Help understanding proof with conjugacy classes Can someone please "walk" me through the intuitive understanding of this proof? In particular, I don't understand where all of the conclusions comes from...I have been trying for awhile to understand the relationship between conjugacy classes and Centralizer...? And this problem seems to involve my question.
Problem: Let $G$ be a finite group, $H$ a subgroup of $G$ of index $2$. Let $x \in H.$ Prove that if $C_{G}(x)$ is not contained in $H$, then $cl_{H}(x)=cl_{G}(x)$.
Proof: Since $y \in cl_{H}(x)$ we know that $y=hxh^{-1}$ for some $h \in H$. But, then $h \in G$ as well, so $y \in cl_{G}(x).$ Thus, $cl_{H}(x) \subset cl_{G}(x)$. And, since $G$ is finite, it will be sufficient to show that $|cl_{H}(x)|=|cl_{G}(x)|$ in order to prove the required statement. 
According to the fundamental counting principle, $|cl_{G}(x)|=|G:C_{G}(x)|$ and $|cl_{H}(x)|=|H:C_{H}(x)|$. And, note that $C_{H}(x)=H \cap C_{G}(x)$.
We also know - by hypothesis - that $H$ is a subgroup of index $2$, so that means $H$ is normal in $G$ and is a maximal subgroup. In particular, $HC_{G}(x)$ is a subgroup of $G$. If $C_{G}(x)$ is not contained in $H$, then $H$ is a proper subgroup of $HC_{G}(x)$, and since $H$ is a maximal subgroup in $G$, this implies that $G=HC_{G}(x)$. Hence,
$|G|=|HC_{G}(x)|=\frac{|H||C_{G}(x)|}{|H \cap C_{G}(x)|}=\frac{|H||C_{G}(x)|}{|C_{H}(x)|}$,
and therefore, $|G|/|C_{G}(x)|=|H|/|C_{H}(x)|$. Therefore,
$|cl_{G}(x)|=|G:C_{G}(x)|=|H:C_{H}(x)|=|cl_{H}(x)|$.
Thus, $cl_{H}(x)=cl_{G}(x)$.
 A: Instead of writing down formal proofs for each statement, I tried to give some intuition and name/reference so that you can search further. I hope it will be useful, excuse if there is any argument that misleads.


Since $y\in cl_H(x)$ we know that $y=hxh^{−1}$ for some $h∈H$. But, then $h∈G$ as well, so $y\in cl_G(x)$. Thus, c$l_H(x)\subset cl_G(x)$. 

I believe this is clear. If $x$ is conjugate to $y$ in $H\leq G$, it is conjugate to $y$ in $G$ by the same element.

And, since $G$ is finite, it will be sufficient to show that $|cl_H(x)| = |cl_G(x)|$ in order to prove the required statement.

For any two finite set, if they have the same size(cardinality), and one contained in the other then they are equal.

According to the fundamental counting principle, $|cl_G(x)|=[G:C_G(x)]$ and $|cl_H(x)|=[H:C_H(x)]$.

First of all, note that centralizer is a subgroup. Here or anywhere else you can find proofs.
Intuition is, since $x$ is fixed by all the elements of centralizer, cosets of $C_G(x)$ will correspond to conjugacy class. i.e. suppose $y=gxg^{-1}$ and $c\in C_G(x)$,
\begin{equation}
(gc) x (gc)^{-1}= gxg^{-1} = y
\end{equation} 
In other words, all of $gC_G(x)$ will send $x$ to $y$. Since $C_G(x)$ is a subgroup, each coset will represent another element. Hence count number of cosets.
You can find proofs under the name orbit-stabilizer theorem.

And, note that $C_H(x) = H\cap C_G(x)$.

This is clear.
(If $h\in H$ and fixes $x$, then $h\in C_H(x)$. And read it backwards.)

We also know - by hypothesis - that $H$ is a subgroup of index $2$, so that means $H$ is normal in $G$ and is a maximal subgroup. 

Let me try to quickly prove it. $H$ is subgroup of index $2$ means $G=H\cup gH$ for some $g\in G$. 
Let $x\in G$, if $x\in H$, $xHx^{-1}=H$. Suppose not. Then $xH = gH$ and $Hx = gH$. Hence $xH = Hx\implies xHx^{-1}=H$. In words, there is no way for left and right cosets to be different.
$H$ is maximal because if there exists $H\leq S\leq G$ then $[G:H]=[G:S][S:H]$, but $2 = [G:S][S:H]$ implies either $S=H$ or $S=G$. 

In particular, $HC_G(x)$ is a subgroup of $G$.

This is true because $H$ is normal. You can find some discussion here. 

If $C_G(x)$ is not contained in $H$, then $H$ is a proper subgroup of $HC_G(x)$, and since $H$ is a maximal subgroup in $G$, this implies that $G=HC_G(x)$.

I believe this should be clear by now.

\begin{equation}
|G| = |HC_G(x)| = \frac{|H||C_G(x)|}{|H\cap C_G(x)|}
\end{equation}

This is called product formula. Let me try to give intuition. Argument is somewhat similar to orbit-stabilizer.Say $S,T$ subgroups of $G$. Note that $T\cap S$ is a subgroup, say subgroup of $T$.  Think about cosets of $S$. For $x\in T\cap S$, $xS = S$, hence $x$ 'stabilize' S. Hence each coset of $T\cap S$ in $T$ will give distinct coset of $S$. You may think like
\begin{equation}
|TS| = [T:T\cap S]|S| = \frac{|T|}{|T\cap S|}|S|
\end{equation}

and therefore, $|G|/|C_G(x)|=|H|/|C_H(x)|$. Therefore,
  $|cl_G(x)|=|G:C_G(x)|=|H:C_H(x)|=|cl_H(x)|$.
  Thus, $cl_H(x)=cl_G(x)$.

This I believe also clear.
