# On left braces in which every subbrace is an ideal

A. Ballester-Bolinches Departament de Matemàtiques, Universitat de València;
Dr. Moliner, 50; 46100 Burjassot, València, Spain<EMAIL_ADDRESS>ORCID 0000-0002-2051-9075<EMAIL_ADDRESS>ORCID 0000-0002-2321-8139;
<EMAIL_ADDRESS>ORCID 0000-0003-4101-8656 R. Esteban-
Romero00footnotemark: 0 L. A. Kurdachenko Department of Algebra and Geometry,
Oles Honchar Dnipro National University, Dnipro 49010, Ukraine;
<EMAIL_ADDRESS>ORCID 0000-0002-6368-7319Part of the research of this
author has been carried out in the Departament de Matemàtiques, Universitat de
València; Dr. Moliner, 50; 46100 Burjassot, València, Spain V. Pérez-
Calabuig-2-2footnotemark: -2

###### Abstract

The aim of this paper is to introduce and study the class of all left braces
in which every subbrace is an ideal. We call them Dedekind left braces. It is
proved that every finite Dedekind left brace is centrally nilpotent.
Structural results about Dedekind left braces and a complete description of
those ones whose additive group is elementary abelian are also shown. As a
consequence, every hypermultipermutational Dedekind left brace whose additive
group is elementary abelian is multipermutational of level $2$. A new class of
left braces, the extraspecial left braces, is introduced and plays a prominent
role in our approach.

_Mathematics Subject Classification (2020):_ 16T25, 16N40, 81R50

_Keywords:_ Dedekind left braces, Yang-Baxter equation, central nilpotency,
elementary abelian, extraspecial left braces.

##  1 Introduction

In his inspiring paper [13], Drinfeld posed the question of study the so-
called set-theoretical solutions of the Yang-Baxter equation (YBE) which has
given rise to an intensive research by means of algebraic and combinatorial
methods (see [14, 16, 24], for example). The main problem in this context is
to find and classify all set-theoretical solutions that satisfy some
prescribed properties which are determined by the action of algebraic
structures. Rump’s _left braces_ stand out among them as a first exponent of a
new algebraic structure introduced in [25] to deal with the classification of
all involutive non-degenerate set-theoretical solutions of the YBE (solutions,
for short).

A left brace is a nonempty set $A$ endowed with two group structures $(A,+)$
and $(A,\cdot)$, such that $(A,+)$ is abelian and

$a\cdot(b+c)=a\cdot b+a\cdot c-a,\ \text{for every $a,b,c\in A$.}$

In particular, a left brace is called _abelian_ if both operations coincide.
Every left brace provides a solution of the YBE, and viceversa, every solution
of the YBE is a restriction of the solution associated to a left brace. Then,
it is abundantly clear that the more we know about the algebraic structure of
left braces the more we go forward in the classification problem of solutions.

The illustrative survey of Cedó [7] features a selection of the most important
algebraic developments on left braces with a great impact on the study of
classes of solutions. In particular, the class of _multipermutational
solutions_ sticks out as it admits an algebraic treatment by means of left-
ordered groups and nilpotent notions of left braces (see [8, 9, 18, 17, 28]
for further information). Moreover, this class has been object of an intensive
study since it covers almost all solutions: “among all involutive solutions
with at most ten elements, more than 95% are multipermutation” (in [22]).
Consequently, the classification problem of solutions undoubtedly calls for an
analysis of subfamilies of multipermutation solutions. In this light, it is
particularly studied the case of multipermuation solutions of level $2$ (v.
gr. [19, 20, 21, 26]).

Like every algebraic structure, the algebraic study of left braces is based on
interactions between substructures (subbraces) and quotients (provided by the
so-called ideals). In this paper, we introduce the brace theoretic analogue of
Dedekind groups, namely _Dedekind left braces_ or left braces for which every
subbrace is an ideal (see Section 3). One of our main motivations comes from
the well-known fact that Dedekind groups form a family of groups with
nilpotency class $2$ (it follows from the structural theorems proved by
Dedekind [12] and Baer [1]). A first glance to examples of Dedekind left
braces in Section 3 shows that a structural theorem for them is a challenging
open problem. In this direction, our first main result shows that every finite
Dedekind left brace is centrally nilpotent (see Theorem 12 and Section 2 for
preliminary notions on nilpotency of left braces). As a consequence, every
solution associated with a finite left brace is multipermutational.

The fact that the subbrace structure coincides with the ideal structure of a
left brace not only has an algebraic interest by their own, but also could
shed light on the study of simple and soluble left braces and associated
simple and soluble solutions (see [4] and [10] for further information).
Delving into the study of Dedekind left braces, one catches a glimpse of the
importance of getting a tighter grip of the abelian nature of the left brace
structure. According to the classification of finite abelian groups,
elementary abelian groups can be considered as atomic abelian structures.
Therefore, it is natural to think about them as a first remarkable family for
which a structural theorem can be proved. Section 4 is devoted to prove such
structural theorem: see Theorems 27 and 33, and Corollary 34. These results
hinge on two main facts. On one hand, we introduce extraspecial left braces
which play a key role in the proof of our results. On the other hand, we prove
that hypermultipermutational Dedekind left braces whose additive group is
elementary abelian have multipermutational level $2$. It follows that finite
Dedekind left braces whose additive group is elementary abelian have central
nilpotency length $2$. Consequently, remarkable families of left braces whose
associated multipermutation solutions have level $2$ are completely described.

##  2 Preliminaries

In this section, we fix notation and give some background concepts and results
on braces. Although many of them can be considered folklore, others are new.
We also include a subsection on results of bilinear forms which are crucial to
prove our main results.

In this paper, products on left braces will be denoted by juxtaposition and
come before sums on left braces. Left braces $(A,+,\cdot)$ will be referred
simply as $A$, whenever the operations $+$ and $\cdot$ are clear from the
context. We shall denote with $(A,+)$ (resp. $(A,\cdot)$) the additive (resp.
multiplicative) group of a left brace $A$, and we denote by $0$ the common
identity element in $(A,+)$ and $(A,\cdot)$. If $A$ is finite, we denote
$\pi(A)$ the set of primes dividing the order $|A|$ of $A$.

Following Rump [25], left braces can be seen as a generalisation of Jacobson
radical rings by means of the so-called _star operation_

$\ast\colon(a,b)\in A\times A\longmapsto a\ast b=ab-a-b\in A.$

We write star products preceding both products and sums in left braces.
Moreover, there exists an action of the multiplicative group on the additive
group, usually denoted as the homomorphism $\lambda\colon
a\in(A,\cdot)\mapsto\lambda_{a}\in\operatorname{Aut}(A,+)$, such that
$ab=a+\lambda_{a}(b)$ and $a\ast b=\lambda_{a}(b)-b$ links all operations in
$A$. In particular, the following equalities are easily verified and are
widely used throughout the paper:

$\displaystyle a\ast(b+c)$ $\displaystyle=$ $\displaystyle a\ast b+a\ast c$
(1) $\displaystyle(ab)\ast c$ $\displaystyle=$ $\displaystyle a\ast(b\ast
c)+b\ast c+a\ast c$ (2)

Substructures in left braces are classified according to each embeddings in
their additive and multiplicative groups. Let $A$ be a left brace. A non-empty
subset $S$ of $A$ is called a _subbrace_ if $(S,+)$ and $(S,\cdot)$ are each
subgroups of $(A,+)$ and $(A,\cdot)$, respectively. We write $S\leq A$. A
subbrace $I$ of $A$ is called an _ideal_ if it is $\lambda$-invariant and
$(I,\cdot)\unlhd(A,\cdot)$. Ideals are characterised as those subbraces $I$ of
$A$ such that $A\ast I,I\ast A\subseteq I$. If $I$ is an ideal of $A$ then
$a+I=aI$, and therefore, a quotient left brace structure $(A/I,+,\cdot)$ is
provided. Moreover, it holds $(a+I)\ast(b+I)=(a\ast b)+I$, for every $a,b\in
A$.

Let $S$ be a subset of a left brace $A$. We denote by $\langle S\rangle$ the
_subbrace generated_ by $S$, which is the intersection of all subbraces of $A$
containing $S$, and $\langle S\rangle_{+}$, $\langle
S\rangle_{\boldsymbol{\cdot}}$ denote generated subgroups in the additive and
the multiplicative group of $A$, respectively. If $S$ is a singleton
$\\{a\\}$, we just write $\langle a\rangle$ for the corresponding generated
substructure. If $X$ and $Y$ are subsets of $A$, $X\ast Y=\langle x\ast y\mid
x\in X,\,y\in Y\rangle_{+}$. A left brace $A$ is said to be _abelian_ if
$A\ast A=0$, or equivalently, $ab=a+b=b+a=ba$ for every $a,b\in A$. We say
that a subbrace (resp. ideal, quotient left brace) is abelian if it is abelian
as a left brace structure. For abelian left braces, both group structures
coincide, and therefore, we can refer to them as _isomorphic_ to the common
group structure.

######  Lemma 1.

Let $A$ be a left brace and let $a\in A$ such that $a\ast a=0$. Then, $\langle
a\rangle=\langle a\rangle_{+}=\langle a\rangle_{\boldsymbol{\cdot}}$ is an
abelian subbrace of $A$.

###### Proof.

Since $a\ast a=0$, it holds $a^{2}=2a$. Then, we see that

$a^{3}=aa^{2}=a(a+a)=a^{2}+a^{2}-a=2a+2a-a=3a$

By induction, it follows $a^{n}=na$ for all positive integers $n$. Moreover,

$a=a^{-1}a^{2}=a^{-1}(2a)=a^{-1}(a+a)=a^{-1}a+a^{-1}a-a^{-1}=0+0-a^{-1}=-a^{-1}$

Thus, $a^{-1}=-a$. Then,

$a^{-2}=a^{-1}a^{-1}=a^{-1}(-a)=a^{-1}(0-a)=a^{-1}-a^{-1}a+a^{-1}=-a-0-a=-2a$

By induction, we obtain that $a^{-n}=-na$, for all positive integer $n$.
Therefore, it follows that $(ka)(na)=(k+n)a$ and $(ka)^{-1}=-ka$, for all
integers $k,n$. This shows that $\langle a\rangle=\langle a\rangle_{+}=\langle
a\rangle_{\boldsymbol{\cdot}}$ is an abelian subbrace of $A$. ∎

_Homomorphisms_ between two left braces $A$ and $B$ are mappings conserving
sums and products; $A$ and $B$ are called _isomorphic_ if there exists an
isomorphism (i.e. a bijective homomorphism) between them; and the _direct
product_ $A\oplus B$ is constructed as the direct product of each sums and
products of $A$ and $B$.

The following particular substructures play a key role in the structural study
of a left brace $A$. Given $\emptyset\neq S\subseteq A$, the following sets

$\displaystyle\operatorname{Ann}_{A}^{l}(S)$ $\displaystyle=\\{a\in A\mid
a\ast x=0,\,\text{for all $x\in S$}\\}=$ $\displaystyle=\\{a\in A\mid
ax=a+x,\,\text{for all $x\in S$}\\};$
$\displaystyle\operatorname{Ann}_{A}^{r}(S)$ $\displaystyle=\\{a\in A\mid
x\ast a=0,\,\text{for all $x\in S$}\\}=$ $\displaystyle=\\{a\in A\mid
xa=x+a,\,\text{for all $x\in S$}\\};$ $\displaystyle\operatorname{Ann}_{A}(S)$
$\displaystyle=\operatorname{Ann}_{A}^{l}(S)\cap\operatorname{Ann}_{A}^{r}(S)=$
$\displaystyle=\\{a\in A\mid ax=a+x=xa,\,\text{for all $x\in S$}\\};$

are respectively defined as the _left annihilator_ , _right annihilator_ , and
_annihilator_ of $S$. In case that $S=A$, they are respectively called the
_socle_ $\operatorname{Soc}(A)$, the _fix_ $\operatorname{Fix}(A)$, and the
_centre_ $\zeta(A)$ of the left brace $A$ (see [5], [6] or [11], for example).
It follows that a left brace $A$ is abelian if, and only if, $\zeta(A)=A$.
Moreover, every subbrace of the centre of a left brace is an abelian ideal and
every subbrace of the socle is abelian.

######  Lemma 2.

Let $A$ be a left brace and let $\emptyset\neq S\subseteq A$. Then, the
following hold

  1. 1.

$\operatorname{Ann}_{A}^{l}(S)\leq(A,\cdot)$. Moreover, if $S$ is an ideal,
$(\operatorname{Ann}_{A}^{l}(S),\cdot)\unlhd(A,\cdot)$.

  2. 2.

$\operatorname{Ann}_{A}^{r}(S)\leq(A,+)$.

  3. 3.

$\operatorname{Ann}_{A}(S)$ is a subgroup of the centraliser of $S$ in the
multiplicative group $(A,\cdot)$. Moreover, if $S$ is an ideal, then
$(\operatorname{Ann}_{A}(S),\cdot)\unlhd(A,\cdot)$.

###### Proof.

1\. Let $a,b\in\operatorname{Ann}_{A}^{l}(S)$. For every $x\in S$, it holds

$x=a^{-1}ax=a^{-1}(a+x)=0-a^{-1}+a^{-1}x,$

i.e $a^{-1}x=a^{-1}+x$. Moreover, $(ab)x=a(bx)=a(b+x)=ab+ax-a=ab+x$. Thus,
$(\operatorname{Ann}_{A}^{l}(S),\cdot)\leq(A,\cdot)$.

Assume that $S$ is an ideal. Then, we can consider the restriction of the
$\lambda$-action $\bar{\lambda}\colon a\in
A\rightarrow\left.\lambda_{a}\right|_{S}\in\operatorname{Aut}(S,+)$. Thus,
$a\in\operatorname{Ker}\bar{\lambda}$ if, and only if, $\lambda_{a}(s)=s$ for
every $s\in S$, i.e. $a\in\operatorname{Ann}_{A}^{l}(S)$. Hence,
$\operatorname{Ann}_{A}^{l}(S)=\operatorname{Ker}\bar{\lambda}\unlhd(A,\cdot)$.

2\. It clearly follows from (1).

3\. By definition, $\operatorname{Ann}_{A}(S)$ is a subgroup of the
centraliser of $S$ in the multiplicative group $(A,\cdot)$. Suppose that $S$
is an ideal of $A$. Since $(S,\cdot)\unlhd(A,\cdot)$ it follows that the
centraliser of $S$ in the multiplicative group $(A,\cdot)$ is a normal
subgroup.

Let $b\in A$ and $a\in\operatorname{Ann}_{A}(S)$. Thus,
$bab^{-1}\in\operatorname{Ann}_{A}^{l}(S)$ and $(bab^{-1})x=x(bab^{-1})$, for
every $x\in S$. Hence,

$x(bab^{-1})=(bab^{-1})x=bab^{-1}+x=x+bab^{-1},$

i.e. $bab^{-1}\in\operatorname{Ann}_{A}(S)$. ∎

In particular, it follows that socle and centre of a left brace are ideals
while fix subbrace of a left brace is a left ideal.

Let $A$ be a left brace. The left and right series of $A$ are respectively
defined as

$\displaystyle A^{1}=A\geq A^{2}=A\ast A\geq\ldots\geq A^{n+1}=A\ast
A^{n}\geq\ldots$ $\displaystyle A^{(1)}=A\geq A^{(2)}=A\ast A\geq\ldots\geq
A^{(n+1)}=A^{(n)}\ast A\geq\ldots$

Following [25], $A$ is said to be _left (right) nilpotent_ if the left (right)
series reaches $0$. Then, $A$ is said to be _centrally nilpotent_ or
_nilpotent in the sense of Smoktunowicz_ if it is left and right nilpotent
(see [2, 5, 11] for further information).

Now, we consider

$\displaystyle
0=\operatorname{Soc}_{0}(A)\leq\operatorname{Soc}_{1}(A)\leq\ldots\leq\operatorname{Soc}_{n}(A)\leq\operatorname{Soc}_{n+1}(A)\leq\ldots$
$\displaystyle
0=\zeta_{0}(A)\leq\zeta_{1}(A)\leq\ldots\leq\zeta_{n}(A)\leq\zeta_{n+1}(A)\leq\ldots$

ascending series of ideals defined by the recursive rule:

$\begin{array}[]{ll}\operatorname{Soc}_{n+1}(A)/\operatorname{Soc}_{n}(A)=\operatorname{Soc}(A/\operatorname{Soc}_{n}(A)),&\zeta_{n+1}(A)/\zeta_{n}(A)=\zeta(A/\zeta_{n}(A));\end{array}$

for every non-negative integer $n$. These are respectively called the _socle
(resp. upper central) series_ of $A$. A left brace is said to have _finite
multipermutational level_ if $\operatorname{Soc}_{n}(A)=A$ for some $n$. On
the other hand, we consider

$0=\operatorname{Fix}_{0}(A)\leq\operatorname{Fix}_{1}(A)\leq\ldots\leq\operatorname{Fix}_{n}(A)\leq\operatorname{Fix}_{n+1}(A)\leq\ldots$

a series of left ideals defined by the recursive rule:

$\operatorname{Fix}_{n+1}(A)=\\{a\in A\mid x\ast
a\in\operatorname{Fix}_{n}(A),\,\text{for every $x\in A$}\\}$

for every non-negative integer $n$. This is called the _fix series_ of $A$.

######  Proposition 3 ([3, 5, 11]).

Let $A$ be a left brace. Then, $A$ is left (right, centrally) nilpotent if,
and only if, the fix (socle, upper central) series reaches $A$. Moreover, the
length of this series is called _the left (right, centrally) nilpotent level_
of $A$.

### Bilinear forms

We end the section of preliminaries with a subsection which features some
concepts and results on bilinear forms over finite dimensional vector spaces.
_From now on, $V$ denotes a finite dimensional vector space over a field
$\mathbb{F}$ and $\phi$ a bilinear form on $V$._

Let $U\leq V$. We write

$\begin{array}[]{l}{}^{\perp}U=\\{x\in V\mid\phi(x,u)=0,\,\text{for all
elements $u\in U$}\\},\\\ U^{\perp}=\\{x\in V\mid\phi(u,x)=0,\,\text{for all
elements $u\in U$}\\}.\end{array}$

It turns out that ${}^{\perp}U$ and $U^{\perp}$ are subspaces of $V$, which
are called the _left (resp. right) orthogonal complement_ of $U$ in $V$.

The following result is well-known.

######  Proposition 4.

Given $U\leq V$, it follows that

$\dim_{\operatorname{\mathbb{F}}}U+\dim_{\operatorname{\mathbb{F}}}\,^{\perp}U-\dim_{\operatorname{\mathbb{F}}}(U\cap
V^{\perp})=\dim_{\operatorname{\mathbb{F}}}V$

Moreover, if the restriction of $\phi$ over $U$ is non-degenerate, then
$V=U\oplus\,^{\perp}U$.

A bilinear form $\phi$ is called _strong non-degenerate_ if the restriction of
$\phi$ over every non-zero subspace is non-degenerate.

######  Corollary 5.

Assume that $\phi$ is strong non-degenerate on $V$. Then, $V$ has a basis in
which the matrix of $\phi$ is upper triangular.

###### Proof.

Let $n=\dim_{\operatorname{\mathbb{F}}}V$ and assume the assertion true for
every vector space of dimension less than $n$ with $n>1$. Take $U_{1}=\langle
u_{1}\rangle$ for some $0\neq u_{1}\in V$. Since the restriction of $\phi$
over $U_{1}$ is non-degenerate, Proposition 4 shows that
$V=U_{1}\oplus\,^{\perp}U_{1}$. By induction, there exists a basis
$\\{u_{2},\ldots,u_{n}\\}$ of ${}^{\perp}U_{1}$ in which the basis of
$\phi|_{{}^{\perp}U_{1}}$ is upper triangular. Hence,
$\\{u_{1},u_{2},\ldots,u_{n}\\}$ is a basis of $V$ in which the matrix of
$\phi$ is upper triangular. ∎

######  Corollary 6.

Assume that $\phi(x,x)\neq 0$ for every $0\neq x\in V$. Then, $V$ has a basis
in which the matrix of $\phi$ is upper triangular.

######  Corollary 7.

Suppose that $V$ is a vector space over a finite field
$\operatorname{\mathbb{F}}_{q}$, with $q=p^{t}$ for some prime $p$. Let $\phi$
be a bilinear form on $V$ such that $\phi(x,x)\neq 0$ for every $0\neq x\in
V$. Then, $V$ has dimension at most $2$.

###### Proof.

Assume that $\dim_{\operatorname{\mathbb{F}}_{q}}(V)>2$. By Corollary 6, we
can take a subspace $S\leq V$ with a basis $\\{b_{1},b_{2},b_{3}\\}$ such that
$\phi(b_{2},b_{1})=\phi(b_{3},b_{1})=0$. Set
$\sigma_{i,j}:=\phi(b_{i},b_{j})$, for $1\leq i\leq j\leq 3$. Let
$x=\lambda_{1}b_{1}+\lambda_{2}b_{2}+\lambda_{3}b_{3}\in S$ with
$\lambda_{i}\in\operatorname{\mathbb{F}}_{q}$, $1\leq i\leq 3$. Then, it holds

$\phi(x,x)=\sum_{1\leq i\leq j\leq 3}\lambda_{i}\lambda_{j}\sigma_{i,j}$

By Chevalley’s Theorem (see [23, Corollary 6.6], for example),

$\sum_{1\leq i\leq j\leq 3}\sigma_{i,j}X_{i}X_{j}$

has a non-zero solution. Therefore, there exists $0\neq x\in V$ such that
$\phi(x,x)=0$, a contradiction. ∎

##  3 Dedekind left braces

Our definition of Dedekind left braces is inspired by the correspondent notion
of Dedekind groups.

######  Definition 8.

A left brace $A$ is said to be _Dedekind_ if every subbrace of $A$ is ideal.

It is clear that the property of being Dedekind is inherited in subbraces and
quotients.

######  Remark 9.

Obviously, every brace with no proper subbraces is Dedekind. Theorem A in [4]
shows that finite left braces with no proper subbraces are abelian left braces
isomorphic to a cyclic group of order prime.

Of course, abelian left braces are Dedekind, but there are left braces whose
multiplicative group is isomorphic to the additive group which are not
Dedekind: the left brace corresponding to SmallBrace(8, 7) in the Yang–Baxter
library [29] for GAP [15] has both additive and multiplicative groups
isomorphic to the direct product of cyclic groups
$\operatorname{C}_{4}\times\operatorname{C}_{2}$ but it is not Dedekind as it
has a subbrace of order $2$ which is not ideal.

This example also shows that having both the additive and multiplicative
groups Dedekind it is not sufficient for a left brace to be Dedekind. This
situation changes if we assume the additive group to be cyclic.

######  Proposition 10.

Let $A$ be a left brace with a cyclic additive group and a Dedekind
multiplicative group. Then, $A$ is Dedekind.

###### Proof.

Since $(A,+)$ is cyclic, it follows that every subgroup of $(A,+)$ is
$\lambda$-invariant. Thus, every subgroup of $(A,+)$ is a subbrace of $A$.
Since both $(A,+)$ and $(A,\cdot)$ are Dedekind groups, we conclude that every
subbrace is an ideal. ∎

The condition on the multiplicative group can not be removed in general: there
exists a left brace with additive group isomorphic to the cyclic group
$\operatorname{C}_{6}$ and multiplicative group isomorphic to the symmetric
group $\text{Sym}(3)$ such that the cyclic subgroup of order $2$ in
$\operatorname{C}_{6}$ is a subbrace but not an ideal.

Moreover, Proposition 10 can not been extended for one-generated braces. In
[4], one-generated left braces $A$ with $\operatorname{Soc}_{2}(A)=A$ are
shown to admit useful descriptions. However, this condition it is not
sufficient to yield Dedekind left braces.

######  Example 11.

Let $A$ be the left brace corresponding to SmallBrace(8, 7) in the Yang–Baxter
library [29] for GAP [15]. It has additive group $(A,+)$ isomorphic to
$\operatorname{C}_{4}\times\operatorname{C}_{4}$, and multiplicative group
$(A,\cdot)$ isomorphic to $\operatorname{C}_{8}\times\operatorname{C}_{2}$
which is a Dedekind group. Moreover, $\operatorname{Soc}_{2}(A)=A$ but $A$ is
not a Dedekind left brace since it has a subbrace of order $2$ which is not an
ideal. Indeed, it is a minimal counterexample. In addition, the left brace $A$
corresponding to SmallBrace(81,167) serves also as a minimal counterexample of
odd order.

The previous examples evince that there is a long way off to find a
characterisation for Dedekind left braces. In this section, we present a first
remarkable result for the case of finite left braces.

######  Theorem 12.

Every finite Dedekind left brace is centrally nilpotent.

We work towards a proof of Theorem 12.

######  Proposition 13.

Let $A$ be a finite Dedekind left brace. For every $p\in\pi(A)$, the Sylow
$p$-subgroup $I_{p}$ of $(A,+)$ is an ideal of $A$. In particular,
$A=\bigoplus_{p\in\pi(A)}I_{p}$.

###### Proof.

Let $p\in\pi(A)$ and let $(I_{p},+)$ be the Sylow $p$-subgroup of $(A,+)$.
Since it is a characteristic subgroup, it is $\lambda$-invariant, and
therefore, a subbrace of $A$. Hence, $I_{p}$ is an ideal as $A$ is Dedekind. ∎

######  Proposition 14.

Let $A$ be a finite Dedekind left brace of order $p^{k}$ for some prime $p$
and positive integer $k$. Then every subbrace $S$ of order $p$ is contained in
$\operatorname{Soc}(A)$.

###### Proof.

Let $A$ be a minimal counterexample. According to Remark 9, the set $\\{S\leq
A\mid|S|=p\\}$ is not empty. Let $S\leq A$ such that $|S|=p$ and
$S\lneq\operatorname{Soc}(A)$. Then, $S$ is an abelian ideal isomorphic to a
cyclic group of order $p$, i.e. $S=\langle a\rangle_{+}$ for some $a\in A$.
Thus, $S$ is $\lambda$-invariant and we can restrict the $\lambda$-action to
$\bar{\lambda}\colon
a\in(A,\cdot)\rightarrow\left.\lambda_{a}\right|_{S}\in\operatorname{Aut}(S,+)$.
It follows that $\operatorname{Ker}\bar{\lambda}=A$ as
$|\operatorname{Aut}(S,+)|=p-1$. Thus, $\lambda_{b}(x)=x$, or equivalently
$b\ast x=0$, for every $b\in A$ and every $x\in\langle a\rangle_{+}$.
Therefore, $S\leq\operatorname{Fix}(A)$ and it also holds that

$I:=\operatorname{Ann}_{A}(S)=\\{b\in A\mid x\ast b=0,\,\text{for every $x\in
S$}\\}=\operatorname{Ann}_{A}^{r}(S).$

By Lemma 2, $I$ is a subbrace of $A$, and therefore, an ideal of $A$.

Let $b\in A\setminus I$. Then, $0\neq a\ast b=ka$ with $0<k<p$. Since $\langle
b\rangle$ is an ideal of $A$, it holds $a\ast b\in S\cap\langle b\rangle$.
Thus $S\cap\langle b\rangle\neq\\{0\\}$, and therefore, $S\leq\langle
b\rangle$. Suppose that $\langle b\rangle\lneq A$. By the minimality of $A$,
$S\leq\operatorname{Soc}(\langle b\rangle)$. Therefore, $a\ast b=0$, a
contradiction. Hence, we have that $\langle b\rangle=A$, for every $b\in
A\setminus I$.

By the minimality of $A$, there exists $0\neq b+I\in A/I$ such that
$b+I\in\operatorname{Soc}(A/I)$. In particular, $b\ast b\in I$. Since $b\ast
a=0$, it holds $a+b=b+a=ba$. Therefore, applying $S\leq\operatorname{Fix}(A)$,
we get

$\displaystyle(a+b)\ast(a+b)$ $\displaystyle=(ba)\ast(a+b)=$
$\displaystyle=b\ast(a\ast(a+b))+a\ast(a+b)+b\ast(a+b)=$
$\displaystyle=0+a\ast a+a\ast b+b\ast a+b\ast b=ka+b\ast b$ (3)

Thus, we obtain

$(a+b)(a+b)=2(a+b)+(a+b)\ast(a+b)=(k+2)a+b\ast b+2b.$ (4)

Now, it holds that

$\displaystyle\Big{(}(a+b)(a+b)\Big{)}\ast(a+b)=\Big{(}(ba)(ba)\Big{)}\ast(a+b)=$
$\displaystyle=(ba)\ast\Big{(}(ba)\ast(a+b)\Big{)}+2\Big{(}(ba)\ast(a+b)\Big{)}=\quad\text{(by\leavevmode\nobreak\
\eqref{eq:igual1})}$ $\displaystyle=(ba)\ast(ka+b\ast b)+2ka+2(b\ast b)=$
$\displaystyle=(ba)\ast(ka)+(ba)\ast(b\ast b)+2ka+2(b\ast b)=\
\text{($ka\in\operatorname{Fix}(A)$ and $b\ast b\in I$)}$
$\displaystyle=b\ast(b\ast b)+2ka+2(b\ast b)$

On the other hand, by (4), it also holds

$\displaystyle\Big{(}(a+b)(a+b)\Big{)}\ast(a+b)=\Big{(}2b+b\ast
b+(k+2)a\Big{)}\ast(a+b)=$
$\displaystyle=\Big{(}b^{2}+(k+2)a\Big{)}\ast(a+b)=\
\text{($(k+2)a\in\operatorname{Fix}(A)$)}$
$\displaystyle=\Big{(}b^{2}\big{(}(k+2)a\big{)}\Big{)}\ast(a+b)=$
$\displaystyle=b^{2}\ast\Big{(}\big{(}(k+2)a\big{)}\ast(a+b)\Big{)}+\big{(}(k+2)a\big{)}\ast(a+b)+b^{2}\ast(a+b)$
$\displaystyle=0+\big{(}(k+2)a\big{)}\ast(a+b)+b^{2}\ast(a+b)=$
$\displaystyle=\big{(}(k+2)a\big{)}\ast b+b^{2}\ast b=$
$\displaystyle=\big{(}(k+2)a\big{)}\ast b+b\ast(b\ast b)+2(b\ast b)$

Hence, we have that $2ka=\big{(}(k+2)a\big{)}\ast b$.

Since $S$ is abelian, it follows that $(k+2)a=a^{k+2}$. Moreover, by
induction, we claim that $a^{n}\ast b=n(a\ast b)$ for every positive integer
$n$. Indeed, if $a^{n}\ast b=n(a\ast b)$ for some $n\geq 1$, then

$a^{n+1}\ast b=a\ast(a^{n}\ast b)+a^{n}\ast b+a\ast b=0+n(a\ast b)+a\ast
b=(n+1)(a\ast b)$

Thus, it follows that

$2ka=\big{(}(k+2)a\big{)}\ast b=a^{k+2}\ast b=(k+2)(a\ast
b)=(k+2)ka=k^{2}a+2ka$

Hence, $k^{2}a=0$, and therefore, $p|k^{2}$; a contradiction. ∎

######  Corollary 15.

Let $A$ be a finite Dedekind left brace of order $p^{k}$ for some prime $p$
and positive integer $k$. Then, $A$ is centrally nilpotent.

###### Proof.

By Proposition 14, $A$ is right nilpotent. Moreover, Proposition 4.4 of [11]
shows that finite left braces of order a power of a prime are left nilpotent.
Therefore, $A$ is centrally nilpotent. ∎

###### Proof of Theorem 12.

By Theorem 4.13 in [2], it follows that a left brace $A$ is centrally
nilpotent if, and only if, the Sylow $p$-subgroups $I_{p}$ of $A$ are
centrally nilpotent ideals and $A=\bigoplus_{p\in\pi(A)}I_{p}$. The result
follows from Proposition 13 and Corollary 15. ∎

In case of left braces with an elementary abelian additive group we can say
more.

######  Proposition 16.

Let $A$ be a finite Dedekind left brace. If $(A,+)$ is elementary abelian,
then $\operatorname{Soc}_{n}(A)=\zeta_{n}(A)$ for every positive integer $n$.

###### Proof.

Clearly $\zeta(A)=\zeta_{1}(A)\leq
S:=\operatorname{Soc}_{1}(A)=\operatorname{Soc}(A)$. Since $S$ is an ideal, by
Lemma 2, $\operatorname{Ann}_{A}^{l}(S)$ is the kernel of the restriction of
the $\lambda$-action

$\bar{\lambda}\colon
a\in(A,\cdot)\rightarrow\left.\lambda_{a}\right|_{S}\in\operatorname{Aut}(S,+),$

i.e.
$\operatorname{Ker}\bar{\lambda}=\operatorname{Ann}_{A}^{l}(S)\unlhd(A,\cdot)$.
Moreover, $\big{(}A/\operatorname{Ann}_{A}^{l}(S),\boldsymbol{\cdot}\big{)}$
is isomorphic to a subgroup of

$F=\\{\varphi\in\operatorname{Aut}(S,+)\mid\varphi(H)=H,\,\text{for every
$H\leq(S,+)$}\\}\leq\operatorname{Aut}(S,+)$

Since the exponent of $S$ is $p$, Theorem 1.5.6 of [27] shows that
$F\cong\operatorname{Aut}(\operatorname{C}_{p})$. Thus,
$A=\operatorname{Ann}_{A}^{l}(S)$, i.e $as=a+s=s+a=sa$, for every $a\in A$ and
every $s\in S$. Hence, $\zeta(A)=S$. Using similar arguments and induction, we
can conclude $\operatorname{Soc}_{n}(A)=\zeta_{n}(A)$, for every positive
integer $n$. ∎

######  Remark 17.

The socle series admits also an extension by means of ordinals. Let $A$ be a
left brace. We can consider the _ordinal socle series_

$0=\operatorname{Soc}_{0}(A)\leq\operatorname{Soc}_{1}(A)\leq\ldots\operatorname{Soc}_{\alpha}(A)\leq\operatorname{Soc}_{\alpha+1}(A)\leq\ldots\leq\operatorname{Soc}_{\gamma}(A)$

an ascending series of ideals, defined by the recursive rule

$\begin{array}[]{l}\operatorname{Soc}_{\alpha+1}(A)/\operatorname{Soc}_{\alpha}(A)=\operatorname{Soc}(A/\operatorname{Soc}_{\alpha}(A))\
\text{for all ordinals $\alpha$}\\\
\operatorname{Soc}_{\lambda}(A)=\bigcup_{\mu<\lambda}\operatorname{Soc}_{\mu}(A)\
\text{for the limit ordinals $\lambda$.}\end{array}$

The last term $\operatorname{Soc}_{\infty}(A)=\operatorname{Soc}_{\gamma}(A)$
of this series is called the _upper hypersocle of $A$_. A left brace is said
to be _hypermultipermutational_ if it coincides with its upper hypersocle.

######  Corollary 18.

Let $A$ be a Dedekind hypermultipermutational left brace. If $(A,+)$ is
elementary abelian, then $\operatorname{Soc}_{\alpha}(A)=\zeta_{\alpha}(A)$
for every ordinal $\alpha$. In particular, $A$ is hypercentral.

###### Proof.

By the definition of the ordinal socle series of $A$, each quotient is an
abelian left brace. Thus, each factor is isomorphic to an abelian $p$-group.
Therefore, it follows that $(A,\cdot)$ is also a $p$-group. Then, we can argue
as in Proposition 16 using transfinite induction. ∎

### Extraspecial left braces

We introduce here a family of Dedekind left braces. These will play a key role
in Section 4. The following definition of extraspecial left braces is
motivated by the corresponding one in groups.

######  Definition 19.

A non-abelian left brace $E$ is called _extraspecial_ if its additive group is
elementary abelian and there exists $c\in\zeta(E)$ such that $0\neq E/\langle
c\rangle_{+}$ is an abelian left brace. An extraspecial brace $E$ is called
_strong extraspecial_ if $a\ast a\neq 0$ for every element $a\notin\langle
c\rangle_{+}$.

######  Remark 20.

Let $E$ be an extraspecial brace such that $(E,+)$ is an elementary abelian
$p$-group, for some $p$ prime. Observe that the element $c\in\zeta(A)$
satisfies $C:=\langle c\rangle_{+}$ is an abelian ideal which is isomorphic to
a cyclic group of order $p$. Then, $C$ can be embedded in the field of
$p$-elements $\operatorname{\mathbb{F}}_{p}$ as $kc\mapsto k$, for every
$0\leq k<p$. Moreover, $E/C$ can also be regarded as a vector space over the
field $\operatorname{\mathbb{F}}_{p}$. _From now on
$\operatorname{\mathbb{F}}_{p}$ denotes the field of $p$-elements with $p$ a
prime_.

######  Remark 21.

If $E$ is an extraspecial left brace, $E\neq\operatorname{Soc}(E)$, and then
$\operatorname{Soc}_{2}(E)=E$, since $a\ast b\in\langle
c\rangle_{+}\leq\operatorname{Soc}(E)$ for every $a,b\in E$. Thus, $E\ast
E=\langle c\rangle_{+}$. Moreover, if $E$ is strong extraspecial,
$\zeta(E)=\operatorname{Soc}(E)$, as $a\notin\zeta(E)$ implies that $a\ast
a\neq 0$. Therefore, $\operatorname{Ann}_{E}(\operatorname{Soc}(E))=E$.

It follows that (strong) extraspecial left braces are characterised by the
existence of (strong non-degenerate) bilinear forms.

######  Theorem 22.

Let $E$ be a (strong) extraspecial left brace and let $C:=\langle
c\rangle_{+}\leq\zeta(E)$ such that $E/C$ can be regarded as a vector space
over $\operatorname{\mathbb{F}}_{p}$ for some prime $p$. The map $\phi\colon
E/C\times E/C\rightarrow\operatorname{\mathbb{F}}_{p}$ defined as:

$(xC,yC)\mapsto k_{x,y},\ \text{where $x\ast y=k_{x,y}c$,}\ \text{for every
$xC,yC\in E/C$},$

is a (strong non-degenerate) bilinear form.

Conversely, let $V$ be a vector space over $\operatorname{\mathbb{F}}_{p}$,
let $C=\langle c\rangle_{+}$ be an additive cyclic group of order $p$ and let
$\phi$ be a (strong non-degenerate) bilinear form on $V$. The group
$A=(V,+)\oplus C$ endowed with the product

$(x,k)(y,t):=(x+y,k+t+\phi(x,y)c),\quad\text{for every $x,y\in V$, $k,t\in
C$}$

turns out to be a (strong) extraspecial left brace.

###### Proof.

Recall that Remark 21 states that $x\ast y=k_{x,y}c\in C$, for every $x,y\in
E$. First, we check that $\phi$ is well-defined. Let
$x_{1}C,x_{2}C,y_{1}C,y_{2}C\in E/C$ such that $x_{1}C=x_{2}C$ and
$y_{1}C=y_{2}C=y_{2}+C$. Then, $x_{2}=x_{1}c$, $y_{2}=y_{1}+c^{\prime}$, for
some elements $c,c^{\prime}\in C$. Since $C\leq\zeta(E)$, it holds

$\displaystyle x_{2}\ast y_{2}$
$\displaystyle=(x_{1}c)\ast(y_{1}+c^{\prime})=(x_{1}c)\ast y_{1}+(x_{1}c)\ast
c^{\prime}=(x_{1}c)\ast y_{1}=$ $\displaystyle=x_{1}\ast(c\ast y_{1})+c\ast
y_{1}+x_{1}\ast y_{1}=x_{1}\ast y_{1}.$

Now, we see that $\phi$ is bilinear. Given $xC,zC\in E/C$, we have that
$xC+zC=(x+z)C=xzC$, as $E/C$ is abelian. Thus, $x+z=xzc$, for some element
$c\in C$. Applying that $z\ast y\in\zeta(E)$, it holds

$\displaystyle k_{x+z,y}c$ $\displaystyle=(x+z)\ast y=(xzc)\ast
y=(xz)\ast(c\ast y)+c\ast y+(xz)\ast y=$ $\displaystyle=(xz)\ast y=x\ast(z\ast
y)+z\ast y+x\ast y=x\ast y+z\ast y=k_{x,y}c+k_{z,y}c$
$\displaystyle=(k_{x,y}+k_{z,y})c.$

Similarly,

$k_{x,y+z}c=x\ast(y+z)=x\ast y+x\ast z=k_{x,y}c+k_{x,z}c=(k_{x,y}+k_{x,z})c.$

Moreover, if $E$ is strong extraspecial, then $x\ast x\neq 0$ for every
$x\notin C$, which means $\phi(x+C,x+C)\neq 0$. Thus, $\phi$ is strong non-
degenerate.

Conversely, assume that $(A,+)=(V,+)\oplus C$, where $(V,+)$ is an elementary
abelian $p$-group and $C=\langle c\rangle_{+}$ is a cyclic group of order $p$,
and the product

$(x,k)(y,t):=(x+y,k+t+\phi(x,y)c),\quad\text{for every $x,y\in V$, $k,t\in
C$}$

is defined for some bilinear form $\phi\colon V\times
V\rightarrow\operatorname{\mathbb{F}}_{p}$.

Only the associative law of the product and the left brace distributivity law
are in doubt. Let $(x,k),(y,t),(z,s)\in A$. Then

$\displaystyle\big{(}(x,k)(y,t)\big{)}\big{(}z,s\big{)}$
$\displaystyle=\big{(}x+y,k+t+\phi(x,y)c\big{)}\big{(}z,s\big{)}=$
$\displaystyle=\big{(}x+y+z,k+t+\phi(x,y)c+s+\phi(x+y,z)c\big{)}=$
$\displaystyle=\big{(}x+y+z,k+t+s+\phi(x,y)c+\phi(x,z)c+\phi(y,z)c\big{)};$
$\displaystyle\big{(}x,k\big{)}\big{(}(y,t)(z,s)\big{)}$
$\displaystyle=\big{(}x,k\big{)}\big{(}y+z,t+s+\phi(y,z)c\big{)}=$
$\displaystyle=\big{(}x+y+z,k+t+s+\phi(y,z)c+\phi(x,y+z)c\big{)}=$
$\displaystyle=\big{(}x+y+z,k+t+s+\phi(y,z)c+\phi(x,y)c+\phi(x,z)c\big{)};$

i.e. the product is associative.

Furthermore,

$\displaystyle\big{(}x,k\big{)}\big{(}(y,t)+(z,s)\big{)}$
$\displaystyle=(x,k)(y+z,t+s)=$
$\displaystyle=\big{(}x+y+z,k+t+s+\phi(x,y+z)c\big{)}=$
$\displaystyle=\big{(}x+y+z,k+t+s+\phi(x,y)c+\phi(x,z)c\big{)};$

and

$\displaystyle(x,k)(y,t)+(x,k)(z,s)-(x,k)=$
$\displaystyle=\big{(}x+y,k+t+\phi(x,y)c\big{)}+\big{(}x+z,k+s+\phi(x,z)c\big{)}-(x,k)=$
$\displaystyle=\big{(}x+y+z,k+t+s+\phi(x,y)c+\phi(x,z)c\big{)}.$

Hence, $(A,+,\cdot)$ is a left brace. ∎

The following example provides three families of finite extraspecial left
braces.

######  Example 23.

Let $p$ be a prime and $\mathbb{F}_{p}$ the field of $p$ elements. We define
the following sets with their corresponding addition and product:

  1. 1.

Let $\operatorname{E}_{0}(m,p)=\mathbb{F}_{p}\oplus\mathbb{F}_{p}$, with
$0<m<p$. For every $k_{1},k_{2},t_{1},t_{2}\in\mathbb{F}_{p}$, we define

$\displaystyle(k_{1},k_{2})+(t_{1},t_{2})$
$\displaystyle:=(k_{1}+t_{1},k_{2}+t_{2}),$
$\displaystyle(k_{1},k_{2})(t_{1},t_{2})$
$\displaystyle:=(k_{1}+t_{1},k_{2}+t_{2}+mk_{1}t_{1})$

By Theorem 22, $\operatorname{E}_{0}(m,p)$ is a strong extraspecial left
brace, since the map
$\phi\colon\operatorname{\mathbb{F}}_{p}\times\operatorname{\mathbb{F}}_{p}\rightarrow\operatorname{\mathbb{F}}_{p}$,
given by $\phi(k,t)=mkt$ for every $k,t\in\operatorname{\mathbb{F}}_{p}$, is a
strong non-degenerate bilinear form. Moreover, given
$(k_{1},k_{2}),(t_{1},t_{2})\in\operatorname{E}_{0}(m,p)$, it holds

$(k_{1},k_{2})\ast(t_{1},t_{2})=(0,mk_{1}t_{1})$

Therefore, $\zeta(\operatorname{E}_{0}(m,p))=\langle(0,1)\rangle_{+}$.

  2. 2.

Let
$\operatorname{E}_{1}(m,p)=\mathbb{F}_{p}\oplus\mathbb{F}_{p}\oplus\mathbb{F}_{p}$,
with $0\leq m<p$. For every $k_{1},k_{2},t_{1},t_{2}\in\mathbb{F}_{p}$, we
define

$\displaystyle(k_{1},k_{2},k_{3})+(t_{1},t_{2},t_{3})$
$\displaystyle:=(k_{1}+t_{1},k_{2}+t_{2},k_{3}+t_{3})$
$\displaystyle(k_{1},k_{2},k_{3})(t_{1},t_{2},t_{3})$
$\displaystyle:=(k_{1}+t_{1},k_{2}+t_{2},k_{3}+t_{3}+mk_{1}t_{1}+k_{2}t_{2})$

By Theorem 22, $\operatorname{E}_{1}(m,p)$ is an extraspecial left brace,
since the map
$\phi\colon\operatorname{\mathbb{F}}_{p}^{2}\times\operatorname{\mathbb{F}}_{p}^{2}\rightarrow\operatorname{\mathbb{F}}_{p}$,
given by

$\phi((k_{1},k_{2}),(t_{1},t_{2}))=mk_{1}t_{1}+k_{2}t_{2}$

is a bilinear form. Moreover, it holds

$(k_{1},k_{2},k_{3})\ast(t_{1},t_{2},t_{3})=(0,0,mk_{1}t_{1}+k_{2}t_{2})$

If $m=0$,
$\zeta(\operatorname{E}_{1}(m,p))=\langle(1,0,0),(0,0,1)\rangle_{+}$.
Otherwise, $\zeta(\operatorname{E}_{1}(m,p))=\langle(0,0,1)\rangle_{+}$.
Hence, $\operatorname{E}_{1}(m,p)$ is strong if, and only if, $m\neq 0$ and
the polynomial $mX^{2}+1$ has no roots in $\operatorname{\mathbb{F}}_{p}$.

  3. 3.

Let
$\operatorname{E}_{2}(m,p)=\mathbb{F}_{p}\oplus\mathbb{F}_{p}\oplus\mathbb{F}_{p}$,
with $0\leq m<p$. For every $k_{1},k_{2},t_{1},t_{2}\in\mathbb{F}_{p}$, we
define

$\displaystyle(k_{1},k_{2},k_{3})+(t_{1},t_{2},t_{3}):=(k_{1}+t_{1},k_{2}+t_{2},k_{3}+t_{3})$
$\displaystyle(k_{1},k_{2},k_{3})(t_{1},t_{2},t_{3}):=(k_{1}+t_{1},k_{2}+t_{2},k_{3}+t_{3}+mk_{1}t_{1}+k_{1}t_{2}+k_{2}t_{2})$

By Theorem 22, $\operatorname{E}_{2}(m,p)$ is an extraspecial left brace,
since the map
$\phi\colon\operatorname{\mathbb{F}}_{p}^{2}\times\operatorname{\mathbb{F}}_{p}^{2}\rightarrow\operatorname{\mathbb{F}}_{p}$,
given by

$\phi((k_{1},k_{2}),(t_{1},t_{2}))=mk_{1}t_{1}+k_{1}t_{2}+k_{2}t_{2}$

is a bilinear form. Moreover, it holds

$(k_{1},k_{2},k_{3})\ast(t_{1},t_{2},t_{3})=(0,0,mk_{1}t_{1}+k_{1}t_{2}+k_{2}t_{2})$

Therefore, $\zeta(\operatorname{E}_{2}(m,p))=\langle(0,0,1)\rangle_{+}$.
Hence, $\operatorname{E}_{2}(m,p)$ is strong if, and only if, $m\neq 0$ and
the polynomial $mX^{2}+X+1$ has no roots in $\operatorname{\mathbb{F}}_{p}$.

Indeed these are the unique families of finite strong extraspecial left
braces.

######  Theorem 24.

Let $E$ be a finite strong extraespecial left brace. Then, $E$ is isomorphic
to either $\operatorname{E}_{0}(m,p)$, $\operatorname{E}_{1}(m,p)$ or
$\operatorname{E}_{2}(m,p)$ for some $p$ prime and some $0<m<p$.

###### Proof.

From Theorem 22 and Corollary 7 it follows that $E$ has order at most $p^{3}$
for some $p$ prime. Moreover, $|E|=p$ is not possible, since $E$ would be an
abelian left brace.

Assume that $|E|=p^{2}$. Then, $\zeta(E)=\langle c\rangle_{+}$ for some $c\in
E$, and $(E,+)=\langle b\rangle_{+}\oplus\langle c\rangle_{+}$, for some $b\in
E\setminus\zeta(E)$. Since $E$ is strong, $0\neq b\ast b\in\zeta(E)$. Write
$b\ast b=mc$ with $0<m<p$. Thus, for every $x=k_{1}b+k_{2}c,y=t_{1}b+t_{2}c$,
$x\ast y=(mk_{1}t_{1})c$. By Theorem 22, $E$ is isomorphic to
$\operatorname{E}_{0}(m,p)$.

Assume that $|E|=p^{3}$. Let $C=\langle c\rangle_{+}\leq\zeta(E)$ such that
$E/C$ is abelian. Since $E$ is strong, without loss of generality, we can
assume that there exists $b\in E\setminus C$ such that $b\ast b=c$. By Theorem
22, $E/C$ can be regarded as a vector space so that

$(xC,yC)\mapsto k_{x,y},\quad\text{where $x\ast y=k_{x,y}c$,}$

defines a strong non-degenerate bilinear form $\phi\colon E/C\times
E/C\rightarrow\operatorname{\mathbb{F}}_{p}$.

According to Corollary 6, we can find $a\notin C$ such that $E/C=\langle
aC\rangle_{+}\oplus\langle bC\rangle_{+}$, and $0\neq\phi(aC,aC)=m_{1}$,
$\phi(aC,bC)=m_{2}$, $\phi(bC,bC)=1$ and $\phi(bC,aC)=0$, with
$m_{1},m_{2}\in\operatorname{\mathbb{F}}_{p}$. If $m_{2}\neq 0$, without loss
of generality, we can assume that $m_{2}=1$. Hence, we can write $E/C=\langle
aC\rangle_{+}\oplus\langle bC\rangle_{+}$ so that $a\ast a=mc$, $a\ast
b=\varepsilon c$ and $b\ast b=c$, with $0\neq
m\in\operatorname{\mathbb{F}}_{p}$ and $\varepsilon\in\\{0,1\\}$.

Therefore, for every $x=k_{1}a+k_{2}b+k_{3}c,y=t_{1}a+t_{2}b+t_{3}c\in
E\setminus C$, with $k_{i},t_{i}\in\operatorname{\mathbb{F}}_{p}$, $1\leq
i\leq 3$,

$x\ast y=(mk_{1}t_{1}+\varepsilon k_{1}t_{2}+k_{2}t_{2})c$

Hence, $E$ is either isomorphic to $\operatorname{E}_{1}(m,p)$ or
$\operatorname{E}_{2}(m,p)$ by whether $\varepsilon=1$ or not. ∎

######  Corollary 25.

$\operatorname{E}_{0}(m,p)$ is a Dedekind left brace for every prime $p$ and
$0<m<p$. Moreover, $\operatorname{E}_{1}(m,p)$ and $\operatorname{E}_{2}(m,p)$
are Dedekind if, and only if, $m\neq 0$ and the polynomials $mX^{2}+1$ and
$mX^{2}+X+1$ have respectively no roots in $\mathbb{F}_{p}$.

###### Proof.

Example 23 and Theorem 27 yield the result for the family
$\operatorname{E}_{0}(m,p)$ and also yield the sufficient condition for the
families $\operatorname{E}_{1}(m,p)$ and $\operatorname{E}_{2}(m,p)$. For the
necessary condition we can argue on the family $\operatorname{E}_{2}(m,p)$, as
the argument for $\operatorname{E}_{1}(m,p)$ is equivalent.

Assume that $\operatorname{E}_{2}(m,p)$ is Dedekind and that there exists
$0\neq k\in\mathbb{F}_{p}$ such that $mk^{2}+k+1=0$. Consider
$x=(k,1,0)\in\operatorname{E}_{2}(m,p)\setminus\zeta(\operatorname{E}_{2}(m,p))$.
According to Example 23, $x\ast x=(0,0,0)$. By Lemma 1, $\langle
x\rangle=\langle x\rangle_{+}$.

Take $y=(1,0,0)$. Then, $x\ast y=(0,0,mk)$ and $mk\neq 0$, as $m,k\neq 0$.
Thus, $x\ast y\notin\langle x\rangle$ and so, $\langle x\rangle$ is not an
ideal; a contradiction. ∎

######  Example 26.

For specific primes $p$, Corollary 25 allows to establish conditions on the
parameter $m$ to characterise Dedekind left braces. For example, for $p=5$,
$\operatorname{E}_{1}(m,p)$ is Dedekind if, and only if, $0<m<p$ and
$m\notin\\{1,4\\}$.

##  4 Dedekind left braces whose additive group is elementary abelian

In this section we present a structural theorem for Dedekind left braces whose
additive group is elementary abelian.

The following theorem provides a sufficient condition in terms of strong
extraspecial left braces.

######  Theorem 27.

Let $A$ be a left brace whose additive group is elementary abelian. Suppose
that $A=E\oplus Z$ where $E$ is a strong extraspecial brace and
$Z\leq\zeta(A)$. Then, $A$ is a Dedekind left brace.

###### Proof.

Let $c\in\zeta(E)$ such that the quotient $E/\langle c\rangle_{+}$ is abelian.
By previous Remark 21, $\zeta(A)=\zeta(E)\oplus Z=\operatorname{Soc}(A)$, and
$\operatorname{Soc}_{2}(A)=A=\operatorname{Ann}_{A}(\operatorname{Soc}(A))$.
Let $a,a^{\prime}\in A$ and $e,e^{\prime}\in E$, $z,z^{\prime}\in Z$ such that
$a=e+z$ and $a^{\prime}=e^{\prime}+z^{\prime}$. It holds that

$a\ast a^{\prime}=(e+z)\ast e^{\prime}+(e+z)\ast z^{\prime}=(e+z)\ast
e^{\prime}=(ez)\ast e^{\prime}=e\ast e^{\prime}\in\langle c\rangle_{+}$

Thus, $A\ast A=E\ast E=\langle c\rangle_{+}$ and $A/\langle c\rangle_{+}$ is
abelian.

Let $a\in A$. If $a\in\zeta(A)$, then $\langle a\rangle=\langle a\rangle_{+}$
is an ideal of $A$. If $a\notin\zeta(A)$, then $\langle c\rangle_{+}=\langle
a\ast a\rangle_{+}\leq\langle a\rangle$. Since $A/\langle c\rangle_{+}$ is
abelian, $\langle a\rangle/\langle c\rangle_{+}$ is an ideal of $A/\langle
c\rangle_{+}$ and so, $\langle a\rangle$ is an ideal of $A$. Hence, every
subbrace of $A$ is ideal. ∎

The rest of the section is devoted to show the structural part of the converse
of this theorem. To this aim, it is imperative to study the first two terms of
the socle series of a Dedekind left brace. Moreover, Theorems 22 and 24, and
Corollary 25 play also a prominent role.

_From now on, for an arbitrary left brace $A$, $S$ denotes
$\operatorname{Soc}(A)$ and $C$ denotes $\operatorname{Soc}_{2}(A)$_.

######  Lemma 28.

Let $A$ be a Dedekind left brace. Suppose that $(C,+)$ is an elementary
abelian $p$-group for some prime $p$. Let $d\in C\setminus S$. Then, the
following statements hold

  1. 1.

$\langle d\ast d\rangle=\langle d\ast d\rangle_{+}$ and $\langle
d\rangle=\langle d\rangle_{+}\oplus\langle d\ast d\rangle_{+}$;

  2. 2.

$d\ast a\in\langle d\ast d\rangle_{+}$ for every $a\in A$;

  3. 3.

$\langle d\ast d\rangle_{+}=\langle c\ast c\rangle_{+}$ for every $c\in
C\setminus S$.

###### Proof.

1\. Since $d\in C$, we have that $u:=d\ast d\in S$. If $u=0$, then $\langle
d\rangle=\langle d\rangle_{+}$ by Lemma 1. If $u\neq 0$, then $u\ast u=0$ and
by Lemma 1, it holds $U:=\langle u\rangle=\langle u\rangle_{+}$ is an ideal.
Using again Lemma 1, it follows $\langle d+U\rangle=\langle
d+U\rangle_{+}=\langle d\rangle_{+}+U$. Therefore, $\langle
d\rangle\leq\langle d\rangle_{+}+U$. The other inclusion is trivial. Since $C$
is elementary abelian and $d\in C\setminus S$, $\langle
d\rangle_{+}\cap\langle u\rangle_{+}=\\{0\\}$ and the equality holds.

2\. We have that $\langle d\rangle$ is an ideal. Thus, $d\ast a\in\langle
d\rangle\cap S$, for every $a\in A$. Assume that $x=kd+tu\in\langle
d\rangle\cap S$, for some $0\leq k,t<p$. Thus, $kd\in\langle d\rangle_{+}\cap
S$. Therefore, $k=0$ as $d\notin S$. Hence, $\langle d\rangle\cap S=\langle
u\rangle_{+}$ as desired.

3\. Let $c\in C\setminus S$. Then, $v:=c\ast c\neq 0$, otherwise $c\in S$ by
item $2$. Thus, $\langle c\rangle=\langle c\rangle_{+}+\langle v\rangle_{+}$.
Suppose that $\langle u\rangle_{+}\neq\langle v\rangle_{+}$. It follows that
$\langle u\rangle_{+}\cap\langle v\rangle_{+}=0$, as $(C,+)$ is elementary
abelian. Therefore, $d\ast c=c\ast d=0$ by item $2$.

Set $e:=c+d$. Since the quotient left brace $C/S$ is abelian,
$c+d+S=cd+S=cdS$, so that $c+d=cdu$, for some $u\in S$. It holds that

$\displaystyle e\ast e$ $\displaystyle=(c+d)\ast(c+d)=(cdu)\ast c+(cdu)\ast
d,$

where

$\displaystyle(cdu)\ast c$ $\displaystyle=(cd)\ast(u\ast c)+(cd)\ast c+u\ast
c=(cd)\ast c=$ $\displaystyle=c\ast(d\ast c)+c\ast c+d\ast c=c\ast c=v$

and

$\displaystyle(cdu)\ast d$ $\displaystyle=(cd)\ast(u\ast d)+(cd)\ast d+u\ast
d=(cd)\ast d=$ $\displaystyle=c\ast(d\ast d)+c\ast d+d\ast d=c\ast u+u=$
$\displaystyle=cu-c-u+u=-c+cu=\lambda_{c}(u)$

Since $\langle u\rangle_{+}$ is an ideal, $\lambda_{c}(u)=ku$, for some
$0<k<p$. Thus, $e\ast e=v+ku\neq 0$, as $\langle v\rangle_{+}\cap\langle
u\rangle_{+}=0$. By the first item, $\langle e\rangle=\langle
e\rangle_{+}\oplus\langle v+ku\rangle_{+}$, which is an ideal. Thus, $d\ast
e,c\ast e\in\langle e\rangle\cap S=\langle v+ku\rangle_{+}$. On the other
hand, it holds

$\displaystyle d\ast e$ $\displaystyle=d\ast(c+d)=d\ast c+d\ast d=u,$
$\displaystyle c\ast e$ $\displaystyle=c\ast(c+d)=c\ast c+c\ast d=v.$

This contradicts the fact that $\langle u\rangle_{+}\cap\langle
v\rangle_{+}=0$. Hence, $\langle u\rangle_{+}=\langle v\rangle_{+}$. ∎

######  Corollary 29.

Let $A$ be a Dedekind left brace. Suppose that $(C,+)$ is an elementary
abelian $p$-group for some prime $p$. Then, $C\ast C$ is an ideal of $A$ of
order $p$.

###### Proof.

By definition, $C\ast C=\langle d\ast u\mid d,u\in C\rangle_{+}\subseteq S$.
Since $S$ is abelian, $C\ast C\leq S$ and thus, $C\ast C$ is an ideal, as $A$
is Dedekind.

Let $d\in C$. If $d\in S$, then $d\ast a=0$ for every $a\in A$. Suppose that
$d\notin S$. By Lemma 28, $d\ast u\in\langle d\ast d\rangle_{+}$, for every
$u\in C$, and $\langle d\ast d\rangle_{+}=\langle c\ast c\rangle_{+}$, for
every $c\in C\setminus S$. Hence, $C\ast C=\langle d\ast d\rangle_{+}$ has
order $p$. ∎

######  Corollary 30.

Let $A$ be a Dedekind left brace. Suppose that $(C,+)$ is an elementary
abelian $p$-group for some prime $p$. Then, $C=E\oplus Z$ where $E$, $Z$ are
ideals of $A$ satisfying

  1. 1.

$Z\leq S$;

  2. 2.

$B:=C\ast C=E\ast E$ has order $p$ and $C/B=\operatorname{Soc}(A/B)$;

  3. 3.

$u\ast u\neq 0$ for all $u\in E\setminus B$.

###### Proof.

By Corollary 29, $B:=C\ast C\leq S$ is an ideal of order $p$. Since $S\leq C$,
$(S,+)$ is elementary abelian and $S=B\oplus Z$ for some subbrace $Z$ of $S$.
Thus, $Z$ is also an ideal of $A$. Since the quotient $C/B$ is abelian, $S/B$
has a complement in $C/B$, say $E/B$, where $E$ is also an ideal of $A$. Thus,
we can write $C/B=S/B\oplus E/B$ and it follows $C=E\oplus Z$.

Let $u\in E\setminus B$. If $u\ast u=0$ then $u\in S$ (otherwise, Lemma 28
leads to a contradiction). Thus, $u\in S\cap E=B$, a contradiction. Therefore,
$u\ast u\neq 0$ and again by Lemma 28, $E\ast E=C\ast C$.

Finally, let $a\in A$ and $d\in C\setminus B$. If $d\notin S$, by Lemma 28,
$d\ast a\in\langle d\ast d\rangle_{+}=C\ast C$. Hence,
$C/B\leq\operatorname{Soc}(A/B)$. On the other hand, let
$u+B\in\operatorname{Soc}(A/B)$. Then, for every $a\in A$, $u\ast a\in B$.
Since $B\leq S$, it follows that $u+S\in\operatorname{Soc}(A/S)=C/S$. Hence,
$C/B=\operatorname{Soc}(A/B)$. ∎

The following theorem shows that $\operatorname{Soc}_{2}(A)=A$ holds for
hypermultipermutational braces with elementary abelian additive group (see
Remark 17).

######  Theorem 31.

Let $A$ be a hypermultipermutational Dedekind left brace whose additive group
is an elementary abelian $p$-group for some prime $p$. Then,
$A=\operatorname{Soc}_{2}(A)$.

###### Proof.

Suppose that $A\neq\operatorname{Soc}_{2}(A)$. Then,
$\operatorname{Soc}_{3}(A)\neq\operatorname{Soc}_{2}(A)$. Without loss of
generality, we may suppose that $A=\operatorname{Soc}_{3}(A)$. Call
$S=\operatorname{Soc}_{1}(A)$ and $C=\operatorname{Soc}_{2}(A)$. By Corollary
30, $B=C\ast C\leq S$ has order $p$ and $C/B=\operatorname{Soc}(A/B)$. Hence,
$\operatorname{Soc}_{2}(A/B)=A/B$. Moreover, Corollary 18 yields $S=\zeta(A)$
and $\operatorname{Soc}(A/B)=\zeta(A/B)$. Thus, $B\leq\zeta(A)$.

Applying Corollary 30 to $A/B$, we can write $A/B=(E/B)\oplus(Z/B)$, where $E$
and $Z$ are ideals of $A$ such that

  * •

$Z/B\leq\zeta(A/B)$;

  * •

$D/B:=(A\ast A)/B=(E\ast E)/B\leq\zeta(A/B)$ has order $p$ and

$(A/B)/(D/B)$ is abelian, or equivalently, $A/D$ is abelian;

  * •

$u\ast u\notin B$ for every $u\in E\setminus D$.

Let $a\in A\setminus D$ and set $d:=a\ast a\neq 0$. Thus, $d\in D\leq C$ and
$\langle d+B\rangle_{+}=D/B$. If $d\ast d=0$ then $d\in S$ (otherwise, Lemma
28 yields a contradiction). Thus, $D=\langle d\rangle_{+}\oplus B\leq S$.
Therefore, $\operatorname{Soc}_{2}(A)=A$ as $D=A\ast A$, a contradiction.
Then, we can assume that there exists $a\in A\setminus D$ such that $a\ast
a=d\in D\leq C\setminus S$ and $d\ast d\neq 0$. Set $0\neq b:=d\ast d\in C\ast
C=B$. Thus, $\langle b\rangle_{+}=B\leq\zeta(A)$, as $B$ has prime order.
Hence, it follows that $\langle a\rangle=\langle a\rangle_{+}\oplus\langle
d\rangle_{+}\oplus\langle b\rangle_{+}$ is a subbrace, and therefore, an ideal
of $A$.

Take $0\leq\alpha,\beta<p$ such that $a\ast d=\alpha b=b_{1}$ and $d\ast
a=\beta b=b_{2}$, and let $u:=a+d+b$. We know that $uu=u\ast u+2u$. Since
$d+B\in\zeta(A/B)$, it follows that $a+d+B=ad+B=adB$. Thus, there exists $w\in
B\leq\zeta(A)$ such that $a+d=adw$. Then, it holds that

$\displaystyle u\ast u$ $\displaystyle=(a+d+b)\ast(a+d+b)=(adw)\ast(a+d+b)=$
$\displaystyle=(ad)\ast(a+d+b)=(ad)\ast a+(ad)\ast d+(ad)\ast b=$
$\displaystyle=a\ast(d\ast a)+d\ast a+a\ast a+a\ast(d\ast d)+d\ast d+a\ast d=$
$\displaystyle=a\ast b_{2}+b_{2}+d+a\ast b+b+b_{1}=$
$\displaystyle=b_{2}+d+b+b_{1}=d+(\alpha+\beta+1)b$

Thus, $uu=u\ast u+2u=2a+3d+(3+\alpha+\beta)b$.

Observe that $a^{2}=2a+a\ast a=2a+d$, and similarly, $d^{2}=2d+b$ so that

$d^{3}=d(2d+b)=2d^{2}-2d+db=4d+2b-2d+d+d\ast b+b=3d+3b$

Since $d+B\in\zeta(A/B)$ and $B\leq\zeta(A)$, there exists $w_{1}\in B$ such
that

$\displaystyle 2a$ $\displaystyle=a^{2}-d=a^{2}+(p-1)d=a^{2}d^{p-1}w_{1}$
$\displaystyle 3d$ $\displaystyle=d^{3}-3b=d^{3}+(p-3)b=d^{3}b^{p-3}$

Moreover, we claim that $d^{n}\ast a=nb_{2}$ and $d^{n}\ast d=nb$, for every
positive integer $n$. Indeed, if we assume that $d^{n}\ast a=nb_{2}$ and
$d^{n}\ast d=nb$, for some $n\geq 1$, we get

$\displaystyle d^{n+1}\ast a$ $\displaystyle=d\ast(d^{n}\ast a)+d^{n}\ast
a+d\ast a=0+d^{n}\ast a+d\ast a=(n+1)b_{2}$ $\displaystyle d^{n+1}\ast d$
$\displaystyle=d\ast(d^{n}\ast d)+d^{n}\ast d+d\ast d=0+nb+b=(n+1)b$

Since $w_{1},b_{2},b\in B\leq\zeta(A)$, we obtain

$\displaystyle(2a)\ast a$ $\displaystyle=(a^{2}d^{p-1}w_{1})\ast
a=(a^{2}d^{p-1})\ast a=$ $\displaystyle=a^{2}\ast(d^{p-1}\ast a)+d^{p-1}\ast
a+a^{2}\ast a=$ $\displaystyle=a^{2}\ast((p-1)b_{2})+(p-1)b_{2}+a\ast(a\ast
a)+2a\ast a=$ $\displaystyle=-b_{2}+b_{1}+2d=2d+(\alpha-\beta)b$ (5)
$\displaystyle(2a)\ast d$ $\displaystyle=(a^{2}d^{p-1}w_{1})\ast
d=(a^{2}d^{p-1})\ast d=$ $\displaystyle=a^{2}\ast(d^{p-1}\ast d)+d^{p-1}\ast
d+a^{2}\ast d=$ $\displaystyle=a^{2}\ast((p-1)b)+(p-1)b+a\ast(a\ast d)+2a\ast
d=$ $\displaystyle=-b+2b_{1}=(2\alpha-1)b$ (6) $\displaystyle(3d)\ast a$
$\displaystyle=(d^{3}b^{p-3})\ast a=d^{3}\ast a=3b_{2}=3\beta b$ (7)
$\displaystyle(3d)\ast d$ $\displaystyle=(d^{3}b^{p-3})\ast d=d^{3}\ast d=3b$
(8)

Now, since $d+B\in\zeta(A/B)$, we can write
$uu=2a+3d+(3+\alpha+\beta)b=(2a)(3d)w_{2}$, for some $w_{2}\in B$. Then,
applying Equations (5)–(8) and the fact that $w_{2},b,b_{1},b_{2}\in\zeta(A)$,
we obtain

$\displaystyle(uu)\ast u$ $\displaystyle=\big{(}(2a)(3d)w_{2}\big{)}\ast
u=\big{(}(2a)(3d)\big{)}\ast u=$ $\displaystyle=(2a)\ast\big{(}(3d)\ast
u\big{)}+(3d)\ast u+(2a)\ast u$ $\displaystyle=(2a)\ast\big{(}(3d)\ast
a+(3d)\ast d+(3d)\ast b\big{)}+$ $\displaystyle+(3d)\ast a+(3d)\ast d+(3d)\ast
b+(2a)\ast a+(2a)\ast d+(2a)\ast b=$ $\displaystyle=(2a)\ast(3\beta
b+3b)+3\beta b+3b+2d+(\alpha-\beta)b+(2\alpha-1)b=$
$\displaystyle=2d+(2+3\alpha+2\beta)b$

Thus, $(uu)u=(uu)\ast u+uu+u=3a+6d+(6+4\alpha+3\beta)b$. On the other hand,
similarly we see

$\displaystyle u\ast(uu)$ $\displaystyle=(adw)\ast(uu)=(ad)\ast(uu)=$
$\displaystyle=a\ast\big{(}d\ast(uu)\big{)}+d\ast(uu)+a\ast(uu)=$
$\displaystyle=a\ast\Big{(}d\ast(2a)+d\ast(3d)+d\ast\big{(}(3+\alpha+\beta)b\big{)}\Big{)}+$
$\displaystyle+d\ast(2a)+d\ast(3d)+d\ast\big{(}(3+\alpha+\beta)b\big{)}+$
$\displaystyle+a\ast(2a)+a\ast(3d)+a\ast\big{(}(3+\alpha+\beta)b\big{)}=$
$\displaystyle=2b_{2}+3b+2d+3b_{1}=2d+(3+3\alpha+2\beta)b$

and so, $u(uu)=u\ast(uu)+uu+u=3a+6d+(7+4\alpha+3\beta)b$. Therefore,
$(uu)u-u(uu)=b\neq 0$, and we arrive to a contradiction. Hence, the theorem
holds. ∎

######  Corollary 32.

Let $A$ be a finite Dedekind left brace whose additive group is elementary
abelian. Then, $A$ is centrally nilpotent of level $2$.

###### Proof.

It follows from Theorems 12 and 31. ∎

Now, we can state the converse of Theorem 27.

######  Theorem 33.

Let $A$ be a hypermultipermutational Dedekind left brace such that $(A,+)$ is
an elementary abelian group. Then, $A$ is an extraspecial brace satisfying

  1. 1.

$A=E\oplus Z$ where $E$ and $Z$ are ideals of $A$, with $E$ strong
extraspecial;

  2. 2.

$A\ast A=E\ast E$ has order $p$;

  3. 3.

$Z\oplus(E\ast E)=\zeta(A)=\operatorname{Soc}(A)$.

###### Proof.

It follows from Theorem 31 and Corollary 30. ∎

In the finite case, applying Theorem 24 and Corollary 25, we can complete the
description of Theorem 33.

######  Corollary 34.

Let $A$ be a finite Dedekind left brace such that $(A,+)$ is an elementary
abelian $p$-group for some $p$ prime. Then, $A$ is an extraspecial brace $A$
satisfying

  1. 1.

$A=E\oplus Z$ where $E$ and $Z$ are ideals of $A$, and $E$ is a strong
extraspecial left brace of order at most $p^{3}$;

  2. 2.

$A\ast A=E\ast E$ has order $p$;

  3. 3.

$Z\oplus(E\ast E)=\zeta(A)=\operatorname{Soc}(A)$.

Furthermore, the following hold

  1. 1.

if $|E|=p^{2}$, then $E$ is isomorphic to $\operatorname{E}_{0}(m,p)$ with
$0<m<p$;

  2. 2.

if $|E|=p^{3}$, then either $E$ is isomorphic to $\operatorname{E}_{1}(m,p)$
with $0<m<p$ such that $mX^{2}+1$ has no roots in
$\operatorname{\mathbb{F}}_{p}$, or $E$ is isomorphic to
$\operatorname{E}_{2}(m,p)$ with $0<m<p$ such that $mX^{2}+X+1$ has no roots
in $\operatorname{\mathbb{F}}_{p}$.

## Acknowledgements

The third author is very grateful to the Conselleria d’Innovació,
Universitats, Ciència i Societat Digital of the Generalitat (Valencian
Community, Spain) and the Universitat de València for their financial support
and grant to host researchers affected by the war in Ukraine in research
centres of the Valencian Community. He is sincerely grateful to the first and
second authors for their hospitality, support and care. The third author would
like also to thank the Isaac Newton Institute for Mathematical Sciences,
Cambridge, for support and hospitality during the Solidarity Supplementary
Grant Program. This work was supported by EPSRC grant no EP/R014604/1. He is
sincerely grateful to Agata Smoktunowicz.

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