Strong quadratic modules

G. Stroth

One of the main tools in the quite recent theory of parabolic systems is the knowledge of GF (p)-modules V for the groups involved admitting quadratically acting p-groups.

Let G be a finite simple group, V a GF (p)G-module, p a prime. We call V quadratic if there is a nontrivial p-subgroup E of G such that [V, E, E] = 0. For p ≥ 5 the groups G possessing such a module have been studied by J. Thompson [Th]. For p = 3 results have been obtained by Ch.Ho [Ho]. The corresponding modules for the groups of Lie type have been classified by A.A. Premet and I.D. Suprunenko [PS]. Similar results have been obtained by U. Meierfrankenfeld [Mei]. If p = 2 the definition of a quadratic module is not any restriction at all as for any involution x we have [V, x, x] = 0. So in this case we have to add a further assumption. In [MeiStr1] and [MeiStr2] the case of G a sporadic simple group, an alternating group or a group of Lie type over a field of odd characteristic was investigated under the additional assumption that |E| ≥ 4. Again this seems not to be the right condition for groups of Lie type over a field of characteristic 2, as root groups tend to act quadratically on many modules. So a classification, which might be possible, would contain a long list of modules defined over GF (q) (if G is defined over GF (q)) and so would be just too long for interesting applications. This paper is a first step in obtaining a satisfactory result for quadratic modules for Chevalley groups over GF (q), q = 2^{n}. We are going to prove the following theorem.

Theorem. Let G = G(q) be a Chevalley group over GF (q), q = 2^{n}. Let R be a long root group and Q = Z_{2}(O_{2}(N_{G}(R))). Let V be a nontrivial irreducible GF (2)G-module with [V, R, Q] = 0. Then one of the following holds

(i) G ≃ (S)L(n, q) or (S)U(n, q) and V = V (λ) for some fundamental weight λ.

(ii) G ≃ Ω^{±}(2n, q) or Sp(2n, q) and V is the natural module or a spin module.

(iii) G ≃ E_{6}(q) and V = V (λ_{1}) or V (λ_{6}).

(iv) G ≃ E_{7}(q) and V = V (λ_{7}).

(v) G ≃ ^{2}E_{6}(q) or F_{4}(q) and V = V (λ_{4}).

(vi) G ≃ G_{2}(q) or ^{3}D_{4}(q) and V is the natural module.

In section 1 we prove the Theorem, and then in sections 2 and 3 we deduce the following corollaries. In fact corollary 1 had originally inspired the work of this paper.

Corollary 1. Let G = G(q) be a Chevalley group over GF (q), q = 2^{n}, and V be a nontrivial irreducible GF (2)G-module. Let X be some nontrivial elementary abelian 2-subgroup of G such that [V, X, X^{g}] = 0 whenever [X, X^{g}] = 1, g ∈ G. Then V is one of the modules in the theorem or G ≃ (S)U(3, q) and V is a basic module (8-dimensional module) or G ≅ Sz(q) and V is the natural module.

A question like the one of corollary 1 recently occured in a paper of Timmesfeld [Ti]. If E is a quadratic fours group we get the following corollary.

Corollary 2. Let G = G(q) be a Chevalley group over GF (q), q = 2^{n}, and V be a nontrivial irreducible GF (2)G-module. Let E be a fours group such that [V, E, E] = 0. If E intersects some long root subgroup R of G nontrivially, but E ⊈ R, then either V is one of the modules in the theorem or G ≅ Sp(2n, q) or F_{4}(q) and V = V (λ) for some fundamental weight λ.

I do not know whether G = F_{4}(q) and V = V (λ_{2}) really occurs.

For a Chevalley group G let q(G) be the minimum value of n such that the following is true: If V is an irreducible module and A is some elementary abelian subgroup of G with [V, A, A] = 0 and |A| = 2^{n}, then V is one of the modules in corollary 2. It would be interesting to determine q(G).

0. Preliminaries

(0.1) Lemma. Let G be a finite group and ⟨x, y⟩ ≤ G be a fours group acting nontrivially on some irreducible GF (2)G–module V. Let x ∼ y ∼ xy and G = ⟨ I C_{G}(s), I C_{G}(t)⟩ for any pair s, t from the triple x, y, xy such that ⟨s, t⟩ = ⟨x, y⟩. Then [V, x] ∩ [V, y] ≠ 0.

Proof. As V is an irreducible module we have I C_{V} (x) ≠ I C_{V} (xy). So [ I C_{V} (xy), x] = [ I C_{V} (xy), y] ≠ 0.

(0.2) Lemma. Let W be a 2–group which carries the structure of a 2–dimensional vectorspace over GF (q), q = 2^{n}. Let V be a GF (2)W –module on which q of the q + 1 onedimensional subspaces of W act quadratically. Then any one-dimensional subspace of W acts quadratically.

Proof. Let W = W_{1}W_{2} where W_{1} = ⟨x_{1}, …, x_{n}⟩, W_{2} = ⟨y_{1}, …, y_{n}⟩ and W_{1}, W_{2} are one– spaces which act quadratically. We may assume n ≥ 2.

Let w_{1}, w′_{1} ∈ W_{1}, w_{2}, w′_{2} ∈ W_{2}, v ∈ V. We first show

\begin{matrix}[v, w_{1}w_{2}, w′_{1}w′_{2}] = [v, w_{1}, w′_{2}] + [v, w′_{1}, w_{2}] \\ = [v, w′_{2}, w_{1}] + [v, w_{2}, w′_{1}] \end{matrix}  (1)

For this we use freely the following general facts

[v, ab] = [v, a] + [v, b] + [v, a, b] and [v, a, b] = [v, b, a] for commuting a, b.

We have

z = [v, w_{1}w_{2}] = [v, w_{1}] + [v, w_{2}] + [v, w_{1}, w_{2}] = [v, w_{2}] + x = [v, w_{1}] + y

where x ∈ [V, W_{1}] and y ∈ [V, W_{2}].

Now

[v, w_{1}w_{2}, w′_{1}w′_{2}] = [z, w′_{1}w′_{2}] = [z, w′_{1}] + [z_{1}, w′_{2}] + [z, w′_{1}, w′_{2}]

We have

[z, w′_{1}] = [v, w_{2}, w′_{1}] as [x, w′_{1}] = 0 by quadratic action
[z, w′_{2}] = [v, w_{1}, w′_{2}] as [y, w′_{2}] = 0 by quadratic action
[z, w′_{1}, w′_{2}] = [[v, w_{2}], w′_{1}, w′_{2}] = [[v, w_{2}], w′_{2}, w′_{1}] = 0 by quadratic action

This establishes (1).

For proving the lemma it is enough to show that for x_{1}y, x_{2}z ∈ W_{0}, W_{0} a one-space in W, y, z ∈ W_{2}, we get [v, x_{1}y, x_{2}z] = 0 for any v ∈ V.

Let y = ∏ y_{i}^{m_{i}} then set \overline{y} = ∏ x_{i}^{m_{i}}. So to x_{1}y, x_{2}z there correspond y_{1}\overline{y} and y_{2}\overline{z}. Looking at W as a two dimensional vectorspace over GF (q) implies that there is an automorphism ρ of order q − 1 (the multiplication by field elements) which acts on W_{1} and W_{2} in the same way. Now the q +1 one dimensional subspaces of W are just the q +1 ⟨ρ⟩-invariant subgroups of order q. As x_{1}y and x_{2}z are in W_{0} there is some power of ρ mapping x_{1} onto x_{2} and y onto z. But then the same power maps y_{1} onto y_{2} and ȳ onto z̄, so y_{1}ȳ, y_{2}z̄ are contained in some subspace W_{3}. We may assume that W_{3} acts quadratically. Denote by W_{4} = ⟨x_{1}y_{1}, …, x_{n}y_{n}⟩, a diagonal of W_{1}W_{2}. We may also assume that W_{4} acts quadratically. As y\overline{y}, z\overline{z} are in W_{4} we get with (1)

0 = [v, y\overline{y}, z\overline{z}] = [v, y, \overline{z}] + [v, \overline{y}, z] for any pair y, z ∈ W , v ∈ W.  (2)

Hence in our situation we have

[v, y_{1}, \overline{z}] = [v, x_{1}, z] and [v, \overline{y}, y_{2}] = [v, y, x_{2}] , v ∈ V  (3)

and

0 = [v, y_{1}\overline{y}, y_{2}\overline{z}] =_{(1)} [v, y_{1}, \overline{z}] + [v, \overline{y}, y_{2}]  (4)

So by (2) and (3) we have

[v, x_{1}, z] = [v, y, x_{2}], v ∈ V  (5)

This gives [v, x_{1}y, x_{2}z] = [v, x_{1}, z] + [v, y, x_{2}] = 0 by (5).

(0.3) Lemma. ([S]). Let G be a Chevalley group over GF (q) and V be an irreducible GF (2)G–module. Then I C_{V} (K) is an irreducible GF (2)P –module for any parabolic P of G, where K = O_{2}(P).

(0.4) Lemma. (a). Let G be a Chevalley group over GF (q), P_{1}, …, P_{n} be the set of minimal parabolics of G containing a given Sylow 2–subgroup. Set K_{i} = O_{2}(P_{i}), i = 1, …, n. Let V be an irreducible module over GF (q). Then V is uniqueley determined by the action of P_{i} on I C_{V} (K_{i}), i = 1, …, n.

(b) If V = V (λ), λ = ∑_{i=1}^{n} a_{i}λ_{i}, λ_{i} fundamental weights, then (a) says that whenever P_{i} acts

nontrivially on I C_{V} (K_{i}), we get a_{i} ≠ 0, otherwise a_{i} = 0.

(c) Let V be as in (b). If all I C_{V} (K_{i}) are trivial for all but one i = i_{0} and I C_{V} (K_{i_{0}}) is a natural module, then V = V (λ) for some fundamental weight λ.

Proof. (a) is [RS], while (b) and (c) are easy consequences from (a).

A consequence of (0.4) we will use quite often is

(0.5) Lemma. Let G be a Chevalley group over GF (q), G ≅ L̸_{n}(q), and R be some long root subgroup. Let Q = O_{2}(N_{G}(R)), S ∈ Syl_{2}(N_{G}(R)) and P_{i} the minimal parabolic with S ⊆ P_{i}, P_{i} ⊈ N_{G}(R), O^{2′}(P_{i}/O_{2}(P_{i})) ≅ L_{2}(q). Let V be some irreducible nontrivial module for G over GF (q) with [V, R, Q] = 0. Suppose that there is some g ∈ G such that R^{g} ⊆ Q, R^{g} ≠ R and (RR^{g}) ⊆ x^{G} for x ∈ R^{♯}. Then O^{2′}(P_{i}) has a fixed point on V. Furthermore if V =

V (λ), λ = ∑_{i=1}^{n} a_{i}λ_{i}, λ_{i} fundamental weights, then the coefficient a_{i} of λ_{i} corresponding to P_{i} is zero.

Proof. As G ≅ L̸_{n}(q), we have that N_{G}(R) is a maximal parabolic of G. Now we are going to apply (0.1) to ⟨x, x^{g}⟩, x ∈ R^{♯}. As I C_{G}(x) = I C_{G}(R) we get G = ⟨ I C_{G}(x), I C_{G}(x^{g})⟩. So by (0.1) 0 ≠ [V, x] ∩ [V, x^{g}] = U. Now ⟨Q, Q^{g}⟩ centralizes U and so S⟨Q, Q^{g}⟩ = O^{2′}(P_{i}) has a fixed point on V, i.e. a_{i} = 0 by (0.4).

(0.6) Lemma. Let G ≃ Sz(q), V an irreducible GF (2)G–module. Then [V, Z(S), S] ≠ 0, for S ∈ Syl_{2}(G).

Proof. Suppose false. Then [V, Z(S), Z(S)] = 0. Any two conjugates of Z(S) generate G. By [BHu;XI,§ 3] there is some involution i ∈ Z(S) inverting some element ω of order 5 in G. As G = ⟨Z(S), Z(S)^{ω}⟩ = ⟨i, Z(S)^{ω}⟩ we get

V = [V, Z(S)] ⊕ [V, Z(S)^{ω}] = I C_{V} (Z(S)) ⊕ I C_{V} (Z(S)^{ω})

and

I C_{V} (i) ∩ I C_{V} (Z(S)^{ω}) = 0.

So I C_{V} (i) = I C_{V} (Z(S)) and I C_{V} (i) ∩ I C_{V} (ω) = 0. As i normalizes ⟨ω⟩ we get I C_{V} (ω) = 0. By [Mar] V is the natural module. But then [V, Z(S), S] ≠ 0, [BHu;XI,§ 3].

(0.7) Lemma. Let G = G_{2}(q), q = 2^{n}, V be a nontrivial GF (2)G–module, R a long root subgroup of G. Let G_{1} = N_{G}(R), G_{2} be the other minimal parabolic containing a Sylow 2–subgroup S of G_{1}. Set V_{i} = I C_{V} (O_{2}(G_{i})). If [V_{2}, O^{2′}(G_{2})] = 0 and [V, R, R] = 0, then [V_{1}, S, S] = 0.

Proof. Set Q = ⋂_{g∈G_{2}} O_{2}(G_{1})^{g}. Then |Q| = q^{3}. Set V_{3} = I C_{V} (Q).

We have V_{1} ⊆ V_{3}. As ⟨G_{1}, G_{2}⟩ = G, V_{1} ≠ V_{3}. We study the action of G_{2} on V_{3}. Let g ∈ G such that S = O_{2}(G_{2})R^{g}. By assumption [V_{3}, R^{g}, R^{g}] = 0. As S/Q is isomorphic to a Sylow 2–subgroup of L_{3}(q), there is an elementary abelian group W of order q^{2} in S/Q such that S/Q = W (O_{2}(G_{2})/Q) and R^{g}Q/Q ⊆ W. As W carries the structure of a 2–dimensional vectorspace over GF (q) and R^{g}Q/Q corresponds to a one–space, we get with (0.2) that W ∩ O_{2}(G_{2})/Q acts quadratically on V_{3}. Now also O_{2}(G_{2})/Q carries the structure of a 2–dimensional vectorspace over GF (q) and W ∩ O_{2}(G_{2})/Q is a one–space. Now any one–space in O_{2}(G_{2})/Q acts quadratically on V_{3} as all one–spaces in O_{2}(G_{2})/Q are conjugate under G_{2}. There is a subgroup T of O_{2}(G_{2}) such that T Q/Q is a one–space in O_{2}(G_{2})/Q and T O_{2}(G_{1}) = S. Hence [V_{1}, T, T] ⊆ [V_{3}, T, T] = 0 and so [V_{1}, S, S] = 0.

For what follows we fix the following: Let M be some irreducible GF (2)–module for G = G(q). Then M ⊗GF (q) is a direct sum of algebraic conjugates of some irreducible GF (q)–module V. If M ⊗GF (q) is an algebraic conjugate of a fundamental module for some fundamental weight λ we write M = V (λ).

As the assumptions of the theorem and the corollaries are independent of the field, it is enough to prove the theorems for V to be an irreducible GF (q)–module, where G is defined over GF (q).

1 Some special modules

Hypothesis 1: Let G be a Chevalley group over GF (q), q = 2^{m}. Let R be some root group and Q = Z_{2}(O_{2}(N_{G}(R))). In case G ≅ Sp(2n, q) or F_{4}(q) we assume in addition R to be long. Let V be a nontrivial irreducible GF (q)G–module. Assume [V, R, Q] = 0.

In this chapter we always work under hypothesis 1. Remember that if G is not L(2, q), Sp(4, q) or ^{2}F_{4}(q) then R = Q′. Furthermore in all cases but ^{2}F_{4}(q) we have Q = O_{2}(N_{G}(R)).

(1.1) Lemma. Let H be a maximal parabolic of G such that [V, R, O_{2}(H)] = 0. If there is some g ∈ G such that R^{g} ⊆ H − O_{2}(H) and [R^{g}, S] ⊆ O_{2}(H) for some S ∈ Syl_{2}(H) and [ I C_{V} (O_{2}(H)), R^{g}] ≠ 0, then for the minimal parabolic P of G containing S but P ⊈ H we get

[ I C_{V} (S), O^{2′}(P)] = 0. So if V = V (λ), λ = ∑_{i=1}^{n} a_{i}λ_{i}, λ_{i} fundamental weights, then a_{i} = 0, where

P = P_{i}.

Proof. Set X = I C_{V} (O_{2}(H)). Then by assumption [R^{g}, X] ≠ 0. As [S, R^{g}] ⊆ O_{2}(H), we have that [R^{g}, X] is invariant under S. This gives I C_{V} (S) ∩ [R^{g}, X] ≠ 0. By (0.3) I C_{V} (S) ⊆ [R^{g}, X]. So I C_{V} (S) is centralized by ⟨S, O_{2}(H)^{g}⟩ ⊈ H.

(1.2) Lemma. Let G = SL(n, q), n ≥ 2, then V = V (λ) for some fundamental weight λ.

Proof. Let first n = 2. Then by hypothesis 1 we have [V, S, S] = 0, for some Sylow 2– subgroup S of G. So the assertion follows with [Hig].

The general case we treat by induction on n. We have n ≥ 3. Let G_{1} be the stabilizer of a point and G_{n−1} be the stabilizer of a hyperplane incident to the point for G_{1} in the natural representation of G. Then N_{G}(R) = G_{1} ∩ G_{n−1}. Let K_{i} = O_{2}(G_{i}) and H_{i} a complement of K_{i} in O^{2′}(G_{i}), i = 1, n − 1. Set V_{i} = I C_{V} (K_{i}). As there are conjugates of R^{g} contained in H_{1}, H_{n−1} respectively, with Q^{g} ∩ H_{i} = Z_{2}(O_{2}(N_{H_{i}}(R^{g}))) we get by induction that V_{i} are trivial modules or fundamental modules.

Let V = V (λ), λ = ∑_{i=1}^{n−1} a_{i}λ_{i}. Now as V_{1} is trivial or fundamental, we get with (0.4) that for

j > 1 there is at most one a_{j} ≠ 0. Similarly for V_{n−1}. Together we have either just one a_{i} is nonzero hence 1 and so V = V (λ), λ a fundamental weight, or λ = a_{1}λ_{1} + a_{n−1}λ_{n−1}. In the latter [V, R, H_{1} ∩ H_{n−1}] = 0. Application of (1.1) now yields the contradiction a_{1} = 0 = a_{n−1}.

(1.3) Lemma. Let G be Ω^{+}(2n, q), n ≥ 4, then V is the natural or spin module.

Proof. Let G_{1}, G_{n−1}, G_{n} be the three maximal parabolics with connected diagram, i.e. O^{2′}(G_{1}/K_{1}) ≃ Ω^{+}(2n − 2, q), O^{2′}(G_{n−1}/K_{n−1}) ≃ O^{2′}(G_{n}/K_{n}) ≃ SL(n, q), where K_{i} = O_{2}(G_{i}). Set V_{i} = I C_{V} (K_{i}).

As there are conjugates of R in G_{i} intersecting K_{i} trivially, i = 1, n, n − 1, we may apply (1.2) or use induction. Let V = V (λ), then this puts restrictions on the coeffizients other than

a_{i}, for λ = ∑_{i=1}^{n} a_{i}λ_{i}. So we have that V_{1} is either trivial, a natural module or a spin module.

Suppose first that V_{1} is trivial. Then by (1.2) both V_{n−1}, V_{n} are natural modules and so by (0.4) V is the natural module.

So we may assume that V_{1} is nontrivial. Let g ∈ G such that R^{g} ≤ Q − R. Then by (0.5) O^{2′}(P_{2}) has a fixed point on V and a_{2} = 0. This shows that V_{1} has to be a spin module. So we may choose notation such that a_{n} = 0. Then we get with (1.2) that V_{n} is the natural module and so V_{n−1} is a trivial module. Hence we have that V is a spin module.

(1.4) Lemma. Let G be Ω^{−}(2n, q), n ≥ 3. Then V is the natural module or the spin module.

Proof. We first treat the case n = 3. Let G_{1}, G_{2} be the two maximal parabolics containing a Sylow 2–subgroup S, where we choose notation such that G_{2} = N_{G}(R). As R acts quadratically and S = R^{g}O_{2}(G_{2}) for some g ∈ G, we get that V_{2} = I C_{V} (O_{2}(G_{2})) is either a trivial O^{2}(G_{2}/O_{2}(G_{2})) module or the natural SL(2, q)–module.

Suppose first that V_{2} is not a trivial module. We may assume R^{g} ≤ O_{2}(G_{1}). Now we have W = [V, R]∩[V, R^{g}] ≠ 0. This gives that W is centralized by ⟨Q, Q^{g}⟩. Thus O^{2′}(G_{1}) = ⟨Q, Q^{g}⟩S has a fixed point on V, where S ∈ Syl_{2}(G_{1}). So we have that V is the natural module.

Suppose now that V_{2} = [V, R] is a trivial module for O^{2′}(G_{2}). Then [V, R, O_{2}(G_{1})] = 0 and so [V, O_{2}(G_{1}), O_{2}(G_{1})] = 0, as O_{2}(G_{1}) is generated by G_{1} – conjugates of R. Now there is some h ∈ G such that O_{2}(G_{1})(O_{2}(G_{1})^{h} ∩ G_{1}) = S. Hence a Sylow 2–subgroup of SL(2, q^{2}) acts quadratically on V_{1}, which gives that V_{1} is the natural SL(2, q^{2})–module. So by (0.4) V is the spin module.

So assume now n ≥ 4. Let G_{1} be the maximal parabolic such that O^{2′}(G_{1}/K_{1}) ≃ Ω^{−}(2n − 2, q), K_{1} = O_{2}(G_{1}). Set V_{1} = I C_{V} (K_{1}). Then by induction V_{1} is trivial, a natural module or a spin module for G_{1}. Let G_{n−1} be the maximal parabolic containing a Sylow 2–subgroup of G_{1} such that O^{2′}(G_{n−1}/K_{n−1}) ≃ SL(n − 1, q), K_{n−1} = O_{2}(G_{n−1}). Set V_{n−1} = I C_{V} (K_{n−1}). Then by (0.3) and (1.2) V_{n−1} is either trivial or a fundamental module. In particular (0.4) puts

restrictions on the coefficients a_{i}, V = V (λ), λ = ∑_{i=1}^{n} a_{i}λ_{i}.

Suppose first that V_{1} is trivial. Then we have that V_{n−1} has to be the natural module. So just a_{1} is nonzero and hence V is the natural module by (0.4).

So let V_{1} be nontrivial. Let g ∈ G such that R^{g} ≤ Q − R. Then by (0.5) we have a_{2} = 0. Hence V_{1} is the spin module. So a_{i} = 0 for i > 1, i ≠ n − 1. If now V_{n−1} is a trivial module we get a_{1} = 0 and so V is the spin module. So assume that V_{n−1} is nontrivial. Then a_{1} ≠ 0 and so V_{n−1} has to be the natural module. Now by (1.1) I C_{V} (S) is centralized by O^{2′}(P_{n−1}), so a_{n−1} = 0, contradicting the fact that V_{1} is a spin module.

(1.5) Lemma. Let G be Sp(2n, q), n ≥ 2, and R be a long root group. Then V is the natural module or the spin module.

Proof. Let first n = 2. Let G_{1} = N_{G}(R). Then as S = QR^{g} ∈ Syl_{2}(G_{1}) for suitable g ∈ G, we get that I C_{V} (Q) is either a trivial or a natural SL(2, q)–module for O^{2′}(G_{1}/Q) ≃ SL(2, q).

If I C_{V} (Q) is the natural module, we get that 0 ≠ [V, R] ∩ [V, R^{g}] is centralized by ⟨Q, Q^{g}⟩, and so I C_{V} (S) is centralized by S⟨Q, Q^{g}⟩ = O^{2′}(G_{2}), where G_{2} is the other parabolic containing S. Hence V is the natural module by (0.4).

Let I C_{V} (Q) be a trivial module. Then [V, R, O_{2}(G_{2})] = 0 and so [V, O_{2}(G_{2}), O_{2}(G_{2})] = 0, as O_{2}(G_{2}) is generated by G_{2} – conjugates of R. As S = O_{2}(G_{2})(O_{2}(G_{2})^{h} ∩G_{2}) for suitable h ∈ G, we get that V_{2} = I C_{V} (O_{2}(G_{2})) is the natural module for O^{2′}(G_{2}/O_{2}(G_{2})) ≃ SL(2, q). Then V is the spin module.

Let now n ≥ 3. Let G_{1}, G_{n} be the maximal parabolics containing a given Sylow 2–subgroup S such that O^{2′}(G_{1}/K_{1}) ≃ Sp(2n − 2, q) and O^{2′}(G_{n}/K_{n}) ≃ SL(n, q), where K_{i} = O_{2}(G_{i}), i = 1, n. Set V_{i} = I C_{V} (K_{i}), i = 1, n. By induction we have that V_{1} is a trivial module, a natural

module or a spin module. By (1.2) V_{n} is a fundamental module. Let V = V (λ), λ = ∑_{i=1}^{n} a_{i}λ_{i}. If

V_{1} is trivial we get with (0.4) a_{i} = 0 for i ≠ 1. So a_{1} ≠ 0 and V is the natural module.

So suppose that V_{1} is nontrivial. Choose R^{g} ≤ Q − R. Then (0.5) yields a_{2} = 0. In particular we get that V_{1} is a spin module, implying a_{n} ≠ 0. If V_{n} is trivial, we get that V is the spin module. So assume that V_{n} is the natural module. But as in (1.4) we get that I C_{V} (S) is centralized by O^{2′}(P_{n}) and so a_{n} = 0, a contradiction.

(1.6) Lemma. Let G be SU(n, q), or Sp(2n, q), n ≥ 3. Then V = V (λ) for some fundamental weight λ.

Proof. Suppose first G ≅ SU(n, q). Let n = 3. Then we have Q ∈ Syl_{2}(G). Hence by (0.3) I C_{V} (Q) is an irreducible module for the torus. This gives that I C_{V} (Q) is one dimensional over GF (q^{2}) and as [V, R] ≤ I C_{V} (Q) we get |[V, R]| ≤ q^{2}. As there are g, h ∈ G such that G = ⟨R, R^{g}, R^{h}⟩, we get |V | ≤ q^{6} and so V has to be the natural module. For n = 4, we have U(4, q) ≃ Ω^{−}(6, q) and the assertion follows with (1.4).

Assume n ≥ 5 or G a symplectic group. In the latter the assertion follows for R long with (1.5). So we may assume that R is short. Let G_{1} = N_{G}(R). Then by induction V_{1} = I C_{V} (Q) is a trivial module for G_{1} or some fundamental module. Let V_{1} be nontrivial. There is some R^{g} ≤ G_{1} − Q such that [S, R^{g}] ≤ Q, where S ∈ Syl_{2}(G_{1}). Now we get with (1.1) that I C_{V} (S) is centralized by ⟨S, Q^{g}⟩ = O^{2′}(P_{1}). Further as V_{1} is fundamental we get with (0.4) that V = V (λ), λ = Σa_{i}λ_{i}, where a_{i} = 0 for all but one i, i > 1. As I C_{V} (S) is centralized by O^{2′}(P_{1}) we get a_{1} = 0 and so V is fundamental. If V_{1} is trivial, then let G_{m} be the other maximal parabolic containing S with connected diagram. As [V, R, I C_{G}(R)] = 0, we get [V, R, O_{2}(G_{m})] = 0. Now R ⊆ O_{2}(G_{m}) and so [V, ⟨R^{G_{m}}⟩, ⟨R^{G_{m}}⟩] = 0. Hence a Sylow 2–subgroup of P_{1}/O_{2}(P_{1}) acts quadratically on I C_{V} (O_{2}(P_{1})) = W. So W is the natural module by [Hig]. Let V = V (λ), λ = Σa_{i}λ_{i}. We have by (0.4) a_{1} ≠ 0 and a_{i} = 0 for i > 1. So V is the natural module.

(1.7) Lemma. Let G be E_{n}(q), n = 6, 7, 8, with Dynkin diagram

0_{4} 0^{1}  0^{2}  0^{3}  0^{5}  0  0^{n}^{· · · · · ·}

Then one of the following holds (i) n = 6 and V = V (λ_{1}) or V (λ_{6})

(ii) n = 7 and V = V (λ_{7}).

Proof. (i) Let G = E_{6}(q). Then N_{G}(R) = G_{4}. Furthermore O^{2′}(G_{4}/Q) ≃ SL(6, q). So by (1.2) V_{4} = I C_{V} (Q) is trivial or a fundamental module. Let R^{g} ≤ Q − R. Then by (0.5) I C_{V} (S)

is centralized by O^{2′}(P_{4}) for S ∈ Syl_{2}(P_{4}), so if V = V (λ), λ = ∑_{i=1}^{6} a_{i}λ_{i}, then a_{4} = 0. Now some

a_{i}, i ≠ 4, has to be nonzero.

We have that O^{2′}(G_{1}/K_{1}) ≃ O^{2′}(G_{6}/K_{6}) ≃ Ω^{+}(10, q), where K_{i} = O_{2}(G_{i}), i = 1, 6. Now by (1.3) V_{i} = I C_{V} (G_{i}) is trivial, a natural module or a spin module for O^{2′}(G_{i}/K_{i}), i = 1, 6. Now V_{1} implies that a_{3} = a_{5} = 0 and at most one of a_{2} or a_{6} is nonzero. Furthermore V_{6} implies that a_{2} = a_{3} = 0 and at most one of a_{1} or a_{5} is nonzero. This shows that a_{2} = a_{3} = a_{4} = a_{5} = 0. As V_{4} is fundamental we get exactly one of a_{1} or a_{6} is nontrivial, the assertion.

(ii) Let G ≃ E_{7}(q). Then G_{1} = N_{G}(R). We have O^{2′}(G_{1}/Q) ≃ Ω^{+}(12, q). So by (1.3) V_{1} = I C_{V} (Q) is a trivial, natural or spin module for Ω^{+}(12, q). Application of (0.5) shows that

[ I C_{V} (S), O^{2′}(P_{1})] = 0. Let V = V (λ), λ = ∑_{i=1}^{7} a_{i}λ_{i}, λ_{i} fundamental weights. We have a_{1} = 0

by (0.5) and in any case a_{6} = 0. Further exactly one of a_{i}, i ≠ 1, 6, is nonzero. Hence V_{1} is nontrivial. As there is some conjugate R^{g} ≤ G_{7} − O_{2}(G_{7}), we may apply (i). As a_{1} = a_{6} = 0 we get [ I C_{V} (S), O^{2′}(G_{7})] = 0 and so a_{i} = 0 for i ≠ 7 and then V = V (λ_{7}).

(iii) Let now G ≃ E_{8}(q). Then G_{8} = N_{G}(R). By (0.5) we again have that I C_{V} (S) is centralized by O^{2′}(P_{8}). By (ii) we have that V_{8} = I C_{V} (Q) is a trivial O^{2′}(G_{8})–module. Let

V = V (λ), λ = ∑_{i=1}^{8} a_{i}λ_{i}, λ_{i} fundamental weights. By (0.5) a_{8} = 0. By (ii) a_{i} = 0 for i ≠ 7. With

(1.3) applied to V_{1} = I C_{V} (O_{2}(G_{1})) we get a_{7} = 0, a contradiction.

(1.8) Lemma. Let G be ^{2}E_{6}(q) or F_{4}(q) with Dynkin diagram

0^{1}  0^{2}  0^{3}  0^{4}

Choose notation such that N_{G}(R) = G_{1}, then V = V (λ_{4}).

Proof. By (0.5) we have that [ I C_{V} (S), O^{2′}(P_{1})] = 0, S ∈ Syl_{2}(G_{1}). So V_{1} is a nontrivial module. Furthermore there is some conjugate R^{g} ∈ G_{1} − Q, such that R^{g} ≤ Z(O_{2}(G_{4})). This gives O^{2′}(G_{4}) = ⟨Q, Q^{g}⟩S. As V_{1} ⊆ [V, R] we have [V, R] ∩ [V, R^{g}] ≠ 0 and so we get that I C_{V} (O_{2}(G_{4})) is a trivial module. Now by (1.5), (1.6) we have that V_{1} is the natural module, so V = V (λ_{4}).

(1.9) Lemma. Let G be G_{2}(q). Then V is the natural module.

Proof. The assertion is clear for q = 2. So assume q ≥ 4. Let G_{1} = N_{G}(R) and G_{2} be the other maximal parabolic containing a Sylow 2–subgroup S of G. Let V_{i} = I C_{V} (O_{2}(G_{i})), i = 1, 2. Then by (0.5) we have that V_{2} is a trivial module. By (0.7) we have that I C_{V} (Q) is the natural G_{1}/Q–module. Now (0.4) yields the assertion.

(1.10) Lemma. Let G be ^{3}D_{4}(q). Then V is the natural module.

Proof. Let G_{1}, G_{2} be the two minimal parabolics containing a Sylow 2–subgroup S of G. Choose notation such that G_{2} = N_{G}(R). Let K_{i} = O_{2}(G_{i}) and V_{i} = I C_{V} (K_{i}), i = 1, 2. Then by (0.5) we have that V_{1} is a trivial module.

Choose conjugates R^{g}, R^{h} such that ⟨R, R^{g}⟩ ≤ G_{1}, R^{h} ∩ K_{1} = 1, R^{h} ≤ K_{2}. Choose x ∈ R, y ∈ R^{g}. We have

[ I C_{V} (xy), x] = [ I C_{V} (xy), y]

and so

| I C_{V} (xy) : I C_{V} (⟨x, y⟩)| = | I C_{V} (xy) : I C_{I C_{V} (xy)}(x)| =
|[ I C_{V} (xy), x] ∩ [ I C_{V} (xy), y]| ≤ |[V, x] ∩ [V, y]| ≤ |[V, R] ∩ [V, R^{g}]|

As xy ∼ x we get

| I C_{V} (x) : I C_{V} (⟨x, y⟩| ≤ |[V, R] ∩ [V, R^{g}]| (1)

Now choose u ∈ R^{h}. As O^{2′}(G_{1}) acts 2-transitively on R^{G_{1}} and [[V, R]∩[V, R^{g}], O^{2′}(G_{1})] = 0, we get that [V, R] ∩ [V, R^{g}] = [V, (R^{g})^{u}] ∩ [V, R^{g}] = [V, R^{g}]^{u} ∩ [V, R^{g}]. This shows |[[V, R^{g}], u]| = |[V, R] : [V, R^{g}] ∩ [V, R]|. As R^{h} ∼ R^{g} in G_{2}, we get for x ∈ R by (1) using [V, R^{g}, x] = 0:

|[V, R] ∩ [V, R^{g}]| ≥ | I C_{V} (x) : I C_{V} (⟨x, u⟩| ≥ |[V, R^{g}] : I C_{[V,R^{g}]}(u)| =

|[[V, R^{g}], u]| = |[V, R] : [V, R^{g}]∩[V, R]|. (2)

As [V, R] ∩ [V, R^{g}] is normalized by G_{1} and centralized by O^{2′}(G_{1}), we get that |[V, R] ∩ [V, R^{g}]| ≤ V_{1}. As V_{1} is trivial for O^{2′}(G_{1}) and irreducible for a torus we get |V_{1}| ≤ q^{3} and so |[V, R] ∩ [V, R^{g}]| ≤ q^{3}. Furthermore [V, R] is a nontrivial module for L_{2}(q^{3}) which shows |[V, R]| ≥ q^{6}. Hence by (2) |[V, R]| = q^{6} and |[V, R] ∩ [V, R^{g}]| = q^{3}. If q ≠ 2, then we have I C_{[V,R]}(S) = I C_{[V,R]}(s) for any s ∈ S^{#}. If q = 2 the same is true as any involution in S inverts an element of order 7 in SL(2, 8) ≃ G_{2}/K_{2}. So S acts quadratically on [V, R]. Hence V_{2} is the natural SL(2, q^{3})–module. So by (0.4) we get the assertion.

(1.11) Lemma. G ≄ ^{2}F_{4}(q).

Proof. Suppose false. The proof follows [MeiStr;(2.17)] where the case of the Tits group has been treated. Let G_{1}, G_{2} be the two minimal parabolics containing a Sylow 2–subgroup S. Choose notation such that G_{1} = N_{G}(R). Let Γ be the coset graph with respect to G_{1} and G_{2}. Let (1, 2, 3, 4, 5) be a path of length 4 in Γ with “G_{1} = G_{1}” and “G_{2} = G_{2}”. Set Z_{2} = ⟨R^{G_{2}}⟩ and U_{1} = ⟨Z_{2}^{G_{1}}⟩, then U_{1} = Q. We know that [V, R, U_{1}] = 0. As U_{1} ≰ O_{2}(G_{3}) ≱ U_{5} we get ⟨U_{1}, U_{5}⟩O_{2}(G_{3}) ≥ O^{2}(G_{3}). As U_{1} ∩ U_{5} is centralized by ⟨U_{1}, U_{5}⟩, we see |U_{1} ∩ U_{5}| ≤ q. Now |U_{1} ∩ O_{2}(G_{3})| = q^{4}, which gives U_{1} ∩ O_{2}(G_{3}) ≰ U_{3}. So we get ⟨U_{1}, U_{5}⟩ ≥ O^{2}(G_{3}) and (O^{2}(G_{3})∩G_{1})R = G_{1}∩G_{2}. Moreover ⟨U_{1}, U_{5}⟩ centralizes [ I C_{V} (U_{1}), Z(O_{2}(G_{5}))] and hence G_{1}∩G_{2} centralizes [U, Z(O_{2}(G_{5}))] for any chief factor U of G_{1} in I C_{V} (U_{1}). As O^{2′}(G_{1}/O_{2}(N_{G}(R))) ≃ Sz(q) we get with (0.6) that [U, O^{2′}(G_{1})] = 1. Hence we have that [ I C_{V} (Q), O^{2′}(G_{1})] = 1.

Choose now R^{g} ≤ Q − R. Then (0.5) shows that I C_{V} (S) is centralized by O^{2′}(G_{2}). But then I C_{V} (S) is centralized by G, a contradiction.

Now (1.1) – (1.11) and (0.6) prove the Theorem.

2 Strong quadratic modules

In this chapter we are going to prove Corollary 1. So we assume

Hypothesis 2: Let G be a Chevalley group over GF (q), q = 2^{n}, V a nontrivial irreducible GF (q)G–module and A be a nontrivial 2–subgroup of G such that [V, A, A^{g}] = 0 for any g ∈ G with [A, A^{g}] = 1.

Choose A maximal satisfying hypothesis 2. Let h ∈ G such that [A, A^{h}] = 1. Set B = ⟨A, A^{h}⟩. Then for g ∈ G with [B, B^{g}] = 1 we get [V, ⟨A, A^{h}⟩, ⟨A^{g}, A^{hg}⟩] = 0. By maximality of A this implies A = B, i.e. A is weakly closed in I C_{G}(A) with respect to G.

For the remainder of this chapter let G, V be as in hypothesis 2 and A ≤ G be maximal satisfying hypothesis 2.

(2.1) Lemma. Let G ≄ Sp(4, q), G_{2}(2) and R be a long (if G ≃ F_{4}(q) any) root group. Suppose there is some quadratic fours group W, (with ⟨W^{N_{G}(R)}⟩ nonabelian in case G ≃ SL(n, q)), contained in N_{G}(R) such that one of the following holds

(i) |W ∩ Z(O_{2}(N_{G}(R)))| = |W ∩ R| = 2.

(ii) W ∩ Z(O_{2}(N_{G}(R))) = 1 and there is some g ∈ G such that w_{1}^{g} = w_{1}, w_{2}^{g} = w_{2}r, for 1 ≠ r ∈ R and W = ⟨w_{1}, w_{2}⟩.

Then (G, V) satisfies the assertion of corollary 1.

Proof. Suppose (ii), then ⟨w_{1}, r⟩ satisfies (i). So we may assume that we have (i). Set Q = Z_{2}(O_{2}(N_{G}(R))). Now we get ⟨W ^{I C_{G}(R∩W)}⟩ ≥ Q and as [V, R ∩ W, W ^{I C_{G}(R∩W)}] = 0 we get [V, R ∩ W, Q] = 0. Now [V, R, Q] = [V, (R ∩ W)^{N_{G}(R)}, Q^{N_{G}(R)}] = 0. Application of the theorem gives that (G, V) is as in the corollary 1.

(2.2) Lemma. Let G ≄ SL(n, q), U(3, q), Sp(4, q), Sz(q) or G_{2}(2). Let R be a long root group. If A ∩ R ≠ 1, then corollary 1 holds.

Proof. As G ≅ S̸L(2, q), U_{3}(q), or Sz(q), we have R is not weakly closed in I C_{G}(R). So A ≰ R. If A ≰ Z(O_{2}(N_{G}(R))), then A contains a fours group satisfying (2.1)(i). (Notice G ≅ S̸L(n, q)).

So (2.1) yields the assertion. Let A ⊆ Z(O_{2}(N_{G}(R)). So we have Z(O_{2}(N_{G}(R))) ≠ R and then G ≃ Sp(2n, q) or F_{4}(q). As N_{G}(R) acts indecomposably on Z(O_{2}(N_{G}(R))) and A is weakly closed in I C_{G}(A) we get A = Z(O_{2}(N_{G}(R))). In case of F_{4}(q) there is another root group L such that L ∩ A ≠ 1. Now the same argument shows A = Z(O_{2}(N_{G}(L))). As Z(O_{2}(N_{G}(R))) ≠ Z(O_{2}(N_{G}(L))) in F_{4}(q) we get a contradiction.

We are left with G ≃ Sp(2n, q), n ≥ 3, |A| = q^{3}. Let L be a short root group L ≤ A. Then ⟨A^{N_{G}(L)}⟩ = O^{2}(N_{G}(L)). This gives [V, L, O^{2′}(N_{G}(L))] = 0. By (1.6) we get that V is the natural module and we are done.

For the remainder of this chapter we fix the following notation: Let R be a long root group centralized by A and Q = O_{2}(N_{G}(R)). Let S be a Sylow 2-subgroup of N_{G}(R) containing A.

(2.3) Lemma. If G ≄ (S)L(n, q), ^{2}F_{4}(q), Sp(2n, q), F_{4}(q), (S)U(3, q), Sz(q) or G_{2}(2), then corollary 1 holds.

Proof. Let A ≤ Q. In all Chevalley groups not excluded by assumption we have [A, Q] ≤ R and if a ∈ A − R, then [a, Q] = R. So as A is weakly closed in I C_{G}(A), we get either A ⊆ R or R ≤ A. Now the assertion follows with (2.2).

Let now A ≰ Q. Suppose A ∩ Q = 1. Then [N_{Q}(A), A] = 1. As I C_{G}(Q) ⊆ Q, we have N_{Q}(A) ≠ Q. Now for x ∈ Q − N_{Q}(A), [x, A] ⊆ N_{Q}(A), we have [A^{x}, A] = 1. But A is weakly closed in I C_{G}(A), a contradiction. So A ∩ Q ≠ 1.

Let 1 ≠ x ∈ A ∩ Q. Set Y = I C_{Q}(x). Choose a ∈ A^{♯}, g ∈ Y with a^{g} = ay, y ≠ 1. Then ⟨x, y⟩ acts quadratically on V. If x ∈ R, then (2.2) yields the assertion. Suppose x ∉ R. If y ∉ Z(Y), then ⟨x, y⟩ ∩ Z(O_{2}(N_{G}(R))) = 1. Furthermore there is some u ∈ Y with y^{u} = yr, r ∈ R^{♯}, as R = O_{2}(N_{G}(R))′. So (2.1)(ii) applies and corollary 1 holds. We are left with y ∈ Z(Y). Let Y_{1} = [a, Y] ⊆ Z(Y). We have |Q : Y | = q and |Q : I C_{Q}(t)| = q for any t ∈ Q − Z(Q). If |Y_{1}| > q, then there is some t ∈ Q − Y with x^{t} = xr, r ∈ R^{♯}, but y_{1}^{t} = y_{1} for some y_{1} ∈ Y_{1}^{♯}. So either (2.1)(ii) applies to ⟨y_{1}, x⟩ or y_{1} ∈ R and so (2.1)(i) applies as x ∉ R. In any case we get the assertion.

So as [Q, A] ≤ Y, we get |[Q/R, a]| ≤ q^{2} for any a ∈ A − Q. In particular G ≅ E̸_{n}(q), n = 6, 7, 8, ^{2}E_{6}(q), or ^{3}D_{4}(q), by using Chevalley commutator formulas. So we have G ≅ (S)U(n, q), G_{2}(q) or Ω^{±}(2n, q).

Let G ≃ (S)U(n, q). Then Q/R is the natural module for SU(n − 2, q). So A is a group of transvections on the natural module implying |A : A ∩ Q| ≤ q and there is some g ∈ G such that AQ ≤ QR^{g}. Furthermore by (2.1) we may assume A ∩ Q ≤ Z(Y), otherwise ⟨x, y⟩, x ∈ A ∩ Q, y ∉ Z(Y) satisfies (2.1)(ii).

We have [Z(Y), R^{g}] = 1. Let a ∈ A − Q, then a = uv, u ∈ Y, v ∈ R^{g}, [u, v] = 1. Suppose u ∉ Z( I C_{Q}(R^{g})). Then there is some s ∈ I C_{Q}(R^{g}) with u^{s} = ur, r ∈ R^{♯}. Now a^{s} = ar but x^{s} = x. Hence ⟨x, a⟩ satisfies (2.1)(ii) and so we have the assertion. Assume now u ∈ Z( I C_{Q}(R^{g})). Then uv is in a conjugate of R^{g} and so we have the assertion with (2.2).

Let G ≅ G_{2}(q) and L ≠ N_{G}(R) be the other parabolic containing S. Set E = O_{2}(L). As [E, O^{2′}(L)] ≠ 1, we get as before that A ∩ E ≠ 1. As Z(E)^{♯} ⊆ R^{G}, we get with (2.2) that A ∩ Z(E) = 1 or the conclusion holds. So assume A ∩ Z(E) = 1. Then [A, Z(E)] = 1 and so A ⊆ E. As E′ = Z(E) this contradicts the fact that A is weakly closed in I C_{G}(A).

So assume finally that G ≅ Ω^{±}(2n, q). Let L be the parabolic with O^{2′}(L/E) ≅ Ω^{±}(2n−2, q), E = O_{2}(L). Again A ∩ E ≠ 1. If A ⊆ E, then by weak closure A = E, which with (2.2) yields the assertion. Let A ≰ E. As E is the natural module for Ω^{±}(2n − 2, q), there is some t ∈ A − E and s ∈ E such that [t, s] = f is contained in a some long root group F. So ⟨f, e⟩, e ∈ E ∩ A is a quadratic fours group. As e ∉ T and Z(O_{2}(N_{G}(F))) = F, ⟨f, e⟩ satisfies (2.1)(i). So we get the assertion.

In what follows we analyze the cases left in (2.3) one by one.

(2.4) Lemma. If G ≅ Sp(2n, q), corollary 1 holds.

Proof. Suppose false. Let L be a short root subgroup contained in Z(S) and M = O_{2}(N_{G}(L)). Suppose first n = 2. Obviously we may assume q ≠ 2. Then A ≤ Q or M as any elementary abelian subgroup of S is in Q or M. As N_{G}(Q) and N_{G}(L) act irreducibly on Q, M, respectively, we get A = Q or A = M by weak closure. Now the assertion follows with the theorem.

So we have n ≥ 3. Let A ≤ Q. If A ∩ R ≠ 1 the assertion follows with (2.2). So we may assume A ∩ R = 1. Now [A, Q] ⊆ R. If A ⊈ Z(Q), then ⟨A^{Q}⟩ = ⟨A, R⟩, contradicting A^{Q} = A and R ∩ A = 1. So A ⊆ Z(Q). If q > 2, then N_{G}(Q) acts indecomposably on Z(Q). As A is weakly closed in Z(Q), we get A = Z(Q), contradicting A ∩ R = 1. So we have q = 2 and A is a complement of R in Z(Q), i.e. |A| = 4. Let U be the maximal parabolic containing S such that O^{2′}(U/O_{2}(U)) ≅ L_{n}(2). Then ⟨A^{U}⟩ is the natural module (n ≥ 3), a contradiction to the fact that A is weakly closed in I C_{G}(A).

So we may assume A ≰ Q. We have O^{2′}(N_{G}(R)/Q) ≅ Sp(2n − 4, q) × L_{2}(q). By (2.2) we may assume A ∩ R = 1. Suppose A ∩ Z(Q) ≠ 1. Let a ∈ A ∩ Z(Q)^{♯}, x ∈ A − Q. Then ⟨a, x⟩ acts quadratically on V. So [V, a, x^{I C_{N_{G}(R)}(a)}] = 0. Hence [V, a, Q] = 0 because any normal subgroup of I C_{G}(Z(Q)) not contained in Q has to contain Q. So there is a fours group ⟨ba, a⟩, ba ∈ R acting quadratically on V. Let a ∈ L. If A ≤ M, then by weak closure A = M. But then A ∩ R ≠ 1, a contradiction. So we have A ≰ M. But as A ∩ M ⊈ L by weak closure, we get some t ∈ M ∩ A, t in a long root group, which with (2.2) yields the assertion.

So we are left with a ∉ L. Then a is an involution of type c_{2}. So we may assume b ∈ L. As a ∈ N_{G}(L) we get a^{I C_{G}(b)} ≥ M and so [V, b, M] = 0. Now [V, b, M ∩ Q] = 0. As [V, a, Q] = 0, we get [V, ab, M ∩ Q] = 0. But then [V, ab, (M ∩ Q)^{I C_{G}(ab)}] = 0 and so [V, ab, Q] = 0, and the assertion holds by the theorem.

So we just have to treat the case A ∩ Z(Q) = 1. Now weak closure implies A ⊆ I C(Z(Q)). As in (2.3) we get that |[Q/Z(Q), t]| ≤ q^{2} for any t ∈ A − Q. This shows |A : A ∩ Q| ≤ q and A induces transvections on both Sp(2n − 4, q)-modules involved in Q/Z(Q). Now there is some ρ ∈ N_{G}(R), o(ρ) = q + 1, with [A, ρ] ⊆ Q. As A ⊆ N(M), we get A ⊆ N(M^{ρ}) and so A ∩ M ≠ 1 ≠ A ∩ M^{ρ}. But M ∩ M^{ρ} = (Q ∩ M) ∩ (Q ∩ M^{ρ}) ⊆ Z(Q). This shows that there is some ⟨x, y⟩ ⊆ A, ⟨x, y⟩ ∩ Z(Q) = 1, x ∈ M, y ∈ M^{ρ}. Hence there is some g ∈ M ∩ Q with y^{g} = yr, r ∈ R^{♯}. Application of (2.1)(ii) gives the assertion.

(2.5) Lemma. G ≅ F_{4}(q), corollary 1 holds.

Proof. Let A ≤ Q. If A ≰ Z(Q), then ⟨A^{Q}⟩ = AR. As A is weakly closed in I C_{G}(A) we get R ⊆ A. Now (2.2) yields the assertion. So assume A ⊆ Z(Q). By weak closure we get A = Z(Q) and so again R ≤ A and (2.2) yields the assertion. So we may assume A ≰ Q. Now we show that we may assume that A induces transvections on Z(Q). For this choose t ∈ A − Q such that |[Z(Q), t]| ≥ q^{2}. As A ∩ Z(Q) ≠ 1, there is a foursgroup ⟨a, b⟩ ≤ Z(Q), a ∈ A, such that ⟨a, b⟩ acts quadratically on V, b is a root element, and ⟨a, b⟩ ≤ O_{2}( I C_{G}(b)) − Z( I C_{G}(b)). But now (2.1)(i) yields the assertion.

So assume that A induces transvections on Z(Q). This shows |A : A ∩ Q| ≤ q and |[A, Q/Z(Q)]| = q^{4}. If |A ∩ Q : A ∩ Z(Q)| > q, then there is some foursgroup ⟨x, y⟩ ⊆ A ∩ Q such that ⟨x, y⟩ ∩ Z(Q) = 1 and for some h ∈ Q, x^{h} = x, y^{h} = yr, r ∈ R^{♯}, as |Q : I C_{Q}(t)| = q for any t ∈ Q − Z(Q). Now (2.1)(ii) yields the assertion. So assume |A ∩ Q : A ∩ Z(Q)| ≤ q. Let a ∈ A ∩ Q − Z(Q). Then the action of A on Q shows that there is some g ∈ I C_{Q}(a) − Z(Q), such that b^{g} = bc, for b ∈ A − Q and c not in the root group determined by a. But then ⟨a, c⟩ acts quadratically on V, and (2.1)(i) yields the assertion. So assume A∩Q = A∩Z(Q). Let L be the other type of root groups such that LR ⊆ [Z(Q), A]. Set M = O_{2}(N_{G}(L)). Then as A induces transvections on Z(Q), we get AQ ⊆ Z(M)Q. Now (Z(M)Q)′ ⊆ Z(M), so [A, Q] ⊆ Z(M). As |[a, Q/Z(Q)]| = q^{4} for a ∈ A − Q, we get that |[a, Q]L/L| = q^{5} and [a, Q, a] = 1. But then [a, Q]L/L is centralized by a and [[A, Q], A] = 1. Hence ⟨A^{Q}⟩ is abelian. Now weak closure gives |A ∩ Z(M)| ≥ q^{5}. But then as |Z(M) : Z(M) ∩ Q| = q, we get |A ∩ Q| = |A ∩ Z(Q)| ≥ q^{4}. So |A ∩ Z(Q) ∩ Z(M)| ≥ q^{4}, but this contradicts |Z(Q) ∩ Z(M)| = q^{2}.

(2.6) Lemma. If G ≅ SL(n, q), corollary 1 holds.

Proof. Let Q = E_{1}E_{2}, E_{i} ≤ N_{G}(R), E_{i} elementary abelian, i = 1, 2. Then A ≤ N_{G}(E_{1}). As in all cases before we get A ∩ E_{1} ≠ 1 by weak closure. But all elements in E_{1}^{♯} are conjugate and so we may assume A ∩ R ≠ 1.

If A ⊆ E_{1} we get by weak closure A = E_{1} and then the assertion with [MeiStr1;(1.6)]. So assume A ⊈ E_{1}. Choose x ∈ R ∩ A, y ∈ A − E_{1}. Then ⟨x, y⟩ is a quadratic fours group and U = ⟨⟨x, y⟩^{N_{G}(R)}⟩ ≥ E_{1}. So U′ ≠ 1. Now we may apply (2.1) (i) to ⟨x, y⟩ which implies the assertion.

(2.7) Lemma. G ≄ ^{2}F_{4}(q).

Proof. Suppose false. By (2.2) A ∩ R = 1. So by weak closure we have A ≰ O_{2}(N_{G}(R)). Let t ∈ A − O_{2}(N_{G}(R)). Then |[t, Q]| = q^{2}. As A is weakly closed in I C(A) we get that either |A∩Q| ≥ q^{2} or there is some a ∈ A∩Q, b ∈ Q such that t^{b} = tr, where r is not in the root group belonging to a. Hence ⟨a, r⟩ is a quadratic fours group. So in both cases we get a quadratic fours group in Q which is not contained in a root subgroup, which contradicts (2.1).

(2.8) Lemma. If G ≅ Sz(q), then V is the natural module.

Proof. We have A = R. As G = ⟨R, R^{g}⟩ for any g ∈ G − N_{G}(R), we get V = [V, R] ⊕ [V, R^{g}]. By [Mar] V is the natural module.

(2.9) Lemma. If G ≅ (S)U(3, q), then V is a basic module as in corollary 1.

Proof. We have R = A. Let X = ⟨R, R^{g}⟩ ≃ SL(2, q) for some g ∈ G. Then we know that V involves just natural X - modules by (2.6). As any irreducible G-module is a restriction of some irreducible SL(3, q^{2})-module, we may assume that V is irreducible for H = (S)L(3, q) and X ≤ G ≤ H. By Steinberg’s tensor product lemma any irreducible H-module is a tensorproduct of algebraic conjugates of the basic modules. We may assume that all this basic modules are nontrivial. Then V involves tensorproducts of algebraic conjugates of the natural X-module. The quadratic action of X now gives that the tensor product just has one factor, i.e. V is an algebraic conjugate of a basic module. Over GF (2) this means that V is a basic module.

(2.10) Lemma. If G ≅ G_{2}(2), then V is the natural module.

Proof. We have a quadratic fours group. It is easy to see that V has to be the natural 6-dimensional module.

Now Corollary 1 follows from (2.3) - (2.10).

3 Quadratic fours groups

Hypothesis 3: Let G be a Chevalley group over GF (q), q = 2^{n}, V a nontrivial irreducible GF (q)G-module. Let R be a long root subgroup of G and E be a fours group with |E ∩ R| = 2. Assume [V, E, E] = 0.

For the remainder of this section we are working under hypothesis 3. We are going to prove Corollary 2.

We fix the following notation:

(i) ⟨e⟩ = E ∩ R.

(ii) X = ⟨E ^{I C_{G}(e)}⟩.

(3.1) Lemma. If X ⊇ Z_{2}(O_{2}(N_{G}(R))), then the conclusion of corollary 2 holds.

Proof. As [V, e, X] = 0 and ⟨e^{N_{G}(R)}⟩ = R, the assertion follows from the main theorem.

(3.2) Lemma. If G ≅ S̸L(n, q), Sp(2n, q), F_{4}(q) or G_{2}(q) the conclusion of corollary 2 holds.

Proof. In all other cases I C_{G}(R) acts irreducibly on Z_{2}(O_{2}(N_{G}(R)))/R and Z_{2}(O_{2}(N_{G}(R)))/R is the only minimal normal subgroup of I C_{G}(R)/R. So Z_{2}(O_{2}(N_{G}(R))) ⊆ X. The assertion follows with (3.1).

(3.3) Lemma. Let G ≅ SL(n, q), then V = V (λ) for some fundamental weight λ.

Proof. By (3.1) we may assume O_{2}(N_{G}(R)) ⊈ X. Let O_{2}(N_{G}(R)) = E_{1}E_{2}, E_{i} elementary abelian of order q^{n−1}, E_{i}◁N_{G}(R), i = 1, 2. Then we may assume X ⊆ E_{1}. Assume first X = E_{1}. As X = ⟨e^{N_{G}(x)}⟩, we get 0 = [V, ⟨e^{N_{G}(R)}⟩, X] = [V, X, X]. As X is weakly closed in I C_{G}(X) with respect to G, the assertion follows with corollary 1.

So X ≠ E_{1}. This is only possible for n = 3. We have that R acts quadratically on V. Furthermore S = E_{1}R^{g} ∈ Syl_{2}(N_{G}(E_{1})) for suitable g ∈ G. Set V_{1} = I C_{V} (E_{1}).

If [R^{g}, V_{1}] ≠ 0, then I C_{V} (S) is centralized by ⟨S, X^{g}⟩. As X^{g} ⊆ E_{1}^{g} and E_{1}^{g} ∩ N_{G}(E_{1}) = R, we get ⟨S, X^{g}⟩ = O^{2′}(N_{G}(E_{2})). As by [Hig] V_{1} is a natural SL(2, q)-module, we get with (0.4) that V is the natural module, as V_{1}, V_{2} are like in the natural module.

If [V_{1}, O^{2′}(N_{G}(E_{1}))] = 0, then I C_{V} (E_{2}) has to be a natural module by [Hig] as S = E_{2}R^{h} for suitable h ∈ G, S ∈ Syl_{2}(N_{G}(E_{2})). Now again by (0.3) V is a natural module.

(3.4) Lemma. Let G ≅ Sp(2n, q), then V = V (λ) for some fundamental weight λ.

Proof. By (3.1) we may assume X = Z(O_{2}(N_{G}(R))). Let f ∈ X, f ∈ R̃, R̃ a short root subgroup. Then [V, ⟨e, f⟩, ⟨e, f⟩] = 0. Set Y = ⟨⟨e, f⟩^{I C_{G}(f)}⟩. As e ∈ O_{2}( I C_{G}(R̃)), we get Y = O_{2}( I C_{G}(R̃)). Furthermore ⟨f^{N_{G}(R̃)}⟩ = R̃ and so [V, R̃, Y] = 0. Now the assertion follows with (1.6).

(3.5) Lemma. Let G ≅ F_{4}(q), then V = V (λ) for some fundamental weight λ.

Proof. By (3.1) we may assume that X = Z(O_{2}(N_{G}(R))). So we may assume E = ⟨e, f⟩, e ∈ R, f ∈ R̃, R̃ ⊆ X, R̃ a short root subgroup. As G possesses an outer automorphism α with R^{α} = R̃, we may also assume that ⟨E ^{I C_{G}(f)}⟩ = Z(O_{2}(N_{G}(R̃))). So E ⊆ Z(O_{2}(N_{G}(R̃))) ∩ Z(O_{2}(N_{G}(R))) = RR̃. Furthermore we have ⟨e^{N_{G}(R)}⟩ = R and ⟨f^{N_{G}(R̃)}⟩ = R̃.

This gives

(∗) [V, R, Z(O_{2}(N_{G}(R)))] = 0 = [V, R̃, Z(O_{2}(N_{G}(R̃)))]

Now we fix a Sylow 2-subgroup S of N_{G}(R) with Z(S) = RR̃. Let G_{1}, G_{2}, G_{3}, G_{4} be the four maximal parabolic subgroups of G containing S. Set K_{i} = O_{2}(G_{i}) and V_{i} = I C_{V} (K_{i}).

Let V_{i} be a nontrivial module for O^{2′}(G_{i}) for some i. There is some g ∈ G such that for T = R^{g} or R̃^{g} the following holds:

(i) T ⊆ G_{i}\O_{2}(G_{i}).

(ii) [T, S] ⊆ O_{2}(G_{i}).

(iii) [V_{i}, T] ≠ 0.

By (0.3), (ii) and (iii) I C_{V} (S) ⊆ [V_{i}, T] and so by (∗) I C_{V} (S) is centralized by ⟨S, Z(O_{2}(N_{G}(T)))⟩. As Z(O_{2}(N_{G}(T))) ⊈ G_{i}, we get that I C_{V} (S) is centralized by O^{2′}(P_{i}), where P_{i} is the minimal parabolic subgroup containing S but not contained in G_{i}. So we have

(∗∗) If [V_{i}, O^{2′}(G_{i})] ≠ 0 for some i ∈ {1, 2, 3, 4}, then [ I C_{V} (S), O^{2′}(P_{i})] = 0, where S ⊆ P_{i}, P_{i} ⊈ G_{i}.

From (∗∗), we get that there is exactly one j ∈ {1, 2, 3, 4} such that [ I C_{V} (S), O^{2′}(P_{j})] ≠ 0. As there is some h ∈ G such that for U = R^{h} or R̃^{h} we have S = O_{2}(P_{j})U and [V, U, U] = 0 by (∗), we get with [Hig] and (0.3) that I C_{V} (O_{2}(P_{j})) is the natural module for O^{2′}(P_{j}/O_{2}(P_{j})) ≅ SL(2, q). Now V = V (λ_{j}) by (0.4).

(3.6) Lemma. Let G ≅ G_{2}(q), then V is the natural module.

Proof. By (3.1) we may assume X ⊆ O_{2}(N_{G}(R)), X ≠ O_{2}(N_{G}(R)). Let G_{1}, G_{2} be the two maximal parabolics containing S ∈ Syl_{2}(N_{G}(R)), where G_{1} = N_{G}(R). There is some g ∈ G such that R^{g}O_{2}(G_{2}) = S. As R ⊆ X, we have [V, R, R] = 0.

Suppose first that [ I C_{V} (O_{2}(G_{2})), O^{2′}(G_{2})] ≠ 0. Then I C_{V} (S) ⊆ [ I C_{V} (O_{2}(G_{2})), R^{g}] and so I C_{V} (S) is centralized by S and X^{g}. But X^{g} ⊈ G_{2} and so ⟨S, X^{g}⟩ = O^{2′}(G_{1}) centralizes I C_{V} (S). By (0.3) and [Hig] I C_{V} (O_{2}(G_{2})) is the natural O^{2′}(G_{2})/O_{2}(G_{2}) ≅ SL(2, q) module. So by (0.4)(c) V = V (λ_{2}).

Let f ∈ Z(O_{2}(G_{2})) − R, then f ∼ e ∼ ef. By (0.1) W = [V, f] ∩ [V, e] ≠ 0. Let g ∈ G_{2} with e^{g} = f, then W is centralized by ⟨X, X^{g}⟩.

We have that W is invariant under O_{2}(G_{2}) and so W ∩ I C_{V} (O_{2}(G_{2})) ≠ 0. This shows that there is some v ∈ I C_{V} (O_{2}(G_{2}))^{♯} such that v is centralized by some element ν ≠ 1 of odd order in O^{2′}(G_{2}). (Notice that ⟨X^{N_{G}(S)}⟩ = O_{2}(G_{1}) and so X ⊈ O^{2}(G_{2})). But I C_{V} (O_{2}(G_{2})) is the natural module, a contradiction.

So we have that [ I C_{V} (S), O^{2′}(G_{2})] = 0. As R ⊆ X, we get with (0.6) that I C_{V} (O_{2}(G_{1})) is the natural module for O^{2′}(G_{1})/O_{2}(G_{1}) and so by (0.3) V is the natural module for G.

Now Corollary 2 follows from (3.1) - (3.6).

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