16.3. SHOWING L IS STANDARD IN G 1179as E is of order 4, so by hypothesis K is a component of the centralizer of an
involution e E NE(fi). Thus as e centralizes K, which is a full diagonal subgroup
of IiJi =I, e centralizes I, contrary to E faithful on I.
Therefore (t, I) is described in case (3) of 16.1.7. Then as K centralizes the
subgroup E of order 4 faithful on I, we conclude from 16.l.4 and 16.1.5 (applied
to I described in 16.1.7.3) that E ~ E4, and I/0 2 (J) ~ M 12 , J 2 , HS, or Ru. By
16.3.4, 02(I) # 1. We may assume that conclusion (2) fails, so I/0 2 (I) ~ HS,
with E = (s, t) a 4-group such that E(Cr(s)) ~ A 8 and E(Cr(t)) ~ A 6. But this
contradicts the hypothesis that K is a component of Cr(e) for each e EE#. D
LEMMA 16.3.6. Assume E :::; Tc is of order 4, with L a component of Ca(e)
for each e EE#. Then Lis a component of Ca(i) for each involution i E Ca(EL).
PROOF. Assume otherwise. Let i be a counterexample to the lemma, set Gi :=
Ca(i), and take E(i) :::; Ti E Syb(Na, (L)). As T E Syb(Na(L)) by Theorem
16.2.4, we may assume Ti:::; T, so that Ti= CT(i). Then i E CT(L) =Tc.
As L is a component of Ca ( e) for each e E E#, and we are assuming the lemma
fails, L <I:= (LE(G,)), where I, E, and Lare described in 16.3.5.2 with L, I in
the roles of "K, I". In particular 02(I) # 1. Set R := CT 0 (i); as Ti= CT(i),
R =CT, (L), so z ER. Also TL:::; Ti since i E Ca(L), so CTLTc(i) = TLR·
Let Ro := CR(I). By 16.3.5.2, E ~ E 4 is faithful on I and AutE(I) =
CAut(I)(AutL(I)), so R = RoE with En Ro = 1. Next R < Tc: for otherwise
TLTc :::; Ti, so that IT : Til :::; IT : TLTcl :::; IOut(L)l2 :::; 2 by inspection of the
cases in 16.3.5.2, contrary to 16.1.8 with z in the role of "t".
Now pick the counterexample i so that R is maximal. As R < Tc, there is
y E NTo (R) - R with y^2 E R. Suppose X := Ro n R'6 # l. Then as R normalizes
Ro and y normalizes R, R also normalizes Rg, and hence normalizes X. Thereforethere is an involution i 1 in X central in R(y), contrary to the maximality of R.
Therefore RonRg = 1, so Ro is isomorphic to a subgroup of R/ Ro = RoE /Ro ~
E 4 , and in particular ~(Ro)= 1. As 02 (J) # 1, from (5b) and (7b) of I.2.2, non-
2-central involutions of I/Z(I) lift to.4-elements of I, so either Cr(L) ~ Qs, or
I/0 2 (I) ~ M 12 and Cr(L) ~ Z 4. In either case there is e EE# inducing an innerautomorphism on I, so that e = r 0 f with ro E Ro and f E Cr(L); then f is of
order 4 with j2 E Z(I), so r 0 is also of order 4, contr~dicting ~(Ro)= 1. D
We are now ready to state the main result of this section:
THEOREM 16.3.7. Lis standard in G.
Until the proof of Theorem 16.3.7 is complete, assume Lis a counterexample.Thus by 16.3.2, there is an involution t E Ca(L) such that L is not a component
of Gt:= Ca(t). Recall Tc= CT(L) E Syb(Ca(L)), so we may assume t E Tc.
LEMMA 16.3.8. Tc is dihedral or semidihedral oforder at least 8. In particular
CT 0 (t) = (z, t)..
PROOF. As Lis not a component of Gt, t -=f. z, so ITcl > 2. Since T normalizesTc by Theorem 16.2.4, we may choose E :::; Tc of order 4 with E :'SI T, and
set S := CT 0 (E). Then IT : CT(e)I :::; 2 for each e E E#, and hence L is a
component of Ca(e) by 16.1.8. Hence Lis a component of Ca(s) for each s Es#