If $G$ is abelian and $\varphi:G\to H$ is a non-trivial homomorphism, then does $H$ have to be Abelian?

Question

If $G$ is abelian and $\varphi:G\to H$ is a non-trivial homomorphism, then does $H$ have to be Abelian?

My guess is yes since if $a,b\in G$ and $\varphi (a*b)=\varphi(a)*\varphi(b)$ and $\varphi(b*a)=\varphi(a*b)$ then $\varphi(a)*\varphi(b)=\varphi(b)*\varphi(a)$, so $H$ must be abelian too.

Is this sound reasoning?

Answer 1

Consider $G=\Bbb Z_2$ and $H=\Bbb Z_2\times S_3$. Define

$$\begin{align} \varphi: G&\to H,\\ g&\mapsto (g,e). \end{align}$$

Then $\varphi$ is a homomorphism, $G$ is abelian, but $H$ is nonabelian.

Exercise: Generalise $\varphi$, $G$, and $H$ to find other counterexamples.

Hint: Let $\varphi:G\to H$ be a homomorphism that is not surjective. Then ${\rm Im}(\varphi)$ is indeed abelian if $G$ is abelian. What can you say about $(H\setminus {\rm Im}(\varphi))\cup\{e\}$?