Find the factor groups of the $\mathbb{Z}/252\mathbb{Z}$ which are projective $\mathbb{Z}/252\mathbb{Z}$-modules

Question

I'm reading about projective modules and saw an algebra exam problem that and uses a bunch of concepts I am familiar with (question in the title) but I am not sure of the best method to solve it. Should I begin by considering some epimorphism from a $\mathbb{Z}252$-module to some factor group of $\mathbb{Z}252$ and see what orders the kernel of said epimorphism can take and still trivially intersect another subgroup with an order that divides 252? I am still new to modules so suspect I am missing some obvious points. I know there is also the definition of a projective module being the direct summands of free-modules but for some reason free-moduels are tougher for me to grasp than projective-modules (which are tough anyway).

Answer 1

The factor groups of $R=\mathbb{Z}/n\mathbb{Z}$ are the $\mathbb{Z}/d\mathbb{Z}$'s, $d\mid n$ (up to isomorphism)

Assume that $\mathbb{Z}/d\mathbb{Z}$ is projective, and consider the surjective $R$-linear map $f: \mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/d\mathbb{Z}$. The kernel of $f$ is isomorphic to $\mathbb{Z}/(n/d)\mathbb{Z}$, and we have an exact sequence $ 0\to \mathbb{Z}/(n/d)\mathbb{Z}\to \mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/d\mathbb{Z}\to 0$, which splits since $\mathbb{Z}/d\mathbb{Z}$ is projective.

Hence, we have $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/d\mathbb{Z} \times \mathbb{Z}/(n/d)\mathbb{Z}$ as $R$-modules and thus as abelian groups.

This will happen if and only if $d$ and $n/d$ are coprime (if they are not coprime, the product won't have an element of order $n$, and if they are , this CRT).

Conversely, if $d$ and $n/d$ are coprime , then you can show easily that the isomorphism given by CRT is also an isomorphism of $R$-modules, so that $\mathbb{Z}/d\mathbb{Z}$ is a direct factor of the free $R$-module $R$, hence projective.

So we proved:

Thm. The projective factor groups of $\mathbb{Z}/n\mathbb{Z}$ are the $\mathbb{Z}/d\mathbb{Z}$'s, where $d\mid n$ and $d$ is coprime to $n/d$.

For $n=252$, this gives $d=1,4,9,7,28,36,63,252$.