i3.2. SOME PRELIMINARY RESULTS ON A5 AND A 6 877permutation module for L. In particular if L/02(L) = A5, then 02,z(L) s CL(V).
Therefore as Out(L) is a 2-group and Vis T-invariant, Mv =LT~ An or Sn from
the structure of NAut(L)(V). We also adopt the notational conventions of section
B.3; in particular, {1, 2, 3, 4} is an orbit under T.
By B.3.3, if Zv =/= 1 then n = 6, Vis the core of the permutation module for L
on n := {1, ... , n }, and Zv is generated by en. In any event V is the irreducible
quotient of the core of the permutation module modulo (en).
When n = 6 we can also view Vas a 4-dimensional symplectic space over F 2
for L ~ Sp4(2)'. When n = 5, V = V since Vis projective for L ~ A 5 (cf. Ll.6.1),
and we can view Vas a 4-dimensional orthogonal space for L ~ n4 (2). Thus we
can speak of isotropic or singular vectors in V, nondegenerate subspaces of V, etc.
We also adopt the following notational conventions:
NOTATION 13.2.1. For 1 s i s 4, let Vi be the preimage in V ofan i-dimensional
subspace of V stabilized by T. Set Mi := NLT(Vi), Li := 02 (Mi), and let Ri be
the preimage in T of 02(Mi)· Notice for i < 4 that IRi : 02(LiT)I s 2, with
equality iff L/02(L) ~ A.6 and T 1:. L, in which case 02(LiT) = 02(Li)02(LT).
When L/02(L) ~ A 6 , define Lo := 02 (0 2 ,z(L)), and for i = 1, 2, set Li,+ :=
02 ([Li,TnL]); observe !Lo: 02(Lo)I = 3 =!Li,+: 02(Li,+)I.
13.2.1. Results on A 6 • We collect a number of results on A 6 into a single
lemma. The first few are easy calculations invoving only L and V, which do not
require Hypothesis 12.2.3.
LEMMA 13.2.2. Assume n = 6 and set Q := 02 (LT). Then
( 1) L is transitive on V#.
(2) Each v EV# is in the center of a Sylow 2-subgroup of LT.
(3) If L/02(L) ~ A5, then Li= Li,+Lo for i = 1, 2.
(4) If L = [L, J(T)], then Li= [Li, J(T)].
(5) LT controls G-fusion in V.
(6) m2(Ri) = m2(Q), so Vs J(Ri).(7) Either there is a nontrivial characteristic subgroup of Baum(Ri) normal in
LT (and hence Na(Baum(Ri)) s M), or Lis an A5-block.
(8) If L/0 2 (L) ~ A5, then J(02(LiT)) = J(02(LT)), so every nontrivial
characteristic subgroup of Baum(02(LiT)) is normal in LT.
(9) If L/02(L) ~ A5, then Na(Li) s M 2: Na(Lo).
{10) One of the following holds:
(I) Some nontrivial characteristic subgroup of Baum(T) is normal in LT.
(II) L is an A 6 -block, and A(02(LT)) ~ A(T).
(III) L2 = [L2, Ji(T)].
PROOF. Parts (1) and (3) are easy calculations, and (2) follows from (1) since
the elements of Vi are central in T. If Zv = 1,, then (5) follows from (1). By
Burnside's Fusion Lemma A.1.35, Na(T) controls fusion in Z, while Na(T) s M
by Theorem 3.3.1. Therefore if Zv =/= 1, then M = Mv controls fusion in Vi, so as
Mv =LT, (5) follows from (1) in this case also.
Next we establish (4) and (6), which will follow fairly easily from B.3.4. First
Ri contains no strong FF*-offenders by parts (1) and (2ii) of B.3.4, so by B.2.4.3,
m 2 (Ri) = m2(Q) and A(Q) ~ A(Ri). Then as VS 51i(Z(Q)), VS J(Q) S J(Ri),
completing the proof of (6).