Number of subgroups of prime order I've been doing some exercises from my introductory algebra text and came across a problem which I reduced to proving that:

The number of distinct subgroups of prime order $p$ of a finite group $G$ is either $0$ or congruent to $1\pmod{p} $. 

With my little experience I was unable to overcome this (all I was able to conclude is that these groups are disjoint short of the identity), and also did not find any solution with a search on google (except for stronger theorems which I am not interested in because of my novice level).
I remember that a similar result is widely known as one of Sylow Theorems. This result was proven by the use of group actions. But can my problem be proved without using the concept of group actions? Can this be proven WITH the use of that concept?
EDIT: With help from comments I came up with this:
The action Derek proposed is well-defined largely because in a group if $ab = e$ (the identity), then certainly $ba = e$. By Orbit-Stabilizer Theorem we can see that all orbits are either of size 1 or $p$ (here I had most problems, and found out cyclic group of order $p$ acts on the set of solutions in the same way).
The orbits of size 1 contain precisely the elements $(x,x,x....,x)$ for some element x in G. In addition, orders of all orbits add up to $|G|^{(p-1)}$ because the orbits are equivalence classes of an equivalence relation.
But certainly $(e,e,e....,e)$ is in an orbit of size 1, and that means there has to be more orbits of exactly one element, actually $p-1 + np$ more for some integer $n$. These elements form the disjoint groups I am looking for. if $p-1$ divides $(p-1 + np)$, it's easy to check the result is 1 mod p.
Could someone check if I understood this correctly?  

Here's another approach. Consider the solutions to the equation $x_1x_2\cdots x_p=1$ in the group $G$ of order divisible by $p$. Since there is a unique solution for any $x_1,\ldots,x_{p-1}$, the total number of solutions is $|G|^{p-1}$, which is divisible by $p$. If $x_1,x_2,\ldots,x_p$ is a solution, then so is $x_2,x_3,\ldots,x_p,x_1$, and so we have an action of the $p$-cycle $(1,2,3,\ldots,p)$ on the solution set.
Since $p$ is prime, the orbits of this action have size $p$ if $x_1,x_2,\ldots,x_p$ are not all equal, and size 1 if they are all equal. So the number of solutions of $x^p=1$ is a multiple of $p$. Now use Steve D's hint to complete the proof.
Incidentally there is a theorem of Frobenius that says that for any $n>0$ and any finite group of order divisible by $n$, the number of solutions of $x^n=1$ is a multiple of $n$.  
