610 4. PUSHING UP IN QTKE-GROUPS
By A.4.4.3, a 2 ,F* (M) 'f:_ H, so in view of 4.2.9, there is K E C(M) with
K/a 2 (K) quasisimple and Ki H.
LEMMA 4.2.10. {1) Lh :S: Kn H < K.
{2) mv(K) = 1 for each odd prime p E 7r(L), and K :::l M.
PROOF. First by 4.2.6.1, M = Na(L) since ME M. Then Lh :S: CH(L/a2(L))
by 4.2.7 and 1.2.1.2, and hence Lh ::::; CM(L/a2(L)). Similarly LI K as Ki H,
so K :S: CM(Lja 2 (L)) by 1.2.1.2. Hence by 1.2.1.1, KLh :S: ( C(CM(L/a2(L)))) =:
Ko :::l M.
Let p E 7r(L) be an odd prime. As M is an SQTK-group, mv(M) ::::; 2, so as
Ko ::::; CM(Lja 2 (L)), mv(Ko) :S: l. Thus Lh :S: av' (Ko) =: Ki, and Ki E C(Ko).
If K I Ki then K acts on LLh = H 0 , so that K ::::; Na(Ho) = H by 4.2.8,
contradicting K i H. Therefore Lh ::::; Ki = K, and then (1) holds as Ki H.
Further as K =Ki, mv(K) = 1 and K =av' (Ko) by earlier observations, so (2)
holds as Ko :::l M. D
. We are now in a position to complete the proof of Theorem 4.2.5. First K :::l M
by 4.2.10.2, so Q acts on K. Set (KQ)* := KQ/CKQ(K/a2(K)) and J := Lh.
Then K* and the action of Q* on K* are described in C.2.7. Now J::::; Kn MH
by 4.2.10.1, while by 4.2.6.2 and 4.2.7, MH = NH(L) = NH(J). Hence J* :::l
(Kn MH)*. As J* is not solvable, inspecting the list of possibilities in C.2.7.3,cases (a)-(d) and (f) are eliminated, as are the cases in (h) where the parabolic is
solvable. The condition in 4.2.10.2 that mv(K) := 1 for each odd prime p E 7r(J*)
then eliminates the remaining cases. This contradiction completes the proof of
Theorem 4.2.5.NOTATION 4.2.11. Assume Hypothesis 4.2.1, set M+ :=(LT), and let I be the
set of subgroups I of M such that
L ::::; Ia2( (L, T) ), LT =LI, and a2(I) # l.
LEMMA 4.2.12. Assume Hypothesis 4.2.1, IE I, and HE M(I) - {M}. Let
a2(I)::::; R+ E Syb(CM(M+/a2(M+))). Then
{1) M_ := M+CM(M+/a2(M+))I E :E(M+) and IE μ(M+,M-).
(2) Assume I E μ* and set LH := (L n I)^00 • Then M+ = L, LH E C(H n M)
is normal in HnM, [!li(Z(R+)),LH] = [Di(Z(R+)),L] = [R 2 (LT),L], and LH :S:
KE C(H) with Ki M, K/a 2 (K) quasisimple, and K is described in one of cases{1)-(9) of Theorem C.4.8.
PROOF. Set R := a2(I). Since TE Syb(G), we may assume that R::::; TnI E
Syb(I). By 4.2.2, Hypothesis 4.1.1 is satisfied. By construction, M_ E :E(M+)·
By definition of I E I in Notation 4.2.11, LI = LT, 1 I R, and L _::::: IR+,
where R+ := a 2 ((L,T)). By A.4.2.4, R+ = CT(M+/a 2 (M+)). As LT= LI,
M+::::; IR+, and hence M+::::; Ia2(M+), so R = a2(I)::::; CT(M+/a2(M+))::::; R+
and M_ = CM(M+/a2(M+))I. Thus I E 'f], and as H E M(I) - {M}, I E μ.
That is, (1) is established.
Assume IEμ* and set V+ = [Di(Z(R+)),M+J, MH :=Mn H, LH := (L n
H)^00 , and Mo := a^2 (M+ n H). As Hypothesis 4.2.3 holds, by 4.2.4 we may apply
4.1.4 and 4.1.5. By 4.1.5.3, V+ = [V+, M 0 ]. Also by 4.2.4, Mo = (L1IfH) and the
quintuple H, LH, MH, R, V+ satisfies Hypothesis C.2.8.