Computing $\mathrm{Hom}_{\mathbb Z}(\mathbb Z_n,G)$ by left-exactness

Question

In chapter 2 of Rotman's An introduction to homological algebra, we are set the exercise to show,

$$\mathrm{Hom}_{\mathbb Z}(\mathbb Z_n, G) = \{g\in G : ng = 0\}$$

where $G$ is an abelian group, hinting to use the left-exactness of $\mathrm{Hom}$. As far as I understand, this means choosing a useful exact sequence,

$$0 \to A \to B \to C \to 0$$

and using the induced sequence,

$$0\to \mathrm{Hom}_{\mathbb Z}(\mathbb Z_n,A) \to \mathrm{Hom}_{\mathbb Z}(\mathbb Z_n,B) \to \mathrm{Hom}_{\mathbb Z}(\mathbb Z_n,C).$$

Obviously either $A, B$ or $C$ I am guessing must be chosen as $G$, but I don't know how to proceed from here and which modules to choose for the computation.

Is my approach correct and if so how do I proceed? A hint is appreciated rather than a full answer.

Answer 1

Try instead $\mathrm{Hom}(-,G)$ instead. In particular, you have

$$0 \to \mathbb Z \to \mathbb Z \to \mathbb Z/n\mathbb Z \to 0.$$

Then $$0 \to \mathrm{Hom}(\mathbb Z/n\mathbb Z,G) \to \mathrm{Hom}(\mathbb Z,G) \to \mathrm{Hom}(\mathbb Z,G)$$ is exact.

Of course, this is really \begin{align}0 \to \mathrm{Hom}(\mathbb Z/n \mathbb Z,G) \to G \to G.\end{align} what are the maps?