Artin's theorem and Brauer's theorem on Characters

Question

Artin: Let $\chi$ be rational valued complex character of (finite) group $G$. Then $\chi$ can be written as $\mathbb{Q}$-linear combination of characters $1_H^G$ for some cyclic subgroups $H$ of $G$.

In the theorem of Brauer, the subgroups $H$ are allowed to be elementary subgroups it is quite general than Artin's theorem in following sense:

Brauer: Every irreducible complex character $\chi$ of $G$ can be written as $\mathbb{Z}$-linear combinations of characters $\lambda_H^G$ for some subgroups $H$ of $G$, which are elementary subgroups, and $\lambda$ is a linear character of $H$.

Q. In the theorem of Brauer, comparing with Artin,

1) subgroups $H$ are allowed to be elementary (which include cyclic subgroups of $G$ also).

2) For elementary subgroups $H$, we consider $\lambda_H^G$, with $\lambda$ a linear character of $H$, not necessarily $1$.

I was wondering, what will happen if we consider subgroups $H$ to be elementary but the characters of $G$ to be $1_H^G$ (instead of $\lambda_H^G$); these induced characters are rational valued; so by integral combination of such characters of $G$, can we get all the rational valued characters of $G$?

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A (sub)group $K$ is said to be $p$-elementary, for a prime $p$, if $K=C_m\times P$ where $C_m$ is cyclic group of order coprime to $p$ (may be trivial) and $P$ is a $p$-group (may be trivial). We say that $K$ is elementary (sub)group if it is $p$-elementary for some prime $p$.

Answer 1

The answer seems to be negative:

If we take the characters $\{ (1_H)^G \,\, | \,\, H \mbox{ is elementary subgroup of $G$} \}$, then an integer valued character of $G$ may not be integral combination of these specific induced characters.

Consider the quaternion group $G=Q_8=\langle x,y\rangle$. All subgroups of $Q_8$ are elementary subgroups: $$1, \,\, \langle x^2\rangle,\,\, \langle x\rangle,\,\, \langle y\rangle,\,\, \langle xy\rangle,\,\, G.$$ Consider trivial characters of these subgroups, and induce to $G$, and denote them, respectively by $\psi_1, \ldots, \psi_6.$

Consider $\chi$ the degree $2$ complex irreducible character of $G$, it is integer values.

Suppose $\chi=a_1 \psi_1 + \cdots + a_6\psi_6$ with $a_i\in\mathbb{Z}$. Then $$(*) \,\,\,\,\,\,\,2=\chi(1)=a_1 \psi_1(1)+\sum_{i>1} a_i\psi_i(1) \mbox{ and } -2=\chi(x^2)=a_1 \psi_1(x^2)+\sum_{i>1} a_i\psi_i(x^2)$$ Note for $2\le i\le 6$, $\psi_i(1)=\psi_i(x^2)$; whereas, $\psi_1(1)=8$ and $\psi_1(x^2)=0$.

Subtracting equations in (*) and noting that the five-term sums in both are same, we get $4=8a_1$, so $a_1\notin \mathbb{Z}$.