Showing a group is abelian of order $\det(M)$

Question

I'm having some difficulty with this question, I don't really know how to tackle it, if anyone can help me I would be very grateful.

Let $M \in \Bbb Z^{n \times n} $ be of matrix-rank n with rows $\underline m_1,\underline m_2,\cdots,\underline m_n$. Let $S:=\langle \underline m_1,\underline m_2,\cdots,\underline m_n\rangle \le \Bbb Z^n$ and $A:=\Bbb Z^n/S$. Show that A is a finite abelian group of order $\det(M)$.

Answer 1

Hint for the first part:

It is a finitely generated abelian group, and tensoring with $\mathbf Q$ above $\mathbf Z$, you obtain $$S\otimes_{\mathbf Z}\mathbf Q\simeq \mathbf Z^n\otimes_{\mathbf Z}\mathbf Q\simeq \mathbf Q^n,$$ hence $\;A\otimes_{\mathbf Z}\mathbf Q=\{0\}$.