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 