the number of subgroups of $S_p$ for prime p which have order p I want to do this without using Sylow theorems. I was wondering if it is possible.
 A: Groups of prime order are cyclic. Every cyclic group of prime order $p$ has $p-1$ generators.
Therefore, the function
$$ \{\textrm{elements of order }p\}\to\{\textrm{subgroups of order }p\}$$
given by $\pi\mapsto\langle \pi\rangle$ is a $(p-1)$-to-$1$ mapping.
Any permutation has a disjoint cycle representation, and its order is the lcm of the cycle lengths, so any element in $S_p$ of order $p$ must be a $p$-cycle $(a_1\,a_2\,\cdots\,a_p)$. Thus the elements of order $p$ are precisely those elements with this cycle type.
A general fact about permutations is how conjugation affects cycles:
$$ \pi(a_1\,a_2\,\cdots\,a_p)\pi^{-1}=(\pi(a_1)\,\pi(a_2)\,\cdots\,\pi(a_p)). \tag{$\circ$}$$
(This of course applies with any length cycle, and hence with whole disjoint cycle representations.)
Thus, the collection of elements of order $p$, i.e. of all permutations with this cycle type, form a single conjugacy class in $S_p$. One canonical representative is $t=(1\,2\,\cdots\,p)$. Any conjugacy class can be thought of as an orbit of a group action if we interpret $S_p$ as acting on itself by conjugation, so by the orbit stabilizer theorem the size of the conjugacy class is the index of the stabilizer, in other words the centralizer $C_{S_p}(t)$.
Note that $(a_1\,a_2\,\cdots\,a_p)=(b_1\,b_2\,\cdots\,b_p)$ if and only if $b_1,\cdots,b_p$ are the same numbers as $a_1,\cdots,a_p$ but possibly cycled. The permutations which cycle the entries of $t$ are precisely the elements of $\langle t\rangle$. Therefore we have
$$[S_p:\langle t\rangle]=(p-1)!$$
elements of order $p$, in which case there are
$$\frac{(p-1)!}{(p-1)}=(p-2)! $$
subgroups of order $p$.
