Pontryagin Dual of a Finite Abelian Group 
Let $M$ be a finite abelian group. I want to show that the Pontryagin dual is a finite abelian group, and in particular I am interested in computing the elementary divisors/invariant factors of it.

Any help is appreciated.
 A: Actually there's not much to say. Since a finite abelian group is a direct sum of cyclic modules of prime power order, you just have to compute $\operatorname{Hom}_{\mathbb{Z}}(M,\mathbb{Q}/\mathbb{Z})$ for a finite cyclic group of prime power order $M$ and it turns out that this is isomorphic to $M$. (Prime power order is not necessary for the isomorphism, but it makes the proof easier.)
Therefore, for a finite abelian group $M$, we have
$$
\operatorname{Hom}_{\mathbb{Z}}(M,\mathbb{Q}/\mathbb{Z})\cong M
$$
Sketch of the proof in the prime power order case. If the group is $\mathbb{Z}/p^n\mathbb{Z}$, a homomorphism $\mathbb{Z}(p^n)=\mathbb{Z}/p^n\mathbb{Z}\to\mathbb{Q}/\mathbb{Z}$ has its image contained in the $p$-torsion part of $\mathbb{Q}/\mathbb{Z}$, which is the Prüfer $p$-group $\mathbb{Z}(p^\infty)$. The image of the homomorphism is thus contained in the subgroup annihilated by $p^n$ which is isomorphic to $\mathbb{Z}(p^n)$. Since for a cyclic group $G$ we have $\operatorname{Hom}_{\mathbb{Z}}(G,G)\cong G$, we are done.
A: Since $M$ is finite and abelian, there exists $n$ such that $nx=0, x\in M$ where $0$ is the neutral element of $M$. Let $f\in Hom_Z(M,Q/Z)$, you must have $nf(x)=0$, thus $f(x)$ is the class of $m/n$. This implies that $Hom_Z(M,Q/Z)$ is finite. 
