For isomorphism $\phi:G\to H$, show that $ab=ba$ for $a,b\in G$ iff $\phi(a)\phi(b)=\phi(b)\phi(a)$.

Question

Could someone please verify my following proof?

For isomorphism $\phi:G\to H$, show that $ab=ba$ for $a,b\in G$ iff $\phi(a)\phi(b)=\phi(b)\phi(a)$.

Proof:

Let $ab=ba$ for all $a,b\in G$. Then $G$ is abelian. Then $H$ is abelian. Since $\phi$ is bijective, $H=\phi(G)=\left \{ \phi(g) : g\in G \right \}$. Therefore, $\phi(a)\phi(b)=\phi(b)\phi(a)$ for all $\phi(a),\phi(b)\in H$.

Let $\phi(a)\phi(b)=\phi(b)\phi(a)$ for all $\phi(a),\phi(b)\in H$. Since $\phi$ is bijective, $H=\phi(G)=\left \{ \phi(g) : g\in G \right \}$. Then $H$ is abelian. Then $G$ is abelian. Then $ab=ba$ for all $a,b\in G$.

Answer 1

If $\phi:G\to H$ is any homomorphism, and $a,b\in G$ are a particular commuting pair of elements: $ab=ba$ (we don't need to assume that the whole $G$ is Abelian), then we already have $$\phi(a)\phi(b)=\phi(ab)=\phi(ba)=\phi(b)\phi(a)\,.$$ If $\phi$ is an isomorphism, the converse also holds for the particular commuting pair $\phi(a),\phi(b)$: specifically, we can apply the same argument for the homomorphism $\phi^{-1}$.

Of course, your weaker interpretation of (the quantifiers in) the statement is also valid, and your proof is correct.