2.4. THE CASE WHERE r3 IS NONEMPTY 539to show that s^0 n T ~ s, for the involution we have been denoting bys: For then
s^0 n Tc ~ s^0 n Tc n s = s^0 n R = 0
using 2.4.21.2. Then as T/Tc is cyclic, s ¢ 02 (G) by Generalized Thompson
Transfer A.l.37.2, as usual contrary to the simplicity of G.Thus it remains to show that s^0 n T ~ S. By 2.4.23, Rs is the strong closure
of Q in Ts E Syh(Gs)· As Z(L) = 1, Rs is of type L3(2n/^2 ) by 2.4.12. Finally
by 2.4.14, for each involution i ET - S, CR(i) is of type L 2 (2n) or U 3 (2n/^2 ), and
in either case, D1(CR(i)) ::; z. To show that s^0 n T ~ S, we must show that
i ¢ s^0 for each such i; so we assume that that i E s^0 , and it remains to derive a
contradiction.Assume first that CR(i) is of type L 2 (2n). Then i centralizes Z of order 2n,
whereas for each g E G with zg n Rs =f=. 1, I c zg ( s) I = 2n/^2 , contrary to i E s^0 and
2.4.21.1.Therefore R;, := CR(i) is of type U3(2n/^2 ). Set Zi := Cz(i) = Z(R;,). Then
i9 = s for some g E G, and for suitable c E Gs, Rfc ::; Ts as Ts E Syh(Gs) by
2.4.23.4. Then z;c ::; Rs by 2.4.23.2. Interchanging U and ux if necessary, we
may assume that zyc ::; Us. Indeed we claim zyc = Zs: For assume otherwise.
By 2.4.18.1, zyc and Zs are TI-subgroups of Gs of order q^112 , so Us= Zs x zyc,
and hence Rfc::; CT.(Us/Zs) = Rs(s). Then zyc = iJJ(Rn ::; iJJ(Rs(s)) =Zs, a
contradiction establishing the claim that zyc = Zs.By the claim, Rfc ::; CTs(Zs)· But R(x) = CT(Z) with T/R(x) cyclic, so
R(x, s) = CT( Zs) as Z is a TI-subgroup. Thus ICT. (Zs) : Rs(s) I ::; 2, so as
IRil = q^312 = IRsl, also ICT.(Zs): Rfc(s)i::; 2, and hence IUs(s): Us(s) nRfc(s)I::;
2. Now Us(s) is elementary abelian of order 2q, while D 1 (Rfc(s)) = Zfc(s) is
elementary of order 2q^112 , so 2q ::; 4q^112 , and hence we conclude q = 4. Therefore
Ts = (s) x Rs(x), with x an involution by 2.4.24.3, and Rs = UsU': 95 Ds, so
Rs(x) 95 D 16. This is impossible, as the group Ri of type U3(2) is isomorphic to
Q 8 , and Z2 x D16 contains no such subgroup.
This contradiction finally completes the treatment of the case s = 1 of Theorem
2.4.7.The cases= 2.
So we turn to the cases= 2. Here we will produce members of I'o other than
H = L 0 S, which we use to obtain a contradiction.As s = 2, Lo = L 1 L2 with L = L1, and we set Ui := U(Li), so that Uo =
(U^8 ) = U 1 U 2 • By 2.4.5.1, Hypotheses C.5.1 and C.5.2 are satisfied with Sin the
roles of both "TH" and "R", for any subgroup Mo of T with IMo : SI = 2. Observe
U 0 , Baum(S) play the roles that "U, S" play in section C.5. Further as I Mo : SI = 2,
the hypotheses of C.5.6.7 are satisfied by 2.4.5.2.Recall from the beginning of this subsection 2.4.1 that R = J(S), and also
that D is defined there; and from the opening few pages of this section 2.4 that
Q = 02 (H) = 02(L 0 S). By 2.4.5.2, Q = UoC E A(S), where C := Cs(Lo), and
U 0 i Q. As s = 2, case (iii) of C.5.6.7 holds; hence there are two S-invariant