16.4. INTERSECTIONS OF NG(L) WITH CONJUGATES OF CG(L) 1193particular Tc does not centralize R. For 1 "I= r E R, observe by 16.4.2.5 that
Cr 0 (r) :::; Nr 0 (K') = Cr 0 (R) = S, so that Cr 0 (r) = S. By parts (4) and (6) of
I.7.7, S :::;J Tc and R is abelian. ·
Set Ro := Rn LK, U := fh(R), and recall K' E .6. 0 so that U :::; R 0. In
particular Ro "I= 1. For r E Ro, we can write r = s(r)l(r) with s(r) E Tc and
l(r) EL. Observe that l(r) is determined up to an element of L n K = Z(L), and
then s(r) is determined by l(r). Also s(r)Z(L) ~ Cr 0 (s(r)) :::; Cr 0 (r). Then as
Cr 0 (r) = S < Tc, s(r)Z(L) ~ S, but s(r) ~ Z(Tc). Indeed s(r) := s(r)Z(L)
is a uniquely determined element of S := S/Z(L), and s : Ro -+ S is a group
homomorphism. Also Rn L = 1 as Cr 0 (r) = S < Tc for r E R#, so ker(s) =Ron Z(L) = 1 and hences is injective. For R 1 :::; R 0 , let s(R1) be the preimage in
S of s(R 1 ). As s(r) tj. Z(Tc) for each r ER#, s(U) n Z(Tc) = 1.
As R is abelian, U is elementary abelian. Suppose s(U) is elementary abelian.Then since s(U)/Z(L) ~ s(U) ~ U ~ fh(S) ::=: s(U), we conclude that Z(L) = 1
and the map s : U -+ D 1 (S) is an isomorphism. So as S :::;J Tc, we have a
contradiction to s(U) n Z(Tc) = 1.
Thus for some u E U#, we may choose s( u), and hence also l ( u), of order at
least 4. Thus 1 "I= l(u)^2 E Z(L), so Z(L) "I= 1. Therefore L* is not L 3 (4), since
Z(L) is elementary abelian, and from I.2.2.3b, involutions in L* lift to involutions
in L. Hence by inspection of the remaining groups in 16.1.2.1, Out(L) has Sylow2-groups of order at most 2, so IR : Ro I :::; 2. Therefore
IRI = ISi:::: ls(Ro)llZ(L)I = IRollZ(L)I:::: IRllZ(L)l/2.
So as IZ(L)I ::=: 2, all inequalities are equalities, and hence S = s(Ro) with Z(L) =
(l(u)2) oforder 2. Thus there is 1 "I= v EU with E := (s(v))Z(L) a normal subgroup
of Tc of order 4. Then ITc: Cr 0 (s(v))I = ITc: Cr 0 (v)I = 2, so S = Cr 0 (s(v)) is
of index 2 in Tc. Thus RS :::'.] RTc, so as Nr 0 (R) = S ~ R is abelian, while Risa
TI-set in TcR under Na(TcR) by I.7.2.3, SR= Rx Rt fort E Tc - S. ThereforeD 1 (S) = Cuut(t) = [U, t], and hence s(w) E Z(Tc) for each w E U, contrary to
s(U) n Z(Tc) = 1. This contradiction completes the proof. D
The next lemma summarizes some fundamental properties of members of .6.o;in particular it shows that .6. 0 defines a symmetric relation on K^0.
LEMMA 16.4.11. (1) R centralizes Tc~ R. In particular, Tc:::; H'.
(2) KE .6.o(K').
(3) If we choose g as in 16.4.2.4, then R =Tl;.
PROOF. By 16.4.10, RE Syb(K'), so that R ~Tc. Then since R ~ Cr 0 (R)
by 16.4.2.1, Tc centralizes R, and so (1) holds using 16.4.2.5.
By 16.4.3.1, K E .6.(K'). Suppose that K ~ .6.o(K'). Then by 16.4.9.1, ITcl =
2, and by 16.4.9.1, z induces an outer automorphism on L'. Applying 16.4.8 with
the roles of K and K' reversed, we conclude that Lz := 02 (Cu(z)) :::; L. As
Ca(r) :::; Na(L'), we conclude that Lz :::;J Lr. Comparing the fixed points of outer
·and inner automorphisms of order. 2 in 16.1.4 and 16.1.5, we conclude L* ~ M12
and L; = L~ ~ A 5. As r induces an inner automorphism on L, if Z(L) i= 1, then
I.2.2.5b says that the projection TL of r on Lis of order 4, so r = rcrL with re E Tc
of order 4, contradicting ITcl = 2. Thus Z(L) = 1, so CL(r) ~ Z2 x S5. HoweverCL(r) :::; CH1(z) by 16.4.2.5, and as z induces an outer automorphism on L', no
element of CH1(z) induces an outer automorphism on Lz. Therefore (2) holds.