free $\mathbb Z$-module of rank $2$

Question

Let $M$ be a free $\mathbb{Z}$-module of rank $2$ and $p$ be a prime. Determine the number of submodules $N$ of $M$ such that $M/N \cong \mathbb{Z}/p \mathbb{Z}$.

The answer may be $2$, but I cannot find a proof.

Answer 1

There is a minimal submodule $K$ such that $M/K$ is of exponent $p$, and this is clearly $K = p M$.

So the question reduces to counting the submodules $N/K$ of $M/K$ such that $(M/K)/(N/K) \cong M/N$ is isomorphic to $\mathbb{Z}/p \mathbb{Z}$.

Since $M / K \cong \mathbb{Z}/p \mathbb{Z} \oplus \mathbb{Z}/p \mathbb{Z}$, a vector space of dimension two over the field $\mathbb{Z}/p \mathbb{Z}$, this number is readily seen to be $p+1$.