1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

14.7. FINISHING Ls(2) WITH (vGi) ABELIAN 1075

Therefore 11 appears in one of the cases (2)-(13) of F.6.18. Further the subcase

of case (2) of F.6.18 with 02 (1+) ~ 31+^2 is eliminated, since in that case there is

no subnormal subgroup of J+ isomorphic to S 3. Thus if l+ is solvable, then by

F.6.18, I' ~ S3 x S3, so there is a normal subgroup Kt of l+ contained in J(I)+

isomorphic to S3. Then as Y = [Y, T], either y+ = 02 (Kr)+ or y+ centralizes Kj.

Similarly either Lt= 02 (Kr)+ or Lt centralizes Kj. Therefore as y+ does not

act on Lt, we conclude using F.6.6 that 02 (!) = (Y, Ll) centralizes Kt, impossible

as 02 (Kj) t Z(Kj).

Therefore I' is nonsolvable, so as l+ has a subnormal subgroup isomorphic to

S3, L3(2) or A5, it follows from F.6.18 that I'~ L3(2). Thus Kr= 02 (1) = (Y,L1)

and l = KrT. But now E4 ~ E = [E, Y] :<:::; [Vr, Kr], so as LlT centralizes Vi and

Kr= 02 (!), Vr = [Vr, Kr] = (Vj_Kr) is of rank 3 by H.5.5. This is impossible, since

we saw earlier that (Z~^1 ) is of rank 4. D

LEMMA 14.7.71. (1) H has two noncentral 2-chief factors, both isomorphic to

tJ, one on U and one on QH/CH(U).

(2) L 1 has five noncentral 2-chief factors, one in 02 (L1), and four in S.

(3) [Q, L] :<:::; S.

PROOF. The proof is similar to some of the analysis in the second subsection,

but is substantially easier. First Ll has one noncentral chief factor on 02(Li), two

on tJ, and hence also two on QH/CH(U) by the duality in 14.5.21.1. Thus Ll has

at least five noncentral 2-chief factors.

Next as Ua :<:::; S by 14.7.70.4, using 14.7.66 we have

02(Li) = (U~L^1 ) :<:::; S. ()

Set Qx := [QH, K]CH(U). As [Qx/CH(U), SJ= [Qx/Cx(U), 02(Li)] is of corank

2 in Qx, with (S, Qx] :<:::; S, and as m(Qx/CQx(V)) = 2 by the duality in 14.5.21.1,

we conclude

Thus one noncentral 2-chief factor for Ll in Q lies in S*, two lie in U :<:::; S, and by

(**) a fourth factor also lies in S. Now if (1) holds, then Ll has four noncentral

2-chief factors in QH, so £ 1 has exactly five noncentral 2-chief factors by (*). Then

as L 1 has at least four noncentral chief factors on S, (2) holds, and of course (3)

follows from ( 2).

Thus it remains to prove (1), so we must show that [CH(U), K] :<:::; U. But

K = [K, UaJ, so it suffices to show [CH(Ua), Ua] :<:::; U.

Now CH(U) :<:::; CH(Za) :<:::; Na(Ua), so [Ua, CH(U)] :<:::; Cu,JU). We will show

that m(Cu°'(U)/U n Ua) :<:::; 1; then as m([W, Ua]) ~ 2 for any nontrivial H-chief

factor Won CH(U)/U since U~ :<:::; K* by 14.7.66, our proof will be complete.
By 14.7.69, m(U~) = 1, and by 14.7.70.3, m(Ua/U n Ua) = 2. So indeed
m(Cu°' (U)/U n Ua) :<:::; 1, as desired. D

LEMMA 14.7.72. (1) S = 02(L) = (02(L), L].

(2) S/V is the Steinberg module for L/S.

(3) Uo = V1.

(4) V = Z(S) = (S) = (S, SJ.