TITLE: Shortest way of proving that the Galois conjugate of a character is still a character
QUESTION [14 upvotes]: Let $G$ be a finite group and $\chi$ a character of $G$. The values of $\chi$ generate an abelian Galois extension $K$ of $\mathbb{Q}$, and so one can consider the conjugate $\sigma(\chi)$ of $\chi$ by any element $\sigma$ of the Galois group. What's the shortest way to prove that $\sigma(\chi)$ is also a character of $G$?
This follows from the fact that the corresponding representation $V$ is realizable over a finite extension of $K$, but this fact is somewhat annoying to prove, and on an exam I don't want to prove it from scratch if I can avoid it. Are there any shorter arguments?
Motivation: I'm looking at a representation theory exam question (for practice) which asks me to fill out a character table. If I could assume the above statement, I could conclude that the missing characters are integer-valued.

REPLY [9 votes]: It's a bit late, but you could argue like this. By Schur's Lemma, when $\chi$ is an irreducible complex character of the finite group $G$, and $\sigma$ is a representation 
affording $\chi$, then for each $z \in Z(\mathbb{C}G$, we have $z\sigma = \lambda(z)I$
for some scalar $\lambda(z).$ Taking traces tells us that $\lambda(z) = \frac{\chi(z)}{\chi(1)}.$ Also, $\lambda$ defines an algebra homomorphism from $Z(\mathbb{C}G)$ to $\mathbb{C}.$ Since the dimension of $Z(\mathbb{C}G)$ is $k= k(G)$, the number of 
conjugacy classes of $G$, and since there are $k$ different irreducible characters of $G$
(I am assuming the orthogonality relations as given), which give (using the orthogonality relations) $k$ distinct algebra homomorphisms from $Z(\mathbb{C}G)$ to $\mathbb{C}$, we see that all such algebra homomorphisms come from irreducible characters.
We use the natural "integral" basis of $Z(\mathbb{C}G)$ (that is, the basis of conjugacy class sums) to show that for $\lambda$ as above, $\lambda^{\sigma}$ is also an algebra homomorphism for each automorphism $\sigma \in {\rm Gal}(\mathbb{Q}[\omega]/\mathbb{Q}$,
where $\omega$ is a primitive complex $|G|$-th root of unity. Notice that $\lambda(C)
\in \mathbb{Q}[\omega]$ for each class sum $C$. Furthermore, for class sums 
$C_{r}$ and $C_{s}$, there are integers $a_{rst}$ such that 
$\lambda(C_r C_s) = \sum_{t=1}^{k}a_{rst}C_t$. That $\lambda$ is an algebra homomorphism
is encapsulated precisely by the fact that we have $\lambda(C_r) \lambda(C_s) = \sum_{t=1}^{k}
a_{rst}\lambda(C_t)$. Since the $a_{rst}$ are rational integers, we can apply $\sigma$
to this equation to conclude that $C_r \to \lambda^{\sigma}(C_r)$ for each class sum,
(and extending by $\mathbb{C}$-linearity to $\mathbb{C}$-combinations of class sums)
is an algebra homomorphism from $Z(\mathbb{C}G)$ to $\mathbb{C}$. Associated to this is,
as discussed above, is a complex irreducible character $\mu$ of $G$ such that 
$\frac{\mu(C)}{\mu(1)} = \frac{\chi^{\sigma}(C)}{\chi(1)}$ for each class sum $C$.
Hence $\mu(g) = \mu(1)\frac{\chi^{\sigma}(g)}{\chi(1)}$ for each $g \in G.$
Since $\{ \chi(g): g \in G\}$ is $\sigma$-stable, we have $\sum_{g \in G}|\chi^{\sigma}(g)|^{2} = |G|$, so that $\chi(1) = \mu(1)$ and $\chi^{\sigma} = \mu.$