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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.2012-10-14

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