14.6. ELIMINATING L 2 (2) WHEN (vG1) IS ABELIAN 1021Assume the hypotheses of (6), and let Zi := Oi(Z(Ti)). As R S Ti by hy-
pothesis, and HE He, Zi SW, so that [02,F•(H),Zi] = 1 by (5). As Ti :::;I T,
Tacts on Zi, so Hi := 02,F•(H)T S No(Zi). Now Hi E H(T,M) by (4), soGi = !M(H 1 ) by (1), and then (6) follows. D
LEMMA 14.6.2. If 1 -=/:- X = 02 (X) S 02,F•(H) with QH S NH(X), then
Zs [UH,X].
PROOF. As CH(UH) = QH while x = 02 (X) -I-1, Ux := [UH,X] -I-1. As
QH acts on X, Ux is normal in QH, but Ux is not central in QH by 14.6.1.5.
Then 1 -=/:- [Ux, QH] S Ux n Z using 14.5.15.1, so as IZI = 2 we conclude that
ZS Ux. D14.6.1. Preliminary results on suitable involutions in UH. In the proof
of Theorem 14.6.18 and also at the end of the section, we will need to control
the centralizers of involutions in UH which satisfy certain special conditions (cf.
14.6.17.3 and 14.6.24.1). Thus we are led to define U(H) to consist of those u
satisfying
(UO) u E UH,(Ul) Tu:= CT(u) E Syh(CH(u)), and To:= CT( ii) is of index 2 in T,
(U2) [0 2 (Gi),u] -=/:-1 #-[02(Gi),uut] fort ET-To, and
(U3) T = No 1 (To).
LEMMA 14.6.3. Assume u E U(H). Then(1) IT: Tul = 4, ITo: Tul = 2, To= NT(Tu), and To= 02(Gi)Tu = QHTu.
(2) No(To) = T.
(3) CQH(u) i CQH(V) and L = [L, Co 2 (0 1 )(u)] = [L, CQH(u)].
(4) No(Tu) =To, Tu E Syl2(Co(u)), and u .;:_ z^0.
PROOF. Set Qi := 02(Gi). First [Qi, u] -=/:- 1 by (U2) and u E UH S U by
(UO), so [QH,u] = [Qi,u] =Vi is of order 2 by 14.5.15.1. Hence CQ 1 (u) =Qi nTu
is of index 2 in Qi, Tu is of index 2 in To, and To= QiTu = QHTu as CT(u) =To
and CT(u) =Tu by (Ul).
Pick t E T - To. If t normalizes CQ 1 ( u), thenCQ 1 (u) = CQ 1 (u)t = CQ 1 (ut).Therefore for x E Qi - CQ 1 (u), z = [x,u] = [x,ut], and hence Qi= (x,CQ 1 (u))
centralizes uut, contrary to (U2). Thus t does not normalize CQ 1 (u), so as NT(Tu)
normalizes Tun Qi = CQ 1 (u), t .;:_ NT(Tu)· As ITo : Tul = 2 we conclude that
To = NT(Tu), and as IT: Toi = 2 by (Ul), IT: Tul = 4, completing the proof of
(1).
As QH S To by (1), and To :::;I T by (Ul), we may apply 14.6.1.6 to conclude
that No(To) S Gi. Then as No 1 (To) = T by (U3), (2) holds. By (1), To = NT(Tu),
so To E Syh(No(Tu)) by (2)..
As [U, Qi] = Vi by 14.5.21.2, and U = (V^01 ), also [V, Qi] = Vi, so that
CQ 1 (V) is of index 2 in Qi since m(V) = 2. Suppose that CQ 1 ( u) S CQ 1 (V). Then
·CQ 1 (u) = CQ 1 (V), as both are of index 2 in Qi, so (u)Cu(Qi) = VCu(Qi) by
the duality in 14.5.21.l. Thus fort E T - To, (ut)Cu(Qi) = VCu(Qi), so that