1036 i4. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTY[K,ui] =/= 1 =/= [K,uiui]. Further in all cases of 14.6.14, KS 02 (02,F*(Gi)) =: X,
so [X, ui] =/= 1 =/= [X, uiui]. Now by 14.6.1.5 applied to Gi in the role of "H", X
centralizes Oi(Z(Qi)), so as F*(Gi) =Qi, Qi does not centralize ui or uiui. Thus
(U2) holds, completing the proof that ui E U(H).
Assume for the moment that Zry S UH,i, and if case (2) of 14.6.13 holds, assume
further that u; S Hi; we will show that these assumptions lead to a contradiction.
Let Z 1 = (u 1 /. We claim first that u 1 E U(H). Suppose that case (2) of 14.6.13
holds, so that u; S Hi by assumption. By 14.6.15, p = 3, so u;(T* n H2) and T 0
are Sylow in H 0 , and therefore conjugating in Hi, we may take T 0 = u;(T* nH2).
Then fi 1 is centralized by T 0 , so by the previous paragraph, u 1 E U(H), establishingthe claim in this case. Suppose instead that case (1) of 14.6.13 holds. Then each
member of UH,i - Z is conjugate to an element in Z(To), so as before u 1 E U(H),
completing the proof of the claim. But then by by the claim, we may apply 14.6.3.4
to conclude that u 1 ¢:. zG, contrary to ( ury / = Z 1 = Z9b. Thus the hypotheses of
the first sentence of this paragraph lead to a contradiction.
If case (2) of 14.6.13 holds, that result shows we may choose 'Y so that u; S Hiand Z 1 S UH,i, contrary to the previous paragraph. Thus case (1) of 14.6.13 holds,
establishing (1). Then (2) follows from the previous paragraph.Next by (1), H has two orbits on UK: the singular and nonsingular vectors,
with tif:,i U Uf:, 2 the set of nonsingular vectors. Thus (3) follows from .the first
paragraph.Choose u as in (3). By 14.6.3.4, Tu E Syl2(CG(u)), and by 14.6.3.1, IT: Tul =
- But if w E UK with w singular, then ICH(w)I = JTJ/2 > JTuJ, so that w ¢:. uG.
Therefore UG n UK= UH, so using A.1.7.1:
CG ( u) is transitive on the G-conjugates of UK containing u. ( *)
As case (1) of 14.6.13 holds, [UH,i, UJi.] =/= 1 for some g E G with UJi. S NH(UH,i)
and UH S H9. In particular by (1) we may take u E [UH,i, UJi.] S UJc By (*),
Uk: = UJ'< for some h E CG ( u). Therefore as [U H,i, UJ'<] =/= 1, while UK S (VG^1 / and
(VG^1 / is abelian, h ¢:. Gi. Thus CG ( u) 1:. Gi. Finally as u E U H,i, u is centralized
by K2, so CH(u) 1:. M. Thus CG(u) is in the set I= I(T,u) defined in the first
subsection, so ( 4) holds.
As V =: (v/ s Z(T), v = fiifi2c, where (fii/ = ·cu-H,i (To) and c E Cu-H (H).Therefore (V
02
CH^2 l/ = (uic, UH,2l is of rank 4 since z s UH,2 by 14.6.12.3, so (5)
holds, completing the proof of 14.6.17. DWe are now in a position to derive a contradiction, and hence establish Theorem
14.6.11.
Let To and u be defined as in 14.6.17. By 14.6.17.4, CG(u) E I, so I* isnonempty, and if Tu = Tr := T n I for some I E I, then also CG(u) E I
by 14.6.4. By 14.6.17.1, JT : QHJ > 4 and m(UH/CuH(QH)) = 4. Thus the
hypotheses of 14.6.9 are satisfied for any I E I*, so by that result JTJ > 211 ,
and for any such I, setting Lr := 02 (L n I) we have LT = LrTr02(LT). Pick
IE I*, choosing I:= CG(u) if Tr= Tu for some IE I*. Seth:= 02 (H 2 )Tr,
Ii := LrTr, and Io := (Ii, I2l· Observe H 2 has a noncentral 2-chief factor on UH
and on QH/CQH(UH) by the duality in 14.5.21.1. Therefore Io EI* by 14.6.6.6.
Further 02 (H) centralizes u by Coprime Action; so if Tr= Tu, then 02 (H 2 )Tu = h
centralizes u, while Ii S CG(u) by our choice of I, so that Io S CG(u). Thus Io