Absolute values of complex irreducible characters of finite groups

Question

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?)

Answer 1

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.