see $C_{N}(u) = P_u C_{T}(u)$ and $C_{N}(P_u) = Z(P_u) C_{T}(P_u)$ for some $p$-complement $T$ of $N$. Hence $$e_u = |C_{N}(u) : P_u C_{N}(P_u)| = |C_{T}(u): C_{T}(P_u)|.$$ Now suppose that $e_u \neq 1$. Then $O_{p'}(C_{G}(P)) \subseteq C_{T}(P_u) \neq  C_T(u)$. So by applying Lemma 3 for $X = C_{T}(u)$, $u \in C_{P}(T)$ and hence $C_{N}(u) = P_u T$. (So we have $ u^N = u^P$. ) Since $\langle x^{p^{n-1}} \rangle $ is a normal subgroup of $P$ of order $p$, $\langle x^{p^{n-1}} \rangle \subseteq C_{P}(u) = P_u$. &gt;From (2) and (3), $C_{T}(x^{p^{n-1}})= O_{p'}(C_{G}(P))$, hence $C_{T}(P_u) = O_{p'}(C_{G}(P))$. Therefore we see $e_ u = e$. Moreover $C_{\tilde{P}}(u)$ is a $p$-hyperfocal subgroup of $C_{G}(u)$ because $[C_{T}(u), C_{\tilde{P}}(u)] = C_{\tilde{P}}(u)$.

Finally suppose $e_u = 1.$ Then $O_{p'}(C_{G}(P)) = C_{T}(u)$ by the argument as in the case $e_u \neq 1$, and hence $C_{N}(u) = P_{u}O_{p'}(C_{G}(P))$. (Then we see $|u^{N}| = e |u^{P}|$.) $\blacksquare$

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Let $\Pi \subseteq P$ be a set of representatives for the conjugacy classes of $p$-elements of $G$. Let $\eta$ be a $G$-stable generalized character of $P$, that is, if $u$ and $u'$ in $P$ are $G$-conjugate, then $\eta(u) = \eta(u')$. By a theorem of Brauer on generalized characters ([@NT], Theorem 3.4.2) and 