This is a standard result in representation theory of finite groups. A poof of the column orthogonality of the character table can be found here. By orthogonality of columns I mean that $$\sum_{V}\overline{\chi}_V(g)\chi_V(h)=\begin{cases} \vert G \vert/\vert c_g\vert& \text{if }g\sim h\\ 0& \text{otherwise. }\end{cases} $$ Here $G$ is a finite group, and $c_g$ is the conjugacy class of $g\in G$. Also, $V$ is supposed to be running in the irreducible representations of $G$.The prove mentioned before, although clever, it seems a bit artificial. In the sense that treating a character table as a matrix is not obvious at all. I think there must be a more pedestrian way to deduce this result from the row orthogonality $$\sum_g\overline\chi_V(g)\chi_W(g)=\begin{cases}\vert G\vert&\text{if }V\simeq W\\ 0& \text{otherwise.} \end{cases}$$ There must be an algebraic manipulation of equation that lead to this result. Is there one?

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    $\begingroup$ What is wrong with treating it as a matrix? Anyway, what is a rigorous definition for the character table itself? Answer: you define it as a matrix. If $\chi_1,...,\chi_r$ are the irreducible characters and $g_1,...,g_r$ are representatives for the conjugacy classes of $G$ then you simply define a matrix by $(A)_{ij}=\chi_i(g_j)$. This is absolutely well defined. So the character table is a matrix by definition. (if you want to define it formally) $\endgroup$
    – Mark
    Sep 1, 2020 at 21:45
  • $\begingroup$ @Mark Yes. It is well defined, but I mean that it is not obvious, right away, that we should construct a table, let alone a matrix. If all we knew was that characters of different irreducible representation are orthonormal; how could we come up with the other result? $\endgroup$ Sep 1, 2020 at 22:02
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    $\begingroup$ I don't know other proofs. I mean, multiplying matrices is exactly the "algebraic manipulation" that you ask about. When you multiply matrices, the corresponding equations give you the result. I think it is a pretty natural idea to construct a matrix when we work with finite orthogonal sequences. $\endgroup$
    – Mark
    Sep 1, 2020 at 22:10

1 Answer 1


Here is an alternative proof:

Lemma: The characters of irreducible representations form a basis for the space $\mathbb C_{class}(G)$ of class functions $f:G\to \mathbb C$.

Proof: We prove that whenever a class function $f$ is orthogonal to every character of irreducible representations, then it is the zero map. Let $\rho:G\to\text{GL}(V)$ be an irreducible representation of $G$. If $f\in \mathbb C_{class}(G)$, then the map $\phi=\sum_{g\in G}f(g)\rho(g^{-1})$ is a representation homomorphism. By Schur's Lemma $\phi=\lambda I$ for some $\lambda\in \mathbb{C}$. The trace of $\phi$ vanishes by assumption. Thus $\phi$ is the zero map in any irreducible representation and therefore in any representation of $G$. Setting the representation $\rho $ as the regular representation we have $\sum_{g\in G}f(g)\rho(g^{-1})=0$, and all group actions in the regular representation are linearly independent. Then $f=0$. $\blacksquare$

I sketch what is needed to prove your result, you can fill in the details.

  1. Fix $g\in G$, let $f(h)=1$ if $h\sim g$ and $f(h)=0$ otherwise.
  2. Write $f=\sum_{i}a_i\chi_i$, we consider the irreducible representations $V_i$ with characters $\chi_i$.
  3. Compute $a_i$ by means of the Hermitian product (use orthonormality).

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