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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?)

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Answers:

  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.
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Jyrki Lahtonen
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  • @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. – P Vanchinathan Nov 10 '16 at 07:07
  • @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. – Jyrki Lahtonen Nov 10 '16 at 07:12
  • Anyway, $\Bbb{Q}(\root3\of2)/\Bbb{Q}$ is not Galois, so $z=\root3\of2$ is not an element of any cyclotomic field. – Jyrki Lahtonen Nov 10 '16 at 07:14
  • Embarrassingly simple. Thanks. – P Vanchinathan Nov 10 '16 at 07:15