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

Then by Conjugates of Elements in Centralizer, $$x$$ 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.