i3.3. STARTING MID-SIZED GROUPS OVER F 2 , AND ELIMINATING U 3 (3) 887PROOF. As KE .C1(G, T), MK= !M(KT) by 13.3.2.2,; then as Hf:._ MK,
02((K,H)) = 1. (*)
Let MK := MK/CMK(K/02(K)). As K/02(K) is quasisimple and T :S: MK,
K = [K,TnK]. Suppose (1) fails, so that K :S: Y. Then K = [K, (TnK)*] =
[Y, (T n K)] = [Y,T n K]*, and as Y is T-invariant, [Y,T n K] ::; Y. Thus
K = (KnY)02(K), so as Tacts on Y, K :S: Y :S: H, contrary to(*) as 02 (H) -=f=. 1.
Thus (1) holds.
Let Yo := 0{^2 ,^3 }(Y); then YO* < MK: by (1). But by 13.3.4, the proper over-
groups of T in MK = Auta(K) are {2, 3}-groups, so we conclude that Y 0 * = 1.
Then Yo :S: Ca(K/02(K)), so K normalizes 0{^2 ,^3 }(YQ0 2 (K)) = Y 0. However if
Yo -=/=-1, then 02(Yo) -=/=-1 by 1.1.3.1, contrary to (*). Thus (2) holds. D
LEMMA 13.3.9. Assume L ~ A 6 , H E 1i(T, M), and Y = 02 (Y) <I H with
Y :S: CM(v) for some v E Vi -Cv(L). Then either
(1) Y = 1, or
(2) Y =Li. Further if L/02(L) ~ A5 then Y =Li, while if L/02(L) ~ A5
then Y = Li,+.PROOF. As in the proof of the previous lemma, with L in the role of "K" ,
02((L,H)) = 1.
By hypothesis Y = 02 (Y), and as Y centralizes v, Y :S: Mv by 12.2.6. Therefore
Y ::; 02 (Mv) = L by 12.2.10.2; and by 13.3.8.1, Y < L. By 13.3.8.2, Y is a
{2, 3}-group. By 1.1.3.1, 02(Y) -=/=-1.
If Y = 1, then L normalizes 02 (Y0 2 (L)) = Y, and hence (1) holds by (*).
Thus we may assume that Y-=/=-1, so that Y =Li for i = 1 or 2 by 13.3.4.3. Then
as Y centralizes v, i = 1. Further Yi := B(Y) ::; B(M) = L by 12.2.8, so Yi ::; Li.
Suppose first that Cy 1 (V) i 02(Yi). Then by 13.3.7, L/02(L) ~ A.6 and
L 0 :S: Yi. Now Lo :S: Yi :S: Li, so Yi = Lo or Li, and in either case H :S: Na (Yi) :S: M
by 13.2.2.9, contrary to (*). Therefore Cy 1 (V) :S: 02(Yi). So as Y is a {2, 3}-
group, Cy(V) ::; 02 (Y), and hence Y = Yi is of order 3. Therefore Y = Yi artd
[Y: 02 (Y)[ = 3, so (2) holds. D
LEMMA 13.3.10. (1) If L ~ A 5 then J(Ri) = J(02(LT)), B := Baum(Ri) =
Baum(0 2 (LT)), and C(G, B) :S: M.
(2) If L ~ A 6 or U 3 (3) then either there is a nontrivial characteristic subgroup
of B := Baum(Ri) normal in LT (so that Na(B) :S: M), or Lis an A5-block or a
G 2 (2)-block. Moreover if Lis a G2(2)-block, then Na(B) :S: M.
(3) If L ~ A 6 then either some nontrivial characteristic subgroup of B :=
Baum((T n L)0 2 (LT)) is normal in LT (so that Na(B) :S: M), or L is an A5-
block.
(4) If L ~ L 3 (2), then either some nontrivial characteristic subgroup of B :=
Baum(Ri) is normal in LT (so that Na(B) :S: M), or Lis an L3(2)-block.
PROOF. Part (1) follows from 13.2.4.1, and part (3) follows from case (b) of
C.1.24; Lis not a A 6 -block since V/Cv(L) is the A5-module by 13.3.2.3. Similarly
the first sentence in (2) follows from 13.2.2 when L ~ A 6 , and from C.1.37 when
L 9:'. U 3 (3). When L 9:'. L 3 (2), C.1.37 also establishes (4).
Thus it only remains to establish the final sentence of (2), so we assume that
Lis a G 2 (2)-block, but that Na(B) f:._ M, and it remains to derive a contradiction.