Suppose $H$ is a finite group acting smoothly on a smooth connected manifold $M$. The action is trivially proper, as $H$ is discrete. If the action of $H$ were also known to be free, i.e. $h\cdot p\neq p$ for all $p\in M$ and non-identity $h\in H$, then it would follow that the quotient space $M/H$ by the action of $H$ would have a unique structure of a manifold making the quotient map a a smooth normal covering.

My question is, without knowing a priori that the action of $H$ is free, does the finiteness of $H$ ensure that the action is free? If not, can we say something maybe less strong but to the same effect, namely something admitting a smooth normal covering map?

My idea was to somehow define define a different action of $H$ on $M$ which was smooth and free, thereby obtaining a quotient map which, if I was clever enough, intertwined the $H$ actions, but I am not even sure this is possible. Any hints?

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    $\begingroup$ You cannot expect freeness of the action, consider for instance the action of $\Bbb Z/ 2\Bbb Z$ on $\Bbb R^2$ given by $1\cdot (x,y)=(y,x)$. This is as smooth an action as they come, and it fails to be free, as the main diagonal of $\Bbb R^2$, the set of all pairs $(x,x)$, is fixed. The quotient space of this homeomorphic to $\Bbb R\times [0,+\infty[$, so it is a manifold with boundary. $\endgroup$ – Olivier Bégassat Dec 6 '12 at 4:17
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    $\begingroup$ There is a topological obstruction. The Euler characteristic is multiplicative in covers, so if the Euler characteristic of $M$ is not divisible by the order of $G$ then this never happens even for merely continuous actions. $\endgroup$ – Qiaochu Yuan Dec 6 '12 at 4:58

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