2.4. THE CASE WHERE r 0 IS NONEMPTY
(2) CR(K) = Ca 2 (K) is cyclic of order 4, and G2/CR(K) S:f Aut(A 6 ).
(3) Cr(Lo) = 1 and [T[ = 28.
549PROOF. We claim that z2 is the unique involution in CR(K). Assume the claim
fails, and let z2 -/=-r E CR(K) be an involution. Recall R::::; G 2.
Under this assumption, we establish a second claim: namely that K :::] Gr :=
Co(r). First K is a component of Cor(z2) using 2.4.33, so by I.3.2, there is a
2-component Kr of Gr such that either K ::::; Kn or K ::::; KrK;^2 with Kr -/=-K;.^2 -
and in the latter case, Kr/0 00 (Kr) S:f K. As K S:f A 6 by 2.4.35, the former caseholds by 1.2.1.3. As Kr is a 2-component of Gr, Kr E C(Gr) and 02(Kr) S Z(Kr)·
As m2(Kr) ~ m2(K) > 1 and 02(Kr) S Z(Kr), Kr is quasisimple by 1.2.1.5.
Now as m3(Kr) ~ m3(K) = 2, Kr:::] Gr using 1.2.1.3, so our second claim
holds if K = Kr. Thus we may assume that K < Kn and it remains to derive
a contradiction. We verify the hypotheses of 1.1.5 for Gr in the role of "H": LetCR(r) S Tr E Syl2(Gr), and Tr S T^9 , so that z9 E Z(T9) ::::; Tr, and hence z9 E
Z(Tr)i thus z^9 , Tn M^9 play the roles of "z, S,M". As r E 02 (Gr n Af9), trivially
Co 2 (M9) (02(Gr n M^9 )) ::::; Gr. This completes the verification of the hypotheses of
1.1.5. As K S:f A6 is a component of CKr(z2), we conclude from inspection of thelist of 1.1.5.3 that one of the following holds:
(i) z 2 induces a field automorphism on Kr S:f Sp 4 (4).
(ii) z2 induces an outer automorphism on Kr S:f L4(2) or L 5 (2).
(iii) z 2 induces an inner automorphism on Kr S:! HS.
Recall that \T: R[ = 4, while [R: CR(r)[::::; 2 by 2.4.30, and z 2 E Z(R). Thus
[Tr: Crr(z2)[ S [Tr: CR(r)[ <IT: CR(r)[ S 8,
where the strict inequality holds since r is not 2-central in G, as Gr rf.11.e. Since z 2centralizes K but not Kr, we conclude (ii) holds, with Kr S:f L4(2) S:f A 8 • Now L is
an A4-subgroup of Kr fixing 4 of the 8 points permuted by Kr, so it centralizes an
A 4 -subgroup Lr of Kr· Then using A.3.18 and the fact that z1 = Zz E 02(L),
Ko:= (LnL2) S 0
31
(Co(L))::::; 0
31
(G1) = K^8 •Now K^8 S:f A 6 with z 2 E L2 ::::; K^8 and z 2 induces an outer automorphism on Lr.
Thus (z2)Lr S:f 84, so (z 2 )Lr is a maximal subgroup of K^8 • It follows that Ks =
Ko ::::; Ca(L), so m 2 ,3(LK0) = 3, contradicting G quasithin. This contradiction
establishes the second claim, namely that K = Kr is a normal component of Gr
for each involution r E CR(K).
Set Er := (z2, r). Using 2.4.35.2, CKss 1 (z2) is a maximal subgroup of K^8 S1,
which does not contain CK•s 1 (a) for any involution a ~ z 2 Co(K^8 ). Thus in the
notation of Definition F.4.41, K^8 S1 = rl,Er(K^8 S1), so K^8 S No(K) using the
second claim. Then as m 2 , 3 (No(K)) ::::; 2 since G is quasithin, K =Ks. This is
impossible as z 1 E K but z 2 = zf centralizes K. This contradiction completes the
proof of the first claim that z 2 is the unique involution in CR(K).
By 2.4.35, CR(K) is of index 2 in CR(L) S:f CQ(Lo) x Ds, so by the uniqueness
of z 2 , CR(K) is cyclic of order 4 and Cq(Lo) = 1. Then Cr(Lo) = Cq(Lo) = 1.
Therefore R S:f Ds x Ds by 2.4.30.4, so [Tl = 4IR[ = 2^8 by 2.4.30.5, completing the
proof of (3).As RK S:f Ds and RK n RK: :::] S but Z(RK) = (z 1 ) 1:. Z(S), we conclude
RK n RK: = 1. Thus R ~ RKRK = RK x RK:; so as [R\ = [RKl^2 , R = RK x RK,
and (1) holds.