1158 15. THE CASE .Cr(G, T) =^0
LEMMA 15.4.6. If Mis maximal in M(T) with respect to :S and [V(M), J(T)] =
1, then M is the unique maximal member of M(T) under :S.
The two groups which satisfy Hypothesis 15.4.l appear in conclusion (2) of the
next result. The second subsection completes the treatment of Hypothesis 15.4.1,
by eliminating the case where IZI > 2, and the case where IZI = 2 but conclusion
(1) of the result fails.
LEMMA 15.4.7. Assume Z is of order 2. Then either
(1) There exists a nontrivial characteristic subgroup C2 := C2(T) of Baum(T)
such that
Mt= !M(NG(C2)),
and Mt is the unique maximal member of M(T) under :S, or
(2) G ~ L3(2) or As.
PROOF. Let 8 := Baum(T) and choose Ci := Ci(T) for i = 1, 2 as in the
Glauberman-Niles/Campbell Theorem 0.1.18. Thus 1 -=J. C 2 char 8, and 1 -=J. C1 :::;
Z, so as Z is of order 2 by hypothesis, C 1 = Z.
Assume (2) fails; we claim that:
()For each ME M(T), M = NM(C2)CM(V(M)).
Assume() fails and set V := V(M). If [V, J(T)] = 1, then 8 = Baum(CT(V))
by B.2.3,5, so ()holds by a Frattini Argument, contrary to our assumption. Thus
[V, J(T)] -=J. 1.
By 15.4.2.2, M is solvable, so by Solvable Thompson Factorization B.2.16,
J(M) = Y = Y1 x · · · x Yr with Yi ~ 83 and V = V1 x · · · x Vr x Cv(J(M)),
where 1!i := [V, Yi] ~ E4, and Y and Yi denote the preimages in M of Y and Yi.
As Mis an SQTK-group, r:::; 2 by A.1.31.1. As IZI = 2, Cv(J(M)) = 1 and Tis
transitive on {Y 1 , ... , Yr}. Thus if r = 1, then m(V) = 2 and M = Y = GL(V),
while if r = 2 then m(V) = 4 and M is the normalizer Ot(V) of Y in GL(V).
In either case as C1 = Z, CM(C1) = CM(Z) = CM(V)T. Thus as () fails,
M > NM(C2)CM(V) = NM(C2)CM(C1) = (NM(C2),CM(C1)). Therefore as Mis
solvable, we conclude from 0.1.28 that there is an A 3 -block A4 ~ X :SI :SI M such
that X = [X, J(T)].
Let Xo := (XM). By the previous paragraph, either r = 1 and X 0 = X, or
r = 2 and Xo = X1 x X2 with X = X 1 and X 2 = xt for suitable t ET-NM(X).
Let H E M(XoT). As His solvable by 15.4.2.2, applying 0.1.27 to H, X in the
roles of "G, K", we conclude that Xo :SI H, so H = NG(Xo) as H E M. Thus
H = !M(XoT), so M = H = NG(Xo) = !M(XoT) as XoT :::; M E M. Next
XoT/CT(Xo) ~ 84 or 84 wr Z2. As IZI = 2, Z:::; 02(Xo), so CT(Xo) = 1. Then
as ME He, CM(Xo) = 1, so M = XoT ~ 84 or 84 wr Z2. In the second case,
Theorem 13.9.1 supplies a contrad1ction, so suppose the first case holds. Then
T ~ Ds, so as F*(CG(Z)) = 02(CG(Z)) by 1.1.3.2, we conclude T = Ca(Z).
Further Thompson Transfer shows that each noncel).tral involution of T is fused
into Z(T), so that G has one conjugacy class of involutions. Thus (2) holds by
I.4.1.2, contrary to our assumption; this completes the proof of the claim (*).
Pick"Mt E M(NG(C2)). By'(*), for each ME M(T), M ;S Mt· Thus Mt
is the unique maximal member of M(T) under :S-in particular Mt is uniquely
determined since :Sis a partial order (cf. A.5.5, and in particular A.5.4). Thus (1)
holds. D