Let $G$ be a finite group. Several recent papers (see e.g. <http://www.jstor.org/discover/10.2307/2695441>) deal with the following notion: $G$ is called a *group with perfect order subsets* or briefly, a *POS-group* if the number of elements of any possible order in $G$ is a divisor of $|G|$. Note that the symmetric group $S\_n$ is not a POS-group for any $n\ge 4$ by <http://arxiv.org/pdf/1007.0568.pdf>.


Inspired by the above notion, we will call $G$ a *group with perfect subgroup order subsets* or briefly, a *PSOS-group* if the number of subgroups of any possible order in $G$ is a divisor of $|G|$. Obviously, every finite cyclic group is a PSOS-group. Also, there are many examples of non-cyclic PSOS-groups, such as the dihedral groups $D\_{2n}$ with $n$ odd.


My question is whether $S\_n$ is a PSOS-group, more precisely which are the positive integers $n$ such that $S\_n$ is a PSOS-group? 



A question on the number of subgroups of symmetric groups
The answer is yes for $n=1,2,3$ only.


The group $S\_n$ has an elementary abelian subgroup $H$ of order $2^{\lfloor n/2 \rfloor}$ generated by the transpositions $(1,2), (3,4), \ldots,$. You can check that $H$ has at least $2^{\lfloor n/4 \rfloor \lfloor (n+2)/4 \rfloor}$ subgroups of order $2^{\lfloor n/4 \rfloor}$.


Now, for $n \ge 82$, $2^{\lfloor n/4 \rfloor \lfloor (n+2)/4 \rfloor} \gt n!$, so $S\_n$ cannot be a PSOS-group.


For $4 \le n \le 81$, you can check rather tediously that the number of subgroups of $S\_n$ of order 2 does not divide $n!$.


(Of course I have only counted those subgroups of order $2^{\lfloor n/4 \rfloor}$ that lie in a specific subgroup $H$, so it is a gross underestimate of the total number!)


