Show that commutativity is a structural property

[Mr Davis 97]

I am trying to prove that if two groups are isomorphic then one is abelian iff the other is abelian. This is a simple task, but I am a little confused about how to write it up.

Suppose that $\phi : G \to H$ is an isomorphism. Let $a,b \in G$. Then $ab = ba \implies \phi (a) \phi (b) = \phi (b) \phi (a)$. Here is where my question lies. To show that $H$ is abelian, I need to show that any two arbitrary elements commute. Why does $\phi (a) \phi (b) = \phi (b) \phi (a)$ tell me that any two arbitrary elements of $H$ commute? Does it have something to with $\phi$ being a bijection?

 

[fresh_42]

I am trying to prove that if two groups are isomorphic then one is abelian iff the other is abelian. This is a simple task, but I am a little confused about how to write it up.

Suppose that $\phi : G \to H$ is an isomorphism. Let $a,b \in G$. Then $ab = ba \implies \phi (a) \phi (b) = \phi (b) \phi (a)$. Here is where my question lies. To show that $H$ is abelian, I need to show that any two arbitrary elements commute. Why does $\phi (a) \phi (b) = \phi (b) \phi (a)$ tell me that any two arbitrary elements of $H$ commute? Does it have something to with $\phi$ being a bijection?

Yes. $\phi$ covers all elements of $H$:
$ab = ba \Longleftrightarrow \phi (a) \phi (b) = \phi (b) \phi (a) \Longleftrightarrow a'b'=b'a'$
You simply already have written all elements of $H$ since $H=\operatorname{im} \phi = \{\,\phi(a)\,|\,a\in G\,\}$.