564 2. CLASSIFYING THE GROUPS WITH IM(T)I = 1So assume that case (i) holds. Suppose first that L < Lo. We saw that
L ~Ki~ L 2 (p) for a suitable prime p, so Lo= L x Li with Ki= CL 0 (i) a full
diagonal subgroup of L 0. By 1.2.2, Lo = 031 (Gj), sot acts on Lo and then on
CL 0 (i) =Ki, contrary to our assumption that case (i) holds. Thus L =Lo, and
by an earlier remark, L = [L, i] is L 3 (4) or J 2. But then tacts on L by 1.2.1.3, soK+ = KiKf ::; L, a contradiction as L3(4) and Jz contain no such subgroup. This
completes our proof that K+ ::;I Gj·
We showed that in case (ii), that Ki/Z(Ki) is not HS; hence in either case (i)
or (ii), IR: Roi = 2, so R =Ro x (t). As i E Z(Ns(K)) and Ku(t) centralizes K,
SnKu(t)::; Cs((i,K)) = R. Therefore SnKu(t) =Sox (t), where So:= R 0 n(SnKu)(t). Thus S 0 (t) is Sylow in Ku(t), so from the structure of Aut(K) ~ PGL2(P)
for p ?: 7 a Fermat or Mersenne prime, and using the second claim,
Ku= (CKu(j): j an involution of So) ::; Na(K+)·Therefore Ku::; (Na(K+)nCa(K))^00 ::; Ca(K+) from the structure of CAut(K+)(K).
But now as m 3 (K+) = 2 in cases (i) and (ii), m 2 , 3 (K+Ku) > 2, contradicting G
quasithin. This contradiction completes the proof of (2), which we saw suffices to
establish 2.5.19. DLEMMA 2.5.20. K =Ko.
PROOF. Assume K < K 0. By 2.5.16 and 1.2.1.3, K ~ L 2 (p) with p ?: 7 a
Fermat or Mersenne prime, and Ko= KKu for u ES - Ns(K). By 2.5.19.1, K is
a component of Ca(i) for each i E Cs(K).
We claim that Ko = 031 (Na(Ku)). For let i be an involution in Kun S =
S''k Then as Ku ~ Lz(p) has one class of involutions, by a Frattini Argument,
Na(Ku) =Ku Ii where Ii := Ca(i) n Na(Ku). Further we just saw that K is a
component of Ca(i), and hence K is a component of h As K ~ L 2 (p) has no outer
automorphism of order 3, 031 (N 1 i(K)) = K0^31 (CI;(K)) = K0^31 (C 1 JK 0 )). As G
is quasithin and m2,3(Ko) = 2, 031 (C1i (Ko))= 1, so 031 (NI;(K)) =Kand hence
K = 031 (Ii) as K is subnormal in h Thus 031 (Na(Ku)) = Kuo^3 ' (Ii) =Ku K,
establishing the claim.
Then as u interchanges K and Ku, also Ko = 031 (Na(K)), so that Ku =
031 (Ca(K)) and hence Na(K) = Na(Ku). Thus Ca(i) n Na(K) = Ca(i) n
Na(Ku) =Ii, so that 031 (Ca(i))nNa(K)) = 031 (Ii)= K. We saw K is subnormal
in Ca(i), so
0
31(Ca(i)) = K,
and hence Ca(i)::; Na(K) = Na(Ku). Thus ifthere is an involution i E KunKug,
then K = 031 (Ca(i)) = KB, so g E Na(K) = Na(Ku); that is, Ku is tightly
embedded in G. Then as SK is nonabelian, I.7.5 says that distinct conjugates
of SK in T commute. Suppose Si< ::; T with SK =/- Si< =f. S''f(. Then Si< ::;
Ca(SKS"f{) ::; Na(Ku) = Na(K 0 ) since Ku is tightly embedded. Then since
the center of a Sylow 2-subgroup of Auts(K) is elementary abelian, <P(Si) ::;
<P( Cr(SKS"f{) n Na(Ko) ) ::; Cr(Ko), and then KKu =Ko::; 031 (Ca(<P(Si<))) =
Ku^9 , a contradiction. Therefore {SK, S"f{} = S<fr n T, so T permutes the set 6. of
groups· 031 (Ca (j)) for j an involution in SK U S"f{. We showed 6. = { K, Ku}, so
Tacts on Ko. Therefore H = K 0 S::; K 0 T::; M = !M(T), contradicting Hi. M.
This completes the proof of 2.5.20. D