Enumerating Palindromes and Primitives in Rank Two Free Groups

TL;DR

This paper introduces a new method to enumerate primitive and palindrome words in rank two free groups, providing explicit representations and improved generation criteria, along with a new proof of existing primitive word results.

Contribution

It develops an iteration scheme that uniquely identifies palindromes or products of two palindromes in rank two free groups, enhancing previous enumeration methods.

Findings

E_{p/q} is a palindrome if pq is even.

E_{p/q} is a product of two palindromes if pq is odd.

Pairs (E_{p/q}, E_{r/s}) generate the group when |ps - rq|=1.

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.