On the generic groups of p-type

U. Meierfrankenfeld B. Stellmacher Draft of May 23, 2002

Contents

1 Random Observations 2 2 Preliminaries 5 3 CS generated modules 13 4 Groups with m_{2′}(G) ≤ 3 15 5 Subnormal Subgroups 23 6 Nice Modules 23 7 An interesting choice of an amalgam for generic p-type groups 47 8 Some general amalgam results 48 9 Amalgams involving uniqueness groups 61 10 Connected parabolics not normalizing Z 70 11 The case b = 1 with G_{α} connected and G_{β} minimal 75 12 Elementary results on p-connected groups 77 13 Establishing Geometries 79 14 Large Alternating Groups 84 15 Tits Chamber Systems 87

dgpty

Definition 0.1 Let G be a finite group, p a prime dividing. the order of G and S ∈ Syl_{p}(G). Then G is of generic p-type provided that

(a) If L is a p-local subgroup of G with S ≤ L, then F^{∗}(L) = O_{p}(L).

(b) G is generated by the p-locals containing S.

(c) all p-locals of G are K-groups.

dqt

Definition 0.2 1. A quasisimple group K is called a C_{2} - group if and only if K is a quasisimple group of Lie type in characteristic 2 or K = P SL(2, q) for q a Fermat or Mersenne prime or q = 9 or K = P SL(3, 3), P SL(4, 3), P SU(4, 3), 2U(4, 3) or G_{2}(3) or K/Z(K) = M_{11}, M_{12}, M_{22}, M_{23}, M_{24}, J_{2}, J_{3}, J_{4}, HS, Suz, Ru, Co_{1}, Co_{2}, F i_{22}, F i_{23},

F i′_{24}, F_{3}, F_{2}, or F_{1} except 2A_{8}, Sp(4, 3) and [X]L_{3}(4) for exp X = 4 are not C_{2} groups.

2. L_{2}(G) = {K : for some involution x of G, K is a component of C_{G}(x)/O_{2′}(C_{G}(x))}

3. G is of even type if and only if the following conditions hold:

(a) Every element of L_{2}(G) is a C_{2} - group

(b) O_{2′}(C_{G}(x)) = 1 for every involution x of G; and

(c) m_{2}(G) ≥ 3.

4. Let G be of even type and let S be a Sylow 2−subgroup of G. Then σ(G) = {p : p is an odd prime and m_{p}(M) ≥ 4 for some maximal 2-local M of G with

|S : S ∩ M| ≤ 2}.

5. G is of quasithin type if G is a simple group of even type with σ(G) empty.

dhead

Definition 0.3 Head(P) =^{def} O^{p}(P)O_{p}(P)/O_{p}(P).

1 Random Observations

Let G be a finite group, S the Sylow 2-subgroup of G and B the intersection of the maximal 2-locals containing M.

Borel

Lemma 1.1 Let G be a finite group such that F^{∗}(G) is the direct product of simple groups of simple groups of Lie type in characteristic 2. Suppose that all the 2-locals of G containing S are of characteristic 2-type. Then S acts transitive on the set of components of G, B = N_{G}(S ∩ F^{∗}(G)) and BF^{∗}(G) = G.

Remark: False for D_{4}(q).3 and D_{4}(q).Sym(3)

Proof: Let E_{1}, …, E_{n} be the components of G, E = F^{∗}(G) = E_{1}E_{2} … E_{n} and T = E ∩ S. Suppose that S does not act transitively on the set of components of G. Then ⟨E_{1}, S⟩ is contained in a 2-local which is not of 2-type, a contradiction.

Let M be any maximal 2-local of G containing S. As M is of 2-type and C_{E}(O_{2}(M)) ≠ 1 we conclude O_{2}(M) ∩ E ≠ 1.

Let Q_{i} be the projection of O_{2}(M) ∩ E onto E_{i} and Q = Q_{1} · Q_{2} · . . . · Q_{n}. Then Q is a 2 group normalized by M and so O_{2}(M) ≤ Q ≤ O_{2}(M), Q = O_{2}(M) and M = N_{G}(Q).

Suppose now that n = 1.

Let M_{i} = N_{M}(E_{i}) and M_{i}^{∗} a maximal 2-local subgroup of E_{i} containing M ∩ E_{i}. Then ⟨M_{i}^{∗M}⟩ ∩ E_{i} = M_{i} and so ⟨M_{i}^{∗}, M⟩ is contained in a 2-local of G. Thus M_{i}^{∗} = M ∩ E_{i}.

TO BE CONTINUED

Remark 1.2 It seems that in groups of charateristic 2-type, B-irreducible subgroups actually have B as a maximal subgroup. For example if G has a parabolic P with P/O_{2}(P) ≅ Sym(5) then the the inverse image of the Sym(4) seems always to be in the Borel group.

2closed

Lemma 1.3 For L ∈ L(= L(S) put Z_{L} = ⟨Ω_{1}(Z(S))^{L}, C_{L} = C_{L}(Z_{L}) and L^{∗} = N_{L}(S ∩ C_{L}). Let R ⊆ L put R = ⟨L^{∗} | L ∈ R⟩.

(a) L = L^{∗}C_{L} for all L ∈ L.

(b) Let L ∈ L and P ∈ calN(L, S). Then P ≤ L^{∗} or O^{2}(P) ≤ C_{L}.

(c) Let L ∈ calL. Then O_{2}(L^{∗}) = S ∩ C_{L}

(d) If R ∈ L, then C_{R} is 2-closed and R = R^{∗}.

(e) Let R = L

e.1. Suppose R ∈ L. Then for all L ∈ L, L = (R ∩ L)(C ∩ L).

e.2. Suppose that R ∉ L. Then there exists R_{i} ⊆ L, i = 1, 2 so that R_{i} ∈ L but O_{2}(⟨R_{1}, R_{2}⟩) = 1.

Proof: (a) follows by the Frattini argument.

To prove (b) let L ∈ R. Then L^{∗} ≤ R, Z_{L} ≤ Z_{R}, and S ∩ C_{R} ≤ C_{L}. Thus S ∩ C_{R} = (S ∩ C_{L}) ∩ C_{R} and S ∩ C_{R} is normalized by L^{∗}. As R is generated by the L^{∗}’s, L ∈ R, S ∩ C_{R} is normal in R and so also in C_{R}. Thus C_{R} is 2-closed and R = R^{∗}.

(c) and (d) are obviuos.

(e.1) follows since from (a) as L^{∗} ≤ R ∩ L and C_{L} ≤ L ∩ C.

For (e.2) let for R_{1} be maximal in L with R_{1} ∈ L and let R_{2} = {L} for some L ∈ L\R_{1}.

∗gomi

Lemma 1.4 Let R = R_{L} and suppose that R ∈ L.

(a) N_{G}(Z_{L}) is the unique maximal 2-local of G containing R.

(b) Let L ∈ N (R, S) with O^{2}(L) ⊴ ⊴R and P ∈ N (S) with P ≰ R. If ⟨P, L⟩ ∈ L, then O^{2}(P) ∩ S ≤ O_{2}(L).

(c) Let P ∈ calN(S) so that P does not normalize Z_{R}. Then there exists L ∈ N (R, S) with O^{2}(L) ⊴ ⊴R and ⟨L, P ⟩ ∉ calL.

Proof: (a) Let R ≤ M ∈ L. Then M^{∗} ≤ R and so R = M^{∗} and Z_{M} = Z_{M^{∗}} = Z_{R}. Thus M ≤ N_{G}(Z_{R}).

(b) Let M = ⟨P, L⟩. As P ≰ R, O^{2}(P) ≤ C_{M}. By 3.6 [Z, L] ≠ 1 and so S ∩ O^{2}(P) ≤ S ∩ C_{M} ≤ O_{2}(L).

(c) As O_{2}(R) is the intersection of the O_{2}(L)’s, L as in the statement of (c) we conclude that O^{2}(P)∩S ≤ O_{2}(R). Hence O_{2}(R) is a Sylow 2-subgroup of O^{2}(P)O_{2}(R). By (a) ⟨P, R⟩ is not a 2-local and we conclude that ⟨Ω_{1}(O_{2}(R)_{2}^{O}(P)⟩ is an FF-module for O^{2}(P)O_{2}(R). But this contradicts [Z, P] = 1.

L+

Lemma 1.5 Let N^{+}(S) = {L ∈ N (S) | [Z, L] ≠ 1} and for L ∈ L put L^{+} = ⟨N^{+}(L, S)⟩. Then

(a) O_{2}(L^{+}) = S ∩ C_{L} = O_{2}(L^{∗})

(b) L = L^{+}(L ∩ C).

(c) Z_{L} = Z_{L^{+}}.

Proof: Put T = S ∩ C_{L} and R = N_{L}(T). Then by 3.6 F_{2}^{∗}(R) ≤ R^{+} an so O_{2}(L^{+}) = O_{2}(F_{2}^{∗}(R). As O_{2}(L/C_{L}) = 1, O_{2}(F_{2}^{∗}) = T. So (a) holds.

For (b) suppose first that C_{L} ≠ O_{2}(L). By the Frattini argument, L = RC_{L} and by induction R = R^{+}(R ∩ C). Hence L = R^{+}C_{L}(R ∩ C) = L^{+}(L ∩ C).

So suppose that C_{L} = O_{2}(L).Then R = L. Let E = S ∩ F_{2}^{∗}(L) and H = N_{L}(T). By the Frattini argument, L = F_{2}^{∗}(L)H and by induction, H = H^{+}(H ∩ C). Hence L = F_{2}^{∗}(L)H^{+}(H ∩ C) = L^{+}(L ∩ C)..

(c) follows directly from (b)

Rgomi

Lemma 1.6 Let N^{+}(S) = {L ∈ N (S) | [Z, L] ≠ 1} and D = ⋂{O_{2}(L^{∗}) | L ∈ L}.

(a) Let P ∈ N^{+}(S) with P ≰ N_{G}(D). Then there exists L ∈ N^{+}(S) so that ⟨P, L⟩ ∉ L.

(b) Let R^{+} = N^{+}(S)⟩ and suppose that R^{+} ∈ L.

(b.a) For all L ∈ L, L = (L ∩ R^{+})(L ∩ C).

(b.b) Suppose that R^{+} ≤ L ∈ L. Then R^{+} = L^{+}, Z_{L} = Z +R^{+} and O_{2}(R^{+}) = C_{L} ∩S.

(b.c) O_{2}(R) = D.

(b.d) N_{G}(Z_{R^{+}}) is the unique maximal 2-local of G containing R^{+}.

Proof: Suppose (a) is false. Let L ∈ calN^{+}(S) and put M = ⟨L, S⟩. By assumption M ∈ calL and so by 1.3b, M = M^{∗}. Let Y ∈ N (M) with O^{2}(Y) ⊴ ⊴M. Then by 3.6, [Z, Y] ≠ 1 and so Y ∈ N^{+}(S). Hence the Gomi argument implies that P normalizes D.

(b.a) follows directly from 1.5b

Since

N^{+}(S) ⊆ N^{+}(R^{+}, S) ⊆ N^{+}(L, S)N^{+}(S),

R^{+} = L^{+}. Thus by 1.5a, O_{2}(R) = C_{L} ∩ S. Furthermore, by 1.5c, Z_{L} = Z_{L^{+}} = Z_{R^{+}} (b.c) follows from 1.5a. (b.d) follows directly from (b.b)

dpred

Definition 1.7 Let L ∈ L(S). Then a p-reduced normal subgroup of L is a elementary abelian normal p-subgroup Y of L so that O_{p}(L/C_{L}(Y)) = 1,( i.e all normal subgroups of L which act unipotently on Y already centralize Y.

bpred

Lemma 1.8 Let L ≤ L(S).

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

(b) Let R ∈ (L, S) 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) Let S_{L} = C_{S}(Y_{L}) and L^{f} = N_{G}(S_{L}). Then S_{L} = O_{p}(L^{f}) and Y_{L} = Ω_{1}Z(S_{L}).

Proof: (a) Let Y_{L} be the subgroup generated by the p-reduced normal subgroups of L. Let N be a normal subgroup acting unipotently on Y_{L}. Then N also acts unipotently on all the generators of Y_{L}. Hence N centralizes all the generators of Y_{L} and so Y_{L}. Thus Y_{L} is p-reduced.

(c) Let Y = ⟨X^{L}⟩ and C = C_{L}(Y). Let N/C = O_{p}(L/C). Then N = (N ∩ S)C and in particular, N = (N ∩ L)C. As X is p reduced, N ∩ L centralizes X. The same is true for C and so also for N. Since N is normal in L and Y = ⟨X^{L}⟩, N centralizes Y. Thus N = C and Y is p-reduced.

(b) Put C = C_{L}(Y_{L}). By Frattini, L = L^{f}C. Since O_{p}(L/C) = 1 we conclude O_{p}(L_{f}) ≤ C Hence O_{p}(L_{f}) ≤ C ∩ S = S_{L} and so O_{p}(L_{f}) = S_{L}). Let X = Ω_{1}(Z(S_{L})). Then clearly Y_{L} ≤ X and L_{f} normalizes Y. Put Y = ⟨Y^{L}⟩ = ⟨Y^{C}⟩. Clearly X is p-reduced for S_{L} and so by (b) applied to C, Y is p-reduced for C. Let N be a normal subgroup of L acting unipotently on Y. Since Y_{L} ≤ Y and Y_{L} is p-reduced for L, N ≤ C. As Y is p-reduced for C, N centralizes C and so Y is p-reduced for L. By maximality of Y_{L} we get Y ≤ Y_{L}. But Y_{L} ≤ X ≤ Y and so Y_{L} = X = Y.

2 Preliminaries

e

Lemma 2.1 Let r and s be positive real numbers and put e = \frac{rs^{2}−r−s}{s^{2}}.

(a) Suppose that s > 1. Then e > O if and only if r > \frac{s}{s^{2}−1}. In particular e > 0 if r ≥ 2 and s ≥ 1.3.

(b) e ≤ 1 if and only if (r − 1)(s − 1) ≤ 1.

Proof: (a) is easily computed and for (b) note that the following are equivalent:

e ≤ 1 , rs^{2} − r − s − s^{2} ≤ 0, (rs − r − s)(s + 1) ≤ 0, rs − r − s ≤ 0, (rs − r − s) + 1 ≤ 1 and (r − 1)(s − 1) ≤ 1.

G = Qt

Lemma 2.2 Let P ∈ P(S) be of weak L_{2}(2)^{k} type. Put Δ = {L_{i} | 1 ≤ i ≤ k} and let Q⊴S such that

(i) |Z_{P} /C_{Z_{P}} (A)| < |A/C_{A}(Z_{P})|^{2} for some A ≤ Q with [Z_{P} , A] ≠ 1.

(ii) Q contains an involution t acting fixed point freely on Delta.

Then O^{2}(P) ≤ ⟨C_{Q}(Δ)^{e}, t⟩ for some e ∈ P.

Proof: Let Δ{L_{i} | 1 ≤ i ≤ k}. Choose A as in (i) with |A| minimal. Then it easy to see that A acts trivially on Δ. Next let T be maximal in C_{Q}(Δ) so that T fullfills |Z_{P} /C_{Z_{P} (T)}| < |T/C_{T} (Z_{P})|^{2}. By [CD] T is unique and so T ⊴ S. Let E = O^{2}(P)C_{P} /C_{P}. Then S acts irreducible on E and E = E_{1} × … × E_{k} with |[Z_{p}, E_{i}]| = 4. We claim that each of the E_{i} is a Wedderburn component for T on E. Indeed, let E^{∗} be a Wedderburn component for T on E and suppose that E^{∗} = E_{1} … E_{t}. Then k = lt for some integer l, C_{T} (E^{∗}) = C_{T} (E_{1}), |T/C_{T} (E^{∗})| = 2 and |T/C_{T} (Z_{P})| = |T/C_{T} (E)| ≤ 2^{l}. On the otherhand Z_{P} /C_{Z_{P}} (T) = 2^{k}. Thus k < 2l and as l divides k, l = k.

We conclude that:

(1) Each T invariant subspace in E is a sum of some of the E_{i}’s.

As t acts fixed point freely on Δ, t inverts an element e ∈ O^{2}(P) with projects nontivially on each of the E_{i}’s. Thus (1) implies

(2) E = ⟨\overline{e}^{T} ⟩.

Let L = ⟨T^{e}, t⟩. Then T^{e^{−1}} = (T^{e})^{t} ≤ L and so also [T, e] ∈ L. Since C_{E}(T) = 1, \overline{e} ∈ [T, \overline{e}] and (2) implies that E ≤ \overline{L}. Hence P = LS and O^{2}(P) ≤ ⟨T^{P} ⟩ = ⟨T^{L}⟩ ≤ L. As T ≤ C_{Q}(Δ) the lemma is proved.

s

Lemma 2.3 Let H be a group, V, B and Z_{i} ∈ I subgroups of H and s a positve real number. Supppose that

(i) V = ⟨Z_{i} | i ∈ I⟩ and for all i ∈ I, Z_{i} ⊴ V.

(ii) For all i in I and D ≤ B, B normalizes Z_{i} and |D/C_{D}(Z_{i})|^{s} ≤ |Z_{i}/C_{Z_{i}}|.

Then |B/C_{B}(V)|^{s} ≤ |V/C_{V} (B)|.

Proof: Without loss I − {1, …, n}. Let B_{1} = B and B_{i+1} = C_{B_{i}}(Z_{i}). Then B_{n+1} = C_{B}(V). Moreover, by (ii) applied to D = B_{i},

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

Thus

|B/C_{B}(V)|^{s} ≤ ∏_{i=1}^{n} |Z_{i}/C_{Z_{i}}(B_{i})|  (2).

As by defintion B_{i+1} centralizes Z_{i} we get

|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})|  (3).

Thus

∏_{i=1}^{n} |Z_{i}/C_{Z_{i}}(B_{i})| ≤ |C_{V} (B_{i+1})/C_{V} (B_{i})| = |V/C_{V} (B)|.  (4).

The lemma now follows from (2) and (4).

qrc

Lemma 2.4 Let V = ⟨W_{i} | i ∈ I⟩, where W_{i} is a normal subgroup of V for all i ∈ I. Let B be a subgroup of A normalizing all the W_{i}’s. If A ≠ B define r by |A/B|^{r} = |V/C_{V} (A)| and t by |V/C_{V} (A)|^{t} = |A/C_{A}(V)|. Let I = {1, 2, … n} and define A_{0} = B and inductively A_{i} = C_{A_{i−1}}(V_{i}). Choose notation so that B = A_{0} > A_{1} > … > A_{k} = C_{A}(V). Define s_{i} by |A_{i−1}/A_{i}|^{s_{i}} = |W_{i}/C_{W_{i}}(A_{i−1})| and s = min_{i=1}^{k} s_{i}. Then

(a) |B/C_{B}(V)|^{s} ≤ |V/C_{V} (B)|.

(b) If A ≠ B, then trs ≤ r + s.

(c) Suppose that A ≠ B and equality holds in (b). Then

(c.a) s_{i} = s for all 1 ≤ i ≤ k. (c.b) C_{V} (B) = C_{V} (A). (c.c) |B/V_{B}(V)|^{s} = |V/C_{V} (B)|.

Proof: (a) follows from 2.3. Note that |A/B|^{rt} = |V/C_{V} (A)|^{t} = |A/C_{A}(V)| = |A/B||B/C_{B}(V)| and therefore

|B/C_{B}(V)| = |A/B|^{rt−1}. Suppose that A ≠ B. By (a) we conclude

|A/B|^{r} = |V/C_{V} (A)| ≤ |V/C_{V} (B)| ≤ |B/C_{B}(V)|^{s} = |A/B|^{(rt−1)s}

and so (rt − 1)s ≤ r and rts ≤ r + s.

(c) follows by investigating the places where ” < ” was used.

msn

Lemma 2.5 Let H be a finite group, P a p-subgroup of H and suppose that P is subnormal in all proper subgroups of H containing P, but is not subnormal in H. Then A is contained in a unique maximal subgoup of H.

Proof: Suppose that A is contained in two distinct maximal subgroups M_{1} and M_{2}. Choose the M_{i}’s so that M_{1} contains a Sylow p-subgroup of H and so that |M_{1} ∩ M_{2}|_{p} is maximal. Let D be a Sylow p-subgroup of M_{1} ∩ M_{2} and put B_{i} = ⟨A^{h} | h ∈ H, A^{h} ≤ M_{i}⟩. Then by asumption B_{i} ≤ O_{p}(M_{i}) ≤ M_{j}.

Suppose that D is not a Sylow p-subgroup of M_{2}. Then M_{M_{2}}(D) ≰ M_{1} and |N_{M_{2}}(D) ∩ M_{1}|_{2} > |D|, a contradiction. Thus D is a Sylow p-subgroup of M_{2} and so B_{2} ≤ D and N_{G}(D) normalizes B_{2}. Thus N_{G}(D) ≤ M_{2} and so D is also a Sylow p-subgroup of M_{1}. Hence B_{1} ≤ D and B_{1} = B_{2}, a contradiction.

almp

Lemma 2.6 Let H be a finite group, p a prime, S a Sylow p-subgroup of H and suppose that S lies in a unique maximal subgroup M of H. Let P ≤ S and suppose that P ≰ O_{p}(H). Then there exist a subgroup L of H and h ∈ H so that

(a) P ≤ L and P ≰ O_{p}(L)

(b) M^{h} ∩ L is the unique maximal subgroup of L containing P.

(c) S^{h} ∩ L is a Sylow p-subgroup of L.

Proof: If M is the unique maximal subgroup of H containg P, then the lemma holds with L = H and h = 1. Hence there exists a proper subgroup K of H such that P ≤ K and K ≰ M. Choose K so that |M ∩ S|_{p} is maximal and then with K minimal. Let T = M ∩ K and R = ⟨P^{G} ∩ T ⟩. Let S^{∗} ∈ Syl_{p}(M) with T ≤ S^{∗}. Then M is the unique maximal subgroup of H containing S^{∗} and so T ≠ S^{∗}. Thus T < N_{S^{∗}}(T) ≤ N_{H}(R) and |M ∩ K|_{p} < |M ∩ N_{H}(R)|_{p}. Thus by the choice of K, N_{H}(R) ≤ M. In particular, N_{K}(R) ≤ K ∩ M and so T is a Sylow p-subgroup of K. Hence O_{p}(L) ≤ T ≤ M. If R ≤ O_{p}(K), then R ⊴ K, contradiction. P^{∗} ∈ P^{H} ∩ T with P^{∗} ≰ O_{p}(K). By the minimal choice of |K|, M ∩ K is the unique maximal subgroup of K containing T and so we can apply induction. Thus there exists L^{∗} ≤ K with P^{∗} ≤ L^{∗}, P^{∗} ≰ O_{p}(L^{∗}) and h^{∗} ∈ K so that (M ∩ K)^{h^{∗}} ∩ L^{∗} is the unique maximal subgroup of L^{∗} containing P^{∗}. Let x ∈ H with P^{∗x} = P and put h = h^{∗}x and L = L^{∗}x. The clearly (a) and (b) hold.

anc

Lemma 2.7 Remark: Quadratic groous normalize components

mqm

Lemma 2.8 Let A ≤ H and V a faithful GF (p)H-module. Suppose that

(i) A is contained in a unique maximal subgroup of H.

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

(iii) A ≰ O_{p}(H)

(iv) One of the following holds:

1. V = ⟨Z^{H}⟩ for some Z ≤ V with [Z, A] = 1.

2. V = C_{V} (A)[V, H].

Let t ∈ A \ O_{p}(H). Then each of the following holds:

(a) Then C_{V} (t) = C_{V} (A).

(b) |V/C_{V} (A)| ≥ |A/A ∩ O_{p}(H)|^{c}, where c is the number of non-trivial chief-factors for H on V.

(c) [V, t] ∩ C_{V} (H) = 1 and |[V, t]^{2}| = |V/C_{V} (H).

(d) Suppose that (iv)1 holds and O_{p}(L) normalizes Z. Then one of the follwing holds:

1. [V, A ∩ O_{p}(H)] ≤ C_{V} (H).

NI2 p = 2, H/O_{p}(H) ≅ Dih(2r^{k}), r an odd prime C_{H}([V, A∩O_{p}(H)]) ≰ O_{p}(H).

(e) [V, H] ∩ C_{V} (H) ≤ [V, A]

(f) W = C_{W} (H)[W, H] for each H-section on V. In particular, H has no central chieffactor on V/C_{V} (H).

Proof: Note first that (iv)1. implies (iv)2. So we assume from now on that (iv)2. holds. Let M be the unique maximal subgroup of H containg A and N = Core_{M}(G). By a Frattini argument, N is p-closed with O_{p}(H) as the Sylow p-subgroup. Hence t ∉ N and so there exists h ∈ H with t ∉ M^{h}. Put B = A^{h}. Then H = ⟨t, B⟩ and so [V, H] = [V, t][V, B]. By (iv)2. we conclude and (ii) we conclude

V = C_{V} (A)[V, B] = C_{V} (t)[V, B].

Thus

C_{V} (B) = [V, B](C_{V} (A) ∩ C_{V} (A) = [V, B]C_{V} (H).

Hence also

C_{V} (A) = [V, A]C_{V} (H)

and so by (iv)2.,

V = C_{V} (H)[V, H].

That is (f) holds for W = V. Moreover, C_{V} (t) = C_{V} (A)(C_{V} ([V, B]) ∩ C_{V} (t)) = C_{V} (A) and so (a) holds. Let Y = [V, A] ∩ C_{V} (H) = [V, B] ∩ C_{V} (H) = [V, A] ∩ [V, B]. Then [V, A] = [V, H] ∩ [V, A] = [V, t]([V, A] ∩ [V, B] and so [V, A] = [V, t]Y. On the otherhand,

|[V, t]| = |[V, B, t]| = |[V, B]/([V, B] ∩ C_{V} (t)) = |[V, B]/Y | = |[V, A]/Y |

and so [V, A] = [V, t] ⊕ Y. In particular [V, t] ∩ C_{V} (H) = 1. Moreover |[V, H]| = |[V, t]^{2}|Y |. C_{[V,H]}(A) = [V, A] and so C_{[V,H]}(H) = Y. Thus (c) and (e) hold. Let W be an nontrivial chief-factor for H on V. Since H = A⟨t^{H}⟩, A/O_{p}(H)/O_{p}(H) acts faithfully on W. Also W = [W, A] ⊕ [W, B] and so |W/C_{W} (A)| = |[W, A]|. Let x ∈ W \ C_{W} (A). By (a) |AO_{p}(H)/O_{p}(H)| = |[x, A]| ≤ |[W, A]| = |W/C_{W} (A)|. Thus (b) holds. Clearly (iv)2 is inherited by quotients of V so it is enough to verify (f) for H-submodules W of V. By (d) applied to V/[W, H], W ≤ [V, A][W, H] and so W = ([V, A] ∩ W)[W, H] fulfills (iv)2. Thus (f) holds.

It remains to prove (d). Let h ∈ H \M. As A is quadratic, A centralizes [Z^{h}, A∩O_{p}(H)]. As O_{p}(H) normalizes Z^{h}, also A^{h} centralizes [Z^{h}, A ∩ O_{p}(H). Since M ≠ M^{h}, H = ⟨A, A^{h}⟩ and [Z^{h}, A ∩ O_{p}(H) ≤ C_{V} (H).

Y = ⟨Z^{h} | h ∈ H \ M⟩.

Then [Y, A ∩ O_{p}(H)] ≤ C_{V} (H).

Suppose first that |AO_{p}(H)/O_{p}(H) ≥ 3. We claim then that B normalizes Y. For this let h ∈ H \ M and b ∈ B. We need to show that Z^{hb} ≤ Y. If hb ∉ M, this is true by definition of Y. So suppose that hb ∈ M. Since |AO_{p}(H)/O_{p}(H) ≥ 3 there exists c ∈ B with c ∉ O_{p}(H) ∪ O_{p}(H)b. If hc ∈ M, then b^{−1}c ∈ B ∩ M. But by 2.9 (10), b^{−1}c ∈ O_{p}(H), a contradcition. Thus hc ∉ M. Similarly hbc ∉ M. Thus Z^{h}Z^{hbc}Z^{hc} ≤ Y. Since Z^{h}Z^{hbc}Z^{hc} = Z^{h}[Z^{h}, bc][Z^{h}, c], the quadratic action of B implies that ⟨bc, c⟩ normalizes Z^{h}Z^{hbc}Z^{hc}. Hence Z^{hb} ≤ Y as claimed.

Suppose next that |AO_{p}(H)/O_{p}(H)| = 2. Then p = 2 and H/O_{2}(H) ≅ Dih(2r^{k}). If k = 1, then M = AO_{p}(H) normalizes Z and so V = ZY and again d1 and as a matter of fact also d2 holds. So suppose k > 1 and define L as in d2. Then L ≤ M. Also let H^{∗} be minimal with A ≤ H^{∗} and H^{∗}O_{p}(H) = M. Let V^{∗} = ⟨Z^{H^{∗}} = Z^{M}. Then V = V^{∗}Y.

Also A ∩ O_{p}(H) ≤ O_{p}(H^{∗}) and so by induction R =^{def} C_{H^{∗}}([V^{∗}, A ∩ O_{p}(H)] O̸_{p}(H^{∗}). Since [V, A ∩ O_{p}(H)] = [V^{∗}, A ∩ O_{p}(H)][Y, A ∩ O_{p}(H] we have [V, A ∩ O_{p}(H), R] = 0. Since R ≰ O_{p}(H), d2 holds in this case.

mq

Lemma 2.9 Let H be a finite group, p a prime, A a p-subgroup of H and V a faithful GF (p)H-module. Suppose that A ≰ O_{p}(H), that A acts quadratically on V and that A lies in a unique maximal subgroup of H. Then one of the following holds for \overline{H} = H/O_{p}(H):

1. \overline{H} ≅ SL_{2}(p^{k}).

2. p = 2 and \overline{H} ≅ Sz(2^{k}).

3. p = 2 and \overline{H} ≅ Dih(r^{k}), r an odd prime.

Proof: Let M the the unique maximal subgroupp of H containing A and D = ⋂_{h∈H} M^{h}. Note that M contains a Sylow p-subgroup S of H and so O_{p}(H) ≤ D. Replacing V by the direct sum of the H-composition factors on V and H by \overline{H} we may assume that O_{p}(H) = 1. Moreover, if |A| = 2, 3. holds so we may assume |A| > 2.

Let T be an A invariant Sylow p-subgroup of D. Then H = DN_{H}(T). If H = N_{H}(T) we get N_{H}(T) ≤ M and so H = DM ≤ M, a contradiction. Hence T ⊴ H and so T ≤ O_{p}(H) = 1. Thus D is a p′-group. Let R be a maximal subgroup of H and suppose that D ≰ R. Then H = DR and so R contains a Sylow p-subgroup of H. Hence A ≤ R^{h} for some h ∈ H and thus R^{h} ≤ M. But then H = DR = DR^{h} ≤ M, a contradiction. Thus D ≤ R. It follows that

mq − 1

(1) D ≤ Φ(H) and D is a nilpotent p′ group.

Let N be a normal subgroup of H. If H ≠ NA then NA ≤ M and so N ≤ D. Put L = O^{p}(H) and suppose that L ≤ D. Then H = DS ≤ M, a contradiction. Thus L ≰ D, H = LA. Hence:

mq − 2

(2) Each normal subgroup of H is either contained in D or contains L. In particular, L/D is characteristicly simple.

Since H acts faithfully on [V, O^{p}(H)] and on V/C_{V} (O^{p}(H)) we may assume that

mq − 3

(3) V = [V, H] and C_{V} (H) = 0.

Let 1 ≠ a ∈ A and pick g ∈ H with a ≰ M^{g}. Then H = ⟨a, A^{g}⟩ and so by (3) V = [V, a] + [V, A^{g}] and C_{V} (a) ∩ C_{V} (A^{g}). Since A is quadratically on V we also have

[V, a] ≤ [V, A] ≤ C_{V} (A) ≤ C_{V} (a).

We conclude that

mq − 4

(4) [V, a] = [V, A] = C_{V} (A) = C_{V} (a) and |V | = |[V, A]|^{2}

With a similar argument:

mq − 4′

(5) C_{V} (b) = [V, b] for each non-trivial quadratic element b in H.

We may assume without loss that A is a maximal quadratic subgroup of H and so

mq − 5

(6) A = C_{H}([V, A]) ∩ C_{H}(V/[V, A])

From (4) and (6) we conclude that

mq − 6

(7) C_{H}(a) ≤ N_{H}(A) and A ∩ A^{h} = 1 for all h ∈ H \ N_{H}(A).

Let h ∈ H with A ∩ M^{h} ≠ 1 and let b ∈ A ∩ M^{h}. Choose k ∈ M^{h} so that ⟨b, A^{hk}⟩ is a p-group. Then C_{V} (b) ∩ C_{V} (A^{hk}) ≠ 0 and so also V_{V} (A) ∩ C_{V} (A^{hk}) ≠). Thus H ≠ A, A^{hk}⟩ and so M = M^{hk} = M^{h}. We proved

mq − 7

(8) Let h ∈ H. Then h ∈ M or A ∩ M^{h} = 1.

If p is odd, then by (5)

dim[V, A] = min{dim[V, b] | 1 ≠ b ∈ H, [V, b, b] = 0}

Hence by the work of Thompson and Ho, H ≅ SL_{2}(p^{k}) or p = 3 and H ≅ 2.Alt(5). But in latter case, A lies in more than one maximal subgroup of H, a contradiction.

Thus we may assume from now on that

mq − 8

(9) p = 2 and |A| ≥ 4.

In particular, by (7)

O_{p′}(H) = ⟨C_{O_{p′}(H)}(a) | 1 ≠ 1 ∈ A⟩ ≤ C_{H}(A).

and we conclude:

mq − 7

(10) D = Z(H) and L = E(L) = E(H).

Note that the exceptionell case in 2.7 is not possible and so A normalizes the components of L and thus

mq − 8

(11) L is quasisimple.

None of the groups in ?? is a minimal parabolic and so L is an alternating group or a Lie type in characteristic 2. Since S lies in a unique maximal subgroup of H we get L ≅ Alt(2^{k} + 1), L_{2}(2^{k}), SU_{3}(2^{k}), Sz(2^{k}), SL_{3}(2^{k}) or Sp_{4}(2^{k}). In the last two cases A has to induce a graph automorphism on L, which contradicts the quadratic action of A on V. If L ≅ Alt(2^{k} + 1), A either is contained just has one non-trivial orbit and that one has lenght four or all orbits of A have length at most 2. Since A lies in a unique maximal subgroup of H we conclude that L = H ≅ Alt(5) ≅ SL_{2}(4). If L ≅ SU_{3}(2^{k}), A lies in the normalizer of a Sylow 2-subgroup and in a SL_{2}(2^{k}), a contradiction, which completes the proof of the lemma.

factorize

Lemma 2.10 Let G be a finite group, M ≤ G, p a prime with F^{∗}(M) = O_{p}(M) and T ∈ Syl_{p}(M). Let Z_{M} = ⟨Ω_{1}(Z(T))^{M}⟩, C_{M} = C_{M}(Z_{M}) and J(M) = ⟨J(T)^{M}⟩.

(a) C_{M} ≤ N_{G}(Z_{T})

(b) Z_{M} is a faithful J(M)C_{M}/C_{M}-module and J(M)C_{M}/C_{M} = P^{∗}(J(M)C_{M}/C_{M}), Z_{M}).

(c) M/J(M) ≅ N_{M}(J(T)/N_{J(M)}(J(T)

(d) Suppose that T is normal in a Sylow p-subgroup S of G. Then N_{G}(Z(T)) ∈ L(S) and N_{G}(J(T)) ∈ L(S).

Proof: Obvious.

vqnhg

Lemma 2.11 Let G be a finite group, N ⊴ H ≤ G, p a prime, S ∈ Syl_{p}(H) , V an elementary abelian normal p-subgroup of H, and C_{S}(V) ≤ Q ≤ S ∩ N Suppose that A(Q)^{G}∩ ⊈ N, then there exists an elementary abelian subgroup A of S with H ≰ N, [V, A] ≠ 1 and |V/C_{V} (A)| ≤ |A/C_{A}(V)|.

Proof: Let D ∈ A(Q) and g ∈ G with D^{g} ≤ H and D^{g} ≰ N. As S is a Sylow p-subgroup of H there exists h ∈ H with D^{gh} ≤ S. Put A = D^{gh}. As N is normal in H, A ≰ N. Since C_{N}(V) ≤ Q ≤ N, [V, A] ≠ 1. Moreover, V C_{A}(V) ≤ Q and so |V C_{A}(V)| ≤ |A|.

i2lt

Lemma 2.12 Let L be an alternating group or simple group of Lie-type in characteristic 2. Let H ≤ L with |L|_{2}/|H|_{2} ≤ 2. Then all non abelian composition factors of H are alternating or a simple groups of Lie type.

Proof: Let T ≤ Syl_{2}(H), and S ≤ Syl_{2}(L) with T ≤ S. Then S′ ≤ T.

Suppose first that L = Alt(Ω_{)}. If H is intransitive or imprimitive we are done by induction. So suppose that H is primitive. If H has a non-trivial abelian normal subgroup A, then H = H_{i}A for any i ∈ Ω. Thus T_{i} has index two in a Sylow 2-subgroup of L_{i} and again we are done by induction.

Hence we may assume that H has no non-trivial solvable normal subgroup. Since |S/T | ≤ 2, T contains an element x of cycle type (2, 2). Since x ∉ O_{2}(H), 1 ≠ x · x^{h} has odd order for some h ∈ H. Its is now straight forward to verify the lemma.

So suppose L is a group of Lie type. and not an alternating group. If O_{2}(H) ≠ 1, then H is contained in a parabolic subgroup of L and the lemma follows by induction. Hence we may assume that O_{2}(H) = 1.

If S is abelian, L ≅ L_{2}(q) and the result is readily verified in this case.

So we may assume that S is not abelian. In particular, S′ and so also H contains a long root group R with R ≤ Z(S). As R ≰ O_{2}(H), there exists h ∈ H with X =^{def} ⟨R, R^{h}⟩ ≅ SL_{2}(q), where q = |R|. Let r be the highest ling root in the root system associate to L. Without loss ω_{r} ∈ X ≤ H. It is now easy to verify that L = ⟨S′ω_{r}⟩ and so L ≤ H, a contradiction.

Remark: this is rather scetchy

3 CS generated modules

In this section G is a finite group, p a prime and V a ( finite dimensional) GF (p)G-module.

dcvg

Definition 3.1 (a) _{G}V = ⟨C_{V} (S) | S ∈ Syl_{p}(G)⟩.

(b) V is called CS-generated provided that V = _{G}V.

ec

Lemma 3.2 Let L ⊲ ⊲G. Then _{G}V ≤ _{L}(V).

Proof: Let S ∈ Syl_{p}(G). Then S ∩ L ≤ Syl_{p}(L) and C_{V} (S) ≤ C_{V} (S ∩ L).

dualtrans

Lemma 3.3 Let p be a prime, G a finite group, L a normal subgroup of G, S ∈ Syl_{2}(G). Then S normalizes a complement to C_{V} (L) in C_{V} (S ∩ L).

Proof: Remark: This is a standard result in cohomology, the map π below is called the corestriction map, a reference should be included

Let T = S ∩ L, X a set of right coset representatives for T in L and define

\begin{aligned}π : & C_{V} (T) & → & V \\ & v & → & ∑_{x∈X} v^{T x} \end{aligned}

Then clearly π(v) = π(v^{l}) for al l ∈ L and so π(C_{V} (T) ≤ C_{V} (L). On the otherhand π restricted to C_{V} (L) is just multiplication by L/T. Thus π |_{C_{V} (L)} is an isomorphism and C_{V} (T) = C_{V} (L) ⊕ ker π. Moreover, it follows immediately from the definition of π that for all v ∈ C_{V} (T) and s ∈ S, π(v^{s}) = π(v)^{s}. Thus S normalizes ker π.

cc

Lemma 3.4 Let L ⊲ ⊲G with [C_{V} (S), L] = 1, then [C_{V} (L ∩ S), L] = 1.

Proof: Clearly we may assume that L ⊴ G. By 3.3 there exists an S invariant complement D to C_{V} (L) in C_{V} (S ∩ L). Moreover, C_{D}(S) ≤ C_{V} (S) ≤ C_{V} (G) ≤ C_{V} (L) and so C_{D}(S) = 0. This implies D = 0 and C_{V} (S ∩ L) = C_{V} (L)

c2

Lemma 3.5 Let L be subnormal subgroup of G. If [C_{V} (S), L] = 1 then [_{G}V, L] = 1.

Proof: By 3.4 C_{V} (S ∩ L) ≤ C_{V} (L). So L centralizes _{L}V and hence the lemma follows from 3.2.

c

Lemma 3.6 Let L ⊲ ⊲G. Then L ∩ C_{G}(_{G}V) = C_{L}(_{L}V).

Proof: Let L^{∗} = C_{L}(_{L}V and L_{∗} = C_{G}(_{G}V). By 3.2 L^{∗} ≤ L_{∗}. Moreover, L_{∗} is subnormal in G and centralizes C_{V} (S). Thus by 3.4 L_{∗} centralizes C_{V} (L_{∗} ∩ S). By 3.2 _{L}V ≤ _{L^{∗}}V = C_{V} (L_{∗} ∩ S) and so L_{∗} ≤ L^{∗}.

[V, L]

Lemma 3.7 Let L ⊴ G with G = LC_{G}(L). If V is CS-generated then [V, L] is a CSgenerated G-module and V = [V, L]_{G}C_{V} (L)

Proof: Let S ∈ Syl_{p}(G), T = S ∩ L, R = S ∩ C_{G}(L) and put W = _{L}C_{V} (R). Then by Gaschütz theorem W = [W, L]C_{W} (L). Moreover, C_{W} (T) = C_{[W,L}(T)C_{W} (L). It follows that [V, L] = ⟨C_{[W,T} (T)^{G}⟩ and [V, L] is a CS geneated G-module. Moreover, V = ⟨W^{G}⟩ = [V, L]⟨C_{W} (L)^{G}⟩ and so V = [V, L]_{G}C_{V} (L).

gaschuetz

Lemma 3.8 Suppose that G = Π_{i∈I}L_{i} for some subgroups L_{i} ≤ G such that [L_{i}, L_{j}] = 1 whenever i, j ∈ I, i ≠ j. For Δ ⊆ I let L_{δ} = ⟨L_{i} | i ∈ Δ and

V_{δ} = [_{G}C_{V} (L_{I\Δ}, L_{i_{1}}, L_{i_{2}}, … L_{i_{r}}]

where r = |Δ| and Δ = {i_{1}, … l_{r}}. (Note that by the Three Subgroup Lemma this defintion is independent form the order in which the i_{j}’s are chosen). Also put V_{∅} = C_{V} (G).

Suppose that V is a CS-generated GF (p)G-modules. Then

∗  V = ∑_{Δ⊆I} V_{Δ}.

Moreover, each of the V_{Δ}’s is CS-generated as G-module.

Proof: By 3.7 The V_{δ}^{s} are CS-generated as G-module and it remains tp prove (*). For this we may assume without loss that V is not the direct sum of two proper CS-generated G-submodules. Let Δ = {i ∈ I | [V, L_{i}] ≠ O and let i ∈ Δ. 3.7 implies V = [V, L_{i}]_{G}C_{V} (L_{i}) with both summands CS generated. Hence V = [V, L_{i}] and V = V_{δ}.

4 Groups with m_{2′}(G) ≤ 3

gwm3
epc2r3

Lemma_{QT} 4.1 Let p be an odd prime, P a p group of exponent p, class at most two and rank at most three. Then P ≅ E_{p^{i}}, i ≤ 3, Ex(p^{1+2i}), i ≤ 2 or C_{p} × Ex(p^{1+2}).

Proof: [As, 3.1,3.2]

l3p

Lemma_{QT} 4.2 Let p be an odd prime, G a irreducible subgroup of GL_{3}(p) and Λ = Z(GL_{3}(p)) Then there exists an irreducible normal subgroup H of G so that one of following holds.

1. H = SL(V) ≅ SL_{3}(p).

2. H = Ω(V, q) for some non degenerate quadratic form q on V.

3. H ≅ Alt(5), p^{2} ≡ 1mod10 and G ≤ Λ × H.

4. H ≅ L_{3}(2), p^{3} ≡ 1mod7 and G ≤ Λ × H.

5. H ≅ 3.Alt(6), p ≡ 1, 19mod30 and G ≤ ΛH.

6. H is cyclic of order dividing p^{3} − 1 but not p − 1 and H = G or |G/H| ≅ C_{3}.

7. H ≅ Ex(3^{1+2}) and GΛ/HΛ ≤ SL_{2}(3).

8. G is monomial

Proof: [As, 3.12]

msp4p

Lemma_{QT} 4.3 Let p be an odd prime, V a four dimensional non-degenerate symplectic space over GF (p) and G a maximal subgroup of Sp(V). Then one of the following holds.

(a) G is the normalizer of a singular 1-space in V and G ∼ Ext(p^{1+2}) : (C_{p−1} × SL_{2}(p)).

(b) G is the normalizer of a singular 2-space in V and G ∼ E_{p^{3}} : GL_{2}(p)

(c) G ∼ SL_{2}(p^{2}).2 and G′ fixes a non-degenerated 2-dimensional sympectic form over GF (p^{2}) on V.

(d) G ≅ SL_{2}(p) ≀ C_{2} and G fixes a decompostion of V into the orthorgonal sum of two non-degenerated 2-dimensional subspaces.

(e) G ∼ GL_{2}(p).2 and G fixes a decomposition of V into the direct sum of two singular 2-spaces.

(f) G ∼ GU_{2}(p).2 ∼ (C_{p+1} · SL_{2}(p)).2 and the subgroup of index 2 fixes a non-degenerate 2-dimensional unitary form over GF (p^{2}) on V.

(g) G ≅ SL_{2}(p) and V is the third symmetric power of the natural module for G.

(h) G ∼ Ext_{−}(2^{1+4}).Alt(5)(.2).

(i) G ∼ 2.Alt(6)(.2) and V is the half-spin module for G

(j) p = 7, G ∼ 2.Alt(7) and V is the half-spinmodule for G

Proof: See [Mi, Theorem 10]. We remark that this list can be easily checked if one is only interested in K-groups. Namely let W be the natural Ω_{5}(p) module for P Sp_{4}(p), H = Sp_{4}(p) and \overline{H} = H/Z(H). We may assume that G acts irreducible on W.

If Sol() ≠ 1 let A be a minimal solvable normal subgroup of \overline{G}. If A is cyclic, |A| divides p^{5} − 1 and |H|. Hence |A| divides p − 1 and A acts as a scalar on W, a contradiction. So A is not cyclic and it is now easy to see that (h) holds.

If Sol(\overline{G}) = 1, let E be a component of G. Since O_{2}^{±}(p) is solvable, [W, E]C_{W} (E)/C_{W} (E) is at least three dimensional. It follows that C_{H}(G) is solvable and so EZ(H) = F^{∗}(G) and E acts irreducibly on W. If Z(H) ≰ E, m_{2}(Z(H)E) ≥ 3, a contradiction to m_{2}(Z(H)) = 2. Thus Z(E) = Z(H). Let V be the natural Sp_{4}(p) module for H. If E does not act irreducible on V then since V ⋀ V = W ⊕ GF (p), E is not irreducible on W. So E acts irreducible on W. Using the list of finite simple groups its now easy to verify that one of (g),(i) or (j) holds or that E ≅ 2.Alt(5). But in the latter case, G is contained in a subgroup of type (i) or (j).

sp4p

Lemma_{QT} 4.4 Let p be an odd prime, V a four dimensional non-degenerate symplectic space over GF (p) and G ≤ Sp(V) with O_{p}(G) = 1.

(a) If G = O^{p′}(G) ≠ 1, then one of the following holds:

1. G ≅ Sp_{4}(p), SL_{2}(p^{2}), SL_{2}(p) × SL_{2}(p) or SL_{2}(p)

2. p = 7 and G ≅ 2.Alt(7).

3. p = 5 and G ∼ 2.Alt(5), Ext_{−}(2^{1+4}).Alt(5), Ext_{(}2^{1+4}).C_{5}.

4. p = 3 and G ∼ 2.Alt(5), Ext_{−}(2^{1+4}).Alt(5), Ext_{(}2^{1+4}).C_{3}.

(b) If G is quasisimple then one of the following holds:

1. G ≅ Sp_{4}(p), SL_{2}(p^{2}) or SL_{2}(p).

2. G ≅ 2.Alt(5) or 2.Alt(6).

3. G ≅ 2.Alt(7) and p = 7.

Proof: [As, 3.13]

mp3Q

Lemma_{QT} 4.5 Let p be an odd prime, G a group with F^{∗}(G) = O_{p}(G) =^{def} Q, m(Q) ≤ 3 and G^{∗} = G/Q.

(a) If G = O^{p′}(G) ≠ Q, then one of the following holds:

1. G^{∗} ≅ SL_{2}(p) or SL_{3}(p).

2. G^{∗} ≅ SL_{2}(p) × SL_{2}(p), SL_{2}(p^{2}), or Sp_{4}(q) and m_{p}(G) > 3.

3. p = 7 and G^{∗} ≅ 2.Alt(7).

4. p = 5 and G ∼ SL_{2}(5), Ext_{−}(2^{1+4}).Alt(5) or Ext_{−}(2^{1+4}).C_{5}.

5. p = 3, G ∼ 2.Alt(5) or Ext_{−}(2^{1+4}.Alt(5) and m_{3}(G) > 3.

6. p = 3 and G ∼ Ext_{(}2^{1+4}).C_{3}

(b) If G^{∗} is quasisimple then one of the following holds:

1. G^{∗} ≅ Sp_{4}(p), or SL_{2}(p^{2}) and m_{p}(G) > 3.

2. G^{∗} ≅ L_{2}(p), SL_{2}(p)orSL_{3}(p)

Remark: SL_{3}(p) also should have m_{p}(G) > 3

3. G^{∗} ≅ Alt(5), 2.Alt(5) or 2.Alt(6). Moreover, if p = 3 then m_{3}(G) > 3.

4. G^{∗} ≅ L_{3}(2) and p^{3} ≡ 1mod7

5. G^{∗} ≅ 3.Alt(6) and p ≡ 1, 19mod30

6. G ≅ 2.Alt(7) and p = 7.

Proof: By [As, 3.13] we only need to show that m_{p}(G) > 3 in a.5, b.1 and for p = 3 in b.3. As in Aschbacher’s proof let G be a minimal counterexample and D a critical subgroup of Q. As G^{∗} = O^{p}(G^{∗}), G = O^{3}(G).

Let t be an involution in G with t^{∗} ∈ Z(G^{∗}). By minimality G = DC_{G}(t) and without loss D = [D, t]. It follows that D ≅ Ext(p^{1+4}). In particular, as m(Q) ≤ 3, Ω_{1}(C_{Q}(D)) = Z(D). As G acts irreducible on D/Z(D), Q = DC_{Q}(D). Since G centralizes Ω_{1}(C_{Q}(D)), G = O^{3}(G) centralizes C_{Q}(D).

Considering the p-part of the Schur multiplier of G^{∗} we see that C_{G}(t)′ ≅ G^{∗} or p = 3 and C_{G}(t)′ ≅ 3.SL_{2}(3^{2}). In any case there exists X ≤ C_{G}(t)′ so that X is an elementary abelian p-group and XD′/D′ ≅ C_{p}. Moreover [D, X, X, X] ≤ D′ and so [Y, X] ≤ D′ for some Y ≤ D with Y ≅ E_{p^{3}}. Since Y = [Y, t] × D′ we have [Y, X] = 1 and so Y X ≅ E_{p^{4}}.

dcr

Definition 4.6 Let p be an odd prime, Q a p-group and H a group acting on Q.

(a) CR_{Q}(H) is the set of maximal, H-invariant, class 2 and exponent p, normal subgroups of Q.

(b) We say that Q is H-homogeneous of rank n provide that there exists A ∈ CR_{H}(Q) so

that A ≅ E_{p^{n}} and H acts irreducible on A.

homo

Lemma 4.7 Let p be an odd prime, Q a p-group, H a group acting on Q. Let D ∈ CR_{Q}(H) and T = C_{Q}(D). Then CR_{T} (H) = {Z(D)}. For i ≥ 0 put T_{i} = Ω_{i}(T). Then T_{i+1}/T_{i} = Ω_{1}((T/T_{i}) = Ω_{1}(Z(T/T_{i})) ∈ CRT/T_{i}(H) and if i ≥ 1, T_{i+1}/T_{i} is isomorphic to HQ-submodule T_{i+1}^{p}T_{i−1}/T_{i−1} of T_{i}/T_{i−1}.

Proof: Let A = Z(D). Clearly A ≤ Ω_{1}(Z(T)). Let A^{∗} ∈ CR_{T} (H). Then DA^{∗} has class two und exponent p and so by maximality of D, A^{∗} ≤ D∩T ≤ A. By maximality of A^{∗}, A ≤ A^{∗} and so A = A^{∗} and CR_{T} (H) = {A}. Let C/A ∈ CR_{Q/A}(H) and B/A = Z(C/A). Then B is of class two and Ω_{1}(B) = A by maximality of A. As p is odd the map

ϕ : B/A → A
bA → b^{p}

is a HQ-homomorphism. As Ω_{1}(B) = A, ϕ is one to one thus B/A ≅ B^{p} as HQ-module. Let c, e ∈ C The c^{p} ∈ A ≤ Z(T) and so c^{p} = (c^{p})^{e} = (c^{e})^{p} Put d = cc^{−e}. As c^{e} ∈ cB,⟨c⟩B has class two and p is odd, d^{p} = c^{p}(c^{e})^{−p} = 1. It follows that d ∈ Ω_{1}(B) = A. Hence cA = c^{e}A for all e ∈ C and so cA ∈ Z(C/A) = B. Thus C = B and B/A ∈ CR_{Q/A}(H). Since T centralizes B^{p} ≤ A, T/Acentralizes B/A. The lemma now follows by induction on |T |.

ccr

Corollary 4.8 Let p be an odd prime, Q a p-group, H a group acting on Q and D ∈ CR_{Q}(H). Then C_{H}(D)/C_{H}(Q) is p-group.

Proof: Note first that C_{H}(D) centralizes Q/C_{Q}(D) and Z(D). Let T and T_{i} be as 4.7. Then by 4.7, C_{H}(D) centralises all factors of the normal series

1 = T_{0} ≤ T_{1} ≤ T_{2} · T_{k} = T ≤ Q.

Thus C_{H}(D)/C_{H}(Q) is a p-group. Lemma 4.9 Let p be a prime with p ≥ 5, A ≅ C_{p^{2}} × C_{p^{2}} and t ∈ Aut(A) with t^{p} = 1. Then t centralizes Ω_{1}(A). In particular, Aut(A) has no subgroup isomorphic to SL_{2}(p). Proof: Identify t which its image in the ring End(A). Since |A| = p^{4} we have (t − 1)^{4} = 0 and since p ≤ 4 we get

(1) (t − 1)^{p} = 0 Since |A^{p}| = p^{2} we have

(2) p(t − 1)^{2} = 0 Since t^{p} = 1 we have

(3) t^{p} − 1 = 0 Consider the polyomial f(x) = x^{p−1} + x^{p−2} + … + x + 1 ∈ Z[x]. Since f(x) ≡ (x −

1)^{p−1}modp, f(x) = (x − 1)^{p−1} + p · g(x) for some g(x) ∈ Z[x]. Write g(x) = h(x)(x − 1) + d for some h(x) ∈ Z[x], d ∈ Z. Then p = f(1) = p · d and so d = 1 and f(x) = (x − 1)^{p−1} + p · h(x)(x − 1) + p. Since f(x)(x − 1) = x^{p} − 1 we obtain

(4) x^{p} − 1 = (x − 1)^{p} + h(x)p(x − 1)^{2} + p(x − 1) Substituting t for x in (4) and using (1) to (3) we obtain

(5) 0 = p(t − 1)

Hence t centralizes A^{p} = Ω_{1}(A).

pe = 3

Lemma_{QT} 4.10 Let G be a finite, perfect K-group with O_{2}(G) = 1 and m_{2′}(G) ≤ 3.

(a) G is the central product of its Sol-components.

(b) If G is a Sol-component of G then one the following holds:

(b1) G is quasisimple and if G/Z(G) is a group of Lie type in characteristic 2 or an alternating group then G/Z(G) is one of the following: Alt(n), 5 ≤ n ≤ 11; L_{n}(q), n ≤ 4; L_{n}(2), n ≤ 7; Sp_{2n}(q), n ≤ 3; G_{2}(q); U_{n}(q), n ≤ 4; Sz(q); Ω_{8}^{−}(q); 3D_{4}(q); 2F_{4}(q).

(b2) F^{∗}(G) = F (G). Let p be a prime dividing |[F (G), G]| and put Q = [O_{p}(G), G]. Then one of the following holds:

1. G/F (G) ≅ 2.Alt(5) or SL_{2}(p), and Q ≅ Ext(p^{1+2}) or Q is of G homogenous of rank 2.

2. G/F (G) ≅ SL_{3}(p); L_{3}(2) ( p^{3} ≡ 1mod7); L_{2}(p); (2.)Alt(5); or (2.)3.Alt(6) ( p ≡ 1, 19mod30 and Q is G-homogenous of rank 3.

3. G/F (G) ≅ SL_{2}(p), 2.Alt(5), (3.)2.Alt(6) or 2.Alt(7) ( and p = 7) and Q ≅ Ext(p^{1+4}).

(c) Let E be quasisimple so that E/Z(E) is alternating or a group of Lie type in characteristic 2. Suppose that G is a central product of r copies of E with r ≥ 2. Then r ≤ 3 and one of the following holds:

(b1) E/Z(E) ≅ L_{2}(q),L_{3}(2) or Sz(q). (b2) E ≅ 3.Alt(6) or SL_{3}(4), r = 2 and |Z(G)| = 3.

Proof: (a) Let L be a Sol-component of G.

Suppose first that L does centralize all its distinct conjugates under G. Then |L^{G}| ≤ 3 and as Sym(3) is solvable, G normalizes L. As L is a K-group, Out(L/Sol(L)) is solvable and so G = LC_{G}(L/Sol(L)). Bu induction C_{G}(L/Sol(L)^{∞} is the central product of its Sol-components.

Hence we may in any case assume that there exist distinct Sol-components L_{1} and L_{2} of G with [L_{1}, L_{2}] ≠ 1. Note that [L_{1}, L_{2}] ≤ Sol(G) and by induction G = L_{1}L_{2}. Moreover, L_{i} is normal in G. If [F (G), L_{1}, L_{2}] = 1 and [F (G), L_{2}, L_{1}] = 1 we get [L_{1}, L_{2}] ≤ C_{G}(F^{∗}(G) ≤ F (G) and so [L_{1}, L_{2}] = [L_{1}, L_{2}, L_{2}] = [L_{1}, L_{2}, L_{1}, L_{2}] ≤ [F (G), L_{1}, L_{2}] = 1, a contradiction. Hence we may assume that [O_{p}(G), L_{1}, L_{2}] ≠ 1 for some odd prime p. Put Q = O_{p}(G) and D ∈ CR_{Q}(G). Then [D, L_{1}] ≠ 1 ≠ [D, L_{2}]. We conclude that D ≅ Ext(p^{1+4}) and [D, L_{1}, L_{2}] = 1. Moreover, [D, Q] ≤ D′, Q = C_{Q}(D)D, C_{Q}(D) is cyclic and so [C_{Q}(D), G] = 1. Thus [Q, L_{1}, L_{2}] = [D, L_{1}, L_{2}] = 1, a contradiction.

(b) If E(G) ≠ 1, then G is clearly a component of G and it is now easy to verify that (b1) holds.

So suppose that E(G) = 1. Then by definition F^{∗}(G) = F (G). Let p and Q be as in (b2). Let D ∈ CR_{Q}(G), D^{∗} = D/D′ and \overline{G} = G/C_{G}(D^{∗}). Let R be minimal in G with respect to D ≤ R and G = RC_{G}(D). Then C_{R}(D)D/D is nilpotent and so C_{R}(D) is nilpotent. In particular, F^{∗}(R/O_{p′}(R)) is a p-group.

Assume that Sol(\overline{G}) ≠ O_{p}(\overline{G})Z(\overline{G}). Then its easy to see that D ≅ Ext(p^{1+4}) and \overline{G} ∼ Ext_{−}(2^{1+4}).Alt(5). Moreover, by 4.55, applied to R/O_{p′}(R), p > 3.

Assume that O_{p}(\overline{G} ≠ 1. Then D ≅ E_{p^{3}}, C_{p} × Ext(p^{1} + 2) or Ext(p^{1+4}. Mostly without loss,(TO BE CONTINUED) G = R and O_{p′}(G) = 1.

Suppose that D ≅ Ext(p^{1+4}) and let A/D′ be a minimal G invariant subgroup of D/D′. If |A/D′| = p we get conclude that [A, G′] = 1 and so [A, G] = 1 and [A, D] = 1, a contradiction. Hence |A/D^{p}rime| = p^{2} and \overline{G} ∼ p^{3}SL_{2}(p) or p^{3}2.Alt(5). Let t be an involution in which inverts A/D′. Then C_{G}(t) ∼ p^{1+3}SL_{2}(p) or p^{1+3}2.Alt(5) and so contains a normal E_{p^{4}}, a contradiction.

Suppose that D ≅ C_{p} × Ext(p^{1+2}). Then G = G′ centralizes Z(D) and Z(D)/D′ and so Z(D) ≤ Z(G). By 4.7 we conclude that G also centralizes C_{Q}(D) and so C_{Q}(D) = C_{Q}(G). Let t be an involution in G inverting D/Z(D). Then Q/C_{Q}(D) has order p^{4} and is inverted by t. Thus Q/C_{Q}(D)is abelian and Q′ ≤ Z(G). In particular Q has class two and so Ω_{1}(Q) = D. Let x, y ∈ Q so that t inverts x and y and Q = C_{Q}(D)D⟨x, y⟩. Since t inverts x^{p}, x^{p} ∈ D and since x^{p} ≠ 1, we conlude that D = ⟨x^{p}, y^{p}⟩Z(D) and so Q = C_{Q}(D)⟨x, y⟩. Hence Q′ = ⟨[x, y]⟩ is cyclic and so Q′ ∩ D = D′. Thus [Q, D] ≤ D′ and [D^{∗}, Q] = 1, a contradiction.

Thus D ≅ E_{p^{3}} and so Q/C_{Q}(D) ≅ E_{p^{2}}. We will use 4.7 without further reference. In particular we are done if G normalizes a hyperplane in Q. So suppose |C_{D}(G)| = p. Let T and T_{i} be as in 4.7. Let t be an involution in G inverting D/C_{D}(G). Assume first that T = D. The t inverts Q/C_{D}(G) and thus Z(Q) = Q′ = C_{D}(G). It follows that Q is extra special, a contradiction to D ∈ CR_{Q}(G). Thus T ≠ D. Let A/D = C_{T/D}(G). Note that C_{Q}(t) = C_{T} (t) is cyclic and A = C_{A}(t)D. Thus t inverts Q/C_{Q}(A). It is now easy to see in Aut(A) that C_{Q}(A) = T and A = C_{A}(G)D. If T_{2} ≠ A put B = T_{2} otherwise let B = Q. Note that since G is perfect, Q = [Q, t] and T = [T, t]Q′. But |Q′[T, t]/[T, t] ≤ p and so if A = T_{2}, A = T. Hence in any case |B/A| = p^{2}, [B, Q]D = A and t inverts B/C_{A}(G). In particular, B′ ≤ C_{A}(G). Since t centralizes Hom(B/A, A/C_{A}(G)), [B, Q] ≤ C_{A}(G). If Q/C_{Q}(B) has exponent p we conclude that [B, Q] has exponent p and [B, Q] ≤ D, a contradiction. Thus Q/C_{Q}(B) ≅ C_{p^{2}} × C_{p^{2}} and hence Q^{p} = T. Hence [B, T] = C_{D}(G),

Assume that Sol(\overline{G} = Z(\overline{G}. Then as G is a Sol-component, \overline{G} is quasisimple.

Remark: Lots of case with L/F (G) ≅ 2.Alt(5) or Ext_{−}(2^{1+4} need to be worked into the statement of the theorem,4.9 has to be used to exclude smilar cases for SL_{2}(p) TO BE CONTINUED

P arAlt

Lemma 4.11 Let G ≅ Sym(Ω) or Alt(Ω), |Ω| = n finite, and H a maximal subgroup of G such that |G/H| is odd.

(a) For an integer k let b_{2}(k) = {2^{i} | a_{i} ≠ 0} where k = ∑_{i=1}^{n} a_{i}2^{i} with a_{i} ∈ {0, 1}. Then one of the following holds.

1. H = N_{G}(Λ) where Λ ⊂ Ω and b_{2}(|Λ|) ⊆ b_{2}(Ω)

2. H = N_{G}(Π), where Π is a partition of Ω into m parts of size l and l is a power of 2 dividing n.

3. G = Alt(7) and H ≅ L_{3}(2).

4. G = Alt(8) and H ∼ 2^{3} : L_{3}(2).

(b) If G = Alt(7), then H = L_{3}(2), Alt(6), Sym(5) or Sym(3) ∧^{|}Sym(4).

(c) If G = Sym(7), then H = Sym(6), Sym(5) × C_{2} or Sym(3) × Sym(4).

(d) If G = Sym(9) then H = Sym(8).

(e) If G = Sym(10), then H = Sym(8) × C_{2} or C_{2} ≀ Sym(5).

(f) If G = Sym(11), then H = Sym(8) × Sym(3), Sym(9) × C_{2} or Sym(10).

(g) If G = Alt(n), n ≥ 9, then H = H^{∗} ∩ Alt(n) for some maximal subgroup H^{∗} of Sym(n) which contains a Sylow 2-subgroup of Sym(n).

Proof: Remark: Maybe we should find a reference, below is a the sketch of aproof

If G = Sym(Ω), this easliy follows since the subgroup of H generated by the 2-cycles in H is a direct product of natural embedded symmetric groups. So we may assume that G = Alt(Ω) and N_{Sym(Ω)}(H) ≤ Alt(Ω). Moreover, we may assume that H acts primitively

on Ω. Let X ⊂ Ω with |X| = 4 and A_{X} =^{def} O_{2}(Alt(X)) ≤ H. Let h ∈ H.

If |X ∩ X^{h}| = 3, then ⟨A_{X}, A_{X}^{h}⟩ = Alt(X ∪ X^{h}) and so H = G, a contradiction. If |X ∩ X^{h}| = 1, then |X ∩ X^{a}| = 3 for all a ∈ A_{X}^{h}, a contradiction to by previuos case.

Thus |X ∩ X^{h}| ∈ {0, 2, 4} for all h ∈ H.

Let V be the power set of Ω viewed as a vector space over GF (2) and endowed with the natural symmetric form. It follows that U =^{def} ⟨X^{H}⟩ is a singular subspace of V and all sets in U have size divisible by 4. Moreover if |X ∩ X^{h}| = 2, then X + X^{h} is in ⟨A_{X}, A_{X}^{h}⟩ conjugate to X and X^{h}. Since X ∩ X^{h} is not a set of imprimitivety, there exists l ∈ H with |X ∩ X^{h} ∩ X^{l} ∩ X^{hl}| = 1 It follows that |X ∩ X^{h} ∩ Y = 1 or some Y ∈ {X^{l}, X^{hl}. Let Z = X ∪ X^{h} ∪ Y. Since |X ∩ Y | = |X^{h} ∪ Y | = 2 we get |Z| = 7. Put L = ⟨A_{X}, A_{X}^{h}, A_{Y} ⟩ then L ≅ L_{3}(2). If n ≤ 7 we are done. If n ≥ 8, there must exists k ∈ H with Z ∩ X^{k} ≠ ∅ and X^{k} ⊄ Z. Since X^{k} is perpendicular to ⟨X^{L}⟩ we get that |Z ∩ X^{k}| = 3 ( and indeed Z ∩ X^{k} = Z \ X^{r} for some r ∈ L. Let W = Z ∪ X^{k} and K = ⟨L, A_{X}^{k}. Then K ≅ 2^{3} : L_{3}(2). We n = 8 then K = H an we are done. If n ≥ 9 then there exists s ∈ H with W ∩ X^{s} ≠ ∅ and X^{s} ⊄ W. Since K acts transitively on W, we conclude that X^{s} intersects each subset of sixe seven in W in O or 3 elements, a contradiction, which completes the proof of the lemma.

5 Subnormal Subgroups

OqRcQ

Lemma 5.1 Let G be a finite group, L a subnormal subgroup of G, Q a normal q-subgroup of G and R a subgroup of G which centralizes L and N_{Q}(L). Then O^{q}(R) centralizes Q.

Proof: Without loss R = O^{q}(R). Suppose the lemma is false and let X be minimal in Q such that L and R normalize X, and R does not centralize X. Then [X, R, R] ≠ 1 and so X = [X, R]. As O^{q}(L) is subnormal in Q^{q}(L)X and X is a q-group we conclude that [X, O^{q}(L)] ≤ L. Thus R centralizes [X, O^{q}(L)] and hence [X, Q^{q}(L)] ≠ X. But this implies [X, L] ≠ X and so by minimal choice of X, [X, L, R] = 1. The three subgroup lemma implies [X, R, L] = 1 and thus [X, L] = 1 and X ≤ N_{Q}(L). We conclude that [X, R] = 1 and the lemma is established.

pi

Lemma 5.2 Let G be a finite group, π a set of primes and L a subnormal subgroup of G such that L = O^{π}(L). Then E_{π}(N_{G}(L)) = E_{π}(G).

Proof: Note first that N_{G}(L) = N_{G}(LO_{π}((G))), E_{π}(G/O_{p}(G)) = E_{π}(G)/O_{π}(G) and E_{π}(N_{G}(L)/O_{π}(G)) = E_{π}(N_{G}(L)/O_{p}(G). Thus we may assume that O_{π}(G) = 1.

Put H = N_{G}(L). Since E(G) normalizes L we have E(G) ≤ E(H). Let R be the group generated by O_{π}(H) and the π-components of H which are not contained in E(G). Then R centralizes E(G) and F (G) ∩ H. By the previous lemma applied with Q a Sylow subgroup of F (G) we conclude that R centralizes F (G) and F^{∗}(G). Thus R ≤ F^{∗}(G) and since E_{π}(H) = E(G)R, E_{π}(H) = E_{π}(G) = E(G).

OqF

Corollary 5.3 Let G be a finite group, p, q distinct primes and L a subnormal subgroup of G such that L = O^{p}(L) and L/O_{p}(L) is a q-group. Then O^{q}(F_{p}^{∗}(N_{G}(L)) = O^{q}(F_{p}^{∗}(G)).

Proof: Apply the previous lemma with π = q′.

USN

Lemma 5.4 Let G be a finite group and L a subgroup of G such that L = O^{p}(L), O_{p}(L) ≠ 1 and L_{/}O_{p}(L) is either quasi-simple or a q-group. Then L is subnormal in at most one maximal p-local subgroup of G containing N_{G}(L).

Proof: Let M_{1} and M_{2} be maximal p-locals of G containing N_{G}(L). By the previous lemma E_{p}(M_{1}) = E_{p}(N_{G}(L) = E_{p}(M_{2}). As O_{p}(L) ≠ 1, O_{p}(E_{p}(N_{G}(L)) ≠ 1 and so N_{G}(E_{p}(N_{G}(L))) is a p-local containing M_{1} and M_{2}. Thus M_{1} = M_{2}.

6 Nice Modules

NM

Definition 6.1 Let H be group and V a faithful GF (p)H-module. Then davh

1. a_{V} (H) is defined by |V/C_{V} (H)|^{a_{V} (H)} = |H|.

2. qa_{V} (H) = min{a_{V} (A) | 1 ≠ A ≤ H, [V, A, A] = 1]}, where qa_{V} (H) = ∞ if H has no nontrivial quadratic subgroups.

3. ra_{V} (H) is the minimum of the qa_{W} (H), where W runs through the non-trivial composition factor for H on V

4. Let a be a positive real number. Then V is called an F a module if qa_{V} (H) ≤ a and an F^{∗}a module if qa_{V} (H) < a.

5. An F F -module is an F 1-module.

oqu

Lemma 6.2 Let G be a finite group, p an odd prime, S ∈ Syl_{2}(G) and V a faithful GF (2)- module. Suppose that

(i) G = O_{p}(G)S.

(ii) [V, S, S] = 0.

The there exists a set of hyperplanes H of S and G-submodules V_{H}, H ∈ H so that

(a) V = C_{V} ([O(G), S]) ⊕ oplus_{H∈H}V_{H}

(b) For all H in H, H centralizes V_{H}.

Proof: We may assume without loss that V is not the direct sum of two proper Gsubmodules. Put P = O_{p}(G) and Q = [P, S]. If Q = 1 we are done. So suppose Q ≠ 1 and let E be a normal subgroup of G in Q minimal with respect to [E, Q] ≠ 1. Let F = C_{E}(QS). Then by minimality of E, G acts irreducibly on E/F. In particular, [E, P] ≤ F, S inverts E/F and |E/F | = p. Since F ≤ Z(Q) ∩ E ≤ Z(E), E is abelian. Then also [Ω_{1}(E), S] ≠ 1 and hence E is elementary abelian. Let T = C_{S}(E). Then |S/T | = 2.

Suppose first that F = 1. Then E = [E, S] ≤ ⟨S^{E}⟩ ≤ C_{G}([V, T]). Since C_{V} (E) = 0, T = 1 and the lemma holds.

Suppose next that F ≠ 1 and ley D be the set of all hyperplanes D in E with C_{V} (D) ≠). Then

V = ⊕_{D∈D}C_{V} (D).

As V is indecomposable, G acts transitively on D. Moreover, T is a Sylow 2 subgroup of C_{G}(E) and so G = N_{G}(T)C_{G}(E). In particular, N_{G}(T) acts transitively on D. We may assume that [C_{V} (D), T] ≠ 0 for some D ∈ D and so [C_{V} (D), T] ≠ 1 for all D ∈ D. As [C_{V} (D), T, S] = 0, S normalizes C_{V} (D) and D. Since F ≠ 1 and F ⊴ G, F ∉ D. Hence E = F D and [E, S] = [D, S] ≤ D. It follows that [E, S] ≤ ⋂_{D∈calD} D, contradicting the minimal choice of E.

F F MP

Lemma 6.3 Let H be finite group such that the Sylow subgroup is contained in a unique maximal subgroup of H. Let V be a faithful GF (2) FF-module for H. Then H has a normal subgroup L = L_{1} × L_{2} × … × L_{k} such that

(a) L_{i} ≅ SL_{2}(q) or Sym(q + 1), q power of 2.

(b) Put \overline{V} = V/C_{V} (L) and V_{i} = [V, L_{i}]. Then \overline{V} = \overline{V_{1}} ⊕ \overline{V_{2}} ⊕ … ⊕ \overline{V_{k}} and \overline{V_{i}} is a natural SL_{2}(q)-module for \overline{L_{i}}.

(c) H = LS and S transitively permutes the L_{i}’s.

F ∗ 2

Lemma 6.4 Let H be finite simple group such that the Sylow subgroup is contained in a unique maximal subgroup of H. Let V be a faithful faithful GF (2) F^{∗}2-module for H. Then either V is an F F -module or H has a normal subgroup L = L_{1} × L_{2} × … × L_{k} such that Remark: maybe we should do all F2 modules, even the non-quadratic ones

(a) L_{i} ≅ Alt(q + 1), SL_{3}(q) or O_{4}^{±}(q), q a power of two.

(b) Put \overline{V} = V/C_{V} (L) and V_{i} = [V, L_{i}]. Then \overline{V} = \overline{V_{1}} ⊕ \overline{V_{2}} ⊕ … ⊕ \overline{V_{k}} and either L_{i} ≅ Alt(q + 1) and |\overline{V_{i}}| is natural module or L_{i} ≅ SL_{3}(q) and \overline{V_{i}} is the direct sum of a natural module and its dual.

(c) H = LS and S transitively permutes the L_{i}’s.

(d) If L_{i} ≅ SL_{3}(q), then some element of N_{H}(L_{i}) induces a graph automorphism on L_{i}. dtendec

Definition 6.5 Let K be a field, H a group and V a KH-module. Then a tensor decomposition of V for H is a tuple (F, V_{i}, i ∈ I) such that

(a) F ≤ End_{K}(V) is a field with K ≤ F.

(b) H acts F -semilinear on V.

(c) Put E = C_{H}(F) ( the largest subgroup of H acting F -linear on V). Then V_{i} is an F E-promodule.

(d) As F E-modules, V and ⨂_{F} {V_{i} ∈ I} are isomorphic.

qtp

Lemma 6.6 Let Q be a group with |Q| ≥ 3. 1 ≠ Z ≤ Z(Q), K a field with charK = p, p a prime, V a faithful KQ-module with [V, Z, Q] = 0 and (F, V_{i}, i ∈ I) a tensor decomposition of V for Q. Then Q acts F -linear and one of the follwing holds:

1. There exists i ∈ I so that [V_{i}, Z, Q] = 0 and Q acts trivially on all other V_{j}’s.

2. p = 2, Q is F -linear and there exist i, j ∈ I, a_{k} ∈ End_{F} (V_{k}) with a_{k}^{2} = 0 (k=i,j)and a monomorphism λ : Q → (F, +) so for q ∈ Q,

(a) For k = i, j, q acts on V_{k} as 1 + λ(q)a_{i}.

(b) Q centralizes all V_{s}’s with s ≠ i, j.

Proof: Note first that as Z acts quadratically on V, Z is an elementary abelian p-group. Also [V, Z, Q] = 0 and [Q, Z] = 1. So the three subgroup lemma implies that [V, Q, Z] = 1.

Suppose that Q does not act F -linear. Note thet z induces some field automorphism σ on F. Let F_{σ} be the fixed field of σ in F. As z is quadratic on V, f − f^{σ} ∈ F_{σ} for all f ∈ F. It easy to see that this implies F = F_{σ} or p = 2 and F_{σ} has inded two in F. Moreover, [V, z] is an F_{σ}-subspace centralized by Q. So Q is F_{σ} and F_{σ} ≠ F. Since [V, C_{Q}(F)] is an F -spave centralizes by z, C_{Q}(F) = 1. Thus |Q| = 2 in contradcition to the assumptions.

Suppose from now on the Q is F -linear. Since Z is a p-group, we mau assume that the V_{i}’s are actually F Z-modules and not only promodules. If Q acts trivially on some V_{k}, V is a direct sum of copies of the F Q-module ⨂_{F} {V_{i} | i ∈ I − k}. So the latter has the same properties as V. Thus we may assume fom now on that Q acts non-trivially on each V_{i}. If |I| = 1, then 1. holds

Suppose next that |I| = 2 and say I = {1, 2}. Note that

[C_{V_{1}}(Z) ⊗ V_{2}, Z] = C_{V_{1}} ⊗ [V_{2}, Q].

Q acts as scalars on [V_{2}, Z] and [V_{1}, Z]. Hence we may choose the promodules V_{1} and V_{2} so that [V_{i}, Z, Q] = 0 for i = 1, 2. For q ∈ Q let q_{i} be the endomorposim q − 1 of V_{i}. Then z_{i}q_{i} = 0. Moreover, in End_{F} (V_{1} ⊗ V),

z − 1 = (1 + z_{1}) ⊗ (1 + z_{2}) − 1⊗ = z_{1} ⊗ 1 + 1 ⊗ z_{2} + z_{1} ⊗ z_{2}.

Thus [V, z, q] = 0 implies

z_{1} ⊗ q_{2} = −q_{1} ⊗ z_{2}

If z_{1} = 0 then as V is faithful, z_{2} ≠ 0. Thus the previuos equation implies q_{2} = 0 for q, a contradcition to the assumption that Q does not centalize V_{2}. Hence both z_{1} and z_{2} are not zero. Choosing q = z we see that p = 2. Hence for arbitray q, q_{1} = λ(q)z_{1} and q_{2} = λ(q)z_{2} for some λ(q) ∈ F. Thus 2. holds in this case.

Suppose now that |I| ≥ 3. Say 1, 2 ∈ I and but W = ⨂_{F} {V_{i} | i ∈ I \ {1, 2}. Then V ≅ (V_{1} ⊗ V_{2}) × W. Then by the prviuos case Q acts faithfully on V_{1} ⊗ V_{2} z − 1 and q − 1 are linear dependent on V_{1} ⊗ V_{2}. Let λ = λ(q) be as above. Then on v_{1} ⊗ v_{2}

q − 1 = (1 + λz_{1}) ⊗ (1 + λz_{2}) − 1 ⊗ 1 = λ(z_{1} ⊗ 1 + 1 ⊗ z_{2} + λz_{1} ⊗ z_{2}).

On the otherhand z − 1 = z_{1} ⊗ 1 + 1 × z_{2} + z_{1} ⊗ z_{2} and we conclude that λ = 0, 1 and so |Q| = 2, a contradiction.

Definition 6.7 Let H be a finite group, F a finite field, V a finite dimensional F H-module and s a postive real number.

(a)

P_{s}(H, V) = {A ≤ H | |A|^{s}|C_{V} (A)| ≥ |B|^{s}|C_{V} (B)| for all B ≤ A}

.

(b)

P_{s}^{∗}(H, V) = {A ∈ P_{s}(H, V) | |A|^{s}|C_{V} (A)| > |B|^{s}|C_{V} (B)| for all C_{A}(V) < B < A}

.

(c) PQ_{s}(H, V) = {A ∈ P_{s}(H, V) | [V, A, A] = 0

(d) PQ_{s}^{∗}(H, V) = {A ∈ P_{s}^{∗}(H, V) | [V, A, A] = 0

bpgv

Lemma 6.8 Let H be a finite group, F a finite field, V a finite dimensional F H-module, s a postive real number and A ≤ H.

(a) A ≤ P_{s}(H, V) if and only if |W/C_{W} (A)| ≤ |A/C_{A}(W)|^{s} for all W ≤ V.

(b) A ∈ P_{s}^{∗}(H, V) if and only if |V/C_{V} (A)| ≤ |A|^{s} and for each W ≤ A one of the following holds:

1. [W,A]=0.

2. C_{A}(W) = C_{A}(V).

3. |W/C_{W} (A)| < |A/C_{A}(W)|^{s}.

(c) Let A ∈ P_{s}(H, V) and W an F A-submodule in V. Then A ∈ P_{s}(N_{H}(W), W).

(d) Let A ∈ P_{s}^{∗}(H, V) and W an F A-submodule in V. Then A ∈ P_{s}^{∗}(N_{H}(W), W).

Proof: (a) Suppose first that A ∈ P_{s}(H, V) and let W be a F -subspace of V. Let B = C_{A}(W) . Then W ≤ C_{V} (B). Since A ∈ P_{s}(H, V) we have |C_{V} (B)/C_{V} (A)| ≤ |A/B|^{s} and thus

bpgv − 1

(1)

|W/C_{W} (A)| ≤ |C_{V} (B)/C_{V} (A)| ≤ |A/B|^{s} = |A/C_{A}(W)|^{s}.

.

Suppose next that |W/C_{W} (A)| ≤ |A/C_{A}(W)|^{s} for all W ≤ V and let B ≤ A. Put W = C_{V} (B). Then B ≤ C_{A}(W) and

bpgv − 1

(2)

|C_{V} (B)/C_{V} (A)| ≤ |W |/|C_{W} (A)| ≤ |A/C_{A}(W)|^{s} ≤ |A/B|^{s}.

(b) Suppose first that A ∈ P_{s}^{∗}(H, V) and let W be a F -subspace of V. Let B = C_{A}(W) Then W ≤ C_{V} (B). If A = B, then 1. holds. If B = C_{A}(V), then 2. holds. So assume C_{A}(V) < B < A. Then by minimalty of |A| the middle ”≤” in (2) becomes a ”<” and so 3.holds.

Suppose next that |V/C_{V} (A) ≤ |A/C_{A}(V)|^{s} and that 1.,2. or 3. holds for each W ≤ V. Let B < A. Put W = C_{V} (B). If 1. holds then, C_{V} (A) = C_{V} (B) and so clearly |A|^{s}|C_{V} (A)| > |B|^{s}|C_{V} (B)|. If 2. holds then B ≤ C_{A}(V) and so |A|^{s}|C_{V} (A)| ≥ |V ||C_{A}(V)|^{s} ≥ |C_{V} (B)||B|^{s}. If 3. holds then the middle ”≤” in ?? becomes a ”<” and (b) is proved.

Finally (c) follows from (a), and (d) from (c) and (b).

bqpgv

Lemma 6.9 Let H be a finite group, F a finite field V a finite dimensional F H-module and s a postive real number with s ≤ 2. Let A ∈ PQ_{s}(G, V)

(a) Suppose that Δ is a System of imprimitivity for A on V and U ∈ Δ.

(a.a) One of the following holds:

1. A normalizes U.

2. |F | = 2 = |U| and s ≥ 1.

3. |F | ∈ {2, 4}, |U| = 4 and s = 2.

(a.b) If in addition A ∈ P ∗ (H, V) and either (a.a.2) with s = 1 or (a.a.3) holds, then |A| = 2 and A centralizes ∠Δ \ U^{A}⟩.

(b) Suppose that V = ⊗_{i=1}^{n}V_{i} for some F H-module V_{i}, 1 ≤ i ≤ n and that [V_{1}, A] ≠ 0 ≠ [V_{2}, A] and dim_{F} V_{i} > 1. Then n = 2, s = 2, dim_{F} V_{1} = 2 = dim_{F} V_{2}, C_{A}(V_{1}) = C_{A}(V_{2}) = C_{A}(V) and |A/C_{A}(V)| = q.

Proof: (a) Let W = ⟨U^{A}⟩ and suppose that A does not normalize U. Since A acts on W, we get char F =2, [U, N_{A}(U)] = 0 and |U^{A}| = 2. Thus |A/C_{A}(W)| = 2. Hence by 6.8c, W/C_{W} (A) ≤ 2^{s}. Since U ∩ C_{W} (A) = 0 we get |U| ≤ 2^{s} and so 2. or 3. holds. Suppose that A ∈ P∗(G, V) and either 2. with s = 1 or 3. holds. Then |W/C_{W} (A)| = |A/C_{A}(W)|^{s}. Thus by 6.8b, C_{A}(V) = C_{A}(W). Since |V/C_{V} (A)| ≤ |A/C_{A}(W)|^{s} we conclude V = W C_{V} (A) and so (a) is proved.

(b) If |A| ≥ 3, this follows this is an easy consequence of 6.6. If |A| = 2 we get |V/C_{V} (A)| ≤ 2^{s} ≤ 4 and again (b) is easily verified.

lbfs

Lemma 6.10 F a finite field , A a finite group, V a n-dimensional F A-module with [V, A] ≠ 0 = [V, A] and s defined by |V/C_{V} (A)| = |A/C_{V} (A)|^{s}. Then s ≥ \frac{1}{dim_{F} [V,A]} ≤ \frac{1}{n−1}.

Proof: We may assume that A acts faithfully on V. Let m = dim_{F} V/C_{V} (A) and k = dim[V, A]. Then A ≤ |F |^{km} and so

|V/C_{V} (A)| = |F |^{m} ≤ |A|^{s} ≤ |F^{kms}.

Thus m ≤ kms and s ≤ \frac{1}{k} ≤ \frac{1}{n−1}.

s2

Lemma 6.11 Let H be a finite group, p a prime and V an irreducible, faithful GF (p)Hmodule. Let s be a positive integer with s ≤ 2 and L = ⟨PQ_{s}^{∗}(H, V)⟩. Suppose that L ≠ 1 and that L acts irreducible on V. Let A ∈ PQ_{s}^{∗}(H, V) and F = End_{L}(V), then one of the following holds:

1. p = 2, 3, L ≅ SL_{2}(p), |A| = p, |F | = p, dim_{F} V = 2 and s ≥ 1.

2. p = 2, L ≅ Dih(D_{10}, |A| = 2, |F | = 4, dim_{F} V = 2 and s = 2.

3. p = 2, L ≅ SU_{3}(2)′, |A| = 2, |F | = 4,dim_{F} V = 3 and s = 2.

4. p = 2, 3, L ≅ SL_{2}(p) ∗ SL_{2}(p), |A| = p, |F | = p, dim_{F} V = 4 and s = 2.

5. p = 2, L ≅ SL_{2}(F) × SL_{2}(F), |A| = |F |, |F | ≥ 4, dim_{F} V = 4 and s = 2.

6. p = 2, L ≅ O_{+}^{4}(F), |A| ≤ 2|F |, |V/C_{V} (A)| = |F |^{2}, |F | ≥ 4, dim_{F} V = 4 and s ≥ \frac{4}{3}.

7. p = 3, L ∼ Ext_{−}(2^{1+4}).Alt(5), |A| = 3, |F | = 3, dim_{F} V = 4 and s = 2.

8. p = 2, L ≅ Sym(5) or Sym(3) ∧^{|}Sym(5), |A| = 2 or A ≤ L′, F | = 2, dim_{F} V = 4,

s = 2 and |End_{L′}(V)| = 4.

9. p = 2, s = 2, F ≤ 4. There exists a system of imprimitivity Δ for L on V with L/C_{L}(Δ) = Sym(Δ). Let U ∈ Δ, then |U| = 4. If A ≤ C_{L}(Δ) then |A| = 2. C_{L}(Δ) is a Sym(Δ) invariant subgroup of Sym(3)^{Δ}. If |F | = 2 then C_{L}(Δ) induces Sym(3) on U and if |F | = 4 then C_{L}(Δ) induces C_{3} on U.

10. Let K = E(L). Then K is quasi simple, K acts irreducible on V, F = End_{K}(V). Moreover, L acts primitively and tensor indecomposable on V.

11. s > 1. There exists a central extension L^{∗} so that V ≅ V_{1} ⊗ V_{2} for some faithful F L^{∗} modules V_{1} and V_{2}. Let {i, j} = {1, 2}, P_{i} = {A ∈ PQ_{s}^{∗}(H, V) | [V_{j}, A] = 0} and L_{i} = ⟨P_{i}⟩. Then PQ_{s}^{∗}(H, V) = P_{1} ∪ P_{2}, L = L_{1}L_{2} and [L_{1}, L_{2}] = 1. Let K_{i} = E(L_{i}) Then V_{i} is an irreducible F K_{i} module module and F = End_{K_{i}}(V_{i}). P_{i} ∈ PQ∗_{\frac{s}{n_{j}}} (L_{i}, V_{i}).

Let A_{i} ∈ P_{i}, n_{i} = dim F V_{i} and and let s_{i} be defined by |V_{i}/C_{V_{i}}(A_{i})| = |A_{i}|^{s_{i}}. Then s_{i} ≤ \frac{s^{2}}{n_{i}+s} ≤ \frac{4}{n_{i}+2} and \frac{n_{j}}{s} + 1 ≤ n_{i} ≤ s(n_{j} − 1).

Proof: We will first prove:

s2 − 2

(1) Suppose V can be regarded as a vector space over a field F so that L acts F -semilinear but not F -linear on V. Then |A| = p = 2, |F | = 4 or 16, |V | = 4 or 16 and L is one of Dih(6), Dih(10), Sym(3) × Sym(3), Sym(5) or Sym(3) ∧^{|}Sym(5). Moreover if s ≠ 2, then s ≥ 1, |F | = |V | = 4 and L ≅ Sym(3).

Choose A ∈ PQ^{∗}(L, V) which does not act F -linear on A. Since A acts quadratically on V we conclude that |A| = 2. Moreover, |V | = |V/C_{V} (A)|^{2} ≤ |A|^{2s} = 2^{2s} ≤ 16. Thus L ≤ ΓGL_{2}(4). (1) now follows by inspecting the irreducible subgroups of ΓGL_{2}(4) ≅ Sym(3) ∧^{|}Sym(5) generated by involutions.

s2 − 3

(2) Suppose there exist a central extension L^{∗} of L, a field F and F L^{∗}-moduln V_{1} and V_{2} so that V ≅ V_{1} ⊗_{F} V_{2} as GF (p)L^{∗} modules. Then one of the following holds:

1. s = 2, p = 2, dim_{F} V_{i} = 2, |A| = |F | for all A ∈ PQ_{s}^{∗}(L, V) and L ≅ SL_{2}(F)×SL_{2}(F)

2. s > 1. Let {i, j} = {1, 2}, P_{i} = {A ∈ PQ_{s}^{∗}(H, V) | [V_{j}, A] = 0} and L_{i} = ⟨P_{i}⟩.

Then PQ_{s}^{∗}(H, V) = P_{1} ∪ P_{2}, L = L_{1}L_{2} and [L_{1}, L_{2}] = 1. P_{i} ∈ PQ∗_{\frac{s}{n_{j}}} (L_{i}, V_{i}). Let

A_{i} ∈ P_{i}, n_{i} = dim F V_{i} and and let s_{i} be defined by |V_{i}/C_{V_{i}}(A_{i})| = |A_{i}|^{s_{i}}. Then s_{i} ≤ \frac{s^{2}}{n_{i}+s} ≤ \frac{4}{n_{i}+2} and \frac{n_{j}}{s} + 1 ≤ n_{i} ≤ s(n_{j} − 1).

Suppose first that there exists A ∈ PQ_{s}^{∗}(H, V) with [V_{1}, A] ≠ 0 ≠ [V_{2}, A]. Using 6.9b it is then easy to see that refs2-31. holds. So suppose that no such A exists. Then clearly PQ_{s}^{∗}(H, V) = P_{1} ∪ P_{2}, L = L_{1}L_{2} and [L_{1}, L_{2}] = 1.

Note that V is as an L_{i} module the direct sum of n_{j} copies of V_{i}. Hence for all B ≤ L_{i},

|C_{V} (B)| = |C_{V_{1}}(B)|^{n_{j}} and so (|B|^{\frac{s}{n_{j}}} |C_{V_{1}}(B)|)^{n_{j}} = |B|^{s}|C_{V} (B)|. Thus P_{i} ∈ PQ∗_{\frac{s}{n_{j}}} (L_{i}, V_{i}).

Moreover, we see that s_{i}n_{j} ≤ s. Thus s_{i} ≤ \frac{s}{n_{j}}. By 6.10 we have s_{i} > \frac{1}{n_{i}−1} and so \frac{s}{n_{j}} ≥ s_{i} ≥ \frac{1}{n_{i}−1} and thus n_{i} ≥ \frac{n_{j}}{s} + 1. Hence also n_{j} ≥ \frac{n_{i}}{s} + 1 = \frac{n_{i}+s}{s}. Therfore s_{i}(\frac{n_{i}+s}{s}) ≤ s_{i}n_{j} ≤ s and s_{i} ≤ \frac{s^{2}}{n_{i}+s}. Hence refs2-32 holds.

s2 − 4

(3) If V is tensor-decomposable as L-module, then 4.,5. or 11. holds.

In case (2)1, 4. or 5. holds. So suppose (2)2. holds. Since P_{i} ≤ PQ_{\frac{s}{n_{i}}} (L_{i}, V_{i}) can imply induction to (L_{i}, V_{i}). Moreover, either \frac{s}{n_{i}} < 1 or \frac{s}{n_{i}} = 1 and n_{i} = 2. If n_{i} = 2, then s_{i} = 1 and s_{i}n_{j} ≤ s implies n_{j} = 2. It follows that 4. or 11 holds in this case.

We may and do assume form now on that V is tensor indecomopsable.

Suppose that L acts irreducible but does not primitively on V and let Δ be a system of imprimitivity for L on V. Since L acts irreducble on V, L acts transitively on Δ. Thus there exists U ∈ Δ and 1 ≠ A ∈ PQ_{s}^{∗}(H, V) so that A does not normalizes U. If |U| = 2, L centralizes the sum of the non-zero elements in ⋃ Δ, a contradiction to the irreducible action of L. Hence by 6.9a we conclude that |U| = 4, s = 2, |A| = 2 and A centralizes ⟨Δ \ U^{A}. In particular, A acts a 2-cycle on Δ and we conclude that L/C_{L}(Δ) = Sym(Δ).

Thus

s2 − 1

(4) If L acts irreducible but not primitively on V, then p = 2, s = 2 and L is a subgroup of SL_{2}(2) ≀ Sym(n), where n = dim V/2.

Suppose next that L acts irreducible and primitively on V.

Let K be a normal subgroup of L minimal with respect to [K, L] ≠ 1. As L acts primitively, V is a as K-module isomorphic to the direct sum of isomorphix irreducible GF (p)Kmodules. In particular KC_{GL(V)}(K) acts irreducible on V and so F =^{def} End_{KC_{GL(V )}(K)}(V) is a field. By (1) we may assume that L acts F -linear on V. As V is tensor indecoposable we conclude that K acts irreducible on V. If K is cyclic, we conclude that V is 1-dimensional over F and so L is cyclic, a contradicion, since O_{p}(H) = 1. Thus K is not cyclic and we may assume that all cylic normal subgroup of L are contained in Z(L). In particular C_{L}(K) ≤ Z(L).

Assume that K is a q-group for be a prime q. Then q ≠ p. Pick A ∈ PQ^{∗}(L, V) with [K, A] ≠ 1. Then p = 2 or 3. Moreover, [K, A] ≰ Z(K) and so 1 ≠ [A, K, K] ≤ Z(L).

Suppose that p = 2, then by 6.2 and the irreducible action of K, A is cyclic. But then |A| = 2 and so |[V, A]| = |[V/C_{V} (A)| = 2^{r} ≤ 2^{s} ≤ 4 for some integer r ≤ s ≤ 2. Hence there exist 1 ≠ k ∈ [A, K, K] with |V | = |[V, k]| ≤ 2^{4r}. Also note that since Z(K) ≠ 1, |F | ≥ 4 and so dim_{F} V ≤ 2r. Since K is non-abelian and acts irreducible on V, we conclude that r = 2 and

s2 − 5

(5) |A| = 2 = p, s = 2, K ≅ Ext(3^{1+2}), |V | = 2^{6}, and L = KA ≅ SU_{3}(2)′

Suppose next that p = 3. Then q = 2 and [K, A] is extraspecial. If A is not cyclic we obtain a contradiction to 6.9b applied to an irreducible submodule for [K, A]A in V. Hence A is cyclic and similarly [K, A] ≅ Q_{8}. Moreover |C_{V} (A)|^{2} = |V | and so |V | ≤ 3^{2s} ≤ 3^{4}. As L is irreducible and tensor indecomospable on V one of the following holds:

s2 − 6

(6) 1. |A| = p = 3, s ≥ 1, |V | = 3^{2} and L ≅ SL_{2}(3).

2. |A| = p = 3, s = 2, |V | = 3^{4} and L ∼ Ext_{−}(2^{1+4}).Alt(5).

Suppose next that K is not nilpotent. Then K = E(K) and L acts transitively on the components of L.

Assume that K is not quasisimple. Then there exist a component R of K and A ∈ PQ_{2}^{∗}(L, V) so that A does not normalize R. Since A acts quadratically this implies p = 2, R ≅ SL_{2}(F) and |R^{A}| = 2. Moreover, using 6.9b we get:

s2 − 7

(7) Put q = |F |. Then p = 2, s ≥ \frac{4}{3}, q > 2, |A| ≤ 2q, dim_{F} V = 4, |V/C_{V} (A)| = q^{2}, and L ≅ Ω_{4}^{+}(F) ∼ SL_{2}(F) × SL_{2}(F) : 2.

Assume finally that K is quasi simple. Then

s2 − 8

(8) K = E(L) is quasi simple, C_{L}(K) = Z(L), L acts irreducibly, primitively, tensor indecompsable and F -linear on V.

Lieq

Lemma 6.12 F 2-modules for groups of Lie type and maybe also the non-quadratic F 2modules

P GSym

Lemma 6.13 Let Ω be a finite set, G = Sym(Ω), and V (Ω) = GF (2)[Ω] the natural permutation module GF (2)G-permutaion module. Define V_{O}(Omega) = [V (Ω), G], \overline{V (Ω)} = V (Ω)/C_{V (Ω)}(G) and \overline{V_{0}(Ω)} = V_{0}(Ω)/C_{V_{0}(Ω)}(G). Let V be one of the modules, V (Ω), V_{0}(Ω), \overline{V (Ω)} and \overline{V_{O}(Ω)}.

(a) Let A be a non-trivial elementary abelian subgroup of G with |V/C_{V} (A)| ≥ |A|. Then there exists commuting transpositions t_{1}, t_{2}, … t_{k} so that one of the following holds

1. A = ⟨t_{1}, t_{2}, …, t_{k}⟩.

2. |Ω| = 2k, V = V_{0}(Ω) or \overline{V_{O}(Ω)} and A = ⟨t_{1}t_{2}, t_{2}t_{3}, …, t_{i−1}t_{i}, t_{i+1}, t_{i+2}, …, t_{k}⟩,

where 1 ≤ i ≤ k.

3. |Ω| = 2k + 4, V = V_{0}(Ω) or \overline{V_{O}(Ω)} and A = ⟨t_{1}, t_{2}, …, t_{k}, (ab)(cd), (ac)(bd)⟩,

where a, b, c, d are the four common fixed points of t_{1}, …, t_{k}.

4. |Ω| = 4|, V = \overline{V (Ω)} and A ≤ Alt(Ω).

5. |Ω| = 8, V = \overline{V_{O}(Ω)}, |A| = 8 and A acts regularly on Ω.

(b) Suppose |Ω| ≠ 8 and let H ≤ G with H = ⟨P(H, V). Let Ψ an orbit for H on Ω. Then one of the following holds:

1. H/C_{H}(Ψ) = Sym(Ψ).

2. H/C_{H}(Ψ) = Alt(Ψ).

3. |Ψ| is even and H/C_{H}(Ψ) = N_{Sym(Ψ)}(Δ) ≅ C_{2} ≀ Sym(|P si|/2), where Δ is a partion of Ψ into sets of size 2.

4. |Ψ| = 4 and H/C_{H}(Ψ) ≅ E_{4}.

5. |Ψ| = 6 and H/C_{H}(Ψ) ≅ Alt(5).

6. |Ψ| = 8 and H/C_{H}(Ψ) ∼ 2^{3} : L_{3}(2).

Proof: (a) By induction on |A|, V and |Ω|. Suppose that A ∉ P(G, V) and let 1 ≠ B ≤ A with B ∈ P(A, V) with |B||C_{V} (B)| > |A||C_{A}(V)| ≤ |V |. Then by induction Ω = 2k and B = ⟨t_{1}, t_{2}, …, t_{k}⟩. But then A ≤ C_{G}(B) = B and so A = B, a contradiction.

Hence A ∈ P(G, V). Let B = C_{V} ([V, A]). Then 1 ≠ B ∈ P(G, V). Suppose B ≠ A and apply (a) to B. In case (a3) A ≤ C_{G}(B) ≤ A, a contradiction. In case (a1) and (a2), C_{G}(B) = ⟨t_{1}, t_{2}, … t_{k}⟩ × Sym(Ω′). If |Ω| = 2k, then C_{G}(B) acts quadratically on V, a contradiction to A ≠ B. Thus |Ω| ≠ 2k and A = B × D, where D = B ∩ Sym(Ω′). We may view V_{O}(Ω′) as a subspace of V. Then A ≤ P(A, V_{O}(Ω′) and so D ∈ P(Sym(Ω′, V_{O}(Ω_{′}). In particular we can apply (a) to D. Since C_{D}([V, A]) = 1 we get that C_{D}(V (Ω′) = 1. But this implies that (a3) with k = 0 holds for D on V_{O}(Ω′). Thus also (a3) holds for A on V.

So we may assume that [V, A, A] = 0. Suppose that A has an orbit of length larger then four on Ω. If |Ω| = 4, (a3) or (a4) holds. So assume |Ω| > 4. If A has an orbit of lenght less then four on Ω then [V_{Ω}, A, A] has an element of lenght four, a contradiction to [V, A, A] = 0. Thus all orbits of A have length at least four. Moreover, [V (Ω), A, A] has an element of lenght four and [V_{Ω}, A, A] has an element of length eight. We conclude that |Ω| = 8 and V = \overline{V_{0}(Ω)}. If A has an orbit of lenght eigth on Ω, (a5) holds. So suppose that A has two orbits of length four. If 1 ≠ a ∈ A acts trivially on on of the orbits of A on Ω, then [V, a, A ≠ 0. Thus |A| = 4, but |V/C_{V} (A)| = 8, a contradiction.

Hence we may assume that all the orbits of A on V have length at most 2. If A has a fixed point on Ω we are done by induction. Hence we may assume that A acts fixed point freely on Ω. Suppose that there exists v ∈ V (Ω) with 0 ≠ [v, A] ≤ C_{V (Ω)}(G). Then it os easy to see that C_{A}(v) = 1 and so |A| = 2 and |Ω| = 2. So we may assume that no such v exists. Hence |V/C_{V} (A)| ≥ 2^{k−1}, where k = Ω|/2 and thus |A| ≥ 2^{k−1} and (a2) holds.

(b) Let A ∈ P(H, V) so that A does not act trivially on Ψ.

Suppose first that some element of H induces a transposition on Ψ. If H acts primitively on Ψ, (b1) holds. So suppose that Δ is a system of imprimitivity for H on Ψ. Since A is generated by elements of support less or equal to four, we conclude that elements of Δ have size two and A on its action on Δ is generated by transopsition. As H acts transitively on Δ, H/C_{H}(Δ) = Sym(Δ). Moreover, all the transposition in H act trivially on Δ and so C_{Sym(Ψ)}(Δ) ≤ H/C_{H}(Ψ) and (b3) holds.

So suppose that no element of H induces a transposition on Ψ.If A fulfils (a3) or (a4) then |Ψ| = 4 and (b4) holds.

So we may assume that A fulfils (a2). Then Ψ = Supp(⟨t_{1}, t_{2}, … t_{k}⟩ and we may assume without loss that Ψ = Ω = {1, …, 2k} and t_{i} = (2i − 1, 2i). It is easy to see that k ≥ 3. Suppose that Δ is a system of imprimitivity for H on Ψ and without loss that A acts non trivially on Δ. Let D ∈ Δ. Then |D| = 2 and say D = {1, 3}. Then |D^{t_{1}t_{3}} ∩ D| = 1, a contradiction.

Thus A acts primitively in Ψ. Hence if H contains a 3-cycle, (b2) holds. So we may assume that H contains no three cycle. Let A^{∗} ∈ P(H, V) with A ≠ A^{∗} and so that A^{∗} does not normalize A. Let a ∈ A and a ∈ A^{∗} with |Supp(a)| = |Supp(a^{∗})| = 4 and A ≠ A^{a^{∗}}. If |Supp(a) ∩ Supp(a^{∗})| = 1, then (aa^{∗})^{2} is a three cycle, a contradiction. Hence |Supp(a) ∩ Supp(a^{∗})| ≠ 3, for all such a and a^{∗}.

Suppose a^{∗} = (1, 2)(3, 5). Then (12)(34)a^{∗} is a three cycles, a contradiction.

Suppose that a^{∗} = (1, 3)(2, 5). If k ≥ 4 we obtain a contradiction by choosing a = (34)(78). Thus k = 3, A^{∗} = ⟨(1, 3)(2, 5), (1, 3)(4, 6)⟩ and ⟨A, A^{∗}⟩ ≅ Alt(5). It follows that H = ⟨A, A^{∗}⟩ and (b5) holds.

Up to conjugation under N_{Sym(Ψ)}(A) we now may assume that a^{∗} = (1, 3)(5, 7). If n ≤ 5 we obtain a contradiction by choosing a = (1, 2)(9, 10). Thus k = 4. By the previous case neither (13)(26) nor (13)(28) can be in A^{∗} and we conclude that the orbits of A^{∗} on Ψ are 13, 24, 57 and 68. In particular, A and A^{∗} normalize {1, 2, 3, 4} and ⟨A, A^{∗}⟩ ∼ 2^{4}Sym(3). It is now readily verified that (b6) holds.

LP GV

Lemma 6.14 Let G be a finite group with F^{∗}(G) quasisimple. Let V be a faithful GF (p)Gmodule and A a G invariant subset of P(G, V). Let S ∈ Syl_{p}(G) and put J = J_{A}(S) = ⟩A ∩ S⟨. L ≤ G with L = N_{G}(O_{p}(L) and J ≤ L and suppose that K is p-component of L so that J does not normalize K. Then p = 2, ⟨A⟩ ≅ O_{2n}^{+}(2^{k}), n ≥ 3, k ≥ 2 and K/O_{2}(K) ≅ SL_{2}(2^{k}) all non-trivial composition factors for ⟨A⟩ on V are natural O_{2n}^{+}(2^{k})- modules. In particular, if n = 3, then P(O_{p}(L), V) = 1 .

Remark: If n > 3, then it can be shown that K is not subnormal in C_{G}(C_{V} (S), where S ∈ Syl_{p}(L).

Proof: Let H = F^{∗}(G). We may assume without loss that H centralizes all proper G-submodules in V. That is V = [V, H] and G actss irreducible on V/C_{V} (H). In particular by the Three Subgroup Lemma, O_{p}(G) = 1.

If p = 2 and H/Z(H) is an alternating group we obtain a contradiction from 6.13. So we may assume that:

LP GV − 1

(1) H is a group of Lie type in characteristic p.

We may assume without loss that H centralizes all proper G-submodules in V. That is V = [V, H] and G acts irreducibly on V/C_{V} (H). In particular by the Three Subgroup Lemma, O_{p}(G) = 1.

If O_{2}(L) ∩ H = 1, then [O_{2}(L), K] = 1 and so by the P × Q-lemma, [C_{V} (O_{2}(L), K] ≠ 1. But L ∩ A ⊆ P(L, C_{V} (O_{2}(L)) and K maps onto a component of L/C_{L}(C_{V} (O_{2}(L)), a contradiction.

Hence O_{2}(L) ∩ H ≠ 1. Let M = N_{G}(O_{2}(L) ∩ H). Then L ≤ M and N_{O_{2}(M)}(O_{2}(L)) ≤ O_{2}(L) and so O_{2}(M) ≤ O_{2}(L). Hence O_{2}(M) ∩ H = O_{2}(L) ∩ H and M ∩ H is a parabolic subgroup of H. We have proved:

LP GV − 3

(2) There exists a parabolic subgroup M of G with L ≤ M and O_{2}(M) ∩ H = O_{2}(L) ∩ H.

It follows immediately from (2) that

LP GV − 4

(3) H has rank at least three.

Note that C_{V} (H) = 0 unless H ≅ Sp_{2n}(q) and V/C_{V} (H) is a natural Sp_{2n}(q)-module. In which case we have C_{V} (X)C_{V} (H)/C_{V} (H) = C_{V/C_{V} (H)}(X) and so P(G, V) ⊂ P(G, V/C_{V} (H)). Hence we may assume without loss that C_{V} (H) = 0 and so V is irreducible as G-module.

LP GV − 2

(4) One of the following holds

1. ⟨A⟩ = H

2. p = 2, ⟨A⟩ =≅ O_{2n}^{±}(2^{k}), n ≥ 3 and V is a natural Ω_{2n}^{±}(2^{k}) module for H.

Let P ∈ ∩P (G, S) so that [C_{V} (O_{2}(P)), O^{2}(P)] ≠ 1. Then J induces inner automorphisms on Head(P) and (4) follows from the structure of P and V.

Suppose that O_{2}(M) = O_{2}(L). Then L = M is a parabolic of G and so the p−componets of L are normal in H ∩ L. Using (4),we conclude that the lemma holds. So we may assume that

LP GV − 5

(5) O_{2}(M) ≠ O_{2}(L) and O_{2}(L) ≰ H.

Note that [O_{2}(L), L ∩ H] ≤ O_{2}(L) ∩ H ≤ O_{2}(M) and so L/O_{2}(M) = C_{M/O_{2}(M)}(O_{2}(L)). In particular, [J ∩ H, O_{2}(L) ≤ O_{2}(M). Without loss S ≤ M and S ∩ L ≤ Syl_{p}(L). Since J ≰ O_{2}(M) there exists P ∈ P(M, S) with J ⋬ P. Then J ≰ O_{2}(P) and [J ∩ H, O_{2}(L) ≤ O_{2}(P). Let \overline{P} = P/O_{2}(P)

Suppose that J ≤ H. Then N_{P} (S∩H) normalizes J and we conclude that Z(\overline{S ∩ P)} ≤ \overline{J}, or p = 2 and \overline{P} ≅ Sym(3) ≀ C_{2}. As O_{2}(L) centralizes \overline{J} and O_{2}(L) ≰ H one of the following has to hold

LP GV − 6

(6) 1. p = 2, H ≅ SL_{n}(q), O_{2}(L) induces a graph automorphism on H and \overline{P ∩ H} ≅ L_{2}(q) or SL_{3}(q)

2. p = 2 H ≅ SU_{n}(q), O_{2}(L) induces a field automorphism of order two on H and \overline{P ∩ H} ≅ L_{2}(q) or SU_{3}(q)

3. p = 2 and O_{2}(L)H ≅ O2n^{±}(q).

4. p = 2 and G = O_{2}(L)H = Aut(L_{n}(2)).

In case (6)1 or (6)2, P is uniquely determined. Let R be the maximal parabolic of M with P ≰ M. Then we conclude that J ⊴ R and so [J, [R, O_{2}(L)] ≤ O_{2}(M). By the structure of M this implies J ≤ O_{2}(M), a contradiction. In case (6)3 it is easy to see that L is the normalizer of a non-singular isotropic space and so all p-components of L are normal in L. In case (4), since J does not normalize K and J ≤ H, M most have parbolic E with E/O_{2}(E) ≅ L_{3}(2)≀C_{2} and J ≰ O_{2}(E). Let T be a 2-componet of E. As [J, O_{2}(L)] ≤ O_{2}(E) and O_{2}(E) does not normalizes T, T ∩J ≤ O_{2}(E). Hence [T ∩S, J] ≤ O_{2}(E) and J i normal in both minimal parabilocs of E, a contradiction.

We have proved:

LP GV − 7

(7) J ∉ H, p = 2 and JH ≅ O_{2n}^{±}(q).

If O_{2}(L) ≤ JH we are done by the argument in (6)3 we are done. So suppose O_{2}(L) ≰ JH. Then O_{2}(L) induces field automorphisms on H and on Head(P). In particular q ≥ 2. If J ≤ HO_{2}(P), we get that \overline{S ∩ P)} = \overline{J ∩ H}, a contradiction. Thus J ≰ HO_{2}(P) and so P is uniquely determined. But now the argument in (6)1&2 yields a contradiction.

qusp

Lemma 6.15 Let H be a finite group such that L = F^{∗}(H) is quasi simple but neither a group of Lie type in charcateristic 2 nor an alternating group. Let V be a faithful irreducible GF (2)H-module and 1 ≠ A ≤ G with [V, A, A] = 1 and let B be a maximal quadratic subgroup of H containing A. Moreover assume that there exists at least one fours group in H acting quadratically on V.

(a) One of the following holds.

Remark: Information should be written down more clearly

1. L ≅ Mat_{12} and V is 10-dimensional.

1.1. |B| = 4, A ≤ L, N_{L}(A) ∼ 2^{5}.Sym(3) ∼ N_{L}(B), [V, B] = C_{V} (B) is 5-

dimensional and either 1.1.1. A = B 1.1.2. |A| = 2 and [V, A] is 4-dimensional.

1.2. |B| = 4, B ≰ L, N_{L}(B) ∼ C_{2} × Sym(5), C_{V} (B) = [V, B] is 5-dimensional

and either 1.2.1. A ≰ L and C_{V} (A) = C_{V} (B) = [V, B] = [V, A] 1.2.2. A = B ∩ L and [V, A] is 4-dimensional.

2. L ≅ 3.Mat_{22} and V is 12-dimensional.

2.1. |A| = 2, A ≤ L and [V, A] is 4-dimensional. 2.2. |A| = |B| = 2, |A| ≰ L and [V, A] = C_{V} (A) is 6-dimensional. 2.3. |A| ≥ 4, |B| = 8, B ≤ L, N_{L}(B) ∼ C_{3} × 2^{3}.L_{3}(2) and C_{V} (A) = C_{V} (B) =

[V, B] = [V, A] is 6-dimensional. 2.4. |A| ≥ 4, |B| = 16, B ≤ L, N_{L}(B) ∼ 2^{4} : 3.Alt(6) and C_{V} (A) = C_{V} (B) =

[V, B] = [V, A] is 6-dimensional.

3. L ≅ Mat_{22} and V is 10 dimensional.

3.1. |A| = |B ∩ L| = 2 and [V, A] is 4-dimensional. 3.2. |A| = 2, |B| = 4, A ≰ L, C_{L}(A) ∼ 2^{3}.L_{3}(2) and [V, A] is 3-dimensional. 3.3. |A| = |B| = 4, A ≰ L, N_{L}(A) = N_{L}(A ∩ L) and C_{V} (A) = C_{V} (B) = [V, B] =

[V, A] is 5-dimensional.

4. H ≅ Mat_{24} and V is 11-dimensional.

4.1. |A| = 2, |B| = 4, N_{G}(A) ∼ 2^{1+3+¯3}.L_{3}(2) and [V, A] is 4-dimensional. 4.2. |A| = |B| = 4, N_{G}(A) ∼ 2^{8}.(Sym(3) × Sym(3)) ≤ 2^{6} : (Sym(3) × L_{3}(2))

and either

V is the Golay code module and C_{V} (A) = [V, A] is 6-dimensional or V is the Todd module and C_{V} (A) = [V, A] is 5-dimensional

4.3. |A| ≤ 4, |B| = 4, N_{L}(A) ≤ N_{L}(B) ∼ 2^{2+4} : 3 : Sym(5) ≤ 2^{6} : 3.Sym(6) and either V is the Golay code module and C_{V} (A) = C_{V} (B) = [V, B] is 6-dimensional or V is the Todd module and [V, A] = C_{V} (B) = [V, B] is 5-dimensional

5. L ≅ 3.U_{4}(3), V is 12-dimensional.

5.1. |A| = 2, A ≤ L and [V, A] is 4-dimensional.

5.2. |A| = |B| = 2, A inverts Z(L) and [V, A] = C_{V} (A) is 6-dimensional. 5.3. |A| = 2, A ≰ L, C_{L}(A) ≅ C_{3} × U_{4}(2) and |[V, A]| = 4. 5.4. |A| = 2, A ≰ L, |B| = 2^{5} and C_{V} (A) = [V, A] = C_{V} (B) = [V, B] is 6-

dimensional and C_{L}(A) ∼ 2^{4}(Sym(3) × Sym(3). 5.5. |B ∩ L| = 16, N_{L}(B) ∼ 2^{4} : 3.Alt(6) and either

C_{V} (A) = [V, A] = C_{V} (B) = [V, B] is 6-dimensional or |A| = 4, |A ∩ L| = 2 and [V, A] = [V, A ∩ L] is 4 dimensional.

6. L ≅ J_{2} and V is 12-dimensional. 6.1. |A| = 2, |B| = 4, N_{L}(A) ∼ 2^{1+4}Alt(5) and [V, A] is 4-dimensional. 6.2. |A| = |B| = 4, N_{L}(A) ∼ 2^{6}.Sym(3) and [V, A] = C_{V} (A) is 6-dimensional. 6.3. |B| = 4, N_{L}(A) ≤ N_{L}(B) ≅ Alt(4) × Alt(5) and C_{V} (A) = [V, A] = C_{V} (B) =

[V, B] is 6-dimensional. 6.4. |A| = |B| = 2, A ≰ L and [V, A] is 6-dimensional.

7. G ≅ Co_{1} and V is 24-dimensional. 7.1. |A| = 2, |B| = 4, N_{L}(A) ∼ 2^{1+8}Ω_{8}(2) and [V, A] is 8-dimensional. 7.2. |A| = |B| = 4, N_{L}(A) ∼ 2^{14}.Sym(3) × Alt(8) and [V, A] = C_{V} (A) is 12-

dimensional. 7.3. |B| = 4, N_{L}(A) ≤ N_{L}(B) ∼ (Alt(4) × G_{2}(4)).2 and C_{V} (A) = [V, A] =

C_{V} (B) = [V, B] is 12-dimensional. 7.4. |A| = |B| = 2, N_{L}(A) ∼ 2^{11}Aut(M_{12}), and [V, A] is 12-dimensional.

8. G ≅ Co_{2} and V is 22-dimensional. 8.1. |A| = 2, |B| = 4, N_{L}(A) ∼ 2^{1+8}Sp_{6}(2) and [V, A] is 6-dimensional. 8.2. |A| = 2, |B| = 4, N_{L}(A) ∼ 2^{1+4+6}Alt(8) and [V, A] is 8-dimensional. 8.3. |A| = |B| = 4, N_{L}(A) ∼ 2^{15}.L_{3}(2) and [V, A] = C_{V} (A) is 11-dimensional. 8.4. |A| = |B| = 2, N_{L}(A) ∼ 2^{10}Aut(Alt(6)), and [V, A] is 11-dimensional.

9. L ≅ 3.Sz and V is 24-dimensional. 9.1. |A| = 2, |B| = 4, N_{L}(A) ∼ 2^{1+6}Ω_{6}(2) and [V, A] is 8-dimensional. 9.2. |A| = |B| = 4, N_{L}(A) ∼ 2^{14}.Sym(3) × Alt(5) and [V, A] = C_{V} (A) is 12-

dimensional. 9.3. |B| = 4, N_{L}(A) ≤ N_{L}(B) ∼ (Alt(4) × L_{3}(4)).2 and C_{V} (A) = [V, A] =

C_{V} (B) = [V, B] is 12-dimensional. 9.4. |A| = |B| = 2, A ≰ L and [V, A] is 12-dimensional.

(b) Suppose in addition that q ≤ 2, where |A|^{q} = |V/C_{V} (A)|. Let c be the case in (a) fulfilled by A and a = |A|. Then (c, a, q) is one of the following Remark: this doesn’t look very nice

1. (2.3, 8, 2).

2. (2.4, 8, 2) or (2.4, 16, \frac{3}{2}).
3. (5.3, 2, 2).
4. (5.5.1, 8, 2), (5.5.1, 16, \frac{3}{2}) or (5.5.1, 32, \frac{6}{5}).
5. (5.5.2, 4, 2)

Inparticular, L ≅ Mat_{22}, 3.Mat_{22} or 3.U_{4}(3); and q ≥ \frac{3}{2} unless L ≅ 3.U_{4}(3) and

|A| = 32.

Proof: This can be verified using [MS] and [At].

dpromo

Definition 6.16 Let H be a group and F a field. Then an F H promodule for H is a pair (ϕ, V) there V is a vector space over F and ϕ : H → GL_{K}(V) is a map so that the induced map ϕ^{∗} : H → P GL_{K}(V) is a homomorphism.

V ZQ

Lemma 6.17 Let p a prime and H be a finite group p-connected group with O_{p}(H) = 1. Let S ∈ Syl_{p}(H) and Z and Q non-trivial normal subgroups subgroups of S with Z ≤ Z(Q) and |Q| ≥ 3. Let L = O^{p}(H).

(a) Suppose p = 2 and H is a transitive subgroup of Sym(Ω) such that Z acts trivially all Q orbits of size larger than two. Then one of the following holds:

1. The exists a system of blocks D for H on Ω such that (a) If Δ ∈ D, then Q normalizes Δ, Q = ZC_{Q}(Δ) and |Q/C_{Q}(Δ)| = 2.

(b) For Δ ∈ D let L_{Δ} = C_{L}(⋃ D − Δ). Then L = ×_{Δ∈D}L_{Δ}.

2. L ≠ O(L). Let D be the set of orbits of O(H) on |Ω|. Then H/O(H) acts faithfully on H. Let Δ be an orbit for L on D and for X ≤ H let X^{Δ} = N_{X}(Δ)/C_{X}(Δ).Then

(a) Q normalizes Δ.

(b) L^{Δ} = F^{∗}(H^{Δ}) is simple.

(c) 1 ≠ Z^{Δ} ≤ Z(Q^{Δ}), Z^{Δ} and Q^{Δ} are normal in S^{Δ}, S^{Δ} is a Sylow 2-subgroup of H^{Δ}, |Q^{Δ}| ≥ 4, and each orbit for Q^{Δ} on Δ is either centralized by Z^{Δ} or has size at most 2.

(d) One of the following holds:

1. H^{Δ} = Alt(Δ) or Sym(Δ).

2. Δ can be viewed as projective space over the field with two elements so that H^{Δ} = P GL(Δ). Moreover if K is a component of L/O(L), then N_{S}(K) induces only inner autmorphism on K.

3. |Δ| = 6 and H^{Δ} ≅ Alt(5) or Sym(5).

4. |Δ| = 10 and H^{Δ} ≅ Sym(6) or Aut(Alt(6)).

5. |Δ| = 12 and H^{Δ} = Mat_{12} or Δ = 24 and H^{Δ} ≅ Aut(Mat_{12}.

6. |Δ| = 22 and H^{Δ} = Mat_{22} or Aut(Mat_{22}.

7. |Δ| = 24 and H^{Δ} = Mat24. Remark: This needs careful checking

(b) Let K be a field with charK = p and suppose that H is an irreducible subgroup of GL_{K}(V) with [V, Z, Q] = 0. Let W a Wedderburn componet for L on V. For X ≤ H let X^{W} = N_{X}(W/C_{X}(W). Then one of the following holds.

1. p = 2 and there exists a system of blocks D for H on V such that (a) If U ∈ D, then Q normalizes U, Q = ZC_{Q}(U) and |Q/C_{Q}(U| = 2.

(b) For U ∈ D let L_{U} = C_{L}(⋃ D − U). Then L = ×_{U∈D}L_{U}.

2. p = 2 and there exists a system D of H-blocks on V with C_{H}(D) = O(H) and so that the action of H/O(H) on D is described as in (a)2.

3. L = E(L) and (a) Q normalizes W.

(b) L acts irreducible on W.

(c) 1 ≠ Z^{W} ≤ Z(Q^{W}), Z^{W} and Q^{W} are normal in S^{W}, S^{W} is a Sylow 2subgroup of H^{Δ}, |Q^{W} | ≥ 3, [W, Z, Q] = 0 and F^{∗}(H^{W}) = L^{W}.

(d) One of the follwing holds.

1. L^{W} is quasi-simple.

2. p = 2, L^{W} = L_{1}L_{2}, where L_{1}L_{2} are the components of L^{W}. Q normalizes L_{1} and L_{2} and as L^{W} Q^{W} module W = W_{1} ⊗_{F} W_{2} for some faithul F L_{i}Q^{W} modules W_{i}. Moreover Q^{W} acts linear dependently on W_{i}.

3. p = 2, L^{W} Q^{W} ≅ L_{2}(q) ≀ C_{2} and W is the natural Ω_{4}^{+}(q)-module for

L^{W} Q^{W} .

4. One of the following holds:

1. p = 2, L = O_{3}(L), L^{W} ≅ Ext(3^{1+2}),Z^{W} ≅ C_{2}, Q^{W} ≅ C_{4} or Q_{8} and |W | = 2^{6}.

2. p = 3, L = O_{2}(L), L^{W} ≅ Q_{8}, Z^{W} = Q^{W} ≅ C_{3} and |W | = 3^{2}.

5. p ∈ {2, 3}. Let {2, 3} = {p, q} and M = O_{q}(H)^{W} /Z(O_{q}(H)^{W}. Then (a) O_{q}(L)^{W} ≅ Ext(q^{1+2n}) or C_{4} ∘ Ext(2^{1+2n}), n ≥ 2

(b) Z^{W} ≅ C_{p} and Q^{W} ≅ C_{3}, C_{4} or Q_{8}.

(c) L acts irreducible on M.

(d) |[M, Q]| = q^{2}.

(e) O_{q}(H) acts irreducible on W.

(f) Conjecture If p = 2, then L/C_{L}(M) = Sp_{2n}(3) and if p = 3, then L/C_{L}(M) ≅ Ω_{2n}^{±}(2), Alt(2n+1), Alt(2n+2), Sp_{2n}(2) or SU_{n}(2^{2}). Also there are restricions on n from the fact that Q is normal in S.

Proof: (a) The proof is divided into a series of steps

V ZQ − 1

(1) Let Δ be a block for Q on Ω.

(a) One of the following holds:

1. Q normalizes Δ.

2. Z centralizes Δ and so also ⋃ Δ^{Q}.

3. |Δ^{Z}| = |Δ^{Q}| = 2 and N_{Q}(Δ) centralizes Δ and so also ⋃ Δ^{Q}.

(b) One of the following holds:

1. Q normalizes Δ.

2. N_{Z}(Δ) centralizes Δ and so also ⋃ Δ^{Q}.

Clearly (a) implies (b). For (a) suppose that Z does not centralize Δ. If Z normalizes Δ then Z has a non-trivial orbit on Δ and Q has to normalize that orbit. Since Δ is a block, Q normalizes Δ in this case. If Z does not normalize Δ, pick z ∈ Z with Δ ≠ Δ^{z}. Then Δ ∪ Δ^{z} is a union of non-trivial z orbits and so Q normalizes Δ ∪ Delta^{z}. Let ω ∈ Δ. Then N_{Q}(Δ) normalizes Δ ∩ {ω, ω^{z}} = {ω}. Hence 3. holds in this case.

V ZQ − 2

(2) Let Δ be an L-invariant H-block. Then

(a) Ω = ⋃ Δ^{S}.

(b ) Z does not centralize Δ.

(c) If Z normalizes Δ and |Q/C_{Q}(Δ)| = 2, then (a)1. in the lemma holds.

(d) If Q does not normalize Δ, then (a)1. in the lemma holds.

Since H = LS,(a) holds. Since Z ⊴ S, (a) implies (b). If the assumptions of (d) hold, then by (b) and (1)(a), also the assumptions of (c) are with Δ replaced by Δ^{Z}. So it remains to prove (c). By (b) and (1)(a), Q normalizes Δ. Let D = Δ^{H}, Q_{D} = C_{Q}(D) and note that Q/Q_{D} = 2. Let Γ be the union of the blocks in Δ^{H} centralized by Q_{D}. We claim thhat Γ is a H-block. Otherwise there exists s ∈ S with Q_{D}^{s} ≠ Q_{D} and a block in Δ^{H} centralized by Q = Q_{D}Q_{D}^{s}, a contradiction to (b). Hence Γ is a block and replacing Δ by Γ we may assume Γ = Δ. Define L_{Δ} as in (a)1. of the lemma. Let R = ⟨L_{Δ} | Δ ∈ D. Then R is a normal subgroup of H and R = ×_{Δ∈D}L_{Δ}. It remains to show that R = L. Let D = {Δ, Δ_{1}, Δ_{2}, …, Delta_{n}}. Put L_{0} = L and inductively for 1 ≤ i ≤ n, L_{i} = [L_{i}, Q_{Δ_{i}}]. We claim that L = L_{i}C_{L}(Δ). This is obvious for i = 0 Since H is 2 connected, L = [L, Q] and so by induction, L = [L_{i−1}, Q]C_{L}(Δ). Since Q = Q_{Δ}Q_{Δ_{i}} and Q_{Δ} ≤ C_{L}(Δ) we conclude, L = [L_{i−1}, Q_{Δ_{i}}]C_{L}(Δ) = L_{i}C_{L}(Δ). Thus L = L_{n}C_{L}(Δ). But L_{n} ≤ L_{i} for all i and L_{i} centalizes Δ_{i}. Thus L_{n} ≤ L_{Δ} and so L = L_{Δ}C_{L}(Δ) But this clearly implies L = R completing the proof of (2).

V ZQ − 2a

(3) Let F ≤ Q with |F/F ∩ Z| ≤ 2. Then an orbit for F on Ω has length at most for 2. In particular, F is elementary abelian.

Either Z ∩ F acts trivially on a given F -orbit or not. In both cases the orbit has size at most two.

V ZQ − 3

(4) Let P be a subgroup of odd order in H normalizes by Q. Let Δ be an orbit for P Q on Δ such that P acts transitively and Z non-trivially on Δ. Then |Q/C_{Q}(Δ)| = 2.

By the Sylow theorem and the Frattini argument, Q fixes a point ω ∈ Δ. Also P = [P, Q]C_{P} (Q) and replacing P by [P, Q] and Δ by ω^{[P,Q]} we may assume that P = [P, Q]. Let R be a maximal Q invariant normal subgroup of P. If R is transitive on Ω, then by induction on |P |, Z centralizes P. Hence Z/C_{Z}(Δ) acts semiregulary on Δ and all orbits of Z on Ω have size two. Also Q and hence [R, Q] normalizes all orbits of Z. Thus [R, Q] centralizes Δ. Since P = [P, Q], [R, P centalizes Δ and so R/C_{R}(Δ) acts regularly. But then R centralizes Δ, a contradition. So R is not transitive. Let D be the set of orbits for R on Δ. Then the abelian group M =^{def} P/R acts regularly on D and D and and M are ismorphic as Q-sets. Suppose that Z centralizes M, then P = C_{P} (Z)R and M acts non-trivially on each member of D. But then Q normalizes each member of D. Thus Z acts non-trivially on M and D. Similarly, if C_{Q}(M), acts non-trivialy on Δ, Z is forced to act trivially on D. Thus Q/C_{Q}(Delta) acts faithfully on M and D. Let z ∈ Z \ C_{Z}(M). Since z ∈ Z(Q) and Q acts irrducibly on M, z inverts M. Let m ∈ M^{#}. Then Q normalize {m, m^{−1} and as Q is irreducible, M = ⟨m⟩ and |Q/C_{Q}(M) = Q/C_{Q}(Δ)| = 2.

V ZQ − 5

(5) Suppose (a)1. does not hold and let D be the set of orbits for O(H) on Ω. Then H/O(H) acts faithfully on D.

Suppose not. Then since H is 2-connected, L centralizes D .Let Δ ∈ D. By (2), Q normalizes Δ. Also Z acts non-trivially on Δ and O(G) acts transitively. Thus by (4), |Q/C_{Q}(Δ)| = 2 and by (2) (a)1. holds.

We assume form now on that (a)1. does not hold. Replacing Ω by the set of orbits of O(H) on Ω and H by H/O(H) we also may assume that O(H) = 1. Thus L = ×_{i=1}^{m}L_{i} for some non-abelian simple groups L_{i}. Let Δ be an orbit for L on Ω. We wish to whow that a2 holds. a2a and a2c follow from (2). Let M = L^{Δ}. Then M = ×_{i=1}^{n}E_{i}, where {E_{1}, …, E_{n}} consists of whose L_{i}^{Δ}(≅ L_{i}) which act non-trivially on Δ. Suppose for a contradiction that n ≥ 2. Let 1 ≠ z ∈ Z ∩ Z(S). Then z centralizes the Sylow 2-subgroup M ∩ S of M and so z normalizes all L_{i} and E_{i}. If Q does not normalize the componets of M, then |[S ∩ M, Q]| ≥ |S ∩ M_{i}| ≥ 4 and so |M ∩ Q| ≥ 4. So replacing Q by (M ∩ Q)Z in this case, we may assume that Q does normalize the components of M.

Let E = E_{1} and F = C_{M}(E_{1}). Since z ∈ Z(S), E = [E, z]. Suppose that C_{Q}(E)^{Δ} ≠ 1 and pick t ∈ C_{Q}(E)^{Δ} with |t| = 2. Then z normalises all the non-trivial orbits for t on Ω. Since E centralizes t, the same is true for E = [E, t]. But the E = E′ centralizes each non-trivial orbit of t, a contradiction. Thus C_{Q}(E)^{Δ} = 1.

Suppose that E does not act transitively on Δ. Since M acts transitively, M does not normalize any orbit of E. As M = [M, z] there exists an orbit Γ for E on Δ with Γ ≠ Γ^{z}.

Thus by (1), P = C_{Q}(Γ) has index two in Q. But then [E, P] centralizes Δ and so [E, P] = 1 and P^{Δ} ≤ C_{Q}(E)^{Δ} = 1, a contradiction to |Q/P | = 2.

Thus E acts transitively on Δ. By symmetry also F is transitively on Δ and so E is regular. Let F be a group of order four in Q^{Δ} with z^{Δ} ∈ F. Let ω ∈ Δ. Let F = {1, f_{1}, f_{2}, f_{3}} and ω^{f_{i}} = ω^{e_{i}} for some e_{i} ∈ E. Let E_{i} = {e ∈ E | e^{f_{i}} = e_{i}^{−1}}. Note that E_{i} is a coset of the proper subgroup C_{E}(f_{i}) in E. Let e ∈ E. By (3), there exists f_{i} ∈ F with ω^{e} = ω^{ef_{i}} = ω^{f_{i}e^{f_{i}}} = ω^{e_{i}e^{f_{i}}}. As E is regular we get e_{i}e^{f_{i}} = 1 and so e ∈ E_{i}. Thus E = E_{1} ∪ E_{2} ∪ E_{3} is covered three proper cosets. But this implies that E has a subgroup of index two or three, a contradiction as E is non-abelian simple. Thus a2c holds.

To prove a2d we assume without loss that Δ = Ω so L = F^{∗}(H) is simple. Let V = GF (2)Ω be the permutation module associate to Ω. Then [V, Z, Q] = 0 and so V is a faithful GF (2)H-module with a quadratic fours group. Hence by 6.15, L is a group of Lie type in characteristic 2, or L = Mat12, Mat22, Mat24, J_{2}, CO_{1} or Co_{2}. Let 1 ≠ z ∈ Z and R = ⟨Q^{C_{H}(z)}. Then R normalizes all non trivial orbits of z on Ω and [V, z, Q] = 0.

Suppose that L is one of the sporadic groups. Then H has a unique class of 2-central involution. If L is J_{2}, C0_{1} or CO_{2} we get that O_{2}(C_{L}(z)) ≤ R and so V, z, O_{2}(C_{L}(z))] = 1, a contradcition. Hence L = Mat12, Mat22 or Mat_{24}. TO BE CONTINUED

(b) Again we divide the proof into a series of steps and use a similar strategy as in the proof of (a)

V ZQ − 11

(6) Let U be a block for Q on V.

(a) One of the following holds:

1. Q normalizes U.

2. Z centralizes U and so also ∑ U^{Q}.

3. p = 2, |U^{Z}| = |U^{Q}| = 2 and N_{Q}(U) centralizes U and so also ∑ U^{Q}.

(b) One of the following holds:

1. Q normalizes U.

2. p = 2 and N_{Z}(U) centralizes U and so also ∑ U^{Q}.

Clearly (a) implies (b). For (a) suppose that Z does not centralize U. If Z normalizes U, then 0 ≠ [U, Z] ≤ U and Q centralizes [U, Q]. Since U is a block, Q normalizes U in this case. If Z does not normalize U, pick z ∈ Z with U ≠ U^{z}. Since z ∈ Z(Q), U + U^{z} is a block for Q.Also Q centralizes [U, z] and so normalizes U + U^{z}. As a N_{Q}(U) module, U ≅ U + U^{z}/U^{z} = [U, z] + U^{z}/U^{z} ≅ [U, z]/[U, z] ∩ U^{z} and so N_{Q}(U) centralizes U. Hence 3. holds in this case.

V ZQ − 12

(7) Let U be an L-invariant H-block. Then

(a) V = ∑ U^{S}.

(b ) Z does not centralize U.

(c) If Z normalizes U and |Q/C_{Q}(U)| = 2, then (b)1. in the lemma holds.

(d) If Q does not normalize U, then (b)1. in the lemma holds.

The proof is essentially the same as the one for (2).

V ZQ − 13

(8) Suppose exists an H-block which is not L-invariant, then (b1) or (b2) in the lemma holds.

Let calD be a block system for H on V with L acting non-trivially on D and let D be maximal with this property. Then p = 2, C_{H}(D ≤ O(H) and we can apply (a) to H/C_{H}(D) and D. In case (a)1., (b)1. holds. In case of (a).2 the maximality of D implies that O(H) acts trivially on D. Thus (b)2. holds.

We assume from now on without loss that neither (b)1. nor (b)2. hold.

V ZQ − 15

(9) Let W be a Wedderburn component for L on V. Then Q normalizes Q and W is irreducible as L-module.

By (7)d, Q normalizes W. As V is irredicble for H, W is irreucible for N_{H}(L). As W is L-homogenous and N_{H}(L)/L is a p-group, L is irreducible on W.

V ZQ − 16

(10) Suppose that L = E(L). Then (b3) holds.

If Q/C_{Q}(W)| = 2, then (b1) holds. Hence ((b3a),(b3b) and (b3c) holds It remains to verify (b3d). Let L_{1}, L_{2}, … L_{n} be the components of L/C_{L}(W). If n = 1, (b3d1) holds. Put F = End_{K}L(W) and let P the largest subgroup of Q normalizing the components of L^{W}. As in part (a), P^{W} has order at least three and (Z ∩ P)^{W} ≠ 1. Then W has a tensor compostion (F, W_{i}, 1 ≤ i ≤ n), where W_{i} is an C_{LP} (F) module centralized by all L_{j}, j ≠ i. Then by 6.6,p = 2, n = 2 and P^{W} acts linearly dependently on W_{1} and W_{2}. If Q = P, (b3d2) holds. So suppose that |Q/P | = 2 and let q ∈ Q \ P. Note that Q is F -linear. Let 1 ≠ z ∈ P Z. Let U be an irreducible F U subspace in W with U ≠ U^{z}. Then U = W_{1} ⊗ a_{2} for some a_{2} ∈ W_{2}. Also U^{q} is an irreducible F L_{2}P subspace and so U^{q} = a_{1} ⊗ W_{1} for some w_{1} ∈ W_{2}. Similarly U^{z} = b_{1} ⊗ W_{2} and U^{zq} = W_{2} ⊗ W_{1}. Thus (U + U^{z}) ∩ (U + U^{z})^{q} = (F a_{1} + F b_{1}) ⊗ (F a_{2} ⊗ F b_{2}). On the otherhand , q centralizes [U, z] ≤ U + U^{z} and we conclude that dim_{F} U = 2. We conclude that W_{1} and W_{2} are 2-dimensional and by say Dicksson’s theorem, (b3d3) holds.

V ZQ − 14

(11) Suppose that W is tensor decomposable for LQ. Then (b3) holds.

By 6.6, p = 2 and Q is elementary abelian and C_{L^{W}} (q) = C_{L^{W}} (Q for all 1 ≠ q ∈ Q. Thus O(H)^{W} ≤ Z(L^{W}) and so L = E(L). So the claim follows from (10).

Suppose from now on that W is tensor indecomposable. Let M be a normal subgroup of H minimal with respect to [M, L] ≠ 1. Note that M/C_{M}(L) is characteristicly simple. Hence either M = E(M) or M is a q-group for some prime q. If M = E(M), it is easy to see that M is not a p′ group and so M = L since H is p-connected. So in view of (10) we may assume that M is a q- group.

V ZQ − 17

(12) M acts irreducible on W and M^{W} ≅ Ext(q^{1+2n}) or C_{4} ∘ Ext(2^{1+2n}), n ≥ 2.

If M is not homogenous on W. Then L acts non-trivially on the Wedderburn components of M on V, a contradiction to (8). Hence M is homogenous. As W is tensor indecoposable, this implies that M is irreducible on W. Let F = End_{KM}(W). Then by 6.6, Q and so also L = [L, Q] is F -linear on W. Thus [Z(M^{W}), L = 1, C_{L}(M) = Z(L) and C_{M}(L) = Z(M). By a standard argument the structure of M^{W} is as described.

V ZQ − 18

(13) One of the following holds:

1. p = 2, q = 3 and [M^{W} , Q]Q^{W} ≅ SU_{3}(2) or Ext(3^{1+2})C_{4}

2. p = 3, q = 2 and [M^{W} , Q]Q^{W} ≅ SL_{2}(3).

Let P = [M^{W} , Q], R = P Q^{W} and Y and irreducible R-submodule in W. Then P and so also R acts faithfuly on Y. Then P is extra-special. Let 1 ≠ z ∈ Z^{W}. Then as z acts quadratically on W, Hall-Higmann implies p = 2, or p = 3 and q = 2. Suppose that P ≠ [P, z]. Then [P, z] and C_{P} (z) are normal in R and P = [P, z] ∘ C_{P} (z). But then Y is tensor decomposable for R. Then the argument in (11) gives a contradiction. Thus P = [P, z]. A be a maximal abelian z-invariant normal subgroup of P. Let A = {D ≤ A|A = Z(P)D, D ∩ Z(P) = 1}. Then P acts transitively on A and z fixes a unique member of D, namely [A, z]. Also Y ⨁_{D∈A} C_{Y} (D). If p = 3 we conclude that |A| = 1 and so |P | = 8 and 2. holds. So suppose p = 2. Let |P | = q^{1+2n}. Then |A| = q^{1+n}, |A| = q^{n} we conclude that dim_{F} [Y, z] = \frac{q^{n}−1}{2}, dim_{F} C_{Y} (z) = \frac{q^{n}+1}{2} and dim_{F} C_{Y} (z)/[Y, z] = 1. Let q ∈ Q^{W} \ ⟨z⟩. If |q| = 2, we may assume that q normalizes A. But then [Y, z, t] = 0 implies that t normalizes all the orbits of z on A, a contradicition. Thus |q| = 4 and we may assume q^{2} = z. Since [Y, q, t] = 0, |[Y, q] + [Y, z]/[Y, z] has dimension at most 1 over F. Hence there exists an q invariant F -hyperplane U in Y with [U, q] ≤ [Y, t] ≤ C_{U}(q). Thus [U, q, q] = 0 and [U, q^{2}] = 1. Thus Y/C_{Y} (z) = 1 is 1-dimensional. So \frac{q^{n}−1}{2} = 1. q^{n} = 3 and |P | = 3^{3}. Hence 1. holds in this case.

V ZQ − 19

(14) Either L acts irreducible on M^{W} /Z(M^{W}) or (b4) holds.

Let Z(M^{W}) < P ≤ M^{W} be minimal with respect to being L-invariant. Put \overline{M} = M^{W} /Z(M^{W}). If Q does not normalize P, then by (13), |\overline{U}| ≤ q^{2}. Thus L/C_{L}(U) is a solvable {p, q} group. Since H is p-connected we conclude that L/C_{L}(U) is a p′ group and so a q-group. Since L is irreducible on U we conclude [U, L] = 1. Since H is irreducible on M/Z(M) we conclude [M, L] ≤ Z(M). Thus O^{q}(L) ≤ C_{L}(W) ≤ Z(L) and L = O_{q}(L)Z(L). Since [Z(L), Q] = 1, p-connectivity of H implies, L = O_{q}(L). Thus (b4) holds in this case.

So we may assume that Q normalizes U. If U is abelian, then by (13), Q centralizes U and so also L centalizes U, a contradiction. Hence U is not abelian and M^{W} = P C_{M^{W}} (P). Thus 6.17-14 implies P = M^{W}.

V ZQ − 20

(15) If L acts irreducible on M^{W} /Z(M^{W}) then (b5) holds.

This follows form (13).

V ZQm

Lemma 6.18 Let p be a group, H a finite p-minimal group with O_{p}(H) = 1. Let S ∈ Syl_{p}(H) and Z and Q non-trivial normal subgrous of S with Z ≤ Z(Q). Let R be maximal in Q with [V, R] ≤ [V, Z]. Let V be a faithful GF (p)H-module so that

(i) [V, Z, Q] = 0.

(ii) V = [V, O^{p}(H)].

(iii) V/C_{V} (O^{p}(H)) is irreducible as H-module.

Then |Q/R| ≤ V/C_{V} (Z). Moreover if T ⊴ S with Z ≤ T. Then either T ≤ R or [V, T] = [V, Q]

Proof: Remark: Some parts of the proof are still very sketchy, also the proof is a lot longer than it should be and to much of a case by case analysis Let Y = C_{V} (L) and V̄ = V/Y. Then V̄ is irreducible as H-module.

Let C = C_{H}(V̄).Then C ∩ L] centralizes U and V/U and so C ∩ L is a p-group. Since O_{p}(H) = 1 we conclude C ∩ L = 1. Thus O^{p}(C) = 1, C is p-group and C = 1.

Hence H acts faithfully on V̄ and we can apply 6.17(b) to V̄.

Let W be a LQ submodule in V minimal with respect to [W, L] ≠ 0. Then W = [W, L]. For X ∈ LQ let X/C_{X}(W). Let 1 ≠ z ∈ Z(S) ∩ Z.

V QZm − 1

(1) Suppose that |Q^{W} /Z^{W} | ≤ W̄ /CW̄ (Z) and [W, T] ∈ {[W, Z], [W, Q]. Then the lemma holds.

Since V̄ is irreducible and H = LS, V̄ = ⟨W̄^{S}⟩ Thus there exists s_{i} ∈ S, 1 ≤ i ≤ k with V̄ = ⊕_{i=1}^{k}W̄^{s_{i}}. Then V = [V, L] = [∑_{i=1}^{k} W^{s_{i}}, L] = ∑_{i=1}^{k} W^{s_{i}}. Let P = ⋂_{i=1}^{k} ZC_{S}(W^{k}). Then P ≤ R and

|Q/R| ≤ |Q/P | ≤ |Q^{W} /Z^{W} |^{k} ≤ W̄ /CW̄ (Z)^{k} = |V̄ /CV̄ (Z) ≤ V/C_{V} (Z)

Also [W, T] = [W, Z] implies [V, T] = [V, Z], while [W, T] = [W, Q] implies [V, T] = [V, Q]

V QZm − 2

(2) C_{LQ}(W̄) = C_{LQ}(W).

Let B = C_{LQ}(W̄). Then B ∩ L centralizes Y and W + Y/Y and so acts as a p-group on W. Since no composition factor of L on L is a p-group, B ∩ L centralizes W. Thus [B, L, W] = 0 and [W, B, L] = 0. Thus by the three subgroup lemma [W, L, B] = 0. As W = [W, L] we conclude [W, L] = 0 and so (2) holds.

V ZQm − 3

(3) If |QW̄ | ≤ p^{2}, the lemma holds.

By ??, |Q^{W} | ≤ p^{2}.Also Z^{W} ≠ 1 and Z does not centralize W̄. Thus (3) follows from ??.

V ZQm − 4

(4) If O_{p′}(L) ≠ 1, then Y = 0.

By Mascke, V = C_{V} (O_{p′}(L))⊕[V, O_{p′}(L)]. Also Y ≤ C_{V} (O_{p′}(L)) and as V̄ is irreducible, V = Y + [V, O_{p′}(L))]. Thus V = [V, L] = [V, O_{p′}(L)), L]] = [V, O_{p′}(L)] and (4) holds.

Suppose first that 1. in 6.17(b) holds for V̄. Then |QW̄ | = 2 and we are done by (3).

Suppose next that 2. in 6.17)(b) holds. Let D/Y ∈ D and Δ = D^{L}. Without loss W ≤ ∑ Δ. Since H is p-minimal we conclude from 6.17(a2) that L^{Δ} ≅ Alt(n) with n = 2^{k} + 1, k ≥ 2 or n = 6. If n ≤ 6 it is easy to see that Q^{Δ} ≤ 4 and so also |Q^{W} | ≤ 4. So we may assume that m = 2^{k} + 1, k ≥ 2. Let E ∈ Δ with E ≠ E^{z}. Then N_{Q}(E) centralizes E. Let M = N_{LQ}(E). Then M^{Δ} ≅ Alt(2^{n}) or Sym(2^{n}) and so M^{E} = ⟨N_{Q}(E)^{M}⟩O(L). Hence M = C_{M}(E)O(L). If O(L) centralizes E. Then V̄ is a permutation module for L, a contradiction to CV̄ (L) = 0. Thus O(L) ≠ 1 and by (4), Y = 0. It follows that [D, Z] = [D, Q]. Let F be the unique fixed point for z on Δ. Since F and E are conjugate under L, all p-elements in N_{LQ}(F) act trivially on F. So [F, Q] = 0 and [V, Z] = [V, Q].

Suppose that 3. in 6.17(b) holds. By (3) we may assume that |Q^{W} | > p^{2}. Then pminimality and quadratic action implies that the components for L are one of SL_{2}(q), SU_{3}(q), Sz(q),Alt(q+1), Sp_{4}(q)′ or L_{3}(q) Here q is a power of p, p = 2 in the last four cases, and a graph automorphism is induced on the components in the last two cases.

If 3d2 or 3d1in 6.17(b) holds then Y = 0. Let F = End_{L}(W). Then |Q^{W} | ≤ 2 · |F |, |W/C_{W} (Z)| ≥ |F |^{2} and [W, T] = [W, Q] if |T^{W} | ≥ 4. Thus we are done by ??.

So suppose that L^{W} is quasi simple. If Q^{W} is not elementary abelain then W is a strongly quadratic module in the sense of Stroth and so W̄ is the natural module. Because of the graph automorphism, L = Sp_{4}(q)′ is impossible in this case. Thus Y = 0 and the lemma is readily verifed in this case.

So suppose that Q^{W} is elementary abelian. Then its is easy to check that |CW̄ (Z)|^{2} = W̄ and |Q^{W} | ≤ |W̄ /CW̄ (Z)|. In particular, Q acts quadratically on W. Let J ≤ H^{W} minimal with Q^{W} ≤ J and Q^{W} ≰ O_{p}(J). Suppose first that O_{p}(J) = 1. Then ( for example by 2.9), J ≅ SL_{2}(q̃) or Sz)q̃). Thus there exists j ∈ J with J = ⟨T^{W j}, T^{W} ⟩. Thus [W, J] = [W, T]^{j} + [W, T] and [W, Q] = ([W, T]^{j} ∩ [W, Q]) + [W, T]. But [W, T]^{j} ∩ [W, Q] ≤ C_{W} (J) ∩ [W, T]^{j} ≤ [W, T] and so [W, Q] ≤ W. So we may assume that O_{p}(J) ≠ 1 and J is not generated by two conjugate of T^{W} in J. In particular, L^{W} ≅ Sp_{4}(q). We conclude that either [W, T] ≤ [W, Z] or Y ∩ W ≤ [W, T]. In the latter case, [W, Q] ≤ [W, T] and the lemma holds in this case.

Suppose finally that 4. or 5. in 6.17(b). In view of (3) we may assume that Q^{W} ≅ Q_{8}. So p = 2 Also by (4), Y = 0. Let X = ⟨Q^{O_{3}(L)}⟩. Then X^{W} ≅ SU_{3}(2) and W is a direct sum of natural modules for X^{W}, Again it is easy to verify the assumptions of ?? and the lemma is proved.

7 An interesting choice of an amalgam for generic p-type

groups

gpt
hgpt

Hypothesis 7.1 p is a prime, G is a finite groupe of generic p-type and S ∈ Syl_{p}(G).

dcalw

Definition 7.2 (a) W is the set of sets {M_{1}, M_{2}} such that (a) M_{i} ∈ L(J(S)

(b) O_{p}(⟨M_{1}, M_{2}⟩) = 1.

(b) Define an partial ordering ” ≤ ”l on W by defining (H_{1}, H_{2}) < (M_{1}, M_{2}) if and only

if one of the the follwing holds.

1. Some Sylow p subgroup of H_{1} ∩ H_{2} is properly contained in a Sylow p -subgroup of M_{1} ∩ M_{2}.

2. H_{1}∩H_{2} and M_{1}∩M_{2} have a common Sylow subgroup T and C_{H_{1}∩H_{2}}(Ω_{1}(Z(T)) < C_{M_{1}∩M_{2}}(Ω_{1}(Z(T))

3. H_{1} ∩ H_{2} < M_{1} ∩ M_{2}.

4. H_{1} ∩ H_{2} = M_{1} ∩ M_{2} and (possible after interchanging M_{1} and M_{2} and H_{1} and

H_{2}, M_{1} < H_{1} and M_{2} ≤ H_{2}. ” ≤ ” is defined as ” < ” or ” = ”

(c) W^{∗} is the set of maximal elements of A under the order defined in (b).

We leave it as an easy exercise to the reader to verify that (W, ≤) is a partially ordered set. Lemma 7.3 Let (M_{1}, M_{2}) ∈ W^{∗}, M_{12} = M_{1} ∩ M_{2}, T ∈ Syl_{p}(M_{12}) and put Z_{0} = Ω_{1}Z(T)). Then

(a) For i = 1, 2, |M(M_{i})| = 1.

(b) Suppose R is a p-subgroup of M_{1} with T < R. Then M(R) = M(M_{1}) and T ∈ Syl_{p}(M_{2}).

(c) Suppose that T ∉ Syl_{p}(G). Then C(G, T) ∈ L, C(G, T) lies in a unique maximal p-local M of G, |M(S)| = 1 and either T is a Sylow p-subgroup in M_{1} and M_{2}, or M = M_{i}^{∗} for some i.

(d) M_{12} is a maximal subgroup of M_{1} and of M_{2}.

(e) One of the following holds:

1. C_{M_{1}}(Z_{0}) = C_{M_{12}}(Z_{0}) = C_{M_{2}}(Z_{0}).

2. There exists {i, j} = {1, 2} so that

(a) C_{M_{i}}(Z_{0}) ≰ M_{j}, M(M_{i}) = M(C_{M_{i}}(Z_{0}) = M(C_{G}(Z_{0})).

(b) C_{M_{j}}(Z_{0}) ≤ M_{i}.

Proof: (a) Suppose M_{1} is contained in two distinct maximal p-locals L_{1},L_{2}. Then M_{1} ∩ M_{2} < M_{1} ≤ H_{1} ∩ H_{2}. But this contradicts the maximal choice of (M_{1}, M_{2}).

(b) Let M ∈ M(R). Then T is properly contained in a Sylow M_{1} ∩ M and so by that maximality of (M_{1}, M_{2}), M_{1} ≤ M. If T is not a Sylow p-subgroup of M_{2}, then we conclude M(M_{1}) = M(N_{L}(T)) = M(M_{2}), a contradcition. Thus (b) holds.

(c) Assume without loss that T < S. Then by maximality N_{S}(T) lies in a unique p-local subgroup M of G. Clearly C(G, T) ≤ M and it is easy to see that (c) holds.

(d) Let M_{12} < L_{1} ≤ M_{2} and put M = ⟨L_{1}, M_{2}⟩. If M ∈ L, then (M, M_{2}) ∈ W and M_{12} < L_{1} ≤ M ∩ M_{1}, a contradiction to the maximality of (M_{1}, M_{2}). Thus O_{p}(M) = 1 and (L_{1}, M_{2}) ∈ W. Also L_{1} ∩ M_{2} = M_{12}, L_{1} ≤ M_{1} and M_{2} ≤ M_{2}. So by maximality L_{1} = M_{1}.

(e) Suppose that C_{M_{1}}(Z_{0}) ≰ M_{2} and let M ∈ M(C_{M_{1}}(Z_{0}). Suppose that M_{1} ≰ M. Since T ≤ M_{1} ∩ M, maximality implies that T is a Sylow p-subgroup of M_{1} ∩ M. But then part 2. of the definition of ”¡” gives a contradiction. Thus (ea) holds. Clearly (ea) implies (eb).

cjt

Lemma 7.4 Let M ∈ L(S) and 1 ≠ x ∈ Z_{M} ∩ ZJ(S) Suppose that Z_{M} ≰ O_{p}(C_{G}(x)0. Then TO BE CONTINUED

Proof: Assume without loss that M is a maximal p-local. Put Q = C_{S}(Z_{M}). Note that C_{G}(x) ∈ L(B(S). Pick L ∈ L(Q) so that Z_{M} ≰ O_{p}(L), |L|_{p} is maximal and |L| is minimal. Let T be a Sylow p-subgroup of |L| with Q ≤ T. Let R be an T invariant subgroup of L with [R, Z_{M} ≰ O_{p}(R). Then by minimality of L, L = RS. In particular, L ∈ N (T). Also

Z_{M} ≤ D = =^{def} ⋂{O_{p}(P) | P ∈ M(L, T)}.

Case 1 T is not a Sylow p-subgroup of G.

Let C be a non-trivial characteristic subgroup of T. Then N_{G}(C) has a larger p-part then L and so by choice of L, Z_{M} ≤ O_{p}(N_{G}(C)). In particular, C is not normal in L. In particular, [Z_{L}, Z_{M}] ≠ 1.

Suppose that F^{∗}(L) is not a p-group. Then no element of O_{p}(L) is of p-type. Pick E ∈ L with Q ≤ L, F^{∗}(E) is not a p-group, |E|_{p} maximal and |E| minimal. Then Z_{M} ≰ O_{p}(E). Let R be a Sylow p-subgroup of E containing Q and R ⊲ R^{∗} for some p-group R^{∗}. Let 1 ≠ r ∈ R ∩ Z(R^{∗}). Then Q ≤ C_{G}(r) and C_{G}(r) has larger p-part then E. Thus r is of p-type and so r ≰ O_{p}(E). Thus [O_{p}(E), O^{p}(E)] = 1. TO BE CONTINUED

8 Some general amalgam results

geamre
amalgam

Hypothesis 8.1 1. G is a group.

2. p is a prime.

3. G_{1} and G_{2}b are finite subgroups of G.

4. G = ⟨G_{1}, G_{2}⟩

5. S ≤ G_{1} ∩ G_{2} so that S is a Sylow p-subgroup of G_{1} and G_{2}

6. Both F^{∗}(G_{1}) and F^{∗}(G_{2}) are p-groups.

Let O_{S}(G) be the largest subgroup of S which is normal in G. Let Z = Ω_{1}Z(S). Let Γ = Γ(G; G_{1}, G_{2}) be the coset graph for G with respect two G_{1}, G_{2}. In equal the vertices are the right cosets of G_{1} and G_{2} in G and two cosets are adjacent if they are distinct and have non-empty intersection. For γ ∈ Γ, let G_{γ} be the stabilizer of γ ∈ G, Q_{γ} = O_{p}(G_{γ}), Z_{γ} = Ω_{1}(Z(T)) | T ∈ Syl_{p}(G_{γ}), △(γ) is the set of neighbors of γ, G_{γδ} = G_{γ} ∩ G_{δ}. G_{γ}^{(1)} = ⋃_{δ∈△(γ)} G_{γδ}, V_{γ} = ⟨Z_{δ} | δ ∈ △(γ), C_{γ} = C_{G_{γ}}(Z_{δ}), E_{γ} = O^{p}(G_{γ}), Q_{γ}^{∗} = [Q_{γ}, E_{γ})], X_{γ} = Ω_{1}Z(Q_{γ}), X_{γ}^{∗} = C_{Q_{γ}}(Q_{γ}^{∗}), Y_{γ} is the largest p-reduced normal subgroup of G_{γ}

For γ ∈ Γ let b_{γ} = min{d(γ, δ) | Z_{γ} ≰ G_{δ}^{(1)}. Let b = min_{γ∈Γ} b_{γ} = min{b_{G_{1}}, b_{G_{2}}. Let α, α′ ∈ Γ with d(α, α′) = b and Z_{α} ≰ G_{α′}^{(1)}. Let

(α, α + 1, α + 2, …, α + b) = (α′ − b, …, α′ − 1, α′)

be a shortest path form α to α′. Put β = α + 1. Without loss {G_{α}, G_{β}} = {G_{1}, G_{2}}. Let q_{δ} = qa_{Z_{δ}}(G_{δ}), r_{δ} = min{r | |AQ_{β}/Q_{β}|^{r} = |V_{β}/C_{V_{β}}(A)} for some A ≤ S with

A ≰ Q_{β} and [V_{β}, A, A] = 1. Let c_{β} the number of non-trivial chief factors for G_{β} on V_{β}.

connected

Definition 8.2 Let H be a group and T a subgroup of H.

1. H is connected with respect to T if T is not normal in H and for each normal subgroup N of H, either N ∩ T is normal in H or H = NT.

2. H is p-connected if H is connected with respect to some Sylow p-subgroup of H.

3. H is p-minimal with H is not p-closed and a Sylow p-subgroup of H lies in a unique

maximal subgroup of H.

rrc

Lemma 8.3 If G_{β} is connected then, r_{β} ≥ ra_{V_{β}}c_{b}.

Proof: A ≤ G_{β} with [V_{β}, A, A] = 1 and put r = ra_{V_{β}}. Let U be a non-trivial chief factor for G_{β} on S Then as G_{β} ∈ N^{∗}(S), C_{A}(U) = A ∩ Q_{β}. So by definition of ra_{V_{β}}(S), |AQ_{β}/Q_{b}|^{r} ≤ |U/C_{U}(A)|. Multiplying together these inequalities over all such U in a chief series we obtain |AQ_{β}/Q_{b}|^{rc_{β}} ≤ |V/C_{V} (A)| and so r_{b} ≥ rc_{β}.

QRC

Lemma 8.4 Suppose that b ≥ 2 and allow for the case that O_{S}(G) ≠ 1.

(a) Suppose that q_{α} > 1 and [V_{β}, J(S) ≠ 1. Then b is odd or ∞ and (q_{α} − 1)(r_{β} − 1) ≤ 1.

(b) Suppose that C_{α} ∩ Q_{β} is not normal in G_{α} and put Q = ⟨C_{α} ∩ Q_{b}^{G_{β}}⟩. Then Q acts quadratically on Z_{α}, |[Z_{α}, Q]| ≤ |Q/C_{Q}(Z_{α})|, Z_{α} is an F F module and [C_{Z_{α}}(Q), E_{β}] = 1.

Proof: (a) If b is even,8.17 shows that Z_{α} or Z_{α′} is F F, a contradiction to q_{α} > 1. Thus b is odd or ∞. In particular, b ≥ 3 and V_{β} is abelian.

Since [V_{β}, J(S) ≠ 1, there exists A ∈ A(S) with [V_{β}, A] ≠ 1. By the Thompson replacement lemma we may assume that [V_{β}, A, A] = 1. Suppose A ≤ Q_{β} and let δ ∈ △(β). Then q_{δ} > 1 implies [Z_{δ}, A] = 1 and [V_{β}, A] = 1, a contradiction. Thus A ≰ Q_{β}. Put B = A∩Q_{β}. We will apply 2.4 with I = △(β) and W_{i} = Z_{i} for i ∈ I. Define r, t and s as in the 2.4. Since A ∈ A(S), |V_{β}/C_{V_{β}}(A)| ≤ |A/C_{A}(V_{β}) and so t ≥ 1. Also s ≥ q_{a} > 1 and r ≥ r_{β}. By 2.4b to obtain trs ≤ r + s, rs ≤ r + s, (s − 1)(r − 1) ≤ 1 and (q_{α} − 1)(r_{β} − 1) ≤ 1.

(b) Let D = C_{Z_{α}}(E_{α}). If D = Z_{α}, then Z_{α} and Q = C_{α} ∩ Q_{β} are normal in G_{b} in contrast to our assumptions. Thus Z_{α} ≠ D and we can choose D ≤ E ≤ Z_{α} with E ⊴S and |E/D| = p. Let W = ⟨E_{β}^{G}⟩. Note that [E, Q] ≤ D and so is centralized by E_{b} and normalized by S. Thus [E, Q] ⊴ G_{β}, [E, Q] = [W, Q] Since [W, E_{β}] ≠ 1 and c_{β} = 1, [V_{β}, E_{b}] ≤ W and so V_{β} = Z_{α}W. Hence [V_{β}, C_{Q_{β}}(Z_{a})] ≤ [W, Q and so [Z_{a}, Q] ≤ [V_{β}, Q] = [W, Q] = [E, Q] ≤ Z_{α}. [C_{α} ∩ Q_{β} centralizes D, Q centralizes D and [E, Q]. Hence [E, Q] = {[e, q] | q ∈ Q}, where e ∈ E \ D. Thus |[E, Q]| = |Q/C_{Q}(e) ≤ |Q/C_{Q}(Z_{α})|. If C_{Z_{α}}(Q) ≠ D, we can choose [E, Q] = 1 and we get [Z_{α}, Q] = 1 and so Q = C_{α} ∩ Q_{β} is normal in G_{β},a contradiction.

ocf

Lemma 8.5 Suppose that b is odd, b ≥ 3 and L ≤ G_{α′} with

(i) L = (G_{α′−1} ∩ L)O^{p}(L).

(ii) G_{α′} = ⟨G_{α′−1}, L⟩.

(iii) L has at most one non-central composition factor on ⟨Z_{ap−1}^{L}⟩.

Then one of the following holds

1. [Z_{ap−1}, [Q_{α′}, O^{p}(L)] ≠ 1 and Z_{α} is an F F -module for G_{α}/C_{α}.

2. [Z_{ap−1}, [Q_{α′}, O^{p}(L)] = 1 and

(a) V_{β} = Z_{α}C_{V_{b}}(Q_{b}).

(b) C_{a} ∩ Q_{β} ⊴ G_{β}.

(c) C_{V_{b}}(Q_{b}) is an F F module for ⟨Q_{a}^{G_{β}}⟩.

Proof: Let V = ⟨Z_{ap−1}^{L}⟩ and Q = [Q_{α′}, O^{p}(L). Then by (i), V = ⟨Z_{ap−1}^{O^{p}(L)}⟩ and we may assume without loss that L = O^{p}(L). Note also that Q_{α′} normalizes Z_{α′} and V.

Suppose first that [Z_{α′−1}, Q] ≠ 1. If [V, Q, L] ≠ 1, then by (iii), V = Z_{α′−1}[V, Q] and so V = Z_{α′−1}, a contradiction to (ii). Thus [V, Q, L] = 1 and by [St1] (1) holds.

So we may assume that Q centralizes Z_{α′−1} and V. Hence (iii) implies that [V, Q_{ap}, L] = 1 and [V, L, Q_{ap}] = 1. Thus V = Z_{α′−1}C_{V} (Q_{α′}) and so L normalizes Z_{ap−1}C_{V_{α′}} (Q_{α′}).

Therefore (ii) implies that G_{α′} normalizes Z_{ap−1}C_{V_{α′}} (Q_{α′}) and so V_{α′} = Z_{ap−1}C_{V_{α′}} (Q_{α′}). Thus C_{Q_{α′}} (V_{ap}) = C_{α′−1}∩Q_{α′} and (a) and (b) are proved. Moreover we get [V_{β} ∩Q_{α′}, V_{α′}] = 1 and [V_{α′} ∩ Q_{β}, V_{β} = 1. Hence (c) follows from 8.17.

q < 2

Lemma 8.6 Suppose that G_{β} is a mimimal parabolic and allow for the case that O_{S}(G) ≠ 1. Then one of the following holds:

1. S centralises Z_{α}.

2. Z_{α} ≰ Q_{β}.

3. q_{a} ≤ 2

4. Z_{α} is the dual of an F F -module

5. There exists a non-tivial characteristic subgroup C of B(S) with C ⊴ G_{β} and G_{α} = N_{G_{α}}(C)C_{α}. Moreover, either C = J(S) or Q_{b}^{∗} ≤ B(S) ≤ C_{α}.

6. Put G_{β}^{∗} = B(S)O^{2}(G_{β}). T O_{2}(G_{β}^{∗}) ≤ B(S) ≤ C_{α} and non-trivial characteristic subgroup of B(S) is normal in G_{β}^{∗}. Moreover, Z ⊴ G_{β}.

7. Z and Z_{α} are normal in G_{β} and centralized by E_{β}. Futhermore, S ∩ C_{α} is a Sylow p-subgroup of C_{G_{β}}(Z_{a}).

Proof: Without loss Z_{a} ≤ Q_{b}. If [J(S), Z_{α}] ≠ 1, r(S, Z_{1}) ≤ 1. So we may assume that J(S) ≤ C_{α}. Thus Z_{α} ≤ C_{S}(J(S)) and B(S) ≤ C_{α}. Hence

q < 2 − 1

(1) G_{α} = N_{G_{α}}(B(S))C_{a} = N_{G_{α}}(C)C_{a} for any characteristic subgroup C of B(S).

If E_{β} centralizes V_{β}, then 7. holds. So suppose [V_{β}, E_{β}] ≠ 1. If J(S) ⊴ G_{β}, 5. holds. Hence we may assume that J(S) ≰ G_{β}. in particular, [V_{β}, J(S)] ≠ 1. By 6.3, r_{V_{b}}(G_{β}) ≥ 1. If c_{β} ≥ 2, then 8.3 implies r_{β} ≥ 2. By refQRCa, (q_{α} − 1)(r_{b} − 1) ≤ 1 and so 3. holds. If c_{β} = 1, then 8.4b implies that 4. holds or C_{α}capQ_{β} is normal in G_{β}. So suppose the latter.

Since J(S) ≤ C_{α}, J(S) centralizes Q_{β}/Q_{β} ∩C_{α}. Since J(S) ≰ Q_{β}, E_{β} ≤ ⟨J(S)_{β}^{G}⟩ and so E_{β} centralizes Q_{β}/Q_{β} ∩ C_{α}. Thus Q_{β}^{∗} ≤ C_{α} ∩ Q_{β} and [V_{β}, Q_{b}^{∗}] = 1. Thus [C_{Q_{β}}(Q_{β}^{∗}), E_{β}] ≠ 1 and by Thompson’s P × Q-lemma, [X_{β}, E_{β}] ≠ 1. Thus by 8.10 ( and the remark following 8.10), O_{p}(E_{b}) ≤ B(S). Now either there exists a non-trivial charcteristic subgroup of B(S) which is normal in G_{β}^{∗} or there does not. In the first case (1) implies that 5. holds and in the second 6. holds.

Lemma 8.7 Suppose b > 1, s_{Z_{α}}(S) ≥ 1, C_{G_{β}}(V_{b}) is p-closed and [V_{β} ∩ Q_{α′}, V_{α′} ∩ Q_{β}] = 1. Then V_{β} is F 2 for G_{β}.

Proof: We may assume without loss that V_{β}Q_{α′}/Q_{α′} ≥ V_{α′}Q_{β}/Q_{a}p. Since s_{Z_{α}}(S) ≥ 1 we can apply 2.3 with s = 1, V = V_{β} and B = V_{α′} ∩ Q_{β} and conclude

|B/C_{B}(V_{b})| ≤ |V_{b}/C_{V_{b}}(B)|

By assumption V_{β} ∩ Q_{α′} ≤ C_{V_{b}}(B) and so

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

Hence V_{β} is F^{2}.

mqrc

Lemma 8.8 Let (P_{0}, P_{1}, P_{2}) be an amalgam over S. Let Z_{0} = ⟨Z^{P_{0}}⟩. For i = 1, 2 put L_{i} = ⟨P_{0}, P_{i}⟩ and Z_{i} = ⟨Z^{L_{i}}⟩. Suppose that

(i) P_{1} and P_{2} are in P(S).

(ii) For {i, j} = {1, 2}, O^{2}(P_{i}) ≰ O_{2}(P_{j}).

(iii) For i = 1, 2, Z ≤ O_{S}(L_{i})

The one of the following holds for some i ∈ {1, 2}

1. J(S) ⊴ P_{0}.

2. J(S) ⊴ P_{i}, [Z_{0}, O^{2}(P_{i})] ≠ 1 and r(S, Z_{i}) ≤ 1.

3. Z_{i} ≰ Q_{j}

4. r(S, Z_{j}) ≤ 2 or r^{∗}(S, Z_{j}) ≤ 2

Proof: Without loss J(S) ⊲ P_{O} and since J(S) is not normal in all the P_{i}’s we may assume that J(S) ⊴ P_{1}. If [Z_{0}, O^{2}(P_{1})] ≠ 1 we conclude that [Z_{1}, J(S)] ≠ 1 and 2. holds. So we also may assume that [Z_{0}, O^{2}(P_{1})] = 1. Then Z_{0} is not normal in P_{2} and hence [Z_{0}, O^{2}(P_{2})] ≠ 1.We apply 8.6 to G_{α} = L_{2} and G_{β} = P_{1}. As J(S) ⊴ P_{1} = G_{β} we conclude that either 3. holds or 4. holds or [Z_{2}, Q_{1}^{∗}] = 1. In the latter case Q_{1}^{∗} ≰ O_{2}(P_{2}) implies [Z_{2}, O^{2}(P_{2}) = 1, a contradiction to [Z_{0}, O^{2}(P_{2})] ≠ 1.

cb

Lemma 8.9 Let L be a subgroup of G_{β} which acts transitively on △(β). Put D_{β} = ⋂_{δ∈△(β)} Z_{δ} and l minimal with [Z_{α}, Q_{β}, l] ≤ D_{β}. Suppose that V_{β} ≤ Q_{β}. Then for all 0 ≤ i < l, L acts non-trivially on [V_{β}, Q_{β}, i]/[V_{β}, Q_{β}, i + 1].

Proof: Put Z_{i} = [Z_{α}, Q_{β}, i] and V_{i} = [V_{β}, Q_{β}, i]. As L acts transitively on △(β), V_{i} = ⟨Z_{i}^{L}⟩. Let i be so that L acts trivially on V_{i}/V_{i+1}. Then V_{i} = Z_{i}V_{i+1} and so V_{i}/Z_{i} = [V_{i}/Z_{i}, Q_{β}]. Hence V_{i} = Z_{i} and Z_{i} ≤ D_{β}. Thus i ≥ l.

Baumann

Lemma 8.10 Let G be a finite group, p a prime, p-subgroup of G, V = ⟨Ω_{1}(Z(O_{p}(G)), B(S) = C_{S}(Ω_{1}(Z(J(S)), J(G) = ⟨J(S)^{G}⟩, B(G) = ⟨B(S)^{G}⟩, \overline{G} = G/C_{G}(V), and Ṽ = V/C_{V} (O^{p}(B(G)) and suppose that each of the following holds:

(i) C_{G}(V) is p-closed.

(ii) If A ∈ P(\overline{G}, V) then |Ṽ /CṼ (A)| ≥ |A|.

(iii) If U is an F F -module for G/O_{p}(G) module with Ṽ ≤ U and U = C_{U}(B(S))Ṽ, then U = C_{U}(O^{p}(J(G)))Ṽ.

Then O_{p}(B(G)) ≤ B(S).

Proof: and Y = Ω_{1}ZJ(S). Let A ∈ A(S). Then \overline{A} ∈ P(\overline{G}, V) and so by (ii), |Ṽ /CṼ (A)| ≥ |\overline{A}|. By (i), |\overline{A}| = |A/A ∩ Q| and so V (A ∩ Q) ∈ A(S). Thus Y ≤ V (A ∩ Q) ≤ Q. Put W = ⟨Y^{G}⟩V. We conclude that W ≤ Ω_{1}ZJ(Q) and so W is elementary abelian and (A ∩ Q)V centralizes W. Hence W ≤ (A ∩ Q)V and W = V (A ∩ W) = V C_{W} (A). It follows that A centralizes W/V. Since A was arbitray in A(S), J(G) centralizes W/V. As Y = Ω_{1}ZJ(S∩J(G)), Sylow’s theorem implies that J(G) acts transitively on Y^{G}. Thus W = Y V and so [W, Q] = [Y, Q] ≤ Y. Hence [W, Q] ≤ C_{W} (B(G)). Let D = C_{W} (O^{p}(B(G)) and U = W/D. Then O_{p}(G) centralizes U. Since Ṽ ≅ V D/D and U = Y V/D, we can apply (iii) to conclude that W = DV and U ≅ Ṽ. Since A ∈ A(S), |W/W ∩A| ≤ |A/C_{A}(W)| = |A/A∩Q|. One the otherhand by (i), |A/A ∩ Q| ≤ |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, B(G)] = 1. Therefore [W, O_{p}(B(G)] ≤ [D, B(G)][V, Q] = 1 and so O_{p}(B(G)) ≤ C_{S}(Y) = B(S).

Remark 8.11 Assume (i) in 8.10. Then (ii) and (iii) hold as well unless \overline{J(G)} has a component K with K ≅ Alt(2n), n ≥ 3; SL_{n}(q), n ≥ 3; SU_{n}(q), n ≥ 6; Sp_{2n}(q), n ≥ 2; Ω_{2n}^{+}(q), n ≥ 3; or Ω_{2n}^{−}(q), n ≥ 4 ; and some compostion factor for K on V is a natural module.

pump

Lemma 8.12 pushing up minimal parabolics, odd elements

pusym

Lemma 8.13 pushing up sym(10) over ⟨(12), (34), (56), (78), (9, 10)⟩

trpu

Lemma 8.14 some trivial pushing up result, at least including L_{5}(2) over the O_{2} of a point stabilizer, saying that b = 4 and non trivial center; or b = 2 and O_{2} basicly a natural module

qaniqb

Lemma 8.15 Suppose that G_{α} is a p-minimal. Then Q_{α} ≰ Q_{β}.

Proof: This follows from 8.12Remark: This needs some thought

znnab1

Lemma 8.16 Suppose that each of the follwing holds:

(i) α, β = {γ, δ}.

(ii) G_{γ} is p-minimal and [X_{γ}, E_{γ}] ≠ 1.

(iii) G_{δ} is p-connected or C_{S}(X_{δ}) = Q_{δ}.

Then one of the following holds.

(a) [X_{δ}^{∗}, E_{δ}] = 1 and Z ⊴ G_{δ}.

(b) J(S) ≰ Q_{δ} and X_{δ} is an F F -module for G_{δ}.

(c) (a) J(S) ⊴ G_{δ}.

(b) O_{p}(B(G_{γ})) ≤ B(S) = B(Q_{δ}).

(c) E_{γ} is a SL_{2}(p^{r})^{k}-block, Alt(2^{r} + 1)^{k}-block or SL_{2}(3^{r})^{k}- double block.

(d) If G is finite and S ∈ Syl_{p}(G), then G contains a p-local R with B(S) ≤ R and C_{R}(O_{p}(R)) ≰ O_{p}(R).

Proof: We may assume that [X_{δ}^{∗}, E_{δ}] ≠ 1. Then by Thompsons’s A×B-lemma, [X_{δ}, E_{δ}] ≠ 1. Hence if G_{δ} is p-connected, C_{S}(X_{δ}) = Q_{δ}. Thus by (ii) C_{S}(X_{δ}) = Q_{δ}.

If J(S) ≰ Q_{δ}, then (b) holds.

So suppose J(S) ≤ Q_{δ}. Then X_{δ} ≤ ZJ(S) and so B(S) ≤ C_{S}(X_{δ}) ≤ Q_{δ} and B(S) = B(Q_{δ}). By 8.10, O_{p}(B(G_{γ})) ≤ B(S). Thus (ca) and (cb) hold in this case.

Since G_{γ} is p-minimal, G_{γ} = B(G_{γ})S. Let R be normal subgroup of B(G_{γ}). Let U be unique maximal subgroup of G_{γ} containing S. Let C be a non-trivial characteristic subgroup of B(S). Then C is normal in G_{δ} and so C is not normal in G_{γ}. Since S ≤ N_{G_{γ}}, this implies N_{G_{γ}} ≤ U. Let W = W_{γ} =⟩Ω_{1}Z(J(S))^{G_{γ}}. Then W is an F F -modules for B(G_{γ}) and O_{p}(B(G_{γ})) centralizes V. Hence W/C_{W} (E_{γ}) is a natural SL_{2}(p^{r})^{k} or Sym(2^{r} + 1)^{k} module for B(G_{γ}). Let E be minimal with B(S) ≤ E, and O^{p}(E) maps onto on normal SL_{2}(q)′’s or Alt(q + 1)’s. Then E ≰ U and so C ⋬ E. Hence by 8.12 O^{p}(E) is an L_{2}(p^{r})- block, Alt(2^{r} + 1) block or SL_{2}(q)-double block. It is now easy to see that O^{p}(E) is normal in E_{γ} and that (cc) holds.

Suppose now that G is finite and S ∈ Syl_{p}(G). Assume first that E_{γ} is a SL_{2}(p^{r})^{k}- or Alt(2^{r} + 1)^{k}-block. Then there exists λ ∈ △(δ) with [W_{γ}, W_{λ}] ≠ 1. Then W_{λ} ≤ B(Q_{δ}) = B(S) ≤ B(G_{γ})). Suppose that [X_{δ}, Q_{γ}] ≠ 1TO BE CONTINUED

P F F

Lemma 8.17 Let λ, μ ∈ Γ and F_{λ}, F_{μ} normal p-subgroups of G_{λ} and G_{μ}, respectively. Suppose that

(i) F_{λ} ≤ G_{μ} and F_{μ} ≤ G_{λ}.

(ii) [F_{λ}, F_{μ}] ≠ 1.

(iii) For ρ ∈ {λ, μ}, C_{G_{ρ}}(F_{ρ}) is p-closed

(iv) [F_{λ}, F^{μ} ∩ Q_{λ}] = 1 and F_{μ}, F_{λ} ∩ Q_{μ}] = 1.

Then one of the follwing holds

1. F_{λ} is an F^{∗}1 module for G_{λ}.

2. F_{μ} is an F^{∗}1 module for G_{μ}.

3. Both F_{λ} and F_{μ} are F F -modules. Proof: By (iii) and (iv) F_{λ} ∩ Q_{μ} = C_{F_{λ}}(F_{μ}) and F_{μ} ∩ Q_{λ} = C_{F_{μ}}(F_{λ}). |F_{λ}/F_{λ} ∩ Q_{μ}| is

either less, larger or equal to F_{μ}/F_{μ} ∩ Q_{λ}. In the first case |F_{λ}/C_{F_{λ}}(F_{μ})| < F_{μ}Q_{λ}/Q_{λ}| and 1. holds. Similarly the second case implies 2. and the third 3.

vbvap

Lemma 8.18 Supposse that b ≥ 3, b is odd and r_{α} ≥ 1.

(a) (r_{a} − 1)(r_{b}1) ≤ 1.

(b) Suppose that equality holds in (b). Then (b.a) |V_{α′}Q_{β}/Q_{a}| = V_{β}Q_{α′}/Q_{α′}| (b.b) C_{V_{α′}} (V_{β} ∩ Q_{α′}) = C_{V_{α′}} (V_{β}). (c.b) Let δ ∈ △(β) with [Z_{δ}, V_{α′}] ≠ 1. Then V_{α′} ∩ Q_{β} ≰ Q_{δ} and |(V_{α′} ∩ Q_{b})Q_{α}/Q_{a}|^{s} =

|Z_{δ}/C_{Z_{δ}}(V_{α′})|. (c.d) |V_{β}Q_{α′}/Q_{α′}|^{r} = |V_{α′}/C_{V_{α′}} (V_{β})|. Proof: By 2.4 we have

vbvap1

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

and

vbvap2

(2) |V_{β} ∩ Q_{α′}/C_{V_{β}}(V_{α′})|^{r_{α}} ≤ |V_{α′}/C_{V_{α′}} (V_{β} ∩ Q_{α′})|. Suppose first that V_{α′} ≤ Q_{β}. Since r_{α} ≥, (1) implies |V_{α′}/C_{V_{α′}} (V_{β})| ≤ |V_{β}/C_{V_{β}}(V_{α′})|. If

V_{α′} Q̸_{β} the situation is symmetric in α′ and β and we may assume in any case that

vbvap3

(3) |V_{α′}/C_{V_{α′}} (V_{β})| ≤ |V_{β}/C_{V_{β}}(V_{α′})|

TO BE CONTINUED

rsc

Lemma 8.19 Suppose that r_{β} ≥ 1, s_{α} ≥ \frac{3}{2} and s_{α}^{∗} > 1. Then

(a) \frac{3}{2} ≤ s_{α} ≤ 2.

(b) 1 ≤ r_{β} ≤ \frac{3}{2}.

(c) c = 2 or 3.

(d) If c = 3, then s_{α} = \frac{3}{2} and r_{β} = 1.

(e) If r_{β} = \frac{3}{2}, then c = 2, s_{α} = \frac{3}{2} and (s_{α} − 1)(r_{β}c_{β} − 1) = 1.

(f) If s_{α} = 2, then c = 2, r_{β} = 1 and (s_{α} − 1)(r_{β}c_{β} − 1) = 1.

(g) [Z_{α}, Z_{α′}] = 1.

Proof: As s_{α}^{∗} > 1, 2.4 implies c_{β} ≥ 2. All but the last statement are now an immediate consequece of 8.4. The last statement follows from 8.17.

p3/2

Lemma 8.20 Suppose that b is odd and β^{+}, β^{−} ∈ Γ_{2} with d(β^{+}, β^{−}) = b−1 For ϵ ∈ {+, −} let Λ^{ϵ} ⊆ △(β^{ϵ}). Define V^{ϵ} = ⟨Z_{λ}|λ ∈ Λ^{ϵ}⟩ and B = V^{ϵ} ∩ ⋂_{λ∈Λ^{−ϵ}} G_{λ}. Finally, let s be a postive real number so that for all ϵ ∈ {+, −}, all λ ∈ Λ^{−ϵ}, and all A ≤ B^{ϵ}, |Z_{λ}/C_{Z_{λ}}(A)|^{s} ≤ |A/C_{A}(Z_{λ})|. Then

(a) (aa) |B^{+}/C_{B^{+}}(V^{−})| ≤ |V^{−}/V_{V^{−}}(B^{+})|^{\frac{1}{s}} ≤ |V^{−}/C_{V^{−}}(B^{+})|^{\frac{1}{s}} (ab) |V^{+}/C_{V^{+}}(V^{−})| ≤ |V^{+}/B^{+}||B^{+}/C_{B^{+}}(V^{−})| (ac) |V^{+}/C_{V^{+}}(V^{−}) ≤ |V^{+}/B^{+}||V^{−}/C_{V^{−}}(V^{+})|^{\frac{1}{s}}.

(b)(b.a) |V^{+}/C_{V^{+}}(V^{−})|^{\frac{s^{2}−1}{s^{2}}} ≤ |V^{+}/B^{+}||V^{−}/B^{−}|^{\frac{1}{s}}.

(b.b) |B^{+}/C_{B^{+}}(V^{−})|^{\frac{s^{2}−1}{s}} ≤ |V^{+}/B^{+}|^{\frac{1}{s}} |V^{−}/B^{−}|.

(c) Suppose s > 1 and V^{+} = B^{+}, then |V^{+}/C_{V^{+}}(V^{−})| ≤ |V^{−}/B^{−}|^{\frac{s}{s^{2}−1}} .

(d) Suppose s > 1 and that r is a positive real number with |V^{−}/B^{−}|^{r} ≤ |V^{+}/C_{V^{+}}(V^{−})|. Put e = \frac{rs^{2}−r−s}{s^{2}}. (d.a) |V^{−}/B^{−}|^{e} ≤ |V^{+}/B^{+}|. (d.b) |B^{−}/C_{B^{−}}(V^{+})| ≥ \frac{|V^{−}/B^{−}|^{r}}{|V^{+}/B^{+}|} (d.c) If e > 0, then |B^{+}/C_{B^{+}}(V^{−})|| ≤ |V^{+}/B^{+}|^{\frac{rs}{rs^{2}−r−s}}

(e) Suppose s > 1 and r is a positive integer so that for ϵ ∈ {+, −}, |V^{ϵ}/B^{ϵ}|^{r} ≤ |V^{−ϵ}/C_{V^{−ϵ}}(V^{ϵ})|. Put e = \frac{rs^{2}−r−s}{s^{2}} and suppose that e > 0. (e.a) |V^{−}/B^{−}|^{e} ≤ |V^{+}/B^{+}||V^{−}/B^{−}|^{\frac{1}{e}} (e.b) If V^{−} ≠ B^{−}, then V^{+} ≠ B^{+} and e ≤ 1. Proof: The first inequa lityin (aa) follows from 2.3 while the second is obvious. (ab)

is obviuos and (ac) follows from (aa) and (ab).

Interchanging ” + ” and ” − ” in (ac) and substituting the result into (ac) we obtain |V^{+}/C_{V^{+}}(V^{−})| ≤ |V^{+}/B^{+}||V^{−}/B^{−}|^{\frac{1}{s}} |V^{+}/C_{V^{+}}(V^{−})|^{\frac{1}{s^{2}}} .

Thus (b.a) holds. Simimalrly interchanging ” + ” and ” − ” in (ac) and substituting the result into (ab) one obtains (bb).

(c) follows easily from (b.a). (ea) follows from (da) and using symmetry in ” + ” and ” − ”. (eb) follows from (eb). So it remains to prove (d). By assumption |V^{−}/B^{−}|^{r} ≤ |V^{+}/C_{V^{+}}(V^{−})|. As s > 1 we can raise this inequality to the \frac{s^{2}−1}{s^{2}} power and obtain

|V^{−}/B^{−}|^{\frac{r(s^{2}−1)}{s^{2}}} ≤ |V^{+}/C_{V^{+}}(V^{−})|^{\frac{s^{2}−1}{s^{2}}} .

Thus (da) follows from (ba). For (db) note that

|V^{−}/B^{−}|^{r} ≤ |V^{+}/C_{V^{+}}(V^{−})| ≤ |V^{+}/B^{+}||B^{+}/V_{V^{+}}(V^{−}).

Finally (d.c) follows from (d.a), (b.b) and a simple computation.

LLp

Lemma 8.21 Suppose b > 1 and G_{β} is p-minimal. Let M_{αβ} be the unique maximal subgroup of G_{β} containing G_{αβ}. Put β^{+} = β, β^{−} = α′. Then one of the follwing holds

1. For each ϵ ∈ {+, −} there exists L^{ϵ} ≤ G_{β^{ϵ}} and μ^{ϵ} ∈ △(β^{ϵ}) so that for V^{ϵ} = ⟨Z_{μ^{ϵ}}^{L^{ϵ}} each of the following holds.

(a) V^{−ϵ} ≰ 0_{p}(L_{ϵ}).

(b) V^{−ϵ} ≤ G_{μ^{ϵ}} and G_{β^{ϵ}μ^{ϵ}} contains a Sylow p-subgroup of L^{ϵ}

(c) L^{ϵ} ∩ M_{β^{ϵ}μ^{ϵ}} is the unique maximal subgroup of L^{ϵ} containg V^{−ϵ}.

(d) [V^{−ϵ}, Z_{μ^{ϵ}}] = 1.

2. There exists ϵ ∈ {+, −}, L^{ϵ} ≤ G_{β^{ϵ}}, μ^{ϵ} ∈ △(β^{ϵ} and μ ∈ △(β^{−epsilon}) so that with V^{ϵ} = ⟨Z_{μ^{ϵ}}^{L^{ϵ}} each of the follwing holds.

(a) V_{ϵ} ≤ G_{μ}, Z_{μ} ≤ L^{ϵ} and Z_{μ} ≰ 0_{p}(L_{ϵ}).

(b) Z_{μ} ≤ G_{μ^{ϵ}} and G_{β^{ϵ}μ^{ϵ}} contains a Sylow p-subgroup of L^{ϵ}

(c) L^{ϵ} ∩ M_{β^{ϵ}μ^{ϵ}} is the unique maximal subgroup of L^{ϵ} containg Z_{μ}.

(d) [Z_{μ}, Z_{μ^{ϵ}}] = 1.

3. There exist μ^{+} ∈ △(β^{+}) and μ^{−} ∈ △(β^{−}) so that Z_{μ^{+}} ≤ G_{μ^{−}}, Z_{μ^{−}} ≤ G_{μ^{+}} and [Z_{μ^{+}}, Z_{mu^{−}}] ≠ 1.

Proof: Suppose that 3. does not hold. For ϵ ∈ {+, −} choose L^{ϵ} ≤ G_{β^{ϵ}} and μ^{ϵ} ∈ △(β^{ϵ}) so that |L^{+}||L^{−}| is minimal with respect to

(i) For all ϵ, V^{−ϵ} ≤ L^{ϵ} ∩ G_{β^{ϵ}μ^{ϵ}}.

(ii) For all ϵ, G_{β^{ϵ}μ^{ϵ}} ∩ L^{ϵ} contains a Sylow p-subgroup of L^{ϵ} and M_{β^{ϵ}μ^{ϵ}} ∩ L^{ϵ} the unique maximal subgroups of L^{ϵ} containg that Sylow p-subgroup.

(iii) For at least one ϵ, V^{−ϵ} ≰ O_{p}(L^{ϵ}).

Note that (i),(ii) and (iii) are fulfilled with L^{ϵ} = G_{β^{ϵ}}, μ^{+} = α + 2 and μ^{−} = α′ − 1 and so we can make such a minimal choice.

Case 1 For some ϵ ∈ {+, −} and some μ ∈ μ^{ϵL^{ϵ}}, [V^{−ϵ}, Z_{μ}] ≠ 1 and V^{−ϵ} ≤ G_{μ}.

For ease of notation we assume without loss that ϵ = −.

LLp − 11

(1) In case 1, Z_{μ} ≰ O_{p}(L^{+}) and [Z_{μ^{+}}, Z_{μ}] = 1.

Suppose Z_{μ} ≤ O_{p}(L^{+}) and pick ρ ∈ μ^{+L^{+}} with [Z_{ρ}, Z_{μ}] ≠ 1. Then Z_{μ} ≤ G_{ρ}, Z_{ρ} ≤ G_{μ} and so 3. holds, contrary to our assumption. As Z_{μ} ≤ G_{μ^{+}}, the same argument shows [Z_{μ^{+}}, Z_{μ}] = 1.

LLp − 12

(2) In case 1, 2. holds.

By 2.6 there exists L ≤ L^{+} and h ∈ L^{+} such that Z_{μ} ≤ L, Z_{μ} ≰ O_{p}(L), (G_{β^{+}μ^{+}}∩L^{+})^{h}∩L contains a Sylow p-subgroup of L, and (M_{β^{+}μ^{+}} ∩ L^{+})^{h} ∩ L is the unique maximal subgroup of L containing Z_{μ}. Thus 2. holds with ϵ = +, L in place of L^{ϵ}.

Case 2 Case 1 does not hold.

LLp − 13

(3) In case 2, for all ϵ, V^{−ϵ} ≰ O_{p}(L^{ϵ}) and [V^{−ϵ}, Z_{μ^{ϵ}}] = 1.

If the first statement is false pick μ ∈ μ^{ϵL^{ϵ}} with [Z_{μ}, V^{−ϵ}] ≠ 1, if the second statement is false put μ = μ^{ϵ}. Then in any case V^{−ϵ} ≤ G_{μ} and the assumption of Case 1 are fulfilled.

LLp − 2

(4) In case 2. 1. holds.

We prove is basicly the same as for (2). By 2.6 there exists L ≤ L^{ϵ} and h ∈ L^{ϵ} such that V^{−ϵ} ≤ L, V^{ϵ} ≰ O_{p}(L), (G_{β^{ϵ}μ^{ϵ}} ∩L_{ϵ})^{h} contains a Sylow p-subgroup of L, and (M_{β^{ϵ}μ^{ϵ}} ∩L_{ϵ})^{h}∩L is the unique maximal subgroup of L containing V^{ϵ}. Hence (i), (ii) and (iii) are still fulfilled if we replace L^{ϵ} be L, μ^{ϵ} by μ^{ϵh} and leave L^{−ϵ} and μ^{−ϵ} as they are. Thus the minimal choice of |L^{+}||L^{−}| implies L = L^{ϵ} and so 1. holds holds.

znnab

Lemma 8.22 Assume that each of the following holds for each {γ, δ} = {α, β} and each critical pair (α, α′)

(I) Z_{αβ} ⋬ G_{γ}.

(ii) If N ⊲ G_{γ} with N ∩ O_{p}(G_{α}β) ≰ Q_{γ} then G_{γ} = NG_{αβ}.

(iii) Let O = O_{γδ} = {A ≤ Q_{δ} | |Z_{γ}/C_{Z_{γ}}(A) ≤ |AQ_{γ}/Q_{γ}| ≠ 1, [Z_{γ}, A, A] = 1}. Then Z_{γ}/C_{Z_{γ}}(A) = |AQ_{γ}/Q_{γ}| for all A ∈ O.

(iv) Either O = ∅ or A_{γδ} =^{def} ⋂_{A∈O}[Z_{γ}, A] ≠ 1.

(v) Z_{β}Z_{α} ⋪G_{α}

(vi) One of the following holds

(vi.1) If α − 1 ∈ △(α) such Z_{α′} does normalize Z_{α−1}Z_{α}, then Z_{α−1} ≰ Q_{α′−1}. (vi.2) There exists α − 1 ∈ △(α) with G_{α} = ⟨G_{αα−1} and Z_{α−1} ≰ Q_{α′−1}.

Then

(a) O_{αβ} ≠ ∅ ≠ O_{αβ}.

(b) If b ≥ 2, then A_{βα} ⊴ G_{α}.

(c) b ≤ 2.

Proof: By (iii), Z_{α′} ∈ O_{αβ}. By (ii) and (vi), there exists α − 1 ∈ △(α) so that Z_{α′} does not normalize Z_{α−1}Z_{α}. Hence by (vi) we may choose α − 1 so that Z_{α−1} ≰ Q_{α′−1}. In particular,

znnab − 1

(1) Z_{α′−1} ∈ O_{α−1α}

Thus (a) holds.

Let H = N_{(}G_{α})(Z_{α}Z_{α} − 1)), G = {g ∈ G_{α} | Z_{α′}^{g} H̸} and T = ⟨Z_{α′}^{g} | g ∈ G⟩. Let g ∈ G. Then g ∈ G or Z_{α′}^{g} ≤ H. Hence ⟨H, T ⟩ ≥ G_{α−1α}Z_{α′}^{G_{α}}⟩ = G_{α}, where the last euality follows from (ii). Since both H and T normalize T, we conclude that T = ⟨Z_{α′}^{G_{α}}⟩ and inparticular

znnab − 2

(2) G_{α} = G_{α−1α}⟨Z_{α′}^{g} | g ∈ G}.

Suppose now that b > 1 and A_{βα} ⋪G_{α}. Then by (2) we may assume that Z_{α′} does not normalize A_{α−1α}. But (1) and the definition of A_{α−1α} imply A_{α−1α} ≤ [Z_{α−1}, Z_{α′−1}]. Hence A_{α−1α} ≤ Z_{α′−1} and b > 1 provides the contradiction, [A_{α−1α}, Z_{α′}] = 1. Thus (b) holds.

Suppose now that b > 2. Then by (b) applied to (α − 1, α′ − 1) in place of (α, α′), A_{αα−1} ⊴ G_{α−1}. Hence by (2) we may now assume that Z_{α′} does not normalize A_{αα−1}. On the otherhand by (1) there exist α − 2 ∈ △(α − 1) so that Z_{α′−2} ∈ O_{α−2α−1} Hence

A_{αα−1} = A_{α−2α−1} ≤ [Z_{α} − 2, Z_{α′} − 2] ≤ Z_{α′−2}.

Since b > 2 we conclude [A_{αα−1}, Z_{α′}] = 1, a contradiction and so also (c) is established.

vvi1

Lemma 8.23 Suppose that (i) to (v) in 8.22 holds. Suppose in addition that

(a) If A ∈ Q and B is an elementary abelian subgroup of Q_{δ} with [Z_{γ}, A, B] = 1 and A ≤ B. Then [Z_{γ}, B] ≤ [Z_{γ}, A][C_{Z_{γ}}(A), B]

(b) If A ∈ Q then there exists λ ∈ △(γ) with G_{γ} = ⟨G_{λγ}, A⟩.

Then (vi.2) in 8.22 and so also the conclusions of 8.22 hold.

Proof: By (b) there exists α − 1 ∈ △(a) with G_{α} = ⟨G_{α−1α}, Z_{α}⟩. Suppose that Z_{α−1} ≤ Q_{α′−1}. Then by (a) applied with γ = α′, A = Z_{α} and B = Z_{α−1}Z_{α}, we conclude that

[Z_{α′}, Z_{α−1}Z_{α}] ≤ [Z_{α′}, Z_{α}][Z_{α′} ∩ Q_{α}, Z_{α−1}] ≤ Z_{α−1}Z_{α}.

Thus Z_{α−1}Z_{α} is normalized by ⟨G_{α−1α}, Z_{α}⟩ = Gα, a contradiction to (v).

cznnab

Lemma 8.24 Suppose that (i) to (v) in 8.22 holds. In addition assume that for each A ∈ Q and each elementary abelian subgroup B each Q_{δ} with [Z_{γ}, A, B] = 1 and A ≤ B the following statements hold

(a) |B/C_{B}(C_{Z_{γ}}(A) ≤ |C_{Z_{γ}}(A)/C_{Z_{γ}}(B).

(b) If [C_{Z_{γ}}(A), B] = 1 then [Z_{γ}, B] ≤ [Z_{γ}, A].

(c) Suppose that [C_{Z_{γ}}(A), B] ≠ 1. Then for elementary abelian subgroup C of Q_{δ} with B ≤ C and [Z_{γ}, B, C] = 1, [C_{Z_{γ}}(A), C] ≤ [C_{Z_{γ}}(A), B]

(d) There exists λ ∈ △(γ) with L_{γ} = ⟨O_{p}(G_{λγ}), A⟩.

Then the conclusions of 8.22 hold.

Proof: We may assume that (vi.2) in 8.22 does not hold. Thus by (d) we can choose a critical pair (α, α′) and α − 1 ∈ △(a) with G_{α} = ⟨G_{α−1α})Z_{α}⟩ and Z_{α−1} ≤ Q_{α′−1}. If [Z_{α′}, Z_{α−1}Z_{α}] ≤ [Z_{α′}, Z_{α}] we get that Z_{α−1}Z_{α} is normalized by ⟨G_{α−1}, Z_{α′}⟩ = G_{α}, a contradiction to (v). Then by (b) we may assume that [Z_{α′} ∩ Q_{α}, Z_{a−1}] ≠ 1. Put X = Z_{α′} ∩ Q_{α}. Then by (a) [X ∈ Q_{α−1α} and so 8.22(a) holds.

Moreover, A_{α−1α} ≤ [Z_{α} − 1, X] ≤ Z_{α′} and so A_{α−1α} is normalized by G_{α−1}α and Z_{α′} and so 8.22b holds.

Suppose that b > 2. By (d) there exists α − 2 ∈ △(α − 1) with G_{α−1} = ⟨G_{α−2α−1}, X⟩. If Z_{α−2} ≰ Q_{α′−2}, then A_{α−2α−1} ≤ [Z_{α} − 2, Z_{α′} − 2] ≤ Z_{α′} − 2. As b > 2 we get that G_{αα−2}, X and Z_{α′} normalize A_{α−2α−1}. But then A_{α−2α−1} is normal in G_{α−1} and G_{α}.

Hence Z_{a−2} ≤ Q_{α′−2}. If Z_{α−2} ≰ Q_{α′−1}, then since also Z_{α′−1} ≤ Q_{α−1} we conclude from 8.22(iii) that Z_{α′} − 1 ≤ calQ_{α′−2α′−1}. But then A_{α−2α−1} ≤ [Z_{α} − 2, Z_{α′} − 1] a we get the same contradiction to the previous paragaph.

Thus Z_{α−2} ≤ Q_{α′−1} and so by (c) applied with C = Z_{α−2} and γ = α′ we conclude that [Z_{α−2}, X] ≤ [Z_{α−1}, X] ≤ Z_{α−1}. Hence Z_{α−2}Z_{α−1} is normalized by G_{α−2α−1} and X, a contradiction to 8.22(v).

znnabmp

Lemma 8.25 Suppose that G_{α} and G_{β} are minimal parabolics and Z ⋬ G_{α} and Z ⋬ G_{β}. Then b ≤ 2 or Z_{α}Z_{β} ⊴ G_{α}

Proof: We assume without loss that G_{αβ} is Sylow 2-subgroup of G_{α} and G_{β}. Put T_{αβ} = ⟨Z_{α′}^{G_{αβ}}⟩Q_{a} and Z_{αβ} = C_{Z_{α}}(T_{αβ}. Note that T_{αβ} only depends on α and β but not on Z_{α′}. Let α − 1 ∈ △(a) with Z_{αβ} ∩ Z_{αα−1} ≤ D_{α}. For O ≤ i ≤ b, put W_{i} = ⟨Z_{α′−i}^{G_{αβ}}⟩. Then W_{b} = Z_{α} and W_{0}Q_{α} = T_{αβ}. Put T = T_{α−1α} and supose that W_{1}Q_{α−1} ≠ T. Then there exists a U ⊴ T so that Z_{α} = ⟨U^{G_{αα−1}}⟩ and [W_{1}, U] = 1. Hence U ≤ Q_{α′−1}. It is now easy to see that Z_{α′} ∩ Q_{α} ≤ T_{α−1α} and so [U, Z_{α′}] ≤ [Z_{α}, Z_{α′}][U, Z_{α′} ∩ Q_{α}] ≤ Z_{α}[U, T]. Hence [U, W_{0}] ≤ Z_{α}[U, T] and W_{O}. Let L = ⟨T, W_{O}. Then O^{2}(L) centralizes UZ_{α}/Z_{α}. As Z_{β} = ⟨U^{G_{α−1α}}⟩ we conclude that G_{α} normalizes Z_{β}Z_{α}. Remark: It is easy to see that V_{α}/Z_{α} is an FF-module. This will kill any problem O_{2Φ} might cause, also this shows that basicly T_{αβ} = T_{βα}

Hence W_{1}Q_{α−1} = T. Choose α−i−1 ∈ △(α−i) with Z_{α−i−1α−i}∩Z_{α−i+1α−i} ≤ Z(G_{α−i}. Then a similar argument shows inductively that W_{i}Q_{α−i} = T_{α−iα−i+1}. Hence Z_{α}Q_{α−b} = T_{α−bα−b+1}. Therfore we may assume that Z_{α′}Q_{a} = T_{αβ}. The above argument now shows that Z_{α′−1}Q_{α−1} = T and we conclude that if b > 1, then [Z_{α−1}, T] = [Z_{α−1}, Z_{α−1}] ≤ D_{α}. Moreover, if b > 2, [Z_{α} −2, Z_{α′} −2] ≤ D_{α} −1∩D_{α}, a contradiction and the lemma is proved.

Lemma 8.26 Let M_{i} ∈ calL(S), 1 ≤ i ≤ 3 and suppose that that

(i) For i = 2, 3, O^{2}(M1i ∩ S ≤ Q_{23}

(ii) O^{2}(M_{1}) ∩ S = (O^{2}(M12 ∩ S)(O^{2}(M13 ∩ S).

Then Q_{23} is a Sylow 2-subgroup of O^{2}(M_{1})Q_{23} and Q_{1} ∩ Q_{23} = O_{2}(O^{2}(M_{1})Q_{23}) is normal in M_{1}

Proof: Let L = O^{2}(M_{1})Q_{23}. Then by (ii) and (i)

Q_{23} ≤ L ∩ S = (O^{2}(M_{1}) ∩ S)Q_{23} = (O^{2}(M12 ∩ S)(O^{2}(M13 ∩ S)Q_{23} = Q_{23}

. Since L ⊴ LS = M_{1}, O_{2}(L) ≤ Q_{1}.

Hence O_{2}(L) = Q_{1} ∩ L = Q_{1} ∩ Q_{23}.

9 Amalgams involving uniqueness groups

minparun

Hypothesis 9.1 (i) Hypothesis 8.1 holds with G finite.

(ii) G_{α} is a minimal parabolic.

(iii) E_{β}B(S) lies in a unique maximal p-local M_{β} of G.

(iv) Q_{β}^{∗} ≤ O_{p}(M_{b}).

(v) G_{β} = E_{β}G_{αβ}

(vi) M_{αβ} = =^{def} M_{β} ∩ G_{α} is the unique maximal subgroup of G_{α} containing S.

(vii) G_{β} ∈ CL(S).

Put Q_{αβ} = O_{2}(M_{αβ}), X_{b} = Ω_{1}(Z(Q_{b})) and X_{β}^{∗} = Ω_{1}(C_{Q_{β}}(Q_{b}^{∗})) Put D_{β} = ⋂_{δ∈△(β)} Z_{δ} and R = [Z_{α}, Z_{α′}]. The next two lemmas reveal how the assumptions on E_{β} can be used

Qab

Lemma 9.2 (a) Q_{β}^{∗} ≤ O_{2}(M_{β}) ≤ Q_{αβ}.

(b) Let γ ∈ △(β) and R_{α} be a normal subgroup of G_{α}. Then

R_{γ} ∩ Q_{β} ≤ (R_{α} ∩ Q_{β})Q_{β}^{∗} ≤ (R_{α} ∩ Q_{β})Q_{αβ} ≤ (R_{α} ∩ Q_{β})O_{2}(M_{β}) ≤ R_{α}Q_{αβ}.

(c) Let γ ∈ △(β). Then Q_{γ} ∩ Q_{β} ≤ Q_{α}O_{2}(M_{β}) ≤ Q_{αβ}.

(d) Let R_{αβ} be a normal subgroup of G_{αβ} contained in Q_{β}. Then for all γ ∈ △β,

R_{αβ} ≤ ⟨R_{αβ}^{G_{β}}⟩ ≤ O_{2}(M_{β})R_{γα}.

Proof: By hypothesis, Q_{β}^{∗} ≤ O_{2}(M_{β}). As G_{αβ} contains a Sylow 2-subgroup of M_{β}, O_{2}(M_{β}) ≤ G_{αβ} and (a) holds.

Since E_{β} acts transitively on △(β) we have R_{γ} ∩ Q_{β} ≤ (R_{α} ∩ Q_{β})[Q_{β}, E_{β}] and so (b) follows from (a).

Since Q_{α} ≤ Q_{αβ}, (c) follows from (b) applied to R_{α} = Q_{α}. As R_{αβ} ≤ ⟨R_{γβ}^{E_{β}}⟩ ≤ [Q_{β}, E_{β}]R_{γα} ≤ O_{p}(M)R_{γα}, (d) holds.

Mtrick

Lemma 9.3 Suppose 1 ≠ D ≤ ZJ(S) and E_{β}] ≤ N_{G}(D). Then

(a) N_{G_{α}}(D) ≤ M_{αβ}

(b) Let δ ∈ Γ such that d(β, δ) = b − i with 1 ≤ i < b . Suppose that N_{G_{δ}}(D) normalizes no non-trivial 2-subgroup of G_{δ}/Q_{δ}. Then

(ba) V_{β}^{(i+1)} ∩ G_{δ} ≤ Q_{δ}

(bb) V_{β}^{(i)} ≤ Q_{δ}.

(bc) If N_{G_{δ}}(D) contains a Sylow p−subgroup of G_{δ}, then V_{β}^{(i+1)} ≤ Q_{δ}.

(c) If b is odd and b ≥ 3, then E_{α′} does not normalize D.

(d) Suppose that b is even, b ≥ 3 and E_{α′−1} normalizes D, then

(da) V_{β}^{(3)} ∩ G_{α′−1} ≤ Q_{α′−1} ≤ G_{α′}.

(db) If G_{α′−1} normalizes D, then V_{β}^{(3)} ≤ Q_{α′−1} ≤ G_{α′}.

Proof: As B(S) and E_{β} normalize D, N_{G}(D) ≤ M_{β}. Thus (a) holds.

For (b) let γ ∈ △(β) with d(γ, δ) = b − i − 1. Then by 9.2(d) V_{β}^{(i+1)} ≤ V_{γ}^{(i−1)}O_{2}(M_{β}).

By minimality of b, V_{γ}^{(i)} ≤ Q_{δ}. Since N_{G_{δ}}(D) ≤ M_{β}, N_{G_{δ}}(D) normalizes the 2-group G_{δ} ∩ O_{2}(M_{β}). Thus by assumption, G_{δ} ∩ O_{2}(M_{β}) ≤ Q_{δ}. Hence V_{β}^{(i+1)} ∩ G_{δ} ≤ V_{γ}^{(i−1)}(O_{2}(M_{β}) ∩ G_{δ}) ≤ Q_{δ}. So (ba) holds. Clearly (ba) implies (bb). In case (bc) O_{2}(M_{β}) ≤ G_{δ} and so V_{b}^{(i+1)} ≤ G_{δ}.

Suppose b is odd and E_{α′} centralizes D. Then by (bb) applied with δ = α′ and i = 1, V_{β} ≤ Q_{α′}, a contradiction.

(d) follows from (ba) and (bc) applied with δ = α′ − 1 and i = 2.

uznn

Lemma 9.4 Suppose that [Z, E_{β}] ≠ 1. Then Z_{β} is an F F -module.

Proof: 8.16

1za

Lemma 9.5 Suppose that [Z_{α}, Z_{α′}] ≠ 1 and [Z, E_{β}] = 1.

(a) Let L_{α} = ⟨Z_{α′}^{G_{α}}⟩Q_{α}. Then L_{α}/C_{α} ≅ SL_{2}(q)^{k}, where k is a postive integer and q a power of 2.

(b) Z_{α} is a natural module for L_{α}/C_{a}.

(c) Z_{α′}Q_{α} is a Sylow p-subgroup of ⟨Z_{α′}^{L_{α}}⟩Q_{α}.

Proof: As [Z_{α}, Z_{α′}] ≠ 1 we may assume that Z_{α′} acts as an offending subgroup on Z_{α}. Since [Z, E_{β}] = 1, C_{Z_{α}}(L_{α}) = 1.Moreover, by 9.2c Z_{α′} ≤ Q_{αβ}, which excludes the possibilty that Z_{α} is a a natural Sym(q + 1)^{k}-modules for q ≥ 4. Thus the lemma follows from 6.3.

Define Z_{αβ} = C_{Z_{a}}(S ∩ L_{α}) and Z_{β}^{∗} = ⟨Z_{αβ}^{G_{β}}⟩. In the next two lemmas we will assume [Z_{α}, Z_{α′}] ≠ 1. Let V be an irreducible L_{α} submodule in Z_{α} not centralized by Z_{α′} and similarly choose V′ ≤ Z_{α′}. Put R = [V, V′].

zb∗

Lemma 9.6 Suppose that [Z_{α}, Z_{α′}] ≠ 1 and [Z, E_{β}] = 1. Then one of the following holds:

1. Z_{αβ} is normal in G_{β}.

2. Z_{αβ} ≤ X_{β}^{∗} and [X_{β}^{∗}, E_{β}] ≠ 1.

3. q = 2 and k ≥ 2. Moreover, if U_{αβ} be maximal in Z_{αβ} with [U_{αβ}, G_{αβ}] ≤ Z_{β} and U_{β} = ⟨U_{αβ}^{G_{β}}⟩, Then U_{β}/Z_{β} is an FF-module for G_{β}/Q_{β}

Proof: We may assume that Z_{αβ} is not normal G_{β} and so is not centralized by E_{β}. Suppose first that q > 2 or k = 1. Then Q_{β}^{∗} ≤ Q_{αβ} ≤ L_{α} and so Z_{αβ} ≤ X_{β}^{∗}. Thus

[X_{β}^{∗}, E_{β}] ≠ 1 and the P × Q lemma implies [X_{β}, E_{β}] ≠ 1.

So suppose now that q = 2 and k > 1. Let α − 1 ∈ △(α) with ⟨G_{αα−1}, V′⟩ = G_{α}. By 9.5c, [Z_{αβ}, Z_{α′}] = 1 and so

zb ∗ −1

(1) (a) Z_{β}^{∗} ≤ Q_{δ} for all δ ∈ Γ with d(β, δ) < b.

(b) [Z_{α′−1}^{∗}, V′] = 1, even if b = 2.

In particular, [Z_{β}^{∗}, Z_{α′}] = 1 and as S acts transitively on the L_{α}^{(i)} and normalizes C_{Q_{β}}(Z_{b}^{∗}) we conclude

zb ∗ −1a

(2) (a) S ∩ L_{α} = C_{Q_{β}}(Z_{β}^{∗})Q_{α}.

(b) Z_{β}^{∗} ∩ Z_{α} = Z_{αβ}.

By definition of U_{αβ} we have [U_{αβ}, Q_{β}] ≤ Z_{β} and thus

zb ∗ −2

(3) [U_{β}, Q_{β}] ≤ Z_{β}.

In particular, D =^{def} [U_{α−1}, U_{α′−1} ∩ Q_{α−1}] ≤ Z_{α−1}. On the otherhand, by (1)a,U_{α−1} ≤ Z_{α−1}^{∗} ≤ Q_{α′−2} ≤ G_{ap−1} and so D ≤ U_{α′−1} ≤ Z_{ap−1}^{∗} and so by (1)c, [D, V′] = 1. Hence by choice of α − 1, D is centralized by G_{α} and G_{α−1}. Thus

zb ∗ −3

(4) [U_{α−1}, Z_{α′−1}^{∗} ∩ Q_{α−1}] = 1.

Suppose that U_{α−1} ≤ Q_{α′−1}. As [R, U_{α−1}] = 1 we conclude that [U_{α−1}, V′] ≤ R ≤ Z_{α}. Thus

U_{α−1}Z_{α} ⊴ ⟨G_{α−1α}, V′⟩ = G_{α}.

Hence also [U_{α−1}, Q_{α}] ⊴ G_{α}. By (4), Z_{α} ≰ U_{α−1} and since Z_{α} is the unique minimal normal subgroup of G_{α} in Q_{α} we conclude that [U_{α−1}, Q_{α}] = 1. Thus [U_{β}, Q_{α}] = 1. Since E_{β} ≤ ⟨Q_{α}^{G_{β}}⟩T we get [U_{β}, E_{β}] = 1. Note also that [U_{αβ} ≤ ZJ(S) and that there exists 1 ≠ D ≤ U_{αβ} with C_{G_{α}}(D) ≰ M_{αβ}. Hence we obtain a contradiction to 9a. We proved

zb ∗ −4

(5) (a) [U_{β}, E_{β}] ≠ 1.

(b) U_{α−1} ≰ Q_{α′−1}.

If [U_{α−1} ∩ Q_{α′−1}, U_{α′−1}] = 1, then 8.17 and (4) imply that 3. holds. Thus we may assume:

zb ∗ −5

(6) Z_{α′−1} = [U_{α−1} ∩ Q_{α′−1}, U_{α′−1}] ≤ U_{α−1}

Suppose that b = 2. Then by (6) and (2)b, Z_{β} = Z_{α′−1} ≤ U_{α−1} ∩ Z_{α} ≤ Z_{α−1}^{∗} ∩ Z_{α} = Z_{αα−1}. But this contradicts the choice of α − 1. Hence

zb ∗ −6

(7) b ≥ 4.

By (6), there exists λ ∈ △(α′ − 1) and t ∈ U_{α−1} ∩ Q_{α′−1} with [t, U_{α′−1λ}] = Z_{α′−1}.

Suppose t normalizes one of the Z_{λ}^{(i)} and let X be the sum of the Z_{λ}^{(j)}, j ≠ i. Then U_{α′−1λ} = U_{α′−1λ} ∩ Z_{λ}^{(i)} ⊕ U_{α′−1λ} ∩ X, t centralise U_{α′−1λ} ∩ Z_{λ}^{i} and so Z_{α′−1} = [U_{α′−1λ}, t] ≤ [X, t] ≤ X, a contradiction.

zb ∗ −7

(8) t acts fixed-point freely on {L_{λ}^{(i)} | 1 ≤ i ≤ k}.

Thus by 2.2 and (2)a there exists μ ∈ △(λ) with O^{2}(G_{λ}) ≤ ⟨C_{Q_{μ})}(Z_{μ}^{∗}), t⟩. As t centralizes Z_{α}, (8) implies that Z_{α} ≤ Q_{λ}. Moreover, U_{μ} ≤ Q_{α+2} ≤ G_{β} and so [V_{β}, U_{μ}] ≤ U_{μ} ∩ V_{β}. Since b ≥ 4, we conclude from (1)a that U_{α−1} and so also t centralizes [V_{β}, U_{μ}]. Since C_{Q_{λ}}(O^{2}(G_{λ})) = 1 the choice of μ implies [V_{β}, U_{μ}] = 1 and so

zb ∗ −8

(9) U_{μ} ≤ Q_{β} ∩ Q_{α} ≤ G_{α−1}.

Since d(μ, α′) = 3 < b, (2) implies [⟨U_{μ}^{G_{λ}}⟩, V′ = 1. Thus [t, U_{μ} ∩ Q_{α−1}] ≤ Z_{α−1}(V′) = 1. From C_{U_{μ}}(t) ≤ C_{Q_{λ}}(O^{2}(G_{λ})) = 1 we get

zb ∗ −9

(10) U_{μ} ∩ Q_{α−1} = 1

Thus

|U_{α−1}/C_{U_{α−1}}(U_{μ})| ≤ |U_{α−1}| = |U_{μ}| = |U_{μ}Q_{α−1}/Q_{α−1}|

and 3. holds.

pred

Lemma 9.7 Suppose that [Z_{α}, Z_{α′}] ≠ 1 and Z_{αβ} is normal in G_{β}. Then b = 2, E_{β} centralizes Z_{αβ} and G_{a} is of L_{2}-type.

Proof: By8.15 Q_{α} ≰ Q_{β}. As Q_{α} centralizes Z_{αβ} and E_{β} ≤ ⟨Q_{α}^{G_{β}}⟩ we conclude that E_{β} centralizes Z_{αβ}. Note that V ∩ Z_{αβ} ≠ 1 and so by 9, C_{G_{α}}(V ∩ Z_{αβ} ≤ M_{αβ}. Thus k = 1 and G_{α} is of L_{2}-type. It remains to show that b = 2.

Suppose that b > 2. Let α − 1 ∈ △(α) with ⟨G_{α−1α}, V ⟩ = G_{α} and note that R = Z_{β}^{∗} = Z_{α′−1}^{∗} is normalized by G_{β} and G_{α′−1}. Hence 9(d) implies that V_{α−1} ≤ G_{α′}. As V_{α−1} centralizes R we conclude that [V_{α−1}, Z_{α′}] ≤ R and G_{α} normalizes V_{α−1}, again a contradiction.

2za

Lemma 9.8 Suppose that [Z_{α}, Z_{α′}] = 1, b > 1 and r_{β} > 1. Then there exists a normal subgroups L_{α} of G_{α} and normal subgroups L_{α}^{(i}, 1 ≤ i ≤ k of L_{α} such that

(a) C_{α} ≤ L_{α} and C_{α} ≤ L_{α}^{(i)}

(b) \overline{O^{2}(L_{α})} = \overline{L_{α}^{(1)}} × … × \overline{L_{α}^{(k)}}

(c) G_{α} = L_{α}S, S transitively permutes the L_{α}^{(i)}’s and L_{α} is the largest subgroup of G_{α} normalizing all the L_{α}^{(i)}’s.

(d) Put Z_{α}^{(i)} = [Z_{α}, L_{α}^{(i)}]. Then Z_{α} = Z_{α}^{(1)} ⊕ … ⊕ Z_{α}^{(k)}.

(e) One of the following holds

1. \overline{L_{α}^{(i)}} ≅ SL_{2}(q), q a power of 2 and Z_{α}^{(i)} is a natural SL_{2}(q)-module for L_{α}^{(i)}.

2. \overline{L_{α}^{(i)}} ≅ C_{3}, |Z_{α}^{(i)}| = 4 and s_{Z_{α}}(O_{2}(M_{β})) < 2.

3. \overline{L_{α}^{(i)}} ≅ SL_{3}(q), q a power of 2; Z_{α}^{(i)} is direct sum of a natural SL_{3}(q)-module for

L_{α}^{(i)} with its dual; some element of S induces a graph automorphism on \overline{L_{a}^{(i)}} and c_{β} = 2

Proof: Suppose first that c_{β} = 1. Then the lemma holds by 8.4 and 6.3, where the Sym(q + 1) case is excluded as in 9.5.

So suppose that c_{β} ≥ 2. Then r_{β}c_{β} − 1 > 1 and so by 2.4a, r_{α} < 2. Thus we can apply 6.4 with the Sym(q + 1)-case excluded as usual. Note that in case (e3) we actually have r_{a} = \frac{3}{2}. As r_{β} > 1, 2.4 implies c_{β} = 2 and all parts of the lemma are proved.

Put Z_{αβ} = C_{Z_{α}}(L_{α} ∩ S) and Z_{β}^{∗} = ⟨Z_{αβ}^{G_{β}}⟩.

l3q.2

Lemma 9.9 Suppose that [Z_{α}, Z_{α′−1}] = 1, b > 1 and the conclusions of 9.8 hold for case e3 hold. Then Q_{β}Q_{α}/Q_{α} ≤ Z(S ∩ L_{α}/Q_{α}), [X_{β}, E_{β}] ≠ 1 and X_{β} is an F F -module.

Proof: Suppose that E_{β} centralizes Z_{αβ} and let D be the intersection of Z_{αβ} with one of the irreducible L_{α} submodule in Z_{α}. Then D ≠ 1, N_{G_{α}}(D) ≰ M_{αβ} and E_{β}B(S) centralizes D, a contradiction to 9a.

Thus E_{β} does not centralize Z_{αβ}.

Recall that c_{β} = 2 in case 9.8e3. Thus 8.9 applied to L = E_{b} shows that [Z_{α}, Q_{β}, 2] ≤ D_{β}. By 8.15 Q_{α} Q̸_{β}. Hence E_{β} ≤ ⟨Q_{a}^{G_{β}}⟩ and so [D_{β}, E_{β}] = 1. In particular Z_{αβ} ≰ D_{β} and so Z_{αβ} ≰ [Z_{α}, Q_{β}, 2]. As S normalizes [Z_{α}, Q_{β}, 2] we conclude from the action of S on Z_{α} that [Z_{α}, Q_{β}, 2] < Z_{αβ}. Since Q_{β} is normal in S this implies that Q_{β} ≤ L_{α} and then that Q_{β} acts quadratically on each of the irreducible L_{α} submodules in Z_{α}. As S normalizes Q_{β} and induces a graph automorphism on the L_{α}^{(1)} we get Q_{β}Q_{α}/Q_{a} ≤ Z(S ∩ L_{α}/Q_{α}) and Z_{αβ} ≤ X_{β}. Hence [X_{β}, E_{β}] ≠ 1 and so by ?? X_{β} is an F F -module.

l2k

Lemma 9.10 Suppose that [Z_{α}, Z_{α′−1}] = 1, b > 1 and the conclusions of 9.8 hold for case e1 or e2 hold. Then one of the following is true:

1. k = 1, [Z_{αβ}, E_{β}] = 1 and V_{β} is an F F -module for G_{β}

2. k = 1, b = 3 and V_{β} is an F 2-module.

3. [Z_{αβ}, E_{β}] ≠ 1 and X_{β} is an F F -module.

4. q = 2, k ≥ 2 and [Z_{αβ}, E_{β}] ≠ 1. Let U_{αβ} be maximal in Z_{αβ} with [U_{αβ}, Q_{β}] ≤ Z_{β} and put U_{β} = ⟨U_{αβ}^{G_{β}}⟩. Then U_{β} is an F F -module for G_{β}.

Proof: By 9a, [Z_{αβ}, E_{β}] = 1 implies, k = 1. Suppose that q > 2 or k = 1. Then Q_{β}^{∗} ≤ O_{2}(M_{β}) ≤ Q_{αβ} ≤ L_{α} and so Z_{αβ} ≤ X_{β}^{∗}. So if

in addition [Z_{αβ}, E_{β}] ≠ 1, then ?? implies that 3. holds. Hence we may assume from now on that

l2k − 1

(1) One of the following holds:

(Case 1) k = 1 and [Z_{aβ}, E_{β}] = 1.

(Case 2) q = 2, k ≥ 2 and [Z_{αβ}, E_{β}] ≠ 1.

Put D_{β}^{∗} = Z_{αβ} ∩ D_{β} and note that in case Case 1, D_{β}^{∗} = Z_{αβ} while in case Case 2 9a implies D_{β}^{∗} = Z_{β}. In Case 1 let U_{αβ} = Z_{α} and in Case 2 let U_{αβ} be maximal in Z_{αβ} with

[U_{αβ}, Q_{β}] ≤ D_{β}^{∗}. Put U_{β} = ⟨U_{αβ}^{G_{β}}⟩. It follows easily from the definitions and 9.2c that:

2z ∗ b (2) (a) [U_{αβ}, E_{β}] ≠ 1

(b) [U_{β}, Q_{β}^{∗}] ≤ [V_{β}, O_{2}(M_{β})] ≤ D_{β}^{∗} ≤ Z_{α}

(c) [U_{β}, Q_{β} ∩ Q_{α+2}] ≤ D_{β}^{∗}.

By 9d applied with D = D_{β}^{∗} ∩ D_{α′−1}^{∗} we get

zbza (3) D_{β}^{∗} ∩ D_{α′−1}^{∗} = 1

By (2)c, [U_{β} ∩ Q_{α′}, U_{α′} ∩ Q_{β}] ≤ D_{β}^{∗} ∩ D_{α′−1}^{∗} = 1 and so

vqvq (4) [U_{β} ∩ Q_{α′}, U_{α′} ∩ Q_{β}] = 1

We may and do assume from now on that U_{β} is not an F F -module and and will show that 2. holds.

Suppose that U_{α′} ≤ Q_{β}. As b ≥ 3, U_{α′} acts quadratically on Z_{α}. Let V be an irreducible L_{α} submodule in Z_{α} with V ≰ Q_{α′}. Assume first that U_{α′} normalizes V. Then

|V/C_{V} (U_{α′}) = q ≥ |U_{α′}/C_{U_{α′}} (V)|.

If q =2, this clearly implies that U_{α′} is an F F -module. If q > 2 we are in Case 2 and so V ≤ U_{β} and by (4), U_{β} ∩ Q_{α′} ≤ C_{V} (U_{α′}). Hence |V Q_{α′}/Q_{α′}| ≥ q. Again U_{α′} is an F F -module, a contradiction.

Thus U_{α′} does not normalizes V and quadratic action implies |U_{α′}/C_{U_{α′}} (V)| ≤ 2, again a contradiction. Thus

sym

(5) U_{α′} ≰ Q_{β} and the situation is symmetric in β and α′.

Suppose that [U_{β}, U_{α′} ∩ Q_{β}] = 1 = [U_{α′}, U_{β} ∩ Q_{α′}]. Then by 8.17 we get that U_{β} is an F F -module. Thus

vvqa

(6) D_{β} = [U_{β}, U_{α′} ∩ Q_{β}] ≤ U_{α′} or D_{α′} ≤ [U_{α′}, U_{β} ∩ Q_{α′}] ≤ U_{β}

Hence we may assume [U_{β}, U_{α′} ∩ Q_{β}] ≠ 1 and so

vvq

(7) D_{β}^{∗} = [U_{β}, U_{α′} ∩ Q_{β}] ≤ U_{α′}.

Pick μ ∈ △(β) and t ∈ U_{α′} ∩ Q_{β} with [U_{μβ}, t] ≠ 1. Then by (4), Z_{μ} ≰ Q_{α′} and we may assume that μ = α. Hence

vbq

(8) There exists t ∈ U_{α′} ∩ Q_{β} with [U_{αβ}, t] ≠ 1. In particular, t ∉ Q_{α}

In particular, by 9.2c, O_{2}(M_{β}) ≰ Q_{α}, as O_{2}(M_{β}) is normal in M_{αβ} we conclude ( compare also (8) in 9.6).

QMbQS

(9) (a) In case 1, O_{2}(M_{β})Q_{a} = S ∩ L_{α}.

(b) In Case 2, t acts fixed point freely on {L_{α}^{(i)} | 1 ≤ i ≤ k}.

In particular, ( also use 2.2 in Case 2) there exists α − 1 ∈ △(α) with

O2G

(10) E_{α} ≤ ⟨O_{2}(M_{α−1}) ∩ Lα, t⟩.

By (4) and (8) we have |U_{β}Q_{α′}/Q_{α′}| ≥ |U_{αβ}Q_{α′}/Q_{α′}| = |U_{αβ}/C_{U_{αβ}}(t)| ≥ q. We record

vbqa

(11) |U_{β}Q_{α′}/Q_{α′}| ≥ q.

Define Y_{α} = ⋂_{δ∈△(α)} U_{δ}Z_{α}. Suppose now that [U_{α−1}, V_{α′−2}]=1. Then U_{α−1} ≤ Q_{α′−2} ∩ Q_{α′−1}. Put A = U_{α−1} ∩

(U_{β}Q_{α′}). Then A ≤ U_{β}(U_{β}U_{α−1} ∩ Q_{α′}) ≤ U_{β}(Q_{α′−1} ∩ Q_{α′}). Thus by (2)

[A, t] ≤ [U_{β}, t][Q_{α′−1} ∩ Q_{α′}, t] ≤ D_{β}^{∗}D_{α′}^{∗}.

Let X be maximal in A with [X, t] ≤ D_{β}^{∗}. As |D_{α′}^{∗}| = q we have |A/X| ≤ q. Since D_{β}^{∗} ≤ X, t normalizes X. By (2), O_{2}(M_{α−1}) also normalizes XZ_{α}. As E_{α} is transitive on △(α) we conclude from (10) that XZ_{α} ≤ Y_{α}. Put a = |U_{α−1}/A|. Then |U_{α−1}Y_{a}/Y_{a}| ≤ |U_{α−1}/A||A/X| ≤ aq. Hence

|U_{β}Y_{a}/Y_{α}| ≤ aq.

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

|U_{α′}/C_{U_{α′}} (U_{α−1}U_{β})| ≤ |U_{α′}Y_{α′−1}/Y_{α′−1}| = |u_{β}Y_{a}/Y_{α}| ≤ aq.

On the otherhand by definition of a, an isomorphism theorem and (11)

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

By the last two equations, U_{α′} is an F F -module, a contradiction. Hence

va − 1va − 2

(12) [U_{α−1}, V_{α′−2}] ≠ 1

Suppose that V_{α′−2} ≤ Q_{a−1}. Then by (5), V_{α−1} ≤ Q_{α′−2}. Note that by (10), C_{D_{α−1}^{∗}}(t) =

1. Thus

1 ≠ [U_{α−1}, V_{α′−2}] ≤ D_{α−1}^{∗} ∩ D_{α′−2}^{∗} ≤ C_{D_{α−1}^{∗}}(t) = 1

a contradiction to (12). Thus

va − 1qa − 1

(13) V_{α′−2} ≰ Q_{α−1}

In particular, (α′ − 2, α − 1) has the same properties as (β, α′) and we conclude from

(5) that

va − 1qa − 2

(14) U_{α−1} ≰ Q_{α′−2}

Suppose that 1 ≠ x ≤ D_{α′−2}^{∗} ∩ U_{α−1}. As t centralizes x, x ∈ X ≤ Y_{α} and so E_{α} normalizes xZ_{α}.

Suppose first that [x, Q_{α}] ≠ 1. Since E_{α} normalizes [x, Q_{a}], Z_{α}^{(i)} ≤ [x, Q_{a}] for some i. Put L = O^{p}(L_{α}^{(i)}) and Q = [Q_{α}, L]. Then [x, Q_{α}, L] = Z_{α}^{(i)} and [x, L, Q_{α}] = 1. Thus be the three subgroup lemma, [x, Q] = Z_{α}^{(i)} = [x, L]. Since [x, Q, Q] = 1 we colcude that xQ = x^{Q} = x^{L} and so by the Frattini argument, L = C_{L}(x)Q. Since x ≤ D_{α′−2}, x is centralised by E_{α′−2} and the Thompson subgroup of G_{α′−1α′−2}. By the proof of (ba), Mtrick t ∈ V_{α′} ∩ G_{α} ≤ V_{α′−2}^{(3)} ∩ G_{α} ≤ Q_{p}(M_{ap−2} ∩ G_{α}. As C_{L}(x) normalizes Q_{p}(M_{ap−2} ∩ G_{α} we get [t, L] ≤ Q_{α}. In case 1 this is impossible since t ∉ Q_{a} and in Case 2 this contradicts ??b.

Suppose next that [x, Q_{α}] = 1, but x ∉ Z_{α}. Then its is easy to see that q > 2 and C_{E_{a}}(x)Q_{α}/Q_{α} is ismorphic to D_{2·q±1} and again C_{E_{a}}(x) normalizes no non-trivial 2-subgroup in G_{a}/Q_{a} and we get the same contradiction as above.

Hence x ∈ Z_{α} and so D_{α′−2}^{∗} ≤ Z_{α}. Note that t centralizes D_{α′−2}^{∗}. In Case 2 we have n x ∈ Z_{α}, [x, O_{2}(M_{α−1}∩L_{α})] ≤ Z_{α−1} and s_{Z_{α}}(O_{2}(M_{α−1}∩L_{α}) < 2 implies, [x, O_{2}(M_{α−1}∩L_{α})] = 1. Hence by (10), [x, E_{α}] = 1 a contradiction to C_{Q_{α}}(E_{α}) = 1.

In case Case 1 we conclude that D_{α′−2}^{∗} = D_{β}^{∗}. If b > 3, 9bb implies that V_{α−1} ≤ V_{β}^{(3)} ≤ Q_{α′−2}, a contradiction. We have proved

dcu

(15) If D_{α′−2}^{∗} ∩ U_{α−1} ≠ 1, then b = 3 and Case 1 holds.

Assume that b > 3. Then t centralizes [U_{α′−2} ∩Q_{α−1}, U_{α−1}] and as by (10) C_{D_{α−1}^{∗}}(t) = 1 we get [U_{α′} − 2 ∩ Q_{α−1}U_{α−1}] = 1. Thus by (6) and ?? that D_{α′−2}^{∗} = [U_{α−1} ∩ Q_{α′−2}, U_{α′−1}] ≤ U_{α−1} a contradiction to (15). Thus

l2kb3

(16) b = 3.

Suppose that k > 1. By (6) applied to (α − 1, β) in place of (β, α′) we get Z_{α−1} = D_{α−1}^{∗} ≤ U_{β} or Z_{β} = D_{b}^{∗} ≤ U_{α′−1}. In the first case [Z_{α−1}, O_{2}(M_{β}) ≤ Z_{β} and as above so [Z_{α−1}, O_{2}(M_{β} ∩ L_{α}) = 1. But this implies Z_{α−1} ≤ Z_{αβ} and Z_{αα−1} = Z_{αβ} a contradiction to (10). The second case yields the same contradicion.

Thus k = 1 and so V_{β} = U_{β}. By (4) and ??, V_{β} is F 2 and so 2. holds.

We remark that an example for case 2 of the previous theorem occurs in ^{2}F_{4}(q). In that example V_{β} is exactly F 2 ( that is not F^{∗}2)

10 Connected parabolics not normalizing Z

UII
HUII

Hypothesis 10.1 (a) Hypothesis 8.1 holds.

(b) C_{G_{α}}(Y_{α}) is p-closed.

(c) G_{β} is p-minimal.

(d) Y_{α} is neither an F F nor an dual F F -modul.

Remark: ”b” in this section is defined with respect to Y_{γ} not Z_{γ}

dmab

Definition 10.2 M_{αβ} is the unique maximal subgroup of G_{β} containing S.

zc

Lemma 10.3 b is odd, Z ⊴ G_{β} and [X_{β}, E_{β}] = 1.

Proof: By 8.17 b is odd and as p-minimal groups have no F 1^{∗}-module, Z ⊴ G_{β}. Since Y_{α} is not F F, J(S) ≰ Q_{β}. If [X_{β}, E_{β}] ≠ 1, we conclude that X_{β} is F F. As G_{β} is p-minimal this gives the contradiction, Z ⋬ G_{β}.

qbniqa

Lemma 10.4 Q_{β}^{∗} ≰ Q_{α} and Q_{α} ≰ Q_{β}.

Proof: Suppose that Q_{β}^{∗} ≤ Q_{α}. Then [V_{β}, Q_{β}^{∗}] = 1 and so by Thompson’s P × Q-Lemma, [X_{β} ∩ V_{β}, L_{β}] ≠ 1, a contradiction to 10.3. The second statement holds since

Z_{α} ≤ Q_{α′−2} ∩ Q_{α′−1} ≤ Q_{α′−1}^{∗}Q_{α′}.
sb

Lemma 10.5 (a) r_{V_{β}}(G_{β}) ≤ 1.

(b) c_{β} ≥ 2.

(c) q_{α} ≤ 2.

Proof: (a) holds since G_{β} is p-minimal. Since Q_{α} ≰ Q_{β} and Q_{β}^{∗} ≰ Q_{α}, Q_{α} ∩ Q_{b} is not normal in G_{β}. Thus by 8.4b, (b) holds. Hence by 8.4a also c. is true.

nLLp

Lemma 10.6 Suppose that b > 1.

(a) 8.21.1 or 8.21.2 holds.

(b) For each ϵ in 8.211. or 2., L^{ϵ} has at least two non trivial chief-factors on V^{ϵ}.

(c) In case ?? q_{α} < \frac{1+\sqrt{17}}{4}.

(a) Suppose that 8.21.3 holds. Then by 8.17 one of Z_{μ^{+}} and Z_{μ^{−}} is F F. But then Z_{α} is F F, a contradiction.

(b) Suppose L^{ϵ} has at most one non-central chief factor on V^{ϵ}. Since L^{ϵ} and G_{β^{ϵ}} are pminimal, 2. implies L^{ϵ} = O^{p}(L^{ϵ})(G_{β^{ϵ}μ^{ϵ}} ∩ L^{ϵ}) and G_{β^{ϵ}} = ⟨G_{β^{ϵ}μ^{ϵ}}, L^{ϵ}⟩. Thus we can apply 8.5 to (μ^{ϵ}, β^{ϵ} in place of (α, β). Since by assumption α is not a dual FF- module we conclude that V_{β} ≤ Z_{α}X_{β}. But then [V_{β}, Q_{α}] ≤ X_{β} and so [V_{β}, E_{β}] ≤ X_{β} and [V_{β}, E_{β}] = 1, a contradiction.

(c) Suppose that q_{a} ≥ \frac{1+\sqrt{17}}{4}. Put Λ^{+} = μ^{+L^{+}} and Λ^{−} = \{μ\}. Abusing notation define V^{+}, V^{−},B^{+} and B^{−} as in that lemma. Note that V^{+} is the same V^{+} as defined before, but V^{−} now is Z_{μ}. Also B^{+} = V^{+} and B^{−} = Z_{μ} ∩ O_{p}(L^{+}). In particular, V^{−} ≠ B^{−} and V^{+} = B^{+}. We wish to apply 8.20e with r = 2 and s = q_{α}. By ?? and since L^{+} is p-minimial, |Z_{μ}/B^{−}|^{2} ≤ |V^{+}/C_{V^{+}}(Z_{μ})|. Also |V^{+}/B^{+}|^{2} = 1 ≤ |Z_{μ}/C_{Z_{μ}}(V^{+})| and so the asumptions of 8.20e are indeed fullfiled for this choice of r and s. Also e > 0 by 2.1a. Thus 8.20e gives the contradiction V^{+} ≠ B^{+}.

zf1

Proposition 10.7 There exists 1 ≠ x ∈ Z_{α} and λ ∈ Γ with d(α, λ) = b and Z_{α} ≰ O_{p}(C_{G_{λ}})(x).

Proof: Suppose the lemma is false. Then by 10.3 b > 1 and we can apply 8.21. In case 8.21.1 we assume without loss that α ∈ μ^{+L^{+}} with Z_{α} ∉ O_{p}(L^{−}). Put Q = O_{p}(L^{+}).

In case 8.21.2 we assume ϵ = − and α = μ. Put Q = G_{α} and V^{+} = Z_{α}.

In each case note that by 8.21 the assumptions of 2.8 with H = L^{−}, V = V^{−}, A = Z_{α} and Z = Z_{μ^{−}} are fulfilled.

znn − .5

(1) V^{−} ∩ Q ≤ G_{α} and C_{V^{−}}(Z_{α}) = C_{V^{−}}(V^{+}) ≤ V^{−} ∩ Q

In case 8.21.2 there is nothing to prove. So suppose 8.21.1 holds. Then O_{p}(L^{+}) ≤ G_{α} and so the first statement holds. The second follows from 2.8a.

zn1 − 1

(2) [Z_{α} ∩ O_{p}(L^{−}), V^{−} ∩ Q] = 1

Suppose 1 ≠ x ∈ [Z_{α} ∩ O_{p}(L^{−}), V^{−} ∩ Q]. Then x ∈ Z_{α} . Thus by 2.8d, Z_{α} ≰ O_{p}(C_{L}(x)) and so also Z_{α} ≰ O_{p}(C_{G_{α′}} (x), a contradiction.

Since L^{−} has at least two non-central chief-factors on V^{−} and as Z_{α} is not F F we now compute |V^{−}/V^{−}∩Q|||V^{−}∩Q/C_{V^{−}}(V^{+})| = |V^{−}/C_{V^{−}}(V^{+})| = |V^{−}/C_{V^{−}}(Z_{α})| ≥ |Z_{α}/Z_{α}∩O_{p}(L^{−})|^{2} ≥ |Z_{α}/C_{Z_{α}}(V^{−}∩Q)|^{2} ≥ |V^{−}∩Q/C_{V^{∩}Q}(Z_{α})|^{2} = |V^{−}∩Q/C_{V^{−}}(V^{+})|^{2}

(1)

Hence

|V^{−}/V^{−} ∩ Q| ≥ |V^{−} ∩ Q/C_{V^{−}}(V^{+})|.  (2)

In case of 8.21.2 we conclude V^{−} = V_{V^{−}}(Z_{α}), a contradiction. Thus

zn1 − 3

(3) 8.21.1 holds.

In particular, the sitution is symmetric in + and − and Q = O_{p}(L^{+}). Since by 8.21.1, L^{+} has two non-central chief factor on V^{+},

|V^{+}/C_{V^{+}}(V^{−})| ≥ |V^{−}Q/Q|^{2} = |V^{−}/V^{−} ∩ Q||V^{−}/V^{−} ∩ Q|

and so by (2)

|V^{+}/C_{V^{+}}(V^{−})| ≥ |V^{−}/V^{−} ∩ Q||V^{−} ∩ Q/C_{V^{−}}(V^{+})| = |V^{−}/C_{V^{−}}(V^{+})|

But the same inequality holds with the roles of + and − are interchanged. Hence equality holds here and also in (1). But has Z_{α} is not F F this is only posibble if V^{−} ∩ Q centralizes Z_{α}. But then all the numbers compared in (1) are equal to 1 and so V^{−} = C_{V^{−}}(V^{+}), a contradiction which completes the proof of ??.

ffb1

Theorem 10.8 Suppose G is of generic p-type, S ∈ Syl_{p}(G) and V is a maximal member of {Y_{L} | L ∈ L(S). Then either V is an F F -or dual F F -module for S or V ≰ O_{p}(C_{G}(Z)).

Proof: Let M = N_{G}(V) and L = N_{G}(C_{S}(V)). Then M is the unique maximal p-local of G containing L. Let G_{α} = L and H a p-minimal member of L(S) not contained in M. Suppose that V is neither F F nor dual F F for S. Then the assumptions of this section are fulfilled. Hence by ?? there exists a p-local subgroup H with O_{p}(L) ≤ H and V ≰ O_{p}(H). Choose such an H with |H ∩ M|_{p} maximal and then |H|-minimal. Let R be a Sylow p-subgroup of H ∩ M with O_{p}(L) ≤ R. Since O_{p}(L) is a Sylow psubgroup of C_{G}(V), O_{p}(L) = C_{R}(V) ⊴ R and so R ≤ L. Without loss R ≤ S.

Since O_{p}(L) ≤ R and V is not F F, J(R) ≤ O_{p}(L). Hence L ≤ N_{G}(J(R)) and so N_{G}(J(R)) ≤ M. Thus N_{H}(J(R)) ≤ M and in particular, N_{H}(R) ≤ M. Thus R is a Sylow p-subgroup of H.

Let W = Z_{H} and suppose that [W, V] ≠ 1. Since W ≤ O_{p}(H) ≤ R ≤ S, |V/C_{V} (W) > |W/C_{W} (V). Thus V is F ∗1 on W. By the minimality of H, V ≤ O_{p}(P) for all P ∈ L(H, S) with P ≠ H and contradiction to ??

Hence V centralizes W By minimiality of H, H = ⟨V^{H}⟩R and so Ω_{1}(Z(R)) = W ≤ Z(H). Thus V ≰ O_{p}(N_{G}(W). By maximaliy if |H ∩ M|, R is a Sylow p-subgroup of M ∩ N_{G}(W). Thus N_{S}(R) ≤ N_{S}(W) ≤ R, R = S and W = Z. Thus the theorem is proved

CV anV

Lemma 10.9 There exists 1 neqA ≤ SC_{α}/C_{α} with

(a) [Z_{α}, A, A] = 1

(b) |Z_{α}/C_{Z_{α}}(A)| ≤ |A^{2}|.

(c) ⟨C_{Z_{α}}(a) | a ∈ A^{#}⟩ ≠ Z_{α}.

(d) If 8.212 holds,then |Z_{α}/C_{Z_{α}}(A)| < |A^{\frac{3}{2}} |.

Remark: We proof contains more information than stated in the lemma Proof: Let L^{ϵ}, μ^{ϵ} and μ as in 8.21.

In case of 8.211. may assume without loss that |V^{+}/C_{V^{+}}(V^{−})| ≤ |V^{−}/C_{V^{−}}(V^{+})|.Pick μ ∈ μ^{+L^{+}} with Z_{μ} ≰ O_{p}(L^{−}) and put B^{−} = V^{−} ∩ O_{p}(L^{+})

In case of 8.212 we assume without loss ϵ = −. Put V^{+} = Z_{μ} and B^{+} = Z_{μ} ∩ O_{p}(L^{−}). In general pick t ∈ Z_{μ} \ O_{p}(L^{−}). By 8.21 the assumptions for 2.8 are fulfilled with

H = L^{−}, A = V^{+},V = V^{−} and Z = Z_{μ^{−}}. We conclude that C_{V^{−}}(t) = C_{V^{−}}(V^{+}). Thus

⟨C_{Z_{μ}}(a) | a ∈ B^{−} \ C_{μ}⟨≤ Z_{μ} ∩ O_{p}(L^{−}).

Suppose now that 8.211. holds and define s by |B^{−}/C_{B^{−}}(Z_{μ})|^{s} = |Z_{μ}/C_{Z_{μ}}(B^{−})|. Note that that C_{B^{−}}(Z_{μ}) ≤ C_{B^{−}}(t) ≤ C_{B^{−}}(V^{+}). Let c be the number of non-central chief-factors for L^{+} on V^{+}. By 2.8 |V^{−}/B^{−}|^{c} ≤ |V_{+}/C_{V^{+}}(V^{−}). Then by 2.4b, (with A = V^{−}, V = V^{+}, ”s = s”, t ≥ 1, r ≥ c ≥ 2) we get that s ≥ 2. Thus the lemma holds in this case with A = B^{−}C_{μ}/C_{μ}( and μ in place of α).

Suppose next that 8.212 holds. As L^{−} has at least two non-trivial chief-factors on V^{−}, we conclude from ?? that

|Z_{μ}/B^{+}|^{2} ≤ |V^{−}/C_{V}^{−}(Z_{μ}).

On the otherhand has Z_{α} is not F F, 2.4a implies

|B^{+}/C_{B^{+}}(V^{−})| < |V^{−}/V_{V^{−}}(B^{+})| ≤ (V^{−}/C_{V^{−}}(Z_{μ}).

Combining the last two inequalites we get |Z_{μ}/C_{Z_{μ}}(V^{−})| ≤ |V^{−}/C_{V^{−}}(Z_{μ})|^{\frac{3}{2}} . Hence the lamma holds also in this case with A = V^{−}C_{μ}/C_{μ}.

zair

Lemma 10.10 Either Z_{α} is irreducible as G_{α} module or some non-trivial chief-factor for G_{α} on Z_{α} is F F.

Proof: Since [Z, E_{β}] = 1, C_{Z_{α}}(E_{α}). Since Z_{α} is CS-generated, we conclude Z_{α} = [Z_{α}, E_{α}]. So if G_{α} a unique non-central chief-factor, Z_{α} is irreducible. If Z_{α} has more than one noncentral chief-factor, then as Z_{α} is F 2 and G_{α} is p-connected, at least one chief-factor is F F.

nospor

Proposition 10.11 Let U be a non-trivial chief-factor for E_{α} on Z_{α}. Let E = E_{α}/C_{E_{a}}(U). Then one of the following holds:

1. E is solvable and one of the following holds:

1.1. p = 2, E ≅ C_{3} and |U| = 2^{2}.

1.2. p = 3, E ≅ Q_{8} and |U| = 3^{2}.

1.3. p = 2, E ≅ C_{5} and |U| = 2^{4}.

1.4. p = 2, E ≅ Ext(3^{1+2}) and |U| = 2^{6}.

1.5. p = 3, E ≅ Ext(2_{+}^{1+4}) and |U| = 3^{4}.

2. E is perfect but Sol(E) ≰ Z(E) and one of the folloing holds.

2.1. p = 2, E = (C_{3} ≀ Alt(n))′, n ≥ 5 and |U| = 2^{2n}.

2.2. p = 3, E = Ext(2_{−}^{1+4}).Alt(5) and |U| = 3^{4}.

3. E is quasisimple and one of the following holds.

3.1. E is group of Lie type in charcateristic p.

3.2. p = 2 and E/Z(E) is an alternating group.

3.3. p = 2, E ≅ 3.U_{4}(3) and |U| = 2^{12}.

4. E = E_{1}E_{2} for some components E_{1}, E_{2} of E, E_{1} and E_{2} are isomorphic groups of Lie type in charactaristic p, U = U_{1} ⊗ U_{2} for some U_{i} module E_{i} such that (E_{1}, U_{1}) and (E_{2}, U_{2}) isomorphic. Moreover, if n is the dimension of U_{i} over End_{E_{i}}(U_{i}) then U_{i} is a quadratic F \frac{2}{n}-module for E_{i}.

Proof: Let W be a non-trivial chief-factor for G_{α} on Z_{α}. By 10.9 Z_{α} is quadratic F 2 and since G_{α} is p-connected, W is quadratic F_{2}. Let Let H = G_{α}/C_{G_{α}}(W) and L = ⟨PQ_{2}^{∗}(G_{α}/C_{G_{α}}(V), V). As p-connected O^{p}(H) ≤ L. Let V be a Wedderburn component for H on W. Since N_{H}(V) is irreducible on V and N_{H}(V)/L is a p-group, V is irreducible for L. Hence we can apply 6.11 to \overline{L} = L/C_{L}(V). In particular we see that ( except in case 6.114 with p = 2) O^{p}(L) is irreducible on V and clearly any chiefactor for E_{α} on Z_{α} arises in such a way. Moreover, since G_{α} is p-connected, Case 8 of 6.11 does not arise and in case 9, C_{L}(Δ) is a 3-group. Thus it remains to show that in cases 10, 11 the componets of L are groups of Lie type or E(L)/Z(E(L)) is an alternating group. But this is clear in case 11 and so we may assume that E(L) is quasi simple and neither an alternating group, a group of Lietype in characteristic p nor 3.U_{4}(3)

Then G_{α} has no F F -module and so W is the unique non-trivial composition factor for G_{α} on Z_{α} and as Z ⊴ G_{β} we get that Z_{α} is irreducible. We conclude that E_{α}C_{α}/C_{a} the central product of its components L^{(}i), 1 ≤ 1 ≤ n and Z_{α} the the direct sum of the Z_{α}^{i} = [Z_{α}, L^{(}i)]. By 6.15b L^{(}i) is isomorphic to 3..Mat_{22}

Let A be as in 10.9 and put X = ⟨C_{Z_{α}}(a) | a ∈ A^{#}⟩ ≠ Z_{α}. Pick V = Z_{α}^{(}i) so that V ≰ X and pick t ∈ V \ X. Then C_{A}(t) = 1 and so A acts faithfully, quadratic and F 2 on V. Thus by 6.15b, A ≥ 2^{3} and 6.152.3 or 2.4 hold. Let a ∈ A^{#}. Then C_{V} (a) ≠ C_{V} (A) and so C_{V} (A) < X ∩ V < V. Since X ∩ V is invariant under N_{G_{α}}(A) we conclude that case 2.4 with |A| = 2^{3} holds. Note that V is actually a 6-dimensional space GF (4). Each a ∈ A^{#} C_{V} (a)/C_{V} (a) is 1-dimensional over GF (4) and differnt a’s give different 1-spaces. Hence X/C_{V} (A) contains 7 different GF (4)-1-spaces and so X = V, a contradiction.

11 The case b = 1 with G_{α} connected and G_{β} minimal

b1c
hyb1c

Hypothesis 11.1 (a) Hypothesis 8.1 holds, except for the S ≤ G_{α} ∩ G_{β} we only assume Q_{α} ≤ S and S ∈ Syl_{p}(G_{β}).

(b) G_{α} is p-connected.

(c) b = 1, that is Z_{α} ≰ Q_{β}.

dV Mab

Definition 11.2 (a) V is a normal subgroup of G_{β} minimal with respect to [V, E_{β}] ≠ 1.

(b) M_{αβ} is the unique maximal subgroup of G_{β} containing S.

qmzff

Lemma 11.3 Suppose that G_{β} is p-minimal. Then either [Q_{α}, E_{α}] ≤ Z_{α} or Q_{a}/Z_{a} has a unique non-central chief-factor and that chief-factor is F F.

Proof: Let D = [V, Q_{β}]. Then [D, E_{β}] = 1. Also note that V = [V, E_{β}] and since E_{β} ≤ ⟨Z_{α}^{G_{β}}⟩ we conclude that V = ⟨[V, Z_{α}]^{G_{β}}⟩. Thus D = ⟨[V, Z_{α}, Q_{β}]^{G_{β}}⟩. Since [V, Z_{α}, Q_{β}] is normalized by SE_{β} = G_{β} we conclude that D = [V, Z_{α}, Q_{β}] ≤ Z_{α}. Let \overline{V} = V/D. Then [V, Z_{α}, Q_{α}] ≤ [Z_{α}, Q_{α}] = 1. So let R be maximal in Q_{α} with [\overline{V} , R] ≤ [\overline{V} , Z_{α}]. Then by 6.18,

|Q_{α}/R| ≤ |\overline{V} /C\overline{V} (Z_{α})| ≤ |V/C_{V} (Z_{α}) = |V Q_{α}/Q_{α}|

Also [R, V] ≤ [V, Z_{α}]D ≤ Z_{α}. Let Q̃_{α} = Q_{α}/Z_{α} , we conclude

|Q̃_{α}/CQ̃_{α}(V)| ≤ |V Q_{α}/Q_{α}|.

.

Futhermore, [V, Z_{α}] ≠ 1 and so V ≰ Q_{α}. It remains to show that G_{α} has at most one non-central chief-factor on Q̃_{α}. So suppose [Q̃α, E_{α}] ≠ 1 and let P be a normal subgroup of G_{α} minimal with respect to [P, E_{α}] ≰ Z_{α}. Then [P, V] ≰ Z_{α}] and so P ≰ R. By 6.18, we conclude [\overline{V} , P] = [\overline{V} , Q_{α}] and so [Q_{α}, V] ≤ [P, V_{α}] ≤ P. Hence [Q_{α}, E_{α}] ≤ P and the lemma is proved.

zaf2

Lemma 11.4 Z_{α} is a cubic F 2-module for G_{α}.

Proof: Remark: 1. There should be a much nicer proof which does not go through the list of finite simple groups

2. The structure of L has determined in proof should be recorded as an independent lemma

Assume that Z_{α} is not F F and let L be minimal such that

(i) Z_{α} ≤ L.

(ii) Z_{α} ≰ O_{p}(L).

(iii) G_{α} ∩ L contains a Sylow p-subgroup T of L.

(iv) C_{L}(O_{p}(L)) ≤ O_{p}(L).

By minimality of L, L = ⟨Z_{α}^{L}⟩. Let R be a normal subgroup of L with L ≠ RZ_{α}. Then again by minimality Z_{α} ≤ O_{p}(RZ_{0}). Thus [R, Z_{α}] ≤ O_{p}(R) ≤ O_{p}(L) and [R, L] ≤ O_{p}(L). In particular L is p-connected. Let V be a non-central chief-factor for L on O_{p}(L). Since O_{p}(L) ≤ T ≤ G_{α}, Z_{α} acts quadratically on O_{p}(L) and so also on W. Let L̃ = L/C_{L}(W). If |Z̃_{α}| = 2, then L/O_{2}(L) is a dihedral group. If |Z̃_{α}| ≥ 3, we can apply 6.17 to L̃ and W. So in any case we conclude that one of the following holds ( where we used the minimalty of L to rule out some of the cases)

1. p = 2 and L̃ ≅ Dih(2r), r and odd prime.

2. F^{∗}(L̃) is quasisimple.

3. p = 3 and L̃ ≅ SL_{2}(3).

Suppose first that Z_{α} lies in a unique maximal subgroup M of L. Put

Put A = Z_{α}, B = A ∩ O_{p}(L) and Q = ⟨B^{L}. Let l ∈ L \ M. Then L = ⟨A, A^{l}⟩ and so as [Q, A] ≤ B, Q = BB^{l}. Moreover, B ∩ B^{l} = C_{B^{l}}(L) = C_{B^{l}}(A). And so

B^{l}/C_{B}^{l}(A) = B^{l}/B ∩ B^{l} = |Q/B| = |Q/C_{Q}(A)| ≥ |AQ/Q| = |A/B|

where the last inequality holds has L is F ∗ 1-modules.

Now |B/C_{B}(B^{l})| ≤ |B/C_{B}(A^{l})| = |B/B ∩ B^{l} = |B^{l}/C_{B^{l}}(A).

Hence B^{l} is F 2 on A. Since [A, B^{l}] ≤ Q and B^{l} is quadratic on Q, B^{l} is cubic on A. Thus the lemma holds in this case.

So we may assume form now on that A lies in more one one maximal subgroup of L. In particular, K = F^{∗}(L̃) is quasi simple. Let T ≤ M < L. Then by minimality of L, A ≤ O_{p}(M) ≤ T. Put Q_{M} = ⟨A^{M}⟩. If Q_{M} is not abelian, then [A, A^{m}] ≠ 1 for some m ∈ M. But then A is F F on A^{l} or A^{l} is F F on A, a contradiction. Hence Q_{M} is abelain for all such M and so acts quadratically on Q. Let 1 ≠ ã ∈ Ã ∩ Z(T̃). We conclude

zaf2 − 1

(1) Ã lies in an abelian normal subgroup of CL̃(ã) which acts quadratically on Q.

Suppose next K is not a group of Lie type in characteristic p. Then p = 2 or 3. If p = 3, then |Ã| = 3 and Ã lies subgroup of L is morphic to SL_{2}(3), a contradiction to the minimality of L. So p = 2. Since |Ã| ≥ 2, 6.15 and (1) apply L̃ ≅ 3.Mat_{22}, Aut(Mat(22)) or Mat_{24}. But in each of these cases there exists a overgroup of M̃ which does not have a non trivial quadratic normal subgroup.

We conclude

zaf2 − 2

(2) L is a group of Lie type in characteristic p of rank at least two.

Suppose that Ã is contained in a root group X if Ã. Then X ≤ T and X the Sylow subgroup of some (S)L_{2}(q) in L̃. But this contradcits the minimal choice of L. Hence Ã is not contained in a root group. By (1) and as A is contained in O_{p}(M) for all T ≤ M ≤ L we conclude that p = 2, L ≅ Sp_{2n}(q) or F_{4}(q) and A ≤ Z(T). The minimality of L implies L ≅ Sp_{4}(q). But Sp_{4}(q) has no module on which the O_{p}’s of both parabolic acts quadratically.

12 Elementary results on p-connected groups

p − con
dcn

Definition 12.1 N (S) is the set of all p-connected L ∈ L(S) wh

Remark: change this to N^{∗} and use N for P ∪ E

NS

Lemma 12.2 Let L ∈ L(S). Put E = O^{2}(L). Then L is in N (S) if and only if one of the following holds:

1. L is solvable, E/O_{2}(E) has odd order and for all maximal S invariant normal subgroups N of E, C_{S}(E/N) = O_{2}(L).

1. E is perfect, and E/O_{2,2′}(E) is the direct product of simple groups which are transitively permuted by S.

Proof: It is trivial to verify that (1) and also (2) imply L ∈ N (S). So assume now that L ∈ N (S) and let K be the unique maximal normal subgroup of E with K/O_{2}(K) of odd order. Note that O_{2}(E) ≤ K and by the odd order theorem, K is solvable .

Suppose first that K = E. Let and let N be a maximal S invariant normal subgroup of E. Then NC_{S}(E/N) is normalized by ES = L. Since E N̸C_{S}(E/N) we conclude that C_{S}(N) ≤ O_{2}(L). Thus (1) holds in this case.

Suppose next that E ≠ K and let E^{∗}/K be a minimal L invariant subgroup of E/K. Then E^{∗}/K does not have odd order, S ∩ E^{∗} ≰ K, S ∩ E^{∗} ≰ O_{2}(L) and so E ≤ E^{∗} and E = E^{∗}. As E = O^{2}(E), E/K is not a 2-group and so E/K is not solvable. Thus E/K is the direct product of simple groups transitively permuted by S. Since E′ ∩ S ≰ O_{2}(L), E = E′.

The following is an extended version of a lemma from [St2] which describes the structure of rank 2 groups.

gsr2

Lemma 12.3 Let P_{1}, P_{2} ∈ N (S). Put L = ⟨P_{1}, P_{2}⟩. Let L_{0} be a normal subgroup of L maximal with respect to O^{2}(P_{i}) ≰ L_{0} for i = 1 and i = 2. Let L_{1}/L_{0} be a minimal normal subgroup of L/L_{0}. Then Remark: change L_{1}, L_{0} notation

(a) S ∩ N = O_{2}(L) and L_{0}/O_{2}(L) has odd order.

(b) Let O^{2}(P_{i}) ≤ L_{1} for at least one i ∈ {1, 2}.

(c) If O^{2}(P_{j}) ≰ L_{1}, then P_{j} ≤ N_{L}(L_{1} ∩ S) and O_{2}(O^{2}(P_{i})) ≤ O_{2}(P_{j}).

(d) Suppose that L_{1}/L_{0} = E_{1} × E_{2} × … × E_{r} is the direct product of alternating groups or simple groups of Lie type in characteristic 2. Then P_{j} acts transitively on the E_{l}’s and one of the following holds:

(d.1) O^{2}(P_{j}) ≰ L_{1} and O^{2}(P_{i})L_{0}/L_{0} is the product of some of the E_{l}’s.

(d.2) O^{2}(P_{j}) ≰ L_{1}, E_{1} ≅ D_{4}(q) and some element on P_{j} induces a graph automor-

phism of order 3 on E_{1} (d.3) O^{2}(P_{j}) ≤ L_{1}, j = 1, 2, L = L_{1}S = ⟨E_{1}^{S}⟩S and E_{1} = ⟨E_{1} ∩ P_{1}, E_{1} ∩ P_{2}⟩.( modulo

L_{0})

Proof: As O_{2}(L)L_{0} ∩ P_{i} = O_{2}(L)(L_{0} ∩ P_{i}) ≤ O_{2}(P_{i}) the maximality ofL_{0} implies O_{2}(L) ≤ L_{0}. Let N be a normal subgroup of L and k ∈ {1, 2}.

We next prove that

gsr2 − 1

(1) Suppose that S ∩ N ≤ O_{2}(P_{k}). Then P_{k} normalizes S ∩ N.

Indeed this is clear as S ∩ N = O_{2}(P_{k}) ∩ N in this case.

gsr2 − 2

(2) If O^{2}(P_{k}) ≰ N, then P_{k} normalizes S ∩ N

As O^{2}(P_{k}) ≰ N we have S ∩ N ≤ O_{2}(P_{k}) and so (2) follows from (1).

By definition of N (S) and O^{2}(P_{i}) ≰ L_{0} we have S ∩ L_{0} ≤ O_{2}(P_{i}). By (1) applied to N = L_{0} and k = 1, 2 we conclude that L_{0} ∩ S is normal in L = ⟨P_{1}, P_{2}⟩ and so (a) holds.

(b) follws from the maximal choice of L_{0}. The first part of (c) follows from (2) while the second follows from the first.

To prove (d) we assume without loss that L_{0} = 1. Note that P_{i} ∩ L_{1} is a parabolic subgroup of L_{1} and P_{i} = (P_{i} ∩ L_{1})S. Thus either P_{i} normalizes S ∩ L_{1} or we may choose notation so that P_{i} = ((P_{i} ∩ E_{1}) × … (P_{i} ∩ E_{l}))S, where P_{i} ∩ E_{1} is a parabolic of E_{1} with O^{2′}(P_{i} ∩ E_{1}) = P_{i} ∩ E_{1}.

Suppose now that O^{2}(P_{j}) ≰ L_{1}. Pick E_{1} so that S ∩ N_{L}(E_{1}) is a Sylow 2-subgroup of N_{L}(E_{1}). Then as L_{1} ∩ S is not normal in L, (c) implies that P_{i} does not normalise L ∩ S. If E_{1} ≤ P_{i}, (d.1) holds. So we may assume that P_{i} ∩ E_{1} is a proper parabolic subgroup of E_{1}. Suppose that (d.2) does not hold and that E_{1} is a group of Lie type in characteristic two.

Then no element of odd order in N_{P_{j}}(E_{1}) induces a non-trivial graph automorphism on E_{1} and so O^{2}(N_{G}(P_{j})) normalizes P_{i} ∩ S. Hence N_{G}(P_{j}) = O^{2}(N_{G}(P_{j}))(N_{S}(E_{1})) normalizes P_{i} ∩ E_{1} and so L ≠ ⟨(P_{i} ∩ E_{1})^{P_{j}}P_{j} = ⟨P_{1}, P_{2}⟩, a contradiction. If E_{1} is an alternating of degree at least six, then N_{Aut(E_{1})}(S ∩E_{1}) is a 2-group and we obtain a similar contradiction.

So assume now that O^{2}(P_{j}) ≤ L_{1} for j = 1, 2. Then it is easy to verify that (d.3) holds.

13 Establishing Geometries

EG

Throughout this section we assume

(i) U_{0}, U_{1} ∈ N (S)

Remark: redefine N as P ∪ E ?

(ii) all non-abelian composition factors of elements of L(S) are alternating groups, rank one group of Lie type over GF (q), G_{2}(q)’s or classical groups over GF (q), where q is a power of two.

(ii) U_{0} ≰ U_{1} and U_{1} ≰ U_{0}.

compfact

Lemma 13.1 Let H ≤ G with F^{∗}(H) = O_{2}(H) and |S/S ∩ H| ≤ 2. Then all non-abelian composition factors of elements of L(S) are alternating groups, rank one group of Lie type over GF (q), G_{2}(q)’s or classical groups over GF (q), where q is a power of two.

Proof: By 2.10 we may assume that H ≤ L^{∗} ∈ L(S). Hence the claim follows from 2.12.

O2Ln1

Lemma_{QT} 13.2 Put L = ⟨U_{1}, U_{2}⟩ and suppose that L ∈ L(S). Then the L_{0} and L_{1} in 12.3 and {i, j} = {0, 1} can be chosen so that one of the following holds

1. [O^{2}(U_{0}), O^{2}(U_{1}) ≤ Q.

2. L is not solvable and L ∈ N (S).

3. O^{2}(L)O_{2}(L)/O_{2}(L) is a p-group for some prime odd p.

4. L_{i} is a {2, p}-group for some prime p, O^{2}(P_{i}) ≤ L_{1} and L_{1}/L_{0} is an elementary abelian p-group. Moreover, there exists an odd prime q ≠ p so that the image of O^{2}(P_{j}) in Aut(L_{1}/L_{0}) has one of the following shapes: cyclic q group with q | p^{4} − 1; homocyclic q group of rank 2 with q | p − 1; Ext(3^{1+2}) with p ≠ 3; Ext_{−}(2^{1+4}).5; Ext_{−}(2^{1+4}).Alt(5); Alt(4), 2.Alt(n), n = 4, 5; 2.Alt(4)×2.Alt(4); 2.Alt(5)×2.Alt(5), p ≡ 0, 1, 4(5); 2.Alt(6); 2.Alt(7) ( with p = 7); Alt(5); L_{3}(2) or 3.Alt(6).

5. U_{i} induces Sym(3) on the set of components of L_{1}/L_{0}, U_{j} is the product of one or two 2-components of L_{1} and U_{i}/O_{2}(U_{i}) ≅ Dih_{2·3^{l}}.

6. O^{2}(U_{i}) acts trivially on the set of components of L_{1}/L_{0}, U_{i}/O_{2}(P_{i}) is a dihedral group, U_{i} normalizes O^{2}(U_{j}), and O^{2}(U_{j}) = E_{2}(L_{1}). Moreover, O_{2}(U_{j}) = O_{2}(L).

Remark: The case that O^{2}(U_{i}) ≤ L_{1} for i = 0 and 1 and L_{1}/L_{0} is a direct product of perfect simple groups still needs some attention: one needs to show that L_{1}/L_{O} is ”central” ( and this should be possible) and also things L/O_{2}(L) ≅ C_{3} × Alt(5).2 arise here, this is covered by case 6. But O^{2}(U_{i}) induces inner automorphism on O^{2}(U_{j}). So this probably should be listed as a seperate case, but it is also kind of the same as 1.

Proof: Remark: numbering and notation needs to be updated

We use the results and notation of 12.3. As m_{2′}(L) ≤ 3, case d.2 in 12.3 is not possible. Put D = C_{L}(L_{1}/L_{0}).

Suppose first that L_{1}/L_{0} is not solvable. Then O^{2}(U) ≤ L_{1}. If D ≠ L_{0} we get D ∩ L_{1} = L_{0} and by maximality L_{O}, O^{2}(P) ≤ D. Thus O^{2}(U), O^{2}(P) ≤ L_{O}. In this case we replace L_{1} by O^{2}(P)L_{O}. So we may assume that D = L_{0}. As m_{2′}(L) ≤ 3, r ≤ 3

Assume in addition that O^{2}(P) ≤ L_{1}. As P is solvable, d.1 is impossible. Thus d.3 holds. Moreover, L = L_{1}S and so O^{2}(L) ≤ L_{1} thus 4. holds in this case.

So assume that O^{2}(P) ≰ L_{1}.

If O^{2}(P) does not act trivially on the set of components of L_{1}/L_{0} we conclude that r = 3 and P induces Sym(3) on the set of components of L_{1}/L_{0}. As e(G) ≤ 3 and L_{1}/L_{0} has three components, [L_{1}^{∞}, L_{0}] ≤ O_{2}(L)]. Thus 5. holds.

So suppose that O^{2}(P) acts trivially on the set of components of L_{1}/L_{0}. The S acts transitively thereon and r ≤ 2. If r = 2, then O^{2}(U) = E_{2}(L_{1}). Since e(G) ≤ 3 we have E_{1} is L_{2}(q), Sz(q), L_{3}(4), L_{3}(2), Alt(6), Alt(7). But in the last three cases Out(E_{1} is a 2-group, a contradiction. In the first two cases, Out(E_{1}) is cyclic and so P L_{1}/L_{1} is a dihedral group. If E_{1} ≅ L_{3}(4), then O^{2}(U)O_{2}(L)/O_{2}(L) ≅ SL_{3}(4)∗SL_{3}(4). Since the action of Aut(L_{1}/L_{O}) on Out(L_{1}/L_{0}) on the 3-part of the Schur multiplier respectively the outer automorphisms of L_{1}/L_{O} are isomorphic we conclude that S does not act irreducibly on O_{3}(Out(L_{1}/L_{0}) and so O^{2}(P)L_{1}/L_{1} ≅ C_{3} and so again P/O_{2}(L) is a dihedral group. Thus 6. holds

If r = 1 we conclude that P L_{1}/L_{1} is isomorphic to a subgroup of Out(E_{1}) and so Out(E_{1}) is not abelian. Hence E_{1} ≅ U_{3}(q), U_{4}(q), L_{3}(q) and P/O_{2}(P) is a dihedral group and 6. holds.

Assume now that L_{1} is solvable.

Suppose that L_{2}/L_{O} is a minimal normal subgroup of L/L_{0} different from L_{1}/L_{O}. Then we may choose notation so that O^{2}(P) ≤ L_{1} and O^{2}(U) ≤ L_{2}. Then [O^{2}(P), O^{2}(U)] ≤ L_{0}, L_{1} = O^{2}(P)L_{O} and L_{2} = O^{2}(U)L_{0}.

Suppose that O^{2}(U) ≤ L_{1}. Then by assumption L_{1}/L_{O} is an elementary abelian 3group.

TO BE CONTINUED

Corollary 13.3 Assume that

(i) U_{0} ∈ P(S)

(ii) If U_{1} ∈ P(S) and U_{1} is solvable then U_{1} is a {2, 3}-group.

(iii) L =^{def} ⟨U_{0}, U_{1}⟩ ∈ L(S).

Then one of the following holds

TO BE CONTINUED

O2L1

Lemma_{QT} 13.4 Suppose that

(i) E ∈ E(S) \ P(S).

(ii) O_{2}(⟨U_{1}, E⟩) = 1.

(iii) For all U^{∗} ∈ N (E, S) with U^{∗} ≠ E, ⟨U_{1}, U^{∗}⟩ ∈ L(S)

(iv) There exists a maximal element U_{1} ∈ N (E, S) so that one of the cases 3-6 in 13.2 holds.

Then one of the following holds for L(1) = ⟨U_{0}, U_{1}⟩.

1. U_{1} is solvable.

2. Head(U_{1}) ≅ L_{2}(q)^{r}, r ≤ 2, q ≥ 4; U_{O}/O_{2}(U_{O}) ≅ D_{2·3^{k}}, Head(L_{1}(1) ≅ L_{2}(q)^{3} and O^{2}(U_{O}) transitively permutes the three 2-components of L(1)

3. O^{2}(U_{1})/O_{2}(U_{1}) ≅ Alt(5), Head(E) ≅ U_{4}(2) and O^{2}(U_{0}) ≤ O_{2,p}(L(1)), p a prime with p > 3. Moreover, if TO BE CONTINUED

4. Put R_{1} = O^{2}(U_{1})O_{2}(U_{1}). Then

(a) U_{O} normalizes R_{1} and no non-trivial characteristic subgroup of R_{1} is normal in E.

(b) One of the following holds

1. Head(E) ≅ U_{4}(2), U_{O}/O_{2}(U_{O}) ≅ D_{2·3^{k}} and Head(U_{1}) ≅ Alt(5).

2. There exists a maximal element U_{2} of N (E, S) which fulfils 3. with U_{2} in place of U_{1}.

Remark: Case 4b1 is impossible by a trivial pushing up argument ( or by quoting pushing up)

Proof: Let N be the set of proper maximal elements U^{∗} ∈ N (E, S). We assume without loss that U_{1} is not solvable.

By 8.2 there exists U_{2} in N so that ⟨U_{1}, U_{2}⟩ = E. Under all these U_{2}’s with pick one which ( possibly trivial) 2-component K with K/O_{2}(K)| maximal.

In particular O^{2}(E) = ⟨O^{2}(U_{1}), O^{2}(U_{2})⟩. For i = 1, 2 let L(t) = ⟨U_{O}, U_{i}. We will apply 13.2 to L(1) and L(2). We write Case t(i) if Case t in 13.2 holds for L(1). For i = 0, 1, 2 put Q_{i}^{∗} = [O_{2}(U_{i}), O^{2}(U_{i})]. The next two statement follow immediately from 13.2 applied to L(1).

O2L1 − 1

(1) U_{O} is solvable and O^{2}(U_{1})/O_{2}(O^{2}(U_{1})) ≅ A̸lt(n) for 8 ≤ n ≤ 11.

O2L1 − 2

(2) One of the following holds:

• Case 4(1) with (i, j) = (O, 1) and Head(U_{1}) ≅ Alt(5), Alt(6), 3·Alt(6), Alt(7) or L_{3}(2)

• Case 5(1) with (i, j) = (O, 1) and Head(U_{1}) ≅ L_{2}(k)^{r} or L_{3}(2)^{r}, with r ≤ 2.

• Case 6(1) with (i, j) = (O, 1)

By 8.2, the second statement in (1) and as U_{1} is not solvable we can choose U_{2} so that U_{1} ∩ U_{2} is a maximal parabolic of U_{1}.

Remark: this needs to be proved very carefully for the the symmetric groups

Next we prove

O2L1 − 3

(3) In Case 1(2), 5(1) holds.

As we are in case 1(2), [O^{2}(U_{O}), O^{2}(U_{2})] is a 2-group. Hence also [O^{2}(U_{O}), U_{1} ∩ O^{2}(U_{2})] is a 2-group. On the other hand in case 4(1), U_{1} ∩ O^{2}(U_{2}) acts fixed point freely on L_{1}(1)/L_{0}(1), a contradiction. In case 6(1) O^{2}(U_{0}) normalizes O^{2}(U_{1}) and O^{2}(U_{2}), again a contradiction. Thus case 5(1) holds.

O2L1 − 4

(4) In Case 4(1), Case 4(2) holds.

By (3) we may assume that Case 2(2),3(2), 5(2) or 6(2) holds. As P_{O} is solvable, we get in case 2(2), 3(2) and 5(2) that P_{0} is a 2, 3-group a contradiction. Hence Case 6(2) holds, Head(U_{O}) is cyclic and O^{2}(P_{0}) induces field or diagonal automorphism of odd order larger than 3 on O^{2}(U_{2})/O_{2}(O^{2}(U_{2}). But this contradicts the structure of U_{1} and E.

O2L1 − 5

(5) If Case 4(1) and Case 4(2) holds, 3. holds

Considering the action of Q_{2}^{∗} on L_{1}(1)/L_{0}(1) we see that [O^{2}(U_{O}), Q_{2}^{∗}] = O^{2}(U_{0}) Remark: more details please . Hence O^{2}(U_{2}) ≰ L_{1}(2) and so O^{2}(U_{O}) ≤ L_{1}(2). Moreover, Q_{2}^{∗} ≰ O_{2}(L(2). Hence either U_{2} is solvable or acts as Ext2^{1+4}.A_{5} on L_{1}(2)/L_{0}(2). In the latter we get L_{1}(2) ≤ P_{0} ≤ L(2) and then L_{1}(1) = L_{2}(1), a contradiction. Thus U_{2} is solvable and so U_{2}/O_{2}(U_{2}) ≅ Sym(3) or Sym(3) ≀ C_{2}.

In the latter case, [L_{1}(2)/L_{0}(2), Q_{2}^{∗} ≠ 1 implies that S acts irreducible on [L_{1}(2)/L_{0}(2)]. But then L_{1}(2) ≤ P_{0} ≤ L_{1}(1), a contradcition.

Thus U_{2}/O_{2}(U_{2}) ≅ Sym(3) and as U_{1} is not solvable we conclude that Head(E) ≅ U_{4}(2). Hence 3. holds.

O2L1 − 6

(6) In case 5(1), 2.holds.

We may assume that Head(U_{1}) ≅ L_{3}(2)^{r}, r = 1, 2. If r = 1 and U_{1} induces no graph automorphism on Head(U_{1}), then Head(E) ≅ L_{4}(2), Sp_{6}(2), Ω_{8}^{−}(2) or (3.)Alt(7). If r = 1 and U_{1} induces a graph automorphism on Head(U_{1}), then Head(E) ≅ L_{5}(2). If r = 2 then now element of U_{1} induces a graph automorphism on Head(U_{1}) and Head(E) ≅ L_{6}(2), L_{7}(2) or 3.(Alt(7) × Alt(7). Let K be the normaliser in U_{1} of some 2-component of U_{1} and P ∈ P(K, S ∩ T). Then |S/S ∩ P | ≤ 2. Let H_{O} = N_{L(1)}(O^{2}(P), H_{1} = N_{E}(O^{2}(P)) and H = ⟨H_{1}, H_{2}. Then Head(H_{O}/O^{2}(P)) ≅ L_{3}(2) × L_{3}(2). Moreover we can and do choose P so that H_{1} ≰ L(1) and so H ≠ H_{1}. As m_{3}(H) ≤ 3 and O^{2}(P)O_{2}(H)/O_{2}(H) is a normal subgroup of order three in H. By 4.10 we conclude that H^{∞}/O_{2,2′}(H^{∞}) ≅ L_{3}(2) × L_{3}(2) or L_{3}(2) × Alt(7). In the first case each minimal parabolics of H is either contained in H_{0} or is solvable and not a {2, 3}-group, a contradition to H_{1} H̸_{O}. In the second case H has a 2-component R with Head(R) ≅ 3.Alt(7), O_{2,3}(R) ≤ P and Head(R ∩ H_{1}) ≅ C_{3} × L_{3}(2). It follows that P ∩ K induces a group of automorphisms on 3.Alt(7)(= Head(R)) which inverts the central three but centralizes an L_{3}(2) subgroup, a contradiction.

O2L1 − 7

(7) In case 6(1),4. holds.

By case 6(1) O_{2}(U_{1}) = O_{2}(L(1)) and U_{0} normalizes O^{2}(U_{1}). Thus the first statement in 4. holds. As U_{1} induces diagonal or field automorphism of odd order on Head(U_{1}), E is not a group of Lie type in over the field of 2-elements, except maybe U_{4}(2).

Suppose first that U_{2} is solvable. Then Head(E) ≅ U_{4}(2), Head(U_{1}) ≅ Alt(5) and so 4b1 holds.

Suppose next that U_{2} is not solvable.In case 1(2) or 6(2), P_{O} normalizes O^{2}(U_{2}), a contradiction as P_{0} already normalizes O^{2}(U_{1}). Suppose Case 2(2) holds. As U_{O} is solvable, we conclude that Head(L_{1}(2)) ≅ U_{4}(2). Let Q = O_{2}(U_{2}). In E we see that Q induces inner automorphism on Head(U_{1}), in L(2) we see that Q inverts Head(U_{0}) and in L(1) we see that every element that inverts Head(U_{0}) induces an outer automorphism on Head(U_{1}), a contradiction.

Hence we may assume that one of 4(2) or 5(2) holds. In particular, U_{2} in place of U_{1} fulfils the assumption of this lemma and so by (4) and (6) applied with U_{1} and U_{2} interchanged 5(2) we get that case 5(2) holds and Head(U_{2}) ≅ L_{2}(q)^{r}, r ≤ 2. Thus 4b2 holds. Remark: I forget to think about 3.Alt(6) for Head(U_{1}). This might arrise for Head(E) = 3.Alt(7)

nl2ws3

Lemma_{QT} 13.5 Retain the assumptions of 13.4 and assume that 13.4.2 holds. Then one of the following holds:

a. troet

Proof:

nl2ws3 − 1

(1) (a) If r = 1, then Head(H) ≅ (3.)Alt(7) ( with q = 4); Alt(10) (with q = 4); (S)L_{3}(q); Sp_{4}(q); G_{2}(q); U_{4}(q); U_{4}(\sqrt{q}); or L_{4}(q) ( with S inducing a graph automorphismus).

(b) If r = 2 then Head(H) ≅ 3.(L_{3}(4) × L_{3}(4)) (with q = 4), 3.(Alt(7) × Alt(7) with q = 4) or L_{4}(q)( with S inducing a graph automorphism).

(c) Let H ∈ L(S) with L_{1}(1)S ≤ H. Then Head(H^{∞}) = H_{1} · H_{2} · H_{3}, where S normalizes H_{1} and interchanges H_{2} and H_{3}, for 1 ≤ i ≤ 2, H_{i}/O(H_{i}) ≅ (2.)Alt(5) and O(H_{0}) and O(H_{1}) have coprime order.

This follows easily from 4.10

Let K_{1}, K_{2}, K_{3} be three different 2-components of L(1) with K_{1} ≤ U_{1}. Put K = K_{1}K_{2}K_{3}. Let {i, j, k} = {1, 2, 3}. Put H^{i} = N_{G}(K_{i}) and K_{j}^{i} = ⟨K_{j}^{H^{i∞}}. As L(1) ≤ H^{i} and H^{i} contains a Sylow 2-subgroup of G we can apply (1)c and conclude that K_{k}^{i} normalizes K_{j}^{i}

and K_{j}). Hence K_{k}^{i} ≤ H^{j} and K_{k}^{i} ≤ K_{k}^{j}. By symmetry K_{k}^{j} ≤ K_{k}^{i} and so K_{k}^{∗} def^{=} K_{k}^{j} = K_{k}^{i}.

In particular K_{i}^{∗} normalizes K_{j}^{∗} and the K_{j}^{∗}’s are pairwise isomorphic. By (1)c applied to K_{1}^{∗}K_{2}^{∗}K_{2}^{∗}S we conclude that O_{2}2′(K_{i}^{∗}) = O_{2}(K_{i}) and so K_{i}^{∗} = K_{i}. It follows that

nl2ws3 − 2

(2) Put L = N_{G}(K). Then L is the unique maximal 2-local of G containg KS. Moreover, C_{L}(K/O_{2}(K)/O_{2}(L) is coprime to |L_{2}(q)|

Remark: the same argument works for any group with three 2-componets which are conjugate in G so we should make an extra lemma and use it in the L_{3}(2) ≀ Sym(3) case

Suppose that Head(E) ≅ Alt(7)orAlt(10). Then Head(U_{2}) ≅ Alt(6) or Alt(8) respectively and U_{1} ∩U_{2}/O_{2}(U_{1} ∩U_{2}) ≅ Sym(3). Hence we see in L(1) that U_{0} does not normalize U_{1} ∩ U_{2} and Head(⟨U_{O}, U_{1} ∩ U_{2}⟩ ≅ C_{3} ≀ C_{3}. Hence U_{O} does not normalize U_{2}. It follows that case 2(2) holds and Head(L(2)) ≅ Alt(7), Sp_{6}(2), L_{6}(2), Alt(9), Alt(10) or Alt(11). But this contadicts the stucture of ⟨U_{O}, U_{1} ∩ U_{2}⟩.

Suppose that q = 4 and Head(E) ≅ .(L_{3}(4) × L_{3}(4)) or 3.(Alt(7) × Alt(7) and let K_{1} be a 2-component of U_{1}. Then N_{G}(K_{1}) involves L_{3}(4) respectively Alt(7), a contradiction to (1).

Let L = KO_{2}(L(1), T = L ∩ S and B = N_{L}(T). Note that B normalizes K_{1}. Let F = ⟨B, E⟩.

Suppose that F ∉ L(S). TO BE CONTINUED

14 Large Alternating Groups

In this section we assume that G is a quasi thin group, and that there exists an amalgam (P, E) so that P ∈ P(S), E ∈ E(S), Head(E) ≅ Alt(n), n = 10, 11 Remark: we should at least also allow E/O_{2}(E) ≅ Sym(9)

notA11

Lemma_{QT} 14.1 Suppose n = 11 and let U ≤ calL(E, S) with Head(U) ≅ Alt(10). Then (P, U) is an amalgam.

Proof: Let L = ⟨P, U⟩ and suppose that L ∈ L(S). Then by 13.2, [O^{2}(P), O^{2}(U)] is a 2-group or L ∈ N (S). In the second case we get that Head(L) ≅ Alt(11) and so O_{2}(L) = O_{2}(U) = O_{2}(E) a contradiction. Thus [O^{2}(P), O^{2}(U)] is a 2-group. As m_{3}(O^{2}(U)) = 3 we conclude that P is a 3′ group. Let T be a Sylow 2-subgroup of O^{2}(P). Then clearly U normalizes T and so T ≤ O_{2}(U) and O_{2}(U) is a Sylow 2-subgroup of O_{2}(U)O^{2}(P). As O_{2}(U) = O_{2}(E), no non-trivial characteristic subgroup of O_{2}(U) is normal in O_{2}(U)O^{2}(P). Hence O_{2}(U)O^{2}(P) has a non-trivial irreducible F F -module and so is not a 3′ group, a contradiction.

notA10

Lemma_{QT} 14.2 Suppose E/O_{2}(U) ≅ Sym(9), Alt(10) or Sym(10) and let U ≤ calL(E, S) with U/O_{2}(U) ≅ Sym(8). Then (P, U) is an amalgam.

Proof: Let L = ⟨P, U⟩ and suppose that L ∈ L(S). Then by 13.2, [O^{2}(P), O^{2}(U)] is a 2-group or L ∈ N (S).

Suppose that O_{2}(E) ≤ O_{2}(L). Then O_{2}(U) ≠ O_{2}(E) and E/O_{2}(E) ≅ Sym(10). Let R ≤ E with O_{2}(L) ∈ Syl_{2}(R) and R/O_{2}(E) ≅ Sym(3). Let C be a characteristics subgroup of O_{2}(L) normal in R. Then C is normal in L and in ⟨U, R⟩ = E. Hence C = 1 and so by 8.12 O^{2}(P) normalizes Ω_{1}(Z(O_{2}(E)), a contradiction.

notA10 − 1

(1) O_{2}(E) ≰ O_{2}(L).

Let U^{∗} ∈ L(U, S) with U^{∗}/O_{2}(U^{∗}) ≅ Sym(3) and Let Q/O_{2}(U) be the unique elementary abelian, normal subgroup of order 16 in U^{∗}/O_{2}(U). Then N_{E}(Q)/Q ≅ Sym(5). Let C be a characteristic subgroup of Q normal in L. Then C is normal in ⟨U, N_{E}(U)⟩ = E and so C = 1. We proved

notA10 − 2

(2) O_{2}(L) < Q and no non trivial characteristic subgroup of Q is normal in L.

Remark: (2) and its set up makes no sense for the Sym(9) case, some fixing necessary

Suppose that L ∈ calN(S). Then Head(L) ≅ Alt(m), 9 ≤ m ≤ 11 or L/O_{2}(L) ∼ L_{6}(2).2.

If Head(L) ≅ Alt(m), m = 9 or 11, L cannot be generated by U and a minimal parabolic unless m = 9 and P = L. We conclude P/O_{2}(P) ≅ Sym(9) and O_{2}(E) ≤ O_{2}(U) ≤ O_{2}(L), a contradiction

If Head(L) ≅ Alt(10), the situation is symmetric in E and L. L(1) = ⟨N_{E}(Q), N_{L}(Q). Then Q = O_{2}(L(1)) and 13.4 provides a contradiction. Remark: One has to make sure that the possibility of two different complements Sym(5) to a group of odd order was really ruled out

If L/O_{2}(L) ≅ L_{6}(2).2,

O_{2}(U) = [O_{2}(U), U]O_{2}(L) ≤ O_{2}(E)O_{2}(L) ≤ O_{2}(U)

and so O_{2}(U) = O_{2}(E)O_{2}(L). If E/O_{2}(E) ≅ Sym(9) or Alt(10), then O_{2}(L) ≤ O_{2}(U). Hence no non-trivial characteristic subgroup of O_{2}(U) is normal in L and we conclude that [J(U), ⟨Ω_{1}(Z(O_{2}(U))^{L}⟩ = 1, a contradiction. Thus E/O_{2}(E) ≅ Sym(10). Let V = Ω_{1}(Z(O_{2}(L)). Then by (2), C_{S}(V) = O_{2}(L). On the otherhand, L/O_{2}(L) has no faithful module with respect to it O_{2}(U)/O_{2}(L) contains an offending subgroup. Hence J(O_{2}(U) ≤ O_{2}(L) and so J(O_{2}(U) ≰ O_{2}(E). It follows that there exists a conjugate of J(O_{2}(U)) under E which is contained in U but not in U′O_{2}(U). Hence by 2.11 there exists an offender for L on V which is not contained in L′O_{2}(L), a contradiction.

We have proved that [O^{2}(U), O^{2}(P)] ≤ O_{2}(U). Put P^{0} = O^{2}(P)Q. As O^{2}(P) ∩ S ≤ O_{2}(U) ≤ Q, S ∩ P^{0} = Q. Put U_{1} = N_{E}(Q) and L(1) =^{def} ⟨P, U_{1}⟩ By 8.12 we conclude that

notA10 − 3

(3) [O_{2}(P), O^{2}(P)] ≤ O_{2}(L(1)).

By a similar argument O_{2}(L) = O_{2}(U) leads to a contradiction and so O_{2}(L) ≠ O_{2}(U). In particular, E/O_{2}(E) ≅ Sym(10). As U normalizes O^{2}(P), U_{1} does not. So by 13.4, L(1) ∈ N (S). By (3), the compoents of Head(L(1)) cannot be groups of Lie type in characteristic 2 and thus are alternating groups. Furthermore, as m_{3}(L) ≤ 3 and m_{3}(U) = 2, m_{3}(P) ≤ 1. This leads to Head(L(1) ≅ (3.)Alt(7) or Alt(11). In particular P/O_{2}(P) ≅ Sym(3). In the second case N_{(L(1)}(O^{2}(P)) involves Sym(8) and we obtain a contradiction by considering ⟨N_{(L(1)}(O^{2}(P)), U⟩ (note here that U ≰ L(1) as already U_{1} ≤ L(1). Thus Head(L(1)) ≅ (3.)Alt(7). By (1), O_{2}(E) inverts Head(P). Thus L/O_{2}(L) ≅ Sym(3) × Sym(8). As U^{∗} ≤ U_{1} ≤ L(1) we get L(1)/O_{2}(L(1)) ≅ (3.)Sym(7). The 3.Sym(7) case is exclude by considering N_{G}(O^{2}(P)). Thus L(1)/O_{2}(L(1)) ≅ Sym(7).

In L we see that O_{2}(L) = O_{2}(U) ∩ O_{2}(P), in L(1) that O_{2}(L(1)) = O_{2}(U_{1}) ∩ O_{2}(P) and in E that O_{2}(U) ≤ O_{2}(U_{1}). Hence O_{2}(L) ≤ O_{2}(L(1)). Moreover, in L we see that |O_{2}(E)O_{2}(L)/O_{2}(L)| = 2 and in L(1) that |O_{2}(E)O_{2}(L(1))/O_{2}(L(1))| = 2. It follows that F = =^{def} O_{2}(E) ∩ O_{2}(L) = O_{2}(E) ∩ O_{2}(L(1)). Thus F is normalized U and U_{1} and so F is normal in E. Note that O^{2}(U) ∩ O_{2}(E) ≤ O_{2}(O^{2}(U)) ≤ O_{2}(L) and so O^{2}(U) ∩ O_{2}(E) ≤ F. Hence by the ” Satz von Gaschütz, O^{2}(E) ∩ O_{2}(E) ≤ F. Put E^{∗} = O^{2}(E)O_{2}(L). Since O_{2}(L)∩O^{2}(E)O_{2}(E) = F we conclude that O_{2}(E^{∗}) = F ≤ O_{2}(L). Now the same argument as in the proof of (1) gives a contradiction, which completes the proof of the lemma.

We remark that Sym(14) has parabolics C_{2} ≀ Sym(7), Sym(8) × C_{2} ≀ Sym(3) and Sym(10) × C_{2} ≀ C_{2}, intersecting in the same way has the groups in the last case we ruled out. But of course these parabolics in Sym(14) are not of 2-type and so do not furnish a counter example.

notAlt9

Lemma_{QT} 14.3 Suppose E/O_{2}(E) ≅ Alt(9) and let U ≤ calL(E, S) with U/O_{2}(U) ≅ Alt(10). Then one of the following holds

1. (P, U) is an amalgam.

2. Let L = ⟨P, U⟩. Then L/O_{2}(L) ≅ L_{5}(2), [O_{2}(L), O^{2}(L) is a natural module and [Z, E] = 1.

Proof: We may assume that L ∈ L(S).As above [O^{2}(U), O^{2}(P) is not a 2-group and L/O_{2}(L) ≅ A̸lt(9). This leaves the possiblity L/O_{2}(L) ≅ L_{5}(2). Note that O_{2}(L) ≤ O_{2}(E) and so no non-trivial characteristic subgroup of O_{2}(E) is normal in L. Let Z_{1} = Ω_{1}(Z(O_{2}(L) and Z_{2} = Ω_{1}(O_{2}(E) and note that Z_{2} = C_{Z_{1}}(O_{2}(E)). Suppose that [Z_{2}, E] ≠ 1. Then [Z_{2}, E] ≠ 1. As Z_{1} is an FF-module, all non-trivial composition non-trivial factors of L in Z_{1} are isomorphic natural modules. Hence Z_{2} is as U module the direct sum of isomorphic natural modules and trivial modules. Let d be an element of order three in U acting fixed point freely on the natural module for U, then it is easu to see that C_{Z_{2}}(d) = C_{Z_{2}}(U) = C_{Z_{1}}(E) and so d acts fixed point freely on Z_{2}/C_{Z_{2}}(E). It follows that Z_{2} involves a spinmodule for E and so also two non-isomorphic natural modules for U, a contradiction. Remark: u se the easier alt 7 argument

Hence [Z_{2}, E] = 1. It follows that Z_{1} is a natural module for L and so by 8.14 and as C_{Z_{1}}(E) we get [O_{2}(E), O^{2}(E) = Z_{1} and so (2) holds

15 Tits Chamber Systems

In this section we us the following assumptions and notations:

(i) I is a finite set with |I| ≥ 3,

(ii) For i ∈ I, P_{i} ∈ P(S).

(iii) For J ⊂ I put J′ = I \ J, P_{J} = ⟨P_{j} | j ∈ J⟩ and M_{J} = P_{I′}

(iv) Define a graph on I by considering i and j to be adjacent if and only if [O^{2}(P_{i}), O^{2}(P_{j})] is not a 2-group.

(v) If J ⊂ I is connected with |J| ≥ 2, then P_{J} ∈ E(S) and for all j ∈ J, S ∩ P′_{J} ⋬ P_{j}.

(vi) Let i ∈ I. Then Head(M)i is a central extension of a groups of Lie type in characteristic two.

(vii) Let J be a proper subset of J. Q_{J} = O_{2}(P_{J}) and Z_{J} = ⟨Z^{P_{J}}. Then C_{P_{J}} (Q_{J}) ≤ Q_{J}).

(viii) ⟨P_{i}| ∈ I ≰ L(S).

zni2p

Lemma 15.1 Suppose there exists two distinct i, j in I with Z ⋬ P_{i} and Z ⋬ P_{j}. Then one of the following holds: TO BE CONTINUED

Proof: Suppose first that there exists k ∈ I \ {i, j} so that k′ is connected. Apply 8.6 with to G_{α} = M_{k} and G_{β} = P_{k}. As P_{i} does not centralize Z, 8.61 does not hold. By the stucture of M_{k}, 8.61 implies C ⊴ M_{k} and P_{k}, a contradiction.

In case (6) 8.12 implies that [Q_{k}.O^{2}(P_{k})] ≤ Z_{k}. let k ≠ r so that r is connected. Then [Q_{k}, O^{2}(P_{k})] ≤ Z_{k} ≤ Z_{r′} ≤ Q_{r′} a contradiction to (v) and (vi).

Hence we mau assume that q(M_{k}, Z_{k′} ≤ 2. As two parabolics of M_{k} act non-trivially on Z we get from 6.12 that M_{k} is of type L_{n}(q), k′ is a string with i and j as endpoints and M_{k} has exactly two non-central composition factors on Z_{k′}. Moreover these composition factors are natural modules dual to each other. Is is easy to see that Z ⊴ P_{k}. Let J = I \ {i, j, k}. Assume that k is adjacent to some element of J. Then we can apply 8.22 to G_{α} = M_{i}, G_{β} = M_{j} and G_{αβ} = M_{ij}. Thus TO BE CONTINUEDAssume that k is not adjacent to an element of J and without loss that k is adjacent to i. Then we can apply 8.22 to G_{α} = M_{i}, G_{b} = M_{k} and G_{αβ} and we conclude that J = ∅. Thus TO BE CONTINUED

Remark: the effect of graph automorphisms needs to be worked in, Z_{α}Z_{β}⊴G_{β} needs to be ruled out

Suppose next that no such k exists. Then clearly I is a string with i and j as the end notes. Then we can apply 8.22 to G_{α} = M_{i}, G_{β} = M_{j} and G_{αβ} = M_{ij}. Thus TO BE CONTINUED

References

[As] M. Aschbacher, Finite Groups of Rank 3.I., Invent. Math.63, (1981), 357-402.

[At] J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker and R. A. 1985.n, The Atlas of Finite Group, Oxford University Press, London,(1985)

[Bl] D. Bloom, The subgroups of P SL_{3}(q), for q odd. Trans.Amer.Math.Soc. 127,(1967), 150-178.

[CD] A. Chermak, A. Delgado, A measuring argument for finite groups, Proc. Amer. Math. Soc.107,(1989) 907-914.

[MS] U. Meierfrankenfeld and G. Stroth, Quadratic GF(2)-modules for sporadic simple groups and alternating groups, Com. Alg. 18(7) (1990),2099-2139.

[Me] U. Meierfrankenfeld, Quadratic pairs in odd charctaristic, Preprint, East Lansing, (1986).

[Mi] H. Mitchell, The subgroups of the quaternary abelian groups. Trans.Amer.Math.Soc. 15, 379-396(1914).

[St1] B. Stellmacher, On the 2-local structure of finite groups, in Groups, Combinatorics, and Geometry, M. Liebeck and J.Saxl (eds), Cambridge University Press, (1992),495-506.

[St2] B. Stellmacher, An application of the amalgam method: the 2-local structure of Ngroups of characteristic 2-type, preprint, Kiel (1995).

Department of Mathematics

Michigan State University

East Lansing, Michigan 48824 meier@math.msu.edu

Mathematisches Seminar Universität Kiel Ludewig-Meyn Str. 4 24118 Kiel