# Character table: sum of elements in rows

• Apr 10th 2011, 11:49 AM
bluepidgeon
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.
• Apr 10th 2011, 12:02 PM
Drexel28
Quote:

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 $\displaystyle \chi\in\text{irr}(G)$ then $\displaystyle \displaystyle \sum_{g\in G}\chi(g)\in\mathbb{N}\cup\{0\}$, right? But, think about it if we let $\displaystyle \chi^{\text{triv}}$ denote the trivial character $\displaystyle g\mapsto 1\in\mathbb{C}$ then by the orthogonality relations for irreducible characters we have that $\displaystyle \displaystyle \left\langle \chi,\chi^{\text{triv}}\right\rangle=\delta_{\chi, \chi^{\text{triv}}$. But, by definition $\displaystyle \displaystyle \left\langle\chi,\chi^{\text{triv}}\right\rangle=\ frac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\chi^{\text{triv}}(g)}=\frac{1} {|G|}\sum_{g\in G}\chi(g)$ so that $\displaystyle \displaystyle \sum_{g\in G}\chi(g)=\delta_{\chi,\chi^{\text{triv}}}|G|$, which is what we want, right?
• Apr 10th 2011, 03:10 PM
NonCommAlg
Quote:

Originally Posted by Drexel28
I'm a little bit confused, about the hint. What you're basically saying is that if $\displaystyle \chi\in\text{irr}(G)$ then $\displaystyle \displaystyle \sum_{g\in G}\chi(g)\in\mathbb{N}\cup\{0\}$, right? But, think about it if we let $\displaystyle \chi^{\text{triv}}$ denote the trivial character $\displaystyle g\mapsto 1\in\mathbb{C}$ then by the orthogonality relations for irreducible characters we have that $\displaystyle \displaystyle \left\langle \chi,\chi^{\text{triv}}\right\rangle=\delta_{\chi, \chi^{\text{triv}}$. But, by definition $\displaystyle \displaystyle \left\langle\chi,\chi^{\text{triv}}\right\rangle=\ frac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\chi^{\text{triv}}(g)}=\frac{1} {|G|}\sum_{g\in G}\chi(g)$ so that $\displaystyle \displaystyle \sum_{g\in G}\chi(g)=\delta_{\chi,\chi^{\text{triv}}}|G|$, which is what we want, right?

but $\displaystyle \sum_{g \in G} \chi(g)$ is not the sum of elements in the row of the character table corresponding to $\displaystyle \chi$ because your sum is over all elements of $\displaystyle G$ and not over representative elements of each conjugacy class. this problem can be resolved by looking at the character of permutation representation, acting on $\displaystyle G$ by conjugation, rather than the trivial character.