Show that $\lvert \mathbb Z^n / N\rvert = \lvert \det (A) \rvert$.

Question

Let $N$ be a rank $n$ submodule of $\mathbb Z^n$, and let $A$ be the matrix with rows being the generators of $N$. Show that $\lvert \mathbb Z^n / N\rvert = \lvert \det (A) \rvert$.

So this is a homework problem, and I am a little confused. Shouldn't it be the case that $\mathbb Z^n$ is the unique (up to isom.) free module of rank $n$? Then, wouldn't this imply that $N = \mathbb Z^n$? I think I am missing something here, but even assuming that this is indeed the case, then we would be trying to prove that $\lvert \det (A) \rvert = 1$ for every invertible matrix $A$ with entries in $\mathbb Z$, which I don't think is true.

Answer 1

First of all one does not necessarily have that $N \cong \Bbb Z^n \implies N = \Bbb Z^n$, consider for example $\Bbb Z$ and $2\Bbb Z$ (the even numbers). Both modules are isomorphic but one is a proper subset of the other. Now consider $N$ as a sublattice of $\Bbb Z^n$ and let let the matrix $A$ determine a basis of $N$ then $A$ can be replaced by its Smith normal form (a diagonal matrix), as explained in the answer to this question. The unit of the lattice $N$ is a polyhedron whose dimensions are described by the diagonal elements $d_1, \ldots, d_n$, so one can chose a representative of the quaotient $\Bbb Z^2/N$ by a vector $(z_1, \ldots, z_n)$ where $0 \leq z_i < d_i$, so their number equals $d_1d_2\ldots d_n$, which equals the determinant of $A$. The situation is analoguous to $\Bbb Z/n\Bbb Z$ in more than one dimension.