1068 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cf(G, T) IS EMPTYPROOF. Suppose first that U~ is not an FF*-offender on UH. As m(U~) 2:
m(U/D), Ua does not centralize D, so that Za = [D, Ua] S U using F.9.13.6.
Therefore by 14.7.60.1, we may take ZaUo = V2Uo, and by 14.7.60.3, U~ is B* :=
B(V 1 , V2) or A:= A(V 1 , l/2). As U~ is not an FF-offender on U, U~ = B*, so
- • - - -- -J
[Ua, U] = U(Vi, V2) by 14.7.55.4. Also by 14.7.60.3, either D SU or DU_= V 2.
- • - - -- -J
The first case is impossible, as m(U/D) S m(U~) = 2, whereas m(U/U-) = 3.
Thus DU_= -- V-J_ - - -J_ * -
2 , and hence Za = [D,Ua] = [V 2 ,B] = l/2, so that Za S V2 S V.Therefore Vi S Ua and m(U~) = 2 = m(U/D) by 14.7.58.2. Then as D lies in
the hyperplane fJ[r = "V;.l of U, while m(U/U)"= 3, we obtain m(DU_/D) = 1,
and in particular [! i D. Since u = [U, Ua]Uo and [U, Ua] s u n Qa = D,
we conclude there is u 0 E U 0 - D. But this is impossible, as then [Ua, uo] S V1,
whereas no nontrivial element of Ga/Qa induces a transvection on Ua/Za.
Thus U~ is an FF-offender on U, so H ~ G2(2) by 14.7.55.5. As U~ S Ri
by Notation 14.7.1, (1) follows from 14.7.57.3. Then (2) follows from 14.7.55.4.
Suppose Di U_. Then as c;;(U:) = Co(A*) = [;_, 1-/=-[D, Ua], so as in the
previous paragraph, Za SU by F.9.13.6. However this contradicts 14.7.60.3a since
U~ =A*. Therefore D s U_, so as 3 = m(U~) 2: m(U/D) 2: m(U/U_) = 3, we
conclude (3) and (4) hold. Then (4) implies (5), completing the proof. DSet H-;!; := Ha/Qa and let W denote the preimage of W(Vi, V2) = V:f in U.
LEMMA 14.7.62. (1) V1 i Ua and Za i U.
(2) u+ = A*(Za, V2 a) for a suitable conjugate V2 a of Vz in Ua containing Za.
(3) w+ = B*(Za, V2,a)· ,
PROOF. Recall that U S G 7 S Ca(A 1 ), so that A 1 S Cu'Y(U) S QH and
hence also Za S QH.
Suppose first that Vi S Ua. Then by 14. 7.61.2, Un Ua = U _ is of codimension
3 in Uai so as m(U~) = 3, QH n Ua = Un Ua. Thus Za S QH n Ua S U. Then
CQH (Za) is of index 2 in QH by 14.5.21.1, with [CQH (Za), Ua] s QH n Ua s u, soU~ centralizes a hyperplane of QH /CH(U). But this is impossible since by 14.5.21.1,
· QH/CH(U) is H-dual to U, and no member of H acts as a transvection on U.
Therefore Vi i Ua. Then by the symmetry in 14.7.61.5, Za i U, so (1) holds.
By 14. 7 .61.1, u~ is an FF -offender on f)' so by symmetry u ID is also an
FF-offender on Ua/Za. In~icular (2) holds.
As V1 i Ua, Co( a)= Cu( a) for each a E Ua, so as each w E Wis centralized
by some 1-/=-b EB by 14.7.55.4, m(Ua/Cua(w)) S 2. Thus w+ is a hyperplane
of u+ such that m(Ua/Cua (w+)) S 2 for each w E W, so (3) follows from 14.7.55.4.
D
We now enter the last stages of our proof of Theorem 14.7.53.
From 14.7.62, in the symmetry between 'Yl and a appearing in 14.7.61.5, thetuple H, U, V1, l/2, W, U~, B*, ')'1 corresponds to the tuple Ga, Ua, Za, V2,a, B,
u+, w+, a, where B is the preimage in Ua of B*.
Now since V1 i Ua, using 14.7.55.4 we see that
:F := {[U, b] : 1 -/=-b* EB*}
consists of three 4-subgroups, with
(a) V(Vz) = {FVi: FE :F}: