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

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Jul 21, 2014 at 12:31	comment	added	Geoff Robinson		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)"
Jul 21, 2014 at 11:51	comment	added	Dietrich Burde		Yes, see math.stackexchange.com/questions/870642/….
Jul 21, 2014 at 10:41	history	edited	Andreas Thom	CC BY-SA 3.0	added 138 characters in body
Jul 21, 2014 at 10:38	comment	added	Andreas Thom		You are right, maybe something like $n! 2^n$ is the right bound then.
Jul 21, 2014 at 10:31	comment	added	eins6180		Thanks for your answer. I think the Weisfeiler bound was improved by Collins in 2007 to $(n+1)!$ for $n$ large enough, see here. But anyway, I am not sure that I am convinced that the same bound should hold for virutally abelian subgroups. Aren't the crystallographic groups already a counterexample? What about $\mathbb{Z}^n \rtimes P$, where $P$ is the hyperoctahedral group?
Jul 21, 2014 at 10:30	comment	added	abx		Oops -- of course, sorry. I overlooked the $n+1$. But you still need an argument to pass from finite to discrete.
Jul 21, 2014 at 10:26	comment	added	Andreas Thom		Reducing to $GL(n)$ is possible I think, but requires some additional argument -- reducing to $GL(n+1)$ is trivial.
Jul 21, 2014 at 10:06	comment	added	abx		Is it clear that you can reduce to $GL$? A semi-direct product of abelian group is not necessarily abelian...
Jul 21, 2014 at 9:58	history	answered	Andreas Thom	CC BY-SA 3.0