The index of the centraliser of an element of a group equals the order of its conjugation class. Let $G$ be a group, and $x \in G$. The conjugation class of $x$ is:
$$ Cl(x) = \{ y \in G \mid \exists g \in G, y=gxg^{-1} \} $$
And the centraliser of $x$ in $G$ is:
$$ C_G (x) = \{ g \in G \mid gx=xg \} $$
Show that $|Cl(x)| = |G:C_G (x)|$.
I have been working in this proposition for many hours. It appears in the notes of an introduction course in group theory. I have tried to build an isomorfism from $Cl(x)$ to $G/C_G (x)$ but I think $C_G (x)$ it's not a normal group, so I don't know how to keep working.
 A: To expand on my comment, we consider the map $\phi : G / C_G (x) \to Cl(x)$ defined by $\phi (g C_G(x)) = gxg^{-1}$.
Note, we do not require that $G / C_G(x)$ is a group. Rather, $G / C_G(x) = \{g C_G(x) : G \in G\}$ is merely the set of left cosets of $C_G(x)$ in $G$.
All we have to do is show that $\phi$ is a bijection. We have to check 3 things:

*

*$\phi$ is well defined. That is, if $g_1 C_G(x) = g_2 C_G(x)$, then $\phi(g_1 C_G(x)) = \phi (g_2 C_G(x))$.

*$\phi$ is a surjection.

*$\phi$ is an injection.

I'll show 1, as that is usually the part people overlook. If  $g_1 C_G(x) = g_2 C_G(x)$ then $g_1 = g_2 y$ for some $y \in C_G(x)$. We have
\begin{align}
\phi(g_1 C_G(x)) &= g_1 x g_1^{-1} = (g_2 y) x (g_2 y)^{-1}
\\&= g_2 y x y^{-1} g_2^{-1} = g_2 x yy^{-1} g_2^{-1}
\\&= g_2 x g_2^{-1} = \phi (g_2 C_G(x))
\end{align}
which shows that $\phi$ is well defined.
Proving that $\phi$ is an injection essentially amounts to reversing the argument for 1.
Finally, $\phi$ is obviously a surjection (why?).
Having established 1, 2, and 3, we conclude $\phi$ is a bijection, i.e. $| G / C_G (x) | = |Cl (x)|$.

What is the moral of this problem? To show that two sets have the same size, exhibit a bijection. For undergraduate math, the necessary bijection will usually be the only bijection that makes sense.
