Let $G$ be a finite group and let $\chi$ be a rational character of $G$. A theorem due to Artin states that $|G| \chi$ is a $\mathbb{Z}$-linear combination of characters of the form $(1_H){\uparrow}^G$, where $H \leq G$ is a cyclic subgroup.
It's a fact (see, for instance, exercise 5.20 of the same book by Isaacs) that the set $P(G)$ of $\mathbb{Z}$-linear combinations of characters of the form $(1_H){\uparrow}^G$, for $H \leq G$ (not necessarily cyclic here) is a ring.
The problem I'm trying to solve is as follows:
Let $n_G(\chi)$ the the smallest non-zero natural number such that $n_G(\chi)\chi \in P(G)$ and let $N = \ker \chi$, so that $\chi$ may be considered a character of $G/N$. Show that $n_G(\chi) = n_{G/N}(\chi)$.
The following hint is also provided:
Hint: Writing $n\chi = \sum a_H (1_H){\uparrow}^G$, show $n\chi = \sum a_H (1_{NH}){\uparrow}^G$ by obtaining $(1_H){\uparrow}^G = (1_{NH}){\uparrow}^G + \xi_H$, where $N$ isn't contained in the kernel of any of the constituents of $\xi_H$.
My attempt
Assuming the hint as given: It's trivial (from the hint) that $n_{G/N}(\chi) \leq n_G(\chi)$. Furthermore, if $m\chi = \sum a_\overline{H} (1_\overline{H}){\uparrow}^G$, the correspondence theorem yields $m\chi = \sum a_{NH} (1_{NH}){\uparrow}^G$. Thus, $n_{G/N}(\chi) \geq n_G(\chi)$ and we are done.
Proving the hint: I first tried to show that every constituent of $(1_{NH}){\uparrow}^G$ is also a constituent of $(1_{H}){\uparrow}^G$. This was relatively simple: $$\langle (1_{NH}){\uparrow}^G, \eta \rangle > 0 \iff \langle 1_{NH}, \eta {\mid}_{NH} \rangle > 0 \iff \eta {\mid}_{NH} = 1_{NH} + \sum \eta_i$$
Restricting $\eta$ to $H$, we get at least one component of $1_H$, which gives me the result. That said:
How do I know all these constituents appear in $(1_H){\uparrow}^G$ the same number of times they do in $(1_{NH}){\uparrow}^G?$
In principle, some of the $\eta_i$ may be different from $1_{NH}$ but also restrict to $1_H$ – and this can't happen if the hint is true…
Also:
How do I show that every constituent of $(1_H){\uparrow}^G$ containing $N$ in its kernel has to be a constituent of $(1_{NH}){\uparrow}^G$?
The idea used before doesn't work here, since inducing $\psi{\mid}_H$ to $NH$ doesn't necessarily yield the restriction $\psi{\mid}_{NH}$.
Thanks in advance!
Best Answer
By Problem 5.2 of the book, $((1_H)^{HN})_N=(1_{H\cap N})^N$. This shows that $1_{HN}$ is the only constituent of $(1_H)^{HN}$ with $N$ in its kernel. Now you can use linearity and transitivity of induction, $$(1_H)^G-(1_{HN})^G=((1_H)^{HN}-1_{HN})^G=\sum_{\psi\ne 1_{HN}}\psi^G$$ where $\psi$ are certain non-trivial characters of $HN$, which do not have $N$ in their kernel.
EDIT (answer to question in comment): Using the hint you get $n\chi=\sum a_H(1_{HN})^G+\sum a_H\psi_H$, where $N$ is not contained in the kernel of any constituent of any $\psi_H$. On the other hand, $N$ is contained in the kernel of every constituent of $(1_{HN})^G$ (and of $\chi$). Since the irreducible characters form a basis of the space of complex class function, it follows that $\sum a_H\psi_H=0$.