I had few questions on complex irreducible characters of finite groups which are mostly on their arithmetic nature. I will also mention here that I am considering only $\mathbb{C}$-irreducible characters of finite groups.

If $\chi$ is an irreducible $\mathbb{C}$-character of a finite group $G$, then one can see that $|\chi(g)|\leq |G|$ for any $g\in G$. My question is about opposite side of this fact. To avoid triviality, we do not consider zero character values.

Question 1. Is there lower bound on $\{|\chi(g)|:g\in G\}\setminus \{0\}$?

For second question, it is well known that character values are algebraic integers, and so are their absolute values (am I right?). But, absolute values are also real numbers. This forced me to consider the question:

Question 2. Consider those real numbers which are absolute values of irreducible $\mathbb{C}$-characters of finite groups. Is this set dense in $\mathbb{R}$?

The third question came because of the very basic property of characters.

Question 3. Given any algebraic integer, does there exists a finite group which takes this value for some irreducible character? (In other words, does any algebraic integer sits in character table of some finite group?)



  1. No. A way of seeing this is that all integers of cyclotomic fields occur as character values. Among them are numbers of the form $2-2\cos(\pi/n)$ for any positive integer $n$, and those become arbitrarily close to zero.
  2. Yes. All the algebraic integers of all cyclotomic fields occur (see the above answer). Those form a dense set (consider integer multiples of the numbers I used in part 1).
  3. No. Character values of finite groups are sums of roots of unity. Those reside inside abelian Galois extensions of $\Bbb{Q}$. This means that algebraic integers like $\root3\of2$ cannot occur as values of characters of a finite group.
  • $\begingroup$ @Jyrkii Lahtonen: Great answer ( I followed the link in the answer to your first part of the question). For third part I am not able to see why cube root of 2 is not in any cyclotomic extension. Being real this boils down to a statement on values of $f(\zeta +\bar\zeta)$ . I am not able to proceed. $\endgroup$ – P Vanchinathan Nov 10 '16 at 7:07
  • $\begingroup$ @PVanchinathan: If $z$ is an element of some cyclotomic field $K=\Bbb{Q}(\zeta_n)$, then $\operatorname{Gal}(K/\Bbb{Q})$ is abelian (the Galois group $G$ is isomorphic to $\Bbb{Z}_n^*$ actually). By Galois theory $\Bbb{Q}(z)$ is then the fixed field of some subgroup $H\le G$. Because $G$ is abelian we see that $H\unlhd G$. Therefore $\Bbb{Q}(z)/\Bbb{Q}$ is Galois, and its Galois group $G/H$ is abelian. This is the easy direction of Kronecker-Weber. $\endgroup$ – Jyrki Lahtonen Nov 10 '16 at 7:12
  • $\begingroup$ Anyway, $\Bbb{Q}(\root3\of2)/\Bbb{Q}$ is not Galois, so $z=\root3\of2$ is not an element of any cyclotomic field. $\endgroup$ – Jyrki Lahtonen Nov 10 '16 at 7:14
  • $\begingroup$ Embarrassingly simple. Thanks. $\endgroup$ – P Vanchinathan Nov 10 '16 at 7:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.