1124 15. THE CASE .Cr(G, T) = 0PROOF. Let SS T1 E Syl2(I). By 15.3.6.2, Na(S) SM, so as IT: SI = 2,
T 1 SM. Thus either (1) holds, or T1 E Syl 2 (M), and in the latter case, 15.3.2.4
supplies a contradiction as Y+ -=f. L Thus (1) is established.
We next prove (2)-(6).Suppose first that case (1) of Hypothesis 15.3.10 holds. Then (3) holds vac-
uously, and (4) holds as M = Na(Y) by 15.3.7. Also V = (ZY) S U1, so (2)
holds as M = Na(V). Further from 15.3.7 and the structure of Min 15.3.2.1, (6)
and the first three statements of (5i) hold. As R :::) YT, M = !M(Na(R)) by
15.3.7, so that C(G, R) s M. Also R = Cr(V) as 02(YS) = 1. As Y :::) M1
and R = 02 (YS) with SE Syl 2 (I), RE B 2 (M 1 ) and RE Syb((RM^1 )) by C.1.2.4.Then as N 1 (R) S M1, R = 02 (N1(R)), completing the proof that conclusion (i) of
( 5) holds in this case.Now suppose that case (2) of Hypothesis 15.3.10 holds, so that Y+ = 02 (Cy(Vi)).
ThenM = !M(YS) by Hypothesis 15.3.10, so that Yi. I. Thenas IY: Y+02(Y)I =
3, Y+ = 02 (Y n I). Further as V 1 , 112, and Y+ are normal in YS, the remainderof (3) and also (4) follow as M = !M(YS). Then as V 2 S (zY+) S U1, (2) follows
from (3). We next prove (5) and (6). First suppose that conclusion (1) of 15.3.7
holds. Then Y = 0
31
(M), so Y+ = 031 (M 1 ) as Y+ = 02 (Y n I). Thus (6) holds,and visibly conclusion (ii) of (5) holds. Therefore we may assume that conclusion
(2) of 15.3.7 holds. Then Y* ~ 3i+^2 and Y+/0 2 (Y+) ~ E 9 , with Y+S/R ~ 83 x 83
using the structure of M in 15.3.2.1. This time Y = B(M), so (6) holds. As
Cr(V) = Cs(Y+), R = Cr(V) = 02(YS), so C(G,R) s Mas M = !M(YT).As Y+ :::) M1 by (6) and R = 02(Y+S) with S E Syh(I), R E B2(M1) and
R E Syb((RM^1 )) by C.1.2.4. Then as N1(R) S M1, R = 02(N1(R)), so that
conclusion (i) of (5) holds.It remains to prove (7)-(12).
As IT : SI = 2 and F*(M) = 02 (M), 1.1.4.7 implies (7). In case (1) of
Hypothesis 15.3.10, V = [V, Y] so that O(I) S C1(V) S M1 by A.l.26.1, and hence
(8) follows from (7). In case (2) of Hypothesis 15.3.10, 112 = [112, Y+J, and (8) follows
similarly from (7) as C1(l12) S M by (3).Next X := Y+C1(U1) S M1 by (2), so Y+ = B(Y+C1(U1)) by (6); then (9)
follows from (4) as Y+ is nontrivial on 1 #-[112, Y+] S U1 by construction. By (5),
Cr(V) SR SS. So if [V, J(T)] = 1, then Baum(R) = Baum(S) = Baum(Cr(V))
and C(J, Baum(S)) .:::; M1 by 15.3.5 and B.2.3.5, so that conclusion (i) of (10)
holds. Then I= J(I)N1(J(S)) = J(I)M1 by a Frattini Argument, so as Ii. M,
J(I) i. M, and hence J(I)* i. Mj by (2), establishing (11). So suppose insteadthat [V, J(T)] #-1. Then case (2) of 15.3.5 holds, so that J(T) = S. But by (5),
Y+ = fY+, SJ and Cr(V) SR SS, so Y+ = [Y+,J(S)], and hence conclusion (ii) of
(10) holds. Thus if J(I) S Mj, then Y+ = B(J(I)) ::::J I* by (6), contrary to (9).
This completes the proof of (10) and (11).
Assume the hypotheses of (12), with 1 -=f. C char S and C :::) (L, S). Then
(0^2 (Y n L), T) S Na(C), so as 02 (Y n L) i. CM(V) by hypothesis, Na(C) S M
by 15.3.2.4, a contradiction. 0
LEMMA 15.3.12. Assume IE H+ and there is LE C(J) with m 3 (L) 2: 1. Also
assume m 3 (Y+) = 2. Then(1) YL := 02 (Y+ n L) -=f. 1. Further R normalizes L and 02 (L)0 2 (YL) SR.