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Since a(a)=P(a)=v(a), it follows that $(v(a)) is a well defined element of U(U). Thus *(#") =$-i(fi(v))7 (^*))$- 1 (u(v)- 1 ) =$-1(fi(v))v'(a)$-1(u(v)-1) =$-1(u(v)«l'-1(*(vr ( a ) ) * - 1 ^ ) - 1 ) =$-1(u(v)$(a(a))u(v)-1) =$-'(fi(v)[$ -Tor-'av)^)-1) =#" '(u(v )u(v r V " ^7)u(v )u(v)"') =# _1 (a) Now consider a relation of the form x*~xctx*=b*. In U(P) this implies the relation x~lax=b. Vtf^cfx*) =f>-1(u(v))jc-1$-1(u(x)-1)

If x and y are units, then xy eAvcGv. If exactly one of x and y are units, we may assume x is the unit. Since y is cyclic by definition, (xy)~l~y~l~y~xy. Thus xy is cyclic, so xy e V. Suppose both x and y are maximal. Then x and y are cyclic and x,yeV. If xy is maximal, then 1 1 1 _1 xy—x—y" —^" ^" , so xyeGy. If xyeU, then (xy)y =xeK, so xy stabilizes V. thus xy eAyCGy. Hence (iii) follows. Finally, suppose H is a subgroup of P and xeHC\V. If yeH is maximal, then y _ 1 x e / / =>y~x, s o y e F .

Moreover, Vy=zVc, so Vy and Vc are in the same orbit. Since x~lyeP, xeVy. Hence x and Gc are in the same orbit, so xeVc by construction of the generating set. Recall that a is cyclic maximal and aiQ. In particular, Va # VC9 because GceG. Nevertheless, x~laeP, so x~a, so Va = Vx = Vc, a contradiction. Therefore z must be maximal. 3: Suppose z is maximal. We may assume y is maximal, for otherwise we could cut along y. PREGROUPS 35 Now y~z~x and y l~c, z l~c l. Thus Vc, Vc-i9 Vx, and Vx-i are all in the same =* -class.