# Bound on the index of an abelian subgroup in discrete subgroup of the euclidean group?

$\DeclareMathOperator{\isom}{Isom}$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 $m$?

The index is the order of the crystallographic point group, in the cocompact case. It is then a finite subgroup of $GL(n,\mathbb{Z})$. There are several bounds known for the maximal order of such subgroups. Friedland proved in $1997$ that $$m\le 2^nn!$$ for $n\ge n_0$ and gave conditions where equality is attained (The maximal orders of finite subgroups of $GL(n,\mathbb{Q})$. Rockmore in $1995$ used a different description: For every $\epsilon>0$ there exists a constant $c(\epsilon)$ such that $m\le c(\epsilon)(n!)^{1+\epsilon}$.
• @eins6180: Every discrete group of Euclidean isometries acts cocompactly on some subspace of $R^n$: This is the content of one of Bieberbach's theorems. A proof should be in the book J.Wolf, "Spaces of constant curvature". Jul 19 '14 at 16:47
• There a confusion in the statement that every discrete group of Euclidean isometries acts cocompactly on some subspace of $\mathbf{R}^n$: the missing adjective is "proper", and possibly the action has a finite kernel. Hence there is a difference between the class (C) of cocompact discrete subgroups of isometries of Euclidean spaces, and the class (D) of discrete subgroups of isometries : namely (D) consists of virtually abelian f.g. groups, while (C) consists of those elements in (D) having no nontrivial finite normal subgroup. E.g., a nontrivial finite group is in (D) but not (C).