890 i3. MID-SIZE GROUPS OVER F2
Furthermore 031 (H) = K, again using the fact that m 3 (K 2031 (H)) S m3(Na(l)) =
- Thus Ki SK, so as Ki ::::! MH, we conclude n = 1 in case (ii). Next Ki= [Ki, t]
for some t E NrnL (l). Thus if K is an L 2 (2m )-block, t induces a field automorphism
on K/0 2 (K) and [MK, t] S CL(l), so [MK, t] is a {2, 3}-group; we conclude in case
(i) that K is an L2(4)-block.
NextE15 ~ z.l = [l.l,Ki] S Z(Q) = Z(02(KiTH)). (*)
Assume K/0 2 (K) ~ L 3 (2). Then by(*), case (2) of C.l.34 occurs, with 02 (K) the
direct sum of two isomorphic natural modules for K/02(K). Hence Ki has exactly
three noncerrtral 2-chief factors, two of which are in lj_ S V. Thus as 02(Ki) = 1,
Ki has one noncentral chief factor on Q/V, impossible as Ki has more than one
noncentral chief factor on each nontrivial irreducible on Q/V under L ~ U3(3).
Therefore K is an L 2 (4)-block, and Q is Sylow in QK. Now Z(Q) s
CKQ(02(KQ)) = Z(0 2 (KQ)) as KQ E He. Hence
E15 ~ z.l = [l.l,Ki] s [Z(Q),Ki] s Z(Q) n U(K).
However this is impossible as Ki has at most one noncentral chief factor on Z( Q) n
U(K), since Q E Syb(KQ) and K is an L 2 (4)-block. This contradiction finally
establishes the claim that H = Ca(l) s M.
Now ( 2) follows directly from the claim. Further if r ( G, V) = 3, then there isa totally isotropic 3-subspace U of V with Ca(U) f:. M. By 7.3.3 in [Asc87], L
has two orbits on such subspaces, represented by V3 and Y where N Mv (Y) ~ L 3 (2)
is faithful on Y. Thus CM(Y) = CM(V), so as r(G, V) > 1 by (1), we conclude
Ca(Y) SM by E.6.12. So U E Vl, and (3) follows.
Assume the hypotheses of (4). Then interchanging V and V9 if necessary, we
may assume m(V9) ;:::: m(V/Cv(V9)). We apply B.4.6.13 much as in the proof
of 13.3.10: First T contains no strong FF*-offenders, so that V9 E P(T, V) by
B.1.4.6; then there is a unique conjugacy class of FF*-offenders in LT represented
by the subgroup "Ai" of that lemma, so we may assume that V9 =Ai and hence
Cv(V9) = [V, V9] E vp. Thus we may take V3 = [V, V9]. By hypothesis, [V, V9] s
V n V9, and V n V9 s Cv(V9) =Va, so V 3 = V n V9. Also m(V/(V n V9)) = 3 =
m(V9 / (V n V9)), so we have symmetry between V and V9. We conclude V3 E V!L
9
,and hence we may take g E G 3. Let U 5 be the preimage in V9 of V9 n L. By (3)
and (4) of B.4.6,
U5 = {u E V^9 : Cv(u) >Va}
and similarly
V5 = Vi.l = {v EV: Cvg(v) >Va},
so U5 E vtNLg(Vg)' and hence we may take Vt = U5. Then as Vi = [U5, V5],
V{ = Vi, so g E Gi n G3. But AutLT(V3) is the stabilizer in GL(V3) of Vi, so
Gin G3 = NLT(V3)Ca(V3). Then as LT normal~zes V, we may take g E Ca(V3),
and hence Ca(Vi) f:. M as CM(V3) ::::; Mv by 12.2.6. This completes the proof of
W. D
During the remainder of this subsection, we will assume the following hypoth-
esis:
HYPOTHESIS 13.3.13. Cv(L) = 1, so that V is not a 5-dimensional module
when L ~ A5. Set Qi:= 02(G1).