13.7. FINISHING THE TREATMENT OF Aa WHEN (vG1) IS NONABELIAN 943PROOF. Assume otherwise. Observe there is X1 E 8yl 3 (L 1 ) with V 1 V{ =
Cv(X1), and we can take g E NL(X 1 ). Thus X 1 :::; H n H9, so X 1 acts on DH,
and as UH= [UH,L1] is of rank 4, X1 is irreducible on UH/V 3. Thus X 1 has two·
nontrivial chief factors on UH = U"}[. By (7) and (9) of 13.7.10, Vs :::; Z1 =EH,
so Vs:::; UH n EH= DH. Then as DH <UH by 13.7.11.1, we conclude DH= Vs,so that Ll is irreducible on UH/DH. Then X 1 is also irreducible on U'fI / Df:J ~
U'fI* :::; 02(Li), so Ll has a noncentral 2-chief factor not in QH. Also L 1 has two
noncentral chief factors on each of Uj; and QH /He by 13. 7.4.2, so L 1 has at least
five noncentral 2-chief factors, with [He, Ll] :::; UH in case of equality. Therefore by
13.7.10.8, m(A*) 2". 3, and [He,L1]:::; UH in case of equality. Then as m(Uj;) = 4,inspecting the subgroups H of GL4(2) of 2-rank at least 3 with 02 (H) = 1, we
conclude H ~ L4(2) or 85, with m(A) = 3 in the second case. The first case is
impossible by 13. 7.6.1 and 13. 7.9.2. In the second, [He, L 1 ] :::; UH, so [VH, HJ :::; UH
in view of 13.7.3.2, contrary to 13.7.7. DLEMMA 13.7.14. F(H) is centralized by each minimal FF-offender B* on Uj;
contained in A*.PROOF. Set P := {C E P(H,Uj;): [F(H),C] =/= 1} and suppose B E P
with B :::; A. Then by B.1.9, there is a normal subgroup N of H such that
N* = Hi x · · · x H;, Ht ~ L2(2), with s = 1 or 2 since m3(H) :::; 2, U"j; :=
[Uj;, N*] = ut EB··· EB u; with ui+ := [Uj;, H;J ~ E4 affording the natural modulefor Ht, and
P = LJ 8yl2(Ht).
iIn particular we may take B E 8yl2(Hi)· Then [Ut, B] = [Uj;, B]:::; [Uj;, A]:::;
v 3 + by 13.7.12.2. As s :::; 2 and Ll = 0^2 (L1), Ll acts on Ht for each i, and hence
also on ui+. Then v 3 + = [Ut, B, Li] :::; ut, so v 3 + = ut as both are of rank 2.
However as A :::; Q, B :::; A :::; R]' :::) L]'T, so L]' acts on R]' n Hi = B and hence
also on [Ut, B] of rank 1. This is impossible as Li is irreducible on v 3 + = ut. D
Since 02 (H*) = 1, by 13. 7.14 some member B* of P* (H*, Uj;) contained inA acts nontrivially on E(H). So there is K E C(H) with K* quasisimple and
[K,B] =/= 1. Let Ko:= \KT) and UK:= [UH,K]. By B.1.5.4, B acts on K, so
K = [K,B*].
LEMMA 13.7.15. (1) K* ~ Ln(2), An, 8L3(4), or 8p4(4).
(2) U"}( E Irr+ (K, Uj;).
(3) Ko =K.
(4) Uj; = U"}(.
PROOF. By B.1.5.1, AutB(U°j() is an FF-offender on U°j(. Therefore by B.5.6
and B.5.1.1, either U"}( E Irr +(K*, Uj;), or one of conclusions (ii)-(iv) of B.5.1.l
holds.
In the first case, (2) holds, with K* and UK := U"}(/Cu+ K (K) described in B.4.2.
However by 13.7.12.2, m([UK,B]) :::; 2, so we conclude K ·is one of the groups
listed in (1) in this case: Recall B.4.6.13 eliminates K ~ G 2 (2)', and K is not A 6
since m([UK, B]) = 4 for the unique FF-offender B* in B.4.2.8.
So assume that the second case holds. As m([U"}(, B*]) :::; 2, U"}( has exactly two
chief factors U1 and U2, and B* induces a group of transvections with fixed center