$N\trianglelefteq G$. Now let $A$ be a conjugacy class in $G$ such that $A\subseteq N$.
- Prove $A$ is a union of conjugacy classes in $N$ and they all have the same size.
- Denote this number of classes as $k$, now prove $k=[G:C_G(x)N]$ for each $x\in A$.
This question alreay appears on stackexchange but my proof seems different. (Or perhaps I just wrote it down in to much detail) Is the following correct?
First statement
Choose an $x$ in $N$ and let $A=\{x^g: g\in G\} = x^G$ be a conjugacy class in $G$.
Now let $N$ act on $A$ by conjugation, such that $(x^g)^n = x^{gn}$. Notice how this element is once again an element of $A$.
Now consider the orbits resulting from this action. These orbits partition $A$ so $A$ is a union of conjugacy classes in $N$. Now we prove that they have the same size
Choose a $x^g$ and $x^h \in A$. I wish to show $|(x^g)^N|=|(x^h)^N|$. By the orbit-stabilizer formula $$ |N| = |N_{x^g}| \cdot |(x^g)^N| \qquad |N| = |N_{x^h}|\cdot |(x^h)^N|. $$
Since $N_{x^g} = g^{-1}N_x g$ and $N_{x^h} = h^{-1}N_x h$ it follows that $|N_{x^g}| = |N_{x^h}|$. Because $\iota_g, \iota_h\in \mathrm{Inn}(N)$, this implies $|(N_x)^g| = |N_x| = |(N_x)^h|$. Then this results in $|(x^g)^N| = |(x^h)^N|$.
Second statement
Here Conjugacy class of $G$ splitting as Conjugacy class of $H\unlhd G$ seems to provide adequate inspiration.