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.

  1. 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, $|Z_G|$.

  2. 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.

  • $\begingroup$ I am only interested in the simplest case when everything is done using complex numbers $\endgroup$
    – thedude
    Commented Nov 10, 2020 at 15:14
  • $\begingroup$ Are you familiar with Wedderburn's theorem? If $\pi_1,...,\pi_r$ are all the irreducible representations, and their dimensions are $n_1,...,n_r$ then $\mathbb{C}G\cong M_{n_1}(\mathbb{C})\times...\times M_{n_r}(\mathbb{C})$ as algebras. Then your theorem simply follows from computing the dimension of the center of these $2$ algebras and comparing the answers. $\endgroup$
    – Mark
    Commented Nov 10, 2020 at 15:29
  • $\begingroup$ @Mark I saw mention of this theorem, but it was over my head. Can you recommend an accessible account? $\endgroup$
    – thedude
    Commented Nov 10, 2020 at 15:34
  • $\begingroup$ You can try reading this article, it is about Wedderburn's theorem. I used it when I learned representation theory. Though I didn't read it from start to end, so can't say how good it really is. But it's the best I could find. sites.lafayette.edu/bloomjs/files/2015/07/Wedderburn-Theory.pdf $\endgroup$
    – Mark
    Commented Nov 10, 2020 at 15:40
  • $\begingroup$ I don't why you are trying to prove this yourself. It is a standard theorem which one normally learn when studying representation theory. (And you ought to say irreducible complex representations, because the result is not true over all fields.) $\endgroup$
    – Derek Holt
    Commented Nov 10, 2020 at 15:42

1 Answer 1


(1) Is wrong. You are failing to take into account there can be linear combinations of non-fixed-points which are fixed points of the conjugation action. Thus, $|Z(G)|$ is just a lower bound for the multiplicitiy of the trivial rep in the conjugation rep. Indeed, the trivial-rep isotypic component (i.e. sum of all the trivial subreps) of the conjugation rep admits a basis of the form $\{\sum_{g\in C}g\mid C\}$ as $C$ runs over the conjugacy classes of $G$. Since central elements form singleton conjugacy classes, these are the single-summand elements of this basis. But all you get is that the number of conjugacy classes equals itself.

You will basically have to learn Artin-Wedderburn if you want to understand this fact. You should already know there are algebra homomorphisms $\mathbb{C}[G]\to\mathrm{End}(V)$ for each irrep $V$. This means you can put them together into a homomorphism $\mathbb{C}[G]\to\bigoplus\mathrm{End}(V)$ (the sum is over irreps). This must have trivial kernel, since any element of the kernel would act as $0$ on any irrep, hence on any rep (by semisimplity), hence on the regular rep $\mathbb{C}G$ hence it times $1$ would be $0$ within $\mathbb{C}[G]$. To see that the map is also onto (and hence an algebra isomorphism), we can show the dimensions match.

To do this, let's find the multiplicity of any irrep $V$ within $\mathbb{C}G$. That equals the dimension of the hom space $\hom_G(\Bbb CG,V)$ (which you can show using semisimplicity and the distributivity of $\mathrm{hom}$). But any equivariant map $\mathbb{C}G\to V$ is determined by where $1$ is sent to, which is arbitrary, hence this hom-space is naturally isomorphic (as a vector space) to $V$ itself, so the dimension (hence the multiplicity) is $\dim V$ itself. As a result, we get $\dim\mathbb{C}[G]=\sum(\dim V)^2$.

On the one hand, the multiplicity of the trivial rep in the conjugaction action is the number of conjugacy classes. (More generally, if $\Omega$ is a $G$-set, the multiplicity of the trivial rep in the permutation rep $\mathbb{C}\Omega$ equals the number of orbits, by using a basis like the one I described for the conjugation rep. Can you show this? It is a nice exercise, and relevant to Burnside's counting lemma.) On the other hand, the trivial representation under the conjugation action corresponds to the center $Z(\mathbb{C}[G])$ of the group algebra, which has the same dimension as $\bigoplus Z(\mathrm{End}(V))$, which equals the number of irreps.


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