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When can one conclude that a character table has non-real entries?

In particular, by constructing the character table for $\mathbb{Z}/3\mathbb{Z}$ or $A_4$ how does one determine that some of the entries will be nonreal? Is the reason that the same complex values in the table for $\mathbb{Z}/3\mathbb{Z}$ also appear in the table for $A_4$ because $A_4 / ({1,(12)(34),(13)(24),(14)(23)})\cong \mathbb{Z}/3\mathbb{Z}$?

Here's what I have for $\mathbb{Z}/3\mathbb{Z}$.

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Using short orthogonality relations I conclude that $$1+a^2+b^2=1+c^2+d^2+=1+a^2+c^2=1+b^2+d^2=3,$$ and $$1+a+b=1+c+d=1+ac+bd=0,$$ I don't see how to conclude from this that $a=d=\omega$ and $b=c=\bar{\omega}=\omega^2$, where $\omega=e^{2\pi i/3}.$

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    $\begingroup$ The orthogonality relations should be $1+|a|^2+|b|^2=3$ (with modulus !). $\endgroup$ – Watson Jun 7 '16 at 9:26
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A question from the Cambridge Part II Maths course on Representation Theory:

Let $x$ be an element of order $n$ in finite group $G$. Say, without detailed proof, why:

  1. if $\chi$ is a character of $G$, then $\chi(x)$ is a sum of $n$th roots of unity;
  2. $\tau(x)$ is real for every character $\tau$ iff $x$ is conjugate to $x^{-1}$;
  3. $x$ and $x^{-1}$ have the same number of conjugates in $G$.

Prove that the number of irreducible characters which take only real values is equal to the number of self-inverse conjugacy classes.

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Characters of degree 1 are homomorphisms from the group to the multiplicative group of nonzero complex numbers. It follows that if, say, $g$ is a group element of order 3, then $(\chi(g))^3=1$, so $\chi(g)$ is 1, $\omega=e^{2\pi i/3}$, or $\omega^2$. Once you have found all the linear characters with $\chi(g)=1$, if you have any more characters they must have $\chi(g)=\omega$ or $\chi(g)=\omega^2$.

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If an element of the group and its inverse are in the same conjugacy of of a group then the value of any character at that element will be a real number.

In the symmetric groups $S_n$ any element and its inverse have the same cycle decomposition structure and hence they are conjugates.

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  • $\begingroup$ Is the converse of your first statement also true? $\endgroup$ – The Substitute Jun 7 '16 at 6:38

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