i2.6. ELIMINATING As ON THE PERMUTATION MODULE 835that Cv(Ao) = VnNa(VY) for all such A 0. That is, A E A 3 -k(Mv, V). Therefore
· k =f. 0 by (1). The remaining statement in (3) holds in view of E.3.34.2.
Hence we have reduced to the case k = 1, so that m(A) = 5, and A E
A2(Mv, V). Now by (2), there exists fJ of corank 3 in A satisfying [£(.A, fJ)[ > 2.
Consider any Ai E £(.A, D); thus the preimage Ai in VY has rank 3. Since
Cv(A1) > Cv(A) = VnNa(VY) from the previous paragraph, Cv(Ai) i Na(VY).
We conclude from Theorem 12.6.2 that the 3-subspace Ai is totally singular in
VY; in particular, D is a totally singular 2-subspace of VY. But then D lies injust two totally singular 3-subspaces of VY, whereas there are at least 3 choices for
Ai E £(.A, D). This contradiction shows that k =f. 1, and so completes the proof of
W- DLEMMA 12.6.29. If Ho E H(T) with n(Ho) = 1 or Ho solvable, then Ho ::::; M.
PROOF. Assume that n(Ho) = 1. We may apply 12.6.20.3 and 12.6.28.4 to see
that min{r(G, V), w(G, V)} > 1, so Ho ::::; M E.3.35.1. Recall also that if Ho is
solvable, then n(Ho) = 1 by E.1.13 DAs H i M, n(H) > 1 and H is not solvable by 12.6.29. Thus H^00 =f. 1.
Suppose H^00 ::::; M. Then as CL(z) is solvable, H^00 ::::; CM(V) ::::; CM(L/0 2 (£)), so
L normalizes (H^0002 (L))^00 = H^00 • But then H::::; Na(H^00 ) ::::; M = !M(LT), a
contradiction. We conclude H^00 i M, so that by 1.2.1.1, there exists K E C(H)
with K i M. As usual by 1.2.1.3, Li = 02 (Li) normalizes K. Let Ko := (KT),so that Ko ::;I H by 1.2.1.3.
Notice that KoLiT E Hz.
For the rest of the section, we assume H = KoLiT, where KE C(Hi) for some
Hi EHz withK-j;M.
Let MH :=Mn H. Notice that L]':::::; 02,3(MH).LEMMA 12.6.30. (1) Hypothesis F.9.8 is satisfied for each H2 E Hz with Vs in
the role of "V+ ". In particular it holds for H = KoLiT.
(2) q(H*,UH)::::; 2.
(3) K/0 2 (K) is quasisimple, and K 0 and its action on UH are described in (4)or (5) of F.9.18.
PROOF. By 12.6.24. 7, Hypothesis F.9.1 is satisfied, while F.9.8.fholds by 12.6.27,
and case (i) of F.9.8.g holds by 12.6.24.6. Thus (1) holds. Then (1) and F.9.16.3
imply (2).
Suppose K/0 2 (K) is not quasisimple. Then K ::;I H by 1.2°.l.3, and by 1.2.1.4,X := Bp(K) =f. 1 for some prime p > 3. By 12.6.29, X::::; M = Na(L). By 1.3.3,
X E B(G, T); so X ::;I LXT by 1.3.4 since L cannot play the role of "L" in that
result. Thus H::::; Na(X)::::; M = !M(LT), a contradiction. Therefore K/0 2 (K)is quasisimple, so as H = K 0 LiT, (3) follows from F.9.18. D
LEMMA 12.6.31. One of the following holds:
(1) H* ~ Aut(L 3 (4)) or 8L3(4) extended by a 4-group. Further MH is the
product of T with a Borel subgroup of 02 (H).
(2) H is of index at most 2 in 85 wr Z2, and MH is the product of T with a
Borel subgroup of K 0.