---
abstract: 'In this paper, we investigate the block that has an abelian defect group of rank $2$ and its Brauer correspondent has only one simple module. We will get an isotypy between the block and its Brauer correspondent. It will generalize the result of Kessar and Linckelmann ([@KL]).'
author:
- Xueqin Hu
date: 'School of Mathematics and Statistics, Central China Normal University, Wuhan 430079, China'
title: '**Blocks with abelian defect groups of rank $2$ and one simple module**'
---

Introduction
============

Let $p$ be a prime and $\mathcal{O}$ a complete discrete valuation ring having an algebraically closed residule field $k$ of characteristic $p$ and a quotient field $\mathcal{K}$ of characteristic $0$. We will always assume that $\mathcal{K}$ is big enough for the finite groups below.

Let $G$ be a finite group and $b$ a block of $\mathcal{O}G$ with a defect group $P$. Denote by $\mathrm{Irr}_\mathcal{K}(G,b)$ and $\mathrm{IBr}(G,b)$ the set of irreducible ordinary characters in $b$ and the set of irreducible Brauer characters in $b$ respectively. Set $l_G(b)=|\mathrm{IBr}(G,b)|$. Let $c$ be the Brauer correspondent of $b$ in $N_G(P)$. In [@KL], Kessar and Linckelmann investigated the block $b$ under the assumptions that $l_{N_G(P)}(c)=1$ and $P$ is elementary abelian of rank $2$. They showed that the inertial quotient of $b$ is abelian and there is an isotypy between $b$ and $c$ all of whose signs are positive.

In this note, we will generalize these results to the blocks with defect groups of rank $2$.

\[MT\] Keep the notation as above. Assume that $P$ is abelian of rank $2$ and $l_{N_G(P)}(c)=1$. Then the inertial quotient of $b$ is abelian and there is an isotypy between $b$ and $c$.

These results are well-known when either $p$ is $2$ or the inertial quotient of $b$ is trivial. Therefore, we may assume that $p$ is odd and the inertial quotient of $b$ is non-trivial throughout this paper.

The structure of the block $c$
==============================

Keep the notation as above. In this section, we will investigate the structure of the inertial quotient of $b$ and irreducible ordinary characters of the block $c$.

Given a positive integer $a$, denote by $C_a$ the cyclic group of order $a$. We will use $[-\,\ ,\,\ -]$ to represent the commutator. Assume that $P=C_{p^n}\times C_{p^m}$ for some positive integers $n,m$. We will fix a maximal $b$-Brauer pair $(P,b_P)$. For any $Q\leq P$, denote by $(Q,b_Q)$ the unique $b$-Brauer pair contained in $(P,b_P)$. Let $E$ be the inertial quotient of $b$ associated with $(P,b_P)$, namely, $E=N_G(P,b_P)/C_G(P)$.

\[E is abelian\] The inertial quotient $E$ is abelian if $l_{N_G(P)}(c)=1$.

Let $\Phi(P)$ be the Frattini subgroup of $P$. So $P/\Phi(P)$ is $C_p\times C_p$. Set $H$ to be $N_G(P,b_P)$. Then $\Phi(P)\unlhd H$ and denote $H/\Phi(P)$ by $\bar{H}$. For any subset $X$ of $\mathcal{O}H$, $\bar{X}$ denotes the image of $X$ under the canonical map $\mathcal{O}H\longrightarrow\mathcal{O}\bar{H}$.

Since $l_H(b_P)=1$, $l_{\bar{H}}(\bar{b}_P)=1$ and $\bar{b}_P$ is a block of $\bar{H}$ with defect group $\bar{P}=C_p\times C_p$. Let $\hat{C}$ be the subgroup of $H$ such that $\hat{C}/\Phi(P)=C_{\bar{H}}(\bar{P})$. Hence, $\hat{C}=\{x\in H~|~[P,x]\subseteq\Phi(P)\}$. It is clear that $P=[P,\hat{C}]\times C_P(\hat{C})$. So $P=C_P(\hat{C})$ since $[P,\hat{C}]\leq\Phi(P)$. This means $C_{\bar{H}}(\bar{P})=\bar{C}_G(P)$. Hence, $(\bar{P},\bar{b}_P)$ is a maximal $\bar{b}_P$-Brauer pair of $\mathcal{O}\bar{H}\bar{b}_P$. By [@KL Proposition 5.2], $N_{\bar{H}}(\bar{P},\bar{b}_P)/C_{\bar{H}}(\bar{P})$ is abelian. It is evident that $E$ is isomorphic to $N_{\bar{H}}(\bar{P},\bar{b}_P)/C_{\bar{H}}(\bar{P})$. We are done.

By [@DJ Lemma 2] and the structure of blocks with normal defect groups, $E$ is a direct product of two isomorphic groups. Next, we will show that $E$ acts diagonally on $P$. This can be deduced from the following general fact.

\[diagonal action\] Let $D$ be an abelian $p$-group of rank $2$ and $F\leq\mathrm{Aut}(P)$ an abelian $p^\prime$-group which is a direct product of two isomorphic subgroups. Then we have the decompositions $F=F_1\times F_2$ and $D=D_1\times D_2$ such that $F_1$ acts faithfully on $D_1$ and centralises $D_2$ and $F_2$ acts faithfully on $D_2$ and centralises $D_1$ and $F_1\cong F_2$. In particular, $F_1$ and $F_2$ are cyclic groups of order dividing $(p-1)$.

We will exhibit it by induction on $|D|$. When $D$ is elementary abelian, it is actually done in [@KL Proposition 5.3]. We may assume that $n\geq 2$ or $m\geq 2$. Let $\Phi(D)$ be the Frattini subgroup of $D$. So $D/\Phi(D)$ is $C_p\times C_p$. Let $\pi$ be the canonical map from $F$ to $\mathrm{Aut}(D/\Phi(D))$. For any subset $X$ of $F$, $\bar{X}$ denotes the image of $X$ under $\pi$. It is clear that $\pi$ is injective. So there exist two subgroups $F_1$ and $F_2$ of $F$ and two subgroups $D_1$ and $D_2$ of $D$ containing $\Phi(D)$ satisfying the properties $\bar{F}=\bar{F}_1\times \bar{F}_2$ and $D/\Phi(D)=D_1/\Phi(D)\times D_2/\Phi(D)$ and $\bar{F}_1$ acts faithfully on $D_1/\Phi(D)$ and centralises $D_2/\Phi(D)$ and $\bar{F}_2$ acts faithfully on $D_2/\Phi(D)$ and centralises $D_1/\Phi(D)$ and $\bar{F}_1\cong \bar{F}_2$. Hence, $D_1$ and $D_2$ are $F$-stable and they fulfill

\(i) $D_1=[D_1,F_1]\cdot\Phi(D)$ and $[D_1,F_2]\subseteq\Phi(D)$ and $F_1$ acts faithfully on $D_1$;

\(ii) $D_2=[D_2,F_2]\cdot\Phi(D)$ and $[D_2,F_1]\subseteq\Phi(D)$ and $F_2$ acts faithfully on $D_2$;

\(iii) $D_1\cap D_2=\Phi(D)$ and $D_1/\Phi(D)\cong C_p\cong D_2/\Phi(D)$ and $D=D_1\cdot D_2$.

Suppose that $\Phi(D)$ is cyclic. Then $D=C_p\times C_{p^m}$ with $m\geq 2$ and $\Phi(D)=C_{p^m-1}$. Since $D_2=[D_2,F_1]\times C_{D_2}(F_1)$ and $[D_2,F_1]\subseteq\Phi(D)$, $\Phi(D)=[D_2,F_1]\times C_{\Phi(D)}(F_1)$. Then either $[D_2,F_1]=1$ or $C_{\Phi(D)}(F_1)=1$ by the assumption that $\Phi(D)$ is cyclic. If $[D_2,F_1]=1$, then $\Phi(D)\leq D_2\leq C_D(F_1)$. Clearly, $D=[D,F_1]\times C_D(F_1)$ and $D_2$ is a maximal subgroup of $D$. Thus, $D_2=C_D(F_1)$ and $[D,F_1]=[D_1,F_1]$. Since $F_1$ and $F_2$ commute with each other and $[D_1,F_2]\subseteq\Phi(D)\subseteq C_D(F_1)$, $[[D_1,F_1],F_2]=1$. So $[D_1,F_1]\leq C_P(F_2)$. Since $F_1$ acts faithfully on $D_1$ and $D_1=[D_1,F_1]\times C_{D_1}(F_1)$, $F_1$ acts faithfully on $[D_1,F_1]$. Thus, the decompositions $F=F_1\times F_2$ and $D=[D_1,F_1]\times D_2$ are what we want. We may assume that $C_{\Phi(D)}(F_1)=1$. Then $\Phi(D)=[D_2,F_1]$ and $D_2=\Phi(D)\times C_{D_2}(F_1)$. If $C_{\Phi(D)}(F_2)=1$, we can get $\Phi(D)=[D_1,F_2]$ and $D_1=\Phi(D)\times C_{D_1}(F_2)$ similarly. Then $D=C_{D_1}(F_2)\times C_{D_2}(F_1)$ which is impossible. So $C_{\Phi(D)}(F_2)\neq1$. Then replacing $D_2$ by $D_1$ in the previous argument, we can obtain the decompositions that we need.

Suppose $\Phi(D)$ is of rank $2$. Then both $D_1$ and $D_2$ are of rank $2$. Let $K$ be subgroup of $F$ consisting of automorphisms acting trivially on $D_1$. Then $D=[D,K]\times C_D(K)$ and $D_1\leq C_D(K)$. Hence, $K$ has to be trivial since $D_1$ has rank $2$. This means $F$ acts faithfully on $D_1$. By induction, we have $D_1=D_{11}\times D_{12}$ and $F=F_{11}\times F_{12}$ such that $F_{11}$ acts faithfully on $D_{11}$ and centralises $D_{12}$ and $F_{12}$ acts faithfully on $D_{12}$ and centralises $D_{11}$ and $F_{11}\cong F_{12}$. Then $D=[D,F_{11}]\times C_{D}(F_{11})$ and $D_{11}=[D_{11},F_{11}]\leq[D,F_{11}]$ and $D_{11}\leq C_{D}(F_{12})$. In particular, $C_{[D,F_{11}]}(F_{12})\neq 1$. But $[D,F_{11}]$ is cyclic. Then $[D,F_{11}]\leq C_D(F_{12})$ and moreover $C_D(F_{12})=[D,F_{11}]\times(C_D(F_{11})\cap C_D(F_{12}))$. But $C_D(F_{12})$ is also cyclic. We have $[D,F_{11}]=C_D(F_{12})$. Similarly, we can prove that $[D,F_{12}]=C_D(F_{11})$. Then the decompositions $D=[D,F_{11}]\times [D,F_{12}]$ and $F=F_{11}\times F_{12}$ are what we want. We are done.

Hence, by Lemma \[diagonal action\], we have $E=E_1\times E_2$ and $P=P_1\times P_2$ such that

\(i) $E_1$ acts faithfully on $P_1$ and centralises $P_2$;

\(ii) $E_2$ acts faithfully on $P_1$ and centralises $P_1$;

\(iii) $E_1\cong E_2$ are cyclic groups of order $l$, which $l$ is a positive integer dividing $(p-1)$.

We can easily describe the source algebra of the block $c$ by the structure theory of blocks with normal defect groups and the structure of inertial quotient $E$. It is well-known that there exists a central extension

$$\xymatrix@C=0.5cm{
  1 \ar[r] & Z \ar[r] & \tilde{E} \ar[r] & E \ar[r] & 1 }$$ with $Z$ cyclic $p^\prime$-group such that there is an irreducible ordinary character $\theta$ of $Z$ which is covered by a unique irreducible character of $\tilde{E}$. Let $e_\theta\in\mathcal{O}Z$ be the central idempotent corresponding to $\theta$. Set $N=P\rtimes\tilde{E}$. Then $\mathcal{O}Ne_\theta$ is the source algebra of the block $c$. Note that $e_\theta$ is still a block of $C_N(R)$ for any $R\leq P$. The following lemma gives some information about the degrees and number of irreducible ordinary characters of $\mathcal{O}Ne_\theta$, which is similar with [@KL Proposition 5.3]. We will skip the proof.

\[irreducible character of c\] Set $A$ to be $\mathcal{O}Ne_\theta$. Then the degree of an element of $\mathrm{Irr}_\mathcal{K}(A)$ is either $l$ or $l^2$ and $\mathrm{Irr}_\mathcal{K}(A)$ has $p^n+p^m-1$ elements of degree $l$ and $\frac{p^n-1}{l}\cdot\frac{p^m-1}{l}$ elements of degree $l^2$.

The extension of local system
=============================

Keep the notation as above. In this section, we will use the so-called $(G,b)$-local system introduced by Puig and Usami in [@PU] to prove the main theorem.

First, let us recall some notation and state the definition of $(G,b)$-local system under our setting (see [@PU]).

Let $\mathcal{CF}_\mathcal{K}(G)$ be the vector space of $\mathcal{K}$-valued class functions of $G$ and $\mathcal{BCF}_\mathcal{K}(G)$ be the vector space of $\mathcal{K}$-valued class functions on the set $G_{p^\prime}$ of $p^\prime$-elements of $G$. It is clear that the set of irreducible ordinary characters of $G$ is a $\mathcal{K}$-basis of $\mathcal{CF}_\mathcal{K}(G)$ and the set of irreducible Brauer characters of $G$ is a $\mathcal{K}$-basis of $\mathcal{BCF}_\mathcal{K}(G)$. For $\chi,\chi^\prime\in\mathcal{CF}_\mathcal{K}(G)$, we denote by $\langle\chi,\chi^\prime\rangle$ the inner product of $\chi$ and $\chi^\prime$.

Let $u$ be a $p$-element of $G$. we have the well-known surjective $\mathcal{K}$-linear map $d_G^u:\mathcal{CF}_\mathcal{K}(G)\longrightarrow
\mathcal{BCF}_\mathcal{K}(C_G(u))$ defined by $d_G^u(\chi)(s)=\chi(us)$ for any $\chi\in\mathcal{CF}_{\mathcal{K}}(G)$ and $s\in C_G(u)_{p^\prime}$. It has a section $e_G^u:\mathcal{BCF}_\mathcal{K}(C_G(u))\longrightarrow
\mathcal{CF}_\mathcal{K}(G)$ such that for $\varphi\in\mathcal{BCF}_\mathcal{K}(C_G(u))$, $e_G^u(\varphi)(g)=0$ if the $p$-part of $g$ is not conjugate to $u$ in $G$.

For the block $b$, let $\mathcal{CF}_\mathcal{K}(G,b)$ be the subspace of $\mathcal{CF}_\mathcal{K}(G)$ generated by the elements in $\mathrm{Irr}_\mathcal{K}(G,b)$ and $\mathcal{L}_\mathcal{K}(G,b)$ the group of generalized characters in $b$. Also, let $\mathcal{CF}_\mathcal{K}^\circ(G,b)=
\mathcal{CF}_\mathcal{K}(G,b)\cap\mathrm{Ker}(d_G^1)$ and $\mathcal{L}_\mathcal{K}^\circ(G,b)=
\mathcal{L}_\mathcal{K}(G,b)\cap\mathrm{Ker}(d_G^1)$.

(Puig-Usami [@PU 3.2]) With the above notation and assumption. Let $X$ be an $E$-stable non-empty set of subgroups of $P$ and assume that $X$ contains any subgroup of $P$ containing an element of $X$. Let $\Gamma$ be a map over $X$ sending $Q\in X$ to a bijective isometry $$\Gamma_Q:\mathcal{BCF}_{\mathcal{K}}(C_N(Q),e_\theta)\longrightarrow
  \mathcal{BCF}_\mathcal{K}(C_G(Q),b_Q).$$ If $\Gamma$ satisfies the following conditions, then $\Gamma$ is called a $(G,b)$-[[local system]{}]{} over $X$.

\(i) For any $Q\in X$, any $\eta\in\mathcal{BCF}_\mathcal{K}(C_N(Q),e_\theta)$ and any $s\in E$, we have $\Gamma_Q(\eta)^s=\Gamma_{Q^s}(\eta^s)$.

\(ii) For any $Q\in X$ and any $\eta\in\mathcal{L}_\mathcal{K}(C_N(Q),e_\theta)$, the sum $$\sum\limits_{u}e_{C_G(Q)}^u(\Gamma_{Q\cdot\langle u\rangle}
  (d_{C_N(Q)}^u(\eta)))$$ where $u$ runs over a set of representatives $U_Q$ for the orbits of $C_E(Q)$ in $P$, is a generalized character of $C_G(Q)$.

Let $\Gamma$ be a $(G,b)$-local system over $X$. Such $\Gamma$ always exists by [@PU 3.4.2]. For any $Q\in X$, we have a map $\Delta_Q:
\mathcal{CF}_\mathcal{K}(C_N(Q),e_\theta)\longrightarrow
\mathcal{CF}_\mathcal{K}(C_G(Q),b_Q)$ defined by $$\Delta_Q(\eta)=
\sum\limits_{u\in U_Q}e_{C_G(Q)}^u(\Gamma_{Q\cdot\langle u\rangle}
  (d_{C_N(Q)}^u(\eta))).$$ Then by [@PU 3.3 and 3.4] $\Delta_Q$ gives a perfect isometry between the block $e_\theta$ of $C_N(Q)$ and the block $b_Q$ of $C_G(Q)$ and $\Delta_Q(\lambda\ast\eta)=\lambda\ast\Delta_Q(\eta)$ for any $\lambda\in\mathcal{CF}_\mathcal{K}(P)^{C_E(Q)}$ and $\eta\in\mathcal{CF}_\mathcal{K}(C_N(Q))$. Here, $\mathcal{CF}_\mathcal{K}(P)^{C_E(Q)}$ denotes the set of $C_E(Q)$-stable elements of $\mathcal{CF}_\mathcal{K}(P)$ and $\ast$ denotes the $\ast$-construction of charaters due to Brou$\acute{\mathrm{e}}$ and Puig (see [@BP]). Hence, if $X$ contains the trivial subgroup $1$ of $P$, then $\Delta_1$ induces a perfect isometry between the block $e_\theta$ of $N$ and the block $b$ of $G$. Moreover, this is an isotypy in the sense of [@B] by [@WZZ Proposition 2.7].

In [@PU], Puig and Usami developed a criterion for the extendibility of the $(G,b)$-local system. With the notation above. Suppose that $1\not\in X$ and let $Q$ be a maximal subgroup of $P$ such that $Q\not\in X$. Denote by $X^\prime$ the union of $X$ and the $E$-orbit of $Q$. For any subset $Y$ of $\mathcal{O}C_N(Q)$, denote by $\bar{Y}$ the image of $Y$ under the canonical map from $\mathcal{O}C_N(Q)$ to $\mathcal{O}C_N(Q)/Q$. We have the similar notation for $\mathcal{O}C_G(Q)$. So $\bar{e}_\theta$ and $\bar{b}_Q$ are the blocks of $\bar{C}_N(Q)$ and $\bar{C}_G(Q)$ respectively. Set $\Delta_Q^\circ=
\sum\limits_{u\in U_Q-Q}e_{C_G(Q)}^u\circ\Gamma_{Q\cdot\langle u\rangle}
  \circ d_{C_N(Q)}^u$ (see [@PU 3.6.2]). By [@PU Proposition 3.7 and Remark 3.8], $\Delta_Q^\circ$ induces a bijective isometry $$\bar{\Delta}_Q^\circ:
\mathcal{CF}_\mathcal{K}^\circ(\bar{C}_N(Q),\bar{e}_\theta)\cong
\mathcal{CF}_\mathcal{K}^\circ(\bar{C}_G(Q),\bar{b}_Q)$$ such that $\bar{\Delta}_Q^\circ(\mathcal{L}_\mathcal{K}^\circ(\bar{C}_N(Q),\bar{e}_\theta))=
\mathcal{L}_\mathcal{K}^\circ(\bar{C}_G(Q),\bar{b}_Q)$. Clearly, $\bar{\Delta}_Q^\circ(\lambda\ast\eta)=
\lambda\ast\bar{\Delta}_Q^\circ(\eta)$ for $\lambda\in\mathrm{Irr}_\mathcal{K}(\bar{P})^{C_E(Q)}$ and $\eta\in\mathcal{L}_\mathcal{K}^\circ(\bar{C}_N(Q),\bar{e}_\theta)$ (see [@W05 Case 2.2]) and $\bar{\Delta}_Q^\circ$ is $N_E(Q)$-stable. The following is the key criterion of extendibility.

([@PU Proposition 3.11])\[extendibility\] With the notation above, the $(G,b)$-local system $\Gamma$ over $X$ can be extended to a $(G,b)$-local system $\Gamma^\prime$ over $X^\prime$ if and only if $\bar{\Delta}_Q^\circ$ can be extended to an $N_E(Q)$-stable bijective isometry $$\bar{\Delta}_Q:\mathcal{CF}_\mathcal{K}(\bar{C}_N(Q),\bar{e}_{\theta})\cong
 \mathcal{CF}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$$ such that $\bar{\Delta}_Q(\mathcal{L}_\mathcal{K}(\bar{C}_N(Q),\bar{e}_\theta))=
\mathcal{L}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$.

In order to prove Theorem \[MT\], it suffices to show that there is a $(G,b)$-local system over the set of all the subgroups $P$. Hence, by Proposition \[extendibility\], we can assume that there is a $(G,b)$-local system $\Gamma$ over $X$ such that $1\not\in X$ and $Q$ is a maximal subgroup of $P$ such that $Q\not\in X$.

\[MT’\] With the notation above and assumptions of Section $2$. Then $\bar{\Delta}_Q^\circ$ can be extended to an $N_E(Q)$-stable bijective isometry $$\bar{\Delta}_Q:\mathcal{CF}_\mathcal{K}(\bar{C}_N(Q),\bar{e}_{\theta})\cong
 \mathcal{CF}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$$ such that $\bar{\Delta}_Q(\mathcal{L}_\mathcal{K}(\bar{C}_N(Q),\bar{e}_\theta))=
\mathcal{L}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$.

By the structure of $E$ and $P$, $C_E(Q)$ has only three possibilities: $1$, $E$ and $E_1$ or $E_2$. So we will divided the proof into $3$ cases.

   Assume that $C_E(Q)=1$.

Then the blocks $e_\theta$ of $C_N(Q)$ and $b_Q$ of $C_G(Q)$ are nilpotent. By the same argument as in [@PU 4.4], $\bar{\Delta}_Q^\circ$ can be extended to an $N_E(Q)$-stable bijective isometry $\bar{\Delta}_Q$.

   Assume that $C_E(Q)=E$.

Then $Q$ has to be trivial subgroup of $P$ and $N_E(Q)=E$. So $\bar{C}_N(Q)=N$ and $\bar{C}_G(Q)=G$ and we have a bijective isometry $$\bar{\Delta}^\circ:\mathcal{CF}_\mathcal{K}^\circ(N,e_\theta)\longrightarrow
  \mathcal{CF}_\mathcal{K}^\circ(G,b)$$ such that $\bar{\Delta}^\circ(\mathcal{L}_\mathcal{K}^\circ(N,e_\theta))=
  \mathcal{L}_\mathcal{K}^\circ(G,b)$.

The following technique we adopt to extend $\bar{\Delta}^\circ$ is essentially due to Kessar and Linckelmann (see [@KL Theorem 4.1]).

By Lemma \[irreducible character of c\], we have the following disjoint union $$\mathrm{Irr}_\mathcal{K}(N,e_\theta)=\Lambda_1\cup\Lambda_2,$$ where $\Lambda_1$ consists of irreducible ordinary characters of dimension $l$ and $\Lambda_2$ consists of irreducible ordinary characters of dimension $l^2$. Hence, $|\Lambda_1|=p^n+p^m-1$ and $|\Lambda_2|=\frac{p^n-1}{l}\cdot\frac{p^m-1}{l}$. We can assume that $n\geq 2$. Then $|\Lambda_1|>2$ and $|\Lambda_2|>2$. Choose an element $\psi_i\in\Lambda_i$ and set $\Lambda_i^\prime=\Lambda_i-\{\psi_i\}$ for $i=1,2$. Since $l_{N}(e_\theta)=1$, it is easy to see $$\mathcal{B}=\{\psi_1-\psi_1^\prime\,|\,\psi_1^\prime\in\Lambda_1^\prime\}
  \cup\{\psi_2-\psi_2^\prime\,|\,\psi_2^\prime\in\Lambda_2^\prime\}
  \cup\{\psi_2-l\psi_1\}$$ is a $\mathbb{Z}$-basis of $\mathcal{L}_\mathcal{K}^\circ(N,e_\theta)$. Since $p$ is odd, $|\Lambda_i^\prime|\geq 3$ for $i=1,2$. So by the same argument in [@PU 4.4], for any $i=1,2$, there exists a subset $\Omega_i=\{\chi_{\psi_i},\chi_{\psi_i^\prime}\,|\,
  \psi_i^\prime\in\Lambda_i^\prime\}$ of $\mathrm{Irr}_\mathcal{K}(G,b)$ and $\delta_i\in\{\pm 1\}$ such that $\bar{\Delta}^\circ(\psi_i-\psi_i^\prime)=
  \delta_i(\chi_{\psi_i}-\chi_{\psi_i^\prime})$. Since $\langle\psi_1-\psi_1^\prime,
  \psi_2-\psi_2^\prime\rangle=0$ for any $\psi_1^\prime\in\Lambda_1^\prime$ and $\psi_2^\prime\in\Lambda_2^\prime$, $\{\chi_{\psi_1},\chi_{\psi_1^\prime}\,|\,
  \psi_1^\prime\in\Lambda_1^\prime\}$ and $\{\chi_{\psi_2},\chi_{\psi_2^\prime}\,|\,
  \psi_2^\prime\in\Lambda_2^\prime\}$ have trivial intersection. Denote $\psi_2-l\psi_1$ by $\mu$. Then $\langle\mu,
  \psi_1-\psi_1^\prime\rangle=-l$ for all $\psi_1^\prime\in\Lambda_1^\prime$. Thus $$\label{equation E}
    \begin{array}{ll}
    \bar{\Delta}^\circ(\mu)=
    \delta_1(a-l)\chi_{\psi_1}+
    \delta_1a\sum\limits_{\psi_1^\prime\in\Lambda_1^\prime}\chi_{\psi_1^\prime}+
    \Xi
    \end{array}$$ for some integer $a$ and some element $\Xi\in\mathcal{L}_\mathcal{K}(N,e_\theta)$ not involving any of elements in $\Omega_1$. Since $\langle\mu,
  \psi_2-\psi_2^\prime\rangle=1$ and $\bar{\Delta}^\circ(\psi_2-\psi_2^\prime)=
  \delta_2(\chi_{\psi_2}-\chi_{\psi_2^\prime})$, $\Xi$ must involve one of the two characters occuring in $\bar{\Delta}^\circ(\psi_2-\psi_2^\prime)$ for any $\psi_2^\prime\in\Lambda_2^\prime$. Taking norms on both sides in equation (\[equation E\]), we have $$\label{inequation norm}
    \begin{array}{ll}
      &1+l^2\geq(a-l)^2+(p^n+p^m-2)a^2=
      (p^n+p^m-1)a^2-2la+l^2\\
      \Longleftrightarrow
      &1\geq(p^n+p^m-1)a^2-2la
    \end{array}$$

Suppose that $a\leq0$. Since $a$ is integer and $p^n+p^m-1,l$ are positive integers, $a$ has to be $0$.

Suppose that $a>0$. Since $p^n+p^m-1>2l$, $(p^n+p^m-1)a^2-2la>(p^n+p^m-1)(a^2-a)$. This forces $a=1$. Hence, $a=0$ or $1$. Notice that $\Xi\neq 0$. This implies (\[inequation norm\]) is a proper inequality. So $a$ must be $0$. Then equation (\[equation E\]) becomes $$\bar{\Delta}^\circ(\mu)=-\delta_1l\chi_{\psi_1}+\Xi.$$ Comparing norms, we have $\langle\Xi,\Xi\rangle=1$.

For any $\psi_2^\prime,\psi_2^{\prime\prime}\in\Lambda_2^\prime$, $$\langle\bar{\Delta}^\circ(\mu),
  \delta_2(\chi_{\psi_2}-\chi_{\psi_2^\prime})\rangle=
  \langle\mu,
  \psi_2-\psi_2^\prime\rangle=1$$ and $$\langle\bar{\Delta}^\circ(\mu),
  \delta_2(\chi_{\psi_2^{\prime}}-\chi_{\psi_2^{\prime\prime}})\rangle=
  \langle\mu,
  \psi_2^\prime-\psi_2^{\prime\prime}\rangle=0.$$ Then $\Xi=\delta_2\chi_{\psi_2}$. But $\bar{\Delta}^\circ(\mu)(1)=0$. This forces $\delta_1=\delta_2$. Since $\mathcal{B}$ is a $\mathbb{Z}$-basis of $\mathcal{L}_\mathcal{K}^\circ(N,e_\theta)$, $\mathrm{Irr}_\mathcal{K}(G,b)=\Omega_1\cup\Omega_2$. Hence, we get a bijective isometry $\bar{\Delta}$ from $\mathcal{L}_\mathcal{K}(N,e_\theta)$ to $\mathcal{L}_\mathcal{K}(G,b)$ mapping $\psi_i$ and $\psi_i^\prime$ to $\chi_{\psi_i}$ and $\chi_{\psi_i^\prime}$ respectively, where $i=1,2$. In particular, $l_G(b)=l_N(e_\theta)=1$. Clearly, it is an extension of $\bar{\Delta}^\circ$. Since $\bar{\Delta}^\circ$ is $E$-stable and $l_G(b)=l_N(e_\theta)=1$, $\bar{\Delta}$ is also $E$-stable.

   Assume that $C_E(Q)=E_i$ for some $i=1,2$.

We can assume that $C_E(Q)=E_1$ and then $1\neq Q\leq P_2$ and $N_E(Q)=E$. It suffices to prove that $\bar{\Delta}_Q^\circ$ can extend to an $E_2$-stable bijective isometry $\bar{\Delta}_Q:\mathcal{L}_\mathcal{K}(\bar{C}_N(Q),\bar{e}_\theta)\longrightarrow
  \mathcal{L}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$.

By [@W14 Theorem 1], $|\mathrm{Irr}_\mathcal{K}(\bar{C}_N(Q),\bar{e}_\theta)|=
  |\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)|$ and $l_{\bar{C}_N(Q)}(\bar{e}_\theta)=l_{\bar{C}_G(Q)}(\bar{b}_Q)$ since the block $\bar{b}_Q$ of $\bar{C}_G(Q)$ has a cyclic hyperfocal subgroup. It is clear $C_N(Q)=(P_1\rtimes\tilde{E}_1)\times P_2$ and $C_N(Q)\unlhd N$, where $\tilde{E}_1$ is the preimage of $E_1$ in $\tilde{E}$. Hence, $E_1$ is the inertial quotient of the block $e_\theta$ of $C_N(Q)$ and $P_1$ is a hyperfocal subgroup with respect to $E_1$. By [@W14 Theorem 1], $l_{C_N(Q)}(e_\theta)=l$. We will claim that $N$ acts transitively on $\mathrm{IBr}(C_N(Q),e_\theta)$. Indeed, this holds because $l_N(e_\theta)=1$ by the assumption and $N/C_N(Q)\cong E_2$ is a cyclic group of order $l$.

Denote by $\mathrm{Irr}_\mathcal{K}(\tilde{E}_1)_\theta$ the subset of $\mathrm{Irr}_\mathcal{K}(\tilde{E}_1)$ consisting of characters covering $\theta$. Then $|\mathrm{Irr}_\mathcal{K}(\tilde{E}_1)_\theta|=l$ and we set $\mathrm{Irr}_\mathcal{K}(\tilde{E}_1)_\theta=\{\tau_i\,|\, i=1,2,\cdots,l\}$, which is transitively acted by $N$. Hence, we can write $\mathrm{Irr}_\mathcal{K}(\tilde{E}_1)_\theta$ as $\{\tau^a\,|\,a\in E_2\}$ for any $\tau\in\mathrm{Irr}_\mathcal{K}(\tilde{E}_1)_\theta$. By Clifford theorem, we have $\mathrm{Res}_Z^{\tilde{E}_1}(\tau_i)=\theta$ for any $i$ and $\mathrm{Ind}_Z^{\tilde{E}_1}(\theta)
  =\sum\limits_{i=1}^l\tau_i$. Let $M$ be a representative of $\tilde{E}_1$-orbit of $\mathrm{Irr}_\mathcal{K}(P_1)-\{1_{P_1}\}$, where $1_{P_1}$ is the trivial character of $P_1$. Then $$\mathrm{Irr}_\mathcal{K}(\bar{C}_N(Q),\bar{e}_\theta)=
  \{\tau_i\bar{\zeta}_j\,|\, \bar{\zeta}_j\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2),
     i=1,2,\cdots,l\}\cup
  \{\mathrm{Ind}_{P_1\times Z}^{P_1\rtimes\tilde{E}_1}(\xi\theta)\bar{\zeta}_j\,|\,
     \xi\in M,\bar{\zeta}_j\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}.$$ We will write $\mathrm{Ind}_{P_1\times Z}^{P_1\rtimes\tilde{E}_1}(\xi\theta)$ and $\bar{\chi}\cdot 1_{\bar{P}_2}$ as $\mathrm{Ind}(\xi)$ and $\bar{\chi}$ respectively for simplicity. Here, $\bar{\chi}$ is an element of $\mathcal{CF}_\mathcal{K}(\overline{P_1\rtimes\tilde{E}_1})$. Clearly, $\mathrm{Ind}(\xi)$ is $N$ and $E_2$-stable for any $\xi\in M$. Similar to the argument of [@W05 Case 2], $$\{(\sum\limits_{i=1}^l\tau_i-\mathrm{Ind}(\xi))\bar{\zeta}\,|\,\xi\in M,
    \bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}\cup
    \{\tau_i-\tau_i\bar{\zeta}\,|\,i=1,2,\cdots,l,
         1_{\bar{P}_2}\neq\bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}$$ is a $\mathbb{Z}$-basis of $\mathcal{L}_\mathcal{K}^\circ(\bar{C}_N(Q),\bar{e}_\theta)$.

   Assume that $\bar{P}_2=1$, i.e., $Q=P_2$.

Set $H=N_G(Q,b_Q)$. Then $H=C_G(Q)N_G(P,b_P)$ and $b_Q$ is still a block of $H$. Let $d$ be the Brauer correspondent of the block $b_Q$ of $H$ in $N_H(P)$. Then $l_{N_H(P)}(d)=1$ by the assumption. We claim that $l_H(b_Q)=1$.

Indeed, considering the canonical map from $\mathcal{O}H$ to $\mathcal{O}(H/Q)$, denote by $\bar{X}$ the image of $X$ under this canonical map for any subset $X$ of $\mathcal{O}H$. Then $\bar{b}_Q$ is still a block of $\bar{C}_G(Q)$ and $\bar{H}/\bar{C}_G(Q)$ is a cyclic group of order $l$. By [@KR Lemma 3.5], $\mathrm{Br}_{\bar{P}}(\bar{b}_Q)=\overline{\mathrm{Br}_P(b_Q)}$. Since $l_{N_H(P)}(d)=1$, $\bar{d}$ is still a block of $\bar{N}_H(P)$. Therefore, $\mathrm{Br}_{\bar{P}}(\bar{b}_Q)=\bar{d}$ is a block of $\bar{N}_H(P)$. Suppose that the blocks of $\bar{H}$ covering the block $\bar{b}_Q$ of $\bar{C}_G(Q)$ have the same defect group $\bar{P}$. Then $\bar{b}_Q$ is a block of $\bar{H}$ since $\mathrm{Br}_{\bar{P}}(\bar{b}_Q)$ is a block of $\bar{N}_H(P)$ and $N_{\bar{H}}(\bar{P})=\bar{N}_H(P)$. Hence, it has a defect group $\bar{P}$ which is cyclic by our assumption. In particular, we have $l_H(b_Q)=l_{\bar{H}}(\bar{b}_Q)=l_{N_H(P)}(d)=1$ since $N_{\bar{H}}(\bar{P})=\bar{N}_H(P)$. Consequently, the argument follows from the lemma below.

[*[**Lemma 3.4**]{} Let $L$ be a normal subgroup of $K$ such that $K/L$ is a cyclic $p^\prime$-group. Let $i$ be a $K$-stable block of $L$ with defect group $D$. For any block $e$ of $K$ covering $i$, $e$ has defect group $D$.*]{}

[*Proof.*]{} We will prove it by induction on $K/L$. Let $M\leq K$ such that $M$ contains $L$ and $|M/L|$ is a prime. Then $M\trianglelefteq K$ and $K/M$ is still a cyclic $p^\prime$-group. Denote by $M[i]$ the subgroup of $M$ consisting of elements acting on $\mathcal{O}Li$ as inner automorphisms. Therefore, $M[i]=M$ or $L$. Let $f$ be a block of $M$ covered by $e$. So $f$ covers the block $i$ of $L$. If $M[i]=M$, then $\mathcal{O}Mf$ and $\mathcal{O}Li$ are source algebra equivalent by [@K90 Theorem 7]. In particular, the block $f$ has defect group $D$. If $M[b]=L$, then $f=i$ by [@D Theorem 3.5] and certainly they have the same defect group. In conclusion, $D$ is a defect group of the block $f$. Let $K_f$ be the stabilizer of $f$ in $K$. Then blocks of $K_f$ covering $f$ have defect group $D$ by induction. So is $e$.

Moreover, we claim that there is a regular $E_2$-orbit of $\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$, namely, $H$ acts transitively on it.

Indeed, since the block $\bar{b}_Q$ of $\bar{H}$ has a cyclic defect group, it must be nilpotent. By [@P11 Theorem 3.13], the block $\bar{b}_Q$ of $\bar{C}_G(Q)$ is basic Morita equivalent to its Brauer correspondent. Note that the block $\bar{b}_Q$ of $\bar{C}_G(Q)$ is not nilpotent since $l>1$. This implies that every irreducible Brauer character of the block $\bar{b}_Q$ of $\bar{C}_G(Q)$ can be uniquely lifted to an irreducible ordinary character by the theory of cyclic blocks.

On the other hand, since $l_{C_G(Q)}(b_Q)=l$ and $l_H(b_Q)=1$ and $H/C_G(Q)\cong E_2$ has order $l$, $H$ acts transitively on $\mathrm{IBr}(C_G(Q),b_Q)$. Combining this with the argument above, there exits a regular $H$-orbit of $\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$. We are done.

   Assume that $|M|=1$.

Then $\mathrm{rank}_\mathcal{O}
  (\mathcal{L}_\mathcal{K}^\circ(\bar{C}_N(Q),\bar{e}_\theta))=1$ and $\mathcal{L}_\mathcal{K}^\circ(\bar{C}_N(Q),\bar{e}_\theta)=
  \mathbb{Z}(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)$. Since there is a regular $E_2$-orbit of $\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$, $\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)=\{\chi_0\}\cap
  \{\chi_1,\chi_2,\cdots,\chi_l\}$ such that $\chi_0$ is $E_2$-stable and $E_2$ acts regularly on $\{\chi_1,\chi_2,\cdots,\chi_l\}$. Then we have $$\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)=
  \delta_0\chi_0-\sum\limits_{i=1}^l\delta_i\chi_i$$ for some $\delta_0,\delta_i\in\{\pm1\},i=1,2,\cdots,l$. Since $\bar{\Delta}_Q^\circ$ is $E_2$-stable, we have $\delta_1=\delta_2=\cdots=\delta_l=\delta_0$. If we write $\{\tau_1,\tau_2,\cdots,\tau_l\}$ and $\{\chi_1,\chi_2,\cdots,\chi_l\}$ as $\{\tau^a\,|\,a\in E_2\}$ and $\{\chi^a\,|\,a\in E_2\}$ respectively, then we can define a bijective isometry as below $$\bar{\Delta}_Q:\mathcal{L}_\mathcal{K}(\bar{C}_N(Q),\bar{e}_\theta)\longrightarrow
  \mathcal{L}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$$ $$\mathrm{Ind}(\xi)\mapsto\delta_0\chi_0$$ $$\tau^a\mapsto\delta_0\chi^a.$$ It is evident that it is an extension of $\bar{\Delta}_Q^\circ$ and $E_2$-stable. We are done for this case.

   Assume that $|M|\geq2$.

Then there are at least two different $\xi_1,\xi_2\in M$. So $\mathrm{Ind}(\xi_1)-\mathrm{Ind}(\xi_2)\in
  \mathcal{L}_\mathcal{K}^\circ(\bar{C}_N(Q),\bar{e}_\theta)$ and $\langle\mathrm{Ind}(\xi_1)-\mathrm{Ind}(\xi_2),
  \mathrm{Ind}(\xi_1)-\mathrm{Ind}(\xi_2)\rangle=2$. Then there exist $\chi_1\neq\chi_2\in
  \mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$ such that $$\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi_1)-\mathrm{Ind}(\xi_2))=
  \delta(\chi_1-\chi_2)$$ for some $\delta\in\{\pm1\}$. Since $\bar{\Delta}_Q^\circ$ is $E_2$-stable, we have ${^a}(\delta\chi_1-\delta\chi_2)=\delta(\chi_1-\chi_2)$ for any $a\in E_2$. This means that $\chi_1$ and $\chi_2$ are both $E_2$-stable.

If there is a $\xi_3\in M$ different from $\xi_1$ and $\xi_2$, then there is a $\chi_3\in\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$ different from $\chi_1$ and $\chi_2$ such that $$\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi_1)-\mathrm{Ind}(\xi_3))=
  \delta\chi_1-\delta\chi_3~\mathrm{or}~
  -\delta\chi_2+\delta\chi_3$$ and $\chi_3$ is $E_2$-stable; then we may choose the notation in such a way that $$\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi_1)-\mathrm{Ind}(\xi_2))=
  \delta(\chi_1-\chi_2)
  ~\mathrm{and}~
  \bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi_1)-\mathrm{Ind}(\xi_3))=
  \delta(\chi_1-\chi_3)$$ for some $E_2$-stable elements $\chi_1,\chi_2,\chi_3$ of $\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q).$

If $|M|\geq 4$, then for any $\xi\in M-\{\xi_1,\xi_2,\xi_3\}$, there is a unique $\chi\in\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)-
  \{\chi_1,\chi_2,\chi_3\}$ such that $$\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi_1)-\mathrm{Ind}(\xi))=
  \delta(\chi_1-\chi)$$ and $\chi$ is $E_2$-stable.

In conclusion, we have an injective isometry $$\Phi:\mathbb{Z}\{\mathrm{Ind}(\xi)\,|\,\xi\in M\}\longrightarrow
  \mathcal{L}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$$ mapping $\mathrm{Ind}(\xi)$ to $\delta\chi_\xi$ such that $$\Phi(\mathrm{Ind}(\xi)-\mathrm{Ind}(\xi^\prime))=
  \bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\mathrm{Ind}(\xi^\prime))$$ and $\chi_\xi$ is $E_2$-stable for any $\xi,\xi^\prime\in M$.

Denote $\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)-
  \{\chi_\xi\,|\,\xi\in M\}$ by $\Omega$. Then $|\Omega|=l$ and $E_2$ acts on $\Omega$. Since there is a regular $E_2$-orbit of $\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$, $E_2$ acts regularly on $\Omega$. This means that $\Omega$ can be represented as $\{\chi^a\,|\,a\in E_2\}$ for some $\chi\in\Omega$.

Now we fix an element $\xi$ of $M$. Suppose that $\chi$ does not get involved in $\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)$. Then there is $\xi^\prime\in M$ such that $\langle\chi,\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi^\prime)-
  \sum\limits_{i=1}^l\tau_i)\rangle\neq 0$ since $\{\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i\,|\,
  \xi\in M\}$ is a $\mathbb{Z}$-basis of $\mathcal{L}_\mathcal{K}^\circ
  (\bar{C}_N(Q),\bar{e}_\theta)$. Hence, $\chi$ has to get involved in $\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)-
  \bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi^\prime)-\sum\limits_{i=1}^l\tau_i)$ which is $\delta(\chi_\xi-\chi_{\xi^\prime})$. This is impossible. So $\chi$ must get involved in $\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)$ for any $\xi\in M$. Since $\bar{\Delta}_Q^\circ$ and $\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i$ are $E_2$-stable, $\chi^a$ has to get involved in $\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)$ for any $a\in E_2$ and $\xi\in M$. Since $\langle\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i,
   \mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i\rangle=1+l$ and $\langle\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i,
    \mathrm{Ind}(\xi)-\mathrm{Ind}(\xi^\prime)\rangle=1$, $\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)=
    \delta\chi_\xi-\sum\limits_{a\in E_2}\delta_a\chi^a~\mathrm{or}~
    -\delta\chi_{\xi^\prime}-\sum\limits_{a\in E_2}\delta_a\chi^a$, where $\delta_a\in\{\pm1\}$ for any $a\in E_2$. Note that the last situation can happen if and only if $|M|=2$. By switching $\chi_\xi$ and $\chi_{\xi^\prime}$ if necessary, we can assume that $\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)=
    \delta\chi_\xi-\sum\limits_{a\in E_2}\delta_a\chi^a$. Since $\bar{\Delta}_Q^\circ$ is $E_2$-stable and $E_2$ acts regularly on $\Omega$, $\delta_a$ is equal to $\delta$ for any $a\in E_2$. Then we can define an $E_2$-stable bijective isometry as follows $$\bar{\Delta}_Q:\mathcal{L}_\mathcal{K}(\bar{C}_N(Q),\bar{e}_\theta)\longrightarrow
  \mathcal{L}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$$ $$\mathrm{Ind}(\xi)\mapsto\delta\chi_\xi$$ $$\tau^a\mapsto\delta\chi^a.$$ It is clear that $\bar{\Delta}_Q$ is an extension of $\bar{\Delta}_Q^\circ$.

   $\bar{P}_2>1$, namely, $Q$ is a non-trivial proper subgroup of $P_2$.

Then $\mathrm{Ind}(\xi)-\mathrm{Ind}(\xi)\bar{\zeta}\in
  \mathcal{L}_\mathcal{K}^\circ(\bar{C}_N(Q),\bar{e}_\theta)$ for any $\xi\in M$ and $1_{\bar{P}_2}\neq\bar{\zeta}\in
  \mathrm{Irr}_\mathcal{K}(\bar{P}_2)$.

Now we fix an element $\xi\in M$. Since $p$ is odd, $|\bar{P}_2|\geq 3$. Then there are at least two elements $\bar{\zeta}$ and $\bar{\zeta}^\prime$ of $\mathrm{Irr}_\mathcal{K}(\bar{P}_2)$ different from $1_{\bar{P}_2}$. With the same argument in the first three paragraphs in Case 3.1.2, we can get a subset $\{\chi_\xi,\chi_{\bar{\zeta}}\,|\,
  1_{\bar{P}_2}\neq\bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}$ of $\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$ such that $$\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\mathrm{Ind}(\xi)\bar{\zeta})=
  \delta(\chi_\xi-\chi_{\bar{\zeta}})$$ for any $1_{\bar{P}_2}\neq\bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)$, where $\delta\in\{\pm1\}$.

Given any $1\neq a\in E_2$ and $1_{\bar{P}_2}\neq\bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)$, ${^a}(\mathrm{Ind}(\xi)\bar{\zeta})=\mathrm{Ind}(\xi)({^a}\bar{\zeta})$. Since $\bar{\Delta}_Q^\circ$ is $E_2$-stable, this means ${^a}\chi_\xi-{^a}\chi_{\bar{\zeta}}=\chi_\xi-\chi_{{^a}\bar{\zeta}}$. Hence, we have $\chi_\xi$ is $E_2$-stable and ${^a}\chi_{\bar{\zeta}}=\chi_{{^a}\bar{\zeta}}$. On the other hand, $$(\mathrm{Ind}(\xi)-\mathrm{Ind}(\xi)\bar{\zeta})\bar{\zeta}=
  (\mathrm{Ind}(\xi)-\mathrm{Ind}(\xi)\bar{\zeta}^2)-
  (\mathrm{Ind}(\xi)-\mathrm{Ind}(\xi)\bar{\zeta}).$$ Since $\bar{\Delta}_Q^\circ$ is compatible with $\ast$-structure, using $\bar{\Delta}_Q^\circ$ on both sides in the above equality, we can get $$\delta(\chi_\xi-\chi_{\bar{\zeta}})\ast\bar{\zeta}=
  \delta(\chi_{\bar{\zeta}}-\chi_{\bar{\zeta}^2}).$$ Therefore, $\chi_{\bar{\zeta}}=\chi_\xi\ast\bar{\zeta}$ for any $1_{\bar{P}_2}\neq\bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)$.

Suppose that there is another element $\xi^\prime$ of $M$ different from $\xi$. Similarly, we can get a subset $\{\chi_{\xi^\prime}\ast\bar{\zeta}\,|\,
  \bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}$ of $\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$ such that $\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi^\prime)-
  \mathrm{Ind}(\xi^\prime)\bar{\zeta})=
  \delta^\prime(\chi_{\xi^\prime}-\chi_{\xi^\prime}\ast\bar{\zeta})$ for any $1_{\bar{P}_2}\neq\bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)$ and $\chi_{\xi^\prime}$ is $E_2$-stable, where $\delta^\prime\in\{\pm 1\}$. Assume that $\{\chi_\xi\ast\bar{\zeta}\,|\,
  \bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}\cap
  \{\chi_{\xi^\prime}\ast\bar{\zeta}\,|\,
  \bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}\neq\emptyset$. Then there is $\bar{\zeta}_0\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)$ such that $\chi_\xi=\chi_{\xi^\prime}\ast\bar{\zeta}$. If $\bar{\zeta}_0=1_{\bar{P}_2}$, then $\chi_\xi=\chi_{\xi^\prime}$. This implies that $\mathrm{Ind}(\xi)-\mathrm{Ind}(\xi)\bar{\zeta}=
  \pm(\mathrm{Ind}(\xi^\prime)-\mathrm{Ind}(\xi^\prime)\bar{\zeta})$ for any $1_{\bar{P}_2}\neq\bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)$. This is impossible. Then $\bar{\zeta}_0$ is non-trivial. But it implies that $\chi_{\xi^\prime}\ast\bar{\zeta}_0^2=\chi_{\xi^\prime}$ since $\langle\mathrm{Ind}(\xi)-\mathrm{Ind}(\xi)\bar{\zeta}_0,
  \mathrm{Ind}(\xi^\prime)-\mathrm{Ind}(\xi^\prime)\bar{\zeta}_0\rangle=0$. It is well-known that $\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\backslash\{1_{\bar{P}_2}\}$ acts freely on irreducible ordinary characters of height zero in the block $\bar{b}_Q$ of $\bar{C}_G(Q)$ (see [@R §1]). Hence, $\bar{\zeta}_0^2=1_{\bar{P}_2}$ since the defect group of the block $\bar{b}_Q$ of $\bar{C}_G(Q)$ is cyclic. But it is impossible because $p$ is odd. Then $$\{\chi_\xi\ast\bar{\zeta}\,|\,
  \bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}\cap
  \{\chi_{\xi^\prime}\ast\bar{\zeta}\,|\,
  \bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}=\emptyset$$ for any different $\xi,\xi^\prime\in M$. It is clear that $\chi_\xi\ast\bar{\zeta}$ is an irreducible ordinary character in the block $\bar{b}_Q$ of $\bar{C}_G(Q)$ by [@BP Corollary]. Then we get an injective isometry $$\Psi:\mathbb{Z}\{\mathrm{Ind}(\xi)\bar{\zeta}\,|\,
  \xi\in M,\bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}
  \longrightarrow
  \mathcal{L}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$$ mapping $\mathrm{Ind}(\xi)\bar{\zeta}$ to $\delta_\xi(\chi_\xi\ast\bar{\zeta})$ such that $\Psi(\mathrm{Ind}(\xi)-\mathrm{Ind}(\xi)\bar{\zeta})=
  \bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\mathrm{Ind}(\xi)\bar{\zeta})$ and $\chi_\xi$ is $E_2$-stable for any $\xi\in M$ and $1_{\bar{P}_2}\neq\bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)$, where $\delta_\xi\in\{\pm 1\}$.

At the same time, $\tau-\tau\bar{\zeta}\in
  \mathcal{L}_\mathcal{K}^\circ(\bar{C}_N(Q),\bar{e}_\theta)$ for any $\tau\in\mathrm{Irr}_\mathcal{K}(\tilde{E}_1)_\theta$ and $1_{\bar{P}_2}\neq\bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)$. Take an element $\tau$ of $\mathrm{Irr}_\mathcal{K}(\tilde{E}_1)_\theta$. With the same arguments as above, we can get an element $\chi_\tau$ of $\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$ and $\delta_\tau\in\{\pm1\}$ such that $\bar{\Delta}_Q^\circ(\tau-\tau\bar{\zeta})=
  \delta_\tau(\chi_\tau-\chi_\tau\ast\bar{\zeta})$ for any $1_{\bar{P}_2}\neq\bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)$. Choosing any $1\neq a\in E_2$ and $1_{\bar{P}_2}\neq\bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)$, since $\bar{\Delta}_Q^\circ$ is $E_2$-stable, we have $$\delta_{{^a}\tau}(\chi_{{^a}\tau}-\chi_{{^a}\tau}\ast{^a}\bar{\zeta})=
  \bar{\Delta}_Q^\circ({^a}\tau-{^a}\tau({^a}\bar{\zeta}))=
  {^a}(\bar{\Delta}_Q^\circ(\tau-\tau\bar{\zeta}))=
  \delta_\tau({^a}\chi_\tau-{^a}\chi_\tau\ast{^a}\bar{\zeta}).$$ Then $\chi_{{^a}\tau}={^a}\tau$ or $\chi_{{^a}\tau}={^a}\chi_\tau\ast{^a}\bar{\zeta}$. If $\chi_{{^a}\tau}={^a}\chi_\tau\ast{^a}\bar{\zeta}$, then ${^a}\chi_\tau=\chi_{{^a}\tau}\ast{^a}\bar{\zeta}$. Therefore, $\chi_{{^a}\tau}=\chi_{{^a}\tau}\ast{^a}(\bar{\zeta}^2)$, which is impossible. Hence, $\chi_{{^a}\tau}={^a}\chi_\tau$ and $\delta_{{^a}\tau}=\delta_\tau$ for any $a\in E_2$ since $E_2$ acts transitively on $\mathrm{Irr}_\mathcal{K}(\tilde{E}_1)_\theta$. And we denote $\delta_\tau$ by $\delta$. By the facts that $\langle\tau-\tau\bar{\zeta},
   \tau^\prime-\tau^\prime\bar{\zeta}^\prime\rangle=0$ and $\langle\mathrm{Ind}(\xi)-\mathrm{Ind}(\xi)\bar{\zeta},
  \tau-\tau\bar{\zeta}^\prime\rangle=0$ for any $\tau\neq\tau^\prime\in\mathrm{Irr}_\mathcal{K}(\tilde{E}_1)_\theta$ and $\xi\in M$ and $\bar{\zeta},\bar{\zeta}^\prime\in
  \mathrm{Irr}_\mathcal{K}(\bar{P}_2)-\{1_{\bar{P}_2}\}$, we can get $$\{\chi_\tau\ast\bar{\zeta}\,|\,
  \bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}\cap
  \{\chi_{\tau^\prime}\ast\bar{\zeta}\,|\,
  \bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}=\emptyset$$ and $$\{\chi_\xi\ast\bar{\zeta}\,|\,
  \xi\in M, \bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}\cap
  \{\chi_\tau\ast\bar{\zeta}\,|\,
  \tau\in\mathrm{Irr}_\mathcal{K}(\tilde{E}_1)_\theta,
  \bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)\}=\emptyset.$$ Hence, we have a well-defined $E_2$-stable bijective isometry as below $$\bar{\Delta}_Q:\mathcal{L}_\mathcal{K}(\bar{C}_N(Q),\bar{e}_\theta)\longrightarrow
  \mathcal{L}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)$$ $$\mathrm{Ind}(\xi)\bar{\zeta}\mapsto\delta_\xi\chi_\xi\ast\bar{\zeta}$$ $${^a}\tau\bar{\zeta}\mapsto\delta{^a}\chi_\tau\ast\bar{\zeta}.$$ It suffices to show that $\bar{\Delta}_Q$ is an extension of $\bar{\Delta}_Q^\circ$, namely, $$\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)=
  \delta_\xi\chi_\xi-\delta\sum\limits_{i=1}^l\chi_{\tau_i}$$ for any $\xi\in M$.

Choose an element $\xi$ of $M$. Since $\langle\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i,
  \tau-\tau\bar{\zeta}\rangle=-1$, then at least $\chi_\tau$ and $\chi_\tau\ast\bar{\zeta}$ must get involved in $\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)$ for any $\tau\in\mathrm{Irr}_\mathcal{K}(\tilde{E}_1)_\theta$ and $\bar{\zeta}\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)-\{1_{\bar{P}_2}\}$.

Keep the notation as above. Suppose that there are $\tau$ and $\bar{\zeta}$ such that $\chi_\tau\ast\bar{\zeta}$ gets involved in $\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)$. Since $\langle\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i,
  \tau\bar{\zeta}-\tau\bar{\zeta}^\prime\rangle=0$ for any $\bar{\zeta}^\prime\in\mathrm{Irr}_\mathcal{K}(\bar{P}_2)$ different from $\bar{\zeta}$ and $1_{\bar{P}_2}$, $\chi_\tau\ast\bar{\zeta}$ must get involved in $\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)$ for any $\bar{\zeta}$. At the same time, since $\bar{\Delta}_Q^\circ$ is $E_2$-stable and $\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i$ is $E_2$-stable, we have ${^a}(\chi_\tau\ast\bar{\zeta})$ must get involved in $\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)$ for any $a\in E_2$ and $\bar{\zeta}$. Then there are at least $l\cdot(|\bar{P}_2|-1)$ different irreducible characters involved in $\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)$. This is impossible since $\langle\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i,
  \mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i\rangle=1+l$ and $|\bar{P}_2|-1\geq 2$ and $l>1$.

So for any $\xi\in M$, $\bar{\Delta}_Q^\circ(\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i)=
  a_\chi\chi-\delta\sum\limits_\tau\chi_\tau$. Here, $a_\chi\in\{\pm 1\}$ and $\chi$ is an element of $\mathrm{Irr}_\mathcal{K}(\bar{C}_G(Q),\bar{b}_Q)-
  \{\chi_\tau\,|\,
  \tau\in\mathrm{Irr}_\mathcal{K}(\tilde{E}_1)_\theta\}$. Since $\langle\mathrm{Ind}(\xi)-\sum\limits_{i=1}^l\tau_i,
  \mathrm{Ind}(\xi)-\mathrm{Ind}(\xi)\bar{\zeta}\rangle=1$ for any $\bar{\zeta}\neq 1_{\bar{P}_2}$, we have $a_\chi=\delta_\xi$ and $\chi=\chi_\xi$. We are done.

Then the proof of Theorem \[MT\] will follow by Theorem \[MT’\] and [@PU 3.4.2].

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