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:

The conjugacy classes of $S_3$ are:

Proof
We have that Symmetric Group on 3 Letters is Isomorphic to Dihedral Group D3.

This is the group presentation of $D_3$:

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:

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.