984 i4. L 3 (2) IN THE FSU, AND L 2 (2) WHEN L.'.r(G, T) IS EMPTY
(1) K/0 2 (K) ~ L 3 (4), and some element of T induces a graph automorphism
on K/02(K).
(2) B =Bi is of order 3 and B :S OM(V).
(3) K = 031 (Mg'°).
PROOF. Assume QM:= 02 (M) i 8; in particular, 8 < T, so one of the cases
of 14.2.9 holds. Now QM= [QM,B]OqM(B) by Ooprime Action, and using A.1.6,
[QM,B] ::; [0 2 (BT),B] ::; 8. Thus if OT(B) ::; 8, then QM ::; 8, contrary to
assumption; so Or(B) i 8, and then of the cases in 14.2.9, only conclusion (1) of
the present result can hold.
As K/0 2 (K) ~ L 3 (4) by (1), B =Bi is of order 3. By 14.2.11, B :S OM(V), so
(2) holds. By 14.2.9, K =Kc E C(Mc)· By A.3.18, OMc(K/02(K)) is a 3'-group,
so (3) holds. D
LEMMA 14.2.13. 02(Y) :'S 8.
PROOF. Assume not. Then as 02 (Y) ::; 02(M), 02(M) i 8, so conclusions
(1)-(3) of 14.2.12 are satisfied. In particular KE C(Mc) and K/02(K) ~ Ls(4).By 14.2.5, Mc E 1-lz· Let U := (VMc).
We first show that U is abelian. Suppose not. Let y E Y be of order 3 and
set I := (UY). We appeal to 12.8.9; recall Vz = V, and Y, I play the roles
of "0^2 (P), Iz''. Thus by 12.8.9.2, 02(I) = U1Uf, where U1 := Un 02(I). By
12.8.9.1, Y = 02 (I) and Tacts on I. Thus Tacts on 02(I) = U1Uf, so that as
U :'S QH by 14.2.6.2, Uf is a normal elementary abelian subgroup of T. Thus as
KT::; Aut(Ls(4)), we conclude Uf::; K. But then 02(Y) :S 02(I) = U1Uf :S
8, contrary to our hypothesis.
Therefore U is abelian. So by 12.8.6.5, Hypothesis F.9.8 is satisfied, for each
HE 1-lz, with Z, Vin the roles of "Vi, V+". As K ~ Ls(4) and T is nontrivial
on the Dynkin diagram of K, H has no FF-modules by Theorem B.4.2, so we
conclude from (3) and (4) of F.9.18 that there is J E Irr +(K, UH) with I :SI H
and q(AutH(i), f) ::; 2. This contradicts B.4.2 and B.4.5. D
Set 82 := 02 (Y)(T n K). We begin to verify the hypotheses of F.1.1 with Ki,
Y82, 8 in the roles of "Li, L2, 8": By 14.2.13, 02(Y)::; 8, while TnK::; 8
by definition, so that 82 ::; 8 .and 8i := 8 n Ki E 8ylz(Ki). By construction
02(Y) :S 82, so that 8 n Y82 = 82 E 8yl2(Y82). Thus hypothesis (b) of F.1.1
holds. By definition, 8 acts on Ki. As 8 acts on Kand Y, 8 acts on Y8 2. Thus
hypothesis (a) of F.1.1 holds. Next Kif0 2 (Ki) is a Bender group by construction,
and so satisfies (c) of F.1.1. Since Y/0 2 (Y) ~ Z 3 ~ L2(2)', to verify (c) for Y8 2
we must show:
LEMMA 14.2.14. y = [Y, 82].
PROOF. If not, then 82 :SI YT, so Theorem 3.1.1 applied to 82 , YT in the
roles of "R, Mo" says 02 ((YT,H)) # 1, contrary to 14.2.2.3. DNext NK 1 (8i) = 8iBi =: Oi lies in Mand so normalizes Y, and hence nor-
malizes Y8 2 by construction, and 02 := Nys 2 (82) = 82 normalizes Ki. Thus (d)
of F.1.1 holds with Oi, 02 in the roles of "Bi, B2" ; and (f) of F.1.1 also follows