number of subgroups of order $ n $ in $S_n$ 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.
 A: 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. 
