Number of conjugates of $x \in G$ in $H \triangleleft G$ 
Let $G$ be a finite group, let $H$ be a normal subgroup of prime index, and let $x \in H$ satisfy $C_H(x) < C_G(x)$. If $y \in H$ is conjugate to $x$ in $G$, then $y$ is conjugate to $x$ in $H$.

I have to prove this assertion. My attempt is:
(all the letters used here are elements of $\mathbb{Z}$)
let $|G|=n$, by Lagrange theorem $n=p \cdot h$ where $p$ is prime and $h= |H|$.
 Let $|C_G(x)|=m$, cleary by hypothesis, $m>1$ and is not a prime (it has a proper subgroup), let $|C_H(x)|=b$, again $b|m , \ b|h$ so I put $h=b \cdot a$ and $m=b \cdot c$. The last observation I did is,by Lagrange $m= b \cdot c| h \cdot p \Longrightarrow b \cdot c | h$ so exists $k \in \mathbb{Z}$ s.t. $b \cdot c \cdot k = b \cdot a = h$ and $a = c \cdot k$
By the Orbit-Stabilizer Theorem (for finite groups) we have
 $|x^G| = \dfrac{|G|}{|C_G(x)|} = \dfrac{h \cdot p}{b \cdot c} =\dfrac{b \cdot a \cdot p}{b \cdot c} = k \cdot p$
and
$|x^H| = \dfrac{|H|}{|C_H(x)|} = \dfrac{b \cdot a}{b} = a = c \cdot k$
My problem is I can't prove that $p=c$ in other words that  $[C_G(x) : C_H(x) ]=p$. (tried to use 2nd isomorphism theorem, but got a circular argument) and I'm out of ideas:(
I hope I'm on the right way to prove that otherwise every advice is welcomed :)
 A: There is some mistake in the line "by Lagrange $m=b\cdot c|h\cdot p\Longleftarrow b\cdot c|h$ so exists $k$ s.t. $b\cdot c\cdot k=b\cdot a=h$". Also, I'm not sure what this is trying to accomplish.
Anyway, you can prove the theorem using a counting argument as follows (with your notation):$$|x^G|={b\cdot a\cdot p\over b\cdot c}={a\cdot p\over c}, |x^H|={b\cdot a\over b}=a.$$
We want to show $|x^G|=|x^H|$ which is equivalent to showing $c=(C_G(x):C_H(x))=p.$
Define $\varphi :C_G(x)\to G/H$ by $g\mapsto gH$. This map is a homomorphism with kernel $C_G(x)\cap H=C_H(x)$. By the first isomorphism theorem, $C_G(x)/C_H(x)$ is isomorphic to a subgroup of $G/H$ and by Lagrange $c=|C_G(x)/C_H(x)|$ divides $|G/H|=p$. But since $C_H(x)$ is a proper subgroup of $C_G(x), c\neq 1$ so $c=p$. $\square$
Here is a proof that better shows, why the theorem works:
Since $(G:H)$ is prime, $H$ is a maximal subgroup of $G$ (for any $K$ with $H\le K\le G$ we have $p=(G:H)=(G:K)(K:H)$, so either $(G:K)=1$ or $(K:H)=1$). Thus, for any subgroup $K$ of $G$ with $K\not \subseteq H$, we have $G=KH$. Note that $C_G(x)\not \subseteq H$, because $C_G(x)\cap H=C_H(x)\neq C_G(x)$, thus $G=C_G(x)H$. Now suppose $y$ is a conjugate of $x$ in $G$, then for some $g\in G$,$$g^{-1}xg=y.$$
Write $g=kh$ with $k\in C_G(x), h\in H$, then $$y=g^{-1}xg=h^{-1}k^{-1}xkh=h^{-1}k^{-1}kxh=h^{-1}xh.$$ (The third equality holds, because $k\in C_G(x)$, i.e. $xk=kx$) Therefore, $y$ is a conjugate of $x$ in $H$. $\square$
A: Your idea is in the right direction. Using an isomorphism theorem gives
$$C_G(x) / C_H(x) = C_G(x) / C_G(x) \cap H \cong C_G(x)H/H$$
Here $C_G(x)H/H$ is a subgroup of $G/H$. Can you see now why $[C_G(x) : C_H(x)] = p$?
