538 2. CLASSIFYING THE GROUPS WITH IM(T)I = 1
possibilities in Theorem C (A.2.3) that Ks = K ~ Sp4(q^112 ), and x induces an
outer automorphism on K nontrivial on the Dynkin diagram. Then CT.(Ks) =
02(KsTs) = (s). Finally Out(Ks) is cyclic and Rs E Syl2(Ks), so Ts/Rs(s) is
cyclic. Further (cf. 16.1.4 and its underlying reference) a Sylow 2-subgroup of
Out(Ks) is generated by the image of any 2-element nontrivial on the Dynkin
diagram of K, so ITs: Rs(s)I = n2 = IOut(K)'2, completing the proof of (5). DLEMMA 2.4.24. Let TB := NT(B). Then
(1) Tis the split extension of R by TB.
(2) Z(L) = 1.
(3) TB = (x) x F, where x is an involution such that CR(x) = Z, R(x) =CT(Z), and F is cyclic and induces field automorphisms on L/Q.
PROOF. Part (1) is one of the conclusions of 2.4.8.5. By 2.4.12.1, TB = F(x),
where F := N 8 (B) is cyclic and induces field automorphisms on L/Q, and x E
TB - S. By 2.4.22.1, TB is noncyclic. Choose s E F.
Suppose first that Z ( L) = 1. By 2.4.17 .1, either CT ( Z) = R, or there exists
some involution x E TB - S with Z = CR(x) such that CT(Z) = R(x). In the
former case, TB is cyclic by 2.4.17.2, contrary to the previous paragraph, so the
latter must hold. Then [x, F] ::::; Cp(Z) = 1, so TB = (x) x F, establishing (3).
Since (2) holds by assumption, the lemma holds in this case. Thus we may assumethat Z(L) "I- 1 and it remains to derive a contradiction.
Suppose first that n/2 is odd. Then IS : RI = 2 since R < S by 2.4.22,
so IT : RI = 4 using 2.4.8.1. Hence TB ~ T/R ~ E4, since TB is noncyclic,
so there is an involution x in T - S and by 2.4.14, CR(x) is of type Sz(q), so
V := D 1 (CR(x)) ~ Eq and V# ~ ~- It will suffice to show that Vis the strongclosure of u in a Sylow 2-subgroup Tx of Ca(x) containing CT(x): For by 2.4.23.2,
Rs is the strong closure of u in a Sylow 2-group of Ca(s), and hence is nonabelian
by 2.4.12.2. So as Vis the strong closure of u in Tx, it follows thats rf. x^0. Further
x^0 n R = 0 by 2.4.21.2, so as all involutions in S -R are fused to s by 2.4.12.2, we
conclude that x^0 n S = 0. Then x rf. 02 (G) by Thompson Transfer, for the usual
contradiction to the simplicity of G.So it remains to show that V is strongly closed in Tx. Now conjugates of u
generate R by 2.4.20; so by 2.4.21 and 2.4.20.4, R is the strong closure of u in
Ca(u). Therefore as V = D 1 (CR(x)) and V# ~ u^0 , Vis strongly closed in Tx. Aswe mentioned, this completes the elimination of the case n/2 odd.
Therefore n/2 is even, so q > 4. Thus by 2.4.23.5, Ts/ Rs (s) is cyclic of order
n2 2: 4, and n2 = IOut(Ks)l2· Let tRs(s) denote the involution of Ts/Rs(s); then
this involution lies in the cyclic subgroup of index 2 in Ts/ Rs(s) inducing field auto-
morphisms, so any preimage t of tRs(s) induces an involutory field automorphism
on Ls/Us. Thus t induces a field automorphism of order 4 on L/Q, sot is not an
involution. Since s E TB, TB/(s) ~ T/R(s) ~ Ts/Rs(s) using 2.4.23, sos is the
unique involution in TB· Also TB is not quaternion since TB/ (s) is cyclic. Therefore
TB is cyclic, contrary to our earlier reduction. This contradiction completes the
~~ D
We can now finally eliminate the case where the numerical parameter we de-
noted earlier by "s" has the value 1: Let Tc := CT(Z). By 2.4.24.2, Z(L) = 1.
Then Z = Z(R) :'SJ T, so Tc :'SJ T. By 2.4.24.3, there is an involution x E T - S
such that Z = CR(x), Tc= R(x), and.T/Tc ~ TB/(x) ~Fis cyclic. It will suffice