Conjugacy Classes of the Symmetric Group, S3

# Conjugacy Classes of the Symmetric Group, S3

Let $S_3 = \{ (1), (12), (23), (13), (123), (132) \}$ be the symmetric group on the $3$-element set $\{ 1, 2, 3 \}$. We now examine the conjugacy classes of $S_3$.

First, let's look at $Z(S_3)$. Note that $(12)(13) = (132) \neq (123) = (13)(12)$ and $(12)(23) = (123) \neq (132) = (23)(12)$, and so $(12), (23), (13) \not \in Z(S_3)$. Furthermore, $(123)(12) = (13) \neq (23) = (12)(123)$ and $(132)(12) = (23) \neq (13) = (12)(132)$. So $(123), (132) \not \in Z(S_3)$. So the only trivial conjugacy class is $[(1)] = \{ 1 \}$.

Now observe that for the element $(12)$ we have that:

(1)
\begin{align} \quad (12)(12)(12)^{-1} &= (12)(12)(21) = (12) \\ \quad (13)(12)(13)^{-1} &= (13)(12)(31) = (23) \\ \quad (23)(12)(23)^{-1} &= (23)(12)(32) = (13) \\ \quad (123)(12)(123)^{-1} &= (123)(12)(321) = (23) \\ \quad (132)(12)(132)^{-1} &= (132)(12)(231) = (13) \end{align}

Therefore the conjugacy class of $(12)$ is $[(12)] = \{ (12), (13), (23) \}$. The remaining elements in $S_3$ are $(123)$ and $(132)$. Since neither of these elements have trivial conjugacy classes, it must be that $[(123)] = \{ (123), (132) \}$.

We can partition $S_3$ into its conjugacy classes as:

(2)
\begin{align} \quad S_3 = [(1)] \cup [(12)] \cup [(123)] = \{ (1) \} \cup \{ (12), (13), (23) \} \cup \{ (123), (132) \} \end{align}

And The Class Equation for $S_3$ is:

(3)
\begin{align} \quad S_3 = 1 + (2 + 3) \end{align}