10Z4 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTY
Hz/Oz(Hz) ~ 83 by the hypothesis of (6). Then since To = QiTu by 14.6.3.1, and
Qi is normal in H, we conclude Iz/0 2 (fz) ~ 83.
Suppose first that Oz(Io) -/= 1. Since L = L10z(L), we have Ii 1:. G1, while
Hz 1:. M by hypothesis, so ]z 1:. M since we saw Hz = oz(Hz)To. Thus Io E I,
and indeed as T1 ::::; Io, Io E I* and T1 0 = T1 by 14.6.4, so that T1 E 8ylz(Io) by
(2). We conclude that (I 0 ,1 1 ,Jz) is a Goldschmidt triple in the sense of Definition
F.6.1, so that (6) holds in this case.
So we suppose instead that Oz(Io) = 1, and it remains to derive a contradiction.
By construction, Hypothesis F.1.1 is satisfied with I 1 , Iz, T1 in the roles of "L1, Lz,
8". So by F.1.9, a:= (Ii, T1,Iz) is a weak BN-pair of rank 2, and as T1 plays the
role of "B/' for j = 1, 2, a appears on the list of F.1.12. Since Ii/Oz(Ii) ~ 83, and
Iz/Oz(Iz) has at least two noncentral chief factors by hypothesis, it follows that a
is of type Gz(2)', Gz(2), M1z or Aut(M1z). But then IT1I ::::; 27 , so as IT: T1I ::::; 4
by (1) and 14.6.3.1, !Tl::::; 29 , contrary to the hypothesis for (6). This contradiction
completes the proof of (6).
Finally observe that as T1 = Tu or To by (1), Na(T1) ::::; T by (2) or (4) of
14.6.3. Thus (7) holds since I 1:. M. This completes the proof of (7), and hence of
14.6.6. D
LEMMA 14.6.7. Assume IE I*. Then
(1) The hypotheses of 1.1.5 are satisfied with I, G 1 in the roles of "H, M".
{2} F*(Iz) = Oz(Iz)·
(3) O(I) = 1.
(4) If K is a component of I, then K 1:. Iz and (K, T1) EI*.
PROOF. As u E UH::::; U::::; Oz(G1), u E Oz(I n G1). Therefore
Co 2 (G 1 )(0z(InG1))::::; Co 2 (G 1 )(u)::::; Tu::::; I,
so (1) holds; hence we may apply 1.1.5. Then 1.1.5.1 implies (2). In view of
(2), to prove (3) it suffices to show that O(I) ::::; G 1. But as L is transitive on
V#, V::::; Oz(Ca(v)) for each v EV# since V::::; Oz(G 1 ) by 14.5.15.1. Therefore
[V,Co(JJ(v)]::::; O(I) n Oz(Ca(v)) = 1. Then using Generation by Centralizers of
Hyperplanes A.1.17, O(I)::::; C1(V)::::; Gi, establishing (3).
Suppose K is a component of I. By 1.1.5.3, Ki Iz. Further if K::::; M, then
as m(V) = 2 by 14.2.1.4, K ::::; C1(V) ::::; Iz, contrary to the previous remark; so
also K i M. Thus (K, T1) E I, so that (K, T1) E I* by 14.6.4, completing the
proof of ( 4). D
LEMMA 14.6.8. Assume I E I and F(I) -/= Oz(I). Then mz(I/Oz(I)) 2
m(UHOz(I)/Oz(I)) 2: m(UH/CuH(QH)).
PROOF. By 14.6.7.3, O(I) = 1, so as F*(I)-/= Oz(I) by hypothesis, we conclude
there is a component K of I. By 14.6.7.4, z does not centralize K, so that Zn
Oz(I) = 1 as Z has order 2. Set P := CQH(u). By 14.5.15.1, [UH n Oz(I), P] ::::;
Zn Oz(I) = 1. So since Z i 02(1), using the duality in 14.5.21.1 we obtain
Therefore
UH n 02(I) < CuH(P) = (u)CuH(QH)·
m2(I/02(I)) 2 m(UH02(1)/02(I)) = m(UH/(UH n 02(I)))
> m(UH/CuH(P)) = m(UH/CuH(QH))-l,