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Given that $N \lhd G$ and $x \in N$, there are two cases. If $C_G(x) \nsubseteq N$ we have that the orbit $G(x) = \{gxg^{-1} \mid g \in G\}$ is equal to the orbit $N(x)$. If $C_G(x) \subseteq N$ then the orbit $G(x)$ is the union of $p$ orbits in $N$ with equal number of elements; in other words, there are $x_1,\cdots,x_p \in N$ so that $\displaystyle G(x) = \bigcup_{i=1}^p N(x_i)$.

How can I show that?

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Please edit the condition $[G : N]=p$ into your question. Also if $p$ is prime, which I assume is the case, please indicate that too. –  peoplepower Oct 14 '12 at 22:19
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1 Answer

Hint: In the first case, $C_G(x)\not\subseteq N$, if a normal subgroup has prime index, then it is maximal, so express $G$ as a product. When $y$ is conjugate to $x$ write the expression $y={}^gx$ in a different way.

When $C_G(x)\subseteq N$, consider the fact that $C_N(x)=C_G(x)$ and put it together with a different form of $C_G(x^g)$. Finally, what can be determined from examining the action of $N$ on $G(x)$.

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thanks for the answer. What do you mean by "express G as a product" ? and what means "$y={}^gx$ ? –  André Oct 15 '12 at 8:33
    
By product, I mean look at a certain internal product $HN$, where $H$ is some subgroup and $N$ is some normal subgroup. –  peoplepower Oct 15 '12 at 22:16
    
@André The notation ${}^gx$ is the conjugate $gxg^{-1}$. –  peoplepower Oct 15 '12 at 22:17
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