Why is every discrete subgroup of a Hausdorff group closed? 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.
 A: 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 $U_x \cap H = \emptyset$. 
A: 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.
