Let $\chi(g)$ is a nonnegative real number. Show that if $\chi$ is irreducible, then $\chi$ is the trivial character

Question

I am working on a problem from my abstract algebra class:

Let $\chi$ be a character of a group $G$ with the property that $\chi(g)$ is a nonnegative real number for all $g \in G$. Use the row orthogonality relations to show that if $\chi$ is irreducible, then $\chi$ must be the trivial character.

My proof so far:

Row orthogonality tells us $$\sum_{i=1}^k \frac{\chi_r\left(g_i\right) \overline{\chi_s\left(g_i\right)}}{\left|C_G\left(g_i\right)\right|} = \delta_{r s} .$$ If $\chi$ is irreducible, then $\delta_{rs} = 1$, so \begin{align*} \sum_{i=1}^k \frac{\chi_r\left(g_i\right) \overline{\chi_s\left(g_i\right)}}{\left|C_G\left(g_i\right)\right|} &= 1 \\ \sum_{i=1}^k \chi_r\left(g_i\right) \overline{\chi_s\left(g_i\right)} &= \sum_{i=1}^k \left|C_G\left(g_i\right)\right| \end{align*} so for each $i = \{1, \dots, k\}$ we have that \begin{align*} \chi_r\left(g_i\right) \overline{\chi_s\left(g_i\right)} &= \left|C_G\left(g_i\right)\right|. \end{align*}

I think I am heading in the right direction but I can't seem to pull out that $\chi$ is trivial.

Answer 1

Edit: I'll rewrite this in a much simpler way - basically what MarkH wrote in the comment. What I wrote last night was extremely convoluted.

Since $\chi$ is irreducible, $\chi$ is one of your $\chi_r$'s.

Let $\chi_{r_{\text{trivial}}}$ be the trivial character.

There are two possibilities:

• Possibility (a): $\chi$ is the trivial character. In this case, the orthogonality relations tells us that $\sum_{i=1}^k \frac{\chi \left(g_i\right) \overline{\chi_{r_{\text{trivial}}}\left(g_i\right)}}{\left|C_G\left(g_i\right)\right|} = 1 .$

• Possibility (b): $\chi$ is some irreducible character, different from the trivial character. In this case, the orthogonality relations say that $\sum_{i=1}^k \frac{\chi \left(g_i\right) \overline{\chi_{r_{\text{trivial}}}\left(g_i\right)}}{\left|C_G\left(g_i\right)\right|} = 0 .$

We need to rule out possibility (b).

To rule out possibility (b), we can use the following facts:

• $\chi_{r_\text{trivial}}(g_i) = 1$ for all $g_i$.
• $\chi (g_i) \geq 0$ for all $g_i$ (by assumption)
• $\chi (g_i) > 0$ for the $g_i$ corresponding to the identity element.

These three facts imply that it is impossible for the equation $\sum_{i=1}^k \frac{\chi \left(g_i\right) \overline{\chi_{r_{\text{trivial}}}\left(g_i\right)}}{\left|C_G\left(g_i\right)\right|} = 0 $ to hold.