i2.7. THE TREATMENT OF A 6 ON A 6-DIMENSIONAL MODULE 843by (4). Finally if 2 E 02, we may take 2 to act on Vt by (3), so 1 =/:-CVt(z) S U,
contradicting Ca(U) i Min view of 12.7.8. This completes the proof of (5), and
hence of 12.7.10. D
LEMMA 12.7.11. Wo := Wo(T, V) S CT(V), so that w(G, V) > 0 and Na(Wo)
SM.
PROOF. Suppose A:= VB ST with A=/:-1. By 12.7.10.2, s(G, V) > 1, so by
E.3.10, A E A2(f', V), and hence A= R2 by 12.7.3. Now m(A/CA(V)) = 2, so
by 12.7.10.2, VS Ca(CA(V)) S Na(A). Thus V2 = [V, A] SA by 12.7.2.4. This
contradicts 12.7.9.3, as t E V2. Hence Wo S CT(V) = 02(LT), and Na(Wo) SM
by E.3.34.2. D
LEMMA 12.7.12. CT(L) = 1.
PROOF. If CT(L) =/:-1, also Cz(L) =/:-1; so by 12.2.9.1, Ca(Z) S M. But by
12.7.10.2, s(G, V) > 1, so by 12.4.1 there is g E G with VB S T and [V, VB] =/:-1,
contrary to 12.7.11. D
LEMMA 12.7.13. (1) Hypothesis G.2.1 is satisfied for each H E H(P 1 T) n
Na(V1), with Pi in the role of "Li".
(2) VS 02(Na(Vi)).
PROOF. By 12.7.2.3, P 1 is irreducible on V/Vi, so (1) holds. Then (2) follows
from G.2.2.1. D
Much of the rest of the section is devoted to the proof of the following result,
which identifies the remaining group in the conclusion of Theorem 12.7.1.
THEOREM 12.7.14. If Vi 02(Gz), then as Gt SM, G is isomorphic to M24.
Until the proof of Theorem 12.7.14 is complete, assume V i 02(Gz)· Recall
the subgroup Li defined in 12.7.2.5. Set K := (V^0 z), U := (V{), Gz := Gz/(z),
H := KLiT, Qz := 02(H), and H* := H/CH(U). As Vi is LlT-invariant, U ::! H.
As Vi 02(Gz), 02 (K) =/:- 1 and Vi 02(H), so Ki M by 12.2.6.
LEMMA 12.7.15. {1) (U) S (z), and U E R2(H), so that 02(H*)'= 1.
(2) Either
(a)U=Ri,or
{b) Mv ~ 86 and 0 is either Z(LiT) of order 2 or 02(L1T) of order 8.
{3) If U is abelian, then 0 = Ri.(4) m(V) = 2 or 4 and V = [V*, Li], so Li/02(Li) ~ Z3.
PROOF. Observe that Hypothesis G.2.1 is satisfied with (z), Vi, Gz, in the
roles of "V 1 , V, G 1 "; hence (1) holds by G.2.2. As
CH(U) S CH(V1) S Na(Vi),
and V i 02 (H), V does not centralize U by 12.7.13.2. Thus V =/:-1, so 0 =/:-1.
However U j H, so 0 j L 1 T, and hence (2) follows. Further if U is abelian then
0 S C.M(Vi), so (3) holds. As V/V1 = [V/Vi, Ll], V = [V, Ll], so as V =/:- 1 and
Li/0 2 (Li) ~ Z3, (4) holds. D