connected nilpotent subgroup of Isom(R^n)

Question

Is it true that every connected nilpotent Lie subgroup of Isom(R^n), the isometry group of R^n, is actually abelian?

Any reference on it?

Answer 1

This result is well known, and it is worth, I think, to look into a book on geometry for a detailed proof and additional background. One possible reference is the book on Three-Dimensional Geometry and Topology, Volume 1 by Thurston, namely Corollary $4.1.13$ and its proof. There it says "to complete the proof we must show that any connected, closed nilpotent subgroup $N\subseteq Isom(E^n)$ is abelian." The proof given there uses induction, and of course a result on subgroups of $O(n)$, and is certainly worth reading.