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
We have that Symmetric Group on 3 Letters is Isomorphic to Dihedral Group D3.
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.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 6.5$. Orbits: Example $119$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.17$