How many subgroups of order 17 does $S_{17}$ have?

How many subgroups of order 17 does $S_{17}$ have ?

My attempt :

An order 17 group is of prime order, hence cyclic and each element in it is a generator and of order 17.
In $S_{17}$ group we can get an order 17 element only through a 17-cycle.

Number of elements of order 17 in $S_{17}$ is $\frac{17!}{17} = 16!$.

Now given that two sylow 17 subgroups have only a trivial intersection. We can conclude that 16 of these elements fall into each sylow 17 subgroup.

Hence the number of sylow 17 subgroups would be $\frac{16!}{16} = 15!$

Your answer looks correct.

It is true that there are $16!$ elements of order $17$, but if this is a homework question a marker might want you to elaborate on why that is.

An alternative (but not necessarily better) proof is as follows:

Consider the set of subgroups of $S_{17}$ of order $17$. As you noted, these subgroups must be cyclic. In particular they must be transitive.

For each subgroup $G$ fix $g\in G$ with $g(1)=2$. $g$ is the only such element of $G$ and generates $G$ so uniquely defines $G$.

Write $g=(1,2,x_3,\ldots,x_{17})$. There are $15!$ choices for the $x_i$ so $15!$ such subgroups.

Prove that the number of $p-$Sylow subgroups in the symmetric group $S_p$ is $(p − 2)!$.

Proof : Any $p-$Sylow subgroup is cyclic of order $p$ and has precisely $p − 1 $generators. Moreover, if two $p-$Sylow subgroups share a generator, they are identical. So, the elements of order p are partitioned according to which p-Sylow subgroup they belong to. We need to count the number of elements of order exactly p. This is precisely the number of distinct $p-$cycles, which is $p!/p = (p−1)!$. Grouping them into distinct $p-$Sylow subgroups (with $p−1$ in each clump), we see that the number of $p-$Sylow subgroups is $(p − 1)!/(p − 1) = (p − 2)!.$

Now take take $p= 17$ ,then

Number of $p-$Sylow subgroups in the symmetric group $S_{17}$ is $(17 − 2)!=15!$.