15.3. THE ELIMINATION OF Mr/CMf(V(Mr)) = 83 wr Z 2 1143are equivalent. Further Na(Zs) = NM(Zs) by 15.3.46.5, so as N!VJ(Zs) = f' is a
2-group, (4) implies (5).Thus we may assume that (3) fails, and it remains to derive a contradiction.
Hence case (2) of 15.3.7 holds, so that Y/0 2 (Y) ~ 31+^2 , Y = e(M), Cy(V) =
02,z(Y), and Yo:= 02 (Cy(V)) # 1. Hence:(b) Yo= {)(MH)·
Let D := D1(QH). Now [QH, Yo] :'S: QHnYo :'S: 02(Yo), so Q :'S: D1(CT(Yo/02(Yo))) =
Z(f'). Thus D :'S: S, so as UH= (Zff):
(c) UH :'S: Z(D).
(d) VH is elementary abelian.
For [V, QH] ::::; VnQH::::; D::::; CH(UH) by (c), so VH centralizes QH/CH(UH)· Now
Hypothesis F.9.1 holds by 15.3.48.1, and UH is abelian by (c), so we may applyF.9.7 to conclude that QH/CH(UH) is H-isomorphic to the dual of UH. So as H*
is faithful on UH, VH::::; QH. In particular VH normaliz~s V, so V commutes with
.each H-conjugate of V by 15.3.46.4, and hence VH is abelian, establishing (d).
We next extend Hypothesis F.9.1 to:
(e) Hypothesis F.9.8 holds.
For suppose Z ::::; V n V^9 for some g E G. As Mc = Ca(Z), and M controlsG-fusion in V by 15.3.46.1, we conclude from A.1.7.1 that Mc is transitive on
{U E v^0 : Z::::; U}. Thus we may take g E Mc, and then [V, Vg] = 1 by (d) applied
to Mc in the role of "H". Thus condition (f) of Hypothesis F.9.8 holds. Further
case (ii) of condition (g) of Hypothesis F.9.8 holds by 15.3.47.We now adopt the notation of the latter part of section F.9 and obtain:
(f) [EH, V 1 ] = 1 = [E 1 , VH]· In particular, CvH(U 1 /Z 1 ) = CvH(U 1 ).
For as VH is elementary abelian by (d), E 1 = V 1 n QH ::::; D, and so [E 1 , UH] = 1
by (c). Thus as Zs ::::; UH, E 1 ::::; CT(Zs). Suppose E 1 does not centralize V.
Then 1 # [V, E 1 ] ::::; V 1 , so as V 1 is abelian, V 1 E v^0 n Ca([V, E 1 ]) ~ Ca(V) by
15.3.47, contradicting our assumption that [V, E 1 ] # 1. Thus E 1 centralizes V.
But as E 1 ::::; QH, E 1 normalizes each H-conjugate of V, so this argument gives
the second equality in (f). Before completing the proof of (f), we recall [V, U 1 ] # 1
since V 1:. G~
1
), so as [D 1 , VJ :::; [E 1 , VJ = 1:
(g) D 1 < U,.By (g) we have symmetry between l'l and 1' as discussed in the first paragraph
of Remark F.9.17, so that the remaining equality in (f) follows from that symmetry.
Further by F.9.16.4, we can choose 1' so that 0 < m(u;) ;:: m(UH /DH), and hence
by (f):
(h) u; and V_; are quadratic FF*-offenders on UH.
Choose h EH with ')'o = ')' 2 h, set a:= ')'h, and observe VO: ::::; 02(Y 0 *T*)-sincefrom the proof of 15.3.48, Yo plays the role of "£ 1 ''. Let JH := (V!). We show:
(i) JH is the product of C-components L of JH with L =[£,Yo].For if JH is not the product of members of C(JH ), then by (h) and Theorem B.5.6,
there is£ subnormal in J'H with L ~ 83, 03(£) = [03(£), VO:], and [UH, L]
of rank 2. Further Yo acts on L as there are at most two H-conjugates of L in
Theorem B.5.6 and Yo= 02 (Y 0 ). As 03 (£) = [03(£), VO:] and VO:::::; 02 (Y 0 T),
03 (L) # Y 0 . Hence Y 0 centralizes L/0 2 (£) so that L normalizes 02 (Yo0 2 (L)) =