If $f:\mathbb{Z}^m\to\mathbb{Z}^m$ is a module homomorphism, then $\mathbb{Z}^m/\operatorname{im}(f)$ is a finite abelian group

Question

If $f:\mathbb{Z}^m\to\mathbb{Z}^m$ is an injective group homomorphism, then is $\mathbb{Z}^m/\operatorname{im}(f)$ a finite abelian group?

I think yes. Let the homomorphism be given by a $m\times m$ matrix. Then, by existence of smith normal form over $\mathbb{Z}$, I think $\mathbb{Z}^m/\operatorname{im}(f)$ would be a direct product of cyclic groups of orders given by the entries in the diagonal of the smith normal form. Is this right? If so, how do we prove it? Any hints? Thanks beforehand.

Answer 1

The Smith normal form works fine (and with $f(x)=Ax$ it proves that $|\Bbb{Z}^n/Im(f)| = |\det(A)|$) but it is easier to find a non-zero matrix $B\in M_n(\Bbb{Z})$ such that $AB=d\ I$.