I am trying to prove that the number of irreducible representations of a finite group $G$ equals the number of conjugacy classes in it.
It seeems the action of the group on itself by conjugation $g(x)=gxg^{-1}$ is related to this. Let us call this the conjugation representation. It is not irreducible, so we consider its decomposition into irreducibles.
If $x$ is such that $gx=xg$ for every $g$, than this leads to a trivial representation. So the multiplicity of the trivial representation, $m_1$, equals the number of such $x$'s. This is the order of the center of the group, $|Z_G|$.
The character $\chi(g)$, in the conjugation representation, equals the order of the centralizer of $g$. This is $|G|/|C_g|$ where $C_g$ is the conjugacy class of $g$. Then the multiplicity of representation $R$ is given by $m_R=\frac{1}{|G|}\sum_{C}|C|\chi^R(C)\chi(g)=\sum_{C}\chi^R(C)$. When $R$ is the trivial representation, we get that $m_1$ equals the number of conjugacy classes.
Fine. We are almost there. It remains to prove that the order of the center of the group equals the number of irreducible representations. What is the simplest way of proving this?
PS: Wait a second, this cannot be. The center of the permutation group $S_n$ is trivial, $|Z_{S_n}|=1$, but the number of conjugacy classes/irreps is not 1. I am confused.