subgroup of topological group is either clopen or has empty interior

Theorem - Every subgroup of a topological group is either clopen or has empty interior.

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Proof: Let $G$ be a topological group and $H\subseteq G$ a subgroup. Suppose the interior of $H$ is nonempty, i.e. there is a non-empty open set $U$ of $G$ such that $U\subseteq H$. Translating $U$ around $H$ we can see that $H$ is open: if $u\in U$ then for every $h\in H$ the set $h{u}^{-1}U$ is open in $G$, is contained in $H$ and contains $h$, which implies that $H$ is open in $G$.

Let us now see that $H$ is closed. Let $\overline{H}$ denote the closure of $H$ and let $H^2$ be the set of elements of the form $h_1{h}_2$ where $h_1,h_2\in H$. Of course, since $H$ is a subgroup of $G$, we have that $H^2=H$. Also, since $H$ is open we know that$H\subseteq \overline{H}\subseteq H^2$ (see this entry (http://planetmath.org/BasicResultsInTopologicalGroups) - 5). Hence $\overline{H}=H$, i.e. $H$ is closed.

We have proven that a subgroup of a topological group must be clopen or it must have empty interior. Since this two topological properties can never be satisfied simultaneously, we have that every subgroup of a topological group is either clopen or it has empty interior. $\mathrm{\square }$