536 2. CLASSIFYING THE GROUPS WITH IM(r)I =^1so Z(I) = [I,I] ::; R*. Now we saw that involutions of Z(I) lie in~' so we may
assume that u E Z(I). Thus Z(R) ::; Ca(u) SM by 2.4.20.2. By 2.4.16.2, Z(R)
is generated by a pair of conjugates of Z, so Z(R) ::; R ::; S by 2.4.20.6. As
a E Z(R), this contradicts our assumption that a t'J_ S. Therefore a= u^9 , and so
T E T^0. Let Q E A(T) and S = Nr•(Q). Then IT : S*I = 2, so arguing
much as before, a E Z(T)::; Z(R) and Z(I) = [I,I]::; S. Then as S/R* is
cyclic by 2.4.12.1, either Z(I) is noncyclic so that Z(I) nR* =f. 1, or Z(I) is of order
2 so that q = 4. In the former case we obtain a contradiction as before, and in the
latter T / R is of order at most 4 and hence abelian, so again [I,I] S R*, for the
same contradiction. This completes the proof of the claim.
We now summarize the remaining possibilities: If Z(L) =f. 1 then a E S =Nr(Z) by the claim, so that I is of type Sp 4 (q^112 ) by 2.4.12.2. So assume that
Z(L) = 1. Then a= z9 and T = Nr(Z), so again a normalizes Z. If a t'J_ S, then
by 2.4.14, either I= Z is of type L 2 (q), or I is of type U 3 (q^112 ). Finally if a ES,
then I is of type L 3 (q^112 ) by 2.4.12.2.Assume that a centralizes Z. Then by the previous paragraph, Z(L) = 1,
a= z9 E T-S, and I= Z is of type L 2 (q). Since Z is a TI-subgroup by 2.4.18.1 and
a= z9 centralizes Z, [Z, zg] = 1 by I.6.2.1. Thus aZ ~ V := zzg ~ E 2 2n. However[R, a] is transitive on aZ by 2.4.14. Thus for r E [R, a], ar E aZ ~ V::; Ca(V), so
again by I.6.2.1, zgr s Ca(V). Then as m(V) = 2n = m 2 (T), zgr s V, so [R, a]
normalizes (zg[R,a]) = V. Notice VE A(G) = Q^0 in view of 2.4.8.3, and of course
z E zG n v. Now l[R, a]VI = q^3 = IRI and by 2.4.17.1, R is Sylow in GQ n Ca(Z),so that [R, a]V = Rh for some h E G. By 2.4.14, R is transitive on a[R, a], so for
s E R, as E Rh. Thus as is contained in some conjugate of Z contained in Rh, so
as Z is a TI-subgroup of G, zgs ::; Rh. Then V = (zg[R,a]) ::; (zgR) =: Xis a
subgroup of Rh normalized by R. It follows that R normalizes Rh: for if X < Rh,
then V = J(X), so that R normalizes [R, a]V =Rh. So as Rh = J(Th) is weaklyclosed in Th, R = Rh. But then a E V ::; Rh = R, contradicting our observation
that a t'J_ S.
Therefore [a, Z] =f. 1, so from our earlier summary, I is of type Sp 4 (q^112 ),
U 3 (q^112 ), or L3(q^112 ). In each case [I,I] = Z(I). Furthermore setting Za := Cz(a),
either Za S [I, I], or q = 4 and I~ Z2 x Ds is of type Sp4(2).
Suppose first that a= z9. Assume Z(L) = 1. Then by 2.4.17.2, T*/Cr·(Z9)
is cyclic, and using the previous paragraph, Za ::; [I, I] ::; Cr* (Z9). Thus as
1 # Za S Z, [Z^9 ,Z] = 1 by I.6.2.1, contradicting [a,Z] =f. 1. Thus Z(L) =f. 1 so
T* E s^0 by 2.4.20.5, and T* / R* is cyclic by 2.4.12.1. Hence [I, I] s R* =Cr· (Z9).Thus if Za S [I, I], we get the same contradiction as above, so from the previous
paragraph, q = 4 and [I,I] =: (u)::; R* = Cr·(Z9). Then a E Z9::; (Z^0 nC 0 (u)),
as Ca(u) SM by 2.4.20.2, so a ER using 2.4.20.6. Again this contradicts [a, Z] =f.
1, so a tJ_ z^0.Therefore a = u^9 , so that Z ( L) =f. 1 by our previous summary; and it also now
follows from our remarks at the start of the proof that R is the weak closure of z
in T. From our summary, a E S and I is of type Sp 4 (q^112 ). We may assume z EI.
Then I= (z^0 n I) S (z^0 n Ca( a)) =: Y. Since Ca(a) s M9 by 2.4.20.2, and Ris the weak closure of z in T, we conclude from 2.4.20.4 that z E Y ::; R9. But
then z is contained in a conjugate of Zin R9 = R*, so as Z is a TI-subgroup of G,
ZS R^9 S Ca(a), again contradicting [a, Z] =f. 1. This finally completes the proof
of (1), and hence of 2.4.21. D