Prove $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},A)\cong A[n]$ by using left-exactness of Hom? Let $A$ be an abelian group, then $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},A)\cong A[n]$, where $A[n]=\{a\in A:na=0\}$. How do I prove this by using left-exactness of Hom? (This is Exercise 2.7 from Rotman's An Introduction to Homological Algebra.)
I tried applying the functor $\operatorname{Hom}_\mathbb{Z}(-,A)$ to the short exact sequence $0\to\mathbb{Z}\xrightarrow{\mu_n}\mathbb{Z}\xrightarrow{\pi}\mathbb{Z}/n\mathbb{Z}\to0$, where $\mu_n$ is multiplication by $n$ and $\pi$ is the canonical projection, but it didn't work.

When I do that I get
$$0\to\text{Hom}(\Bbb Z/n\Bbb Z,A)\to A\to A$$
is exact where the $A\to A$ map is multiplication by $n$. That gives
me $\text{Hom}(\Bbb Z/n\Bbb Z,A)\cong A[n]$ straight away.
