14.5. STARTING THE CASE (vG1) ABELIAN FOR L 3 (2) AND L 2 (2) 1015(VJ/^1 ) is abelian, so we have symmetry between LT, V and H =KT, VH; that is,
Hypothesis 14.5.1 holds with H, VH in the roles of "LT, V".
Now Y 1:. M = !M(LT), so that 02( (LT, H)) = 1. Hence Hypotheses F.7.1
and F.7.6 are satisfied with LT and H in the roles of "G 1 " and "G 2 ", so we can
form the coset geometry r of Definition F.7.2 with respect to this pair. Similarly
we can form the dual geometry r' where the roles of LT and H are reversed. Let
'Yo :=LT, 'Y1 := H, and for g, h E (LT, H) let V 709 :=VB and V 71 h := V]I. Also for
O" EI' let Qa := 02(Ga)· Observe G 70 m = LTnH = XT and kerxr(Gi) = 02 (Gi)
for i = 1, 2, is the centralizer in Gi of V or VH, respectively, so
Qa = G~l) = CaJVa)·
Next as usual choose a geodesica := ao, ... , ab =: /3
in r of minimal length b, subject to Va 1:. Qf3. Then b = min{b(r, V), b(I'', VH )},
so by F.7.9.1, Va :S G(3 and Vf3 :S Ga, and hence
1 I [Va, V(3]::::: Van Vf3. (*)Thus by 14.5.2 and the corresponding result for VH, /3. is not conjugate to a, sob is
odd. Replacing r by r' if necessary, we may assume Va = V, and we may assume
z E V n Vf3 by transitivity of L on V#. As H is also transitive on Vff, Vf3 = VJ} for
some g E G1 by A.1.7.1, so
w(3a1) = wff 1) = ( wr) 01) = w 2 a1) = wa1)since V = (Vf^1 ). Then as (V^01 ) is abelian, Vf3 centralizes V = Va, contrary to
(*). D
LEMMA 14.5.9. [VH, J(T)] = 1 and J(T) :s:] H.
PROOF. If J(T) centralizes VH, then J(T) = J(QH) by 14.5.7.2 and B.2.3.5,
so the lemma holds. Thus we assume [VH, J(T)] I 1, and derive a contradiction.
By 14.5.8, we may apply Solvable Thompson Factorization B.2.16 to conclude that
J(H) =Ki x · · · x K;, with Kt~ 83 and Vi:= [VH,Ki] ~ E4. Notices :S 2 by
A.1.31.1. As X = [X, T] either X = 02 (Ki) for some i, or [X, J(H)] = 1. The
same holds for Y as Y = [Y, T]. Thus if X = 02 (Ki), then Y normalizes X,
contrary to 14.5.6. Therefore X centralizes J(H)*, so that J(H) n X ::=:;; 02 (X).
Similarly J(H) n Y :S 02(Y). Then we may apply F.6.8 to J(H), to conclude that
J(T) :ST n J(H) :S QH :S CH(VH), contrary to our assumption. D
LEMMA 14.5.10. J(T) = J(02(XT)) 1:. 02 (LT) and X = [X, Jl(T)].
PROOF. By 14.5.9 and 14.5.7.2, J(T) :S Cr(VH) = QH :S 02 (XT), so J(T) =
J(02(XT)) by B.2.3.3. If J(T) 1:. 02(LT), then Ji(T) 1:. R2 by 14.3.9.3, and
hence the lemma holds. On the other hand if J(T)·= J(0 2 (LT)) then by 14.5.9,
H:::; Na(J(T)) :SM= !M(LT), contradicting 14.5.4.1. D
LEMMA 14.5.11. (1) H* is a {2, 3}-group.
(2) 031(H)) :S CH(VH), so H* is a quotient of H+.
PROOF. Assume [031(H),X] I 1. Then as 031(H) is solvable of odd order
by (2) and (4) of 14.5.7, [R,X*] I 1 for some prime p > 3 and some supercritical