1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

570 2. CLASSIFYING THE GROUPS WITH JM(T)J = 1

hence e inverts vx and centralizes Sa. Therefore ex inverts V and centralizes S(;,

so ex ¢_ SK as SK centralizes V.

As SK ::::J S and x acts on S, SK n S'f< ::::l S. However tx = ZK by 2.5.25, so

Z(S) n SK n S'k::::; Z(SK) n Z(S'K) = W) n (t) = 1,

and hence SK n S'K = 1. Thus [SK, ex] ::::: [SK, S'K] ::::: SK n S'f< = 1, so e"' E

D1(C3(SK)) = So(z). Hence as e EK centralizes Sa,

s 0 ::::: c 8 (i~x) = C 8 (Sa) =Sa x SK ~ Z2 x Dg.

Thus as S(; ~ S(; by 2.5.5.1, and Z 2 x Dg contains no Qg or Zg subgroups, we

conclude from 2.5.27 that Sa= V ~ Z4, and hence ISi = 27.

Next A:= EX Ex= (t, tx, e, ex)~ Em, and

NH(A) =(ex, V) x NK(E) ~ Dg x S4,

as ex inverts V and centralizes NK(E). It follows that NHx(A) ~ Dg x S4

and I := (NH(A),x) acts on A. Now Ns(A) = Ns(E) E U*(H) by 2.5.23, so

(Ns(A),NH(A)) E U(I) ~ U(Na(A)) from the definitions in Notation 2.3.4. As

T n I contains (Ns(A), x) of order 27 = ISi where SE Syb(H) for HE I'*, and U

has maximal order in U, from the maximality of these groups in the definition of

I' in Notation 2.3.5, also Na(A) EI' ~ I' 0. This is impossible: for z EA, so that

-A E Si(G) by 1.1.4.2; hence Na(A) E He, so that Na(A) E r 0 , contradicting our
hypothesis in this section that ro = 0.
This contradiction completes the proof of ThE)orem 2.1.1.