2.5. ELIMINATING THE·SHADOWS WITH rQ EMPTY 567Suppose first that i either centralizes Kx or induces an outer automorphism on
Kx. If i induces an outer automorphism on Kx, then Kx \ i) ~ PG L 2 (p), so in either
situation, i centralizes an Eq2-subgroup of K+, where q is an odd prime divisor of
p + c:, p = E mod 4, and E = ±1. This is impossible, as K = 02 (Gt) ~ L 2 (p) is
of q-rank 1. Therefore i induces a nontrivial inner automorphism on Kx. Then
t E CKx(i) ~ Dp-E centralizes CK(i) ~ Dp+E, sot E ( CKx(i)) n Ca(CK(i))):::; Ca;(Kg),
since the centralizer in Aut(Kg) of a Dp+csubgroup of Kg is of order 2. Then
as K = 02 (Gt), Kg = K, a contradiction as i centralizes Kg but not K. This
establishes the claim that I= tG n M+.
We've shown that tG n M+ = I, so tG n M+ = tM+. We also showed thatGt :::; M+, so by 7.3 in [Asc94], t fixes a unique point in the representation of G
by right multiplication on G / M+. Therefore as T is nilpotent (cf. the proof of2.2.2), T:::; M+. Further M+ is the unique fixed point of each member of sf::, so
SK U S'K is strongly closed in T with respect to G. Thus the hypotheses of 3.4 in[Asc75] are satisfied with SK, S'K, M+ in the roles of "A 1 , A 2 , H", so that result
says G = M+, contradicting G simple.
We have reduced to the case where Sc ~ Q 8. In particular H ~ M 10. Now
ScSc = Sc x Sc by 2.5.5.2. Since SK ~ Ds, Be 1:. K, so H = KBc· Thus
[8 0 , SK] is the image of the cyclic subgroup Y of index 2 in SK. Then as 80 :::l S,
Y = [8 0 ,SK] :::; 80 , so 80 = Y\v) for v E 80 - K. Then as [Sc,8 0 ] = 1 andv induces an outer automorphism on K with v^2 = ZK E Y :::; K, it follows that
H =Sc x S 0 K, so S =Sc x S 0 SK with S 0 SK a Sylow 2-subgroup of M1 0. Since
ZK = tx, Cs(S 0 ) =Sc x Z(S 0 ) = Sc\tx), and henceICs(Sc)I = 16 < 32 = ICs(Sc)I,
a contradiction as x acts on S. This finally completes the proof of 2.5.21. D
In view of 2.5.21, it only remains to eliminate the case H ~ Aut(A 6 ). In
particular K ~ A 6 and SK ~ Ds.LEMMA 2.5.22. (1) If zg E S for some g E G, then K = [K, zg], and zg induces
an automorphism in 85 on K.
(2) H =Gt and CH(z) = S.
PROOF. Assume zg E S for some g E G. Then as Cat (zg) E He by 1.1.3.2,
CK(zg) E He using 1.1.3.1. Further K =Ko :::] Gt by 2.5.20, so since H ~ Aut(A 6 )
by 2.5.21, (1) follows.
Let C := Cat(K). By 2.5.18, Gt= KBC and CE He. Thus R := 02(Gt):::; Sc
and R:::] S. By 2.5.5.1, S 0 nS 0 = 1, so R ~ R,x :::] S. As Sis Sylow in H ~ Aut(A 6 ),it follows that either
(i) R is abelian and m(R) :::; 2, or
(ii) [R, R] =: Y is the cyclic subgroup of index 2 in SK, and either m(R/Y) :::; 2
or R= S.We conclude that Aut(R) is a {2, 3}-group, and hence C is a {2, 3}-group. However
as His an SQTK-group, C is a 3^1 -group, so as F*(C) = 02 (C), C is a 2-group.
Thus Gt= KBC= KS= H,'so CH(z) =Sas K ~ A5. D