4.2. PUSHING UP IN THE FUNDAMENTAL SETUP 609(1) holds, and NH(V) s MH and V = [V, Mo] by 4.1.5. As V s Z(R+) and
02(MoR) s R+, V s Z(02(MoR)). Thus part (3) of Hypothesis C.2.8 holds,completing the verification of that Hypothesis, and establishing (2). D
THEOREM 4.2.5. Assume Hypothesis 4.2.3 and HE M(I) - {M}. Then
02,F•(H) i M.
The proof of Theorem 4.2.5 involves a short series of reductions, culminating in4.2.10. Until it is complete, assume I, Hafford a counterexample; that is, assume
02,F• (H) S M.
By 4.2.4, the quintuple H, MH := H n M, LH := (L n H)^00 , R, V :=
[01(Z(R+)), M+] satisfies Hypothesis C.2.8, so we can apply results in the latter
part of section C.2 to this quintuple.
LEMMA 4.2.6. (1) M+ = L.
(2) H+ := LH E C(H), and MH = H n M = NH(LH) is of index 2 in H.
PROOF. As we are assuming 02 ,F· (H) s M, we may apply C.2.13. Since
M # H E M we have MH < H, so case (1) of C.2.13 does not hold. Thus
case (2) of C.2.13 holds, so that (2) holds. By Hypothesis 4.2.3, LI = LM, while
LI~ LMH = {L} by (2), so (1) holds. D
We now reverse the roles of H, M-applying suitable results on pushing up to
M instead of H.
Set Q := 02(MH ). By assumption 02,F• (H) s M, so Q = 02 (H) by A.4.4.1.
Now as H E M, H. = Na(02(H)), and C(M, Q) = MH by A.4.4.2. Thus Q E
B 2 (M) and Q is Sylow in (QMH) = Q, so the triple Q, MH, M satisfies Hypothesis
C.2.3 in the roles of "R, MH, H". Therefore we can apply the results from Section
C.2 based on Hypothesis C.2.3 to this triple. Further as Q E B2(M),
02(M) s Q
by C.2.1.2.
LEMMA 4.2.7. (1) L = LH E C(H).
(2) LH = {L,Lh} for each h EH -M.
PROOF. By 4.2.6.1, M+ =LE C(M). By 4.2.4,· we may apply 4.1.4. Then by
4.1.4.3, L = LH0 2 (L), so as 02(L) s 02 (M) :":'.: Q :":'.: H, L = LH. Thus (1) holds,
and then (2) follows from 4.2.6.2. D
In the remainder of the proof of Theorem 4.2.5, let h denote an element of
H - M. Set Ho := (LH). Then Ho s NH(L) = MH :":'.:Musing 4.2.6.2 and 4.2.7.1.
As Ho '.5) H and H E M, we have:
LEMMA 4.2.8. H = Na(Ho).LEMMA 4.2.9. 02,F(M) S H.
PROOF. Recall that Q, M satisfy Hypothesis C.2.3 in the roles of "R, H". We
may assume that 02 ,F(M) i H, so by C.2.6, there is a subnormal A4-block Y of
M with Yi H. As m3(M) s 2, Ho s 02 (M) :":'.: NM(Y), so as Aut(Y/02(Y)) is
a 2-group, [Y,H 0 ] s 02(Y) s 02(M) s Q. But then Y acts on 02 (HoQ) =Ho, so
Y s H by 4.2.8. This contradicts Yi H, completing the proof. GJ