Brauer induction

If $G$ is trivial, the assertion is immediate. Otherwise, for every prime $p\mid \mathrm{\mid }G\mathrm{\mid }$, write $\mathrm{\mid }G\mathrm{\mid }=p^{n_p}{l}_p$ with $p\nmid l_p$. By [F1], $l_p{1}_G\in I_{{\mathcal{E}}_p}(G)\subseteq I_{\mathcal{E}}(G)$. The integers $l_p$ have greatest common divisor $1$: for each prime divisor $q$ of $\mathrm{\mid }G\mathrm{\mid }$, the particular integer $l_q$ is prime to $q$. Bézout therefore gives $1_G\in I_{\mathcal{E}}(G)$.

By [F2], every $\chi \in R(G)$ satisfies $\chi =\chi 1_G\in I_{\mathcal{E}}(G)$. Thus it is an integral sum of characters ${\mathrm{Ind}}_H^G\theta$ with $H$ elementary and $\theta \in R(H)$.

By [F3], each such $\theta$ is an integral combination of characters induced from linear characters of elementary subgroups $K\le H$. Transitivity [F4] changes ${\mathrm{Ind}}_H^G{\mathrm{Ind}}_K^H\lambda$ into ${\mathrm{Ind}}_K^G\lambda$, which is exactly the claimed form. ∎