We know that if we want to construct a space with a given fundamental group $G$ ,we can use cells and attaching maps, or fundamental domains and attaching maps, as in : How to determine space with a given fundamental group.
But there is a different way: if $M$ is, I think a manifold, and $G$ acts proper-discontinuously on $M$, then the quotient $ M/G$ has $G$ as a fundamental group. EDIT: As Seirios points out, $M$ may have to be simply-connected.
Now, can we always do this for any group $G$, i.e., can we always construct a space with fundamental group $G$ by defining a proper discontinuous action $M/G$? Can we always find $M$ so that the action is of the necessary type? Thanks.
You don't need any hypotheses on $M$. For any topological group $G$, we can construct a weakly contractible space $EG$ on which $G$ acts freely. The resulting quotient $BG$ is called the classifying space of $G$, and when $G$ is discrete it is an Eilenberg-MacLane space $K(G, 1)$, and in particular has fundamental group $G$. Both $EG$ and $BG$ can be constructed using the bar construction.
If you want $M$ to be a reasonable manifold then there are some size conditions on $G$: in particular, if you want $M/G$ to be a closed manifold then $G$ must be finitely presented. With this caveat, as Mike Miller says in the comments, it's known that any finitely presented group occurs as the fundamental group of some closed $4$-manifold, and you can take $M$ to be its universal cover.