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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

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