Finite groups of local characteristic p The non-central rank 2 Case

Andrew Chermak and Ulrich Meierfrankenfeld

Department of Mathematics Kansas State University Manhattan KS, 66502 chermak@math.ksu.edu

Department of Mathematics Michigan State University East Lansing M 48824I meier@math.msu.edu

August 17, 2002

1 Introduction

Let G be a finite group, fix a prime divisor p of |G|, and fix a Sylow p-subgroup S of G. Recall that a p-local subgroup of G is, by definition, the normalizer in G of a nonidentity p-subgroup of G, and recall the notation: for any subgroup X of G, O_{p}(X) is the largest normal p-subgroup of X. We most often write Q_{X} for O_{p}(X). We say that X is of characteristic p if C_{X}(Q_{X}) ≤ Q_{X}. G is of local characteristic p if all p-local subgroups of G have characteritic p. This paper is part of the program to understand the finite groups of local characteristic p. See [MSS] for an overview of the this program.

We denote by Z_{X} the subgroup generated by the elements of order p in Z(Q_{X}). Throughout, we write Z for Z_{S}. We also remind the reader that the Thompson Subgroup of S, denoted J(S), is the subgroup of S generated by the elementary abelian subgroups of S of maximal order. The Baumann Subgroup of S, denoted B(S), is the group C_{S}(Z_{J(S)}). Set J(G) = ⟨J(S)^{G}⟩ and B(G) = ⟨B(S)^{G}⟩. We remind the reader also that, for any group X, O^{p}(X) is the subgroup of X which is generated by the set of elements of X of order prime to p.

We say that G is p-minimal if S is not normal in G and S is contained in unique maximal subgroup of G. A subgroup X of G is called a parabolic subgroup if X contains a Sylow p-subgroup of G. Denote by P the set of subgroups P of G such that P is a parabolic subgroup of G, P is p-miminal and Q_{P} ≠ 1. We refer to the members of P as the minimal parabolic subgroups of G. P(S) is the set of minimal parabolics of G containing S. We shall work with the following hypothesis.

Hypothesis A The following hold.

(i) G is a finite group, p is a prime and S ∈ Syl_{p}(G).

(ii) All p-local subgroups of G containg B(S) have characteristic p.

(iii) There exist minimal parabolic subgroups P and P̃ containing S, with O_{p}⟨P, P̃⟩) = 1, and with both [Z_{S}, P] ≠ 1 and [Z_{S}, P̃] ≠ 1.

(iv) Z_{P} ZP̃ is neither normal in P nor in P̃.

Let X be a group and V an 𝔽_{p}X module. Let 𝔽 be a field in characteristic p and CL_{𝔽}(W) a classical group defined over the 𝔽-vector space W. We say that X is a natural CL_{𝔽}(W)-module for X provided that there exists a onto homomorphism Φ : X → CL_{𝔽}(W) such that V and W are isomorphic as 𝔽_{p}X-module (where we view W as an 𝔽_{p}X-module via the homorphism: X →^{Φ} CL_{𝔽}(W) → GL_{𝔽}(W) → GL_{𝔽_{p}}(W). Note that a given group X can have more than one isomorphism class of natural CL_{𝔽}(W)-modules. Namely X might have more than one factor group isomorphic to CL_{𝔽}(W); or different choices of Φ, with the same kernel can give different isomorphism types of natural modules. Let I be a set and W the power set of I. Let W_{e} be the set of all finite even size subsets of I. Note that W and W_{e} are 𝔽_{2} vector spaces, where the addition is given by the symmetric diffence. If I ∈ W_{e} put N = W_{e}/{∅, I} and if I ∉ W_{e} put N = W_{e}. Note that N is an 𝔽_{2} Sym(I)- and an 𝔽_{2} Alt(I)-module. We say that V is a natural Sym(I)− (Alt(I)-) module for X provided that p = 2 and there exists an onto homomorphism Φ : X → Sym(I), (Alt(I)), such that V and N are isomorphic 𝔽_{2}X-modules.

Theorem B Assume Hypothesis A. Then for all X ∈ {P, P̃}, B(X)/Q_{X} ≅ SL_{2}(q), q a power of p, and one of the following holds:

1. Q_{X} is the natural SL_{2}(q)-module for B(X)/Q_{X}.

2. p = 2 and Q_{X} is the natural O_{3}(q)-module for B(X)/Q_{X}.

3. p = 3 and

(a) Z(Q_{X}) = Z(B(X)) × [Z(Q_{X}), B(X)].

(b) Z(B(X)) = Φ(Q_{X}) has order q.

(c) [Z(Q_{X}), B(X)] and Q_{X}/Z(Q_{X}) are natural SL_{2}(q)-modules for B(X)/Q_{X}.

(d) Q_{X}/Φ(Q_{X}) is indecomposable as B(X)/Q_{X} module.

Theorem C Assume Hypothesis A with p = 2, and let q be as in Theorem B. Then one of the following holds.

1. G is isomorphic to a subgoup of P ΓL_{3}(q) which contains L_{3}(q).

2. G is isomorphic to a subgroup of ΓSp_{4}(q) which contains Sp_{4}(q).

3. q = 2 and G is isomorphic to the alternating group Alt(6).

4. q = 4 and G is isomorphic to the Mathieu group M_{23}.

Theorem D Assume Hypothesis A, and let q be as in Theorem B. Let M(S) be the set of maximal p-local subgoups of G containing S. Then one of the following holds.

1. M(S) = {N_{G}(B(P)), N_{G}(B(P̃)} and P and P̃ are not conjugate in G.

2. p ≠ 2, M(S) = {N_{G}(B(P)), N_{G}(B(P̃), N_{G}(B(S))} and P and P̃ are conjugate via N_{G}(B(S)).

3. q = 4 and M(S) = {X, Y } with X ∼ 2^{4} : Alt(7) and Y ∼ 2^{4} : ΓL_{2}(4).

4. q = 3, 5, 7 or 13 and if X ∈ M(S) then either X = N_{G}(B(S)) or B(S) ∼ q^{2} : SL_{2}(q). The precise structure of X/Q_{X} and the size of M(S) can be obtained from 4.4.

As in [MSS] we say that Q! holds in G provided there exists a maximal p-local subgroup C̃ in G such that C̃ is of characteristic p, N_{G}(Z_{S}) ≤ C̃ and with Q = O_{p}(C̃),

C_{G}(x) ≤ C̃ for all 1 ≠ x ∈ Z_{Q}.

Theorem E Assume that A(i), A(ii), A(iii) and Q! holds in G. Then all of Hypothesis A holds

Some of the groups in Theorem C will be identified by means of an involution-counting method of Bender [Be1] Here is the form of Bender’s result that we need (and which is a little bit weaker than the result stated in [Be1]).

Proposition 1.1 (Bender) Let G be a group of even order, and denote by J the set of involutions in G. Let X be a subgroup of G such that |J| > |G : X|, and set

J_{n} = {t ∈ J − X : |Xt ∩ J| = n},

b_{n} = |{Xt : t ∈ J_{n}}|, and

f = (|J|/|G : X|) − 1. Then

b_{1} < f^{−1}[|J ∩ H| + ∑_{n≥2}(n − 1)b_{n}] − 1 − ∑_{n≥2} b_{n}.

2 Minimal parabolics

Since the members of P play such an important role in this paper, we devote our first section to two general results concerning the structure of groups p-mimimal groups.

Recall that for any group X the Frattini subgroup Φ(X) is, by definition, the intersection of all the maximal subgroups of X. We write Φ_{p}(X) for the inverse image in X/Q_{X}) of Φ(X/Q_{X}).

Lemma 2.1 Let X be p-mimimal and S ∈ Syl_{p}(X).

(a) For any normal subgroup N of X, either S ∩ N ≤ Q_{X} or N ≥ O^{p}(X).

(b) We have Φ_{p}(X) = Φ_{p}(O^{p}(X))Q_{X} = ∩{M^{x}}_{x∈X}.

(c) Let X_{0} be a subgroup of X such that Φ_{p}(X)X_{0} ≥ O^{p}(X). Then X_{0} ≥ O^{p}(X)O_{p}(X).

Proof: Without loss, we may assume that Q_{X} = 1. (a) is well known, see for example [PPS]. Set K = ∩{M^{x}}_{x∈X}. Then KS ≤ M, so (a) implies that K ∩ S = 1. Now let M_{1} be an arbitrary maximal subgroup of X. If K ≰ M_{1} then M_{1}K = X, and M_{1} contains a conjugate of S, by Sylow’s theorem. But in that case we have M_{1} = M^{x} for some x ∈ X, and so K ≤ M_{1} after all. This shows that KΦ(X), and hence K = Φ(X). Now set M_{0} = M ∩ O^{p}(X). Then ∩{M_{0}^{x}}_{x∈X} = K ∩ O^{p}(X) = K, and so K ≤ Φ(O^{p}(X)). Since S ≠ Q_{X} we have X ≠ S, and then X ≠ Φ(O^{p}(X))S. Thus Φ(O^{p}(X)) ≤ M, and so N ≥ Φ(O^{p}(X)). Equality then holds, and we have (b).

Finally, let X_{0} be as in (c). As Φ(X)X_{0} = Φ(O^{p}(X))X_{0} we then have

O^{p}(X) = O^{p}(X) ∩ Φ_{p}(X)X_{0} = Φ_{p}(O^{p}(X))(O^{p}(X) ∩ X_{0}) = O^{p}(X) ∩ X_{0}.

Thus (c) holds. □

Lemma 2.2 Let H be a finite group, let S be a Sylow p-subgroup of H, and denote by P_{H}^{∗}(S) the set p-minimal parabolics of H containing S Then H = ⟨P_{H}(S)⟩N_{H}(S).

Proof: Well known, see for example [PPS]. □

3 F F -modules

Let V be a module over the field 𝔽_{p} of p elements, for a group X. Denote by A(X, V) the set of all non-identity, subgroups A of X such that [V, A, A] = 0 ≠ [V, A] and such that |A/C_{A}(V)| ≥ |V/C_{V} (A)|. Set J(X, V) = ⟨A(X, V)⟩. An element of A(X, V) is said to be a quadratic offender on V, and V is said to be an F F -module for X if J(X, V) ≠ 1. Let A_{∗}(X, V) be set of all A ∈ A(X, V) such that AC_{X}(V)/C_{X}(V) is a minimal element of A(X/C_{X}(V), V).

Proposition 3.1 Let X be a mimimal parabolic with O_{p}(X) = 1, and let V be a faithful F F -module for 𝔽_{p}X.

Then there exists a uniquely determined X-invariant set K(X, V) of subgroups L_{1}, ⋯, L_{r} of X, such that the following hold.

(a) J(X, V) = O^{p}(X)J(S, V) = L_{1} × ⋯ × L_{r}, and S acts transitively on K(X, V).

(b) V = [V, L_{1}] + ⋯ + [V, L_{r}] + C_{V} (J(X, V)), and [V, L_{i}, L_{j}] = 0 for i ≠ j.

(c) One of the following holds.

1. L_{1} ≅ SL(2, p^{n}) for some n, and [V, L_{1}]/C_{[V,L_{1}]}(L_{1}) is a natural module for L_{1}, or 2. p = 2, L_{1} ≅ Sym(2^{n} + 1) for some n, and [V, L_{1}] is isomorphic to natural module

for L_{1}.

(d) Let A ∈ A(X, V). Then A is the direct product of its subgroups A ∩ X_{i}. If X_{i} ≅ SL(2, q) and A ∩ X_{i} ≠ 1 then A ∩ X_{i} is a Sylow p-subgroup of X_{i}, while if p = 2 and X_{i} ≅ Sym(2^{n} +1) then A∩X_{i} is generated by a set of pairwise commuting transvections on [V, X_{i}].

(e) For any A ∈ A(X, V) we have |A| = |V/C_{V} (A)|.

(f) Let S ∈ Syl_{p}(X). Then S acts tranisitively on A_{∗}(S, V).

Proof: This is Theorem 3 in [Ch]. See [He] for a proof that does not depend on a K-group assumption. □

Corollary 3.2 Let X and V satisfy the hypotheses of 10.1, and let S be a Sylow p-subgroup of X. Then two conjugates of J(S, V) suffice to generate J(X, V).

Proof: Let L ∈ K(X, V), and let A ∈ A(S, V) be chosen as large as possible for the condition that [A, L^{∗}] = 1 for all L^{∗} ∈ K(X, V) − {L}. Then 3.1 shows that either L is isomorphic to SL(2, p^{n}) and A = S ∩ L is a Sylow p-subgroup of L, or L ≅ Sym(2^{n} + 1) and A is generated by a maximal set of pairwise commuting transvections in L. In the SL(2, p^{n}) case it is evident that two conjugates of A generate L. In the Sym(2^{n} + 1) case we may identify A with the group generated by the transpositions (2i − 1 2i), 1 ≤ i ≤ 2^{n−1}. There is then a conjugate A^{g} of A in L which is generated by the transpositions (2i 2i + 1), 1 ≤ i ≤ 2^{n−1}, and evidently ⟨A, A^{g}⟩ = L. Set B = ⟨A^{S}⟩. Then 3.1(d) shows that B ∈ A(S, V), and indeed B = J(S, V), and since S acts transitively on K(X, V), the foregoing argument shows that two conjugates of B generate ⟨K(X, V)⟩. That is, in view of 3.1(a), two conjugates of J(S, V) suffice to generate J(X, V). □

Definition 3.3 Let X be a p-minimal with O_{p}(X) = 1, V a faithful F F -module for 𝔽_{p}X, S ∈ Syl_{p}(G), T = J(S, V), and A ∈ A(T, V). Define (a) K(A, T, V) := ⟨F ∈ A_{∗}(X, V) | F ≰ M, [A, F] ≠ 1⟩, where M is the unique maximal

subgroup of X containing N_{X}(T).

(b) U(A, T, V) := [V, K(A, T, V)].

(c) U′(A, T, V) := U(A′, T, V), where A′ = ⟨A_{∗}(T, V) − A_{∗}(A, V)⟩.

(d) The subgroups U(A, T, V), A ∈ A_{∗}(T, V) are called the T -components of V.

Lemma 3.4 Let X be a p-minimal with O_{p}(X) = 1, V a faithful F F -module for 𝔽_{p}X, S ∈ Syl_{p}(G), T = J(S, V). and A ∈ A(T, V). Set K = K(A, T, U), U = U(A, T, V) and U′ = U′(A, T, V).

(a) Suppose that A ∈ A^{∗}(T, V).

1. If the V -components of X are isomorphic to SL_{2}(q), then K is the unique V - component of X containg A and U/C_{U}(K) is the natural SL_{2}(q)-module for K.

2. If the V -components of X are isomorphic to Sym(2^{k} + 1), then K ≅ Sym(3) and |U| = 4.

(b) [V, J(X, V)] = ∑_{E∈A_{∗}T,V)} U(E, T, V).

(c) U(A, T, V) = ∑_{E∈A_{∗}(T,V)} U(E, T, V).

(d) [V, J(X, V)] = U + U′ and A centralizes U′.

(e) Let B ≤ C_{S}(A), then [U, B] ≤ [U, A].

(f) Suppose C_{V} (X) = 1, W is a T -component of V and b ∈ T with [W, b] ≠ 1. Then [W, b] = [W, T].

(g) Suppose that A ∈ A^{∗}(T, V) and set L = N_{X}([U, A]). Then O_{p}(L) = A and O^{p}(L) ≤ ⟨T^{L}⟩.

Proof:

Proof: (a) L be the V -component containing A. If L ≅ SL_{2}(q), then any F ∈ A^{∗}(X, V) with [F, A] ≠ 1 is contained in L. Thus K = K(A, T, V) = L and (a) holds in this case. Suppose now that L ≅ Sym(m) and identify L with Sym(m) and [V, L] with the set of even subsets of {1, …, m}. Say T ∩ L fixes m, T ∩ L = {(1, 2), (3, 4), …, (m − 2, m − 1)} and A = ⟨(1, 2)}. Let M be the unique maximal subgroup of X containing N_{G}(T). Then M ∩L is exactly the stabilizer of m in L. The only 2-cycles which neither fix m nor commute with (1, 2) are (1, m) and (2, m). Thus K = K(A, T, V) = ⟨(1, m), (2, m)⟩ ≅ Sym(3) and U = U(A, T, V) = ⟨{1, m}, {2, m}⟩. So (a) holds.

(b) Let L be a V -component of X. It suffices to show that

[V, L] = ∑_{E∈A_{∗}(T ∩L,V)} U(E, T, V).

If L ≅ SL_{2}(q) this follows from (a). If L ≅ Sym(m) then

∑_{E∈A_{∗}(T ∩L,V)} U(E, T, V) = ⟨{1, m}, {2, m}⟩+⟨{3, m}, {4, m}⟩+…+⟨{m−2, m}, {m−1, m}⟩

which equals [V, L] and again (b) holds.

(c) Let F ∈ A_{∗}(X, V) with F ≰ M. We claim that there exists a unique E ∈ A_{∗}(T, V) with [F, E] ≠ 1. Indeed F and E must be in the same V -component L of X. If L ≅ SL_{2}(q), then E = T ∩ L. If L ≅ Sym(m) and say F = ⟨(1, m)⟩(with the same set up as in (b)) then E = ⟨(1, 2)⟩.

If follows from the claim that K(A, T, V) = ⟨K(E, T, V) | E ∈ A_{∗}(A, V)⟩. Clearly this implies (c).

(d) follows from (b), (c) and the definition of U and U′.

(e) Replacing B by AB we may assume that A ≤ B. We will proof by induction on |A| that B = C_{B}(U)A.

Let E ∈ A_{∗}(A, V). Since B ≤ S ≤ N_{G}(T) and B centralizes E, B normalizes L := K(E, T, V) and W := [V, L] = U(E, T, V). Then L ≅ SL_{2}(q) (where q = 2 in the Sym(m) case), W = [W, L], W/C_{W} (L) is the natural SL_{2}(q)-module for L, and E ∈ Syl_{p}(L). Since no outer automorphism of L centralizes E we conclude that B ≤ C_{X}(L)L. Hence B ≤ C_{X}(L)E and B = C_{B}(L)E. Since L is irreducible on W/C_{W} (L), [W, C_{B}(L)] ≤ C_{W} (L). From W = [W, L] and the three subgroup lemma we conclude that C_{B}(L) centralizes W. Thus B = C_{B}(W)E. Set Ã = ⟨A_{∗}(A, V)−E⟩ and Ũ = U(Ã, T, V). Note that Ã ∈ A(T, V), A = ÃE, Ã = C_{A}(W) and U = Ũ + W. Hence by induction on |A|, C_{B}(W) = C_{B}(Ũ)Ã and so B = C_{B}(U)A. Thus [B, U] ≤ [A, U] and (e) holds.

(f) Chooose E ∈ A_{∗}(T, V) with W = [V, L] where L = K(E, T, V). Since C_{V} (X) = 1, we have C_{V} (J(X, V)) = 1 and so also C_{W} (L) = 1. Hence W is a natural SL_{2}(q)-module for L. Note that T = EC_{T} (W) and (f) from the structure of a natural SL_{2}(q)-module.

(g) Readily verfied. □

Lemma 3.5 Let p be a prime, H a finite group, X ≤ H, T ∈ Syl_{p}(H) and V a faithful 𝔽_{p}H module. Suppose that

(i) N_{H}(X) contains a Sylow p-subgroup of H.

(ii) X ≅ SL_{2}(q), q a power of p and X is the natural SL_{2}(q)-module for X.

(iii) N_{H}(C_{V} (T)) ≤ N_{H}(T).

Then one of the following holds:

1. X ⊴ H and H is isomorphic to a subgroup of ΓL_{2}(q).

2. (a) q = 4.

(b) N_{H}(X) ≅ Sym(5)

(c) H ≅ Alt(7)

(d) N_{H}(T) ≅ Sym(4) ∧^{∣} Sym(3)

(e) For all z ∈ V^{#}, C_{H}(z) ≅ L_{3}(2) and V is indecomposable as an C_{H}(z)-module.

Proof: Note first that N_{H}(X) is isomorphic to a subgroup of ΓL(2, q).

Suppose that C_{V} (T) ∩ C_{V} (T^{g}) = 0 for all g ∈ H − N_{H}(T) and let h ∈ H. Since X is transitive on V^{#} there exists x ∈ X with C_{V} (T)^{x} ∩ C_{V} (T)^{h} ≠ 1. Then T^{h} = T^{x}, T^{H} = T^{X} and X = ⟨T^{X}⟩ = ⟨T^{H}⟩ is normal in H. Thus (1) holds.

Suppose next that there exists g ∈ H with T ≠ T^{g} and C_{V} (T) ∩ C_{V} (T^{g}) ≠ 1. Choose such a g with C_{V} (T) ∩ C_{V} (T^{g}) maximal. Set U = C_{V} (T) ∩ C_{V} (T^{g}).

By (iii) C_{V} (T) ≠ C_{V} (T^{g}). Thus there exists v ∈ C_{V} (T) − C_{V} (T)^{g}. Since T^{g} acts transitively on v + C_{V} (T)^{g} we can choose h ∈ T^{g} with 0 ≠ [v, g] ∈ C_{V} (T) ∩ C_{V} (T^{g}). It follows that ⟨v⟩ + U ≤ C_{V} (T) ∩ C_{V} (T^{h}). The maximality of U implies T = T^{h}. (i) and the Fratti-argument, N_{H}(X) contains a Sylow p-subgroup of N_{H}(T). Thus T ⟨h⟩ is under N_{H}(T) conjugate to a subgroup of N_{H}(X) ∩ N_{H}(T). Since C_{V} (h) = C_{V} (T^{g}) ≠ C_{V} (T), h ∉ T. Since h acts quadratically on V we conclude that p = 2. Let R = C_{T} (h). Then R ≤ C_{H}(h) ≤ N_{H}(C_{V} (h) = N_{H}(C_{V} (T^{g})) and so by (iii), R ≤ N_{G}(T^{g}). Put S = RT^{g}. Note that R ∩ T^{g} = 1. Since S is isomorphic to a subgroup of ΓL_{2}(q), S/T is cyclic. Thus |R| = 2, q = 4 |S| = 8, S ∈ Syl_{2}(H) and N_{H}(X) has order at least 120. Let N = C_{H}(U). Since no element in S acts as a transvection on V, N acts faithfully on V/U. Hence N is isomorphic to a subgroup of SL(3, 2). Set L = ⟨T, T^{g}⟩. Then L ≤ N and L normalizes the subgroup C_{V} (T)C_{V} (T)^{g}/U of otdet four in V/U. We conclude that L ≅ Sym(4). Sym(4) is a maximal subgroup of SL(3, 2) ( for example by Lagrange Theorem) and so either L = N or N ≅ L(3, 2). Since H is transitive on V, |H| = 15|N| and so either |H| = 360 or 2520. In the first case N_{H}(X) has index three in H and we get a contradiction. So H has order 2520 and H has index eight in L_{4}(2). Thus L_{4}(2) ≅ A_{8} and H ≅ Alt(7). Moreover, N ≅ L_{3}(2). Since |C_{V} (S)| = 2, N does not normalize a complement to U in V. Since N acts transitively on (V/U)^{#}, H acts transitively on the 35 2-dimensional subspaces of V. Thus N_{H}(T) = N_{H}(C_{V} (T)) has index 35 in H. Thus all parts of (2) hold. □

Lemma 3.6 Let X ≅ SL_{2}(q), q = p^{k}, p an odd prime and V a 𝔽_{p}X-module. Suppose that

(i) There exists a submodule W in V such that both W and V/W are natural SL_{2}(q)- modules for X.

(ii) Let T ∈ Syl_{p}(X). Then [V, X, X] ≠ 0 = [V, X, X, X].

Then p = 3 and V is indecomposable.

Proof: Replacing V by an appropriate section of V ⊗_{𝔽_{p}} 𝔽_{q} we may assume that V is a four dimensional 𝔽_{q}X-module. Note that V/W is as an 𝔽_{q}X isomorphic to W^{σ} for some σ ∈ Aut(𝔽_{q}). Let B = N_{X}(T), U = C_{W} (V). and R/W = C_{V/W} (T). If [R, T] ≰ U, we get [V, T, T, T] ≠ 0. Thus [R, T] = U and there exists a 1 dimensional B-submodule D/U in R/U with D ≠ V. Suppose that [D, T] ≠ 0. Then the action of T ≅ Hom(D/U, U) as a 𝔽_{p}B/T module. Thus there exists μ ∈ Aut(𝔽_{q}) with k^{2μ} = kk^{−σ} for all k ∈ 𝔽_{q}^{#}. Thus k^{2μ+σ−1} = 1. Let q = p^{n}, σ = Frob(p^{m}) and μ = Frob(p^{l}), where 0 ≤ m, l < n. Then p^{n} − 1 divides 2p^{m} + p^{l} − 1. Since p > 2 we get p = 3 (and m = l = n − 1). So we are done in this case.

Thus we may assume that [D, T] = 1. Thus C_{V} (T) is 2-dimensional. We may assume that the same is true for the dual of V and so also [V, T] is 2-dimensional. Since [V, T, T] ≠ 1 we have [V, T] ≠ C_{V} (T) and so [V, T]/U, C_{V} (T)/U and W/U are three distinct B-submodules in R/U. It follows that V/U and R/V are isomorphic 𝔽_{q}B/T modules. Hence k^{−1} = k^{σ} for all k ∈ 𝔽_{q}. Thus p^{n} − 1 divides p^{m} + 1, a contradiction. □

Lemma 3.7 Let X ≅ SL_{2}(q), q = p^{k}, p a prime, V a 𝔽_{p}X-module and T ∈ Syl_{p}(X) Suppose that V/C_{V} (X) is natural SL_{2}(q)-module, V = [V, X] and C_{V} (X) ≠ 1. Then

(a) p = 2, q > 2, V is isomorphic to a quotient of the natural O_{3}(q)-module and C_{V} (X) ≤ [V, T].

(b) If |C_{V} (X)| < q, then C_{V} (X)^{#} = [V, T] − ({[v, t] | v ∈ V, t ∈ T }.

(c) If |C_{V} (X)| = q then C_{V} (X) and [C_{V} (T), N_{X}(T)] are only subgroup of order q in [V, T] intersecting {[v, t] | v ∈ V, t ∈ T } trivially.

Proof: (a) is well known and easy to proof. See for example [Ch]. Given that V is a quotient of the natural O_{3}(q) module, (b) and (c) are easy to compute. □

4 The Sylow p-subgroup of L_{3}(q)

In later section we will need some detailed information about the automorphism group of a Sylow p-subgroup of SL(3, p^{n}). This will be the subject of this section.

Lemma 4.1 Let (V, s) be a non-degenerated symplectic space over the field F. Let ΓSp(𝔽, V, s) be set of all α ∈ ΓL(𝔽, V) such that there exists σ ∈ Aut(𝔽) and λ ∈ 𝔽 with s(v^{α}, w^{α}) = λs(v, w)^{σ} for all v, w ∈ V.

(a) Let α ∈ ΓSp(𝔽, V, s). Then λ and σ as above are uniquely determined by α.

(b) Let g ∈ GL_{𝔽_{p}}(V). Then g ∈ ΓSp(𝔽, V, s) if and only if s(v^{g}, w^{g}) = 0 for all v, w ∈ V with s(v, w) = 0.

Proof: Well known and easy to prove. □

Lemma 4.2 𝔽 be a field, V a vectors space over 𝔽 and s a non-degenerate symplectic form on V. Set T := T (s) := {(v, k) | v ∈ V, k ∈ 𝔽} and define (v, k) ∗ (w, l) = (v + w, s(v, w) + k + l). Then

(a) (T, ∗) is a group.

(b) Define

Φ : Hom_{ℤ}(V, 𝔽) ⋊ ΓSp(𝔽, V, s) → Aut(T), (α, δ) → Φ(α, δ)

where

Φ(α, δ) : T → T, (v, k) → (v^{α}, λk^{σ} + v^{δ})

Then Φ is a well defined monomorphism.

(c) If char K = 2, then T is an elementary abelian 2-group and if char K ≠ 2, then T′ = Z(T) = {(0, k) | k ∈ 𝔽}.

(d) Suppose that char K ≠ 2. Then Φ is an isomorphism.

(e) Suppose that char K = 2. For l ∈ 𝔽 and (v, k) ∈ T define l · (v, k) = (lv, l^{2}k) and q((v, k)) = k. Then V is a vector sppace over 𝔽, q is a quadratic form on V and Φ(ΓSp(𝔽, V, s)) = ΓO(𝔽, V, q).

Proof: (a) (0, 0) is the identity, (−v, −k) is the inverse of (v, k) and for the associative law note that

(u, j) ∗ (v, k) ∗ (w, l) = (u + v + w, s(u, v) + s(u, w) + s(v, w) + j + k + l)

for either of the two ways to compute the left hand side.

(b) We first verify that Φ(α, δ) is an automorphism of T:

\begin{aligned}(v, k)^{Φ(α,δ)} ∗ (w, l)^{Φ(α,δ)} & = & (v^{α}, λk^{σ} + v^{δ}) ∗ (w^{α}, λl^{σ} + w^{δ}) \\ & = & (v^{α} + w^{α}, s(v^{α}, w^{α}) + λk^{σ} + λl^{σ} + v^{δ} + w^{δ}) \end{aligned}

and

\begin{aligned}((v, k) ∗ (w, l)^{Φ(α,δ)} & = & (v + w, s(v, w) + k + l)^{Φ(α,δ)} \\ & = & ((v + w)^{α}, λ(s(v, w) + k + l)^{σ} + (v + w)^{δ}) \end{aligned}

By the properties of α and δ, the right hand sides of the last two displayed equations are equal. Hence Φ(α, δ) is a homomorphism. That Φ is a homomorphism we leave as an easy exercise to the reader. Since Φ is a homomorphism and Φ(id_{V} , 0) = id_{T} each Φ(α, δ) is invertible and Φ(α, δ) ∈ Aut(T). This completes the proof of(b).

(c) We compute

[(v, k), (w, l)] = (−v, −k) ∗ (−w, −l) ∗ (v, k) ∗ (v, l)
= (−(v + w), s(v, w) − (k + l)) ∗ (v + w, s(v, w) + k + l)
= (0, 2s(v, w)).

This easily implies (c).

(d) Let W = T/Z(T). The map T → V, (v, k) → v is a homomorphism and by (c) has kernel Z(T) and we obtain an isomorphism Ξ : W → V, (v, k)Z(T) → v. Let ϕ ∈ Aut(W) and \overline{ϕ} ∈ End_{F_{p}}(W), the map induced by ϕ. View W as a symplectic space over 𝔽 such that Ξ becomes an 𝔽 linear isometry. Note that two vectors v, w ∈ W are perpedicular if and only if their inverse image in T commutes. Thus 4.1 implies that \overline{ϕ} ∈ ΓSp(W, 𝔽_{q}). Adjusting ϕ be an element of Φ(ΓSp(V, F_{q})) we may assume that \overline{ϕ} = 1. Define δ : V → 𝔽 by ϕ(v, 0) = (v, v^{δ}). Then ϕ = Φ(id_{V} , δ) and (d) is proved.

(e) Readily verified. □

Lemma 4.3 Let q be a power of p, N = 𝔽_{q}^{2} ⋊ ΓL(2, 𝔽_{q}), S ∈ Syl_{p}(N) and T = B(S).

(a) Let A, B ∈ A(T). Then |A| = q^{2}, and if A ≠ B then A ∩ B = Z_{T}.

(b) We have |A(T)| = 2 if p = 2, and |A(T)| = q + 1 if p is odd.

(c) Assume p is odd and set D = Aut(T), and W = T/Z_{T}, Then W and Z_{T} may be regarded as semilinear 𝔽_{q}D modules in such a way that

(a) T ≅ T (s) for some non-degenerate F_{q}-bilinear symplectic form on W.

(b) C_{D}(W) = O_{p}(D)

(c) D/O_{p}(D) ≅ ΓL(2, q) induces that the full group of 𝔽_{q}-semilinear automorphisms on W.

(d) Choose notation such that S fixes (0, 1) ∈ 𝔽_{q}^{2} = O_{p}(N). Let v, w ∈ 𝔽_{q}^{#} and σ ∈ Aut(𝔽_{q}). Denote with σ Diag(v, w) the element (a, b) → (va^{σ}, wb^{σ}) of ΓL(2, q). Then g normalizes T and the image of g in D/O_{p}(D) ≅ ΓL(2, q) is σ Diag(wv^{−1}, v).

(e) The 1-dimensional 𝔽_{q}-subspaces of W are given by {A/Z_{T} | A ∈ A(T)}

Proof: If p = 2 then T has exactly two maximal elementary abelian subgroups. Both of order q^{2} and (a) and (b) hold in this case. So suppose that p is odd. Let V = 𝔽_{q}^{2}, and s the synplectic form s((k_{1}, l_{1}), (k_{2}, l_{2}) = k_{1}l_{2} −l_{1}k_{2}. Recall the definition of T (s) and Φ from 4.2. For k, l ∈ 𝔽_{q} let a(k, l) = ((0, k), l) ∈ T (s) and A := {a(k, l) | k, l ∈ 𝔽_{q}} ≤ T (s). View A as a vector space over 𝔽_{q} by ja(k, l) = a(jk, jl). Let b(j) = ((−\frac{1}{2}j, 0), 0) and B = \{b(j) | j ∈ 𝔽_{q}. Then T = AB. Also a(k, l)^{b(j)} = a(jk, k). It follows that T (s) ≅ T and (c:a) holds. (c:b) and (c:c) follows from 4.2.

Let u, v ∈ 𝔽_{q}^{#} and σ ∈ Aut(𝔽_{q}) and let g = σ Diag(u, v) ∈ ΓL(V, 𝔽_{q}). Then s(v^{g}, w^{g}) = uvs(v, w)^{σ} and hence by definition of Φ, ((j, k), l)^{Φ(g,0)} = (uj^{σ}, vk^{σ}, uvl^{σ}). Thus a(k, l)^{Φ(g,0)} = a(vk^{σ}, uvl^{σ}). Hence the automorphism on T induced by σ Diag(v, w) is Φ(σ Diag(wv^{−1}, v), 0) and (c:d) holds.

Note that maximal elementary abelian subgroups of T ≅ T (s) correspond to the maximal isotropic subspaces of V. Thus (c:e) holds. □

|A^{+}(T)|  N_{G}(T)/C_{G}(T)T  N_{G}(T)/C_{G}(A^{+}(T))  q  N_{G}(A)/C_{G}(A)
1  (C_{q−1} × C_{e}).C_{m}  1  any q  SL_{2}(q)C_{e}.C_{m}
1 + 1  (C_{q−1} × C_{f}).C_{m}  1  any q  SL_{2}(q)C_{f}.C_{m}
2  (C_{q−1} × C_{f}).(C_{2} × C_{m})  C_{2}  any odd q  SL_{2}(q)C_{f}.C_{m}
2 + 2  D_{8}  C_{2} × C_{2}  3  GL_{2}(3)
4  SD_{16}  D_{8}  3  GL_{2}(3)
6  C_{4} ∗ GL_{2}(3)  Sym(4)  5  (C_{4} ∗ SL_{2}(3).2
2 + 2  C_{3} × D_{8}  C_{2} × C_{2}  7  SL_{2}(7).2
4 or 4 + 4  C_{3} × D_{16}  D_{8}  7  SL_{2}(7).2
8  C_{3} × SD_{32}  D_{16}  7  SL_{2}(7).2
3  C_{3} × Sym(3)  Sym(3)  7  SL_{2}(7)
3 or 3 + 3  C_{3} × C_{2} × Sym(3)  Sym(3)  7  SL_{2}(7).2
6 or 2 + 6  C_{6} ≀ C_{2}  C_{2} × Sym(3)  7  SL_{2}(7).2 (or GL_{2}(7))
6  C_{12} ∗ GL_{2}(3)  Sym(4)  13  (C_{4} ∗ SL_{2}(13).2

Lemma 4.4 Let p be a prime, G a finite group, S ∈ Syl_{p}(G) and q a power p. Set T = B(S) and let N be the of all subgrous N of G such that p ∤ |G/N_{G}(N)| and Q_{N} is the natural SL_{2}(q) module for N/Q_{N} ≅ SL_{2}(q)}. Also set A^{+}(T) = {Q_{N} | N ∈ N }. Suppose that

(i) T is isomorphic to a Sylow p subgroup of SL_{3}(q).

(ii) A^{+}(T) ≠ ∅

The following table lists the lengths of the orbits of N_{G}(T) on A^{+}(T), the image of N_{G}(T) in Aut(T)/O_{p}(Aut(T)), the image of N_{G}(T) in Sym(A^{+}(T)), and q. The last entree list the structure of N_{G}(A)/C_{G}(A) assumimg that A = Q_{N} for some N ∈ N with that N_{G}(A) ≤ N_{G}(N). Also e is a divisor of q − 1, f is a divisor of q − 1, with \frac{1−q}{f} ≤ d, d = gcd(q − 1, 3) and m a divisor of n, where q = p^{n}

Proof: Pick N ∈ N (T) and put A = Q_{N}. By (i), A ∈ A(T) and there exists a complement to A in T. Since |N|_{p} = q^{3} = |T |, T is a Sylow p-subgroup of N. Hence by Gaschütz Theorem, there exists a complement K to A in L. By definition of N, N_{G}(T) contains a Sylow p-subgroup of G. So we may assume that S ≤ N_{G}(N). Note that NS/C_{NS}(A) is isomorphic to SL(2, 𝔽_{q})Σ, where Σ is some p-subgroup of Aut(𝔽_{q}). Let B be the image of N_{NS}(T) in Aut(T)/O_{p}(Aut(T). It follows from 4.3 that B = ΣH where H := {Diag(k, k^{−2}) | k ∈ 𝔽_{q}^{#}}. Note that H = B ∩ GL(2, 𝔽_{q}). Also let L^{∗} be the image of N_{G}(T) in Aut(T)/O_{p}(Aut(T)) and L the intersection of L^{∗} with GL_{2}(q). Since S is a Sylow p-subgroup of N_{G}(T), H contains a Sylow p-subgroup of L. Thus L is a p′ group. Let \overline{L} be the image of L is P GL_{2}(q). Then \overline{H} is cylic of order \frac{q−1}{d}. Let R be a maximal p′-subgroup of GL_{2}(q) containg L.

Note also that A is unique determined by H, namely A/Z(T) is the unique 1-dimensional subspace of T/Z(T) which is inverted by the involution in T. In particular, we have a one to one ( but not necessarily onto) map from the elements of A^{+}(T) to the involutions with determined −1 in L.

By Dickson List of maximal subgroups of P GL_{2}(q) any maximal p′ subgroup of P GL_{2}(q) is isomorphic to one of the following groups:

D_{2(q−1)}, D_{2(q+1)}, Sym(4) or Alt(5)

Suppose that \overline{R} ≅ D_{2(q−1)}. Then R normalizes a set \{X, Y \} of 1-dimensional 𝔽_{q} subspaces in T/Z(T) Thus for all h ∈ H, h^{2} normalies X and Y.

Assume that A/Z(T) ∉ {X, Y } for some A ∈ A^{+}(T). Then k^{2} = k^{−4} for all k ∈ 𝔽_{q}^{#} and so k^{6} = 1 and q ∈ \{3, 7\}. If q = 3, then \overline{R} ≅ D_{4}, which is not a maximal 3′ subgroup. Thus q = 7 and R ≅ C_{6} ≀ C_{2}. An elementary but tediuos discussion show that that one of the entrees in the table holds. As an example consider the case that L ≅ C_{6} ≀ C_{2}. Then R has two classes of involutions with determinant -1. One class has size 2 the other size 6. Thus |A^{+}(T))| = 2, 6 or 8. If |A^{+}(T))| = 2, Case 3 holds. If |A^{+}(T)| = 6 or 8, then Case 12 holds. ( Also if |A^{L}| = 2, then (N_{G}(A) ∩ N_{G}(T))/T C_{G}(A) ≅ C_{6} × C_{6} and so last entry in Case 12 is GL(2, 7). If |A^{L}| = 6, then (N_{G}(A) ∩ N_{G}(T))/T C_{G}(A) ≅ C_{6} × C_{2} and so we get SL_{2}(7).2.

Assume that A/Z(T) ∈ {X, Y } for all A ∈ A^{+}(T). Then |A^{+}(T)| ≤ 2 and one of first three case must hold.

Suppose next that L̄ is not contained in a D_{2(q−1)} and that R ≅ D_{2(q+1)}. Since gcd(q − 1, 2(q + 1)) = 2 we again get q = 3 or q = 7. If q = 3 we have R ≅ SD_{16} and it is easy to see that one of first five cases holds. If q = 7, then R ≅ C_{3} × SD_{32}. Since we assume that L̄ is not in a D_{2}q − 1, R = L. Thus R has a unique class of involutions with determinant −1 and Case 9 holds.

So we may assume now that \overline{L} is not contained in a Dihedral Group.

Suppose R̄ ≅ Alt(5). Then all elements in R̄ have even order have order 2. Thus \frac{q−1}{d} = 2 q = 3 or 7. But then 5 does not divide |GL(2, q). a contradiction.

Suppose finally \overline{R} ≅ Sym(4). Since \overline{R} is a p′ group p ≠ 3. Since \overline{R} has no elements of even order larger than 4, \frac{q−1}{d} ≤ 4, q ≤ 3 · 4 + 1 = 13 and q = 5, 7 or 13. If q = 7, R contains no involution with determinant −1, a contradcition.

Thus q = 5 or 13. Then \overline{H} contains a cylcic group of order 4. Since \overline{L} is not a dihedral group we conclude that \overline{H} ≅ Sym(4). Since Φ(H) contains an involution with determinant −1, L = R and Case 6 or 12 holds. □

5 Graphs

In this section we introduce the basic material relating to the “amalgam method” of p-local analysis.

All graphs in this paper are “combinatorial”. This means that a graph Γ consists of a set V (Γ) of vertices, together with a symmetric, irreflexive relation on V (Γ) (called the incidence). The unordered pairs of incident vertices are the edges of Γ. The graph Γ is bipartite if there is a non-trivial partition V (Γ) = V_{1} ∐ V_{2}, such that no vertex in V_{1} is incident with any vertex in V_{2}.

A non-empty sequence π = (α_{0}, α_{1}, ⋯, α_{n}) of vertices is a path if

(1) α_{i−1} is incident with α_{i}, 1 ≤ i ≤ n, and

(2) α_{i−1} ≠ α_{i+1}, 1 ≤ i < n.

The integer n is then the length of π, and we say that π is a path from α_{0} to α_{n}. The path π is a cycle if α_{0} = α_{n}. (The condition (2) in the above definition may appear to some readers to be overly restrictive, but the fact is that we shall never encounter a situation in which a more general notion of “path” would be more convenient.)

We say that Γ is connected if, for any vertices α and α′ of Γ there exists a path from α to α′. If there is a path from α to α′ then the minimum, over all paths π from α to α′, of the length of π, is the distance from α to α′, denoted dist(α, α′). There is a natural equivalence relation on V (Γ) whereby the vertices α and α′ are taken to be equivalent if there exists a path from α to α′. The equivalence classes for this relation are the connected components of Γ.

For any vertex α of Γ and any n ≥ 1, we set

Δ^{(n)}(α)Δ_{Γ}(α) = {δ ∈ V (Γ) : dist(α, δ) ≤ n},

and we also write Δ(α) for Δ^{(1)}(α). The cardinality of Δ(α) is the valence of α, and we say that Γ is regular if the valence is constant over all vertices.

Let Γ be a graph. By an automorphism of Γ we mean a bijective mapping on V (Γ) which preserves the incidence. The inverse of an automorphism is then an automorphism, as is the composition of two automorphisms, and we write Aut(Γ) for the group of all automorphisms of Γ. If Γ is bipartite (via a fixed partition of V (Γ)) then we write Aut^{0}(Γ) for the subgroup of Aut(Γ) which preserves the partition. A homomorphism from a group G into Aut(Γ) is said to be an action of G on Γ (and one says that G acts on Γ). If Γ is bipartite and the image of G lies in Aut^{0}(Γ) then the action is said to be orientation-preserving. Equivalently, an action of G on Γ is a pairing

V (Γ) × G ⟶ V (Γ), (α, g) ⟼ α^{g},

satisfying the familiar conditions: (α^{g})^{h} = α^{gh} and α^{1} = α, and such that α^{g} is incident with β^{g} whenever α and β are incident vertices. If G is a group acting on a graph Γ, and α is a vertex of Γ, we denote the stabilizer in G of α by G_{α}, and we denote the point-wise stabilizer in G of Δ^{(n)}(α) by G_{α}^{(n)}.

For G a group and H a subgroup we denote the set of right cosets of H in G by H\G. Let X_{1} and X_{2} be subgroups of G. Then Γ(G; X_{1}, X_{2}) is the bipartite graph Γ whose vertex set is the disjoint union of the sets X_{1}\G and X_{2}\G, and with incidence given by

X_{1}g ∼ X_{2}h if and only if X_{1}g ∩ X_{2}h ≠ ∅.

Equivalently, X_{1}g ∼ X_{2}g if and only if X_{1}g ∩ X_{2}h is a coset of X_{1} ∩ X_{2}. In particular the (identity cosets of) the groups X_{1} and X_{2} are the vertices of an edge of Γ.

Lemma 5.1 Let G be a group, let X_{1} and X_{2} be subgroups of G, and set Γ = Γ(G; X_{1}, X_{2}). Then the following hold.

(a) G acts on Γ via right translation. This action is orientation-preserving, and is transitive on edges.

(b) (Local transitivity) For any vertex α of Γ the stabilizer G_{α} of α in G acts transitively on Δ(α), and the kernel of this action is G_{α}^{(1)}.

(c) Γ is connected if and only if G = ⟨X_{1}, X_{2}⟩. More generally, the (set-wise) stabilizer in G of the connected component containing the edge {X_{1}, X_{2}} of Γ is ⟨X_{1}, X_{2}⟩.

(d) G acts faithfully on Γ if and only if no non-identity normal subgroup of G is contained in X_{1} ∩ X_{2}.

Let L be a finite group then normal subgroup V of L is said to be p-reduced if V is an elementary abelian p-group and O_{p}(L/C_{L}(V)) = 1. It is an easy excercise to show that L is has a unique maxaimal p-reduced normal subgroup Y_{L}, see for example [PPS]. Moreover, Y_{L} ≠ 1 if and only if Q_{L} ≠ 1.

Definition 5.2 Let p be a prime and let (X_{1}, X_{2}) be a pair of non-trivial subgroups of a group G.

(a) (X_{1}, X_{2}) is an amalgam in G if no nontrivial subgroup of X_{1} ∩ X_{2} is normal in X_{1} and X_{2}.

(b) (X_{1}, X_{2}) has characteristic p if both X_{1} and X_{2} have characteristic p and Q_{X_{1}}Q_{X_{2}} ≤ X_{1} ∩ X_{2}.

(c) (X_{1}, X_{2}) is parabolic if X_{1} ∩ X_{2} is a parabolic subgroup of X_{1} and of X_{2}.

(d) (X_{1}, X_{2}) is p-minimal if it is parabolic and if both X_{1} and X_{2} are p-minimal.

(e) (X_{1}, X_{2}) is called non-central if Y_{X_{1}} ≰ Z(X_{1}) and Y_{X_{2}} ≰ Z(X_{2}).

Whenever (X_{1}, X_{2}) is a characteristic p- amalgam in G with corresponding coset graph Γ = Γ(G; X_{1}, X_{2}), we write Q_{δ} for Q_{G_{δ}}, and Z_{δ} for Y_{G_{δ}},δ a vertex of Γ. For any edge (γ, δ) of Γ, set G_{γδ} = G_{γ} ∩ G_{δ}, Z_{γδ} = Y_{G_{γδ}} and Q_{γδ} = O_{p}(G_{γδ}). Note that (see [PPS], Z_{γδ} ≤ Z_{γ} ∩ Z_{δ}.

Lemma 5.3 Let (X_{1}, X_{2}) be amalgam in G. Let Γ = Γ(G, X_{1}, X_{2}).

(a) ⟨X_{1}, X_{2}⟩ acts faithfully on the connected component Γ_{0} of Γ containing the edge {X_{1}, X_{2}}.

(b) Suppose (X_{1}, X_{2}) has characteristic p. Then there is a smallest integer b = b(X_{1}, X_{2}) such that, for some pair (α, α′) in Γ, we have Z_{α} ≰ G_{α′}^{(1)} and dist(α, α′) = b. Moreover, b ≥ 1.

Proof: This follows immediately from the definitions. □

Definition 5.4 A path π = (α, ⋯, α′) in the coset graph of a characteristic p amalgam is said to be a critical path if π has length b and Z_{α} ≰ G_{α′}^{(1)}. We say also in this case that (α, α′) is a critical pair.

Given a fixed critical path π. Then we denote it vertices as follows:

π = (α, α + 1, …, α + b − 1, α + b) = (α′ − b, α′ − b + 1, …, α′ − 1, α′).

We also write β for α + 1.

Lemma 5.5 Let (X_{1}, X_{2}) be p-minimal, characteristic p amalgam in G. Let (α, α′) be a critical pair and δ ∈ Γ. Then the following hold.

(a) Q_{δ} = O^{p′}(G_{δ}^{(1)}). That is, G_{δ}^{(1)} is p-closed, with unique Sylow p-subgroup Q_{α′}.

(b) If Z_{δ} ≰ Z(G_{δ}), then Z_{δ} = Z_{Q_{δ}} and C_{G_{δ}}(Z_{δ}) is p-closed with unique Sylow p-subgroup Q_{δ}

(c) If Z_{α′} ≰ Z(G_{α′}) then

(a) [Z_{α}, Z_{α′}] ≠ 1.

(b) (α′, α) is a critical pair.

(c) Z_{α} is an F F -module for G_{α}, and Z_{α′} is a quadratic offender on Z_{α}.

(d) Z_{α} ≰ Z(G_{α}) and if b is even Z_{α′} ≰ Z(G_{α′}).

Proof: (a) Let γ ∈ △(δ). By definition of a p-minimal amalgam, (X_{1}, X_{2}) is parabolic and so G_{γδ} contains a Sylow p-subgroup S of G_{δ}. Hence Q_{δ} ≤ S ≤ G_{γδ} ≤ G_{γ} and so Q_{δ} ≤ G_{δ}^{(1)}. If G_{δ} ≤ G_{γ}, then Q_{γ} ≤ O_{p}(⟨G_{γ}, G_{δ}⟩) = 1, a contradiction, since G_{γ} is of characteristic p. Thus G_{δ} ≰ G_{γ} and so O^{p}(G_{δ}) ≰ G_{δ}^{(1)}. Thus 2.1 implies S ∩ G_{δ}^{(1)} = Q_{δ} and so (a) holds.

(b) Since Z_{δ} is p-reduced, Q_{δ} centalizes Z_{δ} and so Z_{δ} ≤ Z_{Q_{δ}}. Suppose that Z_{Q_{δ}} is not p-reduced. Then 2.1 implies that [Z_{Q_{δ}}, O^{p}(G_{δ})] = 1. But then also [Z_{δ}, O^{p}(G_{δ})] = 1 and since Z_{δ} is p-reduced, Z_{d} ≤ Z(G_{δ}) a contradiction. Thus Z_{Q_{δ}} is p-reduced and Z_{δ} = Z_{Q_{δ}}.

Since [Z_{δ}, G_{δ}] ≠ 1 we have O^{p}(G_{δ}) ≰ C_{G_{δ}}(Z_{δ}), then again by 2.1, C_{S}(Z_{δ}) = C_{G_{δ}}(Z_{d}) ∩ S = Q_{δ}.

(c) By definition of a critical pair, Z_{α} ≰ G_{α′}. So by (b) applied with δ = α′, [Z_{α}, Z_{α′}] ≠ 1. So (c:a) holds. Applying (b) this time to α we get Z_{α′} ≰ G_{α}^{(1)}. Thus (c:b) holds. Note that |Z_{α}/C_{Z_{α}}(Z_{α′})| ≤ |Z_{α′}/C_{Z_{α′}} (Z_{α}) or |Z_{α′}/C_{Z_{α′}} (Z_{α})| ≤ |Z_{α}/C_{Z_{α}}(Z_{α′}). In the first case we can apply 3.1 to V = Z_{α} and in the second to V = Z_{α′}. In either case 3.1(e) gives |Z_{α}/C_{Z_{α}}(Z_{α′})| = |Z_{α′}/C_{Z_{α′}} (Z_{α}). Thus (c:c) holds.

(d) Suppose that Z_{α} ≤ Z(G_{α}).Then Z_{α} = Z_{αβ} ≤ Z_{β}. So the mimimality of b gives the contradiction Z_{α} ≤ Z_{β} ≤ G_{α′}^{(1)}. □

6 The main reduction

Throughout this section we assume

Hypothesis 6.1 (i) p is a prime and (X_{1}, X_{2}) is a p-mimimal, characteristic p, noncentral amalgam in the group G.

(ii) Z_{X_{i}}Z_{X_{2}} is neither normal in X_{1} nor in X_{2}.

Set Γ = Γ(G; X_{1}, X_{2}) and set b = b(G; X_{1}, X_{2}). For any directed edge (γ, δ) define

T (γ, δ) = J(Q_{γδ}, Z_{γ})  and  H_{γ} = ⟨T (γ, δ)^{G_{γ}}⟩

Two vertices λ, γ ∈ △(δ) are called opposite (in △(δ)) if

O^{p}(G_{δ}) ≤ ⟨T (δ, γ), T (δ, λ)⟩ (*)

For Σ ⊆ Γ set Z_{Σ} = ⟨Z_{σ} | σ ∈ Σ⟩. Set V_{δ}^{(i)} = Z_{△^{(i)}(δ)} and V_{δ} = V_{δ}^{(1)}.

Lemma 6.2 Let (δ, γ) be an edge with T (δ, γ) ≠ Q_{δ}. Then there exists λ ∈ △(δ) such that λ and δ are opposite.

Proof: By 3.2 we can choose λ ∈ △(δ) such that J(G_{δ}, Z_{δ}) ≤ ⟨T (δ, γ), T (δ, λ)⟩C_{G_{δ}}(Z_{δ}). Note that O^{p}(G_{δ}) ≤ J(G_{δ}, Z_{δ}) and so by 2.1(c), O^{p}(G_{δ}) ≤ ⟨T (δ, γ), T (δ, λ)⟩. □

Lemma 6.3 Let (α, α′) be a critical pair.

(a) Z_{α′} ∈ A(Q_{αβ}, Z_{α}).

(b) A(Q_{αβ}, Z_{α}) = A(S, Z_{α}) for all S ∈ Syl_{p}(G_{αβ}).

(c) Q_{α} ≠ T (α, β) ≤ Q_{β}Q_{α}.

Proof:

(a) Note that Z_{α′} ≤ Q_{β} ≤ Q_{αβ}. By 3.1(e) applied to G_{α′} acting on Z_{α′} |Z_{α′}/C_{Z_{α′}} (Z_{α})| ≥ |Z_{α}/C_{Z_{α}}(Z_{α′})|.

(b) Let \overline{G}_{α} = C_{G_{α}}(Z_{α}). By (a) \overline{Q_{αβ}} contains an element of A_{∗}(\overline{S}, Z_{α}). By 3.1(f), S acts transitively on A_{∗}(\overline{S}, Z_{α}) and since S normalizes Q_{αβ} we get A_{∗}(\overline{S}, Z_{α}) = A_{∗}(\overline{Q_{αβ}}, Z_{α}). (b) now follows from 3.1(d).

(c) The proof of (b) shows that T (α, β) = ⟨Z_{α′}^{S}⟩ ≤ Q_{β}Q_{α}. □

Lemma 6.4 Let (γ, α, β) be a path of length two, Σ ⊆ △^{b−1}(β) and Σ^{∗} ⊆ △^{b−2}(β). Suppose that

(i) γ and β are opposite in △(α).

(ii) Z_{Σ}Q_{α} = T (α, β)

(iii) For each σ ∈ Σ there exists σ^{∗} ∈ Σ^{∗} ∩ △(σ).

Then Z_{Σ^{∗}}Q_{γ} = T (γ, α) ≠ Q_{γ}.

Proof: We may assume without loss that Z_{σ} ≰ Q_{α} for all σ ∈ Σ. Let α′ ∈ Σ. Then (α, α′) is a critical pair. Moreover, by (iii) we may assume that α′ − 1 = α′^{∗} ∈ Σ^{∗}. Set U_{α′} := U(Z_{α}, T (α′, α′ − 1), Z_{α′}), ( see 3.3) and U = ⟨U_{α′} | α′ ∈ Σ⟩. Then by 3.4(d), Z_{α′} = U_{α′}C_{Z_{α′}} (Z_{α}) = U_{α′}(Z_{α′} ∩ Q_{α}) and so by (ii),

1^{∘} T (α, β) = UQ_{α}.

Suppose that [Z_{γ}, Z_{Σ^{∗}}] = 1 or that Z_{Σ^{∗}}Q_{γ} ≠ T (γ, α). In the first case set Y = Z_{γ} and in the second Y = U′(Z_{Σ^{∗}}, T (γ, α), Z_{γ}). Then [Y, Z_{α′−1}] = 1 and so [Y, Z_{α}] = 1 and Y ≤ Q_{α′−1α′} ≤ G_{α′−1α′}. Thus by 3.4(e), [Y, U_{α′}] ≤ [Z_{α}, U_{α′}] ≤ Z_{α}. Since this is true for all α′ ∈ Σ we conclude that

2^{∘} U normalizes Y Z_{α}.

By 6.3(c) T (α, γ) ≤ Q_{γ}Q_{α}. Thus by (1^{∘}) and (i) O^{p}(G_{α}) ≤ ⟨T, U⟩Q_{α}. Since G_{αγ} normalizes Q_{γ} and since G_{α} = G_{αγ}O^{p}(G_{α}) = G_{αγ}⟨Q_{γ}, U⟩ we conclude that ⟨Q_{γ}^{G_{α}}⟩ ≤ ⟨Q_{γ}, U}. Thus

3^{∘} O^{p}(G_{α}) ≤ ⟨Q_{γ}, U⟩.

Note that Q_{γ} centralizes Y and so normalizes Y Z_{α}. From (2^{∘}) and (3^{∘}) we conclude that O^{p}(G_{α}) normalizes Y Z_{α}. By 3.4(b) and 3.1, Z_{γ} = ⟨Y^{G_{αγ}}⟩(Z_{α} ∩ Z_{γ}) and we conclude that Z_{γ}Z_{α} is normalized by G_{αγ}O^{p}(G_{α}) = G_{α}, a contradiction to Hypothesis 6.1(ii). □

Lemma 6.5 There exists a critical pair (α, α′) with Z_{α′}Q_{α} = T (α, β).

Proof: Let (δ, δ′) be any crictical pair. Then V_{δ+1}^{b−1}Q_{δ} = T (δ, δ + 1) ≠ Q_{δ}. Hence we can choose path (α, β, …, δ) of length i ≤ b such that V_{δ}^{b−i}Q_{α} = T (α, β) ≠ Q_{α} and i is maximal.

Suppose that i < b. By 6.2 there exists a vertex γ ∈ △(α) opposite to β. Let Σ = △^{b−i}(δ) and Σ^{∗} = △^{b−(i+1)}(δ). Then we can apply 6.4 and conclude that Z_{Σ^{∗}}Q_{γ} = T (γ, α) ≠ Q_{γ}. Since Z_{Σ^{∗}} = V_{δ}^{b−(i+1)} we obtain a contradiction to the maximal choice of i. Thus i = b and so V_{δ}^{b−i} = Z_{δ}. Thus the conclusion of the lemma holds for α′ = δ. □

Theorem 6.6 Suppose that b > 1 and let (α, β) be an edge.

(a) b = 2.

(b) p = 3.

(c) T (α, β) = T (β, α)

(d) H_{α}/Q_{α} ≅ H_{β}/Q_{β} ≅ SL_{2}(q)^{n}.

(e) Z(H_{α}) = Φ(Q_{α}) is elementary abelian of order q^{n}.

(f) Z_{α} = [Z_{α}, H_{α}] × Z(H_{α}) and [Z_{α}, H_{α}] is a natural SL_{2}(q)^{n}-module for H_{α}.

(g) Q_{α}/Z_{α} is natural SL_{2}(q)-module for H_{α}.

(h) Q_{α}/Z(H_{α}) is indecomposable as H_{α}-module.

Proof: For an edge (γ, δ) define D(γ, δ) = [Z_{γ}, T (γ, δ)]. Note that D(γ, δ) ⊴ G_{γδ}. Let (α, α′) be as in 6.5. By 6.2 there exists α−1 ∈ △(α) opposite to α+1. By 6.4, (α−1, α−1) is critical and Z_{α′−1}Q_{α−1} = T (α − 1, α). Hence D(α − 1, α) = [Z_{α} − 1, Z_{α′−1}] ≤ Z_{α′−1}. Since b > 1 we conclude that Z_{α′} centralises [D(α − 1, α). Thus D(α − 1, α) is normal in ⟨G_{α−1α}, Z_{α′}⟩ = G_{α}. Set D_{α} = D(α − 1, α) and note that D_{α} = D(γ, α) for all γ ∈ △(α). Since Z_{α′} centralizes D_{α}, [D_{α}, O^{p}(G_{α})] = 1.

By 6.2 there exists α − 2 ∈ △(α − 1 opposite to α. By symmetry, D(α − 2, α − 1) = [Z_{α−2}, Z_{α′−2}] is normal in G_{α−1}. We write D_{α−1} for D(α − 2, α − 1). Note that D_{α−1} ≤ Z_{α′−2}.

Suppose that b > 2. Then Z_{α′} centralizes D_{α−1}. Since D_{α−1} is normal in G_{α−1} and we conclude that D_{α−1} is also normal in ⟨G_{α−1,α}, Z_{α′}⟩ = G_{α}, a contradiction.

Thus b = 2. Let β = α+1 and δ = α−1. Since O^{p}(G_{α}) centralizes D_{α}, D_{α} ∩Z(G_{β}) = 1. As D_{α} = [Z_{β}, T (β, α)] we conclude that [Z_{β}, H_{β}] is a natural module for G_{β}.

For γ ∈ Γ define m_{γ} = |T_{γ,λ}/Q_{γ}|, where λ ∈ △(γ). Note that |[Z_{γ}, G_{γ}]| = m_{γ}^{2} and |D_{γ}| = m_{λ} Replacing (α, α′) by (δ, β) if necessary we may assume that m_{α} ≥ m_{β}. Then |V_{α}/V_{α} ∩ Q_{β}| = m_{β} ≤ m_{α} = |V_{β}Q_{α}/Q_{α}|. Also [V_{α} ∩ Q_{β}, V_{β}] ≤ D_{β} ≤ Z_{α} and so the sum of the non-central composition factor for G_{α} on V_{α}/Z_{α} is an F -module for G_{α}/Q_{α}. So by 3.1, G_{α} has a unique non-central composition factor on V_{α} and this compostion factor is an a module for O^{p}(G_{a}) isomomorphic to Z_{α}/C_{Z_{α}}(H_{α}). Moreover, m := m_{α} = m_{β}. Since

Φ(V_{α}) ∩ Φ(V_{β}) ≤ D_{β} ∩ D_{α} = 1, V_{α} ∩ V_{β} is elementary abelian. Let L_{α} = ⟨V_{β}^{G_{α}}⟩ and note that L_{α} = ⟨V_{δ}, V_{β}⟩ Since [V_{α}, V_{β}] ≤ V_{α} ∩ V_{β} we see that L_{α} normalizes (V_{δ} ∩ V_{α})(V_{β} ∩ V_{α}. Hence V_{α} = (V_{δ} ∩ V_{α})(V_{β} ∩ V_{α}). Put E_{α} = ⋂ V_{β}^{G_{α}}. Then E_{α} = V_{δ} ∩ V_{α} ∩ V_{β}, E_{α} ≤ Z_{V_{α}} and |V_{α}/E_{α}| ≤ m^{2}. In particular, V_{α}/E_{α} is a natural SL_{2}(q_{α})^{n_{α}}-module for H_{α} and G_{α} acts irreducibly on V_{α}/E_{α}. Since V_{α} is not abelian we conclude E_{α} = Z_{V_{α}}. Since C_{[Z_{β},L_{β}]}(V_{α}) = D_{α} we have |Z_{β}E_{α}/E_{α}| = m = |C_{V_{α}/E_{α}}(V_{β})|. It follows that Z_{β}E_{α}/E_{α} = C_{V_{α}/E_{α}}(V_{β}). Taking the centralizes of this group in G_{αβ} we obtain T (α, β) = T (β, α). Thus Q_{α} ≤ V_{α}Q_{β}, Q_{α} ∩ Q_{β} ≤ Z_{α}Q_{β} and so Q_{α} = V_{α}(Q_{α} ∩ Q_{α′}). In particular, [Q_{α}, L_{α}] ≤ V_{α}. Since Q_{α} ≤ V_{α}Q_{β}, [Q_{α}, Z_{β}] = D_{α} and so [Q_{α}, V_{α}] = D_{α}. Let D_{α}^{∗} := C_{Q_{α}}(O^{p}(G_{α})). Then Φ(Q_{α}) ≤ D_{α}^{∗} and [D_{α}^{∗}, L_{α}] = 1. So D_{α}^{∗} ≤ Q_{α−2} ∩ Q_{α} ∩ Q_{α′} and this latter group is normalized by Q_{α}⟨Z_{α−2}, Z_{α′}⟩. Thus D_{α^{∗}} = Q_{α−2} ∩ Q_{α} ∩ Q_{α′}, |Q_{α}/D_{a^{∗}}| = m^{4} Q_{α} = V_{α}D_{α^{∗}} and C_{Q_{α}}(V_{α}) = Z_{α}D_{α}^{∗}. Since [D_{α^{∗}}, V_{β}] = 1, D_{α^{∗}} ≤ Z_{β}D_{β}^{∗}. Since [D_{α^{∗}}, V_{α}] = 1 and C_{Z_{β}}(V_{α}) = D_{α}Z(H_{β}) we conclude that D_{α^{∗}} ≤ D_{α}D_{β}^{∗}. Thus D_{α}^{∗} = D_{α}(D_{α}^{∗} ∩ D_{β}^{∗}) = D_{α}. Thus |Q_{α}| = m^{5} = |Q_{β}|.

Let K and L be distinct Z_{α}-components of G_{α}. Let Q_{K} = [Q_{α}, O^{p}(K)]. Since Φ(Q_{α}) = D_{α} = Z(H_{α}), [Q_{K}, Q_{L}, K] = 1. Since Q_{α}/Z_{α} is a natural module, [Q_{L}, K] ≤ Z_{α}. So [Q_{L}, K], Q_{K}] = 1 and by the Three Subgroup Lemma, [Q_{K}, K, Q_{L}] = 1. Since Q_{K} = [Q_{K}, K] we conclude that [Q_{K}, Q_{L}] = 1. Since Q_{α} is generated by the various Q_{K}, we conclude that Q_{K} is not abelian.

Suppose that p = 2. Since Q_{K}E_{α}/E_{α} is a natural SL_{2}(q_{α}) or Sym(2^{k} + 1)-module for K, it is easy to see that every element in Q_{K} is conjugate to an element in [Q_{K}, V_{β}]E_{α}. Thus all elements in Q_{K} are involutions. This implies that Q_{K} is abelian, a contradiction.

Hence p is odd. Since Q_{K} ≰ Q_{β}, [Q_{K}, V_{β}] ≰ Z(V_{β}). Thus 1 ≠ [Q_{K}, V_{β}, V_{β}] ≤ D_{β} and so V_{β} does not act quadratically on Q_{K}D_{α}/D_{α}. But [D_{β}, V_{β}] = 1 and so V_{β} acts cubic on Q_{K}D_{α}/D_{α}. By 3.6 we conclude that p = 3. Let x ∈ Q_{K} − Q_{β}. Then 1 ≠ [x, Z_{β}] ≤ [x, Q_{α}] = [x, Q_{K}]. Since [x, Z_{β}] has order at least q_{β} and [x, Q_{K}] has order at most q_{α} we get q_{β} ≤ q_{α}. By symmetry, q_{β} ≤ q_{α} and so q_{α} = q_{β}. As m_{α} = m_{β}, n_{α} = n_{β} and all parts of 6.6 are proved. □

Theorem 6.7 Suppose that b = 1. Then the following holds for all δ ∈ Γ and all directed edges (α, β) in Γ.

(a) Q_{δ} is elementary abelian.

(b) T (α, β) = T (β, α) and m := |T (α, β)/Q_{α}| = |T (β, α)/Q_{β}|.

(c) Q_{δ} = [Q_{δ}, H_{δ}]Z(H_{δ}) and |Z(H_{δ})| ≤ m.

(d) One of the following holds

1. H_{α}/Q_{α} ≅ H_{β}/Q_{b} ≅ SL_{2}(q)^{n} and m = q^{n}. Moreover one of the following holds.
1. For all δ ∈ Γ, [Q_{δ}, H_{δ}] is the natural SL_{2}(q)^{n}-module for H_{δ}/Q_{δ}.
2. p = 2, q > 2 and for all δ ∈ Γ, Q_{δ} is the natural Ω_{3}(q)^{n}-module for H_{δ}/Q_{δ}.

2. p = 2, H_{δ}/Q_{δ} ≅ Sym(2^{k_{δ}} + 1)^{n_{δ}}, [Q_{δ}, H_{δ}] is the corresponding natural module for

H_{δ}/Q_{δ}, m = 2^{n_{δ}2^{k_{δ}−1}} and n_{α}2^{k_{α}} = n_{β}2^{k_{β}}.

Proof: By edge tranisitivity and b = 1, (a, β) is a critical pair. Since Z_{α} is an offender on Z_{β} and G_{αβ} normalizes Z_{α} we get that Z_{α}Q_{β} = T (β, α). Since Q_{α} centralizes Z_{α} we get Q_{α} ≤ Z_{α}Q_{β}. Thus Q_{α} = Z_{α}(Q_{α} ∩ Q_{β}) and so Φ(Q_{α}) = Φ(Q_{α} ∩ Q_{β}) = Φ(Q_{β}) and we conclude that Q_{α} and Q_{β} are elementary abelian. Thus Q_{δ} = Z_{δ} and T (α, β) = Z_{α}Z_{β} = T (β, α).

Moreover, m = |Z_{β}Q_{α}/Q_{α}| = |Z_{α}/C_{Z_{α}}(Z_{β})| = |T (β, α)/Q_{β}|. Since H_{α} is generated by Q_{α} and two conjugates of Q_{β}, |Q_{α}/Z(H_{α})| ≤ m^{2} and so Q_{α} = [Q_{α}, H_{α}]Z(H_{a}). In particular Q_{α}/Z(H_{α}) is the appropriate natural module.

Since Z(H_{α}) ∩ Z(H_{β}) = 1 we have |Z(H_{α})| ≤ |C_{Q_{β}/Z(H_{β})}(Q_{α})| = m. Thus (a),(b) and (c) are proved.

Suppose that Q_{α} components of G_{α} are symmetric groups. Then there exists x ∈ Z_{α} with |[x, Z_{β}]| = 2 and so also the Q_{β} components of G_{β} are symmetric groups.

If [Q_{δ}, H_{δ}] ∩ Z(H_{δ}) = 1 for all δ ∈ Δ we conclude that (d:1:1) or (d:2) holds. So suppose that [Q_{α}, H_{α}] ∩ Z(H_{α}) ≠ 1, then p = 2 and the Q_{α}-components of G_{α} are SL_{2}(q)’s with q > 2. Let U be a G_{α}-component of Q_{α}. Then there exists a unique G_{β}-component V of Q_{β} with [U, V] ≠ 1. Let A := [U, V] ∩ Z(H_{α}) and B := [U, V] ∩ Z(H_{β}). Note that A = U ∩Z(H_{α}). Suppose that |A| < q. Then by 3.7(b) A^{#} = [U, V]−{[u, v] | u ∈ U, V ∈ V }. But this forces A = B = 1, a contradiction the the assumption. Thus |A| = q and so AZ(H_{β}) = [U, V]Z(H_{β}). This implies ([Q_{α}, H_{α}] ∩ Z(H_{α}))Z(H_{β}) = [Q_{β}, T (β, α)]Z(H_{β}) and so (d:1:2) holds.

7 Theorem B

In this section we prove Theorem B. So suppose that Hypothesis A holds and observe that Hypothesis A implies Hypothesis 6.1. So we can apply the result of the section 6 . We also continue to use the notation of section 6

Lemma 7.1 For all α ∈ Γ, G_{α} has a unique Z_{α} component K and K ≅ SL_{2}(q).

Proof: Let (α, α′) be a critical pair with Z_{α′}Q_{α} = T (α, β), where β = α + 1. Let V by a T (α′, α′ − 1) components of Z_{α′}. Then it is easy to see that V ∈ A_{∗}(T (αβ), Z_{α}). Set U = U(A, T (α, β), Z_{α}) and U′ = U′(V, T (α, β), Z_{α}). Then U is a minimal offender on Z_{α′} and V = U(U, T (α′, α′ − 1), Z_{α′}). Set V′ = U′(U, G_{αα′−1}, Z_{α′}, R = [U, V] and L = N_{G}(R). Pick S ∈ Syl_{p}(G_{αβ}). Then B(S) ≤ T (α, β) ≤ L and Hypothesis A implies that L has characteristic p. Set Q = O_{p}(L) and F = G_{αβ} ∩ Q. If b = 2, then H_{β} ≤ L and so F ≤ Q_{β}. Hence in any case F ≤ G_{α′} and so F ≤ O_{p}(N_{G_{α′}} (R)) and so by 3.4(g), F ≤ UQ_{α′} and so [F, V′] = 1. Since T (α, β) = Z_{α′}Q_{α} = V′V Q_{α} = V′O_{p}(H_{α} ∩ L) we conclude from 3.4(g) that [O^{p}(G_{α} ∩ L), F] = 1. Set n := |A_{∗}(S/Q_{α}, Z_{α})|. Then |S/S ∩ L| ≤ n. Let T ∈ Syl_{p}(L) with S ∩ L ≤ T. Since S ∈ Syl_{p}(G), |T | ≤ |S| and so |T/S ∩ L| ≤ n. Since Q ≤ T and Q ∩ S ≤ F, |Q/F | ≤ n. Since C_{G}(Q) ≤ Q and O^{p}(G_{α} ∩ L) centralizes F, we conclude that H_{α} ∩ L/O_{p}(H_{α} ∩ L) acts faithfully on an elementary abelian p-group of order at most n. (Namely the sums of the non-trivial factors of composition series for H_{α} ∩ L on Q and through F.) But H_{α} ∩ L/O_{p}(H_{α} ∩ L) is the direct product of n − 1 non-cyclic groups and so any faithful 𝔽_{p}-representation has order at least p^{2(n−1)}. Hence n = 1. □

Lemma 7.2 Suppose b = 1. Then one of the following hold.

1. For all δ ∈ Γ, Q_{δ} is a natural SL_{2}(q)-module for H_{δ}.

2. p = 2 and for all δ ∈ Γ, Q_{δ} is a natural O_{3}(q) for H_{δ}.

Proof: Suppose neither (1) nor (2) holds. Then by 7.1 and 6.7 H_{δ}/Q_{δ} ≅ SL_{2}(q), [Q_{δ}, H_{δ}] is a natural SL_{2}(q)-module and 1 < |Z(H_{δ})| ≤ q. If q = 2, this means that Q_{δ} is the O_{3}(2)-module. So q > 2. Let (α, β) be a critical pair, S ∈ Syl_{p}(G_{αβ}) and T = Z_{α}Z_{β}. Set L = N_{G}(T). Then L normalizes T′ = [Z_{α}, Z_{β}] and Z(T) = Z(H_{α})×T′ = Z(H_{β})×T′. Since S is a Sylow p-subgroup of L and S normalizes a complement to T′ in Z(T) we conclude from Gaschütz Theorem that L normalizes a complement D to T′ in Z(T). Since q > 2 we have Z(H_{a}) = C_{Z(T)}(L∩H_{α}) = D = C_{Z(T)})(L∩H_{β}) = Z(H_{β}) and so Z(H_{α}) = Z(H_{β}) = 1. □

8 Theorem E

In this section we prove Theorem E. So we assume that A(i), A(ii), A(iii), and Q! hold in G. We wish to show that Z_{P} ZP̃ is neither normal in P nor in P̃. So we also assume that Z_{X}Z_{Y} ⊴ X for some ordering (X, Y) of {P, P̃)} and will derive a contradiction. Note that (X, Y) is a p-minimal, characteristic p, non-central amalgam and we can use the amalgam method.

Lemma 8.1 Let (α, α′) be a critical pair. Then α, α′ ∈ G/X and b is even.

Proof: Suppose that α ∈ G/Y. Then (α, β) is conjugate to (Y, X) and so Z_{α}Z_{b} is normal in G_{β}. Thus b > 1 and Z_{α}Z_{β} = Z_{α+2}Z_{β} ≤ Q_{α′}, a contradiction. So α ∈ G/X. By 5.5(b) (α′, α) is a critical pair. Thus also α′ ∈ G/X and b is even. □

Lemma 8.2 X ≰ C̃ and Z(X) = 1.

Proof: Suppose that X ≤ C̃. Since X contains a Sylow p-subgroup of C̃, Q = O_{p}(C̃) ≤ O_{p}(X). Let (α, α′) be a critical pair. By 8.1 α = Xa and α′ = Xb for some a, b ∈ G. By 5.5(b), 1 ≠ [Z_{α}, Z_{α′}] ≤ Z_{α} ∩ Z_{α′} ≤ Z_{Q^{a}} ∩ Z_{Q^{b}}. Thus by [MSS, Lemma 2.4.2], Q^{a} = Q^{b}. But Z_{Q_{X}} ≤ Z_{Q} ≤ Q and so

Z_{α} ≤ Z_{Q^{a}} = Z_{Q^{b}} ≤ Q^{b} ≤ Q_{α′}

a contradiction.

Thus X ≰ C̃. Note that Z(X) ≤ Z(Q). If Z(X) ≠ 1, then by Q!, X ≤ C_{G}(Z(X)) ≤ C̃, a contradiction. □

Lemma 8.3 Z_{Y} < Z_{X} and Q_{X} < Q_{Y}.

Proof: Suppose that Z_{Y} ≰ Z_{X}. Then [Z_{Y} , Q_{X}] ≠ 1. Since [Z_{Y} , Q_{X}] = [Z_{Y} Z_{X}, Q_{X}] and since both Z_{Y} Z_{X} and Q_{X} are normal in X we conclude that [Z_{Y} , Q_{X}] is normal in X. If Q_{Y} ≤ Q_{X} then Z_{X} ≤ Z_{Y} and Z_{X}Z_{Y} = Z_{Y} is normal in X and Y, a contradiction. Hence Q_{Y} ≰ Q_{X}, since [Z_{Y} , Q_{X}] ≤ Z_{Y} we have that Q_{Y} and so also ⟨Q_{Y}^{X}⟩ centralizes [Z_{Y} , Q_{X}]. From 2.1 we conclude that O^{p}(X) centralizes [Z_{Y} , Q_{X}]. But then Z(X) ≠ 1, contrary to 8.2. □

Lemma 8.4 b = 2.

Proof: Suppose that b > 2. Let g ∈ G with (α + 2)^{g} = α. Set δ = α^{g} ∈ △^{2}(α), γ = β^{g} and T := V_{α}^{(b−2)}. Since [Z_{α}, Z_{α′}] ≠ 1 we have [Z_{α}, V_{a+2}^{(b−2)}] ≠ 1 and so [Z_{δ}, T] ≠ 1. Note that by 5.5(b), T Q_{δ}/Q_{d} ≤ J(Q_{δα}/Q_{δ}, Z_{δ}). Since [Z_{δ}, T] ≠ 1, there exists a T (δ, β) component V of Z_{δ}, with A := [V, T] ≠ 1.

Suppose that O^{p}(G_{α}) normalizes A. Then since Q_{δ} ≤ Q_{γ} ≤ G_{α} and since O^{p}(G_{α}) does not centralize A, we get Q_{δ} ≤ Q_{α}. But then Q_{α} = Q_{δ}, a contradiction to [Z_{d}, T] ≠ 1.

Thus O^{p}(G_{δ}) does not normalize A. Let U be a G_{α′−1α′} component of Z_{α′} with U ≰ Q_{α}. Since O^{p}(G_{α}) ≤ ⟨U^{G_{α}}⟩, there exists h ∈ G_{α} such that U^{h} does not normalize A. Then U does not normalize A^{h^{−1}} and replacing g by gh^{−1} and V by V^{h^{−}1} we may assume that U does not normalize A. Since Z(X) = 1, 3.4(f) implies that A = [V, t] for all t ∈ T with [V, t] ≠ 1.

Suppose that V ≰ Q_{α′−2}. Then A = [V, Z_{α′−2}] ≤ Z_{α′−2}. Since b > 2 we conclude that [A, U] = 1 a contradiction. Thus V ≤ Q_{α′−2} ≤ Q_{α′−1} ≤ G_{α′}. Since [V, Z_{α}] = 1 we get from 3.4(e) that [V, U] ≤ [Z_{α}, U] ≤ Z_{α}. Hence U normalizes V Z_{α}. U also normalizes T and so U normalizes [V Z_{a}, T] = [V, T] = A, again a contradiction. □

Proof of Therorem E

Since b = 2 we have Q_{α} ≤ Q_{β} ≤ G_{αβ}. Also [Q_{α}, Z_{α}] = 1 and so by 3.4(e), [Q_{α}, Z_{α′} ≤ Z_{α}. Thus implies [Φ(Q_{α}), O^{p}(G_{α})] = 1 and since Z(G_{α}) = 1 we get Φ(Q_{a}) = 1. Thus Z_{α} = Q_{α} is a natural SL_{2}(q)^{k} or Sym(2^{l} + 1)^{k} module B(G_{α}). Using that the p-locals subgroups of G containing B(S) are of characteristic p we can argue as in 7.1 and conclude that B(X)/Q_{X} ≅ SL_{2}(q) and Q_{X} is the natural module for B(X)/Q_{X}. Let T = B(S). Choose notation such that G_{α} = X and G_{β} = Y. Then Z_{α}Z_{α′} = T and so T ≤ Q_{Y} and T ⊴ Y. Recall the definition of N in 4.4. Then B(X) ∈ N and so Q_{X} ∈ A^{+}(T). It follows from columns 3 and 4 of the table in 4.4 that O^{p}(Y) acts trivially on A^{+}(T). Thus Y normalizes Q_{X}, a contradiction.

9 The maximal p-locals containing S

In this section we assume Hypothesis A and determine the set M(S) of maximal p-local subgroups of G containing S. Note that Theorem B tells us the structure of P and P̃. Let T = B(S), {X, Y } = {B(P), B(P̃)}. Then T ∈ Syl_{p}(X) ∩ Syl_{p}(Y).

Proposition 9.1 Assume that Case 3 of Theorem B holds. Then one of the following holds:

1. P and P̃ are not conjugate in G and M(S) = {N_{G}(B(P)), N_{G}(B(P̃)}.

2. P and P̃ are conjugate in G, |N_{G}(B(S))/N_{G}(B(S)) ∩ N_{G}(B(P))| = 2 and M(S) = {N_{G}(B(P)), N_{G}(B(P̃)), N_{G}(B(S))}

Proof:

1^{∘} Z_{X} ∩ Z_{Y} = Z(X)Z(Y). Let A = Z_{X}Z_{Y}. Then A = Q_{X} ∩ Q_{Y} is elementary abelian of order q^{4}, A ∈ A(S), and O^{p′}(C_{G}(A)) = A.

Since Z(X)Z(Y) ≤ Z_{X} ∩ Z_{Y} ≤ C_{Z_{X}} (Q_{Y}) we have Z(X)Z(Y) has order q^{2} and so Z_{X}Z_{Y} has order q^{4}. Since ”b=2”, A = Z_{X}Z_{Y} ≤ Q_{X} ∩ Q_{Y}. As Q_{X} ∩ Q_{Y} has order q^{4}, A = Q_{X} ∩ Q_{Y}. From the structure of S it is easy to see that S has no elementary abelian subgroups of order larger that q^{4}. So A ∈ A(S). C_{S}(A) = C_{S}(Z_{X}) ∩ C_{S}(Z_{Y}) = Q_{X} ∩ Q_{Y}. As S is a Sylow p-subgroup of N_{G}(A) also the last statement in (1^{∘}) holds.

2^{∘} T = Q_{X}Q_{Y} ∈ Syl_{p}(X) ∩ Syl_{p}(Y), Z(T) = Z_{X} ∩ Z_{Y} = Z(X)Z(Y) and A = Z_{2}(T).

Here Z_{2}(T) is the second center of T defined by Z_{2}(T)/Z(T) = Z(T/Z(T)).) By (1^{∘}), A ≤ T. Also Q_{X} = ⟨A^{X}⟩ and so Q_{X}Q_{T} ≤ T. Clearly T ≤ Q_{X}Q_{Y} and so T = Q_{X}Q_{T}. Thus Z(T) = Z(Q_{X}) ∩ Z(Q_{Y}) = Z_{X} ∩ Z_{Y} = Z(X)Z(Y). [Z_{X}, T] = [Z_{X}, Q_{Y}] ≤ Φ(Q_{Y}) = Z(Y) ≤ Z(T) ansd so A ≤ Z_{2}(T). [Z_{2}(T), Q_{X}] ≤ Z_{X} ∩ Z_{Y} ≤ Z_{X}. So Z_{2}(T) acts trivially on Q_{X}/Z_{X} and Z_{2}(T) ≤ Q_{X} ∩ Q_{Y} = A.

3^{∘} N_{G}(Q_{X}) = N_{G}(X) and N_{G}(Q_{X})/Q_{X} acts faithfully on Q_{X}/Z_{X}.

Let L be maximal in N_{G}(Q_{X}) with [Q_{X}, L] ≤ Z_{X}. Since S ∩ L = Q_{X}, L/Q_{X} is a p′- goup. Note that Z_{X} = Z(Q_{X}) is normal in N_{G}(Q_{X}). Let \overline{Q}_{X} = Q_{X}/Φ(Q_{X}). Suppose that [\overline{Z_{X}}, L] ≠ 1. Since X normalizes L and X is irreducible on \overline{Z_{X}} we conclude that [\overline{Q_{X}}, L] = \overline{Z_{L}}. By Mascke’s Theorem, \overline{Q_{X}} = \overline{Z_{L}}×C_{\overline{Q}_{X}} (L), a contradiction since \overline{Q}_{X} is indecomposable

on an X-module. Thus L centralizes \overline{Z}_{L} and so O^{p}(L) centralizes \overline{Q}_{X} and Q_{X}. As N_{G}(Q_{X}) has characteristic p, O^{p}(L) = 1 and L = Q_{X}. The commutator map from provides as with a 𝔽_{q} linear symplectic form on Q_{X}/Z_{X} and so Aut(Q_{X})/C_{Aut(Q_{X})(Q_{X}/Z_{X})} ≤ ΓL_{2}(q). Thus N_{G}(Q_{X}) = LN_{G}(X) and (3^{∘}) is proved.

4^{∘} Let X ≤ L ≤ G with Q_{L} ≠ 1. Then L ≤ N_{G}(X).

Replacing L by N_{G}(Q_{L}) we may assume that L is a p-local subgroup. Since B(S) ≤ L, L is of characteristic p, that is C_{G}(Q_{L}) ≤ Q_{L}. Thus Q_{L} ≰ Z_{X}. But Q_{L} ≤ O_{p}(XS) = Q_{X}. Since \overline{Q}_{X} is indecomposable all proper X invariant subgroups of Q_{X} are in Z_{X}. Thus Q_{X} = Q_{L} and (4^{∘}) follows from (3^{∘}).

5^{∘} Suppose Q ≤ G with Q ≅ Q_{X}. If [Z_{X}, X] ≤ Q, then Q ∈ {Q_{X}, Q_{Y}^{X}}. And if A ≤ Q, then Q ∈ {Q_{X}, Q_{Y} }.

To prove the first statement suppose [Z_{X}, X] ≤ Q. Since |Z(Q)| = q^{3} and G has no elementary abelian subgroups of order q^{5}, Z(Q) ∩ [Z_{X}, X] ≠ 1. Since X is transitive on [Z_{X}, X]^{#} we may assume that Z(Q) ∩ Z(Y) ≠ 1. Since ⟨Q, Y ⟩ ≤ N_{G}(Z(Q) ∩ Z(Y) we get from 5 that Q ≤ N_{G}(Y). Since B(N_{G}(Y)) = Y, and since Q_{X} and so als Q is generated by elementary subgroups of maximal order q^{4} we conclude Q ≤ Y.

Suppose that [[Z_{X}, X], Q] ≰ Z_{Y}. Since ⟨[Z_{X}, X]^{Q} ≤ Q_{Y} ∩ Q we conclude that (Q ∩ Q_{Y})Z_{Y} /Z_{Y} contains a 𝔽_{q}-subspace of Q_{Y} /Z_{Y} as a proper subgroup. Thus Z(Y) = Φ(Q_{Y}) = Φ(Q ∩ Q_{Y} ⟩ ≤ Φ(Q). Since both Z(T) and Φ(Q) have order q, Z(Y) = Φ(Q). But Y/Z(Y) has a unique elementary abelian subgroup of order q^{4} namely Q_{Y} /Z(Y) and we conclude that Q = Q_{Y}, a contradiction to [[Z_{X}, X], Q] ≰ Z_{Y}.

Hence [Z_{X}, X], Q] ≤ Z_{Y} and so Q ≤ T. If [Z_{X}, X], Q] ≰ 1 we get again Φ(Q) = Z(Y)) and Q = Q_{Y}. And if [Z_{X}, X, Q] = 1 we get Q ≤ Q_{X} and Q = Q_{X}. So the first statement holds.

To prove the second statment suppose that A ≤ Q and Q ≠ Q_{X}. By the first statement, Q = Q_{Y}^{x} for some x ∈ X. Since A ∩ Q_{Y}^{x} ≤ Q_{Y} ∩ Q_{Y}^{x} = Z_{X} for all x ∈ X − N_{X}(Q_{Y}), we get Q = Q_{Y}.

6^{∘} No elements of Z(T) − (Z(X) ∪ Z(Y)) is conjugate to an element of Z(X) ∪ Z(Y).

Let 1 ≠ z ∈ Z(T) and 1 ≠ x ∈ Z(X). Suppose that x and z are conjugate in G. Note that both x and z are under N_{G}(T) conjugate to an element of Z_{S}. So e may assume that x and z are in Z(S). Thus z = x^{g} for some g ∈ N_{G}(S) ≤ N_{G}(T). Since A ≤ Q_{X} and A = Z_{2}(T) ⊴ N_{G}(T) we get A ≤ Q_{X}^{g}. Thus by (5^{∘}), Q_{X}^{g} ∈ {Q_{X}, Q_{Y} } and so z ∈ Q′_{X} = Z(X) or z ∈ Q′_{Y} = Z(Y).

7^{∘} N_{G}(Z(T)) = N_{G}(T), N_{G}(T) ∩ N_{G}(X) = N_{G}(T) ∩ N_{G}(Y) and |N_{G}(T)/N_{G}(T) ∩ N_{G}(X)| ≤ 2.

By (6^{∘}) N_{G}(Z(T)) normalizes {Z(X), Z(Y)}. By the proof of (6^{∘}), if Z(X) is conjugatte to Z(Y), then Z(X) and Z(Y) are conjugate in N_{G}(T). By (4^{∘}) N_{G}(Z(T) ∩ N_{G}(Z(X)) = N_{G}(X) ∩ N_{G}(T) and so (7^{∘}) holds.

8^{∘} N_{G}(A) = N_{G}(T).

By (5^{∘}), N_{G}(A) ≤ N_{G}({Q_{X}, Q_{Y} }). Since Q_{X}Q_{Y} = T, (8^{∘}) holds.

9^{∘} Let L ∈ M(S) with Q_{L} ≤ Q_{X}. Then L = N_{G}(X).

Note first that Z_{X} ≤ C_{G}(Q_{L}) ≤ Q_{L}.

Suppose that Q_{L} ≤ A. Then A ≤ C_{G}(Q_{L}) ≤ Q_{L} and Q_{L} = A. So by (8^{∘}), L ≤ N_{G}(T) and so T ≤ Q_{L}, a contradiction.

Hence Q_{L} ≰ A. We conclude that [Q_{L}, T]Z_{X} = A and so Φ(Q_{X}) = [Q_{L}, A] = [Q_{L}, [Q_{L}, T] ≤ Φ(Q_{L}).

Thus Φ(Q_{L}) = Φ(Q_{X}). Hence Q_{X} centralizes Q_{L}/Φ(Q_{L}) and Φ(Q_{L}). Since L is of characteristic p, Q_{X} ≤ Q_{L} and Q_{X} = Q_{L}. So (9^{∘}) now follows from (9^{∘})

We are now able to prove the 9.1. Let L ∈ M(S) with T ⋬ L. Then there exists R ∈ P(S) with R ≤ L and T ⋬ R.

Suppose that R ≤ N_{G}(X). Since T ⊴ X ∩ R, then T = O_{p}(X ∩ R) ⊴ R, contrary to our choice of R. Thus T ⋬ X ∩ R and the structure of X implies X = X ∩ R ≤ L. Thus by (4^{∘}) (4^{∘}), L = N_{G}(X).

Suppose next that R ≰ N_{G}(X). Then by (9^{∘}), Q_{X} ≰ Q_{R} and by (4^{∘}) [Z(XS), R] ≠ 1. Thus [Z_{S}, R] ≠ 1 and C_{S}(Z_{R}) = Q_{R}.

Suppose that Q_{X} ≤ Q_{R}. By (9^{∘}), Q_{X} ⋬ R and so ⟨Q_{X}^{R}⟩ = T and T ⊴ R, a contradiction. Thus Q_{X} ≰ Q_{R}. Suppose that Z_{X}Z_{R} is normal in X. Since Z_{X}Z_{R} is abelian we get

Z_{R} ≤ Z_{X}, a contradiction to Q_{X} ≰ Q_{R}.

Hence Z_{X}Z_{R} is not normal in X.

Suppose that Z_{X}Z_{R} is normal in R. Since Q_{R} ≰ Q_{X} we conclue that Z(Y) ≤ [Z_{X}, Q_{R}] ⊴ R. Since Q_{X} centralizes [Z_{X}, Q_{R}] we conclude that O^{2}(R) centralizes Z(Y). So by (4^{∘}) applied with Y in place of X, R ≤ N_{G}(Y), R = Y S and again by (4^{∘}) L = N_{G}(Y).

Suppose that Z_{X}Z_{R} is not normal in R. Hence (XS, R) fulfills all Hypothesis A. So Z_{X} ≤ Q_{R}. By symmetry we may assume Z_{Y} ≤ Q_{R} and so by (5^{∘}) Q_{R} ∈ {Q_{X}, Q_{Y} } and by (3^{∘}) L = N_{G}(X) or N_{G}(Y). □

Lemma 9.2 Assume that Case 2 of Theorem B holds. Then M(S) = {N_{G}(B(P)), N_{G}(B(P̃)}.

Proof:

1^{∘} A(S) = {Q_{X}, Q_{Y} }, T = Q_{X}Q_{Y} and and N_{G}(T) = N_{G}(Q_{X}) ∩ N_{G}(Q_{Y}).

The first statement follows from the structure of X and Y. The second statement now is obvious. For the last note that N_{G}(T) = SO^{2}(N_{G}(T) normalizes Q_{X} and Q_{Y}. Since S normalizes Q_{X}, the second statement follows from the first.

2^{∘} Let R ∈ P(S) with Q_{X} ⋬ R. Then (XS, R) fulils the assumption of Theorem B, Q_{R} = Q_{Y} and Z(R) = Z(Y).

Let L = ⟨R, XS⟩ and that Q_{L} ≠ 1. Then Q_{L} ≤ Q_{X} and since Q_{X} ⋬ R, Q_{L} < Q_{R}. Since L is of characteristic p, [Q_{L}, O^{2}(X] ≠ 1. Since Q_{L} ⊴ X we conclude that q = 2, Q_{L} = [Q_{X}, X] and |Q_{L}| = 4. But then X induces Aut(Q_{L}) on Q_{L} and so L = C_{L}(Q_{L})X = Q_{L}X = X, a contradiction.

Thus Q_{L} = 1. Suppose that Q_{R} ≤ Q_{X}. Then [Q_{X}, Q_{R}] = 1 and since R is of characterictic p, Q_{X} ≤ Q_{R} and Q_{X} = Q_{R}, again a contaradiction.

Thus Q_{R} ≰ Q_{X}. Suppose that Q_{X} < Q_{R}. Then ⟨Q_{X}^{R}⟩ = T,a contradiction to (1^{∘}) and Q_{X} ⋬ R.

Thus Q_{X} ≰ Q_{R}. Suppose that Z_{X}Z_{R} is normal in X. Then Z_{R} ≤ Q_{X} = Z_{X}. Since Q_{X} ≰ Z_{R} we get that [Z_{R}, R] = 1. But then [Z(XS), R] = 1, a contradiction to Q_{L} = 1.

Thus Z_{X}Z_{R} is not normal in X. Suppose that Z_{X}Z_{R} is normal in R. The [Z_{X}, Z_{R}] = 1, Z_{R} ≤ Q_{X} = Z_{X} and Z_{X}Z_{R} = Z_{X} is normal in R and X, a contradiction to Q_{L} = 1.

Thus Z_{X}Z_{R} is not normal in R. Since [Z(XS), R] ≠ 1 we see that (XS, R) fullfills the assumptions Hypothesis A and so we conclude from Theorem B, that Q_{R} ∈ A(S). Thus by (1^{∘}), Q_{R} = Q_{Y}. Also by 3.7(c), Z(R)^{#} = Z(T)−({[x, y] | x ∈ Q_{R}, y ∈ T }∪Z(X)) = Z(Y)^{#} and (2^{∘}) is proved.

3^{∘} N_{G}(Q_{X}) = N_{G}(Z(X)) .

Let R ∈ P(S). If R ≰ N_{G}(Q_{Y}), then (2^{∘}) applied to Y in place of X shows that R ∈ N_{G}(Q_{X}) ∩ N_{G}(Q_{Y}). Suppose next that R ≤ N_{G}(Q_{Y}).

If R ≤ N_{G}(Q_{X}), then R ≤ N_{G}(T). By (1^{∘}) R normalizes Q_{Y} and so by 3.7c, R = SO^{2}(S) normalizes Z(X).

If R ≰ N_{G}(Q_{X}), then by (2^{∘}), O_{p}(XS, R) = 1 and so R ≰ N_{G}(Z(X)).

Finally N_{G}(S) ≤ N_{G}(T) ≤ N_{G}(Q_{X}) ∩ N_{G}(Z(X)).

4^{∘} N_{G}(Q_{X})/Q_{X} acts faithfully on Q_{X}/Z(X).

Let D be maximal in N_{G}(Q_{X}) with [Q_{X}, D] ≤ Z(X). Since D ∩ S = Q_{X}, D/Q_{X} is a 2′ group. If [Z(X), D] = 1, then [O^{p}(D), Q_{X}] = 1 and D is a p-group. So we may assume that [Z(X), D] ≠ 1. Then q ≠ 2 and Q_{X} is indecomposable as an X-module. But by Maschke, Q_{X} = C_{Q_{X}} (D) × [Q_{X}, D], a contradiction.

5^{∘} N_{G}(Z(T)) = N_{G}(T).

Let R ∈ P(S) with Z(T) ⊴ R. Then (2^{∘}) implies Q_{X} ⊴ R and Q_{Y} ⊴ R.

6^{∘} N_{G}(Q_{X}) = N_{G}(X).

Let V = Q_{X}/Z(X) and H = N_{G}(Q_{X}). Then C_{V} (T) = Z(T)/Z(X) and so by (5^{∘}) N_{H}(C_{V} (T)) = N_{H}(T). Hence all the assumption of 3.5 are fullfilled ( with H/Q_{X} in the place of H). Suppose that 3.5(2) holds. Then H/Q_{X} ≅ Alt(7). Note that H^{1}(Alt(7), V) = 1 (for example since the commuting graph on the fixed-point free elements of order three, that is the 3-cycles, is connected). So Q_{X} = Z(C) × [Q_{X}, H], a contradiction to the action of X on Q_{X}. Thus 3.5(1) holds and X ⊴ H.

We are now able to prove 9.2. Let L ∈ M(S). If Q_{X} ⊴ L, then by (6^{∘}), L = N_{G}(X). So suppose that Q_{X} ⋬ L. Since N_{G}(S) ≤ N_{G}(Q_{X}), there exists R ∈ P(S) with R ≤ L and Q_{X} ⋬ L. By (2^{∘}) R ≤ N_{G}(Q_{Y}) and Q_{Y} = Q_{R}. Thus Q_{L} ≤ Q_{Y}, [Q_{L}, Q_{Y}] = 1 and so Q_{L} = Q_{Y}. Hence by (6^{∘}) applied to Y, L = N_{G}(Y). □

Lemma 9.3 Suppose that Case 1 of Theorem B holds. Then one of the following holds.

1. P and P̃ are not conjugate in G and M(S) = {N_{G}(B(P)), N_{G}(B(P̃)}.

2. p is odd, P and P̃ are conjugate in G, |N_{G}(B(S))/N_{G}(B(S)) ∩ N_{G}(B(P))| = 2 and M(S) = {N_{G}(B(P)), N_{G}(B(P̃), N_{G}(B(S))}

3. q = 3, 5, 7 or 13. Let N be define as in 4.4. Then M(S) = {N_{G}(N) | N ∈ N }∪{N_{G}(T)}. The structure of N_{G}(N)/N for N ∈ N and of N_{G}(T)/T is listed in columns 2 and 5 of the lower part of the table in 4.4.

4. After interchanging P and P̃ if necessary,

(a) q = 4.

(b) N_{G}(QP̃) ∼ 2^{4}.Alt(7).

(c) N_{G}(Q_{P})) = N_{G}(B(P)) ∼ 2^{4}.ΓL(2, 4)

(d) For all z ∈ Q# ; P̃, N_{G}(QP̃) ≥ C_{G}(z) ∼ 2^{1+3}L(3, 2).

(e) M(S) = {N_{G}(QP̃), N_{G}(B(P))}.

Proof:

1^{∘} N_{G}(Z_{T}) ≤ N_{G}(T).

Set N = N_{G}(Z_{T}) and N_{0} = ⟨T^{N}⟩. We have [T, T, T] = 1 by Theorem B(3), so [T, O^{p}(N_{0})] = 1. But T is the largest subgroup of S which is invariant under P ∩ T, so Q_{N} ≤ T, and [Q_{N}, O^{p}(N_{0})] = 1. Then O^{p}(N_{0}) = 1 and N_{0} = T ⊴ N. So (1^{∘}) holds.

2^{∘} Let N ∈ P(S) such that neither Q_{X} nor T is normal in N. Then B(N) ∼ q^{2}SL_{2}(q) and Q_{N} ∈ S.

Since X is irreducible on Q_{X} and N does not normalizes Q_{X} we have

O_{p}(⟨XS, N⟩) = 1 (*)

Suppose that Q_{X} ≤ Q_{N}. Since Q_{X} ∈ A(S) and T is generated by any two members of A(S) we get that ⟨Q_{X}^{N}⟩ = T ⊴ R, a contradiction. Thus

Q_{X} ≰ Q_{N} (**)

Next we show

[Z_{S}, N] ≠ 1. (***)

Suppose [Z_{S}, N] = 1]. As T is not normal in N, it now follows from 3.1 that [Z_{N}, O^{p}(N)] = 1. If Q_{N} ≤ T then Z_{T} ≤ Z_{N}, [Z_{T} , O^{p}(N)] = 1, and we contradict (1^{∘}). Thus Q_{N} ≰ T.

Let V = ⟨(Q_{X} ∩ Q_{N})^{N}⟩ and suppose that V ≤ T. Since Q_{N} ≰ T, Z_{T} < [Q_{X}, Q_{N}] ≤ V. This implies C_{S}([Q_{X}, Q_{N}]) ≤ Q_{X} and we conclude that Z_{T} ≤ Z_{V} ≤ Q_{X}. (**) yields [Z_{V} , O^{p}(N)] = 1, a contradiction to (1^{∘}).

Thus V ≰ T and q ≠ p. Let g ∈ N with A := Q_{X}^{g} ∩ Q_{N} ≰ T.

Since S ≠ T, q = p^{pm} for some positive integer m Since S/T is cyclic and A is elementary abelian, |AT/T | = p. Also any an elementary abelian subgroup of T which is centralized by an element in S − T has order at most p^{2m}. Thus A has order at most p^{2m+1}. But A has order at least q^{2}/p^{m}. Thus p^{(2p−1)m} = q^{2}/p^{m} ≤ p^{2m+1}, (2p − 3)m ≤ 1 and so p = 2 and m = 1. Thus q = 4.

Since A ≰ T there exists an involution s ∈ Q_{N} − T. Set R = ⟨Z_{T} , s^{T} ⟩. Then R ≤ Q_{N} and we have R = ⟨Z_{T} , s⟩ ∘ ⟨[T, s]⟩ ≅ D_{8} ∘ D_{8}. Thus Out(R) ≅ O_{4}^{+}(2). As S/R = Q_{P} QP̃/R is elementary abelian of order 4, and N ∈ P, it follows that Q_{N} > R, and that |S/Q_{N}| = 2.

Since Q_{X} ≰ Q_{N} we get S = Q_{X}Q_{N}, Φ(Q_{N}) ≰ Q_{X} and Z_{T} = [Q_{X} ∩ Q_{N}, Q_{N}] ≤ Φ(Q_{N}). Put W = Z_{Φ(Q_{N})}. Then Z_{T} ≤ W and W ∩ Q_{X} ≤ C_{Q_{X}} (Φ(Q_{N})) = Z_{T}. Thus W ∩ Q_{X} = Z_{T}. By symmetry W ∩ Q_{Y} = Z_{T}. But all involutions in T are contained in Q_{X} ∪ Q_{Y} and so W = Z_{T}. Hence N normalizes Z_{T}, a contradiction to (1^{∘}). So (***) holds.

Suppose that Z_{N}Z_{X} ⊴ XS. Then Z_{N} ≤ Q_{X} and [Z_{N}, Q_{X}] = 1. (*) implies [Z_{N}, O^{p}(R)] = 1 a contradiction to (***). Thus Z_{N}Z_{X} ⋬ XS. By (**), Z_{X} = Q_{X} ≰ Q_{N} and so Z_{X}Z_{N} ⋬ R. We verified all the assumptions of Theorem B and (2^{∘}) follows.

3^{∘} If p = 2 then M(S) = {N_{G}(Q_{X}), N_{G}(Q_{Y})}.

Let M ∈ M(S) with Q_{X} ⋬ M. Since |A(S)| = 2, N_{G}(S) ≤ N_{G}(T) ≤ N_{G}(Q_{X}). Thus there exists R ∈ P(S) with R ≤ M and Q_{X} ⋬ R. Then also T ⋬ R and (2^{∘}) implies Q_{R} ∈ A(S) and R acts irreducible on Q_{R}. Thus Q_{Y} = Q_{R} = Q_{M} and M = N_{G}(Q_{Y}). So (3^{∘}) holds.

4^{∘} Suppose XS < H with Q_{H} ≠ 1 and X ⋬ H. Then (4) holds.

Since X is irreducible on Q_{X}, Q_{H} = Q_{X} and we may assume that H = N_{G}(Q_{X}). Since H is of characteristic p, H/Q_{X} acts faithfully on Q_{X}. Since Z_{T} = C_{Q_{X}} (T), (1) implies N_{H}(C_{Q_{X}} (T) = N_{H}(T). 3.5 implies that (4:a) and (4:b) hold. Let 1 ≠ z ∈ Z_{S}. By 3.5(2:e), C_{H}(z) ∼ 2^{1+3}L(3, 2). Thus Q_{Y} ⋬ C_{G}(z) and (3^{∘}) shows C_{G}(z) = C_{H}(z). Thus (4:d) holds.

Suppose Y is not normal in N_{G}(Q_{Y}). Then symmetry in X and Y gives Q_{Y} = O_{2}(C_{G}(z)) = Q_{X}, a contradiction. Thus Y is normal in N_{G}(Q_{Y}). From 3.5(2:d) we

deduce that N_{H}(T)/T ≅ Sym(3) ∧^{∣} Sym(3), and since A(T) = {Q_{X}, Q_{Y} } we have N_{H}(T) ≤ N_{G}(Q_{Y}). Thus N_{G}(Q_{Y})/Q_{Y} ≅ ΓL(2, 4). and (4:c) holds. By (3^{∘}), also (4:e) holds and (4^{∘}) is proved.

5^{∘} If p = 2 then (1) or (4) holds.

This follows from (2^{∘}) and (3^{∘}).

6^{∘} Suppose that p is odd and let M ∈ M(S) with T ⋬ M. Then B(M) ∈ N.

Let R ∈ P(S) with T ⋬ R ≤ M. If Q_{X} ⊴ R, then (4^{∘}) implies X ⊴ R and so X = B(R). Thus XS ≤ M and (4^{∘}) implies X ⊴ M and X = B(M). We may assume that Q_{X} ⋬ R. Then by (2^{∘}) B(R) ∼ q^{2}SL_{2}(q). Thus by (4^{∘}) applied to R in place of XS and M in place of H gives B(R) ⊴ M and so B(R) = B(M). Thus B(M) ∈ N.

9.3 now follows from 4.4. □

Proof of Theorem D

Theorem D is just a combination of 9.1, 9.2 and 9.3.

10 Geometric characterizations of some specific groups

Recall that a rank-2 incidence geometry is a triple G = (Π, Λ, ∼) where Π and Λ are disjoint sets (of points and lines, respectively) and where ∼ is a relation (called the incidence) between Π and Λ. The corresponding incidence graph is the bipartite graph Γ with vertex set Π ∪ Λ and with adjacency of vertices given by incidence.

The following result is well-known, Maybe we should find a reference extends easily (?) to generalized n-gons for n ≥ 3.

Lemma 10.1 Let X and Y be distinct finite subgroups of a group G. Set G_{0} = ⟨X, Y ⟩, and set Γ_{0} = Γ(G_{0}; X, Y). Assume that the following conditions hold.

(i) For any path π = (δ, ϵ ⋯, δ′) in Γ_{0} of positive length n ≤ 3, the point-wise stabilizer G_{π} acts transitively on Δ(δ) − {ϵ}.

(ii) Γ_{0} contains a cycle of length 6.

Then |X : X ∩ Y | = |Y : X ∩ Y | = q + 1 for some prime power q, and Γ(G_{0}, X, Y) is the incidence graph of a desarguesian projective plane of order q.

Proof: As Γ_{0} is a bipartite combinatorial graph, all cycles are of even length, and of length at least 4. Suppose first that there exists a cycle π = (α, β, γ, δ, α)) of length 4. As G_{π} is transitive on Δ(δ) − {γ}, by (1), we then conclude that Δ(δ) = {α, γ}. Applying this to (δ, α, β, γ) in place of π, Δ(γ) = {δ, β}. Thus |Δ(ϵ)| = 2 for every vertex ϵ of Γ_{0}, and then Γ_{0} is itself a 4-cycle. This is contrary to (2), so in fact Γ_{0} contains no 4-cycles.

Fix an edge (α, β) of Γ_{0}. As G_{0} is transitive on edges, there is a cycle σ of length 6 such that (α, β) is an edge of σ. Let π = (α, β, γ, δ) be the path of length 3 in π which begins with the edge (α, β). As G_{π} is transitive on Δ(γ) − {δ}, it follows that every vertex in Δ(γ) is at distance 2 from α. As G_{0} acts transitively on the set of undirected paths of length 3, we conclude that no vertex of Γ_{0} is at distance greater than 3 from α. Thus, for any α′ ∈ α^{G_{0}} − {α}, we have dist(α, α′) = 2. By symmetry, an analogous statement holds for β. Now, call the elements of X\G “points” and the elements of Y \G “lines”, and define incidence of points and lines by non-trivial intersection. Thus, we have arrived at the fact that any two points are incident with a unique line, and any two lines are incident with a unique point. That is, Γ_{0} is the incidence graph of a projective plane (of some finite order q). Then q + 1 = |X : X ∩ Y | = |Y : X ∩ Y |. The transitivity property (1) guarantees that this plane is desarguesian (see [Sa, p.33]), so q is a prime power. □

Proposition 10.2 Let M be a group generated by subgroups L,P and B satisfying the following conditions.

(i) P = (B ∩ P)(L ∩ P).

(ii) B = (L ∩ B)(P ∩ B)(L ∩ B)

(iii) L = (L ∩ P)(L ∩ B)(L ∩ P).

Then

(a) M = LBP = LP L.

(b) Suppose in addition that L ∩ P has index 2 in P, then M is doubly transitive on M/L, the action of L on M/L − {L} is isomorphic to the action of L on L/L ∩ P and |M| = |L| · (|L/L ∩ P | + 1).

Proof: We will first show that LBP = LP L. Indeed

LBP =^{(i)} LB(B ∩ P)(L ∩ P)  = LB(L ∩ P)
=^{(ii)} L(L ∩ B)(P ∩ B)(L ∩ B)(L ∩ P) =^{(i)} L(P ∩ B)(L ∩ B)(L ∩ P)
= L(L ∩ P)(P ∩ B)(L ∩ B)(L ∩ P) =^{(i)} L(P ∩ B)(L ∩ P)(L ∩ B)(L ∩ P)
=^{(iii)} L(P ∩ B)L  = LP L

Since M is generate by L, P and B, (ii) shows that M is generated by L and P. Thus LBP = LP L is invariant under right multiplication by M = ⟨L, P ⟩. Thus (a) holds. (b) follows immediately from (a). □

Proposition 10.3 Let M be a group generated by three proper subgroups L,P and B. Suppose that there exists a Steiner System (Ω, B) of type S(n, l, k), β ∈ Ξ ∈ B, a set Λ and γ ≠ δ ∈ Λ satisfying the following conditions:

(i) k ≥ 2, L acts faithfully and k-transitively on Ω and L preserves B.

(ii) |Λ| = l + 1 and B acts doubly transitively on Λ.

(iii) L ∩ B = N_{L}(Ξ) = C_{B}(γ), L ∩ P = C_{L}(β) and B ∩ P = N_{B}({γ, δ})

(iv) L ∩ P has index two in P.

Set Ω^{∗} = M/L, α = L ∈ Ω^{∗}, Ξ^{∗} = α^{B} and B^{∗} = Ξ^{∗M}. Then (Ω^{∗}, B^{∗}) is a Steiner System of type (n + 1, l + 1, k + 1) and M acts faithfully and k + 1-transitively on M. In particular, |M| = |L|(n + 1).

Proof: We will first verify that the assumptions of 10.2 are fulfilled. Since B is doubly transitive on Λ, B ∩ P = N_{B}({γ, δ}) acts non-trivially on {γ, δ}. Since B ∩ L = C_{B}(γ) we conclude that B ∩ P ≠ P ∩ L and so (iv) implies P = (B ∩ P)(L ∩ P). So 10.2(i) holds. The doubly transitive action of B on Λ ≅ B/B ∩ L implies 10.2(ii). We will show next that L ∩ B ≰ L ∩ P. For this let β ∈ Σ ⊆ Ξ with |Σ| = k. Since (Ω, B) is a steiner system of type S(n, l, k), Ξ is the unique member of B containing Σ. Thus N_{L}(Σ) ≤ N_{L}(Ξ) = L ∩ B. Since k ≥ 2 and L is k-transitive on Ω, N_{L}(Σ) ≰ C_{L}(β) = L ∩ P. Thus indeed L ∩ B ≠ L ∩ P. Thus the the doubly transitive action of L on Ω ≅ L/L ∩ P implies 10.2(iii). By (iv) the extra condition in 10.2(b) is fulfilled.

We verified all the assumptions and we conclude that Ω^{∗} − {α} ≅ L/L ∩ P ≅ Ω as an L set. Thus M is k + 1-transitive on Ω^{∗}. In particular there exists an integer e ≥ 1 such that every k + 1 element subset of Ω^{∗} is contained in exactly e members of B^{∗}. Denote by m the number of pairs (Ξ′, R) where Ξ′ ∈ B and where R is a k + 1-element subset of Ω contained in Ξ′. Then

m = |B^{∗}|\binom{l + 1}{k + 1} = \binom{n + 1}{k + 1}e. (1)

Applying the analogue of (1) to the Steiner system (Ω, B) ( where e = 1) we get

|B|\binom{l}{k} = \binom{n}{k} (2)

Also

|B^{∗}| = |M/B| = \frac{|M/L|}{|B/B ∩ L|}|L/B ∩ L| = \frac{n + 1}{l + 1}|B| (3)

Multiplying (1) with \binom{l}{k} and using (2) and (3) we obtain.

\frac{n + 1}{l + 1}\binom{n}{k}\binom{l + 1}{k + 1} = \binom{l}{k}\binom{n + 1}{k + 1}e

and so e = 1. Hence (Ω^{∗}, B^{∗}) is a Steiner system of type S(n + 1, l + 1, k + 1). Since the kernel of the action of M on Ω^{∗} is contained in L and since L acts faithfully on Ω and so on Ω^{∗}, M is faithful on Ω^{∗}. □

11 L_{3}(2^{n}) and Alt(6)

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

Theorem 11.1 Suppose Hypothesis A holds, with p = 2. Set X = N_{G}(QP̃) and Y = N_{G}(Q_{P}), and assume further that the following hold.

(i) X = C̃ = N_{G}(P̃),

(ii) Y = N_{G}(P).

Then F^{∗}(G) ≅ P SL_{3}(2^{n}) for some n, or G ≅ Alt(6).

We will assume the hypothesis of 11.1 for the remainder of this section. Let G be a minimal counter-example to Theorem 11.1.

Lemma 11.2 We have S = B(S) = Q_{X}Q_{Y}, and G has a single conjugacy class of involutions.

Proof: [OUTLINE. TO BE FIXED UP.] Suppose false. Apply the Thompson Transfer Lemma, and find that G has a subgroup G_{0} of index 2. Minimality of G yields F^{∗}(G_{0}) ≅ P SL(3, q) or G_{0} ≅ Alt(6). But F^{∗}(G) = F^{∗}(G_{0}). If G_{0} ≅ Alt(6) then G ≅ Sym(6), since X and Y are S-invariant. But then contradict Hypothesis A. □

Lemma 11.3 There is a power q = 2^{n} of 2 such that the following hold.

(a) S = Q_{X}Q_{Y}, where Q_{X} and Q_{Y} are elementary abelian of order q^{2}.

(b) X/Q_{X} ≅ Y/Q_{Y} is isomorphic to a subgroup of ΓL(2, q) containing SL(2, q).

(c) X ∩Y has index q+1 in both X and Y, and we have X = (X ∩Y)P̃ and Y = (X ∩Y)P.

(d) For any M ∈ {X, Y } and any involution u ∈ Q_{M}, we have |M : C_{M}(t) = q^{2} − 1, and C_{G}(u) = C_{M}(u).

(e) For any involution t ∈ Q_{Y} − Q_{X}, we have J ∩ Y = t^{Y} ∪ (Q_{Y})^{#}, and |t^{Y} | = q(q^{2} − 1).

For the remainder of this section, let q = 2^{n} be as in 11.3.

Lemma 11.4 The graph Γ(G; X, Y) contains no cycles of length 4.

Proof: □

Denote by J the set of involutions in G, and adopt the notation of 1.1. In particular, we set f = (|J|/|G : X|) − 1. Since X = C̃ by assumption, we have C_{G}(t) = C_{X}(t) for t ∈ J ∩ Q_{X}. As G has a single conjugacy class of involutions, by 11.2, it follows that f = (|X|/C_{X}(t)|) − 1. Then 11.3(d) yields:

(1) f = q^{2} − 2.

Denote by J^{∗} the set of involutions t ∈ G − X such that C_{X}(t) has even order, and let t ∈ J^{∗}. Let u ∈ C_{J∩X}(t). Then u ∉ Q_{X}, and since J ∩ S ≤ Q_{X} ∪ Q_{Y} we may assume, after conjugation in X, that u ∈ Q_{Y} − Q_{X}. By 11.3(d) we then have t ∈ Y. Conversely, for any involution t′ in Y we have C_{Q_{Y}} (t′) ≠ 1, and so t′ ∈ J^{∗}. Thus J^{∗} = ∪{J ∩ (Y^{x} − X)}_{x∈X}, and then ?? yields

(2) |J^{∗}| = ∑_{x∈X} |J ∩ (Y^{x} − X)|.

Notice that S ⊴ X ∩ Y. Thus J ∩ (Y − X) = (J ∩ Y) − S, and so |J ∩ (Y − X)| = |t^{Y} | − |t^{Y} ∩ S| for any t ∈ J ∩ Y − Q_{Y}. Thus |J ∩ (Y − X)| = q(q^{2} − 1) − (q^{2} − q) = q^{2}(q − 1). Then (2) yields:

(3) |J^{∗}| = q^{2}(q^{2} − 1).

Fix t ∈ J ∩ (Y − X), and denote by D the subgroup X ∩ X^{t} generated by the set of elements x of X such that x^{t} = x^{−1}. We have D ≤ X ∩Y ∩X^{t}, by ??, and then D = Q_{Y} D_{0}, where D_{0} is cyclic of order q − 1 and where t inverts every element of D_{0}. The number of elements of D inverted by t is then q + q(q − 2) = q(q − 1), and this is then the number of involutions in Xt. Then (3) yields the following result.

(4) |{Xt : t ∈ J^{∗}}| = q(q + 1), and |J ∩ Xt| = q(q − 1) for any t ∈ J^{∗}.

Let s ∈ J − J^{∗}, and denote by D the set of non-identity elements x ∈ X such that x^{s} = x^{−1}. Let d ∈ D. As C_{G}(s) is 2-closed, C_{G}(d) is of odd order. Since any element of X −P̃H induces a field automorphism on P̃/Q_{X}, we then have d ∈ P̃H, and then d induces a non-identity inner automorphism on P̃/Q_{X}, by —. Set D = ⟨D⟩. Then |D| is odd, and so D is contained in a cyclic subgroup of P̃H of order q − 1 or q + 1. Thus |J ∩ Xs| ≤ q + 1. In the notation of 1.1, we have thus shown that b_{k} = 0 for k > q + 1, k odd.

Set b^{∗} = q(q + 1). Thus b^{∗} = b_{q(q−1)} in the notation of 1.1. Recall from (1) that we have f = q^{2} − 2. For k odd, k > 1, we then have

(5)

f^{−1}(k − 1)b_{k} − b_{k} = [(k − 1)b_{k} − (q^{2} − 2)b_{k}]/(q^{2} − 2)
≤ b_{k}(q − q^{2} + 2)/(q^{2} − 2) ≤ 0,

and so 1.1 yields

0 ≤ b_{1} < f^{−1}[|J ∩ X| + (q^{2} − q − 1)b^{∗}] − 1 − b^{∗}.

Here |J ∩ X| = (q^{2} − 1) + q(q^{2} − 1) = (q^{2} − 1)(q + 1), so with (2),(3), and (4) we have

0 ≤ b_{1}(q^{2} − 2) < (q^{2} − 1)(q + 1) + (q^{2} − q − 1)(q^{2} + q) − (q^{2} − 2)) − (q^{2} − 2)(q^{2} + q)

= (q^{2} − 1)(q + 1) + (q^{2} − 2)(q^{2} + q) − (q − 1)(q^{2} + q) − (q^{2} − 2) − (q^{2} − 2)(q^{2} + q)

= (q^{2} − 1) − (q^{2} − 2) = 1,

and so 0 ≤ b_{1} < 1/(q^{2} − 2). Thus b_{1} = 0. Further, for any odd k with k > 1, (5) yields

(6) 0 < 1 − (k − 1)b_{k} − (q^{2} − 2)b_{k}.

If b_{k} ≠ 0 we have seen that k ≤ q + 1, while (6) yields q^{2} < k + 1. Thus:

(7) Either b_{k} = 0 for all odd k, or else q = 2 and b_{k} = 0 for k ∉ {2, 3}.

Suppose that b_{k} = 0 for all odd k. Then J = (J ∩ X) ∪ J^{∗}, and (3) yields |G : C_{G}(t)| = (q^{2} − 1)(q + 1) + q^{2}(q^{2} − 1) = (q^{2} − 1)(q^{2} + q + 1). Thus |G : X| = q^{2} + q + 1, and |G| = |G^{∗}|, where G^{∗} is a subgroup of P ΓL(3, q) containing P SL(3, q). Further, we have

(8) X\G = {Xt : t ∈ J^{∗}} = {Xg : g ∈ ∪_{x∈X}Y^{x}}.

Now 10.1 implies that Γ(G; X, Y) is the incidence graph of a Desarguesian projective plane of order q. Thus G is isomorphic to a subgroup of P ΓL(3, q). Evidently [G, G] ≥ ⟨P̃, P ⟩, so [G, G] ≅ P SL(3, q) [WHY?], and we have 11.1 in this case.

It remains to consider the case where q = 2 and where b_{3} ≠ 0. Let d be an element of order 3 in X, set K = C_{G}(d), and let t be an involution inverting d, with t ∉ X. As C_{G}(t) is a 2-group, K is of odd order and t inverts every element of K. It follows that K is an abelian Hall subgroup of G, and that K is a T I-set. Further, as t ∉ X we have K ≠ ⟨d⟩. In [Be3], Bender analyzed precisely this situation, and we shall appeal to his result in order to conclude that |K| = 9 and that |G| = 360. Now ?? yields G ≅ Alt(6). This completes the proof of Theorem 11.1.

12 M_{23}

In this section we shall prove the following result.

Theorem 12.1 Assume Hypothesis A, and assume that 9.3(4) holds. Then G ≅ M_{23}.

Proof: Let H = N_{G}(Q_{X}). Note that H/Q_{X} ≅ Alt(7) and so has a unique class of involutions. So every involution in H is conjugate to an involution in Q_{X}Q_{Y}. Furthermore every involution in Q_{X}Q_{Y} is contained in Q_{X} ∪ Q_{Y} and since X and Y are transitive on Q_{X}^{#} and Q_{Y}^{#} we conclude

1^{∘} H has a unique class of involution.

Let z be an involution in G. From (1^{∘}) and 9.3(4:d) , C_{G}(z) has the form 2^{1+3}L_{3}(2). In particular, C_{G}(z) has a unique class of elements of order three and if e ∈ C_{G}(z) has order 3, then (C_{G}(e) ∩ C_{G}(z))/⟨e⟩ is elementary abelian of order four. Moreover, C_{G}(e) acts transitively on the involutions in C_{G}(e). Let H^{∗} = N_{G}(Q_{Y}) and d an element of order three in H^{∗} with Q_{Y} ⟨d⟩ ⊴ H^{∗}. Then C_{H^{∗}}(⟨d⟩ ≅ GL(2, 4). It now follows from just about any theorem or exercise in the universe that C_{G}(d) = C_{Y} (d) ≅ GL_{(}2, 4). We proved:

2^{∘} G has a unique class of subgroups D such that |D| = 3 and |C_{G}(D)| is even. Moreover for any such D, C_{G}(D)/D ≅ Alt(5) and N_{G}(D)/D ≅ Sym(5).

Set L = ⟨X, Y ⟩ and Γ_{0} = Γ(L, X, Y). Let D be a subgroup of X ∩ Y order 3. Note that C_{H}(D) has even order and so we can apply (2^{∘}) to D. Let s and t be involutions in X − Y and Y − X, respectively, which invert H. Since the product of any two distinct involutions in Sym(5) − Alt(5) has order 2 or 3, we conclude from (2^{∘}) that stH/H has order 2 or 3. If [s, t] ∈ H, then t ∈ N_{G}(⟨Q_{Y} , H⟨s⟩) = N_{G}(B(X)) = X, a contradiction. Thus |stH/H| = 3. So Γ_{0} contains a cycle of length 6, and 10.1 implies that Γ_{0} is the incidence graph of a projective plane of order 4, and thus L is isomorphic to a subgroup of P ΓL(3, 4) containing L_{3}(4). As N_{G}(B(X)) = X, and X contains a Sylow 2-subgroup of L we conclude

3^{∘} L is of the form L_{3}(4).2.

Denote by L_{0} the subgroup of index 2 in L. So L_{0} = ⟨B(X), B(Y)⟩ ≅ L_{3}(4). Let Λ_{1} be a set of size 7 with H/Q_{X} ≅ Alt(7) acting faithfully on Λ_{1}. Since X/Q_{H} ≅ Sym(5), X normalizes a subset {γ_{0}, γ_{1}} of Λ_{1}. Since X ∩ Y/Q_{X} ≅ Sym(4), X ∩ Y fixes a point γ_{3} ∈ Λ_{1} \ {γ_{0}, γ_{1}}. Let B_{0} = C_{H}(γ_{0}). Then B_{0}/Q_{X} ≅ Alt(6) and B(X) = B_{0} ∩ X. Let E := C_{H}({γ_{0}, γ_{1}, γ_{2}}) and R = N_{B_{0}}(E) = N_{H}({γ_{1}, γ_{2}}) ∩ C_{H}(γ_{0}). Then E/Q_{X} ≅ Alt(4), R/Q_{X} ≅ Sym(4) and E = O^{2}(X ∩ Y)) = B(X) ∩ B(Y). Also O_{2}(E) = Q_{X}Q_{Y} and so N_{H}(E) ≤ N_{G}(O_{2}(E)) ≤ N_{G}(Q_{Y}) = H^{∗} = N_{G}(B(Y)). Set P_{0} = R B(Y). Then P_{0}/Q_{Y} ≅ Sym(5). Moreover, L_{0} ∩ B_{0} = N_{L}(Q_{X}) ∩ B_{0} = B(X) and L_{0} ∩ P_{0} = N_{L_{0}}(Q_{Y}) ∩ P_{0} = B(Y) ∩ P_{0} = B(Y). Also B_{0} ∩ P_{0} = R and L_{0} ∩ B_{0} ∩ P_{0} = E. Put M_{0} = ⟨L_{0}, B_{0}, P_{0}⟩. We will verify that the assumption of 10.3 are fulfilled. Note that L_{0} acts faithfully on a projective plane of order 4 with points set Ω_{0} and lines B_{0}. Moreover we choose the points such that L_{0} ∩ P_{0} = B(Y) fixes a point β_{0} ∈ Ω_{0}. Then L_{0} ∩ B_{0} = B(X) fixes a line Ξ_{0} ∈ B_{0}. Since L_{0} ∩ B_{0} ∩ P_{0} = E, β_{0} ∈ Ξ_{0}. Note that (Ω_{0}, B_{0}) is a Steiner System of type S(21, 5, 2) and L acts doubly transitively on Ω_{0}. Let Λ_{0} = Λ_{1} \ {γ_{0}} Then L_{0} ∩ B_{0} = B(X) = C_{B_{0}}(γ_{1}) and P_{0} ∩ B_{0} = R = N_{B_{0}}({γ_{1}, γ_{2}}. Finally |P_{0}/L_{0} ∩ P_{0}| = |P_{0}/B(Y)| = 2 and so indeed all the assumptions of 10.3 are fulfilled. Put Ω_{1} = M_{0}/L_{0}, β_{1} = L_{0} ∈ Ω_{1}, Ξ_{1} = β_{1}^{B_{0}} ⊆ Ω_{1} and B_{1} = Ξ_{1}^{M_{0}}. We conclude that (Ω_{1}, B_{1}) is a Steiner System of type S(22,6,3) and M_{0} acts faithfully and 3-transitive on Ω_{1}. Thus for example by [As], M_{0} ≅ M_{22}

We would like to apply 10.3 a second time to identify M_{1} := ⟨M_{0}, H, L⟩. M_{0} acts on the Steiner System (Ω_{1}, B_{1}) and H acts on the set Λ_{1} with the right transitivity properties.

M_{0} ∩ L = N_{M_{0}}(L_{0}) ∩ L = L_{0} ∩ L = L_{0} = C_{M_{0}}(β_{1}), M_{0} ∩ H = N_{M_{0}}(Q_{X}) = B_{0} = N_{M_{0}}(Ξ_{1}) = C_{H}(γ_{0}) and L ∩ H = N_{L}(Q_{X}) = X = N_{H}({γ_{0}, γ_{1}}). Also |L/M_{0} ∩ L| = |L/L_{0}| = 2. Thus the assumption of 10.3 are fulfilled and we conclude that M_{1} acts faithfully and 4-transitive on a Steiner System of type S(23, 7, 4). So by [As] M_{1} ≅ M_{23}.

Suppose that M_{1} ≠ G. By (1^{∘}) applied to M_{1} in place of G, M_{1} has a unique class of involutions. Since H ≤ G^{∗} (1^{∘}) implies that C_{G}(z) ≤ M_{1} for all involutions z ∈ M_{1}. Thus M_{1} is a strongly 2-embedded subgroup of G. But this contradicts [Be4].

Thus G = M_{1} ≅ M_{23} and 12.1 is proved. □

References

[As] Aschbacher’s book on sporadic groups

[Be1] B

[Be2] B2

[Be3] Groups with large subgroups= B???

[Be4] Strongly 2-embedded subgroups

[Ch] small modules for small groups

[Gor] D. Gorenstein, Finite Groups, Chelsea (1980) New York.

[He] Nils Hebbinghaus, Über die Struktur minimalparabolischer Gruppen der lokalen Charakteristik p, Diplomarbeit, (2002) Universitatät Kiel.

[KMR] Kleidman, Meierfrankenfeld, Ryba

[MSS] U. Meierfrankenfeld, B. Stellmacher, G. Stroth, Groups of local characteristic p, an overview (2002), submitted to the Proceedings of the Durham Conference 2001.

[PPS] C. Parker, G. Parmeggiani, B. Stellmacher, The P!-Theorem, J. Algebra ( to appear)

[Sa] Samuel