Determine all group homomorphisms from $\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}$.
Attempt:
Would any group homomorphism be the zero homomorphism or a map $1 \mapsto a$ for any generator $a$ of $\mathbb{Z}/n\mathbb{Z}$? Are these the only group homomorphisms?
I am struggling to come up with a proof for this. I know the kernel of such a map would be $k\mathbb{Z}$ for some $k$.
Does this help at all?
Not quite: $1\mapsto a$ defines a group homomorphism for any $a\in\mathbb{Z}_4$, not only a generator. Note that for $a=0$ this also covers the zero homomorphism.
The group of integers $\mathbb{Z}$ is a free abelian group, meaning that given any abelian group $G$ and any function $f:\{1\}\to G$ we have that $f$ uniquely extends to a group homomorphism $F:\mathbb{Z}\to G$. This group homomorphism is quite easy to define:
$$F(x):=x\cdot f(1)$$
And vice versa every group homomorphism $\mathbb{Z}\to G$ arises in this fashion. Therefore there is a bijection between $Hom(\mathbb{Z},G)$ and $Func(\{1\},G)$.
In particular there are $4$ group homomorphisms $\mathbb{Z}\to\mathbb{Z}_4$ determined by $4$ possible assignements $1\mapsto a$ for $a\in\mathbb{Z}_4$.