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.
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2Since $p$ is prime, saying $M/N \cong {\mathbf Z}/p{\mathbf Z}$ is the same as saying $N$ has index $p$ in $M$. If $N \subset M$ has index $p$ then $p(M/N)$ is 0, so $pM \subset N$. Thus $pM \subset N \subset M$, so counting such $N$ is the same as counting index $p$ subgroups of $M/pM \cong ({\mathbf Z}/p{\mathbf Z})^2$, which is now a linear algebra problem. In a similar way, the index $d$ submodules of $M$ are in bijection with the index $d$ subgroups of $({\mathbf Z}/d{\mathbf Z})^2$. – KCd Jun 13 '14 at 17:04
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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 $p+1$.
Andreas Caranti
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