6134.3. Pushing up L 2 (2n)
In the first exceptional case of Theorem 4.2.13 where L :SI Mand L/0 2 (L) ~
L2(2n) for n > 1, it is possible to obtain a result weaker than Theorem 4.2.13, butstill stronger than M = !M(LT): Namely in Theorem 4.3.2, we show in this case
that at least L is also a uniqueness subgroup. Theorem 4.3.2 will be used in the
Generic Case of the proof of the Main Theorem. Therefore:Throughout this section we assume Hypoihesis 4.2.1, with L/0 2 (L) ~ L 2 (2n),
and L :SI M.
LEMMA 4.3.1. Let S be a 2-subgroup of M, TH E Syb(NM(S)), and assume
that Sn LE Syb(L) and M = !M(LTH)· Then Nc(S) -5:_ M.
PROOF. Assume otherwise, and pick S to be a counterexample to 4.3.1 such
that TH is of maximal order subject to this constraint. We may assume TH -5:. T.
We claim that TH E Syb(Nc(S)). If TH= T this is clear, so we may assume that
TH< T, and hence TH< Nr(TH)· As S -5:_ TH, TH n L =Sn LE Syl2(L) and
by hypothesis M = !M(LTH), so by maximality of [TH[, Nc(TH) -5:. M. Hence
if TH -5: Ts E Syb(Nc(S)), then Nr 8 (TH) -5: Ts n M -5:_ NM(S), so TH= Ts as
claimed.
Observe next that Hypothesis C.5.1 of chapter C of Volume I is satisfied with
LTH, NG ( S), S in the roles of "H, Mo, R". Further we may assume that Hypothesis
C.5.2 is satisfied, or otherwise 02((LTH,Nc(S))) -=f 1, so that Nc(S) -5:. M =
!M(LTH), as desired. Thus we may apply C.5.6.6, and obtain a contradiction to
L :SI M. This completes the proof. · D
THEOREM 4.3.2. M = !M(L).
The proof of Theorem 4.3.2 involves a series of reductions, culminating in 4.3.16.Assume the Theorem fails, and pick I so that L -5:. I -5:_ L02(LT) and I is
maximal subject to M(I) -=f {M}. Set R := 02(I) and R+ := 02(LT), so that
I= LR and R=InR+·
Set V := [0 1 (Z(R+)),L]. Choose HE M(I) - {M}, and set MH := H n M.
Define I as in Notation 4.2.11 and observe IE I. Set M := LCM(L/02(L));
by 4.2.12.1, M E ~(L) and IEμ. Then by maximality of I, IEμ*, so Hypothesis
4.2.3 is satisfied and hence by 4.2.4, the quintuple H, L, MH, R, V satisfies Hy-
pothesis C.2.8. By 4.2.12.2, L -5:_ KE C(H), with K f:. M, K/02(K) quasisimple,
and K appears in one of cases (1)-(9) of Theorem C.4.8. As L/02(L) ~ L2(2n),
case (1) of Theorem C.4.8 holds, so that either V/Cv(L) is the natural module
for L/0 2 (L), or n = 2 and Vis the A 5 -module. Furthermore MH acts on K by
Theorem C.4.8. By 1.2.1.5, either F*(K) = 02(K), or K is quasisimple and hence
a component of H. Therefore K is described in either Theorem C.4.1 or Theorem
C.3.1, respectively. Set MK := Mn K.
Recall from 4.2.4.3 that R+ = 02(LT), and R+ = Cr(L/02(L)) by 1.4.1.2.
Without loss S := T n HE Syb(MH), and we choose HE M(I) - {M} so that
S is maximal. As L -5:_ H and M = !M(LT), T 1:. H, so S < T, and hence also
S < Nr(S).
LEMMA 4.3.3. (a) If S < X -5:_ T, then M = !M(LX).