2021
DOI: 10.48550/arxiv.2101.06132
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A characterization of wreath products where knapsack is decidable

Abstract: The knapsack problem for groups was introduced by Miasnikov, Nikolaev, and Ushakov. It is defined for each finitely generated group G and takes as input group elements g1, . . . , gn, g ∈ G and asks whether there are x1, . . . , xn ≥ 0 withWe study the knapsack problem for wreath products G ≀ H of groups G and H.Our main result is a characterization of those wreath products G ≀ H for which the knapsack problem is decidable. The characterization is in terms of decidability properties of the indiviual factors G … Show more

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Cited by 2 publications
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“…As mentioned in the introduction, the class of knapsack semilinear groups is very rich. Examples of a groups, where knapsack is decidable but solvability of systems of exponent equations is undecidable are the Heisenberg group H 3 (Z) (the group of all upper triangular (3 × 3)-matrices over the integers, where all diagonal entries are 1) [22] and the Baumslag-Solitar group BS(1, 2) [3,25]. These groups are not knapsack semilinear in a strong sense: there are knapsack expressions e such that sol H3(Z) (e) (resp.…”
Section: Knapsack and Exponent Equationsmentioning
confidence: 99%
“…As mentioned in the introduction, the class of knapsack semilinear groups is very rich. Examples of a groups, where knapsack is decidable but solvability of systems of exponent equations is undecidable are the Heisenberg group H 3 (Z) (the group of all upper triangular (3 × 3)-matrices over the integers, where all diagonal entries are 1) [22] and the Baumslag-Solitar group BS(1, 2) [3,25]. These groups are not knapsack semilinear in a strong sense: there are knapsack expressions e such that sol H3(Z) (e) (resp.…”
Section: Knapsack and Exponent Equationsmentioning
confidence: 99%
“…There are groups with a decidable knapsack problem but an undecidable solvability problem for systems of exponent equations. Examples are the discrete Heisenberg group [25] and the Baumslag-Solitar group BS(1, 2) [5,Theorem E.1]. Let us also remark that the variants of these problems, where the variables x i range over Z are not harder, since one can replace a power g x i i with x i ranging over Z by g x i i (g −1 i ) y i with x i , y i ranging over N.…”
Section: Introductionmentioning
confidence: 99%