According to Wikipedia: If G is a finite group and K is the complex number field, the regular representation is a direct sum of irreducible representations, in number at least the number of conjugacy classes of G.

Can anyone prove this? (Only an honors level student, so please try to keep it as simple as possible)

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    $\begingroup$ I am pretty sure that someone can prove this. Have you looked up the standard representation theory literature yet? I would recommend Serres Linear Representations of Finite Groups. As far as I remember, this is proved in the first part of the book. $\endgroup$ Nov 8, 2009 at 22:28
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    $\begingroup$ To echo some of the other commenters: this can be found in many textbooks that were designed for use with honours-level courses. There are introductory books by Ledermann, and James and Liebeck, and no doubt others, which I'd hope you could find in your university library. $\endgroup$
    – Yemon Choi
    Nov 9, 2009 at 0:03
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    $\begingroup$ Remember, MathOverflow is not an encyclopedia. mathoverflow.net/faq#whatnot Also, this is in the Wikipedia article on representations of finite groups en.wikipedia.org/wiki/Representation_of_a_finite_group $\endgroup$
    – Ben Webster
    Nov 9, 2009 at 0:43
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    $\begingroup$ C'mon, let's be easier on the guy. His question is less than perfect, but I get to post a not-so-well-known but readable reference. $\endgroup$ Nov 9, 2009 at 22:29
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    $\begingroup$ And it's not about MO being an encyclopedia -- on the contrary, there are lots of encyclopedic treatments of the representation theory but no half-page introductions known to me. $\endgroup$ Nov 9, 2009 at 22:32

2 Answers 2


If I remember correctly, that statement can be proven via the equivalence between the (group) representations of $G$ and the (algebra) representations of the group algebra $K[G]$. If $K=\mathbb{C}$ is the field of complex numbers (or any other field of characteristic 0, or in general if $\textrm{char} K$ does not divide the order of $G$) then by Maschke's Theorem $K[G]$ is semisimple, and moreover it is Artinian (by finite-dimensionality) so by Artin-Wedderburn it decomposes (as a module over itself) as a direct sum of matrix rings over some division ring over $K$. If $K=\mathbb{C}$, there are no finite dimensional division rings over $\mathbb{C}$ so each summand is a ring of matrices over the complex numbers. The irreducible representations of the group are in one to one correspondence with left ideals of this ring, which are well known (look for instance here. A simple dimension counting argument should convince you that you have at least one ideal for each conjugacy class.

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    $\begingroup$ Once you establish that the group algebra $\mathbb{C}[G]$ is a direct sum of matrix blocks, the irreducible representations are in 1-to-1 correspondence with the blocks. Then the easiest way to answer the counting question is to observe that both the conjugacy classes and the matrix blocks give you a basis of the center $Z(\mathbb{C}[G])$. Two bases of a vector space must have the same cardinality. $\endgroup$ Nov 9, 2009 at 15:49

It's not entirely trivial, but here's the sketch of the proof if you would like to finish it yourself. Note that people usually denote complex numbers by $\mathbb C$:

(1) Every irreducible representation comes as part of regular representation, denoted $\mathbb C[G]$

To prove this, take any non-trivial representation $R$ and consider $\mathrm{Hom}\,(R, \mathbb C[G])$. A careful examination should show it's non-trivial as well.

(2) The character $\chi_R$ is a function on $G$ defined as $\chi _ R(g) = \mathrm{tr}\,g| _ R$. Prove that value of any character on an element of $G$ depends only on element's conjugacy class.

This immediately follows from $\mathrm{tr}\,A = \mathrm{tr}\,BAB^{-1}$ (property of trace).

(3) Prove that any function on $G$ that depends only on a conjugacy class is a linear combination of some characters $\chi_R$.

I'm not sure I remember how to prove this in an elementary way, but it's doable.

(4) Prove your statement from (1)-(3).

Indeed, there must be at least as many irreducible representations as there are conjugacy classes and all of them appear in regular representation by (1), so you have your statement.

There are many references on representation theory, you can basically pick up any book to answer your question. I would like to point out the following very accessible text if you are be interested in learning more:

  • Introduction to representation theory, write-up of lectures by Pavel Etingof to high-school students and MIT undergraduates. They cover more material, finite groups are in Chapter 3.  

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