TITLE: Irreducible characters of the direct product of two groups
QUESTION [8 upvotes]: I am studying representation and character theory because of my field of research. So, my question is not a homework. 
I want to solve a problem of the book "character theory of finite groups" by M. Isaacs. The problem is as follows:

I know that if $A$ is an abelian group, then each character of $A$ in the field of complex numbers is linear. Also I know that if $G=H \times A$ then each irreducible character of $G$ can be written as $\phi \lambda$, where $\phi \in Irr(H)$ and $\lambda \in Irr(A)$. Using these facts, I tried to solve this problem but I could not find any useful ideas.
I will be so grateful for your helpful answers and comments.

REPLY [3 votes]: Okay here is an outline:
First reduce it to the case where $A$ is trivial, this can be done using the facts you mentioned.
Next note that $\chi$ is valued in $\mathbb{Q}[\zeta]$ where $\zeta$ is a primitive $|H|$th root of unity, and $\chi^{(n)}$ is just the image of $\chi$ under the field automorphism defined by $\zeta \to \zeta^n$ (which exists as $(|H|,n)=1$).
Then note that if $\chi$ is irreducible then $\langle\chi,\chi\rangle = \langle\chi^{(n)},\chi^{(n)}\rangle = 1$ since it is invariant under this field automorphism.  
Now show that $\chi^{(n)}$ is in the $\mathbb{Z}$-span of the irreducible characters, as opposed to just the $\mathbb{C}$-span which we know a priori. One way to do this is to use the Newton identities for symmetric polynomials.
We can conclude that $\chi^{(n)}$ is $\pm1$ times an irreducible character.  To verify it is positive, try pairing it with the regular representation (the character of which is fixed by our field automorphism).