Manifold acted on by every finite group Is there a (second-countable, connected) manifold that admits a faithful continuous group action by every finite group?
 A: I will argue that the infinite genus surface, $\Sigma_C$, with the space of ends being a Cantor set has this property(think an regular infinite tree except genus).
Let $\Sigma_g$ be the surface of genus $g$ and $\mathrm{Sym}_n$ symmetric group on $n$ elements. There is a standard construction to show that for any $\mathrm{Sym}_n$ there is a $g$ large enough so that $\mathrm{Homeo} (\Sigma_g$) contains that symmetric groups: consider a Cayley graph for $\mathrm{Sym}_n$, now "thicken" it up to a surface by "thickening" each vertex to a torus, than the edges will correspond to connect sum of two tori(connecting with a tube/annulus). Now an action on the graph can be turned into an action on the surface that has been constructed.
The main fact(and it isn't hard to see in the cases we will be working with) is that infinite type surfaces are basically classified,up to homeomorphism, by the space of ends and some genus and puncture information(see On the classification of noncompact surfaces by Ian Richards for more detail).
On the surface we have constructed, choose a small disk on one of the tori/vertices, and look at the orbit of the disk, cut out the orbit and attach a thickened (under the same process as above) infinite rooted binary tree. Your group still acts on this space, where the group takes the trees you planted to trees, and preserves the thickened Cayley graph. For each $n$ these surfaces are homeomorphic by the "classification of surfaces" above.
A: Yes.  There is a universal finitely presented group $G$ that contains an isomorphic copy of every finitely presented group; this follows from Higman's embedding theorem.  Choose a finite generating set of this group, and construct the Cayley graph.  Choose a nice embedding of this graph in Euclidean 3-space and take a regular neighborhood $N$; let $M$ be the boundary of $N$.  Then $G$ acts faithfully on a manifold homeomorphic to $M$.  You can construct such a homeomorphic copy by replacing the vertices of the Cayley graph by spheres with twice as many holes as the number of generators.  Then attach a cylinder for each edge to the spheres corresponding to the end vertices.  $M$ is a surface of infinite genus, but it is connected and 2nd countable.  Each finite group acts by choosing an embedding into $G$.  Since no vertex sphere is fixed, except by the identity, $F$ acts faithfully.
