TITLE: Can a finite group act freely (as homeomorphisms) on $\mathbb R^n$
QUESTION [11 upvotes]: I am asking if whether or not a finite group acts freely (as homeomorphisms) on $\mathbb R^n$.
To answer in the negative, it suffices to show: for any homeomorphism $f$ such that $f^d=\text{id}_{\mathbb R^n}$, then $f$ has a fixed point.
I am looking for a complete self-contained answer.

REPLY [3 votes]: If $G$ acts freely and continuously on $\mathbb{R}^n$ then the quotient map $\mathbb{R}^n \to \mathbb{R}^n/G=X$ is a covering map of degree $n=|G|$. Hence we have for the Euler characteristic (see https://en.wikipedia.org/wiki/Euler_characteristic#Covering_spaces), 
$$  \chi(X) \cdot  n=\chi(\mathbb{R}^n)=1.$$
 Since the Euler characterstic is an integer, this implies that $G$ is trivial.