Suppose $|G|$ is a prime power $p^n$ and that $N$ is a normal subgroup of $G$. Show that $|y^G|$ is a power of $p$ whenever $y \in G$
Attempt:
Firstly, I assume that $y^G = \{ gyg^{-1} | g \in G \}$' i.e. we're talking about a conjugacy class here. And I know I'm supposed to use the class equation and the fact that $N$ is a normal subgroup in $G$ but I just don't know how. I mean, I saw the Wiki explanation for it but I just don't think I was able to follow the logic in a way I can absorb and repeat.
Could you show me how to answer this question and maybe explain how I can conceptualize conjugation classes in the future?
$|y^G|=[G:C_G(y)]$ by orbit stabilizer theorem and $C_G(y)$ denotes the stabilizer of $y$ in $G$.
Since $C_G(y)$ is a subgroup of $G$ its index is equal to a power of $p$.
http://en.wikipedia.org/wiki/Conjugacy_class