1018 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTYSince V plays the role of "V+" in 14.5.14.2 in the notation of section F.9,
UH = (VH} =: VH, and hence DH = EH. These identifications simplify the
statements of various results in section F .9. In particular:
LEMMA 14.5.17. DH< UH·
PROOF. By F.9.13.5, Vi_ DH, so the remark follows as V $ VH =UH. D
LEMMA 14.5.18 .. {1) If Uy = D-. 0 then UH induces a nontrivial group of
transvections with center Vi on U 'Y •
{2) If m(U;) 2 m(UH/DH), then u; =f. 1 and u; E Q(H*, UH)· In case
2m(U;) = m(UH/CoH(U;)),
then also m(U;) = m(UH/DH), and u; acts faithfully on DH as a group of
transvections with center A 1.(3) q(H*, UH)$ 2.
(4) If we can choose 'Y with D'Y < U'Y, then we can choose 'Y with
0 < m(U;) 2: m(UH/DH),
in which case u; E Q(H*, UH).
{5) Leth E H with 'Yo = "(2h and set a:= "(h. Then Ua $ R1 and if D'Y < U'Y
then U~ E Q(H*,UH)·PROOF. Part (3) holds by F.9.16.3, while (1), (2), and (4) follow from 14.5.17
and the corresponding parts of F.9.16. Assume the hypotheses of (5). By parts
(1) and (2) of F.9.13, Ua $ R 1 , and if u; =f. 1, then since we can choose 'Y so that
u; E Q(H*, UH) in (4), also U~ E Q(H*, UH), completing the proofof (5). 0LEMMA 14.5.19. If K E C(H) then K i. M, so K 0 L1T E 1iz, where Ko :=
(KT}.
PROOF. This follows from 13.3.8.2 applied to L, Ko in the roles of "K, Y". D
LEMMA 14.5.20. Assume Y :SJ H with Y /0 2 (Y) a p-group of exponent p. Then
either
{1) Y n M = 02(Y), or
(2) p = 3, L/02(L) ~ L3(2), L1 $ Y, and one of the following holds:
{i) L1 = Y :SJ H.
{ii) Y/02(Y) ~ 3i+^2 , L1 = 02 (02,z(Y)) = 02 (Y n M), and T is irre-ducible on Y / L1 02 (Y).
{iii) Y/02(Y) ~ Eg and there exists Yo $ H such that L 1 $Yo :SJ Y 0 T
with Yo/02(Yo) ~ Zg and Yo i. M.PROOF. We may assume that (1) fails, so that YM := 02 (Y n M) =f. 1.
Let X be the set of T-invariant subgroups X of H such that 1 =f. X = 02 (X) $
CM(L/02(L)). Then using the T-invariance of X, L normalizes 02 (X0 2 (L)) = X,
so NG(X) $ M = !M(LT). In particular as Hi. M:
For each X EX, NG(X) $ M, so X is not normal in H. (!)