1156 15. THE CASE .Cr(G, T) = (/JBecause of Hypothesis 15.4.1, members of t,(G, T) have few overgroups, so that
t,*(G, T) is nonempty in many situations-cf. 15.4.3.1 and 15.4.12.
15.4.1. Preliminary results, and the reduction to Cc(Z) = T. Recall
B+(G, T) is defined before 3.2.13.LEMMA 15.4.2. (1) .C(G, T) = 0.
(2) Each member of M(T) is solvable..
(3) If XE S(G, T) then [Z, X] =f- 1 and XE Sj(G, T) but X ¢'c B+(G, T). In
particular X/0 2 (X) is a 3-group or a 5-group.(4) Assume Mo E H(T) with M = !M(Mo), and V E R2(Mo) with R :=
CT(V) = 02(CM 0 (V)) and V:::; 02(M). Then q(Mo/CM 0 (V), V):::; 2.
PROOF. Assume .C(G,T) =f-0. Then there is L E .C*(G,T). Setting Lo :=
(LT), Na(L 0 ) = !M((L,T)) by 1.2.7.3. But by Hypothesis 15.4.1.1, .C1(G,T) =
0, so (L, T) :::; Ca(Z) and hence Na(Lo) = !M(Ca(Z)), contrary to Hypothesis
15.4.1.2. Thus (1) holds, and since C(M) ~ .C(G, T) for ME M(T), (1) and 1.2.1.1
imply (2).
Supppose XE S(G, T). By (1), XE S*(G, T), so by 1.3.7, Na(X) = !M(XT).
Thus [Z, X] =f-1 by Hypothesis 15.4.1.2, so X E Sj (G, T). Hence X tfc B+(G, T)
by 3.2.13, completing the proof of (3).
Assume the hypotheses of (4), and let q := q(M 0 /CM 0 (V), V). Pick H E
H*(T, M), and let QH := 02 (H). Observe that Hypothesis D.1.1 is satisfied with
M 0 , H in the roles of "G 1 , 02 ": First, by hypothesis M = !M(Mo), so that
02( (Mo, H)) = 1, and hence part (3) of Hypothesis D.1.1 holds. Second, CT(V) =
02(CM 0 (V)), with 02(CM 0 (V)) = 02(Mo) since V E R2(Mo), so that part (2)
of D.1.1 holds. Finally by 3.1.3.1, H n M is the unique maximal subgroup of Hcontaining T, so that part (1) of D.1.1 holds. Now recall by B.5.13 that if the dual
V* is an FF-module for M 0 /CM 0 (V), then q(M 0 /CM 0 (V), V) :S 2. Thus combining
conclusions (2), (3), and (4) of the qrc-lemma D.1.5 into case (ii) below, one of the
following holds:
(i) ViQH.
(ii) q(Mo/CMo (V), V) :::; 2.
(iii) V:::; Rn QH :::! H, the dual V* is not an FF-module for Mo/CM 0 (V),
U := (VH) is abelian, and H has a unique noncentral chief factor on U.
If (ii) holds, then the conclusion of (4) holds and we are done.
Assume that (i) holds. We verify Hypothesis E.2.8 with H n M in the role of
"M": As Vi QH, Ti QH, so Hi Na(T); hence by 3.1.3.2, His a minimal
parabolic in the sense of Definition B.6.1, and so is described in B.6.8. Therefore by
B.6.8.5, kerHnM(H) is 2-closed with Sylow group QH, so Vi kerHnM(H). Finally
V:::; 02(M) by hypothesis, so that V:::; 02 (H n M). Now q(AutH(V), V) :::; 2 by
E.2.13.2, and again (4) holds.
This leaves case (iii), so as His solvable by (2), RE Sy[z( 02 (H)R) by D.1.4.4.2.
But now 02 ( (M 0 , H)) =f-1 by Theorem 3.1.1, contrary to an earlier observation.
This completes the proof of (4), and hence of 15.4.2. D
Following the notational convention of chapter 1, set I:, f ( G, T) := I:, ( G, T) n X f.
LEMMA 15.4.3. Let X E t,(G, T), with IX : 02 (X)I = p where p is chosen
maximal among such X, and suppose p > 3. Then