TITLE: Why is $\#\{s_\alpha,s_\alpha^{-1}hs_\alpha\in H\} = \frac{|C_G(h)|}{|C_{N_G(H)}(h)|}$?
QUESTION [5 upvotes]: In one literature, I encountered the following problem.
Suppose $G$ is a group and $H$ is its subgroup. $N_G(H)$ is the normalizer of $H$ in $G$. A number $c(h)$ is defined for every $h$ in $H$ this way: 

we first decompose $G$ using coset of $N_G(H)$: $G=\bigcup_{\alpha}s_\alpha N_G(H)$, 
$c(h)$ is defined to be $c(h)=\#\{s_\alpha,s_\alpha^{-1}hs_\alpha\in H\}$, where $\#$ means number of elements in a set.

Then, it is claimed that $$c(h)=\frac{|G|}{\#[h]_G}/\frac{|N_{G}(H)|}{\#[h]_{N_G(H)}}=\frac{|C_G(h)|}{|C_{N_G(H)}(h)|},$$ where $C_G(h)$ ($C_{N_G(H)}(h)$) is centralizer of $h$ in $G$ ($N_G(H)$) and $[h]_G$ ($[h]_{N_G(H)}$) is conjugate class in $G$ ($N_G(H)$) containing $h$.
I cannot prove that these two numbers are equal and cannot find a counter example. Can anyone help me on this?

REPLY [5 votes]: I don't think it is true. Let $G = S_4$, $h=(1,2)(3,4)$, and $H=C_G(h)$. So $|H|=8$.
Then $g^{-1}hg \in H$ for all $g \in G$ and $N_G(H)=H$, so $c(h)=3$. But $|C_G(h)|=|C_{N_G(H)}(h)|=8$.