CHAPTER 14
L3(2) ~n the FSU, and L 2 (2) when .Cr(G, T) is empty
The previous chapter reduced the treatment of the Fundamental Setup (3.2.1)
to the case L ~ £3(2)-which we handle in this chapter. This in turn reduces the
proof of the Main Theorem to the case £ f ( G, T) = 0.
Recall that the case in the FSU where L ~ A5 is actually treated last in the
natural logical order, but because of similarities with the case L ~ A 6 , those cases
were treated together in the previous chapter; this was accomplished by introducing
assumption ( 4) in Hypothesis 13.3.1.
In this chapter it will again be convenient to take advantage of some similarities
in the treatment of two small linear groups: namely between the case L ~ £ 3 (2)
for LE £1(G,T), and suitable LE M(T) such that LT/02(LT) ~ £ 2 (2) acts
naturally on some 2-dimensional member of R2(LT). The latter situation is the
most difficult subcase of the case £1(G, T) = 0, which of course remains after the
Fundamental Setup is treated. As a result, we begin this chapter with several
sections providing preliminary results on the case £1(G, T) = 0, and in particular
on the subcase with L/CL(V) ~ £ 2 (2).
14.1. Preliminary results for the case £f(G, T) empty
As usual, TE Syl2(G) and Z := fh(Z(T)).This chapter includes the beginning of the treatment of the case £ f ( G, T) = 0.
The first few results below are based only on that assumption, but afterwards we
will assume the stronger Hypothesis 14.1.5.
We use the following notation through the section:NOTATION 14.1.1. Let E := fh(Z(J(T))), M E M(T), V E R2(M), and
M := M/CM(V).
Recall from section A.5 of Volume I that for H E H(T), in this section we
deviate from om usual meaning of V(H) in definition A.4.7, instead using the
meaning in notation A.5.1, namely
V(H) := (zH).
Recall the partial ordering on M(T) given by M1 :S M2 whenever
M1 = OM 1 (V(M1))(M1 n M2).
Recall V(H) E R 2 (H) by B.2.14 ..
The first result below does not even require the hypothesis £ f ( G, T) = 0:
LEMMA 14.1.2. Assume J(T):::; CM(V) and set S := Baum(T). Then
(1) V:::; E and S = Baum(CT(V)).
(2) Assume either
975