i4.2. STARTING THE L 2 (2) CASE OF .Cr EMPTY 983each case, we next define a Bender subgroup Ki of K which, together with Y, will
be used to construct our weak BN-pair:NOTATION 14.2.8. One of the following holds:
(1) K/02(K) is L2(2n) or Sz(2n), and we set Ki := K.
(2) K/02(K) is the product of two commuting Bender groups interchanged by
T, and we choose Ki E C(H).
(3) K/02(K) is (S)Ls(2n) or Sp4(2n) for n 2: 2, with T inducing an automor-
phism nontrivial on the Dynkin diagram of K/0 2 (K), and we set Ki :=Pf', where
Pi/02(K), i = 1, 2, are the maximal parabolics of K/0 2 (K) with T n K::; Pi.
Let S := NT(Ki). In each case in 14.2.8, Ki/02(Ki) is a Bender group with
Ki E C(KiS) and Ki i. M. In case (1), K = Ki and S = T, while in cases (2)
and (3), Ki< Kand JT: S[ = 2.
By 14.2.3, H::; Mc= Ca(Z).
LEMMA 14.2.9. Assume S < T. Then Ki is contained in some Kc E C(Mc),
and one of the following holds:
(1) Case (3) of 14.2.8 holds, and K =Kc is of 3-rank 2, with K = (Kf) for
each RE Syl2(Mc) with S::; R. ·
(2) Case (2) of 14.2.8 holds, Ki =Kc, K = (K[), and either K has 3-rank 2,
or Ki/02(Ki) ~ Sz(2n):
(3) Case (2) of 14.2.8 holds, Ki/02(Ki) ~ L2(4), Kc/02(Kc) ~ Ji or L2(P)
for p an odd prime with p^2 = 1 mod 5, S = NT(Kc), and (Kf) is of 3-rank 2 for
each RE Syb(Mc) with S ::; R.
PROOF. The existence of Kc follows from 1.2.4. In case (3) of 14.2.8, K/0 2 (K)
~ (S)L 3 (2n) or Sp 4 (2n), so that KE £(G, T) by 1.2.8.4-except when K/0 2 (K) ~
£ 3 (4), where KE ,C(G, T) by 1.2.8.3, since Tis nontrivial on the Dynkin diagram
of K/02(K). Thus K E C(Mc) by 14.1.6.2, so that K = Kc, and conclusion
(1) holds in this case. In case (2) of 14.2.8, Ki E £( G, T), so by 1.2.8.2, either
Ki E £*(G, T) so that Ki= Kc and (2) holds; or else (3) holds. D
LEMMA 14.2.10. If S < T, then Mc= !M((Ki,Ti)) for each Ti E Syb(Mc)
containing S.
PROOF. By Sylow's Theorem, Ti = TY for some g E Mc. If Ki = Kc, the
result follows from 14.1.6.2 applied to Ti in the role of "T". Thus we may assume
Ki <Kc, so that conclusion (1) or (3) of 14.2.9 holds.
Let Hi:= (Ki, Ti) and Mi E M(Hi). By 14.2.2.5, Mi= Mc or MY, and we
may assume the latter. As case (1) or (3) of 14.2.9 holds, (Kf) is of 3-rank 2 for
each R E Syl2(Mc) containing S, so in particular Hi = (Ki, Ti) is of 3-rank 2.
Then 02 (Hi)::; 02 (Mc n MY)::; CMg(VY) by 14.2.2.1, contrary to 14.2.2.4. D
Let B be a Hall 2'-subgroup of Kn M, and set Bi:= B n Ki.
LEMMA 14.2.11. B acts on Ki, BT= TB, BS= SB, and B::; CM(V).PROOF. As MH = BT, BT =TB. Then as B acts on Ki, BS = SB. As
H::; Mc, B::; 02 (M n Mc) ::; Ca(V) by 14.2.2.1. D
LEMMA 14.2.12. Either 02(M) ::; S, so that S E Syl2(YS), or the following
hold: