TITLE: Number of subgroups in a Bieberbach group.
QUESTION [24 upvotes]: 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...

REPLY [2 votes]: That is not an answer.
I want to give an example where the argument of Erdős does not work directly.
Consider an action of group $\Gamma$ on $\mathbb R^3$ generated by the reflections $r_1, r_2$ and $r_3$ correspondingly in the lines $x=z=0$ and $x+1=z=0$ and $x-y=z-1=0$.
Each of the reflections $r_i$ generate a maxiamal $\mathbb Z_2$-subgroups, all of them are  nonconjugate. 
These groups corespond to three singular circles, say $\Sigma_i$ in the factor $X=\mathbb R^3/\Gamma$. 
($X$ is homeomorphic to $S^3$ and $\Sigma_1$, $\Sigma_2$, $\Sigma_3$ form Borromean rings, but all this is not important.)
Let us try to mimic argument of Erdős.
Take subsets $X_i$ of $X$ of midpoints $m$ between $x\in X$ and a closest $x_0\in\Sigma_i$ to $x$.
As in the argument of Erdős we have $\mathrm{vol}\, X_i>\tfrac{1}{2^3}\cdot\mathrm{vol}\, X$.
BUT $X_1\cap X_3$ has interior points and here argument brakes into parts.
Comments 

Since fixed point sets are 1-dimensional, it would be enough to take $m\in [xx_0]$ such that $\tfrac{|mx_0|}{|xx_0|}=\tfrac1{2\sqrt[3]{2}}$.
But even in this case one has interior points in $X_1\cap X_2$ (the borderline in this example seems to be $\tfrac13$).
There is a natural bisecting hyperplane for any two affine subspaces. We may use it to cut a cylinder domain around each fixed point set of a maximal subgroup. The projection of these cylinders in $X$ gives Voronoi-like domains, but they do not cover whole space in general --- that is OK as far as we have lower bound on their volumes...