15.1. INITIAL REDUCTIONS WHEN .Cr(G, T) IS EMPTY 1091PROOF. As 7t*(T, M) -=!= (i), (1) follows from 15.1.12.4. Next if [V,J 1 (T)] = 1,
then by B.2.3.5, NM(Cr(V)) normalizes J 1 (T) and hence also normalizes E 1 , so
that Nc(E 1 ) ::::; M by 15.1.9.5, contrary to (1). This establishes (2).
By (2), there is A E Al(T) with A-=!= 1. Now m(A)::::; m 2 (M) and m 2 (M)::::; 2
from 15.1.7. As A E A 1 (T), m(V/Cv(A)) ::::; m(.A) + 1, while q(M, V) > 1 by
15.1.8, so that m(V/Cv(A)) = m(A) + 1. Hence for m(.A) = 1 or 2,
_ _ m(V/Cv(A)) _
rA,v - m(A) -^2 or^312 ,respectively. Assume that case (3) of 15.1.7 does not hold. Then by 15.1.8,
q(M, V) = q(M, V) = 2, and the calculation above shows that m(A) = 1 and
rA,v = 2 for each A E Al(T) with A-=!= l, so that A E Q*(M, V). This estabfoihes
(3).
It remains to prove (4). Suppose first that case (3) of 15.1.7 holds. Then
M = Mo and T ~ E4, and Jl(T) -=!= 1 by (2). Therefore either Jl(T) = T,
and hence conclusion (a) of (4) holds, or J 1 (T) is of order 2, and conclusion (b)holds. Thus we may assume that case (3) of 15.1.7 does not hold. Then by (3),
J 1 (T)::::; TnM 0 =: T 0. As case (3) of 15.1.7 does not hold, either case (6) of 15.1.7
holds or ITol = 2. In the latter case, Ji(T) = To so that Ji(M) = Mo, giving
conclusion (a). In the former case, A ::::; Mi for i = 1 or 2 since r A, v = 2, and then
J 1 (T) = (.A,.At) = T 0 , so again conclusion (a) holds. This completes the proof of
· (4). D
LEMMA 15.1.14. Let VE:= Cv(J1(T)). Then
(1) 02 (Cc(Z))::::; Cc(VE).(2) NM(J1(T))::::; Nc(VE)::::; Mc.
(3) Nc(J1(T))::::; Mn Mc.
PROOF. By 15.1.6, 02 (0M(Z)) ::::; CM(V) ::::; CM(VE)· Thus if (1) fails, then
02 (0c(Z)) 1:_ (Mn 02 (Cc(Z))T, 02 (Cc(Z)) n Cc(VE)),
so there exists HE 7t*(T,M) with H::::; Cc(Z) but 02 (H) 1- 02 (Cc(VE)). How-ever since VE ::::; D 1 (Z(J 1 (T))), this contradicts 15.1.12.4, ·so (1) is established.
Then (1) implies (2) since Mc= !M(Cc(Z)). Finally as J(J1(T)) = J(T) by (1)
and (3) of B.2.3,
Nc(J1(T)) = Nc(J(T)) n Nc(J1(T))::::; NM(J1(T))
by 15.1.9.1, so (2) implies (3). D15.1.2. Eliminating some larger possibilities from 15.1.7. Our proof of
Theorem 15.1.3 now divides into two cases:
Case I. M = (OM(Z1), T) for some nontrivial subgroup Z1 of Cv(J1(T)).