# The Order of a Conjugacy Class Divides the Order of the Group

## Problem 455

Let $G$ be a finite group.

The **centralizer** of an element $a$ of $G$ is defined to be

\[C_G(a)=\{g\in G \mid ga=ag\}.\]

A **conjugacy class** is a set of the form

\[\Cl(a)=\{bab^{-1} \mid b\in G\}\]
for some $a\in G$.

**(a)**Prove that the centralizer of an element of $a$ in $G$ is a subgroup of the group $G$.

**(b)** Prove that the order (the number of elements) of every conjugacy class in $G$ divides the order of the group $G$.

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## Proof.

### (a) Prove that the centralizer of $a$ in $G$ is a subgroup of $G$.

Since the identity element $e$ of $G$ satisfies $ea=a=ae$, it is in the centralizer $C_G(a)$.

Hence $C_G(a)$ is not an empty set. We show that $C_G(a)$ is closed under multiplications and inverses.

Let $g, h \in C_G(a)$. Then we have

\begin{align*}

(gh)a&=g(ha)\\

&=g(ah) && \text{since $h\in C_G(a)$}\\

&=(ga)h\\

&=(ag)h&& \text{since $g\in C_G(a)$}\\

&=a(gh).

\end{align*}

So $gh$ commutes with $a$ and thus $gh \in C_G(a)$.

Thus $C_G(a)$ is closed under multiplications.

Let $g\in C_G(a)$. This means that we have $ga=ag$.

Multiplying by $g^{-1}$ on the left and on the right, we obtain

\begin{align*}

g^{-1}(ga)g^{-1}=g^{-1}(ag)g^{-1},

\end{align*}

and thus we have

\[ag^{-1}=g^{-1}a.\]
This implies that $g^{-1}\in C_G(a)$, hence $C_G(a)$ is closed under inverses.

Therefore, $C_G(a)$ is a subgroup of $G$.

### (b) Prove that the order of every conjugacy class in $G$ divides the order of $G$.

We give two proofs for part (b).The first one is a more direct proof and the second one uses the orbit-stabilizer theorem.

#### The First Proof of (b).

By part (a), the centralizer $C_G(a)$ is a subgroup of the finite group $G$.

Hence the set of left cosets $G/C_G(a)$ is a finite set, and its order divides the order of $G$ by Lagrange’s theorem.

We prove that there is a bijective map from $G/C_G(a)$ to $\Cl(a)$.

Define the map $\phi:G/C_G(a) \to \Cl(a)$ by

\[\phi\left(\, gC_G(a) \,\right)=gag^{-1}.\]

We must show that it is well-defined.

For this, note that we have

\begin{align*}

gC_G(a)=hC_G(a) &\Leftrightarrow h^{-1}g\in C_G(a)\\

& \Leftrightarrow (h^{-1}g)a(h^{-1}g)^{-1}=a\\

& \Leftrightarrow gag^{-1}=hag^{-1}.

\end{align*}

This computation shows that the map $\phi$ is well-defined as well as $\phi$ is injective.

Since the both sets are finite sets, this implies that $\phi$ is bijective.

Thus, the order of the two sets is equal.

It yields that the order of $C_G(a)$ divides the order of the finite group $G$.

#### The Second Proof of (b). Use the Orbit-Stabilizer Theorem

We now move on to the alternative proof.

Consider the action of the group $G$ on itself by conjugation:

\[\psi:G\times G \to G, \quad (g,h)\mapsto g\cdot h=ghg^{-1}.\]

Then the orbit $\calO(a)$ of an element $a\in G$ under this action is

\[\calO(a)=\{ g\cdot a \mid g\in G\}=\{gag^{-1} \mid g\in G\}=\Cl(a).\]

Let $G_a$ be the stabilizer of $a$.

Then the **orbit-stabilizer theorem** for finite groups say that we have

\begin{align*}

|\Cl(a)|=|\calO(a)|=[G:G_a]=\frac{|G|}{|G_a|}

\end{align*}

and hence the order of $\Cl(a)$ divides the order of $G$.

Note that the stabilizer $G_a$ of $a$ is the centralizer $C_G(a)$ of $a$ since

\[G_a=\{g \in G \mid g\cdot a =a\}=\{g\in G \mid ga=ag\}=C_G(a).\]

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