748 10. THE CASE LE .Cj(G, T) NOT NORMAL IN M.
It will suffice to show Na(T 1 ) acts on Xi for i = 1and2, since then Na(T1) acts on
(X 1 , X 2 ) = L 0 , so Na(T 1 ) :::; Na(L 0 ) = M, as desired. Therefore we may assume
Na(T 1 ) i Na(Xi) for some i, and we now fix that value of i.
Set K 1 := (Xfj) and KJT := K1T/02(K1) for each j. Notice 02(K1) :::;
02(H1) :::; T 1 using A.1.6. Now Na(T1) :::; G1 so Na(T1) acts on K1. Thus as
Na(T1) i Na(Xi), Xi < Ki, and hence by 10.2.4 either Ki = Ki,1Kf, 1 with
Ki,1 E C(Gi) ands ET - Nr(Ki,1), or KiT ~ Aut(Lk(2)), k := 4 or 5. In either
case, Na(T 1 ) acts on R := T 1 n Ki and 02(Xi(T n Ki)) :::; R.
Suppose first that either Ki ~ Lk(2), or Ki, 1 ~ L2(2m). As Hin Ki is T-
invariant, Hin Ki :::; Ji, where Jt is a T-invariant parabolic of Ki such that Xi is
the characteristic subgroup generated by the elements of order 3 in Ji. Notice
Now when Ki 1 ~ L 2 (2m), TnKi = 02 (Ji), so the inequalities in(*) are equalities,
and then Na(T 1 ) :::; Na~ (R) :::; Na; (Ji) :::; Na(Xi), contrary to our assumption. On
the other hand if Ki ~ Lk(2), then 02 (Jt) is a unipotent radical, and so by I.2.5
is weakly closed in (T n Ki)* with respect to G1; thus Na(T1):::; Na;(02(Ji))):::;
Na(Xi), for the same contradiction.
This leaves the case where Ki 1 ~ L2 (p). If p = ±3 mod 8, then again TnKi =
02 (Xi(T n Ki)) :::; R, so R = T n Ki; and Na(T1) normalizes NK;(T n Ki) =
Xi(T n Ki) and hence also 02 (Xi(T n Ki)) = Xi, for our usual contradiction.
Therefore p = ±1 mod 8, and (T n Ki,1)* is a nonabelian dihedral 2-group. Since
(Xi n Ki,1)* is a T n Ki-invariant A4-subgroup of Ki, 1 , l(T n Ki,1)*1=8.Next R is of index r :::; IT : Til = 2 in T n Ki. Further if r = 2, then
02(Xi(T n Ki)) = J(R), so Na(T1) :::; Na; (02(Xi(T n Ki)) :::; Na(Xi), again
contrary to assumption. Therefore R = T n Ki and there are exactly two subgroups
Y of Ki,1 with Rn Ki,1 :::; Y and Y* ~ 84. So 02 (Na(T1)) acts on both such
subgroups, and in particular on Xi nKi,1· Similarly 02 (Na(T1)) acts on Xi nKt, 1 ,
and hence on the product Xi of these two subgroups, so Na(T1) = T0^2 (Na(T1)):::;
Na(Xi), for our final contradiction. D
LEMMA 10.2.6. (1) M = !M(LoT1).(2) Na(l'i):::; M 2: Na(L).
PROOF. Notice (1) implies (2), so it suffices to prove (1). Suppose that there is
HE M(LoT1) - {M}. Then IT: Tll = 2, and Na(T1):::; M by 10.2.5. By 1.2.7.3,
M = !M(LoT+) for each T+ E Syb(M), so that Ti E Syb(H). Thus by 1.2.4,
Li :::; Ki E C(H), and Ki :::) H by ( +) in 1.2.4. Now from A.3.12, Ki does not
contain Lo = LiL2, so Ki =I K2· Thus as mp(H) :::; 2 for each prime divisor p of
ILi, while L3-i :::; CH(Ki/02(Ki)), we conclude mp(Ki) = 1 for each such prime.
As H =IM= Na(Lo), Lo is not normal in H, so Li <Ki for i := 1 or 2; we fix
this value of i.
Now if L ~ Sz(2n), A.3.12 says Li is properly contained in no Ki withmv(Ki) = 1 for each prime jJ dividing 2n - 1, and similarly Li is proper in no
Ki with m1(Ki) = 1 = m3(Ki) when L ~ L 3 (2). Therefore L ~ L2(2n).
Assume F*(Ki) = 02(Ki)· Set Ho := KiL3-iT1 and R := 02(LoT). Then
LoT1 :::; Mo := Mn Ho. As M =!M(LoT), C(Ho, R) :::; Mo, and by A.4.2.7,
RE B 2 (H 0 ) and RE Syb((RM^0 )). Thus Hypothesis C.2.3 is satisfied, so Ki is