\lambda \in {\rm Irr}(G_{\nu}/\tilde{G}) \Big\}.$$

Generalized decomposition numbers and a proof of Theorem 1
==========================================================

In this section we use notations in Propositions 4 and 5. For $u \in P$, set $b_u = b(C_{G}(u))$. We note that if $C_{G}(u)$ is $p$-nilpotent, then $$d^{u}_{\chi, 1_{C_{G}(u)}} = \chi(u) \ \ (\forall \chi \in {\rm Irr}(b(G)).$$

Let $u \in P$ and assume $e_{u} = 1$. We have $$(\chi_{i}\lambda)(u) = \varepsilon_i \lambda(u),$$ $$({\chi_{\nu}}\lambda_{\nu})\!\uparrow_{G_{\nu}}^G\! (u) = \varepsilon(\eta_{\hat{\nu}}\lambda_{\nu})\!\uparrow_{P_{\nu}}^{P}\!(u),$$ $$(1 \leq i \leq e, \ \lambda \in {\rm Irr}(P/\tilde{P}) = {\rm Irr}(G/\tilde{G}), \ \nu \in {\cal X},\  \lambda_{\nu} \in {\rm Irr}(P_{\nu}/\tilde{P}) = {\rm Irr}(G_{\nu}/\tilde{G}) ).$$

By Lemma 1, $P_u = C_{P}(u)$ is a Sylow $p$-subgroup of $C_{G}(u)$. Here we show that $u$ and $u^{t} $ are not $P$-conjugate for any $u \in P$ with $e_u = 1$ and $t \in T \backslash O_{p'}(C_{G}(P))$ where $T$ is a $p$-complement of $N$. Suppose that $u$ and $u^t$ are $P$-conjugate. Then $\langle t \rangle$ acts on $u^P$ by conjugation. Since $P$ acts on $u^P$ transitively. So by a lemma of Glauberman, there is $u' \in u^P$ such that ${u'}^t = u'$. Hence $u^{v^{-1}tv} = u$ for some $v \in P$. This is a contradiction by Lemma 6. Set $\bar{P} = 