Suppose that $M$ and $N$ are finitely generated free $\mathbb{Z}$-modules, and $f: M \to N$ is a $\mathbb{Z}$-linear map corresponding to a matrix $F$. We would like an upper bound on the order of the torsion subgroup of $N / \mbox{image}(f)$ in terms of the matrix $F$. Is it true that the order of torsion group is bounded by the product of the lengths of the nonzero columns of $f$?
By the way, I realize that if $F$ is in Smith normal form, then this seems to be true. But what if $F$ is not in Smith normal form, and I just want a rough bound on the order of the torsion group, in these kinds of terms?
Here is a sketch of some information that would prove your conjecture.
By a change of basis for both $M$ and $N$, we can ensure that $F$ only has entries down the main diagonal; the torsion subgroup has order equal to the product of those diagonal entries, which is the same as the determinant of the submatrix consisting just of those rows and columns.
We can restrict to a subgroup of $M$ and $N$ of dimension the same as the rank of $F$ in a way that corresponds to throwing away rows and columns of $F$ leaving behind a nonsingular square submatrix. This will not increase the length of any of the columns. This also doesn't change the size of the torsion subgroup, since we're just throwing away the components of $M$ and $N$ that are irrelevant to the torsion.
The determinant of a matrix is the (signed) volume of the paralleliped whose edges are precisely the columns of the matrix. In particular, it cannot be greater than the product of the lengths of the columns.