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?