The problem:
Let $M \leq \mathbb{Z}^n$ be the subgroup generated by the rows of an
$n \times n$ matrix $A$ with entries in $\mathbb{Z}$. Show that $G=\mathbb{Z}^n/M$ is finite if and
only if $\det A \neq 0$, and in that case the order of $G$ is $|\det A|$.
Let's label the rows of $A$ by $v_1, …, v_n \in \mathbb{Z}^n$.
- The if and only if (solution verification)
Firstly, I think I understand how to show the if and only if. Here's what I did, if there are any mistakes please correct me.
If $\det A = 0$, then there will be some $v \in \mathbb{Z}$ such that $v \notin \text{span} \left(v_1, …, v_n \right)$. Then $v+M, 2v+M, 3v+M, …$ are all distinct elements of $G$. Conversely, if $\det A \neq 0$, we can always write any $v$ as some linear combination $v=\sum_{i=1}^n q_i v_i$ where $q_i \in \mathbb{Q}$. $q_i$ will have denominator at most the lowest common multiple of the integers in $v_i$, say $b_i$, otherwise we'll not get an integer. Then we can write any element of $G$ as $v+M=\sum_{i=1}^n \{q_i\} v_i$ where $q_i \in \mathbb{Q}$ where $\{ \cdot \}$ denotes the fractional part. Thus there's at most $\prod_{i=1}^{n} b_i < \infty$ elements in $G$.
- $|G|=|\det A|$ (help)
I'm stuck on this. I don't really get how I can relate anything I've done in the first part of this to the determinant, the use of the lowest common multiple or the product $\prod_{i=1}^{n} b_i$ doesn't seem to lend too easily into thinking about determinants, but perhaps I'm missing something. I'm thinking we could perhaps try to show that the group is isomorphic to the same group with $A$ in Smith Normal Form, which would then lead nicely into thinking of the determinant, since it'll be a diagonal matrix, but I'm not really sure how to approach that.
Best Answer
Smith normal form for $\mathbb Z$-modules can be formulated as the statement that if $M$ is a free $\mathbb Z$ module and $N\leq M$ is a submodule, then there is a basis $\{e_1,\ldots,e_n\}$ such that $N = \text{span}_{\mathbb Z}\{f_1,\ldots,f_k\}$ where $k \leq n$ and $f_i = c_i e_i$ for $c_i \in \mathbb Z$ with $c_1|c_2|\ldots|c_k$.
You can use this formulation to give a proof of both parts of the question.