# 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.

• 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 $p-1$. – Derek Holt Apr 27 '17 at 8:03
• See also this question. For elements of order $n$ in $S_n$ see here. – Dietrich Burde Apr 27 '17 at 8:04
• @DerekHolt so then, for prime $p$, the number is $(p-2)!$ right? – vidyarthi Apr 27 '17 at 8:19
• @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. – Marc van Leeuwen Apr 27 '17 at 8:58
• @vidyarthi You should be able to prove it yourself now. – 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.