620 4. PUSHING UP IN QTKE-GROUPSPROOF. Let X := T+B, Q := 02 (X) and X := X/Q. Then F(X) is of odd
order, so as B is an abelian Hall 21 -subgroup of X, B S Ox• (F(X)) S F(X),
so B = F(X). Thus BQ '.SIX, so by a Frattini Argument (using the transitivity
of a solvable group on its Hall subgroups in P. Hall's Theorem, 18.5 in [Asc86a]),X = QNx(B) ·= QTBB, so that T+ = QTB· Also Q = CQ(B)[Q, BJ by Coprime
Action, with CQ(B) s TB, so T+ = [Q,B]TB· D
LEMMA 4.4.5. Assume parts (1) and (2) of Hypothesis 4.4.1. Then M+ =
LB02(M+)·
PROOF. By 4.4.1.2, [M+, B] S 02 (M+), so M+ acts on B02(M+); hence by
a Frattini Argument, M+ = 02 (M+)CM+(B). Now M+ is perfect by Hypothesis
4.2.1in4.4.1.1, so M+ = 02(M+)CM+(B)^00 = 02(M+)LB· D
In the remainder of this section, we assume we are in a counterexample toTheorem 4.4.3; in particular, GB</:. M.
LEMMA 4.4.6. (1) M = !M(LBTB)·
(2) If L :::) M then M = !M(LB)·
(3) Nc(VB) SM.
PROOF. Set I:= LBTB and VL := [R 2 (LT), L]. Observe that (cf. Notation
4.2.11) I EI: By 4.4.5, L S IR+; LT = LT+ = £TB by 4.4.1.2 and 4.4.4 (since
[0 2 (T+B),B] s R+); and 1 # VB S 02 (I) by 4.4.1.3. Thus if (1) fails, then
by Theorem 4.2.13, L :::) M, and LB/0 2 (LB) e:! L/0 2 (L) appears on the list ofTheorem 4.2.13. Further 4.2.13 says that VL is an FF-module for AutLT(VL), so the
LT-submodule VB is an FF-module for AutLT(VB) by B.1.5. Suppose L/02(L) e:!
Ln(2) for n = 3 or 4. Then case (2) of 4.2.13 holds, so either VL is the sum of one
or more isomorphic natural modules, or VL is the 6-dimensional orthogonal modul<?
for £ 4 (2). Therefore the submodule VB satisfies the same constraints, so conclusion
(i.a) of case (2) of Theorem 4.4.3 holds. Similarly if conclusion (4) or (5) of 4.2.13
holds, then VB = VL and conclusion (i. b) or (i.c) of part (2) of Theorem 4.4.3 holds.
In the remaining cases in Theorem 4.2.13, subcase (i) of case (2) of Theorem 4.4.3
imposes no further restriction on VB; hence subcase (i) of case (2) in 4.4.3 holds.This contradicts our assumption that we are in a counterexample to Theorem 4.4.3,
so we conclude that (1) holds. Under the hypothesis of (2), LT= L, so by Remark
4.4.2, we may take T+ = 1 and I := LB; thus (2) follows from (1). Finally (1)
implies (3), completing the proof of 4.4.6. DLEMMA 4.4.7. (1) 02(GB) = 1.(2) MB is a maximal 2-local subgroup of GB.
PROOF. By 4.4.6.1, M = !M(MB). Hence (2) holds, and as GB
implies (1).
LEMMA 4.4.8. O(GB) S MB.</:. M, (2)
DPROOF. By Hypothesis 4.4.1 and 4.4.5, 1 "I VB= [VB, LB]· As LB is perfect,
m(VB) ::::: 3, and in case of equality, LB acts irreducibly as £3(2) on VB, so VB n
Z*(GB) = 1. Therefore applying A.1.28 with GB in the role of "H", we conclude
that mp(Op(GB)) S 2 for each odd prime p. Thus by A.1.26, VB = [VB,LB] s
Cc(Op(GB)). Hence VB S CvBo(GB)(F(VBO(GB))) S F(VBO(GB)), so VB
02(VBO(GB)) and thus O(GB) S Nc(VB) SM by 4.4.6.3. D