On the subgroup structure of exceptional groups of Lie type

Martin W. Liebeck and Gary M. Seitz

August 7, 1997

Introduction

In this paper we establish results aimed at the study of the subgroup structure of the finite exceptional groups of Lie type, in particular of their maximal subgroups. Let G(r) be an exceptional group of Lie type over a finite field 𝔽_{r} of order r, where r is a power of the prime p. Write G(r) = O^{p′}(G_{σ}), where G is a simple algebraic group over 𝔽̄_{r} and σ is a Frobenius morphism of G.

The reduction theorem [LS1, Theorem 2] determines all maximal subgroups of G(r) which are not almost simple. Thus we concentrate on almost simple maximal subgroups of G(r); let M be such a subgroup. Of particular interest is the so-called “generic case”, in which F^{∗}(M) = X(q), a group of Lie type over a field 𝔽_{q}, also of characteristic p.

Our approach to the generic case is to attempt to show that the embedding X(q) < G(r) lifts to an embedding X̄ ≤ G, where X̄ is a simple closed connected subgroup of G of the same type as X(q). Once this is achieved, we are in a position to apply results from [Se2, LS1] on the subgroup structure of algebraic groups.

The first lifting result for exceptional groups appeared in [ST1], and showed that the finite embedding can be lifted under certain hypotheses, one of the hypotheses being that the characteristic p be suitably large. In this paper we establish lifting results in which there are hypotheses on the size of the ground field 𝔽_{q}, but no assumptions on the characteristic. Consequently, the results hold for all but finitely many possibilities for the group X(q). For example, Theorem 6 below shows that for q sufficiently large, either X(q) is the group of fixed points of an automorphism of G(r), or X(q) = O^{p′}(X̄_{σ}) with X̄ maximal among closed connected M⟨σ⟩-invariant subgroups of G. (In most cases, “sufficiently large” means that q > 9.)

The authors acknowledge the support of NATO Collaborative Research Grant CRG 931394. The second author also acknowledges the support of an NSF Grant.

Once the lifting has been achieved we can apply the results of [Se2, LS1], which determine those subgroups X̄ of G as in the last sentence of the previous paragraph, under some mild characteristic restrictions. These restrictions are much weaker than those required for the result in [ST1]; in particular, p = 0 or p > 7 covers all the restrictions. However, in view of the results obtained in this paper, it becomes a high priority issue to extend the analysis in [Se2, LS1] so as to remove the characteristic restrictions. Such an analysis is under way.

The preceding discussion has been couched in terms of maximal subgroups, but our results apply much more generally. We establish lifting results for all subgroups X(q) of G(r), with q sufficiently large. Namely, in Theorem 1 we show any such subgroup is contained in a closed connected subgroup X̄ of G stabilizing precisely the same subspaces of the Lie algebra L(G) as X(q). We also show (Theorem 4 and Corollary 5) that this connected group X̄ is almost always proper in G, and that, given mild characteristic restrictions, X(q) = O^{p′}(X̄_{δ}) for some Frobenius morphism δ (Theorem 10). At this point the results of [LS2] determine the conjugacy class of X(q) in G(r).

Finally, we use our lifting results to clear up a nagging problem on classical groups. In Theorem 11, we show that an embedding X < Y < C of finite groups of Lie type in the same characteristic, with C classical and X absolutely irreducible on the usual module for C, can usually be lifted to a corresponding embedding of simple algebraic groups; such embeddings of algebraic groups are known [Se1]. See [Se3] for further discussion of this problem.

We now state our results in detail, beginning with our main lifting results. As indicated above, these require no assumption on the characteristic p, but do need an assumption on q, where X = X(q) is as above. In order to specify this assumption we need the following definition.

Definition Let G be a simple adjoint algebraic group, and let Σ = Σ(G) be the root system of G. For a sublattice L of the lattice ℤΣ, let t(L) be the exponent of the torsion subgroup of ℤΣ/L. For α, β ∈ Σ, call the element α − β of ℤΣ a root difference. Define

t(Σ(G)) = max{t(L) : L a sublattice of ℤΣ generated by root differences}.

Theorem 1 Let X = X(q) be a quasisimple group of Lie type in characteristic p, and suppose that X < G, where G is a simple adjoint algebraic group of exceptional type, also in characteristic p. Assume that

q > t(Σ(G)).(2, p − 1) if X = A_{1}(q), ^{2}B_{2}(q) or ^{2}G_{2}(q),
q > 9 and X ≠ A_{2}^{ϵ}(16) otherwise.

Then the following hold:

(i) there is a closed connected subgroup X̄ of G containing X, such that every X-invariant subspace of the Lie algebra L(G) is also X̄-invariant;

(ii) if also X ≤ G_{σ} where σ is a Frobenius morphism, then there is a σ-stable subgroup X̄ containing X such that each X-invariant subspace of each G-composition factor of L(G) is also X̄-invariant.

Theorem 4 below will show that the subgroup X̄ in the conclusion is proper in G, unless X has the same type as G.

Remark If p is a good prime for G, then L(G) is irreducible as a Gmodule, so the subspace invariance conditions in conclusions (i) and (ii) are equivalent; however, for bad primes L(G) can be reducible (see Proposition 1.10 below), in which case (i) is stronger than (ii).

In view of Theorem 1, it would be of interest to know the values of t(Σ(G)), particularly for groups G of exceptional type. Dr Ross Lawther has informed us that he has verified (using a computer) that t(G_{2}) = 12, t(F_{4}) = 68, t(E_{6}) = 124 and t(E_{7}) = 388, while t(E_{8}) is unknown at present.

The bounds in Theorem 1 for groups of type A_{1}, ^{2}B_{2}, ^{2}G_{2} are relatively large, and follow from Corollary 3 below, which we shall deduce from the following general proposition. It should be possible to improve these bounds with additional work.

Proposition 2 Let G be a simple adjoint algebraic group, and let x be a semisimple element of G of finite order greater than mt(Σ(G)), for some positive integer m. Then there is an infinite closed subgroup S of G such that the following both hold:

(i) x ∈ S, and some nontrivial power of x, of order greater than m, lies in S^{0};

(ii) every x-invariant subspace of L(G) is also S-invariant.

Corollary 3 Let X = X(q) be a quasisimple group of Lie type in characteristic p, and suppose that X < G, where G is a simple adjoint algebraic group, also in characteristic p. Assume that q > t(Σ(G))m(X), where m(X) is the order of the Schur multiplier of X/Z(X). Then conclusions (i) and (ii) of Theorem 1 hold.

Note that Proposition 2 and Corollary 3 apply to all types of simple algebraic group, not just exceptional types.

The proofs of Proposition 2 and Corollary 3 are rather short, and are given in §2. Reducing the bounds on q to those in Theorem 1 requires a great deal more effort. After some preliminary results in §3, this is carried out for p ≠ 2 in §§4,5 and in §6 for p = 2. With some further effort, our proof could no doubt be extended to improve the q > 9 bound for subgroups X of rank larger than 2; indeed, when X has rank greater than half the rank of G, [LST, Theorem 2] gives a stronger conclusion assuming only that q > 2.

In order to make use of Theorem 1 (or Corollary 3), we need to know that the subgroup X̄ is proper in G. For this it suffices to show that X̄ acts reducibly on some G-composition factor of L(G). This information is provided by the next result.

Definition We define X and G to be of the same type if X ≅ G_{σ}^{(∞)} for some Frobenius morphism σ; we also say that X is a group of type G.

Theorem 4 Let G be a simple algebraic group in characteristic p, and let X = X(q) be a subgroup of G, where q = p^{e} and X is quasisimple. Suppose that X is not of the same type as G, and that X acts irreducibly on every G-composition factor of L(G).

Then (G, p) = (C_{n}, 2) or (B_{n}, 2), and either X = D_{n}^{ϵ}(q) or n = 3 and X = G_{2}(q); further, X lies in a simple connected subgroup D_{n} or G_{2} of G.

Combining Theorems 1 and 4, we immediately have

Corollary 5 Let G be a simple adjoint algebraic group of exceptional type, and suppose that X = X(q) < G, where q is as in the hypotheses of Theorem 1. Then either X is of the same type as G, or X < X̄ < G, where X̄ is closed and connected, and leaves invariant every X-invariant subspace of L(G).

Using this result we shall derive Theorem 6 below, which concerns the maximal subgroups of the finite exceptional groups of Lie type. In the statement we refer to [LS1, Theorem 1]; in this result, certain positive integers N(X̄, G) are defined for pairs (X̄, G) of simple algebraic groups, as follows:

G = E_{8} E_{7} E_{6} F_{4} G_{2}
X̄ = A_{1}  7 7 5 3 3
A_{2}  5 5 3 3
B_{2}, G_{2}  5 3 3 2
B_{3}  2 2 2
A_{3}, C_{3}, B_{4}  2

For example, N(A_{2}, E_{7}) = 5, and so on. For (X̄, G) not in the table, set N(X̄, G) = 1. We define N(X(q), G) to be equal to N(X̄, G), where X̄ is a simple algebraic group of the same type as X(q).

The statement of Theorem 6 and its corollaries involves groups G, L and L_{1} such that G is a simple adjoint algebraic group of exceptional type in characteristic p, σ is a Frobenius morphism such that L = O^{p′}(G_{σ}) is a finite simple group of exceptional Lie type, and L_{1} is a finite group with socle L (i.e. L ≤ L_{1} ≤ Aut L).

Theorem 6 Let L = O^{p′}(G_{σ}) and L ≤ L_{1} ≤ Aut L, as above. Suppose that H is an almost simple maximal subgroup of L_{1}, and F^{∗}(H) = X(q) with q satisfying the hypotheses of Theorem 1. Then one of the following holds:

(i) X(q) has the same type as G;

(ii) X(q) = O^{p′}(X̄_{σ}) for some maximal closed connected reductive H⟨σ⟩- stable subgroup X̄ of G. (The subgroup X̄ is given by [LS1, Theorem 1] if X̄ is non-simple or if p > N(X(q), G).)

Remarks 1. There is no assumption on p in the theorem.

2. The subgroups satisfying (i) are uniquely determined up to G_{σ}-conjugacy by [LS2, 5.1].

3. Our proof of Theorem 6 shows more generally, that if H is any (not necessarily maximal) subgroup of L_{1} such that F^{∗}(H) = X(q) with q as in Theorem 1, then H ≤ N(X̄_{σ}) for some X̄ as in conclusion (ii).

4. In our proof of Theorem 6 we do not require the full force of the hypothesis on q; indeed, in the proof we only require the existence of a subgroup X̄ satisfying the conclusions of Theorem 1.

We recall one further definition before stating the next result. If G, L, L_{1} are as above, and D is a σ-stable connected reductive subgroup of G of maximal rank (i.e. containing a maximal torus), then we call N_{L_{1}}(D_{σ}) a subgroup of maximal rank of L_{1}. The subgroups of maximal rank which are maximal in L_{1} are determined in [LSS].

Corollary 7 Let L = O^{p′}(G_{σ}) and L ≤ L_{1} ≤ Aut L, as above. Let H be an almost simple maximal subgroup of L_{1}, with F^{∗}(H) = X(q) and q as in Theorem 1, and let G_{σ} = G(q_{1}). Then one of the following holds:

(i) H is a subgroup of maximal rank;

(ii) X(q) has the same type as G;

(iii) q = q_{1} and X(q) = O^{p′}(X̄_{σ}), where X̄ is a simple maximal connected σ-stable subgroup of G not containing a maximal torus. Moreover, assuming that p > N(X(q), G), the possibilities are:

G = G_{2} : X(q) = A_{1}(q) (p ≥ 7)
G = F_{4} : X(q) = A_{1}(q) (p ≥ 13) or G_{2}(q) (p = 7)
G = E_{6} : X(q) = A_{2}^{ϵ}(q) (ϵ = ±, p ≥ 5), G_{2}(q) (p ≥ 11), F_{4}(q) or C_{4}(q) (p ≠ 2)
G = E_{7} : X(q) = A_{1}(q) (2 classes, p ≥ 17, 19) or A_{2}^{ϵ}(q) (ϵ = ±, p ≥ 7)
G = E_{8} : X(q) = A_{1}(q) (3 classes, p ≥ 23, 29, 31) or B_{2}(q) (p ≥ 7).

Assuming p > N(X(q), G), there is just one Aut(G)-class of subgroups X̄ of each isomorphism type occurring, and each such Aut(G)-class gives a unique Aut(G_{σ})-class of subgroups X(q).

Corollary 8 Let L = O^{p′}(G_{σ}) and L ≤ L_{1} ≤ Aut L, as above. There is a constant c (independent of G, L, L_{1}), such that if H is a maximal subgroup of L_{1} with |H| > c, then either

(i) H is the normalizer of a subgroup of the same type as G, or

(ii) H = N_{L_{1}}(X̄_{σ}) for some maximal closed connected H⟨σ⟩-stable subgroup X̄ of G.

Next we state our results on embeddings of arbitrary subgroups X(q) of an exceptional adjoint algebraic group G. Theorem 9 is the analogue of [LS2, Theorem 1] for finite groups. In the statement we use numbers N′(X, G) as defined in [LS2]; these are roughly the same as the numbers N(X, G), with a few differences. The precise definition of N′(X, G) is given by the following table:

G = E_{8} E_{7} E_{6} F_{4} G_{2}
X of type A_{1}  7 7 5 3 3
A_{2}  5 5 3 3
B_{2}  5 3 3 2
G_{2}  7 7 3 2
B_{3}  2 2 2 2
C_{3}  3 2 2 2
A_{3}, B_{4}, C_{4}, D_{4}  2 2 2

Theorem 9 Let X = X(q) < G, with G of exceptional type and q as in Theorem 1, and assume that p > N′(X, G). If X lies in a parabolic subgroup QL of G, with unipotent radical Q and Levi subgroup L, then X lies in a Qconjugate of L. Further, if also X < G_{σ} (where σ is a Frobenius morphism), and QL is σ-stable, then X lies in a σ-stable Q-conjugate of L.

Theorem 10 Let X = X(q) < G, with G of exceptional type and q as in Theorem 1, and assume that p > N′(X, G). Then

(i) X lies in a closed connected simple subgroup X̃ of G of the same type as X, and C_{G}(X)^{0} = C_{G}(X̃)^{0};

(ii) if also X < G_{σ} with σ a Frobenius morphism of G, then X lies in a closed connected semisimple σ-stable subgroup X̄ = X_{1} … X_{t} of G, where each factor X_{i} is simple of the same type as X; moreover, C_{G}(X)^{0} = C_{G}(X̄)^{0}.

The embeddings of the subgroups X̃ and X̄ in the conclusion of Theorem 10 are given by [LS2, Theorems 5 and 7], so Theorem 10 determines the embeddings of X(q) in G and G_{σ}.

For X = X(q) of rank more than \frac{1}{2}rank(G), the conclusion of Theorem 10 is obtained in [LST, Theorem 2] with no assumption on p, and assuming only that q > 2.

Finally, we give a consequence of Theorem 1 to the study of subgroups of finite classical groups.

Theorem 11 Let C be a finite classical simple group in characteristic p, with usual module V. Suppose that X = X(q) and Y are finite simple groups of Lie type in characteristic p, not of the same type, such that X < Y < C and X is absolutely irreducible on V; when Y is of exceptional type, assume also that q is as in Theorem 1 (relative to Y).

Then the embedding X < Y < C lifts to an embedding of closed subgroups X̄ < Ȳ ≤ C̄, where X̄, Ȳ , C̄ are connected simple algebraic groups of the same types as X, Y, C.

Remark Assuming Ȳ , C̄ are not both classical groups on the same natural module, all possibilities for the triple (X̄, Ȳ , C̄) in the conclusion are given by [Se1, Te1].

The layout of the paper is as follows. After the preliminary first section, we give in §2 the proofs of Proposition 2 and Corollary 3. The following four sections are devoted to the proof of Theorem 1. In §3 we describe the strategy and give some preliminary lemmas. The proof is then carried out for q odd in §§4,5, and for q even in §6. In §7 we prove Theorem 4, and §8 contains proofs of Theorem 6 and its corollaries. Theorems 9 and 10 are proved in §9, and Theorem 11 in §10.

We use the following notation throughout the paper. If Y is a connected simple algebraic group over the algebraically closed field K, or a corresponding finite group of Lie type, and λ is a dominant weight, then V_{Y} (λ) denotes the rational irreducible KY -module with high weight λ, and W_{Y} (λ) denotes the corresponding Weyl module. If V_{1}, …, V_{k} are modules, then V_{1}/ … /V_{k} denotes a module having the same composition factors as V_{1} ⊕ … ⊕ V_{k}. We often abbreviate this notation as follows: if μ_{1}, …, μ_{k} are dominant weights for the group Y, and c_{1}, …, c_{k} are positive integers, then

μ_{1}^{c_{1}}/ … /μ_{k}^{c_{k}}

always denotes a KY -module having the same composition factors as the module V_{Y} (μ_{1})^{c_{1}} ⊕ … ⊕ V_{Y} (μ_{k})^{c_{k}}.

When Y is a subgroup of G and V is a KG-module, we use V ↓ Y for the restriction of V to Y. If Y is connected and simple in characteristic p > 0, and q is a power of p, then σ_{q} denotes a standard Frobenius morphism of Y - that is, a morphism inducing the map x_{α}(t) → x_{α}(t^{q}) on root groups. If V is a KY -module, we write V^{(q)} for the KY -module obtained from V by twisting the action of Y by σ_{q} (i.e. changing the action from v → vy to v → vy^{σ_{q}}).

We use T_{i} to denote a torus of rank i, and W (G) for the Weyl group of G. The fundamental roots in a fundamental system for G are denoted α_{1}, …, α_{l}, and the corresponding fundamental dominant weights are λ_{1}, …, λ_{l}. The Dynkin diagram of G is labelled as in [Bo, p.250].

For ϵ = ±, A_{n}^{ϵ}(q) denotes the group A_{n}(q) if ϵ = + and ^{2}A_{n}(q) if ϵ = −; we use similar notation for types D_{n}^{ϵ}, E_{6}^{ϵ}, B_{2}^{ϵ}, G_{2}^{ϵ}, F_{4}^{ϵ}, and also for D_{4}^{ϵ} we allow ϵ = 3 to denote ^{3}D_{4}. Finally, if X is a finite group of Lie type, then rk(X) denotes the rank of the simple algebraic group corresponding to X.

1 Preliminaries

In this section we present some preliminary results which will be used throughout the paper.

The first few results concern representations of groups of type A_{1}. For the group G = SL_{2}(K) over the algebraically closed field K of characteristic p > 0, and for 0 ≤ i ≤ p − 1, we denote simply by i (or sometimes V (i)) the rational irreducible KG-module with high weight iλ_{1}, of dimension i + 1; these are the restricted KG-modules. By Steinberg’s tensor product theorem, every finite-dimensional rational irreducible KG-module is a tensor product i_{0} ⊗ i_{1}^{(p)} ⊗ … ⊗ i_{k}^{(p^{k})} of field twists of restricted modules i_{0}, …, i_{k}, and such a module has high weight ∑ i_{r}p^{r}.

Using [St, Lemma 79], we have the following well known result.

Proposition 1.1 Let V (i) denote the restricted module for G = SL_{2}(K) of high weight i. If i is odd, then the induced group G^{V (i)} = SL_{2}(K) ≤ Sp(V (i)); and if i is even, then G^{V (i)} = P SL_{2}(K) ≤ SO(V (i)).

The next result concerns extensions of modules for SL_{2}(K) and its subgroups SL_{2}(p^{e}), and is taken from [AJL, 3.9 and 4.5].

Proposition 1.2 Let G = SL_{2}(K), and for e ≥ 1 denote by G_{e} a subgroup SL_{2}(p^{e}) of G.

(i) Suppose p > 2 and let 0 ≤ a, b ≤ p − 1. Then Ext_{G_{1}}^{1}(a, b) ≠ 0 if and only if a + b = p − 3 or p − 1.

(ii) Let e ≥ 2, and suppose λ, μ < p^{e} have p-adic expansions λ = ∑_{1}^{e−1} a_{i}p^{i}, μ = ∑_{1}^{e−1} b_{i}p^{i} (where 0 ≤ a_{i}, b_{i} ≤ p − 1). Set a_{e} = a_{0} and b_{e} = b_{0}. Then Ext_{G_{e}}^{1}(λ, μ) ≠ 0 if and only if there exists k ≥ 0 such that

a_{i} = b_{i} for i ∉ {k, k + 1}, a_{k} + b_{k} = p − 2, and a_{k+1} − b_{k+1} = ±1.

Further, Ext_{G_{e}}^{1}(λ, μ) has dimension at most 1, except if e = 2 and (a_{0}, a_{1}) = (b_{1}, b_{0}) = (\frac{1}{2}(p − 3), \frac{1}{2}(p − 1)) or (\frac{1}{2}(p − 1), \frac{1}{2}(p − 3)) (in which case it has dimension 2).

(iii) Let λ, μ be positive integers with p-adic expansions λ = ∑ a_{i}p^{i}, μ = ∑ b_{i}p^{i}. Then Ext_{G}^{1}(λ, μ) ≠ 0 if and only if there exists k ≥ ν_{p}(λ + 1) (where ν_{p}(λ + 1) = max{i : p^{i} divides λ + 1}), such that

a_{i} = b_{i} for i ∉ {k, k + 1}, a_{k} + b_{k} = p − 2, and a_{k+1} − b_{k+1} = ±1.

Further, Ext_{G}^{1}(λ, μ) has dimension at most 1.

Proposition 1.3 Let G = SL_{2}(K), G_{e} = SL_{2}(p^{e}) and let 0 ≤ λ, μ < p^{e}.

(i) The restriction map Ext_{G}^{1}(λ, μ) → Ext_{G_{e}}^{1}(λ, μ) is injective; indeed, this map is an isomorphism unless either e = 1, or λ, μ are as in the exceptional cases at the end of 1.2(ii).

(ii) Suppose that e ≥ 2, and, if e = 2, that λ, μ are not as in the exceptional cases at the end of 1.2(ii). Then there exist λ′, μ′ such that λ′ ↓ G_{e} ≅ λ ↓ G_{e}, μ′ ↓ G_{e} ≅ μ ↓ G_{e} and Ext_{G}^{1}(λ′, μ′) ≅ Ext_{G_{e}}^{1}(λ′, μ′) ≅ Ext_{G_{e}}^{1}(λ, μ). In particular, taking μ = 0, we have H^{1}(G, λ′) ≅ H^{1}(G_{e}, λ′) ≅ H^{1}(G_{e}, λ).

Proof Part (i) is immediate from [CPSK, 7.4], and part (ii) follows from 1.2; note that in (ii) it may be necessary to choose λ′ ≠ λ or μ′ ≠ μ, since by 1.2 there exist λ, μ such that Ext_{G}^{1}(λ, μ) = 0 but Ext_{G_{e}}^{1}(λ, μ) ≠ 0 (the pairs λ, λ′ and μ, μ′ are related by a field morphism of G). ♠

Proposition 1.4 Let G be a connected simple algebraic group over K and let H be a finite subgroup of G. Suppose V is a finite-dimensional rational KG-module satisfying the following conditions:

(i) every G-composition factor of V is H-irreducible;

(ii) for any G-composition factors M, N of V, the restriction map Ext_{G}^{1}(M, N) → Ext_{H}^{1}(M, N) is injective;

(iii) for any G-composition factors M, N of V, if M ↓ H ≅ N ↓ H then M ≅ N (as G-modules). Then G and H fix exactly the same subspaces of V.

Proof The proof goes by induction on dim V. It is clearly sufficient to show that if W is an irreducible H-submodule of V, then G fixes W.

Thus let W be an irreducible H-submodule of V. Define U = ⟨W^{G}⟩, and suppose that W ≠ U. If U is G-irreducible, then by (i) it is H-irreducible, so U = W, contrary to assumption. Therefore U is G-reducible; let W_{0} be a G-irreducible submodule of U. Then W_{0} ≠ W.

Consider V/W_{0}. Now H fixes the subspace (W + W_{0})/W_{0} of this, which is H-irreducible. By induction, G fixes W + W_{0}. Hence U = W + W_{0}. As W, W_{0} are H-irreducible, U = W + W_{0} = W ⊕ W_{0}. Thus U is not Hindecomposable, and so by (ii), U is also not G-indecomposable. Therefore there is a G-module W_{1} such that

U = W ⊕ W_{0} = W_{1} ⊕ W_{0}.

Note that W is H-isomorphic to W_{1}. If W is not H-isomorphic to W_{0}, then W, W_{0} are the only irreducible H-submodules of U, so W_{1} = W and G fixes W, a contradiction. Therefore W is H-isomorphic to W_{0}, and hence by (iii) these two spaces are also G-isomorphic. Now W ⊆ W_{1} ⊕ W_{0}, so W = {w + wϕ : w ∈ W_{1}} for some H-isomorphism ϕ : W_{1} → W_{0}. But ϕ is a G-isomorphism: for let α be any G-isomorphism from W_{1} to W_{0}. Then αϕ^{−1} : W_{1} → W_{1} is an H-isomorphism, so by Schur’s Lemma, αϕ^{−1} = λ.id for some λ ∈ K^{∗}. Hence ϕ = λ^{−1}α, and so ϕ is a G-isomorphism, as claimed. It follows that W is fixed by G, contrary to our assumption that W ≠ U. Therefore W = U, and this completes the proof. ♠

The next result is a consequence of 1.4; it is actually a special case of [CPSK, 7.5].

Corollary 1.5 Let G = SL_{2}(K) and G_{e} = SL_{2}(p^{e}) < G. If V is a finitedimensional rational KG-module such that every composition factor of V ↓ G has high weight less than p^{e}, then G and G_{e} fix exactly the same subspaces of V.

Proof Take H = G_{e} in 1.4. Conditions (i) and (iii) of 1.4 are clearly satisfied, and condition (ii) follows from 1.3(i). ♠

Proposition 1.6 Let G = SL_{2}(K) and Y ≅ SL_{2}(p^{e}), and let T, T_{Y} be Cartan subgroups of G, Y respectively. Let Q be a finite-dimensional rational KG-module, and regard the semidirect product QG as an algebraic group. Suppose that Y < QG, T_{Y} < QT, C_{Q}(T) = C_{Q}(T_{Y}) and that H^{1}(G, Q) ≅ H^{1}(Y, Q). Then there is a closed complement to Q in QG which contains Y.

Proof We may assume that T_{Y} < T. For x ∈ Y, let x = u_{x}r_{x} with u_{x} ∈ Q, r_{x} ∈ G. Then X = {r_{x} : x ∈ Y } is a subgroup of G isomorphic to Y, and the hypotheses hold with X replacing Y. The map γ : X → Q given by γ(r_{x}) = u_{x} is a 1-cocycle. The assumption that C_{Q}(T) = C_{Q}(T_{Y}) implies that the restriction map H^{1}(G, Q) → H^{1}(X, Q) is injective, by [CPSK, 7.3]. Hence by hypothesis, the restriction map is an isomorphism, and γ extends to a rational 1-cocycle γ̄ : G → Q. Then {γ̄(g)g : g ∈ G} is a closed complement to Q in QG containing Y. ♠

Proposition 1.7 Let G = SO_{2n}(K), where K is algebraically closed of characteristic p > 2n, and let u be an element of G of order p. Then there is a closed connected subgroup A ≅ A_{1}(K) of G containing u, such that every composition factor of V_{G}(λ_{1}) ↓ A is restricted.

Proof Let V = V_{G}(λ_{1}). By [SS, IV, 2.14 and 2.15], u is determined up to G-conjugacy by the sizes of its Jordan blocks on V; moreover, the number of Jordan blocks of each even size is even. Let r_{1}, …, r_{k} be the sizes of the odd Jordan blocks of u, and s_{1}, s_{1}, …, s_{l}, s_{l} the sizes of the even Jordan blocks, listed in pairs. Then G has a subgroup Π_{1}^{k}SO_{r_{i}} × Π_{1}^{l}SL_{s_{i}}. Each factor has an irreducible subgroup of type A_{1}(K), and we define A to be a diagonal subgroup of the product of these A_{1}’s such that the composition factors of V ↓ A have high weights r_{i} − 1, s_{i} − 1. Any nontrivial unipotent element of A has the same Jordan block sizes on V as u, hence is G-conjugate to u. This completes the proof. ♠

Next we state two results on modules for groups of type A_{2}. The first is taken from [LS2, 1.9, 1.11], and is a consequence of the sum formula in [An].

Proposition 1.8 Let G = A_{2}(K), and let a, b ∈ {0, 1, …, p − 1}.

(i) If a + b + 2 ≤ p, or if b = 0, then W_{G}(ab) is irreducible.

(ii) If W_{G}(ab) is reducible, then it is indecomposable with two composition factors, V_{G}(ab) and V_{G}(p − b − 2, p − a − 2).

(iii) W_{G}(ab) has dimension \frac{1}{2}(a + 1)(b + 1)(a + b + 2).

The next result is a straightforward application of 1.8.

Proposition 1.9 Let A = A_{2}^{ϵ}(p) with p = 11 or 13, and let S be a subgroup SL_{2}(p) of A centralizing an involution. Suppose V = V_{A}(ab) is a restricted module for A satisfying

(i) a ≥ b and a + b ≥ 9, and

(ii) dim V ≤ 124 if a ≠ b, and dim V ≤ 248 if a = b.

Then the possibilities for a, b and V ↓ S are as in the following table (in the last column, the superscripts denote the multiplicities of each composition factor).

ab  p dim V V ↓ S
12, 0 13  91 12^{1}/11^{1}/10^{1}/....
11, 0 13  78 11^{1}/10^{1}/9^{1}/....
10, 0 11, 13 66 10^{1}/9^{1}/8^{1}/....
90 11, 13 55 9^{1}/8^{1}/7^{1}/....
11, 1 13  102 12^{1}/11^{2}/10^{1}/....
91 13 120 10^{1}/9^{2}/....
11  75 10^{1}/9^{2}/8^{2}/....
81 11, 13 99 9/8^{2}/....
10, 2 13  111 12^{1}/11^{2}/10^{2}/....
82  11  82 10^{1}/9^{2}/8^{2}/7^{1}....
93  13  118 12^{1}/11^{2}/10^{2}/9^{2}/....
73  11  87 10^{1}/9^{2}/8^{2}/....
84  13  123 12^{1}/11^{2}/10^{2}/....
66  13  127 12^{1}/11^{2}/10^{2}/....
64  11  90 10^{1}/9^{2}/8^{2}/....
55  13  216 10^{1}/9^{2}/8^{3}/....
11  91 10^{1}/9^{2}/8^{2}/7^{2}....

Next we prove some results concerning the representation of the simple algebraic group G on its Lie algebra L(G). The first is well known, but we give a proof for completeness.

Proposition 1.10 The G-composition factors of L(G) are as in the following table.

G  p  comp. factors dimensions
of L(G)
A_{n}  p |̸n + 1  λ_{1} + λ_{n}  (n + 1)^{2} − 1
p|n + 1  λ_{1} + λ_{n}/0  (n + 1)^{2} − 2, 1
B_{n}  p ≠ 2  λ_{2}  2n^{2} + n
p = 2, n odd  λ_{1}/λ_{2}/0  2n, 2n^{2} − n − 1, 1
p = 2, n > 2 even λ_{1}/λ_{2}/0^{2}  2n, 2n^{2} − n − 2, 2
p = 2, n = 2  λ_{1}/2λ_{2}/0^{2}  4, 4, 2
C_{n}  p ≠ 2  2λ_{1}  2n^{2} + n
p = 2, n odd  2λ_{1}/λ_{2}/0  2n, 2n^{2} − n − 1, 1
p = 2, n even 2λ_{1}/λ_{2}/0^{2}  2n, 2n^{2} − n − 2, 2
D_{n}  p ≠ 2  λ_{2}  2n^{2} − n
p = 2, n odd  λ_{2}/0  2n^{2} − n − 1, 1
p = 2, n even λ_{2}/0^{2}  2n^{2} − n − 2, 2
E_{8}  any  λ_{8}  248
E_{7}  p ≠ 2  λ_{1}  133
p = 2  λ_{1}/0  132, 1
E_{6}  p ≠ 3  λ_{2}  78
p = 3  λ_{2}/0  77, 1
F_{4}  p ≠ 2  λ_{1}  52
p = 2  λ_{1}/λ_{4}  26, 26
G_{2}  p ≠ 3  λ_{2}  14
p = 3  λ_{1}/λ_{2}  7, 7

Proof For G of exceptional type this is proved in [Se2, 1.2]. Assume now that G is classical.

Consider first G = A_{n}. The highest root vector in L(G) affords weight λ_{1} + λ_{n}, so the composition factor V_{G}(λ_{1} + λ_{n}) of L(G) has all n^{2} + n roots occuring as weights. By [Se1, 8.6], the 0-weight space of V_{G}(λ_{1} + λ_{n}) has dimension n if p |̸n + 1, and dimension n − 1 if p|n. The conclusion follows for A_{n}.

Next consider G = B_{n}, C_{n}, D_{n} with p ≠ 2. The highest long root vector affords weight λ_{2}, 2λ_{1}, λ_{2} respectively; hence L(G) has a composition factor of this high weight. Also, if V = V_{G}(λ_{1}) (of dimension 2n + 1, 2n, 2n respectively), then by [Se1, 8.1], the G-modules ⋀^{2}V, S^{2}V, ⋀^{2}V (respectively) are irreducible, and are isomorphic to V_{G}(λ_{2}), V_{G}(2λ_{1}), V_{G}(λ_{2}). Since ⋀^{2}V, S^{2}V, ⋀^{2}V (respectively) have dimension equal to that of L(G), it follows that L(G) is irreducible of high weight λ_{2}, 2λ_{1}, λ_{2} respectively, as in the conclusion.

In all the remaining cases we have p = 2. We begin with G = D_{n}. We may take n ≥ 4 and G = SO_{2n} = SO(V) (note that the composition factors of L(G) are independent of the form of the group G). The highest root vector affords weight λ_{2}. We work out the dimension of the composition factor V_{G}(λ_{2}), by considering the action of G on ⋀^{2}V (of which V_{G}(λ_{2}) is a section). Observe first that the orbit of λ_{2} under the Weyl group of G has size 2n^{2} − 2n (which is the number of roots of D_{n}); so the dimension of V_{G}(λ_{2}) will be determined once we calculate the dimension of the 0-weight space. Now G has a subgroup GL_{n} containing a maximal torus T. The Weyl group S_{n} of GL_{n} acts on the 0-weight space M of T on ⋀^{2}V as on the natural n-dimensional permutation module over K. Writing V_{i} for an irreducible S_{n}-module of dimension i, we have

M ↓ S_{n} = V_{1} ⊕ V_{n−1}, if n is odd
V_{1}/V_{n−2}/V_{1} (indecomposable), if n is even.

A subgroup GL_{4} of G has a composition factor of high weight λ_{1} + λ_{3} in V_{G}(λ_{2}). Hence by the A_{n} case above, we see that the 0-weight space of V_{G}(λ_{2}) has dimension at least 2; from the S_{n}-action described above, the 0-weight space therefore has dimension at least n−1 if n is odd, and at least n−2 if n is even. Therefore dim V_{G}(λ_{2}) is at least 2n^{2} −n−1 if n is odd, and at least 2n^{2} −n−2 if n is even. As G = SO(V) < Sp(V), we see that G fixes a 1-space ⟨w⟩ in ⋀^{2}V. It follows that dim V_{G}(λ_{2}) = 2n^{2} − n − 1 if n is odd. Finally, dim V_{G}(λ_{2}) = 2n^{2}−n−2 if n is even: for in this case the G-invariant 1-space ⟨w⟩ cannot be a direct summand (as M ↓ S_{n} is indecomposable), so the self-duality of ⋀^{2}V implies that there is a G-invariant subspace W of codimension 1 in ⋀^{2}V, and V_{G}(λ_{2}) ≅ W/⟨w⟩.

Next consider G = B_{n}, with p = 2 and n > 2. The highest long root affords weight λ_{2}, and the highest short root affords λ_{1}. Also V_{G}(λ_{1}) has dimension 2n. Moreover, G fixes the ideal I of L(G) generated by all e_{α} with α a short root; and I ⊆ L(A_{1}^{n}), where A_{1}^{n} is a commuting product of short A_{1}’s. Thus a consideration of weights implies that I ↓ G = λ_{1}/0^{k} for some k. Now the long roots form a subsystem D_{n}, so the composition factor V_{D_{n}}(λ_{2}) of L(D_{n}) is a section of L(G)/I. By [Se1, 4.1(iii)], V_{D_{n}}(λ_{2}) has the same dimension as V_{B_{n}}(λ_{2}), and by the previous paragraph, this dimension is 2n^{2} − n − 1 if n is odd, and 2n^{2} − n − 2 if n is even. The conclusion now follows. The case where G = B_{2}, p = 2 follows from the C_{2} case, given below.

Finally, let G = C_{n} with p = 2. The highest long root affords 2λ_{1}, and the highest short root affords λ_{2}. Again G fixes the ideal of L(G) generated by all e_{α} with α short, which is contained in L(D_{n}). By [Se1, 4.1(iii)], dim V_{C_{n}}(λ_{2}) = dim V_{D_{n}}(λ_{2}), and the conclusion follows as above. ♠

Proposition 1.11 Let G be a simple algebraic group in characteristic p, and let T be a maximal torus of G. Then C_{L(G)}(L(T)) = L(T), unless (G, p) = (C_{n}, 2) (or (B_{2}, 2)). In the exceptional cases, C_{L(G)}(L(T)) = L(A_{1}^{n}) (or L(A_{1}^{2})).

Proof Let A = C_{L(G)}(L(T)). Then A is T -invariant, hence is spanned by L(T), together with those elements e_{α} of L(G) which centralize L(T). Note that if e_{α} ∈ A then e_{−α} ∈ A. We may assume that G ≠ A_{1}.

Suppose A ≠ L(T), so that e_{α} ∈ A for some root α. Conjugating by an element of the Weyl group, we may assume that α is a fundamental root. As h_{α} centralies e_{α}, we have p = 2.

If there is a root β such that (α, β) = −1, then h_{β} does not centralize e_{α}, a contradiction. Therefore α cannot lie in an A_{2} subsystem of Σ(G). Hence there must be more than one root length, and α must be a long root. It follows that G = B_{2} or C_{n}, with α short or long, respectively. The B_{2} case can be treated as C_{2}, so we take G = C_{n}. Now the proof of 1.10 shows that the weight α occurs within a composition factor V_{C_{n}}(2λ_{1}) of L(G). Because of the field twist, L(T) acts trivially on this module. Moreover, the fundamental A_{1}’s corresponding to long roots commute, so this completes the proof. ♠

Proposition 1.12 Let G be a simple algebraic group over K, and let ϕ : G → G be a morphism which is an automorphism of abstract groups.

(i) Suppose that G_{ϕ} is not a finite Suzuki or Ree group, and let V be a G-composition factor of L(G). If M is a subspace of V, then (G_{M})^{ϕ} = G_{M′} for some subspace M′ of V.

(ii) Suppose G_{ϕ} is a finite Suzuki or Ree group, and let V_{1}, V_{2} be the two G-composition factors of L(G). If M is a subspace of V_{i} (i = 1, 2), then (G_{M})^{ϕ} = G_{M′} for some subspace M′ of V_{3−i}.

(iii) Let X be a ϕ-stable subgroup of G, and let M be the collection of all X-invariant subspaces of all G-composition factors of L(G). Then the subgroup ⋂_{W ∈M} G_{W} of G is ϕ-stable.

Proof (i) Let the G-module V have high weight λ and correspond to representation ρ : G → GL(V). We may write ϕ = yτσ, where y, τ, σ are (possibly trivial) inner, graph, field automorphisms, respectively. By 1.10, the representations ρ and τρ of G are equivalent, as they both have high weight λ. Hence if σ is a q-power field automorphism (where q = p^{e} ≥ 1), then the high weight of the representation ϕρ is qλ. There is therefore a q-power field automorphism ω of GL(V) such that the representations ϕρ and ρω of G are equivalent. The automorphism ω is induced by a semilinear transformation V → V which we shall also denote by ω. Then y^{ω} = ω^{−1}yω for y ∈ GL(V). Thus, identifying each g ∈ G with its image gρ ∈ GL(V), there exists x ∈ GL(V) such that g^{ϕ} = g^{ωx} = x^{−1}ω^{−1}gωx, for all g ∈ G.

Writing δ = ωx, this gives δg^{ϕ} = gδ for all g ∈ G, and we have

(vδ)g^{ϕ} = (vg)δ

for all v ∈ V, g ∈ G. If M is a subspace of V, and m ∈ M, g ∈ G_{M}, then (mδ)g^{ϕ} = (mg)δ ∈ Mδ, and hence g^{ϕ} ∈ G_{Mδ}. Therefore (G_{M})^{ϕ} ≤ G_{Mδ}. For the reverse inclusion, write g ∈ G_{Mδ} as g = (g^{ϕ^{−1}})^{ϕ}. Using the displayed equality, we see that g^{ϕ^{−1}} ∈ G_{M}, as required. Part (i) is now established. (ii) Here G_{ϕ} is a finite group of type ^{2}B_{2}, ^{2}G_{2} or ^{2}F_{4}. Let V_{1}, V_{2} be the composition factors of L(G), as in 1.10, and let ρ_{i} : G → GL(V_{i}) (i = 1, 2) be the corresponding representations. As above, consideration of high weights shows that there are automorphisms ω_{i} of GL(V_{i}) such that the representations ϕρ_{i} and ρ_{3−i}ω_{3−i} are equivalent for i = 1, 2. Thus there are invertible linear transformations δ : V_{1} → V_{2}, γ : V_{2} → V_{1} such that

(v_{1}g^{ϕ})δ = (v_{1}δ)g^{ω_{2}}, (v_{2}g^{ϕ})γ = (v_{2}γ)g^{ω_{1}}

for all v_{i} ∈ V_{i}, g ∈ G (where we write just v_{i}g instead of v_{i}(gρ_{i}). A calculation similar to that in part (i) now shows that for subspaces M_{i} of V_{i}, we have

G_{M_{1}} = (G_{M_{1}δω_{2}^{−1}})^{ϕ}, G_{M_{2}} = (G_{M_{2}γω_{1}^{−1}})^{ϕ},

giving (ii).

Finally, (iii) is immediate from (i) and (ii). ♠

The next result is presumably well known, but we have been unable to find a reference for it.

Proposition 1.13 Let G be a simple algebraic group over K = 𝔽̄_{p}, and let σ be a Frobenius morphism of G such that O^{p′}(G_{σ}) = G(q), a finite group of Lie type over 𝔽_{q}. Suppose H is a closed, connected, simple subgroup of G which is σ-stable. Then O^{p′}(H_{σ}) = H(q), a group of Lie type over the same field 𝔽_{q}.

Proof Assume first that G_{σ} is not a Suzuki or Ree group, so that σ is defined over 𝔽_{q}. By [SS, I, 2.9], σ fixes a Borel subgroup B_{H} of H, and a maximal torus T_{H} therein. Let U_{H} = R_{u}(B_{H}). Notice that since σ is defined over 𝔽_{q}, H_{σ} is also not a Suzuki or Ree group; hence there exists a σ-stable long root subgroup U of U_{H}.

We claim that |U_{σ}| = q. To see this, observe first that since the action of σ on U ≅ K^{+} is defined over 𝔽_{q} (see [Bor1, AG 14.4]), we have K[U] = K[x] ≅ K ⊗ 𝔽_{q}[x]. Hence the comorphism σ^{∗} induces the q-power map x → x^{q} on 𝔽_{q}[x]. There is a polynomial f such that σ(u) = f(u) for all u ∈ U, and we have

f(u) = x(σ(u)) = σ^{∗}(x)(u) = x^{q}(u) = u^{q}.

Hence |U_{σ}| = q, as claimed.

We have now shown that a long root subgroup of H_{σ} has order q, from which the conclusion follows.

Finally, assume that G_{σ} is a Suzuki or Ree group. Then σ^{2} is a q-power field morphism of G, and from the above proof we have |U_{σ^{2}}| = q, where U is a σ^{2}-stable long root subgroup of H. As σ squares to σ^{2}, it follows that either σ induces a field morphism of H, or H_{σ} is a Suzuki or Ree group. In the latter case we have the result. The former leads to a contradiction, as here we can choose U to be σ-stable and obtain |U_{σ}| = q^{1/2}. But this is impossible as q is an odd power of p. ♠

We conclude this section by stating a result concerning the centralizers of semisimple elements of prime order in simple algebraic groups; this result is taken from [GL, 14.1].

Proposition 1.14 Let G be a simple algebraic group, and let α_{1}, …, α_{l} be a system of fundamental roots for the root system of G; let α_{0} = ∑ c_{i}α_{i} be the highest root. Suppose x ∈ G is a semisimple element of prime order r. Then C_{G}(x)^{0} is a reductive subgroup of maximal rank in G, and if Δ is the Dynkin diagram of the root system of (C_{G}(x)^{0})′, then one of the following holds:

(i) Δ is obtained by deleting nodes from the Dynkin diagram of G; (ii) Δ is obtained from the extended Dynkin diagram of G by deleting

one node α_{i}, where r = c_{i}.

2 Proof of Proposition 2 and Corollary 3

The proof of Proposition 2 relies upon the following result.

Proposition 2.1 Let G be a simple algebraic group over the algebraically closed field K, let T be a maximal torus of G and let X(T) be the character group of T. For any subgroup L of X(T), define

Ann(L) = {t ∈ T : λ(t) = 1 for all λ ∈ L}

Then the exponent of Ann(L)/(Ann(L))^{0} divides t(L) (the exponent of the torsion subgroup of X(T)/L).

Proof Choose a ℤ-basis f_{1}, …, f_{n} of X(T) such that n_{1}f_{1}, …, n_{r}f_{r} is a ℤ-basis of L, for some positive integers n_{1}, …, n_{r}. Define a morphism h : T → (K^{∗})^{n} by

h(t) = (f_{1}(t), …, f_{n}(t)) for t ∈ T.

Then h has comorphism h^{∗} such that h^{∗}(π_{i}) = f_{i} (where π_{i} is the ith projection map on (K^{∗})^{n}). It follows that h^{∗} is an isomorphism. The surjectivity of h^{∗} implies that h is injective. Since the image of h is closed, dimension considerations imply that h is an isomorphism of algebraic groups. Thus, if

T_{1} = h^{−1}{(α_{1}, …, α_{r}, 1, …, 1) : α_{i} ∈ K^{∗}}, and

T_{2} = h^{−1}{(1, …, 1, α_{r+1}, …, α_{n}) : α_{i} ∈ K^{∗}},

then T = T_{1} ×T_{2} and both T_{1} and T_{2} are tori. Also, X(T) = X(T_{1})×X(T_{2}) and L ≤ X(T_{1}).

Write g_{i} = n_{i}f_{i} for 1 ≤ i ≤ r, and define d : T → (K^{∗})^{n} by

d(t) = (g_{1}(t), …, g_{r}(t), 1, …, 1) for t ∈ T.

Then Ann(L) = ker d, and (ker d)^{0} = T_{2}. Let t ∈ Ann(L) and set e = exp(X(T_{1})/L). As X(T_{1})/L is the torsion group of X(T)/L, we have e = t(L). We need to show that t^{e} ∈ T_{2}. This is equivalent to χ(t^{e}) = eχ(t) = 1 for all χ ∈ X(T_{1}). However, if χ ∈ X(T_{1}), then eχ ∈ L, so that eχ(t) = 1, as required. ♠

Proof of Proposition 2

As in the hypothesis of Proposition 2, let G be a simple adjoint algebraic group, and let x be a semisimple element of G of finite order greater than mt(Σ(G)), for some positive integer m. We aim to find an infinite closed subgroup S of G satisfying conclusions (i) and (ii) of Proposition 2.

Choose a maximal torus T of G containing x, and let X(T) denote the character group of T. Regard the root system Σ(G) of G as a subset of X(T) in the usual way. As G is adjoint, ℤΣ = X(T). Define subgroups L, L′ of ℤΣ as follows:

L′ = ⟨α − β : α, β ∈ Σ(G) and α(x) = β(x)⟩,

L = ⟨α − β, γ : α, β, γ ∈ Σ(G), α(x) = β(x) and γ(x) = 1⟩.

Since γ(x) = 1 implies that γ(x) = γ(x)^{−1}, hence 2γ ∈ L′, we see that t(L) (the exponent of the torsion group of ℤΣ/L) divides t(L′). By the definition of t(Σ(G)), we have t(L′) ≤ t(Σ(G)), and so t(L) ≤ t(Σ(G)).

Let S = Ann(L), the annihilator of L in T. Clearly x ∈ S. By 2.1, the exponent of S/S^{0} divides t(L). Since x has order greater than mt(Σ(G)), this means that some nontrivial power of x, of order greater than m, lies in S^{0}. In particular S^{0} ≠ 1. Thus conclusion (i) of Proposition 2 holds.

To prove (ii) of Proposition 2, write

L(G) = L_{0} ⊕ ∑_{α∈Σ(G)} L_{α}

where L_{0} = L(T), and for α ∈ Σ(G),

L_{α} = {v ∈ L(G) : t(v) = α(t)v for all t ∈ T }.

Let W be an x-invariant subspace of L(G). For a character χ of ⟨x⟩, set W_{χ} = {v ∈ W : x(v) = χ(x)v}; then

W = ∑_{χ} W_{χ},

the sum being over all irreducible characters χ of ⟨x⟩. For nontrivial such χ, we have

W_{χ} ⊆ L(χ) = ∑{L_{α} : α ∈ Σ(G), α(x) = χ(x)}.

If α_{1}, …, α_{k} are the roots such that α_{i}(x) = χ(x), then α_{i} − α_{j} ∈ L for i ≠ j, hence α_{i}(s) = α_{j}(s) for all s ∈ S. It follows that each s ∈ S acts on L(χ) as a scalar multiple of the identity, and hence S fixes W_{χ}. Finally, to see that S fixes W_{1} = C_{W} (x), observe that

W_{1} ⊆ L_{0} + L(1) = L_{0} + ∑{L_{α} : α ∈ Σ(G), α(x) = 1}.

If α(x) = 1 then α ∈ L, hence α(s) = 1 for all s ∈ S; thus S acts as the identity on L_{0} + L(1), so S fixes W_{1}. We conclude that S stabilizes W, completing the proof of Proposition 2.

Proof of Corollary 3

Assume that G is a simple adjoint algebraic group in characteristic p, and that X = X(q) < G, with X quasisimple of Lie type over 𝔽_{q} (q = p^{e}). Assume also that q > t(Σ(G))m(X), where m(X) is the order of the Schur multiplier of X/Z(X).

Observe that X contains a subgroup isomorphic to SL_{2}(q) or ^{2}B_{2}(q), unless X = L_{2}(q) or ^{2}G_{2}(q). Excluding the latter possibilities for X, we see that X has a semisimple element x of order at least q+1, and by assumption this is greater than t(Σ(G)).|Z(X)|; and when X = L_{2}(q) or ^{2}G_{2}(q), X has an element of order (q+1)/(2, p−1) or q+\sqrt{3q}+1 respectively, which is again greater than t(Σ(G)).|Z(X)|. Applying Proposition 2 with m = |Z(X)|, we see that there is an infinite closed subgroup S of G such that x ∈ S, 1 ≠ x^{n} ∈ S^{0} − Z(X) for some n, and every x-invariant subspace of L(G) is S-invariant. Define

X̄ = ⟨X, S⟩^{0}.

Then x^{n} ∈ (X ∩ X̄) − Z(X); hence, as X ∩ X̄ is normal in X and X is quasisimple, we have X < X̄. Also every X-invariant subspace of L(G) is certainly x-invariant, hence is also ⟨X, S⟩-invariant, and therefore is X̄- invariant. This completes the proof of part (i) of Corollary 3.

For part (ii) of Corollary 3, suppose that X ≤ G_{σ}, where σ is a Frobenius morphism. Let M be the collection of all X-invariant subspaces of all Gcomposition factors of L(G), and define

Y = ⋂_{W ∈M} G_{W} .

By 1.12(iii), Y is σ-stable; also Y is clearly closed and contains the subgroup X̄ defined above. Clearly Y fixes every X-invariant subspace of each G-composition factor of L(G), so this completes the proof of part (ii) of Corollary 3.

3 Proof of Theorem 1: strategy and lemmas

We embark upon the proof of Theorem 1. In this section, after some preliminary observations, we explain the strategy of our proof, which is based on the consideration of the centralizer of a suitably chosen element a of the subgroup X. We then present some general lemmas concerning this centralizer.

Assume that G is a simple adjoint algebraic group of exceptional type over the algebraically closed field K of characteristic p, and that X = X(q) is a quasisimple group of Lie type over 𝔽_{q} (q = p^{e}), with X < G. The case where X = A_{1}(q), ^{2}B_{2}(q) or ^{2}G_{2}(q) follows from Corollary 3, so we assume that X is not one of these groups. Suppose further that q > 9 and X ≠ A_{2}^{ϵ}(16) (as in the hypothesis of Theorem 1). We aim to show that X lies in a connected subgroup X̄ of G fixing the same subspaces of L(G) as X. This is enough to complete the proof of Theorem 1, since part (ii) of Theorem 1 follows exactly as at the end of §2.

We begin with a lemma which restricts attention to the case where G = E_{8}.

Lemma 3.1 If Theorem 1 holds for G = E_{8}, then it holds in all cases.

Proof Suppose that Theorem 1 holds for G = E_{8}. We deduce the result for G = E_{7} as follows. Suppose X = X(q) < E_{7}, with q as in the hypothesis of Theorem 3. By the conclusion of Theorem 1 for E_{8}, there is a subgroup X̄ of E_{8} containing X which fixes exactly the same subspaces of L(E_{8}) as X does. Since X fixes L(E_{7}), so does X̄, and therefore X̄ lies in N_{E_{8}}(L(E_{7})); this is equal to A_{1}E_{7} (since it certainly contains A_{1}E_{7}, which is maximal in E_{8}). Passing to the actions of X, X̄ on L(E_{7}), and projecting to the adjoint group E_{7}, we have the conclusion of Theorem 1 for E_{7}. Theorem 1 for G = E_{6}, F_{4}, G_{2} follows using the same argument. ♠

In view of Lemma 3.1, we assume from now on that G = E_{8}.

Observe that X contains a subgroup A isomorphic to A_{2}^{ϵ}(q) (ϵ = ±) or B_{2}(q). There is an element a ∈ A such that C_{A}(a)^{(∞)} = S ≅ SL_{2}(q) generated by long root subgroups of A, and such that a has order divisible by (q − ϵ)/(3, q − ϵ) if A ≅ A_{2}^{ϵ}(q), and divisible by q + 1 if A ≅ B_{2}(q). By [SS, II, 4.4], C_{G}(a) is a connected reductive group; write

D = C_{G}(a)′,

so that D is connected and semisimple, and S < D. Note that o(a) ≥ 6, except when X = L_{3}(13) or U_{3}(11).

The strategy of our proof is based on the following elementary lemma.

Lemma 3.2 Suppose that there is a subgroup Y of X and a closed connected subgroup Z of G such that

(i) X ∩ Z ≰ Z(X), and

(ii) every Y -invariant subspace of L(G) is Z-invariant.

Then the conclusion of Theorem 1 holds.

Proof Take X̄ = ⟨X, Z⟩^{0}. Then X ∩ X̄ contains X ∩ Z, so X ∩ X̄ ≰ Z(X), and hence X < X̄. By (ii), X and X̄ fix the same subspaces of L(G). ♠

In the ensuing proof, we attempt to fulfil the hypotheses of 3.2 using two basic methods. The first method is to try to find a torus T_{k} such that ⟨a⟩∩T_{k} and T_{k} fix the same subspaces of L(G). Then, assuming ⟨a⟩ ∩ T_{k} ≰ Z(X), the hypotheses of 3.2 hold, taking Y = ⟨a⟩ ∩ T_{k}, Z = T_{k}.

This method only works in relatively few cases (usually when D = C_{G}(a)′ has large rank). In other cases, we use the following approach. Observe that S < D, and with a couple of exceptions, D is a product of classical groups. By studying the possible embeddings of S in D, we attempt to find a closed subgroup S̄ ≅ SL_{2}(K) of D containing S, such that every composition factor of S̄ on L(G) has high weight less than q. Then by 1.5, S and S̄ fix the same subspaces of L(G), so the hypotheses of 3.2 hold with Y = S, Z = S̄. It turns out (after much work) that this approach is successful in most cases; the remaining cases are usually dealt with using the representation theory of the subgroup A.

It is convenient to carry out the above strategy by contradiction; thus, in view of 3.2, we assume

(†) there are no subgroups Y ≤ X, Z ≤ G such that Z is closed and

connected, X ∩ Z ≰ Z(X) and every Y -invariant subspace of L(G) is Z-invariant.

We consider now the possibilities for C_{G}(a). Recall that the extended Dynkin diagram of G = E_{8} is

[figure: α_{1}; α_{3}; α_{4}; α_{5}; α_{6}; α_{7}; α_{8}; −α_{0}; α_{2}]

and the highest root α_{0} = 2α_{1} + 3α_{2} + 4α_{3} + 6α_{4} + 5α_{5} + 4α_{6} + 3α_{7} + 2α_{8}.

Lemma 3.3 Suppose that o(a) ≥ 6. Then one of the following holds:

(i) C_{G}(a) = T_{k}D, where T_{k} is a central torus of rank k ≥ 1, and D is a connected semisimple group;

(ii) C_{G}(a) = A_{1}A_{2}A_{5} and o(a) = 6.

Proof This follows easily from 1.14: if (i) does not hold, then every power of a of prime order r has centralizer obtained by deleting a node from the extended Dynkin diagram of G corresponding to a coefficient r in the above expression for α_{0}. Repeating this process, we find that the only possibility with o(a) ≥ 6 is that in conclusion (ii). ♠

Lemma 3.4 Suppose that o(a) ≥ 6 and C_{G}(a) = T_{1}D, with T_{1} a rank 1 torus and D semisimple. Then C_{G}(T_{1}) = L_{i}, a Levi subgroup corresponding to the Dynkin diagram of G with the node α_{i} deleted. Moreover D ≤ L_{i}, and one of the possibilities in Table 1 below holds.

Table 1

L_{i}  D  o(a)  possible q
D_{7}  A_{1}A_{1}D_{5} or A_{3}D_{4} o(a^{2}) ≤ 4 17, 19, 23, 25
A_{7}  A_{7}  o(a) ≤ 6 17, 19
A_{1}A_{6}  A_{1}A_{6}  o(a) ≤ 8
A_{1}A_{2}A_{4} A_{1}A_{2}A_{4}  o(a) ≤ 12
A_{3}A_{4}  A_{3}A_{4}  o(a) ≤ 10
A_{2}D_{5}  A_{2}D_{5}  o(a) ≤ 8 17, 19, 23, 25
A_{2}A_{1}A_{1}A_{3}  o(a^{2}) ≤ 8
A_{1}E_{6} A_{1}E_{6} o(a) ≤ 6 17, 19
A_{1}A_{1}A_{5}  o(a^{2}) ≤ 6
A_{1}A_{2}A_{2}A_{2}  o(a^{3}) ≤ 6
E_{7} A_{1}D_{6} or A_{7} o(a^{2}) ≤ 4 17, 19, 23, 25
A_{1}A_{3}A_{3}  o(a^{4}) ≤ 4
A_{2}A_{5}  o(a^{3}) ≤ 4

Proof Observe that C_{G}(T_{1}) is a Levi subgroup of G containing D, a semisimple group of rank 7; hence C_{G}(T_{1}) = L_{i} for some i.

First suppose that i = 8, so L_{i} = E_{7}. Since the α_{8}-coefficient in the highest root α_{0} is 2, the torus T_{1} has just 5 distinct weights 2,1,0,-1,-2 on L(G); indeed, this can be seen by writing a typical element T_{1}(c) of T_{1} as h_{α_{1}}(c^{2})h_{α_{2}}(c^{3})h_{α_{3}}(c^{4})h_{α_{4}}(c^{6})h_{α_{5}}(c^{5})h_{α_{6}}(c^{4})h_{α_{7}}(c^{3})h_{α_{8}}(c^{2}). If a ∈ T_{1}, then since o(a) ≥ 6, every a-invariant subspace of L(G) is also T_{1}-invariant; hence the subgroups ⟨a⟩ of X and T_{1} of G satisfy (i) and (ii) of 3.2, contrary to our assumption (†). Therefore a ∉ T_{1}, and so a = a_{1}a_{2} with a_{1} ∈ T_{1} and 1 ≠ a_{2} ∈ L_{8} = E_{7} and a_{2} ∉ Z(E_{7}). We have C_{E_{7}}(a_{2}) = D, a proper semisimple subgroup of E_{7} of rank 7. Using 1.14 again, we see that the only possibilities for D are A_{1}D_{6}, A_{7}, A_{1}A_{3}A_{3}, A_{2}A_{5}, and a_{2}^{r} ∈ Z(E_{7}) with r = 2, 2, 4, 3, respectively. Then a^{r} ∈ T_{1}; replacing a by a^{r} and using the previous argument, we conclude that o(a^{r}) < 5. This completes the proof for i = 8, and the other values of i are handled in the same way. ♠

Lemma 3.5 Assuming o(a) ≥ 6, we have SL_{2}(q) ≅ S ≤ C_{X}(a)′ < C_{G}(a)′ = D, and one of the following holds:

(1) D = A_{1}A_{2}A_{5}, o(a) = 6 and q = 17 or 19;

(2) D is as in row 1, 2, 6, 8 or 11 of Table 1;

(3) D = D_{6} = L_{18};

(4) D = E_{6} = L_{78} and p = 2; (5) D is contained in one of the following subgroups of G:

A_{1}A_{3}A_{3}, A_{2}A_{5}  (⊆ L_{8} = E_{7})
A_{1}A_{1}A_{5}, A_{1}A_{2}A_{2}A_{2} (⊆ L_{7} = A_{1}E_{6})
A_{2}A_{1}A_{1}A_{3}  (⊆ L_{6} = A_{2}D_{5})
A_{1}A_{1}D_{4}  (⊆ L_{18} = D_{6})
A_{3}A_{4}, A_{1}A_{2}A_{4}, A_{1}A_{6} (= L_{5}, L_{4}, L_{3} resp.)
A_{2}D_{4}, A_{1}D_{5}  (= L_{16}, L_{17} resp.).

Proof By 3.3, either (1) holds or C_{G}(a) = T_{k}D with k ≥ 1, so assume the latter holds. When k = 1, the possibilities are as listed in Table 1, and visibly either (2) or (5) holds.

Now assume that k = 2, so D ≤ C_{G}(T_{2}) = L_{ij} for some i, j. If i or j is in {3, 4, 5}, or if {i, j} is {1, 6} or {1, 7}, then (5) holds. Also L_{12}, L_{26}, L_{27}, L_{28}, L_{67} and L_{68} lie in conjugates of subgroups listed in (5) of L_{3}, L_{4}, L_{7}, L_{3}, L_{17} and L_{17}, respectively. And if {i, j} = {1, 8} or {7, 8} then L_{ij} = D_{6} or E_{6}, and either D = L_{ij} or D lies in C_{L_{ij}}(a), a proper semisimple subgroup of L_{ij} of rank 6; the only possibilities for the latter are A_{1}A_{1}D_{4}, A_{3}A_{3} (if L_{ij} = D_{6}), or A_{1}A_{5}, A_{2}A_{2}A_{2} (if L_{ij} = E_{6}), all of which occur under (5). If D = L_{18} then (3) holds. Finally, if D = L_{78} = E_{6}, then p = 2, as in (4): for if p ≠ 2 then Z(S) = ⟨t⟩ ≅ Z_{2}, and D ≤ C_{L_{78}}(t), a contradiction.

To complete the proof, observe that if k ≥ 3, then D lies in a Levi subgroup L_{ijk} of co-rank 3. We claim that all such Levi subgroups lie in conjugates of subgroups under (5). To see this, observe that as L_{3}, L_{4}, L_{5}, L_{16} and L_{17} are under (5), we may assume that {i, j, k} does not contain 3, 4, 5, {1, 6} or {1, 7}. Hence ijk = 128, 267, 268, 278 or 678. One checks that in each case L_{ijk} lies in a subgroup under (5), giving the claim. This completes the proof. ♠

Lemma 3.6 Suppose that a ∈ A_{2} = C_{G}(E_{6}). Then p is not 2 or 3; in particular, case (4) of Lemma 3.5 does not occur.

Proof Assume p = 2 or 3. Then q ≥ 16 if p = 2 and q ≥ 27 if p = 3. Hence either o(a) ≥ 26, or o(a) = 15 or 17 (with q = 16), or o(a) = 11 (with q = 32), or o(a) = 21 (with q = 64). Choose a torus T_{2} < A_{2} containing a. We may take T_{2} to consist of diagonal 3 × 3 matrices, and write a = diag (α, β, α^{−1}β^{−1}). The nontrivial composition factors of A_{2} on L(G) are V_{A_{2}}(10), V_{A_{2}}(01) and V_{A_{2}}(11). Hence the eigenvalues of a on L(G) are

1, α, β, αβ, αβ^{−1}, α^{2}β, αβ^{2}, α^{−1}, β^{−1}, α^{−1}β^{−1}, α^{−1}β, α^{−2}β^{−1}, α^{−1}β^{−2}.

If these are all distinct, then T_{2} fixes the same subspaces of L(G) as a, contrary to assumption (†). Therefore two of the above eigenvalues are equal, and so one of the following holds:

(i) α^{k} = 1, β^{k} = 1 or α^{k} = β^{−k}, for some k ∈ {1, 2, 3};

(ii) α^{l} = β^{l}, α^{2l} = β^{−l} or α^{−l} = β^{2l}, for some l ∈ {1, 2};

(iii) α^{2} = β, α = β^{2}, α^{3} = β^{−1}, α^{−1} = β^{3}, α^{3} = β^{−2} or α^{2} = β^{−3}.

Consider (i). Here a^{k} ∈ T_{1}, where T_{1} is a W (A_{2})-conjugate of {diag(1, c, c^{−1}) : c ∈ K^{∗}}. The weights of T_{1} on L(G) are 2,1,0,-1,-2, so T_{1} and a^{k} fix the same subspaces of L(G) provided o(a^{k}) ≥ 5, which is true; this contradicts (†) (note that a^{k} ∉ Z(A), so a^{k} ∉ Z(X)).

In case (ii), a^{l} lies in T_{1}, a W (A_{2})-conjugate of {diag(c, c, c^{−2}) : c ∈ K^{∗}}. Then T_{1} has weights 3,2,1,0,-1,-2,-3. Hence T_{1} and a^{l} fix the same subspaces provided o(a^{l}) ≥ 7, which is true; this again contradicts (†).

Finally in case (iii), a lies in T_{1}, a W (A_{2})-conjugate of {diag(c, c^{2}, c^{−3}) : c ∈ K^{∗}}, which has weights 5,4,3,...,-5. Then T_{1} and a fix the same subspaces since o(a) ≥ 11. ♠

Lemma 3.7 Suppose a ∈ A_{1}A_{1} = C_{G}(D_{6}). Then either q is prime or q = 25; in particular, this holds in case (3) of Lemma 3.5. Further, if q = 25 then D ≠ A_{1}D_{6}.

Proof The argument is similar to that of the previous lemma. Assume that q is not prime and that q ≠ 25. Then either q ≥ 49 or q ∈ {16, 27, 32}; hence either o(a) ≥ 15 or o(a) = 11. We have a ∈ T_{2} < A_{1}A_{1} = C_{G}(D_{6}). Write elements of T_{2} as (c, d) (c, d ∈ K^{∗}), where this represents the element (diag(c, c^{−1}), diag(d, d^{−1})) of A_{1}A_{1} = SL_{2}×SL_{2}. Take a = (α, β). Since the nontrivial composition factors of A_{1}A_{1} on L(G) are 1 ⊗ 0, 0 ⊗ 1, 2 ⊗ 0, 0 ⊗ 2 and 1 ⊗ 1, the eigenvalues of a on L(G) are

1, α, β, α^{2}, β^{2}, αβ, αβ^{−1}, α^{−1}, β^{−1}, α^{−2}, β^{−2}, α^{−1}β^{−1}, α^{−1}β.

As in the previous lemma, two of these must be equal. Hence one of the following holds:

(i) α^{k} = 1 or β^{k} = 1 for some k ∈ {1, 2, 3, 4};

(ii) α = β^{±1} or α^{2} = β^{±2};

(iii) α = β^{±2} or α^{±2} = β;

(iv) α = β^{±3} or α^{±3} = β.

In (ii), a^{2} lies in T_{1} = {(c, c) : c ∈ K^{∗}} or {(c, c^{−1}) : c ∈ K^{∗}}, both of which have weights 2,1,0,-1,-2; hence T_{1} and a^{2} fix the same subspaces of L(G) provided o(a^{2}) ≥ 5, which is true.

In (iii) or (iv), a ∈ T_{1} = {(c, c^{m}) : c ∈ K^{∗}} or {(c^{m}, c) : c ∈ K^{∗}}, where m ∈ {±2, ±3}. If m = ±2 the weights of T_{1} are 4,3,2,...,-4, so T_{1} and a fix the same spaces provided o(a) ≥ 9; this is true by our assumption on q. And if m = ±3, the weights of T_{1} are ±6, ±4, ±3, ±2, ±1; hence T_{1} and a fix the same spaces if o(a) ≥ 13 or o(a) = 11, which is again true.

Finally, consider (i). Here a^{k} lies in T_{1} = {(1, c) : c ∈ K^{∗}} or {(c, 1) : c ∈ K^{∗}}, which has weights 2,1,0,-1,-2; hence T_{1} and a^{k} fix the same subspaces provided o(a^{k}) ≥ 5. By our assumption on q, this is true unless k = 4 and q = 49 (in which case o(a) could be 16). So suppose k = 4 and q = 49, and without loss assume that α^{4} = 1. Recall that a = (α, β) ∈ A_{1}A_{1}. The composition factors of A_{1}A_{1} on L(G) are 0 ⊗ 0, 1 ⊗ 0, 0 ⊗ 1, 2 ⊗ 0, 0 ⊗ 2 and 1 ⊗ 1. Hence we check that T_{1} = {(1, c) : c ∈ K^{∗}} fixes the same spaces as a itself. Finally, if q = 25 and D = A_{1}D_{6}, then o(a) = 8 by 3.4, and we may take a = (α, ±1); as above, a and T_{1} = {(1, c) : c ∈ K^{∗}} fix the same spaces. This completes the proof. ♠

4 Proof of Theorem 1 for p ≠ 2 and q > 13

We continue with the proof of Theorem 1. In this section we handle the case where p ≠ 2 and q > 13. Recall from the discussion following Lemma 3.2, that our basic strategy is to try to find a subgroup S̄ ≅ SL_{2}(K) of D = C_{G}(a)′ such that S̄ contains S and fixes the same subspaces of L(G) as S.

The next lemma provides some useful information for this strategy; it will be used in the following sections also, so we make no assumptions on p and only assume q > 9 in the statement.

To ease notation, we make the following definition. Suppose D = E_{1} … E_{k}, a product of simple factors E_{i}, and write Z = Z(D). Denote by π_{i} the i^{th} projection map D/Z → E_{i}Z/Z. By the projection S_{i} of S in E_{i}, we mean a minimal preimage in E_{i} of the group (SZ/Z)π_{i}. If E_{i} is classical, with usual module V_{i}, we abuse notation by writing V_{i} ↓ S instead of V_{i} ↓ S̃_{i}, where S̃_{i} is the derived group of the preimage of S_{i} in a simply connected cover of E_{i}. And if each of the E_{i} is classical, we usually describe the embedding of S in D by giving the composition factors of V_{1} ↓ S, …, V_{k} ↓ S. We can always adjust these factors by applying (globally) any automorphism of S (this amounts to relabelling the elements of S).

Lemma 4.1 Let E be a simple factor of D of classical type, where D is as in (1), (2), (3) or (5) of Lemma 3.5, and let V be the usual module for E. Suppose that V ↓ S is not completely reducible. Then one of the following holds:

(i) p = 2 and V ↓ S has an indecomposable section 1^{(2^{i})}/0 or 1^{(2^{i})} ⊗ 1^{(2^{j})}/1^{(2^{j})} (with i ≠ j);

(ii) p = 3 and V ↓ S has an indecomposable section 1^{(3^{i})}/1^{(3^{i+1})}, 1^{(3^{i})} ⊗ 1^{(3^{i+1})}/0 or 1^{(3^{i})} ⊗ 1^{(3^{i+1})}/2^{(3^{i+1})};

(iii) p = 5 and V ↓ S has an indecomposable section 3^{(5^{i})}/1^{(5^{i+1})}, 3^{(5^{i})} ⊗ 1^{(5^{i+1})}/0, 2^{(5^{i})}/1^{(5^{i})} ⊗ 1^{(5^{i+1})} or 2^{(5^{i})} ⊗ 1^{(5^{i+1})}/1^{(5^{i})};

(iv) q = p = 11 and V ↓ S has an indecomposable section 4/4, 5/5 or 0/8/0;

(v) q = p = 13 and V ↓ S has an indecomposable section 5/5.

Proof We have dim V ≤ 12; moreover, if dim V ≥ 8 then E = D_{4}, D_{5}, D_{6} or A_{7} (with q prime or q = 25 in the latter two cases). We use Proposition 1.2, which determines the 2-step indecomposables for S = SL_{2}(q).

First assume that q = p. By 1.2(i), if there is an indecomposable W for S with composition factors a/b, then a + b = p − 3 or p − 1, whence dim W ≥ a+1+b+1 ≥ p−1. As q > 9, we have p−1 = q−1 ≥ 10. Therefore E = D_{5} or D_{6}. Thus V ↓ S is self-dual, and it follows that either (iv) or (v) holds, or E = D_{6}, p = 11 and V ↓ S has an indecomposable section a/b/a. The dimension of this section is 2a + b + 3 ≤ 12, so (a, b) = (0, 8) or (1, 7). The first case is in conclusion (iv). And in the second case, the middle factor (of high weight 7) is symplectic, which is impossible.

Now assume q = p^{e} with e ≥ 2. By 1.2(ii), the following is a list of all 2-step indecomposables for S of dimension at most 10 (of dimension at most 12 if q = 25):

p = 2 : 1^{(2^{i})}/0, 1^{(2^{i})} ⊗ 1^{(2^{j})}/1^{(2^{j})}
p = 3 : 1^{(3^{i})}/1^{(3^{i+1})}, 1^{(3^{i})} ⊗ 1^{(3^{i+1})}/0, 1^{(3^{i})} ⊗ 1^{(3^{i+1})}/2^{(3^{i+1})}, 1^{(3^{i})} ⊗ 2^{(3^{i+1})}/1^{(3^{i+1})}
1^{(3^{i})} ⊗ 1^{(3^{j})}/1^{(3^{i+1})} ⊗ 1^{(3^{j})}, 1^{(3^{i})} ⊗ 1^{(3^{i+1})} ⊗ 1^{(3^{j})}/1^{(3^{j})}
p = 5 : 3^{(5^{i})}/1^{(5^{i+1})}, 3^{(5^{i})} ⊗ 1^{(5^{i+1})}/0, 3 ⊗ 1^{(5)}/2^{(5)}(q = 25),

2^{(5^{i})}/1^{(5^{i})} ⊗ 1^{(5^{i+1})}, 2^{(5^{i})} ⊗ 1^{(5^{i+1})}/1^{(5^{i})}, 2 ⊗ 1^{(5)}/1 ⊗ 2^{(5)} (q = 25)

p = 7 : 5^{(7^{i})}/1^{(7^{i+1})}, 4^{(7^{i})}/1^{(7^{i})} ⊗ 1^{(7^{i+1})}, 3^{(7^{i})}/2^{(7^{i})} ⊗ 1^{(7^{i+1})}.

If E = A_{r} then dim V ≤ 8 (and q = 25 if dim V = 8), and the only possible indecomposables appearing in V ↓ S are among those in conclusions (i)- (iii). Otherwise, E = D_{r} with 4 ≤ r ≤ 6 (and q = 25 if E = D_{6}); as V is orthogonal, we see using 1.1 that the last three indecomposables listed for p = 3, the third, fifth and sixth listed for p = 5, and all the indecomposables listed for p = 7, cannot occur in V ↓ S. Hence again one of (i)-(iii) holds. ♠

From now on in this section we shall assume that

p ≠ 2 and q > 13.

Lemma 4.2 Suppose that D is as in (1), (2), (3) or (5) of Lemma 3.5, but D ≠ A_{1}E_{6} (and that p ≠ 2, q > 13). Thus D = E_{1} … E_{k}, a product of classical simple factors E_{i} with usual module V_{i}; let S_{i} be the projection of S in E_{i}. Then one of the following holds:

(i) for each i, V_{i} ↓ S_{i} is completely reducible;

(ii) for some j, V_{j} ↓ S_{j} is not completely reducible; moreover, there is a subgroup S̄_{j} ≅ A_{1}(K) of E_{j} containing S_{j}, and the composition factors of V_{j} ↓ S̄_{j} are as in Table 2 below (up to a twist);

(iii) q = 27, S < D = A_{5} or A_{6} with embedding 1/1^{(3)}/1^{(9)}/0^{m} (where m = 0 or 1);

(iv) q = 27, S < D = A_{1}D_{5} with embedding 1^{(9)}, (1 ⊗ 1^{(3)})^{2}/0^{2};

(v) q = 27, S < D = A_{1}D_{5} with embedding 1^{(9)}, (2^{(3)})^{2}/1 ⊗ 1^{(3)}.

Table 2

p E_{j}  V_{j} ↓ S̄_{j}
5 A_{5+δ} (δ = 0, 1)  3/1^{(5)}/0^{δ}
A_{6+δ} (δ = 0, 1)  2/1 ⊗ 1^{(5)}/0^{δ}
A_{7} (q = 25)  2 ⊗ 1^{(5)}/1
D_{5+δ} (δ = 0, 1)  3 ⊗ 1^{(5)}/0^{2+2δ}
D_{5+δ} (δ = 0, 1)  2^{2}/1 ⊗ 1^{(5)}/0^{2δ}
3 A_{3+δ} (δ = 0, 1, 2, 3) 1/1^{(3)}/0^{δ}
A_{4}  1 ⊗ 1^{(3)}/0
A_{6}  1 ⊗ 1^{(3)}/1^{(3^{i})}/0
D_{4+δ} (δ = 0, 1)  1^{2}/(1^{(3)})^{2}/0^{2δ}
D_{4+δ} (δ = 0, 1)  1 ⊗ 1^{(3)}/0^{4+2δ}
D_{5}  1 ⊗ 1^{(3)}/1^{(3^{i})}/1^{(3^{i})}/0^{2}
D_{5}  1 ⊗ 1^{(3)}/1^{(3^{i})} ⊗ 1^{(3^{l})}/0^{2}

Proof Suppose (i) does not hold, so that some V_{j} ↓ S is not completely reducible. Write E = E_{j}, V = V_{j}. By 4.1, p = 3 or 5 and V ↓ S has a section as in 4.1(ii) or 4.1(iii). Let T be a Cartan subgroup of S.

Assume first that p = 5. Relabelling S, we may take it that V ↓ S has an indecomposable section 3/1^{(5)}, 2 ⊗ 1^{(5)}/1, 3 ⊗ 1^{(5)}/0 or 2/1 ⊗ 1^{(5)}. In the first case, using 1.1 we see that E = A_{5}, A_{6} or D_{6}. Suppose E = A_{5} or A_{6}; then V ↓ S = 3/1^{(5)} or (3/1^{(5)}) ⊕ 0. By 1.3(i), the restriction map Ext_{A_{1}(K)}^{1}(3, 1^{(5)}) → Ext_{A_{1}(q)}^{1}(3, 1^{(5)}) is an isomorphism. It follows that there is a closed subgroup A_{1}(K) of E containing S_{j}, as required for (ii). Now suppose E = D_{6}; then D = E, q = 25 by 3.7, and V ↓ S = 1^{(5)}/3/3/1^{(5)}. Pick an element t ∈ S of order 26; then t lies in a Levi subgroup A_{5} of E, and we may take it that t lies in a 1-dimensional torus T_{1} = {diag(c^{5}, c^{−5}, c^{3}, c^{−3}, c, c^{−1}) : c ∈ K^{∗}} < A_{5}. As the nontrivial composition factors of the restriction of L(G)/L(A_{5}) to A_{5} are just V_{A_{5}}(λ_{i}) (1 ≤ i ≤ 5) (see [LS2, §2]), it follows that t and T_{1} fix the same subspaces of L(G), contrary to (†).

Next, if V ↓ S has an indecomposable section 2 ⊗ 1^{(5)}/1, then E = A_{7} (use 1.1 again), and we obtain (ii) using 1.3(i), as above.

Now assume V ↓ S has an indecomposable section 3 ⊗ 1^{(5)}/0. Then E = D_{5} or D_{6}. In the first case, V ↓ S is an indecomposable module 0/3 ⊗ 1^{(5)}/0, and S_{j} (the projection of S in E) fixes a singular 1-space of V. The stabilizer in E of this 1-space is a parabolic subgroup QL with unipotent radical Q and Levi subgroup L = D_{4}T_{1}, and Q is a KL-module of high weight λ_{1}. If we choose a subgroup S̃ ≅ A_{1}(K) of L such that Q ↓ S̃ = 3 ⊗ 1^{(5)}, then S_{j} < QS̃. Moreover, by 1.3, H^{1}(S, Q) ≅ H^{1}(S̃, Q), and clearly C_{Q}(T) = 0. Therefore by 1.6, S_{j} lies in a closed complement to Q in QS̃, giving (ii). When E = D_{6} we have V ↓ S = 0^{2}/3 ⊗ 1^{(5)}/0^{2}. Since H^{1}(S, 3 ⊗ 1^{(5)}) has dimension 1 by 1.2(ii), in fact V ↓ S = V_{2}⊥(0/3 ⊗ 1^{(5)}/0) where V_{2} is a non-degenerate 2-space, so S_{j} lies in a subgroup D_{5} of E, and now the argument given for E = D_{5} applies.

To complete the case p = 5, suppose V ↓ S has an indecomposable section 2/1 ⊗ 1^{(5)}. Then E = A_{6}, A_{7}, D_{5} or D_{6}. If E = A_{6} or A_{7}, we obtain (ii) using 1.3(i). Now suppose E = D_{5}. Then V ↓ S is an indecomposable module 2/1 ⊗ 1^{(5)}/2. The group A_{1}(K) has a 10-dimensional orthogonal module ⋀^{2}V (4), and ⋀^{2}V (4) ↓ A_{1}(K) = 2/1⊗1^{(5)}/2; moreover this module is indecomposable, since a 1-dimensional unipotent subgroup centralizes only a 1-space of weight 2 vectors in ⋀^{2}V (4). Hence E contains a subgroup S̃ ≅ A_{1}(K) with this representation on the usual module. We may take it that both S_{j} and S̃ lie in the same parabolic subgroup QL stabilizing a totally singular 3-space. The unipotent radical Q has an S̃-series 1 < Q_{1} < Q with Q_{1}, Q/Q_{1} abelian, such that Q_{1} ↓ S̃ = V (2) and (Q/Q_{1}) ↓ S̃ = 2⊗1⊗1^{(5)} = (3 ⊗ 1^{(5)}) ⊕ (1 ⊗ 1^{(5)}). By 1.2, H^{1}(S̃, Q/Q_{1}) is 1-dimensional. Hence, under the action of (Q/Q_{1})Z(L) there are just two classes of closed complements to Q/Q_{1} in (Q/Q_{1})S̃, with representatives S̃ and S̄, where S̄ = L ∩ QS̃. As H^{1}(S̃, Q/Q_{1}) ≅ H^{1}(S, Q/Q_{1}) by 1.3, and C_{Q/Q_{1}}(T) = 0, we deduce from 1.6 that S_{j} therefore lies in a conjugate of either Q_{1}S̃ or Q_{1}S̄. Since also by 1.3, H^{1}(S̃, Q_{1}) ≅ H^{1}(S, Q_{1}) = 0, S lies in a conjugate of S̃ or S̄, giving the conclusion.

Finally, suppose E = D_{6}. If S_{j} lies in a subgroup D_{5} of E, then the conclusion follows as before. Otherwise, V ↓ S = (1 ⊗ 1^{(5)}/2/1 ⊗ 1^{(5)}) ⊕ 0; also q = 25 and D = E, by 3.7. An element t of S of order 26 lies in a 1-dimensional torus

T_{1} = {diag(c^{6}, c^{4}, c^{−4}, c^{−6}, c^{2}, 1, c^{−2}, c^{6}, c^{4}, c^{−4}, c^{−6}, 1) : c ∈ K^{∗}} < D_{6}

(matrices relative to a suitable basis). The nontrivial composition factors of L(G) ↓ E are V_{D_{6}}(λ_{i}) (i = 1, 2, 5, 6). Moreover, T_{1} lies in a Levi subgroup A_{4} of D_{6}, which by [LS2, 2.6] has all its nontrivial composition factors on V_{D_{6}}(λ_{i}) (i = 5, 6) among the modules V_{A_{4}}(λ_{1}), …, V_{A_{4}}(λ_{4}). Hence the eigenvalues on L(G) of a typical element T_{1}(c) of T_{1} displayed above are among c^{±k} with 0 ≤ k ≤ 12. It follows that T_{1} and t fix the same subspaces of L(G), contrary to (†). This completes the proof for p = 5.

Now suppose that p = 3. By 4.1, V ↓ S has an indecomposable section 1 ⊗ 1^{(3)}/0, 1/1^{(3)} or 1 ⊗ 1^{(3)}/2^{(3)}. Consider the first case. Here E = A_{r}, D_{4} or D_{5} (note that E ≠ D_{6} by 3.7). Suppose E = A_{r}. Then r ≠ 7 by 3.4, and r = 4 yields to the usual argument using 1.3(i). If r = 5 then the projection S_{j} of S in E is P SL_{2}(q), so D = A_{1}A_{5}; but then a ∈ C_{G}(A_{1}A_{5}) = A_{2} = C_{G}(E_{6}), contrary to 3.6. The same argument works if r = 6, unless V ↓ S = (1 ⊗ 1^{(3)}/0) ⊕ 1^{3^{(i)}}; but then S < A_{1}A_{4} < A_{6} and we can apply the 1.3(i) argument again.

When E = D_{4} or D_{5}, one of the following holds:

(a) V ↓ S = V_{m}⊥(0/1 ⊗ 1^{(3)}/0), with V_{m} (m = 2 or 4) a non-degenerate m-space on which S acts completely reducibly;

(b) E = D_{5} and V ↓ S = 0/1 ⊗ 1^{(3)}/1^{(3^{i})} ⊗ 1^{(3^{i+1})}/0 (with S fixing a singular 1-space);

(c) E = D_{5} and V ↓ S = (1 ⊗ 1^{(3)}/0/1 ⊗ 1^{(3)})⊥V_{1} (with S fixing a singular 4-space).

In case (a), S_{j} lies in a subgroup of E of type SO_{m}×SO_{6}, and we can use the usual H^{1} argument for the SO_{6} to obtain conclusion (ii) of the lemma. In case (b), observe that S_{j} lies in a subgroup QS̃ of E fixing a singular 1-space, with S̃ ≅ A_{1}(K) and Q ↓ S̃ = (1 ⊗ 1^{(3)}) ⊕ (1^{(3^{i})} ⊗ 1^{(3^{i+1})}). Now by 1.2, H^{1}(S, Q) and H^{1}(S̃, Q) both have dimension 2, and C_{Q}(T) = 0, so conclusion (ii) follows from 1.6. In case (c), the projection S_{j} of S in E is P SL_{2}(q), so D = A_{1}D_{5} (note that D has rank at most 6 by 3.4). Let the projection of S in the factor A_{1} be 1^{(3^{k})}. Pick t ∈ S of order q + 1; then t lies in a rank 1 torus T_{1} in D = A_{1}D_{5}, with

T_{1} = {diag(c^{3^{k}}, c^{−3^{k}}), diag(c^{4}, c^{4}, c^{2}, c^{2}, c^{−4}, c^{−4}, c^{−2}, c^{−2}, 1, 1) : c ∈ K^{∗}}

(matrices in D_{5} relative to a suitable basis). The nontrivial composition factors of L(G) ↓ A_{1}D_{5} are 0 ⊗ λ_{i}, 1 ⊗ λ_{i} (i = 1, 4, 5) and 1 ⊗ 0, together with L(A_{1}) and L(D_{5}). Hence the eigenvalues of a typical element T_{1}(c) of T_{1} shown above on L(G) are

c^{±2.3^{k}}, c^{±3^{k}±6,4,2,0}, c^{±8,6,4,2,0}.

It follows that t and T_{1} fix the same subspaces of L(G), unless q = 27 and 3^{k} = 9; this is the exceptional embedding in conclusion (iv) of the lemma.

Now suppose V ↓ S has an indecomposable section 1 ⊗ 1^{(3)}/2^{(3)}. Here E = A_{6}, A_{7}, D_{5} or D_{6}. In the A_{6}, A_{7} cases, the projection of S in E is P SL_{2}(q), so D has a further factor; but then D has rank at least 7, which contradicts 3.4; and E ≠ D_{6} by 3.7. Therefore E = D_{5}, hence D = D_{5} or A_{1}D_{5}. We have V ↓ S = 2^{(3)}/1 ⊗ 1^{(3)}/2^{(3)}, indecomposable with S fixing a singular 3-space. Pick t ∈ S of order q + 1 ≥ 28. If S < E then t lies in a 1-dimensional torus T_{1} = {diag(c^{6}, c^{−6}, c^{6}, c^{−6}, 1, 1, c^{4}, c^{2}, c^{−2}, c^{−4}) : c ∈ K^{∗}} in a subgroup of type SO_{6} × SO_{4} in E. The nontrivial composition factors of L(G) ↓ D_{5} have high weights λ_{1}, λ_{2}, λ_{4} and λ_{5} (see [LS2, §2]). Hence the highest weight of T_{1} on L(G) is 12. It follows that t and T_{1} fix the same subspaces of L(G), contrary to (†). Hence S ≮ E, so D = A_{1}D_{5}, with embedding 1^{(3^{i})}, 2^{(3)}/1 ⊗ 1^{(3)}/2^{(3)}. Now an obvious adjustment of the previous T_{1} argument goes through, unless q = 27 and 3^{i} = 9, as in conclusion (v).

To complete the proof, suppose V ↓ S has an indecomposable section 1/1^{(3)}. Here E = A_{r}, D_{4} or D_{5}. In the D_{4}, D_{5} cases, we have V ↓ S = (1/1^{(3)}/1^{(3)}/1)⊥V_{k}, where k = 0 or 2 and V_{k} is a nondegenerate k-space. Thus S_{j} < D_{4} ≤ E. As Ext_{S}^{1}(1, 1^{(3)}) has dimension 1 by 1.2, V ↓ S has a submodule 1 ⊕ 1^{(3)} which is totally singular. Hence S_{j} lies in a subgroup QS̃ of E stabilizing this singular 4-space, where S̃ ≅ A_{1}(K) and Q ↓ S̃ = ⋀^{2}(1 ⊕ 1^{(3)}) ≅ (1 ⊗ 1^{(3)}) ⊕ 0^{2}. By 1.2 we have H^{1}(S̃, Q) ≅ H^{1}(S, Q); since also C_{Q}(T) = C_{Q}(T̃) (where T̃ is a Cartan subgroup of S̃), the conclusion now follows from 1.6.

Finally, suppose E = A_{r}. If S has only two nontrivial composition factors on V, then S_{j} < A_{3}A_{r−4} < E, and we can use the usual 1.3(i) argument for the A_{3} factor to obtain the conclusion. So assume that S has more than two nontrivial composition factors on V. In particular, r ≥ 5. If r = 7, or if D has a further factor, 3.4 or 3.6 gives a contradiction. Thus D = A_{5} or A_{6}, and V ↓ S = 1/1^{(3)}/1^{(3^{k})}/0^{m}, where m = r − 5. Let t ∈ S have order q + 1; then t lies in a rank 1 torus T_{1} = {diag(c, c^{3}, c^{3^{k}}, c^{−1}, c^{−3}, c^{−3^{k}}, 1^{m}) : c ∈ K^{∗}} of D, and we check that t and T_{1} fix the same subspaces of L(G) unless q = 27 and 3^{k} = 9, as in conclusion (iii) of the lemma. ♠

We now explain how we propose to use Lemma 4.2. The special configurations in 4.2(iii,iv,v) will be postponed until later (see 4.8). Thus suppose we are in the situation of conclusion (i) or (ii) of 4.2. Consider a module V_{i} such that V_{i} ↓ S_{i} is completely reducible. Write V_{i} ↓ S_{i} = ⨁_{r=1}^{t} W_{i_{r}}, where the W_{i_{r}} are irreducible S_{i}-submodules if E_{i} is of type A_{n}, and are minimal non-degenerate S_{i}-submodules if E_{i} is of type D_{n}. Using 1.1, we see that the action of S_{i} on each W_{i_{r}} extends to an action of A_{1}(K), preserving the quadratic form when E_{i} = D_{n}. Therefore S_{i} lies in a subgroup R_{i} = S̄_{i_{1}} … S̄_{i_{t}} of E_{i}, where each S̄_{i_{r}} ≅ A_{1}(K).

Thus if 4.2(i) or (ii) holds, then S lies in a connected subgroup of D which is a commuting product of A_{1}’s and contains each of the R_{i}. Abusing notation slightly, write S̄_{1} … S̄_{l} for this commuting product, where each S̄_{i} ≅ A_{1}(K). So we have

S < S̄_{1} … S̄_{l} ≤ D.

Obviously S̄_{1} … S̄_{l} is an image of (SL_{2}(K)^{l}. In the ensuing discussion we shall consider subgroups SL_{2}(K) of S̄_{1} … S̄_{l} containing S. When l > 1, there are many such groups, as described below.

Fix matrix groups SL_{2}(q) < SL_{2}(K). For powers p^{i_{1}}, …, p^{i_{l}} < q, the embedding ϕ : x → (x^{(p^{i_{1}})}, …, x^{(p^{i_{l}})}) of SL_{2}(q) in SL_{2}(K)^{l} extends to an embedding ϕ̄ : x̄ → (x̄^{(p^{i_{1}})}, …, x̄^{(p^{i_{l}})}) of SL_{2}(K) in SL_{2}(K)^{l}. Every subgroup of SL_{2}(K)^{l} which is isomorphic to SL_{2}(q) and projects nontrivially to each factor is conjugate in SL_{2}(K)^{l} to Im ϕ for some such ϕ. Notice that it is possible to have such embeddings ϕ, ψ such that Im ϕ = Im ψ but Im ϕ̄ ≠ Im ψ̄: for example, this is the case if q = p^{2} and

ϕ : x → (x, x^{(p)}), ψ : x → (x^{(p)}, x).

In general, Im ϕ = Im ψ means that ψ is obtained from ϕ by applying a fixed field twist to each coordinate (and equating the q-power twist to 1).

Now let ρ be a surjective homomorphism SL_{2}(K)^{l} → S̄_{1} … S̄_{l}. By the previous paragraph, we may assume that S = (SL_{2}(q))ϕρ for some ϕ as above. Then S lies in S̄_{ϕ} = SL_{2}(K)ϕ̄ρ, a closed subgroup of S̄_{1} … S̄_{l} isomorphic to SL_{2}(K). Consequently the first aim of our strategy, to find a subgroup SL_{2}(K) of D containing S, is achieved in the situation of 4.2(i,ii).

Observe that if ψ is another embedding of SL_{2}(q) such that Im ϕ = Im ψ, then S also lies in S̄_{ψ}. Thus if l > 1, then S lies in many subgroups SL_{2}(K) of S̄_{1} … S̄_{l}, obtained from field twists of ϕ. If we can show that one of these subgroups S̄_{ψ} has all its composition factors on L(G) having high weight less than q, it will follow from 1.5 that S and S̄_{ψ} fix the same subspaces of L(G), contradicting our assumption (†), as desired for Theorem 1. In the next few lemmas we show that this can usually be done.

Lemma 4.3 Suppose that case (5) of 3.5 holds (with p ≠ 2 and q > 13), and suppose also that 4.2(i) holds. Then either we can choose S̄ with S < S̄ ≤ D, such that every composition factor of L(G) ↓ S̄ has high weight less than q, or one of the following holds:

(i) q = 25, S < D ≤ A_{1}A_{3}A_{3} with embedding 0^{2}, 3, 3^{(5)} or 1, 3, 3^{(5)}; (ii) q = 25, S < D ≤ A_{3}A_{4} with embedding 3^{(5)}, 4 or 3^{(5)}, 3/0.

Proof We have S < D ≤ Y_{1} … Y_{r}, a product of classical groups Y_{i} as in (5) of 3.5. In view of the natural embedding of D in Y_{1} … Y_{r}, we may replace D by Y_{1} … Y_{r} in the discussion preceding the lemma. We know from that discussion that there is a closed subgroup S̄ ≅ SL_{2}(K) of D containing S. Let V_{i} be the usual module for Y_{i}. As usual, we describe the embedding of S̄ in Y_{1} … Y_{r} by giving the composition factors of V_{1} ↓ S̄, …, V_{r} ↓ S̄. We can certainly choose S̄ so that all these composition factors have high weight less than q.

Suppose first that all the factors Y_{i} are of type A_{m} with m ≤ 4. Projections of S̄ in such factors are given by the following representations (up to field twists):

S̄ → A_{1} : 1 or 0^{2}
S̄ → A_{2} : 2, 1/0 or 0^{3}
S̄ → A_{3} : 3, 2/0, 1/1^{(p^{i})}, 1 ⊗ 1^{(p^{i})}, 1/0^{2} or 0^{4}
S̄ → A_{4} : 4, 3/0, 2/1^{(p^{i})}, 2/0^{2}, 1 ⊗ 1^{(p^{i})}/0, 1/1^{(p^{i})}/0, 1/0^{3} or 0^{5}.

Each composition factor of L(G) ↓ Y_{1} … Y_{r} occurs either in L(Y_{i}) or in a tensor product of modules among the V_{i} and ⋀^{2}V_{i} and their duals (see [LS2, §2]). The highest weight of a composition factor of ⋀^{2}VS̄(a) is 2a − 2, and of ⋀^{2}(1^{(p^{i})} ⊗ 1^{(p^{j})}) (with i > j) is 2p^{i}.

Now Y_{1} … Y_{r} is one of

A_{1}A_{3}A_{3}, A_{1}A_{2}A_{2}A_{2}, A_{1}A_{1}A_{2}A_{3}, A_{3}A_{4}, A_{1}A_{2}A_{4}.

Suppose first that V_{i} ↓ S̄ has a composition factor 3^{(p^{j})} or 4^{(p^{j})} for some i, j. Then p ≥ 5. Applying a field twist, as discussed in the preamble to the lemma, we can take the factor to be 3 or 4. Arguing only from the above observations, we check that the only possibilities where a composition factor of high weight at least q can occur in L(G) ↓ S̄ are:

(a) S̄ < A_{1}A_{3}A_{3}, with embedding 0^{2}, 3, 3^{(p^{i})} or 1^{(p^{j})}, 3, 3^{(p^{i})}: here L(G) ↓ S̄ has possible highest weight 6p^{i} or p^{j} + 4 + 4p^{i};

(b) S̄ < A_{3}A_{4}, with embedding 3^{(p^{i})}, 4 or 3^{(p^{i})}, 3/0: here L(G) ↓ S̄ has possible highest weight 6p^{i} or 4p^{i} + 6.

In both cases, the possible highest weight is at least q only if p = 5; and if q = 5^{e} > 25, we can apply a field twist to take i ≤ e − 2 in all cases, giving highest weight less than q. Therefore a high weight at least q can only occur if q = 25 and conclusion (i) or (ii) of the lemma holds.

We may now assume that no composition factor of V_{i} ↓ S̄ is 3 or 4 (or a twist thereof). From [LS2,2.1 and 2.3] we see that the number of factors V_{i} or ⋀^{2}V_{i} in a tensor product occurring in L(G) ↓ Y_{1} … Y_{r} is at most three; hence, referring to the list of possibilities for Y_{1} … Y_{r} above, we see that the highest weight occurring in L(G) ↓ S̄ is at most 6p^{e−1}, which is less than q if p > 5. Thus we may assume that p = 3 or 5.

For p = 5, a high weight q or more can be achieved only if three factors occur in a tensor product, each factor contributing 2p^{e−1} (since otherwise the highest weight is at most 4p^{e−1}+2p^{e−2} < q). Thus the embedding of S̄ in Y_{1} … Y_{r} has three representations of the form 2^{(p^{e−1})} or 1^{(p^{e−1})} ⊗ 1^{(p^{j})}. This forces S̄ < A_{1}A_{2}A_{2}A_{2}, with embedding (0^{2} or 1^{(p^{i})}), 2^{(p^{e−1})}, 2^{(p^{e−1})}, 2^{(p^{e−1})}. Applying a field twist, this becomes (0^{2} or 1^{(p^{i+1})}), 2, 2, 2, and now L(G) ↓ S̄ has highest weight less than q.

Now consider p = 3. Here o(a) ≥ q − 1 ≥ 26, so we see from 3.2 and 3.4 that the semisimple group D = C_{G}(a)′ cannot have rank 7 or 8. Recall that the product Y_{1} … Y_{r} is as in 3.5(5), with factors A_{m} (m ≤ 4). Consequently D lies in one of the following products:

A_{1}A_{2}A_{3}, A_{1}A_{1}A_{2}A_{2}, A_{2}A_{2}A_{2}, A_{3}A_{3}, A_{2}A_{4}, A_{1}A_{1}A_{1}A_{3}, A_{1}A_{1}A_{4}

Arguing as above, and applying a field twist if necessary, we see that the only case where L(G) ↓ S̄ can have highest weight at least q is S̄ < D ≤ A_{2}A_{2}A_{2}, with embedding 2, 2^{(3)}, 2^{(9)} and q = 27. But in this case a ∈ C_{G}(A_{2}A_{2}A_{2}) = A_{2} = C_{G}(E_{6}), which is not possible by 3.6. This completes the proof when all the Y_{i} are of type A_{m} with m ≤ 4.

Next, suppose that Y_{1} … Y_{r} has a factor A_{m} with m ≥ 5, hence is A_{2}A_{5}, A_{1}A_{1}A_{5} or A_{1}A_{6}. If the usual module for A_{m} has all S̄-composition factors of dimension at most 5, then the arguments for the previous case (all factors A_{n}, n ≤ 4) apply; hence the projection of S̄ in A_{m} is one of the following (up to twists):

\begin{matrix}S̄ → A_{5} : & 5 or 2 ⊗ 1^{(p^{i})} \\ S̄ → A_{6} : & 6, 5/0 or 2 ⊗ 1^{(p^{i})}/0. \end{matrix}

By [LS2,2.1 and 2.3], each composition factor of L(G) ↓ Y_{1} … Y_{r} occurs in L(Y_{i}), in ⋀^{3}V_{i} or in a tensor product of modules among the V_{i} and ⋀^{2}V_{i} (and duals). Moreover, by [LS2, 2.13], the highest weights in ⋀^{3}VS̄(r), ⋀^{2}(2 ⊗ 1^{(p^{i})}), ⋀^{2}(1 ⊗ 2^{(p^{i})}), ⋀^{3}(2 ⊗ 1^{(p^{i})}), ⋀^{3}(1 ⊗ 2^{(p^{i})}) are 3r − 6, 2p^{i} + 2, 4p^{i}, 3p^{i}, 4p^{i} + 1, respectively. Hence, as usual adjusting by a field twist if necessary, we see that L(G) ↓ S̄ has highest weight less than q when p > 3.

Now let p = 3. Then o(a) ≥ q − 1 ≥ 26, so by 3.2 and 3.4, D = C_{G}(a)′ has rank at most 6. Consequently S̄ must lie in a product A_{1}A_{5}, with embedding 0^{2}, 2 ⊗ 1^{(3^{i})} or 1^{(3^{j})}, 2 ⊗ 1^{(3^{i})}. Then L(G) ↓ S̄ has highest weight less than q, unless i = e − 1; but in this case we apply a field twist to adjust the embedding to 0^{2} or 1^{(3^{k})}, 2^{(3)} ⊗ 1, and now L(G) ↓ S̄ has highest weight less than q.

We have now handled all cases where the Y_{i} are all of type A_{m}. It remains to consider Y_{1} … Y_{r} = A_{1}A_{1}D_{4}, A_{2}D_{4} or A_{1}D_{5}. The possible projections of S̄ in the factor D_{m} (m = 4 or 5) are as follows (up to twists):

S̄ → D_{4} : 6/0, 4/2^{(p^{i})}, 4/0^{3}, 3/3, 2/2^{(p^{i})}/0^{2}, 2/1^{(p^{i})}/1^{(p^{i})}/0, 2/0^{5},
1/1/1^{(p^{i})}/1^{(p^{i})}, 1/1/0^{4}, 1 ⊗ 1^{(p^{i})}/1^{(p^{j})} ⊗ 1^{(p^{k})},
1 ⊗ 1^{(p^{i})}/1^{(p^{j})}/1^{(p^{j})}, 3 ⊗ 1^{(p^{i})}, 1 ⊗ 1^{(p^{i})}/0^{4}, 1 ⊗ 1^{(p^{i})}/2^{(p^{j})}/0
S̄ → D_{5} : 8/0, 6/2^{(p^{i})}, 6/0^{3}, 4/4^{(p^{i})}, 4/2^{(p^{i})}/0^{2}, 4/0^{5}, 3/3/0^{2}, 2/2^{(p^{i})}/2^{(p^{j})}/0,
2/2^{(p^{i})}/1^{(p^{j})}/1^{(p^{j})}, 2/2^{(p^{i})}/0^{4}, 2/1^{(p^{i})}/1^{(p^{i})}/0^{3}, 2/0^{7},
1/1/1^{(p^{i})}/1^{(p^{i})}/0^{2}, 1/1/0^{6}, 2 ⊗ 2^{(p^{i})}/0, 3 ⊗ 1^{(p^{i})}/0^{2},

1 ⊗ 1^{(p^{i})}/1^{(p^{j})} ⊗ 1^{(p^{k})}/0^{2}, 1 ⊗ 1^{(p^{i})}/1^{(p^{j})}/1^{(p^{j})}/0^{2}, 1 ⊗ 1^{(p^{i})}/0^{6},

1 ⊗ 1^{(p^{i})}/4^{(p^{j})}/0, 2/2^{(p^{i})}/1^{(p^{j})} ⊗ 1^{(p^{k})}, 2/1^{(p^{i})} ⊗ 1^{(p^{j})}/0^{3}.

By [LS2,2.1 and 2.3],

L(G) ↓ A_{1}A_{1}D_{4} has nontrivial composition factors among 2 ⊗ 0 ⊗ 0, 0 ⊗ 2 ⊗ 0, 0 ⊗ 0 ⊗ λ_{2}, 0 ⊗ 0 ⊗ λ_{1}, 1 ⊗ 1 ⊗ 0, 1 ⊗ 1 ⊗ λ_{i}, 1 ⊗ 0 ⊗ λ_{i}, 0 ⊗ 1 ⊗ λ_{i}, 0 ⊗ 0 ⊗ λ_{i} (i = 1, 3, 4),

L(G) ↓ A_{2}D_{4} has nontrivial composition factors 11 ⊗ 0, 0 ⊗ λ_{2}, 0 ⊗ λ_{j} λ_{i} ⊗ λ_{j} (i = 1, 2, j = 1, 3, 4), and

L(G) ↓ A_{1}D_{5} has nontrivial composition factors 2 ⊗ 0, 0 ⊗ λ_{2}, 0 ⊗ λ_{i} 1 ⊗ λ_{i} (i = 4, 5).

As before, for each embedding of S̄, the highest weight of L(G) ↓ S̄ can be worked out using [LS2, 2.13]. We find that in every case but one, applying a field twist if necessary, L(G) ↓ S̄ has highest weight less than q; the exceptional case is

q = 25, S < A_{1}D_{5}, embedding (0^{2} or 1), 4/4^{(5)}.

But in this case, S also lies in A_{1}A_{3}A_{3}, as in conclusion (i) of the lemma. ♠

Lemma 4.4 Suppose that case (5) of 3.5 holds, and also that 4.2(ii) holds. Then we can choose S̄ with S < S̄ ≤ D, such that every composition factor of L(G) ↓ S̄ has high weight less than q.

Proof We have S < S̄ ≤ D, and by applying a suitable field twist we may take it that D has a factor E with usual module V such that V ↓ S̄ is as in Table 2.

Suppose first that p = 5. Then E = A_{5}, A_{6}, D_{5} or D_{6}; as we are assuming 3.5(5) holds, D lies in A_{2}A_{5}, A_{1}A_{1}A_{5}, A_{1}A_{6} or A_{1}D_{5}. We may take the embedding of S̄ in these groups to be one of the following:

S̄ < A_{2}A_{5} :  projection (0^{3}, 1^{(5^{i})}/0 or 2^{(5^{i})}), 3/1^{(5)} (with 5^{i} < q)
S̄ < A_{1}A_{1}A_{5} : embedding (0^{2} or 1^{(5^{i})}), (0^{2} or 1^{(5^{j})}), 3/1^{(5)} (with 5^{i}, 5^{j} < q)
S̄ < A_{1}A_{6} :  embedding (0^{2} or 1^{(5^{i})}), (3/1^{(5)}/0 or 2/1 ⊗ 1^{(5)}) (with 5^{i} < q)
S̄ < A_{1}D_{5} :  embedding (0^{2} or 1^{(5^{i})}), (3 ⊗ 1^{(5)}/0^{2} or 2^{2}/1 ⊗ 1^{(5)}) (with 5^{i} < q)

In all cases we find that the highest weight of S̄ on L(G) is less than q, as required.

Now suppose p = 3. Then E = A_{r} (3 ≤ r ≤ 6), D_{4} or D_{5}. As in the proof of 4.3, and using 3.4 to rule out all rank 7 possibilities, we see that D lies in one of the following subgroups Y_{1} … Y_{r}:

A_{1}A_{2}A_{3}, A_{3}A_{3}, A_{2}A_{4}, A_{1}A_{1}A_{1}A_{3}, A_{1}A_{1}A_{4}, A_{1}A_{5}, A_{6}, A_{1}A_{1}D_{4}, A_{2}D_{4}, A_{1}D_{5}.

If D ≤ A_{1}A_{2}A_{3}, then the conclusion is clear unless the embedding of S̄ is (0^{2} or 1^{(3^{i})}, 2^{(3^{e−1})}, 1/1^{(3)} (recall q = 3^{e}). Applying a field twist, we can change S̄ to take the embedding to be (0^{2} or 1^{(3^{i+1})}, 2, 1^{(3)}/1^{(9)}; this gives the conclusion unless i + 1 = e, in which case we further twist to replace the 1^{(3^{i+1})} by just 1. Now the weights on L(G) are less than q, as required.

This argument applies to all cases where the factors Y_{i} are all of type A_{k}, except for A_{3}A_{3}, in which case S̄ could be indecomposable on both natural modules for the factors. Here the embedding is 1/1^{(3)}, (1/1^{(3)})^{(3^{i})}, and we can assume that 3^{i} < q. The weights are all less than q unless 3^{i+1} = q. We can now change S̄ by applying a 3-power twist to take the embedding to be 1^{(3)}/1^{(9)}, 1/1^{(3)}, for which the weights are less than q.

It remains to handle the cases where some Y_{i} is D_{4} or D_{5}. For these the weights of S̄ on L(G) are less than q unless possibly S̄ < A_{1}D_{5} with embedding 1 ⊗ 1^{(3)}/1^{(3^{i})}/1^{(3^{i})}/0^{2} or 1 ⊗ 1^{(3)}/1^{(3^{i})} ⊗ 1^{(3^{l})}/0^{2} (i < l) in the D_{5} factor. In the first case, S̄ < A_{1}A_{1}D_{3}, a case handled above. In the second case, if the factor 1^{(3^{i})} ⊗ 1^{(3^{l})} splits off then S̄ < A_{1}A_{1}A_{1}D_{3}, a case already considered. Otherwise, by 1.2, we have l = i + 1. The weights of S̄ on L(G) are less than q unless i+1 = e, in which case we twist to embedding 1^{(3)} ⊗ 1^{(9)}/1 ⊗ 1^{(3)}/0^{2}, for which the weights are all less than q. ♠

Continue to assume that p ≠ 2, q > 13. By 3.5 - 3.7 and the previous two lemmas, we may now assume that either we are in one of the exceptional cases of 4.2(iii,iv,v), 4.3(i,ii), or D and q are as in Table 3 below.

Table 3

\begin{matrix}D & q \\ A_{1}A_{2}A_{5}, A_{1}E_{6}, A_{1}D_{6} & 17, 19 \\ A_{1}A_{1}D_{5}, A_{7}, A_{2}D_{5}, A_{3}D_{4} & 17, 19, 23, 25 \\ D_{6} & q prime or q = 25 \end{matrix}

Lemma 4.5 Suppose that q is prime and D is as in Table 3, with D ≠ A_{1}E_{6}. Then we can choose S̄ ≅ A_{1}(K) such that S < S̄ ≤ D and the highest weight of L(G) ↓ S̄ is less than q, except in the following case:

q = 17, S < D = A_{1}D_{6} or D_{6}, with embedding 10/0 in the D_{6} factor.

Proof Since q = p ≥ 17, it follows from 1.2 that S acts completely reducibly on the usual module for each factor of D. Hence there is a subgroup S̄ ≅ SL_{2}(K) of D containing S (see the preamble to 4.3). We see as in the proof of the previous lemma that either we can take the highest weight of S̄ to be less than q, or D has a factor D_{6} with S projecting via 10/0, as in the conclusion. ♠ Lemma 4.6 If q = 25 and D is as in Table 3, then either we can choose S̄ ≅ A_{1}(K) such that S < S̄ ≤ D and the highest weight of L(G) ↓ S̄ is less than q, or one of the following holds:

(i) D has a factor D_{5} or D_{6}, and the embedding of S in this factor is 4/4^{(5)}/0^{m} (m = 0 or 2);

(ii) S < D = A_{3}D_{4}, with embedding 3^{(5)}, 1^{(5)} ⊗ 3 or 3^{(5)}, 3/3 or 3^{(5)}, 4/2 or 3^{(5)}, 4/2^{(5)} or 3^{(5)}, 4/0^{3};

(iii) S < D = A_{7}, with embedding 3/3^{(5)}.

Proof If S is not completely reducible on the usual module for one of the factors of D, then one of the possibilities given in Table 2 of 4.2 holds, from which we easily see that S̄ exists, with all weights on L(G) less than q. Otherwise, S is completely reducible in all factors of D. The only way the desired S̄ can fail to exist is for some V_{i} ↓ S to have a composition factor 3^{(5)} or 4^{(5)} which cannot be excluded by applying a field twist; all such possibilities are listed under conclusions (i) - (iii). ♠

The next lemma deals with the possibility D = A_{1}E_{6} excluded in 4.5.

Lemma 4.7 Theorem 1 holds if D = A_{1}E_{6}.

Proof Here q = 17 or 19, and by 3.4, o(a) ≤ 6. Consequently the group A must be U_{3}(17) or L_{3}(19). Observe that the projection of S to the factor E_{6} is a group S_{0} ≅ L_{2}(p). Pick u ∈ S of order p, and write u = u_{1}u_{0} with u_{1} ∈ A_{1}, u_{0} ∈ S_{0} < E_{6}.

Assume first that u_{0} is a regular or semiregular element of E_{6}. Now the projection of C_{A}(a) to E_{6} in fact contains P GL_{2}(p). Under these circumstances, [ST2, Theorem 1] implies that there is a subgroup S̄ ≅ A_{1}(K) of E_{6} containing S_{0}. Moreover, by [LT, Table 3], S̄ is determined up to E_{6}-conjugacy; the composition factors of L(E_{6}) ↓ S̄ are as in [Se2, p.65]. It follows that L(E_{6}) ↓ S̄ has a unique composition factor with high weight 16. Hence the same is true of L(E_{8}) ↓ S. Consequently L(E_{8}) ↓ A has a composition factor V_{A}(aa) with 2a ≥ 16. By 1.5, V_{A}(aa) has dimension either (a + 1)^{3} or (a + 1)^{3} − (p − a − 1)^{3}. Since this is at most 248, we must have p = 17 and a = 8, in which case V_{A}(aa) has dimension 217. Now V_{A}(88) ↓ S has two composition factors of high weight 15; moreover, S < A_{1}S̄ < A_{1}E_{6}, and

L(G) ↓ A_{1}E_{6} = L(A_{1})/L(E_{6})/1 ⊗ λ_{1}/1 ⊗ λ_{6}/0 ⊗ λ_{1}/0 ⊗ λ_{6}/(1 ⊗ 0)^{2}/0

(see [LS2, §2]). It follows that the restriction of the A_{1}E_{6}-module 1 ⊗ λ_{1} to S must have a composition factor of high weight 15. By [LS2, 2.5], we have λ_{1} ↓ S_{0} = 16/8/0 or 12/8/4; hence by the previous sentence, in fact λ_{1} ↓ S_{0} = 16/8/0. Then 1 ⊗ λ_{1} ↓ S has two composition factors 15, whence L(G) ↓ S has four composition factors 15. This implies that L(G) ↓ A has a further factor V_{A}(bc) with b + c ≥ 15; but then dim V_{A}(88) + dim V_{A}(bc) > 248, a contradiction.

Now suppose that u_{0} is not regular or semiregular in E_{6}. Then u_{0} lies in the centralizer in E_{6} of a semisimple element, hence lies in a subsystem subgroup Y = D_{5}, A_{1}A_{5} or A_{2}A_{2}A_{2}. Since p ≥ 17, it follows using 1.7 that u_{0} lies in a subgroup S̄ ≅ A_{1}(K) of Y such that S̄ has restricted composition factors on the usual modules for all factors of Y. If we write M for the factor A_{1} of D = A_{1}E_{6}, then u lies in a diagonal subgroup S̃ ≅ A_{1}(K) of MS̄ (where neither projection involves a twist). The restrictions L(E_{8}) ↓ MY are given by [LS2, §2], and we deduce from this that all composition factors of L(E_{8}) ↓ S̃ are restricted. It now follows from [LST, 1.14] that if U is a connected unipotent subgroup of S̃ containing u, then u and U fix the same subspaces of L(E_{8}), contrary to (†). ♠

To complete the proof of Theorem 1 for q > 13, p ≠ 2, all that remains is to handle the exceptional cases in the conclusions of 4.2, 4.3, 4.5 and 4.6.

Lemma 4.8 None of the exceptional cases in 4.2(iii, iv, v) can occur.

Proof Consider first case 4.2(iii); here q = 27 and S < A_{5} < D with embedding 1/1^{(3)}/1^{(9)}. From [LS2, Table 8.1] we have

L(G) ↓ A_{5} = L(A_{5})/λ_{1}^{6}/λ_{2}^{3}/λ_{3}^{2}/λ_{4}^{3}/λ_{5}^{6}/0^{11}.

Hence

\begin{aligned}L(G) ↓ S = & (1 ⊗ 1^{(3)} ⊗ 1^{(9)})^{2}/(1 ⊗ 1^{(3)})^{8}/(1 ⊗ 1^{(9)})^{8}/(1^{(3)} ⊗ 1^{(9)})^{8}/ \\ & 1^{16}/(1^{(3)})^{16}/(1^{(9)})^{16}/2/2^{(3)}/2^{(9)}/0^{31}. \end{aligned}

Recall that S < A = A_{2}^{ϵ}(27) or B_{2}(27). If A = B_{2}(27) then S contains a conjugate b of a (of order 28). However from the above restriction of L(G) to S, we see that dim C_{L(G)}(b) = 34, whereas C_{G}(a) contains A_{5}, of dimension 35, a contradiction.

Therefore A = A_{2}^{ϵ}(27). Since L(G) ↓ S has two factors 1 ⊗ 1^{(3)} ⊗ 1^{(9)}, it follows that L(G) ↓ A has a factor α ⊗ β^{(3)} ⊗ γ^{(9)} and its dual, where α, β, γ ∈ {10, 01}. Moreover,

(α ⊗ β^{(3)} ⊗ γ^{(9)}) ↓ S = 1 ⊗ 1^{(3)} ⊗ 1^{(9)}/1 ⊗ 1^{(3)}/1 ⊗ 1^{(9)}/1^{(3)} ⊗ 1^{(9)}/1/1^{(3)}/1^{(9)}/0.

Now V_{A}(22) ↓ S has three composition factors of high weight 2, hence is not present in L(G) ↓ A. In order to accomodate the further six composition factors 1⊗1^{(3)} in L(G) ↓ S, it follows that L(G) ↓ A must have three factors of the form δ ⊗ ϵ^{(3)} with δ, ϵ ∈ {10, 01}, together with their duals; similarly for the other 2-fold twists.

The composition factors found so far for L(G) ↓ A account for all composition factors of L(G) ↓ S except 1^{2}/(1^{(3)})^{2}/(1^{(9)})^{2}/2/2^{(3)}/2^{(9)}/0^{11}; in order to accomodate these, L(G) ↓ A must have further factors 11/11^{(3)}/11^{(9)}/00^{8}.

Now a is an element of order 26 or 28 in A of the form diag(ω, ω, ω^{−2}) for some ω ∈ K^{∗}. We calculate that dim C_{V_{A}(λ)}(a) is at most 1 if λ = α ⊗ β^{(3)} ⊗ γ^{(9)}, is 0 if λ = δ ⊗ ϵ^{(3)}, and is 4 if λ = 11. Hence from the above we see that dim C_{L(G)}(a) ≤ 22. This is a contradiction, as C_{G}(a) contains A_{5}. This completes the argument for the case 4.2(iii).

The case 4.2(iv) is handled similarly. Here S < D = A_{1}D_{5}, and from [LS2, §2],

\begin{aligned}L(G) ↓ A_{1}D_{5} = & L(A_{1})/L(D_{5})/(1 ⊗ λ_{1})^{2}/1 ⊗ λ_{4}/1 ⊗ λ_{5}/(0 ⊗ λ_{4})^{2}/ \\ & (0 ⊗ λ_{5})^{2}/(0 ⊗ λ_{1})^{2}/(1 ⊗ 0)^{4}/0^{4}. \end{aligned}

From the proof of 4.2, we see that the projection of S in the D_{5} factor is as in case (c) for p = 3, namely (1 ⊗ 1^{(3)}/0/1 ⊗ 1^{(3)})⊥V_{1}, the space 1 ⊗ 1^{(3)} being a singular 4-space. We now calculate that

\begin{aligned}L(G) ↓ S = & (1 ⊗ 1^{(3)} ⊗ 1^{(9)})^{8}/(2 ⊗ 1^{(9)})^{2}/(2^{(3)} ⊗ 1^{(9)})^{2}/2 ⊗ 2^{(3)}/ \\ & (1 ⊗ 1^{(3)})^{16}/2^{7}/(2^{(3)})^{7}/2^{(9)}/(1^{(9)})^{12}/0^{18}. \end{aligned}

If A = B_{2}(27) then as above, S contains a conjugate b of a, and from this restriction we see that C_{L(G)}(b) has dimension 34, which is a contradiction as C_{G}(a)′ = A_{1}D_{5}. Thus A = A_{2}^{ϵ}(27). As in the previous case, L(G) ↓ A must have a composition factor α ⊗ β^{(3)} ⊗ γ^{(9)} with α, β, γ ∈ {10, 01}. But the restriction of this to S has a composition factor 1 ⊗ 1^{(9)}, which is not present in L(G) ↓ S.

Finally, consider 4.2(v). Here we calculate that

L(G) ↓ S = (1 ⊗ 2^{(3)} ⊗ 1^{(9)})^{2}/(1 ⊗ 1^{(3)} ⊗ 1^{(9)})^{2}/(1 ⊗ 2^{(3)})^{4}/
(2^{(3)} ⊗ 1^{(9)})^{4}/(1 ⊗ 1^{(3)})^{6}/(1 ⊗ 1^{(9)})^{4}/(1^{(3)} ⊗ 1^{(9)})^{7}/
2/(2^{(3)})^{8}/(2^{(9)})^{3}/1^{4}/(1^{(3)})^{12}/(1^{(9)})^{8}/0^{8}.

We obtain the usual contradiction if A = B_{2}(27), so assume A = A_{2}^{ϵ}(27). In order to produce (1 ⊗ 2^{(3)} ⊗ 1^{(9)})^{2} for S, L(G) ↓ A must have a factor λ = α ⊗ β^{(3)} ⊗ γ^{(9)} and its dual, where α, γ ∈ {10, 01}, β ∈ {20, 02} (note that L(G) ↓ A cannot have factors α ⊗ 22^{(3)} or α ⊗ 11^{(3)} ⊗ γ^{(9)}, as these are incompatible with L(G) ↓ S). Removing the composition factors of (λ+λ^{∗}) ↓ S from L(G) ↓ S still leaves two factors 2^{(3)} ⊗ 1^{(9)}. To accommodate these, L(G) ↓ A must have δ^{(3)} ⊗ μ^{(9)} and its dual, where δ ∈ {20, 02, 11}, μ ∈ {10, 01}. Removing the restrictions of these to S leaves a further one or three factors 1^{(3)}⊗1^{(9)}. At least one of these must occur in the restriction of a selfdual factor of L(G) ↓ A, which must be 22^{(3)}. Removing all S-composition factors of this from what is left of L(G) ↓ S leaves (1⊗2^{(3)})^{2}/(2^{(3)})^{1}/..... The factors 1 ⊗ 2^{(3)} must occur in the restriction of ϕ ⊗ ψ^{(3)} and its dual, where ϕ ∈ {10, 01}, ψ ∈ {20, 02, 11}; but this restriction has two composition factors 2^{(3)}, which is a contradiction. This completes the proof. ♠

Lemma 4.9 None of the exceptional cases in 4.3 or 4.6 can occur.

Proof Suppose false, so q = 25. By considering the restriction of L(G) to the subsystem subgroup containing D given in 4.3 or 4.6, we see that L(G) ↓ S has a composition factor 4 ⊗ k^{(5)} or 4^{(5)} ⊗ k, where k ∈ {3, 4}. Applying a twist, we may assume the factor is 4 ⊗ k^{(5)}. Recall that X has a subgroup A ≅ A_{2}^{ϵ}(q) or B_{2}(q) which contains S. Let W be a composition factor of L(G) ↓ A such that W ↓ S has 4 ⊗ k^{(5)} as a composition factor.

Observe first that W is not restricted, since the highest weight of V_{A}(ab) ↓ S (a, b ≤ 4) is a + b. Hence W = ab ⊗ cd^{(5)}, with a, b, c, d ≤ 4.

If a + b < 4 then cd^{(5)} ↓ S must have a composition factor l ⊗ k^{(5)} = l^{(25)} ⊗k^{(5)}, with 1 ≤ l ≤ 4. Then 5(c+d) ≥ 25l +5k ≥ 40, forcing c = d = 4. However, 44 (the Steinberg module) has dimension 5^{3} (for A = A_{2}^{ϵ}(25)) or 5^{4} (for A = B_{2}(25)), so ab ⊗ 44^{(5)} has dimension greater than 248, a contradiction.

Therefore a + b ≥ 4. Similarly we see that c + d ≥ k. For A = A_{2}^{ϵ}(q) it is now easy to check that dim W > 248 if W is self-dual (i.e. if a = b and c = d), and that dim(W + W^{∗}) > 248 otherwise. And for A = B_{2}(q) we check that dim W > 248, by counting conjugates of subdominant weights and using [Pr]. ♠

Lemma 4.10 The exceptional case in 4.5 cannot occur.

Proof Suppose false, so that q = 17 and S < D = D_{6} or A_{1}D_{6} with embedding (0^{2}), 10/0 or 1, 10/0. By [LS2, §2], we have

L(G) ↓ A_{1}D_{6} = L(A_{1}D_{6})/(1 ⊗ λ_{1})^{2}/1 ⊗ λ_{5}/λ_{6}^{2}/0^{3}.

Moreover, by [LS2, 2.13], λ_{5} ↓ S ≅ λ_{6} ↓ S = 15/9/5. Therefore L(G) ↓ S has a composition factor with high weight 15 or 16. If A = B_{2}(17) then we can take a to have order 18, and the conclusion follows from the proof of 3.7. Therefore A = A_{2}^{ϵ}(17). Now L(G) ↓ A must have a composition factor V_{A}(ab) with a + b = 15 or 16. By 1.8, this forces a = b = 8, in which case dim V_{A}(ab) = 217. Now V_{A}(88) = W_{A}(88)/W_{A}(77), from which we calculate that

V_{A}(88) ↓ S = 16^{1}/15^{2}/14^{2}/ …

To obtain the factor 16, the embedding of S in A_{1}D_{6} must be 1, 10/0, whence L(G) ↓ S = 16^{1}/15^{2}/14^{3}/ . . .. The factor 14 not occurring in V_{A}(88) ↓ S must occur in the restriction of a self-dual A-composition factor of L(G), which must be V_{A}(77). However, dim V_{A}(88) + dim V_{A}(77) > 248, which is a contradiction. ♠

This completes the proof of Theorem 1 for p ≠ 2, q > 13.

5 Completion of the proof of Theorem 1 for p ≠ 2

In this section we complete the proof of Theorem 1 for p ≠ 2. Since q > 9 by hypothesis, by the previous section it remains only to handle the cases q = 11 and q = 13. Thus let G = E_{8} and suppose that X = X(q) < G with q = 11 or 13. By the hypothesis of Theorem 1, X ≠ L_{2}(q), so X contains a subgroup A isomorphic to one of the following groups:

A_{2}^{ϵ}(11), A_{2}^{ϵ}(13), B_{2}(11), B_{2}(13).

As in the previous section, by 3.2 we may assume

(†) there are no subgroups Y ≤ X, Z = Z^{0} ≤ G, such that every

Y -invariant subspace of L(G) is Z-invariant and X ∩ Z ≰ Z(X).

We deal with the above possibilities for A in a series of lemmas. The most complicated cases occur when A = L_{3}(13) or U_{3}(11), and we leave these until the final lemma.

We begin with a useful fact about self-dual indecomposables for SL_{2}(p).

Lemma 5.1 Let S = SL_{2}(p) (p = 11 or 13), and let V be a self-dual indecomposable KS-module of the form 4/4 (p = 11) or 5/5 (p = 11 or 13). Then S < SO(V) only if V = 5/5 with p = 11; otherwise S < Sp(V).

Proof Suppose S < SO(V) = SO_{2r} (r = 4 or 5). Then S < P_{r} = QL, the stabilizer of a maximal totally singular subspace, with unipotent radical Q; moreover, Q is abelian, and Q ↓ S ≅ ⋀^{2}V (r), which is 6/2 if r = 4 and 8/4/0 if r = 5. Since V is indecomposable, S does not lie in a conjugate of L, so H^{1}(Q, S) ≠ 0. It follows from 1.2(i) that r = 5 and p = 11, as required. ♠

Lemma 5.2 Theorem 1 holds if A = SL_{3}(13) or SU_{3}(11).

Proof Write Z(A) = ⟨t⟩, a group of order 3. We claim that C_{G}(t) = A_{2}E_{6} or A_{8}. Now A < C_{G}(t) = T_{k}E, where E is a semisimple subsystem subgroup of G. Then A < E = C_{G}(t)′, whence t ∈ Z(E). Therefore E has a simple factor of type A_{2}, A_{5}, E_{6} or A_{8}. In the last case the claim holds. If there is an E_{6} factor or a unique A_{2} factor, then C_{G}(t) contains the centralizer of this factor, whence C_{G}(t) = A_{2}E_{6}, as in the claim. If there is an A_{5} factor, this has centralizer A_{2}A_{1} in G, whence C_{G}(t) has an A_{2} factor. It remains to show that C_{G}(t) cannot have more than one A_{2} factor. If there is more than one, then the normalizer of the product of all the A_{2} factors is A_{2}^{4}, whence C_{G}(t) = A_{2}^{4}, contradicting 1.14. Thus C_{G}(t) = A_{2}E_{6} or A_{8}, as claimed.

Suppose first that C_{G}(t) = A_{8}; in fact then, C_{G}(t) = SL_{9}/Z_{3}. But this group cannot contain A, since if it did, the preimage of Z(A) in SL_{9} would be cyclic of order 9, which is clearly impossible.

Now assume C_{G}(t) = A_{2}E_{6}. If A lies in the A_{2} factor, then these groups fix the same subspaces of L(G); hence A projects nontrivially to the E_{6} factor. Choose a ∈ A of order 4 such that C_{A}(a)^{(∞)} = S ≅ SL_{2}(p). The projection of S in E_{6} centralizes an involution, hence lies in a subgroup A_{1}A_{5} or T_{1}D_{5}; hence S lies in T_{1}A_{1}A_{1}A_{5} or T_{1}A_{1}T_{1}D_{5}. The restriction of L(G) to these subgroups is given by [LS2, §2], and for all but one possible embedding of S we conclude that S < S̄ ≅ SL_{2}(K), where all high weights of L(G) ↓ S̄ are less than q, giving the usual contradiction to (†). The one exception is S < A_{1}D_{5} with embedding 1, 8/0 or 0^{2}, 8/0 (note that projection into D_{5} is not the indecomposable 4/4, by 5.1). In this case, let S_{0} be the projection of S in E_{6}, and consider the action of S_{0} on the 27-dimensional module V_{27} = V_{E_{6}}(λ_{1}). Since V_{27} ↓ D_{5} = λ_{1}/λ_{4}/0, there is a composition factor of V_{27} ↓ S_{0} of high weight 10 (see [LS2, 2.13]). Therefore if A_{0} denotes the projection of A in E_{6}, V_{27} ↓ A_{0} must have a composition factor V_{A_{0}}(ab) with a + b ≥ 10, which is impossible by 1.9. ♠

Lemma 5.3 Theorem 1 holds if A = L_{3}(11) or U_{3}(13).

Proof The arguments are very similar to those of the previous lemma, and we give just a sketch. Pick a ∈ A of order 10 or 14 (order 10 if A = L_{3}(11), order 14 if A = U_{3}(13)), such that C_{A}(a)^{(∞)} = S ≅ SL_{2}(p). As usual, by (†) we may suppose that there is no subgroup S̄ ≅ SL_{2}(K) containing S and having all composition factors on L(G) of high weight less than q. If C_{G}(a) = T_{1}D with D semisimple, then 3.4 implies that D = A_{3}A_{4}, A_{1}A_{2}A_{4}, A_{2}A_{1}A_{1}A_{3} or A_{1}A_{1}A_{5}; in each case S̄ as above exists. Otherwise, writing o(a) = 2r, and working in C_{G}(a^{r}) = A_{1}E_{7} or D_{8}, we find that S < C_{G}(a)′ = A_{6}, D_{6}, D_{5} or A_{1}D_{5} (in all other cases, S̄ as above exists).

If S < A_{6}, the non-existence of S̄ as above (having all weights less than q) implies that the embedding is given by the representation 6 of S, and p = 11. But the image of this representation of SL_{2}(11) in A_{6} is L_{2}(11), so this is impossible.

If S < D_{6}, the embedding must be 10/0, 8/2 or 5/5 (an indecomposable extension in the last case, with p = 11 by 5.1). (Note that the embedding 8/0^{3} has image group L_{2}(p) (see [LS2, 2.13]), and 4/4/0^{2} is out by 5.1.) By [LS2, Table 8.1],

L(G) ↓ D_{6} = λ_{2}/λ_{1}^{4}/λ_{5}^{2}/λ_{6}^{2}/0^{6},

and the restriction to S can be worked out from this using [LS2, 2.13]. If the embedding of S is 10/0 then L(G) ↓ S has at least five composition factors of high weight 10, which is not possible by 1.9. If the embedding is 8/2 with p = 13, then L(G) ↓ S has four composition factors of high weight 11, and none of high weight 12; and if the embedding is 8/2 with p = 11, then L(G) ↓ S = 10^{2}/9^{8}/..... Both of these are impossible, by 1.6. Now suppose the embedding is 5/5 with p = 11. We find that L(G) ↓ S = 10^{1}/9^{2}/8^{7}/..... It follows that L(G) ↓ A must have a self-dual composition factor giving a 10 on restriction to S. This must be V_{A}(55); by 1.9, it has dimension 91, and restricts to S as 10^{1}/9^{2}/8^{2}/..... The remaining composition factors of L(G) ↓ A restrict to S to give a further 8^{5}, so it follows that one of them is V_{A}(44), which has dimension 125 by 1.8. There remain composition factors of L(G) ↓ A, with restriction to S giving 8^{4}, and of total dimension 248 − 91 − 125 = 32. This is clearly impossible, by 1.8.

Finally, if S < A_{1}D_{5}, the embedding must be 1, 8/0 (the 4/4 indecomposable with p = 11 is not possible by 5.1). If p = 13 we find that L(G) ↓ S has 11^{2}/10^{6} and no 12; and if p = 11, it has 10^{5}. Both of these are incompatible with 1.9. ♠

Lemma 5.4 Theorem 1 holds if A = B_{2}(11) or B_{2}(13).

Proof Pick an element a ∈ A of order 10 or 12 such that C_{A}(a)^{(∞)} = S ≅ SL_{2}(p). As in the previous proof, we have S < C_{G}(a)′ = A_{6}, D_{6} or A_{1}D_{5}. Since A contains a product of two commuting conjugates of S, there is a conjugate b of a lying in S.

Suppose S < D_{6}; as before, the embedding must be 10/0, 8/2 or 5/5 (p = 11). In the first two cases, C_{V_{D_{6}}(λ_{i})}(b) has dimension at most 4, 14, 0, 0 according as i = 1, 2, 5, 6, and hence C_{L(G)}(b) has dimension at most 36. But C_{L(G)}(a) has dimension at least dim A_{6} = 48, a contradiction. In the last case C_{V_{D_{6}}(λ_{i})}(b) has dimension 0, 14, 8, 0 according as i = 1, 2, 5, 6, so dim C_{L(G)}(b) ≤ 36, again a contradiction.

Similar contradictions are reached when S < A_{6} or A_{1}D_{5}: for the A_{6} case we obtain a group S̄ contradicting (†) (note that the irreducible embedding 6 is not possible as its image in A_{6} is P SL_{2}(q)); and for the A_{1}D_{5} case we obtain S̄ unless the projection of S into the D_{5} factor is 8/0, in which case we obtain a contradiction by considering C_{L(G)}(b) as above. ♠

Lemma 5.5 Theorem 1 holds if A = L_{3}(13) or U_{3}(11).

Proof Pick a ∈ A of order 4 such that C_{A}(a)^{(∞)} = S ≅ SL_{2}(p). Then C_{G}(a) is one of the following groups (cf. [CG, Table 4]):

T_{1}E_{7}, T_{1}A_{1}E_{6}, T_{1}D_{7}, T_{1}A_{7}, T_{1}A_{1}D_{6}, A_{1}A_{7}, A_{3}D_{5}.

We now argue in a series of steps.

STEP 1: If C_{G}(a) = T_{1}E_{7} then the composition factors of L(G) ↓ A are among the modules V_{A}(ab) with ab ∈ {00, 11, 30, 03}.

To see this, observe first that the weights of T_{1} on L(G) are 2, 1, 0, −1, −2, and 2, −2 occur with multiplicity 1; these are the only weights which agree on a, and the sum of the two corresponding weight spaces is the −1-eigenspace for a. It follows that any a-invariant subspace of L(G) containing this eigenspace is also T_{1}-invariant, and so is any a-invariant space which intersects the eigenspace in 0 only. In view of (†), it follows that some composition factor of L(G) ↓ A has −1-eigenspace of dimension 1.

Suppose W = V_{A}(cd) = V_{A}(λ) is a composition factor on which a has no eigenvalues -1. We may assume the composition factor is restricted. Note that Z(A) = 1, so c ≡ d mod 3. Let α, β be fundamental roots of an A_{2} root system. We may assume that c ≥ d, α(a) = i and β(a) = 1. We know that λ(a) = 1, i or −i. If λ(a) = 1 then λ − 2α − bβ cannot be a weight of W for any b ≥ 0, so c ≤ 1 and d ≤ 1. This forces c = d = 0. If λ(a) = i then λ − 3α − bβ cannot be a weight, so c ≤ 2, d ≤ 2, forcing c = d = 1. Finally, if λ(a) = −i then λ − α − bβ cannot be a weight, forcing c = d = 0, a contradiction.

Similarly, we find that the only possible composition factors on which a has exactly one eigenvalue -1 are V_{A}(30) and V_{A}(03). This proves Step 1.

STEP 2: C_{G}(a) ≠ T_{1}E_{7}.

For suppose C_{G}(a) = T_{1}E_{7}. Then by Step 1, the highest weight of S on L(G) is 3. This is less than (p−1)/3, and one now checks that the argument of [ST1, Lemma 2] (second paragraph of proof) provides a rank 1 torus of G which fixes the same subspaces of L(G) as some element of S of order p − 1, contrary to (†).

STEP 3: C_{G}(a) ≠ T_{1}D_{7}.

Suppose C_{G}(a) = T_{1}D_{7}, so that S < D_{7}. Observe first that by [LS2, Table 8.1],

L(G) ↓ D_{7} = λ_{2}/λ_{1}^{2}/λ_{6}/λ_{7}/0.

Arguing as in 4.2 and 4.3, we see that either S < S̄ < D_{7} with S̄ ≅ SL_{2}(K) and all composition factors of L(G) ↓ S̄ of high weight less than p (in which case (†) is violated), or the embedding of S in D_{7} is one of

12/0, 10/0^{3}, 8/4, 8/2/0^{2}, 8/1^{2}/0, 5/5/0^{2}(p = 11)

(where as usual we specify the embedding by giving the composition factors of S on the natural 14-dimensional module for D_{7}). Note that the embeddings 6/6 and 8/0^{5} are not possible, as these correspond to a subgroup P SL_{2}(q) of D_{7} rather than SL_{2}(q).

In the ensuing arguments, observe that since A = L_{3}(13) or U_{3}(11), every composition factor of L(G) ↓ A is of the form V_{A}(ab) with a ≡ b mod 3.

If the embedding is 12/0 then p = 13 and, using the restriction L(G) ↓ D_{7} given above, together with [LS2, 2.13], we see that L(G) ↓ S has three composition factors with high weight 12. Then by Proposition 1.9, L(G) ↓ A must have three composition factors among V_{A}(12, 0), V_{A}(93), V_{A}(66). This forces dim L(G) > 248, a contradiction.

If the embedding is 10/0^{3}, then L(G) ↓ S has at least five composition factors with high weight 10, and again we obtain a contradiction using Proposition 1.9. If the embedding is 8/4 then for p = 13 we find that L(G) ↓ S = 12^{1}/11^{4}/..., and for p = 11, L(G) ↓ S = 10^{2}/9^{4}/8^{4}/7^{6}/..., both of which conflict with 1.9. If the embedding is 8/2/0^{2} or 8/1^{2}/0, we find that L(G) ↓ S = 11^{4}/... or 11^{2}/10^{6}/... (respectively) if p = 13, and is 10^{2}/9^{8}/... or 10^{5}/9^{6}/8^{3}/... if p = 11; all these possibilities conflict with 1.9.

Finally, the embedding 5/5/0^{2} is excluded exactly as in the proof of 5.3. This completes Step 3.

STEP 4: C_{G}(a) ≠ T_{1}A_{7}, T_{1}A_{1}D_{6}, A_{1}A_{7} or A_{3}D_{5}.

The proof of this is entirely similar to that of Step 3: supposing C_{G}(a) is one of the above, we find that either S < S̄ < C_{G}(a)′ with S̄ having all weights less than p, or the embedding of S in C_{G}(a)′ is one of the following:

C_{G}(a)′ embedding S < C_{G}(a)′
A_{7}  7
A_{1}A_{7} 1, 7 or 1, 6/0(p = 11)
A_{1}D_{6} 1, 10/0 or 1, 8/2 or 1, 8/0^{3} or 1, 5/5(p = 11)
A_{3}D_{5} 3, 8/0

(note that in the last two rows we have assumed that S ≰ D_{7}, in view of the argument for Step 3). For each of the above embeddings, it is straightforward to work out the composition factors of L(G) ↓ S, and in all cases we find as in Step 3 that these conflict with 1.9.

STEP 5: C_{G}(a) ≠ T_{1}A_{1}E_{6}.

The argument for this step is similar to the proof of 4.7. Let S_{0} be the projection of S to the factor E_{6}, pick u ∈ S of order p, and write u = u_{1}u_{0} with u_{1} ∈ A_{1}, u_{0} ∈ S_{0} < E_{6}. Note that S_{0} ≅ L_{2}(p).

Assume first that u_{0} is a regular or semiregular element of E_{6}. As in the proof of 4.7, [ST2] then implies that S_{0} < S̄ < E_{6}, where S̄ ≅ A_{1}(K); and by [Se2, p.65] and [LS2, 2.5(ii)] we have, writing V_{27} = V_{E_{6}}(λ_{1}), either

(i) p = 13, L(E_{6}) ↓ S̄ = W (22)/W (16)/W (14)/W (10)/W (8)/W (2) and V_{27} ↓ S̄ = W (16)/W (8)/0 (recall W (r) denotes the Weyl module of high weigth r), or

(ii) L(E_{6}) ↓ S̄ = W (16)/W (14)/W (10)^{2}/W (8)/W (6)/W (4)/W (2) and V_{27} ↓ S̄ = W (12)/W (8)/W (4).

Moreover, by [LS2, §2],

L(G) ↓ A_{1}E_{6} = L(A_{1}E_{6})/1 ⊗ λ_{1}/1 ⊗ λ_{6}/0 ⊗ λ_{1}/0 ⊗ λ_{6}/(1 ⊗ 0)^{2}/0.

Note that S projects nontrivially into the A_{1} factor (since the image of S in the E_{6} factor is P SL_{2}(q)). In case (i), we deduce that L(G) ↓ S has at least 3 composition factors with high weight 10, and none with high weight 12. Using 1.9 we find that this leads to a contradiction. In case (ii) with p = 13, we have L(G) ↓ S = 12^{2}/11^{4}/10^{3}/9^{2}/....; and in case (ii) with p = 11, we have L(G) ↓ S = 10^{2}/9^{4}/8^{5}/..... Both of these restrictions conflict with the possibilities given by 1.9.

Therefore u_{0} is not regular or semiregular in E_{6}, so as in the proof of 4.7, it lies in a subsystem subgroup Y = D_{5}, A_{1}A_{5} or A_{2}A_{2}A_{2}. Using 1.7 for the D_{5} case, we see that there is a subgroup S̄ ≅ A_{1}(K) of Y containing u_{0} and having restricted composition factors on the usual modules for all factors of Y. The restrictions V_{27} ↓ Y are given by [LS2, Table 8.7], and using [LS2, 2.13] we deduce that all composition factors of V_{27} ↓ S̄ are restricted. Therefore, if U is a connected unipotent subgroup of S̄ containing u_{0}, then by [LST, 1.14], u_{0} and U fix the same subspaces of V_{27}. Now S_{0} = A_{1}(p) is not irreducible on V_{27}, hence neither is ⟨S_{0}, U⟩. Thus ⟨S_{0}, U⟩^{0} is a proper nontrivial connected subgroup of E_{6}, hence lies in a maximal connected subgroup, M say. The possibilities for M are given by [Se2, Theorem 1].

Suppose M contains a maximal torus of E_{6}. If M is reductive, then M′ is a subsystem subgroup, hence lies in a subgroup D_{5}, A_{1}A_{5} or A_{2}A_{2}A_{2}. We find in the usual way that either S lies in a subgroup A_{1}(K) having all composition factors on L(G) of high weight less than p, or S_{0} < D_{5} with embedding 8/0. But in this case S < A_{1}D_{5} < D_{7}, which was excluded in Step 3.

Now suppose M is parabolic. Choose a minimal parabolic P = QL of E_{6} containing S_{0}, with unipotent radical Q and Levi subgroup L. The L-composition factors of Q have the structure of KL-modules, with high weights given by [LS2, 3.1]. Using 1.2, we find that either H^{1}(S_{0}, V) = 0 for all such modules V, or L = D_{5} and the embedding of S_{0} in L modulo Q is 8/0. In the latter case, S < QA_{1}D_{5}, so S has the same composition factors on L(G) as a suitable subgroup A_{1}(p) of A_{1}D_{5} < D_{7}, and the proof of Step 3 gives a contradiction. And in the former case, S_{0} lies in a conjugate of L, a case dealt with in the previous paragraph.

Thus M does not contain a maximal torus of E_{6}. The argument of the previous two paragraphs shows that S_{0} lies in no subgroup of maximal rank in E_{6}. Then by [Se2], M = A_{2}, G_{2}, F_{4} or A_{2}G_{2}; the restrictions of L(E_{6}) and V_{27} to each of these subgroups are given in [LS2, §2]. The fact that S_{0} lies in no maximal rank subgroup of E_{6} implies that S_{0} also lies in no maximal rank subgroup of M (since otherwise S_{0} would centralize a nontrivial semisimple element t in the centre of a maximal rank subgroup of M, hence lie in the maximal rank subgroup C_{E_{6}}(t)).

Write R = ⟨S_{0}, U⟩. Observe first that R does not act irreducibly in dimension 27 or 26, so R is not equal to one of the maximal subgroups G_{2}, A_{2} or F_{4} of E_{6}. Also, if R lies in a maximal G_{2}, then by [Se2, pp.193 and 65], R = A_{1} and S_{0} < R has composition factors as in (ii) above, giving a contradiction as before. And if R lies in a maximal A_{2}, then R = A_{1} has all composition factors on L(G) of high weight less than p.

Now suppose R lies in a maximal subgroup A_{2}G_{2} of E_{6}. Then S_{0} < A_{2}G_{2}. Let S_{1} be the projection of S_{0} in the factor G_{2}. Then S_{1} = L_{2}(p) < G_{2} < B_{3}. There is a subgroup S̄_{1} ≅ A_{1}(K) of B_{3} containing S_{1}, with all weights on L(B_{3}) less than p. Hence by 1.5, S̄_{1} fixes all S_{1}-invariant subspaces of L(B_{3}), and in particular fixes L(G_{2}). Therefore S_{1} < S̄_{1} < G_{2}; as S_{1} is not contained in a proper maximal rank subgroup of G_{2}, we have V_{G_{2}}(λ_{1}) ↓ S̄_{1} = 6 and L(G_{2}) ↓ S̄_{1} = 10/2. It follows that S < S̄ < A_{1}A_{2}G_{2} with S̄ ≅ A_{1}(K) having all weights on L(G) less than p.

Finally, suppose R lies in a maximal subgroup F_{4} of E_{6}. By [Se2] again, R lies in a maximal subgroup A_{1} or A_{1}G_{2} of this F_{4}. The subgroup A_{1}G_{2} lies in a maximal A_{2}G_{2} of E_{6}, a case handled in the previous paragraph. Hence R = A_{1}; now R has composition factors as in (i) above, giving a contradiction as before.

This completes the proof of Step 5, and hence of the lemma. ♠

Lemmas 5.2 - 5.5, together with the remarks at the beginning of this section, complete the proof of Theorem 1 for p ≠ 2.

6 Proof of Theorem 1 for p = 2

In this section we complete the proof of Theorem 1 by handling the case where p = 2. Thus we assume that X = X(q) < G with q = 2^{e} ≥ 16, and G = E_{8} (as we may assume by 3.1). As before, by Proposition 2 we can take it that X ≠ L_{2}(q) or ^{2}B_{2}(q). Moreover, X ≠ A_{2}^{ϵ}(16) by the hypothesis of Theorem 1.

Observe that X contains a subgroup A, where A = A_{2}^{ϵ}(q) or B_{2}(q) if q ≥ 32, and A = B_{2}(16) or G_{2}(16) if q = 16. Pick a ∈ A of order divisible by (q − ϵ)/(3, q − ϵ) if A = A_{2}^{ϵ}(q), and of order q + 1 otherwise, such that C_{A}(a)^{(∞)} = S ≅ SL_{2}(q), generated by long root subgroups of A. Note that either o(a) ≥ 17 or o(a) = 11 (which occurs only if A = U_{3}(32).

In view of 3.2, we continue to adopt the assumption (†) made immediately after 3.2.

We begin with several preliminary results which will be required in the proof.

Lemma 6.1 The restrictions to S of V_{A}(10), V_{A}(01) and V_{A}(11) are as in Table 4 below.

Table 4

A  V_{A}(10) ↓ S V_{A}(01) ↓ S V_{A}(11) ↓ S  dim V_{A}(10), V_{A}(01), V_{A}(11)
A_{2}^{ϵ}(q) 1/0  1/0  1^{2}/1^{(2)}/0^{2}  3, 3, 8
B_{2}(q) 1^{2}  1/0^{2}  1^{4}/(1^{(2)})^{2}/0^{4}  4, 4, 16
G_{2}(q) 1^{2}/0^{2}  1^{4}/1^{(2)}/0^{4} (1 ⊗ 1^{(2)})^{2}/1^{10}/ 6, 14, 64

(1^{(2)})^{9}/0^{18}

Proof This is elementary for the high weights 10 and 01, and follows for 11 from the following facts: for A_{2}, the module 11 is the quotient of 10 ⊗ 01 by a trivial submodule 00; for B_{2}, 11 = 10 ⊗ 01; and for G_{2}, 11 is the quotient of 10 ⊗ 01 by a submodule 10 ⊕ 01. ♠

Lemma 6.2 (i) If L(G) ↓ S has a composition factor 1^{(2^{i_{1}})} ⊗1^{(2^{i_{2}})} ⊗1^{(2^{i_{3}})} ⊗ 1^{(2^{i_{4}})}, then A ≠ B_{2}(q) or G_{2}(q); moreover, if this composition factor occurs with multiplicity 1, then A ≠ A_{2}^{ϵ}(q) also.

(ii) If L(G) ↓ S has a composition factor 1^{(2^{i_{1}})} ⊗ … ⊗ 1^{(2^{ir})} with r = 3 or 4, and has no composition factor 1^{(2^{i_{1}})} ⊗ … ⊗ 1^{(2^{i_{r−1}})}, then A ≠ A_{2}^{ϵ}(q) or G_{2}(q).

(iii) If L(G) ↓ S has a composition factor 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})} ⊗ 1^{(2^{i_{3}})} with multiplicity 1, then it has 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})} as a composition factor also.

(iv) If L(G) ↓ S has a composition factor 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})} ⊗ 1^{(2^{i_{3}})} with multiplicity less than 8, or has composition factors 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})} ⊗ 1^{(2^{i_{3}})}, 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})}, the latter with multiplicity less than 8, or has two different composition factors 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})} ⊗ 1^{(2^{i_{3}})} and 1^{(2^{j_{1}})} ⊗ 1^{(2^{j_{2}})} ⊗ 1^{(2^{j_{3}})}, then A ≠ G_{2}(q).

(v) If L(G) ↓ S has composition factors 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})} ⊗ 1^{(2^{i_{3}})} with multiplicity 2, 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})} with multiplicity less than 4, and no composition factor 1^{(2^{i_{2}})} ⊗ 1^{(2^{i_{3}})}, then A ≠ B_{2}(q).

(vi) If L(G) ↓ S has composition factors 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})} ⊗ 1^{(2^{i_{3}})} with multiplicity a, 1^{(2^{i_{1}})} ⊗1^{(2^{i_{2}})} ⊗1^{(2^{j})} (j ≠ i_{3}) with multiplicity b, and 1^{(2^{i_{1}})} ⊗1^{(2^{i_{2}})} with multiplicity less than a + b, then A ≠ A_{2}^{ϵ}(q).

Proof (i) Note first that L(G) ↓ S has no r-fold tensor composition factors with r ≥ 5, since dim L(G) < 3^{5}. By 6.1, L(G) ↓ A must also have a 4-fold tensor composition factor. If A = B_{2}(q) or G_{2}(q), this has dimension at least 4^{4} or 6^{4}, which is greater than 248, a contradiction; and if A = A_{2}^{ϵ}(q) and the 4-fold factor in L(G) ↓ S occurs with multiplicity 1, then L(G) ↓ A has a self-dual 4-fold factor, of dimension at least 8^{4}, again a contradiction.

(ii) Suppose A = A_{2}^{ϵ}(q) or G_{2}(q). By 6.1 and dimensions, L(G) ↓ A must have a composition factor 10^{(2^{i_{1}})} ⊗ … ⊗ 10^{(2^{ir})} (for type A_{2}, each 10 could also be 01); however the restriction of this to S has a composition factor 1^{(2^{i_{1}})} ⊗ … ⊗ 1^{(2^{i_{r−1}})}, contrary to the hypothesis of (ii).

(iii) This follows from (ii) unless A = B_{2}(q). For this case, by 6.1, the multiplicity 1 assumption forces L(G) ↓ A to have a composition factor 01^{(2^{i_{1}})} ⊗ 01^{(2^{i_{2}})} ⊗ 01^{(2^{i_{3}})}; however, the restriction of this to S has a composition factor 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})}.

(iv) Suppose A = G_{2}(q). The hypothesis implies that L(G) ↓ A has a 3-fold composition factor, which by dimensions must be 10^{(2^{i_{1}})} ⊗ 10^{(2^{i_{2}})} ⊗ 10^{(2^{i_{3}})}. The restriction of this to S conflicts with the assumptions in (iv).

(v) Suppose A = B_{2}(q). Again L(G) ↓ A must have a 3-fold composition factor μ_{1}^{(2^{i_{1}})} ⊗ μ_{2}^{(2^{i_{2}})} ⊗ μ_{3}^{(2^{i_{3}})}, where each μ_{i} ∈ {10, 01}. Each possibility for the μ_{i} conflicts with the assumptions on multiplicities.

(vi) This follows from an easy extension of the argument for (ii). ♠

Lemma 6.3 Let e ≥ 5, and take ℤ_{e} = {0, 1, …, e − 1} to act additively on itself (i.e. a ∈ ℤ_{e} sends x → x + a). Let k ∈ {3, 4, 5}, and let i_{1} … i_{k} be an unordered k-tuple of elements of ℤ_{e}. Then i_{1} … i_{k} is in the ℤ_{e}-orbit of a k-tuple j_{1} … j_{k} satisfying one of the following conditions:

(i) j_{r} ≤ e − 3 for all r;

(ii) taking j_{1} ≤ … ≤ j_{k}, we have j_{k} = e − 2, j_{k−1} ≤ e − 3, and, if j_{k−1} = e − 3, then j_{k−2} ≤ e − 4;

(iii) e = k = 5 and j_{1} … j_{k} = 01234, 01223, 00223, 01133 or 01233.

Proof Assume that (i) and (ii) fail. If e = k = 5 and all the i_{j} are distinct, then clearly i_{1} … i_{k} = 01234, as in (iii). So suppose this is not the case. Then the number of distinct i_{j} is less than e. Take i_{1} ≤ … ≤ i_{k}. Applying an element of ℤ_{e}, we can assume that i_{k} < e − 1. Therefore as (i) fails, we have i_{k} = e − 2; we also have i_{1} = 0, since otherwise we can apply the element −1 of ℤ_{e} to obtain conclusion (i).

Thus we have i_{1} = 0, i_{k} = e−2. Applying the element 2 of ℤ_{e}, we obtain (i) again, unless some i_{j} is e − 3 or e − 4. As (ii) fails, we deduce that either

(a) i_{1} … i_{k} contains 0, e − 3, e − 3, e − 2, or

(b) i_{1} … i_{k} contains 0, e − 2, e − 2.

Consider case (a). If k ≤ 4, application of the element 3 of ℤ_{e} gives the tuple 3001, which satisfies (ii). Now suppose k = 5, so the tuple is 0, i, e − 3, e − 3, e − 2. Applying 3 ∈ ℤ_{e} gives the tuple 3, i + 3, 0, 0, 1; this satisfies (i) or (ii) unless either i = e − 4 or e = 5, i = 0. In the latter case the tuple is 33001, in the same orbit as 00223, as in (iii). And if i = e − 4 the tuple is in the orbit of 40112. As (ii) fails, e = 5; then this is in the orbit of 01223, under (iii).

Now consider case (b). If k ≤ 4 the tuple is 0, i, e − 2, e − 2, which is in the same ℤ_{e}-orbit as 2, i + 2, 0, 0; this satisfies (ii) unless i = e − 3, in which case the tuple is in the orbit of 3011, which satisfies (ii).

Now suppose k = 5, so the tuple is 0, i, j, e − 2, e − 2 (i ≤ j), in the same orbit as 2, i + 2, j + 2, 0, 0. As (ii) fails, either j = e − 3, or i = j = e − 4, or e = 5, i = 0, j = 1. In the last case the tuple is 00223, under (iii). If i = j = e − 4 the tuple is in the orbit of 40022; as (i),(ii) fail, we must have e = 5 here, in which case the tuple is in the orbit of 01133, under (iii). Finally, suppose j = e − 3. Then the tuple is in the orbit of 3, i + 3, 0, 1, 1; as (ii) fails, either i = e − 4 or e = 5, i = 0. In the latter case the tuple is 01133, under (iii). In the former, the tuple is in the orbit of 40122; as (ii) fails, e = 5 and this is in the orbit of 01233, under (iii). ♠

Corollary 6.4 Let e ≥ 5, k ∈ {3, 4}, and let i_{1} … i_{k} be an unordered ktuple of elements of ℤ_{e}; if k = 4, assume that i_{1}, …, i_{k} are not all distinct. Then i_{1} … i_{k} is in the ℤ_{e}-orbit of a k-tuple j_{1} … j_{k} satisfying one of the following conditions:

(i) j_{r} ≤ e − 3 for all r;

(ii) e = 5 and j_{1} … j_{k} = 0013, 0023, 0113 or 013;

(iii) e = 6 and j_{1} … j_{k} = 0024 or 024.

Proof Consider first the case where k = 4, and take i_{1} ≤ … ≤ i_{4}. As in the proof of the previous result, we may take i_{4} = e − 2, i_{1} = 0. By 6.3 we may assume that 6.3(ii) holds. Thus i_{1} … i_{4} is either 0, i, e − 3, e − 2 (with i ≤ e − 4) or 0, i, j, e − 2 (with i ≤ j ≤ e − 4). In the first case the tuple is in the orbit of 3, i + 3, 0, 1. This satisfies (i) unless i = e − 4 or e = 5, i = 0; in the latter the tuple is in the orbit of 0023, under (iii), and in the former, the tuple has 4 distinct elements, contrary to hypothesis.

Now suppose the tuple is 0, i, j, e − 2 with i ≤ j ≤ e − 4. This is in the orbit of 2, i + 2, j + 2, 0, which satisfies (i) unless j = e − 4. Since i_{1}, … i_{4} are not all distinct, i must be 0 or e − 4. Hence the tuple is in the orbit of 0244 or 0024. These satisfy (i) unless e = 5 or 6, in which case they lie in the orbits of tuples under (iii). This completes the case k = 4, and the case k = 3 is entirely similar. ♠

As in previous sections, in the course of our proof of Theorem 1 we shall study the embedding of our subgroup S ≅ SL_{2}(q) in C_{G}(a). As before, define D = C_{G}(a)′, a connected semisimple group. By Lemmas 3.4 - 3.7 we know that D is as in 3.5(5). In particular, either D has a factor D_{4} or D_{5}, or D is a product of factors A_{r}. For the D_{4}, D_{5} possibilities we need to study in some detail the embeddings of subgroups SL_{2}(q) in these groups.

For this purpose we now introduce some notation for certain connected semisimple subgroups of D_{4}, D_{5} (in characteristic 2). It is convenient to treat these groups in the forms SO_{8}, SO_{10}. First, we denote by A_{1} a connected fundamental subgroup SL_{2} in SO_{8} or SO_{10}. Thus A_{1}^{4} denotes a subgroup SO_{4}SO_{4} in SO_{8}, naturally embedded. Taking a subgroup SO_{4} of SO_{8} fixing pointwise a non-degenerate 4-space, we write B_{1} for a subgroup SO_{3} of this SO_{4} fixing a nonsingular vector. Then SO_{6} contains a subgroup B_{1}^{2}, SO_{8} a subgroup B_{1}^{3}, and SO_{10} a subgroup B_{1}^{4}; in each case, the subgroup B_{1}^{k} fixes a unique nonsingular 1-space. Note that if v is a nonsingular vector then (SO_{10})_{v} = SO_{9}, and a surjective morphism SO_{9} → Sp_{8} sends B_{1}^{4} → C_{1}^{4} = (Sp_{2})^{4}, naturally embedded in Sp_{8}. Finally, SO_{8} contains a subgroup Sp_{2} ⊗ Sp_{2} ⊗ Sp_{2}, which we denote by C_{1}^{3}; and C_{1}^{2} denotes a subgroup Sp_{2} ⊗ Sp_{2} of this.

Let S_{0} = A_{1}(2^{e}), and fix an injection ι : S_{0} → SL_{2}(K). Whenever S_{0} = A_{1}(2^{e}) maps homomorphically into a commuting product A_{1}^{k}B_{1}^{l}, up to conjugacy the projections of S_{0} to the factors are either trivial, or are field twists of the injection ι. In the following result, when we say S_{0} has “distinct twists on the B_{1} factors”, we mean that the twists involved in all the nontrivial projections to B_{1} factors are all distinct.

Finally, if Y is a connected reductive group, we shall denote by QY a connected group with unipotent radical Q.

Lemma 6.5 Let S_{0} = SL_{2}(2^{e}) (e ≥ 4), and suppose S_{0} < D_{r} with r = 4 or 5.

(i) If r = 4 then S_{0} lies in one of the following connected subgroups of D_{4}:

A_{1}^{4}, QA_{1}^{3}, B_{1}^{3}, QB_{1}^{2}, QC_{1}^{2}, C_{1}^{3}.

In the last four cases S_{0} has distinct twists on the B_{1} and C_{1} factors.

(ii) If r = 5 then S_{0} lies in one of the following connected subgroups of D_{5}:

QA_{1}^{4}, B_{1}^{4}, A_{1}^{2}B_{1}^{2}, QA_{1}B_{1}^{2}, QB_{1}^{3}, QC_{1}^{2}, C_{1}^{3}.

In the third case A_{1}^{2}B_{1}^{2} < SO_{4}SO_{6}. Also, S_{0} has distinct twists on the B_{1} and C_{1} factors in all cases with such factors.

Proof In the course of the proof we shall require information on the possible embeddings of S_{0} in the symplectic groups Sp_{4}, Sp_{6}, Sp_{8}, as well as in SO_{6}, SO_{8}, SO_{10}. It is convenient to start with the smallest dimensions and work our way upwards. For a classical group G = Sp_{2n} or SO_{2n}, we shall denote the natural G-module by V_{2n}. If S_{0} < G, let W be an irreducible S_{0}-submodule of V_{2n}; then W has dimension 1,2,4 or 8, and W is either non-degenerate or totally singular, or, when G is orthogonal, a nonsingular 1-space.

Suppose that S_{0} < Sp_{4}, with natural module V_{4}. As above, let W be an irreducible S_{0}-submodule of V_{4}. If W = V_{4} then V_{4} ↓ S_{0} = 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})}, and S_{0} lies in a connected subgroup Sp_{2} ⊗ Sp_{2} = C_{1} ⊗ C_{1} of Sp_{4}. If W is a non-degenerate 2-space then S_{0} < Sp_{2}⊥Sp_{2} = C_{1}C_{1}. If W is a 1-space, then V_{4} ↓ S_{0} = 0/1^{(2^{i})}/0; we can take this to be indecomposable, as otherwise we are in the previous case. Here S_{0} lies in a parabolic subgroup QC_{1}T_{1} of Sp_{4}, and the unipotent radical Q has a 1-dimensional subgroup Z such that Q/Z is an S_{0}-module of high weight 1; moreover, S_{0} acts trivially on Z. Choose S̃_{0} < C_{1} such that S_{0} ≅ S̃_{0} and QS_{0} = QS̃_{0}. Then H^{1}(S̃_{0}, Q/Z) is 1-dimensional by 1.2, so under the action of (Q/Z)T_{1} there are just 2 classes of complements to Q/Z in (Q/Z)S̃_{0}, both of which give complements to Q in QS̃_{0}. One of these, S̃_{0}, lies in C_{1}; the other, S_{0}, lies in an indecomposable B_{1} contained in SO_{4} < Sp_{4}. Consequently S_{0} < B_{1} < C_{1} ⊗ C_{1} = SO_{4} in this case. The last possibility is that W is a singular 2-space. A graph automorphism τ of Sp_{4} maps the stabilizer of W to the stabilizer of a 1space. Hence by the previous case, S_{0}^{τ} lies in C_{1} ⊗ C_{1}, whence S_{0} lies in C_{1}C_{1}. Summarising, we have shown that if S_{0} < Sp_{4} then S_{0} lies in either C_{1} ⊗ C_{1} or C_{1}C_{1}.

Next suppose S_{0} < SO_{6}, with natural module V_{6}. If W is non-degenerate of dimension 2 or 4, then S_{0} < SO_{4} = A_{1}^{2}. If W is a singular 1-space or 2-space, then S_{0} < QA_{1}^{2} or QA_{1}. The only other possibility is that W is a nonsingular 1-space, in which case V_{6} ↓ S_{0} = 0/V_{4}/0, where V_{4} is symplectic 4-space. Applying the previous paragraph, S_{0} maps into a subgroup C_{1}C_{1} or C_{1} ⊗C_{1} of Sp(V_{4}). Thus S_{0} lies in a subgroup B_{1}^{2} or A_{1}^{2} of SO_{6}. Moreover, if S_{0} < B_{1}^{2} with equal twists on the factors, then V_{6} ↓ S_{0} = 0/1^{(2^{i})}/1^{(2^{i})}/0; as Ext_{S_{0}}^{1}(1^{(2^{i})}, 0) is 1-dimensional by 1.2, it follows that S_{0} fixes a singular 2space, yielding S_{0} < QA_{1}. Summarising, if S_{0} < SO_{6} then either S_{0} < QA_{1}^{2} or S_{0} < B_{1}^{2} with distinct twists on the B_{1} factors.

Next consider S_{0} < Sp_{6}, with natural module V_{6}. If W is non-degenerate then S_{0} < C_{1}C_{2}, so by the Sp_{4} case handled above, S_{0} lies in C_{1}(C_{1} ⊗C_{1}) or C_{1}^{3}. If W is a singular 2-space then S_{0} < QA_{1}C_{1}. Moreover, either V_{6} ↓ S_{0} = 1^{(2^{i})}/1^{(2^{j})}/1^{(2^{i})}, completely reducible by 1.2, or V_{6} ↓ S_{0} = 1^{(2^{i})}/0^{2}/1^{(2^{i})}; hence either S_{0} < A_{1}C_{1} or S_{0} < QA_{1}. Finally, if W is a 1-space then S_{0} < QC_{2}, hence S_{0} lies in Q(C_{1} ⊗ C_{1}) or QC_{1}C_{1}.

Similarly, if S_{0} < Sp_{8} then S_{0} lies one of the subgroups C_{1} ⊗ C_{1} ⊗ C_{1} (with S_{0} irreducible on V_{8}), C_{2}C_{2}, C_{1}C_{3}, QA_{1}C_{2}, QC_{3} or QA_{3} (in the last case W is a singular 4-space), and we can apply the previous conclusions for subgroups of C_{2}, C_{3} to refine this list.

Now we are ready to consider S_{0} < SO_{8}, with natural module V_{8}. If W = V_{8} then S_{0} < C_{1}^{3}, with distinct twists on the factors, and if W is a non-degenerate 4-space then S_{0} < A_{1}^{4}. If W is a singular 4-space, then W ↓ S_{0} = 1^{(2^{i})} ⊗ 1^{(2^{j})} and S_{0} < QC_{1}^{2} < QA_{3}. If W is a singular 2-space then S_{0} < QA_{1}^{3}. And if W is a singular 1-space, then V_{8} ↓ S_{0} = 0/V_{6}/0 with V_{6} an SO_{6}-space, so by the SO_{6} case above, S_{0} lies in QA_{1}^{2} or QB_{1}^{2} (with distinct twists in the latter case).

To complete the analysis of S_{0} < SO_{8}, it remains to handle the case where W is a nonsingular 1-space. Here V_{8} ↓ S_{0} = 0/V_{6}/0 where V_{6} is symplectic 6-space, and S_{0} maps into Sp_{6}. By the Sp_{6} case above, S_{0} maps into a subgroup C_{1}(C_{1}⊗C_{1}), C_{1}^{3}, QA_{1}C_{1}, Q(C_{1}⊗C_{1}) or QC_{1}C_{1} of Sp_{6}. In the first case S_{0} < B_{1}A_{1}^{2} < A_{1}^{4} (in SO_{8}). In the second, S_{0} < B_{1}^{3}; and if S_{0} has equal twists on two of the B_{1} factors here, then V_{8} ↓ S_{0} = 0/1^{(2^{i})}/1^{(2^{i})}/..., and an application of 1.2 shows that S_{0} fixes a 2-space 1^{(2^{i})}, putting us in a previous case. In the third case S_{0} < QA_{1}B_{1}, which lies in QA_{1}^{3} in SO_{8}. In the fourth case S_{0} < QA_{1}^{2}. Finally, in the last case S_{0} < QB_{1}^{2}, and if S_{0} has equal twists in the B_{1} factors then S_{0} < QA_{1}. This completes the proof of part (i) of the lemma.

Now consider S_{0} < SO_{10}, with natural module V_{10}. If W has dimension 8 then S_{0} < C_{1}^{3} (with distinct twists). If W is a non-degenerate 4-space then S_{0} < SO_{4}SO_{6}, so by the SO_{6} case S_{0} lies in QA_{1}^{4} or A_{1}^{2}B_{1}^{2}. If W is a singular 4-space then V_{10} ↓ S_{0} = (1^{(2^{i})} ⊗ 1^{(2^{j})})^{2}/0^{2}, and S_{0} < QC_{1}^{2}. And if W is a singular 1-space, then S_{0} < QSO_{8}, and conclusion (ii) follows from the SO_{8} case (i) (note that the possibility S_{0} < QC_{1}^{3} only arises in the SO_{8} proof when S_{0} is irreducible on V_{8}, in which case V_{10} ↓ S_{0} is completely reducible by 1.2, giving S_{0} < C_{1}^{3} with distinct twists).

Next suppose that W is a nonsingular 1-space, so V_{10} ↓ S_{0} = 0/V_{8}/0 where V_{8} is Sp_{8}-space. By the Sp_{8} case above, S_{0} maps into a subgroup C_{1} ⊗ C_{1} ⊗ C_{1}, C_{2}C_{2}, C_{1}C_{3}, QA_{1}C_{2}, QA_{3} or QC_{3} of Sp_{8}. In the first case we have S_{0} < C_{1}^{3} with distinct twists. Next, if S_{0} maps into C_{2}C_{2} then by the Sp_{4} case above, S_{0} lies in A_{1}^{4}, A_{1}^{2}B_{1}^{2} or B_{1}^{4} in SO_{10}. Moreover we may take it that S_{0} has distinct twists on the B_{1} factors, since equal twists on a pair of B_{1} factors allows us to replace the group B_{1}^{2} by QA_{1}, putting S_{0} inside QA_{1}^{3} or QA_{1}B_{1}^{2}, as in the conclusion of (ii). If S_{0} maps into C_{1}C_{3} then by the Sp_{6} case, S_{0} maps into C_{1}^{2}(C_{1} ⊗ C_{1}), C_{1}^{4}, C_{1}^{2}A_{1}, QA_{1}C_{1}, QC_{1}(C_{1} ⊗ C_{1}), or QC_{1}^{3}. Hence in SO_{10}, S_{0} lies in A_{1}^{2}B_{1}^{2}, B_{1}^{4}, A_{1}B_{1}^{2}, QA_{1}B_{1}, QA_{1}^{2}B_{1} or QB_{1}^{3}. The only subgroup here not in conclusion (ii) is QA_{1}^{2}B_{1}, which lies in QA_{1}^{4}. Next, if S_{0} maps into QA_{1}C_{2} in Sp_{8}, then S_{0} < QA_{1}^{3} or QA_{1}B_{1}^{2} in SO_{10}. If S_{0} < QA_{3} then S_{0} < QC_{1}^{2}. Finally in this case (W a nonsingular 1-space), if S_{0} maps into QC_{3} in Sp_{8}, then S_{0} fixes a singular 1-space in V_{10}, putting us in a previous case.

It remains to handle the case where W is a singular 2-space. Here S_{0} < QA_{1}D_{3}, so by the SO_{6} case, S_{0} lies in either QA_{1}^{3} or QA_{1}B_{1}^{2} (with distinct twists), as in (ii).

This completes the proof of the lemma. ♠

Corollary 6.6 If D = C_{G}(a)′ has a factor D_{4} or D_{5}, then one of the following holds:

(i) S < QA_{1}^{k} < D with k ≤ 5 (the A_{1}^{k} being fundamental A_{1}’s in G); (ii) S < QA_{1}B_{1}^{3} < A_{2}D_{4} or A_{1}D_{5}; (iii) S < QA_{1}^{2}B_{1}^{2} < A_{1}D_{5}; (iv) S < A_{1}B_{1}^{4} < A_{1}D_{5};

(v) S < A_{1}^{3}B_{1}^{2} < A_{1}D_{5}. In cases (ii) - (v), S has distinct twists on the B_{1} factors.

Proof Observe that D ≠ A_{1}A_{1}D_{4}, since otherwise a ∈ C_{G}(A_{1}A_{1}D_{4}) = A_{1}A_{1} = C_{G}(D_{6}), contradicting 3.7. Hence, as D satisfies 3.5(5), we see that D = A_{r}D_{4} or A_{s}D_{5} with r ≤ 2, s ≤ 1. The projection S_{0} of S in the factor D_{4} or D_{5} satisfies 6.5(i) or 6.5(ii). We see from this that either the result holds, or S_{0} lies in a subgroup C_{1}^{3} or QC_{1}^{2} of the factor D_{4} or D_{5}. If S_{0} < C_{1}^{3}, then S_{0} < C_{1}^{3} < D_{4}, and N_{G}(D_{4}) contains a triality automorphism of this D_{4}. This triality sends the subgroup C_{1}^{3} to a subgroup B_{1}^{3} of D_{4}, and hence a conjugate of S lies in a subgroup A_{r}B_{1}^{3} or A_{s}B_{1}^{3} of D, as in (ii) of the conclusion. A similar argument applies if S_{0} < QC_{1}^{2}; here S_{0} < QC_{1}^{2} < D_{4}, and application of triality gives a conjugate of S lying in A_{r}(QB_{1}^{2}) or A_{s}(QB_{1}^{2}), yielding (ii) again. ♠

In the next lemma we consider the other possibilities for D.

Lemma 6.7 Suppose that D is a product of factors A_{m}. Then conclusion (i) or (v) of Corollary 6.6 holds.

Proof Recall that D is as in 3.5(5). Note also that D ≠ A_{3}A_{4} or A_{1}A_{6} by 3.4. Assume that 6.6(i) does not hold. Then D must have a factor A_{3+δ} (δ ≥ 0) with natural module V such that V ↓ S has a composition factor 1^{(2^{i})} ⊗ 1^{(2^{j})}; and D ≠ A_{1}A_{5}, since otherwise a ∈ C_{G}(A_{1}A_{5}) = A_{2} = C_{G}(E_{6}), contrary to 3.6. Moreover, if D has a factor A_{4}, then the natural module for this factor restricts to S as 1^{(2^{i})} ⊗ 1^{(2^{j})}/0; this splits by 1.2, and hence S projects into a subgroup A_{3} of this factor. We conclude that one of the following holds:

(a) S < D = A_{5} or A_{6}, with embedding 1^{(2^{i})} ⊗ 1^{(2^{j})}/1^{(2^{k})}/0^{m} or 1^{(2^{i})} ⊗ 1^{(2^{j})}/0^{2+m} (m ≤ 1); (b) S < A_{1}^{3}A_{3}, with projection 1^{(2^{i})} ⊗ 1^{(2^{j})} in the A_{3} factor;

(c) S < A_{3}A_{3}, with projection 1^{(2^{i})} ⊗ 1^{(2^{j})} in one of the A_{3} factors;

(d) S < D = A_{1}A_{2}A_{3}, with projection 1^{(2^{i})} ⊗ 1^{(2^{j})} in the A_{3} factor. Consider first case (b), S < A_{1}^{3}A_{3}. Now A_{1}^{3}A_{3} < A_{1}^{3}D_{4} in G, and

N_{G}(A_{1}^{3}D_{4}) contains an element τ of order 3 inducing a triality automorphism on the D_{4} factor and permuting the A_{1} factors cyclically. Since the projection of S to the A_{3} factor is 1^{(2^{i})} ⊗ 1^{(2^{j})}, we see that for some k, S^{τ^{k}} projects into a subgroup B_{1}^{2} in the D_{4} factor. Hence S < A_{1}^{3}B_{1}^{2} < A_{1}^{3}D_{4}. Moreover this A_{1}^{3}B_{1}^{2} < A_{1}^{3}D_{3} < A_{1}^{3}D_{4}, so S < A_{1}^{3}B_{1}^{2} < A_{1}^{3}D_{3} < A_{1}D_{5}, as in 6.6(v). Note that the twists on the B_{1} factors are distinct.

Now suppose case (a) holds, and let V be the natural module for D = A_{5} or A_{6}. If V ↓ S is completely reducible then S < A_{1}A_{3} < D, as in case (b) handled above. So assume V ↓ S is not completely reducible. Then applying a twist to the labelling of S, we can take V ↓ S = 1⊗1^{(2^{i})}/1^{(2^{j})}/0^{m} for some i, j. If j ≠ 0, i then by 1.2, V ↓ S = (1 ⊗ 1^{(2^{i})}) ⊕ (1^{(2^{j})}/0^{m}), so S < A_{3}A_{2} < A_{3}A_{3}, as in case (c), which we handle below.

Thus we assume now that j = 0 or i. Applying a twist, we may take j = 0, so V ↓ S = 1⊗1^{(2^{i})}/1/0^{m}. Recall that q = 2^{e} with e ≥ 4, and the nontrivial composition factors of L(G) ↓ D are of the form V_{D}(λ_{r}), together with adjoint modules of the factors ([LS2, §2]). If i ≤ e−3 then we can choose a rank 1 torus T_{1} in D (namely T_{1} = {diag(c^{2^{i}+1}, c^{2^{i}−1}, c^{−2^{i}−1}, c^{−2^{i}+1}, c, c^{−1}, 1^{m}) : c ∈ K^{∗}}) containing an element t ∈ S of order q + 1, such that t and T_{1} fix the same subspaces of L(G), contrary to (†). If i = e − 1 we can twist to take V ↓ S = 1^{(2)} ⊗ 1/1^{(2)}/0^{m} and apply the same T_{1} argument. And if i = e − 2 we twist to V ↓ S = 1^{(4)} ⊗ 1/1^{(4)}/0^{m}, and the T_{1} argument applies unless q = 16. Thus to complete this case (a), assume now that

q = 16, V ↓ S = 1 ⊗ 1^{(4)}/1/0^{m}.

By [LS2, Table 8.1], L(G) ↓ A_{5} = L(A_{5})/λ_{1}^{6}/λ_{2}^{3}/λ_{3}^{2}/λ_{4}^{3}/λ_{5}^{6}/0^{11}. Using [LS2, 2.13] we deduce from this that

\begin{aligned}L(G) ↓ S = & (1 ⊗ 1^{(2)})^{2}/(1 ⊗ 1^{(4)})^{16}/(1 ⊗ 1^{(8)})^{2}/(1^{(2)} ⊗ 1^{(4)})^{8}/ \\ & (1^{(2)} ⊗ 1^{(8)})^{1}/1^{16}/(1^{(2)})^{9}/(1^{(4)})^{16}/(1^{(8)})^{8}/0^{34}. \end{aligned}

As q = 16, we have A = B_{2}(16) or G_{2}(16). If A = B_{2}(16) then S contains a conjugate b of a (of order 17). But from L(G) ↓ S we see that dim C_{G}(b) = 34, whereas dim C_{G}(a) ≥ dim A_{5}T_{3} = 38, a contradiction. Now suppose A = G_{2}(16). Since L(G) ↓ S has composition factor 1^{(2)} ⊗ 1^{(8)} with multiplicity 1, we see from Lemma 6.1 that L(G) ↓ A must have a composition factor 01⊗01^{(4)}; but the restriction of this to S has composition factor 1⊗1^{(8)} with multiplicity 4, which contradicts L(G) ↓ S given above. This completes the argument for case (a).

Next consider case (c), S < A_{3}A_{3}. If S < A_{1}^{2}A_{3} then we are in case (b), so assume this is not so. Then the embedding of S in A_{3}A_{3} is either 1^{(2^{i_{1}})}⊗1^{(2^{i_{2}})}, 1^{(2^{i_{3}})}⊗1^{(2^{i_{4}})} or 1^{(2^{i_{1}})}⊗1^{(2^{i_{2}})}, 1^{(2^{i_{3}})}/0^{2}, and in the latter case the second representation is indecomposable. By [LS2, p.60], the composition factors of L(G) ↓ A_{3}A_{3} are among L(A_{3}A_{3}), λ_{i} ⊗ λ_{j}, λ_{i} ⊗ 0 and 0 ⊗ λ_{j}, and include λ_{1} ⊗ λ_{1}.

Suppose first that the embedding is 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})}, 1^{(2^{i_{3}})} ⊗ 1^{(2^{i_{4}})}. If i_{1}, …, i_{4} are all distinct, then L(G) ↓ S has a composition factor 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})}⊗1^{(2^{i_{3}})}⊗1^{(2^{i_{4}})} (coming from λ_{1}⊗λ_{1}), and either no 1^{(2^{i_{1}})}⊗1^{(2^{i_{2}})}⊗1^{(2^{i_{3}})} or no 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})} ⊗ 1^{(2^{i_{4}})}; this contradicts 6.2(i,ii). Therefore i_{1}, …, i_{4} are not all distinct. If e ≥ 5 we can twist to take i_{1} … i_{4} to satisfy (i), (ii) or (iii) of 6.3. Now S < S̄ < A_{3}A_{3}, where S̄ ≅ SL_{2}(K) has embedding 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})}, 1^{(2^{i_{3}})} ⊗ 1^{(2^{i_{4}})} in A_{3}A_{3}, and it is easily checked that S̄ has all weights on L(G) less than q. Then by 1.5, S and S̄ fix the same subspaces of L(G), contradicting (†). Finally suppose e = 4. We can twist to take the embedding the embedding of S in A_{3}A_{3} to be

1 ⊗ 1^{(2)}, 1^{(2)} ⊗ 1^{(4)} or 1 ⊗ 1^{(4)}, 1 ⊗ 1^{(2)} or 1 ⊗ 1^{(4)}, 1^{(2)} ⊗ 1^{(4)} or
1 ⊗ 1^{(2)}, 1 ⊗ 1^{(2)} or 1 ⊗ 1^{(4)}, 1 ⊗ 1^{(4)}.

Choose S̄ ≅ SL_{2}(K) with S < S̄ < A_{3}A_{3} and S̄ having one of these embeddings in A_{3}A_{3}. Then in the first, second and fourth embeddings, S̄ has all weights on L(G) less than q = 16. In the third case, S̄ could have a composition factor of high weight 16 coming from a composition factor λ_{2} ⊗ λ_{2} for L(G) ↓ A_{3}A_{3}. By [LS2, p.60], the presence of a composition factor λ_{2} ⊗ λ_{2} implies that there are no composition factors λ_{i} ⊗ λ_{2} with i ∈ {1, 3}; so we see that S̄ has highest weight 16 on L(G), and has no composition factor of high weight 1. Hence 1.4 implies that S and S̄ fix the same subspaces of L(G), contradicting (†). The same argument handles the final embedding 1 ⊗ 1^{(4)}, 1 ⊗ 1^{(4)}.

Now suppose that the embedding of S in A_{3}A_{3} is 1^{(2^{i_{1}})} ⊗1^{(2^{i_{2}})}, 1^{(2^{i_{3}})}/0^{2}. Assume first that e ≥ 5. Then we can take i_{1}i_{2}i_{3} (in some order) to satisfy the conclusion of 6.4. If i_{r} ≤ e − 3 for all r, then the T_{1} argument applies (i.e. there is a torus T_{1} in A_{3}A_{3} containing an element t ∈ S of order q + 1 and fixing the same subspaces of L(G) as t, contrary to (†)). Otherwise, by 6.4, we have e = 5 or 6 and {i_{1}, i_{2}, i_{3}} = {0, 1, 3} or {0, 2, 4} respectively. In the first case, either the T_{1} argument applies, or the embedding of S is 1^{(2)} ⊗ 1^{(8)}, 1/0^{2}. From the two possible restrictions L(G) ↓ A_{3}A_{3} given in [LS2, p.60], we deduce that L(G) ↓ S has 8 composition factors 1⊗1^{(2)}⊗1^{(8)}, but no composition factors 1 ⊗ 1^{(8)}; hence A ≠ A_{2}^{ϵ}(q) by 6.2(ii). Thus A = B_{2}(q), whence S contains a conjugate b of a (of order q + 1). Again using [LS2, p.60], we see that C_{G}(b) has dimension 34. Now A_{3}A_{3} lies in a subgroup D_{4}D_{4} of G, which in turn lies in a subgroup D_{8}. Consideration of the eigenvalues of b on the natural D_{8}-module shows that C_{D_{8}}(b)′ is either A_{1}^{5} or A_{1}D_{4}. Since C_{G}(b) has dimension 34, this forces C_{G}(b) = A_{1}D_{4}T_{3}, contrary to the hypothesis of the lemma.

To complete case (c), assume that e = 4 (and the embedding of S in A_{3}A_{3} is 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})}, 1^{(2^{i_{3}})}/0^{2}). If we can twist to take (i_{1}, i_{2}, i_{3}) = (0, 1, 2), (0, 2, 1), (0, 1, 0) or (0, 1, 1), then the T_{1} argument applies. The only remaining possibilities are (i_{1}, i_{2}, i_{3}) = (1, 2, 0) or (0, 2, 0). For A = B_{2}(16) these are handled as in the previous paragraph. For A = G_{2}(16) in the (0, 2, 0) case, we find that L(G) ↓ S has composition factor 1 ⊗ 1^{(2)} with multiplicity 2, conflicting with 6.1. And for A = G_{2}(16) in the (1, 2, 0) case, we work out the composition factors of L(G) ↓ S (using [LS2, p.60]), and show that these conflict with 6.2. This completes case (c).

It remains to consider case (d), S < D = A_{1}A_{2}A_{3}. We can take it that the projections of S to each of the 3 factors is nontrivial (otherwise we are in a previous case). Say the embedding is 1^{(2^{i_{1}})}, 1^{(2^{i_{2}})}/0, 1^{(2^{i_{3}})} ⊗ 1^{(2^{i_{4}})}. Now D lies in a maximal rank subgroup A_{1}A_{2}A_{5} of G, and from [LS2, §2] we have

L(G) ↓ A_{1}A_{2}A_{5} = L(A_{1}A_{2}A_{5})/1 ⊗ λ_{1} ⊗ λ_{1}/1 ⊗ λ_{2} ⊗ λ_{5}/1 ⊗ 0 ⊗ λ_{3}/0 ⊗ λ_{1} ⊗ λ_{2}/0 ⊗ λ_{2} ⊗ λ_{4}.

If i_{1}, …, i_{4} are all distinct then L(G) has a composition factor 1^{(2^{i_{1}})} ⊗ … ⊗ 1^{(2^{i_{4}})} (from 1 ⊗ λ_{1} ⊗ λ_{1}), but no composition factor 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})} ⊗ 1^{(2^{i_{3}})}, contrary to 6.2(i,ii). Thus i_{1}, …, i_{4} are not all distinct.

Assume now that e ≥ 5. Then we can take i_{1} … i_{4} to satisfy the conclusion of 6.4. If i_{r} ≤ e − 3 for all r then the T_{1} argument applies. Otherwise, i_{1} … i_{4} (in some order) is one of 0013, 0023, 0113 (all with e = 5) or 0024 (with e = 6). If i_{2} > 0 then S < S̄ < D, where S̄ ≅ SL_{2}(K) with embedding 1^{(2^{i_{1}})}, 1^{(2^{i_{2}})}/0, 1^{(2^{i_{3}})} ⊗ 1^{(2^{i_{4}})} (the point being that by 1.2, for SL_{2}(K), the indecomposable 1^{(2^{i_{2}})} exists only if i_{2} > 0, in which case Ext_{S}(1^{(2^{i_{2}})}, 0) ≅ Ext_{SL_{2}(K)}(1^{(2^{i_{2}})}, 0)). But now S̄ has all weights on L(G) less than q. Hence i_{2} = 0, and the embedding of S in D is one of

1, 1/0, 1^{(2)} ⊗ 1^{(8)} or 1^{(2)}, 1/0, 1 ⊗ 1^{(8)} or 1^{(8)}, 1/0, 1 ⊗ 1^{(2)} (e = 5), 1, 1/0, 1^{(4)} ⊗ 1^{(8)} or 1^{(4)}, 1/0, 1 ⊗ 1^{(8)} or 1^{(8)}, 1/0, 1 ⊗ 1^{(4)} (e = 5), 1^{(2)}, 1/0, 1^{(2)} ⊗ 1^{(8)} (e = 5), 1, 1/0, 1^{(4)} ⊗ 1^{(16)} or 1^{(4)}, 1/0, 1 ⊗ 1^{(16)} or 1^{(16)}, 1/0, 1 ⊗ 1^{(4)} (e = 6).

Twist the embeddings in the first and third rows by 2^{2}, those in the second row by 2^{3}, and those in the fourth row by 2^{2}; so for example, the embeddings in the first row now read

1^{(4)}, 1^{(4)}/0, 1^{(4)} ⊗ 1 or 1^{(8)}, 1^{(4)}/0, 1^{(4)} ⊗ 1 or 1, 1^{(4)}/0, 1^{(4)} ⊗ 1^{(8)} (e = 5).

Now S < S̄ < D, with S̄ embedded in D with the new labelling, and we see that the highest weight of S̄ on L(G) is less than q. This completes the proof for e ≥ 5.

To complete the proof for case (d), assume finally that e = 4. Then we can take i_{1} … i_{4} (in some order) to be 0012, 0112, 0122 or 0022 (in all other cases the T_{1} argument applies). As above, i_{2} = 0.

For the tuple 0012, the embedding of S is 1, 1/0, 1^{(2)}⊗1^{(4)} or 1^{(2)}, 1/0, 1⊗ 1^{(4)} or 1^{(4)}, 1/0, 1 ⊗ 1^{(2)}. In the last case, the T_{1} argument applies. In the second case, L(G) ↓ S has composition factor 1⊗1^{(2)} ⊗1^{(4)} with multiplicity 3, so A ≠ G_{2}(16) by 6.2(iii); also if b ∈ S of order 17, we calculate from L(G) ↓ D above that C_{L(G)}(b) has dimension 24, so A ≠ B_{2}(16) (as otherwise b is conjugate to a, whereas C_{G}(a) = A_{1}A_{2}A_{3}T_{2}, of dimension 28). In the first case, we twist to take the embedding to be 1^{(2)}, 1^{(2)}/0, 1^{(4)} ⊗ 1^{(8)}. As explained before, we have S < S̄ < D, where S̄ ≅ SL_{2}(K) has this embedding in D. The composition factors of S̄ on L(G) of high weight 16 or more are 1^{(16)} and 1^{(2)} ⊗1^{(16)}, both of which are irreducible for S; moreover, S̄ has no composition factors 1 or 1^{(2)} ⊗ 1. Hence S and S̄ fix the same subspaces of L(G), by 1.4, contradicting (†).

The other tuples 0112, 0122 and 0022 are handled similar fashion. This completes the proof of the lemma. ♠

In view of the lemma just proved, it remains to deal with S as in cases

(i) - (v) of Corollary 6.6. To handle cases (ii) - (v), we require information about the restriction of L(G) to the subgroups A_{1}B_{1}^{4} and A_{1}^{2}B_{1}^{3}, which is provided by the next lemma. In the conclusion, by a “1-fold tensor” we mean a module for the product A_{1}B_{1}^{4} or A_{1}^{2}B_{1}^{3} in which only one of the simple factors acts nontrivially.

Lemma 6.8 (i) For A_{1}B_{1}^{4} < A_{1}D_{5} < G, the nontrivial composition factors of L(G) ↓ A_{1}B_{1}^{4} are as follows:

1-fold tensors 1 or 1^{(2)},

0 ⊗ 1^{(2)} ⊗ 1^{(2)} ⊗ 0 ⊗ 0, …, 0 ⊗ 0 ⊗ 0 ⊗ 1^{(2)} ⊗ 1^{(2)} (one for each pair of B_{1} factors, each with mult. 1), 1 ⊗ 1^{(2)} ⊗ 0 ⊗ 0 ⊗ 0, …, 1 ⊗ 0 ⊗ 0 ⊗ 0 ⊗ 1^{(2)} (one for each B_{1} factor, each with mult. 2), 1 ⊗ 1 ⊗ 1 ⊗ 1 ⊗ 1 (mult. 2), 0 ⊗ 1 ⊗ 1 ⊗ 1 ⊗ 1 (mult. 4).

(ii) For A_{1}^{3}B_{1}^{2} = A_{1}(A_{1}^{2}B_{1}^{2}) < A_{1}(D_{2}D_{3}) < A_{1}D_{5} < G, the nontrivial composition factors of L(G) ↓ A_{1}^{3}B_{1}^{2} are:

1-fold tensors 1 or 1^{(2)} (only 1^{(2)} for B_{1} factors), 0 ⊗ 1 ⊗ 1 ⊗ 0 ⊗ 0, 0 ⊗ 1 ⊗ 1 ⊗ 1^{(2)} ⊗ 0, 0 ⊗ 1 ⊗ 1 ⊗ 0 ⊗ 1^{(2)}, 0 ⊗ 0 ⊗ 0 ⊗ 1^{(2)} ⊗ 1^{(2)}, 1 ⊗ 1 ⊗ 1 ⊗ 0 ⊗ 0, 1 ⊗ 0 ⊗ 0 ⊗ 1^{(2)} ⊗ 0, 1 ⊗ 0 ⊗ 0 ⊗ 0 ⊗ 1^{(2)}, 0 ⊗ 1 ⊗ 0 ⊗ 1 ⊗ 1, 0 ⊗ 0 ⊗ 1 ⊗ 1 ⊗ 1, 1 ⊗ 1 ⊗ 0 ⊗ 1 ⊗ 1, 1 ⊗ 0 ⊗ 1 ⊗ 1 ⊗ 1 (both with mult. 2).

Proof From [LS2, §2] we have L(G) ↓ A_{3}D_{5} = L(A_{3})/L(D_{5})/λ_{2} ⊗ λ_{1}/λ_{1} ⊗ λ_{4}/λ_{3} ⊗ λ_{5}, whence

L(G) ↓ A_{1}D_{5} = (L(A_{3}) ↓ A_{1})/L(D_{5})/(1 ⊗ λ_{1})^{2}/(0 ⊗ λ_{1})^{2}/1 ⊗ λ_{4}/(0 ⊗ λ_{4})^{2}/1 ⊗ λ_{5}/(0 ⊗ λ_{5})^{2}.

Let V = V_{D_{5}}(λ_{1}), the usual 10-dimensional module. Then from the definition of B_{1}^{4} < D_{5}, we have V ↓ B_{1}^{4} = 1^{(2)}/1^{(2)}/1^{(2)}/1^{(2)}/0^{2} (where 1^{(2)} stands for a 1-fold tensor, one for each factor B_{1}), and V ↓ A_{1}^{2}B_{1}^{2} = 1 ⊗ 1/1^{(2)}/1^{(2)}/0^{2} (where 1 ⊗ 1 has the A_{1}^{2} acting, and the 1^{(2)} have a single factor B_{1} acting). It follows, using [LS2, 2.6 and 2.7], that for m = 4, 5, we have

V_{D_{5}}(λ_{m}) ↓ B_{1}^{4} = 1 ⊗ 1 ⊗ 1 ⊗ 1, V_{D_{5}}(λ_{m}) ↓ A_{1}^{2}B_{1}^{2} = 1 ⊗ 0 ⊗ 1 ⊗ 1/0 ⊗ 1 ⊗ 1 ⊗ 1.

Using this and the above description of L(G) ↓ A_{1}D_{5}, we obtain the conclusion. ♠

Lemma 6.9 Case (iv) of 6.6 does not hold.

Proof Suppose 6.6(iv) holds, so S < A_{1}B_{1}^{4} with distinct twists of the B_{1} factors. If S projects nontrivially to each B_{1} factor, say with projections 1^{(2^{i_{1}})}, …, 1^{(2^{i_{4}})}, then we see from 6.8(i) that L(G) ↓ S has a composition factor 1^{(2^{i_{1}})} ⊗ … ⊗ 1^{(2^{i_{4}})}, but no composition factor ⨂_{r∈A} 1^{(2^{ir})} for some 3-set A ⊆ {1, 2, 3, 4}. This contradicts 6.2(i,ii).

Hence S projects trivially to some B_{1} factor. Let 1^{(2^{i_{1}})}, …, 1^{(2^{i_{k}})} be the nontrivial projections of S into the factors of A_{1}B_{1}^{4}; thus k ≤ 4. Then S < S̄ < A_{1}B_{1}^{4}, where S̄ ≅ SL_{2}(K) also has these projections.

If e ≥ 5 then we can twist to take i_{r} ≤ e − 2 for all r. Then from 6.8(i) we see that the weights of S̄ on L(G) are all less than q. Then by 1.5, S and S̄ fix the same subspaces of L(G), contradicting (†).

Now suppose e = 4. If S projects trivially into two or more of the B_{1} factors, then we can twist to take i_{r} ≤ 2 for all r, giving a contradiction as above. Therefore S projects trivially into just one B_{1} factor. We can take the projections into the other B_{1} factors to be 1, 1^{(2)}, 1^{(4)}; then the weights of S̄ on L(G) are all less than q, unless the projection into the A_{1} factor is 1^{(8)}. But then by 6.8(i), L(G) ↓ S has a composition factor 1 ⊗ 1^{(2)} ⊗ 1^{(4)} ⊗ 1^{(8)}, contrary to 6.2(i) (recall that A = B_{2}(16) or G_{2}(16) when q = 16). ♠

Lemma 6.10 Case (v) of 6.6 does not hold.

Proof Assume 6.6(v) holds, so S < A_{1}^{3}B_{1}^{2} with distinct twists on the B_{1} factors. Observe that S < S̄ < A_{1}^{3}B_{1}^{2}, where S̄ ≅ SL_{2}(K).

Suppose first that e ≥ 5. If S has distinct nontrivial twists on all five factors of A_{1}^{3}B_{1}^{2}, we obtain a contradiction using 6.8(ii) together with 6.2(i,ii).

Otherwise, we can twist to take all nontrivial projections of S̄ to be of the form 1^{(2^{i})} with i ≤ e − 2, and then the highest weight of S̄ on L(G) is less than q, giving the usual contradiction.

Now suppose e = 4. Assume first that S projects nontrivially into all five factors of A_{1}^{3}B_{1}^{2}, with projections 1^{(2^{i_{1}})}, …, 1^{(2^{i_{5}})}. If all i_{r} ≤ 2 then, twisting if necessary, S̄ has all weights less than q, so some i_{r} = 3. Then we can twist to take i_{1} … i_{5} (in some order) to be 00123. Now L(G) ↓ S has no 4-fold tensor composition factor, by 6.2(i); it follows from 6.8(ii) that (with the factors ordered as in 6.8(ii)), the projections of S in A_{1}A_{1}A_{1}B_{1}B_{1} are 1, 1^{(2)}, 1^{(4)}, 1, 1^{(8)} or 1, 1^{(2)}, 1^{(8)}, 1, 1^{(4)} or 1, 1^{(4)}, 1^{(8)}, 1, 1^{(2)}. In the first case, by 6.8(ii), L(G) ↓ S has a composition factor 1 ⊗ 1^{(4)} ⊗ 1^{(8)}, but no composition factor 1 ⊗ 1^{(8)}, contrary to 6.2(iii). We obtain similar contradictions in the other cases.

Therefore S projects trivially to at least one of the factors of A_{1}^{3}B_{1}^{2}. The only case where S̄ cannot be chosen to have all weights less than q is that in which S has four nontrivial projections 1, 1^{(2)}, 1^{(4)}, 1^{(8)}. By 6.2(i), L(G) ↓ S has no 4-fold tensor composition factor. Hence by 6.8(ii), we see that the projections of S in A_{1}A_{1}A_{1}B_{1}B_{1} can be taken to be one of the following (up to reordering the 3 A_{1} factors and the 2 B_{1} factors):

(a) 1^{(2^{i_{1}})}, 1^{(2^{i_{2}})}, 1^{(2^{i_{3}})}, 1, 0^{2} (where {i_{1}, i_{2}, i_{3}} = {1, 2, 3}),

(b) 0^{2}, 1^{(4)}, 1^{(8)}, 1, 1^{(2)},

(c) 0^{2}, 1^{(2)}, 1^{(8)}, 1, 1^{(4)}. In cases (a) and (b), the highest weight of S̄ on L(G) is 16, and there are no composition factors of high weight 1, so 1.4 implies that S and S̄ fix the same subspaces of L(G), a contradiction. And in case (c), the composition factors of S̄ having high weight 16 or more are 1^{(16)} and 1^{(2)} ⊗ 1^{(16)}; using 6.8(ii) we check that S̄ has no composition factors 1 or 1^{(2)} ⊗ 1, so again 1.4 gives a contradiction. ♠

The next lemma is useful in dealing with case (ii) of 6.6, in which S < QA_{1}B_{1}^{3}, lying in either A_{2}D_{4} or A_{1}D_{5}. In the conclusion there is a certain subgroup B_{1}B′_{1}B′′_{1} of D_{5}; this is a commuting product of three subgroups SO_{3} of SO_{10}, where the radicals of the associated 3-spaces may be different.

Lemma 6.11 Let S_{0} = SL_{2}(2^{e}) (e ≥ 4). Suppose that S_{0} < QB_{1}^{3} < D_{5}, and S_{0} has three distinct nontrivial twists on the B_{1} factors. Then there is a subgroup B_{1}B′_{1}B′′_{1} ≅ B_{1}^{3} of D_{5} such that QB_{1}^{3} = QB_{1}B′_{1}B′′_{1} and S_{0} < B_{1}B′_{1}B′′_{1}.

Proof Let V_{10} be the usual 10-dimensional D_{5}-module. Now QB_{1}^{3} < QD_{4} < D_{5}, so S_{0} fixes both a singular vector v and a nonsingular vector w in V_{10}. Write V_{8} for the C_{4}-space w^{⊥}/⟨w⟩. By hypothesis, V_{8} ↓ S_{0} has three nontrivial composition factors 1^{(2^{i})}, 1^{(2^{j})}, 1^{(2^{k})}, where i, j, k are distinct. Pulling back to V_{10}, we see that S_{0} fixes just three indecomposable 3-spaces in V_{10} of the form 0/1^{(2^{i})}, 0/1^{(2^{j})} and 0/1^{(2^{k})}; call these W_{i}, W_{j}, W_{k} respectively, and let w_{i}, w_{j}, w_{k} be nonzero vectors fixed by S_{0} in W_{i}, W_{j}, W_{k}.

Now (D_{5})_{W_{i}} is a subgroup B_{1}B_{3} of B_{4} = (D_{5})_{w_{i}}; similarly (D_{5})_{W_{j}} = B′_{1}B′_{3} and (D_{5})_{W_{k}} = B′′_{1}B′′_{3}. The group B_{1} acts trivially on the image of W_{j} mod ⟨w_{i}⟩, and also on ⟨w_{i}⟩, hence acts trivially on W_{j}, and similarly on W_{k}. Hence B_{1} lies in the kernel of the action of (D_{5})_{W_{j}} on W_{j}, which is B′_{3}, and consequently [B_{1}, B′_{1}] = 1. Similarly [B_{1}, B′′_{1}] = [B′_{1}, B′′_{1}] = 1. Finally, (D_{5})_{W_{i},W_{j},W_{k}} = B_{1}B′_{1}B′′_{1}, so this group contains S_{0}. ♠

Lemma 6.12 Case (ii) of 6.6 does not hold.

Proof Suppose 6.6(ii) holds, so S < QA_{1}B_{1}^{3}, lying in either A_{1}D_{5} or A_{2}D_{4}, and S has distinct twists on the B_{1} factors. Moreover, by 6.7, D = C_{G}(a)′ is not a product of factors A_{m}, and hence D has a factor D_{4} or D_{5}.

Suppose first that the projections of S to the factors A_{1}B_{1}^{3} are all nontrivial; say they are 1^{2^{(i_{1})}}, …, 1^{2^{(i_{4})}}. If these are all distinct, then by 6.8(i), L(G) ↓ S has a composition factor 1^{2^{(i_{1})}} ⊗ … ⊗ 1^{2^{(i_{4})}}, but no 1^{2^{(i_{1})}} ⊗ 1^{(2^{i_{3}})} ⊗ 1^{2^{(i_{4})}}, contrary to 6.2(i,ii). Hence i_{1}, …, i_{4} are not all distinct; since the B_{1} twists are distinct, we may take i_{1} = i_{2} (where i_{1} is the twist on the A_{1} factor), and i_{2}, i_{3}, i_{4} are distinct.

If QA_{1}B_{1}^{3} < A_{1}D_{5}, then by the previous lemma we have S < A_{1}B_{1}B′_{1}B′′_{1}, and the conclusion follows from 6.9. Therefore QA_{1}B_{1}^{3} < A_{2}D_{4}, with QA_{1} < A_{2} and B_{1}^{3} < D_{4}. Observe that if i_{1} > 0, then S < S̄ < A_{2}D_{4} with S̄ ≅ SL_{2}(K), by 1.2.

When e ≥ 5, we can take i_{1} … i_{4} to be as in (i), (ii), or (iii) of 6.4. In case (i), the usual T_{1} argument applies: there is a rank 1 torus T_{1} in A_{1}B_{1}^{3} fixing the same subspaces of L(G) as some element of order q + 1 in S. In cases (ii) and (iii) of 6.4, we have (i_{1}, …, i_{4}) = (0, 0, 1, 3), (0, 0, 2, 3), (1, 1, 0, 3) (e = 5) or (0, 0, 2, 4) (e = 6). For the (1, 1, 0, 3) case, we have S < S̄ < A_{2}D_{4} as observed above, and S̄ has all weights on L(G) less than q. In the (0, 0, 1, 3), (0, 0, 2, 3), (0, 0, 2, 4) cases, L(G) ↓ S has composition factors 1 ⊗ 1^{(2)} ⊗ 1^{(8)}, 1 ⊗ 1^{(4)} ⊗ 1^{(8)}, 1 ⊗ 1^{(4)} ⊗ 1^{(16)} (respectively), but no 1 ⊗ 1^{(8)}, 1 ⊗ 1^{(4)}, 1 ⊗ 1^{(4)} (respectively), so A ≠ A_{2}^{ϵ}(q) by 6.2(ii). And for A = B_{2}(q), picking a conjugate b of a lying in S, we find from L(G) ↓ A_{2}D_{4} that C_{L(G)}(b) has dimension 22, contradicting the fact that C_{G}(a)′ has a factor D_{4} or D_{5}.

To complete the case of four nontrivial projections, consider e = 4. Here we can take (i_{1}, …, i_{4}) = (0, 0, 1, 2), (1, 0, 1, 2) or (2, 0, 1, 2). Then L(G) ↓ S has composition factors 1 ⊗ 1^{(2)} ⊗ 1^{(4)}, 1 ⊗ 1^{(4)}, the latter with multiplicity less than 8, so A ≠ G_{2}(16) by 6.2(iv). And for A = B_{2}(16) we find that dim C_{L(G)}(a) = 30 or 26, which is impossible.

Now suppose S has just three nontrivial projections to the factors of A_{1}B_{1}^{3}, say 1^{2^{(i_{1})}}, 1^{(2^{i_{2}})}, 1^{2^{(i_{3})}}. If the projection to the A_{1} factor is trivial, then using the previous lemma we see that either S < B_{1}^{3} < D_{4} or S < B_{1}B′_{1}B′′_{1} < D_{5}, and the conclusion follows from 6.9. Hence the projection to the A_{1} factor is nontrivial; say it is 1^{2^{(i_{1})}}.

When i_{r} ≤ e − 3 for all r, the usual T_{1} argument gives a contradiction. Hence by 6.4, we can take i_{1}i_{2}i_{3} (in some order) to be 012, 013 or 024 with e = 4, 5 or 6, respectively.

For the e = 6 case, we can twist to take (i_{1}, i_{2}, i_{3}) = (4, 2, 0), and now the usual T_{1} argument works.

Now consider the e = 5 case. If i_{1} = 3, the T_{1} argument works; and if i_{1} = 1, we twist to take (i_{1}, i_{2}, i_{3}) = (3, 2, 0), and now the T_{1} argument again works. In the remaining case, (i_{1}, i_{2}, i_{3}) = (0, 1, 3). Here L(G) ↓ S has a composition factor 1 ⊗ 1^{(2)} ⊗ 1^{(8)}, but no 1 ⊗ 1^{(2)}, so A ≠ A_{2}^{ϵ}(q) by 6.2(ii). Thus A = B_{2}(32), and S contains a conjugate b of a of order 33. Now S < QA_{1}B_{1}^{2} < A_{2}D_{4} or A_{1}D_{5}, from which we calculate that C_{L(G)}(b) has dimension 34, whence C_{G}(a)^{0} = A_{1}D_{4}T_{3}. Thus S < QA_{1}B_{1}^{2} < A_{1}D_{4}, where QB_{1}^{2} < D_{4}. The proof of 6.11 shows that in fact S < A_{1}B_{1}B′_{1} < A_{1}D_{4}, where QB_{1}^{2} = QB_{1}B′_{1}. Hence S < S̄ < A_{1}D_{4}, and S̄ has all weights on L(G) less than q.

To conclude the case of three nontrivial projections, consider the e = 4 case, in which {i_{1}, i_{2}, i_{3}} = {0, 1, 2}. We find that L(G) ↓ S has a composition factor 1 ⊗ 1^{(2)} ⊗ 1^{(4)}, but either no 1 ⊗ 1^{(4)} or no 1^{(2)} ⊗ 1^{(4)}, so A ≠ G_{2}(16). Thus A = B_{2}(16), and working with a conjugate b of a lying in S we calculate that C_{L(G)}(b) has dimension 34, 38 or 38, according as i_{1} = 0, 1 or 2; consequently C_{G}(a)′ is A_{1}D_{4}, A_{2}D_{4} or A_{2}D_{4} in the respective cases. Now it follows as before that S < S̄ < C_{G}(a)′, where S̄ has all weights on L(G) less than q.

Finally, when S has fewer than three nontrivial projections, the usual T_{1} argument works in all cases, after twisting suitably. This completes the proof. ♠

We next handle case (i) of 6.6, in which S < QA_{1}^{k} < D with k ≤ 5 (and A_{1}^{k} is a commuting product of fundamental A_{1}’s).

Lemma 6.13 Case (i) of 6.6 does not hold if q ≥ 32.

Proof We begin by discussing the composition factors of L(G) ↓ A_{1}^{5}. Now A_{1}^{5} lies in a maximal commuting product A_{1}^{8} of fundamental A_{1}’s in G, and N_{G}(A_{1}^{8})/A_{1}^{8} ≅ AGL_{3}(2), acting 3-transitively on the set of 8 factors (see [As, Theorem 2]). Further, A_{1}^{8} lies in the subgroup D_{4}D_{4} of G, with each D_{4} containing A_{1}^{4}. As AGL_{3}(2) has only one orbit on 5-sets, we may take our A_{1}^{5} to lie in D_{4}D_{4}, with 4 of the A_{1} factors lying in a D_{4} factor; thus A_{1}^{5} < A_{1}D_{4} < D_{4}D_{4} < G. Now by [LS2, 2.1], L(G) ↓ D_{4}D_{4} = L(D_{4}D_{4})/λ_{1} ⊗ λ_{1}/λ_{3} ⊗ λ_{3}/λ_{4} ⊗ λ_{4}, whence

\begin{aligned}L(G) ↓ A_{1}D_{4} = & (L(D_{4}) ↓ A_{1})/L(D_{4})/(1 ⊗ λ_{1})^{2}/(0 ⊗ λ_{1})^{4}/(1 ⊗ λ_{3})^{2}/ \\ & (0 ⊗ λ_{3})^{4}/(1 ⊗ λ_{4})^{2}/(0 ⊗ λ_{4})^{4}. \end{aligned}

Moreover,

L(D_{4}) ↓ A_{1}^{4} = 1 ⊗ 1 ⊗ 1 ⊗ 1/1^{(2)}/1^{(2)}/1^{(2)}/1^{(2)}/0^{4},
V_{D_{4}}(λ_{1}) ↓ A_{1}^{4} = 1 ⊗ 1 ⊗ 0 ⊗ 0/0 ⊗ 0 ⊗ 1 ⊗ 1,
V_{D_{4}}(λ_{3}) ↓ A_{1}^{4} = 1 ⊗ 0 ⊗ 1 ⊗ 0/0 ⊗ 1 ⊗ 0 ⊗ 1,
V_{D_{4}}(λ_{4}) ↓ A_{1}^{4} = 1 ⊗ 0 ⊗ 0 ⊗ 1/0 ⊗ 1 ⊗ 1 ⊗ 0.

From this we see that L(G) ↓ A_{1}^{5} has no 5-fold tensor composition factors, and has a unique 4-fold tensor composition factor (coming from L(D_{4}) ↓ A_{1}^{4}).

Suppose now that q = 2^{e} ≥ 32, and S < QA_{1}^{k} with nontrivial projections 1^{2^{(i_{1})}}, …, 1^{2^{(i_{k})}} (k ≤ 5) in to the A_{1} factors. If i_{1} … i_{k} is as in (i) or (ii) of 6.3, then the usual T_{1} argument works: an element b of order q + 1 in S lies in a rank 1 torus

T_{1} = {(c^{2^{i_{1}}} , …, c^{2^{i_{k}}}) : c ∈ K^{∗}} < A_{1}^{k}

(where each entry c^{2^{ir}} stands for the matrix diag(c^{2^{ir}} , c^{−2^{ir}}) in the corresponding factor A_{1}); the weights of T_{1} on L(G) are all at most 2^{e−1}, so b and T_{1} fix the same subspaces of L(G), contrary to (†).

Therefore by 6.3, we have e = k = 5, and we may take i_{1} … i_{5} to be one of the following (in some order):

01234, 01223, 00223, 01133, 01233.

Take 1^{(2^{i_{1}})} to be the projection of S into the factor A_{1} of A_{1}^{5} not lying in the D_{4} chosen above (where A_{1}^{5} < A_{1}D_{4}).

If i_{1} … i_{5} = 01234, then from the above description, L(G) ↓ S has a unique composition factor 1^{2^{(i_{2})}} ⊗ … ⊗ 1^{2^{(i_{5})}}, contrary to 6.2(i).

Now suppose i_{1} … i_{5} = 01223. For this case, the above T_{1} argument works unless i_{1} = 1, in which case we calculate that

\begin{aligned}L(G) ↓ S = & (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{4}/(1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{2}/(1^{(2)} ⊗ 1^{(4)} ⊗ 1^{(8)})^{4}/ \\ & (1^{(2)} ⊗ 1^{(8)})^{2}/(1 ⊗ 1^{(2)})^{0}/ … \end{aligned}

If A = A_{2}^{ϵ}(q), the presence of ten 3-fold tensor composition factors in L(G) ↓ S implies the same for L(G) ↓ A, whence dim L(G) ≥ 10.27, a contradiction. And if A = B_{2}(q), the composition factors (1⊗1^{(2)}⊗1^{(8)})^{2}/(1^{(2)}⊗1^{(8)})^{2}/(1⊗ 1^{(2)})^{0} conflict with 6.2(v).

Next consider i_{1} … i_{5} = 00223. The usual T_{1} argument works unless i_{1} = 0, so suppose this is the case. Then

L(G) ↓ S = (1 ⊗ 1^{(4)} ⊗ 1^{(8)})^{4}/(1 ⊗ 1^{(16)})^{1}/(1^{(16)})^{1}/ …

and the (1 ⊗ 1^{(4)} ⊗ 1^{(8)})^{4} are the only 3-fold tensor composition factors appearing. If A = A_{2}^{ϵ}(q), then the single factor 1 ⊗ 1^{(16)} must occur in the restriction of a self-dual composition factor of L(G) ↓ A, which must be 11^{(8)}⊗11^{(16)} (see 6.1); but the restriction of this module to S has composition factor 1^{(16)} with multiplicity 6, a contradiction. And if A = B_{2}(q), then the single factor 1 ⊗ 1^{(16)} must occur in the restriction of a factor 01 ⊗ 01^{(16)} in L(G) ↓ A (see 6.1), but the restriction of this to S has 1^{(16)} with multiplicity 2.

Now suppose i_{1} … i_{5} = 01133. If i_{1} = 1 or 3, we find that L(G) ↓ S = (1 ⊗ 1^{(2)} ⊗ 1^{(16)})^{1}/(1 ⊗ 1^{(16)})^{0}/ … or (1 ⊗ 1^{(4)} ⊗ 1^{(8)})^{1}/(1 ⊗ 1^{(4)})^{0}/ … respectively, contrary to 6.2(iii). Thus i_{1} = 0. Here L(G) ↓ S = (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{8}/(1 ⊗ 1^{(2)})^{0}/ …, so A ≠ A_{2}^{ϵ}(q) by 6.2(ii). Hence A = B_{2}(q), and S contains a conjugate b of a. We calculate that C_{G}(b) has dimension 34. As S < QA_{1}^{5} < C_{G}(a)′, it follows that C_{G}(a)′ = A_{1}D_{4}. Since a subgroup A_{1}^{4} of D_{4} cannot normalize a nontrivial unipotent group therein, it follows that Q = 1, and hence S < S̄ < A_{1}D_{4}, where S̄ ≅ SL_{2}(K) has all weights on L(G) less than q.

To conclude, let i_{1} … i_{5} = 01233. If i_{1} = 3 then L(G) ↓ S has a unique 4-fold tensor composition factor, contrary to 6.2(i). Otherwise, according as i_{1} = 0, 1 or 2, we find that L(G) ↓ S = (1^{(2)}⊗1^{(4)}⊗1^{(16)})^{1}/(1^{(4)}⊗1^{(16)})^{0}/ …, (1 ⊗ 1^{(4)} ⊗ 1^{(16)})^{1}/(1^{(4)} ⊗ 1^{(16)})^{0}/ … or (1 ⊗ 1^{(2)} ⊗ 1^{(16)})^{1}/(1^{(2)} ⊗ 1^{(16)})^{0}/ …, respectively. All of these possibilities conflict with 6.2(iii). This completes the proof. ♠

Lemma 6.14 Case (i) of 6.6 does not hold if q = 16.

Proof As in the previous proof, let S < QA_{1}^{k} (k ≤ 5), with nontrivial projections 1^{2^{(i_{1})}}, …, 1^{2^{(i_{k})}}. If k ≤ 3, we can take i_{1} … i_{k} to be one of 1,11,14,111,112,114,122,124, and the usual T_{1} argument works in all cases.

Now suppose k = 4. Then either the T_{1} argument works, or we can twist to take i_{1} … i_{4} to be one of the following:

0123, 0122, 0112, 0022.

Consider the first case. If S < QA_{1}^{4} with the A_{1}^{4} lying in a subgroup D_{4} of G, then L(G) ↓ S has a 4-fold tensor factor, contrary to 6.2(i). Otherwise, we find that

L(G) ↓ S = (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{2}/(1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{2}/(1 ⊗ 1^{(4)} ⊗ 1^{(8)})^{2}/(1^{(2)} ⊗ 1^{(4)} ⊗ 1^{(8)})^{2}/ … Hence we see using 6.1 that dim L(G) ≥ 4.4^{3} if A = B_{2}(16), and dim L(G) ≥ 2.6^{3} if A = G_{2}(16), both of which are false.

Now suppose i_{1} … i_{4} = 0122. If S < QA_{1}^{4} with the A_{1}^{4} < D_{4}, then L(G) ↓ S = (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{1}/(1 ⊗ 1^{(8)})^{0}/ …, contrary to 6.2(iii). So assume the A_{1}^{4} ≮ D_{4}. Then L(G) ↓ S = (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{4}/ …, ruling out A = G_{2}(16) by 6.2(iv). Thus A = B_{2}(16), and S contains a conjugate b of a. We can choose a subgroup D_{4}D_{4} containing b, and lying in a subgroup D_{8} of G, such that b has weights 4^{2}, −4^{2}, 0^{4} and 3, −3, 1, −1, 4^{2}, −4^{2} on natural modules for the D_{4} factors, and weights 4^{4}, −4^{4}, 3, −3, 1, −1, 0^{4} on the natural D_{8}-module. Hence C_{D_{8}}(b)^{0} = A_{1}A_{1}A_{3}T_{3}. Moreover, from the above description of L(G) ↓ A_{1}^{5} we calculate that dim C_{L(G)}(b) = 24, so it follows that C_{G}(a)^{0} = A_{1}A_{1}A_{3}T_{3}. Now S < QA_{1}^{4} < A_{1}A_{1}A_{3}. Hence Q = 1 and S < S̄ < A_{1}^{4}, where S̄ ≅ SL_{2}(K) has all weights on L(G) less than q.

Next consider i_{1} … i_{4} = 0112. If S < QA_{1}^{4} with the A_{1}^{4} not in D_{4}, then the usual T_{1} argument works. Otherwise, we find that L(G) ↓ S has no 3-fold tensor factors, and has 1 ⊗ 1^{(8)}/1^{0}; this is not possible, by 6.1.

To complete the k = 4 case, let i_{1} … i_{4} = 0022. The T_{1} argument works if S < QA_{1}^{4} with A_{1}^{4} < D_{4}, so suppose this is not the case. If A = B_{2}(16), then we work out C_{G}(a)^{0} as above, and find that it is A_{3}A_{3}T_{2}. Then S < S̄ < A_{1}^{4} < A_{3}A_{3}, where S̄ all weights less than q. Now suppose A = G_{2}(16). Here we find that

L(G) ↓ S = (1 ⊗ 1^{(8)})^{4}/(1^{(2)} ⊗ 1^{(4)})^{4}/(1 ⊗ 1^{(4)})^{16}/(1^{(4)})^{24}/(1^{(8)})^{6}/(1^{(2)})^{6}/1^{24}/0^{32}.

By 6.1, the factors (1 ⊗ 1^{(8)})^{4} must occur in the restriction of a composition factor 10 ⊗ 10^{(8)} or 10 ⊗ 01^{(4)} of L(G) ↓ A.

Suppose there is a factor 10⊗10^{(8)}. This restricts to S as (1⊗1^{(8)})^{4}/(1^{(8)})^{4}/1^{4}/0^{4}. The remaining factors (1 ⊗ 1^{(4)})^{16} in L(G) ↓ S must come from (10 ⊗ 10^{(4)})^{4} in L(G) ↓ A, and this restricts to S as (1⊗1^{(4)})^{16}/1^{16}/(1^{(4)})^{16}/0^{16}. Now the remaining factors (1^{(2)} ⊗ 1^{(4)})^{4} must come from 10^{(2)} ⊗ 10^{(4)}, which restricts to S as (1^{(2)} ⊗ 1^{(4)})^{4}/(1^{(2)})^{4}/(1^{(4)})^{4}/0^{4}. Of the factors in L(G) ↓ S, there remain 1^{4}/(1^{(2)})^{2}/(1^{(4)})^{4}/(1^{(8)})^{2}/0^{8}, of total dimension 32. These factors must come from composition factors of L(G) ↓ A which are twists of 10 (say a of them), 01 (say b of them) and 00. Since 10 ↓ S = 1^{2}/0^{2} and 01 ↓ S = 1^{4}/1^{(2)}/0^{4} (see 6.1), counting 2-dimensional composition factors for S gives 2a + 5b = 12, while counting trivial composition factors yields 2a + 4b = 8. Therefore b = 4, forcing a to be negative, a contradiction. This completes the argument when L(G) ↓ A has a composition factor 10 ⊗ 10^{(8)}. The other case, in which there is a composition factor 10 ⊗ 01^{(4)}, is handled by an entirely similar argument.

The case where k = 4 is now settled.

Assume from now on that k = 5. We have A_{1}^{5} < A_{1}D_{4}; take 1^{(2^{i_{1}})} to be the projection of S into the A_{1} not lying in the D_{4}. Then either the usual T_{1} argument works, or we can twist to take i_{1} … i_{5} to be one of the following (in some order):

00022, 00112, 00122, 01112, 01122, 01222, 00123.

Moreover, either the T_{1} argument works, or we have i_{1} = 0 in the second case, and i_{1} ≠ 2 in the fourth case. Let b be an element of order q + 1 in S. In the following table we present some information for each of the above possibilities, which in almost all cases is sufficient to obtain a contradiction:

i_{1}, i_{2} … i_{5} some comp. factors of L(G) ↓ S C_{G}(b)′
0, 0022  (1^{(2)} ⊗ 1^{(8)})^{1}/(1 ⊗ 1^{(8)})^{2}  A_{1}D_{4}
2, 0002  (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{1}  A_{2}D_{4}
0, 0112  (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{4}  A_{1}A_{4} or A_{1}^{2}A_{2}A_{3}
0, 0122  (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{1}/(1^{(2)} ⊗ 1^{(8)})^{0}
1, 0022  (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{8}/(1 ⊗ 1^{(2)})^{0}  A_{2}D_{4}
2, 0012  (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{4}  A_{1}A_{1}A_{3}
0, 1112  (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{6}/(1 ⊗ 1^{(2)})^{0}  A_{2}D_{4}
1, 0112  (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{2}  A_{1}A_{4} or A_{1}^{2}A_{2}A_{3}
0, 1122  (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{8}/(1 ⊗ 1^{(2)})^{0}  A_{1}D_{4}
1, 0122  (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{1}/(1 ⊗ 1^{(8)})^{0}
2, 0112  (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{4}  A_{2}A_{3}
0, 1222  (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{6}  A_{2}D_{4}
1, 0222  (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{6}  A_{2}D_{4}
2, 0122  (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{1}  A_{1}A_{3}
0, 0123  1 ⊗ 1^{(2)} ⊗ 1^{(4)} ⊗ 1^{(8)}
1, 0023  (1^{(2)} ⊗ 1^{(4)} ⊗ 1^{(8)})^{2}  A_{2}A_{3}
2, 0013  (1^{(2)} ⊗ 1^{(4)} ⊗ 1^{(8)})^{2}  A_{1}A_{3}
3, 0012  (1^{(2)} ⊗ 1^{(4)} ⊗ 1^{(8)})^{2}  A_{1}A_{1}A_{3}

In each case, using 6.1 and 6.2, we see that the information in the composition factors column rules out A = G_{2}(16), and also rules out A = B_{2}(16) in rows 4, 10 and 15 of the table. Therefore A = B_{2}(16) (and we are not in row 4, 10 or 15), and the element b ∈ S is conjugate to a. We know that S < QA_{1}^{5} < C_{G}(a)′. This implies that C_{G}(a)′ is A_{1}D_{4}, A_{2}D_{4} or A_{1}^{2}A_{2}A_{3}; the last case is ruled out by 3.4, and in the first case, S < S̄ < A_{1}D_{4} with S̄ ≅ SL_{2}(K) having all weights on L(G) less than q. Thus C_{G}(a)′ = A_{2}D_{4}. Then S < (QA_{1})A_{1}^{4} < A_{2}D_{4} with QA_{1} < A_{2} and A_{1}^{4} < D_{4}. If i_{1} > 0 then using 1.2 we have S < S̄ < A_{2}D_{4} with all weights of S̄ less than q. The surviving cases are those in rows 7 and 12 of the table. For row 7, twist to take i_{1}, i_{2} … i_{5} = 3, 0001; then S < S̄ < A_{2}D_{4}, where S̄ has highest weight 16 on L(G), and has no composition factor of high weight 1. The conclusion now follows from 1.4. For row 12, we twist to 2, 3000 and apply the same argument. ♠

To complete the proof of Theorem 1, it remains to handle case (iii) of 6.6.

Lemma 6.15 Case (iii) of 6.6 does not hold if q ≥ 32.

Proof Assume 6.6(iii) holds, so S < QA_{1}^{2}B_{1}^{2} < A_{1}(QA_{1}B_{1}^{2}) < A_{1}D_{5}. Moreover, C_{G}(a)′ = A_{1}D_{5}: for otherwise by 6.5 and 6.7, we are in a different case of 6.6, already ruled out. Thus C_{G}(a)^{0} = A_{1}D_{5}T_{2}, of dimension 50.

If S projects trivially to one of the B_{1} factors then S < QA_{1}^{4} (as B_{1} < A_{1}^{2}), contrary to 6.13 and 6.14. Hence S projects nontrivially to both the B_{1} factors (and with distinct twists, by 6.6). Let the nontrivial projections of S to the factors A_{1}(A_{1}B_{1}B_{1}) < A_{1}D_{5} be 1^{(2^{i_{1}})}, …, 1^{(2^{i_{k}})}, written in the order of the factors given (so 1^{(2^{i_{k−1}})}, 1^{(2^{i_{k}})} are the projections to the B_{1} factors). Take i_{k−1} < i_{k}.

If k = 2, then the usual T_{1} argument goes through, so assume k ≥ 3. Also, if k = 4 and i_{1}, …, i_{4} are all distinct, then by 6.8(ii), L(G) ↓ S has a composition factor 1^{(2^{i_{1}})} ⊗ … ⊗ 1^{(2^{i_{4}})} but no composition factor 1^{(2^{i_{1}})} ⊗ 1^{(2^{i_{2}})} ⊗ 1^{(2^{i_{4}})}, contrary to 6.2(i,ii).

Thus if k = 4 then i_{1}, …, i_{4} are not all distinct. When i_{r} ≤ e − 3 for all r, the usual T_{1} argument goes through. As e ≥ 5 by hypothesis, we may therefore apply 6.4 to take one of the following to hold:

(1) e = 5 and i_{1} … i_{k} = 0013, 0023, 0113 or 013 (in some order);

(2) e = 6 and i_{1} … i_{k} = 0024 or 024.

Consider first case (1). We have S < A_{1}(QA_{1}B_{1}^{2}) < A_{1}D_{5}. Given i_{1}, …, i_{k}, the composition factors of L(G) ↓ S can be worked out from the restrictions L(G) ↓ A_{1}D_{5} and V_{D_{5}}(λ_{m}) ↓ A_{1}B_{1}^{2} (m = 1, 4, 5) given in the proof of 6.8(ii). From 6.8(ii), we see that if neither i_{k−1} nor i_{k} is 3, then the usual T_{1} argument goes through. Hence we may take it that i_{k} = 3. In the table below, for each possibility for (i_{1}, …, i_{k}), we give various composition factors of L(G) ↓ S and their multiplicities; we also give the dimension of C_{G}(b), where b is an element of order q + 1 in S. In the left hand column, a symbol “-” indicates a trivial projection.

\begin{matrix}(i_{1}, …, i_{k}) & L(G) ↓ S & dim C_{G}(b) \\ (0, 0, 1, 3) & (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{8}/(1 ⊗ 1^{(8)})^{0}/.... & 24 \\ (0, 1, 0, 3) & (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{4}/(1^{(2)} ⊗ 1^{(16)})^{3}/(1 ⊗ 1^{(2)})^{6}/1^{8}/.... & 20 \\ (1, 0, 0, 3) & (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{4}/(1^{(2)} ⊗ 1^{(16)})^{3}/(1 ⊗ 1^{(2)})^{6}/1^{8}/.... & 20 \\ (0, 0, 2, 3) & (1 ⊗ 1^{(4)} ⊗ 1^{(8)})^{8}/(1 ⊗ 1^{(4)})^{0}/.... & 24 \\ (0, 2, 0, 3) & (1 ⊗ 1^{(4)} ⊗ 1^{(8)})^{4}/(1^{(2)} ⊗ 1^{(4)} ⊗ 1^{(8)})^{2}/(1^{(4)} ⊗ 1^{(8)})^{4}/.... & 16 \\ (2, 0, 0, 3) & (1 ⊗ 1^{(4)} ⊗ 1^{(8)})^{4}/(1^{(2)} ⊗ 1^{(4)} ⊗ 1^{(8)})^{2}/(1^{(4)} ⊗ 1^{(8)})^{4}/.... & 16 \\ (0, 1, 1, 3) & (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{4}/(1 ⊗ 1^{(4)} ⊗ 1^{(8)})^{2}/(1 ⊗ 1^{(8)})^{4}/.... & 16 \\ (1, 0, 1, 3) & (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{4}/(1 ⊗ 1^{(4)} ⊗ 1^{(8)})^{2}/(1 ⊗ 1^{(8)})^{4}/.... & 16 \\ (1, 1, 0, 3) & (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{8}/(1 ⊗ 1^{(2)})^{0}/.... & 32 \\ (0, −, 1, 3) & (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{4}/(1 ⊗ 1^{(2)})^{0}/.... & 34 \\ (−, 0, 1, 3) & (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{8}/(1 ⊗ 1^{(2)})^{0}/.... & 34 \\ (1, −, 0, 3) & (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{8}/(1 ⊗ 1^{(2)})^{0}/.... & 38 \\ (−, 1, 0, 3) & (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{8}/(1 ⊗ 1^{(2)})^{0}/.... & 38 \end{matrix}

If A = B_{2}(q) then S contains a conjugate b of a; but C_{G}(a) has dimension 50, which conflicts with the dimension of C_{G}(b) given in the table.

Therefore A = A_{2}^{ϵ}(q). By 6.2(ii,vi), the information given in the second column of the table rules out all possibilities for (i_{1}, …, i_{k}) except (0, 1, 0, 3) and (1, 0, 0, 3). In these cases we have

L(G) ↓ S = (1 ⊗ 1^{(2)} ⊗ 1^{(8)})^{4}/(1^{(2)} ⊗ 1^{(16)})^{3}/(1 ⊗ 1^{(2)})^{6}/1^{8}/....

The 3-fold tensor factors force L(G) ↓ A to have four composition factors of the form α ⊗ β^{(2)} ⊗ γ^{(8)} with α, β, γ ∈ {10, 01}. As 1^{(2)} ⊗ 1^{(16)} occurs with odd multiplicity, it must occur as a composition factor in a self-dual factor of L(G) ↓ A, which must therefore by 6.1 be 11⊗11^{(8)}. Now (11⊗11^{(8)}) ↓ S and the four factors (α ⊗ β^{(2)} ⊗ γ^{(8)}) ↓ S account for all eight composition factors 1 in L(G) ↓ S. However, two of the composition factors 1 ⊗ 1^{(2)} are still to be accounted for in other composition factors of L(G) ↓ A. By 6.1, there is no irreducible A-module V such that V ↓ S has 1 ⊗ 1^{(2)} but not 1, so this is impossible.

Now consider case (2), in which e = 6 and i_{1} … i_{k} = 0024 or 024. The usual T_{1} argument works unless i_{k} = 4. As above, we calculate the following information.

\begin{matrix}(i_{1}, …, i_{k}) & L(G) ↓ S & dim C_{G}(b) \\ (0, 0, 2, 4) & (1 ⊗ 1^{(4)} ⊗ 1^{(16)})^{8}/(1 ⊗ 1^{(4)})^{0}/.... & 20 \\ (0, 2, 0, 4) & (1 ⊗ 1^{(4)} ⊗ 1^{(16)})^{4}/(1^{(2)} ⊗ 1^{(4)} ⊗ 1^{(16)})^{2}/(1^{(4)} ⊗ 1^{(16)})^{4}/.... & 16 \\ (2, 0, 0, 4) & (1 ⊗ 1^{(4)} ⊗ 1^{(16)})^{4}/(1^{(2)} ⊗ 1^{(4)} ⊗ 1^{(16)})^{2}/(1^{(4)} ⊗ 1^{(16)})^{4}/.... & 16 \\ (0, −, 2, 4) & (1 ⊗ 1^{(4)} ⊗ 1^{(16)})^{8}/(1 ⊗ 1^{(4)})^{0}/.... & 34 \\ (−, 0, 2, 4) & (1 ⊗ 1^{(4)} ⊗ 1^{(16)})^{8}/(1 ⊗ 1^{(4)})^{0}/.... & 34 \\ (2, −, 0, 4) & (1 ⊗ 1^{(4)} ⊗ 1^{(16)})^{8}/(1 ⊗ 1^{(4)})^{0}/.... & 34 \\ (−, 2, 0, 4) & (1 ⊗ 1^{(4)} ⊗ 1^{(16)})^{8}/(1 ⊗ 1^{(4)})^{0}/.... & 34 \end{matrix}

Now we obtain a contradiction as before. This completes the proof of the lemma. ♠

Lemma 6.16 Case (iii) of 6.6 does not hold if q = 16.

Proof Assume q = 16 and 6.6(iii) holds. Then A = B_{2}(16) or G_{2}(16). As in the previous lemma, dim C_{G}(a) = 50.

Again let the nontrivial projections of S to the factors A_{1}(A_{1}B_{1}B_{1}) < A_{1}D_{5} be 1^{(2^{i_{1}})}, …, 1^{(2^{i_{k}})}. As in the previous lemma, the projections to the B_{1} factors are nontrivial and distinct. Twisting if necessary, we can take i_{1}i_{2} ∈ {00, 01, 02, 11, 12, 22, 0−, 1−} and i_{3}i_{4} ∈ {01, 02, 12}. Moreover, the T_{1} argument goes through in all cases where i_{3}i_{4} = 01, except when i_{1}i_{2} = 12 or 22.

For all the possibilities for i_{1} … i_{k}, we calculate L(G) ↓ S and dim C_{G}(b) as in the previous lemma. In all cases we find that dim C_{G}(b) < 50, ruling out A = B_{2}(16) in the usual way. Therefore A = G_{2}(16). If i_{3}i_{4} ≠ 02 or i_{1}i_{2} ∉ {02, 0−}, we find that L(G) ↓ S is either (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{r}/.... with r < 8, or (1 ⊗ 1^{(2)} ⊗ 1^{(4)})^{8}/(1^{(2^{i})} ⊗ 1^{(2^{j})})^{0}/.... for some i, j ∈ {0, 1, 2}. Both possibilities contradict 6.2(ii,iv).

Finally, consider the cases where i_{3}i_{4} = 02 and i_{1}i_{2} = 02 or 0−. Assume first (i_{1}, i_{2}, i_{3}, i_{4}) = (0, 2, 0, 2). Then we find

L(G) ↓ S = (1 ⊗ 1^{(2)})^{2}/(1 ⊗ 1^{(4)})^{12}/(1 ⊗ 1^{(8)})^{6}/(1^{(2)} ⊗ 1^{(8)})^{3}/(1^{(4)} ⊗ 1^{(8)})^{2}/....

The composition factor 1^{(2)} ⊗1^{(8)} appears with odd multiplicity. By 6.1, the only way this can happen is if L(G) ↓ A has a composition factor 01 ⊗ 01^{(4)} (of dimension 196). However, (01 ⊗ 01^{(4)}) ↓ S has 1 ⊗ 1^{(4)} with multiplicity 16, whereas it only occurs with multiplicity 12 in L(G) ↓ S, a contradiction. Similar contradictions are reached for the other possibilities for (i_{1}, …, i_{4}). ♠

This completes the proof of Theorem 1.

7 Proof of Theorem 4

Let G be a simple algebraic group over the algebraically closed field K of characteristic p. Let X = X(q) < G with q = p^{e} and X not of the same type as G, and suppose that X is irreducible on each G-composition factor of L(G). We aim to show that X and G are as in the conclusion of Theorem 4. We may as well assume that G is of adjoint type; as X is irreducible on each G-composition factor of L(G), we then have Z(X) = Z(G) = 1. The G-composition factors of L(G) are given by 1.10.

We now embark upon the proof of Theorem 4. We first dispose of classical groups.

Lemma 7.1 Theorem 4 holds if G is classical.

Proof First observe that X does not lie in a parabolic subgroup of G: for if X < P, a parabolic, then X leaves invariant the series 0 < L(R_{u}(P) < L(P) < L(G), hence cannot be irreducible on all nontrivial G-composition factors of L(G), by 1.10. We now apply [ST1, Theorem 1]; this implies that X lies in a closed, connected, simple subgroup X̄ of G of the same type as X. Then X fixes L(X̄).

We now claim that dim X̄ < dim G−2. For suppose dim X̄ ≥ dim G−2. If a maximal unipotent subgroup of X̄ is also one of G, then X̄ = G (see [Se4, 1.6] for instance), which is not the case. The dimension of a simple algebraic group is equal to the rank of the group plus twice the dimension of a maximal unipotent subgroup; it follows that X̄ is a subgroup of maximal rank in G, and that a maximal unipotent subgroup of X̄ has dimension just one less that a maximal unipotent subgroup of G. It is trivial to check that this cannot happen.

It follows from the previous paragraph and 1.10 that (G, p) must be (B_{n}, 2) or (C_{n}, 2). Moreover, X̄ is irreducible on V_{G}(λ_{1}) and V_{G}(λ_{2}). At this point [Se1, Theorem 1] determines all possibilities for X̄: either X̄ = D_{n}, or n = 3 and X̄ = G_{2}, as in the conclusion of Theorem 2. ♠

In view of 7.1, we assume from now on that G is of exceptional type. Let V be a nontrivial G-composition factor of L(G), so that X is irre-

ducible on V. Write V = V_{G}(λ) with λ as in 1.10(ii) (up to twists).

If P_{X} = Q_{X}L_{X} is a parabolic subgroup of X with unipotent radical Q_{X} and Levi subgroup L_{X}, then by [BT], there is a parabolic subgroup P = QL of G containing P_{X}, such that Q_{X} ≤ Q, the unipotent radical of P. The next result follows from the argument in [ST1, p.565] (third paragraph).

Lemma 7.2 We have C_{V} (Q_{X}) = C_{V} (Q), and P_{X} and P are both irreducible on this space. The high weight of C_{V} (Q) as an L-module is the restriction of λ to L. Assuming this action is nontrivial, we have λ = λ_{i} for some i, and if L_{0} is the simple factor of L involving the root α_{i}, then the possibilities for C_{V} (Q) ↓ L_{0} are as in Table 5 below. Moreover, if (G, L_{0}) = (E_{7}, E_{6}) or (E_{6}, D_{5}), then Q and Q_{X} are abelian.

Table 5

G L_{0}  C_{V} (Q) ↓ L_{0} (up to auts. of L_{0})
E_{8} D_{7} or A_{r} (r ≤ 7)  V_{L_{0}}(λ_{1})
E_{7} E_{6}, D_{5} or A_{r} (r ≤ 6)  V_{L_{0}}(λ_{1})
E_{6} D_{5}, D_{4}  V_{L_{0}}(λ_{4})
A_{r} (r ≤ 4)  V_{L_{0}}(λ_{1})
F_{4} B_{3}, C_{3}(p = 2), or A_{r} (r ≤ 2) V_{L_{0}}(λ_{1})

Lemma 7.3 X is not A_{1}(q).

Proof Suppose X = A_{1}(q), and let t be a generator for a Cartan subgroup of X. By the representation theory of X, C_{V} (t) has dimension 1 unless V is the Steinberg module for X; the latter is impossible as dim V cannot be equal to q (see 1.10). However, t lies in a maximal torus of G, so C_{V} (t) must certainly be bigger than 1, which is a contradiction. ♠

Lemma 7.4 X is not A_{2}^{ϵ}(q).

Proof Suppose that X = A_{2}^{ϵ}(q). Now V is a self-dual X-module, so by [ST1], V lifts to a self-dual A_{2}(K)-module. This is a tensor product of twists of modules of the form V_{A_{2}}(aa), where 0 ≤ a ≤ p − 1.

Now by 1.8, either W_{X}(aa) = V_{X}(aa), of dimension (a + 1)^{3}, or W_{X}(aa) has two composition factors, namely V_{X}(aa) and V_{X}(p − a − 2, p − a − 2); in the latter case, 2a + 2 > p ≥ a + 2 and V_{X}(p − a − 2, p − a − 2) = W_{X}(p − a − 2, p − a − 2), whence dim V_{X}(aa) = (a + 1)^{3} − (p − a − 1)^{3}.

If p = 2 then V is a tensor product of twists of V_{X}(11), which has dimension 8; but from 1.10 we see that dim V is not a power of 8. Hence p > 2.

Suppose first that G = E_{8}. Then dim V = 248 = 8.31, so there is a restricted module V_{X}(aa) of dimension 2^{r}.31 (r ≤ 3). Therefore 2^{r}.31 must be equal to (a + 1)^{3} − (p − a − 1)^{3}. This has a factor 2a + 2 − p, which, since p is odd, forces 2a + 2 − p = 1. Then 2^{r}.31 = (a + 1)^{3} − a^{3}, which is not possible.

Now let G = E_{7}. Then dim V = 133 = 7.19 (recall p ≠ 2). As 133 is not a difference of cubes, we must have V ↓ X ≅ V_{X}(aa) ⊗ V_{X}(bb)^{(p^{i})} with (a+1)^{3} −(p−a−1)^{3} = 7 and (b+1)^{3} −(p−b−1)^{3} = 19. This is impossible.

Next consider G = E_{6} or F_{4}. Here dim V = 52, 77 or 78, none of which is the product of a difference of cubes.

Finally, if G = G_{2}, then as 14 is not a product of differences of cubes, we must have p = 3 and dim V = 7. Subgroups A_{2}^{ϵ}(q) of G_{2} are identified by [LST, Theorem 2]: they lie in maximal rank connected subgroups A_{2}. However these are not irreducible on both V_{G}(λ_{1}) and V_{G}(λ_{2}), so no examples arise here. ♠ Lemma 7.5 Theorem 4 holds if G = G_{2}.

Proof Suppose G = G_{2}. By 7.3, rk(X) ≥ 2 (recall that rk(X) denotes the rank of the simple algebraic group corresponding to X).

Since a parabolic of X lies in a parabolic of G (by [BT]), we deduce that X is of type A_{2}^{ϵ}(q), B_{2}(q), G_{2}(q), ^{3}D_{4}(q) or ^{2}A_{3}(q). The first case is out by 7.4, and X ≠ G_{2}(q) as X is assumed not to have the same type as G; also X ≠ ^{3}D_{4}(q) since^{3}D_{4}(q) has no nontrivial representation of dimension 7. By the argument in [ST1, p.565] (third paragraph), up to twists, V ↓ X is 10, 01 or 11 if X = B_{2}(q), and is 101 or 010 if X = ^{2}A_{3}(q). As dim V = 14 − 7δ_{p,3}, the only possibility is that X = ^{2}A_{3}(q), V ↓ X = 101 with p = 2. But here X has a parabolic subgroup Q_{X}L_{X} with L′_{X} ≅ A_{1}(q^{2}) acting on C_{V} (Q_{X}) as 1 ⊗ 1^{(q)}, which does not agree with the action of a parabolic of G containing Q_{X}L_{X}. ♠

We assume from now on that G ≠ G_{2}.

Lemma 7.6 X is not B_{2}^{ϵ}(q) or G_{2}^{ϵ}(q).

Proof Suppose that X = B_{2}^{ϵ}(q) or G_{2}^{ϵ}(q), and let V ↓ X ≅ V_{X}(ab) ≅ V_{X}(a_{0}b_{0}) ⊗ … with a_{0}, b_{0} ≤ p − 1. At several points in the proof we make use of the sum formula in [An], which can be used to calculate the dimensions of many modules V_{X}(cd) with c, d reasonably small.

First assume p = 2 or 3. By [GS] for X = G_{2}^{ϵ}(q), and by the sum formula [An] for X = B_{2}^{ϵ}(q), the possibilities for dim V_{X}(a_{0}b_{0}) are as follows:

\begin{matrix}X & p & a_{0}b_{0} & dim V_{X}(a_{0}b_{0}) \\ B_{2}^{ϵ}(q) & 2 & 10, 01, 11 & 4, 4, 16 (resp.) \\ G_{2}(q) & 2 & 10, 01, 11 & 6, 14, 64 \\ B_{2}(q) & 3 & 10, 20, 01, 11, 21, 02, 12, 22 & 5, 14, 4, 16, 40, 10, 25, 81 \\ G_{2}^{ϵ}(q) & 3 & 10, 20, 01, 11, 21, 02, 12, 22 & 7, 27, 7, 49, 189, 27, 189, 729 \end{matrix}

As dim V is a product of dimensions in the table, we have a contradiction.

Thus p > 3. At this point we make use of 7.2. Take a parabolic subgroup P_{X} = Q_{X}L_{X} of X, with L′_{X} ≅ A_{1}(q). Then P_{X} lies in a parabolic subgroup QL of G with Q_{X} ≤ Q, and L_{X} is irreducible on the space C_{V} (Q), which is given as L-module in Table 5. We can choose P_{X} so that C_{V} (Q) ↓ L_{X} ≅ V_{L_{X}} (a) or V_{L_{X}} (b).

We claim that one of the following holds:

\begin{matrix}G = E_{8} : & a_{0}, b_{0} ≤ 7 \\ G = E_{7} : & a_{0}, b_{0} ≤ 6 \\ G = E_{6} : & a_{0}, b_{0} ≤ 4 \\ G = F_{4} : & a_{0}, b_{0} ≤ 6. \end{matrix}

To see this we argue as follows. If the claim is false, then from Table 5 we see that L_{0} = D_{7}, E_{6}, D_{5}, D_{5} or D_{4} and L′_{X} is irreducible on V_{L_{0}}(λ), where λ = λ_{1}, λ_{1}, λ_{1}, λ_{4} or λ_{4}, respectively. This is impossible for V_{D_{r}}(λ_{1}) and V_{D_{4}}(λ_{4}), by 1.1, and for V_{D_{5}}(λ_{4}), by [LS2, 2.13]. Finally, if L_{0} = E_{6} then X ≠ G_{2}(q) since Q_{X} is abelian by 7.2, so X = B_{2}(q). We shall see in the next paragraph that V_{X}(ab) is restricted; hence a or b must be 26. But it is easy to see that this forces dim V > dim L(G) = 133, a contradiction. This establishes the claim.

As p > 3, dim V is 248, 133, 78 or 52. The irreducible KX-modules of dimension 8 or less, dividing dim V, are as follows:

X = B_{2}(q) : V_{X}(01) of dimension 4,
X = G_{2}(q) : V_{X}(10) of dimension 7.

Hence, if V_{X}(ab) is not restricted, then it is a tensor product of one of the above modules with a twist of a restricted module V_{X}(cd); the latter module has dimension 62 or 13 (for X = B_{2}(q), G = E_{8} or F_{4}), or dimension 19 (for X = G_{2}(q), G = E_{7}). When X = G_{2}(q), G = E_{7}, we have C_{V} (Q) ↓ L_{X} ≅ V_{L_{X}} (1) ⊗ V_{L_{X}} (c)^{(p^{i})} or V_{L_{X}} (d)^{(p^{i})}, whence c ≤ 2 and d ≤ 6 by 7.2. It is easy to see that no such module V_{X}(cd) has dimension 19: use [GS] for the small weights, and count conjugates for the others. A similar argument applies when X = B_{2}(q). Here c ≤ 7, d ≤ 3. Moreover, since Z(X) = 1 and V ↓ X ≅ 01 ⊗ cd^{(p^{i})}, the module V (cd) must admit the action of the simply connected group of type B_{2}, and hence d must be odd; in particular, the central element −1 has determinant 1 on V (cd), so dim V (cd) is even, hence equal to 62. Now [Pr] shows that all weights of the Weyl module W_{B_{2}}(cd) appear as weights of V_{B_{2}}(cd), and counting conjugates of such weights shows that the only possibilities with dim V_{B_{2}}(cd) ≤ 62 are among the following:

cd = 01, 03, 11, 13, 21, 23, 31, 41.

The dimensions of these modules can be calculated using the sum formula in [An]; we find that none of them is equal to 62.

We conclude that V_{X}(ab) is restricted (that is, a = a_{0}, b = b_{0}). Now suppose X = B_{2}(q). As Z(X) = 1, b is even. The dimensions of V_{X}(ab) with a, b ≤ 7 can be calculated using the sum formula in [An], and we find that the only cases where this dimension is less than or equal to 248 are as follows:

\begin{matrix}ab & dim V_{B_{2}}(ab) \\ 06 & 84 \\ 16 & 231 \\ 04 & 35 \\ 14 & 105 \\ 24 & 220 − 105δ_{p,5} − 71δ_{p,7} \\ 02 & 10 \\ 12 & 35 \\ 22 & 81 − 13δ_{p,5} − 10δ_{p,7} \\ 32 & 154 − 68δ_{p,5} \\ 42 & 260 − 54δ_{p,7} − 81δ_{p,11} \\ 52 & 199 (p = 7) \\ 10 & 5 \\ 20 & 14 − δ_{p,5} \\ 30 & 30 − 5δ_{p,7} \\ 40 & 55 − δ_{p,7} \\ 50 & 91 − 30δ_{p,11} \\ 60 & 140 − 14δ_{p,11} − 55δ_{p,13} \\ 70 & 204 − 5δ_{p,11} − 30δ_{p,13} \end{matrix}

None of these dimensions is equal to dim V, a contradiction.

Now let X = G_{2}(q), and recall that p > 3. Again, application of the sum formula shows that the only cases where V_{X}(ab) (with a, b ≤ 7) has dimension at most 248 are:

\begin{matrix}ab & dim V_{G_{2}}(ab) \\ 40 & 182 − 27δ_{p,11} \\ 30 & 77 \\ 21 & 189 − 64δ_{p,11} \\ 20 & 27 − δ_{p,7} \\ 13 & 244 (p = 13) \\ 12 & 286 − 189δ_{p,5} − 38δ_{p,7} \\ 11 & 64 − 26δ_{p,7} \\ 10 & 7 \\ 03 & 273 − 77δ_{p,5} − 125δ_{p,11} \\ 02 & 77 \\ 01 & 14 \end{matrix}

The only possibility where dim V_{X}(ab) = dim V is

G = E_{8}, p = 7 and V ↓ X ≅ V_{X}(12).

In this case, let α, β be fundamental roots for a G_{2} root system, and let t = h_{α}(−1) ∈ X. We calculate dim C_{V} (t), where V = V_{X}(12). (We shall then obtain a contradiction by comparing this with the known possibilities for dim C_{L(G)}(t).) All the weights appearing in V and their multiplicities are listed in [GS]. The dominant weights which are subdominant to 12 are

12, 40, 21, 02, 30, 11, 20, 01, 10, 00,

and these appear in V_{X}(12) with multiplicities 1,1,2,2,3,4,6,6,8,8, respectively. The orbits of these weights under the Weyl group of X are as follows:

orbit of 12 : 12, 1(−3), 8(−3), 8(−5), 7(−5), 7(−2) and negatives
orbit of 40 : 40, 4(−4), 8(−4) and negatives
orbit of 21 : 21, 2(−3), 7(−3), 7(−4), 5(−4), 5(−1) and negatives
orbit of 02 : 02, 6(−2), 6(−4) and negatives
orbit of 30 : 30, 3(−3), 6(−3) and negatives
orbit of 11 : 11, 1(−2), 5(−2), 5(−3), 4(−3), 4(−1) and negatives
orbit of 20 : 20, 2(−2), 4(−2) and negatives
orbit of 01 : 01, 3(−1), 3(−2) and negatives
orbit of 10 : 10, 1(−1), 2(−1) and negatives

The dimension of C_{V} (t) is equal to the number of weights appearing in V with first coordinate even. Hence we find dim C_{V} (t) = 124. But C_{G}(t) is D_{8} or A_{1}E_{7}, whence dim C_{L(G)}(t) = dim C_{G}(t) = 120 or 136, a contradiction. This completes the proof of the lemma. ♠

Lemma 7.7 Theorem 4 holds if G = F_{4}.

Proof Let G = F_{4}. By 7.3, 7.4 and 7.6, rk(X) ≥ 3. First suppose that q > 2. Then by [LST, Theorem 2] there is a connected simple subgroup X̄ of G containing X and of the same type as X. As X fixes L(X̄), we deduce from 1.10 that L(X̄) has all its X̄-composition factors of dimension 52 − 26δ_{p,2}. By 1.10, there is no such group X̄.

Now let q = 2. The only possible subgroups X having an irreducible module of dimension 26 are D_{4}^{ϵ}(2) and B_{4}(2) ≅ C_{4}(2), the unique such module being V_{X}(λ_{2}). It is enough to rule out X = D_{4}^{ϵ}(2) (since B_{4}(2) contains D_{4}(2) acting irreducibly on V (λ_{2})). So suppose X = D_{4}^{ϵ}(2) and pick a parabolic subgroup P_{X} of X of type Q_{X}D_{3}^{ϵ}(2) if ϵ = ±, and of type Q_{X}L_{2}(8) if ϵ = 3. By [BT], P_{X} < QL, a parabolic of G. If ϵ = ±, this must be a B_{3}- or C_{3}-parabolic. Moreover, by 7.2, L must act irreducibly and nontrivially on C_{V} (Q) = C_{V} (Q_{X}), for V = V_{G}(λ_{i}), i = 1, 4. But this is not the case, as L in fact acts trivially on C_{V} (Q) for one of these values of i.

Finally, suppose ϵ = 3. Then the factor L_{2}(8) acts trivially on C_{V} (Q_{X}) (where V = V_{X}(λ_{2})), so QL is a B_{2}- or A_{1}-parabolic as it must be trivial on both 26-dimensional composition factors of L(G). The 8-dimensional module V_{X}(λ_{1}) restricts to L_{2}(8) as 1^{(2)} ⊗ 1^{(4)}/1/1, hence V_{X}(λ_{2}) ↓ L_{2}(8) has a composition factor 1⊗1^{(2)}⊗1^{(4)}. But on restriction to QL, the modules V_{G}(λ_{i}) (i = 1, 4) have all composition factors of dimension at most 5, which is a contradiction. ♠

From now on, we assume that G = E_{6}, E_{7} or E_{8}. Also, by 7.3 - 7.6, rk(X) ≥ 3.

Lemma 7.8 Let L_{0} be a connected simple group of type A_{r} (r ≤ 7), D_{4}, D_{5}, D_{5}, D_{7} or E_{6}, and let W = V_{L_{0}}(μ), where μ = λ_{1}, λ_{1}, λ_{1}, λ_{4}, λ_{1} or λ_{1}, respectively. Suppose that Y = Y (q) is a quasisimple subgroup of L_{0} such that rk(Y) ≥ 2, Y ≠ G_{2}^{ϵ}(q) and Y is irreducible on W. Then one of the following holds (where V ↓ Y is given up to twists):

Y  ν, where V ↓ Y = V_{Y} (ν)  L_{0}
A_{2}^{ϵ}(q)  10, 01, 20, 02, 11, 22,  A_{2}, A_{2}, A_{5}, A_{5}, A_{7−δ_{p,3}} or D_{4}, E_{6},
α ⊗ β^{(p^{i})} ⊗ γ^{(p^{j})}  E_{6} (resp.)
(α, β, γ ∈ {10, 01})
B_{2}^{ϵ}(q)  10, 01, 02, 20, 11, 01 ⊗ 01^{(p^{i})}, A_{4−δ_{p,2}}, A_{3}, D_{5}, D_{7}, D_{5}, D_{5},
10 ⊗ 10^{(2^{i})}, 10 ⊗ 01^{(2^{i})}(p = 2) D_{5}, D_{5} (resp.)
A_{3}^{ϵ}(q)  100, 001, 010  A_{3}, A_{3}, A_{5}
B_{3}(q)  100, 200(p ≠ 7), 001  A_{6−δ_{p,2}}, E_{6}, A_{7} or D_{4}
C_{3}(q) (q odd)  100, 010(p ≠ 3)  A_{5}, D_{7}
B_{4}(q) (q odd)  0001  D_{5}
C_{4}(q)  1000, 0100, 0001(p = 2)  A_{7}, E_{6}, D_{5}
D_{4}^{ϵ}(q)  1000, 0010, 0001  A_{7} or D_{4}
same type as L_{0} μ  L_{0}

Proof The dimension of W is r+1, 8, 10, 16, 14 or 27 in the respective cases for L_{0}. The Y -modules of dimension up to 10 (self-dual if the dimension is 9 or 10) are well known (cf. [Li, 1.1]), and all possibilities are in the table. So assume dim W = 14, 16 or 27. Let W = V_{Y} (λ). If this is a nontrivial tensor product then the tensor factors have dimension at most 9 and either W is in the table or Y = A_{3}^{ϵ}(q), L_{0} = D_{5}, λ = λ_{4} and W ↓ Y = α ⊗ β^{(p^{i})} with α, β ∈ {100, 001}; however, in the latter case, V_{D_{5}}(λ_{1}) ↓ Y must be either 100/001/000^{2} or 010/000^{4}, and in neither case does the spin module V_{D_{5}}(λ_{4}) restrict irreducibly to Y, so this case does not occur. Now assume V_{Y} (λ) is not a nontrivial tensor product; then we may assume λ is restricted.

Suppose Y is not of type L_{0}. When Y = A_{2}^{ϵ}(q), 1.8 shows that the only possibility is λ = 22 (giving dim V_{Y} (λ) = 27). For other types, we argue by counting conjugates of subdominant weights appearing in V_{Y} (λ) that the only possibilities for (Y, λ), with V_{Y} (λ) having dimension between 11 and 27, are among the following:

\begin{matrix}Y & λ \\ B_{2}(q) & 20, 30, 03, 11, 12 \\ B_{3}(q) & 200, 002 \\ C_{3}(q) & 200, 010, 001 \\ B_{4}(q) & 0100, 0001 \\ C_{4}(q) & 0100 \end{matrix}

For Y = B_{2}(q) the tables in the proof of 7.6 exclude all but λ = 20, 11; for the other cases, [BW] excludes all but (Y, λ) = (B_{3}(q), 200), (C_{3}(q), 010 or 001), (B_{4}(q), 0001) and (C_{4}(q), 0100). All of these except (C_{3}(q), 001) (q odd) are in the conclusion; in the exceptional case, the 14-dimensional module V_{C_{3}}(001) is symplectic rather than orthogonal. Note also that p ≠ 7, 3 for the B_{3}(q), C_{3}(q) cases in the table, respectively, as otherwise V_{Y} (λ) does not have correct dimension. ♠

Lemma 7.9 We have V ≅ V_{X}(δ), where δ is as in Table 6 below (given up to twists).

Table 6

X  δ
A_{3}(q) 111, 202, 222
2A_{3}(q) 0a0 (a ≤ 7), 1a1 (a ≤ 5)
2A_{4}(q) 0110, 1111
B_{3}(q) 011, 020, 110, 111, 102, 202, 220
C_{3}(q) 002, 020, 011, 101, 111, 202, 220, 110(p = 2)
2D_{4}(q) 2010, 2001, 1100
3D_{4}(q) a000, 0b00, 1c00 (a ≤ 4, b ≤ 3, c ≤ 3), 2100,
1010, 1110, 2010, 3010, 3110,
1d10, 1d11, 2d10 (p ≤ 3, d ≤ p − 1)
F_{4}(q) 2000, 0010
2F_{4}(q) 0100, 1100

Proof Recall that V is a self-dual G-module. We show first that if V ↓ X = V_{X}(δ), then the weight δ is restricted. If not, then V ↓ X is a nontrivial tensor product. As dim V is 78−δ_{p,3}, 133−δ_{p,2} or 248, we see that one of the tensor factors must have dimension x ≤ 8. Thus X has a self-dual irreducible V_{X}(χ) of dimension x, from which we deduce that (X, χ, x) = (A_{3}^{ϵ}(q), 010, 6), (B_{3}(q), 100, 7), (B_{3}(q), 001, 8), (C_{3}(q), 100, 6) (C_{4}(q), λ_{1}, 8) or (D_{4}^{ϵ}(q), λ_{i}, 8) (i = 1, 3, 4). We have V ↓ X = χ ⊗ γ^{(r)}, where y = dim V_{X}(γ) is 13 or 22, 11 or 19, 31, 13 or 22, 31, or 31, respectively. Excluding the cases (X, y) = (A_{3}^{ϵ}(q), 22) or (B_{3}(q), 31), we see in the usual way (counting conjugates of subdominant weights, etc.) that the only possibility for a self dual Xmodule of dimension y is X = C_{3}(q), γ = λ_{2} with p = 3 and y = 13; but then xy = 78, whereas dim V = 77 for G = E_{6}, p = 3. For the excluded case with X = A_{3}^{ϵ}(q), we have xy = 132, so G = E_{7}, p = 2. The only possibility for γ is 111; but it is easy to see by counting conjugates of subdominant weights that V_{X}(111) has dimension greater than 22. The other excluded case with X = B_{3}(q) is handled by counting conjugates of dominant weights in the usual way. Thus the weight δ is restricted.

Consider first X = A_{3}(q), and let δ = aba. Choose a parabolic P_{X} = Q_{X}L_{X} of X such that L′_{X} ≅ A_{2}(q) and C_{V} (Q_{X}) ↓ L_{X} ≅ V_{L_{X}} (ab). Then from 7.2 and 7.8, we see that ab is as in the conclusion of 7.8 for A_{2}(q), and so

ab = 00, 10, 01, 20, 02, 11, or 22.

Therefore δ = 101, 010, 202, 020, 111 or 222. Clearly 101, 010 and 020 have dimension less than 77, leaving the possibilities in Table 6 for X = A_{3}(q).

For X = A_{r}(q) with r ≥ 4, we first use the same argument with an A_{r−1}-parabolic: if δ = a_{1} … a_{r}, we see that a_{1} … a_{r−1} is equal to 0 … 0, 10 … 0 or 0 … 01 (or, if r = 4, to 010). As V ↓ X is self-dual, we conclude that a_{1} … a_{r} = 10 … 01. Now choose another parabolic Q_{X}^{(1)}L_{X}^{(1)} of type A_{2} × A_{r−3}; then C_{V} (Q_{X}^{(1)}) has dimension 3(r − 2), and we deduce using 7.2 that 3(r −2) ≤ 8. Hence r = 4 and V ↓ X = V_{X}(1001); but dim V_{X}(1001) ≤ 24, which is a contradiction.

Next let X = B_{3}(q), with δ = abc. From an A_{2}-parabolic and 7.8 we get ab = 00, 10, 01, 20, 02, 11 or 22; and from a B_{2}-parabolic we get bc = 00, 10, 01, 20, 02 or 11. Hence δ is in the following list:

001, 002, 100, 101, 102, 010, 011, 200, 201, 202, 020, 110, 111, 220.

If δ = 001, 002, 100, 101, 010 or 200, then V_{X}(δ) appears in a tensor product V_{X}(λ_{i}) ⊗ V_{X}(λ_{j}) with i, j ∈ {1, 3}, hence has dimension less than 77; and if δ = 201 then p ≠ 2 and the acting group is the simply connected group Spin_{7}(q), with centre of order 2, contrary to the fact that Z(X) = 1. All other possibilities for δ are in Table 6.

Now consider X = B_{r}(q) with r ≥ 4 and δ = a_{1} … a_{r}. Observe that r ≤ 8, as can be seen from an easy inductive argument (embed a B_{r−1}- parabolic of X in a parabolic of G, and repeat). Working with parabolics of types A_{r−1} and B_{r−1} as above, we find a_{1} … a_{r−1} = 00...0, 10...0 or 0...01 (or 010 with r = 4), and a_{2} … a_{r} = 00...0 (or 100, 200, 001 with r = 4, or 0001 with r = 5). Hence δ is one of the following:

10...0, 0001, 00001, 1001, 10001, 0100.

The first three cases have dimension 2r + 1 or 2r, 16 and 32 respectively, so are impossible; and the last has dimension less than 77. If δ = 10001, then an A_{2}B_{2}-parabolic acts on the fixed space of its radical as 10 ⊗ 01, so this fixed space has dimension 12, contrary to 7.2. This leaves only δ = 1001; here also p = 2, since otherwise the acting group has centre of order 2 again. But when p = 2, V_{B_{4}}(1001) ≅ V_{B_{4}}(1000) ⊗ V_{B_{4}}(0001) (see [Bor2, 7.6]), of dimension 8.16, so this case cannot occur either.

The cases X = C_{r}(q) (r ≥ 3) or D_{r}(q) (r ≥ 4) are handled in just the same way as B_{r}(q) (for D_{r}(q) we use two A_{r−1}-parabolics).

For X = F_{4}(q) with δ = abcd, use of a B_{3}-parabolic and 7.2 gives abc = 000, 100, 200 or 001, and use of a C_{3}-parabolic gives bcd = 000, 100, 010 or 001(p = 2). Hence δ = 1000, 1001(p = 2), 2000, 0010 or 0001(p = 2). Of these, the first and last have dimension less than 77, and 1001 ≅ 1000⊗0001 ([Bor2, 7.6]), of dimension 26.26; the rest are in Table 6.

When X = E_{6}(q), E_{7}(q) or E_{8}(q), use of suitable parabolics forces δ = λ_{2}, λ_{1} or λ_{8} respectively, whence X has the same type as G, contrary to our original assumption.

This completes the case where X is of untwisted type.

Now suppose X = ^{2}A_{3}(q) and δ = aba. Choose a parabolic P_{X} = Q_{X}L_{X} of X with L′_{X} ≅ A_{1}(q^{2}). Then C_{V} (Q_{X}) ↓ L_{X} = a ⊗ a^{(q)}. Also Q_{X} is nonabelian, so we deduce from 7.2 that dim C_{V} (Q_{X}) = (a + 1)^{2} ≤ 8, whence a ≤ 1. The usual argument using an A_{1}(q)-parabolic shows that b ≤ 7. Now choose a subgroup B ≅ B_{2}(q) of X. Then V ↓ B has a composition factor V_{B}(b, 2a). Using the list of B_{2}-modules given in the proof of 7.6, we see that either a = 0, b ≤ 7 or a = 1, b ≤ 5, as in Table 6.

Now consider X = ^{2}A_{4}(q), δ = abba. There is a parabolic Q_{X}L_{X} with L_{X} ≅ ^{2}A_{2}(q) and C_{V} (Q_{X}) ↓ L_{X} ≅ V_{L_{X}} (bb). Hence by 7.8, we have b ≤ 2; moreover, if b = 2 then dim C_{V} (Q_{X}) = 27, but this is impossible by 7.2, as Q_{X} is non-abelian. Thus b ≤ 1. As in the previous paragraph, we see that a ≤ 1. Clearly 0000 and 1001 have dimension less than 77, leaving 0110 and 1111, as in Table 6. For X = ^{2}A_{r}(q) with r ≥ 5 and δ = a_{1} … a_{r}, an A_{2}(q^{2}) parabolic Q_{X}L_{X} has C_{V} (Q_{X}) ↓ L_{X} = a_{1}a_{2} ⊗ a_{2}a_{1}^{(q)}, whence a_{1} = a_{2} = 0, and a^{2}A_{r−2}(q) parabolic now gives either a_{i} = 0 for all i, or r = 5, δ = 00100; in the latter case, V_{X}(δ) has dimension 20, a contradiction.

Next let X = ^{2}D_{r}(q) (r ≥ 4). Note again that r ≤ 9. Use of 7.8 with parabolics of types ^{2}D_{r−1} and A_{r−2} yield the following possibilities for δ:

λ_{1}, λ_{r−1}, λ_{r}, λ_{1} + λ_{r−1}, λ_{1} + λ_{r}, 01000, 2000, 2010, 2001, 0100, 1100.

The first three possibilities have dimensions 2r, 2^{r−1}, 2^{r−1} respectively, so these cannot hold. For the next two cases, when r ≥ 6 an A_{r−4}(q) × ^{2}A_{3}(q) parabolic rules them out; for r = 5 they are not self dual; and for r = 4 they have dimension less than 77. Finally, the cases with δ = 01000, 2000, 0100 have dimension less than 77 also. This leaves just the possibilities in Table 6.

Similarly, for X = ^{2}E_{6}(q) we use parabolics of types ^{2}A_{5} and ^{2}D_{4} to force δ = λ_{1}, λ_{6}, λ_{2}, λ_{1} + λ_{2} or λ_{6} + λ_{2}. Of these, only λ_{2} is self dual. But this is the high weight of the adjoint module for X, which means that X and G must have the same type, contrary to assumption.

It remains to deal with X = ^{2}F_{4}(q) or ^{3}D_{4}(q). In the first case, V ↓ X = V_{X}(abcd) ≅ V_{X}(ab00) ⊗ V_{X}(00cd). Each tensor factor is either trivial or has dimension at least 26, so one of them, say 00cd, is trivial. Thus V ↓ X = 1000, 0100 or 1100; the first has dimension 26, and the others are in Table 6.

Finally, let X = ^{3}D_{4}(q), and suppose δ = abcd. Pick a parabolic Q_{X}L_{X} with L′_{X} ≅ A_{1}(q^{3}). Then C_{V} (Q_{X}) ↓ L_{X} = a ⊗ c^{(q)} ⊗ d^{(q^{2})}. Moreover Q_{X} is non-abelian, so we deduce from 7.2 and 1.1 that (a + 1)(c + 1)(d + 1) ≤ 8. Therefore, taking a ≥ c ≥ d, we have

(a, c, d) = (a, 0, 0) (a ≤ 7), (a, 1, 0) (a ≤ 3) or (1, 1, 1).

Now take a subgroup C = G_{2}(q) of X, so that V ↓ C has a composition factor V_{C}(a + c + d, b). This module is restricted unless acd = 210, 110 or 111 and p ≤ 3. These cases are included in Table 6 of the statement. In the restricted case, we see from the list of G_{2}-modules given in the proof of 7.6 that

a + c + d, b = 40, 30, 21, 20, 13, 12, 11, 10, 03, 02, or 01.

Hence abcd is as in Table 6. ♠

The proof of Theorem 4 is completed by the next lemma.

Lemma 7.10 Let X and δ be as in Table 6. Then dim V_{X}(δ) is not equal to 77 − δ_{p,3}, 133 − δ_{p,2} or 248.

Proof For many of the possibilities in Table 6, the dimension of V_{X}(δ) is given by [BW] or [GS]. For several of the remaining cases, counting conjugates of subdominant weights in V_{X}(δ) shows that dim V_{X}(δ) > 248. The final surviving cases are handled using the programme described in [GS] to determine the precise dimension of V_{X}(δ). ♠

This completes the proof of Theorem 4.

8 Proof of Theorem 6 and corollaries

We begin with the proof of Theorem 6. As in the hypothesis of Theorem 6, let G be a simple adjoint algebraic group of exceptional type in characteristic p, and σ be a Frobenius morphism such that L = O^{p′}(G_{σ}) is a finite simple group of exceptional Lie type. Let L_{1} be a finite group with socle L (i.e. L ≤ L_{1} ≤ Aut L). Suppose that H is a maximal subgroup of L_{1} such that F^{∗}(H) = X = X(q), a simple group of Lie type over 𝔽_{q}, with q as in the hypothesis of Theorem 1. Assume further that X is not of the same type as G (otherwise (i) of Theorem 6 holds).

By Corollary 5, there is a proper connected subgroup X̄ of G containing X such that X and X̄ fix the same subspaces of L(G).

Let M be the set of all X-invariant subspaces of all G-composition factors of L(G), and define

Y = ⋂_{W ∈M} G_{W} .

As every automorphism of L extends to a morphism G → G, it follows from 1.12(iii) that Y is H⟨σ⟩-stable. Moreover, by Theorem 4, Y is a proper closed subgroup of G, and X < X̄ ≤ Y. Let Z be a maximal connected H⟨σ⟩-invariant proper subgroup of G containing Y^{0}. By the maximality of H we have H = N_{L_{1}}(Z). Clearly Z is reductive (otherwise H normalizes R_{u}(Z)_{σ}, hence is parabolic). Hence X = O^{p′}(Z_{σ}). Consequently (ii) of Theorem 6 holds.

This completes the proof of Theorem 6.

Remark As remarked after Theorem 6 in the Introduction, the above proof shows more generally, that if H is any (not necessarily maximal) subgroup of L_{1} such that F^{∗}(H) = X(q) with q as in Theorem 1, then H ≤ N(X̄_{σ}) for some X̄ as in conclusion (ii) of Theorem 6.

Proof of Corollary 7

Assume the hypotheses of Corollary 7, so that H is a maximal subgroup of L_{1} as above, and F^{∗}(H) = X = X(q) with q as in Theorem 1. We may take it that K = 𝔽̄_{p}, the algebraic closure of 𝔽_{p}, since the conclusions of Corollary 7 concern finite groups only. (This allows us to apply 1.13 at a suitable point in the argument.)

Assume that H is not a subgroup of maximal rank, and that X(q) is not of the same type as G. By Theorem 6, we have X = O^{p′}(X̄_{σ}), where X̄ is a maximal closed connected reductive H⟨σ⟩-stable subgroup of G. By assumption, X̄ does not contain a maximal torus of G. Moreover, Z(X̄)^{0} = 1, since otherwise H ≤ N(C(Z(X̄)^{0}), a group of maximal rank, which is impossible as H is not of maximal rank by assumption. Therefore X̄ is semisimple. If X̄ is not simple then by [LS1, Theorem 1], X̄ is as in the second column of [LS1, Table II]; from this we see that O^{p′}(X̄_{σ}) is not simple, a contradiction.

Hence X̄ is simple. Therefore, since X = X(q) = O^{p′}(X̄_{σ}), by 1.13 we have q = q_{1} (where G_{σ} = G(q_{1})), as in (iii) of Corollary 7. Moreover, if p > N(X, G), then the possibilities for X̄ are given by [LS1, Theorem 1]; note also from [LS1, Theorem 1] that these maximal H⟨σ⟩-stable subgroups X̄ are also maximal σ-stable.

There remain two points to establish: first, that the fixed point groups X(q) = O^{p′}(X̄_{σ}) are as in conclusion (iii) of Corollary 7; and second, the conjugacy statement at the end of Corollary 7.

The conjugacy statement is straightforward: first, the assertion of the uniqueness of Aut(G)-classes of maximal connected subgroups X̄ is given by [Se2, Theorem 1]. And second, by [SS, I,2.7], the G_{σ}-classes contained in a given G-class of subgroups X̄ correspond to classes in the group H^{1}(σ, N_{G}(X̄)/X̄); however, clearly N_{G}(X̄)/X̄ has order 1 unless X̄ = A_{2}, in which case it has order 2 by Lemma 6.1 below, so the conclusion is immediate.

The statement that the fixed point groups X(q) = O^{p′}(X̄_{σ}) are as in conclusion (iii) of Corollary 7 is clear, except when X̄ = A_{2}, in which case we must establish that both the fixed point groups A_{2}(q) and ^{2}A_{2}(q) arise. This follows from the next lemma.

Lemma 8.1 Let X̄ ≅ A_{2} be a maximal connected subgroup of G = E_{6} or E_{7}. Then N_{G}(X̄) induces a nontrivial graph automorphism on X̄ (and hence N_{G}(X̄)/X̄ has order 2).

Proof We may assume that G is of adjoint type. When G = E_{6}, the conclusion is immediate from [Te2, Claim on p.314]. So assume from now on that G = E_{7}. The class of maximal subgroups A_{2} in E_{7} for p ≥ 7 is constructed in [Se2, 5.8].

Choose a parabolic subgroup QJ_{α}T_{α} of X̄, where J_{α} is a fundamental SL_{2}, T_{α} a rank 1 torus and Q the unipotent radical. Then by [Se2, p.83], C_{G}(T_{α}) = T_{2}A_{1}A_{4}, J_{α} projects onto the factor A_{1}, and J_{α} projects onto a regular A_{1} in the factor A_{4}.

We claim that there is an involution z ∈ N_{G}(T_{2}A_{1}A_{4}) normalizing J_{α} and inverting T_{2}. To see this, we work first in E_{8}, and observe that there is an involution u normalizing a maximal rank subgroup A_{4}A_{4} of E_{8}, inducing a graph automorphism on each factor. Then u normalizes a fundamental A_{1} in one of the factors A_{4}, inducing a nontrivial inner automorphism on this A_{1}. Therefore u ∈ A_{1}E_{7} and is a product yz of elements y, z of order 4, where y ∈ A_{1}, z ∈ E_{7}, and y^{2} = z^{2} is a generator of Z(E_{7}). Then z projects to an involution in the adjoint group of type E_{7}. Now C_{A_{4}A_{4}}(A_{1}) = T_{1}A_{2}A_{4}; therefore u, hence z, normalizes this group, acting as a graph automorphism on the A_{2}, A_{4} factors and inverting the T_{1}. Now z centralizes a subgroup B_{2} of the A_{4} factor, and a subgroup P SL_{2} therein. Adjusting z by a suitable involution of this P SL_{2}, we obtain an element z normalizing a subgroup T_{2}A_{1}A_{4} of T_{1}A_{2}A_{4}, acting in the fashion claimed at the beginning of this paragraph.

Let t be the central involution in J_{α}. Then C_{G}(t) = A_{1}D_{6}, and L(G) ↓ A_{1}D_{6} = L(A_{1}D_{6}) ⊕ (V_{A_{1}}(1) ⊗ V_{D_{6}}(λ_{5})). The restriction of V_{D_{6}}(λ_{5}) to the projection of J_{α} in A_{4} (which lies in D_{6}) is 6^{2}/4^{2}/2^{2}/0^{2}; tensoring with V_{J_{α}}(1), we obtain

V_{A_{1}}(1) ⊗ V_{D_{6}}(λ_{5}) ↓ J_{α} = 7^{2}/5^{4}/3^{4}/1^{4}.

We claim that there is an 8-dimensional J_{α}-submodule isomorphic to 1^{4}. Two of the required four summands are provided by L(Q) and L(Q^{−}). Now consider the action of T = T_{α}; say it acts with weight β on L(Q), hence with weight −β on L(Q^{−}). Now T_{2} centralizes J_{α}, so acts on the homogeneous component A of the socle of L(G) ↓ J_{α} corresponding to high weight 1. It also centralizes T, hence acts on the β and −β weight spaces in A; and z interchanges these. If the β weight space is just 2-dimensional (i.e. equal to L(Q)), then so is the −β weight space, and conversely. But then the Lie algebra span of this part of the socle is just ⟨L(Q), L(Q^{−})⟩ = L(X̄). However, this is not T_{2}-invariant, so this is a contradiction. Therefore the β and −β weight spaces of A have dimension 4, and A ≅ 1^{4}, proving the claim.

Let W_{β}, W_{−β} be the ±β weight spaces in A. These are fixed by T_{2} and interchanged by z. If C_{T_{2}}(W_{β})^{0} ≠ 1, then there is a 1-dimensional torus acting trivially on W_{β}, hence also on W_{−β} (as z inverts T_{2}). But this torus then acts trivially on ⟨L(Q), L(Q^{−})⟩ = L(X̄) (Lie algebra span), which is impossible (as X̄ is maximal). Therefore C_{T_{2}}(W_{β})^{0} = 1. Consequently W_{β} = W_{1} ⊕ W_{2}, a sum of two 2-dimensional T_{2}-weight spaces with different T_{2}-kernels; similarly W_{−β} = W_{3} ⊕ W_{4}.

Now none of the W_{i} is equal to L(Q) or L(Q^{−}) (otherwise we would get a 1-dimensional torus acting trivially on L(X̄) as above). Hence L(Q) is diagonal in W_{1} ⊕ W_{2}. As a J_{α}T_{2}-module, we have W_{β} ≅ M ⊗ N, where M is a natural 2-dimensional module for J_{α} and N is a direct sum of two 1-dimensional T_{2}-modules with T_{2}-kernels having different connected components. As T_{2} has three orbits on the nonzero vectors of N, it follows that T_{2} has three orbits on the set of 2-dimensional J_{α}-invariant subspaces of W_{β}. Since L(Q^{−})^{z} is such a subspace, and is diagonal in W_{1} ⊕W_{2}, there is therefore an element a ∈ T_{2} such that L(Q^{−})^{za} = L(Q). Then za is an involution normalizing the Lie algebra span ⟨L(Q), L(Q^{−})⟩ = L(X̄). Consequently za normalizes X̄, inducing a graph automorphism, and this completes the proof of the lemma. ♠

Proof of Corollary 8

Let L_{1} be as in the first paragraph of this section, and let H be a maximal subgroup of L_{1}. Assume first that H is not almost simple. Then by [LS1, Theorem 2], one of the following holds:

(i) H = N(X̄_{σ}), where X̄ is a maximal connected H⟨σ⟩-stable subgroup of G;

(ii) H is the normalizer of a subgroup of the same type as G;

(iii) H one of the local subgroups 2^{3}.L_{3}(2), 3^{3}.L_{3}(3), 3^{3+3}.L_{3}(3), 2^{5+10}.L_{5}(2), 5^{3}.L_{3}(5) (in G_{2}, F_{4}, E_{6}, E_{8}, E_{8} respectively), given in [CLSS, Theorem 1(II)];

(iv) F^{∗}(H) = Alt_{5} × Alt_{6}. Taking c > 2^{15}|L_{5}(2)|, we see that (i) or (ii) of Corollary 8 holds for |H| > c.

Now assume that H is almost simple, and let X = F^{∗}(H). If X is alternating, sporadic, or of Lie type in p′-characteristic, then H has bounded order (see [LSS, §4] for instance); so we may choose c so that this does not occur for |H| > c. Finally, let X = X(q) with q = p^{e}. The BN-rank of X is at most that of G (since a parabolic of X must lie in a parabolic of G), so again choosing c appropriately, we may assume that q satisfies the hypothesis of Theorem 1. Now the conclusion follows from Theorem 6.

9 Proofs of Theorem 9 and 10

For the proof of Theorem 9, we shall need the information on first cohomology groups given in the next two results. The first is taken from [Ja, 2.3].

Proposition 9.1 (Jantzen) Let Y be a simply connected simple algebraic group over 𝔽̄_{p}, and let σ be a Frobenius morphism of Y such that Y_{σ} is not of type ^{2}B_{2}, ^{2}G_{2} or ^{2}F_{4}. Let Y_{σ} = Y (q), a finite group of Lie type over 𝔽_{q}, with q = p^{r}, and let α_{0} be the highest short root of the root system of Y. Suppose that λ is a dominant weight for Y satisfying

\begin{aligned}⟨λ, α_{0}^{v}⟩ ≤ & p^{r} − 2p^{r−1} − 3, if Y ≠ G_{2} \\ & p^{r} − 3p^{r−1} − 3, if Y = G_{2}, \end{aligned}

and let V = V_{Y} (λ). Then the natural restriction map

H^{1}(Y, V) → H^{1}(Y_{σ}, V)

is an isomorphism.

Proposition 9.2 Let Y be a simple algebraic group in characteristic p, let Y (q) = Y_{σ} be a quasisimple group of Lie type over 𝔽_{q}, q = p^{e}, and let V = V_{Y (q)}(λ). Then H^{1}(Y (q), V) = 0 for all λ in Table 7 below.

Table 7

Y (q)  λ  conditions on p, q
A_{r}^{ϵ}(q)  λ_{1}, λ_{2}, λ_{3}  q > 3, and p > 2 if r = 1
B_{r}(q) (r ≥ 2) λ_{1}, λ_{r}  p > 2 if λ = λ_{1} or r = 2, Y (q) ≠ B_{2}(3)
D_{r}^{ϵ}(q) (r ≥ 4) λ_{1}, λ_{r−1}, λ_{r}
G_{2}^{ϵ}(q)  10  p > 2, q > 3
C_{4}(q)  λ_{1}, λ_{3}  p > 2
F_{4}^{ϵ}(q)  λ_{4}  p ≠ 3, q > 8
E_{6}^{ϵ}(q)  λ_{1}
E_{7}(q)  λ_{7}
A_{2}^{ϵ}(q) 20, 30, 60, 11, 21, 22, 10 ⊗ 10^{(p^{i})}, p ≥ 5 (p ≥ 7 for 60), q ≥ 11
10 ⊗ 01^{(p^{i})}, 10 ⊗ 11^{(p^{i})}, 10 ⊗ 20^{(p^{i})},
10 ⊗ 02^{(p^{i})}
B_{2}(q) 02, 03, 20, 11, 13, 10 ⊗ 10^{(p^{i})}, p ≥ 5, q ≥ 11 (p ≥ 7 for 20,
10 ⊗ 01^{(p^{i})}, 01 ⊗ 01^{(p^{i})}, 01 ⊗ 02^{(p^{i})} p = 3 allowed for λ = 02)
G_{2}(q)  20, 01, 11, 10 ⊗ 10^{(p^{i})}  p ≥ 11
A_{3}^{ϵ}(q)  101, 110, 200, 100 ⊗ 100^{(p^{i})},  p > 2, q ≥ 11
100 ⊗ 001^{(p^{i})}, 100 ⊗ 010^{(p^{i})}
B_{3}(q)  010, 002, 101, 001 ⊗ 001^{(p^{i})}  p > 2, q ≥ 11
C_{3}(q)  100, 010, 001, 110  p > 2 (p > 3 for λ = 010), q ≥ 11
C_{4}(q)  0100, 0010  p > 2, q ≥ 11
D_{4}^{ϵ}(q)  0100, 0011  p > 2, q ≥ 11

Proof For Y (q) as in the first eight rows of the table (i.e. Y (q) = A_{r}^{ϵ}(q), …, E_{7}(q)), the result follows immediately from [JP] and [LS3, 1.8]. For the other cases, our assumptions on p, q imply that H^{1}(Y (q), V) ≅ H^{1}(Y, V) by 9.1. (For cases where λ is non-restricted, we twist V by an automorphism of Y (q) to take i ≤ r/2.) It follows from [LS2, 1.7 and 1.9-1.15] that H^{1}(Y, V) = 0 in all cases. ♠

We shall also need some elementary lower bounds for the numbers t(G) defined in the Introduction.

Proposition 9.3 We have t(G_{2}) ≥ 12, and t(Σ(G)) ≥ 16 for G = F_{4}, E_{6}, E_{7}, E_{8}.

Proof For G = G_{2}, take simple roots α (short) and β (long), and set y = (3α + 2β) − (−3α − β) = 6α + 3β, z = β − (−β) = 2β (the middle terms in these equations are written simply to demonstrate that y and z are root differences); then 12α = 2y − 3z, but 6α and 4α are not in ℤy + ℤz. Hence t(Σ(G)) ≥ 12.

For G = F_{4}, take simple roots α, β, γ, δ with α and β long and β joined to γ. Set x = (α + 2β + 2γ) − (−α − β − 2γ) = 2α + 3β + 4γ, y = (α + β) − (−β) = α + 2β, z = α − (−α) = 2α; then 16γ = 4x − 6y − z, but 8γ is not in ℤx + ℤy + ℤz. Hence t(Σ(G)) ≥ 16.

For G = E_{6}, E_{7} or E_{8} take four simple roots α, β, γ, δ forming an A_{4} subsystem. Set w = (α + β) − (−α) = 2α + β, x = (β + γ) − (−β) = 2β + γ, y = (γ +δ)−(−γ) = 2γ +δ, z = δ −(−δ) = 2δ; then 16α = 8w −4x+2y −z, but 8α is not in ℤw + ℤx + ℤy + ℤz. Hence t(Σ(G)) ≥ 16. ♠

Proof of Theorem 9

Let G be a simple exceptional adjoint algebraic group over the algebraically closed field K of characteristic p, let X = X(q) < G with q as in the hypotheses of Theorem 1, and assume that p > N′(X, G) (as defined before the statement of Theorem 9). Suppose that X < P = QL, a parabolic subgroup of G with unipotent radical Q and Levi subgroup L. We aim to show that X lies in a Q-conjugate of L.

We follow closely the proof of Theorem 1 of [LS2], which appears in [LS2, §3]. Assume first that P is minimal subject to containing X, and choose a subgroup X_{1} of L such that QX = QX_{1}.

Suppose that X ≅ A̸_{1}(q). By Theorem 1, X_{1} < X̄ ≤ L, where X̄ is connected and X_{1} and X̄ stabilize the same subspaces of the Lie algebra of L. Since X̄ lies in no parabolic subgroup of L, it is semisimple, say X̄ = X̄_{1} … X̄_{r} with all X̄_{i} simple. The fact that X_{1} is irreducible on each X̄-composition factor of L(X̄_{i}) implies that all the X̄_{i} are of the same type as X. Therefore X lies in a connected simple subgroup X̄_{0} of L of the same type (a diagonal subgroup of X̄_{1} … X̄_{r}). Hence the possible embeddings of X_{1} in L are given by [LS2, 3.3] (with the same proof). The L′-composition factors within Q have the structure of rational KL′-modules, with high weights given by [LS2, 3.1]. We deduce as in the proof of [LS2, 3.4] that the restrictions to X_{1} of such modules have composition factors with high weights λ all twists of those occurring in the table of Proposition 9.2. Moreover, the conditions on p, q given in the table are satisfied (since we are assuming p > N′(X, G) and q > 9), except in the following cases:

(1) X_{1} ≅ B_{r}(q) (r ≥ 5), p = 2, λ = λ_{1}, L′ = D_{r+1}

(2) X_{1} ≅ C_{3}(q), p = 3, λ = λ_{2}, L′ = A_{5}

(3) X_{1} ≅ F_{4}(q), p = 3, λ = λ_{4}, L′ = E_{6}. Therefore, excluding these cases, the group QX_{1} has just one Q-conjugacy class of complements to Q, and so X is equal to a Q-conjugate of X_{1}, as required. Finally, cases (1),(2),(3) above are dealt with exactly as in the proof of [LS2, 3.6], using 9.1.

Now let X ≅ A_{1}(q). First assume that q = p. By hypothesis, we have q > 2t(Σ(G)); and by 9.3, t(Σ(G)) ≥ 12 if G = G_{2}, t(Σ(G)) ≥ 16 otherwise; so q = p > 24 if G = G_{2}, and q = p > 32 otherwise. By 1.2, we have H^{1}(X, V_{X}(r)) = 0 when r < p − 3. Now [LS2, 2.4],together with the first row of the table of [LS2, 2.13] shows that all X-composition factors within Q have high weights at most 27 (at most 4 if G = G_{2}). Hence H^{1}(X, V) = 0 for all such composition factors V, and we deduce as before that X is a Q-conjugate of X_{1}.

Finally, suppose q = p^{e} with e ≥ 2. By 1.2, if H^{1}(X, V_{X}(r)) ≠ 0 then V_{X}(r) must be a twist of the module V_{X}(p−2)⊗V_{X}(1)^{(p)}. Now the proof of [LS2, 3.5] shows that such a module can appear as an X-composition factor within Q only if

p = 11 or 17, L′ = D_{7} (and G = E_{8}).

The embedding of XQ/Q in D_{7} is such that the action on the natural module is 10⊕2^{(p)}, and the composition factor (p−2)⊗1^{(p)} occurs within the module Q/Q′ with multiplicity 1; moreover, no nontrivial twist of this composition factors occurs within Q/Q′.

Suppose X does not lie in a Q-conjugate of L. By 1.3, the restriction map H^{1}(A_{1}(K), (p − 2) ⊗ 1^{(p)}) → H^{1}(X, (p − 2) ⊗ 1^{(p)}) is injective. Hence there is a connected subgroup F of QL containing XQ′ such that F/Q′ is of type A_{1}. A similar analysis took place in [LS2, 3.7], except that there we were dealing with a connected subgroup of type A_{1} instead of the finite group X. In [LS2, 3.7] it is shown that this connected A_{1} subgroup lies in a Q-conjugate of L. This implies that the group F does not contain a closed complement to Q′. Now Q′ ↓ X = 10 ⊕ 2^{(p)}. Since F contains X, it follows that the restriction map H^{2}(A_{1}(K), 10 ⊕ 2^{(p)}) → H^{2}(X, 10 ⊕ 2^{(p)}) is not injective. However, this map is in fact injective: to see this, we verify the injectivity condition (5.4) of [CPSK] for n = 2, and the isomorphism condition (5.5) for n = 1, upon which the assertion follows from [CPSK, 5.1].

Therefore X lies in a Q-conjugate of L.

We have now proved Theorem 9 (except for the last sentence concerning σ-stability), in the case where P = QL is minimal subject to containing X. The proof for non-minimal parabolics containing X follows as in [LS2, p.43].

We finally deal with the σ-stability statement in the last sentence of Theorem 9. Suppose then that X < G_{σ} and X < P = QL, a σ-stable parabolic. By what we have already proved, we can assume that X < L. Define

Δ = {L^{u} : u ∈ Q, X < L^{u}}.

Observe first that if X < L^{u} (u ∈ Q), then X^{u^{−1}} < L∩QX = X, whence u ∈ N_{Q}(X) = C_{Q}(X). Consequently C_{Q}(X) acts transitively on Δ. Write C = C_{Q}(X).

We claim now that C is connected. To see this, let T = Z(L)^{0}. Then X < C_{G}(T), so T normalizes C. Obviously T acts trivially on the finite group C/C^{0}, so if this group is nontrivial, then T fixes a nonzero vector in some composition factor of Q ↓ T. However C_{Q}(T) = 1, so this is impossible. Hence C is connected, as claimed.

Thus the connected group C acts transitively on the σ-stable set Δ. It now follows from [SS, I,2.7] that Δ contains an element fixed by σ. In other words, X lies in a σ-stable Q-conjugate of L. This completes the proof of Theorem 9.

Proof of Theorem 10

As in the hypothesis of Theorem 10, assume that X = X(q) < G, with q as in Theorem 1 and p > N′(X, G). We first aim to prove part (i) of Theorem 10.

By Corollary 5, there is a proper closed connected subgroup X̄ of G containing X and fixing the same subspaces of L(G) as X. Take X̄ to be minimal subject to these conditions.

Lemma 9.4 X̄ is reductive.

Proof Suppose false, so R_{u}(X̄) ≠ 1. By [BT], X̄ lies in a parabolic subgroup P = QR of G (with unipotent radical Q, Levi subgroup R), such that R_{u}(X̄) ≤ Q. Then by Theorem 9, X lies in a Q-conjugate of R, so we may take it that X < R.

Now X fixes L(R), and hence so does X̄. But N_{QR}(L(R))^{0} = R: for if not, N_{QR}(L(R))^{0} = Q_{0}R with Q_{0} a nontrivial connected subgroup of Q. Then L(Q_{0}) commutes with L(R). However, C_{L(G)}(L(R)) ∩ L(Q) = 0, since if T is a maximal torus of R, then C_{L(G)}(L(R)) ⊆ C_{L(G)}(L(T)) = L(T) by 1.11; this contradicts the previous sentence. Therefore X̄ ≤ N_{QR}(L(R))^{0} = R. But this means that R_{u}(X̄) ≤ R, which is absurd. ♠

By the minimality of X̄ we have X̄′ = X̄, and hence X̄ is semisimple. Write X̄ = X̄_{1} … X̄_{t}, a commuting product of connected simple groups X̄_{i}.

Lemma 9.5 All the X̄_{i} are of the same type as X.

Proof Since X and X̄ fix the same subspaces of L(G), X must be irreducible on each X̄_{i}-composition factor of L(X̄_{i}) (for all i). Hence by Theorem 4, for each i, either X and X̄_{i} are of the same type, or (X̄_{i}, X, p) is (B_{n} or C_{n}, D_{n}^{ϵ}(q), 2) or (B_{3} or C_{3}, G_{2}(q), 2). In each of these exceptional cases, the projection of X in X̄_{i} lies in a connected simple subgroup of X̄_{i} of the same type as X (as this projection is determined up to conjugacy in X̄_{i}). Hence the result follows by the minimality of X̄. ♠

We now have

X < X̄_{1} … X̄_{t}

with each X̄_{i} connected and simple of the same type as X.

We now argue that X < X̃ ≤ X̄_{1} … X̄_{t}, where X̃ is connected and simple of the same type as X. This is clear if t = 1, so suppose t > 1. Let ϕ_{i} : X → X̄_{i}/Z(X̄_{i}) be the i^{th} projection map.

Let Y be a simple algebraic group of the same type as X, and containing X. If Y is classical, then [ST1] implies that each ϕ_{i} extends to a morphism ϕ̂_{i} : Y → X̄_{i}/Z(X̄_{i}); and if not, then as t > 1, we have X = G_{2}(q), and the same conclusion follows from [LS3, 5.1]. Now define

X̂ = {yϕ̂_{1} … yϕ̂_{t} : y ∈ Y },

and take X̃ to be the connected preimage of X̂ in X < X̄_{1} … X̄_{t}. Then X̃ is a closed, connected, simple subgroup of G of the same type as X, and containing X. Moreover X̃ ≤ X̄, so by minimality, X̃ = X̄. This completes the proof of the first part of Theorem 10(i).

For the last part of Theorem 10(i), we must show that C_{G}(X)^{0} = C_{G}(X̃)^{0}. This is obvious if C_{G}(X)^{0} = 1, so assume C_{G}(X)^{0} ≠ 1.

We now argue that C_{G}(X)^{0} is reductive. The proof of this follows closely that of [LS2, Theorem 4.1]. Let C = C_{G}(X)^{0}, and suppose U = R_{u}(C) ≠ 1. Then XC lies in a parabolic subgroup P = QL, with U ≤ Q. Certainly X fixes L(P), hence so does X̃, and therefore X̃ < P. Then by Theorem 9, we may assume that X̃ ≤ L. Now Lemmas 4.2 - 4.4 of [LS2] show that there exists l ∈ L such that w̄_{0}l normalizes X̃ (where w̄_{0} is an automorphism of G normalizing a maximal torus of L and inducing -1 on Σ(G)). Using [LS3, 5.1] we may take it that w̄_{0}l also normalizes X. But now we have 1 ≠ |C_{Q}(X)| = |C_{Q^{w̄_{0}l}}(X^{w̄_{0}l})| = |C_{Q^{−}}(X)|, where Q^{−} is the unipotent radical of the parabolic opposite to P. This gives 1 ≠ C_{Q^{−}}(X) ≤ Q^{−}∩P = 1, a contradiction.

Thus C_{G}(X)^{0} is reductive. Let T be a torus in C_{G}(X)^{0}. Then R = C_{G}(T) is a Levi subgroup of G containing X. Now R = P ∩ P^{−}, the intersection of a parabolic P and its opposite P^{−}. Hence X fixes L(P) and L(P^{−}), and therefore so does X̃. Since N_{G}(L(P)) = P and N_{G}(L(P^{−})) = P^{−}, we deduce that X̃ ≤ P ∩P^{−} = R. Thus X̃ centralizes T. The reductive group C_{G}(X)^{0} is generated by tori, so we conclude that C_{G}(X)^{0} = C_{G}(X̃)^{0}, as required. This completes the proof of Theorem 10(i).

Now we prove part (ii) of Theorem 10. Suppose that X = X(q) < G, with q as in Theorem 1. By Theorem 1, we can choose a σ-stable connected subgroup X̄ of G, minimal subject to containing X and fixing the same subspaces as X of every G-composition factor of L(G).

We claim that X̄ is reductive. Suppose false, so that by [BT], X̄ lies in a σ-stable parabolic P = QR of G, with R_{u}(X̄) ≤ Q; then by Theorem 9, X lies in a σ-stable Q-conjugate of R. If L(G) is G-irreducible, the proof of Lemma 9.4 shows that R_{u}(X̄) ≤ R, a contradiction. Otherwise, by 1.10, we have (G, p) = (E_{7}, 2), (E_{6}, 3), (F_{4}, 2) or (G_{2}, 3). In the first two cases we may assume that G is simply connected, so that G is irreducible on L(G)/Z(L(G)) and X̄ fixes each X-invariant subspace of L(G)/Z(L(G)). As L(R) > L(T) > Z(L(G)) (where T is a maximal torus of R), we conclude that X fixes L(R), and at this point the proof of 9.4 works as before to give the claim. For the F_{4} and G_{2} cases, by the prime restrictions p > N′(X, G), the only possibility is that (G, p, X) = (F_{4}, 2, A_{3}^{ϵ}(q)). Then QL = QB_{3} or QC_{3}. Let I be the 26-dimensional ideal of L(G) generated by short root elements. We may take it that X < L, hence X fixes L(B_{3})∩I or L(C_{3})∩I. Therefore X̄ lies in the stabilizer of one of these subspaces, hence lies in a subgroup B_{3}Ã_{1} or C_{3}A_{1} of F_{4}. Moreover, X, hence X̄, fixes every subspace of L(Ã_{1}) or L(A_{1}), from which it follows that X̄ < B_{3}T_{1} or C_{3}T_{1}, whence X̄ < L.

Thus X̄ is reductive. By minimality, X̄ is semisimple, say X̄ = X̄_{1} … X̄_{t} with each X̄_{i} simple. We claim also that each X̄_{i} is of the same type as X. As in the proof of Lemma 9.5, this follows from Theorem 4, unless for some i, (X̄_{i}, X, p) is (B_{n} or C_{n}, D_{n}^{ϵ}(q), 2) or (B_{3} or C_{3}, G_{2}(q), 2). Assume then that one of the latter two cases holds. The fact that p > N′(X, G) implies that the first case holds with t = 1, n ≥ 4. But then X lies in a σ-stable subgroup D_{n} of X̄, contrary to the minimality of X̄. This proves our claim.

To finish the proof of Theorem 10(ii), we need to show that C_{G}(X)^{0} = C_{G}(X̄)^{0}. We showed for part (i) of Theorem 10 that C_{G}(X)^{0} = C_{G}(X̃)^{0}, where X̃ is a suitable simple connected diagonal subgroup of X̄ of type X. Define CC^{0}(X) = C_{G}(C_{G}(X)^{0})^{0}. This group contains X̃ and is σ-stable. If

Y = ⟨X̃^{σ^{i}} : all i⟩,

then Y lies in X̄ and is σ-stable; hence Y = X̄ by the minimality of X̄, and consequently CC^{0}(X) contains X̄. It follows that C_{G}(X)^{0} ≤ C_{G}(X̄)^{0}. The reverse inclusion is trivial, so C_{G}(X)^{0} = C_{G}(X̄)^{0}, as required.

10 Proof of Theorem 11

In this final section we prove Theorem 11. Let C be a finite classical group in characteristic p with usual module V, and suppose X, Y are simple groups of Lie type in characteristic p, not of the same type, such that X < Y < C and X is absolutely irreducible on V. Suppose also that if Y is of exceptional type, then X = X(q) with q as in the hypothesis of Theorem 11.

First observe that by [ST1], the embedding Y < C lifts to an embedding Ȳ ≤ C̄ of corresponding simple algebraic groups. Let V̄ be the usual module for C̄. If Ȳ is classical, then again by [ST1], X < Y lifts to an embedding X̄ < Ȳ, giving the required conclusion.

Thus we assume that Ȳ is exceptional. By Theorem 1, X < Z ≤ Ȳ for some closed connected subgroup Z of Ȳ fixing the same subspaces of L(Ȳ) as X. By Theorem 4, X is reducible on some Ȳ -composition factor of L(Ȳ), and so Z < Ȳ. Now X, and hence Z, is irreducible on V̄, so Z is semisimple.

At this point we have

X < Z < Ȳ < C̄.

Now we apply [Se1, Theorem 1]; this says that the triple (Z, Ȳ , C̄) is one of those in Table 1 of [Se1]. Inspection of this table (recalling that Ȳ is exceptional) shows that Z must be simple. If Z is classical, then [ST1] again implies that X < X̄ ≤ Z for some simple connected group X̄ of the same type as X, as required. If Z is exceptional and of the same type as X, we take X̄ = Z. Finally, if Z is exceptional and not of the same type as X, we replace Ȳ with Z and repeat the argument; eventually we obtain a suitable group X̄. This completes the proof of Theorem 11.

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Imperial College, London SW7 2BZ, UK

University of Oregon, Eugene, Oregon 97403, USA