# Discrete subgroups of isometry group $\mathbb{R}^n$

Let $G$ be a Hausdorff topological group. We say that a subgroup $S$ of $G$ is discrete if and only if the subspace topology (from $G$) on $S$ is discrete.

Note that isometry group of euclidean space $\mathbb{R}^n$ is displayed by $E(n)$(or affine isometries on $\mathbb{R}^n , \ O(n)\subset E(n)$) and $Sx:=\left\{s(x);\ \forall s\in S\subseteq E(n)\right\}$ is $S$-orbit of any $x\in\mathbb{R}^n$ where S is any subgroup of E(n).

For discrete subgroups S of isometry group of euclidean space $E(n)$ an equivalent condition is: intersection of the $S$-orbit of any $x\in\mathbb{R}^n$ has finite intersection with any compact set of $\mathbb{R}^n$.

Why is there a such the equivalent condition? How can I prove it?

• Affine isometries or $O(n)$? Anyway, recall how you definie the topology on $G$ – Hagen von Eitzen Nov 25 '13 at 17:29

The definition means that $S$ is discrete if for each point $x\in \mathbb{R}^n$ the family $Sx$ is locally finite, i.e., for each point there is a neighbourhood intersecting only finitely many subsets of this family. This is equivalent to the fact each compact set intersects only finitely many subsets of this family.
• What's your mean from the family $Sx$? Is your mean $\mathcal{F}=\{\text{singleton subsets of} \ Sx\}$ for any $x\in \mathbb{R}^n$ that fixed? Why is this family locally finite? I do not figure out The last equivalence. Why are these property only for group action of isometries on $\mathbb{R}^n$? Please describe them more than your posts. – bigli Nov 26 '13 at 5:32
• I mean $Sx=\{s.x \mid s \in S\}$. The word "family" is used in Vinberg's book on discrete groups. Your definition of discrete is just that each Sx is locally finite, see en.wikipedia.org/wiki/Subspace_topology. – Dietrich Burde Nov 26 '13 at 8:36
• Dear Dietrich Burde, My $\mathcal{F}$ is the same with your mean from $Sx=\{s.x\ | s\in S\}$. Why is this family($\mathcal{F}$) locally finite? I do not figure out The last your equivalence in your answer. What's Vinberg's book? What's connection the Wikipedia webpage(in the last your comment) with my problem? – bigli Nov 27 '13 at 3:37