2016
DOI: 10.1142/s0218196716500144
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The power conjugacy problem in Higman–Thompson groups

Abstract: An introduction to the universal algebra approach to Higman-Thompson groups (including Thompson's group V ) is given, following a series of lectures by Graham Higman in 1973. In these talks, Higman outlined an algorithm for the conjugacy problem; which although essentially correct fails in certain cases, as we show here. A revised and complete version of the algorithm is written out explicitly. From this, we construct an algorithm for the power conjugacy problem in these groups. Python implementations of these… Show more

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Cited by 8 publications
(38 citation statements)
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References 23 publications
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“…If an admissible word W has the form (2.1), W = q 1 u 1 q 2 u 2 ...q s , and q i ∈ Q ±1 j(i) , i = 1, ..., s, u i are group words in tape letters, then we shall say that the base of W is the word Q ±1 j(1) Q ±1 j (2) ...Q ±1 j(s) . Here Q i are just symbols which denote the corresponding parts of the set of state letters.…”
Section: S-machinesmentioning
confidence: 99%
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“…If an admissible word W has the form (2.1), W = q 1 u 1 q 2 u 2 ...q s , and q i ∈ Q ±1 j(i) , i = 1, ..., s, u i are group words in tape letters, then we shall say that the base of W is the word Q ±1 j(1) Q ±1 j (2) ...Q ±1 j(s) . Here Q i are just symbols which denote the corresponding parts of the set of state letters.…”
Section: S-machinesmentioning
confidence: 99%
“…The alphabet of tape letters Y of LR(Y ) is Y (1) ⊔ Y (2) , where Y (2) is a (disjoint) copy of Y (1) . The positive rules of LR are defined as follows.…”
Section: Running State Lettersmentioning
confidence: 99%
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