CHAPTER 6
Reducing L 2 ( 2n) to n = 2 and V orthogonal
In this chapter, we continue our analysis of simple QTKE-groups G for which
there exists a T-invariant LE .Cj(G,T) with L/0 2 (L) ~ L 2 (2n). Recall that we
began this analysis in chapter 5. In particular in Theorem 5.2.3 we showed under
these hypotheses, and the hypothesis that n(H) > 1 for some HE 1-{*(T, M), that
either
(I) G is M23 or a group of Lie type of characteristic 2 and Lie rank 2, or
(II) the conclusion of 5.2.3.1 holds; in particular n = 2 and [R 2 (LT), L] is the
sum of at most two As-modules.In Theorem 6.2.20 of this chapter, we complete the reduction to the situation
where n = 2 and [R2(LT), L] is the sum of As-modules by considering the remaining
case where n(H) = 1 for each H E H*(T, M), and [R2(LT), L] is not the sum of
As-modules when n = 2. Section 6.1 carries out the reduction to the subcase n = 2.Then section 6.2 shows that the only quasithin example to arise in this subcase is
M22.This reduction allows us thereafter to regard L/02(L) ~ L2(4) as D4(2). We
treat that case in Part 5, which deals with the situation where there exists L E
.Cj(G, T) with L/0 2 (L) a group of Lie type group defined over F 2.
6.1. Reducing L2(2n) to L2(4)
As mentioned above, we wish to complete the reduction to the situation where
n = 2 and [R 2 (LT), L] is the sum of As-modules, under the hypothesis of chapter
- By Theorem 5.2.3, we may assume Hypothesis 5.1.8 fails. Thus in this section,
we assume the following hypothesis:
HYPOTHESIS 6.1.1. G is a simple QTKE-group, TE Syl2(G), and LE .Cj(G, T)
with L/0 2 (L) ~ L2(2n), L ::::1 ME M(T), and VE R2(LT) with [V,L] #-1. In
addition, assume(1) [V, L] is not the sum of one or two copies of the As-module for L/02(L) ~
As.(2) n(H) = 1 for each.HE H*(T, M).
REMARK 6.1.2. Notice Hypothesis 6.1.1.1 has the effect of excluding cases (2)
and (5) of 5.1.3 plus case ( 4) of 5.1.3 when n = 2. Thus either case (1) or (3) of 5.1.3holds, or n > 2 and case (4) of 5.1.3 holds. Similarly 6.1.1.1 excludes case (3) of
5.1.2; therefore by 5.1.2, either case (3) of 5.1.3 holds, or J(T) :::;; Cr(V) = 02(LT),
so that J(T) ::::] LT and hence M = !M(Na(J(T))).
Throughout this section, define Z := D1(Z(T)), VL := [V,L], and TL:= TnL.
Set Mv := NM(V), and Mv := Mv/CM(V).
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