1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

2.5. ELIMINATING THE SHADOWS WITH rg EMPTY 569

transposition on Ki· As Cs 0 (i) = (i, t), Sc is dihedral or semidihedral by a lemma

of Suzuki (cf. Exercise 8.6 in [Asc86a]), so as Sc is not elementary abelian by

2.5.24, JScJ 2 8. Using 2.5.5.2, S 0 :::; Cs(Sc):::; Cs(i). However, as Cs 0 (i) = (i,t),

KiCs(i) ~ (i) x Sa. Therefore a Sylow 2-subgroup of KiCs(i) nC 0 (t) is isomorphic

to Es x Ds, which contains no Drn or SDrn-subgroup, so Sc 3:! D 8. Hence JS[= 28

since fI ~ Aut(A 6 ) by 2.5.21.

Let V denote the cyclic subgroup order 4 in Sc. By 2.5.22.2 Gt = KS, so

V ::9 Gt, and thus V is a TI-set in G. Hence as V is not elementary abelian,

(VG n T) is abelian by I. 7.5.

Assume VY :::; T for some g E G. Then by the previous paragraph, VY :::;

CT(V) = Cs(V) and hence <I>(VY) :::; <I>(S) :::; SK Sc since S /SK ~ E4. Now no

involution in SK-(z) is a square in S, so no involution in S:KSc-(z)Sc is a square

in S. Hence

(tY) = <I>(VY):::; Di(C(z)sa(V)) = Di(V(z)) = (t,z).

Therefore tY E t^0 n (t, z), so that tY is tor t'n by 2.5.11. Hence VY is either V or

V"'.

Since v^0 n T = {V, V"'}, VV"' ::9 T, so D 1 (VV"') = (t, t"') ::9 T. Then as

SE Sylz(Gt) by 2.5.11.2, [T[ = 2[S[ = 29.

Let Ho := Ki(i, t), Ti E Syl2(Ho), and Ti :::; TY for suitable g E G. As Ki is

a component of Ca(i), Ho i MY by 1.1.3.2. As Ho ~ Z2 x Ss, Ho = (Hi, H2),

where Hi and H2 are the maximal 2-locals of Ho over Ti; thus we may assume

Hi d J::: MY. As I I Ti = 2 s = I T I/ 2, T; y-1 E f3 by 2.3.10, so ( Ti y-1 , Hi y-1) E U ( Hi y-1)

-1

and Hf Er from the definitions in Notation 2.3.4 and Notation 2.3.5. Then by

-1 •

2.3.7.1, Hf E rg, contrary to the hypothesis of this section. This contradiction

finally completes the proof of the claim.

By the claim, K is a component of Ca(i) for each involution i E Sc. Further

K ::9 Ca(i) by 1.2.1.3. Recall t"' = ZK and ScSf; =Sc x S 0 , so for any i E Sc

distinct from t, i"' ~ t"'Sc = zKSc. Therefore from the 2-local structure of Aut(A 6 ),

CK(i"') i SK. Hence as S = CK:s(t"') is a maximal subgroup of KS,

KS= (CKs(t"'), CKs(i"')) :::; Na(K"')

using the claim. By symmetry, K"' acts on K, and K-=/= K"' as t"' centralizes K"'

but not K. Therefore [K, K"'] = 1, a contradiction as m 2 , 3 (KK"') :::; 2 since G is

quasithin. This contradiction completes the proof of 2.5.26. D

LEMMA 2.5.27. Sc~ Z4, Zs, or Qs.

PROOF. By 2.5.26, m2(Sc) = 1; by 2.5.24, Sc is not elementary abelian; and

by 2.5.5.1, Sc 3:! S 0 is isomorphic to a subgroup of S. Thus the lemma holds as the
three groups listed in the lemma are the only subgroups X of SE Syl2(Aut(A 6 ))
of 2-rank 1 with (X) -=/= 1. D

We are now ready to complete the proof of Theorem 2.1.1.

By 2.5.27 there is a cyclic subgroup V of Sc of order 4 normal in S. Let Y