552 2. CLASSIFYING THE GROUPS WITH JM(T)I =^1
(2) IfQ1 E VIH(Hu,2), then (U,HuQ 1 ) EU(H).
(3) If Lis a component of H which is a Bender group and VIH(Hu,2) ~ Qu,
then Qu n LE Syb(L)..(4) There exist~ a component K of H such that K is not a Bender group, and
if HE I', I', then (K, S) EI', I', respectively.
PROOF. By 2.3.8.2, Qu E S:J(G), so Ho:= Nc(Qu) E He. Assume 02(H) S
. k
Qu. By hypothesis, Bi SH= nj=l Nc(Bj), so Bi S 02(H) S Qu. Thus
k
NH(Qu) = n NH 0 (Bi) E Hei=l
by 1.1.3.3. Hence (1) holds.
Assume the hypotheses of (2), and let Q2 := QuQ1 and H2 := HuQ1. As
F(Hu) = 02 (Hu) = Qu since Hu E He, also F(H2) = Q2 = 02(H2); so as
U S Hu S H 2 with U E (3, (U, H2) E U(H), and hence (2) holds.
Assume the hypotheses of (3), and let Lo := (LH). First, 02(H) E VIH(Hu, 2) ~
Qu by hypothesis; so by (1), NH(Qu) E He, and then by 1.1.3.1,
Set Pu:= LoCH(Lo) n Qu, and let PL, P 1 denote the projections of Pu on L, Lo,
respectively. If L < Lo = LL^8 , let PL• be the projection of Pu on L^8 • If P1 = 1 then
AutQu(L 0 )ninn(L 0 ) = 1, so as Lis a Bender group, from the structure of Aut(Lo),02 (F(CL 0 (Qu)))-=/= 1, contrary to(). Thus P1-=/= 1 and P1 E VIH(Hu,2) ~ Qu.
Similarly if L <Lo, P1 S PLPLs E VIH(Hu,2) ~ Qu, and as P1=f1, either PL-=/= 1
or PL· =f 1. Further if PL = 1, then Qu acts on P1 = PL• and hence on L, and
AutQu(L) n Inn(L) = 1 so again 02 (F(CL(Qu)) =f 1, contrary to (). Thus
PL =f 1, and if L <Lo also Py =f 1. Therefore as Lis a Bender group, there is a
unique Sylow 2-group Po of Lo containing P1, so Po E VIH(Hu, 2) ~ Qu and hence
Po = Qu n L 0 , establishing (3).
It remains to prove (4). Let L+ be the product of all Bender-group components
of H, with L+ := 1 if no such components exist. Partially order U(H) by (U1, Hi) S
(U 2 ,H 2 ) if U 1 S U 2 and H 1 S H2, and choose (U,Hu) maximal with respect to
this order. Then by (2) and maximality of (U,Hu), VIH(Hu,2) ~Hu, and hence
VIH(Hu, 2) ~ Qu and in particular 02(H) S Qu. (!)
Observe by (!) that we may apply (1) and (3).
Replacing (U, Hu) by a suitable conjugate under H n M, we may assume Sn
Hu E Syb(Hu n M). Set Q+ :=Sn L+ E Syl2(L+)· Then Q+ = Qu n L+ by
(3), and so Hu S X := NH(Q+)· When L+ =f 1, M+ :=Mn L+ = NL+(Q+) by
2.5.2.1. In any case by a Frattini Argument, H = L+X. Further SE Syb(X) since
SE Syb(H) by 2.5.1.3. Also (U,Hu) E U(X), so XE r. As (U,Hu) E U(X) ismaximal with respect to our ordering and S::::; X, it follows from parts (3) and (4)
of 2.3.7 that XE I'*, r *'when HE I'*, I'*' respectively. Moreover the components
of X are the components of Hnot in L+, so by definition of L+, X has no Bender
components. Thus replacing (H, S, T, z) by (X, S, T, z) ET, and adjoining Q+ toB 1 , •.. , Bk, we may assume L+ = l; that is, that H has no Bender components.
Let H E r *, I'*; it remains to show that there is a component K of H with
(K, S) E r *, I'*, respectively.