1088 15. THE CASE L'.f(G, T) = 0that conclusion (6) holds. In case (2) of 15.1.2, conclusion (3) holds. Cases (4) and
(5) of 15.1.2 are conclusions (7) and (4).
It remains to treat case (1) of 15.1.2. In this case, m(V) = 4 as m(V) > 2,
so Mo is 83 or D 10. If P := 02 (M 0 ) = F*(M), then M :::; Aut(P) ~ 83 orSz(2), respectively, so that conclusion (1) or (5) holds. Thus we may assume that
P < F*(M). Now NI :::; NaL(V)(Mo), and NaL(V)(Mo) is nt(V) or Z4/Z15,
respectively. Since P < F*(M), and 02 (M) = 1 by 15.1.5.1, one of conclusions
(2)-(5) of the lemma holds. Thi$ completes the proof. DOur assumption that G is a counterexample to Theorem 15.1.3 has ruled out
the subcases of 15.1.2 in which M contains an FF*-offender on V; that is we are
left with thoses cases where q(M, V) > 1. Indeed:
LEMMA 15.1.8. One of the following holds:
(1) q(M, V) = q(M, V) = 2.
(2) Case (3) of 15.1. 7 holds, where q(M, V) = 3/2 and q(M, V) = 2.
PROOF. The proof of 15.1.2 showed that one of the following holds:
(i) V = V(M), and (MJ, V) is an indecomposable appearing in one of cases
(1)-(5) of D.2.17.(ii) V = V(M), and the (Mi, \ti) are indecomposable and appear in case (1) or
(2) of D.2.17; hence case (6) of 15.1.7 holds.
(iii) V < V(M), and hence case (3) of 15.1.7 holds.
In cases .(i) and (ii), MJ = Mo by Notation 15.1.4, so we conclude from the
values listed in the corresponding cases of D.2.17 that lj_(Mo, V) = q(Mo, V) = 2-unless case (3) of 15.1.7 holds, where (M, V) appears in case (5) of D.2.17, Mo·=
M, q(Mo, V) = 3/2, and q(M 0 , V) = 2. However by definition of Q*(M, V), if
q(M, V):::; 2, then q(M, V) = q(M 0 , V) and q(M, V):::; q(M, V). Thus the lemma
holds in cases (i) and (ii). In case (iii), conclusion (2) of the lemma holds, again as
(M, V) appears in case (5) of D.2.17. D
Recall that IM(T)I > 1 by Hypothesis 14.1.5.3, so that H*(T, M) is nonempty.
The next few results study properties of members of H*(T, M).
LEMMA 15.1.9. Set R := CT(V). Then
(1) [V, J(T)] = 1 = [V(M), J(T)], so that Baum(T) = Baum(R) and further
C(G,Baum(T)):::; M.
(2) M is the unique maximal member of M(T) under :S.
(3) H*(T, M) ~ Ca(Z):::; Mc.
(4) For each HE H*(T, M), 02 (H n M):::; CM(V).
(5) M = !M(NM(R)).
(6) NM(R) E He, VE R2(NM(R)), R = 02(NM(R)), and NM(R) = M; and
case (II) of Hypothesis 3.1. 5 is satisfied with NM ( R) in the role of "Mo" for any
HE H*(T, M).
(7) Na(T):::; M, and each HE H*(T,M) is a minimal parabolic described in
B.6.8) and in E.2.2 if H is nonsolvable.PROOF. If J(T) does not centralize V(M), then as m(V) > 2 by 15.1.7, 14.1.7