554 2. CLASSIFYING THE GROUPS WITH [M(T)f = iKo := (K^8 ). Set SK :=Sn K, SK 0 :=Sn Ko, Sc:= Cs(Ko), and fl:= H/Sc.
Let x E Nr(S) - S with x^2 E S.
As H E r+, Ko is the product of at most two conjugates of the component K
of H, and H = KoS. Further K is not a Bender group, and SE Syl2(H), so SK E
Syl2(K), SKo E Syl2(Ko), and Sc = 02(H) E Syl2(CH(Ko)). As H E r ~ 1-i,
1 of. S 0. By 2.5.2, z induces an inner automorphism on K with K = [K, z]. Thusz E KoSc - Sc, so z has nontrivial projection in Z(SK) and in Z(SK 0 ).
We begin to generate information about Sc:
LEMMA 2.5.5. (1) Sc n S 0 = 1, so S 0 ~Sc is isomorphic to a subgroup of S.
(2) ScSc =Sc x S 0 , so in particular S 0 ::::; Cs(Sc).
PROOF. Recall Sc= 02(H) :':) S. Then as x normalizes S, S 0 is also normal
in S. As x^2 ES, So:= Sc n S 0 :':) Si:= S(x), and So::::; Sc, so So :':) KoS = H.
Thus if S 0 # 1, then by 2.3.7.2, Na(S 0 ) E ro and S E Syb(Na(S 0 )). This is a
contradiction since S <Si ::::; Na(S 0 ). So So= 1, and hence (1) holds. Then as
both Sc and S 0 are normal in S, (1) implies (2). D
LEMMA 2.5.6. If 1 # E ::::; Sc with E :':) S, then GE .-Na(E) E ro,
SE Syb(GE), and GEE r* if HE r*. Further either(1) K is a component of GE, or
(2) K =Ko~ A5, H/Sc ~ M10, and KE:= (K^0 E) ~ M11.
PROOF. As E ::::; Sc and E :':) S, H = K 0 S::::; GE. Thus by parts (2) and
(4) of 2.3.7, GEE 1-i(H) ~ ro, SE Syb(GE), and GEE r* if HE r*. Next by1.2.4, K ::::; KE E C(GE)· Then by 2.3.7.2, (KE, S) E ro, and (KE, S) tjc 1-ie by
our assumption in this section that I' 0 = 0. As m2(KE/0 2 (KE)) 2':: m2(K) > 1,
KE/0 2 (KE) is quasisimple by 1.2.1.4. So as (KE, S) tjc 1-ie, KE is a component
of GE. Then KE is described in 2.5.2, K is described in one of cases (2)-(4) of
2.5.2, and if K <KE, then the embedding of Kin KE is described in A.3.12. Weconclude that the lemma holds. D
We next show that K is essentially defined over F 2 :
LEMMA 2.5.7. If K/02(K) ~ L3(2n) or Sp4(2n), then n = 1.
PROOF. Assume that n > 1 and set B :=Kn M. By 1.2.1.3, Ko = K, so that
H = KS. By 2.5.2, some element s in S is nontrivial on the Dynkin diagram of
K/02(K) and B is a Borel subgroup of K. Let Ki be a maximal parabolic of Kover B, set Li := K'j'° and V := 02(Li).
We first observe that as case (2) of 2.5.2 holds, either Z(K) = 1, or Z(K) =
02(K) with K/Z(K) ~ L 3 (4). In the latter case, il>(Z(K)) = 1: for otherwise
from the structure of the covering group in I.2.2.3a, Z(S) ::::; Cs(K) =Sc; and as
x E Nr(S), this is contrary to 2.5.5.1. By this observation and the structure of the
covering group in I.2.2.3b when Z(K) of. 1, in each case il>(V) = 1 and V/Cv(Li)
is the natural module for Li/V ~ L 2 (2n).
Recall from Notatiop. 2.5.4 that SK= SnK and SK E Syb(K). Set R := J(S)
and Re :=Sc n R = CR(K). Observe since sis nontrivial on the Dynkin diagram
of K/02(K) that SK= vvs and A(SK) = {V, V^8 } are the maximal elementary
abelian subgroups of SK.
We claim that R =SK Re: For let A E A(S). Suppose first that A::::; N 8 (Li).
As V/Cv(Li) is the natural module for Li/V ~ L 2 (2n), either A centralizes V,