Conjugacy class of $G$ splitting as Conjugacy class of $H\unlhd G$ Assume $H\unlhd G$ and $\mathcal{K}$ is a conjugacy class of $G$ contained in $H$ and let $x\in \mathcal{K}$.
Prove that $\mathcal{K}$ is union of $k$ conjugacy classes of equal size in $H$ where $k=|G:HC_G(x)|$.
I do not really understand the idea behind this : 
$\mathcal{K}$ is a conjugacy class of $G$ means : 
Given $x,y\in \mathcal{K}$ we have $g\in G$ such that $x=gyg^{-1}$
He is saying that $\mathcal{K}$ splits into separate classes when seen as classes over $H$.
Splitting means there are two elements in $\mathcal{K}$ that their conjugating element is not coming from $H$ 
I am not able to proceed further..
I would like to see that if it splits then it splits "Equally"
please help me to clear this.
Thank you
 A: Define an action of $\;G\;$ on $\;H\;$ by conjugation:
$$g\cdot h\mapsto h^g:=g^{-1}hg\;\;,\;\;g\in G\;,\;\;h\in H$$
What's the order of an orbit of an element here?
$$\left|\mathcal O(h)\right|:=\left|\{h^g\;;\;g\in G\}\right|=[G:G_h]\;,\;;\text{with}\;\;G_h:=\{g\in G\;;\;h^g=h\}=:C_G(h)$$
Of course, we have $\;\mathcal K=\mathcal O(x)\subset H\;$ , which btw means also that $\;x\in H\;$ ...
Now let $\;H\;$ act on itself by conjugation, so that
$$\left|\mathcal O_H(x)\right|:=\left|\{x^h\;;\;h\in H\}\right|=[H:H_x]$$
But 
$$y\in H_x\iff x^y=x\iff y\in H\cap C_G(x)$$
and by the (second, third, something) isomorphism theorem:
$$[H:H_x]=\left|H/(H\cap C_G(x))\right|\cong \left|HC_G(x)/C_G(x)\right|=[HC_G(x):C_G(x)]$$
Putting the above together:
$$|\mathcal K|=|\mathcal O(x)|=[G:C_G(x)]=[G:HC_G(x)][HC_G(x):C_G(x)]$$
A: $\newcommand{\Size}[1]{\lvert #1 \rvert}$It is probably useful to generalize this slightly. You have a group $G$ acting transitively on a set $A$, and a normal subgroup $H$ of $G$. Let $a \in A$.
We have
$$
\lvert G \rvert = \lvert a^G \rvert \cdot \lvert G_a \rvert,
\qquad
\lvert H \rvert = \lvert a^H \rvert \cdot \lvert H_a \rvert.
$$
(Here $a^G$ is the orbit of $a$ under $G$, and $G_a$ the stabilizer of $a$ in $G$.)
Also, $H_a = H \cap G_a$, and
$$
\frac{H G_a}{H} \cong \frac{G_a}{H \cap G_a}.
$$
Thus
$$
\frac{\Size{a^G}}{\Size{a^H}}
=
\frac{\Size{G}}{\Size{H}} \cdot \frac{\Size{H_a}}{\Size{G_a}}
=
\frac{\Size{G}}{\Size{H}} \cdot \frac{\Size{H \cap G_a}}{\Size{G_a}}
=
\frac{\Size{G}}{\Size{H}} \cdot \frac{\Size{H}}{\Size{H G_a}}
=
\Size{G : H G_a}.
$$
Note also the following argument. Let $a, b \in A$, with $b = a^g$, for some $g \in G$. Then
$$
b^H = a^{g H} = (a^{g H g^{-1}})^g = (a^H)^g,
$$
(we have used the fact that $H$ is normal in $G$) which shows that $g$ maps the orbit $a^H$ onto the orbit $b^H$, hence they have the same number of elements.
