http://dx.doi.org/10.4153/CMB-1998-010-0
Canad. Math. Bull. 41(1998), 65-70
Published:1998-03-01 Printed: Mar 1998
A. Mohammadi Hassanabadi
Akbar Rhemtulla
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Abstract
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 $n$.
© Canadian Mathematical Society, 2013
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