On totally k-closed nilpotent groups

Abstract

A group $G$ is said to be totally $k$-closed for a positive integer $k$ if, in each of its faithful permutation representations on a set ${\Omega }^k$, $G$ is the largest subgroup of the symmetric group $\mathrm{Sym}(\Omega )$ that preserves every $k$-orbit in the induced action on the set $\Omega \times \cdots \times \Omega ={\Omega }^k$. We prove that for $k\ge 1$, every finite nilpotent group with Sylow subgroups of orders at most $p^k$ for all primes $p$ dividing $|G|$ is totally $k$-closed if and only if it does not contain an elementary abelian subgroup ${\mathbb{Z}}_p^k$ for every prime $p$.