1.3. THE SET 8*(G, T) OF SOLVABLE UNIQUENESS SUBGROUPS OF G 513Next we obtain some restrictions on L. If R :':! LT then any 1 =/= C char R is
normal in LT, so as M = !M(LT) by 1.2.7.3, we conclude Na(C):::; M, establishing
our claim (*). Thus we may assume R is not normal in LT.Suppose next that p is the only odd prime in ?r(0 2 ,F(L)); in particular this
holds in case (c) of 1.2.1.4. Then X = 02 (0 2 ,F(L)), so as R centralizes X/0 2 (X),
[R, L]:::; CL(X/02(X)):::; 02,F(L):::; XR
and hence RX :':! RL. But then R = 02 (RX) :':! LT, contrary to our assumption.
Thus we may assume L is in case ( d) of 1.2.1.4, and hence r = 5 with either p = 5
or p = ±1 mod 5. Further we have shown there is an odd prime q E ?r(0 2 ,F(L))
with p =/= q. Notice that q 2'.'. 5.
For 1 =/= C char R, let Le:= NL(C)^00 , and set Xq := 'Bq(L). Notice Y:::; Le
and 02(Xq) :':! XT, so 02(Xq) :::; R. Therefore RE Syl2(XqR). Then as q 2'.'. 5,
by Solvable Thompson Factorization B.2.16,. XqR = NxqR(J(R))CxqR(D1(Z(R))).
So for Co := J(R) or D1(Z(R)), NxqR(Co) i 02,(Xq)· Therefore as Y is ir-
reducible on Xq/02,(Xq), we conclude Xq :::; Na( Co), so Xq = [Xq, Y] :::; Lc 0 •
Hence ?r(0 2 ,F(Lo 0 )) contains at least two odd primes p and q.
We are now in a position to complete the proof of the claim. Assume (*)
fails. Then there is 1 =/= C char R with N :=· Na(C) i M. As YT :::; Na(R),
NE 7-i(YT); in particular YE L(N,T) as we saw YE L(G,T). So we may apply
1.2.4 to embed Y :::; Yo E C(N), with the inclusion described in A.3.12. Notice inparticular that Yo :':! N, by 1.2.2.b, since Y contains X of p-rank 2. Also Y :::; Le
by the previous paragraph, so Le= [Le, Y]:::; Yo as Le E L(G, T).
Now if X :::;! Yo, then X char Yo using 1.2.1.4, and hence N :::; Na(X) = M,
contrary to our choice of Ni M. Thus we may assume Xis not normal in Yo. As
X :::;! Y, it follows that Y <Ya. In addition the fact that Xis not normal in Yo
means X i:. 000 (Yc), which rules out cases (21) and (22) of A.3.12, leaving only
case (10) of A.3.12 with Yo/02(Yo) ~ L3(p). In particular, p = r = 5, Y =Le,and ?r(0 2 ,F(Lo)) = {2,p}. But we saw earlier that n(02,F(Lo 0 )) contains at least
two odd primes, so we conclude that our counterexample C cannot be the subgroup
Co constructed earlier. That is, Na(R) :::; Na(Co) :::; M.
Let (YcR) := YcR/0 2 (YcR). Then P is the unipotent radical of a maximal
parabolic of Y 0 ~ L 3 (5), so Vly 0 R (P, 2) = 1, giving R* = 1 and hence R :::;
02 (YcR). On the other hand 02(YoR) :::; 02(XT) = R, so R = 02(YcR). But
then Yo :::; Na(R) :::; M, impossible as X is normal in M, but not in Ye. This
establishes (*); namely C(G, R) :::; M.
We now use () and results on pushing up from section 0.2 to complete the proof
of 1.3.8: Assume X tj. S(G, T). Then XT < (K, T·) =: H for some KE L(G, T),
with K/0 2 (K) quasisimple, and H is described in 1.3.4. Asp > 3, H does not
satisfy 1.3.4.4. Now AutTnL(P) is quaternion in cases (c) and (d) of 1.2.1.4, so
AutT(P) is not cyclic and hence 1.3.4.3 does not hold. Thus we have reduced to
cases (1) and (2) of 1.3.4. As Xis not normal in H but X :::;! M, Ki M. Further
from 1.3.4, R acts on Kin each case.
We observe next that the property RE E 2 (H) from Definition 0.1.1 and Hy-
pothesis 0.2.3 of Volume I hold for MH := H n M: Namely C(H, R) :::; MH using
(*);and then by A.4.2.7, R = 02 (NH(R)) and RE Syl 2 ((RMH)).