The Small World Theorem

Ulrich Meierfrankenfeld, Bernd Stellmacher

May 4, 2003

1 Introduction

Assume M(S) ≥ 2 and Q!. We investigate the Structure of E/O_{p}(E).

For L ∈ L define L_{∘} = L^{∘}O_{p}(L). In this section we assume

Hypothesis 1.1 [hypothesis e structure theorem]

ES1 [ES1] M(S) ≥ 2 and Q!.

ES2 [ES2] P ∈ P(S) and P ≰ C̃.

ES3 [ES3] P_{∘}/O_{p}(P) ≅ SL_{2}(q), q a power of 2.

ES4 [ES4] Y_{P} is a natural module for P/O_{p}(P).

ES5 [ES5] N_{P} (S ∩ P_{∘}) ≤ C̃.

ES6 [ES6] ⟨Y_{P}^{E}⟩ is abelian.

ES7 [ES7] 0_{p}(⟨P, E⟩) = 1.

Some remarks on this assumptions. (ES7) follows from E! but not from Q! (example (L_{n}(q). (ES2) to (ES5) follow from P! theorem. But ¬P! have currently been treated only for Y_{M} ≤ Q.

Let L = N_{G}(P^{∘}) and H = E(L ∩ C̃).

By (ES7) O_{p}(⟨H, L⟩) = 1. By part (a) of the preceeding lemma we can apply the amalgam method to (H, L). Let Γ_{0} = Γ(G; L, H) and Γ the connected component of Γ containing L and H. For α ∈ Γ. If α = Lg define E_{α} = P_{∘}^{g}, if α = Hg define E_{α} = (EQ)^{g}, C̃_{α} = C̃^{g} and Q_{α}^{∗} = Q^{g}.

2 Preliminaries

Lemma 2.1 [pushing up sl2q]
Lemma 2.2 [basic q!]

3 A basic lemma

Lemma 3.1 [basic es] Let (α, β) be adjacent vertices with α ∼ H

(a) [a] G_{α} = E_{α}G_{αβ} and G_{β} = E_{β}G_{αβ}.

(b) [c] C_{U}(Y_{β}) = Q_{α}Q_{β} = Q_{a}Q_{β}^{∗} = Q_{αβ} ∈ Syl_{p}(L_{α}), where U ∈ Syl_{p}(G_{αβ}).

(c) [d] G_{αβ} = N_{G_{α}}(Q_{αβ})

(d) [e] Y_{αβ} = Y_{β} = [Y_{α}, x] for all x ∈ Q_{αβ} \ Q_{α}.

(e) [f] Let Ṽ_{β} = V_{β}/Y_{β}. Then C_{G_{β}}(Ṽ_{β})∩C_{G_{β}}(Y_{β}) = Q_{β} In particular G_{αβ} contains a point stablizer for G_{β} on Ṽ_{β}.

Proof: By edge transitivity we may assume that α = L and β = H. By (ES5), and the Frattini Argument, L = P_{∘}(L ∩ C̃) = P_{∘}(L ∩ H). By definition of H, H = (H ∩ L)E. Thus (a) holds.

Note that Y_{H∩L} ≤ Y_{H} and so the definition of E implies [Y_{H}, E] = 1 . Since L = (L∩H)E we get Y_{H∩L} = Y_{H}. Let T = S ∩ P^{∘}. Since N_{P} (T) is a maximal subgroup of L we get N_{P} (T) = H ∩ L and O_{p}(H ∩ L) = T. Since Y_{H∩L} ≤ Y_{L} and Y_{P} is a natutal module for P_{∘} we see that Y_{H} = Y_{H∩L} = C_{Y_{L}}(T) = Ω_{1}Z(T) and C_{S}(Y_{H}) = T. Also since Q ≰ O_{p}(P), O_{p}(H)O_{p}(L)/O_{p}(L) is a non-trivial subgroup of T/O_{p}(L) normalized by H ∩ L. Thus O_{p}(H)O_{p}(L) = T.

For (f) let D = C_{H}(Ṽ) ∩ C_{H}(Y_{H}). Since [Y_{L}, O_{p}(H)] = Y_{H}, O_{p}(H) ≤ D. Note that O^{p}(D) centralizes Y_{P} and so [P^{∘}, O^{p}(D) ≤ O_{p}(P). Since O^{p}(D) = O^{p}(O^{p}(D)O_{p}(P)] we get P^{∘} ≤ N_{G}(O^{p}(D)). Since also H normalizes O^{p}(D) we conclude that O^{p}(D) = 1, D is a p-group and D = O_{p}(H)

G_{αβ} = N_{G}(Q_{αβ})

Suppose that N_{G}(T) ≰ L. Put M = ⟨N_{G}(T), P_{∘}T ⟩. If O_{p}(M) = 1 then pushing up SL_{2}(q) 2.1 and Ω_{1}Z(P^{∘}) = 1 gives [O_{p}(P), O^{p}(P)] ≤ Y_{P}. By ES6, V = ⟨Z_{P}^{H}⟩ ≤ O_{p}(L). Thus V is normal in H and L, a contradiction. The last equality in (e) follows since Y_{P} is a the natural module. □

The following is not needed in the F F -module argumnet:

Lemma 3.2 [more basic]

C_{G_{β}}(x̃) ≤ G_{αβ} for all x ∈ Y_{α} \ Y_{β}.

Proof: Let D^{∗} = C_{H}(x̃) and D = C_{H}(x), T acts transitively on the coset Y_{H}x, D^{∗} = DT. Let x ∈ Y_{H}^{l} for some l ∈ L. Then D ≤ C_{G}(x) and by Q!, D ≤ N_{G}(Q^{l}). Thus D ≤ N_{G}(⟨Q, Q^{l}⟩ = N_{G}(P^{∘}) = L. □

4 The F F -module argument

Let (α, α′) be a critical pair. Let β = α + 1 and α − 1 ∈ Δ(a) with α − 1 ≠ β. Lemma 4.1 b l 2

(a) [a] b > 2.

(b) [b] α ∼ H.

(c) [b] b is odd. Proof: b > 2 follows from (ES6) and α ∼ H follows from Y_{H} ≤ Y_{L}. Suppose that b is even. The by 3.1 Y_{β} = [Y_{α}, Y_{α}] = Y_{α′−1}. Hence by ??(c), E_{β} = E_{α′−1}. Since b > 3, V_{α−1} ≤ Q_{β} and V_{α−1} ≤ N_{G}(Q_{β}^{∗}) = N_{G}(Q_{α′−1}^{∗}). Thus

V_{α−1} ≤ N_{G_{α−2}}(Q_{α′−2}Q_{α′−1}^{∗}) = N_{G_{α′−2}}(Q_{α−2α−1}) = G_{α−2α−1}

As V_{α−1} ≤ Q_{β}, [V_{α−1}, E_{β}] is a p-group and so V_{α−1} ≤ Q_{α′−1} ≤ Q_{α′−1α′} and hence [V_{α−1}, Y_{α′}] ≤ Y_{α′−1} = Y_{β} ≤ Y_{α}. Hence Y_{α′} normalizes V_{α−1}, a contradiction. Thus b is odd. □

Theorem 4.2 [offender on Vbeta] One of the following holds

1. [a] There exists 1 ≠ A ≤ Q_{αβ}/Q_{β} such A is an offender on Ṽ_{β}.

2. [b] b = 3 and there exists an non-trivial G_{αβ} invariant subgroup A of Q_{αβ}/Q_{β} such that A is a quadratic 2F -offender on Ṽ_{β}.

Proof: If Y_{β} ∩ Y_{α′} ≠ 1, then ??(c) implies E_{β} = E_{α′}, a contradiction since [Y_{β}, E_{β}] is a p-group, but [V_{β}, E_{α′}] is not. Hence Step 1 [zbza] Y_{β} ∩ Y_{α′} = 1

Hence by 3.1 [V_{β} ∩ Q_{α′}, V_{α′} ∩ Q_{β}] ≤ Y_{β} ∩ Y_{α′} = 1 and we proved Step 2 [vqvq] [V_{β} ∩ Q_{α′}, V_{α′} ∩ Q_{β}] = 1

Suppose now that b = 3. Since Q_{α′} acts transitively on △(α + 2) \ {α′} get G_{α+2α′} = G_{βα+2α′}Q_{α′}. Hence V_{β}Q_{α′} is normal in G_{α+2α′}. Also [V_{α′}, V_{β}, V_{β}] ≤ [V_{β}, V_{β} = 1. Since G_{α+2} is doubly transitive on △(α + 2),

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

Let δ ∈ △(β). Then no subgroup of Q_{β} is an over-offender on Z_{δ}. This together with Step 2 implies

V_{α′} ∩ Q_{β}/C_{V_{α′}∩Q_{α′−1}}(V_{β}) ≤ |V_{β}/C_{V_{β}}(V_{α′} ∩ Q_{β})| ≤ |V_{β}/V_{β} ∩ Q_{α′}| = |V_{β}Q_{α′}/Q_{α′}| By the lasy two displayed equations, V_{β} is a 2F offender on Ṽ_{α′}. So Case 2. of the

lemma holds.

So we may assume from now on Step 3 [b¿3] b > 3.

Suppose that V_{α′} ≤ Q_{β}. Then by Step 2 [V_{α′}, Y_{α} ∩Q_{α′} = 1. By 3.1f [V_{α′}, Z_{α}] ≠ 1 and since Y_{α} is a natural module

for E_{α} and since V_{α′} ≤ E_{α} we get

|V_{α′}/C_{V_{α′}} (Y_{α′}) ≤ q = |Y_{α}/C_{Y_{α}}V_{α′}|

Thus 1. holds in this case. So we may assume that for all critcial pairs:

Step 4 [sym] V_{α′} ≰ Q_{β} and the situation is symmetric in β and α′.

If [V_{β}, V_{α′} ∩ Q_{β}] = 1 = [V_{α′}, V_{β} ∩ Q_{α′}], then again (1) holds. So we may assume Step 5 [vvqa] Y_{β} = [Y_{β}, Y_{α′} ∩ Q_{β}] ≤ V_{α′} or Y_{α′} ≤ [Y_{α′}, Y_{β} ∩ Q_{α′}] ≤ Y_{β}

By symmetry in α, α′ we may assume Step 6 [vvq] Y_{β} = [V_{β}, V_{α′} ∩ Q_{β}] ≤ V_{α′}.

Pick μ ∈ △(β) and t ∈ V_{α′} ∩ Q_{β} with [Z_{μ}, t] ≠ 1. Then μ ≠ α + 2 and by Step 2, Z_{μ} ≰ Q_{α′} and we may assume that μ = α. Hence Step 7 [vbq] There exists t ∈ V_{α′} ∩ Q_{β} with [Z_{α}, t] ≠ 1. In particular, t ∉ Q_{α}

Note that Step 8 [O2G] O^{p}(E_{α}) ≤ ⟨Q_{α−1}, t⟩.

By Step 2 and Step 7 we have |V_{β}Q_{α′}/Q_{α′}| ≥ |Y_{α}Q_{α′}/Q_{α′}| = |Y_{α}/C_{Y_{α}}(t)| ≥ q. We record Step 9 [vbqa] |V_{β}Q_{α′}/Q_{α′}| ≥ q.

We next show:

Step 10 [vavap-1=1] If [V_{α−1}, V_{α′−2}] = 1 then 1. holds.

Suppose [V_{α−1}, V_{α′−2}] = 1. Then V_{α−1} ≤ Q_{α′−2} ∩Q_{α′−1}. Put A = V_{α−1} ∩(V_{β}Q_{α′}). Then A ≤ V_{β}(V_{β}V_{α−1} ∩ Q_{α′}) ≤ V_{β}(Q_{α′−1} ∩ Q_{α′}). Thus by 3.1

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

Let X be maximal in A with [X, t] ≤ Y_{β}. As |Y_{α′}| = q we have |A/X| ≤ q. Since Y_{β}^{∗} ≤ X, t normalizes X. By 3.1, [XZ_{a}, Q_{α−1}] ≤ [V_{α−1}, Q_{α−1}] = Y_{α−1} ≤ XZ_{α}. So by Step 8, O^{p}(E_{α}) normalizes XZ_{α}. Since O^{p}(E_{α}) is transitively on △(α) we conclude that XZ_{α} ≤ D_{α} := ⋂_{δ∈△(α)} V_{δ}. Put a = |V_{α−1}/A|. Then |V_{α−1}D_{a}/D_{a}| ≤ |V_{α−1}/A||A/X| ≤ aq. Hence

|V_{β}D_{a}/D_{α}| ≤ aq.

Note that V_{α−1} ≤ Q_{α′−2} ∩ Q_{α′−1} ≤ G_{α′}. Since D_{α′−1} ≤ V_{α′−2} we conclude from |V_{β}D_{a}/Y D_{α}| ≤ qa and edge-transitivity that

|V_{α′}/C_{V_{α′}} (V_{α−1}V_{β})| ≤ |V_{α′}D_{α′−1}/D_{α′−1}| = |V_{β}D_{a}/D_{α}| ≤ aq.

On the otherhand by definition of a, an isomorphism theorem and Step 9

|V_{α−1}V_{β}Q_{α′}/Q_{α′}| = |V_{α−1}V_{β}Q_{α′}/V_{β}Q_{α′}||V_{β}Q_{α′}/Q_{α′}| ≥ aq.

By the last two equations 1. holds. So we may assume from now on that

Step 11 [va-1va-2] [V_{α−1}, V_{α′−2}] ≠ 1

Suppose that V_{α′−2} ≤ Q_{a−1}. Then by Step 4, V_{α−1} ≤ Q_{α′−2}. Note that by Step 8, C_{Y_{α−1}}(t) = 1. Thus

1 ≠ [V_{α−1}, V_{α′−2}] ≤ Y_{α−1} ∩ Y_{α′−2} ≤ C_{Y_{α−1}}(t) = 1

a contradiction to Step 11. Thus

Step 12 [va-1qa-1] V_{α′−2} ≰ Q_{α−1}

By Step 4 we get

Step 13 [va-1qa-2] V_{α−1} ≰ Q_{α′−2}

Since b > 3, t centralizes [V_{α′−2} ∩ Q_{α−1}, V_{α−1}]. and so

[V_{α′} − 2 ∩ Q_{α−1}, V_{α−1}] = C_{Y_{α−1}}(t) = 1

Thus by Step 5 and ?? that Y_{α′−2} = [V_{α−1} ∩ Q_{α′−2}V_{α′−1}] ≤ V_{α−1}. Hence there exists 1 ≠ x ≤ Y_{α′−2} ∩ V_{α−1}.

Note that t centralizes x and [x, Q_{α−1} ≤ Y_{α−1} ≤ Y_{α}. So by Step 8, O^{p}(E_{α}) normalizes the coset xY_{α}.

Suppose that [x, Q_{α}] ≠ 1. Let R = O^{p}(E_{α}) and D = [Q_{α}, R]. Since C_{Y_{α}}(R) = 1, the Three Subgroup Lemma implies [x, D] ≠ 1. Since R normalizes [x, D] we get [x, D] = Y_{α}. Thus D acts transitively on Y_{α}x and so by the Frattini argument, R = C_{R}(x)D. Since x ∈ Y_{α′−2}, Q! implies C_{R}(x) ≤ C̃_{α′−2}. Also since [E_{α′−2}, Q_{α′−2}] ≤ Q_{α′−2}^{∗} and E_{α′−2} acts transitively on △α′ − 2 we have

t ∈ V_{α′−1}^{(2)} ∩ Q_{α′−2} ≤ (V_{α′−1}^{(2)} ∩ Q_{α′−2})Q_{α′−2}^{∗} = (V_{α′−3}^{(2)} ∩ Q_{α′−2})Q_{α′−2}^{∗} ≤ (Q_{α} ∩ C̃_{α′−2})Q_{α′−2}^{∗}

The right hand side of this equation is p-group normalized by C_{R}(x) and so ⟨t^{C_{R}(x)}⟩Q_{α}/Q_{α} is a p-group. But this contradicts t ∈ Q_{αβ} \ Q_{α} and O^{p}(E_{α}) ≤ C_{R}(x)Q_{α}.

Thus [x, Q_{α}] = 1, and so x ∈ Ω_{1}Z(Q_{α}) = Y_{α}. Since [x, t] = 1 we conclude x ∈ Y_{β}. Since also x ∈ Y_{α′−2} we conclude that E_{α′−2} ≤ C̃_{β}

Since b > 3,

V_{α−1} ≤ V_{α}^{(2)}Q_{β}^{∗} = V_{α+2}^{(2)}Q_{β}^{∗} ≤ (Q_{α′−2} ∩ C̃_{β})Q_{β}^{∗}

The right hand side is a p-group normalized by E_{α′−2} and we obtain a contradiction to Step 13. □

5 The basic structure of E/Q_{E}.

Let G be a group of local characteristic p. We say that G has rank 2, provided that there exists P, P̃ ∈ P(S) such that P̃ ≤ ES and ⟨P, P̃⟩ ∉ L. We say that H has rank at least three if G has neither rank 1 nor rank 2.

The next lemma shows how P̃! can be used to obtain information about E/O_{p}(E).

Proposition 5.1 [unique component in e] Suppose E!, P!, P̃! and that G has rank at least three. Let L = N_{G}(P^{∘}) and H = (L ∩ C̃)E. Comment:H = C̃?

(a) [a] N_{G}(T) ≤ L for all O_{p}(H ∩ L) ≤ T ⊴ S.

(b) [b] There exists a unique P̃ ∈ P_{H}(S) with P̃ ≰ L. Moreover, P̃ ≤ ES.

(c) [c] P̃/O_{p}(P̃) ∼ SL_{2}(q).p^{k}.

(d) [d] H = K(L ∩ H), L ∩ H is a maximal subgroup of H and O_{p}(H ∩ L) ≠ O_{p}(H).

(e) [e] H has a unique p-component K.

(f) [f] Let Z_{0} = C_{Y_{P}} (S ∩ P^{∘}) and V = ⟨Y_{P}^{H}⟩. Then Z_{0} ⊴ V and V ≤ Q ≤ O_{p}(H).

(g) [g] Let D = C_{H}(K/O_{p}(K). Then D is the largest normal subgroup of H contained in L and D/O_{p}(H) is isomorphic to a section of the Borel subgroup of Aut(SL_{2}(q)

(h) [h] Let \overline{V} = V/Z_{0}. Then

(a) [ha] [\overline{V} , O_{p}(H)] = 1

(b) [hb] C_{H}(\overline{V}) ≤ D and C_{H}(\overline{V}) ∩ C_{H}(Z_{0}) = O_{p}(H).

(c) [hc] Let 1 ≠ X ≤ Y_{P} /Z_{0}. Then N_{H}(X) ≤ H ∩ L.

(d) [hd] H ∩ L contains a point-stabilizer for H on \overline{V}.

Proof: By E!, O_{p}(⟨L, H⟩) = 1 and so E ≰ L. Since {P } = {P^{∘}(S) and N_{G}(S) ≤ C̃, N_{G}(S) ≤ N_{G}(P) ≤ L. Since E = ⟨P_{ES}, N_{E}(S) > there exists P̃ ∈ P_{ES}(S) with P̃ ≰ L. Since rank G is at least three, ⟨P, P̃⟩ ∈ L and so by P̃!, P̃ is uniquely determined.

Let T be as in (a). It follows easily from P! that QO_{p}(L) is a Sylow p subgroup of P^{∘}O_{p}(L). Since QO_{p}(L) ≤ O_{p}(H ∩ L) ≤ T we conclude that T is a Sylow p-subgroup of T P^{∘}. Suppose that M := ⟨P^{∘}T, N_{G}(T)⟩ ∉ L. Then by Pushing up SL2(q) 2.1 and Q! P ∼ q^{2}SL_{2}(q). Since ⟨P, P̃⟩ ∈ L we get P̃ ≤ N_{G}(Y_{P}) and 2.2(dd) gives the contradiction P̃ ≤ N_{G}(P^{∘}) = L. Thus M ∈ L. If P̃ ∈ M, then T ⊴ P̃ and so O_{p}(P^{∘}) ≤ O_{p}(P̃), a contradiction to (P̃ − 2b). Hence P̃ ≰ M. By the uniqueness of P̃ and since S ≤ M, M ≤ L. Thus (a) holds.

Let P^{∗} ∈ P_{H}(S) with P^{∗} ≰ L. Suppose that P̃ ˜P̸. Then P^{∗} ≰ ES and so S ∩ E ≤ O_{p}(P^{∗}) Since EO_{p}(H ∩ L) is normalized by E(L ∩ H) = H we get O_{p}(H ∩ L) ≤ O_{p}(P^{∗}). Thus by (a) P^{∗} ≤ L, a contradiction. So (b) holds. (a) is obvious.

Let H ∩ L < M ≤ H. Then M ≰ L and so P̃ ≤ M. Let R ∈ P(H)(S). If R = P̃, then R ≤ M, if R ≠ P̃, then R ≤ L and again R ≤ M. Since also N_{H}(S) ≤ H ∩ L ≤ M we get M = H. By (a) O_{p}(H ∩ L) ≠ O_{p}(H).

Let N be a normal subgroup of H minimal with respect to N ≰ L. By the uniqeness of P̃, P̃ ≤ NS. Hence O^{p}(P̃) ≤ N and since O^{p}(P̃) ≰ L, N = ⟨O^{p}(P̃)^{H}⟩ ≤ E. Next let F be the largest normal subgroup of H contained in L. Then [O_{p}(H ∩ L), F] ≤ O_{p}(H ∩ L) ∩ F ≤ O_{p}(F) ≤ O_{p}(H). Note that [O_{p}(H), N] is normal in N(H ∩ L) = N and so [O_{p}(H), N] = N and we conclude that [N, F] ≤ O_{p}(H). In particular F ∩ N/O_{p}(N) ≤ Z(N/O_{p}(N).

Suppose that N is solvable. Then the mimimality of N implies that N/O_{p}(N) is a r-group for some prime r ≠ p. In particular H ∩ N < N_{N}(H ∩ N) and the maximality of H ∩ N implies H ∩ N ⊴ H. Thus H ∩ N ≤ F. Suppose that S does not act irreducible on N/H ∩ N. The by coprime action there exists an S-invarinat R ≤ N with R ≰ L and O^{p}(P̃) ≰ R. Then by (b) RS ≤ L, a contradiction. So S acts irreducible on N/H ∩N. Thus N = (H ∩ N)O^{p}(P̃) and N = [N, O_{p}(H)] ≤ O^{p}(P̃). Note that ⟨P^{∘}, N⟩ is normalizes by P^{∘}, H ∩ L and N and so by ⟨L, H⟩, it follows that O_{p}⟨P^{∘}, N⟩ = 1 and so also O_{p}(⟨P, P̃⟩ = 1, a contradiction.

Thus N is not solvable and so the product of p-components.

Let K_{1} be a p-component of N. By minimality of N, N = ⟨K_{1}^{H∩L}⟩ If S does not act tranisitively on the p-components of N, we can choose K_{1} such that O^{p}(P̃) ≰ K_{1}^{S}⟩. But then K_{1} ≤ L, a contradiction. Thus N = ⟨K_{1}^{S}⟩. Suppose that K_{1} ∩ P̃ lies in the unique maximal subgroup of P̃ containing S. Since K_{1} ∩ P̃ is subnormal in P̃ the structure of P̃ implies [K_{1}∩P̃, S] is a p-group. Thus K_{1} ≠ N and K_{1}∩P̃ is a p-group. Hence N ∩S ⊴O_{p}(P̃). Since N_{N}(N ∩ S)/N ∩ S is a p-group we conclude from coprimes action that N ∩ L projects onto N_{K_{1}}(N ∩S)F/F. Thus [K_{1} ∩L)F, N_{K_{1}}(N ∩S)] ≤ [K_{1} ∩L)F, N ∩L] ≤ L. Conjugation with S yields [N ∩L, N_{N}(N ∩S) ≤ N ∩L. Thus H = ⟨P̃, L⟩ ≤ N_{H}(N ∩L) and so N ∩L = F ∩L. Since L contains a Sylow p-subgroup of N, we conclude that NF/F is a p′-group. Let r be a prime divividing the order of NF/F and R/F an S-invariant Sylow p-subgroup of NF/F. Then RS is not contained in L and so P̃ ≤ RS. Thus P̃ = is a {r, p} group and r is unique. Thus NF/F is a r-group, a contradiction.

We proved that K_{1}∩P̃ is not contained in the unique maximal subgroup of P̃ containing S. Since [K_{1} ∩ P̃, (K_{1} ∩ P̃)^{g} is a p-group for all g ∈ P̃ \ N_{H}(K_{1}) the structure of P̃ implies O^{p}(P̃) ≰ K_{1}. Thus S ≤ N_{H}(K_{1}) and so N = K_{1}. Thus (e) is proved.

By P! uniqueness Z_{0} is normal in C̃ and [Z_{0}, Q] = 1. Since Y_{P} is a natural module for P^{∘} we get [Y_{P} , Q] ≤ [V, O_{p}(H) ≤ [V, O_{P} (L ∩ H)] ≤ Z_{0}. So Y_{P} acts trivial an all factors of 1 ≤ Z_{0} ≤ Q and since ℂ is of characteristic p, Y_{P} ≤ Q. This proves (f) and (ha).

To prove (g) note that N ≰ D and so by uniqueness of N, D ≤ L and so D ≤ F. But as seen above F ≤ D and so D = F. Let D_{0} be maximal in D with [P^{∘}, D_{0}] ≤ O_{p}(P^{∘})]. Then H and P^{∘} normalize O^{p}(D_{0}) and so O^{p}(D_{0}) = 1. Thus D_{0} is a p-group and (g) holds.

Note that C_{H}(\overline{V}) ≤ N_{H}(Y_{P})∩H ∩L and So C_{H}(\overline{V}) ≤ D. Let R = O^{p}(C_{H}(\overline{V})∩C_{H}(Z_{0})). Then R) centralizes Y_{P} and [R, P^{∘}] ≤ C_{P^{∘}}(Y_{P}) = O_{p}(P^{∘}) ≤ O_{p}(H ∩ L). But R is normal in H ∩ L and so R = O^{p}(R) = O^{p}(RO_{p}(H ∩ L). Thus H and P^{∘} both normalizes R and so R = 1. Hence (hb) holds.

Let e ∈ Y_{P} \Z_{0} with eZ_{0} ∈ X. Let g ∈ N_{H}(X). Since H ∩L acts transitively on Y_{P} \Z_{0}, there exists h ∈ H ∩ L with e^{gh} = e. Let t ∈ P^{∘} with e ∈ Z_{0}^{t}. Then [e, Q^{t}] = 1 and so gh ≤ C̃^{t}. Thus gh ∈ N_{G}(⟨Q, Q^{t}⟩ = N_{G}(P^{∘}) = L. Hence g ∈ L and (hc) holds.

(hd) follows from (hc). □

6 The Small World Theorem

Given Q! and P ∈ P^{∘}(S). We say that b = 2 for P if b > 1 for P and ⟨Y_{P}^{E⟩} is not abelian. If neither b = 1 nor b = 2 for P we say that b is at least three for P.

Theorem 6.1 (The Small World Theorem) [the small world theorem] Suppose E! and let P ∈ P^{∘}(S). Then one of the following holds:

1. [a] G has rank 1 or 2.

2. [b] b = 1 or b = 2 for P.

3. [c] A rank three situation described below.

Proof: Assume that G has rank at least three and that b is at least three. In the exceptional case of the P!-theorems ( ?? and ?? one easily sees that b = 2 for P. Thus P! holds. Also in the exceptionell case of the P̃! Theorem ?? one gets b = 2 for P. Thus (strong) P̃! holds. We proved

Step 1 [ P! and wP!] P! and P̃! hold.

5.1 gives us a good amount of information about E. We use the notation introduced in 5.1.

Since ⟨H, L⟩ ∉ L, we can apply the amalgam method to the pair (H, L). A non-trivial argument shows

Step 2 [offender on V] One of the following holds:

1. [a] O_{p}(H ∩ L)/O_{p}(H) contains a non-trivial quadratic offender on Ṽ.

2. [b] There exists a non-trivial normal subgroup A of H ∩ L/O_{p}(H ∩ L) and normal subgroups Y_{P} ≤ Z_{1} ≤ Z_{2} ≤ Z_{3} ≤ V of H ∩ L such that:

(a) [a] A and V/Z_{3} are isomorphic as 𝔽_{p}C_{H∩L}(Y_{P})-modules.

(b) [b] |Z_{3}/Z_{2}| ≤ |A|.

(c) [c] [\overline{V} , A] ≤ \overline{Z_{2}} ≤ C\overline{V} (A). In particular, A is a quadratic 2F -offender.

(d) [d] [\overline{x}, A] = \overline{Y_{P}} for all x ∈ Z_{3} \ Z_{2}.

(e) [e] \overline{Z_{1}} is a natural SL_{2}(q)-module for P̃ ∩ C_{H}(Z_{0}).

Using ?? and ?? (and the Z^{∗}-theorem to deal with the case |A| = 2) it is not too difficult to derive

Step 3 [e-structure] K/O_{p}(K) ≅ SL_{n}(q), (n ≥ 3), Sp_{2n}(q)′,(n ≥ 2) or G_{2}(q)′,( p = 2). Moreover, if \overline{W} is a maximal submodule of \overline{V}, then V/W is the natural module for K/O_{p}(K) and H ∩ L contains a point-stabilizer on V/W.

Let M ∈ M(⟨P, P̃⟩) with M^{∘} maximal.

Suppose first that M^{∘} ≰ ⟨P, P̃⟩. Then by the Structure Theorem ?? M^{∘}/O_{p}(M^{∘}) ≅ SL_{n}(q), n ≥ 4 or Sp_{2n}(q), n ≥ 3. Moreover,

R := O^{p}(M^{∘} ∩ C̃ ≤ O^{p}(P̃)^{M∩C̃} ≤ K

In the case of Sp_{2n}(q) we have R/O_{p}(R) ≅ Sp_{2n−2}(q)′. So Step 3 implies K/O_{p}(K) ≅ Sp_{2m}(q) and R ≤ H ∩ L, a contradiction.

Thus M^{∘}/O_{p}(M^{∘}) ≅ SL_{n}(q) and R/O_{p}(R) ≅ SL_{n−1}(q). Let R^{∗} be a parabolic subgroup of KS minimal with RS < R^{∗}. Then R^{∗} ∩ K/O_{p}(R^{∗} ∩ K) ≅ SL_{n}(q) or Sp_{2n−2}(q). Let M^{∗} = ⟨M^{∘}S, R^{∗}⟨. Since R^{∗} ≤ ⟨O^{p}(P̃)^{R^{∗}}⟩S ≤ (M^{∗})^{∘}S the maximality of M^{∘} implies that M^{∗} ∉ L. Now M^{∗} has a geometry of type A_{n} or C_{2n} with all its residues classical we conclude

that M^{∗} has a normal subgroup SL_{n+1}(q) or Sp_{2n}(q). But this contradicts the assumption that V is abelian.

(Remark: One does not have to identify M^{∗} first to obtain this contradiction. Indeed an easy geometric argument shows that M^{∗} has rank at most three on M^{∗}/M^{∗} ∩ C̃. But then V abelian gives the contradiction ⟨Z_{0}^{M^{∗}}⟩ abelian . )

We conclude that M^{∘} ≤ ⟨P, P̃⟩. Let P^{∗} be the unique elemenent of PKS(S) with P^{∗} ≰ N_{G}(O^{p}(P̃)). By maximality of M^{∘} we obtain ⟨P, P̃, P^{∗}⟩ ∉ L.

This is rank three situation eluded to in (c). □

Lemma 6.2 [quadratic normal point stabilizer theorem] Let H be a finite group and V a faithful irreducible 𝔽_{p}H-module. Let P be a point stabilizer for H on V and A ≤ P. Suppose that

(i) [i] F^{∗}(H) is quasi simple and H = ⟨A^{H}⟩

(ii) [ii] A ⊴ P and |A| > 2.

(iii) [iii] A acts quadratically on V.

Then one of the following holds:

1. [a] H ≅ SL_{n}(q), Sp_{2n}(q),or G_{2}(q) and V is the natural module. Moreover,

2. [b] p = 2, H is a group of Lie Typ in char p, and H is contained in a long root subgroup of H.

3. [c] Who knows

References