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 ?