Show that a group with $xa=ax^{-1}$, for all $x\in G$, where $a$ is fixed in $G$, is abelian. Fix $a$ in a group $G$. How can I prove that $G$ is abelian if for every $x\in G$
$$
xa=ax^{-1}?$$
I am thinking that it is enough to show that $x^2=e$, where $e\in G$ is the identity element. I started with
$$
x^2=(ax^{-1}a^{-1})^2=ax^{-1}a^{-1}ax^{-1}a^{-1}=ax^{-2}a^{-1}.
$$
But from here I don't know how to continue.

Hint: Take $x=a$. Then take $x=ya$.
