Proof verification: If $xy=x^{-1}y^{-1}$ for all $x,y\in G$, then $G$ is abelian.

Question

Prove that if $xy=x^{-1}y^{-1}$ for all $x,y\in G$, then $G$ is abelian.

Proof: $G$ is abelian iff $xy=yx$ for all $x,y\in G$.

$xy = x^{-1}y^{-1}=(xy)^{-1}$ (by power laws) $= y^{-1}x^{-1}$ (by inverse laws) $=yx$

Therefore, $G$ is abelian. $\square$

Answer 1

Since that formula holds for all x,y then in holds also for $y=e$ so we have $x^2=e$ (so $x^{-1}=x$) for each x. Take now any a, b in group. Then we have $abab=(ab)^2 =e$ so $ab =b^{-1}a^{-1} =ba$ and we are done.

Answer 2

Through associativity and inverse laws we have:

$x(xy)^2y=xxyxyy=x(xyxy)y=x(x^{-1}y^{-1}x^{-1}y^{-1})y=(xx^{-1})y^{-1}x^{-1}(y^{-1}y)=(e)y^{-1}x^{-1}(e)=y^{-1}x^{-1}=xy$

Where $e$ is the identity element

So $x(xy)^2y = xy$. Using cancellation laws,

$(xy)^2y = y$

$(xy)^2 = (xy)(xy) = e$

We also have

$(xy)(yx) = (x^{-1}y^{-1})(yx) = x^{-1}(y^{-1}y)x = x^{-1}ex = x^{-1}x = e = (xy)(xy)$

By right cancellation,

$yx=xy$