Proving the size of the conjugacy class $x^G$ is given by $|x^G| = \frac{|G|}{|C_G(x)|}$.

Question

I am reading a textbook and am confused of the proof of the following theorem:

Let $x\in G$. The size of the conjugacy class $x^G$ is given by $$|x^G| = \frac{|G|}{|C_G(x)|}$$ where $C_G(x)$ is the set of elements of $G$ commuting with $x$; that is, $$C_G(x)=\{g\in G: xg= gx\}.$$

The first part of proof goes as the following:

For $g$, $h\in G,$ we have $$g^{-1}xg = h^{-1}xh \Longleftrightarrow hg^{-1}x=xhg^{-1} \Longleftrightarrow hg^{-1}\in C_G(x) \Longleftrightarrow C_G(x)g = C_G(x)h$$

My question is how to get the last step; i.e., $C_G(x)g = C_G(x)h$?

Note: The textbook is "Representations and Characters of Group (G. James & M. Liebeck) p.107"

Answer 1

$hg^{-1}$ in $C$ implies $h$ in $Cg$, which implies $Ch=Cg$.