I was trying to solve a statement from Serre's Linear Representation of Finite Groups (taken from the hint of Exercise 10.6(b) in the book) :
Suppose that $H$ is normal in $G$ and that $G/H$ is abelian. Show that $\operatorname{Ind}_H^G(1)\in R'(G)$
The definition of $R'(G)$ is $R'_0(G)+\mathbb{Z}$, where $R'_0(G)$ is the subgroup of $R(G)$ (the space of virtual characters) generated by elements of the form $\operatorname{Ind}_E^G(\alpha-1)$ with $E$ an elementary subgroup of $G$ and $\alpha$ a degree $1$ character of $E$.
Hints from the question and my attempts
The hint suggests that we can try to show that $\operatorname{Ind}_H^G(1)$ is the sum of $(G: H)$ characters of degree $1$ of $G$ each of whose kernel contains $H$.
My attempt so far is that $\operatorname{Ind}_H^G(1)$ is actually $\operatorname{Res}^{G/H}_G(r_{G/H})$, where the restriction map is via the quotient $G\rightarrow G/H$, and $r_{G/H}$ is the regular representation of $G/H$,
so if we take $\chi_i$ to be the irreducible characters of $G/H$, then there are $|G/H|$ many of them and each has degree $1$ (since $G/H$ is abelian), which gives us
$$\operatorname{Res}^{G/H}_Gr_{G/H}=\operatorname{Res}^{G/H}_G(\sum_i\chi_i)=\sum_i\operatorname{Res}^{G/H}_G(\chi_i)$$
so the hint is verified.
Question
I still don't know how to use this hint to show that $\operatorname{Ind}_H^G(1)\in R'(G)$.
I only know if $G$ is assumed to be elementary (since for the purpose of other exercises in that question, this weaker version would suffice), we are done by noticing the calculation :
$$\psi_i=1+(\psi_i-1)=1+\operatorname{Ind}_G^G(\psi_i-1)$$
(where $\psi_i$ denotes $\operatorname{Res}^{G/H}_G(\chi_i)$)
I don't know whether or not that this assumption (that $G$ is elementary) needs to be added for this exercise to be true.