FINITE GROUPS GENERATED BY A PAIR OF

MINIMAL PARABOLIC SUBGROUPS: PART I

Andrew Chermak

Department of Mathematics Kansas State University Manhattan KS, 66502

September, 2000

Section 0: Notation, terminology, and quadratic action

All groups considered here are understood to be finite.

Let X be a group, and let V be a group on which X acts. We write

η(X, V)

for the number of chief factors in any X-chief series through V on which the action of X is non-trivial. If V_{0} is a subgroup of V, and Y is an element or a subgroup of X, we write

C_{V} (Y mod V_{0})

for the set of elements v of V such that [v, Y] ≤ V_{0}.

Suppose further that V is an 𝔽_{p}X-module for some prime p. We say that V is an F 1-module (resp. a quadratic F 2-module, resp. a quadratic F 2^{∗}- module) for X if there exists a subgroup A of X such that [V, A, A] = 1 ≠ [V, A], and such that |A/C_{A}(V)||C_{V} (A)| ≥ |V | (resp. |A/C_{A}(V)|^{2}|C_{V} (A)| ≥ |V |, resp. |A/C_{A}(V)|^{2}|C_{V} (A)| > |V |). Such a subgroup A is then said to be a quadratic F 1-offender (resp. F 2-offender, resp. F 2^{∗}-offender).

When there is a need to be more precise, we employ the notation, due, as far as I know, and in one form or another, to Stellmacher, as follows. Assuming that there exists a non-identity subgroup of X which acts quadratically on V, we define r(X, V) to be the smallest real number r such that there exists a non-identity subgroup A of X with [V, A, A] = 0 and with |A/C_{A}(V)|^{r}|C_{V} (A)| = |V |. We then set

A(X, V) = {A ≤ X : A ≠ 1, [V, A, A] = 0, and |A/C_{A}(V)|^{r(X,V)}|C_{V} (A)| = |V |}.

Typeset by AMS-TEX

Lemma 0.1. Let X be a finite group, p a prime, and V a faithful 𝔽_{p}X-module. Let A and K be subgroups of X, such that 1 ≠ K = O^{p}(K) = [K, A], and such that [V, A, A] = 0. Assume also that A ∈ A(A, V), and set W = [V, K]. Then C_{A}(K) = C_{A}(W), and for any complement A_{0} to C_{A}(K) in A we have r(A_{0}, W) ≤ r(A, V).

Proof. By quadratic action, we have [V, ⟨A^{K}⟩, C_{A}(K)] = 0, and since K = [K, A] ≤ ⟨A^{K}⟩ we then have [W, C_{A}(K)] = 0. Now [V, C_{A}(W), K] = 0, and the Three Subgroups Lemma then yields C_{A}(W) = C_{A}(K).

Choose a complement A_{0} to C_{A}(K) in A, and set r = r(A, V). Suppose that |A_{0}|^{r}|C_{W} (A_{0})| < |W |. We then have

|V | = |A|^{r}|C_{V} (A)| = |A_{0}|^{r}|C_{A}(W)|^{r}|C_{W} (A)||C_{V} (A)/C_{W} (A)|
< |C_{A}(W)|^{r}|C_{V} (A)/C_{W} (A)||W | = |C_{A}(W)|^{r}|W + C_{V} (A)|
≤ |C_{A}(W)|^{r}|C_{V} (C_{A}(W))|.

this is contrary to the assumption that A lies in A(A, V), so in fact |A_{0}|^{r}|C_{W} (A_{0})| ≥ |W |, and the lemma is proved. □

Section 1: P and H

These notes are intended as a preliminary draft of a more extensive work. We assume familiarity with some of the notation and basic concepts associated with the project initiated by Meierfrankenfeld, on groups of characteristic p-type. We begin with a finite group G, a prime p, and a Sylow p-subgroup S of G. The following condition is assumed, throughout.

1.0. For any subgroup L of G containing S, with O_{p}(L) ≠ 1, we have C_{L}(O_{p}(L)) ≤ O_{p}(L).

For any subgroup L of G, we denote by Y_{L} the p-reduced core of L. That is, Y_{L} is the (uniquely determined) normal p-subgroup Y of L such that O_{p}(L/C_{L}(Y_{L})) = 1, and which is largest for this condition. Here are the general properties of the p-reduced core that we will require.

Lemma 1.1. Let X be a group with C_{X}(O_{p}(X)) ≤ O_{p}(X).

(a) We have Y_{X} ≤ Ω_{1}(Z(O_{p}(X))), and if X/O_{p}(X) acts faithfully on Ω_{1}(Z(O_{p}(X))) then Y_{X} = Ω_{1}(Z(O_{p}(X))).

(b) If X ≤ X^{∗}, where C_{X}(O_{p}(X^{∗})) ≤ O_{p}(X^{∗}), then Y_{X} ≤ Y_{X^{∗}}.

We set Z = Ω_{1}(Z(S)), we set C = C_{G}(Z), and we choose a maximal p-local subgroup C̃ of G containing C. Further, we set E = O^{p}(F_{p}^{∗}(C_{G}(YC̃))).

For X a group and R a Sylow p-subgroup of X, we denote by P_{X}^{∗}(R) the set of all subgroups P of X such that R is contained in a unique maximal subgroup of P, and satisfying also the condition that R is not a normal subgroup of P. We set P_{X}(R) = {P ∈ P_{X}^{∗}(R) : O_{p}(P) ≠ 1}. We shall write P for P_{G}(S).

Lemma 1.2. For any finite group X and Sylow subgroup R of X, we have X = ⟨P_{X}^{∗}(R)⟩N_{X}(R), and if O_{p}(X) ≠ 1 then X = ⟨P_{X}(R)⟩N_{X}(R).

Henceforth, all groups under consideration will be subgroups of our fixed finite group G. For any subgroup X of G and p-subgroup R of X we denote by L_{X}(R) the set of subgroups L of X containing R, with O_{p}(L) ≠ 1. We write L for L_{G}(S).

At this point we may state the hypotheses under which we will be working. One of these is that whenever it is inconvenient to do otherwise, we assume that the non-abelian simple sections of elements of L are alternating groups, groups of Lie type, or may be found among the twenty-six “sporadic” groups. We shall make use of this “K-group” hypothesis when we have quadratic action by a group of order bigger than 2 or, at one point (see 3.3, below), to identify simple groups with dihedral or semi-dihedral Sylow 2-subgroups.

1.3 Main Hypothesis.

(1) C_{L}(O_{p}(L)) ≤ O_{p}(L) for any L ∈ L.

(2) C̃ is the unique maximal p-local subgroup of G containing E.

(3) There is a unique member P of P such that P ≰ C̃. Moreover, setting P^{0} = ⟨O_{p}(C̃)^{P} ⟩, we have the following.

(a) P^{0}S = P.

(b) P^{0}/O_{p}(P^{0}) ≅ SL(2, q) for some power q of p, and Y_{p} = Y_{P^{0}} is a natural

SL(2, q)-module for P^{0}.

(c) C_{Y_{P}} (S ∩ P^{0}) ⊴ C̃.

(4) There is at most one member P_{1} of P such that ⟨P, P_{1}⟩ ∈ L and such that Y_{P} is not normal in P_{1}. Moreover, if such a P_{1} exists then, setting L = ⟨P, P_{1}⟩ and L^{0} = ⟨O_{p}(C̃)^{L}⟩, we have the following.

(a) L = L^{0}S.

(b) L^{0}/C_{L^{0}}(Y_{L}) ≅ SL(3, q) or Sp(4, q), and Y_{L} = Y_{L^{0}} is a natural module for

L^{0}/C_{L^{0}}(Y_{L}).

(5) There exists P̃ ∈ P_{ES}(S) such that ⟨P, P̃⟩ ∉ L.

(6) We have Y_{P} ≤ O_{p}(C̃), and if L is as in (4) then also Y_{L} ≤ O_{p}(C̃).

From the outset, we fix P̃ as in 1.3(5) so that, first, the group H := ⟨P ∩ C̃, P̃⟩ is as small as possible, and then, subject to this condition, so that P̃ is as small as possible. Then ⟨P, H⟩ ∉ L.

Our goal is to determine all groups G satisfying the above Hypothesis. In this preliminary set of notes we will fall far short of this goal. For one thing, we will only be dealing, at present, with the case “b ≥ 3” (where b is a parameter to be defined later), and we leave the case b ≤ 2 (which, by 1.3(6) amounts to b = 2) for a second installment. For another thing, we will be content in these notes (so far) to obtain suitable p-local information about G, from which the actual identification of G will follow at a later stage. We mention that, in the case of p = 2, we expect to be able to use (perhaps “steal” would be a better word) large portions of the relevant parts of the “Quasithin”

Project of Aschbacher and Smith, in determining the group G from the local data. In fact, Michael and Steve should really by writing up the stuff I need as a separate paper, as far as I’m concerned. In these notes as they are, we only aim at the following.

Theorem A. Assume b ≥ 3. Then either (N_{H}(P)P, H) is a weak BN-pair, or there are two maximal 2-local subgroups of G having the “same general structure” as maximal 2-local subgroups in the Rudvalis simple group. In particular, in the latter case, we have |S| = 2^{14}.

In fact, in the case where (N_{H}(P)P, H) is not a weak BN-pair we can show that |G| = |Ru|. But, as mentioned, we expect to follow Aschbacher and Smith in their approach to the construction of Ru from the materials at their hands.

In the remainder of this section we will develop basic properties of P and H, and of the coset graph which they determine.

Lemma 1.4. Let K be a subgroup of H containing S. Then one of the following holds. (i) ⟨K, P ∩ H⟩ = H. (ii) K ≤ N_{G}(Y_{P}). (iii) K = ⟨N_{K}(Y_{P}), P_{1}⟩, where P_{1} ∈ P_{K}(S) and P_{1} is the unique element of P

satisfying the conditions in 1.3(4).

Proof. Immediate from 1.3(4) and from the minimality of H. □

Lemma 1.5. We have

N_{P} ((S ∩ P^{0})Q_{P})) = N_{P} (S ∩ P^{0}) = P ∩ C̃ = P ∩ H,

and P ∩ H is the unique maximal subgroup of P containing S.

Proof. We have N_{P} (S) ≤ P ∩C̃ by the E-uniqueness condition 1.3(2). The P -uniqueness condition in 1.3(3), together with 1.1, shows that P ∩ C̃ is a maximal subgroup of P containing S. But 1.3(3b) shows that N_{P} (S ∩ P^{0}) is maximal in P. As P ∈ P, 1.5 follows. □

In all that follows, set H_{0} = ⟨O^{p}(P̃)^{H}⟩. Let us also set

Q_{P} = O_{p}(P), and Q_{H} = O_{p}(H).

Further, set

V = ⟨(Y_{P})^{H}⟩, Ṽ = V/Y_{H}, and \overline{H} = H/C_{H}(Ṽ).

Lemma 1.6. We have Y_{P} = Ω_{1}(Z(Q_{P})) and Y_{H} = C_{Y_{P}} (S ∩ P^{0}) = C_{Y_{P}} (O_{p}(P ∩ H)).

Proof. We have Y_{P} ≤ Ω_{1}(Z(Q_{P})), by 1.1(a). But 1.3(3) shows that C_{P} (Ω_{1}(Q_{P})) = Q_{P}, so the first part of 1.6 follows from the maximality of Y_{P} among the p-reduced subgroups of P.

We have H = H_{0}(P ∩ P̃), and 1.3(5) shows that H_{0} ≤ E. Set Q = O_{p}(C̃). Then Q ≤ Q_{H}, and so Y_{H} ≤ Ω_{1}(Z(Q))) by 1.1. As [YC̃, E] = 1, the Thompson P × Q Lemma shows that also [Ω_{1}(Z(Q)), E] = 1, and so [Y_{H}, H_{0}] = 1. Thus H/C_{H}(Y_{H}) ≅ (P ∩C̃)/C_{P ∩C̃}(Y_{H}). It now follows (from the definition of Y_{H}) that Y_{H} ≤ Z(O_{p}(P ∩C̃)). Setting Z_{0} = C_{Y_{P}} (S ∩ P^{0}), we have Z_{0} = Ω_{1}(Z(O_{p}(P ∩ C̃))), and we conclude from 1.1(a) that Y_{H} ≤ Z_{0}. On the other hand, we observe that Y_{P ∩C̃} = Z_{0}, and so Z_{0} ≤ Y_{H} by 1.1(b). This yields the second part of 1.6. □

Lemma 1.7. We have N_{G}(Y_{P}) ≤ N_{G}(O^{2}(P)).

Proof. Set N = N_{G}(Y_{P}). As P is transitive on (Y_{P})^{♯}, we have N = C_{N}(z)P for any z ∈ Z^{♯}. Then ⟨(O_{2}(C_{G}(z)))^{N}⟩ = ⟨(O_{2}(C_{G}(z)))^{P} ⟩ ≥ ⟨(O_{2}(C̃))^{P} ⟩ = O^{2}(P)O_{2}(C̃). It follows that O^{2}(P) is normal in N. □

Lemma 1.8. Let g ∈ P − H, and set R = (Q_{H})^{g}. Then C_{V} (R) = (Y_{H})^{g}.

Proof. Set U = C_{V} (R)Y_{P}. Then U is Q_{P} -invariant, and we have [U, R][U, Q_{H}] = Y_{P}. Thus U ⊴ P and [U, O^{p}(P)] = Y_{P}. Assume that U ≠ Y_{P} and let U_{0} be a normal subgroup P contained in U, with |U_{0}/Y_{P} | = p. As Y_{P} = Ω_{1}(Z(Q_{P})) we have [U_{0}, Q_{P}] = Y_{P}, and then Q_{P} /C_{Q_{P}} (U_{0}) is isomorphic to Y_{P} as modules for P. As [R, Q_{P}] ≤ R ∩ Q_{P} ≤ C_{Q_{P}} (U_{0}), it follows that R acts trivially on Y_{P}, so R ≤ Q_{P}. As Q_{H} ≰ Q_{P}, by 1.3(3), we have a contradiction. □

Lemma 1.9. Let H_{1} be a subgroup of H containing P̃, and set R = Q_{P} ∩O_{p}(H_{1}). Then R is normal in neither P nor H_{1}.

Proof. Set U = Ω_{1}(Z(O_{p}(C̃))). It is then a fundamental fact, concerning E, that [U, E] = 1. (See Lemma — in the P -Uniqueness paper, for example.)

Set W = Ω_{1}(Z(R)) and set W_{1} = Ω_{1}(Z(O_{p}(H_{1}))). As O^{p}(P̃) ≤ E we have H_{0} ≤ E, and so O^{p}(H_{1}) ≤ E, and [U, O^{p}(H_{1})] = 1. But W_{1} ≤ U, so [W_{1}, O^{p}(H_{1})] = 1. Suppose that R is normal in H_{1}. The Thompson P × Q Lemma then yields [W, O^{p}(H_{1})] = 1. As Y_{P} ≤ W, by 1.3(6), we obtain Y_{P} ⊴ ⟨P, P̃⟩, whereas P̃⟩ ∉ L by assumption. Thus R is not normal in H_{1}. In particular, O_{p}(H_{1}) ≰ Q_{P}.

Suppose next that R is normal in P, and set P_{0} = ⟨O_{p}(H_{1})^{P} ⟩. We have O_{p}(C̃) ≤ O_{p}(H_{1}), and we have O_{p}(C̃) ≤ O_{p}(P ∩H) and O_{p}(C̃) ≰ O_{p}(P) by 1.3(3a). It follows that O_{p}(H_{1}) ∩ O_{p}(P ∩ H) is invariant under P ∩ H, and that O_{p}(H_{1}) is a Sylow p-subgroup of P_{0}. Then P = P_{0}S. As Y_{P} ⊴ ⟨P, P̃⟩ ∉ L, no non-identity characteristic subgroup of O_{p}(H_{1}) is normal in P_{0}. Further, as Z(P) = 1 we have also Z(P_{0}) = 1, and then the Pushing Up Theorem of Niles [Ni] or of Stellmacher [St1] implies that Q_{P} = Y_{P}. As Y_{P} ≤ Q_{H} by 1.3(6), and since P ∩ H acts irreducibly on O_{p}(P ∩ H)/Q_{P}, it follows that Q_{H} = O_{p}(P ∩ H), and that Q_{H} is the product of any two H-conjugates of Y_{P} in Q_{H}. If p = 2 then all of the involutions in Q_{H} lie in two conjugates of Y_{P}, so that |H : N_{H}(Y_{P})| ≤ 2. But Y_{P} ⊴ S, so Y_{P} ⊴ H in this case, contrary to ⟨P, H⟩ ∉ L. Thus p is odd.

Recall the standard notation: V = ⟨(Y_{P})^{H}⟩ and Ṽ = V/Y_{H}. Here we have found that V = Q_{H}. Set Λ = Aut(V)/C_{Aut(V)}(Ṽ). It is well known [REFERENCE ?] that, in odd characteristic, Ṽ admits the structure of a 2-dimensional vector space over 𝔽_{q}, in such a way that Λ is isomorphic to the associated group of all semilinear transformations. Denote by \overline{Λ} the quotient of Λ by the scalar transformations, and let \overline{Λ}_{0} be the linear subgroup of \overline{Λ}. Thus \overline{Λ}_{0} ≅ P GL(2, q).

Let D be a Hall p′-subgroup of P ∩ H, and denote by \overline{D} the image of D in \overline{Λ}. Thus D is cyclic of order q − 1, and since D ∩ Z(O^{p}(P) mod Q_{P}) ≠ 1 the irreducible D-modules Y_{P} /Y_{H} and V/Y_{P} are non-isomorphic. Thus, the image of D in Λ intersects the set of scalar transformations trivially, and so D is isomorphic to \overline{D}. Further, as D acts regularly on Y_{P} /Y_{H}, \overline{D} acts linearly on Ṽ, and so \overline{D} is a cyclic subgroup of \overline{Λ}_{0} of order q − 1.

As P̃ ∈ P, we have Q_{H} ≠ S. As Q_{H} = C_{S}(Y_{H}) it follows that H_{0}Q_{H}/Q_{H} is a p′-group and that S induces a non-identity group of Galois automorphisms on \overline{Λ}_{0} and on Ṽ. In particular we have q − 1 > 5, and then the only p′-subgroups of \overline{Λ}_{0} containing \overline{D} are contained in the normalizer of \overline{D}. As D leaves invariant only two 1-dimensional 𝔽_{q}-subspaces of Ṽ we again arrive at |H : N_{H}(Y_{P})| ≤ 2. As p is odd, P̃ has no subgroups of index 2, and thus P̃ ≤ N_{H}(Y_{P}). As ⟨P, P̃⟩ ∉ L, we have a contradiction at this point. Thus R is not normal in P.

□

Corollary 1.10. We have [Q_{P} , O^{p}(P)] ≰ O_{p}(P̃). □

Lemma 1.11. We have (a) [V, Q_{H}] = Y_{H}.

(b) [C_{H}(Ṽ), S ∩ O^{p}(P)] ≤ Q_{H}.

(c) O^{p}(H)Q_{H}/Q_{H} is a central extension of O^{p}(\overline{H})) by the p′- group C_{H}(Ṽ)/Q_{H}.

Lemma 1.12. Suppose that some element of Q_{H} induces a transvection on a non-central chief factor for P in Q_{P}. Then q = p.

Proof. □

We now form the coset graph

Γ = Γ(P, H).

Thus, Γ is the bipartite graph whose vertices are the right cosets of P and the right cosets of H, with adjacency of vertices given by non-empty intersection. The stabilizer in G of a vertex P g, for g ∈ G, is then the conjugate P^{g} of P (and similarly for vertices Hg). In this way we speak of a vertex α of Γ being of type P (resp. H) if α is of the form P g (resp. Hg) for some g ∈ G. An edge of Γ consists of an un-ordered pair {α, β} of adjacent vertices. If α = P g and β = Hg then the stabilizer in G of the edge {α, β} is (P ∩ H)^{g}.

For any vertex δ of Γ, we denote by Δ(δ) the set of vertices of Γ which are adjacent to δ. For any integer n ≥ 1, we denote by Δ^{(n)}(δ) the set of all vertices γ such that d(δ, γ) ≤ n, where d is the natural metric on Γ. We also set G_{δ}^{(n)} = G_{Δ^{(n)}(δ)}; the point-wise stabilizer in G of Δ^{(n)}(δ).

For any vertex δ of Γ we set

Y_{δ} = Y_{G_{δ}},  and  Q_{δ} = O_{p}(G_{δ}).

For β a vertex of type H we set

V_{β} = ⟨Y_{β} : δ ∈ Δ(β)⟩ and \overline{G_{β}} = G_{β}/C_{G_{β}}(V_{β}/Y_{β}).

and for α a vertex of type P we set

W_{α} = ⟨V_{β} : β ∈ Δ(α)⟩.

Further, for α of type P we set

D_{α} = ∩_{β∈Δ(α)}V_{β}.

We assume that the reader of these notes has a good deal of familiarity and, indeed, facility, with the so-called amalgam method. In particular, we do not wish to develop the notion of critical pair, for the uninitiated, at this time. For a critical pair (α, α′), the distance b = d(α, α′) is an invariant of Γ, which we denote by b(Γ) or by b(P, H). A path of length b from α to α′ will be called a critical path.

The following result is so basic to what follows that it will most often be used without explicit reference. Recall that a group X is said to be p-closed if O_{p}(X) is a Sylow subgroup of X.

Lemma 1.13. Let δ be a vertex of Γ.

(a) If α is of type P then G_{α}^{(1)} is p-closed, and O_{p}(G_{α}^{(1)} = Q_{α}.

(b) If β is of type H then C_{G_{β}^{(1)}}(Y_{β}) is p-closed, and O_{p}(C_{G_{β}^{(1)}}(Y_{β})) = Q_{β}.

Proof. Set R_{P} = ∩{(P ∩ H)^{g}}_{g∈P} and R_{H} = ∩{(P ∩ H)^{g}}_{g∈H}. The Frattini argument yields P = N_{P} (S ∩ R_{P})R_{P}, and since P ∈ P we then have P = N_{P} (S ∩ R_{P}). Thus S ∩ R_{P} = Q_{P}, and we have (a).

Set T = S ∩ C_{R_{H}} (Y_{H}). Then H = N_{H}(T)C_{R_{H}} (Y_{H}). But C_{P ∩H}(Y_{H}) = O_{p}(P ∩ H), as follows from 1.3(3) and from the characterization of Y_{H} given in 1.6. Thus T ⊴ H, so T = Q_{H}, and we have (b). □

Lemma 1.14. Let (κ, λ, μ) be a a path of length 2 in Γ. Then Q_{κ}∩Q_{λ} ≤ [Q_{λ}, O^{p}(G_{λ})]Q_{μ}.

Proof. Set R = [Q_{λ}, O^{p}(G_{λ}](Q_{λ} ∩ Q_{μ}). Then R ⊴ G_{λ}. As G_{λ} is transitive on Δ(λ) we then have Q_{κ} ∩ Q_{λ} ≤ R. □

Lemma 1.15. Let (α, β, ⋯, α′) be a critical path in Γ.

(a) If b is even then b ≥ 2 and [Y_{α}, Y_{α′}] = Y_{β} = Y_{α′−1}.

(b) If b is odd then b ≥ 3 and Y_{β} ∩ Y_{α′} = 1.

Proof. □

Lemma 1.16. Let (β, γ, δ_{)} be a path of length 2 in Γ, with δ of type P. Then Q_{β}(G_{β} ∩ G_{δ}) = G_{β} ∩ G_{γ}.

Proof. Set P = δ and H = β, and let g ∈ P so that γ = β^{g}. Notice that the induced action of P on Δ(δ) is equivalent to the natural doubly transitive permutation representation of P on the cosets of P ∩ H. The two- point stabilizer D := G_{β} ∩ G_{γ} therefore has the property that O_{p}(P ∩ H)D = P ∩ H. But O_{p}(P ∩ H) = Q_{P} Q_{H}, as D acts irreducibly on O_{p}(P ∩ H)/Q_{H}. As Q_{P} ≤ D, the lemma follows. □

Recall that for α of type P we have set D_{α} = ∩_{β∈Δ(α)}V_{β}.

Lemma 1.17. Let α be a vertex of Γ of type P, let σ and λ be distinct neighbors of α, and set Q_{α}^{∗} = (Q_{σ} ∩ Q_{α})(Q_{α} ∩ Q_{λ}). Then the following hold.

(a) V_{σ} ∩ V_{λ} = D_{α}.

(b) [D_{α}, O^{p}(G_{α})] = Y_{α}.

(c) Q_{α}^{∗} ⊴ G_{α}, and [Q_{α}, O^{p}(G_{α})] ≤ Q_{α}^{∗}.

Lemma 1.18. Suppose that b is odd, and let (α, β, ⋯, α′) be a critical path in Γ. Suppose that V_{α′} ≰ Q_{β}, and suppose that [V_{β} ∩ Q_{α′}, V_{α′}] = [V_{β}, V_{α′} ∩ Q_{β}] = 1. Then either V_{α′} is an F 1-offender on V_{β}/Y_{β}, or V_{β} is an F 1-offender on V_{α′}/Y_{α′}.

.

Section 2: Stellmacher’s Basic Amalgam Lemmas

For the remainder of this paper we shall assume:

Hypothesis 2.0. We have b ≥ 3.

Lemma 2.1. Assume that b is even, b ≥ 4, and let (α, β, ⋯, α′) be a critical path in Γ. Let β′ ∈ Δ(α′) − {α′ − 1}. Then V_{β′} fixes the path (β, ⋯, α′), V_{β′} ≰ G_{α}, and V_{β′} is a quadratic F -offender on Ṽ_{β}.

Proof. As b is even, 1.15(a) yields:

(1) [Y_{α}, Y_{α′}] = Y_{β} = Y_{α′−1}.

Then Y_{α′} = Y_{α′−1}Y_{β′}, and so [Y_{α}, Y_{β′}] ≠ 1. In particular, we have Y_{β′} ≠ Y_{α+3}. As V_{α+3} ≤ Q_{β′} and V_{β′} ≤ Q_{α+3}, it now follows from 1.11(a) that

(2) [V_{α+3}, V_{β′}] = 1.

Observe that Y_{β+1} = Y_{α′−1}Y_{α+3} by (1). Then V_{β′} ≤ C_{G_{β+1}}(Y_{β+1}) and so

(3) V_{β′} ≤ Q_{β+1} ≤ G_{β}.

Notice that V′_{β} ∩ Q_{β} ⊴ ⟨Y_{α}, Q_{β′}⟩ ≥ O^{p}(G_{α′}). Then V_{β′} ∩ Q_{β} ≤ D_{α}, and so

(4) |V_{β′}/D_{α′}| ≤ |V_{β′}Q_{β}/Q_{β}|.

Notice also that, by (2), we have [D_{α+2}, V_{β′}] = 1, and so (4) yields

|V_{β}/C_{V_{β}}(V_{β′})| ≤ |V_{β}/D_{α+2}| = |V_{β′}/D_{α′}| ≤ |V_{β′}Q_{β}/Q_{β}|.

But |Ṽ_{β}/CṼ_{β}(V_{β′})| ≤ |V_{β}/C_{V_{β}}(V_{β′})|, so we have shown that either V_{β′} is an F -offender

on Ṽ_{β} or V_{β′} ≤ Q_{β}. Thus, it remains to show that V_{β′} acts quadratically on Ṽ_{β} and that V_{β′} ≰ G_{α}. The required quadratic action follows from the observation that [V_{β}, V_{β′}, V_{β′}] ≤ [W_{α′}, V_{β′}] ≤ Y_{α′−1} = Y_{β}. So assume that we have V_{β′} ≤ G_{α}. Then [Y_{α}, V_{β′}] ≤ Y_{β} = Y_{α′−1} ≤ V_{β′}, whence

V_{β′} ⊴ ⟨Y_{α}, G_{β′}⟩ = ⟨Y_{α}, G_{α′} ∩ G_{β′}, G_{β′}⟩ = ⟨G_{α′}, G_{β′}⟩.

As O_{p}(⟨P, H⟩) = 1, we have a contradiction, and the lemma is proved. □

Here is a “QRC”-argument that will be used in the proof of lemma 2.3. It was shown to me by Ulrich M.

Lemma 2.2. Let V be a p-group, and let {Z_{i}}_{1≤i≤n} be a collection of normal subgroups of V which together generate V. Let s > 0 be a positive real number, and let B be a p-group acting on V. Assume that each Z_{i} is B-invariant, and assume that

|Z_{i}/C_{Z_{i}}(D)| ≥ |D/C_{D}(Z_{i})|^{s}

for every subgroup D of B. Then |V/C_{V} (B)| ≥ |B/C_{V} (B)|^{s}.

Proof. Set B_{1} = B and, inductively, set B_{i+1} = C_{B_{i}}(Z_{i}), 1 ≤ i ≤ n. Then B_{n+1} = C_{B}(V), and

|B_{i}/B_{i+1}|^{s} = |B_{i}/C_{B_{i}}(Z_{i})|^{s} ≤ |Z_{i}/C_{Z_{i}}(B_{i})|.

Then

|B/C_{B}(V)|^{s} = ∏_{i=1}^{n} |B_{i}/C_{B_{i}}(Z_{i})|^{s} ≤ ∏_{i=1}^{n} |Z_{i}/C_{Z_{i}}(B_{i})|.

But also

|Z_{i}/C_{Z_{i}}(B_{i})| = |Z_{i}C_{V} (B_{i})/C_{V} (B_{i})| ≤ |C_{V} (B_{i+1})/C_{V} (B_{i})|

and so we get

∏_{i=1}^{n} |Z_{i}/C_{Z_{i}}(B_{i})| ≤ ∏_{i=1}^{n} |C_{V} (B_{i+1})/C_{V} (B_{i})| = |V/C_{V} (B)|.

Thus |V/C_{V} (B)| ≥ |B/C_{V} (B)|^{s}. □

Lemma 2.3. Assume that we have b ≥ 3,and assume also that Ṽ is not an F 1-module for H via an F 1-offender contained in O_{p}(O^{p}(P))O_{p}(H). Then b is odd, and there exists a critical path π = (α, β, ⋯, α′) such that V_{α′} ≰ Q_{β}, and such that one of the following holds.

(i) V_{β} is a quadratic F 2^{∗}-offender on Ṽ_{α′} and Y_{α′} ≰ V_{β}, (ii) V_{β} is a quadratic F 2^{∗}-offender on Ṽ_{α′} and b = 3, or (iii) V_{β} is a quadratic F 2-offender on Ṽ_{α′}, V_{α′} is a quadratic F 2-offender on Ṽ_{β}, and

b = 3.

Proof. By 2.1, b is odd. Fix a critical path π = (α, β, ⋯, α′) and suppose first that V_{α′} ≤ Q_{β}. Then |V_{α′}/C_{V_{α′}} (Y_{α})| ≤ q. Here Y_{α} ≤ O_{p}(O^{p}(G_{α′−1}))Q_{α′}, so |Y_{α}/C_{Y_{α}}(Ṽ_{α′})| < |Ṽ_{α′}/CṼ_{α′}(Y_{α})|, by hypothesis. Thus, we have Y_{α}/(Y_{α} ∩ Q_{α′})| < q. [Y_{α} ∩ Q_{α′}, V_{α′}] ≤ Y_{β} ∩ Y_{α′} = 1, by 1.15(b), and then [Y_{α}, V_{α′}] = 1 as Y_{α} is a natural module for G_{α}. But Y_{α} ≰ Q_{α′}, so we have a contradiction. Thus:

(1) V_{α′} ≰ Q_{β}.

Notice that the proof of (1) involved also a proof of:

(2) Y_{α} ∩ Q_{α′} = Y_{β}.

Further, it follows from 1.11(a) that we have

(3) [V_{β} ∩ Q_{α′}, V_{α′} ∩ Q_{β}] = 1.

Since V_{β} and V_{α′} act quadratically on each other, V_{β} is not an F -offender on Ṽ_{α′}, and V_{α′} is not an F -offender on Ṽ_{β}. Therefore

(4) [V_{β} ∩ Q_{α′}, V_{α′}] ≠ 1 or [V_{α′} ∩ Q_{β}, V_{β}] ≠ 1.

As (α′, α′ −1, ⋯, β) is a pre-critical path, by (1), we may assume, by (4) and symmetry, that [V_{β}, V_{α′} ∩ Q_{β}] ≠ 1,. Thus, we have [V_{β}, V_{α′} ∩ Q_{β}] = Y_{β}, and Y_{β} ≤ V_{α′}.

Let δ ∈ Δ(β) with [Y_{δ}, V_{α′} ∩ Q_{β}] ≠ 1. Then (3) shows that Y_{δ} ≰ Q_{α′}, and we may therefore take δ = α. Choose t ∈ V_{α′} ∩Q_{β} with [Y_{α}, t] ≠ 1, and choose α−1 ∈ Δ(α)−{β}. Then:

(5) ⟨Q_{α−1}, t⟩ ≥ O^{p}(G_{α}).

Suppose next that [V_{α−1}, V_{α′−2}] = 1. Then V_{α−1} ≤ Q_{α′−2} ∩ Q_{α′−1}. Set A = V_{α−1} ∩ V_{β}Q_{α′}. Then

A ≤ V_{α−1}V_{β} ∩ V_{β}Q_{α′} = V_{β}(V_{α−1}V_{β} ∩ Q_{α′}) ≤ V_{β}(Q_{α′−1} ∩ Q_{α′})

and so [A, t] ≤ [V_{β}, t][Q_{α′−1} ∩ Q_{α′}, t] ≤ Y_{β}Y_{α′}.

Set X = C_{A}(t mod Y_{β}). Then |A/X| ≤ q, and X ⊴ ⟨Q_{α−1}, t⟩ as Y_{β}Y_{α−1} ≤ X. As O^{p}(G_{α}) is transitive on Δ(α), (5) then implies that X ≤ D_{α}. But also [D_{α}, t] ≤ [V_{β}, t] ≤ Y_{β}, so in fact (6) X = D_{α}.

We now have

|V_{α′}/C_{V_{α′}} (V_{α−1}V_{β})| ≤ |V_{α′}/D_{α′−1}|
= |V_{α−1}/D_{α}| = |V_{α−1}/A||A/X| ≤ |V_{α−1}V_{β}Q_{α′}/V_{β}Q_{α′}|q.

But also

|V_{α−1}V_{β}Q_{α′}/Q_{α′}| = |V_{α−1}V_{β}Q_{α′}/V_{β}Q_{α′}||V_{β}/(V_{β} ∩ Q_{α′})| ≥ |Y_{α}/Y_{α} ∩ Q_{α′}| = q, and so we obtain |V_{α−1}V_{β}Q_{α′}/Q_{α′}| ≥ |V_{α′}/C_{V_{α′}} (V_{α−1}V_{β})|. Thus V_{α−1}V_{β} is an F -offender on Ṽ_{α′}, contrary to hypothesis. Thus

(7) [V_{α−1}, V_{α′−2}] ≠ 1.

If now V_{α−1} ≤ Q_{α′−2} then we may apply the result (1) to the pre-critical path (α′ − 2, ⋯, α − 1) and obtain V_{α′−2} ≤ Q_{α−1}. Then (7) yields Y_{α−1} = Y_{α′−2}, and then Y_{α} = Y_{α′−2}Y_{β} ≤ C_{G}(V_{α′}), which is contrary to the case. Thus:

(8) V_{α−1} ≰ Q_{α′−2}.

Suppose now that we have b > 3. Then [V_{α′−2}, V_{α′}] = 1, and so Y_{α−1} ≰ V_{α′−2}. Since both (α − 1, ⋯, α′ − 2) and (α′ − 2, ⋯, α − 1) are pre-critical paths, we are free to replace (β, ⋯, α′ − 2) by one of these, obtaining either Y_{β} ≰ V_{α′} or Y_{α′} ≰ V_{β}. But in fact Y_{β} ≤ V_{α′} as [V_{β}, V_{α′} ∩ Q_{β}] ≠ 1, so we have shown:

(9) If b > 3 then Y_{α′} ≰ V_{β} and [V_{α′}, V_{β} ∩ Q_{α′}] = 1. Suppose that |V_{β}Q_{α′}/Q_{α′}| ≤ |V_{α′}Q_{β}/Q_{β}|. Then (9) yields

|Ṽ_{β}/CṼ_{β}(V_{α′})| ≤ |V_{β}/C_{V_{β}}(V_{α′})| = |V_{β}/(V_{β} ∩ Q_{α′})| ≤ |V_{α′}Q_{β}/Q_{β}|

and V_{α′} is an F -offender on Ṽ_{β}, contrary to assumption. Thus:

(10) If b > 3 then |V_{β}Q_{α′}/α′| > |V_{α′}Q_{β}/qb|.

Notice that 2.2 applies to our situation, with s = 1, V = V_{β}, B = V_{α′} ∩ Q_{β}, and with {Z_{i}}_{1≤i≤n} = {Y_{δ}}_{δ∈Δ(β)}. We therefore have

(11) |V_{β}/C_{V_{β}}(V_{α′} ∩ Q_{β})| ≥ |(V_{α′} ∩ Q_{β})/C_{V_{α′}} (V_{β})|. If now |V_{β}Q_{α′}/Q_{α′}| > |V_{α′}Q_{β}/Q_{β}| then

|V_{α′}/C_{V_{α′}} (V_{β})| = |V_{α′}Q_{β}/Q_{β}||(V_{α′} ∩ Q_{β})/C_{V_{α′}} (V_{β})| <

|V_{β}Q_{α′}/Q_{α′}||V_{β}/C_{V_{β}}(V_{α′} ∩ Q_{β})| ≤ |V_{β}Q_{α′}/Q_{α′}||V_{β}/(V_{β} ∩ Q_{α′})| = |V_{β}Q_{α′}/Q_{α′}|^{2}.

This shows:

(12) If b > 3 then V_{β} is a quadratic F 2^{∗}-offender on Ṽ_{α′}.

Suppose next that we have b = 3. If also |V_{β}Q_{α′}/Q_{α′}| ≥ |V_{α′}Q_{β}/Q_{β}| then the argument for (12) shows that V_{β} is a quadratic F 2-offender on Ṽ_{α′}, while if |V_{β}Q_{α′}/Q_{α′}| ≤ |V_{α′}Q_{β}/Q_{β}| then V_{α′} is a quadratic F 2-offender on Ṽ_{β}. Alternatively, if |V_{β}Q_{α′}/Q_{α′}| < |V_{α′}Q_{β}/Q_{β}| then V_{α′} is a quadratic F 2^{∗}-offender on Ṽ_{β}. In this way we obtain the lemma. □

Lemma 2.4. In outcome (iii) of 2.3 we have Q_{β}V_{α′} ⊴ G_{β} ∩ G_{β+1}.

Proof. Immediate from 1.16. □

Section 3: Quadratic modules

This section contains detailed results on quadratic F 2-modules and F 2^{∗}-modules. Some of these results are stated without proof.

Hypothesis 3.1. X is a finite group with O_{p}(X) = 1, and such that every non-abelian simple section of X is an alternating group, a group of Lie type, or one of twenty-six “sporadic” groups. Further, we are given a faithful 𝔽_{p}X-module V.

Proposition 3.2. Assume hypothesis 3.1, with p = 2, and let t be an involution in X. Assume that F^{∗}(X) is quasisimple, and that X = ⟨t^{X}⟩. Assume further that |V/C_{V} (t)| ≤ 4. Then one of the following holds.

(i) There is a fours group E in X with t ∈ E and with [V, E, E] = 0.

(ii) A Sylow 2-subgroup of X is dihedral or semi-dihedral.

Proof. To be provided. □

Remark: The proof of 3.2 does not require that the simple sections of X be of “known” type. The “known” simple groups with dihedral or semi-dihedral Sylow 2-subgroups are the groups L_{2}(r), r odd; L_{3}(r), r ≡ 3 (mod 4); U_{3}(r), r ≡ 1 (mod 4); Alt(7); and M_{11}. Using this, we may obtain the following result.

Proposition 3.3. Assume hypothesis 3.1. Let X, t, and V be as in 3.2, and set K = F^{∗}(X). Assume that there does not exist a fours group E in X, containing t, and such that [V, E, E] = 0. Then either X ≅ Alt(5) and V is the O_{4}^{−}(2)-module for X, or else X ≅ Sym(5) and V is the ΓL(2, 4)- module for X. In the latter case there exists a fours group in K acting quadratically on V.

Proof. To be provided. □

Corollary 3.4. Assume hypothesis 3.1. Let K be a component of X and let A ≤ X be a quadratic F 2-offender on V, with [K, A] ≠ 1, and set L = ⟨K^{A}⟩. Then one of the following holds.

(i) There exists a subgroup B of LA, acting quadratically on some non-trivial Linvariant section of V and acting faithfully on L, such that |B| > 2.

(ii) p = 2, K = L ≅ SL(2, 4), and |A/C_{A}(K)| = 2.

(iii) p = 2, K ≠ L, and K ≅ SL(2, 2^{n}) for some n.

Proof. Suppose that (i) is false, and let A_{0} be a complement to C_{A}(K) in A. Then |A_{0}| = 2. Suppose that K is A-invariant. Then 0.1 implies that A_{0} induces either a transvection or a 2-transvection on [V, K], and then 3.3 yields (ii). So assume that K is not A-invariant. Theorem 3 in [Chermak] then yields (iii). □

Proposition 3.5. Assume hypothesis 3.1. Set K = F^{∗}(X), and assume that K is quasisimple. Then one of the following holds.

(i) K/Z(K) is of Lie type in characteristic p.

(ii) p = 2 and K/Z(K) is an alternating group.

(iii) p = 2 and K/Z(K) is isomorphic to U_{4}(3).

(iv) p = 2 and K/Z(K) is isomorphic to a sporadic group

Proof. For p = 2 this is a consequence of the work [MS1] of Meierfrankenfeld and Stroth on quadratic fours groups in groups of Lie type, with some help from 3.3. For p > 3 one may appeal to work on so-called quadratic pairs by Thompson, Betty Salzberg [Sa], or by Timmesfeld [Ti]. The case of quadratic pairs for p = 3 was investigated by Meierfrankenfeld in unpublished notes. The author of these notes will produce a paper on quadratic pairs for p = 3, more than sufficient for the purposes here, someday, soon. □

Proposition 3.6. Assume hypothesis 3.1. Let S be a Sylow p-subgroup of X, and assume that S is contained in a unique maximal subgroup of X. Assume further that V is a quadratic F 2-module for X. Let K be a component of X, set U = [V, K], and set Ũ = U/C_{U}(K). Then one of the following holds.

(i) K ≅ SL(2, p^{n}) and every constituent for K in Ũ is a natural module for K. There are at most two such constituents.

(ii) K ≅ SU(3, p^{n}) and U is a natural module for K.

(iii) K ≅ Ω_{4}^{−}(p^{n}) and U is a natural orthogonal module for K.

(iv) p = 2, K ≅ Sz(2^{n}), and Ũ is a natural module for K.

(v) p = 2, K ≅ SL(3, 2^{n}) (resp. Sp(4, 2^{n})), and Ũ is the direct sum of a natural module and its dual (resp. a natural module and its contragredient), for K. Moreover, there exists an element of N_{S}(K) which induces a polarity on the standard Coxeter diagram for K and which interchanges the two irreducible Ksubmodules of Ũ.

(vi) p = 2, K ≅ Alt(2^{n} +1), and every non-trivial constituent for K in U is a natural module (derived from the permutation module) for K. There are at most two such constituents.

(vii) p = 2, K ≅ Alt(9), and Ũ is isomorphic to a spin module for K, of dimension 8.

Proposition 3.7. In Proposition 3.6, suppose that V is a quadratic F 2^{∗}-module for X. Then one of the following holds.

(i) K ≅ SL(2, p^{n}) and Ṽ is a natural module for K.

(ii) p = 2, K ≅ SL(2, 2^{n}), and there exists a quadratic F 2^{∗}-offender A in X such

that ⟨K^{A}⟩ ≅ Ω_{4}^{+}(2^{n}), and U is a natural orthogonal module for ⟨K^{A}⟩. (iii) p = 2, K ≅ Ω_{4}^{−}(2^{n}), U is a natural orthogonal module for K, and K contains

no quadratic F 2^{∗}-offender on U. (iv) p = 2, K ≅ SL(3, 2^{n}), and Ũ is the direct sum of a natural module and its dual,

for K. Moreover, there exists an element of N_{S}(K) which induces a polarity on the standard Coxeter diagram for K and which interchanges the two irreducible K-submodules of Ũ.

(v) p = 2, K ≅ Alt(2^{n} +1), and Ũ is a natural module (derived from the permutation

module) for K. Moreover, if V is an F 1-module for X then either (i) or (v) holds.

Proposition 3.8. In Proposition 3.5, suppose that there is a component K_{1} of X, distinct from K, such that [V, K, K_{1}] ≠ 0. Then KK_{1} ≅ Ω_{4}^{+}(p^{n}), and [V, K] = [V, K_{1}]. Moreover, K_{1} is the unique such component of X distinct from K.

Proposition 3.9. Assume hypothesis 3.1, and assume that X = RA, where R = [R, A] is a normal p′-subgroup of X, and where |A| = p and |V/C_{V} (A)| ≤ p^{2}. Assume also that V = [V, R]. Then one of the following holds.

(i) X ≅ SL(2, 2) and |V | = 4 or 16.

(ii) R is elementary abelian of order 9, |A| = 2, and |V | = 16.

(iii) X is isomorphic to the commutator subgroup of SU(3, 2), and V is a natural

SU(3, 2)-module for X, of order 64. (iv) X is dihedral of order 10, and |V | = 16.

(v) X ≅ SL(2, 3) and |V | = 9 or 81.

(vi) R is a commuting product of two quaternion groups, and |V | = 81.

Proposition 3.10. Assume hypothesis 3.1. Let X = RA, with A a p-group and with R = [R, A] a normal p′-subgroup of X. Assume that A is a quadratic F 2-offender on V. Denote by D the set of non-identity subgroups D of X such that D = [C_{R}(B), A] for some maximal subgroup B of A. Then R is the direct product of the elements of D, and [V, R] is the direct sum of the subspaces [V, D], D ∈ D. Further, for any D ∈ D and for any a ∈ A − C_{A}(D), the triple (D, ⟨a⟩, [V, D]) satisfies the hypothesis of 3.8, in place of (R, A, V). Moreover, A is generated by its elements a such that |V/C_{V} (a)| ≤ p^{2}.

Proposition 3.11. Assume hypothesis 3.1, with X solvable. Let S be a Sylow p-subgroup of X, and assume that S is contained in a unique maximal subgroup of X. Assume further that V is a quadratic F 2-module V for X. Then p = 2 or 3, and there exists an Stransitive set K = {K_{1} ⋯ K_{r}} of subgroups of O_{p′}(X) such that O_{p′}(X) = K_{1} ×⋯×K_{r}, with [V, O_{p′}(X)] = [V, K_{1}] ⊕ ⋯ ⊕ [V, K_{r}], and such that one of the following holds for al i.

(i) K_{i} ≅ O^{p}(SL(2, p)) and [V, K_{i}] is a natural module or a direct sum of two natural modules for K_{i}.

(ii) K_{i} ≅ O^{p}(O_{4}^{+}(p)) and [V, K_{i}] is a natural orthogonal module for K_{i}.

(iii) p = 2, K_{i} is cyclic of order 5, and |[V, K_{i}]| = 16.

(iv) p = 2, K_{i} ≅ O^{2}(SU(3, 2)), and |[V, K_{i}]| = 64.

The next two results come from joint work with Christopher Parker. The reader may wish to recall some of the notation from 1.2.

Lemma 3.12. Assume hypothesis 3.1, and assume that X = DP_{0} where D is a normal r-subgroup of X for some prime r, and where P_{0} ∈ P_{X}^{∗}(S) for some Sylow p-subgroup S of X. Assume further that O^{p}(P_{0}) ≰ C_{P_{0}}(D) and that there exists an element t of S such that [V, t, t] = 0 and such that |V/C_{V} (t)| ≤ p^{2}. Assume also that V = ⟨C_{V} (S)^{X}⟩. Then p = 2 and the following hold.

(a) |V/C_{V} (t)| = 4.

(b) [D, t] is a direct factor of D of order 3, and |[V, [D, t]]| = 16.

(c) Let h be an element of odd order in P_{0} − D, with h^{t} = h^{−1}, and set D^{∗} = D⟨h⟩. Then ⟨t^{D^{∗}}⟩ ≅ SU(3, 2)′, and |[V, [D^{∗}, t]]| = 64.

Proof. We have [D, t] ≠ 1 as [D, O^{p}(P_{0})] ≠ 1. Then the quadratic action of t implies that p = 2 or 3. If p = 3 then there is a subgroup K = ⟨t^{K}⟩ of P with K ≅ SL(2, 3) and with O_{2}(K) ⊈ Φ_{3}(P). On the other hand, if p = 2 then the Baer-Suzuki Theorem implies that there is a subgroup K = ⟨t^{K}⟩ of P with K dihedral of twice odd order, and with O_{2′}(K) ⊈ Φ_{2}(P). Fix K as above, for the two cases, and put J = O_{p′}(K), R = [D, t], and R^{∗} = [DJ, t]. Suppose first that t is a transvection on V, or that R^{∗} is cyclic. Then 3.9 shows that R^{∗} = J, contrary to [D, t] ≠ 1. We conclude that t is a 2-transvection and that R^{∗} is non-cyclic. Then 3.9 implies that R^{∗} ≅ 3^{2} or 3^{1+2} if p = 2, or a commuting product of two copies of Q_{8} if p = 3. Moreover, in the case that R^{∗} ≅ 3^{1=2}, we have [Z(R^{∗}), t] = 1, and thus in any case we have R⟨t⟩ ≅ SL(2, p).

Now C_{D}(t) centralizes R and normalizes R^{∗}. Suppose p = 3. Then R and J are the unique normal quaternion subgroups of R^{∗}, and so C_{D}(t) normalizes J. But the centralizer in Aut(J) of t is ⟨t⟩, so C_{D}(t) centralizes J, and thus J centralizes [D, t]C_{D}(t) = D. As ⟨J^{P_{0}}⟩ contains O^{3}(P_{0}), we have a contradiction at this point, and so p = 2. Now suppose that R^{∗} ≅ 3^{2}. There is then precisely one cyclic subgroup J_{1} of R^{∗} satisfying R^{∗} = R × J_{1} and with |[V, J_{1}] = |[V, R]|. (Here it should be noted that |[V, R]| may be either 4 or 16.) Then J_{1} is D-invariant, so J_{1} centralizes D, and then also O^{2}(P_{0}) centralizes D, as in the case p = 3. So we conclude at this point that R^{∗} ≅ 3^{1+2}. Here [V, R^{∗}] is a natural SU(3, 2)-module for R^{∗}⟨t⟩, as is shown by 3.9, and the Lemma thereby follows. □

Lemma 3.13. Assume hypothesis 3.1, and assume that all of the following conditions hold.

(1) F^{∗}(X) is an r-group for some prime r, r ≠ p.

(2) X ∈ P_{X}^{∗}(S) for a Sylow p-subgroup S of X.

(3) We have V = ⟨C_{V} (S)^{X}⟩.

(4) There exists a non-identity element t of S, acting quadratically on V, such that |[V, t]| ≤ p^{2}.

Set D = ⟨[O_{r}(X), T]^{X}⟩, and denote by B the set of all maximal subgroups B of D such that C_{V} (B) ≠ 0. Then p = 2, r = 3, D is elementary abelian, and X/D induces the full symmetric group on B, with t inducing a transposition.

Proof. Set T = t^{X} and set H = ⟨T ⟩. Then 3.11 implies that p = 2, and 3.12(b) in particular implies that D is an elementary abelian 3-group, with F (X) = D ×C_{F (X)}(H). As X ∈ P_{X}^{∗}(S), by assumption, we have also H ≥ O^{2}(X).

Set \overline{V} = V/C_{V} (D), and let R be a Sylow 2-subgroup of the pre-image in X of O_{2}(X/C_{X}(\overline{V})). Suppose that R ≠ 1. As O_{p}(X) = 1 and X ∈ P_{X}^{∗}(S), the frattini argument then yields O^{2}(X) ≤ C_{X}(\overline{V}), contrary to D ≠ 1. Thus R = 1, and we may therefore replace V with \overline{V} and replace X by X/C_{X}(\overline{V}) while retaining the hypotheses of the lemma. So, we may assume to begin with that C_{V} (D) = 0.

Write [D, t] = ⟨d⟩, and notice that 3.12(c) implies that [V, d] = [V, d, C_{D}(t)], while 3.12(b) yields |[V, d]| = 16. It follows that there are hyperplanes I and J of D with I^{t} = J and with [V, d] = C_{V} (I) × C_{V} (J). As t is a 2-transvection on V, it then follows that t acts trivially on B − {I, J}. Thus:

(1) t acts as a transposition on B.

We also record:

(2) We have [V, d] = C_{V} (I) × C_{V} (J), where I and J are the two elements of B which are interchanged by t.

Denote by F the set of fixed-points for H on B, and set W = ⟨C_{V} (B) : B ∈ B − F⟩. Then W is X-invariant, and (2) shows that W ≥ [V, H] ≥ [V, D_{0}] = V. Then B = B −F, and so

(3) H has no fixed-points on B.

Now (1) and (3) together imply that X acts transitively on B. As X = SH, it follows that S is transitive on the set of H-orbits in B. As ⟨t⟩ is normal in S, by assumption, we conclude that H has just one such orbit. Thus

(4) H is transitive on B.

It now follows from (1) and (4) that X induces the full symmetric group on B. □

Section 4: The structure of H, with b ≥ 3

Recall from Hypothesis 2.0 that we have b ≥ 3.

We begin this section by considering the case where H is solvable. As always, we set \overline{H} = H/C_{H}(Ṽ), and we set H_{0} = ⟨O^{p}(P)^{H}⟩.

Lemma 4.1. Assume that H is solvable and that P is not. Then the following hold. (a) \overline{H}_{0} is an r-group, where \{p, r\} = \{2, 3\}.

(b) [Φ_{p}(H_{0}), S ∩ O^{p}(P)] ≤ O_{p}(H).

(c) H acts irreducibly on H_{0}/Φ_{p}(H_{0}).

(d) (P ∩ H)Φ_{p}(H_{0}) is the unique maximal subgroup of H containing P ∩ H.

(e) Let A be a subgroup of S ∩ O^{p}(P) such that \overline{A} is elementary abelian, and let X_{A} denote the set of elements g ∈ H_{0} such that |A/C_{A}(g mod O_{p}(H))| = p and such that for any a ∈ A with [\overline{g}, \overline{a}] ≠ 1, we have H = ⟨a, (P ∩ H)^{g}⟩. Then H_{0} = ⟨X_{A}⟩C_{H_{0}}(A mod O_{p}(H)).

Proof. Let H_{1} be a normal subgroup of H_{0} such that H_{1}(P ∩ H) is a proper subgroup of H. By our minimal choice of H, we have ⟨P, P_{0}⟩ ∈ L for any P_{0} ∈ P_{H_{1}}(S). For any such P_{0}, P_{0} is solvable, by assumption, and since P is non-solvable it follows from 3.1(4) that P_{0} ≤ N_{G}(O^{p}(P)). As also O^{p}(P) is invariant under N_{G}(S), by P -Uniqueness (i.e. by 1.3(3)) we conclude from 1.2 that H_{1} ≤ N_{G}(O^{p}(P)). But O_{p}(O^{p}(P) ∩ H) = S ∩ O^{p}(P), so [H_{1}, S ∩ O^{p}(P)] ≤ O_{p}(H_{1}). In particular, by taking H_{1} = Φ_{p}(H_{0})(P ∩ H) we obtain (b).

As P̃ ∈ P and P̃ is solvable, S acts irreducibly on O^{p}(P̃)/Φ_{p}(O^{p}(P̃)). As S ∩O^{p}(P) ≰ O_{p}(P̃), by 1.10, it follows that O^{p}(P̃) = [O^{p}(P̃), S ∩ O^{p}(P)], and so also H_{0} = [H_{0}, ⟨(S ∩ O^{p}(P))^{H}⟩]. Thus, there does not exist a normal subgroup H_{1} of H which lies properly between Φ_{p}(H_{0}) and H_{0}, and we obtain (c). Now (d) is an immediate consequence of (c), as is the fact that \overline{H}_{0} is an r-group for some prime r. Then (a) follows from the quadratic action on Ṽ given by 2.3.

Let A be as in (e), and set U = [\overline{H}_{0}/Φ(\overline{H}_{0}), A]. Assume that U ≠ 1. As \overline{A} is elementary abelian, we have U = ⟨u ∈ U : |A/C_{A}(u)| = p. Let g ∈ H_{0} such that g is incident with an element u of U with |A/C_{A}(u)| = p, and let a ∈ A − C_{A}(u). Then ⟨a^{g}, P ∩ H⟩ ≥ ⟨[a, g], P ∩ H⟩ = H, as a consequence of (d). Further, (b) implies that [g, C_{A}(u)] ≤ [C_{H}(Ṽ), C_{A}(u)]. Since [C_{H}(Ṽ), S ∩ O^{p}(P)] ≤ O_{p}(H), by 1.11(b), we see that [g, C_{A}(u)] ≤ O_{p}(H), and thus g ∈ X_{A}. This yields (e). □

Lemma 4.2. If H is solvable then so is P.

Proof. Assume false, so that H is solvable and P is not. Then 4.1(a) implies that q > p.

Define r as in 4.1, so that \overline{H}_{0} is an r-group. Let (α, β, ⋯, α′) be a critical path in Γ, and suppose first that b is odd. Then Y_{β} ∩ Y_{α′} = 1, by 1.15(b). Take H = α′ and P = α′ − 1, and set A = V_{β}. We have A ≤ (S ∩ O^{p}(P))Q_{H}, by 1.14. For any h ∈ H such that [A, h] ≤ Q_{H} we have A ≤ (α′ − 1)^{h}. As A ≰ Q_{H} it then follows from 4.1(e) that there exists g ∈ H_{0} and a hyperplane A_{0} of A such that [A_{0}, g] ≤ Q_{H} and such that, for any a ∈ A − A_{0}, we have H = ⟨a, (P ∩ H)^{g}⟩. Setting μ = (α′ − 1)^{g}, we then have:

(1) There exists μ ∈ Δ(α′) such that |A : A ∩ G_{μ}| = p and such that ⟨a, G_{α′} ∩ G_{μ}⟩ = G_{α′} for any a ∈ A − G_{μ}.

In particular, (1) implies that for any a ∈ A − G_{μ} we have [a, Y_{μ}] ≠ 1. Write A_{0} for A ∩ G_{μ}.

Suppose that Y_{μ} ≤ Q_{β}. Then [A, Y_{μ}] = Y_{β}. As [A_{0}, Y_{μ}] ≤ Y_{α′}, we then have A_{0} = C_{A}(Y_{μ}), and thus |A/C_{A}(Y_{μ})| = p. But p < q, from which it now follows that [Y_{δ}, Y_{μ}] = 1 for all δ ∈ Δ(β). This is contrary to [A, Y_{μ}] ≠ 1, so we conclude that Y_{μ} ≰ Q_{β}.

By analogy with (1), we may now choose δ ∈ Δ(β) so that:

(2) |Y_{μ} : Y_{μ} ∩ G_{δ}| = p, and G_{β} = ⟨G_{δ} ∩ G_{β}, b⟩ for any b ∈ Y_{μ} − G_{δ}.

Now [Y_{δ} ∩ G_{μ}, Y_{μ} ∩ G_{δ}] ≤ Y_{β} ∩ Y_{α′} = 1, and as p < q we then have [Y_{δ}, Y_{μ} ∩ G_{δ}] = 1 and [Y_{δ} ∩ G_{μ}, Y_{μ}] = 1. But [Y_{δ}, Y_{μ}] ≠ 1, as follows from (2), so Y_{δ} ≰ G_{μ}. Then (1) yields G_{α′} = ⟨Y_{δ}, G_{α′} ∩ G_{μ}⟩, and we obtain Q_{α′} ∩ Q_{μ} = C_{Q_{α′}} (Y_{μ} ∩ G_{δ}) ⊴ G_{α′}, contrary to 1.9. This contradiction shows that b is even.

We now have [Y_{α}, Y_{α′}] = Y_{β} = Y_{α′−1}. Also, as b > 2 by assumption, we have b ≥ 4. Let A now denote W_{α}. Then [A, Y_{β}] = 1. As A ≤ Q_{α′−3} ≤ G_{α′−2}, and as Y_{α′−2} = Y_{α′−3}Y_{α′−1} = Y_{α′−3}Y_{β}, we have A ≤ Q_{α′−2}, and so A ≤ G_{α′−1}. We observe also that Φ(A) ≤ Y_{α} ≤ Q_{α′−1}. If A ≤ Q_{α′−1} then [A, Y_{α′}] = [Y_{α}, Y_{α′}] ≤ Y_{α}. But η(G_{α}, A) ≥ 2, as V_{β} is not normal in G_{α}. Thus A ≰ Q_{α′−1}. By analogy with (1) and (2) we may then choose λ ∈ Δ(α′ − 1) so that:

(3) |A : A ∩ G_{λ}| = p, and G_{α′−1} = ⟨a, G_{α′−1} ∩ G_{λ}⟩ for any a ∈ A − G_{λ}.

Set A_{0} = A ∩ G_{λ}. Then [A_{0}, Y_{λ}] ≤ Y_{α′−1} ≤ Y_{α}. By 1.12, no element of Q_{β} − Q_{α} induces a transvection on A/Y_{α}, so we have Y_{λ} ≤ Q_{α}. We observe that, as a consequence of (3), Y_{λ} is not A-invariant, and so [A, Y_{λ}] ≰ Y_{β}. As Y_{λ} ≤ Q_{β} we have [V_{β}, Y_{λ}] ≤ Y_{β}, so we may fix δ ∈ Δ(α) − {β} with [V_{δ}, Y_{λ}] ≰ Y_{β}. In particular, we have [V_{δ}, Y_{λ}] ≠ 1, while Y_{λ} ≤ Q_{α} ≤ G_{δ}. As Y_{δ}Y_{β} = Y_{α} we have [Y_{δ}, Y_{α′}] ≠ 1, and so Y_{δ} ≰ V_{α′−1}. Then Y_{δ} ≰ [V_{δ}, Y_{λ}], so [V_{δ}, Y_{λ} ∩ leqQ_{δ}] = 1. In particular, we have Y_{λ} ≰ Q_{δ}, and by analogy with (1) we may fix γ ∈ Δ(δ) so that |Y_{λ} : Y_{λ} ∩ G_{γ}| = p and so that G_{δ} = ⟨G_{γ} ∩ G_{δ}, Y_{λ}⟩. Then [Y_{γ}, Y_{λ}] ≠ 1. But [Y_{γ}, Y_{λ}∩G_{γ}] = 1 as Y_{δ} ≰ V_{α′−1}. As p < q it follows that Y_{γ} ≰ G_{λ}, and then (3) yields ⟨Y_{γ}, G_{α′−1} ∩ G_{λ}⟩ = G_{α′−1}. Then Q_{α′−1} ∩ Q_{λ} = C_{Q_{α′−1}}(Y_{λ}) ⊴ G_{α′−1}, again contrary to 1.9. This contradiction completes the proof of 4.2. □

Lemma 4.3. Assume that b ≥ 3 and that F_{p}^{∗}(H) = F_{p}(H). Then H is solvable, P is solvable, and either p = 2 or (N_{H}(P)P, H) is a weak BN-pair in characteristic 3.

Proof. Set D = F_{p}(H), X = O^{p}(P̃), and R = O_{p}(O^{p}(P))Q_{H}. As always, set \overline{H} = H/C_{H}(Ṽ). Suppose first that X ≤ D. Then H = (P ∩ C̃)D, H is solvable, and 4.2 implies that P is solvable. By 2.3 there exists a subgroup A of R such that \overline{A} is a nonidentity quadratic F 2-offender on Ṽ. Set R_{0} = ⟨A^{P ∩H}⟩, and suppose that R_{0} ≤ O_{p}(P̃). As [O_{p}(P̃), O^{p}(P̃)] is a p- subgroup of F_{p}(H), we then have \overline{R}_{0}Q_{H} ⊴ ⟨O^{p}(P̃), P ∩H⟩ = H, contrary to A ≰ Q_{H}. Thus we may assume to begin with that A ≰ O_{p}(P̃).

Suppose that p ≠ 2. As \overline{A} acts non-trivially on \overline{D}, \overline{D}\overline{A} involves SL(2, p), so we obtain p = 3, P̃ is a \{2, 3\}-group, and \overline{D} is a 2-group. As P is solvable we get also q = 3, and P ∩ H = S⟨t⟩ where t is an involution satisfying [S, t] ≤ R. Set L = ⟨A^{H}⟩, and set L_{0} = O^{3}(L). Then 3.10 shows that \overline{L}_{0} can be written as \overline{L}_{0} = \overline{K}_{1} ⋯ \overline{K}_{s}, where each \overline{K}_{i} is a normal quaternion subgroup of \overline{L}, and where |[Ṽ, \overline{K}_{i}]| = 9 or 81. Denote by L_{1} the inverse image in L_{0} of Z(\overline{L}_{0}). Thus \overline{L}_{1} is an elementary abelian 2-group, and it then follows from “P_{1}- uniqueness (i.e. from 1.2 and 1.3(4)) that L_{1} ≤ N_{G}(Y_{P}).

Suppose s = 1. As X ≤ L_{0} we then have \overline{H} ≅ GL(2, 3) and (N_{H}(P)P, H) is a weak BN-pair. Thus, we may assume that s > 1. Notice that C_{L_{1}}(Ỹ_{P}) centralizes Ṽ, as Ṽ = ⟨(Ỹ_{P})^{L_{0}}⟩. Thus \overline{L}_{1} acts faithfully on Ỹ_{P}, and since |Ỹ_{P} | = 3 we then have s = 2, Z(\overline{K}_{1}) = Z(\overline{K}_{2}), and Ṽ = [Ṽ, D] is a natural O_{4}^{+}(3)-module for \overline{H}. Here |\overline{A}| = 3 and |Ṽ/CṼ (\overline{A})| = 9, so 2.3 shows that b = 3, and if (α, β, γ, α′) is a critical path in Γ then V_{β} is a quadratic F 2-offender on Ṽ_{α′}. But V_{β}Q_{α′} is normal in (G_{β} ∩ G_{α′})Q_{α′}, and one may observe that (G_{β} ∩ G_{α′})Q_{α′} = G_{γ} ∩ G_{α′} = P ∩ H. But \overline{A} is not normal in P ∩ H since t interchanges \overline{K}_{1} and \overline{K}_{2}. This contradiction proves that either the lemma holds holds or:

(1) We have X ≰ D.

We may now assume that (1) is the case. If D ≤ N_{H}(R) then [D, R] ≤ Q_{H}, and then 1.10 yields [\overline{D}, \overline{O}^{p}(H)] = 1. This implies that O^{p}(H) ≤ D, contrary to (1), so we conclude that D ≰ N_{H}(R). In particular, D ≰ N_{G}(O^{p}(P)). Minimality of H implies that ⟨P, P_{1}⟩ ∈ L for any P_{1} ∈ P_{DS}(S), and then P_{1}-Uniqueness (1.2 and 1.3(4)) yields q = 2 or 3. Notice that, by 3.10, \overline{A} is generated by elements \overline{a} such that |Ṽ/CṼ (\overline{a})| ≤ p^{2}. Then 3.12 yields p = 2, whence also q = 2 and P ∩ H = S. Thus H = P̃, and \overline{D} is a 3-group, with |[\overline{D}, \overline{R}]| = 3. Then |\overline{R}| = 2, so \overline{R} = \overline{A} and |[Ṽ, R]| ≤ 4. We now appeal to 3.13, and find that H/D is a symmetric group of degree 2^{n} + 1 for some n, and that RD/D is generated by a transposition. Then N_{H}(R)D is not maximal in H. This is contrary to P_{1}-Uniqueness, and the lemma is thereby proved. □

Lemma 4.4. Suppose that F_{p}^{∗}(H) ≠ F_{p}(H). Then H = (P ∩ H)E_{p}(H), and O^{p}(P̃) ≤ E_{p}(H). Moreover, H acts transitively on its set of p-components.

Proof. Let K be a p-component of H, and set L = ⟨K^{H}⟩. Suppose that O^{p}(P̃) ≰ L. Then S ∩ L ⊴ P̃, and since P ∩ H_{0} = O_{p}(P ∩ H) we have also S ∩ L ⊴ P ∩ H. Thus S ∩ L ⊴ H, and \overline{L} is a p′-group.

As L ≤ H_{0} we have [L, H_{0}] ≰ Q_{H}, and so [L, O^{p}(P̃)] ≰ Q_{H}. Set R = O_{p}(O^{p}(P)). As R ≰ O_{p}(P̃), by 1.10, [L, R] is not a p-group. Then 1.4 shows that L = ⟨N_{L}(R)⟩P_{1} where P_{1} ∈ P_{LS}(S), and where P_{1} involves SL(2, q). As \overline{L} is a p′-group we then have q ≤ 3, |P_{1} : N_{P_{1}}(R)| ≤ 4, and \overline{P}_{1} ∩ \overline{L} is an r-group, where \{p, r\} = \{2, 3\}. Further, a Frattini Argument implies that \overline{S} normalizes a Sylow subgroup of \overline{L} for every prime divisor of |\overline{L}|, so 1.4 now yields |L : N_{L}(R)| ≤ 4. As no non-abelian simple group has a subgroup of index less than 5, we have a contradiction. Thus O^{p}(P̃) ≤ L, and then L = H_{0}. This yields the lemma. □

In the following lemma there is some non-standard notation. For a group X we write Z_{p}(X) for the complete pre-image in X of Z(X/O_{p}(X)).

Lemma 4.5. Suppose that H is non-solvable. Then the following hold.

(a) O^{p}(P̃) is a product of p-components of H.

(b) If K is any p-component of H then K/Z_{p}(K) ∈ Lie(p) or p = 2 and K/Z_{2}(K) is an alternating group Alt(2^{n} + 1), n ≥ 3.

Proof. As H is non-solvable, 4.3 implies that F_{p}^{∗}(H) ≠ F_{p}(H), and the preceding lemma then yields H = (P ∩ H)E_{p}(H), with O^{p}(P̃) ≤ E_{p}(H).

Set T = C_{S}(Y_{H}). Then T is a Sylow p-subgroup of T H_{0}, and T ≥ O_{p}(O^{p}(P))Q_{H}. Recall the notation given in section 0, and set r = r(T, Ṽ) and A = AF^{∗}(T, Ṽ). Further, set T_{0} = ⟨A⟩. Notice that, by 2.3, we have r ≤ 2. As T_{0} ≰ Q_{H}, we have T_{0} ≰ O_{p}(P̃), and therefore we may fix A ∈ A with A ≰ O_{p}(P̃). As F_{p}^{∗}(H_{0}) = E_{p}(H_{0}) we may choose a p-component K of H such that [\overline{K}, \overline{A}] ≠ 1. Then H_{0} = ⟨K^{H}⟩, by 4.4.

Set L = ⟨K, A⟩, and suppose first that K ≅ SL(2, p^{n}) for some n. Then (ii) holds, and we may assume that (i) is false. Then S ∩ E_{p}(H) ⊴ P̃. As C_{P ∩H}(Y_{H}) = O_{p}(P ∩ H), a Frattini Argument then shows that S ∩ E_{p}(H) ⊴ H, which is absurd. Thus, the lemma holds in this case, and we may assume henceforth that K is not isomorphic to SL(2, p^{n}). Then 3.4 implies that L = KA, K is invariant under ⟨A^{H}⟩, and there exists a subgroup B of L with |\overline{B}| > 2, such that \overline{B} acts faithfully on \overline{K} and quadratically on some non-trivial L-invariant section of Ṽ. Then 3.5 shows yields:

(1) We have \overline{K}/Z(\overline{K}) ∈ Lie(p), or else p = 2 and \overline{K}/Z(\overline{(}K)) is isomorphic to Alt(m) for some m, or to U_{4}(3), or a sporadic group.

Set X = O^{p}(P̃)(S ∩ E_{p}(H)). As O^{p}(P̃) ≤ ⟨A^{H}⟩, K is X- invariant, and indeed every p-component of H is X-invariant. We may therefore assume that K has been chosen so that S ∩ K ≰ O_{p}(X). Then O^{p}(P̃) ≤ [O^{p}(P̃), S ∩ K] ≤ K. Set X_{1} = (S ∩ K)O^{p}(P̃) and set N = N_{P ∩H}(K). As we have already seen, (ii) follows from (i) via 3.6, so we may assume that K ≠ X_{1}. Denote by M the set of maximal subgroups of K containing X_{1}, and let M ∈ M. As H_{0}Q_{H} = E_{p}(H) = ⟨X_{1}, P ∩ H⟩ we have \overline{K} = ⟨(\overline{M})\overline{N}⟩. Thus, \overline{K} has an automorphism ϕ which fixes a Sylow p-subgroup of K, and which moves a maximal subgroup of K containing that Sylow p-subgroup. In the case of the sporadic groups listed in (1), with p = 2, there is no such automorphism, as one may check using [Aschbacher].

Suppose that p = 2 and \overline{K}/Z(\overline{K}) ≅ U_{4}(3). Then \overline{M}/Z(\overline{K}) is of the form 2^{4} : Alt(6), and |N : N_{N}(\overline{M})| = 2. The two maximal subgroups of \overline{M} containing Z(\overline{K})(\overline{S} ∩ \overline{K}) are invariant under all automorphisms of \overline{K} which fix \overline{S} ∩\overline{K}, so we conclude in this case that X_{1} is N-invariant. But K = ⟨(X_{1})^{N}⟩, and so we have a contradiction in this case.

If p = 2 and \overline{K} is an alternating group Alt(m) then N ≤ S, and we again have K = ⟨(X_{1})^{N}⟩ = X_{1}, for a contradiction. Thus, we are now reduced to the case where \overline{K} is a group of Lie type in characteristic p. We shall say that a subgroup \overline{P}_{0} of \overline{K} is “parabolic” if P_{0} ≥ Z(K mod C_{H}(Ṽ)) and \overline{P}_{0}/Z(\overline{K}) is a parabolic subgroup of \overline{K}/Z(\overline{K}) in the ordinary sense. Thus, \overline{M} is a maximal parabolic subgroup of \overline{K}, and as \overline{M} is not N-invariant it follows that some element of N induces a non-trivial automorphism on the Coxeter diagram associated with \overline{K}. Let M_{0} be a parabolic subgroup of M which is minimal subject to containing \overline{X}_{1}. Then K = ⟨(M_{0})^{N}⟩. As P̃ ∈ P, it follows from inspection of the various Coxeter diagrams that \overline{M}_{0} is in fact a minimal parabolic subgroup of \overline{K}, that M_{0} = M is uniquely determined, and that \overline{K}/Z(\overline{K}) is of Lie rank 2. Then S normalizes M_{0}, so p is odd, and \overline{K}/Z(\overline{K}) is isomorphic to L_{3}(p^{n}) or G_{2}(3^{n}) for some n. Moreover, we have |N : N_{N}(M)| = 2.

Set D = N_{H}(S ∩ E_{p}(H)), and suppose next that D ≤ N_{H}(O^{p}(P)). Let N_{1} be a Hall p′-subgroup of N. Then [D∩K, N_{1}] ≤ P ∩K ≤ O_{p}(P ∩H)∩K, and thus \overline{N}_{1} acts trivially on the group \overline{I} = (\overline{D} ∩\overline{K})/(\overline{S} ∩\overline{K}). But in fact, any automorphism of \overline{K} which acts nontrivially on the diagram for \overline{K} also acts non-trivially on \overline{I}, so we have a contradiction in this case. Thus D ≰ N_{H}(O^{p}(P)). Then 1.4 shows that there exists P_{1} ∈ P_{DS}(S) such that ⟨P, P_{1}⟩ ∈ L, with P_{1} ≰ N_{H}(O^{p}(P_{1})). Set J = ⟨P, P_{1}⟩. By 1.3(4) we have J/C_{J}(Y_{J}) ≅ SL(3, q) or Sp(4, q), and since D is solvable, by 1.11(b), we conclude that q = 3, P_{1}/O_{3}(P_{1}) ≅ SL(2, 3), and |P ∩H : S| = 2. As O^{3}(P̃) ≤ K, K is S-invariant, and we may now conclude that N = P ∩ H and that Q_{H}K = E_{p}(H). Then E(\overline{H})\overline{S}/E(\overline{H}) may be identified with a subgroup of Out(\overline{K}) covered by field automorphisms. It follows that [\overline{I}, \overline{S}] ≤ \overline{S} ∩ \overline{K}. As [\overline{P}_{1}, \overline{S}] is not a p-group, we have a final contradiction at this point, proving 4.5. □

Lemma 4.6. There is then a unique maximal subgroup of H containing P ∩ H.

Proof. Immediate from 4.5 and 4.3. □

For the remainder of this section, denote by M the unique maximal subgroup of H containing P ∩ H, and set N = N_{H}(Y_{P}). Also set R = O_{p}(O^{p}(P))Q_{H}, set r = r(R, Ṽ), and fix A ∈ A(R, Ṽ).

Lemma 4.7. Let K be a p-component of H. Then \overline{K}/Z(\overline{K}) is of Lie type in characteristic p. Moreover, the Lie rank of K is at most 2, and if equal to 2 then p = 2 and \overline{K}/Z(\overline{K}) ≅ L_{3}(2^{n}) or Sp(4, 2^{n}) for some n.

Proof. Evidently N ≤ M. Suppose first that this inclusion is proper. Set M_{0} = O^{p}(M ∩ H_{0}). Then M = (P ∩ H)M_{0}, and so M_{0} ≰ N_{G}(Y_{P}). As M_{0}S is a proper subgroup of H, we have ⟨P, P_{1}⟩ ∈ L for every P_{1} ∈ P_{M_{0}S}(S), and the P_{1}!-Theorem then says that there is a unique P_{1} ∈ P_{M_{0}S}(S) such that P_{1} ≰ N_{G}(Y_{P}), and moreover, we have O^{p}(P_{1}/O_{p}(P_{1})) ≅ SL(2, q)′. Suppose now that a p-component K of H is of Lie type in characteristic p. Then M_{0}/O_{p}(M_{0}) is abelian, and hence q = 2, P ∩ H = S, H = ⟨P̃, P ∩H⟩ = P̃, and the lemma holds. In view of —- we may therefore assume that p = 2 and that \overline{K} ≅ Alt(2^{n} + 1) for any 2-component K of H. If n ≥ 4 then M_{0}S is generated by the elements of P_{M0S}(S) − {P_{1}}, by —-, and M = N in that case. Thus, we have n = 3, and \overline{M}_{0} ≅ Alt(8) ≅ L_{4}(2). Again, P_{1}! implies that q = 2 and then that H = P̃. Then S acts transitively on the set of 2-components of H, and since P_{1}/O_{2}(P_{1}) ≅ L_{2}(2) it follows that there is just one 2-component in H.

Suppose that a^{∗} < 2. Then —- shows that [Ṽ, K] is the irreducible permutation module for \overline{K}. Then C_{[Ṽ ,K]}(S ∩ K) = C_{[Ṽ ,K]}(M_{0}), and this group is of order 2. But

also Ṽ = ⟨(Ỹ_{P})^{K}⟩, where |Ỹ_{P} | = 2, so it follows that Y_{P} is M_{0}-invariant in this case. We conclude from —- that a^{∗} = 2, b = 3, and A may be chosen so that \overline{A} is normal in \overline{N}. Here \overline{N} ∩ \overline{M}_{0} is a maximal parabolic subgroup \overline{M}_{1} of \overline{M}_{0}, and inspection of these then shows that \overline{A} = O_{2}(\overline{M}_{1}) (of order 8 or 16). Now — shows that there is no 𝔽_{2}-module for \overline{K} on which \overline{A} is a quadratic F 2-offender. By this contradiction, we may assume henceforth that M = N.

Then —- shows that p = 2, \overline{K} is isomorphic to Alt(2^{n} + 1) for some n, n ≥ 3, and we have M = N. But one may observe that O_{2}(M ∩ H_{0}) = O_{2}(H), and hence C_{S}(Z_{0}) = O_{2}(H). But this means that S ∩ H_{0} = O_{2}(H), which is evidently untrue. □

Lemma 4.8. Assume that H is non-solvable. Then there are normal subgroups L_{i} of H_{0} and subgroups V_{i} of V, 1 ≤ i ≤ s, such that

(1) H_{0} = L_{1} ⋯ L_{s},

(2) [L_{i}, L_{j}] ≤ Q_{H} for all i ≠ j,

(3) V_{i} = [V, L_{i}], and [Ṽ_{i}, L_{j}] = 1 for all i ≠ j,

(4) Set L = L_{1} and set W = V_{1}/C_{V_{1}}(L). Assume that the indexing has been chosen so that [\overline{L}, \overline{A}] ≠ 1, let A_{0} be a complement in A to C_{A}(\overline{L}), and set r_{0} = r(\overline{A}_{0}, W). Then one of the following holds.

(a) \overline{L} is isomorphic to SL(2, p^{n}), and W is a natural module for \overline{L}.

(b) r_{0} = 2, \overline{L} ≅ SU(3, p^{n}) or Sz(p^{n}), and W is a natural module for \overline{L}.

(c) r_{0} = 2, \overline{L} ≅ SL(2, p^{n}), and W involves two non-trivial constituents for the action of \overline{L}, each of which is a natural module.

(d) r_{0} = p = 2, \overline{L} ≅ Sp(4, 2^{n})′ and W is the direct sum of a natural and a contragredient module for \overline{L}.

(e) p = 2, \overline{L} ≅ SL(3, 2^{n}) and W is the direct sum of a natural and a dual module for \overline{L}.

(f) \overline{L} ≅ Ω_{4}^{ϵ}(p^{n}) and W is a natural orthogonal module for \overline{L}.

(5) Ṽ = Ṽ_{1} ⋯ Ṽ_{r}CṼ (H_{0}A).

Proof. Choose K so that [\overline{K}, \overline{A}] ≠ 1, and with K ≤ P̃. Set L = [K, A] and W = [V, L]. Then 0.1 says that C_{A}(\overline{L}) = C_{A}(W̃) and that r(\overline{A}_{0}, W̃) ≤ r(\overline{A}, Ṽ) for any complement A_{0} to C_{A}(\overline{K}) in A. In particular, we have r(\overline{A}_{0}, W̃) ≤ 2. Fix an irreducible K-submodule Û of Ŵ, and denote by X the product of all of the p-components of H which are not contained in L.

Suppose first that L ≠ K, and let a ∈ A_{0} with K ≠ K^{a}. Then Û ∩ (Û)^{a} = 1, and so, in particular, we have |W̃/CW̃(a)| > 4. Then |\overline{A}_{0}| ≠ 2, and we may appeal to Theorem 3 of [P_{G}(V)-paper] for the structure of \overline{L}\overline{A}_{0} and of W̃. Thus, p = 2, \overline{L} ≅ Ω_{4}^{+}(2^{n}) for some n, and W̃ is a direct sum of natural orthogonal modules for \overline{L}. As r(\overline{A}_{0}, W̃) ≤ 2, one easily determines that η(L, W̃) = 1, and thus [W̃, X] = 1.

Suppose next that L = K, and set K^{∗} = N_{S}(K)K. Suppose that η(K^{∗}, W̃) > 1. Then 3.— says that \overline{K} ≅ SL(2, p^{n}), that η(\overline{K}, W̃) = 2, and that the K-irreducible constituents of Ŵ are natural SL(2, p^{n})-modules for \overline{K}. It follows from —– that there exists at most one p-component K_{1} of X such that [W̃, K_{1}] ≠ 1, and that if such a p-component K_{1} exists, then \overline{K}\overline{K}_{1} ≅ Ω_{4}^{+}(p^{n}) and W̃ is a natural orthogonal module for \overline{K}\overline{K}_{1}.

If η(K^{∗}, W̃) = 1 then also η(K, W̃) = 1 and [W̃, X] = 1. With 3.— , the lemma now follows □ Lemma 4.9. Suppose that H is non-solvable, and suppose that M ≠ N. Set U = ⟨(Y_{P})^{M}⟩. Then q = 2, H = P̃, |Y_{P} | = 4, and |U| = 8.

Proof. As M ≠ N, and since M = (M ∩ H_{0})(P ∩ H), it follows from 1.4 that there exists P_{1} ∈ P_{(McapH_{0})S}(S) such that ⟨P, P_{1}⟩ ∈ L, and such that O^{p}(P_{1}/O_{p}(P_{1})) is isomorphic to the commutator subgroup of SL(2, q). On the other hand, (M ∩ H_{0})/O_{p}(M ∩ H_{0}) is abelian, by inspection of the set of outcomes in 4.8(4). Thus q = 2, P ∩ H = S, and H = P̃. Set J = ⟨P, P_{1}⟩. Then 1.3(4) says that J/C_{J}(Y_{J}) is isomorphic to L_{3}(2) or Sp(4, 2), and that Y_{J} is a natural module for J/C_{J}(Y_{J}). Then |U| = 8. □

Lemma 4.10. Assume that H is non-solvable, let K be a p-component of H, and set V̂ = V/C_{V} (H_{0}). Then one of the following holds.

(i) K is the unique p-component of H, M = N, \overline{K} ≅ SL(2, q) or Sz(q), and V̂ is a natural module for \overline{K}. Moreover, if \overline{K} ≅ Sz(q) then b = 3.

(ii) q = 2 and \overline{K} ≅ L_{3}(2) or Alt(6).

(iii) q = 2, M ≠ N, \overline{H} ≅ Sym(5), and V̂ is a natural ΓL(2, 4)-module for \overline{H}.

(iv) q = 2, M = N, b = 3, \overline{H} ≅ Alt(5) or Sym(5), and V̂ is an Ω_{4}^{−}(2)-module for \overline{H}.

Proof. Set U = ⟨(Y_{P})^{M}⟩ and set M_{i} = O^{p}(M ∩ L_{i}), 1 ≤ i ≤ s. If M_{1} = 1 then it follows from 4.8(4) that \overline{L} ≅ L_{3}(2) or Alt(6), so that outcome (ii) of the lemma holds in this case. We may therefore assume that M_{1} ≠ 1.

Suppose first that [U, M_{1}] = 1. Then [U, O^{p}(M ∩ H_{0})] = 1, and since M = (M ∩ H_{0})(P ∩ H) it follows that Y_{P} is M-invariant. Further, as CṼ (M ∩ H_{0}) ≠ CṼ (H_{0}), and since M_{1} ≠ 1, it follows that 4.8(4)(f) holds with p^{n} = 2 and with ϵ = −1. Here R is M-invariant, by 1.7, so we have A ≤ H_{0}Q_{H}, and then r_{0} = 2. As r_{0} ≤ r, by 0.1, 2.3 now implies that b = 3. Thus (iv) holds, and we may assume henceforth that [U, M_{1}] ≠ 1.

Suppose next that we have s > 1, and then suppose further that there exists an element z in ([U, M_{1}] ∩ Y_{P}) − Y_{H}. Then Q_{P} ∩ Q_{H} = C_{Q_{H}} (z) is invariant under ⟨P ∩ H, L_{2}⟩ = H, contrary to 1.9. Thus [Ũ, M_{1}] ∩ Ỹ_{P} = 1. This yields U ≠ Y_{P}, and then 4.9 yields |Ũ| = 4. Since [Ũ, M_{1}] ≠ 1 we obtain Ũ ≤ V_{1}, contrary to s > 1. We have therefore shown that s = 1.

As P ∩H acts irreducibly on Ỹ_{P}, and as Y_{P} is not normal in H, we have C_{Y_{P}} (O^{p}(H)) = Y_{H}. Thus |Ŷ_{H}| = |Ỹ_{H}| = q. Further, we have [Y_{P} , S ∩L] ≤ Y_{H}, since S ∩L ≤ O_{p}(P ∩H).

Suppose that \overline{L} ≅ SL(2, p^{n}) and that V̂ is a natural module for \overline{L}. Supposing that M = N, we obtain q = p^{n}, and so (i) holds. On the other hand, suppose that M ≠ N. Then q = 2, H = P̃, |Ũ| = 4 and \overline{L} ≅ SL(2, 4), by 4.9. Further, there exists P_{1} ∈ P_{M}(S), and therefore \overline{H} ≅ Sym(5). Thus (iii) holds in this case, and we may assume henceforth that the pair (\overline{L}, V̂) does not consist of SL(2, p^{n}) and a natural module. In particular, b is then odd, as follows from 2.1 and 3.9.

Fix a critical path (α, β, ⋯, α′), and take H = β and P = β + 1. As b is odd, 1.15(b) yields Y_{β} ∩ Y_{α′} = 1. Setting V_{0} = V ∩ Q_{α′}, and taking A = V_{α′}, we then have [V_{0}, A ∩ Q_{H}] = 1.

Suppose next that |Y_{α}/(Y_{α}∩Q_{α′})| < q. Then A∩Q_{β} centralizes Y_{α}. As M is the unique maximal subgroup of H containing P ∩ H, we may choose a vertex δ ∈ Δ(α′) so that ⟨Y_{α}, G_{α′} ∩ G_{δ}⟩ = G_{α′}. Suppose further that |Y_{δ}/(Y_{δ} ∩ Q_{β})| < q. Then C_{Q_{α′}} (Y_{δ} ∩ Q_{β}) = Q_{α′} ∩ Q_{δ} which is then normal in G_{α′}, contrary to 1.9. By symmetry, we may then assume:

(1) |Y_{α}/(Y_{α} ∩ Q_{α′})| ≥ q.

Suppose next that r = 2. Then 2.3 implies b = 3, and 1.16 implies that \overline{A} is normal in \overline{P} ∩ \overline{H}. Suppose that \overline{L} ≅ SU(3, p^{n}), Sz(2^{n}), or SL(2, p^{n}), as in 4.8(4b) or 4.8(4c). The quadratic action of V_{β} on V_{α′}/Y_{α′} then yields |V/V_{0}| ≤ p^{n}, and then q ≤ p^{n} by (1). Here [V_{0}, A] ≤ Y_{α′} ≤ Y_{P}. The condition q ≤ p^{n} then eliminates the possibility that Ŵ is a 3-dimensional unitary module over 𝔽_{p^{2n}}, and shows that |Ŷ_{P} | = p^{n}. If \overline{H}_{0} ≅ Sz(2^{n}) we then have outcome (i) of the lemma. So assume that \overline{L} ≅ SL(2, p^{n}) and (as stated in 4.8(4c)) that V̂ involves two natural modules for \overline{L}. Then |[V̂, \overline{A}]| ≥ p^{2n}, and thus [V, A] ≰ Y_{P}. Then also V ∩ A ≰ Y_{P}. On the other hand, we have [V ∩ A, Q_{P}] = [V ∩ A, (Q_{P} ∩ Q_{H})A] ≤ [V, Q_{H}] = Y_{H}, and by symmetry we have also [V ∩ A, Q_{P}] ≤ Y_{α′}. As Y_{β} ∩ Y_{α′} = 1, we have thus shown that V ∩ A ≤ Ω_{1}(Z(Q_{P})), which is to say that V ∩ A ≤ Y_{P}. This contradiction eliminates the case 4.8(4c).

Suppose next that \overline{L} is an orthogonal group Ω_{4}^{ϵ}(p^{n}), and that V/C_{V} (H_{0}) is a natural orthogonal module. Further, assume that \overline{A} ≤ \overline{L}. Then r = 2 and, as we have seen, b = 3 and A is normal in P ∩ H. Quadratic action yields |\overline{A}| ≤ p^{n}. Further, we have |CṼ (\overline{S} ∩ \overline{L})| = p^{n}, and so q ≤ p^{n}. Set W = Y_{α′}/C_{Y_{α′}} (O^{p}(G_{α′})), and denote by W_{0} the image of A ∩ Q_{H} in W. Then |W/W_{0}| ≤ p^{n} and |[W_{0}, V]| ≤ q. It follows that q = p^{n} and that |\overline{A}| = q. Suppose q = 2. As H is non-solvable we then have ϵ = −1 and \overline{L} ≅ Alt(5). Here Ỹ_{P} centralizes \overline{M}, so 1.7 yields M = N, and we have thus returned to outcome (iv) of the lemma. We may therefore assume that q > 2, and then M = N by 4.9. Observe that there exists a subgroup X of C_{M}(Y_{P}) with |\overline{X}| = q − ϵ. It follows that [O^{p}(P), X] ≤ Q_{P}. Since A = V^{g} for some g ∈ O^{p}(P), A is then X-invariant. It follows that ϵ = +1.

As P̃ contains a p-component of H, by 4.5(a), there exists an element f of P ∩ H such that f interchanges the two p-components of H. If p is odd then 1.16 shows that f may be chosen in G_{β} ∩ G_{α′}, so that f normalizes A. But for p odd the only quadratic subgroups of \overline{L} are contained in components of \overline{L}, so we now conclude that p = 2. Then f may be chosen to induce an 𝔽_{q}-transvection on V̂, centralizing Ŷ_{P} in particular. Then f ∈ O_{2}(P ∩ H) = Q_{P} Q_{H}, so we may take f ∈ Q_{P}. Then f normalizes A, and indeed [A, f] ≤ [A, Q_{P}] ≤ A ∩ Q_{H}, since ’ [W, Q_{P}] = [W, O_{2}(G_{γ} ∩ G_{α′})] is the unique 3-dimensional 𝔽_{q}-subspace of W which is invariant under Q_{P}. Thus [\overline{A}, \overline{f}] = 1.

Let D be a subgroup of P ∩ H of order q − 1. Then D acts regularly on Ŷ_{P}, and hence \overline{D} acts as a group of inner automorphisms of \overline{L}. By 1.16 we may take D to normalize A. We recall however that \overline{A} is also invariant under a subgroup \overline{X} of \overline{M} of order q − 1, such that \overline{X} centralizes Ŷ_{P}. It follows that \overline{A} is normal in \overline{M}. But the only proper \overline{M}-invariant subgroups of \overline{S} ∩ \overline{L} are contained in components of \overline{L}, so we have a contradiction at this point.

We are reduced to the case where \overline{A} ≰ \overline{L}. Then [\overline{M}, \overline{A}] is not a p-group. We recall however that A ≤ R and that N normalizes R. Thus M ≠ N, and 4.9 yields q = 2 and |U| = 8. As Y_{P} is not normal in M we then have Û ≠ Ŷ_{P}, and so |Û| = 4. This implies that 2^{n} = 4. Also, we note that [Y_{P} , A] = 1, \overline{A} acts as a group of 𝔽_{4}-linear automorphisms of V̂, and hence |\overline{A} : \overline{A} ∩ \overline{H}| = 2. Then also |\overline{A}| ≤ 8. By symmetry we have also |V/V_{0}| ≤ 8

Suppose that |\overline{A}| > 2, and let \overline{a} be a non-identity element of \overline{A}∩\overline{L}. Then |CV̂ (\overline{a})| = 16, and since [V_{0}, A] ≤ Y_{α′}, of order 2, it follows that |V/V_{0}| ≥ 8. Again by symmetry, we conclude that |\overline{A}| = 8 and that |\overline{A} ∩ \overline{L}| = 4 But CV̂ (\overline{a}) = CV̂ (\overline{b}) for any non-identity element \overline{b} of \overline{A}∩\overline{L}, and this is inconsistent with [V̂_{0}, \overline{A}] being of order at most 2, and with |V̂/V̂_{0}| < 16. We therefore conclude that |\overline{A}| = 2, and also, by symmetry, |V/V_{0}| = 2. Then \overline{A} induces an 𝔽_{4}-transvection on V̂, and similarly for V on W. As V_{0} and A∩Q_{H} are index-2 subgroups of V and A, respectively, we obtain [V_{0}, A] = Y_{α′} and [V, A∩Q_{H}] = Y_{H}. Now 2.3 implies that b = 3.

As \overline{A} is 𝔽_{4}-linear on V̂, and as [V, A] ≥ Y_{H}, we have |[V, A]| ≥ 8. Now [V̂, \overline{Q}_{P}] is the unique Q_{P} -invariant 3-dimensional subspace of V̂, and is therefore equal to CV̂ (\overline{A}). Thus [V, Q_{P} , A] ≤ C_{V} (O^{p}(H)). But also [V, Q_{P} , A] ≤ [V_{0}, A] ≤ Y_{α′}, and since Y_{P} = Y_{H}Y_{α′} we have [Y_{α′}, O^{p}(H)] ≠ 1. Thus [V, Q_{P} , A] = 1, and symmetry yields also [A, Q_{P} , V] = 1. Then [A, V] ≤ Ω_{1}(Z(Q_{P})), by the Three Subgroups Lemma, and so [A, V] ≤ Y_{P}. As |[A, V]| > 4 we have a contradiction, and have thereby succeeded in ruling out the case given by 4.8(4f).

It remains now to consider the cases where p = 2 and \overline{L} ≅ SL(3, 2^{n}) or Sp(4, 2^{n}), with n > 1. Here (1) implies that |Ỹ_{P} | > 2 and so M = N. Thus Ỹ_{P} is invariant under a Borel subgroup of \overline{L}, as well as being invariant under an element of S which induces a diagram automorphism on \overline{L}. It follows that Ỹ_{P} covers CV̂ (\overline{S}∩\overline{L}), and so |Ŷ_{P} | = 2^{2n}. But O^{2}(P ∩H) acts transitively on Ŷ_{P} while preserving any non-trivial irreducible \overline{L}-invariant section of V̂. This is contrary to η(\overline{L}, V) = 2, so we have succeeded in eliminating cases (4d) and (4e) of 4.8. The proof of 4.10 is thereby complete. □

Section 5: Two special cases

Our goal in this section is to show that outcomes (ii) and (iv) in lemma 4.10 can not occur.

Lemma 5.1. Suppose that q = 2, and let K be a 2-component of H. Then \overline{K} is not isomorphic to L_{3}(2) or to Alt(6).

Proof. Suppose false. Choose a critical path (α, β, ⋯, α′) in Γ, and take H = β and P = β + 1. Notice that Ṽ is not an F 1-module for H, as follows from 3.7, and then 2.1 implies that b is odd. Note also that we have G_{β} ∩ G_{β+1} = P ∩ H = S, and hence H = P̃.

Denote by K the set of 2-components of H, and write K = {K_{1}, ⋯, K_{s}}. For any i, 1 ≤ i ≤ s, set V_{i} = [V, K_{i}]. Then 4.8 shows that we have Ṽ = CṼ (H_{0}) × Ṽ_{1} × ⋯ × Ṽ_{r}, and Ṽ_{i} is the direct sum of two non-isomorphic modules for K_{i}, each of dimension 3 if \overline{K} ≅ L_{3}(2), or 4 if \overline{K} ≅ Alt(6). We recall also that N_{S}(K_{i}) contains an element which induces a diagram automorphism on \overline{K}_{i} (viewed as a group of Lie type in characteristic 2) and which interchanges the two irreducible submodules of Ṽ_{i}.

Suppose first that we have b ≥ 5. By 2.3 we may then assume that V_{α′} is a quadratic F 2^{∗}-offender on Ṽ_{β}, and further that Y_{β} ≰ V_{α′}, while Y_{α′} ≤ V_{β}. In particular, 4.8 and 0.1 show that \overline{K}_{i} ≅ L_{3}(2) for all i.

Fix a 2-component K of H such that [\overline{K}, V_{α′}] ≠ 1, and set V_{K} = [V, K]. As V_{α′} acts quadratically on Ṽ_{β} we have |V_{α′}/C_{V_{α′}} (\overline{K})| ≤ 4. As Y_{β} ≰ V_{α′} we have also [V_{K}, C_{V_{α′}} (\overline{K})] = 1. Thus |Ṽ_{α′}/CṼ_{α′}(V_{K})| ≤ 4. But also |[Ṽ_{K}, V_{α′}] ≥ 4 as η(K, V_{K}) = 2,

and so V_{K} ≰ Q_{α′}. It now follows that |V_{K}/(V_{K} ∩Q_{α′})| = 2 and that |Ṽ_{α′}/CṼ_{α′}(V_{K})| = 4. Further, we have \overline{V_{α′}} ≤ C\overline{H}(\overline{K})\overline{K}.

Let Ũ_{1} and Ũ_{2} be the two K-submodules of Ṽ_{K} of order 8, with the indexing chosen so that V_{α′} projects modulo C\overline{H}(\overline{K}) to the unipotent radical of the stabilizer of a point in Ũ_{1} and of a line in Ũ_{2}. Set U_{0} = V_{K} ∩ Q_{α′}. As we have seen, Ũ_{0} is a hyperplane of Ṽ_{K}, and [Ũ_{0}, V_{α′}] = Ỹ_{α′}, of order 2. Then Ũ_{0} = Ũ_{1} × [Ũ_{2}, V_{α′}], and Ỹ_{α′} ≤ U_{1}. Now K is uniquely determined among the 2-components of H by the condition that Y_{α′} lie inside V_{K}. Thus [\overline{H}_{0}, V_{α′}] = \overline{K}, and \overline{K} ≥ \overline{V_{α′}}. In particular, we now have |V_{α′}/C_{V_{α′}} (V_{β})| = 4. Then also |V_{β}/(V_{β} ∩ Q_{α′})| = 2, and [O^{2}(G_{α′}), V_{β}] contains a unique 2-component L of G_{α′}.

Set X_{β} = ⟨(W_{α})^{G_{β}}⟩. As b ≥ 5 we have [W_{α}, X_{β}, W_{α}] ≤ [W_{α}, W_{α}] = 1. Also [V_{β}, X_{β}] = 1 and so, recalling that Y_{α′} ≤ V_{β}, we have [Y_{α′}, X_{β}] = 1. By the definition of b, we have X_{β} ≤ G_{α′−2}, and X_{β} centralizes Y_{α′−2}Y_{α′} = Y_{α′−1}. Since P ∩ H = S is maximal in H, it now follows that X_{β} ≤ Q_{α′−1}, and so X_{β} ≤ G_{α′}. Then L is X_{β}-invariant. Set \overline{G_{α′}} = G_{α′}/C_{G_{α′}} (Ṽ_{α′}). As [W_{α}, X_{β}, W_{α}] = 1, where also Φ(W_{α}) = 1, it follows that either

(i) |W_{α}/C_{W_{α}}(\overline{L})| = 2, or

(ii) |X_{β}/C_{X_{β}}(\overline{L})| ≤ 8.

Set V_{L} = [V_{α′}, L], and suppose that (i) holds. Then W_{α} = C_{W_{α}}(\overline{L})V_{β}, and hence

[V_{L}, W_{α}] ≤ Y_{α′}[V_{L}, V_{β}] ≤ V_{β} ≤ W_{α}.

and hence W_{α} ⊴ ⟨G_{α}, V_{L}⟩. But V_{K} = (V_{K} ∩ Q_{α′})Y_{α}, so |[Y_{α}, V_{L}]| ≥ 4, and thus V_{L} ≰ G_{α}. Then ⟨G_{α}, V_{L}⟩ = ⟨G_{α}, G_{β}⟩, contrary to O_{2}(⟨P, H⟩) = 1. Thus (ii) holds. Then |X_{β}/C_{X_{β}}(\overline{L})V_{β}| ≤ 4. But [C_{X_{β}}(\overline{L})V_{β}, V_{L}] ≤ V_{β}, so X_{β}/V_{β} is an F 1-module for G_{β} via the action of V_{L}. (Indeed, we have |V_{L}/C_{V L}(V_{β})| = 4.) As \overline{G_{β}} has no F 1-modules, we conclude that [X_{β}, O^{2}(G_{β})] ≤ V_{β}, so that W_{α} ⊴ G_{β}, again contrary to O_{2}(⟨P, H⟩) = 1. This proves:

(1) b = 3.

Set γ = β+1. By 1.16 we have Q_{β}V_{α′} ⊴ G_{β}∩G_{γ}, and quadratic F 2-action then implies that \overline{V_{α′}} = Z(\overline{S} ∩\overline{H}_{0}). As b = 3, 1.17 implies that [V_{β}, V_{α′}] ≤ D_{γ} and that [D_{γ}, G_{γ}] ≤ Y_{γ}, where we recall from section 1 that D_{γ} denotes ∩\{V_{δ}\}_{δ∈Δ(γ)}. Thus [Ṽ, Z(\overline{S} ∩ \overline{H}_{0}), \overline{S}] is of order at most 2. As S acts transitively on the set K of 2-components of H, it follows that |K| = 1.

Write V_{0} for V_{β} ∩ Q_{α′} and set V̂ = V/C_{V} (H_{0}). Suppose that \overline{H}_{0} ≅ Alt(6). Then |V/V_{0}| = 4 and |[V_{0}, V_{α′}]| ≤ 2. Further, we have V_{0} ⊴ Q_{β}(G_{β} ∩ G′_{α}) = S, by 1.16, which implies that V̂_{0} intersects each of the two irreducible H_{0}-submodules of V̂ in a hyperplane. But then |[V̂_{0}, V_{α′}]| > 2, and we have a contradiction. As \overline{V_{α′}} ⊴ \overline{S} we conclude:

(2) \overline{H} ≅ Aut(L_{3}(2)) and, setting A = V_{α′}, we have \overline{A} = Z(\overline{S}).

Observe that [V, A] ≥ Y_{β}Y_{α′} = Y_{γ} = Y_{P}. As V = ⟨(Y_{P})^{H_{0}}⟩ and as \overline{A} ≤ \overline{H}_{0}, it follows that Ṽ = [Ṽ, H_{0}]. That is:

(3) C_{V} (H_{0}) = Y_{H}.

Choose λ ∈ Δ(α) − {β}, and set Q_{α}^{∗} = (Q_{λ} ∩ Q_{α})(Q_{α} ∩ Q_{β}). Then Q_{α}^{∗} ⊴ G_{α}, and [O_{2}(G_{α}), O^{2}(G_{α})] ≤ Q_{α}^{∗}. As usual, set D_{α} = ∩{V_{δ}}_{δ∈Δ(α)}. We then claim:

(4) [D_{α}, G_{α}] = Y_{α}.

Indeed, setting X = ⟨(Q_{β})^{G_{α}}⟩, we have [D_{α}, X] = Y_{α} and Q_{α}^{∗} ≤ X. There is a natural embedding of Q_{α}/C_{Q_{α}}(D_{α}) into Hom(D_{α}/Y_{α}, Y_{α}), and so Q_{α}/C_{Q_{α}}(D_{α}) is isomorphic to a direct sum of natural modules for G_{α}/Q_{α}. Thus Q_{α} = C_{Q_{α}}(D_{α})[Q_{α}, O^{2}(G_{α})] = C_{Q_{α}}(D_{α})Q_{α}^{∗}, and this proves (4).

Notice that Ỹ_{P} = CṼ (S). Notice further that C_{Ṽ /Ỹ_{P}} (S) = CṼ (S ∩ H_{0})/Ỹ_{P}. As D_{α}/Y_{P} centralizes S, by (4), we then have:

(5) |D_{α}/Y_{α}| ≤ 2.

Next, since Y_{α} does not induce a transvection on Ṽ_{α′} we obtain |V_{α′}/C_{V_{α′}} (Y_{α})| ≥ 4. We may then choose an element t ∈ V_{α′} ∩ Q_{β} − Q_{α}. Now C_{V_{λ}}(t)Y_{α} ⊴ ⟨Q_{λ}, t⟩, and ⟨Q_{λ}, t⟩ ≥ O^{2}(G_{α}). Therefore C_{V_{λ}}(t)Y_{α} ≤ V_{λ}∩V_{β}. Further, we have V_{λ}∩V_{β} ⊴ O^{2}(G_{α})Q_{β}, so that V_{λ} ∩ V_{β} = D_{α}. With (5) we then have:

(6) |C_{V_{λ}}(t)Y_{α}| ≤ 8, and C_{V_{λ}}(t)Y_{α} ≤ D_{α}.

On the other hand we have |V_{λ}/(V_{λ} ∩ G_{α′})| ≤ 4, so V_{λ}/(V_{λ} ∩ Q_{α′})| ≤ 16, and so |V_{λ}/C_{V_{λ}}(t)| ≤ 32. As Y_{α} ≰ C_{G}(t) we then have |V_{λ}/C_{V_{λ}}(t)Y_{α}| ≤ 16, and as |V_{λ}| = 2^{7} we conclude from (6) that |C_{V_{λ}}(t)Y_{α}| = 8, that C_{V_{λ}}(t)Y_{α} = D_{α}, and that equality holds in each step in achieving this calculation. In particular, we have [V_{λ} ∩ Q_{α′}, t] ≠ 1, and so

(7) [V_{λ} ∩ Q_{α′}, t] = Y_{α′}.

Similarly, we have V_{λ} ∩ G_{γ} ≰ Q_{γ}. Choose t′ ∈ (V_{λ} ∩ G_{γ}) − Q_{γ}. By symmetry, we have C_{V_{α′}} (t′)Y_{γ} = D_{γ}, and [V_{α′} ∩ Q_{λ}, t′] = Y_{λ}. Now

V_{α′} ∩ Q_{λ} ≤ V_{α′} ∩ G_{λ} ≤ C_{V_{α′}} (Y_{λ}) ≤ [V_{α′}, G_{γ} ∩ G_{α′}]
= [V_{α′}, Q_{γ}] ≤ [W_{γ}, Q_{γ}].

As t′ ∈ G_{γ} we then have [t′, V_{α′} ∩ Q_{λ}] ≤ [W_{γ}, Q_{γ}], and so Y_{λ} ≤ [W_{γ}, Q_{γ}].

For any δ ∈ Δ(γ) we have V_{δ}Q_{α′}/Q_{α′} ≤ Z((G_{δ} ∩ G_{α′})/Q_{α′}), and so W_{γ}Q_{α′}/Q_{α′} ≤ Z((G_{δ} ∩ G_{α′})/Q_{α′}). Then [W_{γ}, Q_{γ}] ≤ Q_{α′}, and so Y_{α} ≤ Q_{α′}. As (α, α′) is a critical pair, we have a final contradiction, which then completes the proof of 5.1. □

Lemma 5.2. [Outcome (iv) of lemma 4.10 is out]

Proof. To be provided. (It is in fact very easy, and very short. Maybe set it inside 4.10.) □

Section 6: The Rudvalis Case

We begin by warning the reader that we shall often write Z for Y_{H}.

This section is concerned with outcome (iii) of lemma 4.10. We begin with two lemmas on modules for L_{3}(2).

Lemma 6.1. Let V be a module for L over 𝔽_{2}, where L ≅ L_{3}(2). Assume that there exists a parabolic subgroup P of L and an element x of V such that V = ⟨x^{L}⟩ and dim(⟨x^{P} ⟩) = 2. Then η(L, V) ≤ 2.

Proof. Set S = C_{P} (x). Then S is a Sylow 2-subgroup of L, and so S is contained in a (proper) parabolic subgroup P_{1} of L, P_{1} ≠ P. Set U = ⟨x^{P} ⟩ and set U_{1} = ⟨x^{P_{1}}⟩. Let W be a maximal L-submodule of V.

Observe that dim(U_{1}) ≤ 3. Suppose first that U_{1} = ⟨x⟩. Then |x^{L}| = 7 and so dim(V) ≤ 7. As any non-trivial L-module has dimension at least 3, we have η(L, V) ≤ 2 in this case. Suppose next that dim(U_{1}) > 1, and let y be an element (possibly 0) of C_{U_{1}}(P_{1}). If y ∉ W for any choice of W then V = ⟨y^{L}⟩ and, again, we obtain dim(V) ≤ 7 and η(L, V) ≤ 2. As dim(C_{U_{1}}(P_{1})) ≤ 1 we may therefore take C_{U_{1}}(P_{1}) ≤ W. Setting \overline{V} = V/W, barV is then an irreducible L-module, with dim(⟨\overline{x}^{P} ⟩) = dim(⟨\overline{x}^{P_{1}}⟩) = 2. There are only four isomorphism classes of irreducible 𝔽_{2}L-modules, and the given conditions identify \overline{V} as the adjoint module (which is also the Steinberg module) for L. In particular \overline{V} is projective, so W = 0, and η(L, V) = 1. □

Lemma 6.2. Take L = L_{3}(2) and let W = sl(3, 2) be the 𝔽_{2}L-module consisting of three-by-three matrices of trace zero. Then L has exactly six orbits on the set of non-zero vectors of W. We may denote these orbits by O_{n}, n = 2, 3, 4, 6, 7, 8, where n is the cardinality of C_{L}(w_{n}), w_{n} a representative of O_{n}. The elements w_{n} may be taken as follows.

w_{2} = ( \begin{matrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix} )  w_{3} = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix} )  w_{4} = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} )

w_{6} = ( \begin{matrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} )  w_{7} = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} )  w_{8} = ( \begin{matrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} )

Proof. To determine the cardinalities of the centralizers, notice that w_{3} and w_{7} are elements of L_{3}(2) of order 3 and 7, respectively, and that w_{4} + I and w_{8} + I are elements of order 4 and 2, respectively. A slight knowledge of centralizers in L_{3}(2) thus yields |C_{L}(w_{i})| for i = 3, 4, 7, 8. For w_{2} and w_{6}, some straightforward matrix computation is called for. One observes that the summation of all the numbers 168/n, n = 2, 3, 4, 6, 7, 8, is equal to |W | − 1. □

Lemma 6.3. Suppose that H is non-solvable and that N_{H}(Y_{P}) is not a maximal subgroup of H. Then q = 2, |S| = 2^{14}, and the following hold.

(a) H/O_{2}(H) ≅ Sym(5), and Ṽ is the ΓL(2, 4) module for H/O_{2}(H). Further, C_{H}(V)/V is a trivial module for O^{2}(H) of order 4, and O_{2}(H)/C_{H}(V) is isomorphic to V/Z as modules for H.

(b) Let P_{1} be the maximal subgroup of H containing S, and set L = ⟨P, P_{1}⟩. Then L/O_{2}(L) ≅ L_{3}(2), Φ(O_{2}(L)) = Y_{L}, Y_{L} is a natural L_{3}(2)-module for L/O_{2}(L), and O_{2}(L)/Y_{L} is the adjoint module for L/O_{2}(L).

Proof. By 4.10, we have q = 2, H_{0} is a 2-component of H, \overline{H} is isomorphic to Sym(5), and V/C_{V} (H_{0}) is a natural ΓL(2, 4)-module for \overline{H}. Thus N_{H}(Y_{P}) has index 3 in the maximal subgroup M of H containing P ∩ H, and |⟨(Y_{P})^{M}⟩| = 8. Observe that M ∈ P. For this reason we shall henceforth write P_{1} for M.

Set L = ⟨P, P_{1}⟩. Then 1.3(4) shows that L/C_{L}(Y_{L}) is isomorphic to L_{3}(2), or Sp(4, 2), with Y_{L} a natural L_{3}(2)-module or a natural Sp(4, 2)-module for L. Form the coset graph Γ^{∗} = Γ(L, H), and set b^{∗} = b(L, H). Take α = L and β = H.

For the remainder of this section, take V = V_{β} = ⟨(Y_{P})^{H}⟩ and set W = W_{β} = ⟨(Y_{L})^{H}⟩. Then V_{δ} and W_{δ} are defined for any conjugate δ of β in Γ^{∗}. Also, set U = Uα, β = ⟨(Y_{P})^{P_{1}}⟩. Then conjugation defines Uλ, μ for any edge {λ, μ} of Γ^{∗}.

Notice that since |V | > |Y_{L}|, Y_{L} ∩ V is a proper P_{1}-invariant subspace of V. Then Y_{L} ∩ V = U. Further, as Y_{P} ≰ Z(O_{2}(H)) we have also U ≰ Z(O_{2}(H)). Then O_{2}(H)O_{2}(L)/O_{2}(L) is a non-trivial subgroup of O_{2}(P_{1})/O_{2}(L), and in the case that L/C_{L}(Y_{L}) is isomorphic to Sp(4, 2) it follows that O_{2}(H)O_{2}(L)/O_{2}(L) is not the transvection subgroup of order 2 in O_{2}(P_{1})/O_{2}(L). Thus, in any case we have:

(1) [O_{2}(H), O^{2}(P_{1})] ≰ O_{2}(L).

We now assume that we have:

Case A. L/C_{L}(Y_{L}) ≅ Sp(4, 2).

Notice that, in this case, (1) implies that [Y_{L}, O_{2}(H)] = [Y_{L}V, O_{2}(H)] = U. As U is not normal in H it then follows that Y_{L}V is not normal in H, and so Y_{L}V ≠ W. Thus:

(2) η(H, W) ≥ 2.

We also note the following.

(3) There is a unique element element \overline{t} of O_{2}(P_{1})/O_{2}(L) such that \overline{t} induces a transvection on Y_{L}. Moreover, we have [Y_{L}, \overline{t}] = Y_{H}.

Suppose next that Y_{l}V ⊴ L. As |Y_{L}V/Y_{L}| = |V/(Y_{L} ∩ V)| = 4, we then have [V, O^{2}(L)] ≤ Y_{L}. But evidently [V, O^{2}(P_{1})] = V ≰ Y_{L}. Thus:

(4) Y_{L}V is not normal in L, and if V ≤ O_{2}(L) then η(L, ⟨V^{L}⟩) ≥ 2.

Fix a critical path (α, β, ⋯, α′) in Γ^{∗}, and suppose first that b^{∗} is odd. Then b^{∗} ≥ 3, by 1.3(6). [OR MAYBE BY AN ARGUMENT TO BE SUPPLIED LATER].

Suppose that V_{α′} ≤ Q_{β}. Then [V_{β}, V_{α′}] ≤ Y_{β}, where Y_{β} is of order 2. As V/Z is a ΓL(2, 4)-module for H, no element of H acts as a transvction on V/Z, and so V_{β} ≤ Q_{α′}. Then [V_{β}, V_{α′}] ≤ Y_{β} ∩ Y_{α′}, and so [V_{β}, V_{α′}] = 1, by 1.15(b). Then V_{α′} centralizes the hyperplane Y_{α} ∩ V_{β} of Y_{α}, and so |V_{α′}/C_{V_{α′}} (Y_{α})| ≤ 2. This is contrary to Y_{α} ≰ Q_{α′}, so we conclude that V_{α′} ≰ Q_{β}. In particular, (α′, ⋯, β) is a pre-critical path, and so also V_{β} ≰ Q_{α′}, by symmetry.

Notice that V_{β} ≤ Q_{α′−1} ≤ O_{2}(G_{α′−1} ∩ G_{α′}). For any element x of O_{2}(P_{1}) − O_{2}(H) we have [V, x]Z = U, so now [V_{β}, V_{α′}]Y_{α′} = U_{α′−1,α′}. For the same reason, we have also [W_{β}, V_{α′}]Y_{α′} = U_{α′−1,α′}. In view of (2), it follows that V_{α′} acts as a transvection group on W_{β}/V_{β}. But vap ≤ O_{2}(G_{β} ∩ G_{β+1}) ≤ O^{2}(H)O_{2}(H), and since no element of SL(2, 4) induces a transvection on any SL(2, 4)-module over 𝔽_{2}, we have a contradiction at this point. We conclude that b^{∗} is even, b^{∗} ≥ 2.

Let P_{0} = (P_{1})^{g} be an L-conjugate of P_{1} such that (P_{0} ∩ P_{1})/O_{2}(L) ≅ SL(2, 2). (The existence of such a conjugate P_{0} of P_{1} can be read off easily from a picture of the B_{2} root system.) One then has ⟨t, P_{1}⟩ = L for any t ∈ O_{2}(P_{0}) ∨_{2} (L). By edge-transitivity on Γ^{∗} we may then fix a vertex λ ∈ Δ(α′) such that:

(5) ⟨G_{α′} ∩ G_{λ}, t⟩ = G_{α′}, for any t ∈ O_{2}(G_{α′−1} ∩ G_{α′}) − Q_{α′}.

We next show:

(6) Y_{β} = Y_{α′−1}.

In order to prove (6), suppose first that [V_{β}, Y_{α′}] ≠ 1. As yap ≤ Q_{β} we then have [V_{β}, Y_{α′}] = Y_{β}, and so V_{β} induces a transvection group on Y_{α′}. Then (2) yields [V_{β}, Y_{α′}] = Y_{α′−1}, and so (6) holds in this case. So assume that [V_{β}, Y_{α′}] = 1. Then C_{Y_{α}}(Y_{α′}) = Y_{α} ∩ V_{β} is a hyperplane of Y_{α}, whence also C_{Y_{α′}} (Y_{α}) is a hyperplane of Y_{α′}. Then (3) yields [V_{β}, Y_{α′}] = Y_{β} = Y_{α′−1}, and (6) holds in any case.

The next step will be to show:

(7) We have b^{∗} = 2.

Assume that (7) is false, so that b^{∗} ≥ 4. Suppose that [Y_{β+1}, W_{λ}] ≠ 1. There is then a critical pair (β + 1, μ) with μ ∈ Δ(λ). Applying (6) to this critical pair, we obtain Y_{β+2} = Y_{λ}, and then C_{G}(Y_{λ}) ≥ ⟨Y_{α}, G_{λ}⟩. But (5) then yields C_{G}(Y_{λ}) ≥ ⟨G_{α′}, G_{λ}⟩, contrary to ⟨L, H⟩ ∉ L. We therefore conclude that [Y_{β+1}, W_{λ}] = 1.

Now W_{λ} ≤ Q_{β+1} ≤ G_{β}. Since Y_{α′} ≤ Q_{β} we have |[V_{β}, Y_{α′}]| ≤ 2, and then |V_{β}/(V_{β} ∩ Q_{α′})| ≤ 2. Suppose that V_{λ} ≰ Q_{β}. Then [V_{β}, V_{λ}]Y_{β} = U_{β,β+1} ≥ [V_{β}, W_{λ}]. Since O_{2}(P_{1}) contains no transvections on W/V, it follows that V_{β} ∩ G_{λ} ≤ Q_{λ}. As [Y_{α}, Y_{λ}] ≠ 1, as we have seen, we have Y_{λ} ≰ V_{β}, and then [V_{β} ∩ Q_{λ}, V_{λ}] = 1. Thus V_{λ} centralizes a hyperplane of V_{β}, and so V_{λ} ≤ Q_{β}.

We now have [V_{β}, V_{λ}] ≤ Y_{β} = Y_{α′−1} ≤ Y_{α′}. If V_{β} ≰ G_{λ} it then follows from (5) that Y_{α′}V_{λ} is normal in G_{α′}, contrary to (4). Thus V_{β} ≤ G_{λ}, and since |V_{β}, V_{λ}]| ≤ 2 we obtain V_{β} ≤ Q_{λ} and [V_{β}, V_{λ}] ≤ V_{β} ∩ Y_{λ}. We have already seen that [Y_{α}, Y_{λ}] ≠ 1, so in fact [V_{β}, V_{λ}] = 1. Then V_{λ} induces a transvection on Y_{α}, and [ya, V_{λ}] = Y_{β} ≤ Y_{α′}. This again yields Y_{α′}Y_{λ} ⊴ G_{α′}, contrary to (4), and completing the proof of (7).

Since [Y_{L}, O_{2}(P_{1})] ≤ V, we now have [Y_{α}, Y_{α′}] ≤ Y_{α} ∩ V_{β} ∩ Y_{α′}. But (Y_{α} ∩ V_{β})/Y_{β} and (Y_{α′} ∩ V_{β})/Y_{β} are two different 1-dimensional 𝔽_{4}-subspaces of V/Z (different since G_{α} ∩ G_{β} ≠ G_{β} ∩ G_{α′}, by maximality of P_{1} in H). Therefore [Y_{α}, Y_{α′}] = Y_{β}. This yields:

(8) [W_{β}, W_{b}] = Y_{β}.

Further, Y_{α} ∩ V_{β} is the hyperplane of Y_{α} which centralizes W_{β}. The normal closure of Y_{α} ∩ V_{β} in G_{β} is V_{β}, so we have:

(9) [V_{β}, W_{β}] = 1.

Applying (9) at λ, we have [Y_{α′}, V_{λ}] = 1, so V_{λ} ≤ Q_{α′} ≤ G_{β}. Suppose V_{λ} ≤ Q_{β}. Then [V_{β}, V_{λ}] ≤ Y_{β}, and since V_{β} ≤ Q_{α′} ≤ G_{λ} we get V_{β} ≤ Q_{λ} and [V_{β}, vl] ≤ Y_{β} ∩ Y_{λ} = 1. As in the proof of (7), we then have [Y_{α}, V_{λ}] = Y_{β} and Y_{α′}V_{λ} ⊴ G_{α′}, contrary to (4). We therefore conclude that V_{λ} ≰ Q_{β}.

As b^{∗} = 2 we have W_{λ} ≰ Q_{α′}. Let t ∈ W_{λ} −Q_{α′}. Then [V_{β}, (V_{β})^{t}] ≤ V_{β} ∩(V_{β})^{t} ∩Y_{α′} = (V_{β} ∩ Y_{α′}) ∩ (V_{β} ∩ Y_{α′})^{t}. By our choice of λ in (5), we have O_{2}(G_{α′} ∩ G_{λ}) ∩ G_{β} = Q_{α′}, so t ∉ G_{β}. As t induces a transvection on V_{α′}, by (8), we then have Y_{α′} = C_{Y_{α′}} (t)Y_{β} and (V_{β} ∩ Y_{α′}) ∩ (V_{β} ∩ Y_{α′})^{t} = C_{V_{β}∩Y_{α′}} (t) is of order 4. Then C_{V_{β}∩Y_{α′}} (t) = [V_{β}, V_{λ}] ≤ V_{λ}, and so we have shown that [V_{β}, (V_{β})^{t}] ≤ V_{λ}. But [W_{λ}, V_{β}] = [(W_{λ} ∩ G_{β})⟨t⟩, V_{β}] ≤ V_{β}(V_{β})^{t}, and so [W_{λ}, V_{β}, V_{β}] ≤ [V_{β}, (V_{β})^{t}] ≤ V_{λ}. As |[V_{β}, V_{λ}]| > 2 we have V_{β} ≰ Q_{λ}, and we have shown that V_{β} acts quadratically on W_{λ}/V_{λ}.

Set W_{0} = W_{λ} ∩ Q_{α′}. Then |W_{λ}/W_{0}| = 2 and we have [V_{β}, W_{0}] ≤ U_{β,α′} = [V_{β}, V_{λ}]Y_{β}. Setting Ŵ_{λ} = W_{λ}/V_{λ}, we then see that V_{β} acts on Ŵ_{λ}, with |[Ŵ_{0}, V_{β}]| ≤ 2. It follows that Ŵ_{λ} involves no natural ΓL(2, 4)-module for G_{α′}, and then quadratic action implies that |V_{β}/C_{V_{β}}(Ŵ_{λ})| = 2. Thus V_{β} ∩ Q_{λ} is a hyperplane of V_{β}. We have seen that [V_{β}, V_{λ}] = V_{β} ∩ Y_{α′} ∩ V_{λ} is of order 4, so Y_{β} ≰ [V_{β}, V_{λ}] and Y_{λ} ≰ [V_{β}, V_{λ}]. Then [V_{β} ∩ Q_{λ}, V_{λ}] = 1, which yields V_{λ} ≤ Q_{β} and a final contradiction in Case A. Thus, we are reduced to

Case B: L/C_{L}(Y_{L}) ≅ L_{3}(2).

We may begin by observing that in this case we have U = Y_{L} ≤ V = W. In particular, W is abelian, and so b^{∗} ≥ 3.

Suppose that b^{∗} is even. As V_{β} ≤ Q_{α′−1} and V_{α′−1} ≤ Q_{β}, we have [V_{β}, V_{α′−1}] = [Y_{α}, Y_{α′}] = Y_{β} = Y_{α′−1}. As C_{L}(Z) = P_{1} ≤ H we have Y_{β} ≠ Y_{β+2}, and so b∗ ≥ 6.

Set X = ⟨(V_{β})^{G_{α}}⟩. As V is not normal in L we have X ≠ V_{β} and so η(G_{α}, X) > 1. Now X ≤ Q_{α′−3} ≤ G_{α′−2}, and [X, Y_{α′−1}] = [X, Y_{β}] = 1. It follows that X ≤ G_{α′−1}. If X ≤ Q_{α′−1} then [X, V_{α′−1}] ≤ Y_{β} ≤ Y_{α}, contrary to η(G_{α}, X) > 1. Thus X ≰ Q_{α′−1}. But X ≤ O_{2}(G_{α′−1} ∩ G_{α′−1}) ≤ O^{2}(G_{α′−1})O_{2}(G_{α′−1}), and since Sylow 2-subgroups of O^{2}(\overline{H}) are T I-sets it follows that X ≰ G_{α′}.

Next, observe that V_{λ} ≤ Q_{β+2} ≤ G_{β+1}, and [Y_{β}, V_{λ}] = [Y_{α′−1}, V_{λ}] = 1, so V_{λ} ≤ G_{β}. If V_{λ} ≤ Q_{β} then [V_{λ}, V_{β}] ≤ Y_{β} ≤ Y_{α′} ≤ V_{λ}, and then V_{λ} ⊴ ⟨V_{β}, G_{α′} ∩ G_{λ}⟩ = G_{α′}, whereas V is not normal in L. Thus V_{λ} ≰ Q_{β}, and it follows that C_{V_{β}}(V_{λ}) = Y_{β+1}, of index 4 in V_{β}.

As L/O_{2}(L) ≅ L_{3}(2) we have O_{2}(P_{1}) ∩ (P_{1})^{g} ≰ O_{2}(L) for any g ∈ L. As V_{β} ≤ O_{2}(G_{α′−1} ∩ G_{α′}, and as vb ≰ G_{λ} by definition of λ, we then have |V_{β}/(V_{β} ∩ G_{λ})| = 2. Now [V_{β} ∩ Q_{λ}, V_{λ}] ≤ V_{β} ∩ Y_{λ}. But yl is not centralized by G_{α′}, so [V_{β}, Y_{λ}] ≠ 1, and so [V_{β} ∩ Q_{λ}, V_{λ}] = 1. Thus V_{β} ∩ Q_{λ} = C_{V_{β}}(V_{λ}) which, we have seen, has index 4 in V_{β}. We may then conclude that V_{β} ∩ G_{λ} ≰ Q_{λ}. This yields [V_{β} ∩ G_{λ}, V_{λ}]Y_{λ} = Y_{α′}. As Y_{α′−2} ∩ Y_{α′} = Y_{α′−1}, of order 2, we have [V_{β} ∩ G_{λ}, V_{λ}] ≰ Y_{α′−2}. Now X centralizes Y_{α′−2} and, as b^{∗} ≥ 6, X centralizes V_{β}. As no element of H induces a transvection on V/Z we have the desired contradiction at this point. That is:

(10) b^{∗} is odd.

[AS BEFORE, ASSUME FROM GLOBAL HYPOTHESIS or 1.3(6) THAT b^{∗} ≠ 1.] As V_{β} does not induce a transvection on V_{α′}/Y_{α′} we have V_{α′} ≰ Q_{β}, and since V_{α′} ≤

O^{2}(G_{β})O_{2}(G_{β}) we get also V_{α′} ≰ G_{α}.

Suppose that Y_{α′} ≤ V_{β}. As Y_{α′−1} = [V_{β}, V_{α′}]Y_{α′}, we then have Y_{α′−1} ≤ V_{β}. Set X = X_{α} = ⟨(V_{β})^{G_{α}}⟩, and assume now that b^{∗} ≥ 5. Then X ≤ Q_{α′−3} ≤ G_{α′−2}, and X centralizes Y_{α′−3}Y_{α′−1} = V_{α′−2}. We then have X ≤ Q_{α′−1} ≤ G_{α′}, and so [X, V_{α′}] ≤ Y_{α′−1}V_{β} ≤ X. Thus X is normal in ⟨G_{α}, V_{α′}⟩, and since V_{α′} ≰ G_{α} we get X normal in G_{β}, contrary to ⟨L, H⟩ ∉ L. We may now conclude that yap ≰ V_{β}. As also (α′, ⋯, β) is a pre-critical path, we obtain also Y_{β} ≰ V_{α′}, by symmetry. Then |V_{α′}/(V_{α′} ∩ Q_{β})| = |V_{β}/(V_{β} ∩ Q_{α′})| = 4.

Set D = ⟨(X)^{G_{β}}⟩. Then D ≤ Q_{β} and indeed [D, V_{β}] = 1. Set F = [V_{β}, V_{α′}]. Then F is a complement to Y_{α′} in Y_{α′−1}, contained in C_{V_{α′−2}}(D). As D does not induce a transvection on V_{α′−2}/Y_{α′−2} we get D ≤ Q_{α′−2} ≤ G_{α′−1}. As D ≤ O_{2}(G_{α′−2} ∩ G_{α′−1}), we have |D/(D ∩ G_{α′})| ≤ 2. As X is not normal in G_{β} we have η(G_{β}, X) ≥ 2, and then |D/C_{D}(V_{α′})| ≥ 16, and |[D, V_{λ}]| ≥ 16. Then D ≰ G_{α′}. Suppose that D is abelian. Then Y_{α′} ≰ D, so that D ∩ Q_{α′} ≤ C_{D}(V_{α′}), and then since a Sylow 2-subgroup of G_{α′}/Q_{α′} is non-abelian of order 8 we get |D/C_{D}(V_{α′})| < 16. This contradiction shows that D is non-abelian, and so b^{∗} = 5. Then F ≤ Y_{β+1} ∩Y_{β+3} = Y_{β+2}, which is contrary to |F | = 4. Thus, we have shown:

(11) b^{∗} = 3.

Set γ = β + 1, so that also γ = α′ − 1. Here V_{α′} ≰ Q_{β} since V_{β} is not a transvection group on V_{α′}/Y_{α′}.

We have

V_{β} ∩ V_{α′} ≥ Y_{γ}[V_{β}, V_{α′}] ⊴ ⟨Q_{β}, Q_{α′}⟩Q_{γ} = G_{γ}.

Suppose that |V_{β}| > 32. That is, suppose that there are non-trivial fixed points for H on V/Z. Since V = ⟨(Y_{L})^{H}⟩ where Y_{L} = [Y_{L}, P_{1}], it then follows that |C_{V/Z}(H_{0})| = 2 or 4, and that V = [V, H]. Thus, V/Z is an indecomposable module for H_{0}, and we have |V/[V, O_{2}(P_{1})]| = |V/[V, x]Y_{L}| = 4 for any element x of O_{2}(P_{1}) − O_{2}(H). Thus |V_{β}/[V_{β}, V_{α′}]Y_{γ}| = 4, and since [V_{β}, V_{α′}]Y_{γ} ≤ V_{α′} it follows that Y_{γ} is a proper subgroup of V_{β} ∩ V_{γ}. [SHORTER ARGUMENT FOR THIS ?] Let y ∈ V_{β} ∩ V_{α′} with [y, O^{2}(G_{β})] = 1 and with Y_{γ}⟨y⟩ normal in G_{β} ∩ G_{γ}. Then Y_{γ}⟨y⟩ is normal in ⟨Q_{β}, Q_{α′}⟨Q_{γ} = G_{γ}, and y centralizes [Q_{γ}, O^{2}(G_{β} ∩ G_{γ})]. But the commutator map defines a pairing of ⟨y⟩×O_{2}(H)/C_{O_{2}(H)}(y) onto Y_{L}, as y ∉ Y_{L} = Ω_{1}(Z(O_{2}(L))), and so y does not centralize any conjugate of O^{2}(P_{1}) in L. This contradiction shows that, in fact, we have |V | = 32, and V/Z is irreducible for H.

Observe that |Q_{β}/(Q_{β} ∩ Q_{γ}| ≤ 4, and that [Q_{β} ∩ Q_{γ}, V_{α′}] ≤ [Q_{γ}, V_{α′}] ≤ Y_{γ} ≤ V_{β}. Since no element of Alt(5) induces a transvection on any 𝔽_{2}-module for Alt(5), it follows that η(H, O_{2}(H)/V) ≤ 1. Set Q = O_{2}(H) and set R = C_{Q}(V). If [Q, O^{2}(H)] ≤ R then V ≤ Z(Q) by the Thompson P × Q Lemma, whereas [V, O_{2}(H)] ≥ [Y_{L}, O_{2}(H)] = Z, by (1). Thus η(H, O_{2}(H)/V) = 1, and [R, O^{2}(H)] = V. The commutator map defines a pairing of Q/R × V/Z onto Z, so Q/R is isomorphic to the dual of V/Z as a module for H. We record the results so far as follows.

(12) Both V/Z and O_{2}(H)/C_{H}(V) are natural ΓL(2, 4)-modules for H/O_{2}(H), and [C_{H}(V), O^{2}(H)] = V.

For any vertex δ of Γ and for any positive integer n, define G_{δ}^{(n)} to be the point-wise stabilizer in G_{δ} of all the vertices at distance at most n from δ. Observe that Y_{α}∩G_{α}^{(4)} = 1 as b^{∗} = 3. As Y_{α} = Y_{L} = Ω_{1}(Z(O_{2}(L))), we then have G_{α}^{(4)} = 1. In a similar vein, observe that since [Q_{α}, V_{β}, Q_{α}] ≤ [Y_{α}, Q_{α}] = 1, we have [Q_{α}, Q_{α}] ≤ C_{Q_{α}}(X_{α}) by the Three Subgroups Lemma, and so [Q_{α}, Q_{α}] ≤ G_{α}^{(3)}. We note also that C_{G_{β}}(V_{β}) = G_{β}^{(2)}.

Let us write Q_{L} for O_{2}(L), and Q_{H} for O_{2}(H). Write X = X_{α}. Then Q_{L} = (Q_{L} ∩ Q_{H})X. Now (O_{2}(L) ∩ O_{2}(H))/C_{H}(V) is a 1-dimensional 𝔽_{4}-subspace of Q_{H}/C_{H}(V) as seen in (12). As Q_{H}X = O_{2}(P_{1}) it then follows that Q_{L} ∩ Q_{H} = (X ∩ Q_{H})C_{H}(V). Thus

Q_{δ} = (Q_{α} ∩ Q_{δ})X = G_{δ}^{(2)}X

for all δ ∈ Δ(α), and we have the following result.

(13) Q_{α} = ∩_{δ∈Δ(α)}C_{Q_{α}}(V_{δ})X

We now have

[G_{α}^{(3)}, Q_{α}] = ⋂_{δ∈Δ(α)}[G_{α}^{(3)}, C_{Q_{δ}}(V_{δ})X] = ⋂_{δ∈Δ(α)}[G_{α}^{(3}), G_{δ}^{(2})]

≤ ⋂_{δ∈Δ(α)}[G_{δ}^{(2)}, G_{δ}^{(2)}] ≤ ⋂_{δ∈Δ(α)} G_{δ}^{(3)} ≤ G_{α}^{(4)} = 1.

This shows:

(14) We have G_{α}^{(3)} = Y_{α}.

Suppose next that there exists an element h ∈ C_{H}(O^{2}(H)) − Z. We may then choose h so that [H, h] = Z. By (12) we have h ∈ C_{H}(V), and then [h, X] ≤ [O_{2}(L), X] ≤ Y_{L}. As C_{Y_{L}}(O^{2}(H)) = Z we then have [h, X] ≤ Z. For any μ ∈ Δ(α) we have h ∈ G_{μ}, and since no element of G_{μ} induces a transvection on V_{μ}/Y_{μ} we further have h ∈ Q_{μ}. But then [V_{μ}, h] ≤ Y_{μ} ≠ Y_{α} = Z, and so [V_{μ}, h] = 1. Thus h ∈ G_{α}^{(3)}, and so h ∈ Y_{α} by (14). As C_{Y_{L}}(O^{2}(H)) = Z we have a contradiction. This proves:

(15) C_{H}(O^{2}(H)) = Z.

Here is a further consequence of (13). Namely, we have

[Q_{α}, Q_{β}] ≤ [G_{β}^{(2)}X, Q_{β}] ≤ [G_{β}^{(2)}, Q_{β}]X = [C_{H}(V), Q_{H}]X.

But (15) implies that C_{H}(V)/Z is abelian, so C_{H}(V)/V is abelian, and we then have [C_{H}(V), Q_{H}] ≤ V, by (12). We then conclude that [Q_{α}, Q_{β}] ≤ X, and so X ≥ [Q_{L}, O^{2}(L)]. As X = ⟨V^{L}⟩, where V = [V, O^{2}(P_{1})], we then have X = [Q_{L}, O^{2}(L)]. On the other hand, 6.1 shows that η(L, X/Y_{L}) ≤ 2. If X involves no Steinberg module for L/O_{2}(L) it follows that η(P_{1}, X/Y_{L}) = 2 and that η(P_{1}, O_{2}(P_{1})) = 4. But it is evident from the structure of H that η(P_{1}, O_{2}(P_{1})) = 5. Thus X has a constituent W which is a Steinberg module for L/O_{2}(L). Then η(P_{1}, W) = 3 and it follows that W is the unique non-trivial constituent for L in X/Y_{L}. Projectivity of W now yields W = X/Y_{L}. As X = [Q_{L}, O^{2}(L)] it follows that also Q_{L} = C_{Q_{L}}(X)X. But C_{Q_{L}}(X) = G_{α}^{(3)} ≤ X by (14). Thus:

(16) Q_{L}/Y_{L} is a Steinberg module for L/O_{2}(L), and |S| = 2^{14}.

Finally, we observe from |S| that |C_{H}(V)/V | = 4. This completes the proof of lemma 6.3. □

Lemma 6.4. Set W = Q_{L}/Y_{L}, identify W with sl(3, 2), and identify L/Q_{L} with L_{3}(2), with action on W given by matrix conjugation. Let the orbits for L on W − {0} be given as in 6.2. For any x ∈ Q_{L}, write x̂ for the image of x in W. Then |x| ≤ 2 if x ∈ Y_{L} or if x̂ ∈ O_{7} ∪ O_{8}, and otherwise |x| = 4.

Proof. We have maps S : W ⟶ Y_{L} and B : W × W ⟶ Y_{L}, given by squaring and commutation, respectively. Then, for any x and y in Q_{L}, we have

(1)  S(x̂ + ŷ) = S(x̂) + S(ŷ) + B(x̂, ŷ).

We shall also identify Y_{L} with a natural SL(3, 2)-module, written additively.

Let t ∈ V − Y_{L} with [t, S] ≤ Y_{L}. Then ⟨t̂⟩ = C_{W} (S). We take S to induce conjugation by upper triangular matrices on W. Adopting the notation of 6.2, we then have t̂= w_{8}. As Φ(V) = 1, and as S is constant on L-orbits, we see that S is trivial on O_{8}.

As S(w_{7}) is invariant under C_{L}(w_{7}) (which is of order 7) it follows that S(w_{7}) = (0, 0, 0), and thus S is trivial on O_{7}.

Denote by t̂′ the transpose of t̂. The set of all matrices in W whose lower-left entry is 0 is the unique maximal S-invariant subspace of W, and it therefore contains C_{Q_{L}}(t)/Y_{L}. It follows that B(t̂, t̂′) ≠ (0, 0, 0). Then S(w_{2}) = S(t̂+ t̂′) ≠ (0, 0, 0), by (1). Thus, S is non-trivial on O_{2}.

Next, observe that w_{3} + t̂ ∈ O_{7}. (To see this, either check that, as an invertible matrix, w_{3} + t̂is of order 7, or observe that conjugation of w_{3} + t̂by w_{3} yields w_{7}.) Then (0, 0, 0) = S(w_{3} + t̂) = S(w_{3}) + B(w_{3}, t̂). But evidently w_{3} does not lie in the unique maximal S-invariant subspace of W, so B(w_{3}, t̂) ≠ (0, 0, 0), and so S is non-trivial on O_{3}. Further, as S(w_{3}) is invariant under C_{L}(w_{3}) we obtain S(w_{3}) = (1, 1, 1). For the same reason, we have S(w′_{3}) = (1, 1, 1), where w′_{3} is the transpose of w_{3}.

We next check that w′_{3} + w_{6} ∈ O_{7}. (Indeed, one observes that w′_{3} + w_{6} is the square of the matrix w_{7}.) Suppose that S(w_{6}) = (0, 0, 0). Then (0, 0, 0) = S(w′_{3} + w_{6}) = S(w′_{3})+B(w′_{3}, w_{6}), and thus B(w′_{3}, w_{6}) = (1, 1, 1). Thus B(w′_{3}, w_{6}) is invariant under any element g of L whose matrix for the L-action on W is given by w′_{3}. The matrix of g^{2} is then w_{3}, and setting u = (w_{6})^{g} and v = (w_{6})^{g^{2}}, we have w_{6} = u + v. Then

(1, 1, 1) = B(w′_{3}, w_{6}) = B(w′_{3}, u + v) = B(w′_{3}, u) + B(w′_{3}, v)
= B(w′_{3}, w_{6})^{g} + B(w′_{3}, w_{6})^{g^{2}} = (1, 1, 1) + (1, 1, 1) = (0, 0, 0).

This contradiction shows that, in fact, S is non-trivial on O_{6}.

It only remains to treat O_{4}. For this, define matrices y and g as follows.

y = ( \begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 1 & 0 \end{matrix} ) g = ( \begin{matrix}0 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} )

Then y^{g} = w′_{2}, and we have w_{4} = y + w_{7}. Further, we have

(w_{7})^{g} = ( \begin{matrix}1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{matrix} )

and hence B(y, w_{7}) ≠ (0, 0, 0). Now S(w_{4}) = S(y+w_{7}) = B(y, w_{7}), and so S is non-trivial on O_{4}. □

Corollary 6.5. Set R = C_{H}(V), and let x be an element of order 3 in P_{1}. Then R = C_{R}(x)V, and C_{R}(x) is a quaternion group.

Proof. As V is a natural ΓL(2, 4)-module for H/Q_{H}, we have V = [V, x]Z. By 6.3(a), R/V centralizes O^{2}(H), so R = C_{R}(x)V with C_{R}(x) ∩ V = Z, and with |C_{R}(x)| = 8. We have R ≤ Q_{L} since P_{1}/Q_{L} ≅ Sym(4). As C_{R}(x) − Y_{L} = C_{R}(x) − Z, it now follows from 6.4 that C_{R}(x) − Y_{H} consists entirely of elements of order 4, and hence that C_{R}(x) is a quaternion group. □

Proposition 6.6. Both L and H are maximal 2-local subgroups of G.

Proof. Let L^{∗} be a 2-local subgroup of G containing L. Suppose first that Q_{L} properly contains O_{2}(L^{∗}). Then O_{2}(L^{∗}) = Y_{L} and thus C_{L}(O_{2}(L^{∗})) ≰ O_{2}(L^{∗}), contrary to the Main Hypothesis concerning L(S). Thus O_{2}(L^{∗}) = Q_{L}. But then L^{∗} = C_{L^{∗}}(Y_{L})L, and C_{L^{∗}}(Y_{L})/Q_{L} is of odd order, centralized by ⟨Q^{L}⟩ (where we recall that Q denotes O_{2}(C̃)). Thus, we have L^{∗}/Q_{L} = L/Q_{L}) × O_{2,2′}(L)/Q_{L}. As L acts irreducibly on Q_{L}/Y_{L}, it then follows that L^{∗} = L.

Now let H^{∗} be a 2-local subgroup of G containing H. Suppose first that O_{2}(H^{∗}) does not contain [Q_{H}, O^{2}(H)]. Then O_{2}(H^{∗}) ≤ C_{H}(V), and since H^{∗} is 2-constrained we have V properly contained in O_{2}(H^{∗}). The preceeding lemma then shows that V = Ω_{1}(Z(O_{2}(H^{∗}))) and that Z = Φ(O_{2}(H^{∗}). Then H^{∗} = C_{H^{∗}}(V)H, as Sym(5) is maximal among subgroups of L_{4}(2) having a Sylow 2-subgroup of order 8. Now C_{H^{∗}}(V) ≤ C_{G}(Y_{L}) ≤ L, by what has already been proved, and then C_{H^{∗}}(V) ≤ P_{1} ≤ H. Thus, we may assume that [Q_{H}, O^{2}(H)] ≤ O_{2}(H^{∗}). But this yields V = Z(O_{2}(H^{∗})), and Z = [V, O_{2}(H^{∗})], with the result, as just argued, that H = H^{∗}. □

[NOW COMES THE HARD PART: IDENTIFYING Ru. I have some notes which yield |Ru|. Better yet, I have Michael’s notes which do the same thing, with a clever idea for avoiding some of the computation required for a strict applicaion of the Thompson order formula. He also tells how to get an amalgam for ^{2}F_{4}(2) from (H, L), and with |G| (and much of the local structure for odd primes) in hand, one can obtain ^{2}F_{4}(2) using a result of Bennett and Shpectorov. At that point, he produces Ru as a rank-3 permutation group.]

Section 7: The Solvable Case, p = 2, Part I

In this section we consider the case where p = 2 and where both P and H are solvable. In particular, we then have q = 2, so that H = P̃. Our treatment may be viewed as a re-working of parts of Stellmacher’s N-Group Paper [St2] (specifically, sections 9 and 10). In the case b = 3, we will stick closely to the original (section 10).

This section and the next will be devoted to the proof of the following result.

Theorem 7.0. Assume that b ≥ 3, q = 2, and H = P̃ is solvable. Then (P, P̃) is a weak BN-pair. Moreover, we have b = 3, and P̃/O_{2}(P̃) is isomorphic to either SL(2, 2) or Sz(2).

Remark: From the data given by the Theorem 7.1, we will recover the amalgam for M_{12} or Aut(M_{12}) (when P̃/O_{2}(P̃) is isomorphic to SL(2, 2)), or the amalgam for the Tits group or for ^{2}F_{4}(2) (when P̃/O_{2}(P̃) is isomorphic to Sz(2)). The actual construction of these groups from the local data promises to be an interesting exercise. Or not: it being likely that one has only to copy the relevant portions of the Aschbacher-Smith Opus.

Lemma 7.1. Denote by T the set of elements t of S such that |Ṽ/CṼ (t)| = 2. Let T_{0} be a subset of T , and put T_{0} = ⟨T_{0}⟩. Suppose that Y_{P} ≤ [V, T_{0}]Y_{H}. Then C_{V} (O^{2}(H)) = Y_{H} and T_{0}Q_{H} = ⟨T ⟩Q_{H}. Proof. By assumption we have [V, T_{0}] ≰ Y_{H}, so T_{0} ≰ Q_{H}. Set T = ⟨T ⟩. Then 3.11 implies that O^{2}(\overline{H})\overline{T} is a direct product of groups \overline{K}_{i}, 1 ≤ i ≤ r, where \overline{K}_{i} ≅ SL(2, 2) and where |[Ṽ, \overline{K}_{i}]| = 4 for all i. It follows that [Ṽ, T] ≤ [Ṽ, O^{2}(H)], and thus Y_{P} ≤ [V, O^{2}(H)]Y_{H}, by assumption. This shows that C_{V} (O^{2}(H)) = Y_{H} and that Ṽ is the direct product of the fours groups [Ṽ, \overline{K}_{i}], 1 ≤ i ≤ r. Suppose that \overline{T_{0}} is a proper subset of \overline{T} . Then [Ṽ, \overline{T}_{0}] centralizes at least one of the groups \overline{K}_{i}. As \overline{H} = ⟨\overline{K}_{i}, \overline{S}⟩ for any i, we then have Y_{P} ⊴ H, which one recognizes as a contradiction, and which proves the lemma. □ Lemma 7.2. Suppose that there exists an element t of [Q_{P} , O^{2}(P)]O_{2}(H) such that |Ṽ/CṼ (t)| = 2. Then C_{H}(Ṽ) = O_{2}(H).

Proof. Set N = C_{H}(Ṽ). As H ∈ P (or, more generally, by 1.11(b)) Q_{H} is a Sylow 2-subgroup of N. As N normalizes Y_{P}, 1.7 implies that N ≤ N_{G}(O^{2}(P)). Then [N, t] is a 2-group, and thus [N, t] ≤ Q_{H}. Set Ĥ = H/Q_{H}.

Define subgroups \overline{K}_{i} of \overline{H}, 1 ≤ i ≤ s, as in the proof of Lemma 7.1. Then \overline{K}_{i} = ⟨\overline{t}_{i}, \overline{a}_{i}⟩, where |\overline{a}_{i}| = 3 and where \overline{t}_{i} ∈ \overline{S}. As H ∈ P, \overline{S} acts transitively on \{\overline{K}_{i}\}_{1≤i≤s}. Then each \overline{t}_{i} is the image in \overline{H} of a conjugate t_{i} of t. Set X_{i} = [O^{2}(H), t_{i}]. As [N, t_{i}] ≤ Q_{H}, it follows that |X̂_{i}| = 3. In particular, we have [X̂_{i}, X̂_{j}] = 1 for all i and j. Setting T = ⟨t_{1}, ⋯, t_{s}⟩, we now have [O^{2}(Ĥ), T̂] elementary abelian of order 3^{s}, and where 1 ≠ T̂ ⊴ Ŝ. But H ∈ P, so [O^{2}(Ĥ), T̂] = O^{2}(Ĥ). This completes the proof that N = Q_{H}. □ Lemma 7.3. Let (ρ, σ, τ) be a path in Γ with σ of type P. Set D = V_{ρ}∩V_{τ}, T = Q_{ρ}∩Q_{σ}, and assume:

(i) |V_{τ} /D| = 2, and (ii) there exists t ∈ T − Q_{τ} with [D, t] = 1.

Then H/O_{2}(H) ≅ SL(2, 2), and |V | = 8. Proof. We take S = G_{σ} ∩ G_{τ}, and set \overline{G}_{τ} = G_{τ} /C_{G_{τ}} (V_{τ} /Y_{τ}). Thus, we have \overline{S} ≅ S/Q_{τ}. As S = Q_{σ}Q_{τ} and T ⊴ Q_{σ}, we then have \overline{T} ⊴ \overline{Q}_{σ} = \overline{S}.

We have [D, t] = 1, by assumption, and thus t induces a transvection on V_{τ} /Y_{τ}. Proposition 3.11 then yields the structure of V_{τ} /Y_{τ} as a module for the group X = ⟨t^{G_{τ}} ⟩. Thus, \overline{X} is a direct product of s copies of Sym(3) and, setting Ṽ = V_{τ} /Y_{τ}, we have Ṽ = CṼ (O^{2}(X)) × V_{1} × ⋯ × V_{s}, where each V_{i} is an X-invariant fours group. As t ∈ T, \overline{T} contains all the transvections in \overline{S}. As [D, T] ≤ Y_{ρ}, of order 2, it follows that s ≤ 2. But also D̃ is \overline{S}-invariant, as D ⊴ Q_{σ}. If s = 2 (so that \overline{S} is dihedral of order 8) this condition uniquely determines the hyperplane D̃ of Ṽ, and one sees that in this case |[D, T]| > 2. Thus s = 1, and 8 ≤ |V_{σ}| ≤ 16.

Set W = C_{V_{τ}} (O^{2}(G_{τ})), and suppose that W ≰ D. Then V_{τ} = W D, and so [Ṽ, T] = Ỹ_{ρ}. In this case, Lemma 7.1 yields |V | = 8. On the other hand, suppose that W ≤ D, and suppose further that W ≠ Y_{σ}. Set W_{1} = C_{V_{ρ}}(O^{2}(G_{ρ})). We then have D = W W_{1} and W ∩ W_{1} ≠ 1. But O_{2}(O^{2}(G_{τ})) ≰ Q_{σ}, as Q_{τ} ∩ Q_{σ} is not normal in G_{ρ}. Then [W ∩ W_{1}, O^{2}(G_{σ})] = 1, which is contrary to Z(P) = 1. This contradiction now shows that |V | = 8 in any case. Now P/Q_{P} ≅ SL(2, 2), by Lemma 7.2. □

Lemma 7.4. Suppose that b is even. Then |V | = 8.

Proof. Fix a critical pair (α, α′) and a path π = (α, β, β +1, β +2, ⋯, α′−1, α′) of length b. Fix λ ∈ Δ(α′) − {α′ − 1}. As b is even, we have Y_{β} = [Y_{α}, Y_{α′}] = [Y_{α}, Y_{λ}] = Y_{α′−1}. As Y_{β} ≠ Y_{β+2}, we then have b ≥ 6. In particular, it then follows that [V_{β+2}, Y_{α}] = 1.

It will be convenient to establish, and to list, (and to prove) the following facts.

(1) [V_{β+2}, V_{λ}] = 1.

(2) V_{λ} ≤ G_{β} and V_{λ} ≰ Q_{β}.

In order to prove (1) and (2), notice that we have [Y_{α}, Y_{λ}] ≠ 1. As V_{β+2} ≤ Q_{λ}, we have [V_{β+2}, V_{λ}] ≤ Y_{λ}, while also [V_{β+2}, Y_{α}] = 1. This yields (1). Then [Y_{β}, V_{λ}] = 1, and so V_{λ} ≤ G_{β}. Supposing V_{λ} ≤ Q_{β}, we obtain [Y_{α}, V_{λ}] = [Y_{α}, Y_{λ}] = Y_{α′−1} ≤ V_{λ}, and so V_{λ} ⊴ ⟨Y_{α}, G_{λ}⟩ = ⟨G_{α′}, G_{λ}⟩, which contradicts our basic hypothesis that ⟨P, H⟩ ∉ L. Thus (2) holds.

By Lemma 1.2 in [Bernd’s N-Groups] there exists g ∈ V_{β} − G_{λ} with |V_{λ}/C_{V_{λ}}(g)| ≤ 4. Set D = C_{V_{λ}}(g)Y_{λ}. Then D ⊴ ⟨g, Q_{λ}⟩. Thus D is invariant under O^{2}(G_{α′}), and so D ≤ V_{α′−1} ∩ V_{λ}. Thus |V_{λ}/V_{α′−1} ∩ V_{λ}| = 2, and so also |V_{β}/V_{β} ∩ V_{β+2}| = 2. Now (1) and (2) and Lemma 7.3 together yield |V | = 8. □

We now take up the case where b is odd.

Reminder. For any vertex δ of Γ and for any non-negative integer n, we set V_{δ}^{(n)} = ⟨Y_{γ} : dist(γ, δ) ≤ n⟩.

Lemma 7.5. Assume b odd, b ≥ 5. Let (ρ, σ, τ, λ) be a path of length 3 in Γ, with ρ of type P (and σ of type H). Set B = V_{ρ}^{(b−1)}, and assume that there exists t ∈ B − G_{σ}(τ) with |W_{τ} /C_{W_{τ}} (t)| ≤ 4. Then H/O_{2}(H) ≅ SL(2, 2), and (P, H) is a weak BN-pair.

Proof. Set U = C_{V_{λ}}(t)Y_{λ}, U^{∗} = U(Y_{λ})^{t}, R = N_{Q_{σ}}(U), and R^{∗} = N_{Q_{σ}}(U^{∗}). As t ∉ G_{σ}(τ), by assumption, we have Y_{λ} ≰ U, and so |V_{λ}/U| ≤ 2.

Suppose first that Q_{σ} ∩ Q_{τ} ≰ R. Then ⟨U^{Q_{σ}∩Q_{τ}} ⟩ = V_{λ}. Since (Y_{λ})^{t} ≤ V_{σ} we then have

⟨(U^{∗})^{Q_{σ}}⟩V_{σ} = ⟨U^{Q_{σ}}⟩V_{σ} = ⟨U^{(Q_{σ}∩Q_{τ})Q_{σ}}⟩V_{σ} = ⟨V_{λ}^{Q_{σ}}⟩V_{σ} = W_{τ} .

But this implies that W_{τ} is t-invariant, which is contrary to t ∉ G_{σ}(τ). We therefore conclude that Q_{σ} ∩ Q_{τ} ≤ R.

Suppose next that R ≰ Q_{τ}. Then O^{2}(G_{τ}) ≤ ⟨R, Q_{λ}⟩, and hence U ≤ D_{τ}. Thus |V_{σ}/D_{τ} | ≤ 2, and then 7.3 yields the conclusion of the lemma. Thus, we may assume that R ≤ Q_{τ}, and hence that R = Q_{σ} ∩ Q_{τ}. Then R is not t-invariant, by 1.9, and so R ≠ R^{∗}. As |Q_{σ}/(Q_{σ} ∩ Q_{τ})| = 2, we have thus shown:

(1) R^{∗} = Q_{σ}, R = Q_{σ} ∩ Q_{τ}, and U ≠ U^{∗}.

Let g ∈ R^{∗} −R. As |U^{∗}/U| = 2 we then have |U/(U ∩U^{g})| = 2, and since |V_{λ}/U|/ ≤ 2 we obtain |V_{λ}/(V_{λ} ∩ (V_{λ})^{g})| ≤ 4. As g ∉ G_{λ}, by (1), 1.17 yields

(2) |V_{σ}/D_{τ} | = |V_{σ}/D_{ρ}| ≤ 4.

Recall that B denotes V_{ρ}^{(b−1)}. For any vertex γ with dist(γ, ρ) ≤ b − 1 we have [Y_{γ}, V_{δ}] = 1 for some δ ∈ Δ(ρ), and thus [B, D_{ρ}] = 1. Take P = ρ and H = σ, so that S = G_{ρ} ∩ G_{σ}. We shall also write D for D_{ρ}. We have t ∉ N_{G}(W_{τ}), so W_{τ} ≠ V_{σ}C_{W_{τ}} (t). As |W_{τ} /C_{W_{τ}} (t)| ≤ 4, by assumption, it then follows that |V_{σ}/C_{V_{σ}}(t)| = 2. In particular, t induces a transvection on Ṽ_{σ}. As B ⊴ S, it now follows from 3.11 and from (2) that H/Q_{H} ≅ SL(2, 2) or O_{4}^{+}(2), with Ṽ/CṼ (O^{2}(H)) a natural module for H/Q_{H}. In the SL(2, 2)-case we are done, so we assume henceforth that H/Q_{H} is isomorphic to O_{4}^{+}(2). We next show:

(3) Ṽ is a natural O_{4}^{+}(2)-module for H/Q_{H}.

Suppose (3) is false. Then C_{V} (O^{2}(H)) properly contains Y_{H}, and so there exists x ∈ V with [x, H] = Y_{H}. As [B, D] = 1, it follows from (2) that x ∈ D, and then [x, P] ≤ Y_{P} by 1.17(b). Here x ∉ Y_{P} as Y_{P} is not normal in H. As Y_{P} = Ω_{1}(Z(Q_{P})) and as V is elementary abelian, we then have [x, Q_{P}] = Y_{P} and |S : C_{S}(x)| > 2, for a contradiction. This proves (3).

Notice that D is now identified as [V, S]Y_{H}, of order 8. Since |V/C_{V} (t)| = 2, we then have C_{S}(D/Y_{H}) = O_{2}(H)B, and Q_{H}B/Q_{H} = Q_{H})C_{S}(D)/Q_{H} is the subgroup of S/Q_{H} induced by the elements of S/Q_{H} which induce transvections on Ṽ.

Set J = ⟨(Q_{P} ∩ Q_{H}))^{P} ⟩. Then J ≥ [Q_{P} , O^{2}(P)]. As D ≰ Y_{P} we have Q_{P} /C_{P} (D) isomorphic to Y_{P} as modules for P, and so Q_{P} = C_{Q_{P}} (D)J. As B ≤ J, this says that J induces the whole of the action of Q_{P} on V/Y_{H}. Then, since S = Q_{P} Q_{H}), we obtain S = Q_{H}J. Translating this information to the edge (τ, λ), and setting Q_{τ}^{∗} = (Q_{σ} ∩ Q_{τ})(Q_{τ} ∩ Q_{λ}) we observe that Q_{τ}^{∗} is conjugate to J, by 1.17(c). We have thus shown:

(4) G_{τ} ∩ G_{λ} = Q_{τ}^{∗}Q_{λ} = (Q_{σ} ∩ Q_{τ})Q_{λ}.

As U is normal in Q_{σ} ∩ Q_{τ}, by (1), it now follows from (3) and (4) that

(5) U = [V_{λ}, G_{τ} ∩ G_{λ}] = [V_{λ}, Q_{τ} Q_{λ}].

Note that (5) also holds if λ is replaced by the vertex λ′ ∈ Δ(τ) − {σ, λ}, and U is replaced by C_{V_{λ′}} (t)Y_{λ′}. Set X = [W_{τ} , Q_{τ}]V_{σ}. Then X = ⟨(U)^{G_{τ}} ⟩V_{σ}, by (5). It follows that X is t-invariant, and so X ⊴ ⟨t, G_{σ} ∩ G_{τ} ⟩ = G_{σ}. Thus X ≤ W_{δ} for any δ ∈ Δ(σ), and thus [W_{τ} , Q_{τ}] ≤ W_{γ} for every vertex γ at distance 2 from τ. For every such γ we have [V_{γ}^{(b−3)}, W_{γ}] = 1, and we obtain [W_{τ} , Q_{τ} , V_{τ}^{(b−1)}] = 1. Translating back to the edge (ρ, σ), we then have [V, Q_{H}), B] = 1, and so B induces a transvection on V/Z. This is contrary to B ⊴ S, and the lemma is thereby proved. □

Lemma 7.6. Suppose that \overline{H} ≅ SL(2, 2). Then b ≠ 5.

Proof. We’ll just copy this from [9.11 in Weak (B, N)-pairs of Rank 2]. □

For the remainder of this section we will assume that b is odd, b ≥ 5. (The case b = 3 will be taken up in the next section.) Under this assumption, it follows from 2.3 that there is a subgroup A of [Q_{P} , O^{2}(P)]O_{2}(H) such that A is a quadratic F 2^{∗}-offender on Ṽ. Then 3.11 implies that \overline{H}_{0} is a direct product

\overline{H}_{0} = \overline{L}_{1} × ⋯ \overline{L}_{r}

where |\overline{L}_{i}| = 3 and where |[Ṽ, \overline{L}_{i}]| = 4 for all i, 1 ≤ i ≤ r. For each such i, set Ṽ_{i} = ∏\{[Ṽ, L_{j}]\}_{j≠i}, and set K_{i} = C_{H}(Ṽ_{i}). (Thus, we have \overline{K}_{i} = \overline{L}_{i} unless there exist elements of H which induce transvections on Ṽ.) We set

K = {K_{1}, ⋯, K_{r}}

and for any H-vertex δ = H^{g} of Γ we define K_{δ} to be K^{g}.

The parameter r will be fixed for the remainder of this section.

A path in Γ will always be understood to be non-empty, without stammering. For any path π = (δ, ⋯, δ′) in Γ, we write |π| for the length of π, and we write \overline{π} for the opposite path (δ′, ⋯, δ). If γ and γ′ are vertices of π, with γ preceding γ′, then π(γ, γ′) denotes the sub-path (γ, ⋯, γ′) of π. If π′ is a path whose initial vertex is incident with the terminal vertex of π, then we write π ∘ π′ to denote the path obtained by concatenation of π and π′, in the given order. Let H(π) denote the set of H-vertices of π which are not terminal vertices of π. Then, denote by S(π) the set of vertices δ in H(π) which satisfy the following two conditions, for both ϵ = 1 and ϵ = −1.

(1) There exists K ∈ K_{δ}, and an element g of H such that either g is in K − C_{K}(Ṽ) or g induces a transvection on Ṽ which inverts \overline{K}, and such that δ + ϵ is fused to δ − ϵ by g^{ϵ}.

(2) We have N_{G_{δ}∩G_{δ+ϵ}}(Y_{δ−ϵ}) ≤ G_{δ−ϵ}.

We will refer to the group K in (1) as the linkage group for π at δ.

Lemma 7.7. Assume that b is odd, b ≥ 5.

(a) Let K ∈ K and let g ∈ K − C_{K}(Ṽ). Set T = N_{S}((Y_{P})^{g}). Then there exists g′ ∈ N_{K}(T) such that (Y_{P})^{g} = (Y_{P})^{g′}.

(b) Let (ρ, σ) be an edge of Γ with ρ of type P, σ of type H. Then V_{σ} = ⟨Y_{τ} : σ ∈ S(ρ, σ, τ)⟩Y_{ρ}.

Proof. We have [Ṽ, H_{0}] the direct product of {[Ṽ, K_{i}] : 1 ≤ i ≤ r}, and the action of H on K induces an equivalent action on this set of subgroups of Ṽ. Let y ∈ Y_{P} − Y_{H}. Then T centralizes both ỹ and y^{g}, so T centralizes [ỹ, g], and therefore \overline{K} is T -invariant. Let ϕ be the endomorphism of [Ṽ, H_{0}] which restricts to the identity on [Ṽ, K] and which is the zero mapping on any [Ṽ, K_{i}] for K_{i} ≠ K. Then N_{S}(\overline{K}) centralizes also ϕ(ỹ), and so T centralizes [Ṽ, K]. Setting N = C_{H}(Ṽ), we now have [T, K] ≤ N. As T ≥ O_{2}(H), T is a Sylow 2-subgroup of NT, and the Frattini argument now shows that there exists g′ ∈ N_{NT ⟨g⟩}(T) − NT. This proves (a).

In order to prove (b), take ρ = P and σ = H. Set V_{i} = ⟨(Y_{P})^{K_{i}}⟩. Then V = V_{1} ⋯ V_{r}, and (b) then follows from (a). □

Lemma 7.8. Assume that b is odd, b ≥ 5. Let π = (ρ, σ, τ) be a path in Γ, with σ ∈ S(π), and let A be an elementary abelian subgroup of Q_{ρ}. Assume that there exists a ∈ A − G_{τ} with [V_{σ}, a, A] = 1. Then |A/A ∩ G_{τ} | = 2.

Proof. Let K be the linkage group for π at σ, and let g ∈ K with τ = ρ^{g}. Identify H with σ, and set N = C_{H}(Ṽ). Then [C_{A}(\overline{K}), g] ≤ N, and so C_{A}(\overline{K}) normalizes Y_{τ}. As σ ∈ S(π) we then have C_{A}(\overline{K}) ≤ G_{τ}, and it therefore suffices to show that |A/C_{A}(\overline{K})| ≤ 2. This follows from quadratic action, and from the structure of \overline{H}_{0} and of Ṽ given after 7.6. □

Lemma 7.9. Assume that b is odd, b ≥ 5. Let π = (δ, ⋯, γ) be a path with δ of type P and with γ of type H. There then exists μ ∈ Δ(γ) such that γ ∈ S(π ∘ (μ)). Moreover, if (δ, γ) is a critical pair and |π| = b, then μ may be chosen so that Y_{δ} ≰ G_{μ}.

Proof. Take H = γ and take P = γ − 1, where γ − 1 denotes the vertex of π which is adjacent to γ. Thus, we have S = G_{γ−1} ∩ G_{γ}. Choose K ∈ K, with [\overline{K}, \overline{Y_{δ}}] ≠ 1 if Y_{δ} ≰ Q_{γ}. Set N = C_{H}(Ṽ), choose g ∈ K −N, and set m = (γ−1)^{g}. Set T = N_{S}(Y_{μ}). By Lemma 7(a) we may assume that g normalizes T, and hence that T ≤ G_{μ} ∩ (G_{γ−1})^{g^{−1}}. Thus γ ∈ S(π ∘ (μ)). Now suppose that Y_{δ} ≰ Q_{δ} so that, by choice, [Y_{δ}, K] ≰ N. We then have Y_{δ} ≰ T. Thus, Y_{δ} ≰ G_{μ}, and this completes the proof of the lemma. □

Lemma 7.10. Assume that b is odd, b ≥ 5. Let π = (δ, ⋯, γ) be a path of length b − 1 with both δ and γ of type H. Suppose that V_{δ} ≰ Q_{γ}. There then exists λ ∈ Δ(δ) such that (λ, γ) is a critical pair, and such that δ ∈ S((λ) ∘ π).

Proof. Immediate from Lemma 7.7(b). □

Lemma 7.11. Assume that b is odd, b ≥ 5, and let π = (ρ, σ, τ, λ) be a path in Γ, with σ ∈ S(π). Then the following hold.

(a) Y_{ρ} ∩ V_{λ} = Y_{σ}.

(b) W_{ρ} = ⟨x ∈ W_{ρ} : |V_{λ}/C_{V_{λ}}(x)| ≤ 2⟩.

(c) [W_{ρ}, W_{τ}] ∩ Y_{τ} ≤ Y_{σ}.

Proof. Suppose that (a) is false. As Y_{σ} ≤ V_{λ} we then have Y_{ρ} ≤ V_{λ}, and thus [Y_{ρ}, Q_{λ}] ≤ Y_{λ}. As G_{τ} is 2-transitive on Δ(τ), there exists a vertex α at distance b − 2 from λ such that (α, σ) is a critical pair. As σ ∈ S(π), we have ρ = τ^{x} for some x ∈ G_{σ} − G_{τ}, such that x lies in the linkage group K for π at σ.

Take H = σ and take P = τ, so that S = G_{σ} ∩ G_{τ} = Q_{σ}Q_{τ}. Set R = ⟨(Y_{α})^{Q_{τ}}. Then [Y_{ρ}, R] ≤ [V_{λ}, R] = 1, and since σ ∈ S(π) we then have R ≤ G_{ρ}. Notice that 1 ≠ \overline{R} ⊴ \overline{S}. As \overline{S} is transitive on K, we then have [\overline{R}, \overline{K}] ≠ 1. Then H = ⟨S^{x}, R⟩, and we may conclude that Q_{ρ} ∩ Q_{σ} = C_{Q_{σ}}(Y_{ρ}) is normal in H. This contradicts —, and so (a) is proved.

Suppose next that (b) is false. Then W_{ρ} ≰ Q_{λ}, and so b = 5. Choose γ ∈ Δ(ρ) with V_{γ} ≠ ⟨x ∈ G_{γ} : |V_{λ}/C_{V_{λ}}(x)| ≤ 2⟩. It then folows from Lemma 5(b) that there exists δ ∈ Δ(γ) such that γ ∈ S(δ, γ, ρ) and such that, for some x ∈ G_{δ}, we have |V_{λ}/C_{V_{λ}}(x)| > 2. Here V_{λ} ≤ G_{γ}, V_{λ} acts quadratically on V_{γ}, and V_{λ} ≰ G_{δ}. In particular, V_{λ} does not centralize the linkage group K for (δ, γ, ρ) modulo C_{G_{γ}} (Ṽ_{γ}). 7.6 then yields |V_{λ}/V_{λ} ∩G_{δ}| = 2, and then also [Y_{δ}, V_{λ} ∩G_{δ}] ≠ 1. Thus, we have Y_{γ} = [Y_{δ}, V_{λ} ∩G_{δ}] ≤ V_{λ}, which contradicts (a).

Finally, suppose that (c) is false. Again, we have b = 5, as otherwise b ≥ 7 and [W_{ρ}, W_{τ}] = 1. Write Δ(ρ) = {σ, γ, γ′} and Δ(τ) = {σ, λ, λ′}. If W_{ρ} ≤ Q_{λ} and [W_{ρ}, V_{λ}] ≠ 1 then [V_{γ}, V_{λ}][V_{γ′}, V_{λ}] = Y_{λ}. and so Y_{λ} is contained either in V_{γ} or in V_{γ′}, contrary to (a). Thus, if W_{ρ} ≰ Q_{λ} then [W_{ρ}, V_{λ}] = 1. As [W_{ρ}, W_{τ}] ≠ 1, we may then fix notation (replacing λ by λ′, or γ by γ′, if necessary) so that V_{γ} ≰ Q_{λ}.

By Lemma 7.9 we may choose μ ∈ Δ(λ) so that λ ∈ S(τ, λ, μ), and so that V_{γ} ≰ G_{μ}. Suppose that Y_{μ} ≤ Q_{γ}. Then [V_{γ}, Y_{μ}] = Y_{γ}, and thus Y_{γ} ≤ V_{λ}, which is contrary to (a). We conclude that Y_{μ} ≰ Q_{γ}, and then 7.9 implies that there exists δ ∈ Δ(γ) such that γ ∈ S(δ, γ, ρ) and with Y_{μ} ≰ G_{δ}.

Suppose that [W_{δ}, Y_{λ}] = 1. As σ ∈ S(π) we then have W_{δ} ≤ Q_{τ} ≤ G_{λ}. As [W_{δ}, V_{γ}] = 1, we have [V_{λ}, V_{γ}, W_{δ}] = 1. Then |W_{δ}/W_{δ} ∩ G_{μ}| = 2, by Lemma 8, and then also |W_{δ}/C_{W_{δ}}(Y_{μ})| ≤ 4. Now 7.5 yields r = 1 and |V | = 8. Since also b = 5, we have a contradiction to 7.6. We therefore conclude that [W_{δ}, Y_{λ}] ≠ 1. On the other hand, we have Y_{λ} ≤ Y_{τ} ≤ Y_{σ}[W_{ρ}, W_{τ}], by assumption in (c). Thus Y_{λ} ≤ Y_{σ}[V_{γ}, W_{τ}][V_{γ′}, W_{τ}], and so 1 ≠ [Y_{λ}, W_{δ}] ≤ [V_{γ′}, W_{τ} , W_{δ}].

By (b), above, both W_{τ} and W_{δ} are generated by elements which centralize hyperplanes of V_{γ′}. Since W_{τ} W_{δ} is contained in the 2-group G_{ρ} ∩ G_{γ′} it follows that [V_{γ′}, W_{τ} , W_{δ}] ≤ Y_{γ′}. Thus, [Y_{λ}, W_{δ}] = Y_{γ′}. Write Δ(δ) = {γ, β, β′}. As Y_{λ} ≤ Q_{δ} we conclude that, up to a possible permutation of {β, β′}, we have Y_{γ′} ≤ V_{β}. But this contradicts (a), with (ρ, γ, δ, β) in place of π. The proof of (c) is thereby complete. □

Lemma 7.12. Assume that b is odd, b ≥ 5. There then exists a critical pair (α, α′), a path π = (α, β, ⋯, α′) of length b, and vertices μ ∈ Δ(α′)−{α′ −1} and λ ∈ Δ(μ)−{α′} such that the following hold.

(i) Both β and α′ are in S(π ∘ (μ)).

(ii) Y_{α} is not contained in G_{μ}.

(iii) If W_{μ} ≰ G_{β+1}, then β + 2 ∈ S(π) and Y_{β+2} ≰ [V_{β+2}, W_{μ}].

Proof. Suppose false, and choose a critical path π = (α, β, ⋯, α′) so that |S(π)| is as large as possible. Then β ∈ S(π) by 7.10 (applied to π(β, α′)). By 7.9, we may choose μ ∈ Δ(α′) − {α′ − 1} so that α′ ∈ S(π ∘ (μ)) and so that Y_{α} ≰ G_{μ}.

As the lemma is assumed false, we have W_{μ} ≰ G_{b+1}. Choose λ ∈ Δ(μ) − {α′} so that V_{λ} ≰ G_{β+1}. Suppose next that V_{β+2} ≤ Q_{λ}. Then [V_{β+2}, V_{λ}] = Y_{λ}, and since V_{λ} acts non-trivially on V_{β+2}/Y_{β+2} we conclude that Y_{β+2} ≰ [V_{β+2}, V_{λ}]. As the lemma is false, we then have β + 2 ∉ S(π). Now set π_{0} = (λ, μ) ∘ \overline{π}(α′, β + 2) and apply 7.10 to π_{0}. Thus, there exists γ_{0} ∈ Δ(λ) − {μ} such that (γ_{0}, β + 2) is a critical pair and such that, for π′ = (γ_{0}) ∘ π_{0}, we have λ ∈ S(π′). Notice however that |S(π′)| > |S(π)|, which is contrary to our choice of π. We conclude that V_{β+2} ≰ Q_{λ}.

We now apply 7.10 to the path π_{1} = π(β + 2, α′) ∘ (μ, λ). Thus, there exists a vertex γ_{1} ∈ Δ(β+2) such that (γ_{1}, λ) is a critical pair, and such that, upon setting π^{∗} = (γ_{1})∘π_{1}, we have β + 2 ∈ S(π^{∗}). Observe that |S(π^{∗})| ≥ |S(π)|. Thus, we may replace π by π^{∗}, and since α′ ∈ S(π^{∗}) we then obtain α′ − 2 ∈ S(π). But then maximality of |S(π)| shows that, to begin with, and before making the above replacement, we had α′ − 2 ∈ S(π). By iteration it now follows that S(π) = {β, β + 2, ⋯, α′ − 2}. Moreover, since the lemma fails to hold for π^{∗}, we have (prior to replacement) Y_{β+2} ≤ [V_{β+2}, W_{μ}], and thus (after replacement) Y_{β} ≤ [V_{β}, W_{α′−1}] ≤ W_{α′−1}. Lemma 7.11(c) then yields b > 5, so that [W_{α′−1}, W_{μ}] = 1, whence [Y_{β}, W_{μ}] = 1. But with β +2 ∈ S(π) we then have W_{μ} ≤ Q_{β+1}. Thus W_{μ} ≤ G_{β+1} after all, and the lemma is proved. □

Proposition 7.13. Assume that b is odd, b ≥ 5. Then |V | = 8.

Proof. Assume that |V | > 8. Choose π and μ as in 7.12, and fix t ∈ Y_{α} − Y_{β}. Suppose first that Y_{β} ≤ [V_{β}, V_{α′}]. Then Y_{β} ≤ V_{α′} ≤ W_{μ}, and it then follows from 7.12 that W_{μ} ≤ G_{β}. Suppose that W_{μ} acts quadratically on V_{β}, or that V_{α′} ≰ G_{α}. As β ∈ S(π), 7.8 then implies that |W_{μ}/W_{μ} ∩ G_{α}| ≤ 2, and so |W_{μ}/C_{W_{μ}}(t)| ≤ 4. As 7.5 then yields r = 1, we conclude that W_{μ} is not quadratic on V_{β} and that V_{α′} ≤ G_{α}. In particular, it follows that [W_{μ}, (W_{μ})^{g}] ≠ 1 for some g ∈ V_{β}, and so b = 5. Further, with V_{α′} ≤ G_{α} we have |V_{α′}/C_{V_{α′}} (t)| = 2.

Now V_{β} = (V_{β} ∩ G_{μ})Y_{α}, and W_{m} acts quadratically on the hyperplane V_{β} ∩ G_{μ} of V_{β}. It follows that the W_{μ}-orbits on K_{β} are of length at most 2, and that |W_{μ}/W_{μ} ∩G_{α}| ≤ 4. Thus, we have shown that [t, W_{μ}, W_{μ}] ≠ 1, and |W_{μ}/C_{W_{μ}}(t)| ≤ 8 in the case that Y_{β} ≤ V_{α′} ≤ W_{μ}.

Suppose, on the other hand, that Y_{β} ≰ [V_{β}, V_{α′}]. Then [t, V_{α′} ∩ G_{α}] = 1, and so |V_{α′}/C_{V_{α′}} (t)| = 2, as in the preceding case. Further, we have V_{α′} ≰ Q_{β}. Notice that |W_{μ}/W_{μ} ∩ G_{β}| ≤ 2, by 7.12. As [V_{β}, V_{α′}, W_{μ}] = 1, 7.8 then yields |W_{μ}/W_{μ} ∩ G_{α}| ≤ 4, and so once again |W_{μ}/C_{W_{μ}}(t)| ≤ 8. Moreover, we either have |W_{μ}/C_{W_{μ}}(t)| ≤ 4 or Y_{β} ≤ [t, W_{μ}∩G_{α}]. If now [t, W_{μ}, W_{μ}] = 1 we thus obtain [Y_{β}, W_{μ}] = 1 and |W_{μ}/C_{W_{μ}}(t)| ≤ |W_{μ}/(W_{μ} ∩ Q_{α})| ≤ 4, again contrary to 7.5. We have thus shown that, in any case, we have the following information.

(1) b = 5, [t, W_{μ}, W_{μ}] ≠ 1, |W_{μ}/C_{W_{μ}}(t)| = 8, and |V_{α′}/C_{V_{α′}} (t)| = 2.

Set σ = μ^{t}, and observe that since t induces a transvection on Ṽ_{α′}, we have α′ ∈ S(σ, α′, μ). Further, we have the following consequence of 7.2.

(2) α′ ∈ S(σ, α′, μ).

Suppose next that W_{μ} ≤ Q_{ρ} for all ρ ∈ Δ(σ). Then [W_{μ}, W_{σ}] ≤ Y_{σ}, and then 7.11(c) yields [W_{μ}, W_{σ}] ≤ Y_{α′}. For any ρ ∈ Δ(σ) − {α′} we then have [V_{ρ}, W_{μ}] ≤]yr ∩ Y_{α′} = 1, and so [W_{μ}, W_{σ}] = 1. This implies that [t, W_{μ}, W_{μ}] = 1, contrary to (1). We conclude that there exists ρ ∈ Δ(σ) with W_{μ} ≰ Q_{ρ}. Fix such a vertex ρ and set l = ρ^{t}. Then W_{σ} ≰ Q_{λ}.

Notice that W_{μ} ∩ W_{σ} ≥ C_{W_{μ}}(t)V_{α′}. Then (1) yields:

(3) W_{μ} ∩ W_{σ} is of index at most 4 in W_{μ} and in W_{σ}.

We note also that 7.11(b) yields the following.

(4) W_{σ} = ⟨x ∈ W_{σ} : |V_{λ}/C_{V_{λ}}(x)| ≤ 2⟩.

Suppose that |V_{λ}/C_{V_{λ}}(W_{σ})| = 2. Noting that Y_{λ} ≰ [W_{σ}, V_{λ}] by 7.11(c), we then have |[W_{σ}, V_{λ}]| = 2. Set U = [W_{σ}, V_{λ}]. Since Q_{α′} ∩ Q_{μ} is not normal in G_{α′}, by 1.9, where G_{α′} = ⟨G_{σ} ∩ G_{α′}, G_{α′} ∩ G_{μ}λ⟩, it follows that Q_{α′} ∩ Q_{μ} ≠ Q_{α′} ∩ Q_{σ}, and hence (Q_{α′} ∩ Q_{μ})(Q_{α′} ∩ Q_{σ}) = Q_{α′}. In particular, Q_{α′} ∩ Q_{μ} is transitive on Δ(σ) − {α′}. As W_{σ} ≰ Q_{λ} we then have V_{ρ} ≰ Q_{λ}, and so U = [V_{ρ}, V_{λ}] is t- invariant. As U lies in the center of Q_{α′} ∩ Q_{μ} we then have [U, Q_{α′}] = 1, and then 1.1(a) shows that U = Y_{α′}. Lemma 7.1 then implies that |V | = 8, which is contrary to the hypothesis. Thus |V_{λ}/C_{V_{λ}}(W_{σ})| > 2.

Now (3) implies that W_{μ} = C_{W_{μ}}(W_{σ})V_{λ}. Then [W_{σ}, V_{λ}] = [W_{σ}, W_{μ}] ⊴ Q_{α′}, and so Y_{α′} ≤ [W_{σ}, V_{λ}], by 1.8. Then (4) and Lemma 1 imply that W_{σ}Q_{λ}/Q_{λ} is of order 2^{r}, and (3) then yields r ≤ 2. As |V | > 8, Lemmas 1 and 2 now give:

(5) |V | = 32 and H/Q_{H} ≅ Ω_{4}^{+}(2).

For any edge (γ, δ) of Γ, with γ of type H, set T (γ, δ) = ⟨g ∈ G_{γ} ∩G_{δ} : |[V_{γ}/Y_{γ}, g]| ≤ 2⟩. For any vertices ϕ and ψ in Δ(α′), set F (ϕ, ψ) = [W_{ϕ}, W_{ψ}]. Set F = F (σ, μ). We have seen that |W_{σ}/C_{W_{σ}}(V_{λ})| = 4, that W_{σ} is generated by C_{W_{σ}}(V_{λ}) together with two elements that induce transvections on V_{λ}, and that [W_{σ}, W_{μ}] = [W_{σ}, V_{λ}] ≥ Y_{α′}. Thus F is a fours group containing Y_{α′}, and we have F Y_{λ}/Y_{λ} = [V_{λ}/Y_{λ}, T (λ, μ)]. Setting U = F Y_{λ}, it follows that U ⊴ G_{μ} ∩ G_{λ}. But also F, and hence also U, is normal in Q_{α′}, and so U ⊴ G_{μ}. Then U ≤ V_{α′} ∩ V_{λ}, and U has index 4 in V_{λ}. Since |V_{λ}/C_{V_{λ}}(W_{σ}| = 4, it follows that U = V_{α′} ∩ V_{λ}, and so U = D_{α′}, by 1.17. Similarly, we have F Y_{ρ} = D_{σ}, and 7.11(a) yields F = D_{μ} ∩ D_{σ}.

Write K_{α′} = {K_{1}, K_{2}} and define X_{i} = X_{α′}^{(i)} to be the inverse image in V_{α′} of [V_{α′}/Y_{α′}, K_{i}], (i = 1, 2). As U ⊴ G_{μ} we have U = [V_{α′}, T (α′, μ)], and since Y_{λ} ≰ F it follows that F ≤ X_{i} for some i. For definiteness, take F ≤ X_{1}. Then [F, N_{S}(X_{1})] ≤ Y_{α′}. On the other hand we have U = F Y_{λ} = F Y_{μ}, and so [U, Q_{μ}] = [F, Q_{μ}] = Y_{μ}. As D_{μ} ≠ Y_{μ}, Q_{μ}/C_{Q_{μ}}(F) is a fours group admitting non-trivial action by O^{2}(G_{μ}). Setting R = O_{2}(O^{2}(G_{μ})), we then have [F, R] = Y_{μ}, and thus R ≰ N_{S}(X_{1}). As R ⊴ S, R/R∩Q_{α′} then contains a fours group.

Let us record these results.

(6) We have F (σ, μ) := [W_{σ}, W_{μ}] = V_{ρ} ∩ V_{α′} ∩ V_{λ} = [X_{1}, T (α′, μ)].

(7) R/R ∩ Q_{α′} contains a fours group, where R := O_{2}(O^{2}(G_{μ})).

We have |Δ(α′)| = 9, and we may identify Δ(α′) with the set of singular points in the orthogonal space V_{α′}/Y_{α′}, for the action of G_{α′}. Define a map

B : Δ(α′) × Δ(α′) ⟶ 𝔽_{2}

by the formula

B(ϕ, ψ) = \{ \begin{matrix}0, & if [W_{ϕ}, W_{ψ}] = 1 \\ 1, & if [W_{ϕ}, W_{ψ}] ≠ 1. \end{matrix}

We require the following elementary lemma (for which we need provide no proof).

Lemma 7.14. Let Ṽ be an O_{4}^{+}(2)-space with associated bilinear form B̃, let t be an orthogonal transvection on Ṽ, and let x and v be singular points with x = x^{t} and v ≠ v^{t}. Then the following hold.

(a) B̃(v, v^{t}) ≠ 0.

(b) There exists a singular point y, with y ≠ y^{t}, and such that B̃(x, y) = 0.

Notice that S has two orbits on Δ(α′) − {μ}, each orbit being of length 4. Let τ ∈ Δ(α′) with τ ∉ σ^{S}. Thus, we have τ = μ^{g} where g is an element of order 3 in G_{α′} which is fixed-point-free on V_{α′}/Y_{α′}. For any δ ∈ Δ(τ) − {α′} one may then observe that V_{δ} ∩ V_{α′} ∩ V_{λ} = Y_{α′}, and then for any λ′ ∈ Δ(μ) we have [V_{δ}, V_{λ′}] ≤ Y_{α′}. Lemma 7.1 then implies that [V_{δ}, V_{λ′}] = 1, and thus B(τ, μ) = 0. Since B(σ, μ) = 1, where σ = μ^{t}, it follows from 7.13(a) that B is indeed the bilinear form associated with the orthogonal space V_{α′}/Y_{α′}. Then 7.13(b) says that we may choose τ as above, with B(α′ − 1, τ) = B(μ, τ) = 0, and with t ∉ G_{τ}. In particular, we have [W_{τ} , V_{β+2}] = 1, and so W_{τ} stabilizes the path π(β, α′). As t ∉ G_{τ}, 7.5 shows that |W_{τ} /C_{W_{τ}} (t)| ≥ 8. Then |W_{τ} /W_{τ} ∩ G_{α}| = 4 (so that W_{τ} induces a non-quadratic fours group on V_{β}/Y_{β}) and [Y_{α}, W_{τ} ∩ G_{α}] = Y_{β}. As t ∈ X_{β}^{(i)} for some i, (i = 1 or 2), one may then compute that [t, W_{τ}] = [V_{β}, G_{β} ∩ G_{β+1}], and that [t, W_{τ} , W_{τ}] = Y_{β+1}. On the other hand, 7.13(a) shows that [(W_{τ})^{t}, W_{τ}] ≠ 1, and so [t, W_{τ} , W_{τ}] = [(W_{τ})^{t}, W_{τ}] = F (τ^{t}, τ) ≤ V_{α′}, of order 4. Thus F (τ^{t}, τ) = Y_{β+1} ≤ V_{α′}. This contradicts 7.11(a), thus completing the proof of 7.12. □

[THERE MUST BE AN EASIER ARGUMENT FOR 7.13]

Section 8: The solvable case, b = 3

Lemma 8.1. Suppose that P and H are solvable, with p = 2. Then b = 3, and one of the following holds.

(i) H/O_{2}(H) ≅ SL(2, 2) and |V | = 8.

(ii) H/O_{2}(H) ≅ Sz(2) and |V | = 32.

(iii) \overline{H} is isomorphic to a subgroup of ≅ O_{4}^{+}(2) of index at most 2, and |V | = 32. (iv) O_{3}(\overline{H}) is extraspecial of order 27 and of exponent 3, and

( a) There exists P_{1} ∈ P_{H}(S) such that O_{3}(\overline{P}_{1}) = Z(O_{3}(\overline{H})) and such that

⟨(Y_{P})^{P_{1}}⟩ is invariant under P. ( b) Ṽ is a natural SU(3, 2)-module for O^{2}(\overline{H})O_{2}(\overline{P}_{1}). ( c) O^{2}(\overline{H})O_{2}(\overline{P}_{1}) is isomorphic to a subgroup of SU(3, 2) of index at most

2.

Proof. Suppose b > 3. Then 7.12 yields |V | = 8 and H/O_{2}(H) ≅ SL(2, 2). Now [AN ARGUMENT EXTRACTED FROM THE WEAK BN-PAIRS PAPER] yields a contradiction. Thus, we have b = 3. Set γ = β + 1 = α′ − 1, and take H = β and P = γ. Denote by δ the unique element of Δ(γ) − {β, α′}.

As b > 2 we have V ≤ Q_{P}, and so 1.10 implies that [Q_{H}, O^{2}(H)] ≰ V. Thus, there is at least one non-central chief factor X for H in Q_{H}/V.

Suppose first that Q_{β} ∩ Q_{γ} ≤ V_{β}Q_{α′}. Then [Q_{β} ∩ Q_{γ}, V_{α′}] ≤ V_{β}Y_{α′} = V_{β}. As |Q_{β}/(Q_{β}∩Q_{γ})| = 2, V_{α′} induces a transvection on X. But V_{α′}Q_{β} ⊴ Q_{β}Q_{γ} = Q_{H}Q_{P} = S, and it follows that H/C_{H}(X) ≅ SL(2, 2). Then H/Q_{H} is dihedral of order 2·3^{m} for some m, m ≥ 1. As Ṽ is a quadratic F 2-module for H, it follows from 3.11 that \overline{H} ≅ SL(2, 2). As Ṽ is the closure under H of Ỹ_{P}, we then have |Ṽ| ≤ 8, and then 7.2 implies that H/Q_{H} ≅ SL(2, 2). Suppose that |Ṽ| = 8. As P is doubly transitive on Δ(γ) we have [Ṽ, V_{δ}] = [Ṽ, V_{α′}], and 7.1 implies that neither Y_{β} nor yd is contained in [vb, V_{α′}]. But [V_{β}, V_{α′}] ⊴ Q_{γ}, so Y_{γ} ∩ [V_{β}, V_{α′}] ≠ 1. Then Y_{γ} ∩ [V_{β}, V_{α′}] = Y_{β}. This contradicts 7.2, applied to the action of V_{β} on V_{α′}, so we conclude that |V | = 8 and that outcome (i) of the lemma holds. We may therefore assume that Q_{β} ∩ Q_{γ} ≰ V_{β}Q_{α′}. Then also:

(1) Q_{γ} ∩ Q_{α′} ≰ V_{α′}Q_{β}.

Set A = V_{α′}. As [A, Q_{α′}] = Y_{α′} ≤ Q_{β}, (1) yields:

(2) C\overline{S}(\overline{A}) ≰ \overline{A}.

Suppose that A is an F 1-offender on Ṽ. Then 3.11 shows that \overline{A} is generated by transvections, and since \overline{A} ⊴ \overline{S} it follows that \overline{A} contains all the transvections in \overline{S}. This contradicts (2), and so, in fact, \overline{A} is not an F 1-offender on Ṽ, and similarly V Q_{α′}/Q_{α′} is not an F 1-offender on V_{α′}/Y_{α′}. Then by 1.18 and symmetry, we may assume that [V ∩ Q_{α′}, A] = Y_{α′}. We then have:

(3) There exists v ∈ Ṽ such that [v, A] = Ỹ_{P}.

We note also that A is a quadratic F 2-offender on Ṽ, by 2.3, and then 3.10 shows that \overline{A} is generated by 2-transvections.

By 3.11 there is an \overline{S}-transitive collection D = \{\overline{D}_{i}\}_{1≤i≤r} of subgroups \overline{D} of \overline{H}, such that O^{2}(\overline{H}) = \overline{D}_{1} ×⋯×\overline{D}_{r}, with [Ṽ, O^{2}(\overline{H})] = [Ṽ, \overline{D}_{1}]×⋯×[Ṽ, \overline{D}_{r}], and such that \overline{D}_{i} is isomorphic to ℤ_{3}, ℤ_{5}, or O_{3}(SU(3, 2)). Set U_{i} = [Ṽ, \overline{D}_{i}]. The transitivity of \overline{S} on D then implies that there are non-identity elements u_{i} of U_{i} such that Ỹ_{P} = ⟨u_{1} ⋯ u_{r}⟩. On the other hand, (3) shows that u_{1} ⋯ u_{r} = [v, \overline{a}] for some v ∈ Ṽ and some 2-transvection \overline{a} in \overline{A}. It follows that either r = 1 or that |U_{i}| = 4 and r = 2. In any case we have |\overline{A}| = 2 and \overline{A} induces a 2-transvection on [Ṽ, O^{2}(H)], so (3) implies that [Ṽ, O^{2}(H)] = Ṽ.

If r = 2 we have the desired outcome (iii) of the lemma, so we assume now that r = 1. Then |O^{2}(\overline{H})| > 3, by (2). Suppose that |O^{2}(\overline{H})| = 5. Then (2) implies that \overline{H} ≅ Sz(2), while |V | = 32. Then also H/Q_{H} ≅ Sz(2), by 1.11(b), and we have outcome (ii). Finally, suppose that O^{2}(\overline{H}) is an extraspecial group of order 27 and exponent 3, with |Ṽ| = 64. Denote by H_{1} the inverse image in H of Z(O^{2}(\overline{H})). Then ⟨(Y_{P})^{H_{1}}⟩ is of order 8, and 1.7 then shows that O^{2}(P) is not invariant under H_{1}. By 1.4 there then exists P_{1} ∈ P_{H_{1}S}(S) such that O^{2}(P) is not P_{1}-invariant, and such that ⟨P, P_{1}⟩ ∈ L. Then O^{2}(\overline{P}_{1}) = Z(O^{2}(\overline{H})).

Notice that O_{2}(\overline{P}_{1}) is a subgroup of index at most 2 in a quaternion group. Thus |Ω_{1}(\overline{Q_{γ}})| = 2. For any two distinct vertices δ and δ′ in Δ(γ) we then have [V_{δ}, V_{δ′}] = [V_{δ}, Ω_{1}(\overline{Q_{γ}})], and thus [V_{δ}, V_{δ′}] is G_{γ}-invariant. Thus [V_{β}, V_{α′}] is invariant under both P and P_{1}, and (iv) holds. □

Lemma 8.2. Outcome (iv) of Lemma 8.1 does not hold.

Proof. Suppose false, and let P_{1} be as in 8.1(iv). Set L = ⟨P, P_{1}⟩. As we have seen, we have L ∈ L, and Y_{P} is not normal in L.

Form the amalgam Γ^{∗} = Γ(H, L) and set b^{∗} = b(H, L). It follows from 1.3(4) and from 8.1(iv) that |Y_{L}| = 8 and that L/O_{2}(L) ≅ SL(3, 2). In particular, we have Y_{L} ≤ ⟨(Y_{P})^{H}⟩, and so we have V = ⟨(Y_{L})^{H}⟩. Then b^{∗} ≥ 3, and since b^{∗} ≤ b we conclude that b^{∗} = 3. We may then label a critical path (α, β, γ, α′), and we shall take H = β and L = γ, so that P_{1} = G_{β} ∩ G_{γ}. For δ a vertex with stabilizer H^{g} we write V_{δ} for V^{g}. We note that, since L is doubly transitive on Y_{L}^{♯}, G_{γ} is doubly transitive on Δ(γ), and hence G_{β} is transitive on Δ^{(2)}(β). In particular, any path of length 2 from β is a critical path.

Wishing to avoid needless repetition, we ask the reader to check that the argument at the relevant point in the proof of 8.1, above, shows:

(1) |V/C_{V} (V_{α′})| = 4, C_{V} (V_{α′}) = [V, O_{2}(P_{1})], and [V, V_{α′}] = Y_{L}.

Set X = X_{β} = ⟨C_{V_{δ}}(V) : δ ∈ Δ^{(2)}(β)⟩. Then X_{α′} = Ω_{1}(X_{α′}) and X_{α′} ≤ Q_{γ} ≤ O_{2}(G_{γ} ∩ G_{β}) = O_{2}(P_{1}). Here Ṽ is a natural SU(3, 2)-module for O^{2}(H)O_{2}(P_{1}), and Ω_{1}(O_{2}(\overline{P}_{1})) is of order 2, so [Ṽ, X_{α′}] ≤ [Ṽ, V_{α′}], and then [V, X_{α′}]Y_{L}. Thus [V, X_{α′}] ≤ V_{α′}, and so also [X, V_{α′}] ≤ V, by symmetry. It follows that [X, O^{2}(H)] ≤ V. Further, we have [C_{Q_{H}} (V), V_{α′}] ≤ [Q_{γ}, V_{α′}] ≤ C_{V_{α′}} (V), by (1), and so [C_{Q_{H}} (V), O^{2}(H)] ≤ X. This now shows:

(2) [C_{Q_{H}} (V), O^{2}(H)] = V.

We have [V, Q_{H}] = Y_{H}, so:

(3) Q_{H}/C_{Q_{H}} (V) is isomorphic to the dual of Ṽ as a module for H.

Set D = ⟨[V_{δ}, Q_{γ}] : δ ∈ Δ(γ)⟩. Then (1) shows that D ≤ X ∩ X_{α′}, so [D, V_{α′}] = 1, and so V ≰ D. As η(P_{1}, V) = 3, it follows from (2) that η(P_{1}, D/Y_{L}) = 1. As (V ∩D)/Y_{L} is a P_{1}-invariant subgroup of D/Y_{L} of order 4, it follows that D/Y_{L} is an L-module of order 8, dual to Y_{L}. Thus V ∩ D is a hyperplane of D.

Set W = W_{γ}. Then D ≤ Ω_{1}(Z(W)). We have C_{S}(W) ≤ C_{Q_{H}} (V), and since V ≰ C_{S}(W) we then have η(P_{1}, C_{S}(W)) = 2. Thus [C_{S}(W), O^{2}(L)] = D. Suppose that there exists an element x in Q_{H} such that [x, H] = Y_{H}. From (3) we have x ∈ C_{H}(V), so x ∈ G_{γ}^{(2)} ≤ G_{α′}. As |[x, V_{α′}]| ≤ 2 we have x ∈ Q_{α′}, and since [x, V_{α′}] ≠ Y_{α′} we conclude that [x, V_{α′}] = 1. In this way we have x ∈ G_{γ}^{(4)} and, in particular, x ∈ C_{S}(W). Set R = ⟨x^{L}⟩. Then Y_{L} ≤ R ≤ ⟨x⟩D. As [x, P_{1}] ≤ Y_{H} ≤ Y_{L}, and as C_{D/Y_{L}}(P_{1}) = 1, lemma — [FROM J-MODULES PRELIM LEMMAS] implies that R/Y_{L} is of order 2. But then [x, Q_{L}] = Y_{L}, contrary to |S/C_{S}(x)| = 2. We conclude that no such element x exists, and this yields the following result.

(4) C_{H}(O^{2}(H)) = Y_{H}.

Let d ∈ D − V. In particular, we have d ∈ C_{Q_{H}} (V). If [d, Q_{H}] = Y_{H} then (2) implies that there exists d′ ∈ V d such that [d′, O^{2}(H)] = 1, and we contradict (4). Thus [d, Q_{H}] = V, and it follows from (3) that |[Q_{L} ∩ Q_{H}, d]| ≥ 16. As [d, Q_{L}] ≤ Y_{L}, of order 8, we have a contradiction, and the lemma is proved. □

[THE REMAINDER OF THIS SECTION (AND THE NEXT SECTION) WILL BE A CLOSE COPY OF BERND’S PAPER ON N-GROUPS, SECTION 10, TO PRODUCE THE AMALGAM FOR M_{12}, Aut(M_{12}), the Tits group, OR ^{2}F_{4}(2). AFTER THAT, IMITATE MICHAEL IN ORDER TO PRODUCE THE GROUPS.]