Let $G$ be a finite group, for each irreducible character $\chi$, we define ${\bf Z}(\chi)$ to be the set of all $x\in G$ such that $|\chi(x)|=\chi(e)$ when $e$ is the identity of the group. For every irreducible charcter we know that

$\frac{|G|}{\chi(e)^2} \leq \frac{1}{\chi(e)} \sum_{x\in G} |\chi(x)| \leq \frac{|G|}{\chi(e)}$.

Now suppose that ${\bf Z}(\chi)$ is trivial i.e. it includes only $e$. Can we find a better bound for $\frac{1}{\chi(e)} \sum_{x\in G} |\chi(x)|$?

My firs clue about this question was in observance of affine $p$-groups: they satisfy this condition: (${\bf Z}(\chi)$ is trivial for only non-linear character $\chi_\pi$ of affine $p$-groups). On the other hand, for this non-linear character,

$\frac{1}{\chi_\pi(e)} \sum_{x\in G} |\chi_\pi(x)| =2$

for every $p$.

Subsequently, I tried some more character tables for finite groups that some of their characters, say $\chi$, satisify this condition and I observed that $\frac{1}{\chi(e)} \sum_{x\in G} |\chi(x)|$ is so close to $\frac{|G|}{\chi(e)^2}$ than $\frac{|G|}{\chi(e)}$.


The rightmost inequality can be improved for all non-linear irreducible characters, and if $Z(\chi) = 1$ and $G$ is non-trivial, then $\chi$ must certainly be non-linear. Using Cauchy-Schwarz, we have $$\left(\sum_{x \in G} |\chi(x)|\right)^{2} \leq m(\chi) \left(\sum_{x \in G} |\chi(x)|^{2}\right),$$ where $m(\chi)$ is the number of group elements where $\chi$ does not vanish (so certainly $m(\chi) \leq |G|$, and the inequality is strict when $G$ is non-linear, since by a Theorem of Burnside, each non-linear irreducible character vanishes somewhere). Hence $\sum_{x \in G}|\chi(x)| \leq \sqrt{m(\chi)|G|}$, which removes (at least) a factor of $\chi(1)$ from your inequality.

When $Z(\chi) = 1$, the leftmost inequality becomes strict if $G$ is non -trivial. Added later: it might be worth pointing out that Burnside's argument also implictly shows that $\sum_{x \in G}|\chi(x)| > m(\chi)$ when $\chi$ is a non-linear irreducible character (just notice that if $\chi(x) \neq 0$, then $1 \leq \prod_{y} |\chi(y)|$, where $y$ runs through all the generators of $\langle x \rangle$, and use the arithmetic-geometric mean inequality).

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  • $\begingroup$ Dear Geoff Robinson, First of all, the first inequality sounds a bit strange for me; but I get the last one which says $\|\chi\|_1\leq \sqrt{m(\chi) |G|}$. I have seen that theorem which says that a non-linear character will vanish at least on one conjugacy class. But have you seen ever a theorem that somehow gives an estimation about the numer of $m(\chi)$ or numer of conjugacy classes on them $\chi$ is zero. $\endgroup$ – Mahmood Alaghmandan Jul 20 '11 at 8:56
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    $\begingroup$ @Mahmood: Sorry, yes, you were right (now corrected). To keep the important bit of my deleted comments, John Thompson proved (see Isaacs book) that if $\chi$ is an irreducible character of a finite group $G$, then on at least one third of the group elements, the values taklen by $\chi$ are 0 or roots of unity. $\endgroup$ – Geoff Robinson Jul 20 '11 at 10:29

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