Irr}_{P}(\tilde{P}) \}$. We have $|{\cal X}_1| = \frac{p^{n'} - 1}{e}$. Set $m = \frac{p^{n'} - 1}{e}$.

By the second orthogonality relation for $P$ and the fact that $u$ and $u^t$ are not $P$-conjugate for any $u \in P$ with $e_u = 1$, and for any $t \in T \backslash O_{p'}(C_{G}(P))$, we can show $\sum_{\nu \in {\cal X}_1}|\eta_{\hat{\nu}}(u)|^2 = p^{n'} - e$. Moreover by the definition of $\hat{\nu}$ and the fact $u \not\in C_{P}(E)$ we see $ \sum_{\nu \in {\cal X}_1}{\eta}_{\hat{\nu}}(u) = -1$. In fact by using the second orthogonality relations for $P$ $$|P/\tilde{P}| \sum_{\nu \in {\cal X}_{1}}|\eta_{\hat{\nu}}(u)|^2 
= \sum_{t' \in E} \Big( \sum_{t \in E}\sum_{\nu \in {\cal X}_1} \sum_{\lambda \in {\rm Irr}(P/\tilde{P})}{\hat{\nu}}^t (u^{-1}){\hat{\nu}}^t (u^{t'})\lambda(u^{-1})\lambda(u)$$ $$+  \sum_{t \in E}\sum_{\mu \in {\cal X} \backslash {\cal X}_1} \sum_{\lambda_{\mu}}(\hat{\mu}\uparrow^{P}_{P_{\mu}})^{t}(u^{-1})(\hat{\mu}\uparrow^{P}_{P_{\mu}})^{t}(u^{t'})\lambda_{\mu}(u^{-1})\lambda_{\mu}(u) \Big) = 
|C_{P}(u)| - |P/\tilde{P}|e,$$ where $\lambda_{\mu}$ runs over ${\rm Irr}(P)/ X_{\mu}$, $X_{\mu}$ is the kernel of the restriction map ${\rm Irr}(P/\tilde{P}) \rightarrow {\rm Irr}(P_{\mu}/\tilde{P})$, because $u$ and $u^{t'}$ are not $P$-conjugate if $t' \neq 1$. We also note $\lambda(u^{t'}) = \lambda(u)$. Thus we obtain $\sum_{\nu \in {\cal X}_1}|\eta_{\hat{\nu}}(u)|^2 = p^{n'} - e$. Next we show $ \sum_{\nu \in {\cal X}_1}{\eta}_{\hat{\nu}}(u) = -1$. Write $ u = \tilde{u} r$ ($\tilde{u} \in \tilde{P}$, $r \in C_{P}(E)$). 