908 13. MID-SIZE GROUPS OVER F202 (CK(Z)) :S: M by 13.4.2.2. Then LK :S: B(M) = L by 13.4.2.5, so that LK :S:
02 (CL(Z)) :S: 02 (CL(z)) = Li. Let Le := 02 (0£ 1 (K/02(K)); as LK :S: Li :S:
H :s; Nc(K) and Out(K*) is a 2-group, it follows that Li = LKLe. In each case
LK/0 2 (LK) is an elementary abelian 3-group of rank 1 or 2; similarly Li/02(L1)
is of rank 1 or 2 for L/0 2 (L) isomorphic to A 6 or A 6 , respectively. In particular if
L/02(L) ~ A5, then 3 :s; ILK: 02(LK)I :s; IL1: 02(L1)I = 3, so equality holds and
LK =Li.
Now set R := 02 (L 1 T), S := Baum(R), and TK := R(T n K). Then TK E
Syb(KTK)· Further [T n K, Le] :S: 02(K) :S: R, so as Li= LKLe, 02(LKTK) =
02(L1TK) = R. Also CKh(Z) = LKTK, so R = 02(LKTK) = 02(CKh(Z)).
We are now in a position to complete the proof of (1). We showed that K* ~ A5,
L 3 ( 2), or A 6 ; thus it remains to assume K* ~ A 5 , and derive a contradiction. In
this case we saw that U is the A 5 -module, and we also saw that R = 02(LkT),
so R = TK:. Then R contains no FF* -offenders on U by B.3.2.4, so by B.2.10.1,
S = Baum(R) = Baum(02(KTK)) ::;I KTK.
If C is a nontrivial characteristic subgroup of S normal in LT, then K :s; N c ( C) :s;
M = !M(LT), contrary to K i. M; hence no such C exists. This eliminates the
case L/02(L) ~ A.6, since there 13.2.2.8 shows that each C is indeed normal in LT.
Thus L/02(L) ~ A5, so by an earlier remark LK = L 1 , and hence R = Ri. Now
13.2.2.7 shows that Lis an A 6 -block. Therefore L 1 has exactly two noncentral 2-
chief factors; so also K is an A 5 -block since L 1 = LK. As S = Baum(0 2 (KTK)), S
centralizes 02(K) by C.1.13.c; so by B.2.3.7, each A E A(S) contains 02 (K). Then
[A, K] :s; [02(KS), K] = 02(K) :s; A, so A ::;I KA. However m2(02(LT)) = m2(S)
by 13.2.2.6, so that A(02(LT)) ~ A(S); hence J(02(LT)) ::;I KT, so that K :s; M
for our usui;i-1 contradiction. Therefore K* is not A5, completing the proof of (1).
We next prove (2), so we suppose that L 1 i. K, and derive a contradiction.
If K is A5, then K = B(H) by 12.2.8, and hence L 1 :s; K, contrary to our as-
sumption. Thus K is L3(2) by (1). If L/0 2 (L) ~ A 6 , then we saw earlier
that Li = LK :::; K, contrary to our assumption. Thus L/0 2 (L) ~ A 6. As
Lo and Li,+ are the T-invariant subgroups with images of order 3 in Li/0 2 (L 1 ),
we conclude that {Le,LK} = {Lo,L1,+}· Indeed as Ki. M, while K acts on
02 (Le02(K)) =Le and Nc(Lo) :s; M by 13.2.2.9, we conclude that LK =Lo and
Li,+= Le.
As K ~ L3(2), R = 02(LkTK:) is the unipotent radical of the maximal
parabolic LK:TK: of K* stabilizing Zn U. As L/0 2 (L) ~ A 6 , S ::;I LT by 13.2.2.8,
so no nontrivial characteristic subgroup of S is normal in KT, since K i. M.
Therefore we may apply C.1.37 to conclude that K is an L 3 (2)-block. But then LK
has just two noncentral 2-chief factors, whereas we saw earlier that LK = L 0 , and
Lo has at least three noncentral chief factors on an L-chief section of 02 (L) not
centralized by L 0. This contradiction shows that L 1 :::; K, completing the proof of
(2).
Recall that LK:::; Li, while by (2), Li :S: LK, so Li= LK. Thus Lj0 2 (L) ~ A 6
iff m3(L1) = 2 iff m3(LK) = 2 iff K/02(K) ~ A5. But then by 13.2.2.8 applied to
both LT and KT,