# The average value of irreducible character of a non-trivial finite group

Let $G$ be a non-trivial finite group. Let $\chi$ be an irreducible character of the group $G$. Find $$\frac{1}{|G|}\sum_{g \in G} {\chi( g)}$$

I try. But I think that I am wrong. $$G=C_{i_1}\oplus C_{i_2}\oplus\ldots C_{i_k},$$ as $C_{i}$ is a cyclic group If $\chi$ is 1-dimensional character then $\sum_{g \in C_{m}} {\chi( g)}=0.$ Thus $$\frac{1}{|G|}\sum_{g \in G} {\chi( g)}= \frac{k}{|G|}.$$ Am I on the right path?

There are not answer or hints in our book

• You're not on the right path. I suggest 1) try some examples using whatever character tables you can find 2) think about the definition of the inner product of characters $\langle \chi, \phi\rangle$ – Matthew Towers Jan 28 '14 at 23:11
• "Thus [formula]." Where are you getting the formula? Also, whether or not we know $G$ is abelian (the gray block doesn't say even though you write a factor decomposition of $G$) is important, since it determines how tough the proof may be. – anon Jan 29 '14 at 8:52

Here is an extension of the comment by mt_:

Consider the definition of the inner product on the characters $\langle \chi,\psi\rangle = \frac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\psi(g)}$.

Can you get the sum you are looking as an inner product between suitable characters?

• Using character tables I assume that $\frac{1}{|G|}\sum_{g \in G} {\chi( g)}=0$ as affording irreducible representation is non-trivial and $\frac{1}{|G|}\sum_{g \in G} {\chi( g)}=1$ as affording irreducible representation is trivial. I want get the sum looking as an inner product. Сonsider representation representation $\alpha=\rho \otimes \bar{\sigma}$. we have $$\frac{1}{|G|}\sum_{g \in G} {\chi( g)}= \frac{1}{|G|}\sum_{g\in G}\chi_{\rho}(g)\overline{\chi_{\bar{\sigma}}(g)}=0, \text{as \rho \otimes \bar{\sigma} is n non-trivial}.$$ Does this make sense? – nadia-liza Jan 30 '14 at 8:18
• I don't quite follow what you mean. The sum you are looking at is simply the inner product of $\chi$ with the trivial character. – Tobias Kildetoft Jan 30 '14 at 8:55
• Oh, Yes! I at last understood! – nadia-liza Jan 30 '14 at 8:58

Apply the orthogonality relations.

Alternatively, consider an irreducible representation $\mathscr{X}$ affording $\chi$. Let $A = \sum_{g \in G} \mathscr{X}(g)$. Prove that either $A = 0$ or $\mathscr{X}$ is the trivial representation.

• Using character tables I assume that $\frac{1}{|G|}\sum_{g \in G} {\chi( g)}=0$ as affording irreducible representation is non-trivial and $\frac{1}{|G|}\sum_{g \in G} {\chi( g)}=1$ as affording irreducible representation is trivial. I want get the sum looking as an inner product. Сonsider representation representation $A=\rho \otimes \bar{\sigma}$. we have $$\frac{1}{|G|}\sum_{g \in G} {\chi( g)}= \frac{1}{|G|}\sum_{g\in G}\chi_{\rho}(g)\overline{\chi_{\bar{\sigma}}(g)}=0, \text{as \rho \otimes \bar{\sigma} is n non-trivial}.$$ Does this make sense? – nadia-liza Jan 30 '14 at 8:18