Then $P = \langle \tilde{P}, r \rangle$. $[\tilde{P}, P] = [\tilde{P}, r] = 
\langle [\tilde{P}, r] \rangle$, $r^P = [\tilde{P}, r]r$. Since $e_u = 1$ and hence $u \not\in r^P$, $\tilde{u} \not\in [\tilde{P}, P]$. By the second orthogonality relation for $P/[\tilde{P}, P] \cong (\tilde{P}/[P, \tilde{P}] ) \times C_{P}(E)$, $$0 = \sum_{t \in E} \sum_{\nu \in {\cal X}_1} \sum_{\lambda \in {\rm Irr}(P/\tilde{P})}
({\hat{\nu}}^t \cdot \lambda)(\tilde{u}) + \sum_{\lambda \in {\rm Irr}(P/\tilde{P})} \lambda(\tilde{u})$$ $$= |P/\tilde{P}|\big(\sum_{\nu \in {\cal X}_1}\eta_{\hat{\nu}}(u) +1 \big).$$ Therefore we have $\sum_{\nu \in {\cal X}_1}\eta_{\hat{\nu}}(u) = -1$.

Now from (8), $$p^{n'}-1-\sum_{i=2}^e{x_i}^2=\sum_{\nu\in\mathcal X_1}\left|\eta_{\widehat\nu}(u)-(e-1)+\sum_{i=2}^e\varepsilon_ix_i\right|^2.$$ $${\rm The \ left \ hand \ of} \ (9) = p^{n'}-e-\sum_{i=2}^e\left({x_i}^2-1\right).$$ $${\rm The \ right\  hand\  of }\ (9) = \sum_{\nu\in\mathcal X_1}\left|\eta_{\widehat\nu}(u)+\sum_{i=2}^e\left(\varepsilon_ix_i-1\right)\right|^2$$ $$= \sum_{\nu\in\mathcal X_1}\left|\eta_{\widehat\nu}(u)\right|^2+\sum_{\nu\in\mathcal X_1}\left(\eta_{\widehat\nu}(u)+\overline{\eta_{\widehat\nu}(u)}\right)\sum_{i=2}^e(\varepsilon_ix_i-1)+\sum_{\nu\in\mathcal X_1}\left(\sum_{i=2}^e\left(\varepsilon_ix_i-1\right)\right)^2$$ $$= p^{n'}-e-2\sum_{i=2}^e(\varepsilon_ix_i-1)+m\left(\sum_{i=2}^e\left(\varepsilon_ix_i-1\right)\right)^2.$$ Hence we have $$\begin{aligned}
0 &= & \sum_{i=2}^e\left({x_i}^2-1\right)-2\sum_{i=2}^e(\varepsilon_ix_i-1)+m\left(\sum_{i=2}^e\left(\varepsilon_ix_i-1\right)\right)^2 \\
 &=& \sum_{i=2}^e\left(\left({x_i}^2-1\right)-2(\varepsilon_ix_i-1)\right)+m\left(\sum_{i=2}^e\left(\varepsilon_ix_i-1\right)\right)^2 \\ 
 &=& \sum_{i=2}^e\left({x_i}^2-2\varepsilon_ix_i+1\right)+m\left(\sum_{i=2}^e\left(\varepsilon_ix_i-1\right)\right)^2 \\
&=& \sum_{i=2}^e(\varepsilon_ix_i-1)^2+m\left(\sum_{i=2}^e\left(\varepsilon_ix_i-1\right)\right)^2. \end{aligned}$$ Hence we have $$\varepsilon_ix_i-1=0 \therefore x_i = \varepsilon_i \ ( 2 \leq i \leq e).$$ This completes the proof. $\blacksquare$

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Since $|{\rm IBr}(b(\tilde{G}))| = e < p$ and $G/\tilde{G}$ is a $p$-group, $|{\rm IBr}(b({G}))| = e$. In particular if $e_u = e$ where $u \in P$, then $l(b_u) = e$.

Let $u \in P$ and assume $e_{u} = e$. The 