K)$, the group $L$ canot be equal to $K$. So $G/K f_{J}^{G}$ is a $R$-linear combination of transitive $G$-set $G/L'$ where $|L'|<|K|$. By induction, $G/K$ is a $R$-linear combination of elements of the form $G/I f_{J}^{G}$.

Following [@deiml], let us consider the linear form $\phi_{G}$ on $RB(G)$ defined on a basis element by: $$\phi\big(G/I f_{J}^{G}\big) = \left\{\begin{array}{c}1 \hbox{ if $I=J$,} \\0 \hbox{ if $I\neq J$.}\end{array}\right.$$

\[re1\]

-   If $R=k$ is a field of characteristic $p$, and if $p\nmid |G|$, then the idempotents $f_{J}^{G}$ are the idempotents $e_{J}^{G}$ so it is easy to check that

    $\phi_{G}(X) = \sum_{H\in [s(G)]}\frac{|H|}{|N_{G}(H)|} |X^{H}|$, for $X\in kB(G)$.

-   If $p\mid |G|$, it seems rather difficult to compute the value of $\phi_{G}$ on a transitive $G$-set.

\[indu\] Let $H\leqslant G$ and $J$ be a $p$-perfect subgroup of $H$. Then:

1.  $Ind_{H}^{G}(H/J f_{J}^{H})=G/Jf_{J}^{G}$.

2.  Moreover if $p\mid |N_{H}(J)/J|$ and $p^2\nmid |N_{H}(J)/J|$, let $S_{J}$ be a subgroup of $H$ such that $J\subset S_{J}$ and $O^{p}(S_{J})=J$. Then: $$Ind_{H}^{G}(H/S_{J}f_{J}^{H})=G/S_{J} f_{J}^{G}.$$

Using Lemma \[form\], we have: $$\begin{aligned}
Ind_{H}^{G}\big(H/J f_{J}^{H}\big)&= \frac{|N_{H}(J)|}{|J|}Ind_{H}^{G}(e_{H}^{G})\\
&=\frac{|N_{H}(J)|}{|J|} \frac{|N_{G}(J)|}{|N_{H}(J)|} e_{J}^{G}\\
&=G/J f_{J}^{G}.\end{aligned}$$ For the second part, by Lemma $3.5$ of [@yoshida_idempotent], we have $Res^{G}_{H}(f^{G}_{J})=\sum_{J'} f^{H}_{J'}$ where $J'$ runs the subgroups of $H$ up to $H$-conjugacy such that $J'$ is conjugate to $J$ in $G$. So, we have: $$\begin{aligned}
H/S_{J} Res^{G}_{H}(f^{G}_{J})&=\sum_{J'} H/S_{J} f^{H}_{J'},\end{aligned}$$ but we have: $$\begin{aligned}
H/S_{J}f_{J'}^{H}= \sum_{\underset{O^{p}(K)=J'}{K\leqslant J\hbox{ {\footnotesize up to $H$-conjugacy}}}} |(H/S_{J})^K|e_{K}^{H}. \end{aligned}$$ But $|(H/S_{J})^{K}|=0$ unless $K$ is $H$-conjugate to a subgroup of $S_{J}$. Without lost of generality one can assume $K\subseteq S_{J}$. So the only non zero t