Fundamental Group of a free G-space

Question

It is very well known that any group $H$ can be the fundamental group of a topological space. What happened if we restrict the class of topological spaces to the free equivariant topological spaces? More precisely, for a fixed finite group $G$ what groups $H$ can be the fundamental group of a free finite $G$-simplicial complex?

Answer 1

$EG$ is the contractible, free G-space. Given any space $X$ we may take the product $EG \times X$ to be a G-space by acting on $EG$. This then a free G-space which is homotopy equivalent to $X$. Hence, any fundamental group is realized by a free G-space.

This is the cofibrant replacement of the G-space $X$ with the trivial action, in G-spaces (where a map is a weak equivalence if it is equivariant and the underlying map is a weak equivalence).

Edit: To answer your actual question: all finitely presented groups (as is the case for the nonequivariant case). We can follow the same method described above by taking the product of a finite CW complex with fundamental group $G$ and crossing with a free G-space with a trivial fundamental group.

To construct such a (connected) G-space, start with a copy of G and add segments between any two elements of the group. This has an obvious group action by sending the segment from a to b to the segment from $ga$ to $gb$. Then pick any loop represented by a sequence of edges and attach a disk along this loop. Just as before this has an obvious action by sending the disk attached to $a_1 a_2 a_3 \dots a_n$ to $ga_1 ga_2 ga_3 \dots ga_n$.

Since we killed a generating set of loops, we have constructed a space with trivial $\pi_1$. Similar constructions can be done for any homotopy group.