8.1. ELIMINATING SHADOWS OF THE FISCHER GROUPS 713Thus Hypothesis C.2.8 holds, and we may apply Theorem C.4.8. If Ca(x) i
M, then MH < H. But LH is not listed among the possibilities in C.4.8. Thiscontradiction show that Ca(x) :::::; M. D
COROLLARY 8.1.2. r 2: 6, 7, 8.PROOF. Suppose that U:::::; V with Ca(U) i M. We must show that m(U):::::; 4,
4, 3. By Proposition 8.1.1, Un 02 = 0. During the proof of 7.4.1, we verified thehypotheses of E.6.28; and hence (as observed in the proof of that result), also
hypotheses (1) and (4) of E.6.27 with j = 1. So since the conclusion Ca(U) :::::; M
of that latter result fails, hypothesis (2) or (3) of that result must fail; hence U
centralizes either some (F -1)-offender on V, or some nontrivial element of Mv of
odd order.First we consider the case where U:::::; Cv(A) for some (F - 1)-offender A. By
H.15.2.3, if U:::::; Cv(A) with Un 02 = 0, then m(U) :::::; 4, 4, 3, completfo.g the proof
in this case.So it remains to consider the case where U :::::; W := Cv (y) for some nontrivial
element fj of L of odd order. In the case of M 22 , m(W) :::::; 4 as (3 = 6 in Table
7.2.1. When L is M2 3 or M 2 4, then as U :::::; W with Un 02 = 0, m(U) :::::; 4 or 2 by
H.15.7.3, completing the proof. DUsing this improved bound on r, it is not hard to eliminate the shadows of the
Fischer groups, and isolate the configuration leading to ]4:THEOREM 8.1.3. IfV is the cocode module for L ~ M22, M23, or M24, then L ~
M 24 , and there is a unique solution of the Fundamental Weak Closure Inequality
7.5.1. Indeed that solution satisfies r = 8, m(CA(V)) = 3, w = n(H) = 2, and
A= KT of rank 6, for A aw-offender on V and HE H*(T, M).PROOF. Let A be aw-offender, with A :::::; V^9 for suitable g E G. By 8.1.2,
r;:::: 6, 7, 8, while by Table 7.2.1, w :::::; 2 and m2 :::::; 5, 4, 6. Thus the FWCI is violated
when L ~ M 23. When L ~ M 2 4, the FWCI is an equality, so all inequalities are
equalities, and hence w = 2 and r = 8. Finally when L ~ M22, w 2: 1 by the FWCI.Further m(A) ;:::: r - w ;:::: 4 by 7.5.3.2, and when these inequalities are equalities,
we must have w = 2 and r = 6-since we saw w:::::; 2 and r;:::: 6.
In particular, we have eliminated M2 3. Suppose next that L S:! M24, where
we have shown the FWCI is an equality with r = 8 and w = 2. Let W be thesubspace of V defined in 7.5.3.1, and note W = ~v(A) in the language of H.10.1.
As w = 2 = n
1
, n(H) = 2 by 7.3.4. By 7.5.3.1, m(A) = m 2 = 6. Therefore by
H.14.1.1, A is KT or Ks. If A= Ks, then W = V by H.15.3.3, contrary to 7.5.3.1.
Thus A= KT, so that the Theorem holds in this case.
We have reduced to the case where L ~ M 2 2. This case is a little harder. Recallm 2 :::::; 5, w :::::; 2, and m(A) ;:::: 4, with w = 2 in case m(A) = 4. Let B be the set of
B:::::; A with CA(V):::::; Band m(V9/B) = 5. Then for BE B, m(V9/B) < 6:::::; r,
,so Ca(B) :::::; Na(V9) and hence
W := (Cv(B) : B EB) :::::; Nv(V^9 ).
Further m(B) = m(A) - 5 + w, so m(.B) = 1 if m(A) = 4 (since in that case
we showed w = 2); while if m(A) = 5, then m(.B) = w is either 1 or 2. As