# Symmetric Group on 3 Letters/Conjugacy Classes

## Conjugacy Classes of the Symmetric Group on 3 Letters

Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:

$\begin{array}{c|cccccc} \circ & e & (123) & (132) & (23) & (13) & (12) \\ \hline e & e & (123) & (132) & (23) & (13) & (12) \\ (123) & (123) & (132) & e & (13) & (12) & (23) \\ (132) & (132) & e & (123) & (12) & (23) & (13) \\ (23) & (23) & (12) & (13) & e & (132) & (123) \\ (13) & (13) & (23) & (12) & (123) & e & (132) \\ (12) & (12) & (13) & (23) & (132) & (123) & e \\ \end{array}$

The conjugacy classes of $S_3$ are:

 $\ds$  $\ds \set e$ $\ds$  $\ds \set {\tuple {123}, \tuple {132} }$ $\ds$  $\ds \set {\tuple {12}, \tuple {13}, \tuple {23} }$

## Proof

This is the group presentation of $D_3$:

$D_3 = \gen {a, b: a^3 = b^2 = e, a b = b a^{-1} }$

First we have that Conjugacy Class of Element of Center is Singleton.

From Center of Symmetric Group is Trivial, only $e$ is in the center.

Thus $\set e$ forms one of the conjugacy classes.

We then have:

 $\ds b a b^{-1}$ $=$ $\ds b^{-1} a b$ as $b^2 = e$ $\ds$ $=$ $\ds b^{-1} b a^{-1}$ as $a b = b a^{-1}$ $\ds$ $=$ $\ds a^{-1}$ as $a b = b a^{-1}$ $\ds$ $=$ $\ds a^2$ as $a^3 = e$

and so $a^2$ is in the same conjugacy class as $a$.

From Centralizer of Group Element is Subgroup, the centralizer $\map {C_{S_3} } a$ of $a$ is a subgroup of $S_3$.

Thus $\map {C_{S_3} } a$ also contains $a^2$.

Thus the order of $\map {C_{S_3} } a$ is either $3$ or $6$, from Lagrange's Theorem.

But we have that $b a \ne a b$ and so $b \notin \map {C_{S_3} } a$.

Thus $\map {C_{S_3} } a$ contains $3$ elements.

By the Orbit-Stabilizer Theorem $a$ has $2$ conjugates.

Hence the conjugacy class of $a$ is $\set {a, a^2}$.

By a similar argument, $\map {C_{S_3} } b = \gen b$ and so contains $2$ elements.

Thus $b$ has $3$ conjugates.

As the conjugacy classes form a partition, $\card {\map {C_{S_3} } b} = 3$.

Hence the result.

$\blacksquare$