2.5. ELIMINATING THE SHADOWS WITH rg EMPTY 557PROOF. First either K appears in case (3) or (4) of 2.5.2, or by 2.5.7, K
appears in case (2) with n = 1. Now (1)-(3) and (5) follow by examination of thosegroups. Then Z(SK) is of order 2 by (3), so Z(S) n Ko is of order 2. By 2.5.2,
z induces a nontrivial inner automorphism on K 0 , so Z(S) n Ko = (zK)· Further
Z(S) = Z(SK 0 ), since Sis nontrivial on the Dynkin diagram when K =Ko ~ A 6
by 2.5.7. Then (2) completes the proof of (4). DJust before establishing Notation 2.5.4, we verified that I'* nr+ f-0, and hence
there is a member of T+ with first entry in this set. We now take advantage of this
flexibility:
NOTATION 2.5.9. In the remainder of the section, we choose (H, S, T, z) ET+
with H EI'. Let U(H) denote the pairs (U,Hu) E U(H) with U of maximal
order in U. By definition of I', U(H) f-0.
LEMMA 2.5.10. (1) If (U,Hu) E U(H), then Na(02(Hu)) E He.
(2) If (U,Hu) E U*(H), then U E Syb(No(02(Hu))), so U E Syb(Hu). If
also U ~ S then z E Z(S) ~ Z(U).
PROOF. By 2.3.8.2, N := Na(0 2 (Hu)) E He, establishing (1) and showing
(U, N) E U(N). Then if (U, Hu) E U*(H), U is Sylow in Hu and N by 2.3.2.2 and
maximality of IUI, so the first statement in (2) holds. Finally if U ~ S, then as
U E Syl 2 (N), 02 (N) ~ U =Sn N, and so using (1) we conclude
z E Z(S) ~ CH(U) ~ CH(02(N)) ~ 02(N) ~ U,
completing the proof of (2). D
LEMMA 2.5.11. (1) Z(T) = (z) is of order 2 and Z(S) = (t, z) = (t, tx) =
(t, ZK) ~ E 4 , where t is an involution in Sa and ZK is the projection of z on Ko.
(2) H = KoS ~ Ca(t) EI'*, with SE Syb(Co(t)). In particular, t t;! z^0.
PROOF. By 2.5.8.4, Z(S) = (zK) x Zs,a, where Zs,c := Z(S) n Sa, and
ZK is the projection on SKo of z. In the discussion following Notation 2.5.4 we
observed 1 f- 02 (H) = Sc, so Zs,c f- 1. Then as Zs,c is of index 2 in Z(S)
while Zs,c n Z~,a = 1, we conclude from 2.5.5.l that (t) := Zs,c is of order 2 and
Z(S) = (t, tx). Now (2) follows from 2.5.6.1. Finally as 1 f-z E Z(T) ~ Z(S) from
the definition of T, Z(T) = (z) is of order 2, completing the proof of (1). D
For the remainder of the section, let t be defined as in 2.5.11, and set Gt :=
Ca(t).
LEMMA 2.5.12. Assume K '.Si H, and let (U,Hu) E U*(H) with U ~ S. Then
(1) Hu= NH(E) and U = Ns(E) for some 4-subgroup E of SK·
(2) 02 (Hu) ~ A4 and E = 02(0^2 (Hu)) = CK(E).(3) The map E 1-7 (Ns(E), NH(E)) is a bijection of the set of 4-subgroups of
SK with
{(U', Hu') E U*(H): U' ~ S}.
In particular, Ns(E) E Syl2(NH(E)).
(4) If QE is a 2-group with z E QE :sJ Hu, then No(QE) E I' and U E