Let $G$ be a group. Let $H$=$G\times G$ be the direct product of $G$ with itself. Define $f: H\to G$ to be $f((g,h))=gh$ for any $(g,h)\in H$. Show that $G$ is Abelian if and only if $f$ is a homomorphism.
So there are two parts to this proof. For the part of the proof that assumes $f$ is a homomorphism, I believe this means that $f[(g,h)(g',h')]=f(g,h)f(g',h')$. From here, we simplify the LHS of the equation so that we have $f(gg',hh')=gg'hh$'. Then, we simplify the RHS of the original equation so that $f(g,h)f(g',h')=ghg'h'$. Now, we see that $gg'hh'= ghg'h'$. So from here we see that $g'h=hg'$, which shows that $G$ is Abelian.
I just need help making sure that this proof is correct, and I am also unsure of how to show that if $G$ is Abelian, then $f$ is necessarily a homomorphism. Thanks.