562 2. CLASSIFYING THE GROUPS WITH IM(T)I =^1
Therefore if (i) < 02 (KiCs(i)), then it E t^02 (K;Cs(i)), contradicting it~ t^0. Thus
(i) = 02(KiCs(i)).
Suppose case (a:) or (/3) holds. If 02 (Ki) = 1 then (3) holds, and from the
structure of Aut(Ki), Ki is transitive on involutions in tKi, so tzK E tG, and hence
z = ZK by 2.5.13.3, establishing (4). Thus we may assume that 02(Ki) -:f. 1, so
(i) = 02 (Ki) from the previous paragraph. If (/3) holds, then from the embedding
of Kif ( i) in G 2 ( 4), t acts faithfully on some root subgroup Q / ( i), with Q ~ Qs,
so that ti E tQ, contrary to a remark in the previous paragraph. Thus (a:) holds,
with Kif ( i) ~ L 3 ( 4), so ( 3) holds in this case. Further the field automorphism t
normalizes each maximal parabolic P of Ki over Cs(i) n Ki· From the structure
of the covering group in I.2.2.3b, V := 02(P) is an indecomposable P-module such
that V/(i) is the natural module for P/V ~ L 2 (4). Nowt centralizes X of order 3
in P, and V = [V, X] x (i) with
It follows that tzK Eta, so z = ZK by 2.5.13.3, and hence (4) holds.
Thus it remains to consider the case where ("!) holds. If Ki ~ As, then the
lemma holds, since there we do not assert that tzK E tG. If Ki~ Ls(2) or Sp4(4),
then Ki is transitive on involutions in tKi, so that tzK E tG, and hence z = ZK by
2.5.13.3, so the lemma holds. Thus we have reduced to the case Kif0 2 (Ki) ~HS.
Assume first that Z(Ki) -:f. 1. Then as before, (i) = Z(Ki) = 02(KiCs(i)), so as
we are assuming (t, i) = R and t is inner on Ki in ("!), t E KiCK;Cs(i) (Ki) =Ki.
Thus t E CK; (K) sot is not 2-central in Ki· However, an element of the covering
group Ki projecting on a non-2-central involution of HS is of order 4 by I.2.2.5b.
This contradiction shows that Ki is HS, so that (3) holds. Furthermore if u is the
projection on Ki oft, then UZK E uK; and iuzK E (iu)Ki. Therefore as t = u or
iu, tzK E tK;, and again ( 4) follows from 2.5.13.3. This completes the proof of the
reduction of the proof of the lemma to the proof of (2).
We have shown that it suffices to prove that R = (i, t). Thus we assume that
(i, t) < R, and derive a contradiction. Choose i so that R = Cs(i) n Cs(K) is
maximal subject to Knot being a component of Gi· Further if i E Z(Cs(K)) then
R = Cs(K), and we choose i so that Cs(i) is maximal subject to the constraint
that R = Cs(K).
Recall we showed soon after stating (i) that that assumption implies IR: Roi =
2. Inspecting the groups in cases (o:)-('y) of (ii), we check that either IR: Roi = 2, or
KifZ(Ki) ~HS and IR: Roi= 4. When IR: Roi= 2 we set R2 :=Ro, and when
IR: Roi= 4 we let R2 be the subgroup R 1 of index 2 in R with CK;(R 1 )^00 ~As
discussed earlier. Thus in either case, i E Ro :::; R 2 and IR: R21 = 2.
We next claim that K <Ko and i E Z(Ns(K)). Thus we assume that at least
one of the two assertions of the claim fails, and derive a contradiction. As i ~ Z(S)
there is 8 E Ns(Cs(i))-Cs(i) with 82 E Cs(i). Furthermore we observe when K <
Ko that Cs(i) normalizes K: For otherwise i centralizes some u E Cs(i) - Ns(K)
and K+ -:f. K+.. But in all cases appearing in (i) and (ii), m 3 (K+) = 2; therefore
as K+ and K+. are products of components of Gi, m 3 (K+K+.) > 2, impossible as