1030 14. L 3 (2) IN THE FSU, AND L2(2) WHEN .Cf(G, T) IS EMPTY
Thus T 1 =Tu by 14.6.6.1. Therefore by the hypothesis of part (5), I:; Ca(u).
Further as J 2 :; H, V+ := V 0 (u) :; UH, and then from the discussion above,
V_ := Cv+(Tu) = (z,e5,5,u)..
Suppose that u tj. W. Then W n V = (z, e5,6), so that m(V) = 3. Therefore
as [UH, QH] :; Z by 14.5.15.1, while To = QHTu > Tu by 14.6.3.1, we conclude
V# := Cv(QH) = Cv (To) is a hyperplane ofV withu tj. V#, so that V = V#(u).
Let v be the projection on v# of ei,2,3,4,5,6, and set J := CK(v-); then v =f 1 as
u rf. W. Now J ~ A 6 , so J is contained in neither Mj nor I; which are solvable
from the discussion above, and hence J is contained in neither M nor G1. But then
(To, J):; Ca(v-) EI, contrary to 14.6.4 since To> Tu= T1.
Therefore u E W. Since I:; Ca(u), Cw(K) =f 1, and hence n = 6 and (u) =
Cw(K), so that u = ei,2,3,4,5,6· Let Q1 := 02(I). Since Tu is nontrivial on V by
14.6.3.3, and JT : Cr(V) I = 2, To = TuCr 0 (V), so we may choose t E Cr 0 (V) - Tu.
Since t normalizes Tu and W :::;) Tu, both E :=wt and WWt =WE are normal
in Tu = T1. If E :; Q1, then as [Q1, K] = W since K is a block, J := (K, T1, t)
acts on WE, and J contains KT1 =I and T 0 ;:=: TJ > T1, contradicting 14.6.5.
Thus E f:. Q1, so that E* =f 1. By (Ul), To acts on (z, u), so as V :::;) T, T 0 acts
on Vu := V(u). Therefore as V:; Wand (u) = Cw(K), Vu :; W n B. Similarly
ut E VunZ(Tu), and this latter group is generated by z = ei,2,3,4 and u = ei,2,3,4,5,6·
Therefore as ut tj. zK by 14.6.3.4, we conclude that ut = e 5 , 6.
Notice for v E V# that Wv := (VGK(v)) is a hyperplane of W, and if V =
(v, w), then W = Wv Ww. For example Wz = Vo. Thus E = Wt = W.,;W!.
Now (voa(v)) is abelian by Hypothesis 14.5.1 and the transitivity of L on V#,
and we chose t to centralize V, so w.,; :; (Voa(v)) :; Ca(Wv)· Therefore from
the action of 86 on its permutation module, w; = ((i,j)), where v :=.eo-{i,j}·
Then as w.,;W! = E :::;) Tj, and the only normal subgroup of Tj containing
w;* = ((5, 6)) generated by at most two transpositions is ((5, 6)), we conclude
that E = w_; = ((5, 6)). Thus [W, w_;] = (e 5 , 6 ) = (ut). This is impossible, as
C1(W/(u)) = C1(W), so that C1t(W_;j(ut)) = Cp(W;).
This contradiction completes the proof of (5), and hence of 14.6.10. D14.6.2. Showing O(H/0 2 (H)) = 1. Recall that H denotes a member of
1i(T, M) = 1-lz, and we have adopted Notation 12.8.2. In the remaining two sub-
sections we adopt Notation 14.5.16 and use notation and results from section F.9.
For example I' is the coset geometry determined by LT and H as in section F.7,
with the parameter b, the geodesic ')'1, ... , ')' = 'Yb, the element 9b taking ')'i to ')',
and the subgroups UH, u'Y, DH, D'Y etc., as well as z'Y := zgb defined in section
F.9-where Z'Y was often denoted by A 1 •
This second subsection is devoted to the proof of a key intermediate result:THEOREM 14.6.11. O(H*) = 1 for each HE 1i(T, M).
Until Theorem 14.6.11 is established, assume His a counterexample. Thus H
is a member of 1i(T, M) with O(H*) =f 1, and we must derive a contradiction from
the existence of such an H.
Let P 0 be a minimal normal subgroup of H contained in O(H); then P 0 is an
elementary abelian p-group and P 0 == P* for PE Sylp(P 0 ). Indeed PT E 1i(T, M)