5.1. PRELIMINARY ANALYSIS OF THE L 2 (2n) CASE 631the results of section D.3 rather than those of section 3.2 based on the FSU. By
A.1.42.2, there is IE Irr +(L, V, T). Now Hypothesis D.3.1 is satisfied with Lf', L,
I, VM := (JT) in the roles of "M, M+, V, VM". Hence we may apply D.3.10 to
conclude that I ~ LT.
Suppose first that I < [V, L], and choose an LT-submodule V 1 of V with
[V,L] 1. Vi::'.'. I. As L = F*(Lf') is simple, L-and hence also A-is faithful on V 1
and on V := V/Vi. Thus2 ::'.'. rA,V ::'.'. rA,V 1 +rA,V
in the language of Definition B.1.1. On the other hand, by B.6.9.1, r A,w ::'.'. 1 for
each faithful LA-module W, so r A,v 1 = r A,v = 1. Then by another application of
B.6.9, Vi and V have unique noncentral chief factors, and either both factors arenatural, or n = 2 and at least one is an A 5 -module. Now if a factor is natural,
then A E Syl 2 (L), while if a factor is an A 5 -module, then A 1. L. So if one factor
is an A5-module, then both are A5-modules; then as A5-modules have trivial 1-
cohomology by I.1.6, and we saw Gv(L) = 1, (2) holds. This leaves the case where
both factors are natural modules. Here we choose Vi maximal subject to [if, L] -:/-1,
so as V is an FF-module, V is natural by B.5.1.5. Also V1 is an FF-module, so
Vi/Gv 1 (L) is natural by B.5.1.5; hence as Gv(L) = 1, both Vi= I and V/I are
natural. Further as r A,v = 2 with m(V/Gv(I)) = 2n = 2m(L) for each involution2 EL, A E Syh(L) with Gv(A)) = Gv(a) = [V, a] for each a EA#. Therefore Vis
semisimple by Theorem G.1.3, and hence (1) holds.
Thus we may assume that I= [V, L], and therefore that LT is irreducible on
W := [V,L]/G[V,Lj(L). Then as q(Lf', V)::::; 2, it follows from B.4.2 and B.4.5 that
either W is the natural module, or n is even and W is the orthogonal module. In
the first case (3) holds, so assume the second holds. Then H^1 (L, W) = 0 by I.1.6, so
as Gv(L) = 1, Vis irreducible and hence (4) holds. This completes the proof. D
Recall that by Theorem 2.1.1, there is HE 1-l*(T, M).
LEMMA 5.1.5. Let HE 1-l*(T,M) and DL a Hall 2'-subgroup of NL(T n L).
Then
(1) H n M acts on T n L and on 02 (DLT), and
(2) if n(H) > 1, then H n M is solvable, and some Hall 2'-subgroup of H n M
acts on DL·
PROOF. Let TL:= TnL and B := NL(TL)· Since L/02(L) ~ L2(2n), Bis theunique maximal subgroup of L containing TL. But as M = !M(LT) and H 1. M,
L 1. H, so H n L::::; B; hence H n M acts on 02 (H n L) =TL and on NL(TL) = B.
Thus (1) holds.
Assume n(H) > 1. Then H n M is solvable by E.2.2, so as H n M acts on
B and B is solvable, (H n M)B is solvable. Therefore by Hall's Theorem, a Hall
2'-subgroup DH of H n M is contained in a Hall 2'-subgroup D of (H n M)B,
and DnB is a Hall 2'-subgroup of B .. By Hall's Theorem there is t E TL with
(D n B)t = DL, so as TL ::::; H, the Hall 2'-subgroup Dk of H n M acts on DL,
completing the proof of (2). · D
LEMMA 5.1.6. Let H E 1-l*(T, M), DL a Hall 21 -subgroup of NL(T n L), and
assume 02 ( (DL, H)) = 1. Then n is even and one of the following holds:
(1) n = 2, V is the direct sum of two natural modules for L, and [Z, H] = 1.