Effective group action on an interval

Question

I'm struggling with the following problem.

Let $G$ be a compact Hausdorff topological group acting effectively on $X$ where $X$ is

• $[0,1)$
• $[0,1]$

Prove that in the first case $G = \{e\}$ and in the second case $|G| \leq 2$.

I know that if the action is effective, then the representation of $G$ on homeomorphism of $X$, namely $L: G \rightarrow H(X)$ has trivial kernel.

If I assume that $G \neq \{e\}$, then for $g \neq e$ there is some $x \in X$ such that $gx \neq x$. I know that the map $L_g: X \rightarrow X$ is either increasing or decreasing.

Now I suppose that $L_g$ is increasing and $ x < gx$, therefore we get that $g^n \neq e$ for all $n$. So we obtain an increasing sequence $\{g^nx\}$.

In the case of $X = [0,1]$ we can say that the sequence is monotonic and bounded and therefore it converges to some $a \in X$. So we get $L_g(a) = L_g(\lim g^nx) = \lim g^nx = a$ which is a contradiction. (Edit: I realize that there is no contradiction here)

If $X= [0,1)$ the sequence might not converge in $X$ and I don't know what to do in that case. However, A similar argument works when $L_g$ is decreasing and $x > gx$; even this lately works when $X=[0,1)$.

If I assume $L_g$ increasing and $gx < x$. I can not construct a convergent sequence, but maybe a convergent subsequence. However I can not get a contradiction in that case.

Finally, I am really confusing where to apply exactly the assumption of $G$ being compact. Or maybe if there is an easier way to address the problem.

Answer 1

Here are some hints.

1. For the case $X=[0,1]$ : consider the subset $K$ of $G$ consisting of elements $g$ such that $L_g$ is increasing. Prove that it is a subgroup of $G$, that it has index at most $2$, and that it is compact in $G$.
2. For the case $X=[0,1)$ : prove that the subset $K$ as above is just $G$.
3. In both cases : take a fixed $x\in X$. Prove that the orbit of $x$ under $K$ is compact in $X$, and thus has a maximum and a minimum. Prove that moreover these extrema must be equal to $x$. Conclude that every $g\in K$ acts trivially, and thus that $K$ must be trivial.