902 13. MID-SIZE GROUPS OVER F2
and Q 1 =J Q 2 , conclusion (2) of F.6.18 holds, and by earlier reduction [Ki, K2] =I-1,
so that H 0 ~ E 4 /31+^2. Let X 0 := Z(K 0 ); then Vo= UoFBU1, where U1 := Cv 0 (Xo)
and U 0 := [VO, Xo] is of rank 6k for some k 2 1. Now Z = Zo FB Z1 is of rank 2
where Zi :=Zn Ui; so either m(Zi) = 1 for i = 0 and 1, or U1 = 0 and Z = Zo.
As T* has order 4, 6:::; 6k = m(U 0 ) :::; 4m(Z 0 ) :::; 8; hence k = 1 and Z = Zo with
U 1 = 0, so Vo = U 0 is of rank 6. By Theorem B.5.6, J(T) :::;I Ho. Now conclusion
(liii) of the lemma holds. Hence the proof of (1) is at last complete.
We next prove (2). So assume that the hypotheses of (2) hold. As conclusion
(ii) of (1) holds, Ko :::;I X by 13.4.5.1, so 02 (K 0 ) :::; 02(X). Set Vx := (zx), so that
Vo :::; Vx E R 2 (X) by B.2.14. We must show that Vo = Vx and X = Cx(Vo)Ho. Let
X := X/Cx(Vx); then Cx(Vx):::; Cx(Vo), so H 0 is a quotient of Ho. Furthermore
as conclusion (ii) holds, CK 0 (Vo) = 02(Ko), so that CK 0 (Vo) = CK 0 (Vx) since
02(Ko) :S 02(X), and hence Ko~ K 0.
From the structure of Vo in either case of (ii), 02 (NaL(Vo) (K 0 )) = K 0. Suppose
we have shown that Vo= Vx. Then 02 (X) = K 0 , so that X = KoT =Ho. Hence
X = HoCx(V 0 ), giving the remaining conclusion of (2). Thus it suffices to show
V 0 = Vx, or equivalently that Vx :::; Vo.
By the hypotheses of (2), J 1 (T) 1:. 02 (Ho), so as Ho= KoT, there is A E A.1(T)
with Ko= [Ko, A]. Thus by B.2.4.1,
m(A*) 2 m(Vo/Cv 0 (A)) -1 and m(A) 2 m(Vx/Cvx(A)) -1.
However Vo is not an FF-module for K 0 T* by Theorem B.5.1, so m(A*) <m(Vo/Cv 0 (A)), and hence m(A*) = m(Vo/Cv 0 (A)) - 1. Then by B.2.4.2, B :=
VoCA(Vo) E A.(T); recall CA(Vo) =An Qo, so as Vo :S Cx(Vx),
13:::; An Qo:::; Cx(Ko).
Suppose first that B = 1, so that CA(Vo) = CA(Vx). Then
m(Vo/Cv 0 (A))-l = m(A*) = m(A) 2 m(Vx/Cvx(A))-1,so Vx = VoCvx(A) and hence [Vx,A] :S Va. Therefore as Ko = [Ko, A], Vo =
[Vo,Ko] = [Vx,Ko] is X-invariant; then as Z::::; Vo, Vx = (zx) =Vo, so we are
done.
Thus we may assume instead that B =J 1. Then as B E A.(T) and B centralizesKo, J(Cx(Ko)) =: Y 1:. Cx(Vx). Hence as m3(X) :S 2 with Ko~ L3(2), m3(Y)::::;
1, so we conclude from Theorem B.5.6 that either Y ~ 83 (in which case we set
Yo := Y), or Y = Yo for some Yo E C(X) with m3(Yo) = 1. In either case Yais normal in X. In the latter case since 02 (Y 0 ) :::; 02 (X) ::::; Cx(Vx), we obtain
Ya ~ L3(2) or A5 from 13.4.5.1; then by Theorem B.5.1 and 13.4.5.1, [Vx, Yo] is
either the natural module for Yo or the sum of two natural modules for Ya~ £ 3 (2).
Then Endy- 0 ([Vx, Yo]) is either a field or the ring of 2 x 2 matrices over F2, sothat [Vx, Yo, Ko]= 1. Hence [Z, Yo] :::; [Ko, Vx, Yo]= 1 using the Three-Subgroup
Lemma. So as Yo :::;I X, Yo centralizes Vx = (zx), contrary to Yo =J 1. Thus the
proof of (2) is complete.We next prove (4), so we assume that K 2 = L 2 , and that either conclusion
(ii) or (iv) of (1) holds. Now J(T) :::;I H 0 in either of those cases, so that S :=
Baum(T) = Baum(0 2 (H 0 )) by B.2.3.4. Hence as Ho 1:. M = !M(LT), no nontrivial
characteristic subgroup of Sis normal in LT. Thus conclusion (I) of 13.2.2.10 does