<p>If <InlineEquation ID="IEq1"> <EquationSource Format="TEX">\(G_1\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>G</mi> <mn>1</mn> </msub> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq2"> <EquationSource Format="TEX">\(G_2\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>G</mi> <mn>2</mn> </msub> </math></EquationSource> </InlineEquation> are torsion-free hyperbolic groups and <InlineEquation ID="IEq3"> <EquationSource Format="TEX">\(P&lt;G_1\times G_2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>P</mi> <mo>&lt;</mo> <msub> <mi>G</mi> <mn>1</mn> </msub> <mo>×</mo> <msub> <mi>G</mi> <mn>2</mn> </msub> </mrow> </math></EquationSource> </InlineEquation> is a finitely generated subdirect product, then the conjugacy problem in <i>P</i> is solvable if and only if there is a uniform algorithm to decide membership of the cyclic subgroups in the finitely presented group <InlineEquation ID="IEq4"> <EquationSource Format="TEX">\(G_1/(P\cap G_1)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>G</mi> <mn>1</mn> </msub> <mo stretchy="false">/</mo> <mrow> <mo stretchy="false">(</mo> <mi>P</mi> <mo>∩</mo> <msub> <mi>G</mi> <mn>1</mn> </msub> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation>. The proof of this result relies on a new technique for perturbing elements in a hyperbolic group to ensure that they are not proper powers.</p>

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On the conjugacy problem for subdirect products of hyperbolic groups

  • Martin R. Bridson

摘要

If \(G_1\) G 1 and \(G_2\) G 2 are torsion-free hyperbolic groups and \(P<G_1\times G_2\) P < G 1 × G 2 is a finitely generated subdirect product, then the conjugacy problem in P is solvable if and only if there is a uniform algorithm to decide membership of the cyclic subgroups in the finitely presented group \(G_1/(P\cap G_1)\) G 1 / ( P G 1 ) . The proof of this result relies on a new technique for perturbing elements in a hyperbolic group to ensure that they are not proper powers.