Conjugacy Class Equation/Proof 1

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 from Group equals Center iff Abelian we have:
 * $\map Z G = G$

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.

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

or:
 * $\displaystyle \order G - \order {\map Z G} = \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$

From Size of Conjugacy Class is Index of Normalizer:
 * $\order {\conjclass {x_j} } = \index G {\map {N_G} {x_j} }$

and the result follows.