Enumerating palindromes and primitives in rank two free groups

JOURNAL OF ALGEBRA

卷:332

Article

Gilman, Jane¹  Keen, Linda² 

[1] Rutgers State Univ, Dept Math, Newark, NJ 07079 USA

[2] CUNY, Lehman Coll & Grad Ctr, Dept Math, Bronx, NY 10468 USA

关键词: Palindromes;    Primitive element;    Free group;    Rank two;    Kleinian groups;    Fuchsian groups;    Markoff Numbers;    Farey tesselation;    Conjugacy classes;   

DOI  :  10.1016/j.jalgebra.2011.02.010

来源: Elsevier

【 摘 要 】

Let F be a free group of rank two. An element of F is primitive if it, along with another group element, generates the group. If F = < A, B >, then a word W (A, B), in A and B, is a palindrome if it reads the same forwards and backwards. It is known that in a rank two free group, for any fixed set of two generators a primitive element will be conjugate either to a palindrome or to the product of two palindromes, but known iteration schemes for all primitive words give only a representative for the conjugacy class. Here we derive a new iteration scheme that gives either the unique palindrome in the conjugacy class or expresses the word as a unique product of two unique palindromes that have already appeared in the scheme. We denote these words by E-p/q where p/q is rational number expressed in lowest terms. We prove that E-p/q is a palindrome if pq is even and the unique product of two unique palindromes if pq is odd. We prove that the pair (X, Y) (or (X-1, Y-1)) generates the group if and only if X is conjugate to E-p/q and Y is conjugate to E-r/s where vertical bar ps - rq vertical bar = 1. This improves a previously known result that held only for pq and rs both even. The derivation of the enumeration scheme also gives a new proof of the known results about primitive words. (C) 2011 Elsevier Inc. All rights reserved.

【 授权许可】

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