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I asked the following question on math.SE a couple of days ago. Dietrich Burde gave an answer for the case that the subgroup is not only discrete but also acts cocompactly.

What about the general case? Are there any upper bounds known?

$\DeclareMathOperator{\isom}{Isom}$A discrete subgroup of the group of isometries in euclidean space is almost abelian.

By this I mean that for each $n$ there exists $m$ such that for any discrete subgroup $\Gamma$ of $\isom(\mathbb{R}^n)$ we can find an abelian subgroup $\Gamma' \leq \Gamma$ such that $$[\Gamma : \Gamma'] \leq m,$$ so the index of the abelian subgroup in $\Gamma$ is bounded by $m$.

What is the best known bound on $m$?

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@DietrichBurde: I didn't mean to belittle your answer on math.SE. Apologies if it came across like this. – eins6180 Jul 21 '14 at 13:13
All right. Everything clear then. Isn't the paper of Collins now answering your question ? – Dietrich Burde Jul 21 '14 at 13:19
@DietrichBurde: Not quite, I believe. I think Collins gives the right bound for Jordan's theorem (which only covers finite subgroups of $GL(n,C)$). It doesn't cover the more general case of discrete groups. – eins6180 Jul 21 '14 at 13:31

Since ${\rm Isom}(\mathbb R^n) = \mathbb R^n \rtimes O(n) \subset GL(n+1,\mathbb C)$, the statement can be reduced to subgroups of $GL(n+1,\mathbb C)$. It was proved in

Boris Weisfeiler, Post-classification version of Jordan's theorem on finite linear groups, Proc. Natl. Acad. Sci. USA. Aug 1984; 81(16): 5278–5279.

that any finite subgroup of $GL(n,\mathbb C)$ has an abelian normal subgroup of index at most $(n+1)! n^{a log(n) + b}$. I would expect that the same result holds for virtually abelian subgroups, so that you would also obtain a similar bound in your situation.

Looking at

Eli Aljadeff and Jack Sonn, Bounds on orders of finite subgroups of ${\rm PGL}_n(K)$. J. Algebra 210 (1998), no. 1, 352-360.

it seems that there is further unpulished work by Boris Weisfeiler showing that $(n+2)!$ is enough if $n \geq 64$.

EDIT: It was pointed out in a comment that the passage from finite to virtually abelian is not so straightforward.

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Is it clear that you can reduce to $GL$? A semi-direct product of abelian group is not necessarily abelian... – abx Jul 21 '14 at 10:06
Reducing to $GL(n)$ is possible I think, but requires some additional argument -- reducing to $GL(n+1)$ is trivial. – Andreas Thom Jul 21 '14 at 10:26
Oops -- of course, sorry. I overlooked the $n+1$. But you still need an argument to pass from finite to discrete. – abx Jul 21 '14 at 10:30
Yes, see…. – Dietrich Burde Jul 21 '14 at 11:51
There are exceptions to this bound for small $n.$ W. Feit has itemised them: see my answer on the MO Question:: "Maximal order of finite subgroups of GL(n,Z)" – Geoff Robinson Jul 21 '14 at 12:31

Every discrete group of Euclidean isometries acts properly discontinuously and cocompactly on some subspace of $\mathbb{R}^n$ (a result of Bieberbach). If we are in the crystallographic case, then the index is bounded by the maximal order of a finite subgroup of $GL(n,\mathbb{Z})$. Here we have several results in the literature, e.g., the result of Friedland in $1997$, that $$m\le 2^nn! $$ for $n\ge n_0$. There are exceptions for small $n$, see the comment of Geoff Robinson. There are conditions given in Friedland's paper "The maximal orders of finite subgroups of $GL(n,\mathbb{Q})$", when equality is attained. Rockmore in $1995$ proved the following: for every $\epsilon>0$ there exists a constant $c(\epsilon)$ such that $m\le c(\epsilon)(n!)^{1+\epsilon}$.

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a discrete group $\Gamma$ of isometries of $\mathbf{R}^n$ indees acts properly cocompactly by isometries on $\mathbf{R}^k$ for some $k\le n$ (namely any minimal nonempty invariant affine subspace of $\mathbf{R}^n$ for the given action, but you have to take into account that the new action has a finite kernel, possibly nontrivial, and precisely equal to the maximal finite normal subgroup in $\Gamma$. – YCor Jul 21 '14 at 12:35
@YCor: yes, thank you. I edited the answer, refering only to the crystallographic case. – Dietrich Burde Jul 21 '14 at 12:43

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