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...



Number of subgroups in a Bieberbach group.
Dima, I can not write a comment (yet) so I will start an answer to my
own question.


You may assume that $\Gamma$ acts by isometries, so $A=\mathbb
R^n/\Gamma$ is an Alexandrov space.
For each maximal subgroup $F$ one can take its fixed point set $S\_F$
in $\mathbb R^n$.
$S\_F$ is an (affine) subspace and image (say $E\_F$) in $\mathbb
R^n/\Gamma$ is a singular set (so called extremal subset of Alexandrov
space). The maximality of $F$ implies that $E\_F$ contains no proper
extremal sets (a smaller subset is fixed by bigger group). (In fact
$E\_F$ is a flat manifold and its has a neighborhood which isometric
for a product $E\_F\times Cone$.)


So the question boils down to finding maximal number of such extremal
sets in $A$.
A particular case of such sets are isolated singular points.
Perelman's theorem states that number of "one-point extremal sets" in
an Alexandrov space with curvature $\ge 0$ is at most $2^n$. 
The
proof repeats a proof of Erdős problem: if you have $m$ points in
$\mathbb R^n$ such that all angles in all triangles $\le \pi/2$ then
$m\le 2^n$. We take homothety with coefficient 1/2
for each point, then images of convex hull don't have common internal 
points (otherwise it would occur obtuse angle), then
 compairing volume of convex hull and its images gives
estimate for number of points.


