Normal Group and Conjugacy Class I am trying to solve the following: 

Let $G$ a group and $N$ a normal subgroup of $G$ with index a prime $p$. Consider $C$ a conjugacy class of $G$, which is contained in $N$. Prove that either $C$ is a conjugacy class of $N$ or it is a union of $p$ distinct conjugacy classes of $N$.

[EDIT] Following the suggestion of @stewbasic I have done the following:
After denoting by $X^g=\{g^{-1}xg, x\in X\}$ and by $B$ the set of conjugacy classes of $N$ we can define the action $\cdot:G \times B \to B$ by $g\cdot A= A^g$. It is not that diffcult, but also not completely trivial, to verify that this is in fact an action and that this is trivial over $N$, so we can actually "define" the same action now of $G/N$ on $B$. With this new action I know now that for each $A \in B$ the $|Orb_{G/N}(A)|$ will be either $1$ or $p$ by the Orbit Stabilizer Theorem. If this is $1$ then all other orbits will have the same size and if that is $p$ we will have only one orbit. I think the problem would be done if $C$ given in the statement was a conjugacy class of $N$ but it is a conjugacy class of $G$. Am I missing something?
The thing is: I am studying for my PhD qualyfing exam and I have a lot of difficulties with algebra since I don't like it much. So, I wouldn't like any complete answer because although it is probably a very easy problem, I wanna learn something by thinking of it. So, I would be very thankful if I could have some hints about how to approach this problem or at least some comments about whether my first two approaches could take me somewhere or not!
Thank you so much, friends!
 A: If $X\subseteq G$ and $g\in G$, let $X^g=\{g^{-1}xg\mid x\in X\}$. Let $B$ denote the set of conjugacy classes of $N$.
Hint: Show that $C\mapsto C^g$ defines an action of $G$ on $B$. Show that this action is trivial on $N$, so it induces an action of $G/N$. What are the possible sizes of orbits under this action?
Finally suppose $C$ is a conjugacy class of $G$ contained in $N$. Pick any $c\in C$ and let $C'\in B$ be the conjugacy class of $N$ containing $c$. Show that $C$ is exactly the union of the orbit of $C'$ under the above action.
A: You have to prove that number of conjugacy classes in N equals to
$|G : NS(x)|$, where $S(x)$ is the centralizer of element $x$ of $C$ in $G$. 
To do this show that number of conjugacy classes in $N$ equal to: 
$|G : S(x)|/|N:N \cap S(x)| = |G:NS(x)|\times|NS(x):S(x)|/|N:N \cap S(x)| = |G:NS(x)|$.
(The proof of $|NS(x):S(x)| = |N:N \cap S(x)|$ is very similar to the proof of Second Isomorphism Theorem).
Considering that $p$ is prime, $NS(x)$ is either equal $N$ or $G$. 
