14.1. PRELIMINARY RESULTS FOR THE CASE L.:r(G, T) EMPTY 977LEMMA 14.1.6. (1) M^00 :S CM(V).
(2) .C*(G,T) = C(Mc), so that Mc= !M((K,T)) for each KE C(Mc)·
(3) For each HE H(T), H^00 :S Ca(Z) :S Mc.PROOF. Part (1) follows from 14.1.3, and (3) follows from 14.1.3 applied to
V(H), using Hypothesis 14.1.5.2.
Let L E .C(G, T). Then (L, T) E H(T), so L :S Mc by (3). Therefore if
LE .C*(G,T), then by 1.2.7.3, Na((LT)) = !M((L,T)) =Mc, and hence LE
C(Mc)· Conversely let LE C(Mc) and embed L::::; KE .C*(G, T). We just showed
KE C(Mc), so L =Kand hence LE .C*(G, T). Thus (2) is established. DLEMMA 14.1.7. Assume J(T) 1:. CM(V), and either Mi= Mc or [Z[ = 2. Then
either
(1) m(V) = 2, M = GL(V) ~ £ 2 (2), and En V = Z is of order 2, or
(2) m(V) = 4 and M ~ ot(V). Thus M =(Yi x Y2)(t), where Yi~ L2(2), tisan involution interchanging Y1 and Y;, and V = Vi x V2, where Vi := [V, Yi] ~ E4,
and En V of order 4 contains Z of order 2.
PROOF. Set Y := J(M). By 14.1.6.1, Mis solvable; so by Solvable Thompson
FactorizationB.2.16 Y = Y1X· · ·XYr with Yi~ £ 2 (2) and V = Vi.X· · ·xV,.xCv(Y),
where Vi := [V, Yi]~ E4 for the preimage Yi of Yi. As Mis an SQTK-group, r :S 2
by A.1.31.1. Thus either M = Y x CM(Y), or r = 2 and M = (Y x CM(Y))(t),
where t interchanges :Y 1 and Y;. Then as Endy-(Vi) = F2, CM(Y) centralizes [V, Y].
Next Zn [V, Y] i= 1 and [V, Y] is T-invariant, so by 14.1.5.2,
CM(Y) :S CM([V, Y]) :S CM(Z n [V, Y]) :S Mc.
Suppose that Cv(Y) i= 1. Then c;(Y) i= 1, so [Z[ > 2, and Y :S Ca(Cz(Y)) :SMc. Therefore M = CM(Y)YT :S Mc, and hence M = Mc, contrary to our
hypothesis that Mi= Mc when [Z[ > 2.
Therefore Cv(Y) = 1 so that [V, Y] = V. Then CM(Y) centralizes V, so that
CM(Y) = 1. Hence if r = 1, then (1) holds, so we may assumer= 2. If Y < M,
then (2) holds, so we may assume M = Y = Y 1 x Y2. But then Zi := Zn Vi i= 1,
so [Z[ = 4 and Y3-i :S Ca(Zi) :S Mc, so that M = CM(V)Y1Y2 :S Mc, and thus
M =Mc, again contrary to our choice of M when [Z[ > 2. D
LEMMA 14.1.8. Assume Mi= Mc, X :S M is T-invariant of odd order, and
X f:. Mc.· Then V = [V, X].
PROOF. As Xis of odd order, V = [V,X] x Cv(X). Suppose Cv(X) i= 1.
As Xis T-invariant, Cz(X) i= 1. But then by 14.1.5.2, X :S Ca(Cz(X)) :S Mc,
contrary to hypothesis. D
LEMMA 14.1.9. If M is the unique maximal member of M(T) under :S, then
Mi= Mc.
PROOF. Assume M =Mc. By uniqueness of Mand the definition of :S, for
each M 1 E M(T) we have
Mi= CM 1 (V(M1))(M n Mi) :SM