Let $x$ be an element of a group $G$. Show that the elements of $G$ which commute with $x$ form a subgroup of $G$. Call this subgroup $C(x)$. Prove the size of the conjugacy class of $x$ is equal to the index of $C(x)$ in $G$.
I proved that $C(x)$ is a subgroup of $G$. I got stuck on the second part.
I wanted to use the Orbit-Stabilizer Theorem. We can let $G$ work on it's elements with conjugation so $|G(x)|=|\text{conjugacy class of }x|$ with $x \in G$. Don't really know where to go from here.
I've read this, but I don't think the given answer is the proof that my book means.
If $g\in G$, consider $gxg^{-1}$. It is conjugate to $x$. Now, consider $gh$, with $h\in C(x)$. Then$$(gh)x(gh)^{-1}=ghxh^{-1}g^{-1}=gxg^{-1},$$since $x$ and $h$ commute. So, this defines a map from $G/C(x)$ into the conjugacy classes of $x$, which is surjective (since $g$ is arbitrary).
Let us see that it is one-to-one too. Suppose that $g'\in G$ is such that $gxg^{-1}=g'xg'^{-1}$. Then $(g'^{-1}g)x(g'^{-1}g)^{-1}=x$. Therefore, $g'^{-1}g\in C(x)$, which means that $gC(x)=g'C(x)$.
So, there is a bijection between $G/C(x)$ and the set of those elements of $G$ which are conjugate to $x$.