Topological group with discrete topology

Question

Let $G$ be a topological group. I came to know that if I can show the existence of a homeomorphism of $G$ which moves only finitely many points of $G$, then $G$ has only discrete topology.

How can I prove this ?

Answer 1

Assumption. We assume here that $G$ is a Hausdorff topological group.

Let us first prove the following assertion:

Claim If there exists a continuous $id\ne \psi:G\to G$ such that $\psi(x)\ne x$ for at most finitely many $x\in G$, then $G$ has a finite open subset.

Proof. Consider the function $f(x)=\psi(x)\cdot x^{-1}$. By assumption, $f(x)\ne 1$ for at most finitely many $x\in G$. Also, since $G$ is a topological group, and $\psi$ continuous, it is clear that $f$ is continuous. Since the point $1$ is closed, we have that $U=\lbrace x\mid f(x)\ne 1\rbrace= f^{-1}(G\setminus\lbrace 1\rbrace)$ is finite and open.

Since the group $G$ is Hausdorff (or even T1 would do, I think), you can use this to show that there exists an open singleton in $G$. From here the proof should be obvious (use the fact that group translations are homeomorphisms).

Remark. I don't see where the fact that $\psi$ is a homeomorphism was used. I think that this assumption can be weakened to just continuity, but if anyone sees something I missed please let me know.