1182 16. QUASITHIN GROUPS OF EVEN TYPE BUT NOT EVEN CHARACTERISTIC
while m 2 (Tc) = 2 by 16.3.8; hence
m2(T):::; m2(T/Tc) + m2(Tc) = 3 + 2 = 5,
so as m 2 (T) 2: m 2 (P) 2: m 2 (S 8 ) + 1 = 5, all inequalities are equalities. Hence
m 2 (T) = 5, and for each A E A(T), m(A/A n Tc)= 3 and m(A n Tc)= 2. Thus
ALTc/Tc S=! 86 , so A:::; TcL(u) and hence J(T) = J(To), where To:= TcTL(u).
If (u, t) is not faithful on Y, then w centralizes Tc since we chose IOy(w)I
maximal; therefore T 0 =Tc x TL(w), and (1) follows. Thus we may assume that
(u, t) is faithful on Y, so IYI > 4 and yw = yz. Then we calculate that fh(To) =
(y^2 , t) x TL(w), and then that (2) holds. D
We now complete the proof of Theorem 16.3. 7.
As LS=! A 6 , TL S=! D 8. It follows from 16.3.12 and 16.3.8 that D1(<I>(J(T))) =
(z, v), where (v) = Z(TL), and (z, v) :::; Z(T). On the other hand, by 16.3.11.2,(t) = Op(K), so P = (t) x Q, where Q := (P n K)(z) S=! Ds wr Z2. Thus
J(P) = (t) xS1 xS2, with Si S=! Ds, and f21(<I>(J(P)) = (s1, s2), where (si) = Z(Si),
Si has cycle structure 22 On f2, and S1S2 has cycle structure 24.
Now by 16.3.12, J(T) =Rx TL x (w), where Rand TL are dihedral of orderat least 8. Then by the Krull-Schmidt Theorem A.1.15, either IRI > ITLI and
Na(J(T)) normalizes RZ(J(T)) and TLZ(J(T)), or IRI = ITLI = 8 and Na(J(T))
permutes the pair. Thus Na(J(T)) permutes {z,v} since (z) = D 1 ((RZ(J(T)))
and (v) = D 1 ((TLZ(J(T)))). Next J(P):::; T^9 for some g E G, and m2(P) = 5 =
m 2 (T), so J(P):::; J(T9). Then (s 1 ,s 2 ) = (J(P)):::; (J(T9)) = (z,v)9. This is
impossible as (z, v) :::; Z(T), whereas (s 1 , s 2 ) 1:. Z(P).
This contradiction completes the proof of Theorem 16.3. 7.
LEMMA 16.3.13. (1) LG n Oa(L) = 0.
(2) L is standard in G.(3) If Oa(L) n Na(L^9 ) is of even order for some g E G - Na(L), then L f:_
Na(L^9 ).
PROOF. Observe that (2) is just a restatement of Theorem 16.3.7, and (1) is
a restatement of the condition in the definition of standard form that L commutes
with none of its conjugates.
Assume the hypothesis of (3) and L:::; N := Na(L9). Thus £9 f:. L, and there
is an involution i E ON(L). By Remark 16.3.3, Lis a component of ON(i), so we
may apply I.3.1 with N, (i) in the roles of "H, P", to conclude that L :::; KKi,
where Kand Ki are (not necessarily distinct) 2-components of N. If £9 :::; KKi,
then £9 E {K,Ki}, so as i EN= Na(L9), L:::; KKi = £9, contrary to L f:. £9.
Therefore [U, KKi] = 1 by 31.4 in [Asc86a], so L:::; 00 (£9), contrary to (1). D
16.4" Intersections of NG(L) with conjugates of CG(L)
Recall that in Notation 16.2.2, z is an involution in the center of T, and Lis a
component of Gz = Oa(z). By Theorem 16.3.7, Lis standard in G.
With this setup, we could now finish quickly by quoting some.of the machinery
on standard subgroups and tightly embedded subgroups in the Component Paper
[Asc75] and the Tightly Embedded Subgroup Paper [Asc76], and some of the
classification theorems in the literature based on that theory. But since GLS do
not use this machinery, we will only use some comparatively elementary results