Relationship of size of quotient group to matrix determinant In Alperin's book, Local Representation Theory (p. 169) there is a claim I am finding difficult to verify.
The setup is the following.  Given a free abelian group $V$ spanned by basis elements $\{v_1, \ldots, v_n\}$ and $n$ other elements $\{w_1, \ldots, w_n\}$ spanning subgroup $W$, we want to calculate the size (in number of elements) of $V/W$.
The claim is that this is finite iff the matrix $C$ expressing the $w_i$ in terms of the $v_i$ has nonzero determinant.  Moreover if this is so, its exact size is the modulus of the determinant of $C$.
This is explained only as "by the theory of elementary divisors" and I was wondering if anybody had either a proof of this, or directions to some material towards a proof of this.
Edit: specified "...exact size is the modulus of the determinant..."
 A: Invariant factors refers to Smith normal form (link: wikipedia).
Write $w_i = \sum_{j = 1}^n A_{ji} v_j$, where $A_{ji} \in \mathbb{Z}$. 
The idea is that by changing the generators of $W$ and $V$ suitably, we can find new generating sets $\{ w_1', \ldots, w_n' \}$ of $W$ and $\{v_1', \ldots, v_n'\}$ of $V$ such that for all $i$, $$w_i' = k_i v_i'$$ for some non-negative integers $k_i$ such that $k_i \mid k_{i+1}$. This follows from the Smith normal form applied to the integer matrix $A$. Here we have $|\det A| = k_1 \cdots k_n$.
It is obvious from $w_i' = k_iv_i'$ that $$V/W \cong \mathbb{Z}/k_1 \mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/k_n\mathbb{Z}.$$
Hence $V/W$ is finite if and only if $k_i > 0$ for all $i$; so $V/W$ is finite if and only if $\det A \neq 0$. Finally, when $V/W$ is finite, it has order $k_1 \cdots k_n = |\det A|$.
A: $\mathcal{W} = C \mathcal{V}$ implies $\operatorname{adj}(C)\mathcal{W} = \det(C) \mathcal{V}$. Therefore, $\det(C)v \in W$ for every $v \in V$ and so $\det(C)$ kills every element of $V/W$. Thus, if $\det(C)\ne0$, then $V/W$ is a finitely generated torsion abelian group, and so must be finite. 
