$\mathbb Z/n\mathbb Z$ is not a projective module

Question

I want to show that $\mathbb Z/n\mathbb Z$ is not projective for $n\geq 2$. I choose the exact sequence $\mathbb Z\stackrel{\pi}\rightarrow\mathbb Z/n\mathbb Z\rightarrow 0,$ and from $\mathbb Z/n\mathbb Z$ to $\mathbb Z/n\mathbb Z$ choose the identity map, and let $\phi$ :$\mathbb Z/n\mathbb Z \rightarrow \mathbb Z$, if I can show there is no way the diagram is commuted, then it's done, namely $\text{id}=\phi\circ \pi$ can not be true for any $\phi$, but I am confused here, why it's not. Hope someone can help me with that, thanks in advance.

Answer 1

If $\phi:\mathbb Z/n\mathbb Z\to \mathbb Z$ is a homomorphism, then $\phi(i)$ must have finite order for all $i\in\mathbb Z/n\mathbb Z$, since $i$ has finite order. But the only element of $\mathbb Z$ with finite order is $0$. So $\phi$ must be trivial, hence $\phi\circ \pi\ne \mathrm{id}$.

Answer 2

More directly, an abelian group (i.e., a $\,\Bbb Z-$module) is projective iff it is a free abelian group, and clearly a torsion abelian group cannot be free.