848 i2. LARGER GROUPS OVER F2 IN .Cj(G, T)
- Then H is not solvable by E.1.13, so H is described in E.2.2; in particular
(KH n M)0 2 (H)/0 2 (H) is a Borel subgroup of H/02(H). As V :::; 02(Gz) by
12.7.21, 12.2.11.1 says n(H) :::; 2, proving (1). By 12.2.11.2, (KH n M)/02(H) is a
nontrivial 3-group. Next by 12.2.8,
B(H n M) :::; B(M) = L,where we recall that B(Y) is the characteristic subgroup generated by all elements
of order 3 in a group Y. Thus B(H n M) :::; 02 (CL(z)) =Li, so as ILil3 = 3, we
conclude that B(H n M) = Li. Then inspecting the list of groups in E.2.2 with
n(H) = 2, H n M a {2, 3}-group, and B(H n M)/0 2 (B(H n M)) of order 3, we
conclude that (2) holds and 02 (H n M) = B(H n M), so that (3) holds. D
LEMMA 12.7.24. r(G, V) > 3.
PROOF. Assume otherwise. Then r(G, V) = 3 by 12.7.10.2, and then by12.7.10.5, Co(U) i M where U := Cv(Ii). Now T acts on U, so we may choose
H:::; Co(U)T, so that H = CH(U)T. Hence Li :::; 02 (H) :::; CH(U) by 12.7.23.3,
which is impossible as CLr(U) = Q(i 1 ) by 12.7.10.4. D
We are now in a position to obtain the final contradiction establishing Theorem
12.7.1.
By 12.7.11, No(Wo) :::; M; hence as H i M, Wo i 02(H) by E.3.16, so
that there exists A:= Vg :::; T with Ai 02 (H). Let KH := 02 (H) and H+ :=
H/02(H). In case KJi ~ L2(4), set Hi:= H, Ki:= KH, and Ti:= T. Otherwise
by 12.7.23.2, Kfi ~ L 3 (4). Here as A:::; Q:::; 02 (LiT) by 12.7.11 and LiT = HnM
by 12.7.23.3, A:::; 02(H n M). Therefore we have two subcases: either A+:::; KJi;
or A+ = (a+)Ai<, where Aj( :=A+ n Kif, and a+ induces a graph automorphism
on K"Ji. In the former subcase, replacing A by a suitable conjugate if necessary,
Ai 02(P) for one of the two maximal parabolics P of KH. In this subcase, we let
Hi:= NH(P), Ki:= 02 (Hi), and Ti := T n Hi, and observe that as Ai 02(P),
Cr+(A+):::; Ti. Finally in the latter subcase, CK+(a+) H ~ L 2 (4). In this subcase,
let a+ =/=-b+ E a+(r+ n K+ n Z(Cr+(a+))), Hi the preimage in Hof CH+(b+),
Ki := 02 (Hi), and Ti := T n Hi.
In each case, Ki/02(Ki) ~ L2(4), Ti E Syb(Hi), Ai 02(H1), and Cr+ (A+) :::;
Ti. Alsoineachcase,Ki iMasHnM=LiT. LetQi :=02(Hi),Hi :=Hi/Qi,
B :=An Qi, and D := C2(Qi, V). As A:::; 02(H n M), A :::; 02((Hi n M)) =
(Tin Ki)* E Syb(K)..'). As r(G, V) > 3 by 12.7.24, and Ki i M, we conclude
from 12.7.22.4 that Ki i Co(C 2 (T, V)). As n(H) = 2, Ki E E 2 (H,T,A) in the
sense of Definition E.1.5 by construction. So we apply E.3.17.1 with 0, 2, 2 in the
roles of "i, j, k", to conclude C2(T, V):::; D, so that Ki i Co(D), and Ai Co(D)
by E.1.4. But m(A/B):::; m 2 (Hi) = 2, so D:::; Co(B):::; No(A) as r(G, V) > 3.
Indeed as D centralizes B with m(A/ B) :::; 2, but does not centralize A, we con-
clude from 12.7.10 that m(A/B) = m(A) = 2, and we may take B =Vi and
D:::; R§. As A:::; (T1 n Ki) and m(A) = 2, A= (Tin Ki) E Syb(Ki). Thus
m(D/Cv(A)) 2: 2, as m(W/Cw(A*)) 2: 2 for any nontrivial chief section W for Ki
on D. So as m(R2/Q) = 2, we conclude R§ = DQg and ID: Cv(A)I = 4. Then by
12.7.2.4,
B =Vi= [R§,A] = [D,A]:::; D.
Let k E Ki - M; then Ki= (A,Ak). Now [Bk, A]:::; [D,A] = B, so A acts on
BBk, and by symmetry, so does Ak, so that I:= (A,Ak) acts on U := BBk, and