# When Are Discrete Subgroups Closed?

Let $H\subseteq G$ be a subgroup of a topological group $G$ (henceforth abbreviated "group"). If the induced topology on $H$ is discrete, then we say that $H$ is a discrete subgroup of $G$. A commonplace example is the subgroup $\mathbb{Z}\subseteq \mathbb{R}$: the integers are normal subgroup of the real numbers (with the standard topology).

Observe that in $\mathbb{R}$, the subset $\mathbb{Z}$ is also a closed set. What about an arbitrary discrete subgroup $H\subseteq\mathbb{R}$? In other words, if $H$ is discrete, is $H$ necessarily closed?

The answer is yes! If $G$ is any Hausdorff (equivalently: $T_0$) group then any discrete subgroup is closed. In order to show this, we just need to find an open neighbourhood $U$ of the identity $e$ such that $\overline{U}\cap H$ is closed in $G$, where $\overline{U}$ is the closure of $U$.

But since $H$ is discrete, we just choose a neighbourhood $U$ of the identity such that $U\cap H = \{e\}$, and then take another neighbourhood $V$ of the identity such that $\overline{V}\subseteq U$. Then $\overline{V}\cap H = \{ e\}$, and $\{e\}$ is closed in $G$ because $G$ is Hausdorff.

### No More Hausdorff

What if we drop the condition that $G$ is Hausdorff? Is it possible to find a group $G$ and a discrete subgroup $H\subseteq G$ such that $H$ is not closed in $G$? A first example that comes to mind is any nontrivial group $G$ with the indiscrete topology. Then the subgroup $\{e\}$ is a discrete subgroup but is not closed in $G$.

This is not a very interesting example. How fine can we make the topology be on $G$ while keeping $\{e\}$ a discrete nonclosed subgroup? An interesting topology comes from taking any normal subgroup $N\subseteq G$ and letting the cosets of $N$ be a basis for $G$ (it's not difficult to verify that this actually does give a basis, and that $G$ becomes a topological group with this basis).

This means that we can get some pretty interesting examples: take for instance the finite Abelian group $\mathbb{Z}/(2n)$ where $n\in\mathbb{Z}$ is nonzero. Then the subgroup $\{0,n\}\subseteq\mathbb{Z}/(2n)$ is normal, and so its $n$ cosets form a basis for a topology on $\mathbb{Z}/(2n)$ making it into a topological group.There are quite a few open sets (how many?!) here and yet $\{e\}$ clearly cannot be a closed subgroup, yet it's discrete!

### Some Related Problems

Here are two curious bonus problems:

Problem: can we give $\mathbb{R}$ a topology with infinitely many open sets, making it into a topological group such that $\mathbb{Z}$ is discrete but not closed?

Problem: find a direct proof that every discrete subgroup of $\mathbb{R}$ is closed. Does the same proof work for $\mathbb{R}$ replaced by an arbitrary Hausdorff group?