Let $\Gamma$ be a discrete cocompact subgroup of the euclidean motion group $$ G={\mathbb R}^d\rtimes O(d). $$ Let $\phi:G\to O(d)$ the projection homomorphism. Is it true that $\phi(\Gamma)$ is finite?

$\begingroup$ Aakumadula, why not post this as an answer? It seems sufficiently precise and comprehensive $\endgroup$ – Yemon Choi Mar 6 '13 at 16:13

$\begingroup$ Yemon Choi, I thought this was just a reference. But I see your point, that it will look like an unanswered query. $\endgroup$ – Venkataramana Mar 6 '13 at 17:04
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The answer is yes. This is a theorem of Bieberbach (see Corollary (8.26) of the book "Discrete Subgroups of Lie Groups" by M.S.Raghunathan (Springer Ergebnisse Tract).

$\begingroup$ Interesting (to me!) that this is an established result, since without the cocompactness assumption the conclusion fails: an irrational winding in SO(2) combined with the onedimensional discrete subgroup of translations immediately gives a nondiscrete image in SO(2). $\endgroup$ – paul garrett Mar 7 '13 at 0:47

$\begingroup$ Professor Garrett, I do not understand your remark. If I have an element of the form $(v,r)$ in the semidirect product of ${\mathbb R}^2$ with $SO(2)$, where $r$ is an irrational rotation, there is no guarantee that itw powers (vector component) is discrete: the $n$th power vector component is of the form $(1+r+\cdots +r^{n1})v$, which can well converge. $\endgroup$ – Venkataramana Mar 7 '13 at 1:16

$\begingroup$ Of course, you can take $v\in {\mathbb R}^n$ $n\geq 3$ which is fixed by the rotation $r$, in which case you get a discrete group. When you assume "cocompactness", the vector parts are sufficiently many that the rotation parts do not always fix the vector part. $\endgroup$ – Venkataramana Mar 7 '13 at 1:24

$\begingroup$ @Aakumadula: indeed, this is a funny situation. It is late for me here in my timezone, but (in case you'd not already in your own time thought through the facts) I'll add comments tomorrow which I hope will clarify the thing that'd been in my mind. Irrelevant to the question as posed. yes. $\endgroup$ – paul garrett Mar 7 '13 at 2:00