Proof of the class equation

Proof

We split $G$ into conjugacy classes and sort the classes by size.

1. The conjugacy classes are the orbits of the action by conjugation. Since orbits partition the set, the classes are pairwise disjoint and cover $G$.
2. Let $a \in G$. Then $a = e a e^{-1} \in \operatorname{Cl}(a)$. So $\operatorname{Cl}(a) = \set{a}$ if and only if $g a g^{-1} = a$ for all $g \in G$.
3. As $g a g^{-1} = a$ is equivalent to $g a = a g$, step 2 says $\operatorname{Cl}(a) = \set{a}$ if and only if $a$ lies in the centre $Z(G)$.
4. By step 3, the classes with one element are the sets $\set{z}$ with $z \in Z(G)$. Together they contain exactly $\abs{Z(G)}$ elements.
5. The classes with more than one element are $\operatorname{Cl}(a_1), \dots, \operatorname{Cl}(a_k)$, by the choice of $a_1, \dots, a_k$. Since $G$ is finite, step 1 gives $\abs{G} = \abs{Z(G)} + \sum_{i=1}^{k} \abs{\operatorname{Cl}(a_i)}$.
6. By the class size formula, $\abs{\operatorname{Cl}(a_i)} = [G : C_G(a_i)]$. So $\abs{G} = \abs{Z(G)} + \sum_{i=1}^{k} [G : C_G(a_i)]$.