676 6. REDUCING L2(2n) TO n = 2 AND V ORTHOGONAL(3) <P(E) = 1, E/V ~ Z(LifV), and E = keru(Mv) '.'::! Mv.
(4) <P(R) ~ E, and R/E = Cu(V)/ExCui (V)/ Eis the sum of natural modules
for Lif R with Cu(V)/E = CR/E(U).(5) Mv ~ Nc(R); in particular, R ~ 02(Mv).
PROOF. As we just observed, we may apply G.2.3 with Zs in the role of "Vi";
in that application, L 1 , R, E play the roles of "I, S, 82".
Now Ll = LU by G.2.3.2 and LU = L02(LU) by G.2.3.1, so 02(L1) = Ll n
02 (LT). Hence Un 02 (L 1 ) = Cu(V), so that Cu(V) plays the role of "W". Then
(2) follows from parts (3) and (1) of G.2.3, while (4) follows from G.2.3.6. By
G.2.3.5, E /V ~ Z(Li/V) and <P(E) = 1. Thus it remains to establish the first
statement of (1), the last statement of (3), and (5).Now U = (Vca(Zs)), so as Nc(Zs) = NMv(Zs)Cc(Zs) by 6.1.9.4, Nc(Zs)
acts on U. Next z!fv = Zk, so that Mv = NMv(Zs)L ~ NMv(U)L. Then
as Ll = LU, L 1 '.'::! Mv, completing the proof of (1). Similarly keru(Mv) =keru(L 1 ) ~Un U^1 = E and E '.'::! L 1 by G.2.3.4, so E = keru(L1), completing the
proof of (3). Finally (5) holds as R = 02(L1) and Ll '.'::! Mv by (1). D
During the remainder of the section, Rand E are as defined in 6.1.21.
LEMMA 6.1.22. E < R.PROOF. Assume that R = E. In particular R ~ U, and hence R = Cu(V)
by 6.1.21.2. By 6.1.20.2, we may choose g E I with [V, VB] =f=. 1; then U =
VVBCu(VVB) by 6.1.20.4. Also Cu(VVB) = CR(VB) = CE(VB). By 6.1.21.3,
<P(E) = 1, while by 6.1.20.4, the maximal elementary abelian subgroups of VVB
are V and VB, so the maximal elementary abelian subgroups of U are R = Cu (V)
and RB = Cu(VB). By 6.1.21.5, LT acts on R, so T normalizes both membersof A(U), and hence both R and RB are normal in 02 (I)CT(Zs) = I. But then
I~ Nc(R) ~ M = !M(LT), contradicting Hypothesis 6.1.16. This completes the
~~ D
LEMMA 6.1.23. If So ~ S with RU~ So, then Nc(So) ~ M.
PROOF. By 6.1.21, RU is Sylow in Ll = LU, so that Son L is Sylow in L.
Thus the assertion follows from Theorem 4.3.17. D
Recall H = Nc(Zs). Let H* := H/CH(U) and set q := 2n. By 6.1.21.4 and
6.1.22, R/E is the sum of s 2: 1 natural modules for L 1 /R ~ L 2 (q).
LEMMA 6.1.24. (1) Cu(V) = CR(U).
(2) R* ~ Eqs, and R* = [R*,D] for each 1 =f=. D ~ D£.(3) [R, F(I)J = 1.
(4) 02 (I*) = 1.
PROOF. By 6.1.20.4, U = VBCu(V). Also by 6.1.21.4, CR/E(U) = Cu(V)/E,so that [U, r] i. E for r ER - Cu(V); as ti is abelian by 6.1.17.2, we conclude (1)
holds. By 6.1.21.4, R/ E ~ Eq2s is the sum of s natural modules for Li/ R with
Cu (V) / E the centralizer in R/ E of U, so
R = Cui(V) ~ Cui(V)/E = [R*,DJ ~ Eq•
for each 1 =f=. D ::::; D£. That is, (2) holds.
By 6.1.17.2, ti E R 2 (I). Hence 02 (J) = 1, which proves (4), and also shows
that F(J) ~ O(I). Then as R = [R*, DL] by (2), (3) follows from A.1.26. D