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Proofs

Proofs. Proof of A1: Let a=(yz) -1 and b=(xy)z=[x(yz)]s=(sx)(yz)=(sx)a -1 , where s=(x,y,z). Then x = s -1 (ba). (xz)y = {[s -1 (ba)]z}y = [(ba)z](ys -1 ) = [{[(xy)z]a}z](ys -1 ) = {(xy)[(za)z]}y s -1 = x  {y[(za)z]}y  s -1

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Proofs

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  1. Proofs

  2. Proof of A1: Let a=(yz)-1 and b=(xy)z=[x(yz)]s=(sx)(yz)=(sx)a-1, where s=(x,y,z). Then x = s-1(ba). (xz)y = {[s-1(ba)]z}y = [(ba)z](ys-1) = [{[(xy)z]a}z](ys-1) = {(xy)[(za)z]}y s-1 = x {y[(za)z]}y s-1 = x {[(yz)a]z}y s-1 = [x(zy)]s-1. Therefore, (x,z,y) = s-1 = (x,y,z)-1. 

  3. Proof of A2: [(xy)z]y = x[(yz)y] = [x(yz)]y (x,yz,y)-1= {[x(yz)](x,yz,y)-1}y= {[x(yz)](x,y,yz)}y. Canceling y, [(xy)z]= [x(yz)](x,y,yz), so that (x,y,yz) = (x,y,z). Since zy = yz (z,y) and (z,y) is central, (x,y,zy) = (x,y,z) as well. We get (x, y, zyn) = (x, y, z) by induction on n. The other identity follows by Lemma 2. 

  4. Proof of A7: [(xy)z]w = [x(yz)]w(x,y,z) = x[(yz)w](x,y,z)(x,yz,w) = x[y(zw)](x,y,z)(x,yz,w)(y,z,w) = (xy)(zw)(x,y,z)(x,yz,w)(y,z,w)(x,y,zw)-1 = [(xy)z]w(x,y,z)(x,yz,w)(y,z,w)(x,y,zw)-1(xy,z,w)-1 Canceling [(xy)z]wand bringing the last two associators to the left side of the equation, we get (x,y,zw)(xy,z,w ) = (x,y,z)(x,yz,w)(y,z,w).

  5. Proof that rm=1: Choose j such that 2j>m, and let x = [1,0,1] and y = [1, j, 0]. Then (xy)y = [([1,0,1][1, j,0])[1,j,0] = [rj,j,1][1,j,0] = [r2jz,2j –m,1], and x(yy) = [1,0,1]([1,j,0][1,j,0]) = [1,0,1][z,2j-m,0] = [r2j-mz,2j-m,1]. Since these must be equal, we must have r2j= r2j-m, so rm= 1. 

  6. Next, let x = y = [1,1,0], z = [1,1,1]. Then {(xy)z}y = {([1,1,0][1,1,0])[1,1,1]}[1,1,0] = ([zp1,2*,0][1,1,1])[1,1,0] = [s2*zp1+p2,3*,1][1,1,0] = [rs2*zp1+p2+p3,4*,1] and xf(yz)yg = [1; 1; 0]f([1; 1; 0][1; 1; 1])[1; 1; 0]g = [1; 1; 0][(szp1 ; 2?; 1][1; 1; 0])= [1; 1; 0][rszp1+p2 ; 3?; 1] = [rs1+3?zp1+p2+p3 ; 4?; 1] [Note that 2? +1 ´ 3 (mod m) so that (2? +1)? = 3?, etc. Also note that the exponent of z is the same in both cases.] Thus, f(xy)zgy = xf(yz)yg if and only if s2? = s1+3? . If 8<: m = 2 then 2? = 0; 3? = 1 and so s2 = 1 m = 3 then 2? = 2; 3? = 0 and so s = 1 m > 3 then 2? = 2; 3? = 3 and so s2 = s4 9=; : In all cases, s2 = 1. To see that rn = t2 = 1 is a bit more complicated, because the two conditions are related.

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