614 4. PUSHING UP IN QTKE-GROUPSPROOF. As I= LR::; LS, maximality of S implies (a). Then as S < Nr(S),
(a) and 4.3.1 imply Na(S) ::; M. Therefore as S E Syl2(MH), S E Syl2(H). As
we saw earlier that K is Mwinvariant, K :<:'.'.! H by 1.2.1.3, so H = Na(K) as
HEM. D
LEMMA 4.3.4. R =Sn R+· In particular, S normalizes Rand R = 02(IS).
PROOF. As I::; L(S n R+), this follows from maximality of I. D
We next choose an element t E Nr(S)-S with t^2 ES. If R < R+, then R+ 1:. S
by 4.3.4, so in this case we may choose t so that also t E R+ and t^2 E Sn R+ = R.
By convention, t will denote such an element throughout the proof.
Ast E Nr(S), t normalizes Sn R+ = R. Further t ~ S, so by 4.3.3.a:
LEMMA 4.3.5. M = !M(LS(t)).
LEMMA 4.3.6. F*(K) = 02(K).
PROOF. Assume otherwise. Then from our remarks following the statement of
Theorem 4.3.2, K is a component of H described in Theorem C.3.1. As L/02(L) ~
L 2 (2n), we conclude that either
(i) K/Z(K) is of Lie type and Lie rank 2 over F 2 n, and MK is a maximal
parabolic of K, or
(ii) K/Z(K) ~ M22 or M23, and Lis an L2(4)-block.
Let KS:= KS/CKs(K) and SK := Sn K. Now L::; K::; CH(02(H)) as K isa component of H, and 1 =I-02(H) ::; R by 4.1.4.5, so 1 =I-Ro := CR(L). Recall
from 4.3.4 and our choice oft that S(t) acts on Rand Land hence also on R 0 , so
Na(Ro) ::; M = !M(LS(t)) by 4.3.5. Then [K, Ro] #-1 as Ki. M, so 1 =/:-Ro ::;CR,(L). Inspecting the automorphism groups of the groups in (i) and (ii) (e.g.,
16.1.4 and 16.1.5) for such a 2-local subgroup, we conclude K/Z(K) ~ Sp 4 (2n).
Indeed Z(K) = 1 since the multiplier of Sp 4 (2n) for n > 1 is trivial by I.1.3.Furthermore V = 02(L) is the maximal nonsplit extension of the natural module
for L/02(L) over a trivial module by I.1.6, and Cv(L) is a root subgroup of K.
Since Aut(K) fuses the two K-classes of root subgroups, we may regard Cv(L) as
a short root subgroup of K, and take Z::; Cv(SK) to be a long root subgroup of
K.
Set Gz := Nc(Z). As Z = [Cv(T n L), NL(T n L)] and Tacts on Land V,
Tacts on Z; hence F*(Gz) = 02(Gz) =: Qz by 1.1.4. Let K 2 := NK(Z)^00 whereNK(Z) is the maximal parabolic of K containing SK and distinct from NK(Cv(L)).
As L::; M but K = (L, K2) 1:. M, K 2 1:. M. Further T 1:. Na(K 2 ), or otherwise T
normalizes (L, K2) = K, and hence T::; H by 4.3.3.b, contrary to our observation
just before 4.3.3. We will now analyze G z, and eventually obtain a contradiction
by showing that T :<:::: Na(K2).
First, a Cartan subgroup Y of the Borel group MK n N K ( Z) of K decomposes
as Y = Y1 x Y2, where Y1 := Cy(K2/02(K2)) and Y2 := YnK2 are cyclic of order
2n - 1, Y 1 is regular on z#, and NK(Z) = Y1K2"'.
Next by 1.2.1.1, K2 is contained in the product Li··· Lr of those members Li
of C(Cc(Z)) with Li = [Li,K2]. If r > 1, then for a prime divisor p of 2n - 1,
mp(L1 · · · LrY1) > 2, contradicting Gz an SQTK-group. Therefore K 2 ::; L 1 =:
Kz E C(Ca(Z)). Recall from the remarks after (i) and (ii) above that K ~ Sp 4 (2n)