14.1. PRELIMINARY RESULTS FOR THE CASE £.f(G, T) EMPTY 979Mc = MJ, then Mc is the unique maximal member of M(T) under :S, contrary
to 14.1.9. Thus Mc-=/= MJ, proving (4). So as CM 1 (V(MJ)) S Mc= !M(Ca(Z)),
MJ ::} Mc, and then (1) follows from the first sentence of the proof.
As V(MJ) S VJ, CM,(VJ) S CM 1 (V(MJ)). By a Frattini Argument,
MJ = CM,(VJ)NM 1 (Cr(VJ)) = CM 1 (V(MJ))NM 1 (Cr(VJ)),
so (2) follows from A.5.7.1 and (1).Next CM, (VJ) S Ca(Z) S Mc, while by (1) we may apply A.5.3.3 to each
ME M(T) - {Mc} to conclude that
CM 1 (VJ) S CM, (V(MJ )) S CM(V(M)),
so (3) holds. DLEMMA 14.1.13. Assume T S H S M with R := 02 (H) -=/= 1 and C(M, R) S
H. Then either
(1) 02,F·(M) SH and 02(H) = 02(M), or
(2) IZI > 2, and M =Mc= !M(H).PROOF. Observe that the triple R, H, M satisfies Hypothesis C.2.3 in the roles
of "R, MH, H". Thus we can appeal to results in section C.2, and in particular we
conclude from C.2.1.2 that 02(M) SR.
Suppose L E C(M) with L/0 2 (£) quasisimple and L 1:. H. By 14.1.5.1, L fj.
CJ(G, T), so Lis not a block. Thus RnL fj. Syl2(L) by C.2.4.1, so by C.2.4.2, Rs
NM(L). Then by C.2.2.3, RE !32(LR), so that 02(LR) SR by C.2.1.2. Further
Z(R) s 02(LR) as F*(LR) = 02(LR). Then as L fj. CJ(G, T), L centralizes
n1(Z(02(LR))) 2: n1(Z(R)) =: zR, so
L S CM(ZR) S C(M, R) SH.
Thus we conclude 02,E(M) SH.
Next set F := 02 ,F(M). By C.2.6, RE Syl 2 (FR) and either
(i) FR s H, and hence also 02 ,F·(M) SH, or
(ii) FR= (FR n H)X 0 , where X 0 is the product of A 3 -blocks X subnormal in
M with X = [X, J(R)].If (i) holds, then 02(M) = 02 (M n H) = 02(H) by A.4.4.1, so that conclusion
(1) holds. Thus we may assume (ii) holds. Therefore M = Mc and IZI > 2 by
14.1.10, so it remains to show that Mc = !M(H). Let K := CM(Xo). Then from
(ii) and (*) in the proof of 14.1.10, we see that KT= Ca(Z) and 02 ,F·(K) =
02,F* (M) n H s H, so that 02,F· (K H) s H.
We conclude from A.4.4.1 that 02(KH) = 02 (H) = R. Thus Ca(Z) =KT sC(M, R) S H, so that Mc= !M(H) by 14.1.5.2. This completes the proof that (2)
holds. D
LEMMA 14.1.14. If Ml E M(T)-{M}, then 02(M) < 02(M1nM) > 02(M1).
PROOF. Let H :=Mn Ml, R := 02(H), {M2,M3} = {M,M1}, and assume
that R = 0 2 (M 2 ). Then C(G, R) s M 2 , so that C(M3, R) s H. Thus by 14.1.13,
either R =· 02 (M 3 ), or M3 = Mc = !M(H). In the first case, 02(M2) = R =
02(M3), so Ml= M, contrary to the choice of Ml. In the second case as HS M 2 ,
M = M 1 for the san'le contradiction. · D