Let $G$ and $H$ be cyclic groups. Find a necessary and sufficient condition when $H$ is a homomorphic image of $G$

Question

I think the condition is that $H$ is a homomorphic image of $G$ iff there is an $N\trianglelefteq G$ such that $H\cong G/N$, but I don't know how to prove it.

I saw this question posted elsewhere here, but the answer wasn't complete/they finished explaining in chat so I'm a little lost. Here is the original question for reference:

Necessary and Sufficient Condition for cyclic $H$ to be a homomorphic image of cyclic $G$

Answer 1

What you're probably lacking is just some comfort with the isomorphism theorems. You need to know that if $G$ is a group and we have a homomorphism $f:G \to G'$ then $ker(f)$ is a normal subgroup of $G$ and that $G/ker(f) \cong Im(f)$

So if we have $f: G \to H$ where $f(G) = H$, then the isomorphism theorem tells us that $G/ker(f) \cong H$ and so there exists a normal subgroup $N = ker(f)$ such that $G/N \cong H$.

Answer 2

For $H$ to be a homomorphic image of $G$, there must be a surjective map $G \to H$. Since $G \cong \mathbb{Z}/n \mathbb{Z}$ and $H \cong \mathbb{Z}/ m \mathbb{Z}$ for some $n,m \in \mathbb{N}$, such a map exists only when $m$ divides $n$.

If $G$ is infinite, then $H$ is always the homomorphic image of $G$, since it's a quotient of $\mathbb{Z}$. If $H$ is infinite then $G$ must also be infinite.