744 10. THE CASE L E .Cj(G, T) NOT NORMAL IN M.
LEMMA 10.2.2. Assume HE 1-l*(T, M) with [Z, HJ =f 1, and set W := (ZH).
Then
(1) Lo= [L 0 , J(T)] and one of the first three cases of 10.1.1 holds.If in addition case (1) or {2) of 10.1.1 holds, then:
(2) 02 (H) = [0^2 (H), J(T)] and J(T) 1:_ CT(W), so W is an FF-module for
H/CH(W).
{3) If case {1) of 10.1.1 holds, then Bo:::; Nc(8) and 8 E 8yl2(Lo8).
(4) 02((Na(8),H)) f-1.
PROOF. As [Z, HJ =f 1, [V, J(T)] =f 1 by 3.1.8.3. Thus Lo = [Lo, J(T)], andthen 10.1.2.1 completes the proof of (1). In the remaining assertions we may assume
case (1) or (2) of 10.1.1 holds. Then by 10.1.2.6, LoT is a minimal parabolic
described in E.2.3, and 8 :::; Ti.
In case (1) of 10.1.1, E.2.3.2 says that 8 E 8yl2(Lo8) and 8 :S:: T+ := To02(LoT),
so that 8 = Baum(T+)· But Bo normalizes T+ so Bo :::; Na(8), completing the
proof of (3).Assume [0^2 (H), J(T)] < 02 (H). Then as [Z, HJ =I- 1 we conclude from B.6.8.3d
that 8 = Baum(0 2 (H)) and hence H :::; Na(S). However since we are excluding
case (3) of 10.1.1, Na(8) :::; M by 10.2.1. This contradicts H 1:_ M, so (2) holds.
If J(H)* is the product of copies of L 2 (2m) then by E.2.3.2, 8 E 8yh(0^2 (H)8).
Then using Theorem 3.1.1 as in the proof of 10.2.1, ( 4) follows. Since we may assume(4) fails, we conclude from E.2.3.1 that J(H*) is a product of s:::; 2 copies of 85,
and that no nontrivial characteristic subgroup of 8 is normal in H. Therefore by
E.2.3.3, 02 (H) =Ki x · · · x Ks is the product of A 5 -blocks Ki.
Next 02 (H n M) :::; Lo by 10.1.3, so 02 (H n M) :::; Bo and a Sylow 3-subgroup
P of 02 (H n M) is contained in Po E 8y[s(B 0 ). As 02 (H) is a product of A 5 -
blocks, P centralizes Z, so case (2) of 10.1.1 holds since Cp 0 (Z) = 1 in case (1) by
10.1.2.5. Then since Lo= [Lo, J(T)] by (1), Lof' ~ 85 wr Z2 in view of B.4.2.5, so
Bo E 3(G, T). Since 02 (HnM) is T-invariant and lies in B 0 , while Tis irreducibleon B 0 /02(B 0 ), we conclude 02 (H n M) = B 0 • Therefore P =Po is of order 9, so
s = 2 and 02 (H) =Ki x K2 is the product of two blocks. Therefore 02 (HnM) =
Xi x X2 with Xi := 02 (Ki n M) ~ Z3/Q~. Now as 02 (H n M) = Bo, while Xi
has just two noncentral 2-chief factors, Xi cannot be diagonally embedded in L 0 , so
(interchanging Li and L2 if necessary) Xi= Bon Li. Then Xi is Ti-invariant, and
Li is an A5-block as Xi has two noncentral 2-chief factors. Now Ki is Ti-invariant,
I := (Lr,Ki) :S:: Ca(X2), I is Ti-invariant, and 8 = Baum(Ti) since we saw
8 :S:: Ti. Hence Na(Ti) :S:: M by 10.2.1. Therefore NT(X2) =Ti E 8yl2(Nc(X 2 )),
so Ti E 8yl2(ITi). Hence we can apply 1.2.4 to embed Li :::; Lr E C(I), and
then Ki = [Ki, Xi] :S:: Lr, so Li < Lr since Kin M = Xi. Now Na(X 2 ) is
an SQTK-group, so m3(Na(X2)) :S:: 2 and hence m 3 (Lr) = 1. This rules out thepossibility that 02(Li) :S:: 02(Lr) and Lr/02(Lr) ~ A1 in A.3.14. We now obtain
a contradiction via the argument in the last two paragraphs of the proof of 10.2.1.This contradiction completes the proof 9f (4), and of the lemma. D
PROPOSITION 10.2.3. If HE 1-l*(T, M) with n(H) > 1, then
{1) n(H) = 2.
(2) A Hall 2' -subgroup of H n M is faithful on L 0.
(3) If L ~ L3(2), then To0^2 (HnM) is a maximal parabolic in Lo and H/0 2 (H)
~85 wr Z 2 •