LEMMA 11.4.3. (1) Gz = KCaz(K/02(K))Mz. Therefore if K = L1, then
CaJK/02(K)) i M.
(2) G1 = K(Ca 1 (Vi) n Ca 1 (K/02(K)))M1.PROOF. Recall K = K(V, V 1 ) = K(V, z) is normal in G1 and Gz. Set Y :=
Caz(K/02(K)). From 11.1.2, Out(K/02(K)) is 2-closed, so YKT :::l Gz, and hence
by a Frattini Argument, Gz = YKNaz(T). Now by Theorem 3.3.1, Na(T) ::::; M,
proving the first assertion of (1). So if K = L1, then as Gz i M, Y i M, giving
the remaining assertion of (1).
Now instead set Y := Ca 1 (V 1 ) n Ca 1 (K/0 2 (K)). The same proof shows thatCa 1 (Vi)T = YTKCM 1 (V1). As Cv(L) = 1 by 11.4.1.1, G1 = Ca 1 (V1)M1 by
11.2.3.1, proving (2). D
Since Gz i M, there is H E H*(T, M) n Gz; H has this meaning for theremainder of the chapter.
Since n(H) 2: n in the shadows, our next result eliminates those groups:
LEMMA 11.4.4. n(H) = 1.
PROOF. Assume n(H) > 1; then from E.2.2, 02 (H) = (F) for some I E
£(G, T) with Ii M. By 1.2.4 I::::; Ii E C(Gz), and then by 1.2.1, either [£1,I] ::::;
[K,I1] ::::; 02(K) ::::; 02(L1T), so [L1,I] ::::; 02(L1), or I::::; Ii ::::; K. But by 11.4.1.2,either K = £ 1 or K is described in case (1) or (2) of 11.1.2; in either case L1 is the
unique minimal member of C(G, T) n K. Therefore if I::::; K then I= L1 ::::; M,
contradicting Ii M. Thus [L1,I] ::::; 02(£1).
Let B be a Hall 2'-subgroup of In M. Then B ::::; CM(L1/02(L1)) and B
centralizes z. For each prime divisor p of q - l, L contains each subgroup Bp of B
of order p by 11.0.4, so Bp::::; CL(z) n CL(Li/02(L1)) = L n Rl from the action of
Lon the natural module V. Therefore (IBI, q - 1) = 1.
As B centralizes z, B ::::; Mv by 11.0.3.4. Then as BT = TB, [£1, BJ ::::;02(L1), and (q - 1, IBI) = 1, it follows from the action of NAut(v)(L) on V that
[V, BJ = 1. But then using Remark 4.4.2, Na(B) ::::; M by Theorem 4.4.3; so
H = (H n M, NH(B))::::; M, contradicting Hi M. D
11.5. The' final contradictionWe now work to obtain a contradiction, by analyzing the normal closure (Va^1 )
of V in G 1. The analysis falls into two cases, depending on whether (Va^1 ) is
abelian or not. The strong restriction in 11.4.4 will make weak closure methods
more effective.
LEMMA 11.5.1. L/02(L) is SL3(q) or Sp4(q).
PROOF. In view of 11.0.2, we may assume L ~ G2(q). Hence by parts (5) and
(6) of 11.2.2,Ca(C1(R1, V))::::; M 2: Na(Wo(R1, V)).
Thus it will suffice to find H1 ::::; H with H 1 i M, n(H1) = 1, and Rl E Sy[z(H1):
For since n(H 1 ) = 1 and s(G, V) > 1by11.2.2.3, we may apply E.3.19 to concludethat H1 ::::; M, contrary to our choice of H1.
Suppose first that K = L1. Then by 11.4.3.1, Caz(Li/02(L1)) i M, so we
may choose H with [£ 1 , 02 (H)]::::; 02 (£ 1 ) and set H 1 := R 102 (H); then n(H 1 ) = 1
using 11.4.4. On the other hand if L1 < K, then by 11.4.1.2, K is described in case