1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

15.4. COMPLETING THE PROOF OF THE MAIN THEOREM 1161

By 15.4.11.1, 02(Yo) = Cy 0 (V(M)). IfT acts irreducibly onX, then Yo lies in

B(G, T) or ~(G, T), and hence also in YnM, and Yo satisfies (1) and (3). Otherwise;

X = X1 x X2 where Xi is T-invariant; setting Yi := 02 (Xi) for Xi the preimage in

M of Xi, Yi lies in ~( G, T) and hence also in Y n M, and Yi satisfies (1) and (3).

In particular Y n M =/= 0. By E.2.3, Yi satisfies (4)-(6) for i = 0 or i = 1, 2 in our

two cases.

Now consider any Y E Y n M. By 15.4.11.1, 02 (Y) = Cy(V(M)). Since

Y = [Y, J(T)], Y :::; [03(M), J(T)] by Solvable Thompson Factorization B.2.16.

Hence as Y = 02 (Y) is T-invariant, either Y = Y 0 , or Y = Y 1 or Y 2 , in our two

cases. In particular Y :SI M, so M = Nc(Y) as ME M; similarlyY is normal in

each member of M(YT), so (2) holds. D

LEMMA 15.4.13. For YE Y, YT is not isomorphic to Z 2 x 84.

PROOF. Assume otherwise. Then Z ~ E4 and T ~ Z2 x D 8 with <P(T) of order

2, so 02 (Aut(T)) centralizes z. Thus as Nc(T) controls fusion in Z by Burnside's

Fusion Lemma A.1.35, the three involutions in Z are in distinct G-conjugacy classes.

Pick an involution t E T - 02(YT), and let R := (t)02(Y). Then R ~ D 8 has

two YT-classes of involutions tR and zY for 1 =!= z E Zn R = <P(T). As the three

involutions in Z are in distinct G-classes, at most one of the two involutions in

Z - R can be G-conjugate to t, so that some i E z# satisfies i ¢'. t^0 U z^0. Thus

i^0 n R = 0, so by Thompson Transfer, i ¢'. 02 (G), contrary to the simplicity of

G. D

15.4.2. The final contradiction. In this subsection, we assume that G is

not L3(2) or A5.

LEMMA 15.4.14. (1) For each Y E Y, Cz(Y) is a hyperplane of Z.

(2) m(Z) = 2. Thus Cz(Y) =/= 1.

PROOF. Part (1) follows from 15.4.12.6. By the hypothesis of this subsection,

conclusion (2) of 15.4. 7 does not hold, and by 15.4.11.2, conclusion (1) of 15.4.7 does

not hold. Thus m(Z) > 1 by 15.4.7. Indeed IM(T)I > 1 by Hypothesis 15.4.1.2,

so by 15.4.12 there exist distinct Y, X E Y with !M(XT) =/=. !M(YT), and hence

Cz(Y) n Cz(X) = 1. Thus m(Z) :::; 2 by (1), so (2) holds. D

LEMMA 15.4.15. There exists at most one ME M(T) such that s(Y) = 1 for

some YE YnM.

PROOF. Assume Mi E M(T), i = 1, 2, are distinct, with Yi E Y n Mi such

that s(Yi) = 1. Let Gi := YiT for i = 1, 2, and Go := (G1, G2); then (Go, Gl, G2)

is a Goldschmidt triple. By 15.4.12.2, 02(Go) = 1, so by F.6.5.1, a:= (G1, T, G2)
is a Goldschmidt amalgam, and hence a is described in F.6.5.2. As Gi E H(T),

Gi E He, so a appears in case (vi) of F .6.5.2-namely in one of cases (1), (2),

(3), (8), (12), or (13) of F.1.12. By 15.4.14, Z i Z(Gi) for i = 1 and 2, and
m(Z) = 2. Thus by inspection of the possibilities for a, case (2) of F.1.12 holds;

that is G 1 ~ G 2 ~ Z 2 x 84. However this contradicts 15.4.13. D

Recall that 8 = Baum(T).

LEMMA 15.4.16. Nc(8):::; M for each ME M(T).