= \{  \tilde{\chi}_1 = 1_{\tilde{G}},\tilde{\chi}_2, 
\cdots, \tilde{\chi}_e, \tilde{\chi}_{\mu} \ ( \mu \in \tilde{{\cal X}})\},$ where $$\left\{ \begin{array}{l}
\tilde{\chi}_j (u\rho) = \varepsilon_j = \pm 1 \ ( 1 \leq j \leq e),   \\ 
\tilde{\chi}_{\mu}(u \rho) = \varepsilon \eta_{\mu}(u) \ (\varepsilon = \pm 1)\end{array} \right.$$ for $u (\neq 1) \in \tilde{P}$ and $ \rho \in (C_{\tilde{G}}(u))_{p'}$. Moreover if $P$ is normal in $G$, then $\varepsilon_j = 1$ and $\varepsilon = 1$.

Since $b(\tilde{G})$ has a cyclic defect group $\tilde{P}$, the lemma follows by Lemma 4 and [@Dor], Theorem 68.1. $\blacksquare$

For $\mu \in \tilde{{\cal X}}$, we have $1_{\tilde{G}} * \eta_{\mu} = (e -1)1_{\tilde{G}} - \displaystyle\sum_{i = 2}^{e}\varepsilon_i \tilde{\chi}_i + \varepsilon \tilde{\chi}_{\mu}. $

Let $u \in \tilde{P}\backslash \{ 1\}$ and $\rho \in C_{\tilde{G}}(u)_{p'}$. &gt;From (4), we have $$(1_{\tilde{G}} * \eta_{\mu})(u\rho) = \eta_{\mu}(u),$$ $$((e -1)1_{\tilde{G}} - \sum_{i = 2}^{e}
\varepsilon_i \tilde{\chi}_i + \varepsilon \tilde{\chi}_{\mu})(u\rho) = \eta_{\mu}(u).$$ Therefore $1_{\tilde{G}} * \eta_{\mu} = (e -1)1_{\tilde{G}} - \displaystyle\sum_{i = 2}^{e}\varepsilon_i \tilde{\chi}_i + \varepsilon \tilde{\chi}_{\mu}$ on $p$-singular elements of $\tilde{G}$.

On the other hand for $\tau \in \tilde{G}_{p'}$, by [@NT], Theorem 5.4.5, the equation (4) and [@Dor], Theorem 68.1, (6), $$0 = \sum_{i =1}^{e}\tilde{\chi}_{i}(u)\tilde{\chi}_{i}(\tau) + \sum_{\mu \in \tilde{{\cal X}}}\tilde{\chi}_{\mu}(u)\tilde{\chi}_{\mu}(\tau)$$$$= 1 + \sum_{i =  2}^{e}\varepsilon_i \tilde{\chi}_i(\tau) +  
 \big( \varepsilon 