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

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 $a,b\in G$.

• This is the right idea, a little clumsily written. But there's a sense in which it is too much work. You shouldn't have to make any argument at all, except to practice arguments. See math.stackexchange.com/questions/2039702/… Mar 6 '18 at 21:12
• Are you sure the intent is for all $a,b \in G$ and not for any? Mar 6 '18 at 21:13
• @gt6989b I am not sure what you mean. Don't the universal quantifiers "for all" and "for any" mean the same thing?
– user482939
Mar 6 '18 at 21:39
• I mean you don't need to assume the whole group is Abelian, just that these specific $a,b \in G$ commute Mar 6 '18 at 22:35
• @gt6989b I understand what you mean now...thanks, I will rewrite my proof without the abelian part!
– user482939
Mar 7 '18 at 0:26

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 $\phi^{-1}$.