930 13. MID-SIZE GROUPS OVER F2
as B /0 2 (B) is noncyclic of odd order, while 02 (Mv) ~As had cyclic Sylow groups
for odd primes. This contradiction completes the proof. D
LEMMA 13.6.15. If K+ ~As, then no IE Irr +(K, (zK)) is the As-module.
PROOF. Assume K+j0 2 (K+) ~As and some IE Irr+(K,(zK)) is the A5-
module. Then as K+ = [K+, J(Tv)+], UK =I by Theorem B.5.6.
Further as there are no strong FF* -offenders on I by B.4.2.5, by that result
and B.2.9.1, there is A E A(T) with A+ of order 2 inducing a transposition on K+
and K+ = [K+Lt,A+] = K+. By 13.6.11, [UK,Lv]-=/= 1. Then as Tv acts on Lv,
the projection LI<: of Lt on K+ is a Borel subgroup of K+ wit.h LI<: = [Ll<:, A].
Therefore Lt = [Lt, A+] = Lk as Lv is Tv-invariant and K+ = [K+ Lt, A+].
As Lt= LI<: and UK is the As-module, the Lv-module [WK,Lv] = [UK,LK] is
an indecomposable extension of the trivial module CuK(Lt) by a natural module
for Lt /0 2 (Lt) ~ Z 3. This is a contradiction, as [V, Lv] is an Lv-submodule of
[WK, Lv] of rank 2. D
LEMMA 13.6.16. Assume Ko E .C(Gv, Tv) is Tv-invariant with [z, Ko] -=/= 1.
Then
(1) 02( (Ko, T)) = 1.
(2) If C is a nontrivial characteristic subgroup ofTv, then 000 (Ko)NK 0 (C) <Ko.
(3) Ko= [Ko, J(Tv)].
PROOF. Let H := (Ko, T) and assume 02(H) -=/= 1; then H E 1-l(T). As
IT: Tvl = 2, by 1.2.5 there is K2 E C(Nc(02(H))) containing Ko. As [z,Ko] -=/=
1, K2 E .Ct(G, T). By 13.3.2.2, K2 E .Cj(G, T). We now make a particularly
fundamental use of the special assumption in part ( 4) of Hypothesis 13.3.1 that we
have chosen L with L / 02 ( L) ~ As only in the final case of the FSU when no other
choice was possible: namely by Hypothesis 13.3.1.4, K 2 /0 2 (K 2 ) ~ As. Thus as
As is a minimal nonsolvable group, K2 = 02(K2)Ko, and then as IT: Tvl = 2 and
K2 is perfect, K2 = Ko. As v centralizes Ko, 1 -=/= Cr(K2), contrary to 13.6.3.2,
since by 13.3.2 K 2 satisfies Hypothesis 13.3.1 in the role of "L". This completes
the proof of (1).
Next assume thht C is a nontrivial characteristic subgroup of Tv with Ko =
Ooo(Ko)NK 0 (C). Then there is Ki E .C(NK 0 (C), Tv) with Ko= 000 (Ko)Ki. Re-
placing Ko by Ki, we may assume Ko acts on C. Since Tv is of index 2 in T, C
is normal in T, and hence 1 -=/= C ~ 02 ( (K 0 , T) ), contrary to (1). This establishes
(2).
Finally if J(Tv) ~ 000 (Ko), then Ko = 000 (Ko)NK 0 (J(Tv)) by a Frattini
Argument, contrary to (2). Thus (3) holds. D
LEMMA 13.6.17. Assume a:= z9 with a E Z(Tv)· Then [a,K]-=/= 1.
PROOF. Assurrie K ~·Ga := Cc(a). As a centralizes a subgroup of Tv of
index 2, I02(Ga) : (02(Ga) n Nc(K))I ~ 4. Thus as K = K^00 , K.central-
izes 02(Ga)/(02(Ga) n Nc(K)), and hence K :SI K02(Ga)· Since [z, K] -=/= 1
with z E Uv E RdGv), V(K) -=/= 1 in the language of Definition A.4.7. Then
[Z(02(K02(Ga))), K] -=/= 1 by A.4.9 with K, K02(Ga) in the roles of "X, M".
Then as Ga E 7-le, K does not centralize Za := Z(0 2 (Ga)). By 1.2.1.1, (KT^9 ) =: