I want to prove that, if $x^G$ denotes the conjugacy class of $x$ in a finite group $G$ and $H$ is a subgroup of $G$ then
$$
|\{g\in G : g^{-1}xg\in H \}|=|x^G\cap H|\cdot |C_G(x)|,
$$
where $C_G(x)$ denotes centralizer of $x$ in $G$.
If $x^G\cap H=\emptyset$ then nothing to prove.
But, if $x^G\cap H\neq \emptyset$, I could not see a clever way to get the factor $|C_G(x)|$ in the right side when we try to count number of $g$'s such that $g^{-1}xg\in H$.
Can one help (or suggest any other clever way) to prove above equality?
Best Answer
Fix $h\in H\cap x^G$. Suppose that $g_1^{-1}xg_1=h=g_2^{-1}xg_2$. Then $g_1g_2^{-1}\in C_G(x)$ or equivalently $g_1=cg_2$ with $c\in C_G(x)$. Hope this helps.