\subsection{Topological Dynamics}

\begin{fact}[\cite{801106}]
    \label{fact:topsubgroupclosure}
    Let $H$ be a topological group
    and $G \subseteq H$ a subgroup.
    Then $\overline{G}$ is a topological
    group.

    Moreover if $H$ is Hausdorff and $G$ is abelian,
    then is $\overline{G}$ is abelian.
\end{fact}
\begin{proof}
    Let $g,h \in \overline{G}$. We need to show that $g\cdot h \in \overline{G}$.
    Take some open neighbourhood $g \cdot h \in U \overset{\text{open}}{\subseteq} H$.
    Let $V \overset{\text{open}}{\subseteq} H \times H$
    be the preimage of $U$ under $(a,b) \mapsto a \cdot b$.
    Let $A \times B \subseteq V$ for some open neighbourhoods of $g$ resp.~$h$.
    Take $g' \in A \cap G$ and $h' \in B \cap G$.
    Then $g'h' \in U \cap G$,
    hence $U \cap G \neq \emptyset$.

    Similarly one shows that $\overline{G}$ is closed under inverse images.

    Now suppose that $H$ is Hausdorff and $G$ is abelian.
    Consider $f\colon (g,h) \mapsto [g,h]$\footnote{Recall that the \vocab{commutator} is $[g,h] \coloneqq g^{-1}h gh^{-1}$.}.
    Clearly this is continuous.
    Since $G$ is abelian, we have  $f(G\times G) = \{1\}$.
    Since $H$ is Hausdorff, $\{1\}$ is closed, so
    \[
    \{1\} = \overline{f(G \times G)} \supseteq f(\overline{G \times G}) = f(\overline{G} \times \overline{G}).
    \]
    
\end{proof}