Discrete action on Lie groups

Question

Given a Lie group $G$ and a discrete subgroup $\Gamma$ of $G$. Why is the action of $\Gamma$ on $G$ properly discontinuously?

Answer 1

The fact that $\Gamma$ is discrete is equivalent to saying that there exists a neighborhood $U$ of $1_G$ such that $U\cap \Gamma=\{1_G\}$.

Let $V$ be a neighborhood of $1_G$ such that for every $x,y\in V, xy\in U$ and $x^{-1}\in U$. $V$ exists since the multiplication $G\times G\rightarrow G$ and the inverse are continue.

For every $\gamma\in \Gamma$ distinct of the identity, $\gamma(V)\cap V$ is empty. To see this, suppose that $x\in \gamma(V)\cap V$, $x=\gamma y, y\in V$ and $\gamma=xy^{-1}\in U$. Contradiction. Let $g\in G$ and $\gamma\in \Gamma$ distinct of the identity, $Vg$ is a neighborhood of $g$ $x\in \gamma(Vg)\cap Vg$, $x=\gamma yg=y'g, y,y'\in V$. This implies that $\gamma y=y'$ and $\gamma=y'y^{-1}\in U$, $\Gamma\cap U$ is not empty. This is a contradiction. Thus $\gamma(Vg)\cap Vg$ is empty.