Let $F_n$ be the free group on $n$ generators, and let $u$ and $v$ be two of $F_n$'s generators. My question is,
Do there exist $a, b \in F_n$ such that $(u^{-1}v)^{-1} = b^{-1}a u^{-1}v a^{-1} b$?
Let $F_n$ be the free group on $n$ generators, and let $u$ and $v$ be two of $F_n$'s generators. My question is,
Do there exist $a, b \in F_n$ such that $(u^{-1}v)^{-1} = b^{-1}a u^{-1}v a^{-1} b$?
This is not possible. You can see this very quickly by considering the map $\varphi:F_n\to\mathbb{Z}^n$ sending the generators of $F_n$ to the coordinate vectors (this is the abelianization of $F_n$). Since $\mathbb{Z}^n$ is abelian, $\varphi(b^{-1}au^{-1}va^{-1}b)=\varphi(u^{-1}v)$. But $\varphi(u^{-1}v)\neq \varphi((u^{-1}v)^{-1}))$, since $\varphi(u^{-1}v)\neq 0$ and no nonzero element of $\mathbb{Z}^n$ is its own inverse.