The number of $p$-regular elements in a group I would like to prove the following:

Let $G$ be a finite group, $p$ a prime number and $P$ a Sylow $p$-subgroup of $G$. Let $E$ be the set of all $p$-regular elements of $G$ (i.e. elements whose order is not a multiple of $p$). Let $C$ be the centralizer of $P$ in $G$. 
Then $|E|\equiv|E\cap C|$ (mod $p$).

Unfortunately, I don't know where to start. I have absolutely no intuition on why this statement should be true. 
I'd be glad for anything to start me off.  
 A: Consider the action of $P$ on $E$ defined by $g*x=gxg^{-1}$ for $g\in P$ and $x\in E$. This is well defined, since the property of being $p$-regular is clearly preserved by group isomorphisms. The set of fixed points under this action (i.e. the elements of $E$ left fixed by all elements of $P$) is $C\cap E$ by the very definition of the centralizer.
The proof is finished by the following 
Lemma: Let $P$ be a $p$-group acting on a finite set $E$. Let $E^P$ be the set of fixed points under this action. Then
$$|E|\equiv|E^P|\text{ mod }p.$$ 
Proof: We have to show that the order of $E\backslash E^P$ is a multiple of $p$. By definition, $E^P$ is the union of the one-point orbits of the action. So the order of $E\backslash E^P$ is the sum of the orbit orders $>1$. By the orbit-stabilizer-theorem, the order of an orbit divides the order of $P$. Since $P$ is a $p$-group, it follows that orbit orders $>1$ have to be multiples of $p$. So the order of $E\backslash E^P$ is the sum of multiples of $p$ and therefore itself a multiple of $p$.
