13.7. FINISHING THE TREATMENT OF Aa WHEN (vG1) IS NONABELIAN 945LEMMA 13.7.18. Either(1) K* ~ L2(4) and ·[fK/CuK(K) is the natural module, or
(2) K* ~ A6. and UH/CuH(K) is a natural module on which L 1 has two non-
central chief factors.PROOF. By 13.7.16.3,- rv +
UK/CuK(K) = UK/Cu-:c(K)is an irreducible K-module. Also m([UA:,A])::::; 2 by 13.7.12.2, while B is an FF-
offender on the FF-module UA· Further K* appears in 13.7.15.1, so applying the
remark before 13.7.6 to the restricted list in 13.7.15.1: either U"}(/Cu+ K (K) is the
natural module for K ~ L 2 (4), or K, U"}(/Cu+(K) K is one of the pairs considered
in 13.7.6. In the former case (1) holds, so we may assume the latter. If K* ~ A5,
then (2) holds by 13.7.6.3. Therefore we must eliminate the remaining cases in
13.7.15.1.
Observe that part (1) of 13.7.6 eliminates L 3 (2), parts (1) and (2) of that result
eliminate L5(2), and L4(2) ~As is eliminated by parts (1) and (3) of that result
and 13.7.9.2. The natural module for A 5 is eliminated by part (3) of 13.7.6, and
A 7 is eliminated by parts (3) and (4) and 13.7.9.2. Finally SL 3 (4) and Sp4(4) are
eliminated by part (5) of 13.7.6. D ·
LEMMA 13.7.19. Li::::; K.
PROOF. Assume L 1 1:. K. Then case (1) of 13.7.18 holds by 13.7.18 and A.3.18.
Then as L 1 = [L1,T], while IL1l3 = 3 by 13.7.9.2, we conclude H* ~ I'L2(4)
and either Li = 03 (H), or Li is diagonally embedded in 03(H) x K*. Hence
Ri = (T n K)02(KR 1 ), so m 3 (NH(R1)) > 1. Therefore as L 1 = 0
31(H n M) by
13.7.3.9, and this group has 3-rank 1 by 13.7.9.2, NH(R1) 1:. M. Thus L is an
A 6 -block by 13.2.2.7. Therefore L 1 has just two noncentral 2-chief factors. But
if Li = 03 (H*), then L 1 has two noncentral chief factors on UA:, and hence also
two on QH/Hc by the duality 13.7.4.2. Therefore Li is diagonally embedded, so
Li has one chief factor on 02 (Li), plus one each on UA: and Q H / H c, again a
contradiction. D
LEMMA 13.7.20. UK= UH.
PROOF. This follows from 13.7.19 and 13.7.5.2.LEMMA 13.7.21. K* is not L2(4).
DPROOF. Assume K* ~ L 2 (4). By 13.7.18.1and13.7.20, UH/CuH(K) is the
natural module, while by 13.7.19, L 1 ::::; K, so i/3 = [V 3 , Li] is a complement to
CuH(K) in CuH (T n K) =: W. If CuH (K) = 1, then UH = [UH, Li] is of rank 4,
contrary to 13.7.13. Thus CuH(K)-::/= 1.
By B.4.2.1, (TnK) is the unique FF-offender in T, so A= (TnK)* by
- 7.12.3. But for each 1 =/= a E A, [UA, a] = v 3 +, so Cu+ H (K) = 1 and hence
Uo = CuH(K). Thus V 1 < Uo and Uo::::; [UH,A]V 3. This contradicts 13.7.17.2. D
By 13.7.18and13.7.21, K* ~ A6 and UH/CuH(K) is a natural module on which
L 1 has two noncentral chief factors. Now L 1 has two noncentral chief factors on each