6i7that A(S) = {U,A} is of order 2 with V =Un A of rank 4n. We now obtain a
contradiction similar to that in the L 3 (2n)-case of 4.3.7: Again ut f:. 02 (KS) using
4.3.5 and V = [U,L] by C.4.1. As ut 1:. 02 (KS) and A(S) = {U,A}, ut =A,
while as [U, L] = V is t-invariant, also V = [L, Ut]. This is a contradiction as
[A/U, L] = A/U -::/-1. DSet Q := [02(K),K] and U := Z(Q). By 4.3.10, conclusion (3) of C.1.34
holds; that is, U is the natural module for K / 02 ( K) and Q /U is the direct sum of
two copies of the dual of U. In particular, S is trivial on the Dynkin diagram of
K/0 2 (K), and hence normalizes both maximal parabolics over Sn K.
Set SL :=Sn Land Zs := Cv(SL)· Set Gz := Na(Zs). By C.1.34, Vis anF2n-line in U, so Zs an F2n-point. As SL = T n L and V are T-invariant, Zs is
T-invariant.Set K2 := CK(Zs)^00 , R2 := 02(K2S), and let Y be a Hall 2'-subgroup of
02,21(NK(Zs)). Thus Y is cyclic of order 2n -1, with [K 2 , Y] ::::; 02 (K 2 ), and Y
faithful on Zs. Further Y is fixed point free on the natural module U for K/0 2 (K),so as we saw above just after 4.3.10 that the composition factors of Q are natural
and dual, Q = [Q, Y]. Appealing to 4.3.9, we conclude from C.1.35.3 that:
LEMMA 4.3.11. Q = 02(KS) so 02(KS) = [02(KS), Y].
Next by 1.2.1.1, K2 is contained in the product Li··· Ls of those members Li of
C(Gz) such that K 2 projects nontrivially on Li/0 2 (Li)· Therefore for each primedivisor p of 2n - 1, p divides the order of Li. But if s > 1, then as Y is faithful
on Zs, and Y = 02 (Y) acts on each Li by 1.2.1.3, mp(YLiL 2 ) > 2, contradicting
GzY an SQTK-group. Thuss= 1. Set Kz := L 1. A similar argument shows Kz
is the unique member of C ( G z) of order divisible by p, so that K z ::::] G z. If p = 3
and Kz appears in case (3b) of A.3.18, then m 3 '(YK202,z(Kz)) = 3, contradicting
Gz an SQTK-group. Therefore we may appeal to A.3.18 to obtain:
LEMMA 4.3.12. {1) K2::::; Kz E C(Gz) and Kz :::! Gz.
(2) Forp a prime divisor of2n-1, eithermp(Kz) = 1, or p = 3 and a subgroup
of order 3 in Y induces a diagonal automorphism on Kz/02(Kz) ~ LHq) for q = E
mod 3. ·If T normalizes K 2 , then Tacts on (L, K2) = K, contradicting M = !M(LT).
This shows:
LEMMA 4.3.13. K2 < Kz.
LEMMA 4.3.14. {1) No(R2)::::; NH(K2).
{2) R2 = 02(NL1T(R2)).{3) 02(KzT)::::; R2 and K2 < 02(Kz)K2.
PROOF. Suppose Hi E M(KS). Then as I::::::; KS and K f:. M, Hi E M(I) -
{M}, so the reductions of this section also apply to Hi. In particular by 4.3.3,
Hi= Na(K) = H; that is, H = !M(KS).
Next K 2 is the maximal parabolic over Sn K stabilizing the point Zs ofthe natural module U. Now (KR 2 ,R 2 ) satisfies (MSl) and (MS2) of Definition
C.1.31. If (KR 2 , R 2 ) satisfies (MS3), C.1.34 would apply to R 2 , whereas here
R2 = 02 (CKs(Zs)) which is explicitly excluded in case (3) of C.1.34, which holds
by 4.3.10. Thus (MS3) fails, so there is a nontrivial characteristic subgroup C
of R 2 normal in KS, and hence Na(R2) ::::; Na(C) ::::; H = !M(KS). Then