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

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.

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Projectives are direct summands of free. Is this (torsion)free? – user26857 Jan 4 '13 at 22:29
Z is $\mathbb Z$, right? (Try to use the right mathematical symbols!) – user26857 Jan 5 '13 at 9:49
$\mathbb{Z}$ is indecomposable, because any two nonzero subgroups have non trivial intersection. So the only projective quotients of $\mathbb{Z}$ are $\mathbb{Z}$ and $\{0\}$. – egreg Nov 26 '15 at 21:59

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 $\phi\circ \pi\ne \mathrm{id}$.
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.
@user53800, the result you're asking about is Baer's proof of the Baer-Specker group $\,\Bbb Z^{\Bbb N}\,$. The proof can be found in Kaplansky's book "Infinite Abelian Groups", but also in the following PDF: tinyurl.com/bkjfrxn . This group always intrigued me... – DonAntonio Jan 4 '13 at 23:03