CHAPTER 4
Pushing up in QTKE-groups
Recall that in chapter C of Volume I, we proved "local" pushing up theorems
in SQTK-groups. In this Chapter we use those local theorems to prove "global"
pushing up theorems in QTKE-groups. Let L, V be a pair in the FundamentalSetup (3.2.1), L 0 := (LT), and M := Na(L 0 ). We use L 0 T and our pushing up
theorems to show that large classes of subgroups must be contained in M.
For example, in Theorem 4.2.13 we use the fact that L 0 T is a uniqueness
subgroup to prove roughly that if the pair L, V in the FSU is not too "small", then
each subgroup I of Lo which covers Lo modulo 02(LoT) with 02 (1) =/= 1 is also a
uniqueness subgroup. Then we use Theorem 4.2.13 to prove Theorem 4.4.3, whichshows that for suitable subgroups B of odd order centralizing V, Na(B) ::::; M.
As a corollary, we see in Theorem 4.4.14 that for H E 1-l*(T, M) with n(H) > 1,
a Hall 2'-subgroup of H n M must act faithfully on V. This gives the inequality
n(H)::::; n'(NM(V)/CM(V)), (cf. E.3.38) which is used crucially in many places in
this work.4.1. Some general machinery for pushing up
Our eventual goal is to show roughly in most cases of the FSU that if I is the
set of subgroups I of LoT covering Lo modulo 02(LoT) with 02(I) =/= l, then each
member of I is also a uniqueness subgroup. If some member of I fails to be a
uniqueness subgroup, then we study a maximal counterexample I using the theory
of pushing up from chapter C of Volume I. Our starting point is 1.2.7.3, which saysthat L 0 T is a uniqueness subgroup. We develop some fairly general machinery to
implement this approach. So in this section we assume the following hypothesis
(which we will see in 4.2.2 holds in the FSU):HYPOTHESIS 4.1.1. Assume G is a simple QTKE-group, T E Syb(G), M E
M(T), and M+ = 02 (M+) ::::1 M. Further assume that M = !M(I) for each
subgroup I of M such that
M+Cr(M+/02(M+))::::; I and M = CM(M+/02(M+))I.
Let ~(M+) consist of those subgroups M_ of M containing M+CM(M+/0 2 (M+)).
LEMMA 4.1.2. Let R+ E Syb(CM(M+/0 2 (M+))). Then M = !M(NM(R+)).
PROOF. By hypothesis T is Sylow in M, so as M+ ::::1 M, we may assume
R+ = Cr(M+/0 2 (M+)). Also M+ = 02 (M+), so by A.4.2, M+R+ ::::; Na(R+)·
Now M = CM(M+/02(M+))NM(R+) by a Frattini Argument. So by Hypothesis
4.1.1 with NM(R+) in the role of "I", M = !M(NM(R+)). D
Next we define some more technical notation. We will study overgroups of M+which (in contrast to the subgroups I in 4.1.1) need not cover all of M modulo
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