12.6. ELIMINATING As ON THE PERMUTATION MODULE 825PROOF. Assume J(Tv) ::::; Q. Then J(Tv) = J(Q), so Nc(Tv)::::; Na(J(Tv)) =
Na(J(Q)) ::::; M = !M(LT). Hence as Tv E Syl2(Mv), (1) and (3) hold. Then as
Q::::; 02(LvTv) = Rv, (2) holds. D
LEMMA 12.6.9. If J(Tv) -f:. Q, then J(T) -f:. Q and Lv = [Lv, J(Tv)].
PROOF. Assume J(Tv) -f:. Q. Then by 12.6.7.1, J(T) -f:. Q. So by 12.6.6,
Lv = [Lv, J(Tv)]. D
LEMMA 12.6.10. Let ~(v) be the set of vectors of weight 2 in Vv. Then
L(v) :=(Lu: u E ~(v)) = L.
PROOF. Straightforward. DDuring the remainder of the proof of Theorem 12.6.2, we assume that Gv -f:. M.
In addition when Zv-=!= 1 and Gu 1:. M for some u of weight 2 in V, we choose v to
be of weight 2 rather than 6.
LEMMA 12.6.11. Lv < Kv, so Kv -f:. M.
PROOF. Assume Lv = Kv. Then Lv = 0
31(Gv) by A.3.18. Furthermore
Cav(Vv) permutes {Lu: u E ~(v)}, and hence CaJVv)::::; Na(L(v)), so CaJVv)::::;
Na(L) = M by 12.6.10. We deduce several consequences of this fact: First,
Vv ::::; 02(Lv) ::::; 02(Gv), so 02 (F(Gv)) ::::; CaJVv) ::::; M; then 02 (F(Gv)) ::::;
02 (F*(Mv)) = 1 using 1.1.3.2-that is, Gv E 7-te. Second, suppose that Vv ::::J Gv.
Then as Lv :::1 Gv,
AutaJVv) :=::; NaL(Vv)(AutLv(Vv)) ~ 86 ~ AutLvTv(Vv),
so Gv = LvTvCcv (Vv) ::::; M, contrary to our choice of v with Gv i. M. Therefore
Vv is not normal in Gv.
Suppose first that J(Tv) ::::; Q. Let Hv := CcJLv/02(Lv)). By 12.6.8, S = Tv
and Nc(J(S)) ::::; M. As Out(A5) is a 2-group, Gv = LvBHv, so Hv i. M; then
as Gv -f:. Na(Vv), also Hv i. NaJV,,). Therefore Vv < (Vjiv) =: U. Recall thatthe core V,, of the permutation module for A 6 is generated by Lv-conjugates of a
vector of weight 4 in that module, which is central in Tv = S E Syb ( Gv). Then as
Gv E rte, U::::; D1(Z(02(Gv))) by B.2.14. As Lv = 031 (Gv) and Z(Lv/02(Lv)) = 1,
Hv is a 3'-group. Then we conclude from Theorem B.5.6 that U is not a failure of
factorization module for Hv/CHJU), and hence J(S)::::; CaJU) by B.2.7. Now by
a Frattini Argument, Hv = CHJU)NHv(J(S))::::; Ca(Vv)Nc(J(S))::::; M, contrary
to our remark that Hv 1:. M.
Therefore J(Tv) -f:. Q. Then by 12.6.9, Lv = [Lv,J(Tv)J, SO [R2(Gv),Lv] = Vv
by Theorems B.5.6 and B.5.1. Then Vv :::1 Gv, contrary to an earlier reduction. D
LEMMA 12.6.12. Kv is not quasisimple.
PROOF. Assume Kv is quasisimple. Then m2(Kv) ~ m(V,,) = 5, so Kv/Z(Kv)
is not M23; and if Kv/Z(Kv) ~ M22, then (vo) = Cvv (Lv) ::::; Z(Kv) and Lv is an A5-
block. Next as a 2-local of Kv/Z(Kv) contains a quotient of Lv, as Lv/02(Lv) ~ A5,
and as [0 2 (Lv), Lv] -=!= 1, we eliminate most possibilities for Kv/Z(Kv) in the list
of Theorem C (A.2.3), reducing to Kv/Z(Kv) ~ L5(2), M22, M24, or J4· As
IS: Tvl::::; 4 by 12.6.4 with SnKv E Syh(Kv), we conclude that Kv/Z(Kv) ~ M22·
However Cv(Lv) is of corank 5 in V, so Cv(Kv) ::::; Cv(L~) is of corank at least 5
in V. Hence V/Cv(Kv) is of rank at least 5 in AutaJKv) and centralizes V,,/ (vo),