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Let $G$ be a topological group, acts on a topological space $X$, such that the map $f: G \times X \rightarrow X:(g,x)\mapsto g*x$ is continuous.

We say that this action is $properly\;discontinuous$ if for every element $x\in X$, there exist an open neighbourhood $U_{x}$ for $x$ such that $gU_{x}\cap U_{x}\neq \phi$ and $g\in G$ implies that $g=1_{G}$

I am trying to show that if the action of a group $G$ on $\mathbb{R}$ is properly discontinuous then G is isomorphic to $\mathbb{Z}$.

best regards

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Hint: find the free generator of $G$. – Berci Oct 20 '12 at 0:28

The orbit $\mathcal O=G0$ of $0\in\mathbb R$ is discrete. It follows that it is countable and, morerover, there is either a strictly increasing or decreasing bijection $f:\mathbb Z\to\mathcal O$ or $f:\mathbb N_0\to\mathcal O$. Let me suppose it is increasing; if not, we do something similar.

If $g\in G$, there is a $n_g\in\mathbb Z$ such that $g(f(0))=f(n_g)$. I claim that $$g(f(a))=f(a+n_g)\qquad\text{for all $a$ in $\mathbb N_0$ or $\mathbb Z$,}$$ depending on which case we are in. Indeed, there is an $m$ such that $g(f(1))=f(m)$. If we had $m>1+n_g$, then the image of the interval $[f(0),f(1)]$ under $g$ would contain the interval $[f(n_g),f(n_g+2)]$, and there would exist a point in $(f(0),f(1))$ whose image under $g$ is $f(n_g+1)$. This is absurd. Doing this in general gives my claim.

Now I can define a map $\phi:g\in G\mapsto n_g\in\mathbb Z$. It is a morphism of groups, and it must be injective.

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