1110 i5. THE CASE .Cf(G, T) = 0
M = !M(YiT) by 15.2.6.3, so that Oz((YiT,H)) = 1, and hence part (e) holds. To
verify part (d), we must show that H n M normalizes YiRc. By 15.1.9.3, H::::; Mc,
so H n M acts on Re. By 15.1.9.4, oz(H n M)::::; CM(V), so oz(H n M) acts on
Yi by 15.2. 7.2.
Hence a:= (Yi(H n M), H n M, H) is a weak BN-pair of rank 2 by F.1.9, and
since Ny 1 R,JT) = T, a appears in the list ofF.1.12. Now UH is abelian by 15.1.11.2,
and H has two noncentral 2-chief factors on UH by 15.1.12.1, which are natural
modules for H/Oz(H) ~Sn for n = 3 or 5 by 15.1.12.3. But these conditions are
not satisfied by any member of F.1.12. D
We now define notation which will be in force for the remainder of the section:NOTATION 15.2.9. Pick HE H*(T, M), and let QH := Oz(H), UH := (VH),
and H* := H/Oz(H). Recall in particular by 15.1.9.3 that
H:::;Mc.
By 15.1.12.1, H has exactly two noncentral chief factors Ui and Uz on UH. By
15.2.8, H ~ 83 wr Zz. Thus by 15.1.12.4, m(Ui) = 4 and H = Ot(Ui), so
Ui = Ui,i E9 Ui,z with Ui,j ~ E4, j = 1, 2, the two definite 2-dimensional subspaces
of the othogonal space Ui. Also H = (Hi x H2)(t), where t* is an involution
with Hf= Hz and Hi~ 83. This choice for Hi and Hz is not unique, but 15.1.12
supplies us with a distinguished choice: Pick Hi := CH(Ui, 3 -i)· In particular the
subgroups Hi, i = 1, 2 contain the transvections in H* on Ui. Let Ki := Oz(Hi)
and K := oz(H).
Next let L\ consist of those A E A(H) such that A* is minimal subject to
A i QH. By 15.1.12.2, for each A E L\, A is an FF-offender on Ui and Uz.
From B.2.9.l and the description of FF-offenders in B.1.8.4, A is of order 2 by
minimality of A, so A induces transvections on both Ui and Uz. Thus A lies in
either Hi or Hz, and we can choose notation so that also Hi= CH(Uz, 3 -i)· Then
Uj,i = [Uj, Hi] and UJ,i = Uj,Z·.
For A EL\, let B(A) := AnQH; thus IA: B(A)I = IA*I = 2. Let :E := {B(A):
A EL}.
Observe by 15.2.8 that T =Mn H = NH(V), so IVHI = 9. For h EH, let
L(Vh) := L\ n Th, L\'(Vh) := L\ - L(Vh); :E(Vh) := {B(A) : A E L(Vh)}, and
:E'(Vh) := :E - :E(Vh).
LEMMA 15.2.10. Let fJ E Q(f', V). Then
' (1) fJ ::::; Mo and m(D) = 1.
(2) m([V, DJ) = 2.(3) [V,DJ ::::! T.
PROOF. By 15.2.1, Q(T, V) ~ !li(T) ::::; M 0. Then the lemma follows easily
from 15.2.1: For example (3) follows as T is abelian, and in case (3) of 15.2.1,
m(D) = 1 since fJ acts quadratically on V. D
LEMMA 15.2.11. Let BE :E'(V) and Zs:= [V,B]. Then
(1) BE Q(M, V).