13.6. FINISHING THE TREATMENT OF A 5 933· Now recall there is a E z^0 n Cv(Lv) with a E Z(Tv) ::; Z(0 2 (KTv)) since
F*(KTv) = 02(KTv)· Then
[a, K] ::; [Z(0 2 (K)), K] = U,
by the previous paragraph, so K acts on F := (a)U. Further by B.2.14, P =
UCp(K). Then as we saw earlier that Cu(Lv) = Cu(K), it follows that K central-
izes a, and hence K centralizes a by Coprime Action, contrary to 13.6.17. D
LEMMA 13.6.23. (1) K = 031 (Gv)·
(2) K+ ~ L4(2) and T;j is nontrivial on the Dynkin diagram of K+.
(3) L-:/;T"}( is the middle-node minimal parabolic of K+T"}(, where TK := TvnK.
PROOF. By 13.6.22 and the remarks before 13.6.20, we have reduced to the
cases where K+ ~ Lm(2) form= 4 or 5. Since Lv ::; K by 13.6.20 we conclude as
in the proof of 13.6.22 that L-:/;T"}( is a minimal parabolic of K+.
If Tv is trivial on the Dynkin diagram of K+, then Tv acts on a parabolic
Kt of K+ containing L-:/; with Kt /02(Kt) ~ L3(2). However as [z, Lv] =f. 1,
also [z, Ki] =f. 1. By 13.6.16.3, Kt = [Kt, J(Tv)]+ so that Kf' E Yz and 13.6.22
supplies a contradiction.Hence Tv is nontrivial on the Dynkin diagram of K+. So as Tv acts on the
minimal parabolic LvTK , m = 4 and LvTK is the middle-node minimal parabolic
of K. Thus (2) and (3) hold.
By 1.2.4, K ::; K+ E C(Gv); then K+ E Yz, so by symmetry between K+
and K, K/02(K) ~ L4(2) ~ K+/02(K+), and hence K = K+· Then by A.3.18,K = 031 (Gv), so (1) is established. D
By 13.6.23, K+T;j ~ 88 with L-:/;T"}( the middle-node minimal parabolic of
K+. As [z,Lv] =f. 1 and UK is an FF-module for K+, we conclude from Theorem
B.5.1 that UK/CuK(K) is the 6-dimensional orthogonal module. Thus CK(z) is
the maximal parabolic determined by the end nodes, so using 13.6.23.1 we concludethat
X := 0
31(CK(z)) = 0
31(Cov(z)) = 0
31
(Co(V2)),
and hence that Xis T-invariant and XTv/Rv ~ 83 wr Z2, where Rv := 02(XTv)·As UK/CuK(K) is the orthogonal module, J(Rv) = J(02(KTv)) by B.3.2.4. But
as Tacts on X and Tv, Tacts on Rv, so that J(Rv) ::::! (K, T), contrary to 13.6.16.1.
This contradiction finally completes the proof of Theorem 13.6.7.With Theorem 13.6. 7 now in hand, we can now use elementary techniques such
as weak closure in a fairly short argument to complete the proof of Theorem 13.6.1.LEMMA 13.6.24. If g E G - No(V) with V n V^9 =f. 1, then
(1) V n Vg is a singular point of V, and(2) [V, V9] = 1.
·PROOF. By 13.3.11.1, Gv is transitive on {V"' : v E V"'}, so as Gv ::; Mv by
Theorem 13.6.7.1, Vis the unique member of v^0 containing v. Hence V n V9 istotally singular, so that (1) holds. In particular conjugating in L we may assume
V n Vg = V 1 , and then take g E Co(z) = G1 by 13.3.11.1. Hence (2) follows from
13.6.2. D