716 8. ELIMINATING SHADOWS AND CHARACTERIZING THE .14 EXAMPLE
m(V/Z 1 ) = 3k = 6n and m(.A) = 4n. Therefore m(Z1) = m(V) - 6n = 3n.
By 8.2.2.8, Z 1 = Cv(X). This contradicts H.4.4.4, which says if m(.A) = 4n, no
subgroup of Mv of order 22 n - 1 centralizes a subspace of Cv(A) of rank exactly
3n.
Therefore L is M 22 or M 24 , with k ~ 2. This time m(.A) :::; m2 :::; 6, so by
E.2.14.8, I*~ L 2 (4), m(.A) = 2s for s := 2 or 3, and m(V/Z1) = 2(s + 1). Again
Z1 = Cv(X), contradicting H.16.7, which says there is no subgroup X of order
3 centralizing a subspace of Cv(A) of corank 2(s + 1) in V. So the lemma is
established. D
We can now eliminate the shadows of the groups U 6 (2n) or U1(2n), when
L ~ (S)L 3 (2^2 n) and n > 1. Recall that U 6 (2) can be regarded as a Fischer group
F21·
LEMMA 8.2.4. If L ~ (S)L3(2^2 n) then n = 1, L ~ L3(4), r = 5, k = w = 1,
m(A) = 4, and CA(V) = Z1 is of rank 4.
PROOF. By 8.2.3, k = w = n. By 7.4.l and Table 7.2.1, r 2 4n with equality
only if:
(*) Ca(U) i. M for some U of rank 5n where U is the centralizer of an element
y # 1 of odd order in Mv.So by E.3.28.3, m(.A) 2 r-w 2 3n, and hence by H.4.4.3, m(V/Cv(A)) 2 5n. But
by 8.2.2.7,
m(A) = m(V/Z1) - k 2 m(V/Cv(A)) - n 2 4n,
so as m(.A) :::; m 2 = 4n, we conclude that all inequalities are equalities, so that
m(.A) = 4n and Z1 = Cv(A) is ofrank 4n. Then by the FWCI, r:::; m(.A)+w = 5n.
Assume n = 1. Then from H.4.4.7, L ~ L3(4), and we saw earlier that k =
w = n = 1, m(.A) = 4n = 4, and Z1 = Cv(A) is of rank 4n = 4. The lemma holds
when r = 5, so as 4 :::; r :::; 5, we may assume r = 4, and it remains to derive a
contradiction. Thus (*) holds. By H.4.6.1, (y) = CMv (U) is of order 3, so U is in
the set r of Definition E.6.4. But now E.6.11.2 contradicts the fact that U is not
centralized by an element of Mv of order 15.
Thus we may taken> 1, and it remains to derive a contradiction. As n = k,
there is X of order 2n - 1 in Mv with Cv(X) = Z 1 by 8.2.2.8. However this
contradicts H.4.4.5, completing the proof. D
If L ~ (S)L3(2^2 n) then L ~ L3(4) by 8.2.4 and H.4.4.7. In that event, let ULdenote the unipotent radical of the stabilizer of a line in the natural module for
L3(4). We now obtain the analogue of lemma 8.1.3 in our remaining cases:
PROPOSITION 8.2.5. Let U := Cv(A). Then w = k = n(I) and:
L/V w r A m(.A) m(U)
L3(4)/9 1 5 UL 4 4M22/lO 1 6 Kq 5 4
M24/ll 1 7 Ks 6 4
M24/ll 2 8 Kr 6 3In each case, U :J T, so Na(U) E H(T). Further U = CA(V) = Z1 :::; Z(I) and
so I:::; Oa(U).
PROOF. Recall Z 1 = V n Vh :::; Z(I) by 8.2.2.3, and w = k = n(I) by 8.2.3.