{ "id": "0802.2731", "version": "v5", "published": "2008-02-19T23:02:26.000Z", "updated": "2011-02-12T13:28:15.000Z", "title": "Enumerating Palindromes and Primitives in Rank Two Free Groups", "authors": [ "Jane Gilman", "Linda Keen" ], "comment": "Final revisions, to appear J Algebra", "categories": [ "math.GR", "math.CV", "math.GT", "math.NT", "math.RT" ], "abstract": "Let $F= < a,b>$ be a rank two free group. A word $W(a,b)$ in $F$ is {\\sl primitive} if it, along with another group element, generates the group. It is a {\\sl palindrome} (with respect to $a$ and $b$) if it reads the same forwards and backwards. It is known that in a rank two free group any primitive element is 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. 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 pairs $(E_{p/q},E_{r/s})$ generate the group when $|ps-rq|=1$. This improves the 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 primitives.", "revisions": [ { "version": "v5", "updated": "2011-02-12T13:28:15.000Z" } ], "analyses": { "subjects": [ "20H10", "20F10", "32G15", "30F40", "30F60", "32G15", "11A55", "30B70", "30F10" ], "keywords": [ "free group", "enumerating palindromes", "unique palindrome", "conjugacy class", "iteration scheme" ], "note": { "typesetting": "TeX", "pages": 0, "language": "en", "license": "arXiv", "status": "editable", "adsabs": "2008arXiv0802.2731G" } } }