# Conjectures around number of subgroups of symmetric group [closed]

I am asking if you know unsolved or recently solved conjectures around numbers of subgroups in symmetric or alternating groups. In fact, is there a formula depending of $$n$$ to count subgroups of order $$k$$ in the symmetric group $$S_n$$ ? Particularly, how subgroups $$S_n$$ contains ?

I know it is possible to use GAP to find this on the cases for $$n=1,\dots,15$$, but i don't know if formulas around these questions have already been discovered. If you have references on the topic, don't hesitate.

Thanks

• You have already asked about conjectures on $S_n$, see here. "How subgroups $S_n$ contains"? Every finite group is a contained in some symmetric group $S_n$. On the number of subgroups of a fixed $S_n$, see OEIS. – Dietrich Burde Aug 1 at 18:09
• – MJD Aug 1 at 18:25

In general, it is difficult to find a formula for the number of subgroups of $$S_n$$. However, for $$n=p$$ the answer is easy. If $$n=p$$ is prime then $$S_p$$ contains $$(p-1)!/(p-1) = (p-2)!$$ subgroups of order $$p$$.

For the general case, see the answers at MSE so far:

Enumerating all subgroups of the symmetric group

There are upper bounds for the number by Pyber and Shalev.

See also the paper by Derek Holt for a list of representatives of the conjugacy classes of subgroups of $$S_n$$ for $$n ≤ 18$$, including the $$7274651$$ classes of subgroups of $$S_{18}$$.

• Thanks a lot for your references. I known the case n=p, i also read something about $p$-subgroup of $S_n$. Do you think there exist results like that for particularly $n$ ? I mean for exemple for odd numbers ? – Lazare Aug 1 at 21:14

There is a conjecture of Pyber that the number of subgroups (or conjugacy classes of subgroups if you prefer!) of $$S_n$$ is $$2^{\frac{n^2}{16}+o(n^2)}$$.

It is easy to prove that this is a lower bound: you can do this by looking at elementary abelian two subgroups in which all orbits have length 1 or 2.

I think the best upper bound proven so far is something like $$2^{\frac{n^2}{4}+o(n^2)}$$ but I would need to check.

• When you said "elementary abelian two subgroups" you mean isomorphic to the trivial group and $\mathbb{Z}/2 \mathbb{Z}$ ? Thanks a lot for this conjecture, have you got recent references (articles) for your upper bound? – Lazare Aug 1 at 21:06
• For a prime $p$, an elementary abelian $p$-group is an abelian group $G$ with $g^p=1$ for all $g \in G$. It is isomorphic to a direct product of copies of ${\mathbb Z}/p{\mathbb Z}$. As I said, I am not sure exactly what the best known upper bound is - I will need to hunt around for references. – Derek Holt Aug 1 at 22:02
• Thanks for your answer. If you re-find your references, don't hesitate to make them here. I'm very interested on this upper bound. – Lazare Aug 2 at 13:51
• Another question about this number: There exists a lower-bound (untrivial) for the number of subgroups of $S_n$ ? – Lazare Aug 4 at 13:34
• Yes I answered that question in my answer! It would be a good exercise for you to try and prove that, just to try and get a feeel for the problem. – Derek Holt Aug 4 at 15:02