  If $G$ is a finite group and $x\in G$ then the set of all elements of $G$
having the same order as $x$ is called {\em an order subset of $G$ determined
by $x$} (see [2]). We say that $G$ is a {\em group with perfect order subsets}
or briefly, $G$ is a {\em $POS$-group} if the number of elements in each order
subset of $G$ is a divisor of $|G|$. In this paper we prove that for any $n\geq
4$, the symmetric group $S_n$ is not $POS$-group. This gives the positive
answer to one of two questions rising from Conjecture 5.2 in [3].
