2.5. ELIMINATING THE SHADOWS WITH r 0 EMPTY 551
In addition we define T to consist of the 4-tuples (H, S, T, z) such that HE r 0 ,
S E Syl2(H n M), T E Syb(M) with Z(T) ::::; S < T, and z is an involution inZ(T). For each H E I'o, there exists a tuple in T whose first entry is H, using
2.3.9.3. Throughout this section (H, S, T, z) denotes a member of T.
LEMMA 2.5.1. (1) He(S) ~ M.
(2) H n Mis the unique maximal member of He(S) n H.
(3) SE Syl2(H).PROOF. By 2.3.8.1, S E /3 and S E Syl2(H), so that (3) holds. Suppose there
is X E He(S) with X i M. Then from the definitions in Notation 2.3.4 and
Notation 2.3.5, (S, X) E U(X), so XE r. Then by 2.3.7.4, XE r 0 , contrary to our
assumption in this section that r3 = 0. Thus (1) holds. By 2.3.9.4, H n ME He,so that (1) implies (2). D
From now on we use without comment the fact from 2.5.1.3, that Sis Sylow
in H.LEMMA 2.5.2. Suppose L is a component of H and set ML := Mn L. Then
z induces an inner automorphism on L, L = [L, z] i M, and one of the following
holds:(1) L is a Bender group and ML is a Borel subgroup of L.
(2) L ~ Sp4(2n)' or L3(2n) or L/0 2 (L) ~ £ 3 (4). Further Ns(L) is nontrivial
on the Dynkin diagram of L/Z(L), and ML is a Borel subgroup of L.
(3) L ~ £3(3) or Mu and ML= CL(zL), where ZL is the projection on L of z.
(4) L ~ L2(p), p > 7 a Mersenne or Fermat prime, and ML= Sn L.
PROOF. Observe Lis described in 2.3.9.7, and L = [L, z] i M by 2.3.9.6. If
L ~ £4(2), M22, M23, A1, or A1, then from the description of ML in 2.3.9.7, there is
H 1 E He(S)nH with HinL i ML, contradicting 2.5.1.2. Similarly if conclusion (b)
of 2.3.9.7 holds, then by 2.5.1.2, Sis nontrivial on the Dynkin diagram of L/Z(L),
and ML is as described in (2)-in particular, observe we cannot have L ~ A 6 or
A 6 with z inducing a transposition, since S is nontrivial on the Dynkin diagram,while z E Z(S) as (H, S, T, z) E T. So when L is A5 or A5, z induces an inner
automorphism of L. Indeed as z E LCs(L), and z inverts O(H) by 2.3.9.5, L is
not A 6 for any action of z on L. If conclusion (a) of 2.3.9.7 holds, then by 2.5.1.2,
ML is a Borel subgroup of L, so that (1) holds. The remaining cases (d) and (e) of
2.3.9.7 appear as (3) and (4). Since we have eliminated the case where z inducesan outer automorphism on L/Z(L) ~ A 6 or A1, in each case z induces an inner
automorphism on L by .2.3.9. 7. D
Part ( 4) of the next result produces the component of Hon which the remainder
of the analysis in this section is based. Furthermore it eliminates case (1) of 2.5.2
where the component is a Bender group.
LEMMA 2.5.3. Assume
k
H = n No(Bi) for some' 2-subgroups B1, ... , Bk of H,
i=l