I was recently reading a paper by Cantor and Mills, when I came across the following lemma.
Let $m$ and $t$ be positive integers. Let $X$ be the additive group of all $m$-dimensional column vectors with integer elements, let $Y$ be a finite set of t-dimensional column vectors with integer elements, and let c be the cardinality of $Y$. Suppose that $A$ is an $m$ by $m$ matrix of integers and that $B$ is an $m$ by $t$ matrix of integers. If, for $x\in X$ and $y \in Y$, the column vector $Ax + By$ determines $x$ and $y$ uniquely, then $\vert\text{det}\;A\vert \geq c$.
Proof. Let $G$ be the subgroup of $X$ generated by the columns of $A$. Thus $G$ is the set of all vectors $Ax$ with $x \in X$. By hypothesis, the column vectors of the form $Ax + By$, with $x \in X$ and $y \in Y$, are all distinct. Therefore as $y$ ranges over the $c$ elements of $Y$, By ranges over $c$ distinct cosets of $G$ in $X$. Hence the index $X:G$ of $G$ in $X$ is at least $c$. On the other hand $X: G$ is equal to the absolute value of the determinant of $A$. Thus, $$\vert\text{det}\;A\vert = X:G \geq c$$ and the proof is complete.
I don't understand how the authors get $\vert\det A\vert = X:G$ and I wasn't able to prove it myself. Here's what I tried: We want to find the number of sets $S$ of $m$-dimensional column vectors with integer entries such that for all $c,c' \in S$, there exists $x,x' \in X$ that satisfy $Ax + c = Ax' + c'$. Note that $$Ax + c = Ax' + c' \implies x + \frac{\text{adj}(A) \cdot c}{\det A}= x' + \frac{\text{adj}(A) \cdot c'}{\det A},$$ thus there exists a solution $x,x' \in X$ if each entry of $\text{adj}(A) \cdot c$ and $\text{adj}(A) \cdot c'$ is the same modulo $\det A$. At first, I thought that we could get $\det A$ sets just by assigning one set to each integer from $1, \dots, \det A$. However, not every entry of $\text{adj}(A) \cdot c$ can take on all values $1, \dots, \det A$, and there are plenty of examples of $c,c'$ for which are congruent modulo $\det A$ in one entry but not in another.
I can't seem to make any progress beyond this. I've looked for a proof of the above statement to no avail. Could anyone explain how to prove this statement? Thanks in advance.
