I have to prove the following statement:
Let $d$ be a divisor of $n$. Then $\mathbb{Z}/d\mathbb{Z}$ is a projective $\mathbb{Z}/n\mathbb{Z}$-Module $\iff$ $\text{gcd}(d, \frac{n}{d}) = 1$.
The implication $\impliedby$ follows by the chinese remainder theorem and the fact that projective $R$-Modules are direct summands of free $R$-Modules.
I'm having trouble with the implication $\implies$. Let $c = \frac{n}{d}$ and $N = \mathbb{Z}/d\mathbb{Z}$ be a projective $\mathbb{Z}/n\mathbb{Z}$-Module. Then there exists a $\mathbb{Z}/n\mathbb{Z}$-Module $M$ s.t. $\mathbb{Z}/n\mathbb{Z} \cong M \oplus N$ because $\mathbb{Z}/n\mathbb{Z}$ is free over itself. By doing a change of basis via the canonical homomorphism $\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ we have $\mathbb{Z}/n\mathbb{Z} \cong M \oplus N$ as abelian groups. Now this can only be the case if $M \cong \mathbb{Z}/c\mathbb{Z}$ because otherwise the LHS would be cyclic of order $n$ and the RHS would not. Hence $\text{gcd}(d,c) = 1$.
I was given the hint to use the FT of finitely generated abelian groups but I didn't use this here so I'm not really sure if my proof is correct, corrections and/or hints are appreciated.
From this I have to deduce that if $n = \prod_{i=1}^r p_i^{e_i}$ is the prime factor decomposition of $n$ then every finitely generated projective $\mathbb{Z}/n\mathbb{Z}$-Module $M$ is of the form $$\bigoplus_{i=1}^r (\mathbb{Z}/p_i^{e_i}\mathbb{Z})^{f_i}$$ with suitable $f_i \in \mathbb{N}_0$. I don't really see how this follows from the above statement? Help is greatly appreciated!