I am currently reading the book "Algebra" by Hungerford. I saw some interesting theorem:
If $F$ is a free abelian group of finite rank $n$ and $G$ is a nonzero subgroup of $F$, then there exists a basis $\{x_{1},\ldots,x_{n}\}$ of $F$, an integer $r$ $(1 \leq r \leq n)$ and positive integers $d_{1},\ldots,d_{r}$ such that $d_{1} | d_{2} | \cdots | d_{r}$ and $G$ is free abelian with basis $\{ d_{1}x_{1},\ldots,d_{r}x_{r}\}$.
The proof is also interesting, but this theorem does not give a way of obtaining the basis. It only guarantees the existence of the basis and numbers.
Is there any systematic way of obtaining the basis $\{x_{1},\ldots,x_{n}\}$ and the numbers $d_{1},\ldots, d_{r}$?
I solved this problem when $n=2$.
Let $F=\mathbb{Z} \times \mathbb{Z}$ and $G=a\mathbb{Z} \times b\mathbb{Z}$ with $\gcd(a,b)=n$. By Euclidean algorithm, there exist integers $c,d$ such that $ad-bc=n$. Then $d_{1}=n$, $d_{2}=\frac{ab}{n}$ and $x_{1}=(\frac{a}{n},\frac{b}{n})$, $x_{2}=(c,d)$.
My proof is nothing special. Since $$\det \begin{bmatrix} a/n & b/n\\ c& d \end{bmatrix} =1,$$ the collection $\{(a/n,b/n), (c,d)\}$ forms a basis for $\mathbb{Z}^2$ (see Basis for $\mathbb{Z}^2$). The collection $\{d_{1}x_{1},d_{2}x_{2}\}$ also forms a basis for $G$ because $d_{1}x_{1}=(a,b) \in G$ and $$-(bc/n)nx_{1}+d_{2}x_{2}=(0,b) \in G.$$
Can we generalize this argument to the higher rank? Otherwise, is there any different method?
EDIT: the subgroup $\{(x,y) \in \mathbb{Z}^2: x-y \; \textrm{is} \; \textrm{even} \}$ is not of the form $a\mathbb{Z} \times b\mathbb{Z}$.
