976 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTY(a) Mis maximal in M(T) under :S and V = V(M), or
(b) V = ((V n z)M), and M is the unique maximal member of M(T)
under :S.Then M = !M(NM(S)) and C(G, S):::; M.
PROOF. Part (1) follows from B.2.3.5. Assume one of the hypotheses of (2).
Then M = !M(NM(Cr(V)) = !M(NM(S)) by A.5.7.2, so that C(G,S):::; M. D
The next two preliminary results do assume Lt(G, T) = 0:
LEMMA 14.1.3. Assume Ct(G, T) = 0. Then H^00 :::; CH(U) for each HE 1i(T)
and U E R2(H).
PROOF. By 1.2.1.1, H^00 is the product of groups L E C(H). Then L E £( G, T).
By hypothesis, L 'f:-Ct(G, T), so by 1.2.10, [U, L] = 1. D
LEMMA 14.1.4. Assume Ct(G, T) = 0, Mis maximal in M(T) under.:S, and
J(T):::; CM(V(M)). Then Mis the unique maximal member of M(T) under :S.
PROOF. Let S := Baum(T). By 14.1.2.2, C(G, S):::; Mand M = !M(NM(S)).In particular Nc(J(T)) :::; M.
Let M 1 E M(T) - {M}, and V := V(M 1 ). If [V, J(T)] = 1, then by a Frattini
Argument, Mi= CM 1 (V)NM 1 (J(T)), so we conclude Mi :SM as Nc(J(T)):::; M.
Hence we may assume [V, J(T)] -=/= 1. Set Mi := Mi/CM 1 (V) and I:= J(M1).
By a Frattini Argument, M 1 = INM 1 (J(T)) = I(M 1 nM), so it will suffice to show
that I = CM 1 (V)(I n M). By 14.1.3, Mi is solvable, so by Solvable Thompson
Factorization B.2.16, I* = Ji x · · · x 1; with I[ ~ S 3 , and s :::; 2 by A.l.31.l.Now I :::; 021 (IT*), and by B.6.5, 021 (IT) is generated by minimal parabolics H
above T, so it will suffice to show that H:::; M for those H with H* -=J 1. We apply
Baumann's Lemma B.6.10 to H to conclude SE Syh(0^2 (H)S). Then we apply
Theorem 3.1.1 with NM(S), Sin the roles of "M 0 , R", and as M = !M(NM(S)),we conclude that H :::; M as required. D
We next discuss the basic hypothesis which we will use during the bulk of ourtreatment of the case £ f ( G, T) empty:
The final result in this chapter, Theorem D (14.8.2), determines the QTKE-
groups G in which Lt(G, T) -=J 0. Then in the following chapter we determine
those QTKE-groups G such that £ f ( G, T) = 0. As in the previous chapters on the
Fundamental Setup (3.2.1), we may also assume that IM(T)I > 1, since Theorem
2.1.1 determined the groups for which that condition fails. Finally we divide our
analysis of groups G with Ct(G, T) = 0 into two subcases: the subcase where
IM(Cc(Z))I = 1, and the subcase where IM(Cc(Z))I > 1. The second subcase is
comparatively easy to handle, perhaps because all the examples other than L 3 (2)
and A5 occur in the first subcase.
Thus in this section, and indeed in most of those sections in this and the
following chapter which are devoted to the case Ct(G, T) empty, we assume the
following hypothesis:
HYPOTHESIS 14.1.5. G is a simple QTKE-group, TE Syl 2 (G), and
(1) Ct( G, T) = 0.
(2) There is Mc E M(T) satisfying Mc= !M(Cc(Z)).
(3) IM(T)I > 1.