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:

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

Proof 1

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 is Abelian iff Center Equals Group 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.

$\Box$

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:

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

So:

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

or:

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

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.

$\blacksquare$

Proof 2

Let the distinct orbits of $G$ under the Conjugacy Action be $\operatorname{Orb} \left({x_1}\right), \operatorname{Orb} \left({x_2}\right), \ldots, \operatorname{Orb} \left({x_s}\right)$.

Then from the Partition Equation:

$\left|{G}\right| = \left|{\operatorname{Orb} \left({x_1}\right)}\right| + \left|{\operatorname{Orb} \left({x_2}\right)}\right| + \cdots + \left|{\operatorname{Orb} \left({x_s}\right)}\right|$

From the Orbit-Stabilizer Theorem:

$\left|{\operatorname{Orb} \left({x_i}\right)}\right| \backslash \left|{G}\right|, i = 1, \ldots, s$

The result follows from the definition of the Conjugacy Action.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $25.16 \ \text{(c)}$