# Explicitly describing the subgroups of $\mathbb{Z}^{3}$

I am interested in understanding all the subgroups of $$\mathbb{Z}^{3}:=\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$$.

$$\mathbb{Z}^{3}$$ a free abelian group of rank three, so all subgroups are free abelian of rank at most three. They are thus all isomorphic to either the trivial group, $$\mathbb{Z}$$, $$\mathbb{Z}^{2}$$ or $$\mathbb{Z}^{3}$$ itself. This is easy enough, but I want to understand the explicit subgroups, not just up to isomorphism.

Clearly all subgroups are generated by $$m\leq 3$$ linearly independent vectors in $$\mathbb{Z}^{3}$$, but I can't seem to come up with a nice way to describe these sets of vectors. Any pointers would be much appreciated.

• Look up the classification of finitely generated modules over a PID. It says in this case that each subgroup $H$ of $\mathbf Z^n$ is free of some rank $m \leq n$ and there is a basis $e_1, \dots, e_n$ of $\mathbf Z^n$ and (positive) integers $a_1, \ldots, a_m$ such that $a_1e_1, \ldots, a_me_m$ is a basis of $H$. – KCd May 6 at 4:27
• I don't think there's anything better than saying 3 linearly independent vectors in $\mathbb{Z}^3$. What kind of description are you looking for? Do you have such a description for $\mathbb{Z}^2$ ? – Ted May 6 at 5:04
• I trust, Coffee, that you understand the subgroups of $\bf Z$. Maybe first try to understand the subgroups of ${\bf Z}^2$ before you tackle ${\bf Z}^3$. The ones isomorphic to ${\bf Z}^2$ have generators $(a,b),(c,d)$ where $\delta=ad-bc\ne0$, and they have index $|\delta|$ in ${\bf Z}^2$. – Gerry Myerson May 6 at 5:06
• I have had a look at $\mathbb{Z}^{2}$. As you mentioned, for subgroups of rank $2$ we have generators $(a,b)$ and $c,d$ with $ad-bc\neq 0$. For the rank $1$ subgroups we have a single generator $(a,b)$ with $a,b$ not both zero. I have been unable to generalise this to, for example, the rank $2$ subgroups of $\mathbb{Z}^{3}$. – CoffeeCrow May 6 at 5:28

Let $$G \le {\mathbb Z}^n$$, and suppose that $$G$$ is spanned by the linearly independent vectors $$e_1,\ldots, e_m$$ with $$k \le n$$.
Form an $$m \times n$$ matrix in which these vectors are the rows, then put that matrix into (row) Hermite Normal Form. Then the rows $$f_1,\ldots,f_m$$ of that new matrix provide a canonical set of generators for $$G$$.