2.4. THE CASE WHERE rQ IS NONEMPTY 527Thus 02(H n M):::; 8 and CG(0 2 (H)):::; H, so
Co2(M)(8):::; Co2(M)(02(H)):::; 02(M) n H:::; 02(H n M):::; 8,
establishing (f32). D
NOTATION 2.3.11. Set re = rM := r n He and r 0 = ro,M := ro n He
The proof of Theorem 2.1.l now divides into two cases: Either r 0 is nonempty
or r 0 is empty. In the first case we focus on a member of r 0 ; the structure of suchgroups is described in 2.3.8.4. In the second case 2.3.9 gives us information about
the members of ro, particularly about their components. The two cases are treated
in the remaining two sections of this chapter.2.4. The case where r 0 is nonempty
In this section, we treat the case where r 0 is nonempty. Here by 2.3.8.4, H has
a very restricted structure dominated by x 0 -blocks. We will use this fact to identify
the groups in the conclusion of Theorem 2.1.1 which are not Bender groups, and
eliminate some difficult shadows. The main result of this section is:
THEOREM 2.4.1. If there exists HE ro with F*(H) = 92(H), then G is L2(P),
p > 7 a Mersenne or Fermat prime, Ls(3), or M11.In the remainder of this section, we assume that
HE r 0 , and the pair G, Hafford a counterexample to Theorem 2.4.1.
The groups appearing in the conclusion of Theorem 2.4.1 will emerge during the
proof of 2.4.26.
Choose 8 E 8yl 2 (H) as in 2.3.8, and choose TE 8yl2(M) as in 2.3.9.3; then
8 is Sylow in Hand in H n M, TE 8yb(M), and Z(T):::; 8 < T.
LEMMA 2.4.2. If 8i E 8yb(Hi n M) for Hi E re, then 8i E 8yb(Hi),
(8i, Hi) E U(Hi), and 8i EU.
PROOF. From the definition of r in Notation 2.3.5, U(Hi) contains a member
(U,Hu) with U:::; 8i. Then 8i E 8yb(Hi) and 8i E f3 by 2.3.8.1, so as Hi E He
it follows that (8i, Hi) E U(Hi) and 8i EU. DIn particular
8 EU and (8,H) E U(H)
by 2.4.2. For the remainder of this section, we setQ := 02(H) .and GQ := Na(Q).
LEMMA 2.4.3. (1) 8 E 8yb(GQ) and GQ E r 0. In particular, F*(GQ)
02(GQ) = Q.
(2) Assume Hi Ere and [Hi[2 2: JH[2. Then [HiJ2 = [H[2 and Hi E r 0.PROOF. First 8 E 8yl 2 (GQ) and GQ E r 0 by 2.3.9.1. Then using A.1.6, Q :::;
02 (GQ) :::; 02(H) = Q, so Q = 02(GQ)· As (8, H) E U(H), Q = 02(H) E S:](G)
by 2.3.8.2, so GQ E r 0 , completing the proof of (1).
Next assume the hypotheses of (2). Recall r 0 = r ur from the definitions in
Notation 2.3.5. If HE r, then JHl2 2: JHi[2 by maximality of JHJ2, so as JHil2 2:
IHl2 by hypothesis, Hi Er, so that Hi E r5 in this case. Thus we may assume
HE r*, so 8 is a member of U of maximal order. Choose 8i E 8yb(Hi n M); by