From A First Course In Abstract Algebra by John B. Fraleigh, 7th edition. Section 13, problem 50.
Let $\phi: G\to H$ be a group homomorphism. Show that $\phi[G]$ is abelian if and only if for all $x,y \in G$, we have $xyx^{-1}y^{-1}\in\ker(\phi)$.
[Proof moved to answer]
Best Answer
Suppose $\phi[G]$ is abelian, and let $e^{\prime}$ be the identity element of H.
We have $\phi(xyx^{-1}y^{-1})=\phi(x)\phi(y)\phi(x^{-1})\phi(y^{-1})$. Since $\phi(a^{-1})=\phi(a)^{-1}$, we then have $\phi(x)\phi(y)\phi(x)^{-1}\phi(y)^{-1}$. Since $\phi[G]$ is abelian, we can rearrange this to obtain $\phi(x)\phi(x)^{-1}\phi(y)\phi(y)^{-1}=e^{\prime}e^{\prime}=e^{\prime}$. Thus, $\phi(xyx^{-1}y^{-1})=e^{\prime}$, thus $xyx^{-1}y^{-1}\in Ker(\phi)$.
Now, suppose that $xyx^{-1}y^{-1}\in Ker(\phi)$. Then, $\phi(xyx^{-1}y^{-1})=e^\prime$. By the homomorphism property, then, $\phi(x)\phi(y)\phi(x^{-1})\phi(y^{-1})=e^{\prime}$. Again, inverses are preserved, so we have $\phi(x)\phi(y)\phi(x)^{-1}\phi(y)^{-1}=e^{\prime}$.
But, $\phi(x)^{-1}\phi(y)^{-1}=(\phi(y)\phi(x))^{-1}$, so multiplying the last equation above on the right by $(\phi(y)\phi(x))$ we obtain
$\phi(x)\phi(y)=\phi(y)\phi(x)$, therefore $\phi[G]$ is abelian.
This completes the proof.