14.6. ELIMINATING L 2 (2) WHEN (v^0 1) IS ABELIAN 1037
satisfies the hypotheses of 14.6.10.5, so m((V^12 )) = 3 by that lemma, contrary to
14.6.17.5.
This contradiction completes the proof of Theorem 14.6.11.As a corollaries to Theorem 14.6.11 we have:THEOREM 14.6.18. Each solvable member of H(T) is contained in M.
LEMMA 14.6.19. Let HE H(T,M). Then
(1) 02,21(H) = 02(H).
(2) If KE C(H), then K/02(K) is simple, and hence is described in F.9.18.
PROOF. Part (1) follows from 14.6.1.4 in view of 14.6.18. Then (1) implies
(2). D
14.6.3. The final elimination of U abelian when L/0 2 (L) is L 2 (2).
LEMMA 14.6.20. If HE H(T, M) and KE C(H), then K/0 2 (K) ~ L 3 (2) or
A5, and NT(K) is nontrivial on the Dynkin diagram of K/0 2 (K).PROOF. Let K 0 :=(KT). As L 1 = 1, K 0 T E H(T, M) by 14.5.19, so without
loss H = KoT. By 14.6.19.2, K/0 2 (K) is simple, and is described in (4) or (5)
of F.9.18, so K/0 2 (K) is a group of Lie type and characteristic 2, A1, or M 22. If
K/0 2 (K) ~ A1, then KT is generated by solvable overgroups of T, which lie in
M by 14.6.18, contrary to H 1. M. If K/02(K) ~ M22, solvable overgroups of T
generate a subgroup J of KT with 02 (J)/02(K) ~ A 6 /E 2 4, so that J::::; Kn M;
then J::::; CM(V), impossible as m3(CM(V)) ::::; 1by14.2.2.4. Thus K/02(K) is ofLie type and characteristic 2. Set B := N K (T n K); then B is a Borel subgroup of
K, so BT is solvable, and hence BT::::; M by Theorem 14.6.18.
Suppose first that K = K 0. If K/0 2 (K) is of Lie rank at most 2, then as
B::::; M by the previous paragraph, the lemma follows from 14.3.6.1. Thus we may
assume K/0 2 (K) is of higher Lie rank, and hence K/02(K) is L4(2) or L 5 (2) byF.9.18. Let P be the product of the end-node minimal parabolics of K. Then
PT::::; HnM by Theorem 14.6.18, so P::::; CM(V) by 14.2.2.1, contrary to 14.2.2.4.
Therefore we may assume K < K 0. By F.9.18.5, K/02(K) is either a Bendergroup or L 3 (2). In the former case, since B ::::; M, we contradict 14.3.6.1.ii; so
K/0 2 (K) ~ L 3 (2). Further by Theorem 14.6.18, Ko is not generated byT-invariant
solvable parabolics, so NT(K) is nontrivial on the Dynkin diagram of K/0 2 (K).This completes the proof of the lemma. D
In the remainder of the section, we fix G1 as our choice for HE H(T, M), and
use the symbol H to denote this group. As in the previous subsection, we adopt
the setup of Notation 14.5.16, including the notation reviewed in that subsection
involving the coset geometry r determined by LT and H, the vertex 'Y at distance
b from 'Yo, and the subgroups U =UH, D :=DH, Uy, etc. By Theorem 14.6.18,
G 1 is not solvable, so there is K E C(H). Then K is described in 14.6.20. Set
UK := [U,K]. Recall H* = H/QH; as H = G 1 in this subsection, we do not
require the convention G1 = G1/02(G1) of the previous subsection.
LEMMA 14.6.21. One of the following holds:
(1) H^00 = K, with K/02(K) ~ L3(2) or A5.
(2) H^00 = KKt for some t ET-Na(K), with K/02(K) ~ L3(2).