Show that a group with $xa=ax^{-1}$, for all $x\in G$, where $a$ is fixed in $G$, is abelian.

Question

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.

Answer 1

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

Answer 2

We have $$\begin{align} (xy)a &= a(xy)^{-1} \\ &= ay^{-1}x^{-1} \\ &= yax^{-1}\\ &= yxa. \end{align}$$ Now cancel the $a$.