1152 15. THE CASE .Cf(G, T) = 0
Assume HE 'H*(T, M) with n(H) > 1. Then in view of 15.3.2.6, His described
in E.2.2. In particular Ko := 02 (H) = (KT) for some K E C(H), Ko/02(Ko) is
of Lie type over F 2 n for some n > 1, and setting Mo := Mn Ko, Mo is a Borel
subgroup of K 0. As MH is a 3'-group by 15.3.49.4, n is odd.
By A.1.42.2, we may pick IE Irr +(Ko, UH, T); set Ir:= (IT). We apply parts
(4) and (5) of F.9.18 to the list of possibilities in E.2.2 defined over F 2 n with n
odd. In view of 15.3.64, we may also appeal to Theorems B.4.2 and B.4.5; this
determines the modules from the restrictions given in F.9.18. In particular as n
is odd, there is no orthogonal module for L 2 (2n). We conclude that one of the
following holds:
(i) K/02(K) is a Bender group, and I /01(K) is the natural module for K/02(K).
Further either K =Ko and I= Ir; or K < K 0 , K/02(K) ~ L2(2n) or Sz(2n),
and fort ET-Nr(K), Ir= I+ It and [I, Kt]= 0.
(ii) K/0 2 (K) ~ SL 3 (2n) or Sp 4 (2n), Tis nontrivial on the Dynkin diagram of
K/02(K), and Ir/01T(K) is the sum of a natural module and its conjugate by an
outer automorphism nontrivial on the diagram.
(iii) Ko/0 2 (K 0 ) ~ D;t(2n), and Ir is the orthogonal module.
Now by Theorems B.5.1 and B.4.2, K 0 T/0 2 (K 0 T) has no FF-modules, ex-
cept in (i) with K/0 2 (K) ~ L 2 (2n), where K 0 T/0 2 (KoT) has no strong FF-
modules. We conclude from F.9.18.6 that either Ir= [UH, Ko], or case (i) holds
with K / 02 ( K) ~ L 2 ( 2n), and [UH, Kl/ I is an extension of the natural module for
K/02(K) over a submodule centralized by K. (Recall that n > 1 is odd).
As T centralizes Zs and H = KoT, UH = [UH, Ko]Ou-H(H) by B.2.14. By
15.3.49.1, 02 (M 0 ) centralizes V, and hence Mo centralizes Zs. It follows from the
structure of the modules described in (i)-(iii), that H centralizes Zs. But then K
centralizes Zs by Coprime Action, and so K centralizes UH, contrary to K* =f. 1.
This contradiction completes the proof of 15.3.65. DAs r(G, V) = 2 by 15.3.65.2, there is E4 ~ES V with and GE:= Na(E) 1.
M. Further E is totally singular by 15.3.46.2. Pick E so that TE := Nr(E) E
Syl2(ME), where ME := NM(E). Let YE := 02 (Ny(E)), QE := 02(GE), and
VE:= (VGE).
LEMMA 15.3.66. (1) 'i'E = 'i' n n:t(v).
(2) IT: TEI = 2.
(3) TE is the 4-subgroup of T distinct from S.
(4) ZS E.(5) YETE/02(YETE) ~ 831 V = [V, YE], QE S Oa(E), and 02(YETE)
OyETE(E).
(6) GE= YETEOa(E).
('l) Oa(E) S Mc.
PROOF. As E is a totally singular line in V, AutM(E) = GL(E), so that
QE S Oa(E) and (1) and (5) hold. Then (1) implies (2)-(4), and as YETE induces
GL(E) on E, (6) holds. Finally Oa(E) S Oa(Z) =Mc by (4) and 15.3.4. D
LEMMA 15.3.67. (1) R := Or(V) = Oa(V) and M =YT.
(2) TEE Sy[z(GE) and BE:= Baum(TE) SR, so that O(G,BE) SM.
(3) GE = YEXETE, where XE := 02 (0a(E)) i. M, with XE/02(XE) ~