L_{3}(4), D_{4}(2), or U_{6}(2), and let α be an outer automorphism of K of order 3. Define K̂ and Z_{e} as in 3.12. Then α lifts to an automorphism of K̂ which acts faithfully on Z_{e}.

Proof. The result is contained in [GLS3, Theorem 6.3.1], but we present an altwernative proof here.

Assume that α acts trivially on Z(K^{∗}), let K_{1} be a perfect central extension of K by ℤ_{2}, and view α as an automorphism of K_{1}. Suppose first that K ≅ L_{3}(4). Let P be an αinvariant maximal parabolic subgroup of K. Then P is a semidirect product of SL(2, 4) with the natural SL(2, 4)-module, and α centralizes a complement to O_{2}(P) in P. Let P_{1} denote the pre-image of P in K_{1}. Then O_{2}(P_{1}) is elementary abelian, since P acts transitively on the non-identity elements of O_{2}(P). Further, we may choose K_{1} so that P_{1} has a subgroup isomorphic to SL(2, 4), since SL(2, 4) has no perfect central extension by ℤ_{2} × ℤ_{2}. But now [O_{2}(P_{1}), α](C_{P_{1}}(α))′ is isomorphic to P, and is a complement to Z(K_{1}) in P_{1}. Gaschütz’s Theorem [A2, result(10.4)] then implies that K_{1} splits over Z(K_{1}), and we have a contradiction.

Suppose next that K ≅ U_{6}(2). Then K⟨α⟩ has a subgroup P of the form 2_{+}^{1+8} : (U_{4}(2) × 3). The Schur multiplier of U_{4}(2) contains no fours group, so there is a perfect central extension of K⟨α⟩ by ℤ_{2} in which P lifts to a group P_{1} having a sugroup U_{4}(2)×3. By Lemma 3.2, above, O_{2}(P_{1}) splits over Z(K_{1}), and as in the case of L_{3}(4) we find that [O_{2}(P_{1}), α](C_{P_{1}}(α))′ is a complement to Z(K_{1}) in P_{1}, andr a contradiction is reached as before.

Suppose that K ≅ Sz(8). Let S be an α-invariant Sylow 2-subgroup of K and put U = Ω_{1}(S). Then U = Z(S), and N_{K}(S) acts transitively on the set of non-identity elements of U. Let S_{1} and U_{1} denote the pre-images of S and U, respectively, in K_{1}.

It follows at once that U_{1} is elementary abelian. We have C_{S}(α) ≅ ℤ_{4}, and since all involutions in S_{1} lie in U_{1} it then follows that C_{S_{1}}(α) ≅ ℤ_{4} × ℤ_{2}. Let u ∈ C_{U_{1}}(α) with u ∉ Z(K_{1}). Then C_{S_{1}}(u) is α-invariant, of index at most 2 in S_{1}, and containing C_{S_{1}}(α). We conclude that in fact u ∈ Z(S_{1}), and hence U_{1} = Z(S_{1}). Now let U^{∗} and S^{∗} be the inverse images of U and S in the full covering group K^{∗} of K, and let X be a subgroup of N_{K^{∗}}(S^{∗}) of order 7. It follows from the fore-going that U^{∗} = Z(S^{∗}). Let g ∈ S^{∗} − U^{∗}. Then g^{2} = yz where y ∈ [U^{∗}, X] and where z ∈ Z(K^{∗}). Without loss, we may assume that K_{1} was chosen to begin with so that z projects to the identity element of K_{1}. Taking g_{1} for the image of g, and X_{1} for the image of X in K_{1}, we then have (g_{1})^{2} ∈ [U_{1}, X_{1}], and then Φ(S_{1}) = [U_{1}, X_{1}]. Thus S_{1} splits over Z(K_{1}), with a contradiction as before.

Suppose that K ≅ D_{4}(2). In order to analyze this group we will require the detailed structure of the group P = V : L, where L ≅ Alt(8) and where V ≅ 2^{6} is the unique nontrivial constituent in the permutation module for L over 𝔽_{2}. Here V may be described as follows. Put Ω = {1, 2, ⋯, 8} and let E be the 𝔽_{2}-space of all even-cardinality subsets of Ω, with addition given by symmetric difference. We may then identify V with E/⟨Ω⟩, with natural action by L. We require the following facts.

(1) P is isomorphic to a maximal parabolic subgroup of D_{4}(2).

(2) L ≅ Ω_{6}^{+}(2) and V is isomorphic to the natural orthogonal module for L. Moreover, the singular vectors correspond to the four-element subsets of Ω.

(3) We have H^{1}(L, V) ≅ ℤ_{2}. (Up to isomorphism, E is the unique indecomposable Lmodule of order 2^{7} with quotient module V.)

The next two results are easily computed from the above information.

(4) Let S be a Sylow 2-subgroup of P. Then S has exactly three elementary abelian subgroups of order 2^{6}. They are V, A_{1}, and A_{2}, where |A_{i} ∩ V | = |A_{i} ∩ L| = 8, and N_{L}(A_{i} ∩ L) ≅ 2^{3} : L_{3}(2).

(5) In the semidirect product E : L, the pre-image of each A_{i} is an extraspecial group.

From the D_{4} diagram, and from (4), we obtain the following fact.

(6) Identify S with an α-invariant Sylow 2-subgroup of K, and P with a maximal parabolic subgroup of K. Then α permutes {V, A_{1}, A_{2}} transitively.

With these facts in hand, one may prove that α acts non-trivially on the Schur multiplier of K. For, taking K_{1} as in the previous cases, suppose first that V lifts in K_{1} to a group V_{1} which is elementary abelian. Then (4) and (6) imply that the pre-image L_{1} of L is isomorphic to L × ℤ_{2}. The inverse image P_{1} of P is then isomorphic to E : L, as otherwise Z(K_{1}) has a complement in P. Now (4) and (5) imply that the generalized Fitting subgroups of the pre-images in K_{1} of the remaining two connected maximal parabolics over S are extaspecial. Since α fuses these to V_{1}, we have a contradiction. We therefore conclude that V_{1} is not abelian, and so V_{1} is extraspecial. In the four-fold covering group K^{∗} the pre-image V^{∗} of V is then of the form 2_{+}^{1+6} × 2, by Lemma 3.2. But then, taking K^{∗}/(V^{∗})′ in place of K_{1}, we have a perfect double cover of K in which the pre-image of V is abelian, after all, and so we have a contradiction at this point. □

We end this section with a result which will be useful in determining the possible quadratic subgroups of sporadic groups.

Lemma 3.14. Let G be a finite group, put X = O_{2}(G), and assume that X = F^{∗}(G) is an extraspecial 2-group of width n and sign ϵ. Assume also that either:

(1) G/X ≅ Ω_{2n}^{ϵ}(2), with n ≥ 3 if ϵ = 1, or

(2) G/X ≅ SU(n, 2), with n ≠ 2, and with ϵ = (−1)^{n}. Let G^{∗} be a group having a normal subgroup ⟨t⟩ of order 2, with G^{∗}/⟨t⟩ ≅ G. Then O_{2}(G^{∗}) ≅ X × ⟨t⟩.

Proof. Put M = X/Z(X). Then the squaring map from M into Z(X) defines a quadratic form Q on M, with respect to which M is a non-degenerate orthogonal space over 𝔽_{2}, of Witt index ϵ. If G/X ≅ Ω_{2n}^{ϵ}(2), it follows that M may be identified with the natural G/X-module. If G/X ≅ SU(n, 2), then M may be identified with the natural n-dimensional hermitian module for G/X over 𝔽_{4}, whose hermitian form h satisfies h(v, v) = Q(v) for all v ∈ M. In both the cases (1) and (2), the singular vectors and the non-singular vectors in M with respect to Q each form a single orbit for the action of G/X.

Denote by Z the pre-image of Z(X) in G^{∗}. Also, denote by D the set of subgroups D of G such that [X, D] = [X, D, D] is a quaternion group. Thus, D is a set of groups of order 3, and since n ≥ 3 if G/X ≅ Ω^{+}(2n, 2) it follows that D is non-empty. Fix D ∈ D, and denote by Y the inverse image of ⟨D^{X}⟩ in G^{∗}. Then Y is isomorphic to SL(2, 3) × ℤ_{2}. Denote by z the involution in Z(Y), set G_{0}^{∗} = C_{G^{∗}}(z), and let G_{0} be the image of G_{0}^{∗} in G. Then |G : G_{0}| ≤ 2, and z ≠ t. We now make the following claim.

(*) For any element x^{∗} of O_{2}(G^{∗}) whose image in G has order 4, we have (x^{∗})^{2} = z.

Suppose that (*) is not the case. As G is transitive on the non-singular vectors in M, it follows that G ≠ G_{0}, and that G_{0} has two orbits on D. Let D_{0} and D_{1} be the two orbits for G_{0} on D. Then, for any D_{0} ∈ D_{0} and any D_{1} ∈ D_{1}, we have |[X, D_{0}] ∩ [X, D_{1}]| = 2, and hence [X, D_{0}] commutes with [X, D_{1}]. Thus [X, ⟨D_{0}⟩] commutes with [X, ⟨D_{1}⟩], and so each [X, ⟨D_{i}⟩] is a proper subgroup of X. In particular, it follows that ⟨D_{i}⟩ ≠ O^{2}(G), and hence G/X ≅ Ω_{4}^{+}(2) or SU(2, 2). These two cases are excluded by the conditions placed on n in (1) and (2), so (*) holds.

Now let s ∈ X be an involution. By transitivity of G/X on singular points, we may assume that Y was chosen so that ⟨D^{X}, s⟩ ≅ SL(2, 3) ∘ ℤ_{4} (the central product). Denote by L the pre-image of ⟨D^{X}, s⟩ in G^{∗}. Then O_{2}(L) is not isomorphic to Q_{8} × ℤ_{4}, by (*), and therefore s lifts to an involution in G^{∗}. This shows that {g^{2} : g ∈ O_{2}(G^{∗})} is of cardinality 2, and hence |Φ(O_{2}(G^{∗}))| = 2. This yields the lemma. □

Section 4: Automorphisms, and alternating groups

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

Proposition 4.1. Let G be a minimal counterexample to Theorem A. Then either \overline{H} is a sporadic group, or \overline{H} ∈ Lie(r), r ≠ p, and A induces a group of inner-diagonal automorphisms of \overline{H}.

Lemma 4.2. Assume Hypothesis 1.1, set \overline{G} = G/Z(G), and assume that \overline{H} is a group of Lie type in characteristic r, possibly with r = p. Assume further that G is a minimal counter-example to Theorem A, and let a ∈ A. Then a induces an inner-diagonal automorphism of \overline{H}.

Proof. Denote by α the automorphism of \overline{H} induced by a. By 3.5 we have α = xfg where x is an inner-diagonal automorphism, and where f and g are field and graph automorphisms, respectively. We assume that α ≠ x, and our aim will be to derive a contradiction from this assumption. We proceed by induction on |G|.

Suppose first that \overline{a} is not contained in any r-local subgroup of \overline{G}. Then r ≠ p, and α is not conjugate to f in Aut(\overline{H}). Then 3.6 implies that p = 3, and that \overline{H} ≅ D_{4}(q) or 3D_{4}(q) for some power q of r. As a is contained in an SL(2, 3) subgroup of G, \overline{a} is in a 2-local subgroup of \overline{G}, and so r ≠ 2. Let b be an element of order 3 in C_{H}(a), and set L = O^{r′}(C_{H}(b)). Then 3.8 and 3.7 together imply that L ≠ 1 and that L = L_{1} ⋯ L_{k} is a commuting product of groups L_{i} ∈ Lie(r). Moreover, as r > 3, each L_{i} is quasisimple. If k ≥ 3 and a permutes the factors L_{1}, L_{2}, and L_{3}, then L contains an abelian 3′-subgroup on which a acts non-trivially, and contrary to 2.9. Thus, a fixes each of the factors L_{i}. If [L_{1}, a] = 1 then a is in an r-local subgroup, so in fact [L_{1}, a] ≠ 1. We note that L_{1} ≠ H since O_{3}(G) = 1. By the induction hypothesis, a induces an inner-diagonal automorphism on L_{1}, and then since r ≠ 2, induction in Theorem A implies that r = 5 and L_{1} ≅ SL(2, 5). Now 2.5 shows that either L_{1} or C_{L_{1}⟨a⟩}(L_{1}) contains a quadratic element of order 3. As neither D_{4}(q) nor ^{3}D_{4}(q) occur as outcomes in Theorem A, we conclude, by induction, that C_{L_{1}⟨a⟩}(L_{1}) contains a quadratic element a_{1} of order 3. Now a_{1} lies in an r-local subgroup of G, and we may replace a by a_{1}. That is, we may assume from the beginning, and without loss of generality, that a is in an r-local subgroup of G. As r-local subgroups of \overline{G} are r-constrained, by 3.11, 2.9 implies that r = p, or r = 2 and p = 3.

Write \overline{H} = ^{d}Σ(q) as in section 3, and suppose first that r = p. As O_{p}(G) = 1, 3.12 implies that H ∈ Lie(r), and then any irreducible 𝔽_{p}H-module is the restriction to H of an irreducible module for Σ(q), by [St1, Theorem 13.3]. Here V is irreducible for H, by [Ch, Lemma 1.3], so we may now assume that \overline{H} is a Chevalley group. If g = 1 then \overline{H} may be taken to be L_{2}(q), q = r^{pm}, and we then violate 2.9 via the action of a on a Cartan subgroup of H. On the other hand, suppose that g ≠ 1, so that \overline{H} ≅ D_{4}(q). As a normalizes a Sylow 3-subgroup of H, there is then an a-invariant maximal subgroup M of H with M/O_{3}(M) isomorphic to a commuting product of three copies of SL(2, q), permuted transitively by a. There is a section W of V which centralizes O_{3}(M) and on which (M/O_{3}(M))⟨a⟩ acts faithfully. By 2.9, applied to the action of a on Z(M/O_{3}(M)), M/O_{3}(M) is a central product (with center of order 2). Then a acts on a central product of three quaternion groups in M/O_{3}(M), permuting the factors, and then once again there is an abelian 2-group on which a acts non-trivially. Thus 2.9 is violated in any case, and we conclude that r ≠ p.

We now have r = 2 and p = 3. Suppose next that α = f. There is then an ainvariant subgroup L of H, of Lie rank 1, such that a induces a field automorphism on L. By induction, it follows that \overline{H} itself has Lie rank 1. If H ≅ U_{3}(2^{3m}) then again there is an a-invariant subgroup of H isomorphic to L_{2}(2^{3m}), and on which a induces a field automorphism, contrary to induction. If H ≅ L_{2}(2^{3m}) or Sz(2^{3m}) then a acts non-trivially on the center of a Sylow 2-subgroup of H, contrary to 2.9. Thus Z(H) ≠ 1, and indeed the preceding argument via 2.9 shows that Z(H) contains a non-identity 2-group. Then \overline{H} ≅ Sz(8), by 3.12, and then 3.13 implies that a acts non-trivially on Z(H). Again, this outcome is contrary to 2.9.

We conclude that α ≠ f. Then 3.6 yields either \overline{H} ≅ D_{4}(q) and g ≠ 1, or \overline{H} ≅ ^{3}D_{4}(q). If Z(H) ≠ 1 then 3.12 and 3.13 yield \overline{H} ≅ D_{4}(2), Z(H) ≅ ℤ_{2} × ℤ_{2}, and a acts nontrivially on Z(H), contrary to 2.9. Thus Z(H) = 1. As a lies in an SL(2, 3)-subgroup of G, there exists a maximal 2-local subgroup M of H⟨a⟩ containing a. Set L = O^{2′}(M). Then M ∩H is a parabolic subgroup of H, and [Z(O_{2}(M)), a] = 1. In particular, we have [Z(S), a] = 1 for some Sylow 2-subgroup S of L, and S is also a Sylow 2-subgroup of G. We may then choose M so that M ∩ H is the maximal parabolic subgroup N_{H}(Z(S)), and then L/O_{2}(L) is isomorphic to a direct product of three copies of L_{2}(q) permuted transitively by a (in the D_{4}(q) case), or to L_{3}(q^{3}) (in the ^{3}D_{4}(q) case). A Sylow 3subgroup of L⟨a⟩ is then contained in a complement to O_{2}(L), so in either case we find that a acts non-trivially on an abelian 2-subgroup of L. Again 2.9 yields a contradiction, and the lemma is thereby proved. □

4.3 Lemma. Assume Hypothesis 1.1, and suppose that \overline{G} is an alternating group of degree n. Then |Z(G)| = 2, and if n is not equal to 6 then |A| = 3, and the non-identity elements of A project to 3-cycles in \overline{G}.

Proof. Suppose first that p > 3. There is then a quasisimple subgroup K of G with K/Z(K) ≅ Alt(p) and with K = [K, a]. Then two conjugates of a suffice to generate K⟨a⟩, and then 2.7(a) implies that p = 5 and K ≅ SL(2, 5). If K is contained in an a-invariant subgroup L of G with L/Z(L) ≅ Alt(6) then two conjugates of a will generate L⟨a⟩, which is contrary to 2.7(a). Thus we may assume that n is divisible by 5. If n = 5 then there is nothing more to prove, so we may reduce to the case where \overline{G} ≅ Alt(10) and where \overline{a} is a product of two disjoint 5-cycles. Here a lies in a subgroup L of G of the form SL(2, 5) ∘ SL(2, 5) (central product with amalgamated centers) and two conjugates of a will then generate a subgroup of L isomorphic to Alt(5). This is again contrary to 2.7(a). Thus, we need now only consider the case where p = 3.

By Hypothesis 1.1, 3 does not divide |Z(G)|, and so a classical result of Schur implies that |Z(G)| ≤ 2. Let a ∈ A, a ≠ 1. Let k be the number of 3-cycles in the standard notation for \overline{a}, and suppose first that k > 1. As a lies in no Frobenius subgroup of G of order 21, by 2.9, we then have n = 3k. Suppose k ≥ 3, and let \overline{L} be the stabilizer in \overline{G} of n − 9 points which are permuted by \overline{a} in three 3-cycles. Then \overline{L} has a subgroup \overline{K} isomorphic to SL(2, 8), acting on the nine points of the projective line. Any element of \overline{L} of order 3 is fixed-point-free on these points, so we can choose \overline{L} to be \overline{a}-invariant. Denote by L the pre-image of \overline{L} in G, and set L_{0} = [L, L]. Then L_{0} ≅ SL(2, 8) and L_{0} = [L_{0}, a]. But SL(2, 8) has no subgroup isomorphic to SL(2, 3), so we violate 2.7(a). Assuming now that n ≠ 6, we conclude that every non-identity element of A projects to a 3-cycle in \overline{G}. Then |A| = 3, and since A lies in no subgroup of G which is isomorphic to Alt(4), we obtain |Z(G)| = 2. On the other hand, if n = 6 and Z(G) = 1 then every element of order 3 in G lies in an Alt(4)-subgroup of G. Thus we conclude that, in any case, we have |Z(G)| = 2. □

Notice that 4.1 follows from lemmas 4.2 and 4.3, given our background hypothesis that \overline{H} is a “known” simple group.

Section 5: The case p > 3

Our aim in this section is to give a short proof of the following result (which is is proved also in [Sa]).

Theorem 5.1. Assume Hypothesis 1.1 with p > 3. Then G/Z(G) is a group of Lie type in characteristic p.

We fix notation as in section 1, so that H = F^{∗}(G) is a quasisimple group, and we have \overline{G} = G/Z(H). Assume Hypothesis 1.1, and fix a non-identity element a of A. Take G to be a minimal counter-example to Theorem 5.1. The following result is then immediate.

Lemma 5.2. Let K be a proper quasisimple subgroup of H, with K = [K, a]. Then K/Z(K) is a group of Lie type in characteristic p.

Lemma 5.3. \overline{H} is not a sporadic group.

Proof. Suppose false. Then the outer automorphism group of \overline{H} is of order at most 2, and so G = H. Suppose first that |C\overline{G}(\overline{a})| is even. Let \overline{t} be an involution in C\overline{G}(\overline{a}), and set \overline{C} = C\overline{G}(\overline{t}). As p > 3, 2.9 implies that F^{∗}(\overline{C}) ≠ O_{2}(\overline{C}), and it follows from [GLS3, Table 5.3] that \overline{C} has a component \overline{K} with \overline{a} ∈ \overline{K}. By 5.2, \overline{K}/Z(\overline{K}) is of Lie type in characteristic p, and then [GLS3, Table 5.3] yields p = 5, \overline{K} ≅ Alt(5), and \overline{G} ≅ M_{12}, J_{1}, or J_{2}. The inverse image K of \overline{K} in G is then isomorphic to SL(2, 5), by 2.7, so Z(G) ≠ 1, and so G ≅ 2^{·}M_{12} or 2^{·}J_{2}. In fact, in both these cases the cited table in [GLS3] gives the extra information that K ≅ ℤ_{2} × Alt(5), and so we may obtain a contradiction in this way. Alternatively, one may note that M_{12} has cyclic Sylow 5-subgroups and contains Alt(6), so that we contradict 2.7 in this case. In the case that \overline{G} ≅ J_{2}, we have ⟨\overline{a}⟩ contained in a subgroup isomorphic to Alt(4) × Alt(5) (the unique maximal subgroup of \overline{G} containing \overline{C}), and thus |C\overline{G}(\overline{a})| is divisible by 3. Of the two classes of subgroups of order 5 in \overline{G}, ⟨\overline{a}⟩ is then identified as lying in a subgroup isomorphic to 3^{▪}Alt(6), contrary to 2.7.

We conclude that C\overline{G}(\overline{a}) is of odd order. Now 2.8 implies that G has non-cyclic Sylow p-subgroups. Another trip through the cited table in [GLS3] shows, however, that for any element g of prime order p in a sporadic group X, if X has non-cyclic Sylow p-subgroups then C_{X}(g) is of even order. This yields the desired contradiction. □

Lemma 5.4. G is not isomorphic to SL(2, r^{n}) for any n.

Proof. Suppose false, and put q = r^{n}. From L.E. Dickson’s determination of the subgroups of L_{2}(q) (for which one may see [Suz]), we know that, for p ≠ r, the only possible SL(2, p) subgroups of G are given by p = 5. Here 5 must divide q +1, as otherwise a acts non-trivially on a Sylow r-subgroup of G, in violation of 2.9. Put N = N_{G}(⟨a⟩), and let L be the set of all subgroups L of G containing a and with L ≅ SL(2, 5). Each L ∈ L has precisely 50 elements g such that L = ⟨a, a^{g}⟩. From this we obtain:

|L| = (|G| − |N|)/50 = (q^{3} − 3q − 2)/50

Since each L ∈ L has exactly five Sylow 5-subgroups other than ⟨a⟩ it then follows that G has exactly 1 + (q^{3} − 3q − 2)/10 Sylow 5-subgroups. But this number is then equal to |G : N|, so we obtain:

1 + \frac{1}{10}(q^{3} − 3q − 2) = \frac{1}{2}(q^{2} − q)

which yields q^{3} − 5q^{2} + 2q + 8 = 0. That is: (q − 4)(q − 2)(q + 1) = 0, so that q = 4. As SL(2, 4) does not contain SL(2, 5) we have a contradiction at this point. □

Lemma 5.5. \overline{H} is not a group of Lie type in characteristic r different from p.

Proof. Suppose false. By 4.1, A induces a subgroup of Inndiag(\overline{H}). Let 1 ≠ a ∈ A, and suppose first that |C\overline{H}(\overline{a})| is of even order. Let \overline{t} be an involution in C\overline{H}(\overline{a}). If r = 2 then 3.11 implies that C\overline{G}(\overline{t}) is 2-constrained, and we contradict 2.9. Thus r ≠ 2. If \overline{H} ≅ P SL(2, r^{n}) then H ≅ SL(2, r^{n}) since H involves SL(2, p), and we then contradict 5.4. Thus \overline{H} is not isomorphic to P SL(2, r^{n}), and then by 3.8 and 3.7 there is a subnormal subgroup \overline{K} of C\overline{H}(\overline{t}), with \overline{K} of Lie type in characteristic r. As p > 3 and r ≠ 2 there are no isomorphisms between any members of Lie(r) and Lie(p), and so \overline{K} ∉ Lie(p). If [\overline{K}, \overline{a}] = 1 then \overline{a} is in an r-local subgroup of \overline{G}, and we again contradict 2.9 via the Borel-Tits theorem. Thus, [\overline{K}, \overline{a}] ≠ 1. Then 5.2 implies that ⟨(\overline{K})^{⟨\overline{a}⟩}⟩ is a product of p components of C\overline{H}(\overline{t}), or a commuting product of p copies of SL(2, 3) or of L_{2}(3). Again, the result is that \overline{a} lies in an r-local subgroup of \overline{G}, and a contradiction ensues. We therefore conclude that C\overline{G}(\overline{a}) is of odd order.

Now 2.8 shows that G has non-cyclic Sylow p-subgroups. As p is odd, there is then an elementary abelian subgroup \overline{B} of \overline{G} of order p^{2}, with \overline{a} ∈ \overline{B}. Then 3.9 implies that O^{r′}(C\overline{H}(\overline{b})) ≠ 1 for some \overline{b} ∈ \overline{B}. By 3.7 we then have O^{r′}(C\overline{H}(\overline{b}) = \overline{L}_{1} ⋯ \overline{L}_{m}, where each \overline{L}_{i} is a group of Lie type in characteristic r, and where [\overline{L}_{i}, \overline{L}_{j}] = 1 for all i ≠ j. Here O^{r′}(C\overline{H}(\overline{b})) is \overline{a}-invariant, and since we have already seen that \overline{a} lies in no r-local subgroup of \overline{G}, we conclude that each \overline{L}_{i} is \overline{a}-invariant and that [\overline{L}_{i}, \overline{a}] ≠ 1. Now 5.2 implies that each \overline{L}_{i} is solvable, and since p > 3 it then follows that \overline{L}_{i} has no automorphisms of order p. Then [\overline{L}, \overline{a}] = 1, and we have a contradiction. □

Notice that lemma 4.3, and lemmas 5.2 through 5.5, yield Theorem 5.1.

Section 6: Cross-Characteristic Lie Type Groups, p = 3

In this section we assume Hypothesis 1.1 with p = 3. As always, we set \overline{G} = G/Z(G) and H = E(G) We shall assume further that \overline{H} is a group of Lie type in characteristic different from 3. Indeed, we even wish to assume that there exists no exceptional isomorphism of \overline{H} with a group of Lie type in characteristic 3. Thus, \overline{H} is not isomorphic to Sp(4, 2)′ (≅ L_{2}(9)), G_{2}(2)′ (≅ U_{3}(3)), or U_{4}(2) (≅ P Sp(4, 3)).

By a “parabolic subgroup” of H, we mean the complete inverse image in H of a bona fide parabolic subgroup of H/Z(H). Similarly, we have the notions of “Borel subgroup”, “Cartan subgroup”, and of “root group” in H.

Our goal, in this section, is the following result.

6.1 Theorem. Assume Hypothesis 1.1, with H/Z(H) a group of Lie type, and not isomorphic to a group of Lie type in characteristic 3. Then either G is isomorphic to one of the groups P GU(n, 2), n ≥ 5, or else |Z(G)| = 2, and \overline{G} is isomorphic to one of the groups L_{2}(4), L_{4}(2), Sp(6, 2), D_{4}(2), or G_{2}(4). Moreover, we have |A| = 3 in every case.

For the remainder of this section, let G be a minimal counter-example to Theorem 6.1. Throughout, let r denote the defining characteristic of \overline{H}, r ≠ 3, and fix a non-identity element a ∈ A.

Lemma 6.2. Suppose that the Lie rank of H/Z(H) is equal to 1. Then G ≅ 2^{▪}L_{2}(4).

Proof. By 4.2, A induces inner-diagonal automorphisms on \overline{H}. Thus |Inndiag(\overline{H})| is divisible by 3, and so \overline{H} is not isomorphic to Sz(2^{n}). Also, as r ≠ 3, by assumption, \overline{H} is not a Ree group in characteristic 3. Thus \overline{H} ≅ P SL(2, q) or P SU(3, q) for some q, q = r^{n}.

Suppose first that \overline{H} ≅ P SL(2, q). Then |Inndiag(H) : \overline{H}| ≤ 2, so A ≤ H, and so H = G. As G involves SL(2, 3), we conclude that |Z(G)| = 2. Assuming that \overline{G} is not isomorphic to SL(2, 4), it follows from 3.12 that r is odd. Thus r ≥ 5, and since 2^{·}SL(2, 4) ≅ SL(2, 5) we have q > 5. Put d = q − 1, and let λ be a primitive d^{th} root of unity in 𝔽_{q}. Then take:

a = ( \begin{matrix}−1 & −1 \\ 1 & 0 \end{matrix} ) b = ( \begin{matrix}0 & λ^{−1} \\ −λ & −1 \end{matrix} )

and obtain:

ab = ( \begin{matrix}λ & 1 − λ^{−1} \\ 0 & λ^{−1} \end{matrix} )

Here b is of order 3, so b is conjugate to a in G. By 2.6, we have ⟨a, b⟩ isomorphic to SL(2, 3) or SL(2, 5), and so |ab| ≤ 5. But ab has order q − 1, so we conclude that q ≤ 6, and then q = 5, contrary to our choice of q.

Suppose next that \overline{H} ≅ P SU(3, q). Let L_{0} be a subgroup of H with L_{0} = ⟨a^{L_{0}}⟩ ≅ SL(2, 3). Let t be the involution in L_{0}, and set L = O^{r′}(C_{H}(t)). Then L ≅ SL(2, q), and [L, a] ≠ 1. By what has already been shown in the preceding paragraph, we then have q = 5. Both SU(3, 5) and P GU(3, 5) have extraspecial Sylow 3-subgroups of exponent 3. In particular, all subgroups of order 3 in P GU(3, 5) which are contained in P SU(3, 5) are conjugate. By the Frattini argument, the normalizer in P GU(3, 5) of a Sylow 5-subgroup contains such an “outer” subgroup of order 3, so we must conclude from 2.9 that a ∈ H. One may deduce from the action of SU(3, 5) on its natural module that all subgroups of order 3 in P SU(3, 5) are conjugate. As P SU(3, 5) contains a Frobenius group of order 21, we again contradict 2.9. □

Lemma 6.3. \overline{H} is not isomorphic to P SL_{3}(q) for any power q of r.

Proof. Suppose \overline{H} ≅ P SL_{3}(q), q = r^{n}. As always, we have a subgroup X of G containing a, with X ≅ SL(2, 3). Consider first the case where Z(H) = 1. If r is even, then the centralizer in G of any non-identity 2-subgroup of G is contained in a parabolic subgroup of G, while if r is odd then G has a unique conjugacy class of involutions. In either case we find that C_{G}(Z(X)) is contained in a proper parabolic subgroup P of G. Here O_{r}(P) is abelian, and then 2.9 yields a ∈ C_{G}(O_{r}(P)), whereas O_{r}(P) ≥ C_{G}(O_{r}(P)).

We conclude that Z(H) ≠ 1. As G acts irreducibly on V, |Z(H)| is prime to 3, so 3.12 implies that \overline{H} ≅ L_{3}(4) and Z(H) is a 2-group. Since a centralizes Z(H), we conclude from 3.13 that a ∈ H. As all elements of order 3 in L_{3}(4) are conjugate, and as L_{3}(4) contains a Frobenius group of order 21, we contradict 2.9. This proves the lemma. □

Lemma 6.4. If the Lie rank of \overline{H} is at least 2 then r = 2, and a normalizes a maximal parabolic subgroup of \overline{H}.

Proof. Assume that the Lie rank of \overline{H} is at least 2. Then \overline{a} lies in an r-local subgroup \overline{N} of \overline{G}, by 6.3 and 3.8. By 3.11(a), \overline{N} is r-constrained, so 2.9 implies that r = 2. By 3.11(b) we may choose \overline{N} so that \overline{N} ∩ \overline{H} is a parabolic subgroup of \overline{H}. Let \overline{P} be an a-invariant, proper parabolic subgroup of \overline{H}. As a induces an inner-diagonal automorphism of \overline{H} we may write a = xd where x ∈ P and where d ∈ N_{G}(\overline{S}) where \overline{S} is a Sylow r-subgroup of \overline{P}. Then any maximal parabolic subgroup of \overline{H} containing \overline{P} is a-invariant. □

For the remainder of this section we assume that the Lie rank of G is at least 2. Thus r = 2, by 6.4. Fix a Borel subgroup B of H, and let Σ (resp. Σ^{+}) be the root system (resp. the positive subsystem) associated with H and with B, so that O_{2}(B) is generated by the root groups X_{α}, α ∈ Σ^{+}. If P is a parabolic subgroup of H containing B then the set of simple roots α ∈ Σ^{+} such that X_{−α} ≤ P will be denoted D(P). We take D(P) to have also the structure of a graph, with incidence induced from the Coxeter diagram of Σ, and we say that P is connected if D(P) is connected. More generally, let D_{1}, ⋯, D_{r} be the connected components of D(P), and for each i, 1 ≤ i ≤ r, put

L_{i} = ⟨X_{α_{i}}, X_{−α_{i}} : α ∈ D_{i}⟩ and Λ = Λ(P) = {L_{1}, ⋯, L_{r}}.

We will refer to the members of Λ(P) as the Levi complements of P, relative to Σ.

Lemma 6.5. Assume that the Lie rank of G is at least 2. Then there is a Sylow 2subgroup S of H, and a proper parabolic subgroup P of G containing ⟨N_{H}(S), a⟩, for which the following condition holds.

(*)  For every L ∈ Λ(P) we have 1 ≠ [a, L] ≤ L.

Moreover, we can choose P so that the Lie rank of each L in Λ(P) is equal to 1.

Proof. By 6.4, a normalizes a maximal parabolic subgroup N of H. If [N, a] ≤ O_{2}(N) then a normalizes a Sylow 2-subgroup of N (hence of H), and then a normalizes every parabolic subgroup of H containing S. In particular, there is then a rank-1 parabolic subgroup P of H, invariant under a, and with [O^{2′}(P), a] ≰ O_{2}(P). Thus, the desired conclusion holds in this case, and we may therefore assume that [N, a] ≰ O_{2}(N).

Among all a-invariant parabolic subgroups N with [N, a] ≰ O_{2}(N), choose N so that the Lie rank of N is as small as possible. We then construct the set Λ(N) = {L_{1}, ⋯, L_{r}} of Levi complements in N, relative to a fixed Borel subgroup of N. Then N = O_{2}(N)L_{1} ⋯ L_{r}K, for some Cartan subgroup K of B, and we may assume (possibly after replacing a by a conjugate) that a normalizes L_{1} ⋯ L_{r}K. As a is inner-diagonal, a normalizes each L_{i}, and if [a, L_{i}] = 1 for some i we contradict the minimality of N. This proves the first part of the lemma. But further, if the Lie rank of some L_{i} is bigger than 1, then we may apply induction on the Lie rank, with L_{i}⟨a⟩ in place of G, to conclude that L_{i}⟨a⟩ has a proper parabolic subgroup containing a. From this we again contradict the minimality of N, and thus each L_{i} has Lie rank equal to 1. □

Lemma 6.6. Assume that the Lie rank of \overline{H} is at least 2, and assume that the field of definition for \overline{H} (in the sense of a σ-setup, as in section 3) is larger than 𝔽_{2}. Then G ≅ 2^{·}G_{2}(4), and |A| = 3. Moreover, we have A = Z(R) for some Sylow 3-subgroup R of G, and C_{G}(A) ≅ SL(3, 4).

Proof. By 6.5 there exists a proper parabolic subgroup P of G, and a Levi complement L in P, such that L ≥ [L, a] ≠ 1. Since the field of definition of \overline{G} is larger than 𝔽_{2}, we may apply 6.2 to L⟨a⟩ and obtain L ≅ 2^{·}L_{2}(4). Here L ∉ Lie(2), so Z(L) ≤ Z(H). Then 3.12 and 6.3 yield G ≅ 2^{·}G_{2}(4).

Let R be a Sylow 3-subgroup of G containing A. Then |R| = 27, and R is contained in an SL(3, 4) subgroup X of G_{2}(4). Every element of R − Z(R) is contained in a Frobenius subgroup of X of order 21, so 2.9 implies that A = Z(R) is of order 3. We observe that A is contained in a Cartan subgroup D of X, which is a Cartan subgroup of G. The Chevalley relations imply that X is generated by the set of root subgroups centralized by A, relative to the root system determined by D. Then X = O^{2′}(C_{G}(A)), and C_{G}(A) = XD = X. □

For the remainder of this section we assume that the Lie rank of \overline{H} is at least 2, and that 𝔽_{2} is the field of definition for \overline{H}. Further, we assume that there exists no exceptional isomorphism between \overline{H} and a group in Lie(3). By 6.5, we may fix a parabolic subgroup P of G containing a, such that condition (*) in 6.5 holds, and such that every member of Λ(P) is of Lie rank 1. Let M be the set of of all maximal parabolic subgroups of H containing P, and having the property that every connected component of the diagram D(M) contains at least one vertex of D(P). One readily verifies that M is non-empty, and we fix M ∈ M.

Lemma 6.7. The following hold.

(a) We have [L, a] ≠ 1 for any L ∈ Λ(M).

(b) We have ⟨a^{M}⟩ ≥ O^{2′}(M).

(c) Let S be a Sylow 2-subgroup of M, and suppose that Z(S) ≰ Z(H). Then M = N_{H}(Z(S)), and M = {M}.

Proof. Part (a) is immediate from the definition of M. Then L ≤ ⟨a^{L}⟩ for any L ∈ Λ(M), and (b) follows. Suppose that Z(S) ≰ Z(H). We have [Z(O_{2}(M)), a] = 1 by 2.9, and it follows from part (b) that [Z(O_{2}(M)), O^{2′}(M)] = 1. Then Z(S) ⊴ M, and then since M is a maximal parabolic we have M = N_{H}(Z(S)). This yields (c). □

Lemma 6.8. Suppose that \overline{H} is isomorphic to P SU(n, 2) n ≥ 5. Let ϕ be the canonical homomorphism from GU(n, 2) onto P GU(n, 2), and let U be the natural module for GU(n, 2) over 𝔽_{4}. Then G ≅ P GU(n, 2), |A| = 3, and a = ϕ(a∗) for some element a^{∗} ∈ GU(n, 2) such that C_{U}(a∗) has codimension 1 in U.

Proof. Let S be a Sylow 2-subgroup of M. It follows from 3.14 that Z(S) ≰ Z(G), so 6.7(b) yields M = N_{H}(Z(S)). Then O_{2}(\overline{M}) is extraspecial of width n − 2, and O^{2′}(\overline{M}/O_{2}(\overline{M})) is isomorphic to SU(n−2, 2). Further, as O_{3}(G) = 1 and Z(G) is cyclic, it follows from 3.12 that |Z(G)| ≤ 2, and that Z(G) = 1 if n ≠ 6. Then 3.14 implies that O_{2}(M) = X × Z(G), where X is a central product of n − 2 quaternion groups.

Set Y = [O_{2}(M), a]. We have Φ(Y) ≤ Z(X), so Y is contained in an extraspecial subgroup of O_{2}(M), and then 2.9 implies that Y is a quaternion group. From this we may conclude that M/O_{2}(M)Z(G) is isomorphic to GU(n − 2, 2), and then \overline{G} ≅ P GU(n, 2). In particular, if 3 divides n then a ∉ H, and so 3.13 yields Z(G) = 1. Thus O_{2}(M) = X, and C_{X}(a) is a central product of n − 3 quaternion groups. It follows that O^{2′}(C_{G}(a)) ≅ SU(n − 1, 2). Set G^{∗} = GU(n, 2), let U be the natural module for G^{∗} over 𝔽_{4}, and let a^{∗} be a pre-image of a in G^{∗}. Then O^{2′}(C_{G^{∗}}(a^{∗})) ≅ SU(n − 1, 2), and we may choose a^{∗} so that C_{U}(a^{∗}) has codimension 1 in U.

Suppose that |A| > 3. Then C_{G}(a) = ⟨A^{C_{G}(a)}⟩, and so

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

Then [C_{G}(a), C_{G}(a)] centralizes V, by the Three Subgroups Lemma. But C_{G}(a) is nonabelian, as n > 3. Thus |A| = 3, and all parts of the lemma have been established. □

Lemma 6.9. Assume that \overline{H} is defined over 𝔽_{2}, that \overline{H} is not a unitary group P SU(n, 2) with n ≥ 5, and that \overline{H} cannot be viewed (via an exceptional isomorphism) as a group of Lie type in characteristic 3. Then |Z(G)| = 2, and \overline{G} is isomorphic to Ω_{4}^{−}(2), L_{4}(2), Sp(6, 2), or D_{4}(2).

Proof. As A induces inner-diagonal automorphisms on \overline{H}, it follows that H = G or that \overline{H} ≅ ^{2}E_{6}(2). Suppose first that \overline{G} is isomorphic to L_{n}(2) or Sp(2n, 2). If Z(G) ≠ 1 then 3.12 and 6.3 yield G ≅ 2^{·}L_{4}(2) or 2^{·}Sp(6, 2), and thus the lemma holds in this case. On the other hand, if Z(G) = 1 then C_{G}(Z(S)) is not a maximal parabolic subgroup of G, and we contradict 6.7(b). Thus, we may assume that \overline{G} is not isomorphic to L_{n}(2) or Sp(2n, 2).

Suppose that \overline{G} is an orthogonal group Ω_{2n}^{ϵ}(2), and let U be a natural module for \overline{G} over 𝔽_{2}, of dimension 2n. As G is non-solvable we have n ≥ 2, and n ≥ 3 if ϵ = +1. In view of 6.2, and the isomorphism of Ω_{4}^{−}(2) with SL(2, 4), we need only consider the cases where n ≥ 3. As Ω_{6}^{+}(2) ≅ L_{4}(2), and Ω_{6}^{−}(2) ≅ P Sp(4, 3), we may in fact take n ≥ 4. If Z(H) ≠ 1, then 3.12 yields \overline{G} ≅ D_{4}(2) (which is isomorphic to Ω_{8}^{+}(2)), and then since Z(G) is cyclic, 3.12 yields |Z(G)| = 2. Thus, the lemma holds in this case, and so we may assume that Z(G) = 1.

Let U_{0} be a totally singular subspace of U, of dimension 2, and denote by L the stabilizer in G of U_{0}. Without loss, we may assume that a Sylow 2-subgroup S of L is contained in M. With the aid of Witt’s Theorem on extensions of isometries, we find that L = X(K_{1} × K_{2}), where X = O_{2}(L), K_{1} ≅ Ω_{2n−4}^{ϵ}(2) and K_{2} ≅ L_{2}(2). Further, X is extraspecial, of width 2n − 4, and X/Z(X) is isomorphic, as a module for K_{1}K_{2}, to a tensor product N_{1} ⊗ N_{2}, where N_{i} is a natural module for K_{i} over 𝔽_{2}. In particular, L is a maximal parabolic subgroup of G, and Z(S) = Z(L), so L = M by 6.7(b), and a ∈ L. Now let N_{0} be an irreducible K_{1}-submodule of X/Z(X). Then X/Z(X) = N_{0} ⊕(N_{0})^{g} for any g ∈ K_{1}K_{2} −K_{1}. For any element d of K_{1} of order 3 we then have |[X/Z(X), d]| ≥ 16, and so [X, d] is not a quaternion group. Thus a ∉ K_{1}, by 2.9. But, for any element d of K_{1}K_{2} − K_{1} of order 3, we have |[X/Z(X), d]| ≥ |N_{0}|, where |N_{0}| ≥ 16 as n ≥ 4. As a is conjugate to an element of K_{1}K_{2}, we have a contradiction at this point. Thus, we may assume that \overline{H} is not an orthogonal group.

As \overline{H} is not a unitary group (the case of U_{4}(2) ≅ Ω_{6}^{−}(2) having been treated above), we now conclude that \overline{G} is not a classical group. If \overline{G} ≅ E_{n}(2), (n = 6, 7, 8), then |Z(G)| is odd, and so |M| = 1, by 6.7. Recall, however, that M is the set of maximal parabolic subgroups M of H containing P, where P is a totally disconnected parabolic subgroup of H, and where each connected component of M contains at least one vertex of D(P). One has only to glance at the diagrams for the groups E_{n}(2), however, to see that in fact |M| > 1 for any choice of P. Thus, \overline{G} ≇ E_{n}(2). Suppose that \overline{G} ≅ ^{2}F_{4}(2)′. Then again Z(G) = 1 and M = {C_{G}(Z(S))}. Then M/O_{2}(M) ≅ Sz(2), and so |M| is prime to 3, contrary to a ∈ M. By a similar argument, if \overline{H} ≅ ^{3}D_{4}(2) then |Z(G)| = 1 and a ∈ M = C_{G}(Z(S)), so that M has an a-invariant Levi complement isomorphic to SL(2, 8). But this result is excluded by 6.2. As G_{2}(2)′ may be viewed as a group in characteristic 3, we come finally to \overline{G} ≅ F_{4}(2) or \overline{H} ≅ ^{2}E_{6}(2). Then D(G) is the F_{4} diagram, and since D(P) is totally disconnected we can choose M ∈ M so that D(M) contains a subdiagram of type A_{2}. There then exists L ∈ Λ(M) with L/Z(L) ≅ L_{3}(2) or L_{3}(4). Replacing G by L⟨arangle, we obtain a contradiction from 6.3. □

Lemma 6.10. If \overline{G} ≅ L_{4}(2), Sp(6, 2), G_{2}(4), or D_{4}(2) then |A| = 3.

Proof. Suppose, by way of contradiction, that |A| = 9. If \overline{G} ≅ L_{4}(2) it follows that both classes of elements of order 3 in G are represented in A, and since G contains a Frobenius group of order 21, we contradict 2.9.

Suppose next that \overline{G} ≅ Sp(6, 2), and let U be the natural module for \overline{G} over 𝔽_{2}. If there exists a in A with |[U, \overline{a}]| ≠ 16 then \overline{a} is contained in an L_{3}(2)-subgroup of \overline{G}, contrary to 6.3. On the other hand, we have U = ⟨C_{U}(\overline{a}) : 1 ≠ \overline{a} ∈ \overline{A}⟩, so there exist a, b ∈ A such that |[U, \overline{a}]| = |[U, \overline{b}]| = 4, and with ⟨a, b⟩ = A. Then |[U, \overline{ab}]| = 16, and so we have a contradiction in this case.

Finally, suppose that \overline{G} ≅ D_{4}(2), and let U be a natural O_{8}^{+}(2)-module for \overline{G}. We claim that there exists a ∈ A with 16 ≤ |[U, a]| ≤ 64. Suppose false. Then |[U, a]| = 4 or 2^{8} for every non-identity element a ∈ A. Let a and b generate A. If |[U, a]| = |[U, b]| = 4 then |[U, ab]| = 16, while if |[U, a]| = 4 and |[U, b]| = 2^{8} then either [U, ab] or [U, ab^{2}] is of order 64. The only other case is that in which C_{U}(a) = 0 for every non-identity a ∈ A, which is absurd. The claim is therefore established. Now fix a ∈ A with |[U, a]| = 16 or 64. There is then a non-degenerate a-invariant subspace W of V, of type O_{6}^{+}(2), with |[W, a]| = 16. Let H be the point-wise stabilizer in G of W^{⊥}. Then H ≅ Ω_{6}^{+}(2) ≅ Alt(8), and we have H = [H, a]. Identifying H with Alt(8), and identifying W with the nontrivial irreducible constituent in the natural permutation module for Alt(8), it follows that a induces on W the action of a product of two disjoint 3-cycles. There is then a 7-cycle x in H with x^{a} = x^{2}. This contradicts 2.9, so the lemma is proved. □

Theorem 6.1 follows from lemmas 6.2 through 6.10.

Section 7: Sporadic Groups, p = 3

We continue to assume Hypothesis 1.1, with p = 3. Further, we assume that \overline{H} is among the 26 sporadic simple groups. The index of \overline{H} in its automorphism group is then at most 2, and then since G = ⟨A^{G}⟩ we have G = H. Also, since O_{3}(G) = 1, the only cases in which Z(G) ≠ 1 occur when Z(G) is of order 2 or (in the unique case of M_{22}) of order 4). We will obtain the following result.

7.1 Theorem. Assume Hypothesis 1.1, with \overline{G} a sporadic simple group. Then G ≅ 2^{·}J_{2}, 2^{·}Suz, or 2^{·}Co_{1}, and we have |A| = 3.

We will make free use of the tables in section 5.3 of [GLS3], in which, for each sporadic group X, and each subgroup Y of X of prime order, the normalizer N = N_{X}(Y) is determined, in the sense that a chief series for N is given, along with the action of N on the various chief factors. Also, we will draw on the character tables in the ATLAS of Finite Groups [CCNPW], in order to establish that 2^{·}Co_{1} contains a perfect central extension 6^{·}Suz, and that 2^{·}Suz contains a perfect central extension 6^{·}U_{4}(3).

Six cases may be eliminated right away. Namely, by 2.9, if G has a unique conjugacy class of subgroups of order 3, then G does not contain a Frobenius subgroup of order 21. In this way, we obtain the following result.

7.2. \overline{G} is not isomorphic to M_{22}, M_{23}, J_{1}, HS, Ru, or O′N.

We next observe, that the centralizer of any element of order 3 in any sporadic group is of even order. In particular, |C\overline{G}(a)| is even. For the remainder of this section, we fix an element t of C_{G}(a) with \overline{t} of order 2. Set \overline{C} = C\overline{G}(\overline{t}), and denote by C the inverse image of \overline{C} in G. Also, set K = ⟨a^{C}⟩, and set R = F^{∗}(K). We will proceed by induction on |G|.

Definition 7.3. Let X be a group, and set R_{0} = F^{∗}(X). We say that X is of extraspecial type if the following three conditions hold.

(i) R_{0} = Z(R_{0})E where R_{0} is an extraspecial group of width n ≥ 2.

(ii) X/R_{0} is isomorphic to one of the groups Alt(2n + 1), Alt(2n + 2), GU(n, 2),

Ω_{2n}^{ϵ}(2) (for some sign ϵ), or Sp(2n, 2), and (iii) R_{0}/Z(R_{0}) is a natural (irreducible) 𝔽_{2}-module for K/R_{0}.

Lemma 7.4. The following hold.

(a) If K is quasisimple then K/O_{3}(K) is in the list of quasisimple groups which are outcomes in Theorem 1.2.

(b) If R = Z(R)E is a 2-group where E is extraspecial of width n ≥ 2, then K is of extraspecial type, in the sense of 7.3.

Proof. Part (a) is by induction on |G|. Part (b) is immediate from [Ch, Theorem A]. □

Before going to work with 7.4, it will be convenient to eliminate eight more groups by considering 5-local subgroups.

Lemma 7.5. \overline{G} is not isomorphic to Mc, Co_{3}, Co_{2}, Ly, F_{5}, F_{3}, F_{2}, or F_{1}.

Proof. We first show that in each of the above possibilities for \overline{G} we have |C_{G}(a)| divisible by 5. Indeed, in the cases other than \overline{G} ≅ Mc, F_{3}, or Co_{3}, one checks that the centralizer of every element of order 3 has a subgroup of order 5.

Suppose that \overline{G} ≅ Mc. Then Z(G) = 1, G has one class of involutions, and then C_{G}(t) ≅ 2^{▪}A_{8}. Then C_{G}(⟨a, t⟩) ≅ ℤ_{3} × SL(2, 5), by 4.3, and thus |C_{G}(a)| is divisible by 5 in this case.

Suppose next that \overline{G} ≅ F_{3}. Then Z(G) = 1, G has just one conjugacy class of involutions, and we have C_{G}(t) of the form 2_{+}^{1+8}.Alt(9). Now [Ch] shows that a is incident with a 3-cycle in C_{G}(t)/O_{2}(C_{G}(t)), and so we again get 5 dividing the order of C_{G}(a).

Suppose that \overline{G} ≅ Co_{3} and that 5 does not divide the order of C_{G}(a). Again, we have Z(G) = 1, and we find that C_{G}(a) ≅ ℤ_{3} × L_{2}(8) : 3. In particular, a is not contained in the commutator subgroup of C_{G}(a). Now consider C = C_{G}(t). By 7.4(a), C is not isomorphic to ℤ_{2} × M_{12}. This leaves only the case C ≅ 2^{·}Sp(6, 2). Let U denote the natural Sp(6, 2)-module for C. As 5 does not divide |C_{G}(a)| we have |[U, a]| > 4, and since a is not in the commutator subgroup of C_{G}(a) we have |[U, a]| ≠ 64. This leaves |[U, a]| = 16. But then a lies in a Frobenius 21-subgroup of G, and we have a contradiction via 2.9. Thus, we have found that |C_{G}(a)| is divisible by 5 in all cases under consideration.

Let F be a subgroup of C_{G}(a) of order 5. Then C\overline{G}(\overline{F}) is not 5-constrained, by 2.9. We consult [GLS3, Table 5.3] for the structure of centralizers of elements of order 5. setting D = O^{3′}(C_{G}(F)), we have D ≠ 1. Further, D is not isomorphic to Alt(5) (as follows from 2.4) or to U_{3}(5) (by theorem 6.1), or to HS or F_{5} (by induction in 7.1). But in fact, as one checks, this exhausts the list of possibilities for the structure of C_{G}(F), and so 7.5 is proved. □

7.6. \overline{G} is isomorphic to J_{2}, Suz, or Co_{1}.

Proof. We shall go through the list of groups, and check the conditions in 7.3 against the structure of the centralizers of involutions in the sporadic groups that remain to be considered. In view of 7.2 and 7.5, these are (aside from the three groups mentioned in the statement of the lemma) the nine groups F i′_{24}, F i_{23}, F i_{22}, He, J_{4}, J_{3}, M_{24}, M_{12}, and M_{11}. We note that, among these nine groups, only F i_{22}, and M_{12} have non-trivial Schur multipliers, and in these two cases the multiplier is of order 2.

We begin with G ≅ F i′_{24}. Here there are two classes of involutions, and we find that either K is double cover of F i_{22} or R is an extraspecial 2-group of width 6 with K/R isomorphic to 3U_{4}(3). In both these cases, we violate 7.4.

Suppose next that G ≅ F i_{23}. In view of 7.4(a), K is not a Schur extension of F i_{22} or U_{6}(2). This leaves only the possibility that C is of the form

(1)  (2^{2} × Q_{8}^{(4)})((GU_{4}(2))2).

Then Z(C/O_{2}(C)) is of order 3, acting non-trivially on Z(O_{2}(C)), as follows from the structure of the corresponding involution-centralizer in F i′_{24}. Thus K/R ≅ U_{4}(2). Here Z(C/O_{2}(C)) acts non-trivially on R/Z(R), so that R/Z(R) is the natural unitary module for K/R. This violates 7.4(b).

Suppose that \overline{G} ≅ F i_{22}. By 7.4(a), K is not a Schur extension of U_{6}(2) and examination of the remaining classes of involution centralizers then yields F^{∗}(C) = O_{2}(C). Further, for any involution \overline{s} of \overline{G} such that C\overline{G}(\overline{s}) is 2- constrained, either C\overline{G}(\overline{s}) is of the form

(2)  (2 × Q_{8}^{∗4})U_{4}(2)

or C\overline{G}(\overline{s}) does not contain a Sylow 2-subgroup of G. It follows from 2.9 that [Z(S), a] = 1 for some Sylow 2-subgroup S of G. Since the group in (2) lifts to a subgroup of the group in (1) in F i_{23}, one observes that there exists a 2-central involution s of G with s ∉ Z(G). We may then take s = t, whence \overline{C} is as in (2). This violates 7.4.

In the group He there are are two classes of involutions, and we find that either K is a Schur extension of L_{3}(4) or K is of the form D_{8}^{∗3} : L_{3}(2). Both these possibilities are excluded by 7.4 (or by noticing that in both these groups, each element of order 3 lies in a Frobenius group of order 21).

In J_{4} there are two classes of involutions, and we find that K is of the form Q_{8}^{∗6}(3M_{22}) or 2^{11}M_{22}, in each case violating of 7.4.

If G is isomorphic to J_{1} or J_{3}, we obtain C/O_{2}(C) ≅ Alt(5), and C contains a subgroup isomorphic to Alt(5). This violates 2.7.

In M_{24} there are two classes of involutions, with centralizers of the form (D_{8}^{∗3})L_{3}(2) and (2^{6})Sym(5). Thus, we violate 7.4 if G ≅ M_{24}.

If G is isomorphic to M_{11} then G has a single conjugacy class of elements of order 3, and since M_{11} ≥ M_{10} ≥ Alt(6) ≥ Alt(4), we contradict 2.9.

Finally, suppose that \overline{G} is isomorphic to M_{12}, and let S be a Sylow 3-subgroup of G containing a. Then S is extraspecial of order 27. If a ∈ Z(S) then every elementary abelian subgroup of order 9 in G contains a conjugate of a, and hence a lies in an M_{11}-subgroup of G, contrary to the preceeding paragraph. Thus a ∉ Z(S), and one then has C\overline{G}(\overline{a}) ≅ ℤ_{3} × A_{4}. Let \overline{s} be an involution in C\overline{G}(\overline{a}). If \overline{s} is 2-central then O^{2}(C\overline{G}(\overline{s})) ≅ 2_{+}^{1+4} : ℤ_{3}, whereas C\overline{G}(⟨\overline{a}, \overline{s})⟩) ≅ ℤ_{6} × ℤ_{2}. Thus \overline{s} is not 2-central, and so C\overline{G}(\overline{s}) ≅ ℤ_{2} × Σ_{5}. Then also \overline{a} normalizes a fours group \overline{F} ⊆ E(C\overline{G}(\overline{s})), where every involution in \overline{F} is 2-central. Now 2.9 implies that G ≅ 2^{·}M_{12}. But also, we have N\overline{G}(\overline{F}) ≅ 4^{2} : D_{12}. Let \overline{X} be the normal subgroup of N\overline{G}(\overline{F}) with \overline{X} ≅ ℤ_{4} × ℤ_{4}, let X be the pre-image of \overline{X} in G, and let F be the pre-image of \overline{F} in G. Then X = ⟨x, y⟩, where F = ⟨x^{2}, y^{2}⟩ and F ⟨a⟩ ≅ SL(2, 3). Thus [x^{2}, y^{2}] ≠ 1. But [x, y] ∈ Z(X), so [x^{2}, y^{2}] = [x, y]^{4} = [x^{4}, y] = 1, for a final contradiction. □

7.7. If \overline{G} ≅ J_{2}, Suz, or Co_{1} then |Z(G)| = 2 and |A| = 3, and we have C_{G}(A) ≅ 2^{·}Alt(6), 6^{·}U_{4}(3), or 6^{·}Suz, respectively.

Proof. Suppose first that \overline{G} ≅ J_{2}. There are two classes of involutions in \overline{G}, with centralizers isomorphic to either 2_{−}^{1+4} : Alt(5) or 2^{2} × Alt(5). Thus, \overline{C} has a subgroup L containing a and isomorphic to Alt(5), and so 2.6 implies that Z(G) ≠ 1. There are two conjugacy classes of subgroups of order 3 in \overline{G}, with centralizers 3^{·}Alt(6) and 3×SL(2, 3). Suppose that C\overline{G}(\overline{a}) ≅ 3 × SL(2, 3). In the notation of [GLS3, Table 5.3g] (which is the same as ATLAS notation, cf. [CCNPW]) we then have \overline{a} ∈ 3B, and the table gives [3B, 2C] = 1 where 2C is an outer involution of \overline{G} satisfying C\overline{G}(2C) ≅ L_{3}(2). Thus \overline{a} lies in a Frobenius group of order 21, and we violate 2.9. This shows that \overline{a} is in the class 3A. Now suppose that |A| > 3. A Sylow 3-subgroup S of G is extraspecial of order 27, and N\overline{G}(\overline{S}) contains a dihedral subgroup D of order 8 which acts faithfully on \overline{S}/Φ(\overline{S}). It follows that D acts transitively on the set of maximal elementary abelian subgroups of S, and so A contains representatives from each conjugacy class of subgroups of order 3 in G. But we have seen that A contains representatives of only one class, so in fact |A| = 3.

Suppose next that \overline{G} ≅ Suz. Then \overline{G} has two classes of involutions. One of these has a corresponding centralizer \overline{C}_{0} with O^{3′}(\overline{C}_{0}) ≅ L_{3}(4). It follows from theorem 6.1 that \overline{t} represents the other class, with \overline{C} an extension of an extraspecial group 2_{−}^{1+6} by Ω_{6}^{−}(2). By 3.14, Z(G) is a direct factor of O_{2}(C), and then 2.9 implies that C_{O_{2}(\overline{C})}(\overline{a}) ≅ 2_{+}^{1+4}. Thus C\overline{G}(⟨\overline{a}, \overline{t}⟩) is an extension of ⟨\overline{a}⟩ × C_{O_{2}(\overline{C})}(\overline{a}) by Ω_{4}^{+}(2), and so 2^{7} divides |C\overline{G}(\overline{a})|. This information suffices to single out the conjugacy class of ⟨\overline{a}⟩, and to yield C\overline{G}(\overline{a}) ≅ 3^{·}U_{4}(3). Let \overline{f} be an element of order 5 in C\overline{G}(\overline{a}) and set Y = O^{3′}(C_{G}(f)). Then a ∈ \overline{Y} ≅ Alt(6) or Alt(5), and then 2.6 implies that |Z(Y)| = 2. But Z(Y) ≤ Z(G), and so |Z(G)| = 2.

Suppose that |A| > 3. One checks from the character table for 2^{·}Suz in [CCNPW] that C\overline{G}(\overline{a}) lifts to a completely nonsplit extension 6^{·}U_{4}(3) in G, so C_{G}(a) = ⟨A^{C_{G}(a)}⟩. Then [V, a, C_{G}(a)] = 0, whereas Z(G) is fixed-point-free on V. This contradiction shows that |A| = 3.

Suppose finally that \overline{G} ≅ Co_{1}. Then \overline{G} has three classes of involutions, with corresponding centralizers \overline{C}_{i}, (1 ≤ i ≤ 3), where the groups \overline{C}_{i} have the structure given as follows.

O_{2}(\overline{C}_{1}) ≅ 2^{11} and \overline{C}_{1}/O_{2}(\overline{C}_{1}) ≅ M_{11},
\overline{C}_{2} ≅ 2^{2} × G_{2}(4),
O_{2}(\overline{C}_{3}) ≅ 2_{+}^{1+8} and \overline{C}_{3}/O_{2}(\overline{C}_{3}) ≅ D_{4}(2).

Let C_{i} denote the inverse image of \overline{C}_{i} in G. If C = C_{1} we obtain a faithful quadratic module either for M_{11} or for C_{1}, and we contradict 7.4. Suppose that C = C_{2}. Then 5.1 implies that Z(G) ≠ 1, and it only remains to show that |A| = 3. Further, it follows from 6.6 that C\overline{G}(⟨\overline{a}, \overline{t}⟩) contains a subgroup isomorphic to 2^{2} × SL(3, 4), and this serves to identify ⟨\overline{a}⟩ among the three conjugacy classes of subgroups of order 3 in G, and to yield C\overline{G}(\overline{a}) ≅ 3^{▪}Suz.

On the other hand, suppose that C = C_{3}. We then appeal 3.14 to conclude that O_{2}(C\overline{G}(⟨\overline{a}, \overline{t}⟩)) ≅ 2_{−}^{1+6}, and then also C\overline{G}(⟨\overline{a}, \overline{t}⟩)/O_{2}(C\overline{G}(⟨\overline{a}, \overline{t}⟩)) ≅ Ω_{6}^{−}(2)). This information again serves to identify ⟨\overline{a}⟩, among the three conjugacy classes of subgroups of order 3 in \overline{G}, and we again obtain O^{3′}(C\overline{G}(\overline{a})) ≅ 3^{·}Suz. Let \overline{g} be an element of order 7 in C\overline{G}(\overline{a}). Then O^{3′}(C\overline{G}(\overline{g})) ≅ L_{3}(2) or Alt(7), and then 6.3 and 4.2 yield O^{3′}(C_{G}(g)) ≅ 2^{▪}Alt(7), and Z(G) ≠ 1. Thus, we have shown that, in any case, we have Z(G) ≠ 1, and O^{3′}(C\overline{G}(\overline{a})) ≅ 3^{·}Suz. The character table for 2^{·}Co_{1} in [CCNPW] then yields C_{G}(a) ≅ 6^{·}Suz (with no non-trivial direct factors). As in the case of Suz, we obtain C_{G}(a) ≤ ⟨A^{C_{G}(a)} if |A| > 3, and in that case we contradict the fact that C_{V} (Z(G)) = 1. Thus |A| = 3 and the lemma is proved. □

Notice that results 7.2 through 7.6 yield theorem 7.1. Theorem A is then given by the union of the results 4.1, 5.1, 6.1, and 7.1.

Section 8: Theorem B and Corollary C Hypothesis 8.1. Assume Hypothesis 1.1 and assume also that G is not a group of Lie type in characteristic p.

By Theorem A, Hypothesis 8.1 implies that |A| = p = 3, and G is one of the exceptional groups listed in Theorem A. We aim first of all to determine which subgroups of order 3 in G can possibly be quadratic subgroups, with respect to some irreducible G-module V. Some of these identifications have already been made, in 4.3, 6.6, 6.8, and 7.7. Whenever Hypothesis 8.1 is in effect, let a be a generator of A, set \overline{G} = G/Z(G), and set \overline{C} = C\overline{G}(\overline{A}). We note that, by 2.4, \overline{A} is contained in a 2-local subgroup of \overline{G}, and we may fix a subgroup M of G, containing Z(G)A, such that \overline{M} is a maximal 2-local subgroup of G.

Lemma 8.2. Assume Hypothesis 8.1, and suppose that \overline{G} is isomorphic to D_{4}(2). Then \overline{C} ≅ GU(4, 2), and A is contained in a subgroup L of G of the form (2_{+}^{1+6})L_{4}(2). These conditions determine A up to conjugacy in Aut(G).

Proof. Identify \overline{G} with Ω_{8}^{+}(2) and let U be the natural module for G over 𝔽_{2}. Then |[U, \overline{A}]| = 2^{2k} for some k, 1 ≤ k ≤ 4. The integer k determines the structure of \overline{C}, and we have:

(1) If k = 1 then \overline{C} ≅ 3 × Ω_{6}^{−}(2).

(2) If k = 2 then \overline{C} ≅ GU(2, 2) × Ω_{4}^{+}(2).

(3) If k = 3 then \overline{C} ≅ GU(3, 2) × 3, and \overline{A} ≤ [\overline{C}, \overline{C}].

(4) If k = 4 then \overline{C} ≅ GU(4, 2)

The maximal 2-local subgroup \overline{M} of \overline{G} is a maximal parabolic subgroup. Suppose first that \overline{M} is of the form 2^{6} : Ω_{6}^{+}(2). Then 3.14 implies that M is of the form (2_{+}^{1+6})Ω_{6}^{+}(2), and that [O_{2}(M), A] is a quaternion group. Then C_{O_{2}(M)}(A) is of order 32, and so |\overline{C}| is divisible by 16. In this case we have k = 1 or 4. Let S be a Sylow 2-subgroup of M. There are then three maximal parabolic subgroups of \overline{G} containing \overline{S} and of the form 2^{6} : Ω_{6}^{+}(2). In the full covering group 2^{2}^{·}D_{4}(2) these parabolics lift to subgroups of the form (2×2_{+}^{1+6})Ω_{6}^{+}(2), as follows from 3.14. Since Out(D_{4}(2)) acts faithfully on the Schur multiplier of D_{4}(2), by 3.13, it follows that, in G, two of these maximal parabolics lift to groups which are isomorphic to M, and that one lifts to a group N such that O_{2}(N) is elementary abelian. Let M and M_{1} be the two which are isomorphic to M. Then M and M_{1} are fused in Aut(G), and thus A is determined up to conjugacy in Aut(G) in this case.

On the other hand, suppose that \overline{A} is not contained in a maximal parabolic subgroup of \overline{G} of the form 2^{6} : Ω_{6}^{+}(2). Then k = 2 or 3, and \overline{M} is of the form (2_{+}^{1+8}) : (Sym(3) × Sym(3) × Sym(3)). Set R = [O_{2}(M), A], and let V_{1} be an irreducible RA-submodule of V. Then RA/C_{R}(V_{1}) ≅ SL(2, 3), as follows from Theorem A of [Ch]. Set R_{1} = [C_{R}(V_{1}), A]. If R_{1} = 1 then |\overline{C}| is divisible by 2^{7}, which is contrary to having k = 2 or 3. Thus R_{1} ≠ 1. Let V_{2} be a non-trivial irreducible section for R_{1}A in V. Then R_{1}A/C_{R_{1}}(V_{2}) ≅ SL(2, 3). We have Z(G) ∩ R_{1} = 1, so R_{1} is isomorphic to a subgroup of \overline{R}. As R_{1}/C_{R_{1}}(V_{2}) is a quaternion group, it follows that C_{R_{1}}(V_{2}) is elementary abelian, and then [C_{R_{1}}(V_{2}), A] = 1, by 4.3. Thus, |R/C_{R}(A)| = 16, and so |\overline{C}| is divisible by 32. This is again contrary to k = 2 or 3, and the lemma is thereby proved. □

Lemma 8.3. Assume Hypothesis 8.1, and suppose that \overline{G} ≅ Sp(6, 2). Then \overline{C} ≅ 3 × Sp(4, 2), and this condition determines A up to conjugacy in G.

Proof. There are three conjugacy classes of subgroups of order 3 in \overline{G}, two of which are represented in a subgroup \overline{L} of \overline{G} of the form L_{2}(8) : 3. Theorem 1.1 implies that \overline{A} ≰ \overline{L}, so the conjugacy class of A in G is uniquely determined. Let U be the natural module for \overline{G} over 𝔽_{2}, and let b be an element of order 3 in L − E(L). Then b lies in a Frobenius subgroup of L of order 21, and hence |[U, \overline{b}]| = 16. Let c be an element of order 3 in E(L). Then c is contained in a cyclic group of order 9, and so [U, \overline{c}] = U. Thus, [U, \overline{a}] is of order 4, and the lemma follows. □

Theorem B now follows from the results 4.3 (concerning the alternating groups), 6.6 (concerning 2^{·}G_{2}(4)), 6.8 (concerning the groups P GU((n, 2)), 7.7 (concerning 2^{·}J_{2}, 2^{·}Suz, and 2^{·}Co_{1}), 8.2, and 8.3.

We end this section with the proof of Corollary C. Thus, assume Hypothesis 8.1, and assume that |A|^{2} ≥ |V/C_{V} (A)|. That is, assume that |V/C_{V} (A)| ≤ 9. Denote by L the set of all pairs (L, B) where L is a quasisimple subgroup L of C_{G}(A) and B is a G-conjugate of A contained in L, with B ≰ Z(L). Then [V, A, B] = 0, and so AB acts quadratically on V. This is contrary to Theorem B, so L is empty.

If G ≅ P GU(n, 2) with n ≥ 5 then the conditions given by 6.8 guarantee that L is non-empty. This will also be the case if \overline{G} ≅ Alt(n) with n ≥ 8, by 4.3. If \overline{G} ≅ D_{4}(2) or Sp(6, 2), we again get L non-empty, by 8.2 and 8.3. Suppose that \overline{G} ≅ Co_{1}. Then \overline{C} ≅ 3^{·}Suz, and \overline{A} is not contained in the center of a Sylow 3-subgroup of G. Then \overline{A} is not weakly closed in \overline{C} with respect to \overline{G}, and so L is non-empty in this case as well. Thus, none of these cases occur.

We have 2^{·}Suz ≥ 2^{·}G_{2}(4) ≥ 2^{·}J_{2}, and this descending series of groups corresponds to a descending chain of values for \overline{C}: 3^{·}U_{4}(3) ≥ SL(3, 4) ≥ 3^{·}Alt(6). These conditions guarantee that the class of quadratic elements in 2^{·}Suz restricts to the class of quadratic elements in the groups farther down the chain. Thus, to eliminate these groups it will suffice to eliminate the case \overline{G} ≅ J_{2}. In that case \overline{A} is contained in a subgroup \overline{M} of \overline{G} of the form (2_{−}^{1+4})Alt(5), where the extension is split. Thus, A is contained in a subgroup K of G with K ≅ SL(2, 5) and with Z(K) ≤ Z(G). Then C_{V} (Z(K)) = 0, and V is a direct sum of 2-dimensional subspaces V_{i}, 1 ≤ i ≤ m, where each V_{i} is an irreducible module for a fixed quaternion subgroup K_{1} of K. We may choose K_{1} to be A-invariant, so m ≤ 2. But evidently G ≰ SL(4, 3), so we have a contradiction at this point.

It remains to consider the cases \overline{G} ≅ Alt(n), n = 5 or 7. In these cases, there is an Ainvariant quaternion subgroup K_{1} of G with Z(K_{1}) = Z(G), so we obtain an embedding of G in SL(4, 3). As 7 does not divide the order of SL(4, 3) we conclude that n = 5, and then 2.1 implies that V is a natural SL(2, 9)-module for G. This completes the proof of Corollary C.

Section 9: Examples

As mentioned in the introduction, the classification of the irreducible quadratic modules for 2^{·}Alt(n) is given in [M], where it is shown that all such modules are “spin modules” and that all spin modules are quadratic. In this section we will show, by example, that all of the groups mentioned in parts (a) and (c) of Theorem B have quadratic modules. In order to do this, it will be convenient to have available the information given by the following lemma.

Lemma 9.1. Let X be an extraspecial 2-group, expressed as the central product of subgroups X_{i}, 1 ≤ i ≤ n, where each X_{i} is a quaternion group or a dihedral group of order 8. Let F be a field of characteristic different than 2, and let U be a faithful irreducible module for X over F. Then the following hold.

(a) The module U is uniquely determined up to isomorphism. It has dimension 2^{n}, and it is the tensor product module U = U_{1} ⊗ ⋯ ⊗ U_{n}, where U_{i} is the (unique) faithful 2-dimensional module for X_{i} over F.

(b) We have N_{GL(U)}(X)/C_{GL(U)}(X) ≅ Aut(X).

(c) If the characteristic of F is 3, and a is an automorphism of X such that [X, a] is a quaternion group, then a induces an F -linear automorphism of U with [U, a, a] = 0, and with dim([U, a]) = 2^{n−1}.

Proof. Each X_{i} has four linear characters and one irreducible character of degree 2. Since F is a splitting field for X_{i}, there is then a unique faithful irreducible representation of X_{i} over F, and it has degree 2. Any irreducible representation of X over F factors through a representation of the direct product X_{1} × ⋯ × X_{n}, and is therefore a tensor product of irreducible representations of the groups X_{i}. If the representation is also faithful then each of its tensor factors is faithful, and so (a) holds. Part (b) is immediate from the uniqueness of U. Let a be an automorphism of X such that [X, a] is a quaternion group. Then |a| = 3, and (b) implies that a induces a non-trivial automorphism of U over F. Here [X, a] commutes with C_{X}(a), by the Three Subgroups Lemma, and so we may take [X, a] = X_{1}. The tensor decomposition in (a) then implies that U is a direct sum of isomorphic two-dimensional modules for the group L = ⟨a^{X}⟩ = [X, a]⟨a⟩. Here L ≅ SL(2, 3), and if F has characteristic 3 then a acts quadratically on each irreducible L invariant summand of U. This yields (c). □

Now for the examples.

G = P GU(n, 2) : Let X be the central product of n quaternion groups. Then the semidirect product K = X : GU(n, 2) is contained in a maximal parabolic subgroup of SU(n + 2, 2). It follows from the preceding lemma that there is a quadratic module U for K, of dimension 2^{n} over 𝔽_{3}.

\overline{G} = D_{4}(2) or Sp(6, 2) : Take G = 2^{·}D_{4}(2). Then G is the commutator subgroup of the Weyl group of the E_{8}-root lattice Λ. Set V = Λ/3Λ. Then G acts faithfully on V. Choose a maximal subgroup M of G, of the form (2_{+}^{1+6})Ω_{6}^{+}(2), and let A be a subgroup of order 3 in M, such that [O_{2}(M), A] is a quaternion group. By 9.1, we may identify V with the unique faithful irreducible module for O_{2}(M), and A acts quadratically on V.

We may identify \overline{G} with Ω_{8}^{+}(2), in such a way that \overline{A} centralizes a 6-dimensional non-degenerate subspace of the natural 𝔽_{2}-module U for \overline{G}. Let \overline{G}_{0} be the stabilizer in \overline{G} of a non-singular point in U. Then \overline{G}_{0} ≅ 2 × Sp(6, 2). Let G_{1} be the inverse image in G of the commutator subroup of \overline{G}_{0}. Then A ≤ G_{1}, and since A acts quadratically on V, Theorem 1.2 implies that G_{1} ≅ 2^{·}Sp(6, 2).

\overline{G} = Co_{1}, Suz, G_{2}(4), or J2: Next consider the case where Λ is the Leech lattice and where G = 2^{·}Co_{1} – the automorphism group of Λ. Again, take V = Λ/3Λ. Then G acts faithfully on Λ. Let M be a maximal subgroup of G, such that \overline{M} is of the form (2_{+}^{1+8})D_{4}(2). Then 4.1 implies that Z(G) is a direct factor of O_{2}(M). There then exists a subgroup A of M, of order 3, such that [O_{2}(M), A] is a quaternion group. Let R be a complement to Z(G) in O_{2}(M), chosen so that R is invariant under an elementary abelian subgroup E of M of order 81. Then R is generated by four conjugates of A, and R = [R, E]. Set W = [V, Z(R)]. Then 9.1 implies that dim(W) ≥ 16 and that A acts quadratically on W, with dim([W, A]) = 1/2dim(W). We now have dim(C_{V} (Z(R)) ≤ 8, and evidently neither M/Z(R) nor M/Z(R)Z(G) has a faithful representation of degree 8 over 𝔽_{3}. Thus C_{V} (Z(R)) = C_{V} (R), and then also [V, Z(R)] = [V, R]. This implies that both C_{V} (R) and [V, R] are M-invariant, and so R is normal in M. As Z(G) is not a direct factor of G, it now follows from Gaschütz’s Theorem that M/R is a nonsplit central extension of D_{4}(2), and thus M has a subgroup K with K ≅ 2^{cdot}D_{4}(2), and with A ≤ K. As [V, R] is a quadratic module for K, Theorem 1.3 implies that A is contained in a subgroup N of K of the form (2_{+}^{1+6})L_{4}(2), where [O_{2}(N), A] is a quaternion group. Then 9.1 implies that dim(C_{V} (R)) = 8, that A acts quadratically on C_{V} (R), and dim([C_{V} (R), A]) = 4. Then also dim([V, R]) = 16, dim([V, A]) = 12, and A acts quadratically on V.

We now have C_{G}(A) ≅ 6^{·}Suz, by Theorem 1.2. Then A is not contained in the center of a Sylow 3-subgroup of G, and so there exists a subgroup G_{1} of G with G_{1} ≅ 6^{·}Suz, such that A ≤ G_{1} and A ≰ Z(G_{1}). Set V_{1} = [V, Z(G_{1})]. As Z(G_{1}) is conjugate to A, we have dim(V_{1}) = 12, and V_{1} is then a quadratic module for G_{1}/O_{3}(G_{1}). There are subgroups G_{2} and G_{3} of G_{1}, with G_{2} ≅ 2^{·}G_{2}(4) and with G_{2} ≥ G_{3} ≅ 2^{·}J_{2}. By considering the structure of centralizers of elements of order 3 in the groups G_{i}, one finds that G_{2} and G_{3} contain conjugates of A, and thus V_{1} is a quadratic module for G_{i}, 1 ≤ i ≤ 3. As O_{2}(G_{i}) = Z(G), for all i, all irreducible constituents for G_{i} in V_{1} are non-trivial. As 25 divides the order of G_{3} and does not divide the order of SL(6, 3), we conclude that V_{1} is irreducible for each G_{i}. Then also V is irreducible for G.

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