1144 15. THE CASE .Cr(G, T) =^0Y 0 • Thus L :::; Nc(Yo) = M, contrary to (q) since we just saw 03(L) =J Y 0 .
This contradiction shows that JH is the product of members of C(JH)· Similarly
L = [L,Y 0 ] for each LE C(JH), completing the proof of (i).
Applying (i) to any overgroup of YoT in H we conclude
(j) Each solvable overgroup of YoT in His contained in MH.Pick L E C(JH) and let L 0 := (LT) and Uo := [UH, La]. Then LoYoT E
'H(Y 0 T, M), so replacing H by L 0 Y 0 T, we may assume H = LoYaT. As Zs :S Z(T)
by 15.3.48, UH= UoCuH (H) by B.2.14. Let Z 0 be the projection of Zs on Uo with
respect to this decomposition; thus CH· (Zo) :::; NH( Zs)* :::; M'ff by 15.3.46.5.
By Theorems B.5.1 and B.5.6, L* is A 7 , A. 6 , Ln(2) for n := 4 or 5, or a group
of Lie type of Lie rank 1 or 2 over some F 2 e. Set T 0 := T n La. When L* is of Lie
type, let B denote the Borel subgroup of Lo containing To.
Assume first that Yo i L 0 • Then L* is not A7 or A.6 by A.3.18. Further
T 0 = Y 0 T n Lo is Y 0 T-invariant, so YoT acts on B. Then B :::; MH by (j), so as we
are assuming Yo i L 0 , we conclude from (b) that Bis a 3'-group acting on Ya. As
L 0 = [Lo, Yo] by (i), we conclude from the structure of Aut(L*) for L* of Lie type
that B = T 0 , and so Lis defined over F 2. Then Out(L 0 ) is a 3'-group from the list
of possibilities in Theorem B.4.2, so Yo induces inner automorphisms on L 0 , andthis time we obtain a contradiction from (j) and (b) since the projection of Y 0 * on
L 0 is YoT-invariant and nontrivial. Thus we have shown:
(k) Yo:::; Lo.Suppose that L* is of Lie type, and defined over F 2 e with e > 1; then from
Theorem B.4.2, L* is L2(2e), SL3(2e), Sp4(2e), or G2(2e). Further as To acts on
Yo, Yo is contained in B, and e is even. Then by. (b), e ( N Le; (Yo)) = YQ* is of 3-
rank 1, so we conclude L* = L 0 ~ L 2 (2e). As v,; is an FF*-offender contained in02(Y 0 T), we conclude from Theorem B.4.2 that Uo/Co 0 (L) is the natural module
for L*. But then Zo :S Co 0 (Y 0 ):::; Co 0 (L), contrary to UH= (Zff). Therefore L*
is not of Lie type of F 2 e withe> 1.
Applying (j) and (b) as at the end of the proof of (k), we conclude that L =Loif L* ~ L 3 (2); so using 1.2.1.3, we have reduced to:
(1) Lo= L, L* is Ln(2), 3:::; n:::; 5, A5, A 6 , A7, or G 2 (2)', and either YQ*T 0 is
a minimal parabolic of L* of Lie type, or L* is A 7 or A. 6.
(m) VH > UHV·For suppose that VH = UHV; then because [UH, QH]:::; Vi by 15.3.48.2, [VH, QH] =
[UH, QH][V, QH] :S Vi V = V. Further [V, QH] =J 1 by (2), so Z(T) :S QH as Z(f') is
oforder 2. Thus Zs:::; [V, QH], and hence Zs:::; [VH, QH]:::; V. Therefore [VH, QH]is not totally singular in V, so H:::; Na([VH, QH]) :::; M by 15.3.46.2, contrary to
HE 'H(T,M).
( n) v; is a strong FF* -offender on fj H.Suppose otherwise. By the choice of"(, m(U;) ;:::: m(UH/DH), and u'Y :::; v'Y <
CH (DH) by ( f), so as v; is not a strong offender, we conclude that fJ H = Cu H (V'Y),
u; = V,_;, and m(U;) = m(UH/DH). By the last equality we have symmetry
between "! and "(1 (as discussed in the second paragraph of Remark F .9 .17) so alsoVH = UHCvH(U'Y/Z'Y) by that symmetry. Further CvH(U'Y/Z'Y) = CvH(U'Y) by
(f), so Uy centralizes VH/UH. Hence as L = [L, U'Y], L centralizes VH/UH, so as