TITLE: Subgroups of $\mathbb{Z}^k$ of finite index $n$
QUESTION [6 upvotes]: I want to describe all subgroups  in $\mathbb{Z}^k$ of finite index $n$.
I have solved it for the case $k=2$. In $\mathbb{Z}^2$, each subgroup of index $n$ corresponds to a matrix $\left( \begin{array}{a}
a & b \\
0 & d \end{array} \right)$ with $a, b, d \in \mathbb{Z}$ and $ad=n; a,d>0; 0 \le b<d$. Can someone give a hint how to go about it in the general case? For the case $k=3$, my guess is that each subgroup corresponds to a matrix $\left( \begin{array}{a}
a & b & c \\
0 & d & e \\
0 & 0 & f \end{array} \right)$ where $a, b, c, d, e, f \in \mathbb{Z}$ and $adf=n$; a, d, f >0; $0 \le b <d; 0 \le c <f; 0 \le e<f$. Is this right?

REPLY [5 votes]: If $\Gamma<\Bbb Z^k$ has index $n$, then $n(\Bbb Z^k/\Gamma)=(0)$. Therefore $\Gamma$ contains the subgroup $(n\Bbb Z)^k$.
This shows that there's a bijection between the wanted subgroups and the subgroups of
$$\Bbb Z^k/(n\Bbb Z)^k\simeq(\Bbb Z/n\Bbb Z)^k$$
of index $n$. 
Then one can simply apply the structure theorem for finite abelian groups.