700 7. ELIMINATING CASES CORRESPONDING TO NO SHADOWNote that when L ::::) M, Hypothesis 4.4.1 is satisfied by any abelian subgroup X
of CM(V) of odd order, in view of Remark 4.4.2. Thus the hypotheses of E.6.28
are satisfied, so r ::::: min{ a, ,B} by that result, while column 7 and 8 in Table 7.2.1
give lower bounds on a and ,B, so:PROPOSITION 7.4.1. If L ::::) M then r::::: min{a,,B}, the bound appearing in
the final column of Table 7.2.1.
7.5. Eliminating most cases other than shadows
We begin with the cases which are simplest to eliminate. Recall the Funda-
mental Weak Closure Inequality E.3.29:LEMMA 7.5.1. (FWCI) m2+w2'.'.r.
We add the adjacent columns for w :::; and m 2 :::; in Table 7.2.1, and compare
this sum S with the bound R given by the final column min{ a, ,B} of the Table.We find in the following cases that we get the contradiction S < R to the FWCI,
in view of 7.4.1:
LEMMA 7.5.2. (1) L is not U3(2n), Sz(2n), or M22·(2) If Lo is L 3 (2n) on 3 E9 3, then n = 1.
Certain other cases are not immediately ruled out, but require only a slight
extension of this argument.
For the rest of the section, adopt the notation of the latter part of section E.3:
Let A:= Nvg (V), be a "w-offender" on V; that is m(VB /A)= w with A 1:. Ca(V),
so that A i-1.
LEMMA 7.5.3. (1) Assume the inequality in 7.5.1 is an equality, and let
B := {B :S: A: IE: CA(V)I = 2}, and W := {Cv(B): BE B).
Then m(A) = m 2 , r = m(VB /CA(V)), and W:::; Nv(VB). Further m(V/W)::::: w,and in case of equality, W = Nv(VB) is aw-offender on VB and m(W/Cv(A)) =
m2.
(2) m(A) ::::: r - w.
(3) Cv(A) = Cv(VB).PROOF. By 7.3.4, w < s, so (3) follows from E.3.6. By part (2) of Hypothesis
7.0.2, Hypothesis E.3.24 is satisfied. Thus (1) follows from E.3.31 and (3), and (2)
from E.3.28.3. D
In.certain cases when the FWCI has a unique solution, the embedding of A inMv is determined, which leads to a contradiction:
LEMMA 7.5.4. L is neither M12, nor M 23 on the code module 11.
PROOF. Assume otherwise. From Table 7.2.1 and 7.4.1, the FWCI is an equal-
ity with w = 2. Therefore by 7.5.3.1, m(A) = m 2 = 4 and r = 6 = m(VB /CA(V)).
Define Was in 7.5.3.1, and observe that W:::; Nv(VB) and m(V/W)::::: w = 2 by
that result. But if Mv = M23, then as m(A) = 4, H.16.8 says m(V/W) < 2, a
contradiction.Therefore Mv = M12- Here as m(A) = 4, U = Cv(A) is of dimension at most
3 and m(W) ::::: 8 by H.11.1.4. But then m(W/U) ::::: 5 > 4 = m 2 , contrary to
7.5.3.l. This contradiction completes the proof. D