Number of Conjugates is Number of Cosets of Centralizer

Theorem
Let $G$ be a group.

Let $C_G \left({a}\right)$ be the centralizer of $a$ in $G$.

The number of different conjugates of $a$ in $G$ equals the number of different left cosets of $C_G \left({a}\right)$.

That is: $\left|{\operatorname{C}_a}\right| = \left[{G : C_G \left({a}\right)}\right]$,

where:
 * $\operatorname{C}_a$ is the conjugacy class of $a$ in $G$;
 * $\left[{G : C_G \left({a}\right)}\right]$ is the index of $C_G \left({a}\right)$ in $G$.

Consequently, $\left|{\operatorname{C}_a}\right| \backslash \left|{G}\right|$.

Proof

 * Let $G$ be a group.

Let $x, y \in \operatorname{C}_a$.

By definition of $\operatorname{C}_a$, $x a x^{-1} = y a y^{-1}$.

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

It directly follows that $\left|{\operatorname{C}_a}\right| = \left[{G : C_G \left({a}\right)}\right]$.


 * The last bit follows directly from Lagrange's Theorem.