Conjugacy Class Equation/Proof 1

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Theorem

Let $G$ be a group.

Let $\order G$ denote the order of $G$.

Let $\map Z G$ denote the center of $G$.

Let $x \in G$.

Let $\map {N_G} x$ denote the normalizer of $x$ in $G$.

Let $\index G {\map {N_G} x}$ denote the index of $\map {N_G} x$ in $G$.


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

Let $x_j: j \in \set {1, 2, \ldots, m}$ be arbitrary elements of those conjugacy classes.

Then:

$\ds \order G = \order {\map Z G} + \sum_{j \mathop = 1}^m \index G {\map {N_G} {x_j} }$


Proof

From Conjugacy Class of Element of Center is Singleton, all elements of $\map Z G$ form their own singleton conjugacy classes.


Abelian Group

Suppose $G$ is abelian.

Then by definition of abelian group:

$\map Z G = G$

So there are as many conjugacy classes as there are elements in $\map Z G$ and hence in $G$.

So in this case the result certainly holds.

$\Box$


Non-Abelian Group

Now suppose $G$ is non-abelian.

Thus:

$\map Z G \ne G$

and therefore:

$G \setminus \map Z G \ne \O$

From Conjugacy Class of Element of Center is Singleton, all the non-singleton conjugacy classes of $G$ are in $G \setminus \map Z G$.

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:

$\conjclass {x_1}, \conjclass {x_2}, \ldots, \conjclass {x_m}$

So:

$\ds \order {G \setminus \map Z G} = \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$

or:

$\ds \order G - \order {\map Z G} = \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$


From Size of Conjugacy Class is Index of Normalizer:

$\order {\conjclass {x_j} } = \index G {\map {N_G} {x_j} }$

and the result follows.

$\blacksquare$


Sources

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