# Lattice in motion group

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?

• Aakumadula, why not post this as an answer? It seems sufficiently precise and comprehensive – Yemon Choi Mar 6 '13 at 16:13
• Yemon Choi, I thought this was just a reference. But I see your point, that it will look like an unanswered query. – Venkataramana Mar 6 '13 at 17:04

• Professor Garrett, I do not understand your remark. If I have an element of the form $(v,r)$ in the semi-direct 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 $(1+r+\cdots +r^{n-1})v$, which can well converge. – Venkataramana Mar 7 '13 at 1:16
• 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 "co-compactness", the vector parts are sufficiently many that the rotation parts do not always fix the vector part. – Venkataramana Mar 7 '13 at 1:24