702 7. ELIMINATING CASES CORRESPONDING TO NO SHADOW
prove that r 2: m; indeed recall that we are postponing that proof that r 2: m until
Theorem 7.7.1 in the final section of the chapter.
A second difficulty is that we cannot improve our lower bound on r using E.6.28:
since when V is the 3 EB 3-module for L 3 (2), the elementary groups of rank 1 or
2 in L centralize subspaces of codimensions 2 or 3 in V, respectively, and hence
are (F - 1)-offenders. In the next lemma, we use ad hoc methods to complete the
treatment of the case of L 3 (2) on 3 EB 3.
LEMMA 7.6.1. Lo is not L 3 (2) on 3 EB 3.
PROOF. From Table 7.2.1, n' = 1, so n(H) = 1 for each H E 'H*(T, M) by
7.3.4. Also w > 0 by 7.3.3 and Table 7.2.1, while w S n(H) = 1 by 7.3.4; so in fact
w=l.
Next r 2: 3 as we will show in 7.7.1 in the final section, so as m2 = 2, 7.5.1 is
an equality; hence m(A) = 2 and m(V^9 /CA(V)) = 3 = r by 7.5.3.1.
Suppose first that A 1, L. Then by H.4.3.1, U := Cv(A) has dimension 2, and
(for Bas in 7.5.3.1) Ai := (Cv(B) : BE B) is of dimension 5, while Ai S Nv(V9)
by 7.5.3.1. Also U = Cv(V9) by 7.5.3.3, so that m(AutA 1 (V9)) = 3, contradicting
m2(M) = 2.
Thus A s L. In the notation of 7.7.1 and subsection H.4.1 of chapter H of
Volume I, V =Vi EB Vi with Vi E Irr +(L, V), V2 = V{ fort ET-Nr(Vi), and Vi
has basis denoted by 1, 2, 3. By H.4.3.2, we may take A to be the unipotent radical
of the centralizer of the vector 1 E Vi; then U := Cv(A) = (1) EB (2t, 3t) is of rank
3, and
Ai = (Cv(a) : a E _A#) =Vi EB (2\ 3t)
is of rank 5. So by 7.5.3.1, Ai = Nv(V9); thus we have symmetry between V
and V9, in that Ai is also aw-offender on V9. Set (M9)* := M9 /C 0 (V9). Then
Ai S L^9 * by the previous paragraph, so Ui := Cvg (Ai) is 3-dimensional and
Ui = CA(V).
In particular Zi := [A, Ai] S V n V9, and by H.4.3.2, Zi is generated by the
vector 1 E Vi. Thus
X := (V^9 , V) s Gi := No(Zi) = Co(Zi).
Now A centralizes U and V/U, and by symmetry, Ai centralizes Ui and V9 /Ui.
It follows that X centralizes the quotients in the series
1 < UUi < AAi.
Set X := X/AAi. As V and V^9 have order 2, Xis dihedral; set Y := O(X). A
Hall 2' -subgroup Yo of the preimage of Y centralizes AAi by Coprime Action, and
then as r = 3 while m(V/Ai) = 1 = m(V9 /A), Yo centralizes (V9, V) = X. As Y
is dihedral, Yo= 1, so Xis a 2-group.
We can now finish the proof of the lemma using later Proposition 7. 7.2, which
says that Gi E 'He; we postpone the statement and proof of Proposition 7.7.2 until
the next section, as it is proved in parallel with lemma 7.7.6.
Set Gi := Gi/Z1; then as Gi E 'He, F*(Gi) = 02 (Gi) by A.1.8. Recall
T1 := Cr(Z1) is Sylow in G1 by 3.2.10.4. Now T1 S L02(LT), so Cv- 1 (T1) and
Cv.t(T1) 1 are nontrivial, and by B.2.14 both lie in 02 (Gi). Then as CL(Z 1 ) is
irreducible on Vi = .J01V 1 and (2t, 3t) = A-;:n-v- 2 , it follows that Ai s 02 ( G 1 ).