Class Equation and Conjugacy Class Decomposition

After centralizers and conjugacy classes have been defined, the next natural step is to count the whole finite group by collecting its elements into conjugacy classes. The class equation is a compact expression of this decomposition. It separates central elements, whose conjugacy classes have size one, from noncentral elements, whose conjugacy classes have sizes controlled by centralizers. This result is one of the main bridges from elementary group theory to Sylow theory, because it turns group action ideas into divisibility statements.

Let

G

be a finite group. A \textbf{conjugacy class decomposition} of

G

is a partition of

G

into its distinct conjugacy classes:

\begin{equation*} G=Cl_G(a_1)\cup Cl_G(a_2)\cup \cdots \cup Cl_G(a_r), \end{equation*}

where

a_1,a_2,\ldots,a_r

are representatives of the distinct conjugacy classes and the union is disjoint.

Let

G

be a finite group. A \textbf{conjugacy class decomposition} of

G

is a partition of

G

into its distinct conjugacy classes:

\begin{equation*} G=Cl_G(a_1)\cup Cl_G(a_2)\cup \cdots \cup Cl_G(a_r), \end{equation*}

where

a_1,a_2,\ldots,a_r

are representatives of the distinct conjugacy classes and the union is disjoint.

Let

G

be a finite group. If

a_1,a_2,\ldots,a_r

are representatives of the distinct conjugacy classes of

G

, then

\begin{equation*} |G|=\sum_{i=1}^{r}[G:C_G(a_i)]. \end{equation*}

Given that

G

is a finite group and

a_1,a_2,\ldots,a_r

are representatives of the distinct conjugacy classes of

G

. To prove that

|G|=\sum_{i=1}^{r}[G:C_G(a_i)]

. Since conjugacy is an equivalence relation on

G

, its equivalence classes form a disjoint partition of

G

. Therefore

\begin{equation*} G=Cl_G(a_1)\cup Cl_G(a_2)\cup \cdots \cup Cl_G(a_r). \end{equation*}

Since the union is disjoint,

\begin{equation*} |G|=\sum_{i=1}^{r}|Cl_G(a_i)|. \end{equation*}

For each

i

, the conjugacy class formula gives

\begin{equation*} |Cl_G(a_i)|=[G:C_G(a_i)]. \end{equation*}

Substituting these values gives

\begin{equation*} |G|=\sum_{i=1}^{r}[G:C_G(a_i)]. \end{equation*}

Hence

|G|=\sum_{i=1}^{r}[G:C_G(a_i)]

\square

Let

G

be a finite group. The \textbf{class equation of

G

} is the formula

\begin{equation*} |G|=|Z(G)|+\sum [G:C_G(a)], \end{equation*}

where the summation is taken over one representative

a

from each noncentral conjugacy class of

G

Let

G

be a finite group. Then

\begin{equation*} |G|=|Z(G)|+\sum [G:C_G(a)], \end{equation*}

where the summation is taken over one representative

a

from each conjugacy class not contained in

Z(G)

Given that

G

is a finite group. To prove that

|G|=|Z(G)|+\sum [G:C_G(a)]

, where the summation is over the noncentral conjugacy class representatives. For

a \in G

, we have

a \in Z(G)

if and only if

C_G(a)=G

. Therefore

\begin{equation*} a \in Z(G) \iff [G:C_G(a)]=1. \end{equation*}

By conjugacy class decomposition,

\begin{equation*} |G|=\sum [G:C_G(a)], \end{equation*}

where the summation is over one representative from each conjugacy class. Separate the representatives into central and noncentral representatives. For each central representative

z \in Z(G)

, the conjugacy class is

\{z\}

. Therefore all central conjugacy classes together contribute

|Z(G)|

elements. The remaining conjugacy classes contribute the terms

[G:C_G(a)]

for noncentral representatives

a

. Hence

\begin{equation*} |G|=|Z(G)|+\sum [G:C_G(a)]. \end{equation*}

\square

Let

G=S_3

. Its conjugacy classes are

\begin{equation*} \{e\},\quad \{(12),(13),(23)\},\quad \{(123),(132)\}. \end{equation*}

Since

Z(S_3)=\{e\}

, the class equation is

\begin{equation*} |S_3|=|Z(S_3)|+3+2. \end{equation*}

Therefore

\begin{equation*} 6=1+3+2. \end{equation*}

The class of transpositions has size

3

, and the class of three-cycles has size

2

Let

D_4=\langle r,s \mid r^4=e,\ s^2=e,\ srs=r^{-1}\rangle

be the symmetry group of a square. Its conjugacy classes are

\begin{equation*} \{e\},\quad \{r^2\},\quad \{r,r^3\},\quad \{s,r^2s\},\quad \{rs,r^3s\}. \end{equation*}

The center is

Z(D_4)=\{e,r^2\}

. Therefore the class equation is

\begin{equation*} 8=2+2+2+2. \end{equation*}

Here the term

2

for the center records two singleton conjugacy classes, while the other three terms record the three noncentral conjugacy classes.

This preview turns the class equation into a visible partition of the group. Choose

S_3

D_4

and compare the central singleton contribution with the noncentral conjugacy classes. The bars show how the separate class sizes add back to the group order. Use the labels to identify which classes are central and which are counted by centralizer indices. Try both examples and notice that the class equation is not a new operation; it is a careful way to count every element exactly once.

The central part is counted first because central elements form singleton conjugacy classes. Every other bar represents a noncentral conjugacy class whose size comes from a centralizer index. The visual partition reinforces that the class equation counts the whole group without overlap.

This calculator lets you test whether proposed class sizes add to a finite group order. Enter the group order, the center size, and a comma-separated list of noncentral conjugacy class sizes. The calculator checks the sum and warns if a noncentral class size is

1

, since size-one classes belong in the center. Use the presets for

S_3

and

D_4

before entering your own class equation.

A correct class equation must account for every element exactly once. The calculator checks only the arithmetic, so the mathematical work is still to justify each class size using conjugacy or centralizers.

Let

G

be a finite group with exactly two conjugacy classes. Prove that

|G|=2

Let

G

be a finite group with exactly two conjugacy classes. The identity element satisfies

Cl_G(e)=\{e\}

. Since

G

has exactly two conjugacy classes, there exists

a \in G

with

a \neq e

such that

\begin{equation*} G=Cl_G(e)\cup Cl_G(a). \end{equation*}

Let

|G|=n

. Then

|Cl_G(e)|=1

, so

|Cl_G(a)|=n-1

. By the conjugacy class formula,

\begin{equation*} n-1=[G:C_G(a)]. \end{equation*}

Since

[G:C_G(a)]

divides

|G|

, we have

n-1 \mid n

. Also

n-1 \mid n-1

. Therefore

n-1 \mid 1

. Hence

n-1=1

, and so

n=2

. Hence

|G|=2

\square

Prove that there is no finite nonidentity group in which every nonidentity element commutes with exactly half of the elements of the group.

Let

G

be a finite group with

|G|=n>1

. Suppose that every nonidentity element of

G

commutes with exactly half of the elements of

G

. For each

a \in G

with

a \neq e

, the hypothesis gives

\begin{equation*} |C_G(a)|=\frac{n}{2}. \end{equation*}

Therefore

\begin{equation*} |Cl_G(a)|=[G:C_G(a)]=2. \end{equation*}

Every nonidentity conjugacy class has size

2

, while

Cl_G(e)=\{e\}

has size

1

. Thus the class equation expresses

n

1

plus a sum of terms equal to

2

. Therefore

n

is odd. However,

|C_G(a)|=n/2

is the order of a subgroup for every nonidentity

a

. Hence

n/2

must be an integer, so

n

is even. This contradiction shows that no such finite nonidentity group exists.

\square

[1] Write the class equation of an abelian finite group

G

. [2] Let

G

be a finite group and suppose

Z(G)=G

. What are the conjugacy classes of

G

? [3] Let

G

be a group of order

p^2

, where

p

is prime. Explain why the class equation suggests that

Z(G)

is nontrivial.

[1] If

G

is abelian, then every conjugacy class has one element. Hence

|G|=|Z(G)|=|G|

. [2] Since

Z(G)=G

, every element commutes with every element. Therefore each conjugacy class is a singleton

\{a\}

. [3] In a finite

p

-group, every noncentral conjugacy class size is a power of

p

greater than

1

, hence divisible by

p

. The class equation then gives

|G|\equiv |Z(G)| \pmod p

. Since

p

divides

|G|

, it follows that

p

divides

|Z(G)|

, so

Z(G)

is nontrivial.

It converts conjugacy information into a numerical equation involving the center and centralizers.

a \in Z(G)

, then

gag^{-1}=a

for every

g \in G

, so

Cl_G(a)=\{a\}

The counting formula stated here is for finite groups, because it uses finite cardinalities and subgroup indices as numbers.

BMLabs Mathematics

BMLabs Mathematics

Class Equation and Conjugacy Class Decomposition

Class Equation and Conjugacy Class Decomposition

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