615particular Kz/02,F(Kz) is not SL2(P) for any odd prime p. This rules out cases
(c) and (d) in 1.2.1.4, so that Kz/02(Kz) is quasisimple. Furthermore as Y 1 ::::; Gz
is faithful on Z, Kz ~ Gz by 1.2.2. Similarly as mp(KzY 1 ) ::::; 2, we conclude
from A.3.18 that mp(Kz) = 1 for each prime divisor p of 2n - 1-unless possibly
p = 3 (so that n is even), and a subgroup of Y1 of order 3 induces a diagonal
automorphism on Kz/02(Kz) ~ L3(q), with q = E mod 3. (If case (3b) of A.3.18
were to hold, then m3(Y1K202,3(Kz)) = 3.)
Set U := (Cv(L)Gz). Now Tacts on V and L, and hence on Cv(L), so as
Cv(L) =/= l, Cv(LT) =/= 1. Then as Gz E He, Cv(LT) ::::; D1(Z(Qz)), so as Y is
irreducible on Cv(L) and 02 (K 2 ) = (Cv(L)K^2 ),02(K2)::::; (Cv(L)Gz) = U::::; D1(Z(Qz.)). (*)
In particular U is generated by Gz-conjugates of elements of Z(T), so U E R 2 (Gz)
by B.2.14.Let GZ. := Gz/Caz(U). As Kz/02(Kz) is quasisimple, so is Kz. As V/Cv(L)
is the natural module for L/0 2 (L) ~ L2(2n), Cr(Cv(L)Z) = Cr(V)(T n L) with
Cr(V)(TnL)/Cr(V) ~ E2n, and in fact Cr(V)(TnL) = Cr(V)02(K 2 ). Further
02(K2) ::::; Qz by(*); and also [Qz, V] ::::; Qz n V = 02(K2) n V::::; Cv(T n L), so
that Qz::::; Cr(V)(T n L). Hence
m(02(K2)/Co 2 (K 2 )(V)) = n = m(Qz/CQz(V)) and Qz = 02(K2)CQz(V).
( **)
By (), 02(K2) ::::; U, so as m(V/V n 02(K2)) = n with Cv(U) ::::; Cv(0 2 (K 2 )) =
V n 02(K2), m(V) = n. By(*) and(**), m(U/Cu(V))::::; m(Qz/CQz(V)) = n.
Therefore U is a failure of factorization module for Kz with FF -offender V. In
particular Kz/0 2 (Kz) is not L3(q) with q = E mod 3, since in that event as
U is an FF-module, Theorem B.5.6.l says Kz ~ SL 3 (q), whereas SL 3 (q) is not
isomorphic to L 3 (q) when q = 1 mod 3. This eliminates the exceptional case
in our discussion above, so we conclude that mp(Kz) = 1 for each p dividing
2n -1. Therefore by inspection of the lists in Theorems B.5.1 and B.4.2, Kz = K2,
and U/Z is the natural module for K2 ~ L 2 (2n) or the orthogonal module for
L 2 (4). Thus as 02 (K 2 )/Z is the natural module for K2,, U = 02(K2) by(*), and
as Qz = 02(K2)CQz(V) by (**), we conclude [V, Qz] = [V, U] ::::; U. Then as
K2 = (VK^2 ), [K2, Qz] = U::::; K2,
Kz = (Kfz)::::; (KfzQz) = K2,
and hence K 2 = Kz is normalized by T, contrary to our earlier observation that
T f:. Na(K 2 ). This contradiction completes the proof of 4.3.6. D
By 4.3.6, F*(K) = 02 (K); so as we observed following the statement of The-
orem 4.3.2, K is described in Theorem C.4.1, and as L/0 2 (L) ~ L 2 (2n), one of
cases (1)-(3) of Theorem C.4.1 holds.
LEMMA 4.3.7. K is not a block.PROOF. Assume otherwise. Inspecting cases (1)-(3) of Theorem C.4.1, weconclude that either K is an SL 3 (2n)-block, or n = 2 and K is an A1-block or an
Sp 4 (4)-block. Set U := U(K) in the notation of Definition C.1.7. Now S normalizes
K by 4.3.3.b, so as t normalizes S, S also normalizes ut. Therefore if ut::::; 02(KS),
then as [0 2 (KS),K]::::; U::::; UUt, uut ~ KS(t), forcing K::::; M by 4.3.5, contrary