5.1. PRELIMINARY ANALYSIS OF THE L 2 (2n) CASE 639Now as K ~ .Cj(G, T), K centralizes R 2 (KT) by 1.2.10, so that H = KT
centralizes Z. Then the remaining statement in conclusion (2) follows as NG(K) =
!M(H); and conclusion (2) implies conclusion (3) as Hf:. M =M(LT).Thus it only remains to prove parts (4) and (5) of Theorem 5.1.14. Moreover
throughout the remainder of the proof we can and will appeal to the first three
parts of Theorem 5.1.14.
Set M+ := NG(K); by 5.1.14.1, M+ E M(T). If n is even, define D€ for E = ±1
as in Lemma 5.1.6.LEMMA 5.1.15. One of the following holds:
(1) DL:::; M+.
(2) n = 2 and V is the direct sum of two natural modules for L.
(3) n = 2 or 4 and [V, L] is the natural module for L.
(4) n = 4 or 8, Vis the 0;4(2nl^2 )-module for L, and D_ :::; M+.
PROOF. First if D :::; DL and 02 ( (D, H)) #- 1, then by 5.1.14.1, D :::; M+.
However we may assume conclusion (1) does not hold, so DL i M+ and hence
02((1h,H)) = 1. Cases (1) and (2) of 5.1.6 appear as cases (2) and (3) of 5.1.15.
Case (3) of 5.1.6 cannot occur since there Z(H) = 1, contrary to 5.1.14.2. Finally
in case (4) of 5.1.6, 02((D_,H)) #-1, so D_:::; M+. Thus as D+D-= DL i M+,
02( (D+, H)) = 1, son= 4 or 8 by 5.1.6.4. Hence 5.1.15.4 holds. DWe now begin to make use of the local classification of weak BN-pairs of rank
2 in the Green Book [DGS85]. We recognize weak BN-pairs of rank 2 by verifying
Hypothesis F .1.1.LEMMA 5.1.16. Let CK:= CG(K/02(K)). Then
(1) CK is a 3'-group.
(2) If CK is not solvable, then C'j( /02(C'j() ~ Sz(2k) for some odd k 2: 3,
C'j(f:_M, andDLf::M+.
PROOF. Part (1) follows as His an SQTK-group. Thus it remains to prove (2),
so we assume C'j( "I-1. Hence by 1.2.1, there exists K+ E C (CK). Then any such K+
satisfies K+/0 2 (K+) ~ Sz(2k) for some odd k 2: 3. Further m5(K+) = 1 = m 5 (K),
while m 5 (M+) :::; 2 as M+ is an SQTK-group, so K+ = C'j( by 1.2.1.1, establishing
the first assertion of (2). Further M+ = NG(K+) since we saw M+ EM. Let B+
be a Borel subgroup of K+; then B+ :::; NG(T) :::; M = NG(L) using Theorem 3.3.1.
Now if K+:::; M, then [K+, L] :::; 02 (L), so that L normalizes 02 (K+0 2 (L)) = Kand hence L :::; NG(K+) = M+ contradicting M = !M(LT). Thus K+ f:. M,
proving the second statement of (2).To complete the proof of (2), we suppose by way of contradiction that DL :::;
M+ = NG(K+)· We claim that under this assumption, Hypothesis F.1.1 is satisfied
with L, K+, Tin the roles of "L1, L2, S". Let G+ := (LT,H). As K+ f:. M =
!M(LT), 02 (G+) = 1, establishing hypothesis (e) of F.1.1. We have seen that
B+ :::; M = NG(L), and we are assuming DL :::; NG(K+), so hypothesis (d) of
F.1.1 holds. The remaining conditions in F.1.1 are easy to verify, in particular
since we take S to be the Sylow 2-subgroup T of G; therefore Hypothesis F.1.1 is
satisfied as claimed. We conclude from F.1.9 that a:= (LTB+,DLTB+,DLTK+)
is a weak BN-pair of rank 2. Indeed T ::::) B+T, so by F.1.12.I, a is of type
(^2) F 4 (2k), with n = k-as this is the only type where a parabolic possesses an. Sz(2k)