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So I know that $S_3$ has three conjugacy classes. However, I was reading about Pólya counting today, and am wondering how Pólya counting could be used to derive the number of conjugacy classes for $S_3$. Thanks.

The Pólya enumeration theorem tells us that when a group $G$ acts on a set $X$, the number of orbits of $G$ on $X$ equals the average number of fixed points of $G$. That is,$DeclareMathOperatorFixFix$
$$ #textOrbits = dfrac1 sum_g in G |Fix(g)|.$$

Now when $G$ acts on $G$ itself via conjugation, the orbits are conjugation classes and the set of fixed points of an element $g$ is
$$Fix(g) = , h in G mid ghg^-1 = h, = C_G(g),$$
the centraliser of $g$ — i.e., the set of all elements that commute with $g$.

In the case of $G = S_3$, we know that there are two non-trivial rotations (or cyclic permutations) that commute with all three rotations (including the identity); there are three reflections (or transpositions) each of which commutes only with itself and the identity; and the identity element (as in any group) commutes with all elements (six in this case). Thus, $sum |Fix(g)| = 2 times 3 + 3 times 2 + 1 times 6 = 18$, and $|G| = 6$, so $#textOrbits = 3$.

As a side note, a completely different (and here, shorter) way to count the conjugacy classes in $S_n$ is by using the result that any two permutations in $S_n$ [important!] are mutually conjugate if and only if they have the same cycle structure. That immediately tells us that the identity (as always) is in a conjugacy class by itself; the two cyclic permutations are in one conjugacy class; and the three transpositions are in one conjugacy class. So there are three classes.