1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

992 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTY

Z 3 x Z 63. By 14.3.5, any subgroup of order 3 or 5 permuting with Tis contained in

M, so as H 2 ::;_BT but B </:. M, we conclude that either K/0 2 (K) So!^3 D4(4) with

(B n M)/0 2 (B n M) So! E 9 , or B/0 2 (B) So! E 9 and Tis irreducible on B/02(B).

In the latter case, the irreducible action of T implies that (3) holds. In the former,

m 3 (K n M) = 2. However by 14.2.2.5, K ::; Mc, so 02 (K n M) ::; CM(V) by

Coprime Action, whereas m 3 (CM(V)) ::;_ 1by14.2.2.4. This completes the proof of

(3).

Finally suppose K/0 2 (K) is one of the groups in (4)-(6), and Tis trivial on

the Dynkin diagram of K/0 2 (K) in (4). Then in each case His generated by the

set 1-i1 of T-invariant subgroups H 2 with H2/02(H2) So! L2(2). Thus H ::;_ M by

14.3.5, completing the proof of 14.3.6. D

Next recall from our discussion at the beginning of the section that in case (1)

of Hypothesis 14.3.1, Hypotheses 12.2.3 and 13.3.1 hold, so case (1) of Hypothesis

12.8.1 holds. Further by 14.2.4, case (2) of Hypothesis 12.8.1 holds in case (2) of

Hypothesis 14.3.1. Thus we can appeal to the results in section 12.8, and we adopt

Notation 12.8.2 from that section. In particular Vi is the T-invariant subspace of

V of dimension i for i ::;_ dim(V), Gi := Na(Vi), Li := 02 (NL(Vi)), Ri := 02(LiT),

etc.

Notice Vi = Zn V, and indeed in case (2) of 14.3.1, V 1 = Z by 14.2.1.4, and

so G1 =Mc by 14.2.1.2. Recall Ch := Gi/V1, and by 12.8.3.4,

G1 1:. M, so G1 E 1-i(T, M).

Observe since LT induces GL(V) on V that:

LEMMA 14.3.7. Mv = LCM(V) = L(M n G1). In particular if Mn G1 = LiT

and V :'S] M, then M = LT.

LEMMA 14.3.8. Assume L/0 2 (L) So! Ls(2). If H ::;_ G 1 with HLi = LiH for

i = 1, 2, then H ::;_ M.

PROOF. First V1L^2 H = V1HL^2 = vf^2 , so H acts on (V1L^2 ) = 112. Similarly

v 2 LiH = ~HLi = ~Li, so H acts on (V 2 L^1 ) = V, so H ::;_ Na(V) ::;_ M by 14.3.3.5.

D

LEMMA 14.3.9. Assume L/0 2 (L) So! Ls(2). Then

(1) If J(R1) 1:. 02(LT) then there exists A E A(R 1 ) and gi E L with ABi ::;_ T

but ABi 1:. ~ for i = 1, 2.

(2) If J(T) ::;_ Ri then J(T) :::1 LT.

(3) If J(T) 1:. 02(LT) then J 1 (T) f:. Ri for i = 1, 2.

PROOF. Notice (1) implies (2): For if J(T) ::;_ R 1 , then J(T) J(R 1 ) by

B.2.3.3, so J(R1) ::;_ 02(LT) assuming (1), and hence J(T) = J(R 1 ) = J(0 2 (LT)) :'S]

LT.

Assume J(R1) f:_· 02(LT). Then there is A E A(R 1 ) with A =I-1, and either
A has rank 1, or A= R 1 has rank 2. Since R 1 is not a strong FF*-offender on V,
in the latter case B.2.9.2 says we may make a new choice of A so that A has rank

1. Then there exists gi as claimed. Thus (1) and hence (2) are established, so it
   remains to prove (3).
   Assume the hypothesis of (3), so there is D E A(T) with fJ =I- 1. Now as
   m(fJ) ::;_ 2, we may choose B of index at most 2 in D, with CD(V) ::;_Band B of