Abstract <p>In this paper, we construct a continuum family of nonisomorphic 3-generator groups in which the identity <InlineEquation ID="IEq3"> <EquationSource Format="TEX">\(x^{n}=1\)</EquationSource> <!--ContMath2670002Atabekyan-m3--> </InlineEquation> holds with probability 1, while failing to hold universally in each group. This resolves a recent question about the relationship between probabilistic and universal satisfaction of group identities. Our construction uses <InlineEquation ID="IEq4"> <EquationSource Format="TEX">\(n\)</EquationSource> <!--ContMath2670002Atabekyan-m4--> </InlineEquation>-periodic products of cyclic groups of order <InlineEquation ID="IEq5"> <EquationSource Format="TEX">\(n\)</EquationSource> <!--ContMath2670002Atabekyan-m5--> </InlineEquation> and two-generator relatively free groups satisfying identities of the form <InlineEquation ID="IEq6"> <EquationSource Format="TEX">\([x^{pn},y^{pn}]^{n}=1\)</EquationSource> <!--ContMath2670002Atabekyan-m6--> </InlineEquation>. We prove that in each of these products, the probability of satisfying <InlineEquation ID="IEq7"> <EquationSource Format="TEX">\(x^{n}=1\)</EquationSource> <!--ContMath2670002Atabekyan-m7--> </InlineEquation> is equal to 1, despite the fact that the identity does not hold throughout any of these groups.</p>

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A Continuum of Nonisomorphic 3-Generator Groups with Probabilistic Law\(\boldsymbol{x^{n}=1}\)

  • V. S. Atabekyan,
  • A. A. Bayramyan,
  • V. H. Mikaelian

摘要

Abstract

In this paper, we construct a continuum family of nonisomorphic 3-generator groups in which the identity \(x^{n}=1\) holds with probability 1, while failing to hold universally in each group. This resolves a recent question about the relationship between probabilistic and universal satisfaction of group identities. Our construction uses \(n\) -periodic products of cyclic groups of order \(n\) and two-generator relatively free groups satisfying identities of the form \([x^{pn},y^{pn}]^{n}=1\) . We prove that in each of these products, the probability of satisfying \(x^{n}=1\) is equal to 1, despite the fact that the identity does not hold throughout any of these groups.