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> 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 $H$.
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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 :)