7.6. FINAL ELIMINATION OF Ls(2) ON 3 EB (^3 701)
In the case of A1, we can dig a little deeper to increaser:
LEMMA 7.5.5. Lis not A 7.
PROOF. Assume Lis A1. First r 2 4 by 7.4.1, and by the FWCI 7.5.1, it
suffices to show that r > 5. We appeal to E.6.27 with j = 1: As oo is in thecolumn for a in Table 7.2.1, V is not an (F - 1)-module for Aut.M(L), hence
Ji(M) S CM(V). From the proof of 7.4.1, Ca(X) SM for any 1 =/. X:::;: CM(V) of
odd order. Thus for US V with 02 (CM(U)) S CM(V), E.6.27 says Ca(U):::;: M.
Let U consists of those U1 S V with 02 (CM(U1)) i CM(V); it suffices to show
Ca(U1) SM, for each U1 EU with m(V/U1) < 6. But if U1 EU with m(V/U1) < 6,
then U1 < Us := Cv(s) where s is a 3-element of cycle type 32 in A 7. Thus itwill suffice to show that Ca(U1) S M, for each U1 of codirriension at most 1 in
Us.· Choose a counterexample U1, and let U1 S U2 S V be maximal subject to
Ca(U2) i M. Note that C.M(U 1 ) = (s), and in particular 0
21(CM(U 1 )) :::;: CM(V):
For V = V1 EBVz where {V1, Vz} =Irr +(L, V), so that Us= (UsnVi)EB(UsnVz) and
Us n Vj is a 2-subspace of Vj. If I is an involution in M centralizing U 1 , then i mustact on U1 n Vj =/. 0 and hence on Vj. Thus i centralizes the projection U 1 ,j of U 1 on
Vj, and so for j = 1 or 2, U1,j =Us n Vj. This is impossible as CL(Us n Vj) = (s).
So U 1 , and hence also U2, lies in the set r of Definition E.6.4. Then U 2 satisfies
the hypotheses of E.6.11, so as m(V/U2) < 6 and m(V/U2) 2 r 2 4, we concludefrom E.6.11 that AutcM(u 2 ) (V) contains an element of order 15 or 31, whereas A 7
has no such element. This contradiction shows that Ca(U 1 ) SM, completing the
proof of the lemma. DFinally our weak closure methods provide some numerical information which
will be useful in the next chapter in treating two cases arising in certain shadows:LEMMA 7.5.6. (1) If L is M 22 on the code module then w > 0.
(2) If L is (S)L3(2^2 n) on 9n then w 2 n.
PROOF. Assume that the lemma fails. From Table 7.2.1 and 7.4.1, r 2 4n if
Lis (S)L 3 (2^2 n), while r 2 6 if Lis M 22 on the code module. From Table 7.2.1,
m 2 :::;: 5 when L is M 22 , so 7.5.1 supplies a contradiction to our assumption that
w = 0 in that case. Thus Lis (S)L 3 (2^2 n).
By E.3.10, A E As-w(Mv, V), while by 7.3.2, s = m. Thuss 2 3n by Table
7.2.1, so as w < n, A E A2n+i(Mv, V). By 7.5.3.2, m(A) 2 r - w > 3n. Thus we
have verified the hypotheses of lemma H.4.5.Next if B 1 :::;: A with m(A/ B 1 ) S 3n and B is the preimage in A of B 1 , then
m(Vg /B) S 3n + w < 4n s r, so Cv(B1) = Cv(B) S Na(V^9 ) by E.3.4. Thus
WA = (Cv(B1) : m(A/ B1) s 3n) s Nv(V^9 ).
Therefore [WA, A] S WA n V^9 S Cw A (A), so A is quadratic on WA, contrary to
H.4.5.2. This contradiction completes the proof of (2) and establishes the lemma.
D
7.6. Final elimination of L 3 (2) on 3 EB 3
In this section, we eliminate the case left open in 7.5.2.2. This "small" case