Assume $\Gamma$ be a Bieberbach group which acts on $\mathbb R^n$ (i.e. a discrete subgroup of isometries of $n$-dimensional Euclidean space with a compact fundamental domain). Denote by $M(\Gamma)$ the number of maximal finite subgroups (up to conjugation) in $\Gamma$. Is it true that $M(\Gamma)\le 2^n$?

Things I can do: There is a simple geometric observation (due to Perelman) which shows that if $N(\Gamma)$ is the number of orbits of isolated fixed point of some subgroups of $\Gamma$ then $N(\Gamma)\le2^n$. Clearly, each such point corresponds to a maximal finite subgroup. Thus, $N(\Gamma)\le M(\Gamma)$, but in all examples I know I still have $M(\Gamma)\le 2^n$ (and I believe it is allways true).

The formulation is completely algebraic so maybe it has a completely algebraic solution...