ISOLATED p-MINIMAL SUBGROUPS

ULRICH MEIERFRANKENFELD CHRISTOPHER PARKER PETER ROWLEY

1. Introduction

Suppose that p is a prime, P is a finite group and S ∈ Syl_{p}(P). Then P is p-minimal if S is not normal in P and S is contained in a unique maximal subgroup of P. Now suppose that G is a finite group and S ∈ Syl_{p}(G), if S ≤ H ≤ G and S is not normal in H, then we call P a p-parabolic subgroup of G. In most cases the prime p will be evident from the context in which we are working and in these cases we often simply call P a parabolic subgroup of G. The set of maximal parabolic subgroups of G (containing S) is denoted by M_{G}(S) and the set of p-minimal parabolic subgroups of G (containing S) is denoted by P_{G}(S). Suppose that G is a Lie type group defined in characteristic p. Then the parabolic subgroups of G in the traditional sense are also p-parabolic subgroups in our context, though it should be noted that we do not require that our p-parabolic subgroups contain the Borel subgroup of G. If R is a rank 1 parabolic subgroup in G, then P = O^{p′}(R) is a p-minimal parabolic subgroup of G. Moreover it is easy to see that in the Lie type groups we have G = ⟨P_{G}(S)⟩N_{G}(S) and it turns out that this is a property of p-minimal parabolic subgroups in general (see 2.1). Suppose that P ∈ P_{G}(S) and set

B = B_{P} = ⟨P_{G}(S) \ {P }, N_{G}(S)⟩.

Notice that when P is Lie type group in characteristic p, B_{P} ∈ M_{G}(S) for all P ∈ P_{G}(S). In fact minimal p-parabolic subgroups in the Lie type groups have two further properties. The first is that O_{p}(B_{P}) ≰ O_{p}(P) and the second is that P/O_{p}(P) is a rank 1 Lie type group in characteristic p. We shall see that for p sufficiently large (p ≥ 11) these two properties characterize Lie type groups in characteristic p among the finite simple groups.

We now make these notions more precise. We say that P ∈ P_{G}(S) is isolated with respect to A provided A is a normal p-subgroup of B_{P} and A ≰ O_{p}(P). Notice that A ≤ O_{p}(B_{P}) and so we also have O_{p}(B_{P}) ≰ O_{p}(P). Comment: P is isolated if its isolated with repect to O_{p}(B_{P}). Suppose that P is a p-minimal group, S ∈ Syl_{p}(P) and M is the maximal subgroup of P containing S. Set R = ⋂ M^{P}, the core of M in P, and E = O^{p}(P)R/R. Then, loosely approximating the structure of a rank 1 Lie type group, we say that P is narrow if either E is a simple group or E is elementary abelian and M acts primitively on E. Our first result, proved in Section 2 is a basic structure theorem for groups which possess a narrow, p-minimal, isolated parabolic subgroup.

Theorem 1.1. [Thm1] Suppose that P ∈ P_{G}(S) is narrow and isolated in G. Set Y = ⟨O^{p}(P)^{G}⟩. Then either Y/O_{p}(Y) is quasisimple or Y = O^{p}(P) and G = B_{P} P.

Suppose that G is a p-minimal simple group, then so long Comment: as at the unique maximal subgroup of G is a p-local subgroup, G satisfies the hypothesis of the theorem.

In keeping with the structure of the Lie type groups in characteristic p (perhaps extended by field automorphisms) we wish to restrict our further attentions to those candidates for LS which have P/Q_{P} a rank 1 Lie type group in characteristic p extended by automorphisms of order p Comment: ??? not an arbitray p-group . Thus we let L_{1}(p) consist of groups H with O_{p}(H) = 1 and O^{p}(H) isomorphic to O^{p}(L) for some (either adjoint or universal) rank 1 Lie type group in characteristic p. This has the effect of including some groups which are a little bit smaller than we might expect. For example, as O^{2}(^{2}B_{2}(2)) has order 5, Dih(10) and ^{2}B_{2}(2) are both in H. A similar phenomena occurs with PSU_{3}(2) ≅ 3^{2} : Q_{8}. The extra condition that we will impose is contained in the following definition.

Definition 1.2. [p-restricted] Suppose that p is a prime and G is a group and S ∈ Syl_{p}(G). Then P ∈ P_{G}(S) is p-restricted in G if P isolated in G and P/Q_{P} ∈ L_{1}(p). If G possesses a p-restricted p-minimal parabolic subgroup, then we say that G is p-restricted.

Notice that if P ∈ P_{G}(S) and P/Q_{P} ∈ L_{1}(p), then P is narrow Comment: This is false for 3^{2}C_{4} and 3^{2}D_{8}, I guess we need to add narrow to the definition of L_{1}(p). Thus 1.1 tells us that when O_{p}(G) = 1, then Y = ⟨O^{p}(P)^{G}⟩ is either quasisimple or Y = O^{p}(P). In this latter case we also have that Y is quasisimple or Y is soluble and O^{p}(P) is isomorphic to one of ℤ_{3}, Q_{8}, 2^{2} or 3_{+}^{1+2} or 3^{2}.

Theorem 1.3. [Thm2] Suppose that p is a prime, G is a finite group, X = F^{∗}(G) is a non-abelian simple group and G/X is a p-group. If G is p-restricted, then either X is a Lie type group defined in characteristic p or

(a) [a] p = 2 and X ≅ Alt(12) (see 4.2).

(b) [b] X is a Lie type group in characteristic r with r ≠ p, p ∈ {2, 3} and the possibilities for p, X, P and B_{P} are as listed in Table 1.

(c) [c] X is a sporadic simple group and the possibilities for (X, p) are as follows: (Mat_{12}, 2), (Mat_{12}, 3), (Mat_{22}, 2), (J_{2}, 2), (J_{2}, 3), (Mat_{23}, 2), (HS, 2), (J_{3}, 2), (Mat_{24}, 2), (McL, 3), (He, 2), (Ru, 2), (Suz, 2), (Suz, 3), (O′N, 2), (Co_{3}, 2), (Co_{3}, 3), (Co_{2}, 2), (Co_{2}, 3), (Fi_{22}, 2), (Fi_{22}, 3), (HN, 2), (HN, 5), (Ly, 5), (Th, 2), (Th, 3), (Fi_{23}, 2), (Fi_{23}, 3), (Co_{1}, 2), (Co_{1}, 3), (Co_{1}, 5), (Fi_{24}, 2), (Fi_{24}, 3), (BM, 2), (BM, 3), (BM, 5), (M, 2), (M, 3), (M, 5), (M, 7).

In particular, if p ≥ 11, X is a Lie type group in characteristic p and, if p ≥ 5, then X is a Lie type group or one of the sporadic groups HN, Ly, Co_{1}, BM or M.

Group  Condition on r^{a}  P  B
PGL_{2}(r^{a})  5 (mod 8)  Sym(4)  Dih(2(r^{a} − 1))
PGL_{2}(r^{a})  3 (mod 8)  Sym(4)  Dih(2(r^{a} + 1))
PSL_{2}(7) ≅ PSL_{3}(2)  Sym(4)  Sym(4)
PSL_{2}(9) ≅ Alt(6)  Sym(4)  Sym(4)
PΣL_{2}(9) ≅ Sym(6)  Sym(4) × 2  Sym(4) × 2
PSL_{2}(5) ≅ Alt(5)  PSL_{2}(5)  Alt(4)
PGL_{2}(5) ≅ Sym(5)  PGL_{2}(5)  Sym(4)
PGL_{2}(19)  Dih(40)  Sym(4)
PSU_{3}(r^{a})  3 (mod 8)  2. Sym(4) ∗ 4  (r^{a} + 1)^{2} : Sym(3)
PSU_{3}(r^{a}) : 2  3 (mod 8)  2. Sym(4) ∗ Q_{8}  (r^{a} + 1)^{2} : (2 × Sym(3))
PSU_{3}(3) ≅ G_{2}(2)′  4^{2} : Sym(3)  2. Sym(4) ∗ 4
PSU_{3}(3) : 2 ≅ G_{2}(2)  4^{2} : (2 × Sym(3))  2. Sym(4) ∗ Q_{8}
PSL_{3}(r^{a}) : 2  5 (mod 8)  2. Sym(4) ∗ Q_{8}  (r^{a} − 1)^{2} : (2 × Sym(3))
PSU_{4}(3)  2^{2+2+2}. Sym(3)  2^{1+4}.(Sym(3) × Sym(3))
PSU_{4}(3).2_{1}  2^{2+2+2}. Sym(3).2  2^{1+4}.(Sym(3) × Sym(3)).2
PSU_{4}(3).2_{2}  2^{2+2+2}. Sym(3).2  2^{1+4}.(Sym(3) × Sym(3)).2
PSU_{4}(3).2_{2}  2^{2+2+2}. Sym(3).2  2^{1+4}.(Sym(3) × Sym(3)).2
PSU_{4}(3).4  2^{2+2+2}. Sym(3).4  2^{1+4}.(Sym(3) × Sym(3)).4
PSU_{4}(3).2_{122}^{2}  2^{2+2+2}. Sym(3).2^{2}  2^{1+4}.(Sym(3) × Sym(3)).2^{2}
PSU_{4}(3).2_{133}^{2}  2^{2+2+2}. Sym(3).2^{2}  2^{1+4}.(Sym(3) × Sym(3)).2^{2}
PSU_{4}(3).Dih(8)  2^{2+2+2}. Sym(3).Dih(8) 2^{1+4}.(Sym(3) × Sym(3)).Dih(8)
PSU_{4}(3) : 2_{1}  2^{1+2+1+2}. Sym(3).2  2^{4}. Alt(6).2
PSU_{4}(3) : 2_{1}  2^{1+2+1+2}. Sym(3).2  2^{4}. Alt(6).2
PSU_{4}(3) : 2_{122}  2^{1+2+1+2}. Sym(3).2^{2}  2^{4}. Alt(6).2^{2}
PSU_{6}(3)  ≤ 4^{5}. Sym(6)  \frac{1}{2} GU_{2}(3) ≀ Sym(3) ∩ PSU_{6}(3)
PGU_{6}(3)  ≤ 4^{5}(Sym(6) × 2)  \frac{1}{2} GU_{2}(3) ≀ Sym(3)

PSp_{4}(3) ≅ PSU_{4}(2) PSp_{6}(3)

PΩ_{7}^{+}(3)  \frac{1}{2}O_{1}(3) ≀ Sym(7)) ∩ X
PΩ_{7}^{+}(3) : 2  \frac{1}{2}O_{1}(3) ≀ Sym(7))
PΩ_{8}^{+}(3)  \frac{1}{2}O_{4}^{+}(3) ≀ Sym(2) ∩ X
PΩ_{8}^{+}(3) : 2  \frac{1}{2}O_{4}^{+}(3) ≀ Sym(2)
PΩ_{12}^{+}(3)  \frac{1}{2}O_{4}^{+}(3) ≀ Sym(3) ∩ X
PΩ_{12}^{+}(3) : 2  \frac{1}{2}O_{4}^{+}(3) ≀ Sym(3)
2G_{2}(3) ≅ SL_{2}(8) : 3  SL_{2}(8) : 3  7^{3} : 21
G_{2}(3)  4^{2} : Dih(12)  2^{1+4} : 3^{2}.2
G_{2}(3)  4^{2} : Dih(12).2  2^{1+4} : Sym(3) × Sym(3)
3 D_{4}(3)  4^{2} : Dih(12)  (SL_{2}(3) ∗ SL_{2}(27)).2
E_{7}(3)  2^{3}.(PSL_{2}(3))^{7}.2^{4}.PSL_{3}(2)
E_{7}(3).2  2^{3}.(PSL_{2}(3))^{7}.2^{4}.PSL_{3}(2).2

Table 1: Lie type group exceptions with p = 2

Group  Condition on r^{a}  P  B
PGL_{3}(r^{a})  r^{a} ≡ 4, 7 (mod 9)  3^{2} : SL_{2}(3)  (r^{a} − 1)^{2} : Sym(3)
PGU_{3}(r^{a})  r^{a} ≡ 2, 5 (mod 9)  3^{2} : SL_{2}(3)  (r^{a} + 1)^{2} : Sym(3)
PGL_{3}(7)  3^{2} : SL_{2}(3)  3^{2} : Sym(3)
PSU_{4}(r^{a})  r^{a} ≡ 2, 5 (mod 9) 3_{+}^{1+2} : SL_{2}(3) \frac{1}{(2,r^{a}+1)}(r^{a} + 1)^{3} : Sym(4)
PSU_{4}(2)  3^{3} : Sym(4)  3_{+}^{1+2} : SL_{2}(3)
PSU_{5}(2)  3 × 3_{+}^{1+2}.SL_{2}(3)  3^{4}. Sym(5)
PSU_{6}(2)  3^{5} : Alt(6)  3_{+}^{1+4}(Q_{8} × Q_{8}).3
PSU_{6}(2) : 3  3^{6} : Alt(6)  3_{+}^{1+4}(Q_{8} × Q_{8}).3^{2}
PSp_{4}(2) ≅ PSL_{2}(9)  PSL_{2}(9)  3^{2} : 4
PΩ_{8}^{+}(r^{a})  r^{a} ≡ 2, 5 (mod 9) 3^{1+2}.SL_{2}(3) × 3  O_{2}^{−}(r^{a}) ≀ Sym(4)
PΩ_{8}^{+}(r^{a}) : ⟨τ⟩  r^{a} ≡ 2, 5 (mod 9) 3_{+}^{1+4}.SL_{2}(3)  O_{2}^{−}(r^{a}) ≀ Sym(4).3
2E_{6}(2)  \frac{1}{3}PSU_{3}(2) ≀ Sym(3).3 ∩ X
2E_{6}(2).3  \frac{1}{3}PSU_{3}(2) ≀ Sym(3).3^{2} ∩ X
E_{8}(2)  3^{2}.(PSU_{3}(2)^{4}).3^{2}. GL_{2}(3)

Table 1: Lie type group exceptions with p = 3

For our intended application of 1.3 we will know additional information about the over groups of the p-restricted minimal parabolic subgroup. Indeed we will know for such a maximal p-parabolic subgroup M that M/O_{p}(M) is SL_{n}(p^{b}) or PSL_{n}(p^{b}) perhaps extended by a group of field automorphisms (for some n ≥ 2 and natural number b). Our next result can is designed to make 1.3 more immediately applicable in the circumstances just described.

Corollary 1.4. [L3-restricted] Suppose that p is a prime, G is a finite group, X = F^{∗}(G) is a non-abelian simple group and G/X is a p-group. If P ∈ P_{G}(S) is p-restricted and there is a P_{1} ∈ P_{G}(S) such that O^{p}(⟨P, P_{1}⟩/O_{p}(⟨P, P_{1}⟩) ≅ SL_{3}(p^{b}) or PSL_{3}(p^{b}) for some integer b, then is either X is a Lie type group defined in characteristic p or p = 2 and X is one of the following sporadic simple groups .....

Suppose that p is a prime. Then we use R_{p} to denote the set of group G for which there is P ∈ P_{p}(G) which is narrow and p-restricted. One of our goals is to determine those simple groups which are in R_{p}.

For a group H, we denote the preimage of F (H/O_{p}(H)) by F_{p}(H) and the preimage of Φ(H/O_{p}(H)) by Φ_{p}(H). The remainder of our group theoretic notation is standard as can be found in [1].

2. Groups with isolated p-minimal parabolic subgroups

Recall from the introduction that for a prime p, a group P is called p-minimal if for a Sylow p-subgroup S of P, S is not normal in P and S is contained in a unique maximal subgroup of P. For a group G and S ∈ Syl_{p}(G), we denote the set of p-minimal parabolic subgroups of G (which contain S) by P_{G}(S). For an arbitrary subgroup R of G, we set Q_{R} = O_{p}(R).

The following result is elementary to prove. Comment: See for example [?]

Lemma 2.1. [gen] G = ⟨P_{G}(S)⟩N_{G}(S).

We shall also need the following general result.

Lemma 2.2. [no subnormal] Let M be a maximal subgroup of the finite group H and let N be a subnormal subgroup of H with N ≤ M. Then N ≤ ⋂ M^{H}.

Proof. Suppose that H is a counterexample to the statement Comment: H being a counterexample is not used, remove and select N subnormal in H with N ≤ M of maximal order. Let N = N_{0}⊴N_{1}⊴…⊴N_{k}⊴N_{k+1} = H be a subnormal chain from N to H. By Wielandt’s subnormal lemma, ⟨N^{M}⟩ ⊴ ⊴H. Because of the maximal choice of N, we have ⟨N^{M}⟩ = N ⊴ M. Also by the maximal choice of N, N_{1} ≰ M. Therefore, N ⊴ ⟨N_{1}, M⟩. Since M is a maximal subgroup of H, we have N ⊴ H as claimed. □

Lemma 2.3. [basic p-minimal] Assume that P is p-minimal, S ∈ Syl_{p}(P) and M is the maximal subgroup of P containing S.

(a) [a] ⋂ M^{P} is p-closed, that is S ∩ ⋂ M^{P} = Q_{P}.

(b) [d] If O_{p}(O^{p}(P)) = 1, then ⋂ M^{P} is nilpotent.

(c) [b] If O^{p}(P) is p-closed, then P is a {t, p}-group for some prime t ≠ p.

(d) [c] If N is a subnormal subgroup of P with N ≤ M, then N ∩ S ≤ Q_{P}.

Proof. Comment: remove proof, quote [?] Let F = ⋂ M^{P} and set T = S ∩ F. Plainly Q_{P} ≤ T and, by the Frattini Argument, P = F N_{P} (T). Since N_{P} (T) ≥ S and P is p-minimal, we have P = N_{P} (T). So T ≤ Q_{P} and (a) holds.

Assume that O_{p}(O^{p}(P)) = 1. Then (a) implies that F ∩ O^{p}(P) is a p′-group. Let T ∈ Syl_{t}(F) for some prime t ≠ p. Then T ≤ O^{p}(P) and so T ≤ F ∩ O^{p}(P). The Frattini Argument gives P = N_{P} (T)(F ∩ O^{p}(P)). Since p does not divide |F ∩ O^{p}(P)|, we infer that N_{P} (T) contains a Sylow p-subgroup of P. Thus T is normal in P and F is nilpotent.

Without loss of generality, we now assume that Q_{P} = 1.

For (c) let t be a prime such that t divides |P/M|. Note that t ≠ p. Since O^{p}(P/Q_{P}) is a p′-group, there exists an S-invariant Sylow t-subgroup T of O^{p}(P). Then, as T ≰ M, P = ST as p is p-minimal. Thus (c) holds. Finally (d) follows from (a) and 2.2. □

Definition 2.4. [def:restricted] Comment: already defined in the introduction Suppose that G is a group, S ∈ Syl_{p}(G), P ∈ P_{G}(S), B = B_{P} = ⟨P_{G}(S) \ {P }, N_{G}(S)⟩ and A is a normal subgroup of B. If A ≰ Q_{P}, then we say that P is isolated in H with respect to A.

Note that if P is isolated in H with respect to A, then P is also isolated in H with respect to Q_{B}. Furthermore, we note that if P is isolated in H with respect to A, then certainly A > 1 and so Q_{B} > 1.

The next lemma is the primary structural result about groups which possess an isolated p-minimal subgroup.

Lemma 2.5. [quasi] Suppose that P ∈ P_{H}(S) is isolated in H with respect to A. Set Y = ⟨O^{p}(P)^{H}⟩ and F = ⋂ B^{H}. Then

(a) [a-1] Suppose that S ≤ M ≤ H and M ≰ B, then P ≤ M, Comment: and B ∩ M is a maximal subgroup of M; furthermore, P is isolated in M with respect to A;

(b) [a-2] P ∩ B is the unique maximal subgroup of P containing S;

(c) [a] B is a maximal subgroup of H;

(d) [b] N_{H}(T) ≤ B for all A ≤ T ⊴ S;

(e) [c] if O^{p}(H) is p-closed, then H = BP and ⟨A^{H}⟩ = ⟨A^{P} ⟩;

(f) [d] if R is a normal subgroup of H and R ≰ B, then Y ≤ R;

(g) [e] if R is a proper characteristic subgroup of Y, then Y ≤ F; (h) [e+1] [F, Y] ≤ Q_{H};

(i) [f] either Y = O^{p}(P) or Y Q_{H}/Q_{H} is semisimple; Comment: semiquasisimple and

(j) [g] if KQ_{H}/Q_{H} is a component in Y Q_{H}/Q_{H} and Y ≠ O^{p}(P), then Y = ⟨K^{S}⟩ and K ∩P ≰ B.

Proof. Suppose that S ≤ M ≤ H and M ≰ B. Then, by 2.1, M = ⟨P_{M}(S)⟩N_{M}(S). Since N_{M}(S) ≤ B and P is the unique member of P_{H}(S) which is not contained in B, we have P ∈ P_{M}(S). Since P_{M}(S) \ {P } ⊆ P_{B}(S), we have M ∩ B = ⟨P_{M}(S) \ {P }⟩N_{M}(S). Now let M ∩ B < D ≤ M. Then also P_{D}(S) ≤ P_{M}(S) and so as D > M ∩ B, we must have P ∈ P_{D}(S) and we conclude that D = M. Thus M ∩ B is a maximal subgroup of M. Finally, as A ⊴ B, A ⊴ B ∩ M and A ≰ Q_{P} and so P is isolated in M with respect to A. This proves (a). Parts (b) and (c) follow immediately from (a).

Suppose that A ≤ T ⊴ S. Since A ≰ Q_{P}, P is not in N_{H}(T). Taking M = N_{H}(T), (a) gives N_{H}(T) ≤ B. So (d) holds.

For the proof of the remaining statements Comment: we replace H be H/Q_{H} and so we may assume that Q_{H} = 1.

Suppose that X := O^{p}(H) is p-closed. Then, as Q_{H} = 1, X is a p′-group. Since O^{p}(P) ≤ X, O^{p}(P) is also a p′-group. Therefore, by 2.3(c), P is a {t, p}-group for some prime t ≠ p. Since p and |X| are coprime, for each prime divisor r of |X| there is an S-invariant Sylow r-subgroup S_{r} of X. If r ≠ t, then P ≰ S_{r}S and so S_{r} ≤ B by (a). Hence, by considering |X|, we have H = (X ∩ B)S_{t}S. We now consider S_{t}S, we have that (S_{t} ∩ B)S Comment: = S_{t}S ∩ B is a maximal subgroup of S_{t}S by (a) and so S_{t} ∩ B is a maximal S-invariant subgroup of S_{t}. Since S_{t} is nilpotent, N_{S_{t}}(S_{t} ∩ B) > S_{t} ∩ B and of course N_{S_{t}}(S_{t} ∩ B) is also normalized by S. Hence S_{t} ∩ B is normal in S_{t}. In particular, we have that (S_{t} ∩ B)O^{p}(P) is a subgroup of S_{t} and, as S_{t} ∩ B is a maximal with respect to being S-invariant, we infer that S_{t} = (S_{t} ∩ B)O^{p}(P). Hence

H = (B ∩ X)S_{t}S = (B ∩ X)O^{p}(P)S = (B ∩ X)P = BP.

Finally, we have ⟨A^{H}⟩ = ⟨A^{BP} ⟩ = ⟨A^{P} ⟩ and this completes the proof of (e).

Set Y = ⟨O^{p}(P)^{H}⟩. Suppose R is a normal subgroup of H which is not contained in B. Then, by (a), RS ≥ P and so O^{p}(P) ≤ R. Thus (f) holds. Plainly (g) is a direct corollary of (f).

Set F = ⋂ B^{H}. Then as F ≤ B, F normalizes A. Therefore, A ∩ F ≤ Q_{F} ≤ Q_{H} = 1. Therefore, as [F, A] ≤ A ∩ F, 1 = [F, A] = [F, ⟨A^{H}⟩]. Since ⟨A^{P} ⟩ ≥ O^{p}(P), we have ⟨A^{H}⟩ ≥ ⟨O^{p}(P)^{H}⟩ = Y. Therefore, [F, Y] = 1 and (h) is true.

Suppose that Y is not perfect. Then Y′ < Y hence Y′ ≤ Z(Y) by (g) and (h). In particular, Y is nilpotent and so p-closed. Applying part (e) to Y S, we have that ⟨A^{H}⟩ = ⟨A^{BY} ⟩ = ⟨A^{Y} ⟩ = ⟨A^{P} ⟩ = AO^{p}(P) and so Y = O^{p}(P). If on the other hand, Y is perfect, then Y is Comment: semiquasisimple semisimple by (g) and (h). So (i) holds.

Let K be a component of Y. Then Y is not nilpotent, so Y is semisimple Comment: semiquasisimple by (j) and, in particular, K is normal in Y. Hence, as H = Y B, ⟨K^{B}⟩ = ⟨K^{Y B}⟩ = ⟨K^{H}⟩ and so K ≰ B. Therefore, ⟨K^{S}⟩ ≥ O^{p}(P) by (a). If L is a component of Y not contained in ⟨K^{S}⟩, then O^{p}(P) ≤ ⟨K^{S}⟩ ∩ ⟨L^{S}⟩ ≤ Z(Y) ≤ B, a contradiction. Therefore, K^{H} = K^{B} = K^{S} and the first part of (j) holds. Assume that K ∩P ≤ B ∩P. Since O^{p}(P) ≤ Y, K ≠ Y. Furthermore, as (K ∩P)⊴⊴P, 2.3(d) gives K ∩ S ≤ Q_{P}. Hence Y ∩ S = Y ∩ Q_{P} ⊴ P and, in particular, O^{p}(P) is p-closed. Set R = N_{Y} (Y ∩ S)S. The Comment: Then O^{p}(R) is p-closed and R ≥ P. Therefore, R = (R ∩ B)P and ⟨A^{R}⟩ = ⟨A^{P} ⟩ ≤ P by (e). Suppose that A ≤ N_{H}(K). Then [A, K ∩ R] ≤ K ∩ ⟨A^{R}⟩ ≤ K ∩ P ≤ B which normalizes A. It follows that [A, K ∩ R] is a p-group and so, as K ∩ R is p-closed, [A, K ∩ R] ≤ Q_{P} ∩ K. Comment: [A, K ∩ R] ≤ S ∩ Y ∩ K ≤ Q_{P} ∩ K Therefore, [A, R] = ⟨[A, K ∩ R]^{S}⟩ ≤ Q_{P}. But then ⟨A^{R}⟩ = ⟨A^{P} ⟩ is a p-group, a contradiction. Hence A ≰ N_{H}(K). Therefore, Q_{B} properly permutes the components of Y. Since [K ∩ B, Q_{B}] is a p-group, we get K ∩ B ≤ (K ∩ S)F and then that Comment: Why ? Y ∩ S = O_{p}(Y ∩ B). But then Y ∩ S ⊴ ⟨O^{p}(P), B ∩ Y ⟩ = Y and we conclude that H is p-closed. Now a final application of (e) indicates that Y ≤ ⟨A^{H}⟩ = ⟨A^{P} ⟩ ≤ P. So Y = O^{p}(Y) = O^{p}(P) and thus (g) holds. □

To control the structure of Y in (??) Comment: 2.5 further we have to impose further conditions on the isolated p-minimal subgroup P.

Definition 2.6. [def:narrow] Comment: already defined in the introduction Let P be a p-minimal group and S ∈ Syl_{p}(P). Let M be the maximal subgroup of P containing S, R = ⋂ M^{P} and E := O^{p}(P)R/R. Then P is called narrow provided that either

(a) [a] E is non-abelian and simple; or

(b) [b] E is elementary abelian and B Comment: M acts primitively on E.

Comment: should we really require that P is p-minimal in the definition of narrow?

Lemma 2.7. [qs] Suppose that P ∈ P_{G}(S) is narrow and isolated in G. Then either P/Q_{P} is soluble or O^{p}(P)Q_{P} /Q_{P} is quasisimple. Comment: Only need that P narrow and P is isolated in P, that is Q_{M} ≰ Q_{P}, where M is a maximal parabolic subgroup

Proof. Let M = B ∩ P, then by 2.5(b), M is the unique maximal subgroup of P containing S. Put R = ⋂ M^{P}. Then R is normal in P and contained in B, therefore, [Q_{B}, R] ≤ Q_{B} ∩R ≤ S ∩R = Q_{P}. Thus R/Q_{P} is centralized by ⟨Q_{B}^{P}⟩ ≥ O^{p}(P). Thus RQ_{P} /Q_{P} ≤ Z(O^{p}(P)Q_{P} /Q_{P}. Since P is narrow, we have either P is soluble or O^{p}(P)Q_{P} /Q_{P} is quasisimple. □

Theorem 2.8. [simple] Suppose that P ∈ P_{H}(S) is narrow and isolated in H with respect to A. Assume that Q_{H} = 1 and set Y = ⟨O^{p}(P)^{H}⟩. Then either Y = O^{p}(P) or Y is quasisimple.

Proof. We may assume that Y ≠ O^{p}(P). Set M = P ∩ B. The by 2.5(b) M is the unique maximal subgroup of P containing S. Put R = ⋂ M^{P}. By 2.5(i) and 2.5(j), Y is Comment: semiquasisimple semisimple and for any component K of Y, Y = ⟨K^{S}⟩ = ⟨K^{M}⟩ and K ∩ P ≰ B. If K is normalized by S, then Y = ⟨K^{S}⟩ = K is quasisimple and we are done. Hence we assume that K is not normalized by S and look for a contradiction. Suppose first that O^{p}(P)R/R is a non-abelian simple group. Then, Comment: as O^{p}(P) ≤ Y, (K ∩ P)R/R is normalized by O^{p}(P)R/R, we have (K ∩ P)R ≥ O^{p}(P)R. Now selecting s ∈ S such that K^{s} ≠ K, we have (O^{p}(P)R/R)′ ≤ [K, K^{s}]R/R = (K ∩ K^{s})R/R which is abelian, a contradiction. Therefore, O^{p}(P)R/R is an elementary abelian t-group for some prime t ≠ p and O^{p}(P) is p-closed. Hence P is a {t, p}-group. We have that O^{p}(K ∩ P) ≤ X Comment: X = O^{p}(P)? and so X = ⟨O^{p}(K ∩ P)^{S}⟩. Put R^{∗} = Z(Y) ∩ X and D = O^{p}(K ∩ P). Comment: D = O^{p}(K ∩ Y)R^{∗}/R^{∗}) ??? Then X ∩ Z(Y) is normal in P and contained in B. Therefore R^{∗} ≤ R. Note that, as Y Comment: Y/Z(Y) is semisimple,

X/R^{∗} ≅ XZ(Y)/Z(Y) ≅ ∏_{T ∈D^{M}} T.

For a group L, let Φ_{p}(L) Comment: already defined in the introduction denote the full preimage of Φ(L/O_{p}(L)). Then, as P is narrow and soluble, Φ_{p}(X) ≤ R ∩ X and, as P is p-minimal Maschke’s Theorem Comment: this should be part of 2.3 implies that Φ_{p}(X) = R ∩ X. On the other hand

Φ_{p}(X/R^{∗}) = ∏_{T ∈D^{M}} Φ_{p}(T)

and so we conclude that

XR/R ≅ X/X ∩ R = ∏_{T ∈D^{M}} T/Φ_{p}(T).

Since P is narrow, Comment: M acts primitively on XR/R and it must be that D^{M} = D and so K is normalized by M, a contradiction. □

From Section ?? onwards we will be investigation Comment: investigating specific simple groups with an eye to showing that they have or do not have an isolated narrow pminimal parabolic subgroup. The next few results in this section will be applied to proper subgroups of such groups. We continue the notation from the previous lemmas. In particular, if P ∈ P_{G}(S) is narrow and isolated, then we set Y = ⟨O^{p}(P)^{G}⟩.

Lemma 2.9. [comps] Suppose that G is a group, O_{p}(G) = 1, F (G) = C_{G}(E(G)) and G operates transitively on the components of G. If P ∈ P_{G}(S) is narrow and isolated, then E(G) = Y and, in particular, there is exactly one component in G.

Proof. Assume that P is narrow and isolated in G. Then, if Y ≠ O^{p}(P) or Y = O^{p}(P) and O^{p}(P) is not soluble, 2.8 implies that Y is a component of G. Hence, as by hypothesis G acts transitively on its components, we get Y = E(G). Comment: So we may assume that Thus Y = O^{p}(P) is soluble. Since Y is a normal subgroup of G, Y centralizes E(G). Since G = BY,

Comment: G = ⟨B, Y ⟩ we have that Q_{B}Y ⊴ G and Q_{B}Y centralizes E(G). But then Q_{B}Y ≤ C_{G}(E(G)) = F (G) by assumption. Thus Q_{B}Y is nilpotent and this contradicts Q_{B} ≰ Q_{P}. □

Lemma 2.10. [quot] Suppose that P ∈ P_{G}(S) is isolated in G. Then C_{B}(Y) is normal in G and if X ≤ C_{B}(Y) is normal in G, then

(a) [a] P X/X ∈ P_{G}(SX/X) is isolated in G/X; and

(b) [b] if P is narrow, then P X/X is narrow.

Proof. We first of all note that G = BY. Comment: G = ⟨B, Y ⟩ So C_{B}(Y) = B ∩ C_{G}(Y) is normalized by G = BY. Comment: G = ⟨B, Y ⟩ Now suppose that X ≤ C_{B}(Y) is normal in G. We claim that P X/X is a p-minimal parabolic subgroup which is narrow and isolated in G/X. Suppose that R̄ ∈ P_{G/X}(SX/X) is not contained in B/X. Let R be the full preimage of R in G. Then R ≥ S and R ≰ B so P ≤ R by 2.5(a). Furthermore, as R̄ ∈ P_{G/X}(SX/X) and B ≥ X, B ∩ R is the maximal subgroup R containing SX. Since P ≰ B, we infer that R = P X and consequently R̄ = P X/X is the unique p-minimal parabolic subgroup of G/X not contained in B/X. Let U be the full preimage of O_{p}(\overline{R}) Comment: R̄ and let U_{p} = U ∩ S. We have R = N_{R}(U_{p})U by the Frattini lemma. In particular, as R ≰ B and U ≤ XS ≤ B, N_{R}(U_{p}) ≰ B. Since S ≤ N_{R}(U_{p}), it follows that P ≤ N_{R}(U_{p}) and so U_{p} ≤ Q_{P}. Since Q_{B} ≰ Q_{P}, it follows that Q_{B} ≰ U_{p} = U ∩ S. Therefore, Q_{B}X/Xnot ≤ UX/X Comment: ≰ and P X/X is isolated in G/X. This proves (a).

Suppose that P is narrow. Then P X/X ∈ P_{G}(S) by (a). We have so P X/X ≅ P/P ∩ X and, putting F = ⋂(B∩P)^{P}, we have P ∩X ≤ F. It follows that O^{p}(P)F/F ≅ Ω^{p}(P/X)(F X/X)/(F X/X)

Comment: O^{p} and so P X/X is narrow. □

Corollary 2.11. [quot2]Assume that P ∈ P_{G}(S) is narrow and p-restricted. Let Y = ⟨O^{p}(P)^{G}⟩ and assume that Y = O^{p}(P) is soluble. If the holomorph of Y is soluble but not abelian, then G′ is not perfect.

Proof. We have G/C_{B}(Y) is isomorphic to a subgroup of the holomorph of Y which is soluble but not abelian. □

Lemma 2.12. [Opnormal] Suppose that P ∈ P_{G}(S) is narrow and isolated, L ⊴ H with L soluble and C_{H}(L) ≤ L. Assume also that O_{p}(H) = 1. Then O^{p}(P) ⊴ H and P is soluble.

Proof. Set Y = ⟨O^{p}(P)^{H}⟩. Then Y = O^{p}(P) or Y is quasisimple. Assume that O^{p}(P) is not soluble. Then, because Y ⊴H, O_{p}(Y) = 1 and so when Y = O^{p}(P) we also have that Y is quasisimple by 2.7. Thus in any event Y is quasisimple. Therefore, as L is soluble, L∩Y is soluble and so L∩Y ≤ Z(Y). Hence [L, Y, Y] = 1 and the three subgroup lemma gives Y = [Y, Y] ≤ C_{H}(L) ≤ L, a contradiction. □

Lemma 2.13. [sbnrml2] Suppose that H is a group and A ≤ H. Set L = ⟨A^{H}⟩. Assume that AF (L) ⊴ L, C_{H/F (L)}(L/F (L)) = 1 and that there exists h ∈ H such that [A^{h}, A] ≤ F (L). If Y ⊴ H, then Y/F (Y) is not a non-abelian simple group.

Proof. Suppose that Y/F (Y) is a non-abelian simple group. Since Y ⊴ H, F (Y) ≤ F (H) and since [F (H), L] ≤ F (L), C_{H/F (L)}(L/F (L)) = 1 implies that F (H) = F (L). Now L ∩ Y ⊴ Y and so either Y ≤ L or L ∩ Y = F (Y). In the latter case we have [L, Y] ≤ F (L) which means that Y ≤ F (L), a contradiction. Therefore, Y ≤ L. Since F (L)A is normal in L, we either have AF (L) ∩ Y = F (Y) or AF (L) ≥ Y = Y. In the latter case we have Y ≤ AF (L) ∩ A^{h}F (L) where h is as in the statement of the lemma. This gives

[Y, Y] ≤ [AF (L), A^{h}F (L)] ≤ F (L),

which is a contradiction. Therefore, AF (L) ∩ Y = F (Y) ≤ F (L). But Y ≤ L and so Y normalizes AF (L) and this means that [AF (L), Y] ≤ AF (L) ∩ Y ≤ F (L). Thus [⟨A^{H}⟩, O^{p}(P)] ≤ F (L) and, since C_{H/F (L)}(L/F (L)) = 1, we conclude that Y ≤ F (L) and once again have a contradiction. □

Corollary 2.14. [P soluble] Suppose that H is a group, O_{p}(H) = 1 and A ≤ H. Set L = ⟨A^{H}⟩. Assume that AF (L)⊴L, C_{H/F (L)}(L/F (L)) = 1 and that there exists h ∈ H such that [A^{h}, A] ≤ F (L). If P ∈ P_{H}(S) is narrow and P is restricted in H, then O^{p}(P) ⊴ H and O^{p}(P) ≤ F (L).

Proof. Let M be a maximal subgroup of P containing a Sylow p-subgroup S of P and set R = ⋂ M^{P} Set Y = ⟨O^{p}(P)^{H}⟩. Then by (??) either Y is quasisimple or Y = O^{p}(P). However (??) implies that Y/F (Y) is not simple. So Y = O^{p}(P). Hence O_{p}(P) = 1 and R is nilpotent by (??)d. Since P is narrow (??) implies that O^{p}(P)/F (O^{p}(P)) is abelian and hence p-closed. But then O^{p}(P) is a t-group for some prime t and O^{p}(P) ≤ F (L) as claimed. □

Corollary 2.15. [P soluble2] Suppose that H is a group, O_{p}(H) = 1 and A ≤ H. Set L = ⟨A^{H}⟩. Assume that AF (L)⊴L, C_{H/F (L)}(L/F (L)) = 1 and that there exists h ∈ H such that [A^{h}, A] ≤ F (L). If L is perfect and F (L) = Z(L), then H ∉ R_{p}.

Proof. Assume that H ∈ R_{p}. By the previous corollary we have that O^{p}(P) ≤ F (L). Let B = ⟨P_{H}(S) \ {P }⟩. Then, as H ∈ R_{p}, BO^{p}(P) = H and B < H. As O^{p}(P) ≤ Z(L) ≤ L, L = L ∩ BO^{p}(P) = (L ∩ B)O^{p}(P) and L ∩ B < L. Therefore L′ ≤ ((L ∩ B)Z(L))′ ≤ L ∩ B < L, a contradiction. □

Lemma 2.16. [non-local p-minimal] Suppose G ≥ H ≥ S and Q_{H} = 1. Then H ≥ P.

Proof. Suppose H ≤ B. Then Q_{B} ≤ Q_{H} = 1, a contradiction. Thus H ≰ B and H ≥ P by 2.5(a). □

I don’t think that this next one is used.

Lemma 2.17. [no cubic] Suppose that Q_{B} is abelian and that no non-trivial normal subgroup of S acts non-trivially and cubically on Q_{B}. Then P is not of characteristic p.

Proof. Suppose not and let D be normal subgroup of P in Q_{P} minimal such that [D, O^{p}(P)] ≠ 1. Then D = [D, O^{p}(P)] ≤ O^{p}(P) and so, by the minimality of D, [D, D, D] ≤ [D, D, O^{p}(P)] = 1. In particular, as Q_{B} normalizes D, [Q_{B}, D, D, D] = 1 and so by assumption [Q_{B}, D] = 1. Since Q_{B} ≰ Q_{P}, we get [D, O^{p}(P)] = 1, which is a contradiction. □

We finish this section with one final general lemma. It exploits the fact that isolated p-minimal parabolic subgroups are normalized by N_{G}(S).

Lemma 2.18. [ngs maximal] Suppose that P ∈ P_{G}(S) is isolated in G. Suppose that one of the following holds:

1. [1] N_{G}(S) acts irreducibly on S.

2. [2] N_{G}(S) = B.

3. [3] N_{G}(S) is contained in a unique maximal subgroup of G. Then P ⊴ G

Proof. Suppose (1) holds. Since N_{G}(S) ≤ B we get that Q_{B} = S and so (2) holds.

Suppose (2) holds. Then, by 2.5(c), N_{G}(S) is a maximal subgroup of G and so (3) holds.

So we may assume that (3) holds. Since N_{G}(S) ≤ B, we get that B is the unique maximal subgroup of G containing N_{G}(S). Since N_{G}(S)P ≤ N_{G}(P) and P ≰ B, we have G = N_{G}(P) and so P ⊴ G. □

3. General observations

¿From here on we wish to determine the possibilities for Y in 2.8(??)iven that we know that O^{p}(P/Q_{P}) is a rank 1 Lie type group. So we shall assume the following hypothesis:

Hypothesis 3.1. [hyp] p is a prime, G is a group with Q_{G} = 1 and

(a) [a] X = O^{p}(G) is a non-abelian simple K-group; and

(b) [b] there is a P ∈ P_{G}(S) such that P is narrow and isolated in G (with respect to A) and O^{p}(P/Q_{P}) is a rank 1 Lie type group.

We next establish some elementary consequences of 3.1

Lemma 3.2. [s cyclic] If S is cyclic, then ⟨A^{P} ⟩ ⊴ G.

Proof. Suppose that |SQ_{P} /Q_{P} | = p, then, as A ≰ Q_{P}, S = A = Q_{B} and the result follows from 2.18(??) So assume that |SQ_{P} /Q_{P} | > p, then as S is cyclic, we have P/Q_{P} ≅ Suz(2) or 3_{+}^{1+2}.4. In particular, p = 2 and so transfer (see [?, 37.7]) implies that O^{2}(G) is group of odd order. Therefore the lemma follows from 2.5(e). □

Proposition 3.3. [GL][Gorenstein Lyons] Suppose that X is a finite simple K-group and that S is a Sylow p-subgroup of X. If S is abelian, then N_{G}(S) acts irreducibly on Ω_{1}(S).

Proof. See [3, 12-1,pg 158]. □

This allows us easily to establish the following lemma.

Lemma 3.4. [SinXnotabl] If G satisfies 3.1 and X has abelian Sylow p-subgroups, then G ≠ X.

Proof. Suppose that G = X and X has abelian Sylow p-subgroups. Then, as Q_{B} > 1, (??) and N_{G}(S) ≤ B implies that Ω_{1}(S) ≤ Q_{B} and B = N_{G}(Ω_{1}(S)). If Q_{P} > 1, then as Q_{P} is normalized by N_{G}(S), we have Ω_{1}Z(Q_{P}) = Ω_{1}(S). But then P ≤ B, a contradiction. Thus Q_{P} = 1. Since O^{p}(P/Q_{P}) is a rank 1 Lie type group, we have that S is elementary abelian. Hence 2.18 applies and we have a P ⊴ G, a contradiction. □

Lemma 3.5. [not char] Suppose that P is not of characteristic p and Q_{B} is abelian. Then Q_{B} is elementary abelian.( except maybe for Sz(2) or Ree(3)′).

Proof. Since Q_{B}Q_{P} /Q_{P} is an abelian normal subgroups of (P ∩ B)/Q_{P} we see ( except for Sz(2) that Φ(Q_{B}) ≤ Q_{P} and so Φ(Q_{B}) is normal in P and in B. Hence Q_{B} is elementary abelian □

Lemma 3.6. [irr on z(s)] Suppose that P is not of characteristic p, N_{G}(S) is irreducible on Ω_{1}Z(S) and q > 3. Then Q_{P} = 1 and |Q_{B}| = q.

Proof. By 3.5 Q_{B} is elementary abelian. Suppose q > 3. Then O^{p}(P) ∩ Q_{P} = 1. By irreducibility of N_{G}(S) on Ω_{1}Z(S) we get Z(S) ≤ O^{p}(P) ∩ S and so Ω_{1}Z(S) ∩ Q_{P} = 1. Hence Q_{P} = 1. □

Lemma 3.7. [norm] If P is not soluble, then N_{O^{p}(P)}(S ∩ O^{p}(P)) ≤ B and Q_{B}Q_{P} ⊴ N_{O^{p}(P)}(S ∩ O^{p}(P))S.

Proof. Let R = N_{O^{p}(P)}(S ∩ O^{p}(P)). Since R is normalized by S and, as P is not soluble, RS ≱ P, R ≤ B by 2.5(a). This of course then gives Q_{B}Q_{P} ⊴ RS. □

Lemma 3.8. [QBnX=1] Suppose Q_{B} ∩ O^{p}(P) = 1. Then P is soluble and, in particular, p ≤ 3.

Proof. Assume that Q_{B} ∩ X = 1 and put R = N_{O^{p}(P)}(S ∩ O^{p}(P)). Suppose that P is not soluble. Then by 3.7, R normalizes Q_{B}. Now [Q_{B}, R] ≤ O^{p}(P)∩Q_{B} ≤ X ∩Q_{B} = 1. Therefore, Q_{B}Q_{P} /Q_{P} ≤ Z(O^{p}(RS/Q_{P})). Since, from the structure of P/Q_{P}, C_{P/Q_{P}} (RQ_{P} /Q_{P}) = 1, we infer that Q_{B} ≤ Q_{P}, which is a contradiction. □

Lemma 3.9. [order8] If Q_{B} ≤ Z(S) and Z(S) is cyclic, then |S| ≤ p^{3}.

Proof. As Q_{B} ≤ Z(S), [Q_{P} , Q_{B}] = 1 and so P is not p-constrained. Therefore Ω_{1}(Q_{B}) = Ω_{1}(Z(S))∩ Q_{P} = 1 and so Q_{P} = 1. Since Q_{B}Q_{P} is normalized by N_{O^{2}(P)}(S ∩ O^{2}(P)), the structure of the rank 1 Lie type groups gives O^{p}(P)Q_{P} /Q_{P} is defined over GF(p). But then |S| = |S/Q_{P} | ≤ p^{3}, as claimed. □

4. Alternating groups

It looks like we can do this for narrow p-isolated P and not insist on Lie type. I guess it lets in Sym(p) and Sym(9).

In this section we determine those groups G are p-restricted and have O^{p}(G) = F^{∗}(G) an alternating group of degree at least 5. Among the small alternating and symmetric groups there are a number of isomorphisms with the classical groups and these examples always lead to G being p-restricted for the appropriate prime. Before proving our main result on the symmetric groups we recall these isomorphisms.

Lemma 4.1. [smallalts] We have the following isomorphisms

(a) [a] Sym(3) ≅ SL_{2}(2);

(b) [b] Alt(4) ≅ PSL_{2}(3);

(c) [c] Alt(5) ≅ SL_{2}(4) ≅ PSL_{2}(5);

(d) [d] Alt(6) ≅ PSL_{2}(9);

(e) [e] Sym(6) ≅ Sp_{4}(2);

(f) [f] Alt(8) ≅ PSL_{4}(2);

(g) [g] Sym(8) ≅ O_{6}^{+}(2).

Proof. These facts are well known. But see for example [6, 2.9.1]. □

Lemma 4.2. [altcase] Suppose that p is a prime, G is a group such that F^{∗}(G) = O^{p}(G) ≅ Alt(n) with n ≥ 5 and that G is not isomorphic to a Lie type group defined in characteristic p. If G is p-restricted, then p = 2 and n = 12. Furthermore, in this case if P is p-restricted in G, then B is the stabilizer of a system of imprimitivity with blocks {Ω_{1}, Ω_{2}, Ω_{3}} of size 4 and P, which is isomorphic to a subgroup of index at most 2 in (Sym(2) ≀ Sym(4)) × (Sym(2) ≀ Sym(2)), is contained in Stab(Ω_{1} ∪ Ω_{2}) × Stab(Ω_{3}).

Proof. Let X = F^{∗}(G) = O^{p}(G) ≅ Alt(n) with n ≥ 5, S ∈ Syl_{p}(G) and assume that P ∈ P_{G}(S) is p-restricted in G. Because of 3.2 we may assume that S is not cyclic and so in particular, n > 2p−1. Since, for p ≤ 3, Alt(6) is a Lie type group in characteristic p, we may assume that n ≥ 7. Suppose that n = 7. Then p ∈ {2, 3}. When p = 3, then N_{G}(S) operates irreducibly on S so 2.18 delivers G = P, a contradiction. So assume that p = 2. If G = Alt(7), then 2.16 implies P ≤ Alt(6) ∈ M_{G}(S) and P ≤ Sym(5) ∈ M_{G}(S), which is impossible as Sym(5) is p-minimal. For G = Sym(7), the same argument shows that P ≤ Sym(6) with P = Sym(2) ≀ Sym(3) and B = Sym(5) × Sym(2). But P must also be contained in Sym(3) × Sym(4) and we have a contradiction.

Now we assume that n ≥ 8 and that if n = 8, then p ≠ 2 (as Alt(8) ≅ SL_{4}(2)). We consider G acting on the set Ω = {1, …, n}. Set Z = Ω_{1}(Z(Q_{B})). Suppose first that B operates primitively on Ω. Then Z operates transitively on Ω and, as Z is abelian, Z acts regularly on Ω and, in particular, |Ω| = |Z| = p^{a} for some a ∈ ℕ. As n > p, we have a ≥ 2. Select a p-cycle x ∈ S (or when p = 2 a product of two transpositions). Then

| Fix_{Ω}(x)| = p^{a} − p = |C_{Z}(x)| ≤ p^{a−1}

( 2^{a} − 4 ≤ 2^{a−1} when p = 2) which has no solution in for our values of n = p^{a}. Therefore, B does not act primitively on Ω.

Assume that B is not transitive on Ω and let Ω_{1}, Ω_{2} be proper subsets of Ω fixed by B with |Ω_{2}| = k ≤ \frac{n}{2}. So B ≤ Stab_{G}(Ω_{1}) × Stab_{G}(Ω_{2}). Since B is a maximal subgroup of G we must have B = Stab_{G}(Ω_{1}) × Stab_{G}(Ω_{2}) and k ≠ \frac{n}{2} (otherwise B would not be maximal in G). Since B is a p-local subgroup, we infer that Sym(Ω_{1}) or Sym(Ω_{2}) must be a p-local subgroup. Thus, as n ≥ 8 and k ≠ \frac{n}{2}, we have that k = 2, 3, 4 and Q_{B} = O_{p}(Sym(Ω_{2})). Now let T = ⟨Q_{B}^{g} | g ∈ G, Q_{B}^{g} ≤ S⟩. Then Q_{B} ≤ T ⊴S. By 2.5(d), we have that N_{G}(T) ≤ B, but we have N_{G}(T) ≥ O_{p}(Sym(Ω_{2}))≀Sym([n/|Ω_{2}|]) which plainly does not normalize Q_{B}. Thus B operates transitively but not primitively on Ω. Let B = {Θ_{1}, …, Θ_{r}} be a system of imprimitivity for B on Ω with |Θ_{1}| = k. Then B ≤ Stab_{G}(B) and, since B is a maximal subgroup of G and B is a p-local subgroup of G, we have B = Stab_{G}(B) and k ∈ {2, 3, 4}. Suppose that k = 2 or 4. Then p = 2. Write the 2-adic decomposition of n as n = 2^{a_{1}} + 2^{a_{2}} + … 2^{a_{l}} and notice that as n ≥ 8, a_{1} ≥ 3. For i = 1, …, l, let S_{a_{i}} represent a Sylow 2-subgroup of Sym(2^{a_{i}}). Then take S = S_{a_{1}} × S_{a_{2}} × ⋯ S_{a_{l}} ∩ G as our“standard” Sylow 2subgroup. If k = 2, we note that the subgroup H := Sym(4)≀\underbrace{Sym(2) ≀ ⋯ ≀ Sym(2)}_{a_{1}−2} ×S_{a_{2}} ×… S_{a_{l}} ∩G

is not contained in B and consequently must contain P. Suppose that n ≥ 9 or G = Sym(n). Then H is actually p-minimal and we have that H = P, but H is not narrow. Thus we have that G = Alt(n) and n = 9. So we have that S ≤ Alt(8) and G = P is 2-minimal, a contradiction to the uniqueness of P (as P ≤ H also). Next assume that k = 4. Then set H := Sym(2) ≀ Sym(4) ≀ \underbrace{Sym(2) ≀ ⋯ ≀ Sym(2)}_{a_{1}−3} ×S_{a_{2}} × … S_{a_{l}}. Again we have that H ≰ B and so P ≤ H. Furthermore, if

a_{1} ≥ 4, H is 2-minimal and so H = P contradicting the structure of P. Thus a_{1} = 3 and since k = 4 divides n we infer that n = 12, and we have the example listed in the lemma.

Finally suppose that k = 3. Then let n = b_{1}3^{a_{1}} +⋯+b_{l}3^{a_{l}} be the 3-adic decomposition of n. Let S_{a_{i}} be a Sylow 3-subgroup of Sym(3^{a_{i}}) and S_{a_{l}}^{b_{l}} be a Sylow 3-subgroup of Sym(b_{i}3^{a_{i}}). Then we may suppose that S = S_{a_{1}}^{b_{1}} × ⋯ × S_{a_{l}}^{b_{l}}. We first note that Alt(9) itself is 3-minimal and so it is impossible that G = Alt(9). Then taking H := Sym(9)≀\underbrace{Sym(3) ≀ ⋯ ≀ Sym(3)}_{a_{1}−2} ≀ Sym(b_{1})×S_{a_{2}}^{b_{2}} ×⋯×S_{a_{l}}^{b_{l}} ∩Alt(n)

is not contained in B. Thus H contains P and we have a contradiction via 2.9 if either a_{1} > 3 or b_{1} > 1 and otherwise we use the fact that Sym(9) is not 3-restricted. □

5. Lie type groups in characteristic not p

In this section we begin our investigation of the groups which satisfy 3.1 and have X a Lie type group defined in characteristic r ≠ p. Our objective here is to prove two general results.

Lemma 5.1. [para-arg] Suppose that X is a simple group of Lie type defined over GF(r^{a}) and assume that for every parabolic subgroup R of X, G = N_{G}(R)X. If S ∩ X is contained in a proper parabolic subgroup R of G, then G is not p-restricted.

Proof. Suppose that R is a parabolic subgroup of X and assume that R contains S ∩ X. Then as N_{G}(R)X = G, we have that N_{G}(R) contains S. It follows that S normalizes a standard Levicomplement L of R and thus S ≤ LO_{r}(R) and S ≤ LO_{r}(R)^{x} for x ∈ G with O_{r}(R) ∩ O_{r}(R)^{x} = 1.

Setting H_{1} = SO_{r}(R) and H_{2} = SO_{r}(R)^{x} we have that O_{p}(H_{1}) = O_{p}(H_{2}) = 1 and so 2.16 implies that P ≤ H_{1} ∩ H_{2}. But then O^{p}(P) ≤ O^{p}(H_{1}) ∩ O^{p}(H_{2}) = O_{r}(H_{1}) ∩ O_{r}(H_{2}) = 1 and we have a contradiction. Thus the lemma holds. □

Lemma 5.2. [abelian] Suppose that X = O^{p}(G) is a Lie type group G(r^{a}) with p ≠ r. If X has abelian Sylow p-subgroups and P < G, then p ≤ 3 and G > X.

Proof. If X = G, we simply cite 3.4 to obtain G = P. So assume that G > X and p ≥ 5. Since O^{p}(P) ≤ X, O^{p}(P) has abelian Sylow p-subgroups and so, as p ≥ 5, O^{p}(P/Q_{P}) ≅ SL_{2}(p^{b}) or PSL_{2}(p^{b}) for some b ≥ 1 and P/Q_{P} is the same group extended by a cyclic group of field automorphisms. Since the Sylow p-subgroups of O^{p}(P) are abelian, we infer further that O^{p}(P) ≅ SL_{2}(p^{b}) or PSL_{2}(p^{b}) and that [Q_{P} , O^{p}(P)] = 1. Since N_{G}(S ∩ O^{p}(P)Q_{P})S ≤ B, Q_{B}Q_{P} ∈ Syl_{p}(O^{p}(P)Q_{P}) and Φ(Q_{B}) ≤ Q_{P}. Therefore Φ(Q_{B}) ⊴ ⟨B, P ⟩, so Q_{B} is elementary abelian. Since Q_{B} is normalized by N_{G}(S ∩ O^{p}(P)) and N_{G}(S ∩ O^{p}(P)) acts irreducibly on S ∩ O^{p}(P) and centralizes Q_{P}, we infer that Q_{B} ∩ O^{p}(P) = S ∩ O^{p}(P) > 1. In particular, Q_{B} ∩ X > 1 so Q_{B} ∩ Ω_{1}(S ∩ X) > 1 and 3.3 implies that Q_{B} ∩ X = Ω_{1}(S).

Set S_{0} = S ∩ X and consider S_{0}O^{p}(P). Since S_{0} is abelian and O^{p}(P) ≅ SL_{2}(p^{b}) or PSL_{2}(p^{b}), we infer that Φ(S_{0}) ≤ Q_{P}. But then Φ(S_{0}) ∩ Ω_{1}(S_{0}) ⊴ N_{X}(S_{0}), so 3.3 implies that Ω_{1}(S_{0}) ≤ Q_{P}. But then Ω_{1}(S_{0}) ⊴ ⟨B, P ⟩ = G so S_{0} is elementary abelian.

Since p > 3, G/X does not contain a graph automorphism. If X has a diagonal automorphism of order p, then using p > 3 again, we have that p divides (n, Φ_{1}(r^{a})) and X ≅ PSL_{n}(r^{a}) or p divides (n, Φ_{2}(r^{a})) and X ≅ PSU_{n}(r^{a}). In the first case we have that X contains a monomial subgroup \frac{1}{(n,r^{a}−1)^{2}} (r^{a} − 1)^{n}. Sym(n) and in the second case \frac{1}{(n,r^{a}+1)^{2}} (r^{a} + 1)^{n}. Sym(n) and, as p > 3, these subgroups exhibit that fact that S_{0} is non-abelian. Thus X has no diagonal automorphisms of order p. So G/X must consist of field automorphisms. As p > 3, this means that r^{a} = r^{pa_{0}} for some integer a_{0}. Letting m be the order of r^{a} mod p, [3, pg. 112] presents the relationship

\frac{Φ_{m}(r^{pa_{0}})}{Φ_{m}(r^{a_{0}})} ≡ Φ_{m}(r^{a_{0}})^{ϕ(p)} (mod p).

Finally, [3, 10-1 (2)] shows that the exponent of S_{0} is at least p^{2} (using here that X has no diagonal automorphisms of order p) and so the Sylow p-subgroups of X are not elementary abelian, which is a contradiction. This final contradiction finishes the proof of the lemma. □

We remark here that there are examples of 3-restricted groups G with X = O^{p}(G) a simple Lie type group with abelian Sylow 3-subgroups and |G/X| = 3 (see 6.11).

6. Linear and unitary groups

In this section we investigate the linear and unitary groups. To simplify our notation we use GL_{n}^{+}(r^{a}) and GL_{n}^{−}(r^{a}) to represent the general linear and unitary groups respectively. The notation GL_{n}^{ϵ}(r^{a}) denotes either of the groups. So here ϵ = ±. However we will also write r^{a} − ϵ and in this case we regard ϵ as ±1 according to ϵ = ±. Throughout this section \overline{X} ≅ SL_{n}^{ϵ}(r^{a}) and p is a prime with p ≠ r. Recall that ΓL_{n}^{ϵ}(r^{a}) has \overline{X} as a normal subgroup and includes all the diagonal and field automorphisms of \overline{X}. In the event that \overline{X} ≅ SL_{n}(r^{a}), we denote the inverse transpose automorphism by ι and, for ease of notation, when \overline{X} ≅ SU_{n}(r^{a}) or SL_{2}(r^{a}), we take ι to be the trivial automorphism. Let V represent the natural linear space when \overline{X} ≅ SL_{n}(r^{a}) and the natural unitary space when \overline{X} ≅ SU_{n}(q).

For the remainder of this section we take

\overline{X} ≤ \overline{G} ≤ ΓL_{n}^{ϵ}(r^{a}) : ⟨ι⟩

with \overline{G}/\overline{X} a p-group and \overline{G} non-soluble. Finally set G = \overline{G}/F (\overline{G}), X = \overline{X}F (\overline{G})/F (\overline{G}) and for \overline{S} ∈ Syl_{p}(\overline{G}), S = \overline{S}F (\overline{G})/F (\overline{G}). Our objective is to determine for which values of n and r^{a}, G is p-restricted. The first few lemmas form the base of our final induction arguments.

Our first lemma helps us apply induction.

Lemma 6.1. [subquot] Suppose that \overline{R} ≤ F (\overline{G}) is normal in G and P ∈ P\overline{G}(\overline{S}) is p-restricted in \overline{G}. Then either \overline{G}/ovR is p-restricted or O^{p}(\overline{P}) ≤ \overline{R}.

Proof. If O^{p}(\overline{P}) ≰ \overline{R}, then \overline{R}\overline{S} ≤ B and \overline{R} ≤ C\overline{B}(\overline{Y}). Thus the result follows from 2.10. □

Lemma 6.2. [sl2] If X ≅ PSL_{2}(r^{a}) and P is p-restricted in G, then one of the following holds:

(a) [a] p = 2, r^{a} ≡ 3, 5 (mod 8), G ≅ PGL_{2}(r^{a}), B = C_{G}(Ω_{1}(Z(S))) and P ≅ Sym(4);

(b) [a+1] p = 2, G ≅ PGL_{2}(19), B ≅ Sym(4) and P ≅ Dih(40) with P/O_{2}(P) ≅ Dih(10);

(c) [b] p = 2, G ≅ PSL_{2}(7), PSL_{2}(9) and B ≅ P ≅ Sym(4) or G ≅ PΣL_{2}(9) ≅ Sym(6) and B ≅ P ≅ Sym(4) × 2;

(d) [c] G = P ≅ PSL_{2}(5) ≅ Alt(5) or PGL_{2}(5) ≅ Sym(5);

(e) [d] p = 3, X ≅ ^{2}G_{2}(3) ≅ PSL_{2}(8).3 = PΓL_{2}(8).

Proof. We first consider the case when p = 2. Set S_{0} = S ∩ X. Let t be a central involution in S_{0} and put D = C_{G}(t). We shall often use the following straight forward observation.

1^{∘}. [1] If P ≤ D, then P/O_{2}(P) ≅ Sym(3), Dih(10) or Frob(20) and D∩B is a maximal subgroup of D of index 3 in the first case and 5 in the last two cases.

Suppose first that |S_{0}| = 4. Then r^{a} ≡ 3, 5 (mod 8) and consequently a is odd. Therefore X admits no field automorphisms of order 2 and G is isomorphic to a subgroup of PGL_{2}(r^{a}). If G = X, then 5.2 implies that G = P and in this case (d) holds. Hence we have G ≅ PGL_{2}(r^{a}). So S ≅ Dih(8) and N_{G}(S_{0}) ≅ Sym(4). Assume that P ≤ D. Then, as B is a 2-local subgroup of G, we must have B ≤ N_{G}(S_{0}). Hence (1^{∘}) implies that |D| = 24 if P/O_{2}(P) ≅ Sym(3) and |D| = 40 if P/O_{2}(P) ≅ Dih(10). Since r^{a} ≡ 3, 5 (mod 8) and |D| = r^{a} ± 1, we infer that the second possibility occurs and that G ≅ PGL_{2}(19) which is possibility (b). Next suppose that have P ≰ D. Then B ≥ D and, as N_{G}(S_{0}) ≅ Sym(4) is 2-minimal and is not contained in B, we have P = N_{G}(S_{0}). This delivers the examples listed in part (a). (Here we may mention that if a > 1, then P is contained in the subfield subgroups PGL_{2}(r^{a/b}) whenever b divides a.)

Assume that |S_{0}| ≥ 8 and let F_{1} and F_{2} be non-conjugate fours subgroups of S_{0}.

2^{∘}. [2] Q_{P} ∩ X > 1.

Suppose that Q_{P} ∩ X = 1. Then P ≰ D and so B ≥ D. Additionally, we have O_{2}(O^{2}(P)) ≤ Q_{P} ∩ X = 1 and so, as O^{2}(P/Q_{P}) ∈ L_{1}(p), O^{2}(P) ≅ SU_{3}(2^{n}), PSU_{3}(2^{n}), ^{2}B_{2}(2^{n}), SL_{2}(2^{n}) for some n ∈ ℕ or we have one of our unusual cases O^{2}(P) ≅ 3, 5, 3_{+}^{1+2}, 3^{2}.2, 2.SL_{2}(4). Since S_{0} is a dihedral group and the Sylow 3 and 5-subgroups of X are cyclic, either O^{2}(P) ≅ SL_{2}(4) or P is soluble. In the latter case we have that |S/Q_{P} | ≤ 8 |S| = 2^{3}|Q_{P} | leaving only the extreme case with P/Q_{P} ≅ PSU_{3}(2) ≅ 3^{2} : Q_{8} as a possibility and this contradicts the fact that S/Q_{P} ≅ S_{0} is dihedral. So O^{2}(P) ≅ SL_{2}(4) and again by considering orders we get P/O_{2}(P) ≅ SL_{2}(4) : 2 and |S_{0}| = 8. Since SL_{2}(4) : 2 contains a subgroup P_{1} = Sym(4) and since P_{1} is 2-minimal, B ≥ ⟨P_{1}, D⟩ and hence Q_{B} ∩ X = 1. But then 3.8 implies that P is soluble, a contradiction.

3^{∘}. [3] If P ≰ D, then (c) holds.

Assume that P ≰ D. Then necessarily B ≥ D. By (2^{∘}) we have that Q_{P} ∩ X > 1. Therefore, as P does not centralize t, we have Q_{P} ∩ X is a fours group F of S. Consequently S also normalizes F and so S_{0} is a dihedral group of order 8. Furthermore, as in this case the Sylow 2-subgroups of PGL_{2}(r^{a}) are dihedral of order 32, we infer that G ≤ PΣL_{2}(r^{a}). Letting F_{1} ≠ F also be a fours group of S_{0}, we have H = N_{G}(F_{1}) is also a minimal parabolic subgroup (because H and P are conjugate in PΓL_{2}(r^{a})). Thus H ≤ B also. But then D ≅ Dih(8) or Dih(8) × 2 and we infer that one of the possibilities in part (c) occurs.

We now assume that P ≤ D. Suppose that Q_{B} ∩ X > 1. If |S_{0}| > 8, then ⟨P, B⟩ ≤ C_{G}(t) and we have a contradiction. So |S_{0}| = 8 and Q_{B}∩X is a fours group F_{1} of S_{0}. Furthermore, F_{1} is normalized by S. Let F_{2} be a fours group of S_{0} not equal to F_{1}. Then N_{2} = N_{X}(F_{2})S is 2-minimal and so as N_{2} ≰ D and P ≤ D, N_{2} ≤ B. But then 1 = O_{2}(⟨N_{1}, N_{2}⟩) = Q_{B} ∩ X, a contradiction. Therefore, Q_{B} ∩ X = 1 and in particular, S_{0} ≤ C_{X}(Q_{B}). It follows at once that G is not isomorphic to a subgroup of PGL_{2}(r^{a}) which has dihedral Sylow 2-subgroups. Hence X admits field automorphisms and consequently as |S_{0}| = 8, r^{a} ≡ 7, 9 (mod 16) and a is even. So in fact r^{a} ≡ 9 (mod 16) and hence D ∩ X has order r^{a} − 1. Notice that the centre of the Sylow 2-subgroup of Aut(X) has order 2. Let θ be an automorphism of X which normalizes S and maps Q_{B} to Q_{B}^{θ}. Then D^{θ} = D and N_{G}(Q_{B}^{θ}) is also a 2-local subgroup and which is unequal to B = C_{G}(Q_{B}) (as Q_{B}Q_{B}^{θ} ≥ ⟨t⟩). Therefore, P ≤ N_{G}(Q_{B}^{θ}), But DS is invariant under θ and so O^{2}(P) = O^{2}(P S) ≤ N_{G}(Q_{B}^{θ})^{θ^{−1}} = B, a contradiction.

Assume that p ≥ 3. Since the Sylow p-subgroups of PSL_{2}(r^{a}) are abelian, p = 3 and G > X by 5.2. Now O^{3}(P) ≤ X has cyclic Sylow 3-subgroups. Since O^{3}(P) ∈ L_{1}(3), have O^{3}(P) ≅ Q_{8}, 2^{2} or 2G_{2}(3)′. By considering the Sylow 2-subgroup structure of X we deduce that O^{3}(P) ≅ 2^{2} or r^{a} is a power of 2 and O^{3}(P) ≅ ^{2}G_{2}(3)′. If, in the latter case, r^{a} > 8, then O^{3}(P) is not normalized by N_{G}(S) and we have a contradiction. Thus if O^{3}(P) ≅ ^{2}G_{2}(3)′, we have G = P and we are done. So O^{3}(P) ≅ 2^{2} and P/Q_{P} ≅ PSL_{2}(3). Now Q_{P} ∩ X is cyclic and centralized by O^{3}(P). Since the centralizer of an involution in X is dihedral, we infer that Q_{P} ∩ X = 1. Hence P ∩ X ≅ PSL_{2}(3) and so X has a cyclic Sylow 3-subgroup of order 3. Since X admits field automorphisms of order 3, we in fact have |S ∩ X| is divisible by 9, a contradiction. □

If p is odd then set d = ord(p, ϵr) and if p = 2 put d = 1 if r^{a} ≡ ϵ (mod 4) and otherwise put d = 2. Define \overline{L^{∗}} = GL_{d}^{ϵ}(r^{a}) ≀ Sym(s) × GL_{n−sd}^{ϵ}(r^{a}) where s = [\frac{n}{d}], put \overline{L} = \overline{L^{∗}} ∩ \overline{G} and finally L = \overline{L}/F (\overline{G}). Then we may suppose that S ∩ X ≤ L and that L is normalized by S (notice that the natural realization of L is closed under inverse-transpose operation). We shall use similar notation for other imprimitive subgroups of \overline{G}. Let \overline{K^{∗}} be the base group of \overline{L^{∗}} and \overline{K} = \overline{K^{∗}} ∩ G.

In the next lemma we use the well-known fact that the irreducible section of the natural GF(p) Sym(s)- permutation module has dimension s − 1 if (s, p) = 1 and s − 2 if p divides s. In the latter instance the permutation module is uniserial.

Lemma 6.3. [sln1] Suppose that G ≤ GL_{n}^{ϵ}(r^{a}), n = ds > 2 and assume that L ≤ H. If O_{p}(H) > 1, then either

(a) [a] H > L, n = s = 4, d = 1, r^{a} = 4 + ϵ and \overline{H} ∼ 4 ∗ 2_{+}^{1+4}.(2 × Sym(6)) ∩ \overline{G}; or

(b) [b] H = LS and either d = 1 or d = 2, r^{a} ∈ \{2, 3\} and \overline{L^{∗}} ≅ GL_{2}^{+}(r^{a}) ≀ Sym(s).

Proof. Assume that d = 1 and let R be a normal p-subgroup of L. Put R_{0} = Ω_{1}(R). If \overline{R_{0}} ≤ \overline{K}, then as R_{0} ≠ 1, |R_{0}| = p^{s−1} if (p, s) = 1 and |R_{0}| ≥ p^{s−2} if (p, s) = p. In either case, we have that the homogeneous components of \overline{R_{0}} on V coincide with those of K and consequently H = LS. So we may assume that R_{0} ≰ K. This means that K is a 2-group and s = 4 or 2. The case s = 2 has been ruled out by assumption and so we have that s = 4 and X ≅ PSL_{4}(r^{a}). Now by the definition of d, we have the 4 divides r^{a} − ϵ and consequently, as [K, R_{0}] ≤ K ∩ R_{0} which is elementary abelian, we have that r^{a} = 4+ϵ as claimed in (a). So suppose that d = 2. If r^{a} = 2, then an argument similar to the one above shows that H normalizes O_{3}(L) and we are done. So suppose that r^{a} = 3, let R be a normal 2-subgroup of H. Then R ≤ O_{2}(L) and this is contained in the base group \overline{K} of \overline{L}. Consider \overline{R}_{0} = \overline{R} ∩ Z(\overline{K}). Since Z(\overline{K}) is the natural GF(2) Sym(s) permutation module for \overline{L}, we either have |\overline{R}_{0}| = 2 or |\overline{R_{0}}| ≥ 2^{n−1}. If |\overline{R_{0}}| > 2, then the \overline{K} and \overline{R_{0}} have the same homogeneous components and we are done. So \overline{R_{0}} ≤ Z(G). But then R_{0} ∩Z(K) = 1, a contradiction as Z(K) ≥ Z(S). Finally assume that r^{a} > 3 and that d ≥ 2. Then as L is a p-local subgroup and K is non-soluble, we have that O_{p}(L) ≤ Z(K) and then, as d ≥ 2, we have that d = 2 = p and r^{a} − ϵ is not divisible by 4. We now argue as in the r^{a} = 3 case to obtain our contradiction. □

Lemma 6.4. [QBX] Suppose that n = ds and B ≥ LS, then Q_{B} ∩ X > 1.

Proof. Suppose that Q_{B} ∩ X = 1. Then [Q_{B}, L ∩ X] = 1. Notice that \overline{Q_{B}} induces the same type of automorphism on each SL_{d}^{ϵ}(r^{a}) contained in \overline{K} as it does on X. Suppose that d > 2 or p is odd and d ≥ 2. Then the Sylow p-subgroup structure of Out(X) is identical to that of SL_{d}^{ϵ}(r^{a}) and so in these cases it is impossible for Q_{B} ∩ X = 1. Assume that p = 2 and d = 2. Then the inverse transpose automorphism of X induces an inner automorphism of SL_{2}(r^{a}) but not of GL_{2}(r^{a}) and thus in this case also it is impossible for Q_{B} to centralize K. Assume that d = 1. In this case, as inversion is not a field automorphism, we deduce that the only possibility is that Q_{B} induces a diagonal automorphism of X. But then Q_{B} ≤ L and the result now follows. □

Lemma 6.5. [monomial1] Suppose that t is a prime and L ≤ G is as above with d = 1 so that L ∩ X ∼ \frac{1}{(r^{a}−ϵ,n)}(r^{a} − ϵ)^{n−1}. Sym(n) and n ≥ 4. If Y is a non-trivial elementary abelian t-group of rank at most 2 and Y ⊴ L, then n = 4 and p = 2.

Proof. This is the result of an elementary calculation. □

Lemma 6.6. [monomial2] Suppose that L ∩ X ∼ \frac{1}{(r^{a}−ϵ,n)}(r^{a} − ϵ)^{n−1}. Sym(n) with n ≥ 3. Set T = F (L). If P ∈ P_{LS}(S) is p-restricted in LS, then either

(a) [a] T is a p-group; or

(b) [b] P is soluble and n = 3.

Proof. If P ∈ PLS(S) is p-restricted in LS and T is not a p-group. Then O_{p}(LS) ∩ X ≤ O_{p}(L) which is abelian. Furthermore, E(LS/O_{p}(LS)) = 1, and so O^{p}(P/O_{p}(LS)) is normal and soluble in LS/O_{p}(LS) by ??. Since O^{p}(P) ≤ O^{p}(LS) ≤ X, we have that O^{p}(P/O_{p}(LS)) ≤ T/O_{p}(LS). Because T is abelian, we have O^{p}(P) ⊴ L and, as O^{p}(P) is a t-group for some prime t ∈ {2, 3} of rank at most 2, we either have n ≤ 3 or n = 4 and t = 2 by 6.5. The former case gives (b). In the latter case, we have p = 3 and Q_{B}O^{p}(P)O_{3}(LS)/O_{3}(LS) ≅ PSL_{2}(3). Since also [Q_{B}, C_{B}(Y O_{3}(LS)/O_{3}(LS)] = 1, we obtain a contradiction in this case. □

Lemma 6.7. [monomial3] Suppose that n = p ≥ 5 and d = 1, Then LS contains every proper p-local subgroup of G which contains S.

Proof. We have \overline{L^{∗}} ≅ (r^{a} − ϵ) ≀ Sym(p). Let R be a proper p-local subgroup of G containing S. Set Q_{R} = O_{p}(R) and Q_{L} = O_{p}(L). Now LS/F (L) ≅ Sym(p) × T where T is a cyclic p-subgroup (which may be trivial) consisting of Frobenius automorphisms. In particular, we have S/F (LS) is abelian and so Q′_{R} ≤ Q_{L}. If V |\overline{Q′_{R}} is not homogeneous, then the homogeneous components of \overline{Q′_{R}}

coincide with the homogeneous components of \overline{Q_{L}} and we conclude that R ≤ L as claimed. If on other other hand V |\overline{Q′_{R}} is homogeneous, then \overline{Q_{R}}′ ≤ Z(\overline{L}). It follows that [\overline{Q_{L}}, \overline{Q_{R}}, \overline{Q_{R}}] ≤ Z(\overline{L}). In particular, we see that Q_{R} is not inducing field automorphisms on Q_{L}. Furthermore, since \overline{Q_{L}}/Φ(\overline{Q_{L}}) is isomorphic to a section of the GF(p)-permutation module for Sym(p) of order at least p^{p−1}, p ≥ 5 and [\overline{Q_{L}}, \overline{Q_{R}}, \overline{Q_{R}}] ≤ Z(\overline{L}) imply that Q_{R} ≤ Q_{M}. If V |\overline{Q_{R}} is also homogeneous, then Q_{R} ≤ Z(X) and and R = G, a contradiction. Therefore, V |_{Q_{R}} is not homogeneous and again we have that R ≤ M. □

Lemma 6.8. [L3andU3] Suppose that p = 2, X ≅ PSL_{3}^{ϵ}(r^{a}) and P is p-restricted in G. Then one of the following holds.

(a) [a] G ≅ PSU_{3}(r^{a}) with s ≡ 3 (mod 8), P ≅ 2. Sym(4) ∗ 4 and B ≅ (s + 1)^{2} : Sym(3).

(b) [b] G ≅ PSU_{3}(r^{a}) : 2 with s ≡ 3 (mod 8), P ≅ 2. Sym(4) ∗ Q_{8} and B ≅ (s + 1)^{2} : (2 × Sym(3)).

(c) [c] G ≅ PSU_{3}(3) ≅ G_{2}(2)′, P ≅ 4^{2} : Sym(3) and B ≅ 2. Sym(4) ∗ 4.

(d) [d] G ≅ PSU_{3}(3) : 2 ≅ G_{2}(2), P ≅ 4^{2} : (2 × Sym(3)) and B ≅ 2. Sym(4) ∗ Q_{8}.

(e) [e] G ≅ PSL_{3}(r^{a}) : ⟨ι⟩ with s ≡ 5 (mod 8), P ≅ 2. Sym(4) ∗ Q_{8} and B ≅ (s − 1)^{2} : (2 × Sym(3)).

Proof. Let S ∈ Syl_{2}(G) and assume that P ∈ P_{G}(S) is p-restricted. If r^{a} = 3, we use the Atlas [2] to see that statements (a), (b), (c) and (d) hold. So assume that r^{a} > 3.

Set \overline{L_{1}^{∗}} = GL_{2}^{ϵ}(r^{a})×GL_{1}^{ϵ}(r^{a}), \overline{L} = \overline{L^{∗}}∩G. Let K_{1} be the component of L_{1}. Then K_{1} is S-invariant. By 6.3, O_{2}(⟨K_{1}, L, S⟩) = 1 and so P ≤ K_{1}S or P ≤ L_{1}S (or both). Assume that B ≥ K_{1}S, then Q_{B} ≤ O_{2}(K_{1}S) ≤ Ω_{2}(LS) ≤ Q_{P}, which is a contradiction (note here it could be that r^{a} − ϵ is a power of 2 and that Q_{B} ≰ X induces ι). Thus P ≤ K_{1}S and is p-restricted therein. Using (??) together with s ≡ 1 (mod 4) implies that r^{a} ≡ ϵ(5) (mod 8) and that P ≰ LS. Noting also that when ϵ = +, \overline{S ∩ X} does not operate irreducibly on V, 5.1 implies that G > X. Thus we have the examples in (a), (b) and (e).

So suppose that s ≡ −ϵ (mod 4). Then L ≅ GL_{2}^{ϵ}(r^{a}) contains a Sylow 2-subgroup of X. Suppose that R is a 2-local subgroup of X which contains S ∩ X. Then the decomposition of V restricted to \overline{Ω_{1}(Z(Q_{R}))} is preserved by R and we see that P ∩ X ≤ N_{X}(Ω_{1}(Z(Q_{R}))) ≤ (r^{a} − ϵ)^{2} : Sym(3) or P ≤ LS. The former subgroup does not contain a Sylow 2-subgroup of X and so we infer that every 2-local subgroup of X which contains S ∩ X is contained in LS. Hence we cannot have both Q_{B} ∩ X > 1 and Q_{P} ∩ X > 1. Suppose that Q_{B} ∩ X = 1. Then Lemma 3.7 implies that P is soluble and that P is not 2-constrained. Now L′S < L and contains a Sylow 2-subgroup of G. Thus P ≤ L′S. Then as P is not 2-constrained, Lemma 6.2 implies that s = 19 and G ≅ PGL_{3}^{+}(19) : 2 and that (B ∩ L)Z(L)/Z(L) ≅ Sym(4). But then Q_{B} normalizes L/Z(L) and centralizes (B ∩ L)Z(L)/Z(L). It follows from the structure of Aut(PGL_{2}(19)) that [Q_{B}, L] ≤ Z(L), but then [O^{2}(P), L/Z(L)] = 1, a contradiction.

Next suppose that Q_{P} ∩ X = 1. Then B ≥ N and so |Q_{B} ∩ X| = 2. Since |Q_{B}Q_{P} /Q_{P} | is also cyclic, we infer that P/Q_{P} has Sylow 2-subgroups of order at most 8. Therefore, |S| ≤ 8|Q_{P} | so that |S ∩ X| ≤ 8, which is of course a contradiction. □

Lemma 6.9. [4dim] Suppose that p = 2, and X is isomorphic to PSL_{4}^{ϵ}(r^{a}). If P ∈ P_{G}(S) is 2-restricted, then X ≅ PSU_{4}(3) and the possibilities for P and B and G are as described in Table 1. In particular, if G ≥ PGU_{4}(3), then P ≤ LS.

Proof. Let X = PSL_{4}^{ϵ}(r^{a}). If r^{a} = 3, we inspect the Atlas [2] and see that in the case that X = PSL_{4}(3) there is no candidate for B and so we have that X ≅ PSU_{4}(3) and the possibilities for B and P follow. So assume that r^{a} > 3. Since p = 2, \overline{L_{1}} = GL_{2}^{ϵ}(r^{a}) ≀ Sym(2) ∩ \overline{G} is normalized by S. Using 2.9 and r^{a} > 3 we infer that B ≥ L_{1}S. Assume that s − ϵ ≡ 0 (mod 4). Then d = 1 and \overline{L_{1}} ≠ \overline{L}. By 6.3, B doesn’t contain LS and so P ≤ LS. Since Q_{B} ≤ O_{2}(L), we have Q_{B} ≤ O_{2}(L_{1}) ≤ Q_{P}, a contradiction. So assume that r^{a} + ϵ ≡ 0 (mod 4), then Q_{B} ∩ X = Ω_{1}(Z(S)) has order 2 and again Lemma 3.9 finishes this case. □

Lemma 6.10. [su63] Suppose that X ≅ PSU_{n}(3) with n ∈ {5, 6, 7}. If P ∈ P_{G}(S) is 2-restricted, then X ≅ PSU_{6}(3) and \overline{B} ≥ GU_{2}(3) ≀ Sym(3) ∩ G and P ≤ LS ∼ 4^{5}. Sym(6).

Proof. Suppose first that X ≅ PSU_{5}(3). Then L ≅ 4^{4}. Sym(5) is a 2-minimal minimal parabolic. Thus either L ≤ B or P = L. Let \overline{L_{1}} = GU_{4}(3) × GU_{1}(3) ∩ \overline{G}. Then L_{1} ≅ GU_{4}(3). If B ≥ L_{1}, then, using 6.3, P ≤ LS and Q_{B} ≤ O_{2}(L) ≤ Q_{P}, a contradiction. Therefore, P ≤ L_{1}S and B ≥ L. However, since L_{1}/O_{2}(L_{1}) ≅ PGU_{4}(3), applying 6.9 we obtain P ≤ LS ≤ B, a contradiction .

Next consider X ≅ PSU_{6}(3). Then L∩X ≅ 4^{5}. Sym(6). We set \overline{L_{1}} = (GU_{4}(3)×GU_{2}(3))∩\overline{G} and \overline{L_{2}} = GU_{2}(3) ≀ Sym(3) ∩ \overline{G}. If P ≤ L_{2}S and we 2.12 implies that O^{p}(P)O_{2}(L_{2}S)/O_{2}(L_{2}S) is normal in L_{2}S/O_{2}(L_{2}S) and has order 3. Then Q_{B}O^{p}(P)O_{2}(L_{2}S)/O_{2}(L_{2}S) ≅ Sym(3) and is a direct factor of L_{2}S/O_{2}(L_{s}S), which is a contradiction. Hence B ≥ L_{2}S and by 6.3 B does not contain L and so P ≤ LS.

Suppose finally that X ≅ PSU_{7}(3). Then \overline{L} ≅ 4^{7} : Sym(7). Since Sym(7) is not 2-restricted, we have B ≥ LS. Setting \overline{L_{1}} = GU_{6}(3) × GU_{1}(3) ∩ \overline{G}, we have that P ≤ L_{1}S we are in the PSU_{6}(3) configuration. In particular, P ≤ (L_{1} ∩ L)S ≤ B, a contradiction. □

We next consider the situation when p = 3 and X ≅ PSL_{3}^{ϵ}(r^{a}). However, before initiating the investigation we draw attention to the fact that when r^{a} − ϵ ≡ 3, 6 (mod 9), then 3_{+}^{1+2} : SL_{2}(3) is contained in GL_{3}^{ϵ}(r^{a}) and contains a Sylow 3-subgroup of GL_{3}(r^{a}). (See [4, 6.5.3].)

Lemma 6.11. [3dimchr3] Suppose that p = 3 and X ≅ PSL_{3}^{ϵ}(r^{a}).If P ∈ P_{G}(S) is 3-restricted, then G ≅ PGL_{3}^{ϵ}(r^{a}), r^{a} − ϵ ≡ 3, 6 (mod 9) with r^{a} ≠ 4 and either

(a) [a] P ≅ 3^{2} : SL_{2}(3); or

(b) [b] r^{a} = 7 and P ≅ (3^{2} × 2^{2}) : Sym(3).

Proof. Notice that 3 divides one of r^{a} + 1 and r^{a} − 1. If r^{a} − ϵ is not divisible by 3, then d = 2 and L ≅ GL_{2}^{ϵ}(r^{a}). Since O_{3}(LS) = 1, we infer that P ≤ LS and LS is 3-restricted. Then calling upon 6.2 delivers a contradiction. So we may assume that 3 divides r^{a} −ϵ. In this case \overline{L}^{∗} ≅ GL_{1}^{ϵ}(r^{a})≀Sym(3). Let \overline{K} be the base group of \overline{L^{∗}} and \overline{K_{X}} = \overline{K} ∩ \overline{X}.

If r^{a} = 7 we examine the Atlas [2] and observe the configuration in (a) and (b). So assume that r^{a} ≠ 7. Let t be a prime dividing r^{a} − ϵ and, if l = 2, assume that t^{2} divides r^{a} − 1. Let T ∈ Syl_{t}(K_{X}). Then T normalized by S and, by the choice of t (and using 2.8), P ≰ ST. It follows that ST ≤ B and Q_{B} ≤ O_{3}(ST) = O_{3}(K_{X}S). If P ≤ LS, then Q_{P} ≥ O_{3}(K_{X}S) and we have a contradiction. Therefore P ≰ LS and, by 6.3 B = LS and Q_{B} = O_{3}(LS).

If P is of characteristic 3, then O_{3}(O^{3}(P)) > 1. Let D be a normal subgroup of O^{3}(P) chosen minimal such that [D, O^{3}(O_{3}(P))] ≠ 1. If \overline{D} is abelian, then D ≤ Q_{B} and we have a contradiction as [D, Q_{B}] ≠ 1. Therefore D is non-abelian and hence irreducible on V. In particular, the centralizer of D is cyclic and P/C_{P} (D) ≅ SL_{2}(3). By considering the order of S we now obtain the configuration listed in part (a).

Suppose that P is not of characteristic 3. Then by 3.5 either Q_{B} is elementary abelian or O^{3}(P) ≅ Ree(3)′ ≅ SL_{2}(8). In the former case, it follows that G/X does not have field automorphisms, that |S| = 3^{3} and that X has elementary abelian Sylow 3-subgroups. Therefore, the only possibilities for O^{3}(P) in this case are SL_{2}(9), PSL_{2}(9), 2^{2} or Q_{8}. Since 3 does not divides | Out(PSL_{2}(9))|, the first two cases deliver S elementary abelian of order 27 which is impossible. Therefore, O^{3}(P) is a 2-group and |Q_{P} | = 9. But then Q_{P} is abelian and P is contained in a conjugate of L, a configuration we have already seen off. So assume that Q_{B} is not elementary abelian and that O^{3}(P) ≅ Ree(3)′ ≅ SL_{2}(8). By considering representations of the normalizer of a Sylow 2-subgroup of O^{3}(P), we see that r = 2. Now |S ∩ X| > 27 and so Q_{P} ∩ X > 1. Since [O^{3}(P), Q_{P} ∩ X] = 1, and O^{3}(P) ≰ LS, \overline{O^{3}(P)} is contained in GL_{2}^{ϵ}(r^{a}) × GL_{1}^{ϵ}(r^{a}) in a unique way. But then \overline{S} has to normalize this configuration, a contradiction. □

We next consider the immediate repercussions if 6.11.

Lemma 6.12. [4dim2] Suppose that p = 3, X ≅ PSL_{n}^{ϵ}(r^{a}) with 4 ≤ n ≤ 5 and P ∈ P_{G}(S) is 3-restricted. Then either

(a) [a] G = X ≅ PSU_{4}(r^{a}) with s ≡ 2, 5 (mod 9), P ≅ 3^{1+2} : GL_{2}(3) and B ≅ \frac{1}{(2,s+1)}(s + 1)^{3} : Sym(4);

(b) [b] G = X ≅ PSU_{4}(2), P ≅ 3^{3} : Sym(4) and B ≅ 3_{+}^{1+2} : SL_{2}(3); or

(c) [c] G = X = PSU_{5}(2), P ≅ 3 × 3_{+}^{1+2}.SL_{2}(3) and B = 3^{4}. Sym(5).

Proof. If X ≅ PSL_{n}(r^{a}), then \overline{S}∩\overline{X} cannot act irreducibly on V and consequently S∩X is contained in a parabolic subgroup of X contrary to 5.1. So assume that X ≅ PSU_{n}(r^{a}). We consider the case n = 4 first. Assume that d = 2. Then \overline{L^{∗}} ≅ GU_{2}(r^{a}) ≀ 2, since r^{a} > 3, 2.9 implies that L ≤ B. But then Q_{B} ≤ F (L) and we have that 3 divides r^{a} + 1, which is a contradiction. So d = 1 and we have that \overline{L^{∗}} = (s + 1) ≀ Sym(4). Set \overline{L_{1}} = (GU_{3}(r^{a}) × GU_{1}(r^{a})) ∩ \overline{G}. Assume that B ≥ L_{1}. Then, by 6.3, P ≤ LS. If r^{a} > 2, then Q_{B} ≤ O_{3}(L_{1}) ≤ O_{3}(L) ≤ Q_{P}, a contradiction. Thus if B ≥ L_{1}, then s = 2 and (b) holds. So assume that P ≤ L_{1}. Then 6.11 gives s ≡ 2, 5 (mod 9) as claimed in (a).

Assume now that n = 5. Arguing as in the n = 4 case we have that 3 divides d = 1. Set \overline{L_{1}} = (GU_{3}(r^{a}) × GU_{2}(r^{a})) ∩ \overline{G}. Since Sym(5) is not 3-restricted, we have that B ≥ L from 6.6. Assume that s > 2, then B contains at least one of the components from L_{1} and then B ≥ ⟨LS, L_{1}S⟩, a contradiction to(??). So s = 2 and we have X ≅ SU_{5}(2). Finally, we inspect the subgroup structure in the Atlas [2] to obtain the result as stated in (c). □

The situation in X = SU_{4}(r^{a}) with r^{a} ≡ 2, 5 (mod 9) is more exotic than a first look suggests. In the case that s = 2, we have that U_{4}(2) ≅ Sp_{4}(3). So suppose that s > 2. Then in the monomial group (s + 1)^{3}. Sym(4), there are three subgroups isomorphic to 3^{3} : Sym(4) each containing N_{G}(S) call them P_{1}, P_{2}, P_{3} with P_{i} ∩P_{j} = N_{G}(S) whenever i ≠ j. Let P = 3_{+}^{1+2}SL_{2}(3). Then, up to change of notation we have, we have ⟨P, P_{1}⟩ ≅ ⟨P, P_{2}⟩ ≅ PSp_{4}(3) and ⟨P, P_{3}⟩ ≅ GU_{3}(r^{a}). The embedding of PSp_{4}(3) into SU_{4}(r^{a}) stems from the fact that SU_{4}(r^{a}) ≅ Ω_{6}^{−}(r^{a}) and PSp_{4}(3) has index 2 in the Weyl group of type E_{6}. Finally we note that the two subgroups ⟨P, P_{1}⟩ and ⟨P, P_{2}⟩ are conjugate in GU_{4}(r^{a}).

Because of the situation for SU_{5}(2) we need to individually inspect SU_{6}(2) and SU_{7}(2).

Lemma 6.13. [u62andu72] Suppose that X ≅ SU_{6}(2) or SU_{7}(2) and P is 3-restricted in G. Then X ≅ SU_{6}(2), |G/X| ≤ 3, P ∩ X ≅ 3^{5} : PSL_{2}(9) ≅ 3^{5} : Alt(6) and B ∩ X ∼ 3_{+}^{1+4}.(Q_{8} × Q_{8}) : 3.

Proof. For G ≅ SU_{6}(2), we just inspect the Atlas[2]. So suppose that X ≅ SU_{7}(2). Then L ≅ 3^{6} : Sym(7). Since Sym(7) is not 3-restricted, we have B ≥ LS. Setting \overline{L_{1}} = GU_{6}(2) × GU_{1}(2) ∩ \overline{G}. We have that P ≤ L_{1}S. From the PSU_{6}(2) example we now read that P ∼ 3^{6} : Alt(6) ≤ B, which is a contradiction. □

Lemma 6.14. [dimp2] Suppose that p ≥ 5 and n ≤ p. If X ≅ P̸SL_{n}^{ϵ}(r^{a}), then G is not p-restricted.

Proof. Suppose for a contradiction that P is p-restricted in G. If n < p or d > 1, then X has abelian Sylow p-subgroups and 5.2 delivers a contradiction. Therefore, n = p and d = 1. So \overline{L^{∗}} ≅ (r^{a} − ϵ) ≀ Sym(p). Using p ≥ 5, 6.7 and 3.8 we have that B = LS and Q_{P} = 1 ∩ X. Since Q_{B} is normalized by N_{P} (O^{p}(P)∩S) and Q_{B} ∩X = O_{p}(LS)∩X is abelian, we have that O^{p}(P) ≅ SL_{2}(p^{b}), PSL_{2}(p^{b}) for some a and Q_{B} ∩ O^{p}(P) ∈ Syl_{p}(O^{p}(P)) is elementary abelian. Since Q_{P} ∩ X = 1, we then have b = p − 2 and P ∩ X = O^{p}(P) because SL_{2}(p^{p−2}) has no field automorphisms. This contradicts the fact that X has non-abelian Sylow p-subgroups. □

We at last come to the induction argument.

Lemma 6.15. [LU-generic] Suppose that X ≅ PSL_{n}^{ϵ}(r^{a}) and assume one of the following conditions hold:

(a) [a] if p = 2 and r^{a} > 3 or X ≅ PSL_{n}(r^{a}), then n ≥ 5;

(b) [a1] if p = 2 and X ≅ PSU_{n}(3), then n ≥ 8;

(c) [b] if p = 3 and r^{a} > 2 or X ≅ PSL_{n}(r^{a}), then n ≥ 6;

(d) [b2] if p = 3 and X ≅ PSU_{n}(2), then n ≥ 8;

(e) [c] if p ≥ 5, then n > p. Then G is not p-restricted.

Proof. Furthermore, assume that n as defined in the statement of the lemma and then is chosen minimally so that G is p-restricted. Recall that \overline{L^{∗}} = GL_{d}^{ϵ}(r^{a}) ≀ Sym(s) × GL_{n−ds}^{ϵ}(r^{a}) where s = [\frac{n}{d}] and L is normalized by S.

1^{∘}. [1] ds = n.

Assume that ds < n. Plainly d > 1. Then S ∩ X centralizes an n − ds dimensional subspace of V and so, in particular, it centralizes a 1-dimensional subspace W of V. If ϵ = + and G/X consist of diagonal or graph automorphisms or ϵ = − and W is singular, then S ∩X is contained in a parabolic subgroup of X and we have a contradiction via 5.1. In the other cases we have W is non-degenerate and, setting H̄ = GL_{n−1}^{ϵ}(r^{a}) (fixing W), S∩X ≤ H and H is normalized by S. Because of the choice of n, 6.9, 6.10, 6.12, 6.13 and 6.14 imply that H is not p-restricted. Hence B ≥ H. Since Q_{B} > 1, we infer that p divides r^{a} − ϵ and so d = 1 when p is odd, a contradiction. So p = 2. In this case, since d > 1, |Q_{B}| = 2, and n − 1 is even. 3.7 implies that P/Q_{P} ≅ SL_{2}(2), (a subgroup of) PSU_{3}(2), (a subgroup of) SU_{3}(2) or (a subgroup of) ^{2}B_{2}(2). Since d = 2, Q_{B} ≤ [S ∩ X, S ∩ X, S ∩ X] ≤ Q_{P} (from the structure of a Sylow 2-subgroup in GL_{2}(r^{a})), which is a contradiction.

2^{∘}. [2] s ≥ p.

If s < p, then from the structure of L we see that S ∩ X is abelian. Hence 5.2 implies that p ≤ 3. But then d ≤ 2 and n = ds < dp ≤ 6 so the restrictions on n deliver a contradiction.

3^{∘}. [3] If d > 1, then B ≥ LS and either (a) [i] p = 3, d = r^{a} = 2 and \overline{L^{∗}} ≅ GL_{2}^{+}(2) ≀ Sym(n/2);

(b) [ii] p = 2, d = 2, r^{a} = 3 and \overline{L^{∗}} ≅ GL_{2}^{+}(3) ≀ Sym(n/2).

If B ≥ LS, then by 6.4 Q_{B} ∩ X > 1 and 6.3 implies that one of (a) or (b) holds. So for a contradiction assume that B does not contain LS. Then P ≤ LS. If GL_{d}^{ϵ}(r^{a}), is not soluble, we apply 2.9 to obtain s = 1 and this contradicts (2^{∘}). Therefore GL_{d}(r^{a}) is soluble. Using (??) we have that O^{p}(P) ≤ F_{p}(LS). Suppose that d ≥ 2 and r^{a} = 2. If O_{p}(LS) = 1, then F_{p}(LS) is a 3-group and we have O^{p}(P) is a 3-group from which we infer that p = 2 = r^{a}, a contradiction. Therefore, p = 3, and \overline{L^{∗}}/O_{3}(\overline{L^{∗}}) ∼ 2 ≀ Sym(n/2) and n ≥ 6 by (c) or Q_{8} : 3 ≀ Sym(n/3) and n ≥ 8 by (d). Furthermore, from 2.5(e) we have that LS = (B ∩ LS)O^{p}(P), |Q_{B}O_{3}(LS)/O_{3}(LS)| = 3 and O^{p}(P)O_{3}(LS)/O_{3}(LS) is either elementary abelian of order 4 or isomorphic or Q_{8} and is normal in LS/O_{3}(LS). The only possibility is that n/2 = 3 and that LS/O_{3}(LS) ≅ 2 ≀ Sym(3) ≅ 2 × Sym(4). So L ≅ SL_{2}(2) ≀ Sym(3) ≤ X = G = SL_{6}(2). Now in this case L can be embedded in H ≅ Sp_{6}(2) and so we must have that Sp_{6}(2) is 3-restricted. However, a look in the Atlas [2] confirms that a Sylow 3-subgroup of Sp_{6}(2) is contained in a unique maximal subgroup and consequently Sp_{6}(2) is a 2-minimal parabolic group, a contradiction. Suppose that next that r^{a} = 3. Then just as above we argue that p = 2. Then \overline{L^{∗}}/O_{2}(\overline{L^{∗}}) ∼ Sym(3) ≀ Sym(n/2) and O^{2}(P)O_{2}(L)/O_{2}(L) has order 3. Considering again the centralizer of Q_{B} in L/O_{2}(L), we obtain a contradiction. This proves (3^{∘}).

4^{∘}. [4] If d = 1, then LS ≤ B.

Suppose that B ≱ LS. Then 6.5 gives either O^{p}(P) is soluble and n ≤ 3, which contradicts n ≥ 4, or r^{a} − ϵ is a power of p and LS/O_{p}(LS) is p-restricted. So LS/O_{p}(LS) ∈ R_{p}. If p = 2, then as n ≥ 6, we have LS/O_{2}(LS) ≅ Sym(6), Sym(8) or Sym(12) by 4.1 and 4.2. Suppose that LS/O_{2}(LS) ≅ Sym(6) and define \overline{L_{1}} = GL_{2}^{ϵ}(r^{a}) ≀ Sym(3) ∩ G. Then S ∩ G is contained in L_{1}. If r^{a} > 3, then 2.9 implies that L_{1}S is not 2-restricted. Thus, if r^{a} > 3, then B ≥ L_{1}S and we get Q_{B} ≤ O_{p}(L) ≤ Q_{P}, a contradiction. Therefore, r^{a} = 3. But then L_{1} is soluble and so, if P ≤ L_{1}S, O^{2}(P)(L_{1})O_{2}(L_{1})/O_{2}(L_{1}) has order 3 and L_{1}/O_{2}(L_{1}) = C_{B/O_{2}(L_{1})}(O^{2}(P)O_{2}(L_{1})/O_{2}(L_{1}) × Q_{B}O^{2}(P)O_{2}(L_{1})/O_{2}(L_{1}) has a direct factor isomorphic to Sym(3), a contradiction. Hence B ≥ L_{1}S. Finally consider, \overline{L_{2}} = (GL_{2}^{ϵ}(3)×GL_{4}^{ϵ}(3))∩G. Then as B ≥ L_{1}, P ≤ L_{2}S. And furthermore, letting K be the component in L_{2}, we have P ≤ KS is 3-restricted. It follows that ϵ = − but then we contradict our supposition on the size of n as given in (b).

Next assume that n = 8 and set \overline{L_{1}} = GL_{4}^{ϵ}(r^{a}) ≀ Sym(2) ∩ \overline{G}. By 2.9, B ≥ L_{1}S and so P ≤ LS and Q_{B} ≤ Q_{P}, a contradiction. So suppose that n = 12 and set \overline{L_{1}} = GL_{4}^{ϵ}(r^{a}) × GL_{8}^{ϵ}(r^{a}) ∩ \overline{G}. If P ≤ L_{1}S, then we must have P ≤ GL_{8}(r^{a}) and this contradicts the minimal choice of n. Thus B ≥ L_{1}S and again Q_{B} ≤ Q_{P}, a contradiction. Thus p > 2 and consequently n ≤ 6. Hence p ≥ 5. This then forces n = 5 and L/O_{5}(L) ≅ Sym(5) and this configuration has been considered in 6.14. This proves (3^{∘}).

5^{∘}. [5] p does not divide s.

Suppose that p divides s. Assume additionally that d = 1. Set \overline{L_{1}} = GL_{dp}^{ϵ}(r^{a}) ≀ Sym(s/p) ∩ \overline{G}. Then, as s > p, 2.9 implies that L_{1}S ≤ B. Therefore (4^{∘}) gives B ≥ ⟨L, L_{1}⟩S. Since Q_{B} > 1, this contradicts 6.3. Suppose that d = 2. Then restrictions (b) and (e) imply that n ≥ 8. In particular, n > dp and so s/p > 1. Now we apply the above argument to obtain a contradiction.

Let t = n − d[s/p]. Then 1 ≤ t ≤ p − 1. Let \overline{L_{1}} = GL_{d[s/p]}^{ϵ}(r^{a}) × GL_{t}^{ϵ}(r^{a}). Then P ≤ L_{1}S. If L_{1} has two components, K_{1} and K_{2} say, then at least one of them is contained in B, a contradiction. Therefore, L_{1} has at most one component. Since n > 4, we infer that n has exactly one component say K_{1}. Furthermore, we must have K_{1}S is p-restricted. If t = 1, we have a contradiction from our usual lemmas. Thus t = 2 and r^{a} ∈ {2, 3}. The restrictions on and the minimal choice of n then mean that n = 8. If p = 2, we obtain s = 4 or s = 8 and we contradict (5^{∘}). If p = 3, we have K_{1} ≅ SU_{6}(3) and K_{1}S is 3-restricted. By 6.10 we then have P ≤ L ∩ K_{1}S ≤ B, a contradiction. This completes the proof of the lemma. □

7. Symplectic groups

In this section we suppose that r is a prime with r ≠ p and that X ≅ PSp_{2n}(r^{a}). As usual we have G/X is a r-group. Notice that, as p ≠ 3, G/X never involves the exceptional graph automorphism which only appears in characteristic 2. We define GΓSp_{2n}(r^{a}) to be Sp_{2n}(r^{a}) decorated with its diagonal automorphism of order 2 and its field automorphisms. So [GΓSp_{2n}(r^{a}) : Sp_{2n}(r^{a})] = 2a. Now as in the unitary case we let \overline{X} be Sp_{2n}(r^{a}) and \overline{G} ≤ GΓSp_{2n}(r^{a}) with \overline{G}/\overline{X} a p-group. Then G = G/Z(G) and X = \overline{X}/Z(G). Define d = lcm(2, ord(p, r)) and set s = [\frac{2n}{d}]. Define \overline{L} = Sp_{d}(r^{a}) ≀ Sym(s) × Sp_{2n−ds)}(r^{a}) ∩ \overline{G}. Then L ∩ X contains a Sylow p-subgroup of S ∩ X and L is S-invariant.

Lemma 7.1. [spcases] Assume that X ≅ PSp_{2n}(r^{a}) with n ≥ 2. If X ∈ R_{p}, then one of the following holds

(a) [a] p = 3, X ≅ PSp_{4}(2) and P = X;

(b) [b] p = 2 and X ≅ PSp_{4}(3) ≅ PSU_{4}(2); or

(c) [c] p = 2, X ≅ PSp_{6}(3) and B = ...

Proof. Suppose first that 2n − ds ≠ 0. then S ∩ X centralizes an isotropic vector in V and consequently S ∩ X is contained in a parabolic subgroup of X and this contradicts 5.1. So n − ds = 0. If d = 2n, then X has abelian Sylow p-subgroups. Thus p = 2, 3 by 5.2. But then 2n ≤ d ≤ 2, a contradiction. So we have

1^{∘}. [1] ds = n and s > 1.

Assume that P ≤ LS. Then LS ∈ R_{p} and so 2.9 implies that Sp_{d}(r^{a}) is soluble. That is d = 2 and r^{a} = 2 or r^{a} = 3. Arguing exactly as in 6.15 3 delivers a contradiction unless 2n = 4 and \overline{L} ≅ SL_{2}(2) ≀ 2 or \overline{L} ≅ SL_{2}(3) ≀ 2. Thus we have examples (a) and (b).

Henceforth we assume that B ≥ LS. In particular, as Q_{B} > 1, we have that either p = 3 and G ≅ Sp_{2n}(2) or p = 2. In any case we have that d = 2.

Assume that 2n = 4. If r^{a} ≤ 3, then again possibilities (a) and (b) holds. So assume that r^{a} > 3. Then as Q_{B} > 1, we infer that p = 2 and that |Q_{B}| = 2 contrary to 3.9.

We now assume 2n > 4 and when p = 2 and r^{a} = 3 that 2n > 6. Furthermore, select n minimal so that the above conditions on n are satisfied and G ∈ R_{p}. Write s = 2k + l with l ≤ 1 and set \overline{L_{1}} = Sp_{4}(r^{a}) ≀ Sym(k) × Sp_{2l}(r^{a}) and factor L_{1} as \overline{K_{1}} × \overline{K_{2}} with K_{2} ≅ Sp_{2}(r^{a}). Then as L_{1}S and LS generate G, L_{1}S contains P. Since P ≰ LS, we infer that P S ≤ K_{2}. Then 2.9 implies that k = 1 and that K_{2} ≅ Sp_{4}(r^{a}) ∈ R_{p}. It follows that p = 2 and r^{a} = 3 or p = 3 and r^{a} = 2. The first case fails because of the choice of n, the second case indicates that G ≅ Sp_{6}(2) and we have already seen that this group does not satisfy ℝ_{3}. □

8. The orthogonal groups

Assume first that X ≅ PO_{2n}^{ϵ}(r^{a}) with n ≥ 4 and no triality automorphism.

Suppose that p and r are distinct primes and set d = \frac{1}{2} lcm(2, ord(p, r^{a})).

η = \{ \begin{matrix}+ & ord(p, r) odd and p odd \\ + & p = 2 and r^{a} ≡ 1 (mod 4) \\ − & ord(p, r) even and p odd \\ − & p = 2 and r^{a} ≡ 3 (mod 4). \end{matrix}

Let

s = \{ \begin{matrix}[\frac{n}{d}] & d does not divide n \\ [\frac{n}{d}] & d divides n and η^{[^{\frac{n}{d}}]} = ϵ \\ [\frac{n}{d}] − 1 & d divides n and η^{[^{\frac{n}{d}}]} ≠ ϵ. \end{matrix}

Finally put

\overline{L} = O_{2d}^{η}(r^{a}) ≀ Sym(s) × O_{2(n−ds)}^{θ}(r^{a}).

Then \overline{L} contains a Sylow p-subgroup of O_{2n}^{ϵ}(r^{a}).

Lemma 8.1. [orth1] Suppose that X ≅ PΩ_{2p}(r^{a}), d = 1 and η^{p} = ϵ. If p ≥ 3, then every p-local subgroup of X which contains S ∩ X is contained in L.

Proof. Suppose that R is a p-local subgroup of X containing S ∩ X and put Q_{R} = O_{p}(R). Let Z_{R} = Z(Q_{R}). If Z_{R} ≤ O_{p}(L), then the homogeneous components of \overline{Z_{R}} on V coincide with those of \overline{O_{p}(L)} and we have R ≤ L. So assume that Z_{R} ≰ O_{p}(L). But then Z_{R} operates quadratically on O_{p}(L) and, as Ω_{1}(O_{p}(L)) is the permutation module for Sym(p) and p ≥ 3, we have a contradiction.

□

Lemma 8.2. [Orth2] Suppose that X ≅ PΩ_{2n}^{ϵ}(r^{a}) with n ≥ 4 and G ∈ R_{p}, then one of the following holds:

(a) [a] p = 2, X ≅ PΩ_{8}^{+}(3) and \overline{B} = O_{4}^{+}(3) ≀ Sym(2);

(b) [b] p = 2, X ≅ PΩ_{12}^{+}(3) and \overline{B} ≅ O_{4}^{+}(3) ≀ Sym(3).

(c) [c] p = 3, X ≅ PΩ_{8}^{+}(r^{a}) with r^{a} ≅ 2, 5 (mod 9), \overline{B} ≅ Ω_{2}^{−}(r^{a})≀Sym(4) and P ∼ 3^{1+2}.SL_{2}(3)×3.

Proof. Suppose that n is chosen minimally so that PΩ_{2n−2}^{μ}(r^{a}) ∉ R_{p}.

1^{∘}. [1] n = sd.

Suppose first that p is odd. Then G does not involve the graph automorphism of G and S ∩ X centralizes a non-degenerate subspace of V of dimension 2(n−ds). In particular, S ∩X centralizes a singular vector and hence S ∩X is contained in a parabolic subgroup of G. Suppose then that p = 2. Then we have that \overline{L} = O_{2}^{η}(r^{a}) ≀ Sym(s) × O_{2}^{θ}(r^{a}), η ≠ θ and s = [\frac{n}{d}] − 1. Observe that a Sylow 2subgroup of O_{2}^{θ}(r^{a}) has order 4. Thus we also see that in this case S. Set \overline{L}^{∗} = Ω0_{2}^{η} ≀Sym(s)S. Then L^{∗} leaves invariant two distinct anisotropic 1-spaces. Thus \overline{L}^{∗}S is contained in two proper subgroups of G each isomorphic to O_{2n−1}(r^{a}). Since |Q_{B}| > 2, we have P is contained in the intersection of these two groups and this means that P ≤ L^{∗}S. But then r^{a} = 3 and we have 2n = 8 and \overline{L^{∗}} involves PSU_{4}(3), 2n = 10 and \overline{L^{∗}} involve PΩ_{8}^{+}(3) or 2n = 14 and \overline{L^{∗}} involves PΩ_{12}^{+}(3). In the first case we note that \overline{L_{2}}Ω_{6}^{+}(3) × O_{2}^{−}(3) also contains a Sylow 2-subgroup of G and we obtain B ≥ LS for a contradiction from 6.9. In the second and third cases we use the fact that O_{9}(3) contains a subgroup O_{1}(3) ≀ Sym(9) case O_{13}(3) contains a subgroup O_{1}(3) ≀ Sym(13) and apply 4.2 to see that in each case B must contain this subgroup. This contradicts the structure of B as described in (a) and (b). Thus we have that n = sd.

2^{∘}. [2] s > 2.

For p = 2 or 3, this follows because of the requirements on the size of n. So p ≥ 5 and s ≥ 2 follows from 5.2. So we assume that s = p.

3^{∘}. [3] B ≥ LS.

Suppose on the contrary that P ≤ LS. Assume for a moment that p > 3. Then O^{p}(P) is not soluble. Thus (2^{∘}), 2.9 and ?? together imply that O_{2d}^{ϵ}(r^{a}) is a p-group. But then the only possibility is that p = 2, a contradiction. Thus p ∈ \{2, 3\}. Thus d = 1 and \overline{L} ≅ O_{2}^{η}(r^{a}) ≀ Sym(s). Assume that O_{2}^{η}(r^{a}) is not a 2-group. Then, since the derived subgroup of L is perfect when n ≥ 5, we have that n = 4 from 2.11. Easy TO SEE n ≠ 4. Write a nice argument. Therefore, we have is a 2-group and Sym(n) ∈ R_{2}. Therefore 4.2 implies that n ∈ {4, 6, 8, 12}. Set m = 2, 2, 4, 4 according as n = 4, 6, 8, 12 and define \overline{L_{1}} = O_{2m}^{μ}(r^{a}) ≀ Sym(n/m) where μ = η^{m}. If O_{2m}^{μ}(r^{a}) is not soluble, then, by 2.9, B ≥ L_{1}S and Q_{B} has order 2^{m−1} and is contained in Q_{P}, a contradiction. Therefore, O_{2m}^{μ}(r^{a}) which means that m = 2, r^{a} = 3 and μ = +. Hence X ≅ PΩ_{8}^{+}(3) or PΩ_{12}^{+}(3) which is a contradiction to our choice of n. Hence (3^{∘}) holds.

Since B ≥ LS, we must have that LS is a p-local subgroup. Thus

4^{∘}. [4] L ≅ O_{2}^{η}(r^{a}) ≀ Sym(n).

Suppose for a moment that p ≥ 5. Then 5.2 implies that the Sylow p-subgroups of X are not abelian. Thus n ≥ p. If n = p, then ?? implies that Q_{P} = 1 and the argument in ?? works to give a contradiction. Therefore,

5^{∘}. [5] n > p.

Write n = lp + k where k ≤ p − 1. Then put \overline{L_{1}} = O_{2rp}^{τ}(r^{a}) × ⟨I_{2k}⟩. Then S normalizes L_{1}. Assume that lp ≠ n. Since L_{1}S ≰ B, we have that P ≤ L_{1}S and so L_{1}S/O_{p}(L_{1}S) ∈ R_{p}. If p ≥ 5, this immediately contradicts the minimal choice of n. Therefore, p ≤ 3. If p = 3, then we require lp < 4 which means that n = 4 or n = 5 and that O_{6}^{μ}(r^{a}) ∈ R_{3} and this again contradicts our supposition on n. If n = 5, then set \overline{L_{2}} = ⟨I_{4}⟩ × O_{4}^{−}(r^{a}) and note that as P ≤ L_{1}S, L_{2}S ≤ B, a contradiction. So suppose that p = 2. Then 2n = 12 and r^{a} = 3. But then from the example in O_{12}^{+}(3) we read that P ≤ L ≤ B, a contradiction. Hence n = rp. Now set \overline{L_{2}} = O_{2dp}^{ρ}(r^{a}) ≀ Sym(r) where ρ = η^{p}. Then L_{2} is normalized by S. Plainly L_{1}S ≰ B and so P ≤ L_{1}S. Since r > 1 by (5^{∘}), 2dp = 4 and r^{a} = 3 as well as p = 2. Investigating the structure of \overline{L_{2}} using 2.10 readily reveals a contradiction. This completes the proof of the lemma. □

Lemma 8.3. [Orthodd] Suppose that r is odd and X ≅ PΩ_{2n+1}(r^{a}) with n ≥ 3. If G ∈ R_{p}, then p = 2, X ≅ PΩ_{7}(3) and \overline{B} = O_{1}(3) ≀ Sym(7).

Proof. We have that |O_{2n+1}(r^{a}) : O_{2n}^{ϵ}(r^{a})| = (r^{na} + ϵ)r^{b} for some b. Thus a Sylow p-subgroup of O_{2n+1}(r^{a}) fixes either a plus point or a minus point when acting on V. Since these point stabilizers H have O_{p}(H) of order at most 2, we H ∈ Rp by 3.9. Thus the possibilities for p, r^{a} and n may be read from 6.9 and 8.2. Suppose that n = 3. Then \overline{H} ≅ 2 × O_{6}^{ϵ}(r^{a}). Thus 6.9, 6.12 and 6.14 imply that p is either 2 or 3. If p = 2, then from 6.9 we get ϵ = − and r^{a} = 3 and (??) holds. If p = 3, then again ϵ = − and r^{a} ≡ 2, 5 (mod 9). But then r^{3n} + 1 is divisible by 3, a contradiction.

Assume now that n ≥ 4. Then ?? implies that p = 2 or 3. Assuming that p = 2, the subgroups O_{1}(3) ≀ Sym(9) and O_{1}(3) ≀ Sym(13) yield contradictions. So p = 3 and \overline{H} ≅ O_{8}^{+}(r^{a}) × 2. Set \overline{L} = O_{6}^{−}(r^{a}) × O_{3}(r^{a}). Then L is S invariant. Since r^{a} > 3, O_{3}(r^{a}) is not soluble. Furthermore, writing \overline{L} = \overline{K_{1}} × \overline{K_{2}} with K_{2} ≅ O_{3}(r^{a}), we have that K_{2} ≰ H and so K_{2} ≤ B, but then B ≥ ⟨K_{2}, B ∩ \overline{H}⟩ = G. □

Finally in this section we come to the situation when p = 3, X ≅ PΩ_{8}^{+}(3) and S does not normalize all the parabolic subgroups of X. (So the triality automorphism of X is having an influence. From the examples in ?? we only need consider the case when r^{a} ≢ 2, 5 (mod 9) (otherwise the example in X immediately lifts to an example in G with G/X a 3-group.

Lemma 8.4. [triality] Suppose that X ≅ PΩ_{8}^{+}(3) and G/X is a 3-group. If G ∈ R_{3}, then r^{a} ≡ 2, 5 (mod 9).

9. Exceptional groups of Lie type

In this section we suppose that \overline{X} is a universal group of Lie type defined over a field over GF(r^{a}). We use the notation introduced in [4, page 237] writing

|X| = r^{NA} ∏_{i} Φ_{i}(r^{a})^{n_{i}}

where Φ_{i}(x) is the cyclotomic polynomial for for i-th roots of unity. The product ∏_{i} Φ_{i}(r^{a})^{n_{i}} in the case when X is an exceptional group is conveniently presented in [3, Table 10:2]. We set d = ord(p, r^{a}) if p is odd and, if p = 2, d = 1 when r^{a} ≡ 1 (mod 4) and otherwise d = 2.

Lemma 9.1. [OrdSylow] Let \overline{X} be a universal group of Lie type defined over GF(r^{a}), \overline{S} be a Sylow p-subgroup of \overline{X} with p odd. Then

|\overline{S}| = p^{b}Φ_{d}(r^{a})_{p}^{n_{d}}

where b = ∑_{pd|i} n_{i}. Furthermore, if b = 0, then the Sylow p-subgroups of X are abelian.

Proof. Consult [3, Equation (*) page 113] and [4, Theorem 4.10.2 (c)].

Lemma 9.2. [bigdexcep] Suppose G ∈ Rp and that d = ord(p, r^{a}) > 2. Then d = 4, p = 5 and X ≅ E_{8}(r^{a}).

Proof. Since d.2, p ≥ 5. We show that other than in the d = 4, p = 5, X ≅ E_{8}(r^{a}) cases the Sylow p-subgroups of X are abelian. The result then follows form 5.2. Since d > 2 and p ≥ 5, pd ≥ 20 (note d = 3 and p = 7 gives pd = 21). From [3, Table 10:2] the only exceptional group involving Φ_{k}(r^{a}) with k ≥ 20 is E_{8}(r^{a}). So consider E_{8}(r^{a}). If d = 3, we see no Φ_{21}(r^{a}), if d = 4 we can only have p = 5 and if d ≥ 5, there are also no possibilities. □

Lemma 9.3. [E8p5] Suppose that d = 4, p = 5 and X ≅ E_{8}(r^{a}). Then G ∉ R_{5}.

Proof. In this case we have that 5 divides Φ_{4}(r^{a}) = r2a + 1. Using [3, Table 4-1 (37)] we see that X contains a subgroup L isomorphic to SU_{5}(r^{2a}) and by comparing orders we see that this subgroup contains a Sylow 5-subgroup of X and furthermore L is invariant under S. ¿From 6.15 we have that LS ≱ P and therefore, B ≥ LS and |Q_{B}| = 5. But Z(S) is cyclic and so we have a contradiction with 3.9. □

Lemma 9.4. [SuzandRee] If G ∈ R_{p} and X ≅ _{2}^{B}(2^{a}) or ^{2}G_{2}(3^{a})′, then p = 2 and X ≅ _{2}^{G}(3)′.

Proof. If X ≅ ^{2}B_{2}(2^{a}), then p > 3 and the Sylow p-subgroups of X are abelian. Therefore, 5.2 shows that G ∉ Rp. Let X ≅ ^{2}G_{2}(3^{a}) with a ≥ 3. in this case also the Sylow 2-subgroups are abelian and so 5.2 delivers either a contradiction or X ≅ ^{2}G_{2}(3)′ ≅ SL_{2}(8). □

Guess

Lemma 9.5. [ReeF]If G ∈ R_{p}, then X ≅ ^{2̸}F_{4}(2^{a})′.

Proof. By 9.1 we only need to consider the situation when p = 3 and d = 2. we have |S ∩ X| = 3|Φ_{d}(2^{a})^{2}|_{3}. If r^{a} = 2, we have G = X = _{2}^{F}(2)′ and the Atlas [2] shows us that there are no 3-local subgroups which are maximal subgroups of G. So assume that 2^{a} ≥ 8. Using [7, Table 5.1 and 5.2] we see that S ∩ X is contained in maximal subgroups L_{1} = (2^{a} + 1)^{2}. GL_{2}(3) and L_{2} ≅ SU_{3}(2^{a}) : 2. According to 6.11 L_{2} ∉ R_{3}, so B ≥ L_{2} and P ≤ L_{1}. But then Q_{B} ≤ Q_{P}, a contradiction. □

Lemma 9.6. [F4]If G ∈ R_{p}, then X ≅ F̸_{4}(r^{a}).

Proof. By 9.1 and 5.2 we have that p ∈ {2, 3}. Notice that as p ≠ r, the graph automorphism of F_{4}(2^{a}) makes no appearance in this discussion. We first suppose that p = 2. Then, using ??Table 5.1]LSS, there is a maximal subgroup L ∈ X with L ≅ 2.Ω_{9}(r^{a}). By 8.3, L ∉ R_{p}. Therefore, B = L. But then |Q_{B}| = 2 and ?? provides a contradiction. Next assume that p = 3. Then we set L = (2, q − 1)^{2}.PΩ_{8}^{+}(r^{a}). Sym(3). By 8.2. Notice that L contains S ∩ X and that L is not a 3-local subgroup. Hence, if G ∈ R_{p} then L must also be in R_{p}. But then 8.2 implies that r^{a} ≡ 2, 5 (mod 9) and that P ≤ C_{G}(Z(S)). Furthermore, we note that B ∩ L_{2} ∼ (q + 1)^{4} : W (F_{4}) where W (F_{4}) denotes the Weyl group of type F_{4}. (As a subsystem subgroup the E(L_{1}) is generated by the root spanned by the D_{4} subsystem {} details. ) If r^{a} = 2, we consult the Atlas [2] and see that there is a unique maximal 3-local subgroup and so B = N_{G}(Z(S)). But then we have P ≤ B, a contradiction.

Suppose then that r^{a} > 2 and let L_{2} = 3.(PGU_{3}(r^{a}) × PGU_{3}(r^{a})).3.2 be as in [7, Table 5.2]. Then L_{2} can be chosen to contain S ∩ X. Now P ≤ L_{2} and so P is contained in exactly one of the components of L_{2} and the other component must be contained in B. But then B = G if r^{a} is odd and if r = 2, then B contains a subgroup isomorphic to L_{1} (but this time generated by short root groups). In either case we have a contradiction as Q_{B} = 1. □

Get details for the above argument.

Lemma 9.7. [G2 and 3D4] Suppose that X = O^{p}(G) is isomorphic to either G_{2}(r^{a}) or ^{3}D_{4}(r^{a}), then X ≅ G_{2}(3) or ^{3} D_{4}(3) and the possibilities for G, B, and P are listed in Table 1.

Proof. By Lemma 5.2, we need only consider the cases with p = 2 or p = 3. Suppose that X ≅ G_{2}(r^{a}). If r^{a} = 3, we refer the reader to the Atlas [2] to verify the details needed to confirm that G_{2}(3) is an example. Let Δ be the root system associated with X and let M be the monomial group Φ_{n}(s) : Dih(12) where n is 1 or 2 as appropriate for M to contain a Sylow 2-subgroup, respectively 3, of X. Let {α, β} be a fundamental system for Δ with α a long root. Let T = ⟨h_{α}, h_{β}⟩ be the Torus of order Φ_{1}(s)^{2}. Then N = ⟨X_{α}, X_{−α}, X_{α+2β}, X_{−α−2β}⟩ Then Z(N) = ⟨h_{α}(−1)⟩(= ⟨h_{α+2β}(−1)⟩). In particular, we notice that [T : T ∩N] = 2 and so |T N/N| = 2. Since |X|_{s′} = Φ_{1}(s)^{2}Φ_{2}(s)^{2}Φ_{3}(s)Φ_{6}(s), and |NT |_{s′} = \frac{1}{2}(Φ_{1}(s)Φ_{2}(s))^{2}.2, we infer that NT S contains a Sylow 2-subgroup of G. Since s ≥ 3, NT S contains two normal components F_{1} and F_{2} say, and so at least one of these component must be contained in B. So suppose that F_{1} ≤ B. Then O_{2}(F_{1}S) is isomorphic to a Sylow 2subgroup of SL_{2}(s) and consequently O_{2}(F_{1}S) contains a unique involution, namely t = h_{α}(−1). Since Q_{B} ≤ O_{2}(F_{1}S), we see that B ≤ C_{G}(t). But then Q_{B} = Z(S) is cyclic of order 2 and we have a contradiction via Lemma 3.9.

If X ≅ ^{3} D_{4}(s) and s ≠ 3, then an argument just as above works. CHECK THIS. Need to show that |Z(S)| = |Q_{B}| = 2. Now suppose that p = 3. Then the following groups are overgroups of the Sylow 2-subgroup of X are G_{2}(3), (7 × PSU_{(}3)).2 and (SL_{2}(3) ∗ SL_{2}(27)).2 and taking B = (SL_{2}(3) ∗ SL_{2}(27)).2 and P = 4^{2}.Dih(12) we satisfy the conditions required for X to be P̃- restricted. So this group appears on Table 1. □

Lemma 9.8. [E6]Assume that X ≅ E_{6}(r^{a}) or ^{2}E_{6}(p^{a}) and G ∈ R_{p}, then p = 3 and X ≅ ^{2}E_{6}(2) and B ∩ X ∩ 3.(PSU_{3}(2) × PSU_{3}(2) × PSU_{3}(2)).3. Sym(3).

look at p=5

Proof. Suppose first that d = 2 when X ≅ E_{6}(r^{a}) and that d = 1 when X ≅ ^{2}E_{6}(r^{a}) and as is standard write X ≅ E_{6}^{ϵ}(r^{a}) with ϵ = + when X is not twisted and otherwise ϵ = −. Because of the choice of d we see that p ≤ 3. Suppose that p = 3, then S ∩ X is contained in the subgroup L ≅ F_{4}(q). Using 9.6 we have that B ≥ L and then Q_{B} = 1, a contradiction. So suppose that p = 2 and ϵ = −. Then take L = (4, r^{a} − ϵ).(PΩ_{10}^{ϵ}(r^{a}) × (r^{a} − ϵ)/(4, r^{a} − ϵ)).(4, r^{a} − ϵ) (use root groups X_{α_{3}}, X_{α_{4}}, X_{α_{5}}, X_{α_{2}}X_{−α_{0}} and their negatives so that L is normalized by the graph automorphism. It follows that Q_{B} is cyclic of order 2 and we have a contradiction. If p = 2 and d = 1 we take L as above and see that Q_{B} is cyclic and obtain a contradiction via?? So suppose that p = 3 and d = 1 when ϵ = 1 and d = 2 when ϵ = −. In this case we take L = 3.(PSL_{3}^{ϵ}(r^{a})×PSL_{3}^{ϵ}(r^{a})×PSL_{3}^{ϵ}(r^{a})).3^{2}. Sym(3) and when (r^{a}, ϵ) ≠ (2, −), the components of L are permuted transitively by L. Thus in this case B ≥ L by 2.9 and then |Q_{B}| = 3. Thus (r^{a}, ϵ) = (2, −) and the lemma follows. □

Mention the Fischer groups?

Lemma 9.9. [E7] Assume that X ≅ E_{7}(r^{a}) and G ∈ R_{p}, then p = 2, X ≅ E_{7}(3) and B = 2^{3}.(P SL_{2}(3)^{7}).2^{3}.SL_{3}(2).

Proof. We have to consider p ∈ {2, 3, 5, 7}. According as d = 1, 2 set ϵ = +, − and note that L_{1} = (q−ϵ)^{7} : (2×Sp_{6}(2) contains S∩X. Set L_{2} = (3, r^{a}−ϵ).(E_{6}^{ϵ}(r^{a})×(r^{a}−ϵ)/(3, r^{a}−ϵ)).(3, r^{a}−ϵ).2. Then so long as p ≠ 2, 7, S ∩ X ≤ L_{2} and we easily derive contradictions in these cases??

For p = 7 we set L_{3} = f.PSL_{8}(r^{a}).g.(2 × (2/f)) where f and g are powers of 2. Then ?? shows that L_{2} ≤ B, a contradiction. So suppose that p = 2. Then set L_{3} = 2(PSL_{2}(r^{a}) × PΩ_{12}^{+}(r^{a})).2. Set L_{4} = 2^{3}.(PSL_{2}(r^{a})^{7}).d^{3}.PSL_{3}(2). Assume that r^{a} > 3. Then 6.2 and 8.2 imply that L_{3} = B and that Q_{B} has order 2 contrary to 3.9. Thus p = 3. Suppose that B = L_{3}. Then P ≤ L_{4} and Q_{B} ≤ Q_{2}(L_{4}) ≤ Q_{P} for a contradiction. Therefore, B = L_{4} as claimed. Need to argue P ≰ L_{4} ∩ L_{3} (use ??).

Lemma 9.10. [E8] Assume that X ≅ E_{8}(r^{a}) and G ∈ R_{p}, then p = 3, X ≅ E_{8}(2) and B ∩ X = 3^{2}.(PSU_{3}(2) × PSU_{3}(2) × PSU_{3}(2) × PSU_{3}(2)).3^{2}. GL_{2}(3).

Proof. We have to consider the possibilities p = 2, 3, 5 and 7. Recall also that d ∈ {1, 2}. For p = 2 or 7 the group L = (2, r^{a} −1).PΩ_{16}^{+}(r^{a}).2 contains a Sylow p subgroup of X. Applying 8.2 and 3.9 we obtain a contradiction. For p = 5, we consider the subgroup 5.(PSL_{4}^{ϵ}(r^{a}) × PSL_{4}^{ϵ}(r^{a})).5.4 where ϵ is chosen so that 5 divides r^{a} −ϵ. Then 3.9 and 2.9 delivers a contradiction. So assume that p = 3. Set L_{1} = 3.(PSL_{3}^{ϵ}(r^{a}) × E_{6}^{ϵ}(r^{a})).3.2 L_{2} = 3^{2}.(PSL^{ϵ}(r^{a}) × PSL^{ϵ}(r^{a}) × PSL^{ϵ}(r^{a}) × PSL^{ϵ}(r^{a})).3^{2}. GL_{2}(3) (with ϵ = + if d = 1 and ϵ = − if d = 1). Then 2.9 implies that P ≤ L_{1} and B = L_{2}. Using 9.8 we obtain r^{a} = 2 and this completes the lemma. □

10. Sporadic groups

Before we begin the case by case investigation of the sporadic simple groups we note that, by Lemma ??, we may assume that S is not cyclic.

Lemma 10.1. [m11] X ≠ Mat_{11}.

Proof. Suppose that G ∈ R_{p}. Then p ∈ {2, 3}. Suppose that p = 2. Let H ≤ G with H ∼ Mat_{10}. Then H is a 2-minimal subgroup of G and Q_{H} = 1. Hence by 2.16 H = P, a contradiction. Next we consider p = 3. This time we note that N_{G}(S) is a maximal subgroup of G and apply 2.18 to get a contradiction. □

Lemma 10.2. [m12] Suppose X = Mat_{12}. Then p = 2 or 3 and B is either of the two maximal p-local parabolic subgroups of G.

Proof. For p > 3, S is cyclic and for p ≤ 3, M(S) = P(S) has size two. □

Lemma 10.3. [j1] X ≠ J_{1}.

Proof. For p > 2, S is cyclic and for p = 2, N_{G}(S) acts irreducibly on S. So the lemma follows from 2.18. □

Lemma 10.4. [m22] Suppose X = Mat_{22}. Then p = 2 and B is is either of the two maximal 2-local parabolic subgroups of G.

Proof. For p ≥ 5, the Sylow p-subgroups are cyclic. For p = 3, N_{G}(S) acts irreducibly on S. Thus by 2.18, p = 2. Now B can be either of the maximal 2-local parabolics of G and (by definition) P is the unique minimal parabolic not in B. □

Lemma 10.5. [j2] Suppose that X = J_{2}. The one of the following holds:

(a) [1] p = 3, B ∼ 3.PGL_{2}(9) and P ∼ PSU_{3}(3).

(b) [2] p = 2 and B is either of the two maximal 2-local parabolic subgroups of G.

Proof. For p = 7, S is cyclic and for p = 5, N_{G}(S) acts irreducibly on S. So p ≤ 3. For p = 3, G has a unique maximal 3-local parabolic subgroup M and M ∼ 3.PGL_{2}(9). Thus B = M and B ∼ 3.PGL_{2}(9). Note that G has a parabolic subgroup H ≅ PSU_{3}(3). Then by 2.16 P = H and (a) holds.

For p = 2, P(S) = M(S) has size 2 and (b) holds. □

Lemma 10.6. [m23] Suppose X = Mat_{23}. Then p = 2, B ≅ 2^{4}. Alt(7) and P ≅ 2^{4}. Sym(5).

Proof. For p ≥ 5, the Sylow p-subgroups of G are cyclic. For p = 3, N_{G}(S) acts irreducibly on S. Thus by 2.18, p = 2. Since L = 2^{4}. Alt(7) is a subgroup of G and L/Q_{L} ∈ R_{2} by 4.2 B = L. So P is the unique minimal parabolic subgroup of G not in B and the lemma holds. □

Lemma 10.7. [hs] Suppose X = HS. Then p = 2 and B is any of the two maximal local parabolic subgroups of G.

Proof. For p ≥ 7, the Sylow p-subgroups of G are cyclic. For p = 3, N_{G}(S) acts irreducibly on S.

Suppose p = 5. If G = X, then no 5-local subgroup is maximal in G, a contradiction. If G ≠ X, then N_{G}(S) is maximal in G, again a contradiction.

Thus p = 2. Now B is either one of the maximal local parabolic subgroups and the P is the unique minimal parabolic not contained in B. □

Lemma 10.8. [j3] Suppose X ≅ J_{3}. The p = 2 and B is any of the two maximal local parabolic subgroups of G.

For p > 3, the Sylow p-subgroups are cyclic. For p = 3, N_{G}(S) is maximal. So p = 2, P(S) = M(S) has size 2 and the lemma holds. □

Lemma 10.9. [m24] Suppose X ≅ Mat_{24}. The p = 2 and B is any of the three maximal 2-local parabolic subgroups.

Proof. For p > 3, the Sylow p-subgroups are cyclic. No 3-local subgroup is maximal in G and so p = 2. Now B is one of the three maximal 2-local subgroups containing S and P the unique 2-minimal parabolic subgroup not contained in B. □

Lemma 10.10. [mcl] Suppose X = McL. Then p = 3, B = 3^{1+4}.2. Sym(5) and P = 3^{4}.PSL_{2}(9).

Proof. For p > 5, the Sylow p-subgroups are cyclic. For p = 5, N_{G}(S) is maximal in G. For p = 3, P has a unique 3-minimal parabolic of rank 1 type. Thus the lemma holds in this case.

Suppose p = 2. If G ≠ X, then G has a 2-minimal parabolic subgroup PSL_{3}(4).2^{2} and we get a contradiction to 2.16. Thus G = X. Let S ≤ H ≤ G with H ∼ 2^{4}. Alt(7). Then H ∉ R_{2} by 4.2 and so H = B. But there are two different choices for B, a contradiction. □

Lemma 10.11. [he] Suppose X = He. Then G = X,p = 2 and B ≅ 2^{6}.3. Sym(6)

Proof. For p > 7, the Sylow p-subgroups are cyclic, for p = 7, N_{G}(S) is maximal in G and for p = 5, N_{G}(S) is irreducible on S. Thus p ≤ 3. For p = 3, there exists a minimal parabolic subgroup 2^{6}.3^{3} [≤ 2^{6}3. Sym(6)], a contradiction to 2.16.

Thus p = 2. Suppose G ≠ X. Let H ∈ P_{G}(S). Then H/Q_{H} ≅ Sym(3) ≀ Sym(2) or PSL_{3}(2).2 and so there is no candidate for P.

Thus G = X. Since 2^{1+3+3}.PSL_{3}(2) only contains two of the four members of P_{G}(S), B ≅ 2^{6}. Sym(6) and the lemma is proved. □

Lemma 10.12. [ru] Suppose X = Ru. Then p = 2 and B is any of the two maximal 2-local parabolic subgroups of G .

Proof. For p > 5, the Sylow p-subgroups of G are cyclic. For p = 5, N_{G}(S) is maximal in G. For p = 3, G contains a 3-minimal parabolic subgroup 2^{6}.3^{2} [≤ 2^{6}.G_{2}(2)], contradicting 2.16. So p = 2 and M(S) has size two. □

Lemma 10.13. [suz] Suppose that X ≅ Suz. The one of the following holds.

1. [1] p = 3 and B ≅ 3^{5}. Mat_{11}.

2. [2] p = 2 and B is any of the three maximal 2-local parabolic subgroups of G.

Proof. For p > 5, S is cyclic and for p = 5, N_{G}(S) acts irreducibly on S. So p ≤ 3. For p = 3, G has a maximal parabolic subgroup H ∼ 3^{5}. Mat_{11}. Since H ∉ R_{3} by 10.1 H = B. So (1) holds.

For p = 2 M(S) has size three and (2) holds. □

Lemma 10.14. [on] If X = O′N, then p = 2 then B ∩ X ∼ 4.PSL_{3}(4).2.

Proof. For p > 7 and p = 5, S is cyclic and forp = 3, N_{G}(S) acts irreducibly on S so these cases do not arise. For p = 7, G contains a parabolic subgroup H = PSL_{3}(7).2. Since Q_{H} = 1, H ≠ B and so H ∈ R_{7}. But N_{H}(S) is maximal in H, a contradiction to 2.18 applied to H in place of G.

Thus p = 2. Let H ≤ G with H ∩ X ∼ 4.PSL_{3}(4).2_{1}. Then H is 2-minimal and we conclude that B = H. □

Lemma 10.15. [co3] If X = Co_{3}, then one of the following holds:

1. [1] p = 3 and B ≅ 3^{5}.2. Mat_{11}.

2. [2] p = 2 and B is any of the three maximal local parabolic subgroups of G.

Proof. For p > 5, S is cyclic, so p ≤ 5. For p = 5, G contains a parabolic subgroup McL : 2, a contradiction to 10.10.

For p = 3, G contains a maximal parabolic subgroup H ∼ 3^{5}.2. Mat_{11}. Since H ∉ R_{3}, we conclude that H = B. So (1) holds.

For p = 2, M(S) has size three and (2) holds. □

Lemma 10.16. [co2] If X ≅ Co_{2}, then one of the following holds:

1. [1] p = 3 and B ∼ 3_{+}^{1+4}.2_{−}^{1+4}. Sym(5).

2. [2] p = 2 and B is any of the three maximal 2-local parabolic subgroups of G.

Proof. If p > 5, S is cyclic and for p = 5, N_{G}(S) is a maximal subgroup in G and so p ≤ 3.

Suppose p = 3 and let S ≤ H ≤ G be a local subgroups H ∼ 3_{+}^{1+4}.2_{−}^{1+4}. Sym(5). Then H ∉ R_{3} and so H = B and (1) holds.

For p = 2, M(S) has size two and (2) holds. □

Lemma 10.17. [fi22] Suppose X = Fi_{22}. Then one of the following holds:

1. [1] p = 3 and B ∼ 3_{+}^{1+6}.2^{1+2+2+2}.3^{1+1}.2.

2. [2] p = 2 and B is an arbitrary maximal local parabolic subgroup.

Proof. For p > 5, S is cyclic and for p = 5, N_{G}(S) acts irreducibly on S.

Suppose p = 3. Then G has a maximal parabolic subgroup H with H ∼ 3_{+}^{1+6}.2^{1+2+2+2}.3^{1+1}.2. If H ≠ B we get that O^{p}(P) is normal in H, a contradiction (as can been seen by intersection H with a subgroup Ω_{7}(3). So H = B and (1) holds.

If p = 2 then (2) holds. □

Lemma 10.18. [HN] Suppose X = HN. Then one of the following holds:

1. [1] p = 5 and B‘ ∼ 5_{+}^{1+4}.2_{−}^{1+4}.5.4.

2. [2] p = 2 and B ∩ X ∼ 2_{+}^{1+8}. Alt(5) ≀ Sym(2).

Proof. If p > 5, then S is cyclic so p ≤ 5.

If p = 5 let H be the maximal 5-local parabolic subgroup with H ∼ 5^{1+4}2^{1+4}.5.4. Then, as H is soluble and p ≠ 2, 3, P ≰ H and consequently H = B. Thus (1) holds.

If p = 3, then X contains a subgroup contains a subgroup L ∼ 3^{1+4}.4.Alt(5). Since L ∉ R_{3}, we have that B = X. Let H be the parabolic with H/Q_{H} ≅ 2.(PSL_{2}(3) × PSL_{2}(3)).4 Since N_{H}(S) is a maximal subgroup of H, we get that H ≤ B by 2.18. But then G = B, a contradiction.

Suppose p = 2. Then G has maximal parabolic subgroup H with H ≅ 2_{+}^{1+8}. Alt(5) ≀ Sym(2). using 2.9 we have that H = B and (2) holds. □

Lemma 10.19. [ly] Suppose X = Ly. Then p = 5 and H ∼ 5_{+}^{1+4}.4. Sym(6).

Proof. For p ≥ 7, S is cyclic. So p ≤ 5.

Suppose that p = 5. Let S ≤ H ≤ G with H ∼ 5_{+}^{1+4}.4. Sym(6). Since Sym(6) ∉ R_{5}, we get B = H and the lemma holds in this case.

Suppose p = 3, then G has two maximal parabolic subgroup F, H with F/Q_{F} ∼ 2 × Mat_{11} and H/Q_{H} ∼ 2. Alt(5).Dih_{8}. So by 4.2 and 10.1 neither F nor H are in R_{3}, a contradiction.

Suppose that p = 2. Then G has a parabolic subgroups H ∼ 3. McL .2. Since Q_{H} ≠ 1, H ≰ B, but this contradicts 10.10 □

Lemma 10.20. [th] Suppose X = Th. Then one of the following holds:

1. [1] p = 3 and {B, P } = P(S).

2. [2] p = 2 and B ∼ 2_{+}^{1+8}. Alt(9).

Proof. For p > 7, S is cyclic, for p = 7, N_{G}(S) acts irreducibly on S and for p = 5, N_{G}(S) is maximal in G. Hence p ≤ 3.

For p = 3, (1) holds.

For p = 2, let S ≤ H ≤ G with H ∼ 2_{+}^{1+8}.Alt(9). Then H ∉ R_{2} so B = H and (2) holds. □

Lemma 10.21. [fi23] Suppose that X ≅ Fi_{23}. Then one of the following holds:

1. [1] p = 3 and B ∼ 3_{+}^{1+8}2_{−}^{1+6}.3_{+}^{1+2}.2. Sym(4).

2. [2] p = 2 and B ∼ 2^{11}. Mat_{23} or B ∼ 2^{6+2·4}.(Alt(7) × Sym(3)).

Proof. For p ≥ 7, S is cyclic and for p = 5, N_{G}(S) acts irreducibly on S. So p ≤ 3.

Suppose that p = 3 and let S ≤ H ≤ G with H = 3_{+}^{1+8}.2_{−}^{1+6}.3_{+}^{1+2}.2.Sym(4). Since H is soluble and O^{p}(P) is not normal in H, we get that H = B and (1) holds.

Suppose p = 2 and let S ≤ H ≤ G with H ∼ 2^{11}. Mat_{23}. If H = B (2) holds. So suppose that H ≰ B, then by 10.6, (H ∩ B)/O_{2}(H ∩ B) ≅ Alt(7). Since B is a maximal 2-local subgroup of G, we conclude B ∼ 2^{6+2·4}.(Alt(7) × Sym(3)) and again (2) holds. □

Lemma 10.22. [co1] Suppose X = Co_{1}. Then one of the following holds:

1. [1] p = 5 and B is one of the two maximal 5-local parabolic subgroups of G.

2. [2] p = 3 and B is one of the three maximal 3-local parabolic subgroups of G. (Note here that the two maximal subgroups N(3C^{2}) in the Atlas [2] need to be deleted, see Modular Atlas[5].)

3. [3] p = 2 and B is one of the four maximal 2-local parabolic subgroups of G.

Proof. For p > 7, S is cyclic and for p = 7, N_{G}(S) is irreducible on S. So p ≤ 5 and one of (1), (2) and (3) holds. □

Lemma 10.23. [j4] Suppose X = J_{4}. The p = 2 and B is one of the three maximal 2-local parabolic subgroups of G.

Proof. If p ≤ 5 but p ≠ 11, then S is cyclic and we are done. If p = 11 then N_{G}(S) is maximal so this case falls. If p = 3, then S is contained in H ∼ 2^{11}. Mat_{24} which is not in R_{3} by 10.9. Since Q_{H} = 1 we have a contradiction. So p = 2 and the lemma is proved. □

Lemma 10.24. [fi24] Suppose X = Fi_{24}. The one of the following holds:

1. [1] p = 3 and B ∼ 3^{1+10}.PSU_{5}(2).2.

2. [2] p = 2 and B is any of the four maximal 2-local parabolic subgroups of G.

Proof. For p > 7, S is cyclic and so p ≤ 7. For p = 7, S is contained in He.2 and we obtain a contradiction via 10.11. For p = 5, N_{G}(S) acts irreducibly on S, so this case fails.

Suppose p = 3. If B ∼ 3^{7}.Ω_{7}(3), then P/Q_{P} ≅ Alt(5), a contradiction as Alt(5) ∉ R_{3}. Hence (1) holds in this case.

If p = 2, then (2) holds. □

Lemma 10.25. [bm] Suppose X ≅ BM. Then one of the following holds:

1. [1] p = 5 and B ∼ 5_{+}^{1+4}.2_{−}^{1+4}. Alt(5).4.

2. [2] p = 3 and B ≅ 3_{+}^{1+8}.2_{−}^{1+6}O_{6}^{−}(2).

3. [3] p = 2 and B is any of the four maximal 2-local parabolic subgroups of G.

Proof. If p > 7, then S is cyclic and, if p = 7, N_{G}(S) is irreducible on S. So p ≤ 5.

Suppose p = 5 and let S ≤ H ≤ B with H ∼ 5_{+}^{1+4}.2_{−}^{1+4}. Alt(5).4. Then H ∉ R_{5} and so H = B and (1) holds.

Suppose that p = 3 and let S ≤ H ≤ B with H ∼ 3_{+}^{1+8}.2_{−}^{1+6}.O_{6}^{−}(2). Then H ∉ R_{3} and so H = B and (2) holds.

Suppose that p = 2. The (3) holds. □

Lemma 10.26. [m] Suppose that X ≅ M. Then one of the following holds:

1. [1] p = 7 and B ∼ 7_{+}^{1+4}.6. Sym(7).

2. [2] p = 5 and B ∼ 5_{+}^{1+6}.4.J_{2}.2.

3. [3] p = 3 and B ∼ 3_{+}^{1+12}.2. Suz .2 or B ∼ 3^{2+5+5·2}.(Mat_{11} × GL_{2}(3)).

4. [4] p = 2 and B is any of the five maximal 2-local parabolic subgroups.

Proof. For p > 13, S is cyclic, for p = 13, N_{G}(S) is maximal in G and for p = 11, N_{G}(S) is irreducible on S. Thus p ≤ 7.

For p = 7 choose S ≤ H ≤ G with H ∼ 7_{+}^{1+4}.6. Sym(7). Then H ∉ R_{7} and so B = H. Hence (1) holds.

For p = 5 choose S ≤ H ≤ G with H ∼ 5_{+}^{1+6}.4.J_{2}.2. By 10.5 H ∉ calR_{5} and so B = H. Hence (2) holds.

For p = 3 choose S ≤ H ≤ G with H ∼ 3_{+}^{1+12}.2. Suz .2. If H = B, then (3) holds. If H ≠ B then 10.13 implies that H ∩ B/O_{3}(H ∩ B) ≅ 2. Mat_{11} .2. Thus B ∼ 3^{2+5+5·2}.(Mat_{11} × GL_{2}(3)) and (3) holds.

If p = 2, then (4) holds.

11. The smaller list

In this section we assume in addition that if P̃ ≤ M ≤ S and M/Q_{M} is a classical group extended by field automorphism, then the classical groups is (S)L_{n}(q).

References

[1] Aschbacher, Michael. Finite group theory. Corrected reprint of the 1986 original. Cambridge Studies in Advanced Mathematics, 10. Cambridge University Press, Cambridge, 1993. x+274 pp. ISBN: 0-521-45826-9

[2] Conway, J. H.; Curtis, R. T.; Norton, S. P.; Parker, R. A.; Wilson, R. A. Atlas of finite groups. Maximal subgroups and ordinary characters for simple groups. With computational assistance from J. G. Thackray. Oxford University Press, Eynsham, 1985. xxxiv+252 pp. ISBN: 0-19-853199-0

[3] Gorenstein, Daniel; Lyons, Richard. The local structure of finite groups of characteristic 2 type. Mem. Amer. Math. Soc. 42 (1983), no. 276, vii+731 pp.

[4] Gorenstein, Daniel; Lyons, Richard; Solomon, Ronald. The classification of the finite simple groups. Number 3. Part I. Chapter A. Almost simple K-groups. Mathematical Surveys and Monographs, 40.3. American Mathematical Society, Providence, RI, 1998. xvi+419 pp. ISBN: 0-8218-0391-3

[5] Jansen, Christoph; Lux, Klaus; Parker, Richard; Wilson, Robert. An atlas of Brauer characters. Appendix 2 by T. Breuer and S. Norton. London Mathematical Society Monographs. New Series, 11. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1995. xviii+327 pp. ISBN: 0-19-851481-6

[6] Kleidman, Peter; Liebeck, Martin. The subgroup structure of the finite classical groups. London Mathematical Society Lecture Note Series, 129. Cambridge University Press, Cambridge, 1990. x+303 pp. ISBN: 0-521-35949-X

[7] Liebeck, Martin W.; Saxl, Jan; Seitz, Gary M. Subgroups of maximal rank in finite exceptional groups of Lie type. Proc. London Math. Soc. (3) 65 (1992), no. 2, 297–325.

U. Meierfrankenfeld

C.W. Parker, School of Mathematics and Statistics, University of Birmingham, Edgbaston, Birmingham B15 2TT, UK

E-mail address: C.W.Parker@bham.ac.uk

34 ULRICH MEIERFRANKENFELD CHRISTOPHER PARKER PETER ROWLEY

P.J. Rowley, Department of Mathematics, UMIST, PO Box 88, Manchester M60 1QD, UK