Number of Conjugates is Number of Cosets of Centralizer

Theorem
Let $G$ be a group.

Let $\map {C_G} a$ be the centralizer of $a$ in $G$.

Then the number of different conjugates of $a$ in $G$ equals the number of different (left) cosets of $\map {C_G} a$:


 * $\card {\conjclass a} = \index G {\map {C_G} a}$

where:
 * $\conjclass a$ is the conjugacy class of $a$ in $G$
 * $\index G {\map {C_G} a}$ is the index of $\map {C_G} a$ in $G$.

Consequently:
 * $\card {\conjclass a} \divides \order G$

Proof
Let $x, y \in \conjclass a$.

By definition of $\conjclass a$:
 * $x a x^{-1} = y a y^{-1}$

By Conjugates of Elements in Centralizer, this is the case $x$ and $y$ belong to the same left coset of $\map {C_G} a$.

Hence:
 * $\card {\conjclass a} = \index G {\map {C_G} a}$

It follows from Lagrange's Theorem that:
 * $\card {\conjclass a} \divides \order G$