[([@Wata2014] Lemma 4 (i))]{} Suppose that $\tilde{P} \neq 1$. Let $T$ be a $p$-complement of $N$ and let $X$ be a subgroup of $T$ containing $O_{p'}(C_{G}(P))$ properly. Then $P = \tilde{P}\rtimes C_{P}(X). $ In particular $C_{P}(X) = C_{P}(T)$ and $G = \tilde{G}\rtimes C_{P}(T)$.

Let $u \in P$. Since $\tilde{P}$ is a $p$-hyperfocal subgroup of $G$, $[u, X] \subseteq [P, X] \subseteq \tilde{P}$. Hence $\tilde{P}u$ is $X$-invariant. On the other hand $\tilde{P}$ acts transitively on $u\tilde{P}u$ by right multiplication. By using a lemma of Glauberman ([@Isaacs], 13.8) there is an element of $\tilde{P}u$ fixed by $X$. Since $\tilde{P}$ is cyclic, from (2) and (3), $X/O_{p'}(C_{G}(P))$ acts on $\tilde{P} \backslash \{ 1\}$ fixed-point freely, an element of $\tilde{P}u$ fixed by $X$ is unique. This implies $P = \tilde{P}\rtimes C_{P}(X)$. Since $G = \tilde{G}P$, we have $G = \tilde{G} \rtimes C_{P}(T)$. $\blacksquare$

For $u \in P$, $e_{u} = e$ or $e_{u} = 1$. If $e_{u} = e$, then $C_{N}(u) = P_u T$ for some $p$-complement of $N$ and $C_{\tilde{P}}(u)$ is a $p$-hyperfocal subgroup of $C_{G}(u)$. If $e_{u} = 1$, then $C_{N}(u) = P_u O_{p'}(C_{G}(P))$.

By Lemma 2, $P_u = C_{P}(u)$ is a Sylow $p$-subgroup of $C_{G}(u)$. By using the Schur-Zassenhaus theorem, we can 