Trying to prove that $\hom\left(\Bbb Z_n , \frac{\mathbb{Q}}{\mathbb{Z}}\right) \cong \Bbb Z_n$ Trying to prove that $$\hom\left(\Bbb Z_n , \frac{\mathbb{Q}}{\mathbb{Z}}\right) \cong \Bbb Z_n$$
So I started out trying to understand when two rational numbers $\frac{p}{q}$ and $\frac{r}{s}$ are equivalent mod $\mathbb{Z}$.
$\frac{p}{q} \cong \frac{r}{s} \iff \frac{ps-rq}{qs} \in \mathbb{Z} \iff ps-rq=mqs$ for some $m \in \mathbb{Z}$
but I guess this didn't give me a whole lot of insight. If somebody could explain to me why this is true that would be great! As always multiple perspectives are appreciated!
 A: As always, a homomorphism $\varphi:\mathbb{Z}_n\rightarrow\mathbb{Q}/\mathbb{Z}$ is determined by its value $\varphi(1)$ at $1$. Only restriction is that $n\varphi(1)=0$. For any $0\leq m<n$, the map $1\mapsto m/n+\mathbb{Z}$ gives a homomorphism. I claim that if $\varphi$ is a homomorphism, then it is equivalent to one of the maps $1\mapsto m/n$. Say the order of $\varphi(1)=q+\mathbb{Z}$ is $d$. Then, $d\varphi(1)=0$, i.e. $dq\in\mathbb{Z}$. Hence, $q = m/d$ for some $m$. Write $m=ad+b, 0\leq b<d$, so $m/d+\mathbb{Z}=b/d+\mathbb{Z}$. Write $\frac{b}{d}=\frac{b\frac{n}{d}}{n}$ so we get a number of the form $k/n$, $0\leq k<n$. This shows that every homomorphism $\mathbb{Z}\rightarrow\mathbb{Q}/\mathbb{Z}$ is of the form $1\mapsto k/n$ for $0\leq k<n$. Moreover, all these maps are indeed distinct since for $k\neq s$, $k/n+\mathbb{Z}=s/n+\mathbb{Z}$ implies $(k-s)/n\in\mathbb{Z}$, which is not possible.
This is how we obtain $n$ different homomorphism. I leave the part where one shows the group of homomorphisms is isomorphic to $\mathbb{Z}_n$ to you, you just need to show that $1\mapsto 1/n+\mathbb{Z}$ generates the other maps.
