Suppose that it is known that a finite group $G$ has equally many conjugacy classes as it has irreducible representations $\rho_1,\dots,\rho_k$. How can we then show that $\sum_{\rho}\chi_\rho(g)\overline{\chi_\rho(f)} = \frac{|G|}{|C(g)|}$ if $f = g$ and $\sum_{\rho}\chi_\rho(g)\overline{\chi_\rho(g)} = 0$ if $f \neq g$ for $f, g \in G$? We do also know that the irreducible characters $\chi_{\rho_1},\dots,\chi_{\rho_k}$ form an orthonormal basis for the conjugacy class functions $\mathbb{C}_{\text{class}, G}$ of $G$. Therefore, $\sum_{g \in G}\chi_{\rho_i}(g)\overline{\chi_{\rho_j}(g)} = \delta^i_j$. But then how to extend this to the equality $\sum_{\rho}\chi_\rho(g)\overline{\chi_\rho(f)} = \frac{|G|}{|C(g)|}$?
Edit: Since $G$ is finite it must have a finite number of conjugacy classes and thus a finite number of irreducible characters. Hence by orthonormality of the characters (and reordering of finite sums), we know that $\sum_{g\in G}\sum_{\rho}\overline{\chi_{\rho}(g)}\chi_\rho(g) = \sum_{\rho}\sum_{g\in G}\overline{\chi_{\rho}(g)}\chi_\rho(g) = \#\text{Conjugacy classes of } G$. We know from algebra, that the order of the conjugacy class of a group must divide the order of the group.
Currently, I'm stuck at the proof, as I don't know what is the final straw to put these pieces together to show the claimed result.
Best Answer
What you are asking is just the Schur orthogonality relations. But there are some typos in your statements. The exposition below fixes the typos. Let $\chi_{\rho_1},\cdots,\chi_{\rho_k}$ be the irreducible characters (over $\mathbb C$) and $C(g_1),\cdots,C(g_k)$ be the conjugacy classes of $G$, so $$\bigcup_{\ell=1}^k C(g_{\ell})=G,\qquad (1)$$ where $g_1,\cdots,g_k\in G$ are some chosen representatives. Writing the centralizer of $g$ in $G$ by $C_G(g),$ one then has $$|G|=|C_G(g)|\cdot |C(g)|.$$ Since $\chi_{\rho_i}$’s form an orthonormal basis of the space of class functions, one has $$\frac 1{|G|}\sum_{g\in G}\chi_{\rho_i}(g)\overline{\chi_{\rho_j}(g)}=\delta_{ij},$$ which together with (1) implies that $$\frac 1{|G|}\sum_{\ell=1}^k\sum_{g\in C(g_{\ell})}\chi_{\rho_i}(g)\overline{\chi_{\rho_j}(g)}=\sum_{\ell=1}^k\frac{|C(g_{\ell}|}{|G|}\chi_{\rho_i}(g_{\ell})\overline{\chi_{\rho_j}(g_{\ell})}=\delta_{ij}.\qquad (2)$$ The relation in (2) shows that the $k\times k$ matrix $U=(a_{ij})$ defined by $$a_{ij}=\sqrt{\frac{|C(g_j)|}{|G|}}\chi_{\rho_i}(g_j)$$ is unitary. It follows that the column vectors of $U$ are orthonormal, i.e. $$\frac{\sqrt{|C(g_{\ell})|\cdot|C(g_m)|}}{|G|}\sum_{i=1}^k \chi_{\rho_i}(g_{\ell})\overline{\chi_{\rho_i}(g_m)}=\delta_{\ell m},$$ hence for $f,g\in G,$ one has if $f\not\sim g$ (i.e. not conjugate), then $$\sum_{i=1}^k \chi_{\rho_i}(f)\overline{\chi_{\rho_i}(g)}=0,$$ and if $f\sim g,$ then $$\sum_{i=1}^k \chi_{\rho_i}(f)\overline{\chi_{\rho_i}(g)}=\frac {|G|}{|C(g)|}=|C_G(g)|,$$ as required.