Relationship of size of quotient group to matrix determinant

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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..."

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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 $k_1 \cdots k_n = |\det A|$.

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$\mathcal{W} = C \mathcal{V}$ 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.