1180 16. QUASITHIN GROUPS OF EVEN TYPE BUT NOT EVEN CHARACTERISTICL is not a component of Gt. Therefore Cs(t) = (z), so as Sis of index 2 in Tc,
Cr 0 (t) = (t, z) ~ E 4. Then by a lemma of Suzuki (cf. Exercise 8.6 in [Asc86a]),
Tc is dihedral or semidihedral. As Tc > S ~ E and JEJ = 4, JTcJ > 4. D
Let K := (£E(Gt)). By assumption, K > L, so K, L, and the action of z
on K are described in case (2) or (3) of 16.1.7. To prove Theorem 16.3.7, we will
successively eliminate those possibilities, beginning with the reduction:
LEMMA 16.3.9. One of the following holds, with 02 (K) I= 1 in cases (2)-(4):
(1) K ~ L 4 (2), L ~ A 6 , and z induces a graph automorphism on K.
(2) K/02(K) ~ M12 and L ~ A5.
(3) K/02(K) ~ J2 and L ~ A5 or £3(2).
(4) K/02(K) ~HS and L ~ A6 or As.PROOF. We observe first that either
(i) t fJ. K, so that m2(K(t)) 2m2(K)+1, or
(ii) t E K, so that t E Z(K) and hence Z(K) I= 1.
On the other hand, T normalizes L by Theorem 16.2.4, and m 2 (Tc) = 2 by 16.3.8,
so
m2(K(t)) ~ m2(T) = m2(LT) ~ m2(Aut(L)) +m2(Tc) = m2(Aut(L)) + 2. (!)
We will eliminate those cases in 16.1.7 not appearing in the lemma, primarily by
appeals to(!). Set K := K/Z(K), L := L/Z(L), and m := m 2 (L).
Suppose first that K is not quasisimple, so that case (2) of 16.1.7 holds. By
16.1.2, either L is simple, or L* ~ Sz(8) with m(Z(L)) ~ 2. Furthermore as
Inn(L) ~ Aut(L) ~ Aut(L), using 16.1.4 and 16.1.5, m 2 (Aut(L)) = m; Thus (!)
says that
m2(K(t)) ~ m+2.
Further m2(K) 2 2m. Thus if t fJ. K, then 2m + 1 ~ m + 2 by paragraph one
and (), so that m ~ 1, contrary to L simple. On the other hand if t EK, then
Z(K) I= 1 and hence L ~ K ~ Sz(8). Thus 2m ~ m + 2 by (), contrary to
m = m2(Sz(8)) = 3.
Therefore K is quasisimple, and so K is described in one of the subcases of
part (3) of 16.l. 7.
Suppose first that one of subcases (a)-( d) holds, but that K is neither Sp 4 ( 4)
nor G2(4). By 16.3.4, K is not U 3 (2m). Further either K is simple, and then L
is also simple in each case; or Z(K) I= 1, and then by 16.1.2, K ~ £ 3 (4) and
<P(Z(K)) = 1. Hence when Z(K) -I= 1, involutions in K lift to involutions in K,
and so as <P(Z(K)) = 1, m2(K) = m 2 (K) + m(Z(K)). Therefore in any case,
m2(T) ~ m2(K) + 1 from paragraph one. By inspection, m 2 (K) 2 2m, and
m2(Aut(L)) = m. Thus from(!), 2m + 1~m+2, contradicting m > 1.
The lemma holds if K is £4(2), or if K appears in case (f)-(h) of 16.1.7, using
16.3.4 to conclude that 02 (K) I= 1 in the latter cases. So we may assume that K
is Sp4(4), G2(4), £5(2), or Ru. Now by 16.1.2, either K is simple, or K ~ G 2 (4)
or Ru with Z(K) of order 2. By inspection, m 3 (Aut(L)) = 3 in each case, so (!)
says that
m2(K(t)) ~ 5. (!!)However by inspection, m2(K) 2 6, so if K is simple, then (!!) supplies a contra-
diction. Thus K ~ G2(4) or Ru and JZ(K)J = 2. In the latter case m 2 (K) 2 6