1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

950 13. MID-SIZE GROUPS OVER F2

So assume that [F, U'Y] = [F, V'Y]. Then [F, V'Y] s [F, U'Y]Vi s U'YV1 as UH acts

on U'Y. If V 1 s U'Y, then [F, V'Y] s U'Y, contrary to (1), so (2) holds. D

LEMMA 13.8.10. If m(U;) = 1 and UH< DH, then

(1) m(UH/DH) = 1, so we have symmetry between ')'1 and')' in the sense of

Remark F. 9.17.

(2) Either Vi s u'Y, or UH induces transvections on u'Y with axis D'Y"

PROOF. Assume m(u;) = 1, so in particular case (2) of 13.8.8 holds. Then

1 = m(U;) = m(U'Y/D'Y)::::: m(UH/DH)

and DH< UH by hypothesis, so we conclude that m(UH/DH) = 1, and we have

symmetry between /'l and ')' as discussed in Remark F.9.17. Now by F.9.13.6,

[D'Y, UH] S V1 n U'Y, so (2) follows. D

LEMMA 13.8.11. Assume u; =/= 1 and G'Y = (U~^7 )G'Y,'Yb-i. Then

(1) If either V1 s U'Y, or no element of H induces a transvection on VH/UH,

then u; < v;, so m(V_;) > 1..

(2) If UH does not induce a transvection on U'Y with axis D'Y, then m(V_;) > 1.

PROOF. By hypothesis u; =/= 1 and UH i_ U'Y, so that case (2) of 13.8.8 holds

and DH=!= UH. If u; = v;, then [UH, V'Y] = [UH,U')'], and so 13.8.9.2 supplies a

contradiction with UH in the role of "F"; hence u; < v; so that (1) holds. Assume

the hypotheses of (2) but with m(V,;) = 1. Then as 1 =/= u; s v;, u; = v_; is of

rank 1. Thus V1 s U'Y by 13.8.10.2, contrary to (1); hence (2) holds. D

LEMMA 13.8.12. {1) If KE C(H), then Ki_ M and (K, T)L1E1-iz.

{2) Let X := 02 (0 2 ,p(H) n M). Then one of the following holds:

(a) X = 1.

{b) L/02(L) ~ A6 and X = L1.

{c) L/02(L) ~ A5 and X =Li,+·

PROOF. First Ki_ M by 13.3.9 with (KT) in the role of "Y", so (1) holds.

Now define X as in (2), and assume none of (a)-(c) holds. Then 02 (0 2 ,p(H)) s

M by 13.3.9, so 02,F(H)T E 1-i(T, M). Let F denote a T-invariant subgroup

of 02,F(H) minimal subject to X S F = 02 (F) and FT E 1-i(T, M). Then

X02(F) < F since Fi M, so as F/0 2 (F) is nilpotent, X < 02 (Np(X0 2 (F))).

But also F n M = X, so 02 (Np(X0 2 (F))) = F by minimality of F. Then F

normalizes 02 (X02(F)) = X, again contrary to 13.3.9, now with X, FT in the

roles of "Y, H". D

LEMMA 13.8.13. Each solvable overgroup of L 1 T in G 1 is contained in M.

PROOF. If not, we may choose H solvable, and minimal subject to H E 1-iz.

. Then case (2) of 13.8.8 holds; in particular 1 =/= u; E Q(H, UH) and 1 =/= VC: S Ri.
By 13.8.12, 02 (0 2 ,F(H) n M) = 1 or X, where X := £ 1 if L/0 2 (£) ~ A 6 and
x := L1,+ if L/02(L) ~ A5.
Now as 02 (H) = 1, there exists an odd prime p with [Op(H*), VC:] =!= 1. So

by the Supercritical Subgroups Lemma A.1.21 and A.l.24, there exists a subgroup

P ~ Zp, Ep2, or p1+^2 such that P* :sJ H*, and VC: is nontrivial on P*. If

p s M then p s 02 (02,F(H) n M) s x s L1, so [VC:' P*] s 02(Li) n P* = 1,