Using Pólya counting to find number of conjugacy classes of $S_3$

Question

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.

Answer 1

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,$\DeclareMathOperator{\Fix}{Fix}$ $$ \#\text{Orbits} = \dfrac{1}{|G|} \sum_{g \in G} |\Fix(g)|.$$

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 $\#\text{Orbits} = 3$.

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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.