Conjugacy Class Equation/Proof 1

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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 $G$.


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


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

Abelian Group

Suppose $G$ is abelian.

Then from Group equals Center iff Abelian we have $Z \left({G}\right) = G$.

So there are as many conjugacy classes as there are elements in $Z \left({G}\right)$ and hence in $G$.

So in this case the result certainly holds.


Non-Abelian Group

Now suppose $G$ is non-abelian.

Thus $Z \left({G}\right) \ne G$ and therefore $G \setminus Z \left({G}\right) \ne \varnothing$.

From Conjugacy Classes of Center Elements are Singletons, all the non-singleton conjugacy classes of $G$ are in $G \setminus Z \left({G}\right)$.

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:

$\mathrm C_{x_1}, \mathrm C_{x_2}, \ldots, \mathrm C_{x_m}$


$\displaystyle \left|{G \setminus Z \left({G}\right)}\right| = \sum_{j \mathop = 1}^m \left|{\mathrm C_{x_j}}\right|$


$\displaystyle \left|{G}\right| \setminus \left|{Z \left({G}\right)}\right| = \sum_{j \mathop = 1}^m \left|{\mathrm C_{x_j}}\right|$

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.