824 12. LARGER GROUPS OVER F2 IN .Cj(G, T)
either Rv = Q or Rv = ((1, 2)). By choice of v, Tv := Cr(v) E Syh(Mv) and
IT: Tvl = 4. As CL'r_(v) = LvTv, Rv = 02(LvTv)·
LEMMA 12.6.3. There exists Kv E C(Gv) with Lv S Kv = 0
31
(G:;") :Si Gv,and Kv/02(Kv) quasisimple.
PROOF. By 1.2.1.1, Lv is contained in the product of the C-components of Gv,
so Lv projects nontrivally on Kv/O=(Kv) for some Kv E C(Gv)· As Lv/02(Lv) ~
A6, it follows from A.3.18 that m3(Kv) = 2 and Kv = 0
31
(G:;"). Thus Lv SKv :SI Gv, and as Lv/02(Lv) ~ A6, Kv/02(Kv) is quasisimple by 1.2.1.4. D
Let Tv SSE Syh(Gv), set (KvS)* := KvS/02(KvS), and choose S so that
Ns(Lv) E Syh(NaJLv)). Hence R := Cs(L~/02(L~)) E Syl2(CKvs(L~/02(L~)))
and Rv = Rn Tv. Then:
LEMMA 12.6.4. IS: Tvl SIT: Tvl = 4 ~IR: Rvl· Further 02(KvS) SR.
LEMMA 12.6.5. R normalizes Lv, and therefore [Lv, R] S 02(Lv) and R =Cs(Lv/02(Lv)).
PROOF. Let X be the preimage in KvS of 02(L~). As [R, Lv] S X while
IR: Rvl S 4, (LvX)= = Lv, so the lemma holds. D
Let Vv := [V, Lv]. Then Vv is the 5-dimensional core of the permutation module
for Lv/02(Lv) ~ A 6 • In particular Vv is generated by the Lv-conjugates of a
vector of weight 4 in that module, which is central in Tv E Syh(LvTv), so that
Vv E RdLvTv) and Vv S 01(Z(Rv)) by B.2.14. Let Vo denote the generator of
Cvv (Lv); thus Vo has weight 6 in V and Vv, even though v itself may have weight
2 rather than 6 in V.
LEMMA 12.6.6. If J(T) f:. Q then either
(1) Lv = [Lv, J(Tv)J, or
(2) J(Tv) = J(Q).PROOF. By hypothesis there is some A E A(T) not in Q. Assume first that
A satisfies one of (i)-(iii) in the proof of 12.6.1. Then some L-conjugate of A
centralizes v and is nontrivial on Lv/02(Lv), so that J(Tv) f:. Crv (Vv), Lv =
[Lv, J(Tv)J, and LvTv/02(LvTv) ~ 86, and hence conclusion (1) holds. Thus if
J(Tv) S Crv (Vv), then each A must satisfy (iv), so in particular Zv = 1 and Vv
is a hyperplane of V. But since A satisfies (iv), A centralizes no vector of weight
2, so Ai. f'v and hence J(Tv) S Q. As Q S Tv, we conclude J(Tv) = J(Q) using
B.2.3.3, so that conclusion (2) holds. D
LEMMA 12.6.7. (1) If J(T) SQ, then J(T) = J(Q) = J(Tv)·
(2) If Lv = [Lv, J(Tv)J, then L = [L, J(T)].
PROOF. As Q S Tv ST, (1) holds. Then (1) implies (2).
LEMMA 12.6.8. If J(Tv) SQ, then
( 1) S = Tv and R = Rv.(2) J(Tv) = J(Rv) = J(Q).
(3) Na(J(S)) s M.
D