756 10. THE CASE LE .Cj(G, T) NOT NORMAL IN M ..W 0 ::::; 02 (H) by B.6.8.3d, so Wo = Wo(02(H), V) and then HS Nc(Wo) SM,
contradicting Hi M. Thus W # 1; since s(G, V) = 2, we must have a(H, I) 2:: 2
by E.3.18, contradicting a(H*, I) = 1. D
LEMMA 10.3.2. Cc(z) i M for z E z# n V.
PROOF. Assume that Cc(z) ::::; M. We first prove that V is a TI-subgroup
of G: For as Cv(L 0 ) = 1 by 10.3.1.1, each diagonal involution in V is conjugate
in Lo to z, and hence has centralizer contained in M by hypothesis. By 10.2.9.1,
centralizers of nondiagonal involutions are contained in M. Thus these involutions
are' not 2-central in G, so they are not fused in G to diagonal involutions, and hence
M controls fusion of involutions in V. Therefore Vis a TI-set in G by I.6.1.1.
As Vis a TI-subgroup of G, r(G, V) = m(V) = 6. Let A be aw-offender onV. By 10.2.13, w(G, V) = 2, so as m 2 (Autc(V)) = 4, m(A) = 4 by E.3.28.2. But
as Vis a TI-subgroup of G, I.6.2.2a says that Cv(a) = V n M^9 for each a EA#.
This is impossible as no rank-4 subgroup of M satisfies Cv(a) = Cv(A) for each
a E A#. This contradiction completes the proof. D
By 10.2.13 and 10.3.1.1, the hypotheses of 10.2.11 hold. So for the remainderof the section, we adopt the notation of that lemma; in particular, we study the
group Kz = (X^0 •).
LEMMA 10.3.3. KzT/02(KzT) ~ Ss wr Z2.
PROOF. We first observe that if Y E 1i(T) is generated by Ny(T) and a set
~ of minimal parabolics D such that n(D) = 1 for each D E ~' then Y ::::; M
by Theorem 3.3.1 and 10.2.13. In particular each solvable member H of H(T)
is contained in M by E.1.13 and B.6.5, since H = 0
21
(H)NH(T) by a Frattini
Argument.
Let J := G';'. By a Frattini Argument, Gz = JNc.(T n J), and as Gz/J
and NJ(T n J) are solvable, Na. (T n J) is a solvable member of 1i(T). Therefore
Ne. (Tn J) ::::; M by the previous paragraph, so J 1:. M by 10.3.2. Hence by 1.2.1.1
there is IE C(Gz) with I 1:. M.
Suppose I/02(I) is a Bender group. Then a Borel subgroup of Io:= (IT) lies
in M by the first paragraph, so IoT E H*(T, M). Hence by 10.2.13, n(I) = 2. Then
by 10.2.3.3, IoT/02(IoT) ~ Ss wr Z2 and X ::::; Io, so Io = Kz, and the lemma
holds.
Therefore we may assume I/0 2 (I) is not a Bender group. Suppose next that
X = Kz. As Gz is an SQTK-group, m 3 (IX)::::; 2, so I is a 3'-group. Thus I/0 2 (I)
is a Suzuki group and hence a Bender group, contrary to our assumption.
Thus X < Kz, so by 10.2.4, Kz = (JT), for IE C(Gz) with I 1:. M, and as
I/02(I) is not a Bender group, I/02(I) ~ L2(P) for an odd prime p > 5, L 4 (2),
or Ls(2). But then KzT is generated by NK.T(T) and minimal parabolics D with
n(D) = 1, contrary to an earlier remark. D
We are now in a position to obtain our final contradiction.
By 10.3.3, Kz = (KT) with KE £(G, T) and K/02(K) ~ A 5. In particular
X < Kz, so by 10.2.11.2, Vz E R2(Gz). Thus KE £1(G, T).
·Let M+ E M(KzT) and set Jz := (XM+). As X < Kz .S Jz, by 10.2.4,
Jz = (I'f') for Iz E C(M+)· Furthermore arguing as in the proof of 10.3.3, Jz is not
generated by minimal parabolics D with n(D) = 1, so from 10.2.4, Iz/0 2 (Iz) ~