1038 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN L:r(G, T) IS EMPTY(3) H^00 = KK+ with K, K+ normal C-components of H, and K/02(K) ~
K+/0 2 (K+) ~ L3(2).
PROOF. If H^00 = K then (1) holds by 14.6.20, so assume H^00 > K. Then
by 1.2.1.1, there is K+ E C(H) - { K}, and K+ is also described in 14.6.20. As
m 3 (H) :::; 2, we conclude from 14.6.20 that K/02(K) ~ K+/02(K+) ~ L3(2) andH^00 = KK+. Then by 1.2.1.3, either (2) or (3) holds. D
LEMMA 14.6.22. (1) fh = f~h, 1 + UK,2, where UK,1 is a natural module for
K/0 2 (K) or the 5-dimensional cover of a natural module for K/02(K) ~ A5 1
and UK, 2 = UK, 1 for s E NT(K) acting nontrivially on the Dynkin diagram ofK/02(K).
(2) If there exists K+ E C(H) - {K}, then [UK, K+] = 1.
(3) Z:::; UK,i for i = 1, 2.
PROOF. Let Ko:= (KT) and i E Irr +(Ko, U). As TK := NT(K) is nontrivial
on the Dynkin diagram of K/0 2 (K) by 14.6.20, KTK /0 2 (KTK) has no FF-modules
by Theorem B.5.1. By 14.6.19.2, we may apply F.9.18, so [U, Ko] = (fH) by part(7) of that result. Next as T is nontrivial on the Dynkin diagram of K/02(K),
F.9.18 says [U, Ko] is described in case (iii) of part (4) of F.9.18 if K = K 0 , and in
case (iii.b) of part (5) if K < K 0. Next if 01(K)-=/= l, then i is described in I.1.6.1;
in particular i is 5-dimensional when K* ~ A 6. On the other hand if K* ~ L 3 (2),
then i is the extension in B.4.8.2, and that result says q(H*, UH)> 2, contrary to
part (2) of F.9.18. This completes the proof of (1). Also (2) follows, since UK,l is
not K-isomorphic to UK,2 and EndK(UK,i/OuK,i(K)) is a field by A.1.41. Finally
(3) follows from 14.6.2. DLEMMA 14.6.23. (1) H^00 = 02 (H), so H = H^00 T.
(2) M =LT andT = MnH.
(3) We have
u = ( E9 uK ) + Ou(H).
KEC(H)PROOF. In view of 14.6.19.1, we obtain F(H) = H^00 * from 1.2.1.1. By
14.6.21, Out(H^00 *) is a 2-group, so (1) holds.
Let U 0 := [U, H^00 ]. By 14.6.21 and 14.6.22, Uo = EBKEC(H) UK. Now T
centralizes V of order 2, and U = (VH), while H = H^00 T by (1), so (3) follows
using Gaschiitz's Theorem A.1.39. Further the projection VK of Von UK is of order
2 and centralized by NT(K) for each KE C(H). By 14.6.22.1, OK• (VK) = TnK,
so T = OH(V) by (1). Therefore T =Mn H, so M =LT by 14.3.7. Thus (2)
hclds. D
We next choose an element u E U, which we will show lies in the set U(G 1 )of the first subsection. In cases (1) and (3) of 14.6.21, pick u E UK,l such that
[u,K]-=/= 1 and NT(UK,1) centralizes u. (This choice is possible when UK,l is the
5-dimensional cover of a natural module for K/0 2 (K) ~ A 6 by I.2.3.lia). In case
(2) of 14.6.21, pick u E UK - Z such that NT(K) centralizes u.
LEMMA 14.6.24. (1) u E U(H).
(2) K/02(K) ~ L3(2).
(3) K = H^00 •