Conjugacy Class Equation

Theorem
Let $$G$$ be a group.

Let $$\left|{G}\right|$$ be the order of $$G$$.

Let $$Z \left({G}\right)$$ be the center of $$G$$.

Let $$x \in G$$.

Let $$N_G \left({x}\right)$$ be the normalizer of $x$ in $G$.

Let $$\left[{G : N_G \left({x}\right)}\right]$$ be the index of $N_G \left({x}\right)$ in $G$.

Let $$m$$ be the number of non-singleton conjugacy classes of $$G$$.

Then $$\left|{G}\right| = \left|{Z \left({G}\right)}\right| + \sum_{j=1}^m \left[{G : N_G \left({x_j}\right)}\right]$$.

Proof
From Conjugacy Classes of Center Elements are Singletons, all elements of $$Z \left({G}\right)$$ form their own singleton conjugacy classes.


 * If $$G$$ is abelian, then the result is certainly true, because then $$Z \left({G}\right) = G$$, from Center of Abelian Group is Whole Group and so there are as many conjugacy classes as there are elements in $$Z \left({G}\right)$$ and hence in $$G$$.


 * So, suppose $$G$$ is non-abelian. Thus $$Z \left({G}\right) \ne G$$ and therefore $$G - Z \left({G}\right) \ne \varnothing$$.

From Conjugacy Classes of Center Elements are Singletons, all the non-singleton conjugacy classes of $$G$$ are in $$G - Z \left({G}\right)$$. From the way the theorem has been worded, there are $$m$$ of them.

Let us choose one element from each of the non-singleton conjugacy classes and call them $$x_1, x_2, \ldots, x_m$$.

Thus, these conjugacy classes can be written $$\mathrm{C}_{x_1}, \mathrm{C}_{x_2}, \ldots, \mathrm{C}_{x_m}$$.

So $$\left|{G - Z \left({G}\right)}\right| = \sum_{j=1}^m \left|{\mathrm{C}_{x_j}}\right|$$,

or $$\left|{G}\right| - \left|{Z \left({G}\right)}\right| = \sum_{j=1}^m \left|{\mathrm{C}_{x_j}}\right|$$.

From Size of Conjugacy Class is Index of Normalizer, $$\left|{\mathrm{C}_{x_j}}\right| = \left[{G : N_G \left({x_j}\right)}\right]$$ and the result follows.