9.3. REDUCING TO n(H) = (^1 733)
for KX:T*. Comparing the list in A.3.14 with the list of FF-modules in B.5.6, we
conclude KX: ~ SL3(4), Sp4(4), G2(4), or A1. In the first three cases, D induces
inner-diagonal automorphisms on KX: in a Cartan group stabilizing the parabolic of
KX: normalizing K* and hence K, contrary to an earlier reduction. In the last case
as [Z, K] = 1 we have a contradiction since CKX: ( Cvx (T)) contains no A5-subgroup
when Vx is either of the FF-modules of dimension 4 and 6 for KX: ~ A 7 listed in
B.4.2. This finally establishes 9.3.1. DDefine TK := T n Kand TL := T n Lo E Sylz(Lo).
LEMMA 9.3.2. TLQ = 02(TD) = TK02(HD).
PROOF. First TLQ = 02 (TD) and TD = DT from the structure of L 0 f'.
Also H =KT with D::::; Na(K) by 9.3.1. Then as K/0 2 (K) ~ £ 2 (4), we conclude
HD/ 02 (HD) is a subgroup of 83 x S 5, containing G L2 ( 4). Then from the structureof this group, 02(TD) = TK02(HD). D
Our strategy for the remainder of the section, much as in the proof of 9.3.1,is to construct an overgroup Mo of K and L, and use this overgroup to obtain a
contradiction.
Set Ti:= Nr(L). Then Ti is Sylow in NM(L) of index 2 in M, so IT: Til = 2.In particular Ti contains TLQ, so Ti E Sylz(LDT1) by 9.3.2. Similarly as TLQ =
TK02(HD), Ti is Sylow in KDT1 as well.
Define Mo := (LDT 1 ,K), and Vz := (Vl}. Of course Mo 1. Mas K 1. M.
Observe V 2 is a natural module for L/02(L) ~ L2(4).
LEMMA 9.3.3. 02(Mo) =/= 1.PROOF. Assume otherwise and let S := TLQ. Then Hypothesis F.1.1 is satis-
fied by K, L, Sin the roles of "L 1 , L2, S", and S ::::1 DS so a:= (KDS, DS, LDS)
is a weak BN-pair ofrank 2 described in F.1.12. As K/02(K) ~ L/02(L) ~ L2(4),
the amalgam is one of the untwisted types Az, B2, G2 over F4. As [K, V1] = 1 by
9.3.1, while Vz is the natural module for L/02(L), we conclude a is of type G2(4).
But then 02 (£8) = [0 2 (LS),L] ~ L, which is not the case since TL nLt 1. L. D
LEMMA 9.3.4. Ti E Sylz(Mo).
PROOF. Recall J(T) ::::; T 1 , so if Ti ~To E Syl2(Mo), then To ~ M by 3.2.10.8.
If Ti < To, then To E Syl2(G) as IT : Til = 2, and then LoTo ~ Mo i. M,
contradicting M = !M(LoTo). D
LEMMA 9.3.5. (1/[Vz, K] =/= 1.
(2) [L,K] i. 02(L).
PROOF. First B ::::; K; and B is faithful on Vz as V2 is the natural module
for L/0 2 (£) ~ £ 2 (4) while B = Cv(V 1 ). Thus (1) holds. If (2) fails, then as
[Vi,K] = 1 by 9.3.1, K centralizes Vz = (Vl}, contrary to (1). D
Now Mo E 7i by 9.3.3. As L E C(G, Ti), and Ti E Sylz(Mo) by 9.3.4, L ~
KL E C(Mo) by 1.2.4. Similarly K::::; KK E C(Mo). If KL=/= KK, then by 1.2.1.2
[K, L] ::=; [KK, KL] ~ 02 (M 0 ), contrary to 9.3.5.2; therefore KK =KL =: Ko EC(Mo) and (L, K) ~Ka.
LEMMA 9.3.6. Mo= KoT1 E 'lie, Ko= 02 (Mo), and Z(Mo) = 1.