# In a free group two elements commute if and only if they are powers of a common element

In other words, $uv = vu$ in $F_n$ if and only if $u=w^m$ and $v=w^n$ for some $w\in F_n$.

I would like to prove this without making use of Nielsen-Schreier (every subgroup of a free group is free). We can always find reduced representations $u=t_1^{\epsilon_1}\cdots t_k^{\epsilon_k}$ and $v=s_1^{\eta_1}\cdots s_l^{\eta_l}$ and the statement $uv = vu$ transforms into $$t_1^{\epsilon_1}\cdots t_k^{\epsilon_k}\cdot s_1^{\eta_1}\cdots s_l^{\eta_l}\cdot t_k^{-\epsilon_k}\cdots t_1^{-\epsilon_1}\cdot s_l^{-\eta_l}\cdots s_1^{-\eta_1}\sim 0$$ where $\sim$ denotes the equivalence relation coming from setting $x\cdot x^{-1}\sim 0$. However, the arbitrariness of $u$ and $v$ makes it hard to go on from here.

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What about induction on the lengths of reduced representation, then concentrating only on the first letters..? –  Berci Oct 14 '12 at 11:47
First deal with the two special cases: 1) there is no cancellation when forming the products $uv$ and $vu$; and 2) one of $u,v$ cancels completely when forming these products. If neither of these happens, then $u,v$ have the form $au'a^{-1}$, $av'a^{-1}$, where $a$ is a generator (or its inverse), and $u'$, $v'$ are shorter than $u$ and commute, so you can use induction. –  Derek Holt Oct 14 '12 at 16:32