By Karl Friedrich Geldner
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Additional resources for Avesta, the Sacred Books of the Parsis, Vol. 3
1. Choose some r E R, r fj. R'. Let u == raX (mod R') and v == rby (mod R') for some a, bE ZF (to be specified later). Then, modulo R', we have [raX, r b, yJ == [raX, yJ[raX, rbJY == [ra, yJx [x, yJ[x, rbJY == [x, yJ [ra, yJX[X'Y] [x, rbJY == [x, yJ [ra, yJy-1XY [x, rbJY == [x, yJ[y-l, raJXY[r b, X-I JXY == [x, yJr(a(l-y-1)+b(x-1-1»Xy. Thus, the congruence [raX, rbyJ == [x, yJ (mod R') is equivalent to the congruence r(a(l-y-1)+b(x-1-1» == 1 (mod R'). 1. Nielsen's Commutator Test 23 where a, b E ßF (since R ~ F' implies that ßR lies in ß}).
P(u) = w for some u E V(H). As in the statement ofthe theorem, V is the set oflaws defining the nontrivial variety B. 2. Almost Primitive and Generic Elements 43 of F. If w ~ K, then clearly w ~ V(K). If w E K, then since w is an APE, w is primitive in K since K is a proper subgroup of F. Furthermore, since Bis a nontrivial variety and K is free, we have that K =I- V(K). It follows then from the primitivity of w in K that w ~ V(K). Therefore w E V(F) and for any proper subgroup K of F we have w ~ V(K) and hence w is Bgeneric.
Test Elements in the free solvable group of any bigger rank as weH, but technicaHy it is getting more complicated. We also mention a related result of Timoshenko [382J, who proved that a free metabelian group of rank > 2 does not have any test elements. Now we get to the main results of this section. 1 (). , Z(F/R) has no zero divisors and any two nonzero elements have a common nonzero multiple). Then there exist u,v E F such that [u,vJ == [x,yJ (modR'), while the endomorphism x -+ U, Y -+ v 01 F does not induce an automorphism 01 F/R'.