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?
2026-03-29 17:25:32.1774805132
How to prove column orthogonality of character table
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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.