If $\phi: G \to H$ is an isomorphism, prove that $G$ is abelian if and only if $H$ is abelian I would like to know if my proof is correct. Specifically, I would like you to check that surjectivity is needed for proving the first part and injectivity is needed for proving the second part.


Problem If $\phi: G \to H$ is an isomorphism, prove that $G$ is abelian if and only if $H$ is abelian.

Solution
Let $\phi: G \to H$ be an isomorphism.


*

*If $G$ is abelian, we have $xy = yx$, for any $x,y \in G$. Consider any two element in $H$, say $c$ and $d$. Since $\phi$ is an isomorphism (more specifically a surjection), we have $c = \phi(a)$ and $d=\phi(b)$ for some $a,b \in G$. Hence, $$cd = \phi(a)\phi(b) = \phi(ab) = \phi(ba) = \phi(b)\phi(a) = dc$$
This means $H$ is also abelian.

*If $H$ is abelian, we have $xy = yx$, for any $x,y \in H$. Now consider two elements in $G$, say $a$ and $b$. Since $\phi$ is a mapping, we have $c=\phi(a)$ and $d=\phi(b)$ for some $c,d \in H$. Hence,
$$\phi(ab) = \phi(a)\phi(b) = cd = dc = \phi(b) \phi(a) = \phi(ba)$$
Since $\phi$ is an isomorphism (more specifically injective), we have $ab = ba$. This means $G$ is also abelian.



Thanks
 A: Your proof is very well written. 
You've proven precisely what you needed to prove, and you did so based solely on the definitions of "isomorphism" and "commutativity," working with the tools you've acquired thus far. 
Bravo!
Note: 
When I first encountered "abstract algebra" and what it means for two groups to be isomorphic, I recall completing proofs to verify all the ways in which an isomorphism between groups preserves the group "structures", with commutativity being one such structural property among many.  
My instructor at the time was very demanding, to the finest details, requiring justification for every assertion in a proof. He would have been pleased with your explicit appeal to the fact that, as an isomorphism, $\phi$ is surjective, and also your explicit appeal to the definition of $\phi$ as a mapping, and an injection.
Now, in the event you have such an instructor, my only suggestion would be 


*

*to add something like what I've included using "underbraces" below:


$$cd = \phi(a)\phi(b) = \underbrace{\phi(ab) = \phi(ba)}_{G\;\text{is abelian}} = \phi(b)\phi(a) = dc$$ $$\phi(ab) = \phi(a)\phi(b) = \underbrace{cd = dc}_{H\;\text{is abelian}} = \phi(b) \phi(a) = \phi(ba)$$
