# Closure of a topological group

Let $G$ be a topological group. Define $H(g)=\{g^n\}^{\infty}_{n=-\infty}$ for each $g \in G$. I need to prove that the closure of the set $H(g)$ is a commutative subgroup of $G$. But I am not sure how to interpret the closure of $H(g)$.

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@seaturtles How do you think about the closure of $H(g)$ without knowing about the topology of $G$ ? – the8thone May 19 '14 at 2:58

You have to assume that $G$ is Hausdorff, otherwise this does not work (see here).

Let $K$ be the closure of $H(g)$. Given $g,h\in K$, take nets $(g_\lambda)_{\lambda\in\Lambda}$ and $(h_\mu)_{\gamma\in\Gamma}$ in $H(g)$ such that $g_\lambda\to g$ and $h_\gamma\to h$. Since multiplication is continuous, the net $(g_\lambda h_\gamma)_{(\lambda,\gamma)\in\Lambda\times\Gamma}=(h_\gamma g_\lambda)_{(\lambda,\gamma)\in\Lambda\times\Gamma}$ (which is in $H(g)$, since it is a subgroup) converges to both $gh$ and $hg$, so $gh=hg\in K$ (by the Hausdorff property). This shows that $K$ is a commutative subgroup of $G$.

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Let $G$ be a Hausdorff topological group. Consider the map $c:(x,y)\in G\times G\to xyx^{-1}y^{-1}\in G$. Since $c$ is clearly continuous, the set $X=c^{-1}(\{e\})$ is a closed subset of $G\times G$.

Now let $H\subseteq G$ be an abelian subgroup of $G$. Since $H$ is abelian, we can see at once that $H\times H\subseteq X$. But then $\overline H\times \overline H\subseteq\overline{H\times H}\subseteq X$, because $X$ is closed, and this means that $\overline H$, which is a subgroup of $G$, is also abelian.

Remark. To show that $\overline H$ is a subgroup of $G$ we can proceed similarly, as follows. Let $\alpha:(x,y)\in G\times G\mapsto xy^{-1}\in G$. We know that $\alpha(H\times H)\subseteq H$, so continuity of $\alpha$ implies that $\alpha(\overline H\times\overline H)\subseteq\alpha(\overline{H\times H})\subseteq\overline{\alpha(H\times H)}\subseteq\overline{H}$. Since $e\in\overline H$, this is enough to conclude that $\overline H$ is a subgroup.

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