For anyone familiar with Artin's Algebra book, I just worked through the proof of the following theorem, which can be seen here:
(5.9) Theorem Let $G$ be a group of order $N$, let $\rho_1,\rho_2,\dots$ represent the distinct isomorphism classes of irreducible representations of $G$ and let $\chi_i$ be the character of $\rho_i$.
- (a) Orthogonality relations: The characters $\chi_i$ are orthonormal. In other words $\langle\chi_i,\chi_j\rangle=0$ if $i\ne j$, and $\langle\chi_i,\chi_i\rangle=1$ for each $i$.
- (b) There are finitely many isomorphism classes of irreducible representations, the same number as the number of conjugacy classes in the group.
- (c) Let $d_i$ be the dimension of the irreducible representation $\rho_i$, let $r$ be the number of irreducible representations. Then $d_i$ divides $N$ and
$$N=d_1^2+\dots+d_r^2.$$This theorem will be proved in Section 9, with the exception of the assertion that $d_i$ divides $N$, which we will not prove.
The theorem was contained in the last section, but the proof for part (c) was missing completely. It is mentioned that the divisibility property would not be proved, but the sum of squares formula for $N$ is not verified at all.
In applications on homework this property is used extensively to fill in missing characters for character tables of finite groups, so I would like to understand why it is true.
Can anyone suggest a reference or sketch an argument? Thanks in abundance!
Best Answer
Does the chapter have the following theorem? If so, let $g=h=1$.$$ \sum_{\chi_i}\chi_i(g)\overline{\chi_i(h)}= \begin{cases} |C_G(g)| & \text{if } g\sim h \\ 0 & \text{if } g \not\sim h \end{cases} $$
If not, this is called the second orthogonality relation. The proof is actually fairly simple, and follows from noting that since characters are class functions, we can pick conjugacy class representatives $g_k$ (for each conjugacy class $k$ in the set of conjugacy classes $\mathcal{K}$ of $G$) and rewrite the definition of the inner product as $$\langle \chi_i,\chi_j \rangle = \frac{1}{|G|}\sum_{k\in\mathcal{K}}[G:C_G(g_k)]\chi_i(g_k)\overline{\chi_j(g_k)}.$$ You can find the rest of the argument in detail here.