1104 15. THE CASE .Cr(G, T) = 0Next 02 (P 0 ) = Cs(V 2 ), and as case (4) of 15.1.7 holds, Cs(Vi) = Cr(V), so
Q = 02 (P 0 ). Thus from the structure of the rank-2 parabolics of LS: 02(Y3) :::; Q
so that Q E Syl 2 (QY 3 ), as the nodes determining Po and P 3 are not adjacent in the
diagram of L; but Q rt Syb(P 2 ), as the nodes determining Po and P2 are adjacent.
Therefore as t acts on Q, x;t # Y3, so x;t = Y2.
Let Q 2 := 02 (P 2 ); as Tacts on Y2 and on S, Tacts on Q2. But P2 = CKs(Z),
and K has noncentral 2-chief factors on both 02 (L) and 02(K)02(L)/02(L), so
that K is not an L 3 (2)-block. We conclude from C.1.34 that there is a nontrivial
characteristic subgroup C of Q 2 with C '.::1 KQ 2 , and hence C '.::1 KT. Then
H := (T, K) :::; N c ( C). On the other hand, Yo centralizes Z1 but Vi = [Vi, Yo],
so from our construction in 15.1.16, M = (T, Yo)CM(V). Then by A.5.7.1, M =
!M((T, Yo)). Since Yo:::; K, we conclude H:::; Nc(C):::; M. But then K = K^00 :::;
CM(V) by 14.1.6.1, contrary to V 2 = [V 2 , Y 0 ]. This contradiction completes the
proof of 15.1.29. DObserve that 15.1.28 and 15.1.29 supply a contradiction which establishes The-
orem 15.1.15.
15.20 Finishing the reduction to Mf/CMr(V(Mr)) c:::'. 0!(2)
In this section, we complete the proof of Theorem 15.1.3, begun in section 15.1.
Thus we assume G is a counterexample to Theorem 15.1.3.15.2.1. Preliminary reductions. Recall we are assuming Hypothesis 14.1.5;
in particular by 14.1.5.2,Mc= !M(Cc(Z)).
We continue Notation 15.1.4: namely we set M := MJi and set V := V(M)
unless case (6) of 15.1.2 holds, where we set V := [V(M), MJ]. Also Mo is the
preimage in M of J(M, V).
Since Theorem 15.1.15 eliminated cases (4), (6), and (7) of 15.1.7, we have
reduced to the remaining cases in 15.1.7, which we summarize below for convenience:LEMMA 15.2.1. m(V) = 4, and one of the following holds:
(1) ii!= Mo~ 83.
(2) Mo~ 83 and M ~ 83 x Z3.
(3) ii!= Mo= nt(V).(4) Mo~ D10, T ~ Z2 or Z4, and either F(M) = F(Mo) or F(M) ~ Z15.
Furthermore if V < V(M), then case (3) holds.
LEMMA 15.2.2. If T:::; X:::; M with Mo:::; XCM(V) or Mo:::; XNM(Z n V),
then M = !M(X).PROOF. Let Mi E M(X). By 15.1.5.1, V = ((Zn V)x), and by 15.1.9.2,
Mi,....., < M, so M
1 :::; M by A.5.6. D( ) 2((^0(^2) (Mo)T))
LEMMA 15.2.3. Let Re := 02 M n Mc ' y := 0 Re ' and M* .-
M /02 (M). Then
( 1) 1 # Y = [Y, Re], and one of the following holds:
(i) Y = 02 (M 0 ) ~ Y*. Further if case (3) of 15.2.1 holds, then CM(V) is
a 31 -group.