LOEWY LENGTHS OF CENTERS OF BLOCKS II

Abstract

Let $ZB$ be the center of a $p$-block $B$ of a finite group with defect group $D$. We show that the Loewy length $LL(ZB)$ of $ZB$ is bounded by $|D|/p+p-1$ provided $D$ is not cyclic. If $D$ is nonabelian, we prove the stronger bound $LL(ZB)<\min \{p^{d-1},4p^{d-2}\}$ where $|D|=p^{d}$. Conversely, we classify the blocks $B$ with $LL(ZB)\geqslant \min \{p^{d-1},4p^{d-2}\}$. This extends some results previously obtained by the present authors. Moreover, we characterize blocks with uniserial center.

1 Introduction

The aim of this paper is to extend some results on Loewy lengths of centers of blocks obtained in [9, 12]. In the following we will reuse some of the notation introduced in [9]. In particular, $B$ is a block of a finite group $G$ with respect to an algebraically closed field $F$ of characteristic $p>0$ . Moreover, let $D$ be a defect group of $B$ . The second author has shown in [12, Corollary 3.3] that the Loewy length of the center of $B$ is bounded by

$$\begin{eqnarray}LL(ZB)\leqslant |D|-\frac{|D|}{\exp (D)}+1\end{eqnarray}$$

where $\exp (D)$ is the exponent of $D$ . It was already known to Okuyama [10] that this bound is best possible if $D$ is cyclic. The first and the third author have given in [9, Theorem 1] the optimal bound $LL(ZB)\leqslant LL(FD)$ for blocks with abelian defect groups. Our main result of the present paper establishes the following bound for blocks with nonabelian defect groups:

$$\begin{eqnarray}LL(ZB)<\min \{p^{d-1},4p^{d-2}\}\end{eqnarray}$$

where $|D|=p^{d}$ . As a consequence we obtain

$$\begin{eqnarray}LL(ZB)\leqslant p^{d-1}+p-1\end{eqnarray}$$

for all blocks with noncyclic defect groups. It can be seen that this bound is achieved whenever $B$ is nilpotent and $D\cong C_{p^{d-1}}\times C_{p}$ .

In the second part of the paper we show that $LL(ZB)$ depends more on $\exp (D)$ than on $|D|$ . We prove for instance that $LL(ZB)\leqslant d^{2}\exp (D)$ unless $d=0$ . Finally, we use the opportunity to improve a result of Willems [15] about blocks with uniserial center.

In addition to the notation used in the papers cited above, we introduce the following objects. Let $\operatorname{Cl}(G)$ be the set of conjugacy classes of $G$ . A $p$ -subgroup $P\leqslant G$ is called a defect group of $K\in \operatorname{Cl}(G)$ if $P$ is a Sylow $p$ -subgroup of $\operatorname{C}_{G}(x)$ for some $x\in K$ . Let $\operatorname{Cl}_{P}(G)$ be the set of conjugacy classes with defect group $P$ . Let $K^{+}:=\sum _{x\in K}x\in FG$ and

$$\begin{eqnarray}\displaystyle I_{P}(G) & := & \displaystyle \langle K^{+}:K\in \operatorname{Cl}_{P}(G)\rangle \subseteq ZFG,\nonumber\\ \displaystyle I_{{\leqslant}P}(G) & := & \displaystyle \mathop{\sum }_{Q\leqslant P}I_{Q}(G)\unlhd ZFG,\nonumber\\ \displaystyle I_{{<}P}(G) & := & \displaystyle \mathop{\sum }_{Q<P}I_{Q}(G)\unlhd ZFG\nonumber\end{eqnarray}$$

(here $\unlhd$ means that the subsets are ideals).

2 Results

We begin by restating a lemma of Passman [13, Lemma 2]. For the convenience of the reader we provide a (slightly easier) proof.

Lemma 1. Let $P$ be a central $p$ -subgroup of $G$ . Then $I_{{\leqslant}P}(G)\cdot JZFG=I_{{\leqslant}P}(G)\cdot JFP$ .

Proof. Let $K$ be a conjugacy class of $G$ with defect group $P$ , and let $x\in K$ . Then $P$ is the only Sylow $p$ -subgroup of $\operatorname{C}_{G}(x)$ , and the $p$ -factor $u$ of $x$ centralizes $x$ . Thus $u\in P$ . Hence $u$ is the $p$ -factor of every element in $K$ , and $K=uK^{\prime }$ where $K^{\prime }$ is a $p$ -regular conjugacy class of $G$ with defect group $P$ . This shows that $I:=I_{{\leqslant}P}(G)$ is a free $FP$ -module with the $p$ -regular class sums with defect group $P$ as an $FP$ -basis. The canonical epimorphism $\unicode[STIX]{x1D708}:FG\rightarrow F[G/P]$ maps $I$ into $I_{1}(G/P)\subseteq SF[G/P]$ (recall that $SF[G/P]$ is the socle of $F[G/P]$ ). Thus $\unicode[STIX]{x1D708}(I\cdot JZFG)\subseteq SF[G/P]\cdot JZF[G/P]=0$ . Hence $I\cdot JZFG\subseteq I\cdot JFP$ . The other inclusion is trivial.◻

Lemma 2. Let $P\leqslant G$ be a $p$ -subgroup of order $p^{n}$ . Then

1. (i) $I_{{\leqslant}P}(G)\cdot JZFG^{LL(F\operatorname{Z}(P))}\subseteq I_{{<}P}(G)$ .

2. (ii) $I_{{\leqslant}P}(G)\cdot JZFG^{(p^{n+1}-1)/(p-1)}=0$ .

Proof.

1. (i) Let $\operatorname{Br}_{P}:ZFG\rightarrow ZF\operatorname{C}_{G}(P)$ be the Brauer homomorphism. Since $\operatorname{Ker}(\operatorname{Br}_{P})\cap I_{{\leqslant}P}(G)=I_{{<}P}(G)$ , we need to show that $\operatorname{Br}_{P}(I_{{\leqslant}P}(G)\cdot JZFG^{LL(F\operatorname{Z}(P))})=0$ . By Lemma 1 we have

   $$\begin{eqnarray}\displaystyle & & \displaystyle \operatorname{Br}_{P}(I_{{\leqslant}P}(G)\cdot JZFG^{LL(F\operatorname{Z}(P))})\nonumber\\ \displaystyle & & \displaystyle \quad \subseteq I_{{\leqslant}\operatorname{Z}(P)}(\operatorname{C}_{G}(P))\cdot JZF\operatorname{C}_{G}(P)^{LL(F\operatorname{Z}(P))}\nonumber\\ \displaystyle & & \displaystyle \quad =I_{{\leqslant}\operatorname{Z}(P)}(\operatorname{C}_{G}(P))\cdot JF\operatorname{Z}(P)^{LL(F\operatorname{Z}(P))}=0.\nonumber\end{eqnarray}$$

2. (ii) We argue by induction on $n$ . The case $n=1$ follows from $I_{1}(G)\subseteq SFG$ . Now suppose that the claim holds for $n-1$ . Since $LL(F\operatorname{Z}(P))\leqslant |P|=p^{n}$ , (i) implies

   $$\begin{eqnarray}\displaystyle I_{{\leqslant}P}(G)\cdot JZFG^{(p^{n+1}-1)/(p-1)} & = & \displaystyle I_{{\leqslant}P}(G)\cdot JZFG^{p^{n}}JZFG^{(p^{n}-1)/(p-1)}\nonumber\\ \displaystyle & \subseteq & \displaystyle I_{{<}P}(G)\cdot JZFG^{(p^{n}-1)/(p-1)}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{Q<P}I_{{\leqslant}Q}(G)\cdot JZFG^{(p^{n}-1)/(p-1)}=0.\qquad \nonumber\end{eqnarray}$$

Recall from [9, Lemma 9] the following group

$$\begin{eqnarray}W_{p^{d}}:=\langle x,y,z\mid x^{p^{d-2}}=y^{p}=z^{p}=[x,y]=[x,z]=1,[y,z]=x^{p^{d-3}}\rangle .\end{eqnarray}$$

Note that $W_{p^{d}}$ is a central product of $C_{p^{d-2}}$ and an extraspecial group of order $p^{3}$ . Now we prove our main theorem which improves [9, Theorem 12].

Theorem 3. Let $B$ be a block of $FG$ with nonabelian defect group $D$ of order $p^{d}$ . Then (at least) one of the following holds

1. (i) $LL(ZB)<3p^{d-2}$ .

2. (ii) $p\geqslant 5$ , $D\cong W_{p^{d}}$ and $LL(ZB)<4p^{d-2}$ .

In any case we have

$$\begin{eqnarray}LL(ZB)<\min \{p^{d-1},4p^{d-2}\}.\end{eqnarray}$$

Proof. By [9, Proposition 15], we may assume that $p>2$ . Since $D$ is nonabelian, $|D:\operatorname{Z}(D)|\geqslant p^{2}$ and $LL(F\operatorname{Z}(D))\leqslant p^{d-2}$ . Let $Q$ be a maximal subgroup of $D$ . If $Q$ is cyclic, then $D\cong M_{p^{n}}$ (see [4, Theorem 5.4.4]) and the claim follows from [9, Proposition 10]. Hence, we may assume that $Q$ is not cyclic. Then $LL(F\operatorname{Z}(Q))\leqslant p^{d-2}+p-1$ . Now setting $\unicode[STIX]{x1D706}:=(p^{d-1}-1)/(p-1)$ it follows from Lemma 2 that

$$\begin{eqnarray}\displaystyle JZB^{2p^{d-2}+p-1+\unicode[STIX]{x1D706}} & \subseteq & \displaystyle 1_{B}JZFG^{2p^{d-2}+p-1+\unicode[STIX]{x1D706}}\subseteq I_{{\leqslant}D}(G)\cdot JZFG^{2p^{d-2}+p-1+\unicode[STIX]{x1D706}}\nonumber\\ \displaystyle & \subseteq & \displaystyle I_{{<}D}(G)\cdot JZFG^{p^{d-2}+p-1+\unicode[STIX]{x1D706}}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{Q<D}I_{{\leqslant}Q}(G)\cdot JZFG^{p^{d-2}+p-1+\unicode[STIX]{x1D706}}\nonumber\\ \displaystyle & \subseteq & \displaystyle \mathop{\sum }_{Q<D}I_{{<}Q}(G)\cdot JZFG^{\unicode[STIX]{x1D706}}=0.\nonumber\end{eqnarray}$$

Since $2p^{d-2}+p-1+\unicode[STIX]{x1D706}\leqslant 4p^{d-2}$ , we are done in case $p\geqslant 5$ and $D\cong W_{p^{d}}$ . If $p=3$ and $D\cong W_{p^{d}}$ , then the claim follows from [9, Lemma 11]. Now suppose that $D\not \cong W_{p^{d}}$ . If $\operatorname{Z}(D)$ is cyclic of order $p^{d-2}$ , then the claim follows from [9, Lemma 9 and Proposition 10]. Hence, suppose that $\operatorname{Z}(D)$ is noncyclic or $|\operatorname{Z}(D)|<p^{d-2}$ . Then $d\geqslant 4$ and $LL(F\operatorname{Z}(D))\leqslant p^{d-3}+p-1$ . The arguments above give $LL(ZB)\leqslant p^{d-2}+p^{d-3}+2p-2+\unicode[STIX]{x1D706}$ , hence we are done whenever $p>3$ .

In the following we assume that $p=3$ . Here we have $LL(ZB)\leqslant 3^{d-2}+3^{d-3}+4+\frac{1}{2}(3^{d-1}-1)$ and it suffices to handle the case $d=4$ . By [12, Theorem 3.2], there exists a nontrivial $B$ -subsection $(u,b)$ such that

$$\begin{eqnarray}LL(ZB)\leqslant (|\langle u\rangle |-1)LL(Z\overline{b})+1\end{eqnarray}$$

where $\overline{b}$ is the unique block of $F\operatorname{C}_{G}(u)/\langle u\rangle$ dominated by $b$ . We may assume that $\overline{b}$ has defect group $\operatorname{C}_{D}(u)/\langle u\rangle$ (see [14, Lemma 1.34]). If $u\notin \operatorname{Z}(D)$ , we obtain $LL(ZB)<|\!\operatorname{C}_{D}(u)|\leqslant 27$ as desired. Hence, let $u\in \operatorname{Z}(D)$ . Then $D/\langle u\rangle$ is not cyclic. Moreover, by our assumption on $\operatorname{Z}(D)$ , we have $|\langle u\rangle |=3$ . Now it follows from [9, Theorem 1, Proposition 10 and Lemma 11] applied to $\overline{b}$ that

$$\begin{eqnarray}\hspace{82.79993pt}LL(ZB)\leqslant 2LL(Z\overline{b})+1\leqslant 23<27.\hspace{82.79993pt}\square\end{eqnarray}$$

We do not expect that the bounds in Theorem 3 are sharp. In fact, we do not know if there are $p$ -blocks $B$ with nonabelian defect groups of order $p^{d}$ such that $p>2$ and $LL(ZB)>p^{d-2}$ . See also Proposition 7 below.

Corollary 4. Let $B$ be a block of $FG$ with noncyclic defect group of order $p^{d}$ . Then

$$\begin{eqnarray}LL(ZB)\leqslant p^{d-1}+p-1.\end{eqnarray}$$

Proof. By Theorem 3, we may assume that $B$ has abelian defect group $D$ . Then [9, Theorem 1] implies $LL(ZB)\leqslant LL(FD)\leqslant p^{d-1}+p-1$ .◻

We are now in a position to generalize [9, Corollary 16].

Corollary 5. Let $B$ be a block of $FG$ with defect group $D$ of order $p^{d}$ such that $LL(ZB)\geqslant \min \{p^{d-1},4p^{d-2}\}$ . Then one of the following holds

1. (i) $D$ is cyclic.

2. (ii) $D\cong C_{p^{d-1}}\times C_{p}$ .

3. (iii) $D\cong C_{2}\times C_{2}\times C_{2}$ and $B$ is nilpotent.

Proof. Again by Theorem 3 we may assume that $D$ is abelian. By [9, Corollary 16], we may assume that $p>2$ . Suppose that $D$ is of type $(p^{a_{1}},\ldots ,p^{a_{s}})$ such that $s\geqslant 3$ . Then

$$\begin{eqnarray}\displaystyle \min \{p^{d-1},4p^{d-2}\} & {\leqslant} & \displaystyle LL(ZB)=p^{a_{1}}+\cdots +p^{a_{s}}-s+1\nonumber\\ \displaystyle & {\leqslant} & \displaystyle p^{a_{1}}+p^{a_{2}}+p^{a_{3}+\cdots +a_{s}}-2\leqslant p^{d-2}+2(p-1).\nonumber\end{eqnarray}$$

This clearly leads to a contradiction. Therefore, $s\leqslant 2$ and the claim follows.◻

In case (i) of Corollary 5 it is known conversely that $LL(ZB)=(p^{d}-1)/l(B)+1>p^{d-1}$ (see [7, Corollary 2.8]).

Our next result gives a more precise bound by invoking the exponent of a defect group.

Theorem 6. Let $B$ be a block of $FG$ with defect group $D$ of order $p^{d}>1$ and exponent $p^{e}$ . Then

$$\begin{eqnarray}LL(ZB)\leqslant \biggl(\frac{d}{e}+1\biggr)\biggl(\frac{d}{2}+\frac{1}{p-1}\biggr)(p^{e}-1).\end{eqnarray}$$

In particular, $LL(ZB)\leqslant d^{2}p^{e}$ .

Proof. Let $\unicode[STIX]{x1D6FC}:=\lfloor d/e\rfloor$ . Let $P\leqslant D$ be abelian of order $p^{ie+j}$ with $0\leqslant i\leqslant \unicode[STIX]{x1D6FC}$ and $0\leqslant j<e$ . If $P$ has type $(p^{a_{1}},\ldots ,p^{a_{r}})$ , then $a_{i}\leqslant e$ for $i=1,\ldots ,r$ and

$$\begin{eqnarray}LL(FP)=(p^{a_{1}}-1)+\cdots +(p^{a_{r}}-1)+1\leqslant i(p^{e}-1)+p^{j}.\end{eqnarray}$$

Arguing as in Theorem 3, we obtain

$$\begin{eqnarray}\displaystyle LL(ZB) & {\leqslant} & \displaystyle \mathop{\sum }_{i=0}^{\unicode[STIX]{x1D6FC}}\mathop{\sum }_{j=0}^{e-1}i(p^{e}-1)+p^{j}=e(p^{e}-1)\biggl(\mathop{\sum }_{i=0}^{\unicode[STIX]{x1D6FC}}i\biggr)+(\unicode[STIX]{x1D6FC}+1)\frac{p^{e}-1}{p-1}\nonumber\\ \displaystyle & = & \displaystyle e(p^{e}-1)\frac{\unicode[STIX]{x1D6FC}(\unicode[STIX]{x1D6FC}+1)}{2}+(\unicode[STIX]{x1D6FC}+1)\frac{p^{e}-1}{p-1}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \biggl(\frac{d}{e}+1\biggr)\biggl(\frac{d}{2}+\frac{1}{p-1}\biggr)(p^{e}-1).\nonumber\end{eqnarray}$$

This proves the first claim. For the second claim we note that

$$\begin{eqnarray}\biggl(\frac{d}{e}+1\biggr)\biggl(\frac{d}{2}+\frac{1}{p-1}\biggr)\leqslant (d+1)\biggl(\frac{d}{2}+1\biggr)\leqslant d^{2}\end{eqnarray}$$

unless $d\leqslant 3$ . In these small cases the claim follows from Theorem 3 and Corollary 4.◻

If $2e>d$ and $p$ is large, then the bound in Theorem 6 is approximately $dp^{e}$ . The groups of the form $G=D=C_{p^{e}}\times \cdots \times C_{p^{e}}$ show that there is no bound of the form $LL(ZB)\leqslant Cp^{e}$ where $C$ is an absolute constant. A more careful argumentation in the proof above gives the stronger (but opaque) bound

$$\begin{eqnarray}\displaystyle LL(ZB) & {\leqslant} & \displaystyle \unicode[STIX]{x1D6FC}(p^{e}-1)\biggl(\frac{e(\unicode[STIX]{x1D6FC}-1)}{2}+\frac{1}{p-1}+d-\unicode[STIX]{x1D6FC}e\biggr)+\unicode[STIX]{x1D6FD}(p^{e}-1)\nonumber\\ \displaystyle & & \displaystyle +\,\frac{p^{d-\unicode[STIX]{x1D6FC}e}-1}{p-1}+p^{d-2-\unicode[STIX]{x1D6FD}e}\nonumber\end{eqnarray}$$

for nonabelian defect groups where $\unicode[STIX]{x1D6FC}:=\lfloor (d-1)/e\rfloor$ and $\unicode[STIX]{x1D6FD}:=\lfloor (d-2)/e\rfloor$ . We omit the details.

In the next result we compute the Loewy length for $d=e+1$ .

Proposition 7. Let $B$ be a block of $FG$ with nonabelian defect group of order $p^{d}$ and exponent $p^{d-1}$ . Then

$$\begin{eqnarray}LL(ZB)\leqslant \left\{\begin{array}{@{}ll@{}}2^{d-2}+1\quad & \text{if }p=2,\\ p^{d-2}\quad & \text{if }p>2\end{array}\right.\end{eqnarray}$$

and both bounds are optimal for every $d\geqslant 3$ .

Proof. Let $D$ be a defect group of $B$ . If $p>2$ , then $D\cong M_{p^{d}}$ (see [4, Theorem 5.4.4]) and we have shown $LL(ZB)\leqslant p^{d-2}$ in [9, Proposition 10]. Equality holds if and only if $B$ is nilpotent.

Therefore, we may assume $p=2$ in the following. The modular groups $M_{2^{d}}$ are still handled by [9, Proposition 10]. Hence, it remains to consider the defect groups of maximal nilpotency class, i. e., $D\in \{D_{2^{d}},Q_{2^{d}},SD_{2^{d}}\}$ . By [9, Proposition 10], we may assume that $d\geqslant 4$ . The isomorphism type of $ZB$ is uniquely determined by $D$ and the fusion system of $B$ (see [2]). The possible cases are listed in [14, Theorem 8.1]. If $B$ is nilpotent, [9, Proposition 8] gives $LL(ZB)=LL(ZFD)\leqslant LL(FD^{\prime })=2^{d-2}$ . Moreover, in the case $D\cong D_{2^{d}}$ and $l(B)=3$ we have $LL(ZB)\leqslant k(B)-l(B)+1=2^{d-2}+1$ by [12, Proposition 2.2]. In the remaining cases we present $B$ by quivers with relations which were constructed originally by Erdmann [3]. We refer to [5, Appendix B]. Keep in mind that we need to consider only one quiver for each fusion system.

1. (i) $D\cong D_{2^{d}}$ , $l(B)=2$ :

   
   By [5, Lemma 2.3.3], we have
   $$\begin{eqnarray}ZB=\operatorname{span}\{1,\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D702}^{i}:i=1,\ldots ,2^{d-2}\}.\end{eqnarray}$$
   It follows that $JZB^{2}=\langle \unicode[STIX]{x1D702}^{2}\rangle$ and $LL(ZB)=2^{d-2}+1$ .

2. (ii) $D\cong Q_{2^{d}}$ , $l(B)=2$ : Here [16, Lemma 6] gives the isomorphism type of $ZB$ directly as a quotient of a polynomial ring

   $$\begin{eqnarray}\displaystyle ZB & \cong & \displaystyle F[U,Y,S,T]/\!(Y^{2^{d-2}+1},U^{2}-Y^{2^{d-2}},S^{2},T^{2},SY,\nonumber\\ \displaystyle & & \displaystyle SU,ST,UY,UT,YT)\!.\nonumber\end{eqnarray}$$
   It follows that $JZB^{2}=(Y^{2})$ and again $LL(ZB)=2^{d-2}+1$ .

3. (iii) $D\cong Q_{2^{d}}$ , $l(B)=3$ :

   
   By [5, Lemma 2.5.15],
   $$\begin{eqnarray}\displaystyle ZB & = & \displaystyle \operatorname{span}\{\!1,\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD},(\unicode[STIX]{x1D705}\unicode[STIX]{x1D706})^{i}+(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{i},\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}+\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF},(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2},(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2^{d-2}},\nonumber\\ \displaystyle & & \displaystyle (\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}:i=1,\ldots ,2^{d-2}-1\!\}\!.\nonumber\end{eqnarray}$$
   We compute
   $$\begin{eqnarray}\displaystyle & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})^{2}=(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})^{2}=(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}+\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D706}\unicode[STIX]{x1D6FD}=(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}+(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})(\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})=\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}=\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D706}=\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}+\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF})=\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}=(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{3}=\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D706}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2^{d-2}}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}=\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \!\!\!\!\!\!\!\!((\unicode[STIX]{x1D705}\unicode[STIX]{x1D706})^{2^{d-2}-1}+(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2^{d-2}-1})(\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})=\unicode[STIX]{x1D705}\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FE}+(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2^{d-2}}=(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}+(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2^{d-2}}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}+\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF})=\unicode[STIX]{x1D706}\unicode[STIX]{x1D705}\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}=\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}=\unicode[STIX]{x1D705}\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2^{d-2}}=\unicode[STIX]{x1D706}\unicode[STIX]{x1D705}\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D705}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}+\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF})^{2}=(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}+(\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF})^{2}=(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FF}=(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}+(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2^{d-2}}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}+\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF})(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}+\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF})(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2^{d-2}}=\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF}(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2^{d-2}}=\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D705}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}+\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FF})(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D706}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}=\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}=(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2^{d-2}}=(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2^{d-2}}(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2^{d-2}}=(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2^{d-2}}(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}=\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D705}\unicode[STIX]{x1D702}(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}=\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}=0. & \displaystyle \nonumber\end{eqnarray}$$
   Hence, $JZB^{2}=\langle (\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2}+(\unicode[STIX]{x1D705}\unicode[STIX]{x1D706})^{2},(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}+(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}\rangle$ and $JZB^{3}=\langle (\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{3}+(\unicode[STIX]{x1D705}\unicode[STIX]{x1D706})^{3}\rangle$ . This implies $LL(ZB)=2^{d-2}+1$ .

4. (iv) $D\cong SD_{2^{d}}$ , $k(B)=2^{d-2}+3$ and $l(B)=2$ :

   
   By [6, Section 5.1], we have
   $$\begin{eqnarray}ZB=\operatorname{span}\{1,\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D702}^{i}:i=1,\ldots ,2^{d-2}\}.\end{eqnarray}$$
   As in (i) we obtain $JZB^{2}=\langle \unicode[STIX]{x1D702}^{2}\rangle$ and $LL(ZB)=2^{d-2}+1$ .

5. (v) $D\cong SD_{2^{d}}$ , $k(B)=2^{d-2}+4$ and $l(B)=2$ :

   
   By [6, Section 5.2.2], we have
   $$\begin{eqnarray}\displaystyle ZB & = & \displaystyle \operatorname{span}\{\!1,\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE},(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2},\unicode[STIX]{x1D702}^{i},\nonumber\\ \displaystyle & & \displaystyle \unicode[STIX]{x1D702}+\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}:i=2,\ldots ,2^{d-2}\!\}\!.\nonumber\end{eqnarray}$$
   Since $(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}=\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FE}=(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC})^{2}$ and $(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD})^{2}=\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}=\unicode[STIX]{x1D702}^{2^{d-2}}$ , it follows that
   $$\begin{eqnarray}(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD})^{2}=(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}+(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC})^{2}+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD})^{2}=\unicode[STIX]{x1D702}^{2^{d-2}}.\end{eqnarray}$$
   Similarly,
   $$\begin{eqnarray}\displaystyle & \displaystyle (\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD})\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD})(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD})\unicode[STIX]{x1D702}^{2}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD})(\unicode[STIX]{x1D702}+\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC})=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D702}^{2}=\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D702}^{2}\unicode[STIX]{x1D6FE}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}(\unicode[STIX]{x1D702}+\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC})=\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}\unicode[STIX]{x1D6FC}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}\unicode[STIX]{x1D702}^{2}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}(\unicode[STIX]{x1D702}+\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC})=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \unicode[STIX]{x1D702}^{2}(\unicode[STIX]{x1D702}+\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC})=\unicode[STIX]{x1D702}^{3}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D702}+\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FC})^{2}=\unicode[STIX]{x1D702}^{2}. & \displaystyle \nonumber\end{eqnarray}$$
   Consequently, $JZB^{2}=\langle \unicode[STIX]{x1D702}^{2}\rangle$ and $LL(ZB)=2^{d-2}+1$ .

6. (vi) $D\cong SD_{2^{d}}$ , $l(B)=3$ :

   
   From [5, Lemma 2.4.16] we get
   $$\begin{eqnarray}\displaystyle ZB & = & \displaystyle \operatorname{span}\{\!1,(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{i}+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})^{i},\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D705},(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}},(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2},\nonumber\\ \displaystyle & & \displaystyle \unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}:i=1,\ldots ,2^{d-2}-1\!\}.\nonumber\end{eqnarray}$$
   We compute
   $$\begin{eqnarray}\displaystyle & \displaystyle \!\!\!(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})((\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}-1}+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})^{2^{d-2}-1})=(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}}+\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D706}\unicode[STIX]{x1D6FD}=(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}}+\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})(\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})=\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}}=\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D706}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}=\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}\unicode[STIX]{x1D705}\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FE}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}=\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}=\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}\unicode[STIX]{x1D705}\unicode[STIX]{x1D702}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2}=\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D706}+(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2}=(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}}+(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2}, & \displaystyle \nonumber\end{eqnarray}$$
   $$\begin{eqnarray}\displaystyle (\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}} & = & \displaystyle \unicode[STIX]{x1D705}\unicode[STIX]{x1D706}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE}(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}-1}\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D705}\unicode[STIX]{x1D702}\unicode[STIX]{x1D6FE}(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}-1}=0,\nonumber\\ \displaystyle (\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2} & = & \displaystyle \unicode[STIX]{x1D706}(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}}\unicode[STIX]{x1D705}\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D702}\unicode[STIX]{x1D6FE}(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}-1}\unicode[STIX]{x1D705}=0,\nonumber\end{eqnarray}$$
   $$\begin{eqnarray}\displaystyle (\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}+\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}=0, & & \displaystyle \nonumber\end{eqnarray}$$
   $$\begin{eqnarray}\displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}}(\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}} & = & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}}(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2}\nonumber\\ \displaystyle & = & \displaystyle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2^{d-2}}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}=0,\nonumber\end{eqnarray}$$
   $$\begin{eqnarray}\displaystyle & \displaystyle (\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2}(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2}=(\unicode[STIX]{x1D706}\unicode[STIX]{x1D705})^{2}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702})^{2}=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D706}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D706}\unicode[STIX]{x1D705}\unicode[STIX]{x1D706}\unicode[STIX]{x1D705}\unicode[STIX]{x1D702}=0. & \displaystyle \nonumber\end{eqnarray}$$
   Hence, $JZB^{2}=\langle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{2}+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})^{2},(\unicode[STIX]{x1D705}\unicode[STIX]{x1D706})^{2}+\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}\rangle$ and $JZB^{3}=\langle (\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FE})^{3}+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD})^{3}\rangle$ . This implies $LL(ZB)=2^{d-2}+1$ . ◻

It is interesting to note the difference between even and odd primes in Proposition 7. For $p=2$ , non-nilpotent blocks give larger Loewy lengths while for $p>2$ the maximal Loewy length is only attained for nilpotent blocks.

Recall that a lower defect group of a block $B$ of $FG$ is a $p$ -subgroup $Q\leqslant G$ such that

$$\begin{eqnarray}I_{{<}Q}(G)1_{B}\neq I_{{\leqslant}Q}(G)1_{B}.\end{eqnarray}$$

In this case $Q$ is conjugate to a subgroup of a defect group $D$ of $B$ and conversely $D$ is also a lower defect group since $1_{B}\in I_{{\leqslant}D}(G)\setminus I_{{<}D}(G)$ . It is clear that in the proofs of Theorems 3 and 6 it suffices to sum over the lower defect groups of $B$ . In particular there exists a chain of lower defect groups $Q_{1}<\cdots <Q_{n}=D$ such that $LL(ZB)\leqslant \sum _{i=1}^{n}LL(F\operatorname{Z}(Q_{i}))$ . Unfortunately, it is hard to compute the lower defect groups of a given block.

The following proposition generalizes [15, Theorem 1.5].

Proposition 8. Let $B$ be a block of $FG$ . Then $ZB$ is uniserial if and only if $B$ is nilpotent with cyclic defect groups.

Proof. Suppose first that $ZB$ is uniserial. Then $ZB\cong F[X]/(X^{n})$ for some $n\in \mathbb{N}$ ; in particular, $ZB$ is a symmetric $F$ -algebra. Then [11, Theorems 3 and 5] implies that $B$ is nilpotent with abelian defect group $D$ . Thus, by a result of Broué and Puig [1] (see also [8]), $B$ is Morita equivalent to $FD$ ; in particular, $FD$ is also uniserial. Thus $D$ is cyclic.

Conversely, suppose that $B$ is nilpotent with cyclic defect group $D$ . Then the Broué–Puig result mentioned above implies that $B$ is Morita equivalent of $FD$ . Thus $ZB\cong ZFD=FD$ . Since $FD$ is uniserial, the result follows.◻

A similar proof shows that $ZB$ is isomorphic to the group algebra of the Klein four group over an algebraically closed field of characteristic $2$ if and only if $B$ is nilpotent with Klein four defect groups.