12.6. ELIMINATING A 8 ON THE PERMUTATION MODULE 833Therefore m3(K) = 1. As m(P) = 2 = m 3 (PK), we conclude that PK:= PnK
is of rank 1. Again inspecting the list of Theorem C, and using (*), we conclude
that K is L2(Pe) for some odd prime p with pe = ±1 mod 12. Recalling that
elements of odd order in K inverted by V are of order 3 or 5, we reduce to the case
K ~ L2(Pe), with p =f 23 or 25, and V inverts no element of order p if p > 5, so in
fact pe = -1 mod 12.
Set Ho := (K, LiT), so that Ho E Hz as K i. M. As K* = [K*, V*], V i.
02(Ho), so UHo is nonabelian by 12.6.25. Thus replacing H by H 0 , we may assume
H = (K,L1T).. Next as K ~ L2(Pe), KV is generated by 3 conjugates of V. Thus as
Vs = [UH, V] is of rank 4, for each nontrivial chief section W for K on UH,
m([W,K*]) ::::; 3m([W, V*]) ::::; 12. Thus pe divides [L 12 (2)J, so as pe =f 23 or 25
and pe = -1 mod 12, it follows that pe = 11. But as 11 does not divide JL 9 (2)J,the smallest nontrivial irreducible for K* is of rank at least 10, so we conclude
UK:= [UH,K] E Irr+(UH,K), Vs= [UK, V*], and 10::::; m(UK)· Thus if K =f Kt
for some t ET, then Kt centralizes UK. However as Tacts on V5 and K does notcentralize V5, neither does K*t, a contradiction. Thus T normalizes K, so K :::l H
by 1.2.1.3, and so H = (K, LiT) = KL 1 T = KPT.
Next P = Pc x PK where Pc := Cp(K) and PK = P n K are of order 3. As
H = KPT and 02 (H) = 1, H = KP(JT and P(J = 02 (CH·(K) :::1 H*. In
particular P(J is T-invariant, so V5 = [Vs, Pc] since L 1 T is irreducible on V 5. Then
as P(J and K are normal in H, V5::::; UK, and UH= (V 5 H), UK= UH= [UH, Pc].
As lllcH•(V•)(P,2) s.;; CH·(P) from the structure of Aut(L2(ll)), 02(Li) = 1,
and so 02(L1) ::::; QH. Next [QH, V] ::::; QH n V = V5 ::::; UH, so as K = [K, VJ,
[ Q H, K] ::::; UH. As Pk is inverted by a conjugate of V*, it follows from the firstparagraph of the proof that [UH,PK] ~ Q§. Then as 02 (L 1 )::::; QH by the previous
paragraph, [02(L1),PK] :S [QH,K] :S UH, so that (02(L1),PK] = [UH,PK] ~
Q§ is of order 29. Therefore as [V,PK] ~ E 16 ~ [R 1 ,PK], we conclude that L
is an As-block and 02(L1) = [02(L1),PK] ~ Q^4 - -
8. We saw UH = [UH,Pc], so
UH= [UH,Pc] :S 02(L1). Thus 211 :S [UH[ :S [02(L1)[ = 2^9. This contradiction
completes the proof of 12.6.26. D
12.6.3. Restrictions on H, and the final contradiction. In this section,
we use machinery from section F.9 to handle the case VH abelian. The same
approach will be used many times in the remainder of our treatment of groups over
F2.
LEMMA 12.6.27. If g E G with v n V9 =f l, then [V, V9] = 1.
PROOF. By 12.6.20.1, we may assume z EV n V9. Then by 12.6.20.2, we maytake g E Gz. Applying 12.6.26 to Gz in. the role of "H", we conclude VH =(VG•)
is abelian, so the lemma holds. D
LEMMA 12.6.28. (1) A 3 (Mv, V) = 0.
(2) If BE A 2 (Mv, V), then m(B) 2: 3 and there exists fJ of index 8 in B with
[£(B,fJ)J > 2, where
E(B,fJ) := {E::::; f3: m(E/fJ) = 1 and Cv(E) > Cv(B)}.
(3) Wo := Wo(T, V) centralizes V, so Nc(Wo)::::; M.
(4) W1(T, V) centralizes V, so w(G, V) > l.