Set of discrete orbits under subgroup of $Isom(\mathbb{E}^n)$ is clopen

Question

I'm trying to solve the following exercise (exercise 1.4 from Szczepanski's "Geometry of Crystallographic Groups"):

Let $\Gamma$ be a subgroup of $I(\mathbb{E}^n)$, the group of isometries on $\mathbb{E}^n$. Prove that the following are equivalent:

1. $\forall x \in \mathbb{E}^n$, the orbit $\Gamma \cdot x$ is a discrete subset of $\mathbb{E}^n$
2. $\exists x \in \mathbb{E}^n$, the orbit $\Gamma \cdot x$ is a discrete subset of $\mathbb{E}^n$

Of course, 1. implies 2. I'm having trouble with the reverse implication, however. I'd like to show that the set $\{x \in \mathbb{E}^n \mid \Gamma \cdot x \text{ is discrete}\}$ is an open and closed set. Then this set must be either $\emptyset$ or $\mathbb{E}^n$, and the existence of an $x$ with discrete orbit proves the exercise.

One tool that I have is the projection $$\pi: \mathbb{E^n} \rightarrow \mathbb{E}^n/\Gamma: x \mapsto \Gamma \cdot x,$$ which is continuous, open and closed. I'm not sure if this helps in any way, though.

Answer 1

The statement you are trying to prove is false for all $n\ge 2$. Consider $\Gamma=O(n)$. For $x=0$ the orbit $\Gamma x$ is a discrete subset of $R^n$, but for any $x\ne 0$ the orbit is not discrete.