A positive integer multiple of the trivial character is an integral combination of cyclic permutation characters

For each cyclic subgroup $C\le G$, let ${\eta }_C$ be the class function from [F1], and define $f:={\sum }_{\begin{array}{c}C\le G\\ C\text{ cyclic}\end{array}}{\mathrm{Ind}}_C^G{\eta }_C$. The sum is finite because a finite group has only finitely many subgroups.

By [F2], for each cyclic $C\le G$ one has ${\mathrm{Ind}}_C^G{\eta }_C(g)=\frac{1}{\mathrm{\mid }C\mathrm{\mid }}{\sum }_{\begin{array}{c}x\in G\\ x^{-1}gx\in C\end{array}}{\eta }_C(x^{-1}gx)$. Fix $x\in G$. Among all cyclic subgroups $C\le G$, exactly one of them can make the summand indexed by $x$ nonzero, namely $C=⟨x^{-1}gx⟩$; for that subgroup, the value of ${\eta }_C$ is $\mathrm{\mid }C\mathrm{\mid }$. Therefore the double sum defining $f(g)$ contributes exactly $1$ for each $x\in G$, so $f(g)=\mathrm{\mid }G\mathrm{\mid }$.

Step 2.1 holds for every $g\in G$, hence $f=\mathrm{\mid }G\mathrm{\mid }{1}_G$ as class functions. Expanding each ${\eta }_C$ by [F1] and then using [F3] to replace ${\mathrm{Ind}}_C^G({\mathrm{Ind}}_D^C{1}_D)$ by ${\mathrm{Ind}}_D^G{1}_D$ expresses $f$ as an integral linear combination of characters ${\mathrm{Ind}}_D^G{1}_D$ with $D$ cyclic. Thus $\mathrm{\mid }G\mathrm{\mid }{1}_G$ has the required form.