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I have just begun to learn about topological group recently and is still not familiar with combining topology and group theory together.

I have read a useful property of discrete group on the wikipedia:

every discrete subgroup of a Hausdorff group is closed

But I have no idea how to prove it. I find that it cannot be proved only considering the topological structure, since $\left\{\frac{1}{n}: n=1,2,3,...\right\}$ is a discrete subspace of $\Bbb R$, which is not closed.

I don't know how to use the group structure here. Can you please help? Thanks.

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up vote 15 down vote accepted

Let $H$ be a discrete subgroup of a Hausdorff group $G$. Then, of course, $H=\bigcup_{h\in H}\{h\}$. For each $h\in G$, the sets $\{h\}$ is closed, and the family of closed sets $\mathcal F=\{\{h\}:h\in H\}$ is locally finite: it follows that the union is also closed.

Locally finite means: each point of $G$ has a neighborhood $U$ such that $U$ intersects non-trivially finitely many elements of $\mathcal F$.

Later. I guess a hint was not all that was asked for... Of course, we now have to show that $\mathcal F$ is locally finite... Suppose, to reach a contradiction, that there is a $p\in G$ such that every neighborhood of $p$ contains infinitely many elements of $H$. As $H$ is discrete, there is an open subset $U\subseteq G$ such that $U\cap H=\{e\}$; since $G$ is a topological group, there exists an open set $V$ such that $e\in V$ and $V^{-1}V\subseteq U.$ Now $pV$ is an open neighborhood of $p$ so the hypothesis implies that there exist distinct $h_1$, $h_2\in H$ such that $h_1$, $h_2\in pV$. It follows that $h_1^{-1}h_2\in V^{-1}V\subseteq U$ so that $e\neq h_1^{-1}h_2\in U\cap H$. This is absurd.

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Why is $\mathcal F$ locally finite? – Chris Eagle Mar 28 '11 at 14:41
I am a beginner and not familiar with some concepts. Denote $M=\{1/n: n=1,2,3,...\}\subset R$. Is $M$ locally finite in $R$? – Roun Mar 28 '11 at 15:00
I am sorry about the last question. I read the answer again and I understand the concept of locally finite now. – Roun Mar 28 '11 at 15:03
Thank you very much. I've got the idea now. The key point is that a discrete subgroup contains the identity elements $e$, while subspace may not. – Roun Mar 28 '11 at 15:14
@Roun, well that's not all of it: for example, the set $M'=\{\tfrac{1}{n}-\tfrac{1}{17}:n\in\mathbb N\}$ also contains the identity of the additive group $\mathbb R$, but it is not closed! In the argument I wrote I very much needed the fact that the set $H$ has the property that $h_1^{-1}h_2$ is in $H$ whenever $h_1$ and $h_2$ are; and this property is equivalent to $H$ being a subgroup (provided it is not empty...) – Mariano Suárez-Alvarez Mar 28 '11 at 16:53

the important thing about topological groups is that they are homogeneous in that neighbor hoods of every point look just like neighborhoods of the identity (or any other point). if $H$ is discrete in $G$, then some point (say the identity) has a neighborhood $e\in U$ such that $U\cap H=\{e\}$. consider the neighborhood $hU$ of the some point $h\in H$. then $H\cap hU=\{h\}$ (if not, say $g\in H\cap hU$, then $h^{-1}g\in h^{-1}hU\cap H=U\cap H$, which is false by definition of $U$).

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But that argument only shows that if a subgroup $H\subseteq G$ is such that $e\in H$ is isolated in $H$, then $H$ is discrete; but does not get all the way to proving that $H$ is closed. – Mariano Suárez-Alvarez Mar 28 '11 at 16:54

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