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$\DeclareMathOperator{\Aut}{Aut}\DeclareMathOperator{\Ind}{Ind}$I believe I can prove the following induction theorem (modulo carefully checking a few details), and I would like to know whether this is known, or can be deduced from something known very quickly. If you want, below replace $K$ by the real numbers. If you know the Witt–Berman theorem, you can skip the definition and jump straight to the theorem.

Definition. Let $K$ be a field of characteristic $0$ (for simplicity), let $\bar{K}$ be an algebraic closure of $K$, for any integer $n$ let $\mu_{\bar{K}}(n)$ denote the group of $n$-th roots of unity in $\bar{K}$, let $G_K$ denote the Galois group of $\bar{K}/K$, and let $l$ be a prime number. A finite group is called $K$-$l$-elementary if it is a semidirect product $C\rtimes L$, where $C$ is a cyclic group of order coprime to $l$, $L$ is an $l$-group, and the image of $L\to \Aut C$ is contained in the image of $G_K\to \Aut(\mu_{\bar{K}}(\#C))$, where one fixes an arbitrary isomorphism between the cyclic group $\mu_{\bar{K}}(\#C)$ and $C$. A group is called $K$-elementary if it is $K$-$l$-elementary for some prime number $l$.

For example a $\mathbb{C}$-elementary group is just a direct product of $C$ and $L$ as above. An $\mathbb{R}$-elementary group is a semi-direct product as above where $L$ is only allowed to act on $C$ by $\{\pm 1\}$. A $\mathbb{Q}$-elementary group is any semiderct product of $C$ and $L$ as above.

Theorem. Let $G$ be a finite group, let $p$ be a prime number, let $K$ be a field of characteristic $0$, and let $\chi$ be the character of a $KG$-module such that $\chi$ vanishes on all elements whose order is divisible by $p$ (so-called $p$-singular elements). Then there exist integers $n_H$, as $H$ runs through $K$-elementary subgroups of $G$ of order coprime to $p$, and characters $\psi_H$ of $KH$-modules such that $$ \chi = \sum_H n_H\cdot\Ind_H^G\psi_H. $$

Remarks.

  1. If the condition of $\chi$ vanishing on the $p$-singular elements is omitted, and in the conclusion one does not impose the restriction on $H$ of having order coprime to $p$, then this is the Witt–Berman induction theorem, which has nice expositions in lots of sources, such as Curtis–Reiner, Karpilovsky, and others. I am finding that I still need almost 2 pages to deduce this from Witt–Berman. Yet in J. Queyrut, Anneaux d'entiers dans le même genre, on p. 16 in the phrase starting with "Le fait que..." he seems to be using something like that (in the special case $K=\mathbb{R}$) without any justification or reference, which makes me think that it is either a standard result from the literature, or a much easier consequence of Witt–Berman than my proof suggests.
  2. Suppose that $K$ is a local field of mixed characteristic, let $R$ be the ring of integers, let $k$ be the residue field, and let $p$ be the characteristic of $k$. By way of motivation for the above result, recall that the $KG$-modules whose characters vanish on all $p$-singular elements span exactly the image of the map that is often called $e$ from $K_0(kG)$ to $K_0(KG)$, which takes a projective $kG$-module, lifts it to a projective $RG$-module, and then extends scalars to $K$.
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  • $\begingroup$ I won't write a formal answer, and I don't know if this helps since I don't know your proof. If I were trying too prove this, I would try to reduce it to the case when $G$ itself is a $K$-elementary group by writing the trivial as a combination of things induced from $K$-elementary subgroup, and taking the product with $\chi$. The result is known in the case $K = \mathbb{C}$ or a splitting field of characteristic zero (maybe first due to W.F. Reynolds), and that is more or less how the proof goes. $\endgroup$ – Geoff Robinson May 8 at 14:35
  • $\begingroup$ Dear Geoff, yes, the proof reduces to the case where $G$ itself is $K$-elementary by Witt–Berman, and pretty much my whole proof deals with that case. If you are willing to look at it and tell me if I am missing shortcuts, I will be happy to send you my proof, which at the moment is only fully written out for $K=\mathbb{R}$. $\endgroup$ – Alex B. May 8 at 15:16

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