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abstract: 'Let $G$ be a finite group with a Sylow $p$-subgroup $P$. We prove that the principal $p$-blocks of $G$ and $N_G(P)$ are perfectly isometric under the assumption $G$ has a cyclic $p$-hyperfocal subgroup.'
author:
- Hiroshi HORIMOTO and Atumi WATANABE
title: 'On a perfect isometry between principal $p$-blocks of finite groups with cyclic $p$-hyperfocal subgroups'
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Introduction and preliminaries
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Let $G$ be a finite group and $p$ be a prime number. We denote by $O^{p}(G)$ the subgroup of $G$ generated by the set $G_{p'}$ of $p$-regular elements of $G$. Let $P$ be a Sylow $p$-subgroup of $G$ and set $$\tilde{P}  = \langle \ [\ T, \ O^{p}(N_{G}(T)) \ ] \ | \ T \leq P \ \rangle.$$ We will call $\tilde{P}$ a $p$-hyperfocal subgroup of $G$. By Puig [@Pu5] and [@Pu6], $$\tilde{P} = P \cap O^{p}(G)$$ (see [@BCGLO], Lemma 2.2 or [@Craven], Theorem 1.33 for proofs). Denoting by $b(G)$ the principal block of $G$, $\tilde{P}$ is a hyperfocal subgroup of $b(G)$ in the sense of Puig ([@Pu5] and [@Pu6]). In this paper we prove the following theorem. This study is motivated by Rouquier [@Rou2], A.2.

If $\tilde{P}$ is cyclic, then the principal blocks $b(G)$, $b(N_{G}(\tilde{P}))$ and $b(N_{G}(P))$ are perfectly isometric in the sense of Broué [@Broue].

By [@sasaki], if $p$ is odd and $P$ is metacyclic, then $\tilde{P}$ is cyclic. If $p = 3$ and $G = SL(2, 2^{3^n}) \rtimes C_3$, then $P$ is isomorphic to the metacylic $3$-group $M_{(n + 1 )+ 1}(3)$ (cf. [@HKK], (4.3)).

Now the above theorem is a generalization of [@JWata], Proposition 6. If $\tilde{P} = 1$, then $G$ is $p$-nilpotent, hence the above is clear. We will determine the ordinary irreducible characters in $b(G)$ (Propositions 5 in $\S 2$ below), and the generalized decomposition numbers associated with $b(G)$ (Propositions 6 and 7 in $\S 3$ below), and then we obtain Theorem 1 from Theorem 2 below.

We refer [@NT], [@T] and [@Gorenstein] for the notation and terminology. Let $H$ be a subgroup of $G$. For $x \in G$, we denote by $x^H$ the $H$-conjugacy class containing $x$. For a character $\chi$ of $G$, the restriction of $\chi$ to $H$ is denoted by $\chi \!\downarrow_{H}$. For a character $\zeta$ of $H$, the induced character $\zeta$ to $G$ is denoted by $\zeta\!\uparrow_{H}^{G}$. For characters $\chi$, $\chi'$ of $G$, the innner product of $\chi$ and $\chi'$ is denoted by $(\chi, \chi')$. For a complex number $\alpha$ we denote by $\bar{\alpha}$ the complex conjugate, and for a character ${\chi}$ of $G$ we denote by $\bar{\chi}$ the contragredient character.

Let $({\cal K}, {\cal O}, F)$ be a $p$-modular system such that ${\cal K}$ contains the field ${\bf Q}(\sqrt[|G|]{1})$. For a ($p$-)block $b$ of $G$, we denote by Irr$(b)$ the set of ordinary irreducible characters belonging to $b$ and IBr$(b)$ the set of irreducible Brauer characters belonging to $b$. Here $b$ denotes a block idempotent of ${\cal O}G$. We set $k(b) = |{\rm Irr}(b)|$, the cardinality of Irr$(b)$, and $l(b) = |{\rm IBr}(b)|$. Also we set ${\cal R}_{{\cal K}}(G, b) = \sum_{\chi \in {\rm Irr}(b)}\textrm{{\boldmath $Z$}}  \chi$ and we denote by $b(G)$ the principal block of $G$. We denote by $1_{G}$ the trivial irreducible character and the trivial Brauer character of $G$.

Let $(\pi, {\bf b})$ be a $b$-Brauer element, that is, $\bf b$ is a block of $C_{G}(\pi)$ associated with $b$. For $\chi \in {\rm Irr}(b)$, the class function $\chi^{(\pi, {\bf b})}$ of $G$ is defined as follows: $\chi^{(\pi, {\bf b})}$ vanishes outside of the $p$-section of $G$ containing $\pi$ and $\chi^{(\pi, {\bf b})}(\pi \rho) = \chi(\pi\rho {\bf b})$ for $\rho \in C_{G}(\pi)_{p'}$. Let $Bs({\bf b}) = \{ \varphi^{(\pi)}_j \ | 1 \leq j \leq l({\bf b}) \}$ be a basic set for ${\bf b}$ in the sense of Brauer [@Brauer], that is, $Bs({\bf b})$ is a $\textrm{{\boldmath $Z$}}$-basis of the $\textrm{{\boldmath $Z$}}$-module $$\bigoplus_{\varphi \in {\rm IBr}({\bf b})} \textrm{{\boldmath $Z$}}  \varphi.$$ Then there exist ${d}_{\chi {\varphi}_j^{(\pi)}}^{\pi} \in {\cal O}$, what we call the generalized decomposition numbers with respect to $Bs({\bf b})$, which satisfy $$\chi^{(\pi,{\bf  b})}(\pi \rho) = \sum_{j = 1}^{l({\bf b})} 
{d}_{\chi {\varphi}_{j}^{(\pi)}}^{\pi}
{\varphi}_{j}^{(\pi)}(\rho) \ \ (\forall \rho \in C_{G}(\pi)_{p'}).$$ Set $${D}^{(\pi, {\bf b})}  = \Big( {d}_{\chi {\varphi}_{j}^{(\pi)}}^{\pi}\Big)_
{\chi \in {\rm Irr}(b), \varphi_{j}^{(\pi)} \in Bs({\bf b})}.$$ By [@NT], Theorem 5.4.11, the matrix ${^t}({D}^{(\pi, {\bf b})})\overline{D^{(\pi, {\bf b})}}$ is similar to the Cartan matrix of ${\bf  b}$ [w.r.t.]{} the basic set $Bs({\bf b})$, where $\overline{D^{(\pi, {\bf b})}}$ is the complex conjugate matrix of ${D}^{(\pi, {\bf b})}$ (see [@Navarro], Lemma 7.5). The following plays an important role in this paper.

Let $b$ $(resp. \  b')$ be a block of a finite group $G$ $(resp. \ 
G')$ such that $b$ and $b'$ satisfy [(i) - (iv)]{} below.

[(i)]{} $b$ and $b'$ has a common defect group $P$. Let $(P, b_P)$ $(resp. \ (P, b'_P))$ be a maximal $b$ $resp. \ b')$-Brauer pair,

[(ii)]{} there exists $\Pi \subseteq P$ such that $\{ (\pi, b_{\pi}) \in (P, b_P) \ | \   \pi \in \Pi \}$ $(resp. \ \{ (\pi, b'_{\pi}) \in (P, b'_P) \ |\  \pi \in \Pi \})$ is a set of representatives for the $G$ $(resp. \ G')$-conjugacy classes of $b$ $(resp. \ b')$- Brauer elements,

[(iii)]{} $l(b_{\pi}) = l(b'_{\pi})$ for any $\pi \in \Pi$,

[(iv)]{} For any $\pi \in \Pi \ \backslash \ \{ 1 \}$, there exist basic set $Bs(b_{\pi}) = \{ \varphi_{j}^{(\pi)} \ | \  j = 1, \ 2, \cdots, l(b_{\pi}) \}$ for $b_{\pi}$ and $Bs(b'_{\pi}) = \{ {\varphi'_j}^{(\pi)} \ | \  j = 1, \ 2, \cdots, l(b'_{\pi}) \}$ for $b'_{\pi}$ such that $$d_{\chi_i \varphi_{j}^{(\pi)}}^{\pi} = \varepsilon_i  d_{\chi'_{i} \varphi_{j}'^{(\pi)}}^{\pi}, \ \varepsilon_i = \pm 1 \ (1 \leq i \leq k(b), \ 
1 \leq j \leq l(b_{\pi}))$$ where [Irr]{}$(b) = \{ \chi_i \ | \ i = 1, 2, \cdots, k(b)\}$ and [Irr]{}$(b') = \{ \chi'_i \ | \ i = 1, 2, \cdots, k(b')\}$. Then $$\chi'_i \in R_{{\cal K}}(G', b') \mapsto \varepsilon_{i} \chi_i \in R_{{\cal K}}(G, b)$$ is a perfect isometry from $b'$ to $b$ in the sense of Broué [@Broue].

By the condition (ii) and (iii), $k(b) = k(b')$.

(Separation condition) Let $a'$ be a $p$-element of $G'$ and let $\Pi'_{a'}$ be the set of elements $\pi$ of $\Pi$ which is $G'$-conjugate to $a'$. For $\pi \in \Pi'_{a'}$, let $c'_{\pi a'}\in G'$ be such that $a'^{c'_{\pi a'}} = (a')^{c'_{\pi a'}} = \pi$. Similarly for a $p$-element $a$ of $G$, we define $\Pi_a$ and $c_{\pi a}$ for $\pi \in \Pi_{a}$. Now for $\rho \in G_{p'}$ and ${\rho'} \in C_{G'}(a')_{p'}$, from (1) and by the assumption and [@NT], Theorem 5.4.11 for basic sets, $$\sum_{i= 1}^{k(b)} \varepsilon_{i} \chi_i (\rho)\overline{{\chi}'_i (a' {\rho}'}) = \sum_{i = 1}^{k(b)} \chi_i(\rho) \left( \sum_{\pi \in \Pi'_{a'}} \sum_{j =1}^{l(b_{\pi})} \ \overline{d^{x}_{\chi_i \varphi_{j}^{(\pi)}}}\overline{{\varphi'}_{j}^{(\pi)} ({\rho}'^{c'_{\pi a'}})}\ \right)$$ $$=  \sum_{\pi \in \Pi'_{a'}} \sum_{j = 1}^{l(b_{\pi})} \left( \sum_{i= 1}^{k(b)} \chi_i(\rho) \overline{d^{\pi}_{\chi_i \varphi_{j}^{(\pi)}}}\ \right) \overline{\varphi'^{(\pi)}_{j} ({\rho'}^{{c'}_{\pi a'}})} = 0.$$ (see [@Navarro], Lemma 7.5) Similarly for a $p$-singular element $x \in G$ and $y' \in G'_{p'}$,\
$\displaystyle\sum_{i=1}^{k(b)} \varepsilon_{i}\chi_i(x)\overline{\chi'_{i}(y')} = 0.$ (Integrality condition) Let $\rho \in G_{p'}$ and ${\rho'}\in G'_{p'}$. By the above, $\displaystyle\sum_{i= 1}^{k(b)} \varepsilon_i \overline{\chi'_i({\rho'})}\chi_i$ vanishes on the $p$-singular elements of $G$. Hence $\displaystyle\sum_{i= 1}^{k(b)}\varepsilon_i \chi_i(\rho)\overline{\chi'_i ({\rho'})}$ is divisible by $|C_{G}(\rho)|$ in ${\cal O}$ by [@NT], Theorems 3.6.10 and 3.6.13. Similarly $\displaystyle\sum_{i= 1}^{k(b)}\varepsilon_i \chi_i(\rho)\overline{\chi'_i ({\rho'})}$ is divisible by $|C_{G'}(\rho')|$. Let $a$ (resp. $a'$) be a $p$-element of $G$ (resp. $G'$). For $\rho \in C_{G}(a)_{p'}$ and $\rho'  \in C_{G'}(a')_{p'}$, $$\sum_{i= 1}^{k(b)}\varepsilon_i \chi_i (a\rho) \overline{\chi'_i(a'{\rho}')}$$ $$= \sum_{i = 1}^{k(b)}\left( \sum_{ \pi \in \Pi_a } \sum_{j =1}^{l(b_{\pi})}\varepsilon_i d^{\pi}_{\chi_i \varphi_{j}^{(\pi)}} {\varphi}_{j}^{(\pi)} ({\rho}^{c_{\pi a}}) \right)\left( \sum_{\pi' \in \Pi'_{a'}} \sum_{k =1}^{l(b_{\pi'})}\overline{ d^{\pi'}_{\chi'_i  \varphi'^{(\pi')}_{k}}}\overline{{\varphi'}_{k}^{(\pi')} ({\rho}'^{c'_{\pi' a'}})} \right)$$ $$=  \sum_{\pi \in \Pi_a}\sum_{\pi' \in \Pi'_{a'}}\sum_{k =1}^{l(b_{\pi'})}\left( \sum_{j =1}^{l(b_{\pi})}\left( \sum_{i = 1}^{k(b)} \varepsilon_i d^{\pi}_{\chi_i  \varphi_{j}^{(\pi)}}
\overline{d^{\pi'}_{\chi'_i  \varphi'^{(\pi')}_{k}}}\right) \varphi^{(\pi)}_{j}({\rho}^{c_{\pi a}})\right)\overline{{\varphi'}_{k}^{(\pi')} ({\rho}'^{c'_{\pi' a'}})}.$$ Hence from (1) and by [@NT], Theorem 5.4.11 for basic sets (see [@Navarro], Lemma 7.5) $$\sum_{i= 1}^{k(b)}\varepsilon_i \chi_i (a\rho) \overline{\chi'_i(a'{\rho}')}$$ $$= \sum_{\pi \in \Pi_a \cap \Pi'_{a'}}
\sum_{k = 1}^{l(b_{\pi})}\left( \sum_{j= 1}^{l(b_{\pi})} 
 \left( \sum_{i = 1}^{k(b)} d^{\pi}_{\chi_i \varphi_j^{(\pi)}}
\overline{d^{\pi}_{\chi_i  \varphi_k^{(\pi)}}}\right) \varphi_j^{(\pi)}(\rho^{c_{\pi a}}) \right)\overline{ {\varphi'}^{(\pi)}_{k}(\rho'^{c'_{\pi a'}})}$$ $$= \sum_{\pi \in \Pi_a \cap \Pi'_{a'}}
\sum_{k= 1}^{l(b_{\pi})} \left( \sum_{j = 1}^{l(b_{\pi})} 
c_{\varphi_{j}^{(\pi)}\varphi_{k}^{(\pi)} } \varphi_{j}^{(\pi)}(\rho^{c_{\pi a}})
\right)\overline{\varphi'^{(\pi)}_{k}(\rho'^{c'_{\pi a'}})}\big),$$ here $c_{\varphi_j^{(\pi)} \varphi_k^{(\pi)}}$ is a Cartan integer [w.r.t.]{} the basic set $Bs(b_{\pi})$ in (iv). Thus $\displaystyle\sum_{i= 1}^{k(b)}\varepsilon_i \chi_i (a\rho)\overline{\chi'_i(a'{\rho}')}$ is divisible by $|C_{G}(a\rho)|$ in ${\cal O}$, because $\displaystyle \sum_{j = 1}^{l(b_{\pi})} 
c_{\varphi_j^{(\pi)} \varphi_k^{(\pi)}} \varphi_j^{(\pi)}(\rho^{c_{\pi a}}) $ is divisible by $|C_{G}(\pi \rho^{c_{\pi a}})| = |C_{G}(a\rho)|$ in ${\cal O}$. Similarly $\displaystyle\sum_{i= 1}^{k(b)}\varepsilon_i \chi_i (a\rho)\overline{\chi'_i(a'{\rho}')}$ is by $|C_{G'}(a'\rho')|$ in ${\cal O}$. This completes the proof. $\blacksquare$

Let $P$ be a Sylow $p$-subgroup of $G$ and denote by ${\cal F}_{P}(G)$ the Frobenius category of $G$. The objects of ${\cal F}_{P}(G)$ are subgroups of $P$. For $S, \ T \leq P$, $${\rm mor}_{{\cal F}_{P}(G)}(S, \ T) = \{ \varphi \in {\rm Hom}(S, \ T) \ | \ \varphi = {c_{x}}_{|_S} \ ( \exists  x \in G) \},$$ where $c_x$ is the inner automorphism of $G$ induced by $x$.

[([@Okuyama-Watanabe])]{} If $\tilde{P}$ is abelian, then ${\cal F}_{P}(G) = {\cal F}_{P}(N_{G}(\tilde{P})).$

(Naoki Chigira) $\tilde{P} = P \cap O^{p}(G)$ is strongly closed in $P$ in the sense of [@GLS], §16. Hence the proposition follows from [@GLS], Proposition 16.20. $\blacksquare$

[([@Wata2014], Theorem 3)]{} If $\tilde{P}$ is cyclic, then ${\cal F}_{P}(G) = {\cal F}_{P}(N_{G}(P)).$

Set $\tilde{K} = O_{p'}(C_{{G}}(\tilde{P})) = O_{p'}(C_{\tilde{G}}(\tilde{P})) = C_{\tilde{G}}(\tilde{P})_{p'}$. Then $C_{\tilde{G}}(\tilde{P}) =  \tilde{K} \times \tilde{P}$, and $\tilde{K}$ is normal in $N_{G}(\tilde{P})$. Since $C_{P}(\tilde{P})$ is a Sylow $p$-subgroup of $C_{G}(\tilde{P})$, $C_{G}(\tilde{P}) = C_{\tilde{G}}(\tilde{P}) C_{P}(\tilde{P})  = \tilde{K} C_{P}(\tilde{P})$, by Proposition 1, $$N_{G}(\tilde{P}) = C_{G}(\tilde{P})N_{G}(P) = \tilde{K} N_{G}({P}).$$ This implies $${\cal F}_{P}(N_{G}(\tilde{P})) = {\cal F}_{P}(N_{G}(P)).$$ Note that $[P, K]  \cap P = 1$. This and Proposition 1 complete the proof. $\blacksquare$

The above two propositions are generalized in [@Wata2014]) as follows. Let $b$ be a block (not necessarily the principal block) of $G$ with a maximal $b$-Brauer pair $(P, b_P)$ and a hyperfocal subgroup $\tilde{P}$. Let $b' \in Bl(N_{G}(P))$ be the Brauer correspondent of $b$. If $\tilde{P}$ is cyclic, then the Brauer categories ${\cal F}_{(P, b_P)}(G, b)$ and $ {\cal F}_{(P, b_P)}(N_{G}(P),  b')$ are same.

Set $K = O_{p'}(C_{G}(P))$ and let ${T}$ be a $p$-complement of $N_{G}(P)$. Then $K = T \cap C_{G}(P)$. From (2), $\tilde{T} : = \tilde{K}T$ is a $p$-complement of $N_{\tilde{G}}(\tilde{P})$. Then $$\tilde{T}/\tilde{K} \cong T/K$$ because $T \cap C_{G}(\tilde{P}) \subseteq  C_{G}(P)$.

In the rest of this paper $G$ is a finite group and $P$ is a Sylow $p$-subgroup of $G$, $\tilde{G} = O^{p}(G), \tilde{P} = P \cap O^{p}(G)$ and $N =  N_{G}(P)$. Moreover $e = |E|$, the inertial index of $b(G)$. We assume that $\tilde{P}$ is a cyclic group generated by $x$ and of order $p^{n}$. From (3), $e$ is the inertial index of $b(\tilde{G})$ and hence $e$ divides $p-1$.

If $e = 1$, then $G$ is $p$-nilpotent. Hence we may assume $e \neq 1$ and hence $p \neq 2$ in a proof of Theorem 1. For a $p$-element $u \in G$, we denote by $e_{u}$ the inertial index of $b(C_{G}(u))$ and we call it the inertial index of $u$.

Irreducible characters in $b(G)$
================================

Under our assumption, let $u \in P$. Then $P_u = C_{P}(u)$ is a Sylow $p$-subgroup of $C_{G}(u)$ and $e_u = |C_{N}(u) : P_u C_{N}(P_u)|$.

There exists $a \in G$ such that $P \cap C_{G}(u^a)$ is a Sylow $p$-subgroup of $C_{G}(u^a)$. Then $u^a \in P \cap C_{G}(u^a) \subseteq P$. By Proposition 2, there exists $n \in N$ such that $u^a = u^n$. Since $P \cap C_{G}(u)^n$ is a Sylow $p$-subgroup of $C_{G}(u)^n $, $P_u = P \cap C_{G}(u) = P^{n^{-1}} \cap C_{G}(u) $ is a Sylow $p$-subgroup of $C_{G}(u)$.

By Proposition 2, we have $N_{G}(P_u) = C_{G}(P_u)(N_{G}(P_u) \cap N)$, hence $$\begin{aligned}
C_{G}(u) \cap N_{G}(P_u) &=& C_{G}(P_u)(C_{G}(u) \cap N_{G}(P_u) \cap N)\\
 &=&  C_{G}(P_u)(C_{G}(u) \cap N). \end{aligned}$$ On the other hand we have $C_{G}(u) \cap C_{G}(P_u)P_u = C_{G}(P_u)P_u$. Therefore $$\begin{aligned}
 e_u &=& |C_{G}(u) \cap N_{G}(P_u) : (C_{G}(u) \cap C_{G}(P_u))P_u|\\
&=& |C_{G}(P_u)(C_{G}(u) \cap N) :  C_{G}(P_u)P_u|\\
&=&|C_{G}(u) \cap N : C_{G}(u) \cap N \cap C_{G}(P_u)P_u| \\
&=& |C_{N}(u) :  N \cap C_{G}(P_u)P_u| \\
&=& |C_{N}(u) : C_{N}(P_u)P_u|. \end{aligned}$$$\blacksquare$

The following is special case of [@Wata2014] Lemma 4 (i). For the self-containedness we state it with a proof because this is crucial.

[([@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 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 $|u^{N}| = e |u^{P}|$.) $\blacksquare$

\
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 the second and third main theorem on principal blocks, if $\chi \in {\rm Irr}(b(G))$ then $$\chi * \eta =  \sum_{u \in \Pi }\eta(u)\chi^{(u, b(C_{G}(u)))}$$ is a generalized character belonging to $b(G)$ (c.f. [@BP]).

Let $\tilde{{\cal X}}$ be a set of representatives for the $E$-conjugacy classes of non-trivial linear characters of $\tilde{P}$. For each $\mu \in \tilde{{\cal X}}$, we set $$\eta_{\mu} = 
\sum_{a \in E}\mu^a = \sum_{a \in N_{\tilde{G}}(\tilde{P})/C_{\tilde{G}}
(\tilde{P})}\mu^a.$$ This is $\tilde{G}$-stable.

Suppose that a finite group $G$ has a cyclic Sylow $p$-subgroup $P$. Then the trivial character $1_G$ is not an exceptional character.

We use [@Dor], Theorem 68.1 (cf. [@D]). Set $e = |N_{G}(P): C_{G}(P)|$ and $m  = \frac{|P| -1 }{e}$. Then $e | (p - 1)$. If $m = 1$, then any irreducible character in $b(G)$ is not exceptional. So we may assume $m \neq 1$. Assume that $1_{{G}}$ is exceptional. For some non-trivial linear character $\mu$ of $P$, $$1 = 1_G(\pi) = \varepsilon \sum_{x \in N_{G}(P)/C_{G}(P)} {\mu}(\pi^x) \ (\forall \pi \in {P} \backslash \{ 1 \}),$$ where $\varepsilon = \pm 1$. Then we have $\varepsilon m = \sum_{\pi \in {P} \backslash \{ 1 \}} \mu(\pi) = -1$. This is a contradiction. $\blacksquare$

\

[(Dade [@D])]{} ${\rm Irr}(b(\tilde{G})) = \{  \tilde{\chi}_1 = 1_{\tilde{G}},\tilde{\chi}_2, 
\cdots, \tilde{\chi}_e, \tilde{\chi}_{\mu} \ ( \mu \in \tilde{{\cal X}})\},$ where $$\left\{ \begin{array}{l}
\tilde{\chi}_j (u\rho) = \varepsilon_j = \pm 1 \ ( 1 \leq j \leq e),   \\ 
\tilde{\chi}_{\mu}(u \rho) = \varepsilon \eta_{\mu}(u) \ (\varepsilon = \pm 1)\end{array} \right.$$ for $u (\neq 1) \in \tilde{P}$ and $ \rho \in (C_{\tilde{G}}(u))_{p'}$. Moreover if $P$ is normal in $G$, then $\varepsilon_j = 1$ and $\varepsilon = 1$.

Since $b(\tilde{G})$ has a cyclic defect group $\tilde{P}$, the lemma follows by Lemma 4 and [@Dor], Theorem 68.1. $\blacksquare$

For $\mu \in \tilde{{\cal X}}$, we have $1_{\tilde{G}} * \eta_{\mu} = (e -1)1_{\tilde{G}} - \displaystyle\sum_{i = 2}^{e}\varepsilon_i \tilde{\chi}_i + \varepsilon \tilde{\chi}_{\mu}. $

Let $u \in \tilde{P}\backslash \{ 1\}$ and $\rho \in C_{\tilde{G}}(u)_{p'}$. &gt;From (4), we have $$(1_{\tilde{G}} * \eta_{\mu})(u\rho) = \eta_{\mu}(u),$$ $$((e -1)1_{\tilde{G}} - \sum_{i = 2}^{e}
\varepsilon_i \tilde{\chi}_i + \varepsilon \tilde{\chi}_{\mu})(u\rho) = \eta_{\mu}(u).$$ Therefore $1_{\tilde{G}} * \eta_{\mu} = (e -1)1_{\tilde{G}} - \displaystyle\sum_{i = 2}^{e}\varepsilon_i \tilde{\chi}_i + \varepsilon \tilde{\chi}_{\mu}$ on $p$-singular elements of $\tilde{G}$.

On the other hand for $\tau \in \tilde{G}_{p'}$, by [@NT], Theorem 5.4.5, the equation (4) and [@Dor], Theorem 68.1, (6), $$0 = \sum_{i =1}^{e}\tilde{\chi}_{i}(u)\tilde{\chi}_{i}(\tau) + \sum_{\mu \in \tilde{{\cal X}}}\tilde{\chi}_{\mu}(u)\tilde{\chi}_{\mu}(\tau)$$$$= 1 + \sum_{i =  2}^{e}\varepsilon_i \tilde{\chi}_i(\tau) +  
 \big( \varepsilon \sum_{\nu \in \tilde{{\cal X}}}\eta_{\nu}(u)\big)\tilde{\chi}_{\mu}(\tau)$$ $$=  1 + \sum_{i =  2}^{e}\varepsilon_i \tilde{\chi}_i(\tau) -  
\varepsilon \tilde{\chi}_{\mu}(\tau).$$ From this $$(1 * \eta_{\mu})(\tau) = e = \big((e-1)1_{\tilde{G}} - \sum_{i = 2}^{e}\varepsilon_i \tilde{\chi}_i (\tau) + \varepsilon\tilde{\chi}_{\mu}\big)(\tau).$$ This completes the proof. $\blacksquare$

\
We denote by Irr$_{P}(\tilde{P})$ the set of $P$-invariant linear characters of $\tilde{P}$. Note that if $\nu \in {\rm Irr}_{P}(\tilde{P})$ then $\nu^{a} \in {\rm Irr}_{P}(\tilde{P})$ for $a \in N$. Let $\nu \in $Irr$_{P}(\tilde{P})$. Then $\nu$ is trivial on $[\tilde{P}, P]$. Since $P/[\tilde{P}, P] = (\tilde{P}/[\tilde{P}, P]) \times (C_{P}(T)[\tilde{P}, P]/[\tilde{P}, P])$ by Lemma 2, there is a unique extension $\hat{\nu}$ of $\nu$ to $P$ such that $C_{P}(T) \subseteq {\rm Ker}\  \hat{\nu}$, where $T$ is a $p$-complement of $N$. In fact $C_{P}(U) \subseteq {\rm Ker}\  \hat{\nu}$ for any $p$-complement $U$ of $N$. We call $\hat{\nu}$ the $canonical$ $extension$ of $\nu$. For $a \in N$, $({\hat{\nu}})^a = \widehat{\nu^{a}}$. We set $$\eta_{\hat{\nu}} = \sum_{a \in E}\hat{\nu}^a.$$ This is $N$-invariant, and hence $\eta_{\hat{\nu}}$ is $G$-stable by Proposition 2. Of course $\eta_{\hat{\nu}}\downarrow_{\tilde{P}} = \eta_{\nu}$. Also note $\eta_{\hat{\nu}}(u) = e$ if $u \in C_{P}(T)$.

Let $\nu \in {\rm Irr}_{P}(\tilde{P})$. Under the above notations, $$1_{{G}} * \eta_{\hat{\nu}} = (e - 1)1_{{G}} - \sum_{i = 2}^{e}\varepsilon_i {\chi}_i + \varepsilon {\chi}_{\nu}, \ (\forall \nu \in {\rm Irr}_P(\tilde{P}))$$ where $\chi_i$(resp. $\chi_{\nu}$) is an extension of $\tilde{\chi}_i$ (resp. $\tilde{\chi}_{\nu})$ to $G$.

Since $b(\tilde{G})$ has a cyclic defect group $\tilde{P}$ and since $C_{\tilde{G}}(u)$ is $p$-nilpotent for any $u \in \tilde{P} \backslash \{ 1 \}$, $$(*) \ \ \ (1_{\tilde{G}}^{(1, b(\tilde{G}))}, 1_{\tilde{G}}^{(1, b(\tilde{G}))})  = 1 - \sum_{u}(1_{C_{\tilde{G}}(u)}^{(u, b(C_{\tilde{G}}(u))}, 1_{C_{\tilde{G}}(u)}^{(u, b(C_{\tilde{G}}(u))})$$ $$= 1 - \frac{p^{n} -1}{e}\frac{1}{p^n} = \frac{(e-1)p^n + 1}{ep^{n}},$$ where $u$ runs over a set of representatives for the conjugacy classes of the $p$-elements of $\tilde{G}$. Hence we have $$(1_G^{(1, b(G))}, 1_G^{(1, b(G))}) = (1_G^{(1, b(\tilde{G}))}, 1_G^{(1, b(\tilde{G}))})\frac{1}{|G:\tilde{G}|} = \frac{(e-1)p^n + 1}{e|P|}.$$ Let $\Pi$ be a set of representatives for the $N$-conjugacy classes of $P \backslash \{ 1 \}$. By Proposition 2, $\Pi$ is a set of representatives for the conjugacy classes of $p$-elements of $G$. Let $\Pi_1 = \{ u \in \Pi \ | \  e_u = 1 \}$ and $\Pi_2 = \Pi \backslash \Pi_1 =  \{ u \in \Pi \ | \ e_u = e \}$. Set $b_u = b(C_{G}(u))$ for $u \in \Pi$ and $U = \bigcup_{u \in \Pi_2} u^P$. By Lemmas 1 and 2, $$\sum_{u \in \Pi_1}(1_{G}^{(u, b_u)}, 1_{G}^{(u, b_u)})  =  \sum_{u \in \Pi_1}|C_{P}(u)|^{-1}$$ $$= \frac{1}{e|P|} \sum_{u \in \Pi_1} e |P: C_{P}(u)| = \frac{1}{e|P|}\big(|P| - |U| - 1).$$ Hence we have $$(**) \ \ \ \ \ \ \sum_{u \in \Pi_2}(1_{G}^{(u, b_u)}, 1_{G}^{(u, b_u)})$$ $$= 1 - (1_{G}^{(1, b(G))}, 1_{G}^{(1, b(G))}) 
-  \sum_{u \in \Pi_1}(1_{G}^{(u, b_u)}, 1_{G}^{(u, b_u)})$$ $$=  \frac{(e - 1)|P| - (e - 1)p^n  + |U|}{e|P|}.$$ Since $(\eta_{\hat{\nu}}, 1_{P}) = 0$, $$e + \sum_{u \in \Pi_1}\eta_{\hat{\nu}}(u) e |P :C_{P}(u)| + \sum_{u \in \Pi_2}\eta_{\hat{\nu}}(u) |P : C_{P}(u)| = 0,$$ and hence $$(***) \ \ \ \ \ \ \  \sum_{u \in \Pi_1}\eta_{\hat{\nu}}(u) |C_{P}(u)|^{-1}$$ $$= -|P|^{-1} - \sum_{u \in \Pi_2} |C_{P}(u)|^{-1}   = -|P|^{-1} - |P|^{-1}|U|$$ because $\eta_{\hat{\nu}}(u) = e$ for $u \in \Pi_2$. From $(*), (**)$ and $(***)$ $$( 1_G * \eta_{\hat{\nu}}, 1_G)$$ $$= e(1_{G}^{(1, b(G))}, 1_{G}^{(1, b(G))})  +  \sum_{u \in \Pi_1}\eta_{\hat{\nu}}(u)(1_{G}^{(u, b_u)}, 1_{G}^{(u, b_u)}) +  e\sum_{u \in \Pi_2}(1_{G}^{(u, b_u)}, 1_{G}^{(u, b_u)})$$ $$= \frac{(e - 1)p^n + 1}{|P|}  + (- |P|^{-1} - |P|^{-1}|U|) + \frac{(e - 1)|P| - (e - 1)p^n + |U|}{{|P|}}$$ $$= e - 1.$$ From this we have $$(1_G * \eta_{\hat{\nu}}, 1_G * \eta_{\hat{\nu}} ) =  
(1_G * \eta_{\hat{\nu}}\eta_{\hat{\nu}^{-1}}, 
1_G) =  e + (e-1)^2 = e^2 - e + 1.$$ Moreover, for $\mu, \mu' (\neq) \in {\cal M}$, $$(1_G * \eta_{\hat{\mu'}}, 1_G * \eta_{\hat{\mu}} ) =  (1_G * \eta_{\hat{\mu'}}\eta_{\hat{\mu}^{-1}}, 1_G) =  e(e-1).$$ By [@Dor], Theorem 68.1, $|{\rm IBr}(b(\tilde{G}))| = e < p$. Hence any irreducible Brauer character of $b(\tilde{G})$ is $G$-invariant, hence any $\tilde{\chi}_i$ are $G$-invariant by Lemma 5. On the other hand we see $(1_{G} * \eta_{\hat{\nu}})\downarrow_{\tilde{G}} =  1_{\tilde{G}} * \eta_{{\nu}}$ (cf. Cabanes [@Cabanes], Theorem 1), hence Lemma 6 and the above imply (5). This completes the proof. $\blacksquare$

\
We call $\chi_{i}$ (resp. ${\chi}_{\nu})$ in the above proposition the $canonical$ $extension$ of $\tilde{\chi}_i$ (resp. $\tilde{\chi}_{\nu})$. We set $\chi_1 = 1_{G}$ and call it the canonical extension of $\tilde{\chi}_1 = 1_{\tilde{G}}$. Since $\tilde{\chi}_i$ is $p$-rational from (4), $\chi_i$ is also $p$-rational by Proposition 3 and the uniqueness of $\chi_i$.

Let $G_1$ be a subgroup of $G$ containing $\tilde{G}$ and let $P_1 = P \cap G_1 = \tilde{P} \rtimes (C_{P}(T) \cap G_1)$. Then $P_1$ is a Sylow $p$-subgroup of $G_1$, $\tilde{P}$ is a $p$-hyperfocal subgroup of $G_1$ and $G_1$ also has the inertial index $e$. Let $T$ be a $p$-complement of $N$. Then $P_1$ is normalized by $T$. From (3) $$N_{G_1}(P_1) = O_{p'}(C_{{G_1}}(P_1))TP_1.$$ That is, $O_{p'}(C_{G_1}(P_1))T$ is a $p$-complement of $N_{G_1}(P_1)$ because $T \cap C_{G_1}(P_1) = O_{p'}(C_{G}(P))$. Moreover, for $\nu \in {\rm Irr}_{P}(\tilde{P})$, $$1_{G_1}* \eta_{(\hat{\nu}\downarrow_{P_{1}})} = (1_{G}* \eta_{\hat{\nu}})\downarrow_{G_1}.$$ Therefore the canonical extension of $\tilde{\chi}_i$ (resp. $\tilde{\chi}_{\nu}$) to $G_1$ coincides with $\chi_{i}\!\! \downarrow_{G_{1}}$ (resp. $\chi_{\nu}\!\! \downarrow_{G_{1}}$).

Let $\nu \in $ [Irr]{}$(\tilde{P})$. We denote by $P_{\nu}$ the stabilizer of $\nu$ in $P$ and set $G_{\nu} = \tilde{G}P_{\nu}$. Since $(\eta_{\nu})^v = \eta_{\nu^v}$ and hence $(1_{\tilde{G}}* \eta_{\nu} )^v = 1_{\tilde{G}}* \eta_{\nu^v}$ for $v \in P$, the equation (5) implies $$(\tilde{\chi}_{\nu})^v = \tilde{\chi}_{\nu^v}.$$ Hence $G_{\nu}$ is the stabilizer of $\tilde{\chi}_{\nu}$ in $G$ because $G = \tilde{G}P$. Since $C_{P}(\tilde{P}) \subseteq P_{\nu}$ and $P/C_{P}(\tilde{P})$ is a cyclic $p$-group isomorphic to a subgroup of Aut$(\tilde{P})$, $P_{\nu}$ is normalized by $N$ and hence $G_{\nu}$ is normal in $G$. Let $\hat{\nu}$ be the canonical extension of $\nu$ to $P_{\nu}$ and let $\chi_{\nu}$ the canonical extension of $\tilde{\chi}_{\nu}$ to $G_{\nu}$. For $\lambda \in {\rm Irr}(P_{\nu}/\tilde{P}$) and for $t \in N_{p'}$, $[P_{\nu}, t] \subseteq [P, t] \subseteq \tilde{P}$ and hence $\lambda^t = \lambda$. Therefore $(\eta_{\hat{\nu}}\lambda)\!\uparrow^{P}_{P_{\nu}}$ is $N$-invariant, and hence $G$-stable. Moreover we note that Irr$(P_{\nu}/\tilde{P})$ can be identified with Irr$(G_{\nu}/\tilde{G})$ through the isomorphism $P_{\nu}/\tilde{P} \cong G_{\nu}/\tilde{G}$.

Let $\nu \in $ [Irr]{}$(\tilde{P})$. With the above notations, $ (1_{G_{\nu}} * \eta_{\hat{\nu}})\lambda = 1_{G_{\nu}} * \eta_{\hat{\nu}}\lambda = (e - 1)\lambda - \displaystyle\sum_{i = 2}^{e}\varepsilon_i ({{\chi}_i}\downarrow_{G_{\nu}})\lambda + \ \varepsilon {\chi}_{\nu} \lambda, $ and hence $$1_{G} * ((\eta_{\hat{\nu}} \lambda )\!\uparrow^P_{P_{\nu}}) = (e - 1) \lambda\!\uparrow_{G_{\nu}}^{G} -\sum_{i = 2}^{e}\varepsilon_i (({{\chi}_i \downarrow_{G_{\nu}})\lambda})\!\uparrow_{G_{\nu}}^{G} + \ \varepsilon ({{\chi}_{\nu}
\lambda})\!\uparrow_{G_{\nu}}^{G}$$ for any $\lambda \in $[Irr]{}$(P_{\nu}/\tilde{P}) = $[Irr]{}$(G_{\nu}/\tilde{G})$.

It suffices to show $1_{G} * ((\eta_{\hat{\nu}} \lambda )\!\uparrow^P_{P_{\nu}}) = (1_{G_{\nu}} * (\eta_{\hat{\nu}} \lambda)) \uparrow^G_{G_{\nu}}$. Let $P = \bigcup_{j = 1}^{l} v_j P_{\nu}$ (disjoint) and let $u \in P_{\nu}$. Since $G = \bigcup_{j = 1}^{l}v_j  G_{\nu}$ (disjoint) and $P_{\nu}$ is normal in $P$, for $\rho \in C_{G}(u)_{p'}$, $$(1_{G_{\nu}} * (\eta_{\hat{\nu}} \lambda))\uparrow^G_{G_{\nu}}\!(u\rho) = \sum_{j = 1}^{l} (1_{G_{\nu}} * (\eta_{\hat{\nu}} \lambda))((u\rho)^{v_j})$$ $$= \sum_{j = 1}^{l}( \eta_{\hat{\nu}} \lambda)(u^{v_j}) =  
((\eta_{\hat{\nu}} \lambda )\!\uparrow^P_{P_{\nu}})(u) = (1_{G} * ((\eta_{\hat{\nu}} \lambda)\!\uparrow^P_{P_{\nu}}))(u\rho).$$ This completes the proof. $\blacksquare$

\
For $\nu,$ $ \nu' \in \tilde{{\cal X}}$, $\tilde{\chi}_{\nu}$ and $\tilde{\chi}_{\nu'}$ are $G$-conjugate if and only if those are $P$-conjugate if and only if $\nu $ and $\nu'$ are $(P\rtimes E)$-conjugate. Since $b(G)$ is the unique block of $G$ which covers $b(\tilde{G})$, we have the following.

Let ${\cal X} (\subseteq \tilde{{\cal X}})$ be a set of representatives for the $(P \rtimes E)$-conjugacy classes of [Irr]{} $(\tilde{P}) \backslash \{ 1_{\tilde{P}} \}$. $${\rm Irr}(b(G)) = \bigcup_{i = 1}^e \Big\{ \chi_i \lambda \ | \  \lambda \in {\rm Irr}(G/\tilde{G}) \Big\}
\bigcup_{\nu \in {\cal X}} \Big\{ (\chi_{\nu} \lambda)\!\uparrow_{G_{\nu}}^{G} \ | \ \lambda \in {\rm Irr}(G_{\nu}/\tilde{G}) \Big\}.$$

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 $\bar{P} = P/\tilde{P}$ and let $P_{\nu}$ be the stabilizer of $\nu$ in $P$ for $\nu \in {\cal X}$. Moreover we set $\bar{P}_{\nu} = P_{\nu}/\tilde{P}$ and $x_i = \chi_i(u)$. We recall $x_i$ is a rational integer. From (7), it suffices to show in order to establish the proposition $$x_i = \varepsilon_i \ (1 \leq i \leq e).$$ So we can assume that $P = \langle \tilde{P}, u \rangle$, hence $$G = \langle \tilde{G}, u \rangle.$$ From (7) again and [@NT], Theorem 5.4.11, $$\sum_{\lambda \in {\rm Irr}(\bar{P})}|\lambda(u)|^2  + \sum_{i = 2}^{e} \sum_{\lambda \in {\rm Irr}(\bar{P})}{x_i}^2|\lambda(u)|^2$$ $$+ \sum_{\nu}\sum_{\lambda_{\nu} \in {\rm Irr}(\bar{P}_{\nu})}\left|((\eta_{\hat{\nu}}\lambda_{\nu})\!\uparrow_{P_{\nu}}^P\big)(u) - (e -1)(\lambda_{\nu}\!\uparrow^{P}_{P_{\nu}})(u) + \sum_{i = 2}^{e}\varepsilon_{i} x_i (\lambda_{\nu}\!\uparrow^{P}_{P_{\nu}})(u)\right|^2$$ $$= |C_{P}(u)|$$ where $\nu$ runs over a set $\{ \nu \in {\cal X} \ | \ P_{\nu} =  P \}$. Hence $$1+ \sum_{i = 2}^{e} {x_i}^2 + \sum_{ \nu }\left|\eta_{\hat{\nu}}(u) - (e -1) + \sum_{i = 2}^{e}\varepsilon_{i} x_i \right|^2  = |C_{\tilde{P}}(u)|$$ where $\nu$ runs over $\{ \nu \in {\cal X} \ |\  P_{\nu} = P \}$. Set $p^{n'}  = |C_{\tilde{P}}(u)|$. By Brauer’s permutation lemma ([@NT], Lemma 3.2.18, (i)), the numbers of $P$-invariant linear characters of $\tilde{P}$ is equal to $|C_{\tilde{P}}(u)|$. Let ${\cal X}_1 = \{ \nu \in {\cal X} \ | \ \nu \in {\rm Irr}_{P}(\tilde{P}) \}$. 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$ ($\tilde{u} \in \tilde{P}$, $r \in C_{P}(E)$). 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. $\blacksquare$

\
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 matrix of generalized decomposition numbers of $b(G)$ with respect to a suitable basic set $\{ \varphi_1^{(u)} = 1_{C_{G}(u)}, \varphi_2^{(u)}, \cdots, \varphi_e^{(u)} \}$ for $b_u$ is of the form : $$\left.\begin{array}{|c|cccc|c}
\hline
 & \varphi_1^{(u)} = 1_{C_{G}(u)}&\varphi_2^{(u)} &\cdots &\varphi_e^{(u)} \\
\hline
\chi_{1}\lambda_1&\varepsilon_{1}\lambda_1(u)& 0&\cdots &0  \\
\chi_{2}\lambda_2&0&\varepsilon_{2}\lambda_2(u)& 0& 0\\
 \vdots&\vdots &\vdots &\ddots & \vdots  \\
\chi_{e}\lambda_e& 0& 0&\cdots&\ \varepsilon_e \lambda_e(u) \\ 
(\chi_{\nu }\lambda_{\nu})\!\uparrow^G_{G_{\nu}}&
\varepsilon ({\lambda_{\nu}}\!\uparrow^{P}_{P_{\nu}})(u)&\varepsilon ({\lambda_{\nu}}\!\uparrow_{P_{\nu}}^{P})(u) &\cdots &\varepsilon ({\lambda_{\nu}}\!\uparrow^{P}_{P_{\nu}})(u) \\
\hline
\end{array} \right.$$ where $\lambda_i \in {\rm Irr}(P/\tilde{P}) = {\rm Irr}(G/\tilde{G})$, $\nu \in {\cal X}$ and $\lambda_{\nu} \in {\rm Irr}(P_{\nu}/\tilde{P}) = {\rm Irr}(G_{\nu}/\tilde{G})$.

By Lemmas 2 and 3, $P_u = C_{P}(u)$ is a Sylow $p$-subgroup of $C_{G}(u)$, $u \in C_{P}(T)$ for some $p$-complement $T$ of $N$ and $C_{\tilde{P}}(u)$ is a $p$-hyperfocal subgroup of $C_{G}(u)$. Moreover $l(b_u) = e$ as is stated in the above. Set $p^{n'} = |C_{\tilde{P}}(u)|$ and $m = \frac{p^{n'} -1}{e}$. Now for $\nu \in {\cal X}$, the stabilizer $P_{\nu}$ of $\nu$ in $P$ is normalized by $N$. If $u \not\in  P_{\nu}$, $\eta_{\nu}\!\!\uparrow^P_{P_{\nu}}\!\!(u) = 0$ and $\chi_{\nu}\!\!\uparrow^G_{G_{\nu}}\!\!(u \rho) = 0$ for any $\rho \in C_{G}(u)_{p'}$. Recall that if $u \in P_{\nu}$, then $\eta_{\hat{\nu}}(u) = e$ (see (6) for $G_1 = G_{\nu}$). So from (7) in Proposition 4, in order to establish the proposition, it suffices to show for any $\chi_i$ $$d_{{\chi_i, \varphi_{j}^{(u)}}}^u = \delta_{ij}\varepsilon_i$$ for a suitable basic set $\{ \varphi_{j}^{(u)} \ | \ 1\leq j \leq e \}$ for $b_u$, where $\varphi_1 ^{(u)} = 1_{C_{G}(u)}$. Hence we may assume $$G = \langle \tilde{G}, u \rangle  = \tilde{G} \rtimes \langle u \rangle.$$ (cf. Lemma 2) Set $\bar{C}_{G}(u) = C_{G}(u)/\langle u \rangle = 
(C_{\tilde{G}}(u) \times \langle u \rangle )/\langle u \rangle \cong C_{\tilde{G}}(u)$, $\bar{b}_u = b(\bar{C}_{G}(u))$ and $\overline{C_{P}(u)}  = C_{P}(u)/\langle u \rangle  \cong C_{\tilde{P}}(u)$. Then $\bar{b}_u$ has a cyclic defect group $\bar{P}$, and $\bar{b}_u$ has an inertial index $e$. By [@Broue], Theorem 5.3, there is a perfect isometry from $\bar{b}_u$ to $b(N_{\bar{C}_{{G}}(u)}(\bar{P}))$ which maps $1_{\bar{C}_{G}(u)}$ to $1_{N_{\bar{C}_{G}(u)}(\bar{P})}$. Hence by [@Broue], Theorem 1.5, there exists a basic set $Bs(\bar{b}_u) = \{ \varphi_{j}^{(u)}\ | \ 1\leq j \leq e \}$, where $\varphi_{1}^{(u)} = 1_{\bar{C}_{G}(u)}$, such that the Cartan matrix of $\bar{b}_u$ with respect to $Bs(\bar{b}_u)$ is of the form $${C}^{\bar{b}_u} = \left( \begin{array}{cccc}
{m + 1}&m & \cdots & m\\
m & m+1 &  \cdots &m \\
\vdots&\vdots&\vdots&\vdots\\
m &m & \cdots& m + 1
\end{array} \right)_{e \times e}.$$ Let $C^{b_u}$ be the Cartan matrix of $b_u$ with respect to $Bs(\bar{b}_u)$ which is regarded as a basic set for $b_u$. We have $C^{b_u} = |\langle u \rangle |C^{\bar{b}_u}$ by [@NT], Theorem 5.8.11. Let $$D = \Big( {d}_{\chi {\varphi}_{j}^{(u)}}^{u}\Big)_
{\chi \in {\rm Irr}(b(G)), \varphi_{j}^{(u)} \in Bs(\bar{b}_u)}$$ be the matrix of generalized decomposition numbers of $b(G)$ with respect to $Bs(\bar{b}_u)$. Then we have $$C^{b_u} = {^{t}D\overline{D}}.$$

As in the proof of Proposition 6, let ${\cal X}_1 = \{ \nu \in {\cal X} \ | \ \nu \in {\rm Irr}_{P}(\tilde{P}) \}$. We have $|{\cal X}_1| = m$. Let $\nu \in {\cal X}_1$. Recalling $\eta_{\hat{\nu}}(u) = e$, from (7), $$ed_{\chi_1 \varphi_j^{(u)}}^u = (e - 1) d_{\chi_1 \varphi_j^{(u)}}^u - \sum_{k = 2}^{e} \varepsilon_k d_{\chi_k \varphi_j^{(u)}}^u + \varepsilon d_{\chi_{\nu} \varphi_{j}^{(u)}}^{u},$$ $$\therefore \ \ \ \varepsilon d_{\chi_{\nu} \varphi_j^{(u)}}^u = \sum_{k = 1}^{e}\varepsilon_k d_{\chi_k \varphi_j^{(u)}}^u$$ for $j = 1, 2, \cdots, e$. Note that these are all integers. Since $0 \not\equiv \chi_{\nu}(1) \equiv \chi_{\nu}(u) \ ({\rm mod} \ {\cal P})$ where ${\cal P}$ is the maximal ideal of ${\cal O}$, $\chi_{\nu}(u) \neq 0$. In particular $d^{u}_{\chi_{\nu}\varphi_{j}^{(u)}} \neq 0$ for some $j$. The same holds for $\chi_i$. Now set $$X_{kj}  = \varepsilon_k d_{\chi_k \varphi_j^{(u)}}^u \ ( 1 \leq k, j \leq e).$$ Then $X_{11} = 1$, $X_{1j} = 0 \ \ (2 \leq j \leq e)$. Moreover $\varepsilon d_{\chi_{\nu} \varphi_j^{(u)}}^u = \sum_{k = 1}^{e}X_{kj}$ for $\nu \in {\cal X}_1$. So we can see from (11), $$\sum_{k = 1}^{e}X_{ki}X_{kj} + m\big(\sum_{k= 1}^{e}X_{ki}\big) \big(\sum_{k= 1}^{e}X_{kj }\big) = \left\{ \begin{array}{c}m + 1 \ (i = j) \\
m \  (i \neq j).  
\end{array} \right.$$ Therefore if $m \geq 2$, then $X_{ij} = \delta_{ij}$ by rearranging $\varphi^{(u)}_j$ $(j \geq 2)$. But if $m = 1$, by taking another suitable basic set $\{ {\varphi'}_{j}^{(u)} \ | \ 1\leq j \leq e \}$ for $b_u$ where ${\varphi'}_{1}^{(u)} = 1_{C_{G}(u)}$ if necessary, $\varepsilon_id_{\chi_i {\varphi'}_j^{(u)}}^u = \delta_{ij}$ with respect to $\{ {\varphi'}_{j}^{(u)} \ | \ 1\leq j \leq e \}$. Thus we get (10). This completes the proof. $\blacksquare$

\
[**Proof of Theorem 1.**]{} It is easily seen that $\tilde{P}$ is a hyperfocal subgroup of $N$ and $N_{G}(\tilde{P})$ by Proposition 2 and Proposition 1. Let $\tilde{N} = O^{p}(N) = N \cap \tilde {G}$ and $N_{\nu} = N \cap G_{\nu}$ for $\nu \in {\cal X}$. Then Propositions 3-7 hold for $b(N)$. In fact there exist $e$ irreducible characters $\chi'_i \in b(N)$ and $\chi'_{\nu}$ of $b(N_{\nu})$ corresponding to $\nu \in {\cal X}$ such that for $\lambda_{\nu} \in $[Irr]{}$(P_{\nu}/\tilde{P}) = $[Irr]{}$(N_{\nu}/\tilde{N})$, $$1_{N_{\nu}} * (\eta_{\hat{\nu}}\lambda_{\nu}) = (e - 1)\lambda_{\nu} - 
\sum_{i = 2}^{e} ({\chi'_i}\!\downarrow_{N_{\nu}}\lambda_{\nu}) + \ \chi'_{\nu}\lambda_{\nu},$$ and hence $$1_{N} * ((\eta_{\hat{\nu}} \lambda_{\nu})\uparrow^P_{P_{\nu}}) = (e - 1)(\lambda_{\nu}\!\uparrow_{N_{\nu}}^{N}) - \sum_{i = 2}^{e}({\chi'_i \!\downarrow_{N_{\nu}}\!\lambda_{\nu}})\!\uparrow_{N_{\nu}}^{N} + \ ({\chi'_{\nu}\lambda_{\nu}})\!\uparrow_{N_{\nu}}^{N}.$$ By Proposition 5 for $N$, ${\rm Irr}(b(N)) = \bigcup_{i = 1}^e \Big\{ \chi'_i \lambda \ | \  \lambda \in {\rm Irr}(N/\tilde{N}) \Big\} \bigcup_{\nu \in {\cal X}} \Big\{ (\chi'_{\nu} \lambda_{\nu})\!\uparrow_{N_{\nu}}^{N} \ | \ \lambda_{\nu} \in {\rm Irr}(N_{\nu}/\tilde{N}) \Big\}$ and by Theorem 2, $$\left\{ \begin{array}{l}
\chi'_i \lambda \mapsto \varepsilon_{i} \chi_i \lambda \\
(\chi'_{\nu}\lambda_{\nu})\uparrow^{N}_{N_{\nu}} \mapsto \varepsilon (\chi'_{\nu}\lambda_{\nu})\!\uparrow^{G}_{G_{\nu}}
\end{array} \right.$$ gives a perfect isometry from $b(N)$ to $b(G)$. Similarly $b(G)$ and $b(N_{G}(\tilde{P}))$ also are perfect isometric. $\blacksquare$\

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