13.6. FINISHING THE TREATMENT OF A 5 927Notice cases (c) and (d) of 1.1.5.3 cannot arise: for in those cases z induces
an outer automorphism on K, whereas Lv induces inner automorphisms on K by
A.3.18, so that z E (v)Lv does too. In the remaining cases of 1.1.5.3, z induces an
inner automorphism of K.
Recall there is a E z^0 n Cv(Lv) with a E Z(Tv)· As a E z^0 , F*(CaJa)) =
02(CaJa)) by 1.1.4.3 and 1.1.3.2. Therefore [K, a] =I-1, and then as a E Z(Tv),
a normalizes Kand K = [K, a]. Set X := Nov (K), Tx := Tv n X, and X* .-
X/Cx(K). Since L~ centralizes a* but not z*, a* =I-z*. ThenE4 ~ E := (a,z*)::; Z(T_X),
so neither case (e) or (f) of 1.1.5.3 holds. Thus we have reduced to cases (a) or
(b) of 1.1.5.3, with K* of Lie type over F 2 m for some m. Again as E* ::; Z(T_X),either m > 1 and E ::; K, or X* ~ S 6. Suppose the latter case holds. Then
Lv ::; 031 (Gv) = K by A.3.18, so Lv ~ A4, and hence L is an A 5 -block. By
13.6.3.2, CT(L) = 1. Therefore V = 02 (LT) by C.1.13.c, so V = 02 (M) = Ca(V)
by 3.2.11. Thus as (V^01 ) is abelian by 13.6.2, G 1 ::; Na(V) ::; M, contrary to
13.6.3.1. This contradiction shows that m > 1 with E::; K.
Next we conclude from A.3.18 that one of the following holds:
(i) Lv::; 031 (Gv) = K..
(ii) m3(K) ::; 1.
(iii) K* L~ ~ PGL3(2m) or L~'^0 (2m) and K = 031 (E(Gv)).We claim that Tv normalizes K; assume otherwise and let Ko := (KTv) and
To := Tv n Ko. In (iii), K :::) Gv; and in (i), Tv acts on K since Tv acts on
Lv. Therefore we may assume that (ii) holds. Comparing the groups in (a) or
(b) of 1:1.5.3 to those in 1.2.1.3, we conclude that K* ~ L 2 (2m) or Sz(2m). If
K* ~ L2(2m) then by 1.2.2.a, Lv ::; 031 (Gv) = Ko, so L~ ::; K 0. = K* and L~
centralizes a*, impossible as involutions of K* are centralized only by a Sylow 2-subgroup. Therefore K* ~ Sz(2m), so K 0 /Z(Ko) ~ Sz(2m) x Sz(2m). Let B
be the Borel subgroup of Ko containing T 0 , and set Vo := D1(To). Then (using
I.2.2.4 when Z(Ko) =I- 1 and in particular m = 3) J(Tv) centralizes V 0 , so byB.2.3.5, Baum(Tv) :::) B. Then B::; CMv(v) by 13.6.4 and 13.6.5. Now 02 (B)::;
02 ,^3 (CMv (v)) ::; CM(V). Hence [Lv, 02 (B)] ::; 02(Lv), impossible as as subgroupof order 3 in Lv induces field automorphisms on K. This contradiction completes
the proof of the claim that Tv normalizes K.
Hence Tv = Tx. Also L~ ::; K* in cases (i) and (ii); this is clear in (i), and it
holds in (ii) as Lv = [Lv, Tv] while Out(K) is abelian when m3(K) ::; 1.
Next as Tv acts on Lv, by inspection of the groups in (a) or (b) of 1.1.5.3, L~
is contained in a Borel subgroup B* of K* L~. Hence as
[L~, z*] =I- 1 = [L~, a*]
and E* ::; Z(T_X), we conclude K* ~ Sp 4 (2m), as otherwise Z(T_X n K*) =: R* is
a root subgroup of K*, so CB· (R*) =CB· (z*).
As m > 1, K is simple by I.1.3, so J(Tv) = JK x J(Tc), where TK := Tv n Kand Tc := CTJK). Thus ZJ := D1(Z(J(Tv))) = ZK x Zc, where ZK := Z(TK)
and Zc := D1(Z(J(Tc))). Then using 13.6.4,
S = Baum(T) = Baum(Tv) = TK x Sc,
where Sc := Baum(Tc). Therefore B ::; Na(S), so as before B ::; CMv(v) by
13.6.5. Let Be := 02 (CB(V)). Since IB : 02(B)I > 3 and B = Lv with ILv :