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 \item{a)} $G$ is abelian or a $\BFC$-group of finite exponent bounded by a function of $m$ and $n$. \item{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 $$ X_1\cdots X_n\subseteq \bigcup_{\sigma \in S_n\bs 1} X_{\sigma (1)} \cdots X_{\sigma (n)} $$ 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 $n$.

This note contains the classification of the finite groups $G$ satisfying the condition $N_{G}(H)/C_{G}(H)\cong \Aut(H)$ for every abelian subgroup $H$ of $G$.