Elements of a group commute if and only if the commutator is 1

Question

I am trying to prove that for $x, y \in G$, these elements commute if and only if $[x,y] = x^{-1}y^{-1}xy = 1$. What I have is:

$(\implies)$ If $x,y$ commute then $[x,y] = 1$. Because $x,y$ commute we have $xy = yx$. Then, $x^{-1}y^{-1}xy = x^{-1}xy^{-1}y = 1$.

$(\impliedby)$ If $[x,y] = 1$ then $x,y$ commute. If $[x,y] = 1$, then $x^{-1}y^{-1}xy = 1.$ Then, $(x)(x^{-1}y^{-1}xy) = (x)1 \implies y^{-1}xy = x$. Then, $(y)(y^{-1}xy) = (y)x \implies xy = yx$, which shows that $x,y$ commute. QED

I believe that is it, but I wanted to make sure.

Answer 1

$xy=yx[x,y]$ this may help you shorten your proof a little