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Let $G$ be a finite group. Several recent papers (see e.g. 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

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?

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You checked it only for $n=1,2,3$ or you went up to $n=6$? I vote to close. – Mark Sapir Aug 25 '13 at 12:13
You are right: my question is not clear. I restated it. – Marius Tarnauceanu Aug 26 '13 at 0:15
up vote 5 down vote accepted

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!)

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$S_5$ has 120 elements, 30 is a divisor of 120 and so a "possible order" of a subgroup of $S_5$, the number of subgroups of order 30 in $S_5$ is zero, and zero is not a divisor of 120, so $S_5$ is not PSOS. Is it true that for $n\ge5$, $S_n$ has no subgroup of index 4? If so, that would be another way to show $S_n$ is not PSOS for $n\ge5$. And it's proved at… that for $n\ge5$, $S_n$ has no proper subgroup of index less than $n$, except for $A_n$. – Gerry Myerson Aug 27 '13 at 0:02
Yes, that makes it a lot easier! I thought the question would be more interesting (and more suitable for MO) if we looked for non-divisors other than 0. – Derek Holt Aug 27 '13 at 9:42
@GerryMyerson: Well -- I'm not sure that an order for which there is no subgroup qualifies as "possible order" in the sense of the question ... . – Stefan Kohl Aug 27 '13 at 10:28
I agree that Derek's answer is more interesting, and probably more what OP wanted (and that my answer is easier). – Gerry Myerson Aug 27 '13 at 11:24
$H$ has a subgroup of index 2 contained in $A_n$ so the same argument will work for all sufficiently large $n$, and then it is just a matter of checking small $n$. – Derek Holt Aug 28 '13 at 7:31

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