Enumerating palindromes and primitives in rank two free groups

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 Ep/q where p/q is rational number expressed in lowest terms. We prove that Ep/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 Ep/q and Y is conjugate to Er/s where |ps−rq|=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.