14.7. FINISHING L 3 (2) WITH (VGl) ABELIAN 1071Next suppose g E G with V{ ::::; E. As K centralizes V 1 and D normalizes K, K
centralizes (VP)= E, so K::::; 02 (Ca(V{)) = K9, and hence g E Na(K) =MK.
Thus as U :S 02(K), U^9 ::::; 02(K) ::::; 02(JR 1 ). Also using an earlier remark,CJR 1 (E) = JR1 n CDR 1 (E) = CR 1 (E) = 02(JR1). Therefore we may apply
14.7.31.2 with JR 1 , E in the roles of "Y, Vy", to conclude that W 0 .:9 JR 1. But
then 02 (J) ::::; ND(Wo) ::::; D n M::::; T, a contradiction which completes the proof
of (4), and hence of 14.7.67. .D
LEMMA 14.7.68. (1) zG n Uo = {z}.
(2) Ifu E u'/f with [u,T] = 1, then Ca(u)::::; H, and U 0 is the unique memberof Uff containing u.
PROOF. Assume u satisfies the hypotheses of (2) and set Gu:= Ca(u). Notice
Tu:= CT(u) is of index at most 2 in T and K::::; Gu by Coprime Action.
Suppose first that Gu ::::; H holds; we will show that (1) and the remaining
statement in (2) follow. Assume that u lies in some conjugate ug. Then K9 ::::;
02 (Gu)::::; 02 (H) = K, so that K9 = K. Thus g E Na(K) = H by 14.7.67.4, so in
particular g normalizes U 0 , completing the proof of (2) in this case. Further as z
satisfies these hypotheses in the role of "u", z is in a unique G-conjugate of U 0 , so
zG n U 0 = zNa(Uo) by A.1.7.1. But then as HEM by 14.7.67.4, Na(Uo) = H =
Ca(z) so that (1) also holds.
So to complete the proof of the lemma, we assume Gu 1:. H, and it remainsto derive a contradiction. As K has more than one noncentral 2-chief factor by
14.5.21.1, KTu is not a block, so by C.1.26 there is 1 =/=-C char Tu with C ::::! KTu.
But then as Tu is of index at most 2 in T, H =KT::::; Na(C) so that Na(C) = H
since H E M. Thus if Tu :S To E Syb(Gu), then NT 0 (Tu) :S Na(C) = H, so
that Tu= NT 0 (Tu) and hence Tu= To. Therefore K :S Lu E C(Gu) by 1.2.4, and
Lu ::::) Gu by 1.2.1.3 since T normalizes K. Thus K <Lu as Gu 1:. H = Na(K).
In particular Lu 1:. H since K ::::! H, so as H = Ca(z), [z, Lu]=/=-1. Observe further
as Ua. is elementary abelian and contained in Ri with Ri = Q E Syb(K) that
Lu/0 2 (Lu) does not involve SL2(5) on a group of odd order, and so is quasisimple
by 1.2.1.4.
Observe that the hypotheses of 1.1.6 are satisfied with Gu, H in the roles
of "H, M" j so that we may apply 1.1.5. Suppose first that Lu is quasisimple,
and hence a component of Gu. As u E [U, K] :S Lu :S Gu, Z(Lu) is of even
order. On the other hand z is in the center of the Sylow 2-subgroup Tu of Gu, and
KTu = Ca,, (z). Inspecting the list of possiblities for Lu in 1.1.5.3, we conclude
from this structure of KTu (in particular from the two noncentral 2-chief factors)
that Lu is the covering group of Ru. Next Vis the unique Li-invariant complement
to (u) in (u)V,.so as Lu/(u) ~Ru, NL,,(V) =:Lo satisfies Lo/02(Lo) ~ L3(2}.
Thus Lo ::::; 02 (Na(V)) = L by 14.7.67.5, so as IT: Tul = 2 and L = 02 (L), we
conclude L =Lo. Then as Z(Lo) is of order 2 by I.1.3, (u) = Z(L) n Tu, so (u) is
T-invariant, contrary to Tu E Syb(Gu).
Therefore Lu is not quasisimple, so F*(Lu) = 02(Lu) by 1.2.11. Let Ru :=
02(KTu)· As KTu ::::! H, Ru= 02(H)nKTu ::::! H, so since HEM, C(Gu,Ru) :S
Hu:= H n Gu and Ru= 02 (Hu). Thus Hypothesis C.2.3 is satisfied with Gu, Ru,
Hu in the roles of "H, R, MH". Then as Lu :::] Gu, while Lu 1:. Hand Lu/02(Lu)