If \(M\) is a compact topological manifold, can one say something about its group of homeomorphisms \(\mathrm{Homeo}(M)\)? Of course one can, there is quite a lot to say about it, and there is also quite a lot which we still don’t know.

Today I want to mention the following recent result by Csikós-Pyber-Szabó (arXiv:2204.13375): *Let \(M\) be a compact topological manifold. Then there exists a number \(n(M)\) depending only on the homotopy type of \(M\) such that every finite subgroup \(G < \mathrm{Homeo}(M)\) has a nilpotent normal subgroup \(N \vartriangleleft G\) of index at most \(n(M)\).*

The interesting thing in the statement is of course that the number \(n(M)\) does not depend on \(G\), since otherwise the result would be trivial (take as \(N\) the trivial subgroup).

The result would be also trivial if there are only finitely many conjugacy classes of finite subgroups in \(\mathrm{Homeo}(M)\). Since I am not familiar with homeomorphism groups, it is not clear to me if or when this can actually happen? If I think about, say, non-positively curved groups studied in geometric group theory, then they often have this property (that there are only finitely many conjugacy classes of finite subgroups). So let me therefore ask here (the possibly dumb) question whether we can always find in \(\mathrm{Homeo}(M)\) finite subgroups of arbitrarily high order (for \(M\) of high dimension)?

**edit (May 15th):** I have been told by several people that there is a (meta-)conjecture that a `generic’ manifold does not have any elements of finite order in its homeomorphism group.