Determine $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z}_m)$

Question

Determine $\mathrm{Hom}_{\mathbb{z}}(\mathbb{Z},\mathbb{Z}_m)$

I've already described $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}_n,\mathbb{Z}_m)$ but I am struggling with the simpler version of the problem above. I know I once again look at where 1 is mapped to i.e $\phi(1)=a$ and we know $\phi(0)=0$ so $\phi(0) = \phi(n) = na = 0$ but I'm blanking on what this tells me about a.

Can someone point me in the right direction to finish the proof?

Answer 1

Hint: There's no restriction to $a$ other than being an element of $\Bbb Z_m$.

Prove more generally that there's always a bijection $$\hom(\Bbb Z, G)\simeq G$$ for any group $G$.