How can we find the number of subgroups of order $n$, where $n$ is prime, in the symmetric group $S_n$? Typically, I am interested in, say $S_{17}$. I am stuck at this problem. I think it has something to do with the conjugacy classes of the symmetric group. Any ideas. Thanks beforehand.

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    $\begingroup$ In general this is hard, nut for $n=p$ prime it is easy, because all such subgroups are cyclic of order $p$, so the answer is the number of $p$-cycles divided by $p-1$. $\endgroup$ – Derek Holt Apr 27 '17 at 8:03
  • $\begingroup$ See also this question. For elements of order $n$ in $S_n$ see here. $\endgroup$ – Dietrich Burde Apr 27 '17 at 8:04
  • $\begingroup$ @DerekHolt so then, for prime $p$, the number is $(p-2)!$ right? $\endgroup$ – vidyarthi Apr 27 '17 at 8:19
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    $\begingroup$ @ancientmathematician: I guess you are right, though your example seems more complicated than necessary. Every finite group $G$ acts on itself simply transitively by say left multiplication, giving a permutation group of order $n$ acting transitively on $n$ points; I had overlooked that. If that is how you have your $A_5$s act, it gives an example. Of course the additive group of a finite vector space is just another special case of this. $\endgroup$ – Marc van Leeuwen Apr 27 '17 at 8:58
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    $\begingroup$ @vidyarthi You should be able to prove it yourself now. $\endgroup$ – Derek Holt Apr 27 '17 at 9:15

This question is over 2 years old, and has been discussed in the comments. So that this question does not remain listed as unanswered I will provide an approach.

We consider $S_{p}$, the symmetric group on $p$ letters, with order $p!$.

Subgroups of order $p$ must be cyclic. Notice that a subgroup of order $p$ contains $p-1$ elements of order $p$ and then also the identity element. On the other hand, given an element of order $p$, say $x$, then $\langle x \rangle $ is a group of order $p$. In particular every element of order $p$ is contained in precisely one subgroup of order $p$. (To put this another way, the intersection of any two subgroups of order $p$ is trivial).

So, to find the number of subgroups of order $p$, we can find the number of elements of order $p$ and divide this number by $p-1$ (Since each subgroup is being counted $p-1$ times).

How many elements of order $p$ are there in $S_{p}$? The number of ways we can arrange $p$ letters in a cycle is $p!$, but we are overcounting by a factor of $p$ since all shifts of a cycle are equal (for example in $S_{3}, (1,2,3)=(2,3,1)=(3,1,2)$). Hence the number of elements of order $p$ is $p! / p = (p-1)!$.

Then the number of subgroups of order $p$ is $(p-1)!/(p-1) = (p-2)!$.

For some further reading take a look at: Applying this result in $S_{7}$. Note the top answer here has an interesting aside about Sylow theory.


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