# Thread: Character table: sum of elements in rows

1. ## Character table: sum of elements in rows

Prove that the sum of elements in any row of the character table of a finite group G is a non-negative integer.

This can be proven by considering the character of the permutation representation acquired by letting G act on itself by conjugation.

I don't see the significance of the permutation character here. Any hints would be appreciated.

2. Originally Posted by bluepidgeon
Prove that the sum of elements in any row of the character table of a finite group G is a non-negative integer.

This can be proven by considering the character of the permutation representation acquired by letting G act on itself by conjugation.

I don't see the significance of the permutation character here. Any hints would be appreciated.
I'm a little bit confused, about the hint. What you're basically saying is that if $\chi\in\text{irr}(G)$ then $\displaystyle \sum_{g\in G}\chi(g)\in\mathbb{N}\cup\{0\}$, right? But, think about it if we let $\chi^{\text{triv}}$ denote the trivial character $g\mapsto 1\in\mathbb{C}$ then by the orthogonality relations for irreducible characters we have that $\displaystyle \left\langle \chi,\chi^{\text{triv}}\right\rangle=\delta_{\chi, \chi^{\text{triv}}$. But, by definition $\displaystyle \left\langle\chi,\chi^{\text{triv}}\right\rangle=\ frac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\chi^{\text{triv}}(g)}=\frac{1} {|G|}\sum_{g\in G}\chi(g)$ so that $\displaystyle \sum_{g\in G}\chi(g)=\delta_{\chi,\chi^{\text{triv}}}|G|$, which is what we want, right?

3. Originally Posted by Drexel28
I'm a little bit confused, about the hint. What you're basically saying is that if $\chi\in\text{irr}(G)$ then $\displaystyle \sum_{g\in G}\chi(g)\in\mathbb{N}\cup\{0\}$, right? But, think about it if we let $\chi^{\text{triv}}$ denote the trivial character $g\mapsto 1\in\mathbb{C}$ then by the orthogonality relations for irreducible characters we have that $\displaystyle \left\langle \chi,\chi^{\text{triv}}\right\rangle=\delta_{\chi, \chi^{\text{triv}}$. But, by definition $\displaystyle \left\langle\chi,\chi^{\text{triv}}\right\rangle=\ frac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\chi^{\text{triv}}(g)}=\frac{1} {|G|}\sum_{g\in G}\chi(g)$ so that $\displaystyle \sum_{g\in G}\chi(g)=\delta_{\chi,\chi^{\text{triv}}}|G|$, which is what we want, right?
but $\sum_{g \in G} \chi(g)$ is not the sum of elements in the row of the character table corresponding to $\chi$ because your sum is over all elements of $G$ and not over representative elements of each conjugacy class. this problem can be resolved by looking at the character of permutation representation, acting on $G$ by conjugation, rather than the trivial character.