Published online by Cambridge University Press: 20 November 2018
In this paper we prove the following:
1. Let $m\ge 2,\,n\ge 1$ be integers and let
$G$ be a group such that
${{(XY)}^{n}}\,=\,{{(YX)}^{n}}$ for all subsets
$X,Y$ of size
$m$ in
$G$. Then
a) $G$ is abelian or a
$\text{BFC}$-group of finite exponent bounded by a function of
$m$ and
$n$.
b) If $m\ge n$ then
$G$ is abelian or
$|G|$ is bounded by a function of
$m$ and
$n$.
2. The only non-abelian group $G$ such that
${{(XY)}^{2}}\,=\,{{(YX)}^{2}}$ for all subsets
$X,Y$ of size 2 in
$G$ is the quaternion group of order 8.
3. Let $m$,
$n$ be positive integers and
$G$ a group such that
$${{X}_{1}}\cdot \cdot \cdot \,{{X}_{n}}\,\subseteq \,\bigcup\limits_{\sigma \in {{S}_{n}}\,\backslash \,1}{{{X}_{\sigma (1)}}\cdot \cdot \cdot \,{{X}_{\sigma (n)}}}$$
for all subsets ${{X}_{i}}$ of size
$m$ in
$G$. Then
$G$ is
$n$-permutable or
$|G|$ is bounded by a function of
$m$ and
$n$.