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I'm solving the exercises of chapter 14 in the book Representations and Characters of groups. (Gordon James, Martin Liebeck) Always working with $\mathbb{R}$ or $\mathbb{C}$.

One of them says:

Suppose that $\chi$ is a non-zero, non trivial character of $G$, and that $\chi(g)$ is a non-negative real number for all $g$ in $G$. Prove that $\chi$ is reducible

At this stage of the book, I think that the natural procedure is to calculate $<\chi,\chi>$ and see that it is $\neq 1$

I know that non-trivial means $\exists g \in G$ such that $\chi(g)\neq 1$, but since $\chi(1_G)\in \mathbb{N}$, I don't know what they want to say with the non-zero condition.

Anyway, I've doing the next:

By definition, $<\chi,\chi>=\displaystyle\dfrac{1}{|G|}\sum_{g\in G}\chi(g)\chi(g^{-1})$

I know that there exist a base $\mathcal{B}$ where $[g]_{\mathcal{B}}$ is diagonal, then $\chi(g)$ is sum of $m$th roots of the unity (considering the $order(g)=m$) so $\chi(g^{-1})=\overline{\chi(g)}$

But $\chi(g)$ is a non-negative real number for all $g$ in $G$.

So we obtain $<\chi,\chi>=\displaystyle\dfrac{1}{|G|}\sum_{g\in G}\chi(g)^2$

Since $\chi$ is non-trivial, $\exists g \in G$ such that $\chi(g)\neq 1$, but I'm stuck here. Maybe I'm missing something related to the non-zero condition.

Any advice would be welcome.

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    $\begingroup$ Consider the inner product with the trivial character. $\endgroup$ – Qiaochu Yuan May 21 '18 at 1:37
  • $\begingroup$ @QiaochuYuan Thanks for your hint. $<\chi,\chi_1>=\dfrac{1}{|G|}\sum_{g\in G}\chi(g)\chi_1(g)=\dfrac{1}{|G|}\sum_{g\in G}\chi(g)$ And since $\chi(1_G)\geq 1$ and $\chi(g)\geq 0\ \forall g \in G$, then we have $<\chi,\chi_1> >0$. This means that the trivial character (which is irreducible) is a component of our character $\chi$. But $\chi$ is not trivial, so it must have something else as component. In conclusion, $\chi$ is reducible. $\endgroup$ – G.Jimenez May 21 '18 at 3:33

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