14.7. FINISHING L 3 (2) WITH (vG1) ABELIAN 1073Now F1 is invariant under the symmetry interchanging 'Yl and a, so by this sym-
metry there is a similar partition of F given byF = (F n V^9 ) u (F n ug) u F1,
for g E (LT, H) with Vj_^9 = Za. By 14.7.68.l, zG n ug = {z9}, so as z9 ~ F and
P{! ~ zG, Fv = F n V^9 :::;; V^9. Then as Fv is a hyperplane of V, V = V9 by
14.7.67.2, contrary to our earlier reduction V 1 1:. Ua.
Therefore m(Uo) = 2. This time Pis partitioned by Uo and F 1 , so F has thepartition F = Fo U F1, and again using the symmetry between 'Yl and a as above,
we conclude that F = (F n U8) U F 1 is also a partition, and then that F 0 :::;; ug.
Further Po= Uo:::;; Z(H), so Uo = ug by 14.7.68.2, and hence g E Na(Uo) =Has
HEM by 14.7.67.4, contrary to V{ = Za -=f V].. This contradiction completes the
proof of 14. 7.69. DBy choice of 'Yin Notation 14.7.1, m(U~) ~ m(U / D) > 0, where D :=Un Qa;
so as m(U~) = 1by14.7.69, also m(U/D) = 1. Thus again we have symmetry
between a and ')'1, as discussed in Remark 14.7.39.LEMMA 14.7.70. (1) We may choose a so that Za:::;; V2.(2) m(Uo) :::;; 2.
(3) Un Ua = UoV = [U, Ua]V = [U, Ua]Uo.
(4) b = 3 and Ua E UL.
PROOF. Observe that if (1) holds, then so does (4) by 14.7.3.4. Thus it suffices
to establish (1)-(3).Let F := [U, Ua]· By 14.7.66 and 14.7.69, U~ is a subgroup of Q E Syh(K)
of order 2. Then using Remark 14.7.64, FUo = VU 0 , VU 0 = P x Uo, and Ua
centralizes no F2-hyperplane of U; so 1 -=f [D, Ua], and hence Za = [D, Ua] :::;;
F :::;; U using F.9.13.6. By the symmetry between 'Yl and a discussed above, alsoVi = [Da, U] :::;; F. By 14.7.68.1, Za -f:. Uo.
By Remark 14.7.64, m(Uo) :::;; 2, so that m(Uo) :::;; 3. We now make some
choices: We may conjugate in NH(R1) = L1T and preserve the condition Ua :::;; Rl.
As U~ is of order 2 in Q, conjugating in L 1 , we may assume that U~ :::;; Z(T);
when m(Uo) = 3, we make this choice. When m(Uo):::;; 2, we make a more careful
choice: As Za :::;; F :::;; Uo V, conjugating in L1 we may assume that Za.Uo = ViUo.
As m(Uo) :::;; 2, T centralizes V 2 Uo and hence also Za· Further [Za, QH] = V1
by 14.7.4.1, so by a Frattini Argument, T = Cr(Za). Now as H = Ca(z),
Cr(Za) :::;; Na(Ua), so again U~ :::;; Z(T*). Thus in either case our choice implies
U~:::; Z(T*).
As U~ :::;; Z(T*), Tacts on [U, U~] = F; hence as V1 = [Da, U] :::;; F, T also
acts on F. Recall also that Za :::;; F, so
Suppose first that V 1 = U 0 • Then (2) holds, and by our choice under this
assumption, Za :::;; ViUo = V2, so that (1) holds. Further (3) follows from (*),
completing the proof of the lemma in this case..
Thus we may suppose that V 1 < U 0. Recall Pis a complement to U 0 in VUo.
Further if m(Uo) = 3, then from Remark 14.7.64, P n V = 1, while if m(Uo) = 2