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.


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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    $\begingroup$ Why is $\mathcal F$ locally finite? $\endgroup$ – Chris Eagle Mar 28 '11 at 14:41
  • $\begingroup$ 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 $R$? $\endgroup$ – Roun Mar 28 '11 at 15:00
  • $\begingroup$ I am sorry about the last question. I read the answer again and I understand the concept of locally finite now. $\endgroup$ – Roun Mar 28 '11 at 15:03
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    $\begingroup$ 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. $\endgroup$ – Roun Mar 28 '11 at 15:14
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    $\begingroup$ @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...) $\endgroup$ – Mariano Suárez-Álvarez Mar 28 '11 at 16:53

Here is another solution (note that by neighborhood I mean open neighborhood):

Let $H \leq G$ be a discrete subgroup of the Hausdorff group $G$.

Step 1:

We will show that given a neighborhood $U$ such that $U \cap H = \{e\}$, there exits a neighborhood $V \subset U$ such that $VV^{-1} \subset U$ and $e \in V$.

Let $\sigma: U \times U \rightarrow G$ be the map $\sigma(y_1,y_2)=y_1y_2^{-1}$. By continuity there exits a neighborhood $N \subset U \times U$ of $(e,e)$ such that $\sigma(N)\subset U$. Then $N$ contains an open set of the form $V_1 \times V_2$ where $V_1,V_2 \subset U$ are open and $e \in V_1, V_2$. Take $V=V_1 \cap V_2$. Then $V$ is a neighborhood of $e$ and $V \times V \subset V_1 \times V_2$ and hence $VV^{-1} = \sigma(V\times V) \subset \sigma(V_1 \times V_2) \subset U$.

Step 2:

Let $x \in G-H$. We will find a neighborhood of $x$ contained in $G-H$. Assume $U$ is a neighborhood of $e$ such that $U \cap H= \{e\}$. Let $L_x: G \rightarrow G$ be defined by $L_x(y)=xy$. This is a homeoeomorphism with inverse $L_{x^{-1}}$. Now, let $V \subset U$ be a neighborhood of $e$ with the properties from step 1. Take $W= L_x(V)$; this is a neighborhood of $x$. Suppose now that $h_1, h_2 \in W \cap H$. Then $h_1=xv_1$ and $h_2=xv_2$ for some $v_1,v_2 \in V$. Thus $v_1^{-1}h_1=x=v_2^{-1}h_2 \implies h_1h_2^{-1}=v_1v_2^{-1} \in VV^{-1} \subset U$. Thus $v_1v_2^{-1} \in H \cap U$ so $h_1h_2^{-1}=e$, hence $h_1=h_2$. It follows that $V$ contains at most one element of $H$.

If $V$ contains no element of $H$ then $V$ is the desired subset. Otherwise, if $V \cap H = \{h\}$, since $V$ is Hausdorff, we can separate $x$ and $h$ by open subsets $U_x, U_h \subset V$ (which will also be open in $G$). Then $U_x$ is the desired subset since $U_x \cap H = \emptyset$.


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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    $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Mar 28 '11 at 16:54

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