978 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTYLEMMA 14.1.10. Assume M has a subnormal A3-block X, and 02(M) ::; R::; T
such that X = [X, J(R)]. Then M =Mc and JZJ > 2.PROOF. Let Xo := (X M). Thus as m 3 ( M) ::; 2, ·either Xo = X, or Xo =
X 1 x X 2 with X = X1 while X2 = Xt fort ET - NM(X). Set K := CM(Xo).
As X = [X, J(R)] and 02 (M) ::; R, Autx 0 T(Xo) = Aut(Xo), so as Z(Xo) = 1 we
conclude
M = (K x Xo)T. (*)
Since Zo :=Zn [02(Xo),Xo] #-1, K::; Cc(Zo)::; Mc= !M(Cc(Z)) by 14.1.5.2.
Now if CT(Xo) #-1, then Cz(Xo) #-1, so JZJ > 2 and Xo::; Cc(Cz(Xo))::; Mc=
!M(Cc(Z)). Then M =Mc by(*), so the lemma holds.
Therefore we may assume that CT(Xo) = 1. Then as F*(M) = 02(M), we
conclude from (*) that K = 1. Thus as Autx 0 T(Xo) = Aut(Xo), M = XoT ~ 84
or 84 wr Z 2. In the first case, T ~ D 8 , so G ~ L3(2) or A5 by l.4.3. But then
T = Cc(Z), contrary to Hypothesis 14.1.5. In the second case, Theorem 13.9.1
supplies a contradiction. DLEMMA 14.1.11. There exists a nontrivial characteristic subgroup C2 := C2(T)
of Baum(T), such that for each ME M(T), either
(1) M = CM(V(M))NM(C2), or
(2) M =Mc and [Z[ > 2.
PROOF. Let V := V(M). By 14.1.5.2, Mc = !M(Cc(Z)), so CM(V) ::;
CM(Z) ::; Mc. Let 8 := Baum(T) and choose Ci := Ci(T) for i = 1, 2 as
in the Glauberman-Niles/Campbell Theorem C.1.18. Thus 1 -j. C2 char 8 and1 #-C1 ::; Z. In particular Cc(C1) ::; Mc= !M(Cc(Z)).
Suppose first that [V, J(T)] = 1. Then 8 = Baum(CT(V)) by 14.1.2.1, so (1)
holds by a Frattini Argument since C 2 is characteristic in 8.
Thus we may assume that [V, J(T)] #-1, and that (2) fails, so that one of
the conclusions of 14.1.7 holds. In either case [Z[ = 2, so as 1 -j. C 1 ::; Z
we conclude C1 = Z. Further from the structure of M, CM(C 1 ) = CM(Z) =
CM(V)T::; CM(V)NM(C2). Therefore as we also may assume that conclusion (1)
fails, (CM(C1), NM(C2)) < M. Thus conclusion (2) of C.1.28 holds; in particular,
there is ax-block X of M with X = [X, J(T)] such that X does not centralize V.
Therefore as Mis solvable by 14.1.6.1, we conclude that each such Xis an A 3 -block
of M, and then 14.1.10 contradicts our assumption that (2) fails. D
In the remainder of the section let C 2 := C 2 (T) be the subgroup.defined as in
14.1.11 and its proof.LEMMA 14.1.12. Let Mt E M(Nc(C2)) and V(Mt) ::; Vt E R2(Mt ). Then
(1) Mt is maximal in M(T) under :S, Mt is the unique maximal member of
M(T) - {Mc} under :S, and if [Z[ = 2 then Mt is the unique maximal member of
M (T) under :S.
(2) Mt= !M(NM 1 (CT(Vt))).
(3) CM 1 (Vt)::; M for each ME M(T).(4) Mt#-Mc.
PROOF. If M E M(T) and either M -j. Mc or [Z[ = 2, then by 14.1.11,
M ~ CM(V(M))NM(C2); so as NM(C2) ::; Mn Mt, M ;S Mt. In particular if