870 i3. MID-SIZE GROUPS OVER F2
In particular by the claim, LR centralizes Y / 02 (Y), and hence Y normalizes
(LR0 2 (Y))^00 = LR· As Y :':'.) M, 02(Y) ::::; 02(LRT) = R by a remark in the
first paragraph, so R E Syl2(YR). Now LR/02(LR) ~ L2(P) has 3-rank 1 and
centralizes Y/0 2 (Y), so m 3 (Y) ::::; 1 as LRY is an SQTK-group. In case (2) of
13.1.6, YE £j(G, T), so by 13.1.2.3, Y/02(Y) ~ A5 or £3(2). In case (2) of 13.1.6,
either d = 3 and Y/0 2 (Y) ~ Z 3 , or Y/0 2 (Y) is ad-group ford> 3.
Recall Xis a solvable 31 -group and RE Syb(XR), so we may apply a standard
Thompson factorization theorem 26.18 in [GLS96] to conclude that
XR = NxR(J(R))NxR(ER), where ER:= !ti(Z(Ji(R))).
As XE S(G, T), Tis irreducible on the Frattini quotient of X/0 2 (X), so J(R) or
ER is normal in XR; set J := Jj(R) where j := 0 or 1 in the respective case, so in
either case J(R)::::; J and Nc(J)::::; Mc since Mc= !M(XT) by(+).
Set K := [Y, J]. If K ::::; 02 (Y) then Y normalizes Ri := J0 2 (Y) ::::; R; but
J = Jj(Ri) by B.2.3.3, and hence Y ::::; Nc(J) ::::; Mc, contrary to our choice of
Y 1:. Mc. This contradiction shows that K 1:. 02 (Y). Thus K = Y in case (2)
of 13.1.6, since there YE C(M) with Y/0 2 (Y) quasisimple. In case (1) of 13.1.6,
Y = KNy(J) by a Frattini Argument applied to KJ, and hence Y = K(Y n Mc)·
Thus in either case K = [K, J] and as Y 1:. Mc,
K 1:. Mc, and in particular [Z, K] =f. 1. (!)
As LRT normalizes Y and J, it also normalizes [Y, J] = K and hence normalizers
KR. Further K::::; Y::::; Cc(LR/02(LR)) as we saw after the claim, so [LR, KR] ::::;
02(LR) n KR::::; 02(KR).
As K is subnormal in M, KE He by 1.1.3.1. As RE Syl2(YR), RE Syb(KR).
Thus if we set D := (!ti(Z(R))KR), then DE R 2 (KR) by B.2.14, and Dis LRT-
invariant as Rand KR are £RT-invariant. Set H := LRKR and H := H/CH(D).
We saw that LR E £1(G, T) and R = 02(LRT), so [!ti(Z(R)), LR] =f. 1 by A.4.8.5
with LR, LRT, R, Tin the roles of "X, M, R, S", and hence [D, LR] =f. 1, so
that LR =f. 1. As [Z, K] =f. 1 by (!), [D, K] =f. 1, so that K =f. 1. We have seen
that [LR,KR] ::::; 02(LR) n KR::::; 02(KR), while 02 (KR) = 1 as DE R 2 (KR),
so [LR, KR] = 1. Further 02(LR) centralizes !ti(Z(R)), so H = LR x KR. In
particular F* ( H) = LR x K, so R is faithful on K.
As T acts on D, 1 =f. [D, LR] n Z =: Zo ::::; Z. Then as Cc(Zo) ::::; Mc by ( + ),
K 1:. Cc(Zo) by (!), and hence [D, LR, K] =f. 1. As K = [K, J] and K =f. 1, J =f. 1,
so there is A E Aj ( R) with A =f. 1. As R is faithful on K, so is A.
In a moment we will define a subgroup KB of K, with KB = [KB, A] and KB
nontrivial on [D, LR]· Set Hi := LRKBA· As fI =LR x KA, Hi= LR x KEA.
As LR is simple, we can choose an Hi-chief section Win [D,LR] with LR faithful
on W and KB nontrivial on W. Set Hi := Hi/CH 1 (W) and E := m(A) - m(A);
then Hi = LR x KEA, and we will see that E = 0 or 1.
Assume K is simple. Then as 1 =f. A is faithful on K, K = [K, A]; and as
[D,LR,K] =f. 1, K is faithful on [D,LR]· In this case, we set KB:= K. As A is
faithful on K and K is faithful on W, A is faithful on W, so that m(A) = m(A)
and E = 0 in this case.
So assume K is not simple. Then K is a d-group for d > 3, and as K is ofclass at most 2, exponent d, and d-rank at most 2, we conclude from A.1.24 that
K ~ Ed2 or d1+^2. Therefore A::::; GL 2 (d), so m(A)::::; 2. Now as K = [K,J] and