Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1997. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 24)
KlRALY, B., Residual Lie nilpotence of the augmentation ideal
Residual Lie nilpotence of the augmentation ideal 87 For every k and i C/(6") p *fk{G) E F>{ v}- The class ^{p} is closed with respect to forming subgroups and finite direct products, and by Lemma 2.2. G is discriminated by Vs p\. Consequently we can choose the set fit = {p}. Let us consider the case when G' is a torsion-free group and 1 / g E G' is a generalized torsion element of G. We put ft, — ft g. From Lemma 3.5. (case 2) it follows that N jvi R) =Open From Lemma 3.2. (here we put {G' 0} Q e/ = {(G') p njk(G), k,n = 1, 2,.. .} Pen) and Lemma 3.5. (case 3) we have that G is discriminated by the class VQ,. Let A be an arbitrary subset of ft and let r\ p^\J p(R) ^ 0. If G is not discriminated by the class of groups Dn\Aj then by Lemma 3.2. there exists a set of elements gi , g 2,. •., g n (gi E G) of infinite orders such that oo oo M (si -1)(</2 - l) • • • (jn -1) e f| nn^T'^))pGÍÍ\A A;=l i1 By Lemma 3.5. (case 1) for every element a E n p GA Jp(R) 0.(9 - 1)(<71 - 1)(92 - 1) • • • (g n - 1) € AH (RG). Because A^(RG) = 0 we have that a(g - l)(gi - l)(g 2 - l) • • '(g n - l) = oSince element ^ (i = 1,2, ..., n) has infinite order and so has zero left (and right) annihilator in RG , then for g n we have a(g-l)(g 1-l)(g 2-l)---(g n_ l-l) = 0. Continuing this procedure for i = n — 1, n — 2,..., 1 on the last step we get that a(g - 1) = 0. Since the element g has infinite order, its left annihilator is zero in RG , which implies a — 0. Consequently,, if G is not discriminated by the class of groups 2?n\A, then n pe AJ p(R) = 0. The sufficiency part is proved in Lemma 3.6. Corollary. Let R = Z p, the ring ofp-adic integers. Then A^(Z pG) = 0 if and only if either
