8.3. ELIMINATING L 3 (2) 12 ON 9 725We claim Y:::;; M. IfY is solvable, then n(Y) = 1 by E.1.13, so Y:::;; M by 8.3.6.
So suppose Y is not solvable. Then there is Y 1 E C(Y) with Yi/0 2 (Y 1 ) ~ Sz(2k).
Now a Borel subgroup B of Y1 is solvable, so as before B :::;; M using 8.3.6. Set
H := (Y1, T); then n(H) = k is odd and k ~ 3. If Hi M then as B :::;; M, weget H E 1-l*(T, M), contradicting 7.3.4, which says n(H) :::;; 2. So H :::;; M, and in
particular Y1 :::;; M. These arguments apply to each minimal parabolic H of YS
over S, so as this set of parabolics generates 021 (YS) by B.6.5, 021 (YS) :::;; M.
Finally as S E Syb(YS), by a Frattini Argument YS = 021 (YS)Nys(S) :::;; M,
since we saw Na(S) :::;; M. This completes the proof of the cl~im.
As Gz = KzYzMz, and Yz = XY with X :::;; Kz, we conclude Gz = KzMz,
establishing the first assertion of the lemma.
If Kz = X, then Gz = XMz = Mz:::;; M, contradicting 3.1.8.3.ii, which shows
H :::;; Gz for each H E 1-l*(T, M). Thus X < Kz, so K/02(K) ~ L2(2n) with
n > 2. But now we replace Y1 by K in the argument above, and again obtain a
contradiction to n(H) :::;; 2 in 7.3.4. This completes the proof. D
u.
We can now essentially eliminate the shadows of the linear groups:
LEMMA 8.3.8. Ca(Cv(R)):::;; M.
PROOF. Set U := Cv(R); our proof relies on the following properties:
(a) z EU.
(b) NL 0 (R):::;; NL 0 (U), and there is a subgroup P ~ E 32 of NL 0 (R) faithful on(c) T:::;; Na(U).
Since Ca(U) i M, using (c) we may choose HE 1-l(T, M) with I:= 02 (H) :::;;
Ca(U). By (a), I:::;; Gz, and by (b) and A.1.27, Ca(U) is a 3'-group.
Next Gz = KzMz by 8.3.7. As Ii Mz, the projection Kj of I on (KzT) :=
KzT/0 2 (KzT) is non-trivial. Furthermore Ca(U), and hence also Kj, is a T-
invariant 3'-group. In case (ii) of 8.3.7, Kj(T n Kj) contains a Sylow 2-subgroup
of K; and hence is a parabolic subgroup of K; as this parabolic is a 3'-group,
I :::;; TMz, contradicting I i M. So instead case (i) of 8.3.7 holds, and Kz =
KK^8 with K ~ L 2 (p). Now m 2 (L 2 (p)) = 2, so if P is a 3'-subgroup of K,
then 02 (P) = O(P). Thus as I = 02 (!), the 3'-group Kj is of odd order, so
02 (!) :::;; 02 (KzT), and 02(I) is Sylow in I. Then since X:::;; Kz, by A.1.6 we have
02 (!) :::;; 02 (KzT) :::;; 02 (XT) = S. It follows that S E Syb(IS). But n(I) = 1 as
I is solvable, so I :::;; M by 8.3.6, a contradiction which establishes the lemma. D
Now we achieve our initial goal:PROPOSITION 8.3.9. n(H) = 2 for each HE 1-l*(T, M).
PROOF. Recall n(H):::;; 2 by 7.3.4. As w > 0, Na(Wo):::;; M by E.3.16.1. Also
s = 3 by 8.3.2. Thus if Ca(C 1 (T, V)):::;; M, then n(H) = 2 by E.3.19, so the lemma
holds. However if W1:::;; Ca(V), then Ca(C1(T, V)):::;; M by E.3.16.1.3, so we may
assume that W1 i Ca(V). Then by 8.3.3, W1 = R. Therefore Cv(R) :::;; Cr(W1) =
C1(T, V), so Ca(C 1 (T, V)):::;; M by 8.3.8, completing the proof.· D
LEMMA 8.3.10. (1) Kz = KK^8 with K/02(K) ~ L2(5) ~ L2(4), KzT E
1-l*(T, M), and X(T n Kz) = Kz n Mis a Borel subgroup of Kz.