Exercise 26, § 7

Let $X$ be a set and $g$ an element of the free group $F(X)$. Let $g = \prod_{\alpha=1}^n x_\alpha^{m(\alpha)}$ be the canonical decomposition of $g$ as a product of powers of elements of $X$ (cf. I, p. 84, prop. 7). The length of $g$ is called the integer $l(g) = \sum_{\alpha} |m(\alpha)|$.

a) One says that $g$ is cyclically reduced if one has, either $x_1 \neq x_n$, or $x_1 = x_n$ and $m(1)m(n) > 0$. Now prove that every conjugacy class of $F(X)$ contains one and only one cyclically reduced element; it is the element of minimum length of the class in question. b) An element $x \in F(X)$ is said to be primitive if there do not exist an integer $n \geq 2$ and an element $g \in F(X)$ such that $x = g^n$. Now prove that every element $z \neq e$ of $F(X)$ can be written uniquely in the form $x^n$, with $n \geq 1$ and $x$ primitive. (Reduce to the case where $z$ is cyclically reduced, and observe that, if $z = x^n$, the element $x$ is also cyclically reduced.) Now prove that the centralizer of $z$ is the same as that of $x$; it is the cyclic subgroup generated by $x$.