A Characteristic Subgroup for Pushing Up in Finite Groups

Andy Chermak, Ulrich Meierfrankenfeld

September 19, 2002

1 Introduction

2 The Kieler Lemma and Pointstabilizers

An elementary abelian normal subgroup V of a finite group L is called p-reduced if any subnormal subgroup of L which acts unipotently on V has to act trivially. Note that this is equivalent to O_{p}(L/C_{L}(V)) = 1. Here are the basic properties of p-reduced normal subgroups. Comment:due to Thompson? check history

Lemma 2.1 [YL] Let L be a finite group of characteritic p and T ∈ Syl_{p}(L)

(a) [a] There exists a unique maximal p-reduced normal subgroup Y_{L} of L.

(b) [b] Let T ≤ R ≤ L and X a p-reduced normal subgroup of R. Then ⟨X^{L}⟩ is a p-reduced normal subgroup of L. In particular, Y_{R} ≤ Y_{L}.

(c) [c] Let T_{L} = C_{T} (Y_{L}) and L_{T} = N_{G}(T_{L}). Then L = L_{T} C_{L}(Y_{L}), T_{L} = O_{p}(L_{T}) and Y_{L} = Ω_{1}Z(T_{L}).

(d) [d] Y_{T} = Ω_{1}Z(T), Z_{L} := ⟨Ω_{1}Z(T)^{L}⟩ is p-reduced for L and Ω_{1}Z(T) ≤ Z_{L} ≤ Y_{L}.

Now let L be any finite group and T ∈ Syl_{p}(L). Define P_{L}(T) := O^{p′}(C_{L}(Ω_{1}Z(T))). Then P_{L}(T) is called a point stabilizer of L. We call O_{p}(L) the p-radical of L, or just the radical of L. The radical of P_{L}(T) is called a p-central radical of L. A subgroup R of L is called a radical subgroup of L if R = O_{p}(N_{L}(R)), that is R is the radical of its normalizer. The following lemma ist the principal tool for working with point stabilzers.

Lemma 2.2 [kieler lemma] Let H be a finite group of local characteristic p, T ∈ Syl_{p}(H) and L a subnormal subgroup of H. Then

(a) [a] [Kieler Lemma] C_{L}(Ω_{1}Z(T)) = C_{L}(Ω_{1}Z(T ∩ L))

(b) [b] P_{L}(T ∩ L) = O^{p′}(P_{H}(T) ∩ L)

(c) [c] C_{L}(Y_{L}) = C_{L}(Y_{H})

(d) [d] Suppose L = ⟨L_{1}, L_{2}⟩ for some subnormal subgroups L_{1}, L_{2} of H. Then

(a) [da] P_{L}(T ∩ L) = ⟨P_{L_{1}}(T ∩ L_{1}), P_{L_{2}}(T ∩ L_{2})⟩.

(b) [db] For i = 1, 2 let P_{i} be a point stabilizer of L_{i}. Then ⟨P_{1}, P_{2}⟩ contains a point stabilizer of L.

The proof of the above lemma is elementary and does not require any K-group assumption assumption. Comment: not all parts of this lemma are really needed

Lemma 2.3 [minimal overgroups] Let H be a finite group and F < H.

(a) [a] Let I_{H}(F) be the set of all I with F < I ≤ H such that F lies in a unique maximal subgroup of I. Then H = ⟨I_{H}(F)⟩.

(b) [b] Let J_{H}(F) = {I ∈ I_{G}(F) | F ⋬ I}. Then H = ⟨J_{H}(F)⟩N_{H}(F).

Proof: By induction on |H|. Suppose that F lies in two different maximal subgroups M_{1}, M_{2} of H. By induction, M_{i} = ⟨I_{M_{i}}(F)⟩ = ⟨J_{M_{i}}(F)⟩N_{M_{i}}(F). Thus H = ⟨M_{1}, M_{2}⟩ = ⟨I_{H}(F)⟩ = ⟨J_{H}(F)⟩N_{H}(F).

So suppose F lies in a unique maximal subgroup of H. Then H ∈ I and H = ⟨I⟩. Moreover either F is normal in H or H ∈ J. In any case H = ⟨J ⟩N_{H}(F). □

Lemma 2.4 (Schur multipliers) [schur multipliers]

Proof: [Schur] □

3 Modules

Lemma 3.1 (Point Stabilizer Theorem) [the point stabilizer theorem] Let H be a finite group, V a 𝔽_{p}H-module, L a point stabilizer for H on V and A ≤ O_{p}(L).

(a) [a] If V is p-reduced, then |V/C_{V} (A)| ≥ |A/C_{A}(V)|.

(b) [b] If V is irreducible, F^{∗}(H) is quasi-simple, H = ⟨A^{H}⟩ and A is a non-trivial offender on V, then M ≅ SL_{n}(q), Sp_{2n}(q), G_{2}(q) or Sym(n), where p = 2 in the last two cases.

Proof: [BBSM] □

Lemma 3.2 (FF-modules for miminal parabolics) [ff-modules for miminal parabolics]

Proof: [BBSM] □

Lemma 3.3 [spin module] Let H = Sp_{2n}(q), V a 𝔽_{p}H-module, P a point stabilizer for H on the natural module, T = O_{p}(P), Z = Z(P) and W an 𝔽_{p}T submodule of V. Suppose that

(i) [i] V = ⟨W^{H}⟩.

(ii) [ii] [V, T, T] = 1.

(iii) [iii] [V, Z] ≤ W ≤ C_{V} (T).

Let U = ⋂_{h∈H} W^{h}[V, T]^{h} and \overline{V} = V/U. Let h ∈ H with Z ≰ P^{h}. Then

(a) [a] V = [V, Z]C_{V} (T^{h}) = W [V, T]^{h}, \overline{W} = [\overline{W} , T] = C\overline{V} (T) = C\overline{V} (Z) and \overline{V} = \overline{W} ×\overline{W}^{h}.

(b) [b] If [W, H] ≠ 1, then |\overline{V} | ≥ q^{2^{n}} and |V/C_{V} (T)| ≥ q^{2^{n−1}}.

Proof: Let Y = W [V, T]. Then Y ≤ C_{V} (T). Note that H = ⟨Z, T^{h}⟩. Since [V, Z] ≤ W we conclude that H normalizes W [V, T]^{h} and so by (i), V = W [V, T]^{h}. Also H also normalizes [V, Z]Y^{h} and since W^{h} ≤ Y^{h} we conclude V = [V, Z]Y^{h} = [V, Z]C_{V} (T^{h}). Let X/U = C\overline{V} (Z). Then U ≤ X ∩ Y^{h}. Thus H = ⟨Z, T^{h}⟩ normalizes X ∩ Y^{h} and so X ∩ Y^{h} = U.

Thus \overline{V} = \overline{X} × \overline{Y}^{h}. Since V = [V, Z]Y^{h} we also get \overline{V} = [\overline{V} , Z] × \overline{Y}^{h}. This implies [\overline{V} , Z] = \overline{X} = C\overline{V} (Z).

Note that

[\overline{V} , Z] ≤ [\overline{V} , T] ≤ \overline{Y} ≤ C\overline{V} (T) ≤ C\overline{V} (Z)

Now all the inequalities in the preceeding inequalities have to be equalities. So (a) is proved.

To prove (b) suppose that [W, H] ≠ 1. By (a) also [\overline{W} , H] = 1 and so we may assume that U = 1.

Suppose first that n = 1 and 1 ≠ z ∈ Z. Since H = ⟨z, T^{h}⟩, C_{Y^{h}}(z) ≤ U = 1. Let 1 ≠ y ∈ Y^{h}. We conclude that |[y, Z]| ≥ |Z| = q and so |W | ≥ q and |V | ≥ q^{2}.

Suppose next that n > 1 and let H^{∗} = C_{H}(⟨Z, Z^{h}⟩. Then H^{∗} ≅ Sp_{2n−2}(q) and Z^{∗} := Z^{k} ≤ H^{∗} for some k ∈ H. Then P^{∗} := P^{k} ∩ H^{∗} is a point stabilizer for H^{∗} on its natural module, T^{∗} := T^{k} ∩ H^{∗} = O_{p}(P^{∗}) and Z^{∗} = Z(P^{∗}). Since W = C_{V} (Z) and H^{∗} ≤ C_{G}(Z), W is a 𝔽_{p}H^{∗} submodule of W. Suppose that [W, Z^{∗}, H^{∗}] = 1. Let h^{∗} ∈ H^{∗} with Z^{∗h^{∗}} ≰ P^{∗}. Then [W, Z^{∗}] ≤ [V, Z^{∗}] ∩ [V, Z^{∗h^{∗}}] = 1 and so [W, Z^{∗}] = 1. Thus C_{V} (Z) = W = C_{W} (Z^{∗}) and so P and P^{∗} normalize W, a contradcition since H = ⟨P, P^{∗}⟩. Thus [W, Z^{∗}, H^{∗}] ≠ 1. Let V^{∗} = ⟨[W, Z^{∗}]^{H^{∗}}⟩. Then by induction |V^{∗}| ≥ q^{2^{n−1}}. Since V^{∗} ≤ W and |V | = |W |^{2} we get |V | ≥ q^{2^{n}}. □

We remark that ( for example by [BBSM]), \overline{V} from the preceeding lemma must be a direct sum of spin-modules for H.

Lemma 3.4 (H1 of natural modules) [h1]

Proof: [BBSM] □

4 The Baumann subgroup

For a p-group R we let PU_{1}(R) be the class of all finite groups L containing R such

(a) [a] L is of characteristic p,

(b) [b] R = O_{p}(N_{L}(R))

(c) [c] N_{L}(R) contains a point stabilizer of L.

Let PU_{2}(R) be the class of all finite groups L containg R such that L is of characteristic p and

L = ⟨N_{L}(R), H | R ≤ H ≤ L, H ∈ PU_{1}(R)⟩.

Let PU_{3}(R) be the class of all finite groups L such that

(a) [a]

L is of characteristic p.

(b) [b] R ≤ L and L = ⟨R^{L}⟩

(c) [c] L/Co_{p}(L) ≅ SL_{n}(q), Sp_{2n}(q) or G_{2}(q), where q is a power of p and p = 2 in the last case.

(d) [d] Y_{L}/C_{Y_{L}}(L) is the corresponding natural module.

(e) [e] O_{p}(L) ≤ R and N_{L}(R) contains a point stabilizer of L.

(f) [f] If L/C_{L}(Y_{L}) ≅ G̸_{2}(q) then R is a radical in L. Let PU_{4}(R) be the class of all finite groups L containg R such that L is of characteristic

p and

L = ⟨N_{L}(R), H | R ≤ H ≤ L, H ∈ PU_{3}(R)⟩.

Let B(R) = C_{R}(Ω_{1}Z(J(R))), the Baumann subgroup of R. Recall that a finite group F is p-closed if O^{p′}(F) = O_{p}(F). Lemma 4.1 (Baumann Argument) [baumann argument] Let L be a finite group, R a p-sugroup of L, V := Ω_{1}Z(O_{p}(L)), K := ⟨B(R)^{L}⟩, Ṽ = V/C_{V} (O^{p}(K)), and suppose that each of the following holds:

(i) [i] O_{p}(L) ≤ R and L = ⟨J(R)^{L}⟩N_{L}(J(R)).

(ii) [ii] C_{K}(Ṽ) is p-closed.

(iii) [iii] |Ṽ /CṼ (A)| ≥ |A/C_{A}(Ṽ)| for all elementary abelian subgroups A of R.

(iv) [iv] If U is L/O_{p}(L) module with Ṽ ≤ U and U = C_{U}(B(R))Ṽ, then U = C_{U}(O^{p}(K))Ṽ.

Then O_{p}(K) ≤ B(R) and Ω_{1}ZJ(R) ≤ Ω_{1}Z(O_{p}(K)).

Proof: Let T = O_{p}(L), \overline{L} = L/C_{L}(V) and Y = Ω_{1}ZJ(R). Let A ∈ A(R). Since A ∈ A(R) and V ≤ T ≤ R, |V/C_{V} (A)| ≤ |A/C_{A}(V). By (ii) C_{A}(Ṽ) = A ∩ T and so also C_{A}(V) = A ∩ T. Thus (iii) implies |V/C_{V} (A)| = |\overline{A}| = |A/A ∩ T | and so V (A ∩ T) ∈ A(R) ∩ A(T). Thus Y ≤ V (A ∩ T) ≤ T. Put W = ⟨Y^{L}⟩V. We conclude that W ≤ Ω_{1}Z(J(T)) and so W is elementary abelian and (A ∩ T)V centralizes W. Hence W ≤ (A ∩ T)V and W = V (A ∩ W) = V C_{W} (A). It follows that A centralizes W/V. Since A was arbitray in A(R), ⟨J(R)^{L}⟩ centralizes W/V. Since Y = Ω_{1}ZJ(R), N_{L}(J(R)) normalizes Y. So by (i) also L normalizes Y V. Thus W = Y U and [W, T] = [Y, T] ≤ Y. Since L normalizes [W, T] we get[W, T] ≤ C_{W} (K). Let D = C_{W} (O^{p}(K)) and U = W/D. Then T centralizes U. Since Ṽ ≅ V D/D and U = Y V/D, we can apply (iv) to conclude that W = DV and U ≅ Ṽ. Since A ∈ A(R), |W/W ∩ A| ≤ |A/C_{A}(W)| = |A/A ∩ T |. One the otherhand by (iii), |A/A ∩ T | ≤ |Ṽ/CṼ (A)| = |U/C_{U}(A)| ≤ |W/C_{W} (A)D|. Thus |W/C_{W} (A)| ≤ |W/C_{W} (A)D| and D ≤ C_{W} (A). Hence [D, A] = 1, D ≤ Y and [D, K] = 1. Therefore [W, O_{p}(K)] ≤ [D, K][V, T] = 1 and so O_{p}(K) ≤ C_{R}(Y) = B(R). Hence Y ≤ Ω_{1}Z(O_{p}(K)). □

Lemma 4.2 [pu2(R) in pu4(B(R))] Let R be a p-group. Then PU_{2}(R) ⊆ PU_{4}(B(R)).

Proof: Let L ∈ PU_{2}(R). Since N_{L}(R) ≤ N_{L}(B(R)) we may assume that L ∈ PU_{1}(R). Set P = N_{L}(R). If P < H ≤ L, then clearly H ∈ PU_{1}(R). By 2.3(a) L is generated by the H ≤ L such that P is contained in a unique maximal subgroup of H. If H ∈ PU_{4}(B(R)) for all such H, then by the definition of PU_{4} also L ∈ PU_{4}(B(R)). Hence we may assume from now on that

1^{∘} [1] P is contained in unique maximal subgroup H of L.

Let D be the largest normal subgroup of L contained in P. Then [D, R] ≤ [P, R] ≤ R and so [D, R] ≤ O_{p}(D) ≤ O_{p}(L).

Choose T ∈ Syl_{p}(L) with P_{L}(T) ≤ P. Then R ≤ O_{p}(P_{L}(T) ≤ O_{p}(C_{L}(Ω_{1}Z(T)) and [R, C_{L}(Z_{L})] ≤ O_{p}(C_{L}(Z_{L})) = O_{p}(L) ≤ R. Thus C_{L}(Z_{L}) ≤ N_{L}(R) ≤ P. We proved:

2^{∘} [2] [D, ⟨R^{L}⟩] ≤ O_{p}(L) and C_{L}(Z_{L}) ≤ D

If J(R) ≤ D, then J(R) = J(O_{p}(D)) and so J(R) ⊴ H. Thus [Z_{L}, J(R)] = 1 and so also [Z_{L}, B(R)] = 1. So by (2^{∘}), B(R) ≤ D and B(R) = B(O_{p}(D)). Thus B(R) ⊴ H and so H ∈ PU_{4}(B(R)).

So we may assume that J(R) ≰ D and so by (2^{∘}) [Z_{L}, J(R)] ≠ 1. Let K = ⟨J(R)^{L}⟩, \overline{L} = L/C_{L}(Z_{L}) and Z̃_{L} = Z_{L}/C_{Z_{L}}(O^{p}(K)). By ?? there exists a L-invariant set of normal subgroups K_{i}, 1 ≤ i ≤ l, in K such that

3^{∘} [3]

(i) [3-i] K_{i} = O^{p′}(K_{i}),

(ii) [3-ii] \overline{K} = \overline{K_{1}} × \overline{K_{2}} × … × \overline{K_{l}},

(iii) [3-iii] Z̃_{L} = [Z̃_{L}, K_{1}] × [Z̃_{L}, K_{2}] × [Z̃_{L}, K_{l}],

(iv) [3-iv] \overline{K_{i}} ≅ SL_{n}(q), Sp_{2n}(q), G_{2}(q) or Sym(n), where q is a power of p, p = 2 in the last two cases and n ≡ 2, 3 mod 4 in the last case,

(v) [3-v] [Z̃_{L}, K_{i}] is the natural module for K_{i},

(vi) [3-vi] \overline{J(R)} = (\overline{J(R)} ∩ \overline{K_{1}}) × … × (\overline{J(R)} ∩ \overline{K_{l}})

It is now easy to see that \overline{L} = \overline{K}N\overline{L}(\overline{J(R)}

By (2^{∘}) O_{p}(C_{L}(Z_{L})J(R)) = O_{p}(L)J(R) and so J(R) = J(O_{p}(C_{L}(Z_{L})J(R))). Thus \overline{N_{L}(J(R))} = N\overline{L}(\overline{J(R)} and so

4^{∘} [4] L = KN_{L}(J(R)).

Suppose that K ≤ H. Then by (1^{∘}) and (4^{∘}) J(R) is normal in L and J(R) ≤ O_{p}(L) ≤ D, a contradiction to the assumptions.

Thus K ≰ H. Note that O^{p}(K_{i}) ≤ [K_{i}, J(R)]. From (2^{∘}) and 2.4 we conclude that

5^{∘} [5] C_{K_{i}}(Z_{L}) = O_{p}(K_{i}).

Since J(R), R_{i} ≰ O_{p}(L). Pick j with K_{j} ≰ H. Then by (1^{∘}) L = ⟨K_{j}, P ⟩ = ⟨K_{j}^{P} ⟩P. Thus ⟨K_{j}^{P} ⟩J(R) is normal in L. So P acts transitively on {K_{i} | 1 ≤ i ≤ l}, and L = KP. By (2^{∘}) [C_{L}(Z_{L}), J(R)] ≤ O_{p}(L) and so C_{L}(Z_{L}), K] ≤ O_{p}(L). Hence C_{K}(Z_{L}) is p-closed. Also C_{K}(Z_{L}) = C_{K}(Z̃_{L}).

Note also that B(R) ≤ KO_{p}(L) and so ⟨B(R)^{L}⟩ = K B(R).

Suppose that B(R)O_{p}(L) = O_{p}(P ∩ KO_{p}(L)) or that \overline{K_{j}} ≅ G_{2}(q). Then it is easy to see that the assumptions of 4.1 are fulfiled. We conclude that O_{p}(K B(R)) ≤ B(R). Moreover, either \overline{K_{j}} ≅ G_{2}(q) or B(R) = O_{p}(P ∩ K B(R)). By 2.2(a)

C_{K_{i}}(Ω_{1}Z(T ∩ K_{j} B(R))) = C_{K_{j}}(Ω_{1}Z(T ∩ K_{i})) = C_{K_{j}}(Ω_{1}Z(T))

and we conclude that P ∩ K_{j} B(R) contains a point stabilizer of K_{i}B(R). Suppose in addition that K̄_{j} ≇ Sym(n), n ≥ 7. Then K_{i} B(R) ∈ PU_{3}(B(R)). Also P ≤ N_{L}(B(R)) and L = ⟨P, K_{i} B(R) | 1 ≤ i ≤ l⟩ and so L ∈ PU_{4}(B(R)).

Suppose now that \overline{K_{j}} ≇ G_{2}(q) and either B(R)O_{p}(L) ≠ O_{p}(P ∩ KO_{p}(L)) or K̄_{i} ≅ Sym(n), n ≥ 7. Put q := 2 in the second case. Then \overline{K}_{i} ≅ Sp_{2n}(q) or Sym(n) and |B(R)/O_{p}(K_{i} B(R)| = q. Hence there exists a subgroup D_{i} of K_{i} B(R) with B(R) ≤ D_{i}, D_{i} = ⟨B(R)^{D_{i}}⟩ and D_{i}/O_{p}(D_{i}) ≅ SL_{2}(q). By 4.1 B(R) ∈ Syl_{p}(D_{i}). Thus D_{i} ∈ PU_{3}(B(R)). Moreover, K_{i} = ⟨D_{i}, N_{K_{i}}(B(R))⟩ and so L = ⟨D_{i}, N_{L}(B(R)) | 1 ≤ i ≤ n⟩. Thus again L ∈ PU_{4}(B(R). □

Lemma 4.3 [P(T) in PU4(B(T))] Let P be a finite group of characteristic p. Let T ∈ Syl_{p}(T) and suppose that T lies in a unique maximal subgroup of P. Then either Z_{L} = Ω_{1}Z(L) or P ∈ P_{4}(B(T)).

Proof: Suppose that [J(T), Z_{L}] = 1. Then also [B(T), Z_{L}] = 1 and so by the Frattinargument L = C_{L}(Z_{L})N_{L}(B(T)). Since L is minimal parabolic, L = C_{L}(Z_{L})S or B(T) is normal in L. In the first case Z_{L} = Ω_{1}Z(L) and in the second case L ∈ PU_{4}(T).

So we may assume that [B(T), Z_{L}] ≠ 1. Using 3.2 we can argue just as in 4.2. □

5 The principal amalgam problem

Let R be a group and Σ a set of groups containing R. Then

O_{R}(Σ) = ⟨N ≤ R | N ⊴ L ∀L ∈ Σ⟩

So O_{R}(Σ) is the largest subgroup of R which is normal in all the L ∈ Σ.

Theorem 5.1 [simultanous pushing up] Let R be a finite p-group with R = B(R) and Σ a subset of PU_{3}(R). If O_{R}(Σ) = 1, then one of the following holds

(a) [a] who knows

The proof will be achieved in a long sequence of lemmas. Let G^{∗} be the free amalgamated product of Σ over R. We view L ∈ Σ as a subgroup of G^{∗}. Let Γ be the graph with vertices G^{∗} and edges (L_{1}g, L_{2}g), g ∈ G^{∗}, L_{1} ≠ L_{2} ∈ Σ. Note that G^{∗} acts on Γ by right multiplication. For α ∈ Γ let G_{α} = {g ∈ G^{∗} | α = α^{g}}, Q_{α} = O_{p}(G_{α}), Z_{α} = Z_{G_{α}} and U_{α} = [Z_{α}, G_{α}]. For an edge (α, β) let Q_{αβ} = G_{α} ∩ G_{β} and Z_{αβ} = Ω_{1}Z(Q_{αβ}). Let Δ(α) be the set of neighbors of α and G_{α}^{(1)} = G_{α} ∩ ⋂_{β∈Δ(α)} G_{β}. Let U_{α} = [Z_{α}, G_{α}]. Then by definition of Γ and of PU_{3}(R).

Lemma 5.2 [basics of pushing up]

(a) [a] G_{α} = L^{g} for some L ∈ Σ and g ∈ G^{∗}, and G_{α} is of characteristic p.

(b) [g] C_{G_{α}}(Z_{α}) = Q_{α}.

(c) [b] \overline{G_{α}} := G_{α}/C_{G_{α}}(Z_{α}) ≅ SL_{n_{α}}(q_{α}), Sp_{2n}(q_{α}) or G_{2}(q_{α}), q_{α} a power of p.

(d) [c] Z̃_{α} := Z_{α}/C_{Z_{α}}(G_{α}) is the appropiate natural module.

(e) [d] Q_{αβ} = B(Q_{αβ}) and G_{α} = ⟨Q_{αβ}^{G_{α}}⟩

(f) [e] P_{αβ} := N_{G_{α}}(Q_{αβ}) contains a point stabilizer of G_{α}.

(g) [f] If \overline{G_{α}} ≅ G̸_{2}(q) then Q_{αβ} = O_{p}(P_{αβ}).

Next we show

Lemma 5.3 [more basics of pushing up]

(a) [a] Z_{αβ} ≤ Z_{α} = Ω_{1}Z(Q_{α})

(b) [c] Q_{α} = G_{α}^{(1)}.

(c) [d] One of the following holds:

1. [1] U_{α} ∩ Ω_{1}Z(G_{α}) = 1, that is U_{α} is the natural module.

2. [2] \overline{G_{α}} ≅ Sp_{2n}(q) or G_{2}(q) and U_{α} is a quotient of the natural O_{2n+1}(q)-module for \overline{G_{α}}, (where n = 3 in the G_{2}(q)-case).

(d) [e] For all H ≤ G_{α}, CZ̃_{α}(H) = { \atop C_{Z_{α}}(H)}.

(e) [f] Let T ∈ Syl_{p}(P_{αβ}) and x ∈ Ω_{1}Z(T) with x ∉ Ω_{1}Z(G_{α}). Then C_{G_{α}}(x) = O^{p′}(P_{αβ}).

(f) [g]

(a) [a] If \overline{G_{α}} ≅ SL_{n}(q), then |\overline{Q_{αβ}}| = q^{n} and |Z̃_{αβ}| = q.

(b) [b] If \overline{G_{α}} ≅ Sp_{2n}(q), then |\overline{Q_{αβ}}| = q^{2n−1} and |Z̃_{αβ}| = q.

(c) [c] If \overline{G_{α}} ≅ G_{2}(q), then |\overline{Q_{αβ}}| = q^{3} and |Z̃_{αβ}| = q^{3}.

(g) [h] Q_{αβ} = Q_{α}J(Q_{αβ})

Proof:

(a) follows from 5.2(e),5.2(f)and 3.1.

By 5.2(g) Q_{α} ≤ Q_{αβ} = G_{α} ∩ G_{β}. So (b) holds.

(c) follows from 3.4, and (d) follows from (c) . (e) follows from (d) , and 5.2.

(f) follows easily from 5.2.

To prove ?? put Y := Ω_{1}ZJ(Q_{αβ}). Since Q_{αβ} = B(Q_{αβ}), 4.1 implies Y ≤ Ω_{1}Z(Q_{α}). Thus by (a), Y = C_{Z_{α}}(J(Q_{αβ}). (g) now holds since J(Q_{αβ})Q_{α}/Q_{α} is generated by offenders on Z_{α} and since Q_{αβ} = B(Q_{αβ}) = C_{Q_{αβ}}(Y). □

We say that β ∈ Γ is symplectic if \overline{G_{β}} ≅ Sp_{2n}(q) with n ≥ 2, β is linear if \overline{G_{β}} ≅ SL_{n}(q) and β is a hex if \overline{G_{β}} ≅ G_{2}(q). Let α ∈ Δ(β). Define

X_{αβ} := \begin{cases}[Z_{α}, Q_{αβ}] & if α is symplectic. \\ Z_{α} & otherwise. \end{cases}

Put

A_{αβ} = [X_{αβ}, Q_{αβ}]

Lemma 5.4 [agammadelta] Let (α, β) be an edge in Γ. Then A_{αβ} ≤ Ω_{1}Z(Q_{αβ}) ≤ Ω_{1}Z(Q_{β}) ≤ Z_{β} and A_{αβ} ≰ Z(G_{α}).

Proof: Readily verfied. □

Lemma 5.5 [offenders on xgammadelta] Let (α, β) be an edge in Γ and D = X_{αβ} or D = Z_{α}. Let B ≤ Q_{αβ} be a non-trivial offender on D.

(a) [a] |D/C_{D}(B)| = |B/C_{B}(D)|.

(b) [b] One of the following holds:

1. [1] [D, Q_{αβ}] ≤ [D, B].

2. [2] α is symplectic, D = Z_{α} and [D, C_{Q_{αβ}}(X_{αβ})] ≤ [D, B].

(c) [c] One of the following holds

1. [1] [D, B, Q_{αβ}] = 1.

2. [2] α is symplectic, D = Z_{α}, [X_{αβ}, B] ≠ 1 and [D, Q_{αβ}, Q_{αβ}] = A_{αβ}.

Proof: This follows easily from the action of Q_{αβ} on D. □

Lemma 5.6 [agd in zgd] Let (α, β) be an edge in Γ and suppose that Z_{β} ≤ Q_{α}.

(a) [a] If X_{αβ} ≰ Z_{β} then A_{αβ} ≤ Z(G_{β}).

(b) [b] Suppose α is symplectic and that N is a normal p-subgroup of G_{β} with [X_{αβ}, N] = 1. Then [Z_{α}, N] ≤ Z(G_{β}).

Proof: We prove (a) and (b) at the same time. For the proof of (b) we assume that (a) already has been proved and that [Z_{α}, N] ≠ 1.

For the proof of (a) let D_{αβ} = X_{αβ} and N_{β} = Q_{β}. For the proof of (b) let D_{αβ} = Z_{α} and N_{β} = N. Let R_{αβ} = [D_{αβ}, N_{β}]. From the definition of R_{αβ} we obtain:

1^{∘} [1] R_{αβ} ≤ Z_{αβ}.

Next we show:

2^{∘} [2] Let B ≤ Q_{αβ} and suppose that B is a non-trivial offender on D_{α}. Then R_{αβ} ≤ [D_{αβ}, B] ∩ Z_{αβ}.

By (1^{∘}) we only need to show that R_{αβ} ≤ [D_{αβ}, B]. We apply 5.5(b) with D = D_{αβ}. If 1. holds we have R_{αβ} = [D_{αβ}, N_{β}] ≤ [D_{αβ}, Q_{αβ}] ≤ [D_{αβ}, B] and we are done. Suppose that 2. holds. Then D_{αβ} ≠ X_{αβ} and so we must be in the proof of (b). Hence N_{β} = N ≤ C_{Q_{αβ}}(X_{αβ}) and again R_{αβ} ≤ [D_{αβ}, B].

3^{∘} [3] Let B ≤ Q_{β} and suppose that B is a non-trivial offender on D_{αβ}. Then [D_{αβ}, B, Q_{αβ}] ≤ Ω_{1}Z(G_{β}).

We apply 5.5(c). If 1. holds then [D_{αβ}, B, Q_{αβ}] = 1 and we are done. So suppose 2. holds. Then we are in the proof of (b), [X_{αβ}, B] ≠ 1 and [D_{αβ}, B, Q_{αβ}] = A_{αβ}. Since B ≤ Q_{β}, we get X_{αβ} ≰ Z_{β} and so by (a) A_{αβ} ≤ Ω_{1}Z(G_{β}) and (3^{∘}) is proved.

Since Q_{αβ} = B(Q_{αβ}) and C_{G_{β}}(Z_{β}) = Q_{β} we have [Z_{β}, J(Q_{αβ})] ≠ 1. Thus there exists A ∈ A(Q_{αβ}) with A ≰ Q_{β}. Let a ∈ A with a ∉ Q_{β}. If β is a hex we choose a such that in addition C_{Z_{β}}(a) = Z_{αβ}. Let γ ∈ α^{G_{β}} with Z_{αβ} ∩ Z_{γβ} = Ω_{1}Z(G_{β}) and a ∉ P_{βγ}. The choice of a implies

4^{∘} [4] Z_{γβ} ∩ Z_{γβ}^{a} = Ω_{1}Z(G_{β}).

Suppose first that

(∗) [D_{γβ}, D_{γβ}^{a}] ≠ 1.

Then by 5.5 D_{γβ}^{a} is a non-trivial offender on D_{γβ} and vice versa. So by (2^{∘}) applied to (D_{γβ}^{a}, γ) in place of (B, α)

R_{γβ} ≤ [D_{γβ}, D_{γβ}^{a}] ∩ Z_{γβ}

By (3^{∘}) applied to (D_{γβ}, γ^{a}) in place of (B, α) we have [[D_{γβ}^{a}, D_{γβ}], Q_{γβ}^{a}] ≤ Z(G_{β}). Hence 5.3(f) implies that Q_{αβ} centralizes [D_{γβ}^{a}, D_{γβ}] ∩ Z_{β} and so Z_{β} ∩ [D_{γβ}^{a}, D_{γβ}] ≤ Z_{γβ}^{a}. Thus

R_{γβ} ≤ [D_{γβ}^{a}, D_{γβ}] ∩ Z_{γβ}) ≤ Z_{γβ} ∩ Z_{γβ}^{a} ≤ Ω_{1}Z(G_{β})

and we are done in this case.

Suppose next that

(∗∗) [D_{γβ}, D_{γβ}^{a}] = 1.

Set B := A ∩ Q_{β} and C := C_{B}(D_{γβ}). Then Z_{β}B ∈ A(Q_{β}) ⊆ A(Q_{αβ}). Since by assumption Z_{β} ≤ Q_{α} we get Z_{β} ≤ Q_{γ}. Thus B is an offender on D_{γβ}. Since A is abelian and C ≤ B ≤ A we have B = B^{a} and C = C^{a}. Thus C = C_{B}(D_{γ}^{a}) and C centralizes D_{γ}^{a}. Also Z_{β} ≤ Q_{γ}^{a} and so thus by (**) Z_{β}D_{γβ}C centralizes D_{γβ}^{a}. By 5.5(a) Z_{β}D_{γβ}C ∈ A(Q_{β}) and we conclude that D_{γ}^{a} ≤ Z_{β}D_{γ}C. By symmetry in γ and γ^{a} we conclude Z_{β}D_{γβ}C = Z_{β}D_{γβ}^{a}C. Thus

[D_{γβ}, B] = [D_{γβ}^{a}, B].

Suppose that B does not centralize D_{γβ}. Then by (2^{∘}) applied to γ in place of α, R_{γβ} ≤ [D_{γβ}, B] ∩ Z_{γβ}. From [D_{γβ}, B] = [D_{γβ}^{a}, B] and (3^{∘}) applied to γ^{a} in place of α we get [D_{γβ}, B, Q_{γβ}^{a}] ≤ Z(G_{β}). Now as in the (*)-case R_{γβ} ≤ Z(G_{β}) and we are done.

Suppose next that B centralizes D_{γβ}. Then also Z_{β}B centralizes D_{γβ} and so D_{γβ} ≤ Z_{β}B. Since a centralizes B we conclude that D_{γβ}Z_{β} = D_{γβ}^{a}Z_{β}. Hence

R_{γβ} = [D_{γβ}, N_{β}] = [D_{γβ}Z_{β}, N_{β}] = [D_{γβ}^{a}, U] = R_{γβ}^{a} ≤ Z_{γβ} ∩ Z_{γβ}^{a} ≤ Ω_{1}Z(G_{β})

and we are also done in this final case. □

For adjacent vertices α, β let V_{α}^{β} = ⟨Z_{β}^{G_{α}}⟩.

Lemma 5.7 [qgamma cap qdelta normal] Let (β, α) be an edge of Γ and suppose that V_{α}^{β} and V_{β}^{α} are abelian. Then Q_{α} ∩ Q_{β} is normal in G_{α}.

Proof: As in the proof of 5.6 choose A ∈ A(Q_{αβ}), a ∈ A and γ ∈ α^{G_{β}} with Z_{γβ} ∩ Z_{γβ}^{a} ≤ Ω_{1}Z(G_{β}). Suppose for a contradiction that Q_{α} ∩ Q_{β} is not normal in G_{α}. By conjugation Q_{γ} ∩ Q_{β} is not normal in G_{γ} and so Q_{γ} ∩ Q_{β} ≠ Q_{γ} ∩ Q_{ρ} for some ρ ∈ β^{G_{γ}}. Then [Q_{γ} ∩ Q_{β}, Z_{ρ}] ≠ 1. Let δ ∈ β^{G_{γ}}.

If possible, choose ρ such that in addition [Q_{γ}∩Q_{β}, X_{δγ}] ≠ 1. In this case put D_{δγ} = X_{δγ}. If not possible, put N = ⟨(Q_{γ} ∩ Q_{β})^{G_{γ}}⟩ and D_{δγ} = Z_{δ}. Then [X_{βγ}, N] = 1. Note that Z_{γ} ≤ V_{β}^{γ} ≤ Q_{β}. Since [D_{ργ}, Q_{γ} ∩ Q_{β}] ≠ 1, D_{ργ} ≰ Z_{γ}. Thus we can apply 5.6

with (β, γ) in place of (α, β). We conclude that 1 ≠ R := [D_{ργ}, Q_{γ} ∩ Q_{β}] ≤ Ω_{1}Z(G_{γ}). Since R ≰ Ω_{1}Z(G_{δ}) and ρ ∈ β^{G_{γ}} we get R ≰ Ω_{1}Z(G_{β}). Since Z_{γβ}^{a} ∩ Z_{γβ} = Ω_{1}Z(G_{β}) = Z_{γβ}^{a^{−1}} ∩ Z_{γβ} we have

1^{∘} [1] R ≤ Ω_{1}Z(G_{γ}) and Z_{γβ}^{a^{−1}} ≱ R ≰ Z_{γβ}^{a}.

From the definition of D_{δγ},(1^{∘}), and 5.5(b) we deduce

2^{∘} [2] Let F ≤ Q_{δγ} be a non-trivial offender on D_{δγ}, then R ≤ [D_{δγ}, F].

Let B = A ∩ Q_{β} and C = B ∩ Q_{γ}. Then Z_{β}B and Z_{β}Z_{γ}C are in A(Q_{βγ}).

Next we show

3^{∘} [4] [C, D_{δγ}] = 1.

Assume for a contradiction that [C, D_{δγ}] ≠ 1 and put D = C_{C}(D_{δγ}). By (2^{∘}), R ≤ [C, D_{δγ}] and by 5.5(a) E := Z_{β}Z_{γ}D_{δγ}D ∈ A(Q_{γ}).

We will show that [E, D_{δγ}^{a}] = 1. Since V_{γ^{a}}^{β} is abelian, D_{δγ}^{a} centralizes Z_{β}.

Suppose that [D_{δγ}^{a}, Z_{γ}] ≠ 1. If [D_{δγ}^{a}, X_{γβ}] ≠ 1 put Y = X_{γβ}. Otherwise let Y = Z_{γ}. Then [D_{δγ}^{a}, Y] ≤ Z_{γβ}. Since V_{β}^{γ} is abelian, Y ≤ Z_{γ} ≤ Q_{β} ∩ Q_{γ}^{a}. From 5.5(a) we conclude that Y is an offender on D_{δγ}^{a} (and vice versa). By (2^{∘}) R^{a} ≤ [D_{δγ}^{a}, Y] ≤ Z_{γβ}, a contradiction to (1^{∘}).

Thus [D_{δγ}^{a}, Z_{γ}] = 1 and D_{δγ}^{a} ≤ Q_{β} ∩ Q_{γ}. By symmetry D_{δγ} ≤ Q_{β} ∩ Q_{γ}^{a}. Hence by 5.5(a) D_{δγ} and D_{δγ}^{a} are offenders on each other.

Suppose that [D_{δγ}, D_{δγ}^{a}] ≠ 1. Then by (2^{∘}) R ≤ [D_{δγ}, D_{δγ}^{a}] ≤ Z_{γβ}^{a}, again a contradiction to (1^{∘}).

Thus [D_{δγ}, D_{δγ}^{a}] = 1. Since D centralizes D_{δγ} and since D = D^{a}, D centralizes D_{δγ}^{a}. Thus E centralizes D_{γδ}^{a} and so D_{γδ}^{a} ≤ E. Note that C is a non-trivial offender on D_{δγ} and so by (2^{∘}) R ≤ [C, D_{δγ}]. Since a centralizes C we get

R^{a} ≤ [C, D_{δγ}^{a}] ≤ [C, E] = [C, D_{γδ}] ≤ Z_{γβ}

contradicting (1^{∘}). This completes the proof of (3^{∘}).

4^{∘} [3] D_{δγ} ≤ Z_{β}Z_{γ}C.

Since V_{γ}^{β} is abelian, Z_{γ}Z_{β} centralizes Z_{δ} and so also D_{δγ}.

By (3^{∘}), [C, D_{δγ}] = 1. Since Z_{β}Z_{γ}C ∈ A(Q_{βγ}) we conclude that (4^{∘}) holds.

Suppose that B ≠ C, that is B ≰ Q_{γ}. By (4^{∘}) [B, D_{δγ}] ≤ [B, Z_{γ}] ≤ Z_{γ} and so B ≤ N_{G_{γ}}(D_{δγ}Z_{γ}). Since this is true for all δ ∈ β^{G_{γ}} we conclude that D_{βγ}Z_{γ} is ⟨B^{G_{γ}}⟩Q_{βγ} = G_{γ}. But then D_{βγ}Z_{γ} = D_{ργ}Z_{γ} and Q_{β} ∩ Q_{γ} centralizes D_{ργ}, a contradiction.

Thus B = C. So B centralizes Z_{γ}, Z_{γ} ≤ Z_{β}B and by (2^{∘}) D_{δγ} ≤ Z_{β}B. Since A centralizes B, we conclude that A normalizes Z_{γ}Z_{β} and D_{δγ}Z_{β}. But then A also normalizes Q_{γ} ∩ Q_{β} = C_{Q_{β}}(Z_{γ}Z_{β}) and [Q_{γ} ∩ Q_{β}, D_{δγ}Z_{β}]. Since this latter group is R we get a contradiction to (1^{∘}). □

Lemma 5.8 [zalpha offender] Let (α, β) and (γ, δ) be edges in Γ such that Z_{α}Z_{δ} ≤ Q_{αβ} ∩ Q_{δγ} and [Z_{α}, Z_{δ}] ≠ 1. Then

(a) [a] Z_{α} is an offender on Z_{δ} and vice versa.

(b) [b] |Z_{α}Q_{δ}/Q_{δ}| = |Z_{δ}Q_{α}/Q_{α}|.

(c) [c] G_{α} = ⟨Z_{δ}^{G_{α}}⟩Q_{α}.

Proof: (a) and (b) follows from the fact that Q_{αβ} contains no over-offender on Z_{α}.

Note that O^{p}(G_{α})Q_{α} = G_{α} unless \overline{G}_{α} ≅ SL_{2}(2), SL_{2}(3), Sp_{4}(2) or G_{2}(2). In each of the four exceptionell cases O^{p}(G_{α})Q_{α} has index p in G_{α} and Q_{αβ} ∩ O^{p}(G_{α})Q_{α} contains no non-trivial offender on Z_{α}. Thus (c) follows from (a). □

Lemma 5.9 [critical pairs] Let (α, β) and (γ, δ) be edges in Γ such that Z_{α}Z_{δ} ≤ Q_{αβ}∩Q_{δγ} and [Z_{α}, Z_{δ}] ≠ 1.

Then q := q_{α} = q_{β} and one of the following holds.

1. [1] \overline{G}_{α} ≅ \overline{G}_{δ} ≅ G_{2}(q).

2. [2]

(a) [a] \overline{G}_{α} ≅ Sp_{2n_{α}}(q) and \overline{G}_{δ} ≅ Sp_{2n_{δ}}(q)

(b) [b] |Z_{α}Q_{δ}/Q_{δ}| = |Z_{δ}Q_{α}/Q_{α}| = q.

(c) [c] [Z_{α}, [Z_{δ}, Q_{γδ}]] = 1 and [Z_{δ}, [Z_{α}, Q_{αβ}]] = 1.

3. [3]

(a) [a] \overline{G}_{α} ≅ Sp_{2n_{α}}(q), \overline{G}_{δ} ≅ Sp_{2n_{δ}}(q), n_{α}, n_{δ} ≥ 2, (b) [b] |Z_{α}Q_{δ}/Q_{δ}| = |Z_{δ}Q_{α}/Q_{α}| = q^{2},

(c) [c] [X_{αβ}, X_{δγ}] = 1.

(d) [d] One of the following holds:

1. [1] [X_{αβ}, Z_{δ}] = [X_{δγ}, Z_{α}], U_{α} is the natural module for G_{α} and U_{δ} is the natural module for G_{δ}.

2. [2] q = 2, [X_{αβ}, Z_{δ}] ≠ [X_{δγ}, Z_{α}] and U_{α} ∩ Z(G_{α}) = U_{δ} ∩ Z(G_{δ})

4. [4]

(a) [a] \overline{G}_{α} ≅ SL_{n_{α}}(q) and \overline{G}_{δ} ≅ SL_{n_{δ}}(q)

(b) [b] |[Z_{α}, Z_{δ}]| = q.

5. [5] (After interchanging (α, β) with (δ, γ) if necessary)

(a) [a] \overline{G}_{α} ≅ SL_{n_{α}}(q), n_{α} > 2 and \overline{G}_{δ} ≅ Sp_{2n_{δ}}(q), n_{δ} > 1

(b) [b] |Z_{α}Q_{δ}/Q_{δ}| = |Z_{δ}Q_{α}/Q_{α}| = q,

(c) [c] [X_{δγ}, Z_{α}] = 1

(d) [d] |[Z_{α}, Z_{γ}]| = q

Proof:

Let I_{αδ} = {|[Z_{α}, y]| | 1 ≠ y ∈ Z_{δ}Q_{α}/Q_{α}} and J_{αδ} = {|[x, Z_{δ}]| | x ∈ Z_{α} \ C_{Z_{α}}(Z_{δ})} By 5.3(d) |[Z_{α}, y]| = |Z̃_{α}, y]| and |[x̃, Z_{δ}]| = |[x, Z_{δ}]|, for all y ∈ Z_{δ} and x ∈ Z_{α}. Define

the positive integer k_{αδ} by |Z̃_{α}/CZ̃_{α}(Z_{δ})| = q_{α}^{k_{αδ}} and note that

q_{α}^{k_{αδ}} = |Z_{α}Q_{δ}/Q_{δ}| = |Z_{δ}Q_{α}/Q_{α}| = q_{δ}^{k_{δα}}

Also Z_{δ} is a quadratic offender on Z_{α} and the action of \overline{G_{α}} on Z̃_{α} implies the following table:

\begin{matrix}\overline{G_{α}} & I_{αδ} & J_{αδ} \\ G_{2}(q_{α}) & \{q_{α}^{2}, q_{α}^{3}\} & \{q_{α}^{2}, q_{α}^{3}\} \\ SL_{n_{α}}(q_{α}) & \{q_{α}\} & \{q_{α}\} \\ Sp_{2n_{α}}(q_{α}), k_{αδ} = 1 & \{q_{α}\} & \{q_{α}\} \\ Sp_{2n_{α}}(q_{α}), k_{αδ} > 1 & \{q_{α}, q_{α}^{2}\} & \{q_{α}, q_{α}^{k_{αδ}}\} \end{matrix}

Note that the definitions of I_{αδ} and J_{αδ} imply I_{αδ} = J_{δα}. This allows as to relate \overline{G}_{α} and \overline{G}_{δ}. In particular we see that

q := q_{α} = q_{δ} and k := k_{αδ} = k_{δα}.

If \overline{G}_{α} ≅ G_{2}(q_{α}) we conclude that \overline{G}_{δ} ≅ G_{2}(q_{δ}) So (a) holds in this case.

If \overline{G}_{α} ≅ SL_{n_{α}}(q_{α}) and n_{α} > 2, we get \overline{G}_{α} ≅ SL_{n_{δ}}(q_{δ}) or Sp_{2n_{δ}}(q_{δ}). In the latter cae we get k = 1. In any case since n_{α} > 2 we have that U_{α} is a natural module and so |[Z_{α}, Z_{γ}]| = q. Thus (4) or (5) holds.

Interchanging α and δ if necessary, it remains to investigate the case where \overline{G}_{α} ≅ Sp_{2n_{α}}(q) and \overline{G}_{δ} ≅ Sp_{n_{α}}(q). Then k ∈ \{1, 2\}. If k = 1, (2) holds.

So suppose that k = 2. Then clearly n_{α}, n_{δ} > 2. We will show that (3) holds. We already proved (3)(a) and (b). Also both [X_{αβ}, Z_{δ}] and [X_{δγ}, Z_{α}] have order q. It follows that X_{αβ}Q_{δ}/Q_{δ} is the unique full transvection group in Q_{γδ}/Q_{δ} and thus (3)(c) holds.

If q > 2, then |[X_{δγ}, Z_{α}]| = q implies that U_{α} is a natural module and so also [X_{αβ}, Z_{δ}] = [Z_{α}, X_{δγ}] = U_{α} ∩ Z_{αβ}. Thus (3) holds in this case.

So suppose that q = 2. Note that U_{α}∩Z_{αβ} = [X_{αβ}, Z_{δ}][Z_{α}, X_{δγ}]. If [X_{αβ}, Z_{δ}] = [Z_{α}, X_{δγ}] we conclude that U_{α} is a natural module and (3) holds. If [X_{αβ}, Z_{δ}] ≠ [Z_{α}, X_{δγ}] we get that U_{α} ∩ Z(G_{α}) is the unique subgroup of order two in [X_{αβ}, Z_{δ}][Z_{α}, X_{δγ}] distinct from [X_{αβ}, Z_{δ}] and [Z_{α}, X_{δγ}]. The same is true for U_{δ} ∩ Z(G_{δ}) and again (3) holds. □

Lemma 5.10 [q=2 for g2(q)] Let (α, β, γ, δ) be as in Case 1. of 5.9. Then q = 2 and U_{α} ∩ Z(G_{α}) = U_{δ} ∩ Z(G_{δ}).

Proof: The following argument is taken from [MS].

Let R = [Z_{α}, Z_{δ}] and X = R \ {[x, y] ≠ 1 | x ∈ Z_{α}, y ∈ Z_{δ}]}. Then it is not too difficult to see that X = C_{U_{α}}(G_{α}) = C_{U_{δ}}(G_{δ}). We will compare the actions of U_{α}/X on U_{δ}/X as seen in G_{δ} with the action of U_{δ}/X on U_{α}/X as seen in G_{α}. Let 𝔽_{α} = End_{G_{α}}(U_{α}/X). Then 𝔽_{α} is a field isomorphic to GF (q).

Let

K_{δα} = {C_{U_{δ}}(y) | y ∈ Z_{α}, U_{δ} ∩ Q_{α} < C_{U_{δ}}(y) < U_{δ}}.

and similarly define K_{αδ}. If A ∈ K_{δα} then C_{U_{α}}A) ≠ U_{α} ∩ Q_{δ} and C_{U_{α}}(A)/R is a 1-dim. 𝔽_{α}-subspace of U_{α}/R. Also C_{U_{α}}(A) = C_{U_{α}}(a) for all a ∈ A \ Q_{α}. So C_{U_{α}}(A) ∈ K_{αδ} and the map A → C_{U_{α}}(A) is a bijection between K_{αδ} and K_{δα}. Moreover, \overline{A} is a long root subgroup of \overline{G}_{α}. Let t ∈ Z_{α} with [t, A] ≠ 1.

We show next that

(*) [t, A]X/X is a 1-dim. 𝔽_{α} and 𝔽_{δ} subspace of R/X and a

Clearly it is a 1-dim 𝔽_{δ}- subspace. Let P = C_{G_{α}}(\overline{A}). Then W := U_{α}/C_{U_{α}}(A) is a natural module for P/O_{p}(P) ≅ SL_{2}(q). Let t^{∗} be the image of t in W. Then S := C_{P} (t̃^{∗}) is a Sylow p-subgroup of P and so of G_{α}. Since S centalizes [t, A] we conclude that [t, A]X/X = C_{U_{α}/X}(S), which is a 1-dim. 𝔽_{α}-space. Hence (*) holds.

The preceeding argument also shows that every 1-dim. 𝔽_{α} subspace of [U_{α}, A]X/X is of the form [t, A] for some t ∈ Z_{α}. Also each 1-dim. 𝔽_{α} subspace of R/X is contained in [U_{α}, A]X/X for some A ∈ K_{δα}. Thus (*) implies (**) The 𝔽_{α}- and 𝔽_{δ}-subspaces in R/X coincide.

Let W_{αβ} = [U_{α}, O_{p}(P_{αβ})]X and U_{αβ} = C_{U_{α}}(O_{p}(P_{αβ})). Then U_{αβ}/X is a 1-dim. 𝔽_{α}- subspace of R/X. Moreover, U_{αδ} ≤ [U_{α}, A]X for all A ∈ K_{δα}. Considering the action of U_{α}Q_{δ}/Q_{δ} on U_{δ}/X we conclude that U_{αβ} = U_{γδ}.

Fix z ∈ U_{α} \ W_{αβ} and define Y/U_{δγ} := C_{U_{δ}/U_{δγ}}(z). Then Y/R is 1-dimensional 𝔽_{δ} subspace of U_{δ}/R. Since [z, Q_{αβ}]R = W_{αβ}, the Frattin-argument shows that L := C_{P_{αβ}}(zR/R) has a quotient SL_{2}(q). Since L normalizes Y, we conclude that Y Q_{α}/Q_{a} is a short root subgroup of \overline{G}_{α}.

Hence there exists a subgroup M of \overline{G}_{α} with Y Q_{α}/Q_{α} ≤ M and M ≅ SL_{2}(q). Note that for all t ∈ Y_{α}, [t, Y]X/X is an 𝔽_{δ}-submodule of R/X. Hence [t, Y]X/X is also an 𝔽_{α}-submodule of U_{α}/X. But this implies that U_{α}/X is as an 𝔽_{α}M-module the direct sum three isomorphic natural module. But this implies q = 2. ( For example let P be a mimimal parabolic of G_{α}/Q_{a} with M as a Levi complement, V_{1} = C_{U_{α}/X}(O_{p}(P)) and V_{2} = [U_{α}/X, O_{p}(P)]/V_{1}. Then O_{p}(P)/Φ(O_{p}(P)) is isomorphic to an 𝔽_{p}-submodule of Hom_{𝔽_{α}}(V_{2}, V_{1}). Since V_{2} and V_{1} are isomorphic 𝔽_{α}M modules, we conclude that every composition factor for M in O_{p}(P) is either natural or trivial. Comparing with the action of P on O_{p}(P) we obtain q = 2. Comment: a quote from [BBSM] would be more appropriate □

Lemma 5.11 [b=1 sigma=2] Suppose that |Σ| = 2, Σ = {α, β} and [Z_{α}, Z_{β}] ≠ 1. Then for γ ∈ Σ there exists K_{γ} ≤ Ω_{1}Z(G_{γ}) and L_{γ} ≤ G_{γ} such that G_{γ} = K_{γ} × L_{γ} and one of the following holds.

1. [1] L_{α} ∼ L_{β} ∼ q^{n}SL_{n}(q) and |K_{α}| = |K_{β}| ≤ q.

2. [2] p = 2 and (after interchanging α and β if necessary), G_{α} = L_{α} ∼ q^{1+2n}Sp_{2n}(q), G_{β} = L_{β} ∼ q^{1+2+2·(2n−2)}SL_{2}(q).

3. [3] p = 2, L_{α} ∼ L_{β} ∼ 2^{6}G_{2}(2) and |K_{α}| = |K_{β}| ≤ 2^{3}.

4. [4] p = 2 and G_{α} = L_{α} ∼ G_{β} = L_{β} ∼ q^{1+6+8}Sp_{6}(q).

5. [5] p ≠ 2, L_{α} ∼ L_{β} ∼ q^{2n}Sp_{2n}(q), n ≥ 2 and |K_{α}| = |K_{β}| ≤ q.

6. [6] q = 2, G_{α} ∼ 2^{1+2n}Sp_{2n}(2) and G_{β} ∼ 2^{1+2+1·m+1·m+2·k}SL_{2}(2) for some m, k with m + k = 2n − 2 and k even.

7. [7] q = 2 and O^{2}(G_{α}) ∼ O^{2}(G_{β}) ∼ 2^{1+4}Sp_{4}(2)′.

Proof: By assumption, [Z_{α}, Z_{β}] ≠ 1. Clearly Z_{α}Z_{β} ≤ Q_{αβ} and we can apply 5.9 with (δ, γ) = (β, α).

For {γ, δ} = {α, β} define H_{γ} = ⟨Z_{δ}^{G_{γ}}⟩. Let R = [Z_{α}, Z_{β}], I = {1 ≠ [x, y] | x ∈ Z_{α}, y ∈ Z_{b}} and D_{γ} = C_{Q_{γ}}(O^{p}(G_{γ})).

We devide the proof in a series of Steps.

Step 1 [da cap db] D_{α} ∩ D_{β} = 1.

Proof: This holds since D_{α}∩D_{β} is normalized by G_{α} = O^{p}(G_{α})Q_{αβ} and G_{β} = O^{p}(G_{β})Q_{αβ}. □

We call α non-abelian if α is symplectic and p ≠ 2. Otherwise α is called abelian.

Step 2 [abelian]

(a) [a] α is abelian if only if Q_{αβ}/Q_{α} is elementary abelian.

(b) [b] If α is abelian, then Φ(Q_{β}) ≤ D_{β}.

(c) [c] If α and β are abelian, then Q_{α} ∩ Q_{β} is elementary abelian.

Proof: (a) is obvious. If Φ(Q_{β}) ≤ Q_{α}, then Z_{α} centralizes Φ(Q_{β}) and so Φ(Q_{β}) ≤ D_{α}. Thus (b) holds.

Since Φ(Q_{α} ∩ Q_{β}) ≤ Φ(Q_{α}) ∩ Φ(Q_{β}), Step 1 and (b) imply (c). □

Step 3 [b=1 case 1] Suppose that 5.9(1) holds. Then 5.11(3) holds.

Proof: Note first that Q_{α} ≤ Q_{αβ} = Z_{α}Q_{β}. Thus Q_{α} = Z_{α}(Q_{α} ∩Q_{β}) and Step 2(c)implies that Q_{α} is elementary abelian. Thus by 5.3(a), Q_{α} = Z_{α}. In particular, D_{α} = Ω_{1}Z(G_{α}). By 5.10, q = 2 and

U_{α} ∩ Z(G_{α}) = U_{β} ∩ Z(G_{β}) ≤ D_{α} ∩ D_{β} = 1.

Thus |U_{α}| = 2^{6}.

By [Schur, Schur Multiplier] we get O^{2}(G_{α})/U_{a} ≅ G_{2}(2)′. Since G_{α} = Q_{α}Z_{β}O^{2}(G_{α}) and [Q_{α}, Z_{β}] ≤ [U_{α}, Z_{β}] ≤ U_{α} ≤ O^{2}(G_{α}) we get that G_{α}/O^{2}(G_{α}) is elementary abelian. Hence there exists L_{α} ≤ G_{α} with G_{α} = D_{α} × L_{α} and L_{α} ∼ 2^{6}G_{2}(2). Since D_{α} ≤ Z_{αβ} and D_{α} ∩ D_{β} = 1 we have |D_{α}| ≤ |Z_{αβ}/D_{β}| = 2^{3}. The proof of Step 3 is complete. □

Step 4 [b=1 case 2] Suppose that 5.9(2) holds. Then 1., 2., 4., 5., 6. or 7. in 5.11 holds.

Proof:

Let D_{αβ} = [Z_{α}, Q_{αβ}] and A_{αβ} = [D_{αβ}, Q_{αβ}] ≤ Z_{αβ}.

We show first

1^{∘} [6] [D_{βα}, Q_{α}] ≤ Ω_{1}Z(G_{α}). In particular, either D_{βα} ≤ Z_{α} or A_{βα} ≤ Ω_{1}Z(G_{α}).

Choose δ ∈ β^{G_{α}} with [Z_{δα}, Z_{β}] ≠ 1. If [D_{δα}, D_{βα}] ≠ 1, then

[D_{βα}, Q_{α}] ≤ A_{βα} = [D_{βα}, D_{δα}] ≤ Z_{αβ} ∩ Z_{αδ} ≤ Ω_{1}Z(G_{a})

So suppose that[D_{δα}, D_{βα}] = 1. Then [D_{δα}, Z_{β}] ≤ Z_{αβ} ≤ Z_{α} and so D_{βα}Z_{α} is normal in G_{α} = ⟨Q_{αδ}, Z_{β}⟩. Hence also [D_{βα}, Q_{α}] is normal in G_{α}. Since Q_{αβ} centralizes [D_{βα}, Q_{α}] and G_{α} = ⟨Q_{αβ}^{G_{α}}⟩, the first statement in (1^{∘}) hold. If [D_{βα}, Q_{α}] = 1 then since Ω_{1}Z(Q_{α}) = Ω_{1}Z_{α} we get D_{βα} ≤ Z_{α}. If D_{βα}, Q_{α}] ≠ 1, then A_{βα} = [D_{βα}, Q_{α}] ≤ Ω_{1}Z(G_{α}), completing the proof of (1^{∘}).

Note that

2^{∘} [15] Z_{α} ∩ Q_{β} = D_{αβ}Z_{αβ}.

Next we prove:

3^{∘} [7] If [D_{βα}, Q_{α}] = 1, then Q_{α} ≤ Z_{β}Q_{α} and Z_{β} ∩ Q_{α} ≤ Z_{α} ∩ Q_{β}.

Note that C_{Q_{α}β}(D_{αβ}) = Z_{β}Q_{α} and so the first statement holds. By 5.3, D_{βα} ≤ Z_{α}. Also D_{βα} ≤ Z_{β} ≤ Q_{β} and so by (2^{∘}) also the second statement holds.

4^{∘} [8] If p is odd, then 1. or 5. of 5.11 holds.

If [D_{βα}, Q_{α}] ≠ 1, then by (1^{∘}), R = A_{βα} = [D_{βα}, Q_{α}] ≤ Z(G_{α}) a contradiction. Thus [D_{βα}, Q_{α}] = 1 and by (3^{∘}) Z_{β} ∩ Q_{α} ≤ Z_{α} ∩ Q_{β}. By symmetry in α and β we conclude. Hence Z_{α} ∩ Q_{β} = Z_{β} ∩ Q_{α} = Z_{α} ∩ Z_{β}. Thus |Z_{α} ∩ Z_{β}/Z_{αβ}| = q^{2n_{α}−2} and n_{α} = n_{β}. Since Q_{α} ≤ Z_{α}Q_{β} we get that Q_{α} ∩ Q_{β} is elementary abelian, Q_{α} = Z_{α} and Q_{β} = Z_{β}. Also D_{α} ≤ Z(G_{α}), D_{α} ≤ Z_{αβ} and D_{α} ∩ D_{β} = 1. Thus |D_{α}| ≤ q. Hence 5. holds and (4^{∘}) is proved.

We may assume from now on that p = 2. Set D = D_{αβ}D_{βα} and T = C_{Q_{αβ}}(D). By Step 2(c) Q_{α}∩Q_{β} is elementary abelian. Since C_{Q_{αβ}}(D_{αβ}) = Z_{β}Q_{α} we have T = Z_{α}Z_{β}(Q_{α}∩Q_{β}). Since p = 2 we conclude that

5^{∘} [10] A(T) = {Z_{α}(Q_{α} ∩ Q_{β}), Z_{β}(Q_{α} ∩ Q_{β})}

Let A ∈ A(Q_{αβ}). Then C_{A}(D_{αβ})D_{αβ} is in A(Q_{αβ}) and by 5.5(a), C_{A}(D)D ∈ A(T). So C_{A}(D)D = Z_{γ}(Q_{α} ∩ Q_{β}) for some γ ∈ {α, β}. In particular, C_{A}(D)D ≤ Q_{γ}. Let {α, β} = {γ, δ}. Since E := C_{A}(D_{δγ})D_{δγ} ∈ A(Q_{αβ}), E is an offender on Z_{γ}. Moreover, C_{E}(D) = C_{A}(D)D ≤ Q_{γ} and the action of Q_{γδ} on Z_{γ} implies E ≤ Q_{γ}. Since E ∈ A(Q_{αβ}) we conclude, Z_{γ} ≤ E. Thus [Z_{γ}, A] ≤ [E, A] ≤ [D_{δγ}, A].

Suppose that [Z_{γ}, A] ≠ 1, then also [Z_{γ}, A] ≰ Ω_{1}Z(G_{γ}) and 1 ≠ [D_{δγ}] ≰ Ω_{1}Z(G_{γ}). Hence A_{δγ} = [D_{δγ}] ≰ Ω_{1}Z(G_{γ}) and (1^{∘})( applied to (δ, γ) in place of (α, β))implies D_{δγ} ≤ Z_{γ}] = 1. By (3^{∘}) and (2^{∘}) we get D_{δγ} ≤ D_{γδ}Z_{γδ}, so Z_{γ} ≤ AD_{γδ}Z_{γδ} and thus Z_{γ} = C_{Z_{γ}}(A)D_{γδ}. This implies [Z_{γ}, A] = 1. So [Z_{γ}, A] = 1 and A ≤ Q_{γ}. Hence 6^{∘} [11] A(Q_{αβ}) = A(Q_{α}) ∪ A(Q_{β}).

By 5.3(g), Q_{αβ} = Q_{α}J(Q_{αβ}). Thus Q_{αβ} = Q_{β}J(Q_{α}) and so Q_{α} = J(Q_{α})(Q_{α} ∩ Q_{β}). Since Z_{α}(Q_{α} ∩ Q_{β}) ∈ A(Q_{αβ}) we get Q_{α} = J(Q_{α}). Thus

7^{∘} [12] Q_{α} = J(Q_{α}), Q_{β} = J(Q_{β}) and Q_{αβ} = Q_{α}Q_{β}.

Let A ∈ A(Q_{α}). Then Z_{α} ≤ A and C_{A}(D_{βα})D_{βα} = Z_{α}(Q_{α} ∩ Q_{β}). Thus Q_{α} ∩ Q_{β} = (A ∩ Q_{β})D_{βα} and [Q_{α} ∩ Q_{β}, A] = [D_{βα}, A] ≤ A_{βα} ≤ Z_{β}.

So

8^{∘} [13] [Q_{α} ∩ Q_{β}, Q_{β}] ≤ A_{αβ} and [Q_{α} ∩ Q_{β}, Q_{αβ}] ≤ A_{αβ}A_{βα} ≤ Z_{αβ}

Let Q̂_{β} = Q_{β}/Z_{β}. We conclude that

9^{∘} [14] [{ \atop (Q_{α} ∩ Q_{β})Z_{β}}, Q_{α}] = 1 and [Q̂_{β}, Q_{α}] ≤ { \atop Q_{α} ∩ Q_{β}}

We will now prove

10^{∘} [9] Suppose p = 2, and D_{βα}Z_{α} is normal in G_{α}, then 1.,2.,6. or 7. of 5.11 holds.

Since [Q_{α}, Z_{β}] ≤ D_{βα} and [D_{βα}, Z_{β}] = 1 we get [Q_{α}, O^{p}(G_{α})] ≤ Z_{α}. Let \overline{Q}_{α} = Q_{α}/D_{α}. Then Q_{α} centralizes \overline{Q_{α}}, C\overline{Q_{α}}(O^{p}(G_{α})) = 1 and [\overline{Q_{α}}, O^{p}(G_{α})] = \overline{U_{α}} is a natural module.

Thus the stucture of \overline{Q_{α}} is determined by 3.4. From Q_{β}, Z_{β}] = 1, Q_{α}Q_{β} = Q_{αβ} and (3^{∘}) we get \overline{Q_{α} ∩ Q_{β}} = \overline{D_{αβ}}. Hence Q_{α} ∩ Q_{β} ≤ D_{α}D_{αβ} and so

Q_{α} ∩ Q_{β} = (D_{α} ∩ Q_{β})D_{αβ}

Since [D_{α} ∩ Q_{β}, Q_{β}] ≤ D_{α} ∩ D_{β} = 1 we have D_{α} ∩ Q_{β} ≤ Z_{β}. As Z_{α} centralizes D_{α}, D_{α} ∩ Q_{β} ≤ Z_{β} ∩ Q_{α} = D_{βα}Z_{αβ}. We conclude

Q_{α} ∩ Q_{β} = D_{αβ}D_{βα}Z_{αβ} and T = Z_{α}Z_{β} = U_{α}Z_{β}

Since Q_{β} centralizes D_{βα}, 3.4 implies D_{βα} ≤ D_{α}Z_{αβ} and so

D_{βα}Z_{αβ} = (D_{α} ∩ (D_{βα}Z_{αβ})Z_{αβ}.

Note that r := |Q_{α}/D_{α}U_{α}| ≤ q. Let F = O^{p}(G_{α})∩Q_{αβ}. Then U_{α} ≤ F and |Q_{αβ}/Q_{α}F | = e, where e = 2 if (n_{α}, q) = (2, 2) or (1, 2) and e = 1 otherwise. Since D_{βα} ≤ D_{α}Z_{α}, F centralizes D_{βα} and so F ≤ U_{α}Q_{β} and F = U_{α}(F ∩ Q_{β}). Let F_{1} = C_{F} (D_{αβ}). Since F centralizes D_{βα}, F_{1} ≤ T = U_{α}Z_{β}. Since U_{α} ≤ F_{1}, F_{1} = U_{α}(F_{1} ∩ Z_{β}). We now consider 3 different Case for G_{α}/Q_{α}.

Case 1 of (10^{∘}): G_{α}/Q_{α} ≅ Sp_{2}(2)

Then Q_{α} = D_{a} × U_{α}. Moreover Q_{β} ≤ Z_{β}Q_{α} and Q_{β} = Z_{β}(Q_{α} ∩ Q_{β}) = Z_{β}D_{αβ}. In particular, Q_{β} is elementary abelian and Q_{β} = Z_{β}.

Assume that U_{β} ∩ Z(G_{β}) = 1. Then [D_{α}, Z_{β}] ≤ R ∩ D_{α} = 1 and D_{α} ≤ Z_{β}. Thus D_{α} is abelian and D_{α} is centralized by D_{α}U_{α}Z_{β} = Q_{αβ}. Thus D_{α} ≤ Z_{αβ} and Q_{α} = Z_{α}. Hence Z_{β} ∩ Q_{α} = Z_{αβ} and so G_{β}/Q_{β} ≅ SL_{2}(2). Thus 1. of 5.11 holds in this case.

Assume next that U_{β} ∩ Z(G_{β}) ≠ 1. Then 6. of 5.11 holds with k = 0.

Case 2 of (10^{∘}): G_{α}/Q_{α} ≇ Sp_{2}(2), Sp_{4}(2)

Then F_{1} ∩Z_{β} ≰ Q_{α} ∩Z_{β}. Since D_{α} centralizes F_{1} ∩Z_{β} we conclude that D_{α} ≤ Q_{β}. Since |Q_{αβ}/D_{α}(F ∩ Q_{β})Z_{β} ≤ rq ≤ q^{2} we get |Q_{αβ}/Q_{β}| ≤ q^{2} and so n_{β} = 1. Thus D_{βα} ≤ Z_{αβ} and so Q_{α} ∩ Q_{β} = Z_{α} ∩ Q_{β} = D_{αβ}Z_{αβ} = Z_{α} ∩ Q_{β}. Moreover, Q_{α} ≤ U_{α}Q_{β} and so Q_{α} = U_{α}(Q_{α} ∩ Q_{β}) = Z_{α}.

Assume that D_{αβ}Z_{β} is normal in G_{β}. The situation is then symmetric in α and β and from what we proved so far we conclude that n_{α} = 1 and Q_{β} = Z_{β}. It is now easy to see that 1. or 2. of 5.11 holds.

Assume next that D_{αβ}Z_{β} is not normal in G_{β}. Let W = D_{αβ}Z_{β}, V = ⟨W^{G_{β}}⟩ and U = ⋂_{g∈G_{β}} W^{g}. Since [W, Q_{β}] ≤ Z_{β} ≤ U and [V, Q_{α}] ≤ Q_{α}∩Q_{β} ≤ W we have [V, Q_{αβ}] ≤ W and [W, Q_{αβ}] ≤ U. Thus we can apply 3.3 to V/U and conclude that W = [Z_{α}, V]U. Hence

(∗) Z_{α} ∩ Q_{β} = [Z_{α}, V](Z_{α} ∩ U)

We claim that Z_{α} ∩ U = C_{Z_{α}}(V). Indeed, U ≤ Z(V) and so Z_{α} ∩ U ≤ C_{Z_{α}}(V). For the converse let g ∈ G_{β}. Then [C_{Z_{α}}(V), Z_{α}^{g}] ≤ R^{g} ≤ Z_{α} and so C_{Z_{α}}(V)Z_{β} is normal in G_{β}. Thus C_{Z_{α}}(V) ≤ U. This proves the claim and so

Z_{α} ∩ Q_{β} = [Z_{α}, V]C_{Z_{α}}(V).

Considering the perp of this equation in the sympletic space Z_{α} we have [Z_{α}, V] ∩ C_{Z_{α}}(V) ≤ Z_{αβ}. Let V^{∗} = [V, H_{β}]. Since H_{β} is generated by two conjugates of Z_{α} we derive

V/Z_{β} = V^{∗}/Z_{β} × U/Z_{β}.

Suppose that U ≠ Z(V). Then there exists U < X ≤ Z(V) with [X, Q_{αβ} ≤ U. Then X ≤ W and so X = Z_{β}(X ∩ Z_{α}). Since Z(V) ∩ Z_{α} ≤ U we conclude that X ≤ U, a contradiction. Thus U = Z(V). Since [W, Q_{β}] = A_{αβ} and Φ(Q_{β}) ≤ D_{β} we have A_{β} := A_{αβ} ≤ Z(G_{β}) and A_{β} = [V, Q_{β}]. Hence also [V^{∗}, Q_{β}] = A_{β}. Put D^{∗} = C_{Q_{β}}(V^{∗}). Then Q_{β}/D^{∗} is isomorphic to a submodule of Hom(V^{∗}/U, A_{β}) and since [Q_{β}, O^{p}(G_{β}) ≤ V^{∗} we conclude that Q_{β} = V^{∗}D^{∗}. Note that [D^{∗}, O^{p}(G_{β})] ≤ C_{V} (V^{∗}) = Z(V) = U and so [D^{∗}, O^{p}(G_{β}) = U_{β}. Also [D^{∗}, Q_{β}] ≤ Φ(Q_{β}) ≤ D_{β} and so from 3.4, [D^{∗}, Z_{α}] ≤ [Z_{α}, Z_{β}]D_{β}, Thus [Z_{α}, Q_{β}] = [Z_{α}, D^{∗}V^{∗}] ≤ [Z_{α}, V^{∗}]D_{β}. Hence

(∗∗) Z_{α} ∩ Q_{β} = [Z_{α}, V^{∗}](Z_{α} ∩ D_{β}).

Assume that q ≠ 2 and put H := O^{p}(G_{β}) ∩ Q_{αβ}. Then H centralizes Z_{α} ∩ D_{β} and (**) implies that H ≤ V^{∗}Q_{α} = V^{∗}Z_{α}. Hence H = V^{∗}(H ∩ Z_{α}). Thus Z_{α} = (H ∩ Z_{α})(Z_{α} ∩ Q_{β}).

On the otherhand [H ∩ Z_{α}, D^{∗}] ≤ Z_{αβ}D_{β} ∩ Z_{β} = Z_{αβ} and so D^{∗} ≤ Z_{β}Q_{α} and so D^{∗} = Z_{β}(Z_{α}∩D^{∗}). In particular, Q_{αβ} = Q_{β}Q_{α} = V^{∗}D^{∗}Q_{α} = V^{∗}Q_{α} and so C_{Z_{α}}(V^{∗}) = Z_{αβ}. Thus Z_{α} ∩D^{∗} = Z_{αβ}, D^{∗} = Z_{β} = U and Q_{β} = V. Moreover, |V/Z_{β}| = |Z_{α} ∩Q_{β}/Z_{αβ}|^{2} = q^{2(n_{α}−2)} and 2. of 5.11 holds in this case.

Assume that q = 2. Let |Z_{α} ∩ U/Z_{αβ}| = 2^{m} and |[Z_{α}, V]Z_{αβ}/Z_{αβ}| = 2^{l}. Then using (*), m + l = 2n_{α} − 2, |V/U| = 2^{2l}, |D^{∗}Q_{α}/Q_{α}| = 2^{m} and |D^{∗}/Z_{β}| = 2^{2m}. Note that the square map from V to A_{β} defines a non-degenerate quadratic form on V/U. This form is of +-type and the elements of order three in G_{β} act fixed-point-freely on V/U. This implies that k is even and so 6. of 5.11 holds.

Case 3 of (10^{∘}): G_{α}/Q_{α} ≅ Sp_{4}(2)

If D_{αβ}Z_{β} is normal in G_{β}, then either we are done by interchanging the roles of α and β, or G_{β}/Q_{β} ≅ Sp_{4}(2). In the latter case 7. of 5.11 holds.

So we may assume that D_{αβ}Z_{β} is not normal in G_{β}. Define W,V, U and D^{∗} as above. By 3.3 applied to V/U we conclude that [V/U, Z_{α}] has order at least 2^{2^{n_{β} −1}}. Since |[V/U, Z_{α}]| ≤ |W/Z_{β}| = 4. We conclude that n_{β} ≤ 2. If V/U is irreducible as G_{β} module we get that G_{β} acts transitively on V/U and so all elements in V^{#} are involutions. Thus V is elementary abelian, a contradiction. We conclude that n_{β} = 1, V/U has order 2^{4} and Z_{α} ∩ D^{∗} = Z_{αβ}. Since n_{β} = 1, Q_{α} ≤ Z_{α}Q_{β}. Thus Q_{α} = Z_{α}(Q_{α} ∩ Q_{β}) is elementary abelian and Q_{α} = Z_{α}. As before we conclude that Case 2. of 5.11 holds. Thus (10^{∘}) is proved.

11^{∘} [16] Suppose that p = 2 and that neither D_{βα}Z_{α} is normal in G_{α} nor D_{αβ}Z_{β} is normal in G_{β}. Then Case 4. of 5.11 holds.

For {γ, δ} = {α, β} let W_{γδ} = (Z_{γ} ∩ Q_{δ})Z_{δ}, V_{γδ} = ⟨(W_{γδ})_{δ}^{G}⟩ and U_{γδ} = ⋂_{g∈G_{δ}} W_{γδ}. By 3.3 applied to V/U we conclude that [V_{γδ}/U_{γδ}, Z_{γ}] has order at least 2^{2^{n_{δ}−1}}. Since [V_{γδ}/U_{γδ}, Z_{γ}] ≤ |W_{γδ}/Z_{δ}| = 2^{2n_{γ}−2}. Thus 2^{n_{δ}−2} ≤ n_{γ} − 1. Since this holds for both choices of (γ, δ) and since n − 1 ≤ 2^{n−2} for all positive integers n we conclude that n_{α} = n_{β} ≤ 3. Moreover, all of the above inequalites are equalites and V_{γδ}/U_{γδ} is irreducible as a G_{δ}- module. If n_{d} < 3 we get that G_{δ} acts transitively on V_{γδ}/U_{γδ}, a contradiction as above. Thus n_{α} = n_{β} = 3 and arguing as before we see that(11^{∘}) holds.

The lemma now follows from (4^{∘}),(10^{∘}) and (11^{∘}). □

Step 5 [b=1 case 3] Suppose that 5.9(3) holds. Then Case 5. of 5.11 holds.

Proof: Suppose first that p = 2. Then α is abelian and so X_{αβ} = [Z_{αβ}, Q_{αβ}] ≤ Q_{α}, a contradiction. Thus p is odd. Since [X_{αβ}, Q_{β}] = R ≤ Z_{β} have [Q_{β}, O^{p}(G_{β})] = U_{β} is a natural module for G_{β}. In particular both (Z_{α} ∩ Q_{β})Z_{β} and [Z_{α} ∩ Q_{β}, Q_{β}] are normal in G_{β}. Since R ∩ D_{β} = 1 we conclude that [Z_{α} ∩ Q_{β}, Q_{β}] = 1 and so Z_{α} ∩ Q_{β} = Z_{α} ∩ Z_{β}. Thus Q_{β} ≤ C_{Q_{αβ}}(Z_{α} ∩ Q_{β}) = Z_{β}Q_{α}. Hence Q_{β} = Z_{β}(Q_{α} ∩ Q_{β}). A corresponding statement holds for Q_{α} and so as before Q_{α} = Z_{α} and Q_{β} = Z_{β}. It is now easy to see that Case 5. of 5.11 holds. □

Step 6 [b=1 case 4] Suppose that 5.9(4) holds. Then Case 1. of 5.11 holds.

Proof: Since [Z_{α}, Q_{β}] ≤ R ≤ Z_{β} we get as in the previous step that [Q_{β}, O^{p}(G_{β})] = U_{β} is natural module and then that Q_{α} = Z_{α} and Q_{β} = Z_{β}. Thus Case 1. of 5.11 holds. □

Step 7 [b=1 case 5] 5.9(5) does not occur.

Proof: We choose notation such that \overline{G_{α}} ≅ SL_{n_{α}}(q). As in the two previous steps we get that [Q_{β}, O^{p}(G_{β})] = U_{β}, Z_{α}∩Q_{β} = Z_{α}∩Z_{β} and Q_{β} = Z_{β}(Q_{α}∩Q_{β}). Since |Z_{α}/Z_{α}∩Q_{β}| = q and n_{β} ≥ 2 we conclude that Z_{α} ∩ Z_{β} ≠ Z_{αβ}. Thus R = [Z_{α} ∩ Z_{β}, Q_{αβ}] = [Z_{β} ∩ Q_{α}, Q_{αβ}]. Hence 3.4 implies that U_{β} is a natural module.

Let δ ∈ β^{G_{α}} with Z_{δ} ≰ P_{αβ}. Then [X_{βα}, X_{δα}] ≤ R ∩ Z_{αδ} = 1 and thus [Z_{δ}, X_{δα}] ≤ [Z_{δ}, Z_{α}] ≤ Z_{α}. We conclude that X_{βα}Z_{α} is normal in G_{α}. Hence [Q_{α}, O^{p}(G_{α})] = U_{α} and [X_{βα}, Q_{α}] ≤ R ∩ D_{α} = 1. Then as before Q_{α} = Z_{α} and Q_{β} = Z_{β}. In G_{α} we see that |Z_{α}∩Z_{β}/Z_{αβ}| = q^{n_{α}−2} = |Q_{αβ}/Z_{α}Z_{β}|. In G_{β} we see that |Z_{α}∩Z_{β}| = q^{2n_{β}2} = |Q_{αβ}/Z_{α}Z_{β}| = q^{2n_{α}−2}. Thus n_{α} = 2n_{β}. Since n_{β} ≥ 2, [Z_{α}, Q_{αβ}] ≰ Z_{αβ} and so Q_{αβ}/Z_{αβ} is not abelian. We will show that this contradicts the action of P_{αβ} on Q_{αβ}/Z_{αβ}. Indeed P_{αβ} is transitive on Q_{αβ}/Z_{α} and so if p = 2 all non-trivial elements in Q_{αβ}/Z_{αβ} are involutions. Thus p ≠ 2. Since n_{α} is even, there exists an involution t in G_{α} inverting U_{α}. But then t centralizes Q_{αβ}/Z_{α} and so t centralizes the derived group of Q_{αβ}/Z_{αβ}, again a contradiction. □

5.11 is now an immediate consequence of Step 3 to Step 7 □

Lemma 5.12 [b=2 sigma=2] Suppose that |Σ| = 2, Σ = {α, β} and [Z_{α}, Z_{β}] = 1. Then one of the following holds:

1. [1] Z_{β}Ω_{1}Z_{α} is normal in G_{α}, G_{α} ∼ (2^{1+1} × 2^{3})2.SK_{3}(2) and G_{β} ∼ (2^{2} × 2^{1+2+2}SL_{2}(2).

2. [2] Who knows.

Proof: Again we devide the proof into several steps. Note that Q_{α} ∩ Q_{β} cannot be normal in both G_{α} and G_{β}. Hence by ?? we can choose notation such that V := V_{β}^{α} = V_{β} is not abelian. Pick α′ ∈ α^{G_{β}} with [Z_{α′}, Z_{α}] ≠ 1. Put D_{αβ} = [Z_{α}, Q_{β}], E_{β} := ⟨D_{αβ}^{G_{α}}⟩ and R = [Z_{α}, Z_{α′}].

Just as in Step 1 of 5.11 we have

Step 1 [da cap db 2] D_{α} ∩ D_{β} = 1.

Step 2 [ua cap zga=1] U_{α} ∩ Z(G_{α}) = 1.

Proof: Let A = U_{α} ∩ Z(G_{α}) and suppose that A ≠ 1. Then by 5.3, \overline{G_{α}} ≅ G_{2}(q) or Sp_{2n}(q). In the first case A ≤ A_{αβ} and by 5.6, A_{αβ} ≤ Z(G_{β}). So by Step 1, A = 1. Thus \overline{G_{α}} ≅ Sp_{2n}(q). If [D_{αβ}, V_{β}] = 1 put N = V_{β}. If [D_{αβ}, V_{β}] ≠ 1 but N = E_{β}. Then in any case [D_{αβ}, N] = 1, and N indueces a full transvection group on Z_{α}. Then by 5.6 [Z_{α}, N] ≤ Z(G_{β}). If q ≠ 2 we have A ≤ [Z_{α}, N] and as above A = 1. Hence q = 2 and A ≠ 1 implies n_{α} ≥ 2 and α is symplectic. Thus A ≤ A_{αβ}[Z_{α}, N] and so by Step 1, A_{αβ} ≰ Z(G_{β}). Hence 5.6 implies that D_{αβ} = X_{αβ} ≤ Z_{β}. In particular, R = [Z_{α}, N] ≤ Z(G_{β}). Since n_{α} ≥ 2 we conclude that Z_{α} ∩ Z_{β} ≠ Z_{αβ}.. So

(∗) A_{αβ} = [Z_{α} ∩ Z_{β}, Q_{αβ}]

Moreover, Q_{β} ≤ Z_{α′}Q_{α} and so Q_{β} = Z_{α′}(Q_{α} ∩ Q_{β}) = Z_{α}Z_{α′}(Q_{α} ∩ Q_{β} ∩ Q_{α′}).

Suppose that Q_{α} ∩Q_{β} is normal in G_{α}. [Q_{α} ∩Q_{β}, Z_{α′}] ≤ R ≤ Z_{α} implies [Q_{α}, O^{2}(G_{α})] = U_{α}. Thus Q_{α}/D_{α} is the natural module or the dual of the O_{2n+1}-module for G_{α}. Let γ ∈ α^{G_{β}}.

We claim that exactly one of [Z_{α}, Z_{γ}] ≠ 1 and [Z_{α′}, Z_{γ}] ≠ 1 holds. Indeed if none of the two statements holds, then R ≤ Φ(Q_{γ}) ∩ Z(G_{β}) ≤ D_{γ} ∩ D_{β}, a contradiction to Step 1. If both of the two statement holds, then the images of Z_{α} and Z_{α′} in Q_{γ}/D_{γ} are contained in a common maximal elementary abelian subgroup and hence again R ≤ D_{γ}. This proves the claim.

Define γ, δ ∈ α^{G_{β}} to be equivalent if [Z_{γ}, Z_{δ}] ≠ 1. By the claim this is an equivalence relation on α^{G_{β}} and has exactly two equivalecne classes. Thus both O^{2}(G_{β}) and Q_{αβ} act trivially on the the set of equivalence classes, a contradiction to G_{β} = Q_{αβ}O^{2}(G_{β}) and α′ ∈ α^{G_{β}}.

Thus Q_{α} ∩ Q_{β} is not normal in G_{α}. In particular Q_{α} ≰ Q_{β} and Z_{β} ≰ Z_{α}.

If β is linear with n_{β} ≥ 3 or a hex we get that from (*) that A_{αβ} ≤ A_{βα} and so by 5.6(a) ( applied with the roles of α and β interchanged) A_{αβ} ≤ Z(G_{α}), a contradiction. Thus \overline{G_{α}} ≅ Sp_{2n}(2). Since Z_{β} ≰ Z_{α} we have Q_{α} ≰ Q_{β} and so Z_{α} ∩ Z_{β} ≤ D_{βα}Z_{βα}. Thus (*) implies A_{αβ} = A_{βα} and n_{β} ≥ 2. Using 5.6 we conclude that D_{βα} = X_{βα} ≤ Z_{α}. Thus Z_{α} ∩ Z_{β} has index 2 in Z_{α} and in Z_{β}. Thus |Z_{α} ∩ Z_{β}/Z_{αβ}| = 2^{n_{α}} = 2^{n_{β}} and n_{α} = n_{β}. Also |Q_{α}/Q_{α} ∩ Q_{β}| = 2.

Since Q_{α} ∩ Q_{β} is not normal in G_{α} also Z_{β}Z_{α} is not normal in G_{α}. If V_{α} ≤ Q_{β} we get that [V_{α}, Z_{β}] ≤ R ≤ Z_{α}, a contradiction. Thus V_{α} is not abelian. Moreover, [V_{α}, Q_{α}] = Φ(V_{α}) = [V_{α}, Z_{β}] ≤ Z(G_{α}). Note that [Q_{α}, O^{2}(G_{α})] ≤ V_{α} and so Φ(Q_{α}) ≤ D_{α}. Note that by now the set up is symmetric in α and β and so also Φ(Q_{β}) ≤ D_{β}. Step 1 implies that Q_{α} ∩ Q_{β} is elementary abelian. Let γ ∈ β^{G_{α}} with Q_{α} ∩ Q_{β} ≠ Q_{α} ∩ Q_{γ}. Then (Q_{α} ∩ Q_{β})(Q_{α} ∩ Q_{β}) = Q_{α}, Q_{α} ∩ Q_{β} ∩ Q_{γ} ≤ Ω_{1}Z(Q_{α}) = Z_{α} and |Q_{α}/Z_{α}| ≤ 4. Since n_{α} ≥ 2, this forces Q_{α} ∩ Q_{β} to be normal in O^{2}(G_{α})Q_{αβ} = G_{α}, a contradiction. □

Step 3 [aab in zgb] A_{β} := A_{αβ} ≤ Z(G_{β}).

Proof: If X_{αβ} ≰ Z_{β}, this follows from 5.6(a). So suppose that X_{αβ} ≤ Z_{β}. Then α is symplectic and by 5.6(b)applied with N = V_{β} we conclude that [Z_{α}, V_{β}] ≤ Z(G_{β}). But by Step 2 A_{αβ} ≤ [Z_{α}, V_{β}] and we are done. □

Step 4 [zazb normal in ga] Suppose that Z_{β}Z_{α} is normal in G_{α}. Case 1 of 5.12 holds.

Proof: We first proof that

1^{∘} [1] [Q_{α}, O^{p}(G_{α})] = U_{α}.

Suppose not. Since Q_{α} ∩ Q_{β} is normal in G_{α} and Q_{β} ≰ Q_{α} we get [Q_{α} ∩ Q_{β}, O^{p}(G_{α})] ≰ Z_{α}. By Step 3 we have [Q_{α} ∩ Q_{β}, X_{α′β}] ≤ A_{α′β} ≤ Ω_{1}Z(G_{β}) ≤ Z_{α} and so X_{α′β} ≤ Q_{α}. Thus α′ and α are symplectic. Let γ ∈ △(α) with Z_{α′} ≰ P_{γα} and δ ∈ △(γ). By symmetry X_{δγ} ≤ Q_{γ} ∩ Q_{α} = Q_{β} ∩ Q_{α}. If [X_{δγ}, X_{α′β}] ≠ 1, then [X_{δγ}, X_{α′β}] = A_{β} = A_{γ} ≤ Z_{αβ} ∩ Z_{γα} ≤ Z(G_{α}), a contradiction to Step 1. Thus [X_{δγ}, X_{α′β}] = 1 and so [X_{δγ}, Z_{α′}] ≤ A_{β} ≤ Z_{α}. Put X_{γ} = ⟨X_{δγ} | δ ∈ △(γ)⟩. The X_{γ}Z_{α} is normalized by ⟨Q_{γα}, Z_{α′}⟩ = G_{α}. Also Z_{α′} centralizes X_{γ}Z_{α}/Z_{α} and so [X_{γ}, O^{p}(G_{α})] ≤ Z_{α}. Since [Q_{α} ∩ Q_{β}, Z_{α′}] ≤ X_{α′β} ≤ X_{β} ≤ X_{γ}Z_{α} we conclude that [Q_{α} ∩ Q_{β}, O^{p}(G_{α})] ≤ X_{γ}Z_{α} and (1^{∘}) is proved.

2^{∘} [2] Φ(Q_{α}) ≤ D_{α}.

This follows immediately from (1^{∘}).

3^{∘} [3] Let γ ∈ △(β). Then [Q_{α} ∩ Q_{β}, Z_{γ} ∩ Q_{α}] = 1, Q_{α} ∩ Q_{β} = Z_{α}(Q_{α} ∩ Q_{β} ∩ Q_{γ}) and Q_{β} ∩ Q_{α} is elementary abelian.

Suppose that [Q_{α} ∩ Q_{β}, Z_{γ} ∩ Q_{α}] ≠ 1. The action of Q_{α} ∩ Q_{β} on Z_{γ} implies that A_{γβ} ∩[Q_{α}∩Q_{β}, Z_{γ} ∩Q_{α}] ≠ 1. Since A_{γβ} = A_{β} ≤ D_{β} and Φ(Q_{α}) ≤ D_{α} we get a contradiction to Step 1.

Thus [Q_{α}∩Q_{β}, Z_{γ}∩Q_{α}] = 1 and so Q_{α}∩Q_{β} ≤ Z_{α}Q_{γ}. Hence Q_{α}∩Q_{β} = Z_{α}(Q_{α}∩Q_{β}∩Q_{γ}) and Φ(Q_{α} ∩ Q_{β}) = Φ(Q_{α} ∩ Q_{β} ∩ Q_{γ}) = Φ(Q_{γ} ∩ Q_{α}). Since this is true for all γ ∈ △(β) we get that Φ(Q_{α} ∩ Q_{β}) is normal in G_{β}. By (2^{∘}) Φ(Q_{α} ∩ Q_{β}) is also normalized by O^{p}(G_{α}) and so Φ(Q_{α} ∩ Q_{β}) = 1.

4^{∘} [4] Put F_{β} = C_{Q_{β}}(V_{β}). Then F_{β} = G_{β}^{(2)}, F_{β} is elementary abelian and C_{G_{β}}(V_{β}F_{β}/F_{β}) ≤ Q_{β}.

The first statement follows from 5.2(b) and 5.3(b). The second follows from(3^{∘}). Suppose the last statement is false. Then G_{β} = O^{p}(G_{β})Q_{αβ} normalizes Z_{α}F_{β} and so Z_{α′} ≤ Z_{α}F_{α} ≤ Q_{α}, a contradiction.

5^{∘} [5] D_{α} ≤ F_{β}.

Let γ ∈ △(β). By (3^{∘}) and (1^{∘}) [D_{α} ∩ Q_{β}, Z_{γ}] ≤ D_{α} ∩ [Z_{α}, Z_{γ}] ≤ D_{α} ∩ U_{α} = 1. Hence D_{α} ∩ Q_{β} ≤ F_{β}. Thus [D_{α}, V_{β}] ≤ D_{α} ∩ Q_{β} ≤ F_{β} and (4^{∘}) implies D_{α} ≤ Q_{β}. So D_{α} = D_{α} ∩ Q_{β} ≤ F_{β}.

6^{∘} [6] α is not a hex.

Suppose that α is a hex. Then Q_{αβ} = Q_{α}Z_{α′} and by 3.4, [Q_{α}, Z_{α′}] ≤ D_{α}R ≤ F_{β}. It follows that Φ(Q_{αβ}) ≤ F_{β}. Thus [Q_{αβ}, V_{β}] ≤ F_{β} and (4^{∘}) gives the contradiction Q_{αβ} ≤ Q_{β}.

7^{∘} [7] Q_{α} ≠ D_{α}Z_{α}.

Suppose that Q_{α} = D_{α}Z_{α}. Then by (5^{∘}), Q_{α} ≤ Q_{β} and so by (3^{∘}), Q_{α} is elementary abelian and hence Q_{α} = Z_{α}. Thus Z_{β} ≤ Z_{α} and Z_{β} ≤ C_{Z_{α}}(Z_{α′}). It follows that [Z_{β}, Q_{αβ}] ≤ A_{αβ} ≤ Z(G_{β}) and so also [Z_{β}, G_{β}] ≤ Z(G_{β}) a contradiction.

8^{∘} [8] \overline{G}_{α}/Q_{α} is not a symplectic group.

Otherwise (7^{∘}) and 3.4 imply that p = 2.

Assume first that |V_{β}Q_{α}/Q_{α}| = 2. Then

[Q_{α}V_{β}[Q_{αβ}, O^{2}(P_{αβ})], V_{β}] ≤ D_{α}D_{αβ} ≤ F_{β}

and so [Q_{αβ}, V_{β}] ≤ F_{β}, a contradiction to (4^{∘}).

Assume next that |V_{β}Q_{α}/Q_{α}| > 2. Then by the action of P_{αβ} on Q_{αβ}/Q_{α}, every coset of Q_{α} in Q_{αβ} contains an involution. By 3.4, [Q_{α}/D_{α}D_{αβ}, Q_{αβ}] = 1 and so Q_{αβ}/D_{α}D_{αβ} is elementary abelian. Hence [Q_{αβ}, V_{β}] ≤ F_{β}E_{β} and so Z_{α}F_{β}E_{β} is normal in G_{β}. Thus V_{β} ≤ Z_{α}F_{β}E_{β} ≤ Q_{α}E_{β}. Thus implies [D_{αβ}, V_{β}] = 1, a contradiction to |V_{β}Q_{α}/Q_{α}| > 2. Hence (8^{∘}) is proved.

We are now able to prove Step 4. By (6^{∘}), (7^{∘}) , (8^{∘}) and 3.4 we have G_{α}/ D_{α} ∼ 2^{3+1}L_{3}(2). Note that Q_{α} is not elementary abelian, D_{α}U_{α} has index two in Q_{α}, D_{α}U_{α} ≤ Q_{α} ∩ Q_{β} and Q_{α} ∩ Q_{β} is elementray abelian. Thus Q_{α} ∩ Q_{β} = D_{α}U_{α}. Also

|Q_{β}/F_{β}| = \frac{|Q_{αβ}/D_{β}|}{|Q_{αβ}/Q_{β}||F_{β}/D_{α}|}

.

Suppose that Q_{αβ} ≠ Q_{α}Q_{β}. Then |Q_{αβ}/Q_{α}Q_{β}| = 2, |Q_{α}Q_{β}/Q_{β}| = 2 and |Q_{αβ}/Q_{β}| = 4. Also |V_{β}Q_{α}/Q_{α}| = 2 and so |Z_{α}/Z_{α} ∩ F_{β}| = 2 and |F_{β}/D_{α}| = 4. Thus by (*) |Q_{β}/F_{β}| = 2^{3}. Since |Q_{αβ}/Q_{β}| = 4, G_{β}/Q_{β} ≅ SL_{2}(4) and we conclude that O^{2}(G_{β}) centralizes Q_{β}/F_{β}, a contradiction to (4^{∘}).

Thus Q_{αβ} = Q_{α}Q_{β}. It follows that |Q_{αβ}/Q_{β}| = 2 and G_{β}/Q_{β} ≅ SL_{2}(2). Since Z(G_{β}) ≠ 1 and Z(G_{α}) ∩ Z(G_{β}) = 1, we have |Ω_{1}Z(G_{β})| = 2 and |Z_{β}| = 2^{3}. Hence also |Ω_{1}Z(G_{α})| = 2. Since |G_{α}/C_{G_{α}}(D_{α}| = 2 and D_{α} is elementary abelian we conclude that |D_{α}| ≤ 4. Since Q_{αβ} = Q_{α}Q_{β} we have Z_{β} ≤ D_{α}Z_{αβ} and so by order reasons Z_{β} = D_{α}Z_{αβ} and |D_{α}| = 4. In particular, Q_{β} = C_{Q_{αβ}}(D_{α}).

Suppose that V_{β}Q_{α} ≠ Q_{αβ}. Then |F_{β}/Z_{β}| = 2 and |V_{β}/F_{β}| = 4. But then all elements in V_{β}^{#} are involutions, a contradiction.

Thus V_{β}Q_{α} = Q_{αβ}. Since D_{α}Z_{α} = Z_{β}Z_{α} ≤ V_{β}, we get |Q_{αβ}/V_{β}| = 2, Q_{β} = V_{β} and F_{β} = Z_{β}. The square map from V_{β} to A_{β} defines a non-degenerate quadratic form on V_{β}/Z_{β}. Moreover, [V_{β}/Z_{β}, Q_{αβ}] = Q_{α} ∩ Q_{β}/Z_{β} is a singular subspace of order four for this quadratic form and we conclude that the elements of order three in G_{β} must act fixed-point freely on V_{β}/Z_{β}. Hence the shapes of G_{α} and G_{β} are as claimed. □

Step 5 [yazb not normal in ga] Suppose that Z_{β}Z_{α} is not normal in G_{α}. Then who knows?.

Proof: For (γ, δ) an edge define H_{γδ} = X_{γδ} if X_{γδ} ≰ Z(V_{δ}) and H_{γδ} = Z_{γ} otherwise. Let H_{δ} = ⟨H_{γδ}^{G_{δ}}. Also let Z_{γ}^{∗} be maximal in Q_{γ} with [Z_{γ}^{∗}, O^{p}(G_{δ}) ≤ Z_{α}.

1^{∘} [1] [Z_{α}, H_{β}] = A_{β} ≤ Ω_{1}Z(G_{β}) ≤ Z_{α}.

This follows from that definition of H_{β}, Step 2 and Step 3

2^{∘} [2] Z_{β} ∩ Z(V_{α}) ≤ Z_{α}^{∗}.

Note that Z(V_{α}) ≤ Q_{β} and [Z(V_{α}), Z_{α}] = 1.

Suppose that α is not symplectic. The we conclude that [Z(V_{α}), Z_{α′}] ≤ R ≤ Z_{α}. Hence Z(V_{α}) ≤ Z_{α}^{∗} and we are done in this case.

Suppose next that α is symplectic. Note that [Z(V_{α}), D_{αβ}] ≤ A_{β} ≤ Z_{α}. So if D_{α′β} ≰ Q_{α} we are done again. So assume that D_{α′β} ≤ Q_{α}. Let γ ∈ △(α) with Z_{α′} ≰ P_{αγ}. Put A := [Z_{g} ∩ Z(V_{α}), D_{αβ}] ≠ 1. From the action on Z_{α′} we see that A ≤ A_{β} ≤ Z(G_{β}). From G_{α} = ⟨Q_{γα}, Z_{α′}⟩ we get A_{γα} ∩ A_{β} ≤ Z(G_{α}) ∩ Z(G_{β}) = 1. Hence A ∩ A_{γα} = 1 and so the action of D_{αβ} on Z_{γ} implies A = 1. Hence the action of Z_{γ} ∩ Z(V_{α}) reveals [Z_{γ} ∩ Z(V_{α}), Z_{α′}] ≤ A_{β} ≤ Z_{α}. So (Z_{α} ∩ Z(V_{α}))Z_{α} is normalized by G_{α} = ⟨Q_{γα}, Z_{α′}⟩ and the proof of (2^{∘}) is complete.

3^{∘} [3] V_{α} is not abelian. In particular, the set-up is symmetric in α and β.

By assumption Z_{β}Z_{α} is not normal in G_{α}. In particular, Z_{β} ≰ Z_{α}^{∗} and so (3^{∘}) follows from (2^{∘}).

For α ∈ Σ let

Σ_{1}(α) = {β ∈ Σ | [Z_{α}, Z_{β}] ≠ 1}

and

Σ_{2}(α) = {β ∈ Σ | [Z_{α}, Z_{β}] = 1 ≠ [Z_{α}, V_{β}^{α}]}

Lemma 5.13 Let α ∈ Σ and β ∈ Σ_{1}(α). definitionine L := ⟨G_{α}, G_{β}⟩, L^{∗} := ⟨Ω_{1}Z(R)^{L}⟩, K := O_{R}({G_{α}, G_{β}}) and L̃ := L/K. For {α, β} = {γ, δ}, put K_{γ} = C_{Q_{γ}}(⟨Z_{δ}^{G_{γ}}⟩). Then for γ ∈ {α, β} there exists a normal subgroup L_{γ} of G_{γ} such that (a) [a] [K, L^{∗}] = 1.

(b) [b] K = K_{α} ∩ K_{β} and Φ(K_{α}K_{β}) ≤ K.

(c) [c] G_{α} = K_{α}L_{α} and G_{β} = K_{β}L_{β}.

(d) [d] Interchanging α and β if necessary one of the following holds ( where q is a power of p.

1. [1] L̃_{α} ∼ L̃_{b} ∼ q^{n}SL_{n}(q).

2. [2] p = 2, L̃_{α} ∼ q^{1+2n}Sp_{2n}(q), and L̃_{β} ∼ q^{1+2+2·(2n−2)}SL_{2}(q).

3. [3] p = 2 and L̃_{α} ∼ L̃_{β} ∼ 2^{6}G_{2}(2)

4. [4] p = 2 and L̃_{α} ∼ L̃_{β} ∼ q^{1+6+8}Sp_{6}(q).

5. [5] Who knows.

Proof: Note that K is normal in L and K ≤ R, indeed K is the largest normal subgroup of L contained in R. Let g ∈ K then

[Ω_{1}Z(R)^{g}, K] = [Ω_{1}Z(R)^{g}, K^{g} = [Ω_{1}Z(R), K]^{g} = 1.

Thus (a) holds.

Let H_{γ} = ⟨Z_{δ}^{G_{γ}}⟩, R = [Z_{α}, Z_{β}] and D_{βα} = [Z_{β}, Q_{αβ}.

Note that by (a), K ≤ K_{α} ∩ K_{β} also K_{α} ∩ K_{β} is normalized by

⟨O^{2}(G_{α}), O^{2}(G_{β}), Q_{αβ}⟩ = L.

Thus K = K_{α} ∩ K_{b}. So the first part of (b) holds. By definition [K_{α}, Z_{β}] = 1 and so K_{a} ≤ Q_{β}. Thus Φ(K_{α}) ≤ Φ(Q_{β}) ∩ K_{α}. Note that Φ(Q_{β}) ≤ Φ(Q_{αβ}). Since Q_{αβ}/Q_{α} is elementary abelian, unless α is symplectic, n_{α} > 1 and p ≠ 2, we get

(*) Φ(K_{α}) ≤ K and [Φ(Q_{β}), H_{β}] = 1, unless α is symplectic, n_{α} > 1 and p ≠ 2.

Note that by definition of Σ_{1}(α), [Z_{α}, Z_{β}] ≠ 1. Clearly Z_{α}Z_{b} ≤ Q_{αβ} and we can apply 5.9 with (δ, γ) = (β, α).

Suppose that Case c.1 of 5.9 holds. Then Q_{α} ≤ Q_{αβ} = Z_{α}Q_{β}. Since Q_{α} normalizes Z_{β}, H_{α} is generated by two conjugates of Z_{β}. Thus |Q_{α}/K_{α}| ≤ q^{6} and so Q_{α} = K_{α}U_{α}. By 5.10, q = 2 and U_{α} ∩ Z(G_{α}) = U_{β} ∩ Z(G_{β}). Thus U_{α} ∩ Z(G_{α}) ≤ K and |Ũ_{α}| = 2^{6}. Using [Schur, Schur Multiplier] we get O^{2}(G_{α})/U_{a} ≅ G_{2}(2)′ also by (*) G_{α}/O^{2}(G_{α})K is elementary abelian. Hence there exists L_{α} ≤ G_{α} with O^{2}(G_{α})K ≤ L, G_{α} = K_{α}L_{α} and L_{α} ∩ K_{α} = K. Thus d.3 holds in this case.

Suppose next that Case c.2 of 5.9 holds.

Suppose that n_{β} = 1. Then [Q_{α}, Z_{β}] ≤ [Z_{α}, Z_{β}] ≤ U_{α} and so [Q_{α}, H_{α}] ≤ U_{α}. Also Φ(Q_{α}) ≤ Q_{β} and so [Φ(Q_{α}), H_{α}] = 1. Suppose that also n_{α} = 1. Then H_{α} is generated by two conjugates of Z_{β} and we conclude that |Q_{α}/K_{α}| = q^{2} and Q_{α} = K_{α}U_{α}. Let I = {1 ≠ [x, y] | x ∈ Z_{α}, y ∈ Z_{b}}. If q ≤ |[Z_{α}, Z_{β}]| < q^{2} then U_{α} ∩ Z(G_{α}) = [Z_{α}, Z_{b}] \ I = U_{β} ∩ Z(G_{β}) and thus d.1 holds. If |[Z_{α}, Z_{β}]| = q^{2}, then [Z_{α}, Z_{b}] \ I contains exactly two subgroups of order q and these two subgroups have trivial intersection. Hence either U_{α} ∩ Z(G_{α}) = U_{β} ∩ Z(G_{β}) and d.1 holds; or U_{α} ∩ Z(G_{α}) ∩ U_{β} ∩ Z(G_{b}) = 1 and d.2 holds.

Suppose next that n_{β} > 1 and that D_{βα}Z_{α} is normal in G_{α}. Then A_{α} := [D_{βα}, Q_{αβ} = [D_{βα}Z_{α}, Q_{α}] is normal in G_{α}. Since Q_{αβ} centralizes A_{α} we get A_{α} ≤ Z(G_{α}). Let D_{α} := C_{Q_{α}}(O^{p}(G_{α})). We conclude that D_{αβ} ≤ U_{α}D_{α} and D_{αβ} ≤ D_{α}Z_{αβ}. Note that [Q_{α}, Z_{β}] ≤ D_{βα} and so [Q_{α}, H_{α}] ≤ U_{α}D_{α}.

Note that |RA_{α}/A_{a}| ≥ q and so p = 2 and |U_{β} ∩ Z(G_{β})| =q. By (*) [Φ(Q_{α}), H_{α}] = 1. Thus |Q_{α}/U_{α}D_{α}| ≤ q. Note that O^{2}(G_{α}) ∩ Q_{αβ} centralizes D_{α}Z_{αβ}and so we have O^{2}(G_{α}) ∩ Q_{αβ} ≤ C_{Q_{αβ}}(D_{β}α) = Z_{α}Q_{b}. Note also that Z_{β} ≤ Q_{β}, G_{α} = O^{2}(G_{a})Z_{β} and Z_{α} ≤ Q_{α}. Thus Q_{αβ} = Q_{α}Q_{β}.

If q > 2, then A_{a} ≤ R and we conclude that A_{α} = U_{α} ∩ Z(G_{α}).

Let γ ∈ β^{G_{α}} with [Z_{γα}, Z_{β}] ≠ 1.

Lemma 5.14 [sigma symmetric] Let α, β ∈ Σ and i ∈ {1, 2}. Then α ∈ Σ_{i}(β) if and only if β ∈ Σ_{i}(α).

Proof: For i = 1 this is obvious. Suppose now that β ∈ Σ_{2}(α) but α ∉ Σ_{2}(β). The Z_{a}Z_{β} ≤ Q_{α} ∩ Q_{b}, V_{β}^{α} ≰ Q_{α} and V_{α}^{β} ≤ Q_{β}.

Lemma 5.15 [vdelta non abelian] There exists an edge (γ, δ) in Γ such that ⟨Z_{δ}^{G_{γ}}⟩ is not abelian.

Proof: Suppose not. Let V = ⟨Z_{L}⟩L ∈ Σ and Q = ⋂ O_{p}(L) | L ∈ Σ. Then V ≤ Q and so Q ≠ 1. Let L ∈ Σ. Then Q = ⋂(O_{p}(L) ∩ O_{p}(H) | L ≠ H ∈ Σ and so by 5.7 Q is normal in L. Hence Q is a non-trivial subgroup of R which is normal in all the LΣ, a contradiction.□

Some ideas on the rest of the proof. Define a relation ≈ on Σ by L ≈ H if ⟨Z_{L}^{H}⟩ is not abelian or if Z_{L} = Z_{H}. This should be an equivalence relation and L ≈ H if and only if O_{p}(L) ∩ O_{p}(H) is not normal in L. If L ≉ H we should have [(R ∩ O^{p}(L), O^{p}(H)] = 1. b = 2 ( that is L ≈ H and Z_{L} ≤ O_{p}(H)) seems to occur only for the G_{2}(3^{k}) situation, and 2^{1+4+6}L_{4}(2)

What still needs to be discussed in this section is the consequences of 5.1 for the sets PU_{i}, i = 1, 2, 4. There are some interesting cases: for example an amalgam if Z_{L} is the 6-dimensional module for L/O_{2}(L) ≅ 3Alt(6) then L ∈ PU_{4}(R). Same for Alt(6) or Alt(7) on the four dimensional module.

Also it seems possile to enlarge the set PU_{3} without having to change the ”b < 3” part of the proof of 5.1. Namely can drop the assumption on N_{L}(R) containing a point stabilizor one can allow [Z_{L}, L] to be the four dimensionnal module for SL_{3}(2), This would be usefull for the ¬E! case. Other exceptional F F -modules could be included to. The properties one really needs is: no over-offenders and good commutator control. For example Alt(n) on the natural module should be o.k. This also would be o.k for D_{1}0(q) on the 16-dimensional spinmodule and L_{n}(q), n ≥ 5 on the exterior square. But the choice of a ∈ A will cause some problems. Might not be so important though, maybe we only need ⋂_{a∈A} Z_{γ}^{a} ≤ Ω_{1}Z(G_{δ}).

6 The C(G,T)-Theroem

Suppose that G fullfills CGT. Then S is contained in unique maximal subgroup M of G, but there exists L ∈ L(S) such that L ≰ M and |L ∩ M|_{p} ≠ 1. Choose such an L such that |H ∩ L|_{p} is maximal. Let T be a Sylow p-subgroup of H ∩ T. Without loss T ≤ S. If T = S we get that L ∈ L(S) contradicting our assumption M is the unique maximal p-local subgroup of M. Thus T ≠ S. Let C be a non-trivial characteristic subgroup of S. Then N_{S}(T) ≤ N_{G}(C) and so |M ∩ N_{G}(C)|_{p} > |M ∩ L| Hence the maximal choice of |M ∩ L|_{p} implies N_{G}(C) ≤ M. In particular, N_{L}(C) ≤ M ∩ L. For C = S we conclude that T ∈ Syl_{p}(T). Then we can apply the

Theorem 6.1 (Local C(G,T)-Theorem) [local CGT] Let L be a finite K_{p} group of characteristicp, T a Sylow p-subgroup of L, and suppose that

C(L, T) := ⟨N_{L}(C) | 1 ≠ C a characteristic subgroup of S⟩

is a proper subgroup of L. Then there exists a L-invariant set D of subnormal subgroup of L such that

(a) [a] L = ⟨D⟩C(L, T)

(b) [b] [D_{1}, D_{2}] = 1 for all D_{1} ≠ D_{2} ∈ D.

(c) [c] Let D ∈ D, then D ≰ C(L, T) and one of the following holds:

1. [1] D/Z(D) is the semidirect product of SL_{2}(p^{k}) with a natural module for SD_{2}(p^{k}). Moreover O_{p}(D) = [O_{p}(D), D] is elementary abelian.

2. [2] p = 2 and D is the the semidirect product of Sym(2^{k} + 1) with a natural module for Sym(2^{k} + 1).

3. [3] p = 3, D is the semidirect product of O_{3}(D) and SD_{2}(3^{k}), Z(D) = O_{p}(D) has order 3^{k} and both [Z(O_{3}(D)), D] and O_{3}(D)/Z(O_{3}(D)) are natural SL_{2}(3^{k}) modules for D.

For p = 2 the local C(G, T)-theorem was proved by Aschbacher in [Asch].

For general p by GLS?. For us it will be consequence of the ??.

Back to G. Case 3 can be rules out using that N_{S}(T)/T is odd. Let m = |D| and suppose that m > 1. Let g ∈ N_{S}(T) \ T. Then there exists X, Y ∈ D such that R := [[V, X], [V, Y]^{g}] ≠ 1. Let H = N_{G}(R). Then for all Z ∈ D with D ≠ D, D ≤ N_{G}(R) and since [[V, D], V^{g}] ≠ 1, [V, D] ≰ O_{p}(N_{L^{g}}(R). Thus [V, D] ≰ O_{p}(H). Let U = O_{p}(H). We conclude that [Q ∩ T, D] = 1. Since H is of characteristic p, D acts non-trivially on Q/Q ∩ T.

Let T^{∗} ∈ Syl_{p}(H) with N_{T} (R) ≤ T^{∗}. The maximal choice of |T | implies |T^{∗}/N_{T} (R)| ≤ |T/N_{T} (R)| = T/N_{T} (X). In particular |U/U ∩ T | ≤ |T/N_{T} (X). Thus T does not normalize X. Let e := |T/N_{T} (X)|. Then there are at least e − 1 choices for D, each two of whcih commute and each acting non-trivialy on U/U ∩ T whci has order at most e. This is impossible.

Hence there exists a unique D ∈ D.

Suppose that case 2. holds and n ≥ 3. Then O_{2}(M ∩ L) = O_{2}(L). Let Q = O_{2}(M). Then T ∩ Q ≤ O_{2}(M ∩ L) ≤ O_{2}(L). On the otherhand the maximality of |T | implies N_{Q}(O_{2}(L)) ≤ T. Thus N_{Q}(O_{2}(L)) ≤ O_{2}(L) and so Q ≤ O_{2}(L).

If Q is not elementary abelian that [Φ(Q), D] = 1 implies D ≤ M, a contradiction. Hence Q is elementary abelian.

Since [Q, O_{2}(D)] = 1 and M is of characteritic p we conclude O_{2}(D) ≤ Q. Thus [Q, D] ≤ [O_{2}(L), D] ≤ O_{2}(D) ≤ Q and so D ≤ N_{G}(Q) ≤ M. Thus also L = D(M ∩L) ≤ M, a contradiction.

Suppose that case 2 holds and n = 2. Then we can choose x ∈ [V, D] so that R := [V^{g}, x] has order two. Also C_{D}(x) is divisible by 3 and [V, O^{2}(C_{D}(x))], C_{D^{g}}(x)] is not a 2-group. Argue as above we get C_{D}(x) acts non trivially on Q/Q ∩ T. But |Q/Q ∩ T has order 2 a contradiction.

Thus Case 1. holds. We have proved:

References

[Asch] M. Aschbacher, A Factorization Theorem for 2-constrained Groups, Proc. London. Math. Soc. (3) 43 (1981), 450-477.

[BBSM] B. Baumeister, A. Chermak, U. Meierfrankenfeld, G. Stroth, The Big Book Of Small Modules

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

[MS] U. Meierfrankenfeld, B. Stellmacher, Pushing Up Weak BN-Pairs of rank two, Comm. in Algebra, 21(3), 825-934 (1993).

[Schur] Some Schurmultipliers