14.2. STARTING THE L 2 (2) CASE OF Cf EMPTY 981applying 14.1.17 to the preimage Yo in M of O(M), we conclude Y = Yo and
Cy(V) :S: O(Z(CM(V) R~)). By the first remark, M = YT, so (4) holds by
A.5.7.1. By (2) there is r E Re inverting Y, so as [Cy(V)*, R~] = 1, r inverts y of
order 3 in Y - Cy (V), and Y = [Y, R~] = (y). Therefore Y ~ Z 3 , completing
the proof of (3). As [CM(V), YRc] :S: 02(M), the first two statements in (5) hold,
while the third follows from 14.1.17 .4.
Let KE M(T). By (1) and A.5.3.1, V(K) ~ V; so as [VI = 4, it follows thatV(K) = Z or V. In the latter case K = M by A.5.4; in the former, K :S: Ca(Z) :S:
Mc so that K =Mc, completing the proof of (6) and (7). D
14.2. Starting the L 2 (2) case of Cr empty
We now state Hypothesis 14.2.1, which in effect is the special case of Hypoth-
esis 14.1.5 where V(Mt) is of of rank 2, for Mt the member of M(T) defined in
14.1.12. Namely Hypothesis 14.2.1 implies Hypothesis 14.1.5, and conversely whenHypothesis 14.1.5 holds and V(Mt) is of rank 2, then Hypothesis 14.2.l is satisfied
with Mt in the role of "M" by 14.1.18. Indeed 14.1.18 supplies a normal subgroup
Y of M with YT/02(YT) ~ £ 2 (2) and M =!M(YT). Thus we view Y as a solv-
able analogue of L E Cj(L, T), and then Hypothesis 14.2.l allows us to treat the
case LT/0 2 (LT) ~ £ 2 (2) .in parallel with the final case in the Fundamental Setup
where L/02(£) ~ £3(2).
Thus in this section, and. as appropriate in the later sections of this chapter,
we assume:HYPOTHESIS 14.2.1. G is a simple QTKE-group, TE Syb(G), Z := fh(Z(T)),
and(1) Lt(G, T) = 0.
(2) Mc:= Ca(Z) E M(T).
(3) There exists a unique maximal member M of M(T) under :S.
(4) V := V(M) = (ZM) is of rank 2, and M := M/CM(V) ~ Aut(V) ~ £2(2).
(5) IM(T)I > 1.
We observe that by parts (1), (2), and (5) of Hypothesis 14.2.1, Hypothesis
14.1.5 is satisfied. Indeed by 14.2.1.3 and 14.1.12.1, M is the maximal 2-local Mt
containing Na(C 2 (T)) appearing in 14.1.12. Then by 14.2.1.4, the hypotheses of
14.1.18 are satisfied. As in 14.1.18, we set
Re:= 02(MnMc) and Y := 02 ((R;;r)).
Then applying 14.1.18 we conclude:
LEMMA 14.2.2. (1) T = Cr(V)Rc, and Mn Mc= CM(V)Rc so that 02 (M n
Mc) :S: CM(V).
(2) Y = 02 (M) ~ Z3 and 02(Y) = Cy(V).
(3) M = !M(YT). ,
(4) M/02(M) = YRc/02(M) x CM(V)/02(M) with YRc/02(M) ~ £2(2)
and m3(CM(V)) :S: 1.(5) M(T) = {M,Mc}·
(6) IZI = 2, and hence Cr(Y) = 1.
(1) Na(T) :S: Mn Mc.