Exercise 1.6.3 (Two isomorphic groups are both abelian or both non abelian)

Proof. Let $\phi :G\to H$ be an isomorphism, where $G$ is an abelian group. If $a$ and $b$ are any elements of $H$, since $\phi$ is surjective, there are elements $c,d\in G$ such that $a=\phi (c),b=\phi (d)$. Since $G$ is abelian, we obtain

This shows that $H$ is abelian.

Conversely, suppose that $H$ is abelian. Since ${\phi }^{-1}:H\to G$ is also an isomorphism, the preceding result shows that $G$ is abelian. So $G$ is a abelian if and only if $H$ is abelian.

If $G$ is abelian and $\phi :G\to H$ is a surjective homomorphism, the preceding argument shows that $H$ is also abelian.

If $\phi$ is not surjective, the following counterexample shows that this is false. Consider the map $\phi :\mathbb{Z}\to D_6=⟨r,s\mid r^3=s^2=e,\mathit{sr}=r{s}^{-1}⟩$, where $\phi$ is defined by $\phi (n)=r^n$. Then $\phi$ is a homomorphism (which is not surjective) and $\mathbb{Z}$ is abelian, but $D_6$ is not abelian. □