After reading the exposition about Artin's induction theorem for $\mathbb{Q}$-representations, for instance in Serre's Linear Representations of Finite Groups, 12.5, prop 25, one can be tempted to make the following generalization:
Let $G$ be a finite group. Then $G$ satisfies property (A):
Every character over $\mathbb{Q}$ of $G$ is a $\mathbb{Z}$-linear combination of characters induced from trivial characters of arbitrary subgroups.
In the following, I will say that a given $\mathbb{Q}$-character is nice if it is a $\mathbb{Z}$-linear combination of characters induced from trivial characters of arbitrary subgroups.
Serre immediately mentions that there is no induction theorem because the product $H\times C_3$ of the quaternion group and the cyclic group provides a counterexample. After some computations, this looks wrong to me; let me explain.
If $C$ is a cyclic group of order $n$, its irreducible $\mathbb{Q}$-representations are the $\mathbb{Q}(\zeta_d)$ for $d$ dividing $n$ and $\zeta_d$ a primitive $d$-th root of unity, where the generator acts by multiplication by $\zeta_d$. Since we can decompose the regular representation as the direct sum of these, by Möbius inversion we get that the irreducible $\mathbb{Q}$-characters are nice; so the same holds for every $\mathbb{Q}$-character.
If $G=H\times K$, we have $\mathrm{ind}^{H'}_H(\phi)\cdot\mathrm{ind}^{K'}_K(\psi)=\mathrm{ind}^{H'\times K'}_{H\times K}(\phi\cdot\psi)$ so property (A) holds for a product of groups if it holds for each factor. Therefore it holds for all abelian groups.
We are reduced to studying the irreducible $\mathbb{Q}$-characters of $H=\{1,-1,i,-i,j,-j,k,-k\}$; they are given by the trivial representation, the three sign representations of degree one with respective kernels $<i>$, $<j>$ and $<k>$, and a faithful representation of degree 4. The four first representations come from a cyclic quotient of $H$ so they are nice; finally, since the regular representation is the direct sum of all irreducible representations each with multiplicity one, it follows that the last representation also has a nice character.
Is there something wrong in my reasoning ? Do we know of an actual counter-example ?
Best Answer
The second part of that statement is wrong. I have wrongly assumed that the product of $\mathbb{Q}$-irreducible characters is $\mathbb{Q}$-irreducible. This can fail because the Schur index of a product only divides the product of Schur indexes. The example I talked about was in the old french edition; in the 1977 english edition, it is given with more details as exercise 13.4. Let me elaborate.
Let $\phi$ denote the faithful irreducible complex character of the quaternion group $H$ and $\psi=2\phi$ the faithful $\mathbb{Q}$-irreducible character of $H$ (see Linear representation theory of quaternion group). Let $\chi$ denote the character of the faithful irreducible rational representation of $C_3$ given by $\mathbb{Q}(e^{2i\pi/3})$. Then Serre explains how to construct a faithful $\mathbb{Q}$-irrreducible representation of $H\times C_3$ with character $\phi\cdot\chi$; in particular the character $\psi\cdot\chi=2(\phi\cdot\chi)$ is not $\mathbb{Q}$-irreducible.
Moreover, one computes $<\phi\cdot\chi,\phi\cdot\chi>=2$ and $\phi\cdot\chi$ is of degree 4, hence the multiplicity of $\phi\cdot\chi$ in the regular representation is $$ \frac{<1_{1}^{H\times C_3},\phi\cdot\chi>}{<\phi\cdot\chi,\phi\cdot\chi>}=\frac{(\phi\cdot\chi)(1)}{2}=2 $$ which implies by exercise 13.3 that $\phi\cdot\chi$ is not a $\mathbb{Z}$-linear combination of permutation characters.
For more information about which virtual $\mathbb{Q}$-characters are linear combinations of permutation characters, one can read Bartel, Dokchitser, Rational representations and permutation representations of finite groups