i3.6. FINISHING THE TREATMENT OF A 5 93iKi, Ga in the roles of "X, M". Thus as a E Z(T9) centralizes Ki, 13.6.3.2 applied
to Ki in the role of "L" supplies a contradiction. D
We now begin to eliminate various cases for K+ from the list of possible qua-
sisimple groups in Theorem B.4.2.LEMMA 13.6.18. K+ is not L 2 (2m).
PROOF. Assume otherwise. Then by Theorem B.5.6and13.6.15, UK/GuK(K)
is the natural module for K+ ~ L 2 (2m). Let Ko be a minimal member of C(KTv, Tv)·
Then K 0 = K*, so Ko E Yz, and by minimality of K 0 , Ko is a minimal parabolic
in the sense of Definition B.6.1. Then by 13.6.16.2 and C.1.26, Ko is an L 2 (2m)-
block. Hence replacing K by K 0 , we may assume K is a block. Then by E.2.3.2,
J(Tv) is normal in the Borel subgroup B of KTv over Tv, and S = Baum(Tv) by
13.6.4, so B acts on Sin view of B.2.3.4. Hence B S GMv(v) by 13.6.5. Thus
02 (B) S 02 (0Mv(v)) S LvGM(V), so that [V, 02 ,^3 (B)] = 1. But if n > 2, thenz E CwK(0^2 ,^3 (B)) S GwK(K) as UK/GuK(K) is the natural module for L2(2m),
contradicting [z, K] i 1. Thus n = 2 and BGM(V) = LvGM(V). Then B central-
izes the element a:= z^9 E Gv(Lv) with a E Z(Tv) described before 13.6.5. There-fore as B contains a Borel subgroup of K, and K is an L2(4)-block, K S Ga(a).
This contradicts 13.6.17, completing the proof of the lemma. D
LEMMA 13.6.19. K+ is not SL3(2m), Sp4(2m), or G2(2m) with m > 1.
PROOF. If the lemma fails, then for some maximal parabolic P of K containing
TvnK, Ki:= P^00 does not centralize z. Then Ki E C(Gv, Tv) with K[ /0 2 (K[) ~L 2 (2m). As Ki is not a block, this contradicts 13.6.16.2 in view of C.1.26. D
By Theorem B~5.6, K+ is either a Chevalley group over a field of characteristic
2 in Theorem C (A.2.3), or A5 or A1. Lemmas 13.6.18 and 13.6.19 say in theformer case that K+ is a group over F 2. Therefore the list of B.5.6 is reduced to
K+ ~ L3(2), Sp4(2)', G2(2)', A5, A1, L4(2), or L5(2). We next show:
LEMMA 13.6.20. Lv SK.
PROOF. If m3(K) = 2, then by A.3.18, Lv S B(KLv) = K. Thus we mayassume m 3 (K) = 1, so K+ ~ L3(2). By Theorems B.5.1 and B.5.6, either
UK/ Gu K (K) is a natural module for K+, or UK is the sum of two isomorphic
natural modules for K+. By 13.6.11, [UK, Lv] i 1. So either K+ = [K+, L;)-], or
[K+, L;)-] = 1 and UK is the sum of two isomorphic natural modules for K+, with
Lt S AutK+(UK) ~ L2(2).
Assume first that [K+, Lt] = 1. Then J(Tv)+ is the unipotent radical 02(P+)
of the maximal parabolic p+ of K+ stabilizing a line in each summand of UK, and
s+ = J(Tv)+ by B.2.20. Therefore since S = Baum(Tv) by 13.6.4, p+ = NK(S)+
by a Frattini Argument, while NK(S) S GMv(v) by 13.6.5. But as [K+,Lt] = 1,
02 (NK(S)) acts on 02 (02(K)Lv) = Lv, and hence as NK(S) S Mv, 02 (NK(S))
also acts on [V, Lv] = [z, Lv] ~ E 4. But [z, Lv] contains a point of each summand of
[} K, and so is generated by those two points; whereas we saw that P is the stabilizer
of a line in each summand, so that p+ = Np(S)+ acts irreducibly on each such
line.
Therefore [K+,Lt] i 1. Hence the projection Li<: of Lt on K+ is nontrivial.
So as Li<: is Tv-invariant, it is a maximal parabolic of K+ ~ L3(2), and hence
LK = [LK, Tv n K]. Then as Lv is Tv-invariant, Lv = LK s K, as desired. D