2.5. ELIMINATING THE SHADOWS WITH r 0 EMPTY 553
We first consider the case where E(H) =/=-1; thus there is a component K of
H. As L+ = 1, K is not a Bender group, and so K is described in one of cases
(2)-(4) of 2.5.2. Set Ko := (K^8 ) and Ru := KoCH(Ko) n Qu, and let Ro denote
the projection of Ru on Ko.
We now argue as in the proof of (3) using (!) to conclude that N Ko ( Qu) E He
and Ro ~ Qu. Further z E Qu, so by the initial statement in 2.5.2, we conclude
that Ru 1:. Cs(Ko). Therefore Ro =/=-l. Indeed since 02(NK 0 (Ro)) EVIH(Hu, 2),
02(NK 0 (Ro)) ~ Qu n Ko~ Ro, so that Ro= 02(NK 0 (Ro)). From the description
of Kin cases (2)-(4) of 2.5.2, NK 0 (Ro) E He. Thus if NK 0 (Ro) 1:. M, we can argueas in case (ii) that ( 4) holds.
Therefore we may assume that NK 0 (Ro) ~ M. It follows that 02(M n Ko) ~
02(NK 0 (Ro)) = R 0 • Now from the description of Kand Mn Kin cases (2)-(4)
of 2.5.2, either 02 (M n Ko) E Sylz(K 0 ), or case (3) holds with K = Ko ~ Mnor Ls(3) and 02(M n Ko) = CK(z) is of index 2 in a Sylow 2-group of K. Hence
Ro = 02(M n Ko), and either Ro = Sn Ko E Sylz(Ko), or case (3) holds and
Ro = 02(M n Ko) = 02(CK 0 (z)). In any case Ro :::! Sand Hu ~ Ns(Qu) ~
Ns(Ro). Further Rf! = R~^0 , either by Sylow's Theorem or as Mn and £ 3 (3)
have one class of involutions. Therefore by a Frattini Argument, H = K 0 X 0 ,
where Xo := Ns(Ro). Now (U,Hu) E U(Xo), so that Xo E r, and as usual
X 0 EI'*, I'*, when HE I'*, I'*, respectively. Now (X 0 ,S,T,z) ET and adjoiningRo to B 1 , ... ,Bk, X 0 satisfies the hypotheses for H, so we conclude (4) holds by
induction on the number of components of H.
We have reduced to the case where E(H) = 1, where to complete the proof we
derive a contradiction.
As F*(H) =/=- 02 (H) and E(H) = 1, Y := O(H) =/=-l. By 2.3.9.5, z inverts Y,so Y is abelian. By(!) and (1), 02(Ns(Qu)) = Qu and Ns(Qu) E He. Then
by our maximal choice of (U,Hu), Ns(Qu) =Hu and U E Syl2(Hu) so Qu ~ U.
Then as U ~ S, z E Z(S) ~ Cs(Qu) ~ CNH(Qu)(Qu) = Z(Qu).As E(H) = 1, F*(H) = F(H) = 02(H)Y. Further 02(H) ~ S ~ Cs(z), so
[z, H] ~ CH(0 2 (H)), while as z inverts Y, [z, H] ~ Cs(Y), and Y is abelian, so
[z,H] ~ CH(F*(H)) = Z(F*(H)) = Z(02(H))Y.
Hence setting 02 (H)(z) =: D, DY:::! H, so by a Frattini Argument, H = YNs(D).
As z ED, D E S2(G) by 1.1.4.3, so Na(D) ~ M by 2.5.l.l. Now 02(H) ~ Quby (!), and z E Z(Qu) by the previous paragraph, so D ~ Qu. Hence D =
Qu n DY :::) Hu, so that Hu ~ Na(D) ~ M, contradicting Hu 1:. M. Therefore
( 4) is finally established, completing the proof of 2.5.3. D
In view of 2.5.3.4, we are led to definer+ to consist of those HE I'o such that
H = (K, S), for some component K of Hand SE Syl2(H n M), such that K is
not a Bender group.
We verify that r+ is nonempty: For given any (Ho, S, T, z) E T, we conclude
from 2.3.9.1 that Hi := Na(02(Ho)) E I'o, SE Syl2(H1), and if Ho EI'*, then
also H 1 EI'*. Now applying 2.5.3.4 to the 2-local H 1 , we obtain a component K
of H 1 such that K is not a Bender group, Hz := (K, S) E I'o, and Hz E I'* if
Ho E I'. Thus H 2 E r+, so r+ is nonempty, and since we saw in section 1 that I'
is nonempty, also r+ n I'* is nonempty.
NOTATION 2.5.4. Let T+ consist of the tuples (H, S, T, z) in T such that HE
r+. In the remainder of the section we pick (H, S, T, z) Er+ and let KE C(H) and