Discrete Subgroup of Hausdorff Group is Closed

Theorem
Let $G$ be a Hausdorff topological group.

Let $H$ be a discrete subgroup of $G$.

Then $H$ is closed in $G$.

Proof
Let $g\in \overline H$ be in the closure of $H$. We will show that $g\in H$.

Suppose $g\notin H$.

Let $e$ be the identity of $G$.

Because $H$ is discrete, there exists an open set $U\subset G$ such that $U\cap H = \{e\}$.

Then $V=U\cap U^{-1}$ is an open neighborhood of $e$ in $G$.

By Right and Left Regular Representations in Topological Group are Homeomorphisms, $gV$ is a neighborhood of $g$ in $G$.

By definition of closure, $gV \cap H \neq \emptyset$.

Let $h \in gV \cap H$.

Because $g \neq h$ and $G$ is Hausdorff, there exists a neighborhood $W$ of $g$ in $G$ with $h\notin W$.

Then $gV \cap W$ is a neighborhood of $g$.

Let $k \in gV \cap W \cap H$.

Then $k\neq h$.

Then

Bu assumption, $k^{-1}h=e$, so $k=h$, which is a contradiction.

Thus, by Proof by Contradiction, $g\in H$.

Thus $\overline H = H$.

Thus $H$ is closed.

Also see

 * Group is Hausdorff iff T1
 * Discrete Subgroup of Real Numbers is Closed