#  Central nilpotency of left skew braces and solutions of the Yang–Baxter
equation

A. Ballester-Bolinches Departament de Matemàtiques, Universitat de València,
Dr. Moliner, 50, 46100 Burjassot, València, Spain<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>R. Esteban-
Romero00footnotemark: 0 M. Ferrara Dipartimento di Matematica e Fisica,
Università degli Studi della Campania “Luigi Vanvitelli”, viale Lincoln 5,
81100 Caserta, Italy<EMAIL_ADDRESS>V. Pérez-
Calabuig-1-1footnotemark: -1 M. Trombetti Dipartimento di Matematica e
Applicazioni “Renato Caccioppoli”, Università degli Studi di Napoli Federico
II, Complesso Universitario Monte S. Angelo, Via Cintia, 80126 Napoli, Italy,
<EMAIL_ADDRESS>
###### Abstract

Nipotency of skew braces is related to certain types of solutions of the
Yang–Baxter equation. This paper delves into the study of centrally nilpotent
skew braces. In particular, we study their torsion theory (Section 4.1) and we
introduce an “index” for subbraces (Section 4.2), but we also show that the
product of centrally nilpotent ideals need not be centrally nilpotent (Example
B), a rather peculiar fact. To cope with these examples, we introduce a
special type of nilpotent ideal, using which, we define a good Fitting ideal.
Also, a Frattini ideal is defined and its relationship with the Fitting ideal
is investigated.

A key ingredient in our work is the characterisation of the commutator of
ideals in terms of absorbing polynomials (Section 3); this solves Problem 3.4
of [4]. Moreover, we provide an example (Example A) showing that the idealiser
of a subbrace (as defined in [17]) does not exist in general.

_Mathematics Subject Classification(2022)_: Primary 16T25, 20F18
_Mathematics Subject Classification(2022)_: Secondary 81R50

_Keywords:_ brace, Yang–Baxter equation, nilpotency, Fitting ideal,
_Keywords:_ commutator of ideals, idealiser

##  1 Introduction

The study of set-theoretical solutions of the Yang–Baxter equation (YBE)
provides a common framework for a multidisciplinary approach from different
areas including knot theory, braid theory, or Garside theory among others (see
[8], [11], [14], for example).

A (finite) _set-theoretical solution_ of the YBE is a pair $(X,r)$, where $X$
is a (finite) set and $r\colon X\times X\rightarrow X\times X$ is a bijective
map satisfying the equality $r_{12}r_{23}r_{12}=r_{23}r_{12}r_{23}$, where
$r_{12}=r\times\operatorname{id}_{X}$ and
$r_{23}=\operatorname{id}_{X}\times\,r$. The main challenge in this area is to
classify all set-theoretical solutions with prescribed properties. The
algebraic structure of left skew braces plays a fundamental role in this
classification.

A _left skew brace_ $B$ is a set endowed with two group structures, $(B,+)$
and $(B,\cdot)$, satisfying the following skew distributivity property

$a\cdot(b+c)=a\cdot b-a+a\cdot c\quad\forall\,a,b,c\in B.$

If $(B,+)$ satisfies some property $\mathfrak{X}$ (such as abelianity), we say
that $B$ is a left skew brace of $\mathfrak{X}$ type. Of course, left skew
braces of abelian type are just Rump’s _left braces_ (see [16] and [20]).

A non-degenerate set-theoretic solution of the YBE (solution, for short), i.e.
a solution for which both components of $r$ are bijective, naturally leads to
a left skew brace structure over the group

$G(X,r)=\langle x\in X\,|\,xy=uv,\,\text{if $r(x,y)=(u,v)$}\rangle$

(see [20]) — this is said to be the _structure left skew brace of $(X,r)$_
(note that it is an infinite skew brace). Conversely, every left skew brace
$B$ defines a solution $(B,r_{B})$ of the YBE (see [16]).

Nowadays, we are far from being able to understand arbitrary solutions of the
YBE. But it is becoming more and more clear that almost every solution is
multipermutation — recall that a multipermutation solution is one that can be
retracted into the trivial solution over a singleton after finitely many
identification steps — and nilpotency of left skew braces is precisely
introduced to deal with multipermutation solutions (see [7, 15, 17] for
example). In this paper, we go through a deep study of central nilpotency of
left skew braces, with the aim of providing a rigorous framework that could be
used to show that almost every solution is multipermutation. We hope our paper
could be a reference point for all future work on the argument.

We now highlight some of the main aspects of nilpotency of left skew braces we
deal with (see next sections for the definitions):

  * •

In Section 4.1, we study its torsion theory, providing great extensions of
some of the results in [17] (see in particular Theorem 4.16).

  * •

In Section 4.2, we deal with the problem of defining an index for subbraces,
proving that this is possible in the context of locally centrally-nilpotent
left skew braces.

Also we provide many examples showing how peculiar is the behaviour of
centrally nilpotent left skew braces if compared to that of groups and rings.
Among these examples, two are the most striking (see Section 6):

  * •

Example A shows that, contrary to what claimed in [17], it is not always
possible to define the idealiser of a subbraces of a left skew brace of
abelian type.

  * •

Example B shows that the product of two centrally nilpotent ideals of a left
skew brace (of abelian type) need not be centrally nilpotent.

In order to cope with the above examples, we introduce a new type of
nilpotency for ideals (see Section 5). Using this new concept we are able to
define a well-behaving Fitting ideal for left skew braces. In turns, the
Fitting ideal makes it possible to give a good definition of Frattini ideal
for left skew braces. Using these definitions we are able to prove an analogue
of a celebrated result of Gaschütz that relates the Frattini and the Fitting
subgroups of a group (see Theorem 5.12).

Finally, we note that one of the key steps in our work is showing that the two
known definitions of commutators of ideals (inspired by distinct Universal
Algebra approaches) coincide, thus solving Problem 3.4 of [4] (see Section 3).

##  2 Preliminaries

_From now on, the word “brace” will mean “left skew brace”_

In this section, we fix notation and give some background concepts and results
on braces. Although some of them can be considered folklore, others are new.

Let $(B,+,\cdot)$ be a brace. Both operations in $B$ can be related by the so-
called star product: $a\ast b=-a+ab-b$, for all $a,b\in B$. Indeed, both group
operations coincide if and only if $a\ast b=0$ for all $a,b\in B$; in this
case, $B$ is said to be a _trivial brace_. The following properties of the
star product are essential to our work:

$\displaystyle(ab)\ast c$ $\displaystyle=a\ast(b\ast c)+b\ast c+a\ast c;$ (1)
$\displaystyle ab$ $\displaystyle=a+a\ast b+b;$ (2) $\displaystyle a\ast(b+c)$
$\displaystyle=a\ast b+b+a\ast c-b,$ (3)

for all $a,b,c\in B$. If $X$ and $Y$ are subsets of $B$, then $X\ast Y$ is the
subgroup of $(B,+)$ generated by the elements of the form $x\ast y$, for all
$x\in X$ and $y\in Y$.

For every brace, $0$ denotes the common identity element of both group
operations, and the product of two elements will be denoted by juxtaposition.
As usual, group addition follows group product in the order of operations;
also, the star product comes first in the order of operations.

A _subbrace_ of a brace is a subgroup of the additive group which is also a
subgroup of the multiplicative group. A _homomorphism_ between two braces $A$
and $B$ is a map $f\colon A\rightarrow B$ satisfying that $f(a+b)=f(a)+f(b)$
and $f(ab)=f(a)f(b)$ for all $a,b\in A$. Two braces are said to be
_isomorphic_ if there exists a bijective homomorphism between them.

Let $B$ denote an arbitrary brace. Given two subsets $X,Y\subseteq B$, we
write $\langle X\rangle_{+},[X,Y]_{+}$ for the subgroup generated by $X$ and
the commutator of $X$ and $Y$ in $(B,+)$, respectively, and we write $\langle
X\rangle_{\cdot},[X,Y]_{\cdot}$ for the subgroup generated by $X$ and the
commutator of $X$ and $Y$ in $(B,\cdot)$, respectively. Similarly, if we need
to emphasize that some expressions are related to either the additive or the
multiplicative structure, then we put a $+$ or $\cdot$ symbol, respectively,
next to it.

The so-called $\lambda$-_action_ describes an action of the multiplicative
group of $B$ on the additive one. For every $a\in B$, the map
$\lambda_{a}\colon B\rightarrow B$, given by $\lambda_{a}(b)=-a+ab$, is an
automorphism of $(B,+)$ and the map
$\lambda\colon(B,\cdot)\rightarrow\operatorname{Aut}(B,+)$ which sends
$a\mapsto\lambda_{a}$ is a group homomorphism. For every $a,b\in B$,

$a\ast b:=\lambda_{a}(b)-b=-a+ab-b.$

The following concepts are apt to describe the substructural framework of an
arbitrary brace $B$. _Left ideals_ are $\lambda$-invariant subbraces, or
equivalently subbraces $L$ such that $B\ast L\subseteq L$. A left ideal $S$ is
said to be a _strong left ideal_ if $(S,+)$ is a normal subgroup of $(B,+)$,
and an _ideal_ if $(S,\cdot)$ is also a normal subgroup of $(B,\cdot)$, or
equivalently $S\ast B\subseteq S$. Ideals of braces can be considered as the
true analogues of normal subgroups in groups and ideals in rings, since they
allow to take quotients in a brace. Thus, if $I$ is an ideal of $B$, then
$B/I=\\{bI=b+I:b\in B\\}$ denotes the quotient of $B$ over $I$. Finite sums
and products of ideals coincide and are ideals; moreover, arbitrary
intersections of ideals are ideals. It should also be remarked that, for each
left ideal $L$ and each strong left ideal $I$ of $B$, we have $LI=L+I$ is a
left ideal. Furthermore, for the sake of simplicity, we introduce the
following notations (here, $E$ is a subset of $B$):

  * •

The subbrace $\langle E\rangle$ generated by $E$ in $B$ is the smallest
subbrace of $B$ containing $E$ with respect to the inclusion. In this case,
$E$ is also called a system of generators for $\langle E\rangle$; and if $E$
is finite, we say that $\langle E\rangle$ is finitely generated.

  * •

The ideal $E^{B}$ generated by $E$ in $B$ is the smallest ideal of $B$
containing $E$ with respect to the inclusion. If $E=\\{a\\}$, we usually
denote $\\{a\\}^{B}$ simply by $a^{B}$.

  * •

To denote that $C$ is an subbrace of $B$, we write $C\leq B$. To denote that
$I$ is an ideal of $B$, we write $I\trianglelefteq B$.

  * •

If $C$ is a maximal subbrace of $B$, we also write $C<\\!\\!\\!\cdot\,B$.

  * •

If $C\leq B$, then the maximal ideal of $B$ contained in $C$ is denoted by
$C_{B}$.

The following are examples of, respectively, a left ideal and an ideal of a
brace that play a central role in the study of nilpotency of braces:

$\begin{array}[]{c}\operatorname{Fix}(B)=\\{a\in B\,|\,\lambda_{b}(a)=a,\
\text{for every $b\in B$}\\}\end{array}$

and

$\begin{array}[]{c}\operatorname{Soc}(B)=\operatorname{Ker}\lambda\cap\operatorname{Z}(B,+)=\\{a\in
B\,|\,ab=a+b=b+a,\ \text{for every $b\in B$}\\}.\end{array}$

Also note that $B\ast B$ is always an ideal of $B$.

Minimal and maximal substructures, if they exist, turn out to be essential in
every detailed study of an algebraic structure. Let $S$ be a subbrace (resp. a
left ideal or an ideal) of a brace $B$. Then $S$ is a _minimal_ subbrace
(resp. left ideal or ideal) of $B$ if $S\neq 0$ and $0$ and $S$ are the only
subbraces (resp. left ideals or ideals) of $B$ contained in $S$. Moreover, $S$
is a _maximal_ subbrace (resp. left ideal or ideal) of $B$ if $S$ is the only
proper subbrace (resp. left ideal or ideal) of $B$ containing $S$.

The following commutator of ideals is introduced in [5] and plays a key role
in the study of nilpotency and solubility in braces.

######  Definition 2.1.

Let $I,J$ be ideals of a brace $B$. The _commutator_ $[I,J]^{B}$ is defined as

$[I,J]^{B}:=\langle[I,J]_{+}\cup[I,J]_{\cdot}\cup\\{ij-(i+j):\,i\in I,\,j\in
J\\}\rangle^{B}.$

Clearly, $[I,J]^{B}=[J,I]^{B}$ and $[I,J]^{B}\leq I\cap J$.

A brace $B$ is said to be _abelian_ if $[B,B]^{B}=0$, i.e. if it is a trivial
brace of abelian type. Using this commutator, solubility of braces has been
introduced in [3]. If $I$ is an ideal of $B$, the _commutator_ or _derived
ideal of $I$ with respect to $B$_ is defined as $\partial_{B}(I)=[I,I]^{B}$.
If $B=I$, then we say that $\partial_{B}(B)=\partial(B)$ is the _commutator_
or _derived ideal_ of $B$. By repeatedly forming derived ideals, we
recursively obtain a descending sequence of ideals

$\partial_{0}(I)=I\geq\partial_{1}(I)=\partial_{B}(I)\geq\ldots\geq\partial_{n}(I)\geq\ldots$

with $\partial_{n}(I)=\partial_{B}(\partial_{n-1}(I))$ for every
$n\in\mathbb{N}$. We call this series the _derived series of $I$ with respect
to $B$_. Clearly, $\partial_{n-1}(B)/\partial_{n}(B)$ is an abelian brace for
every $n\in\mathbb{N}$, and $B/\partial(B)$ is the greatest abelian quotient
in $B$.

######  Definition 2.2.

We say that an ideal $I$ of $B$ is _soluble with respect to $B$_, if there
exists a non-negative integer $n$ such that $\partial_{n}(I)=0$. If $I=B$, we
simply say that $B$ is a _soluble brace_ , and the smallest non-negative
integer $m$ for which $\partial_{m}(B)=0$ is the derived length of $B$.

##  3 Unifying Universal Algebra definitions of commutators of ideals in
braces

One of the most fruitful research lines of Universal Algebra is commutator
theory. The definition of commutator of ideals we gave in the previous section
was inspired by a deep study of the so-called _Huq $=$ Smith condition_ for
the category of braces (see [5] for more details). In [4], a definition of
commutator of ideals in braces was introduced by means of another universal
algebra point of view (see [12]). In this section, we show that both
definitions coincide, thus answering a question raised in [4].

We start by briefly recalling the definition of commutator of ideals given in
[4].

######  Definition 3.1.

Let $B$ be a brace. Given $n\in\mathbb{N}$, an $n$-_polynomial_ of $B$ is a
map

$p\colon B\times\stackrel{{\scriptstyle(n)}}{{\cdots}}\times B\rightarrow B$

whose image $p(x_{1},\ldots,x_{n})$ is a finite sequence of sums and products
of the variables $x_{1},\ldots,x_{n}$ (note that also additive/multiplicative
inverses are allowed). $\text{Pol}_{n}(B)$ denotes the set of all
$n$-polynomials of $B$.

A polynomial $p\in\text{Pol}_{n}(B)$ is said to be _absorbing_ if

$p(x_{1},\ldots,x_{i-1},0,x_{i+1},\dots,x_{n})=0,$

for every $1\leq i\leq n$ and $x_{j}\in B$, with $1\leq j\neq i\leq n$.

######  Example 3.2.

If $B$ is a brace, then the maps $p_{1}(b_{1},b_{2})=[b_{1},b_{2}]_{+}$,
$p_{2}(b_{1},b_{2})=[b_{1},b_{2}]_{\cdot}$ and
$p_{3}(b_{1},b_{2})=b_{1}b_{2}-(b_{1}+b_{2})$ are absorbing $2$-polynomials.

######  Definition 3.3.

Let $I,J$ be ideals of a brace $B$. The commutator $\llbracket
I,J\rrbracket^{B}$ is defined as

$\llbracket I,J\rrbracket^{B}=\langle\,p(i,j)\,|\,i\in I,j\in
J,\,p\in\text{Pol}_{2}(B)\text{ with $p$ absorbing}\,\rangle^{B}.$

In [4], Problem 3.4, the following questions about $\llbracket
I,J\rrbracket^{B}$ were raised.

######  Problem 3.4.

Let $B$ be a brace.

  * (1)

Does the equality $\llbracket I,J\rrbracket^{B}=\langle I\ast J+J\ast
I+[I,J]_{+}\rangle^{B}$ hold?

  * (2)

Does the equality $\llbracket I,J\rrbracket^{B}=I\ast J+J\ast I+[I,J]_{+}$
hold?

Our next theorem gives a positive answer to the first question, thus showing
that commutators of ideals defined by means of either the Huq $=$ Smith
condition or absorbing polynomials coincide. For the proof, we first remark
that the following equality holds in braces.

######  Remark 3.5.

Let $B$ be a brace. Then, for every $i,j,x,y,z\in(B,+)$, we have that
$i+x+j+y-i+z-j=[-i,-x]_{+}+x+[-i,-j]_{+}+[-i,-y]^{-j,+}_{+}+[-j,-y]_{+}+y+[-j,-z]_{+}+z$.

######  Theorem 3.6.

Let $I,J$ be ideals of a brace $B$. Then:

  1. (1)

$I\ast J+J\ast I+[I,J]_{+}$ is the least left ideal containing

$X_{I,J}=[I,J]_{+}\cup[I,J]_{\cdot}\cup\\{ij-(i+j):\,i\in I,\,j\in J\\}.$

  2. (2)

$[I,J]^{B}=\langle I\ast J+J\ast I+[I,J]_{+}\rangle^{B}$. Thus,
$[I,J]^{B}=I\ast J+J\ast I+[I,J]_{+}$ if and only if $I\ast J+J\ast
I+[I,J]_{+}$ is an ideal of $B$.

  3. (3)

$[I,J]^{B}=\llbracket I,J\rrbracket^{B}$.

###### Proof.

(1) By [4, Proposition 1.4], $I\ast J+J\ast I$ and $[I,J]_{+}$ are left
ideals. Since $[I,J]_{+}$ is a normal subgroup of $(B,+)$, it follows that
$I\ast J+J\ast I+[I,J]_{+}$ is also a left ideal.

For the inclusion $X_{I,J}\subseteq I\ast J+J\ast I+[I,J]_{+}$, we need to
prove that

$\\{ij-(i+j)\,:\,i\in I,\ j\in J\\},\,[I,J]_{\cdot}\subseteq I\ast J+J\ast
I+[I,J]_{+}$

Let $i\in I$ and $j\in J$. For the former, observe that $ij-j-i=i+i\ast j-i$.
Thus,

$ij-j-i=i+i\ast j-i-i\ast j+i\ast j\in[I,J]_{+}+I\ast J.$

For the latter, using Eq. (2) repeatedly, it follows that

$\displaystyle iji^{-1}j^{-1}$ $\displaystyle=iji^{-1}+(iji^{-1})\ast
j^{-1}+j^{-1}=$ $\displaystyle=ij+(ij)\ast i^{-1}+i^{-1}+(iji^{-1})\ast
j^{-1}+j^{-1}=$ $\displaystyle=i+i\ast j+j+(ij)\ast
i^{-1}+i^{-1}+(iji^{-1})\ast j^{-1}+j^{-1}.$

Applying Eq. (1) repeatedly, the above equation becomes

$\displaystyle i+i\ast j+j+i\ast(j\ast i^{-1})+j\ast i^{-1}+i\ast
i^{-1}+i^{-1}+$ $\displaystyle+(ij)\ast(i^{-1}\ast j^{-1})+i^{-1}\ast
j^{-1}+(ij)\ast j^{-1}+j^{-1}=$ $\displaystyle=i+i\ast j+j+i\ast(j\ast
i^{-1})+j\ast i^{-1}+i\ast i^{-1}+i^{-1}+$
$\displaystyle+i\ast\\!(j\ast(i^{-1}\ast j^{-1}))+j\ast\\!(i^{-1}\ast
j^{-1})+i\ast\\!(i^{-1}\ast j^{-1})+i^{-1}\ast j^{-1}+$
$\displaystyle+i\ast(j\ast j^{-1})+j\ast j^{-1}+i\ast j^{-1}+j^{-1}$

Observe that $i\ast i^{-1}+i^{-1}=-i+0-i^{-1}+i^{-1}=-i$ and $j\ast
j^{-1}=-j-j^{-1}$. Thus, we have

$\displaystyle iji^{-1}j^{-1}$ $\displaystyle=i+i\ast j+j+i\ast(j\ast
i^{-1})+j\ast i^{-1}-i+i\ast(j\ast(i^{-1}\ast j^{-1}))$ ($\star$)
$\displaystyle+j\ast(i^{-1}\ast j^{-1})+i\ast(i^{-1}\ast j^{-1})+i^{-1}\ast
j^{-1}+i\ast(j\ast j^{-1})-j$ ($\diamond$) $\displaystyle-j^{-1}+i\ast
j^{-1}+j^{-1}$

Since $I\ast J,J\ast I\subseteq I\cap J$, Remark 3.5 applied on ($\star$ ‣
3)+($\diamond$ ‣ 3) yields

$iji^{-1}j^{-1}\in I\ast J+J\ast I+[I,J]_{+}+(-j^{-1}+i\ast j^{-1}+j^{-1}).$

But, $-j^{-1}+i\ast j^{-1}+j^{-1}=[j^{-1},-i\ast j^{-1}]_{+}+i\ast
j^{-1}\in[I,J]_{+}+I\ast J$. Hence, $[I,J]_{\cdot}\subseteq I\ast J+J\ast
I+[I,J]_{+}$ follows.

Now, let $L$ be the least left ideal of $B$ containing $X_{I,J}$ (note that
the arbitrary intersection of left ideals is a left ideal). In order to prove
that $I\ast J+J\ast I+[I,J]_{+}=L$, we only need to show that $J\ast
I\subseteq L$.

Let $j\in J$ and $i\in I$. Then

$j\ast i=[-j\ast i,-j]_{+}+(ji-i-j)\in X_{I,J}+(ji-i-j)$

as $j\ast i\in I\cap J$, so it suffices to prove that $(ji-i-j)\in L$. Since
$[j,i]_{\cdot}\in X_{I,J}\subseteq L$ and $L$ is $\lambda$-invariant, it
follows that

$ji=ij[j,i]_{\cdot}=ij+\lambda_{ij}([j,i]_{\cdot})=ij+x$

for some $x\in L\cap I\cap J$. Therefore,

$\displaystyle ji-i-j$
$\displaystyle=ij+x-([-j,-i]_{+}+i+j)=ij+x-(i+j)+[-i,-j]_{+}=$
$\displaystyle=ij-(i+j)+[-(i+j),-x]_{+}+x+[-i,-j]_{+}\in L$

Consequently, $I\ast J+J\ast I+[I,J]_{+}=L$.

(2) By definition, $[I,J]^{B}=\langle X_{I,J}\rangle^{B}$. Then, the previous
statement yields $[I,J]^{B}=\langle I\ast J+J\ast I+[I,J]_{+}\rangle^{B}$.

(3) By Example 3.2, $[I,J]^{B}\leq\llbracket I,J\rrbracket^{B}$. For the other
inclusion, let $p$ be an absorbing $2$-polynomial, and choose $i\in I$ and
$j\in J$. Recall that $p(i,j)$ is a finite sequence of sums and products of
$i$, $j$ and their inverses. Since $[I,J]^{B}=\langle X_{I,J}\rangle^{B}$, we
can express

$p(i,j)+[I,J]^{B}=(q_{i}+q_{j})+[I,J]^{B},$

where $q_{i}\in I$ and $q_{j}\in J$ are such that

$p(i,0)+[I,J]^{B}=q_{i}+[I,J]^{B}\quad\textnormal{and}\quad
p(0,j)+[I,J]^{B}=q_{j}+[I,J]^{B}.$

By definition of absorbing polynomial, $p(i,0)=0=p(0,j)$, so
$q_{i}+q_{j}\in[I,J]^{B}$ and consequently $p(i,j)\in[I,J]^{B}$. ∎

The next example gives a negative answer to Problem 3.4 (2).

######  Example 3.7.

Let

$\begin{array}[]{c}(B,+)=\langle
x,y\,|\,4x=4y=0,\,x+y=y+x\rangle\quad\textnormal{and}\\\\[5.69046pt]
(C,\cdot)=\langle
a,b,c\,|\,c^{4}=1,\,a^{2}=b^{2}=c^{2},\,(ab^{-1})^{2}=1,\,b^{-1}cb=c,\,a^{-1}ca=c\rangle\end{array}.$

We see that $C$ acts on $B$ via

$\begin{array}[]{lll}a(x)=x+2y,&b(x)=x+2y,&c(x)=3x+2y\\\
a(y)=-y,&b(y)=2x+y,&c(y)=2x+3y\end{array}$

Consider the semidirect product $G$ of $B$ by $C$ with respect to this action
$\lambda\colon C\longrightarrow\operatorname{Aut}(B)$. For the sake of
clarity, we use multiplicative notation for $B$ in $G$. Let $D=\langle
xa,yb,xyc\rangle\leq G$. It follows that $G$ is a trifactorised group as
$BD=DC$, $D\cap C=D\cap B=1$. By [1, Lemma 3.2], there exists a bijective
$1$-cocycle $\delta\colon C\rightarrow(B,+)$ given by $D=\\{\delta(c)c:c\in
C\\}$ (see Table 1). This yields a product in $B$ and we get a brace of
abelian type, $(B,+,\cdot)$, which corresponds to SmallBrace(16, 73) in the
Yang–Baxter library [22] for GAP [13].

$g$ | $\delta(g)$ | $g$ | $\delta(g)$ | $g$ | $\delta(g)$ | $g$ | $\delta(g)$
---|---|---|---|---|---|---|---
$1$ | $0$ | $b$ | $y$ | $c$ | $x+y$ | $bc$ | $3x$
$a$ | $x$ | $b^{3}$ | $2x+3y$ | $c^{3}$ | $3x+3y$ | $bc^{-1}$ | $x+2y$
$a^{2}$ | $2x+2y$ | $ab$ | $x+3y$ | $ac$ | $2x+y$ | $abc$ | $2y$
$a^{3}$ | $3x+2y$ | $ab^{-1}$ | $3x+y$ | $ac^{-1}$ | $3y$ | $abc^{-1}$ | $2x$
Table 1: Bijective $1$-cocycle associated with the brace of Example 3.7

Let $I=\langle 2x,y\rangle\leq(B,+)$. Then $\lambda(I)=I$ and $|I|=8$, so $I$
is an ideal of $B$. Since $B$ is of abelian type, we compute

$I\ast I+[I,I]_{+}=I\ast I=\langle u\ast v\,|\,u,v\in I\rangle_{+}=\langle
2x\rangle_{+}=\\{1,2x\\}$

Therefore, $I\ast I$ is not an ideal of $B$, as $\delta(abc^{-1})=2x$ and
$\\{1,abc^{-1}\\}$ is not a normal subgroup of $C$. Hence, $I\ast
I\subsetneq[I,I]^{B}=I$.

##  4 Central nilpotency of braces

The aim of this section is to develop a standard theory of central nilpotency
of braces (for example, canonical lower and upper central series of a brace
are studied and a torsion theory is established). In order to have the
broadest range of applicability for our results, and also to simplify proofs,
we usually deal with concepts that are much more general than central
nilpotency (for example, local central-nilpotency, and hypercentrality).

We start by introducing the basic definitions. Central nilpotency of braces
was first introduced by using a brace-theoretical analogue of the centre of a
group (see [4] and [17]). The _centre_ of a brace $B$ (also known as the
annihilator ideal of $B$) is the ideal of $B$ defined as

$\zeta(B):=\operatorname{Soc}(B)\cap\operatorname{Fix}(B)=\\{a\in
B\,|\,a+b=b+a=ab=ba,\ \forall\,b\in B\\}$

(see [6], where it was first introduced in the context of ideal extensions).
Thus, abelian braces $B$ are precisely those ones satisfying $\zeta(B)=B$.

In [9] and [21], the definition of central nilpotency has been extended to
include more types of braces (see also [10], where similar concepts for braces
of abelian type are considered).

######  Definition 4.1.

Let $B$ be a brace. If $J\leq I$ are ideals of $B$
satisfying$I/J\leq\zeta(B/J)$, we say that $I/J$ is a central factor of $B$.

A $c$-series of $B$ is a chain $\mathcal{I}$ of ideals of $B$ such that
$0,B\in\mathcal{I}$ and whose factors are central, that is,
$I/J\leq\zeta(B/J)$ for all consecutive elements $J\leq I$ of $\mathcal{I}$
(meaning that there is no $C\in\mathcal{I}$ satisfying $J<C<I$). A complete
$c$-series is just a $c$-series containing the unions and the intersections of
all its members. Since every $c$-series can be completed, we usually consider
every $c$-series to be complete. We say that $B$ is $\zeta$-nilpotent if it
admits a $c$-series.

If $B$ has an ascending $c$-series

$0=I_{0}\leq I_{1}\leq\ldots I_{\alpha}\leq I_{\alpha+1}\leq\ldots I_{\mu}=B$

(here $\alpha<\mu$ are ordinal numbers), then $B$ is hypercentral, and the
smallest ordinal number $\mu$ for which such an ascending $c$-series exists is
the hypercentral length $n_{c}(B)$ of $B$. (Note that
$I_{\alpha+1}/I_{\alpha}\leq\zeta(B/I_{\alpha})$ for all ordinals
$\alpha<\mu$.)

If $B$ has a finite $c$-series

$0=I_{0}\leq I_{1}\leq\ldots\leq I_{n}=B,$

then $B$ is centrally nilpotent; in this case, the smallest non-negative
integer for which such a $c$-series exists is referred to as the _class_
$n_{c}(B)$ of $B$. (Note that $I_{i+1}/I_{i}\leq\zeta(B/I_{i})$ for all $0\leq
i<n$.)

Of course, centrally nilpotent braces are hypercentral, and hypercentral
braces are $\zeta$-nilpotent, but the converses do not hold. Moreover,
subbraces of centrally nilpotent (resp. hypercentral, $\zeta$-nilpotent)
braces are centrally nilpotent (resp. hypercentral, $\zeta$-nilpotent). Also,
quotients of centrally nilpotent (resp. hypercentral) braces are still
centrally nilpotent (resp. hypercentral), but the consideration of the
infinite dihedral group shows that quotients of a $\zeta$-nilpotent brace can
be non-$\zeta$-nilpotent. Finally, direct products of hypercentral braces are
hypercentral; direct products of finitely many centrally nilpotent braces are
centrally nilpotent; and restricted direct products of $\zeta$-nilpotent
braces are $\zeta$-nilpotent.

Following [4] and [9], canonical $c$-series are introduced for a brace $B$.

($\blacktriangle$) The upper central series of $B$ is recursively defined by
putting: $\zeta_{0}(B)=0$,
$\zeta_{\alpha+1}(B)/\zeta_{\alpha}(B)=\zeta(B/\zeta_{\alpha}(B))$ for every
ordinal $\alpha$, and
$\zeta_{\lambda}(B)=\bigcup_{\beta<\lambda}\zeta_{\beta}(B)$ for every limit
ordinal $\lambda$. The last term of the upper central series is the
hypercentre $\overline{\zeta}(B)$ of $B$.

($\blacktriangledown$) The lower central series of $B$ is recursively defined
by: $\Gamma_{1}(B)=B$, $\Gamma_{\alpha+1}(B)=\langle\Gamma_{\alpha}(B)\ast
B,B\ast\Gamma_{\alpha}(B),[\Gamma_{\alpha}(B),B]_{+}\rangle_{+}$ for every
ordinal $\alpha$, and
$\Gamma_{\lambda}(B)=\bigcap_{\beta<\lambda}\Gamma_{\beta}(B)$ for every limit
ordinal $\lambda$. Note that since $\Gamma_{\alpha}(B)$ is an ideal for every
$\alpha\leq\mu$, we have that

$\Gamma_{\alpha+1}(B)=\Gamma_{\alpha}(B)\ast
B+B\ast\Gamma_{\alpha}(B)+[\Gamma_{\alpha},B]_{+}=[\Gamma_{\alpha}(B),B]^{B},$

for every ordinal $\alpha<\mu$ (see Theorem 3.6). The last term of the lower
central series is the hypocentre $\overline{\Gamma}(B)$ of $B$.

The following easily provable all-in-one result shows the relations between
the concepts we just introduced (see [18] for the definition of the upper
central series $\\{Z_{\alpha}(G)\\}_{\alpha\in\operatorname{Ord}(G)}$ of a
group $G$).

######  Proposition 4.2.

Let $B$ be a brace.

  * (1)

(see [3, Proposition 17]) If $J\leq I$ are ideals of $B$, then
$I/J\leq\zeta(B/J)$ if and only if $[I,B]^{B}\leq J$. In particular, if
$0=I_{0}\leq I_{1}\leq\ldots\leq I_{n}=B$ is a finite $c$-series, then:

    * (a)

$\Gamma_{j}(B)\leq I_{n-j+1}$ for every $1\leq j\leq n+1$; so
$\Gamma_{n+1}(B)=0$.

    * (b)

$I_{j}\leq\zeta_{j}(B)$ for every $0\leq j\leq n$; so $\zeta_{n}(B)=B$.

    * (c)

$n_{c}(B)$ is the smallest number $n$ such that $\zeta_{n}(B)=B$, and the
smallest number $n$ such that $\Gamma_{n+1}(B)=0$.

  * (2)

$B$ is hypercentral if and only if $B=\overline{\zeta}(B)$. Moreover, in this
case $n_{c}(B)$ is the smallest ordinal $\alpha$ for which
$B=\zeta_{\alpha}(B)$.

  * (3)

$\zeta_{\alpha}(B)\subseteq\operatorname{Z}_{\alpha}(B,+)\cap\operatorname{Z}_{\alpha}(B,\cdot)$
for all ordinal $\alpha$. In particular, centrally nilpotent (resp.
hypercentral) braces are braces of nilpotent (resp. hypercentral) type whose
additive group is nilpotent (hypercentral).

  * (4)

$\partial_{n}(B)\leq\Gamma_{n+1}(B)$ for every $n\in\mathbb{N}$. In
particular, centrally nilpotent braces are soluble with derived length less
than or equal to their class.

The following result generalises Grün’s lemma for groups (see for instance
[18], Part 1, p.48).

######  Theorem 4.3.

Let $B$ be a brace such that $\zeta_{2}(B)>\zeta(B)$. Then either $[B,B]_{+}$
or $[B,B]_{\cdot}$ is a proper subset of $B$.

###### Proof.

We may assume by Grün’s lemma that $Z(B,+)=Z_{2}(B,+)$ and
$Z(B,\cdot)=Z_{2}(B,\cdot)$. By Proposition 4.2,

$\zeta_{2}(B)\subseteq Z_{2}(B,+)\cap Z_{2}(B,\cdot)=Z(B,+)\cap Z(B,\cdot).$

Now, choose $c\in\zeta_{2}(B)\setminus\zeta(B)$ and consider the map

$\varphi_{c}:b\in B\mapsto c\ast b\in\zeta(B).$

Let $b_{1},b_{2}\in B$. Then

$c\ast(b_{1}+b_{2})=c\ast b_{1}+c\ast b_{2}$

because $c\ast b_{2}\in\zeta(B)$, so $\varphi_{c}$ is a homomorphism with
respect to $+$. Since $c\not\in\zeta(B)$, we have that
$c\not\in\operatorname{Ker}(\lambda)$ and consequently $\varphi_{c}(B)\neq 0$.
Thus $[B,B]_{+}$ is properly contained in $B$ and we are done. ∎

In order to see if a brace is centrally nilpotent (resp. hypercentral) or not,
we only need to look at its countable subbraces: this is the content of our
next result.

######  Theorem 4.4.

Let $B$ be a brace. Then:

  * (1)

$B$ is centrally nilpotent of class at most $c$ if and only if its finitely
generated subbraces are centrally nilpotent of class at most $c$.

  * (2)

$B$ is centrally nilpotent if and only if its countable subbraces are
centrally nilpotent.

  * (3)

$B$ is hypercentral if and only if its countable subbraces are hypercentral.

###### Proof.

For each $u,v\in B$, the symbol $u\circ v$ means that we are performing one
(but we do not know which) of the following operations
$[u,v]_{\cdot},[u,v]_{+},u\ast v$. The first statement is a direct consequence
of the fact that $\zeta_{c}(B)$ can be easily characterised as the set of all
elements $b\in B$ such that

$(\ldots((b\circ b_{1})\circ\ldots)\circ b_{c})=0$

for all $b_{1},\ldots,b_{c}\in B$.

(2) Only one implication is in doubt. Thus, assume all countable subbraces of
$B$ are centrally nilpotent but $B$ is not centrally nilpotent. By (1), for
each $c\in\mathbb{N}$, there is a finitely generated centrally nilpotent brace
$B_{c}$ whose class is at least $c$. Then $C=\langle
B_{i}\,:\,i\in\mathbb{N}\rangle$ is a countable subbrace of $B$ that is not
centrally nilpotent, a contradiction.

(3) Only one implication is in doubt. Thus, assume all countable subbraces of
$B$ are hypercentral, and that $B$ is not hypercentral. We may certainly
assume that $\zeta(B)=0$. Let $0\neq x\in B$. Then there are sequences of non-
zero elements

$a_{1},a_{2},\ldots,a_{n},\ldots\quad\textnormal{and}\quad
x=b_{1},b_{2}\ldots,b_{n},\ldots$

of $B$ such that $b_{n+1}=b_{n}\circ a_{n}$ for all $n\in\mathbb{N}$. Let
$C=\langle a_{i},b_{j}\,:\,i,j\in\mathbb{N}\rangle$. Thus, $C$ is countable
and so hypercentral. Now, for each $i\in\mathbb{N}$, let $\alpha_{i}$ be the
smallest ordinal $\beta$ such that $b_{i}\in\zeta_{\beta}(C)$. It follows that

$\alpha_{1}>\alpha_{2}>\ldots>\alpha_{n}>\ldots$

is a strictly descending chain of ordinal numbers, a contradiction.∎

In studying the structure of an arbitrary group, local analogs of nilpotence
are very useful. Similarly, the following definition is crucial for us in
studying centrally nilpotent braces (see [9]).

######  Definition 4.5.

A brace $B$ is locally centrally-nilpotent if every finitely generated
subbrace is centrally nilpotent.

Of course, every subbrace/quotient of a locally centrally-nilpotent brace is
still locally centrally-nilpotent, and also restricted direct products of
locally centrally-nilpotent braces are locally centrally-nilpotent. Moreover,
by [21], Corollary 3.6, every hypercentral brace is locally centrally-
nilpotent, but the converse does not hold. As a consequence of the following
result, we see that every locally centrally-nilpotent brace is
$\zeta$-nilpotent.

######  Theorem 4.6.

Let $B$ be a locally centrally-nilpotent brace.

  * (1)

If $I$ is any minimal ideal of $B$, then $I\leq\zeta(B)$.

  * (2)

If $M$ is any maximal subbrace of $B$, then $M$ is an ideal of $B$.

In particular, every minimal ideal of $B$ has prime order, and $\partial(B)$
is contained in every maximal subbrace of $B$.

###### Proof.

(1) Suppose that $I\not\leq\zeta(B)$. Then there exist elements $b\in B$ and
$x\in I$ such that $\mathcal{S}=\\{[b,x]_{+},[b,x]_{\cdot},x\ast
b\\}\neq\\{0\\}$. Let $c\in\mathcal{S}\setminus\\{0\\}$. Since $I$ is a
minimal ideal of $B$, the ideal generated by $c$ in $B$ is $I$, so there are
elements $y_{1},\ldots,y_{n}\in B$ such that $x$ belongs to the ideal
generated by $c$ in $S=\langle b,c,y_{1},\ldots,y_{n}\rangle$.

Let $J=x^{S}$. Since $S$ is centrally nilpotent, there is a finite chain

$0=J_{0}<J_{1}<\ldots<J_{m}=J$

of ideals of $S$ such that $J_{i+1}/J_{i}\leq\zeta(S/J_{i})$. Choose
$\ell\in\mathbb{N}$ with $c\in J_{\ell}\setminus J_{\ell-1}$; clearly,
$\ell\neq 0,m$ because $c$ is a non-zero element of one of the following
types: $[b,x]_{+},[b,x]_{\cdot},x\ast b$. Now,

$x^{S}\leq c^{S}\leq c^{S}+J_{\ell-1}=\langle c\rangle+J_{\ell-1}\leq
J_{\ell}<J_{m}=J=x^{S},$

a contradiction.

(2) Assume $M$ is not an ideal. If $B\ast B\leq M$, then $(M,+)/(B\ast B,+)$
is a maximal subgroup of the locally nilpotent group $(B,+)/(B\ast B,+)$, so
it is even normal, and it follows that $M$ is an ideal of $B$, a
contradiction. Thus, there exists an element $x\in B\ast B\setminus M$. Since
$M$ is a maximal subbrace of $B$, we have that $B=\langle M,x\rangle$. Then
there is a finite subset $L$ of $B$ such that $x\in\langle L\rangle\ast\langle
L\rangle$. For each $y\in L$, let $B_{y}$ be a finite subset of $M$ such that
$y\in\langle B_{y}\cup\\{x\\}\rangle$. Put

$D=\langle B_{y}\,:\,y\in L\rangle\quad\textnormal{and}\quad E=\langle
D,x\rangle,$

so $E$ is finitely generated and $L\subseteq E$. Now, $E$ is centrally
nilpotent, and $x\not\in D\subseteq M$. Let $N$ be a subbrace of $E$ which is
maximal with respect to containing $D$ but not $x$. Since $E=\langle
D,x\rangle$, we see that $N$ is actually a maximal subbrace of $E$. Since $E$
is centrally nilpotent, there is $n\in\mathbb{N}$ such that $\zeta_{n}(E)\leq
N$ but $\zeta_{n+1}(E)\not\leq N$. Then $N$ is a proper ideal of
$\zeta_{n+1}(E)+N$ and so $N\trianglelefteq E$, since $N$ is a maximal
subbrace of $E$. Therefore $E/N$ is centrally nilpotent, and so $x\in E\ast
E\leq N$, a contradiction. ∎

In [2], chief factors of braces are introduced and shown to play a key role in
its ideal structure. Let $I$ and $J$ be ideals of a brace $B$ such that $J\leq
I$. The quotient $I/J$ is said to be a _chief factor_ of $B$ if $I/J$ is a
minimal ideal of $B/J$. A chain $\mathcal{C}$ of ideals of $B$ is a _chief
series_ of $B$ if $0,B\in\mathcal{C}$ and $I/J$ is a chief factor of $B$
whenever $J<I$ are consecutive terms of $\mathcal{C}$. By Zorn’s lemma, every
brace has a (possibly infinite) chief series. In [2], a brace $B$ is proved to
have a finite chief series if and only if it is noetherian (that is, every
ascending chain of ideals is eventually stationary) and artinian (that is,
every descending chain of ideals is eventually stationary). By Theorem 4.6,
every chief series of a locally centrally-nilpotent brace is a $c$-series, so
we have the following result.

######  Corollary 4.7.

Let $B$ be a locally centrally-nilpotent brace. Then $B$ is $\zeta$-nilpotent.

######  Remark 4.8.

It should be noted that the proof of Theorem 4.6 (1) proves much more than we
stated. In fact, let $\mathfrak{Z}$ be the class of all braces in which every
chief factor is central. Moreover, let L$\mathfrak{Z}$ be the class of all
braces in which every finite subset $F$ is contained in a subbrace
$C_{F}\in\mathfrak{Z}$. The proof of Theorem 4.6 (1) can be modified to show
that L$\mathfrak{Z}=\mathfrak{Z}$.

More in detail, using the notation of the first half of the proof of Theorem
4.6 (1), we get that $S$ is contained in a subbrace $T\in\mathfrak{Z}$. Let
$J=x^{T}$.

Since $[J,T]^{T}$ is an ideal of $T$ containing $c$, we have that $x\in
c^{T}\leq[J,T]^{T}$ and so $J=x^{T}=[J,T]^{T}$. Finally, let $M$ be a maximal
ideal of $T$ contained in $J$ and such that $x\not\in M$. Then $J/M$ is a
chief factor of $T$ and so $[J,T]^{T}\leq M$, a contradiction.

It follows from Corollary 4.7 that any non-zero ideal of a locally centrally-
nilpotent brace contains a (non-zero) central factor of the whole brace. In
case of a hypercentral brace we can say more.

######  Lemma 4.9.

Let $B$ be a hypercentral brace. If $I$ is any non-zero ideal of $B$, then
$I\cap\zeta(B)\neq 0$.

###### Proof.

Let $\alpha$ be the smallest ordinal number such that
$J=I\cap\zeta_{\alpha}(B)\neq 0$. Then $\alpha$ is successor and
$I\cap\zeta_{\alpha-1}(B)=0$. Now, $[J,B]^{B}\leq J\cap\zeta_{\alpha-1}(B)=0$
and so $J\leq\zeta(B)$. ∎

######  Corollary 4.10.

Let $B$ be a brace having a finite chief series (resp. an ascending chief
series) $\mathcal{I}$. Then $B$ is centrally nilpotent (resp. hypercentral) if
and only if every chief factor of $\mathcal{I}$ is central.

Our next two subsections deal with the torsion theory of locally centrally-
nilpotent braces and with the problem of defining a suitable index for
subbraces. Before delving into them, we note that some important results for
nilpotent groups do not hold for braces.

  * •

Bearing in mind the normaliser condition for nilpotent groups, the idealiser
of a subbrace is introduced in [17]: given a subbrace $S$ of a brace $B$, the
idealiser of $S$ is defined as the largest subbrace $N$ of $B$ such that $S$
is an ideal of $N$. It is then stated that every subideal is properly
contained in its idealiser (see Section 4.2 for the definition of subideal).
Example A in Section 6 shows that the idealiser of a subbrace does not exist
in general. We note however that if $C$ is a subbrace of a brace $B$, then one
can define the largest strong left ideal $N_{B}(C)$ of $B$ additively and
multiplicatively normalising $B$ and such that $\lambda_{x}(C)=C$ for every
$x\in N_{B}(C)$ — but such a strong left ideal need not to contain $C$.

  * •

Example B shows that there is no analogue of Fitting theorem for centrally
nilpotent ideals.

  * •

Example C shows that an abelian subideal need not be contained in a centrally
nilpotent ideal.

  * •

The ideal structure of the brace listed as SmallBrace(32, 24003) in the
YangBaxter library [22] for GAP [13] is described in [3]. This brace $B$ has
only a unique maximal subbrace $I$, which is also its only non-zero proper
ideal. Moreover, $\partial(B)=I$. Nevertheless, $B$ is not centrally nilpotent
as it is not even soluble. This shows that a finite brace whose maximal
subbraces are ideals need not be soluble. The same example shows that a finite
brace whose subbraces are subideals need not be centrally nilpotent (see
Example D for more details), although an easy induction shows that they are at
least weakly soluble in the sense explained in [3].

###  4.1 Torsion theory

The aim of this subsection is to establish a torsion theory for locally
centrally-nilpotent braces. We start with some definitions.

######  Definition 4.11.

Let $B$ be a brace. The subset of all periodic elements of $(B,+)$ is denoted
by $T_{+}(B)$, while that of all periodic elements of $(B,\cdot)$ is denoted
by $T_{\cdot}(B)$. Moreover:

  * •

An element $b$ of $B$ is periodic if $\langle b\rangle$ is finite. The order
of $b$ is $|\langle b\rangle|$. If $\pi$ is any set of prime numbers, then $b$
is a $\pi$-element if its order is a $\pi$-number. A $\pi$-subbrace is just a
subbrace containing only $\pi$-elements. Finally, $B$ is periodic if every
element of $B$ is periodic.

  * •

$B$ is torsion-free if every element $b\in B$ is either zero or is non-
periodic.

  * •

$B$ is locally finite if every finitely generated subbrace of $B$ is finite.

  * •

$B$ has finite exponent $n$ if $B$ is periodic and $n$ is the smallest
positive integer such that $b^{n}=nb=0$ for all $b\in B$.

Clearly, every locally finite brace is periodic but the converse does not
hold. The following result shows that in the context of locally centrally-
nilpotent braces we can precisely identify the set of all periodic elements of
$B$.

######  Theorem 4.12.

Let $B$ be a locally centrally-nilpotent brace. Then:

  * (1)

$T_{+}(B)=T_{\cdot}(B)$.

  * (2)

$T_{+}(B)$ is an ideal of $B$.

  * (3)

$T_{+}(B/T_{+}(B))=0$.

  * (4)

If $B$ is periodic, then $B$ is locally finite.

  * (5)

$B$ is locally finite$\iff$ $(B,+)$ is locally finite$\iff$$(B,\cdot)$ is
locally finite.

###### Proof.

The proof of (1)–(3) is an easy consequence of [17], Proposition 4.2. Let us
prove (4). Assume $B$ is periodic and finitely generated. Then $(B,+)$ is a
periodic nilpotent group. Moreover, by Theorem 3.7 of [21], $(B,+)$ is also
finitely generated. Thus $(B,+)$ is finite and (4) is proved. Finally, (5) is
an obvious consequence of [21], Theorem 3.7. ∎

Let $B$ be a brace, and let $p$ be a prime. The Sylow $p$-subbrace of $B$ is
just a maximal element of the set of all its $p$-subbraces with respect to the
inclusion. Our next result shows that the Sylow subbraces of a locally
centrally-nilpotent brace are ideals and that they coincide with the
additive/multiplicative Sylow subgroups.

######  Theorem 4.13.

Let $B$ be a locally finite brace. Then, $B$ is locally centrally-nilpotent if
and only if, for every prime $p$,
$\operatorname{Syl}_{p}(B,+)=\operatorname{Syl}_{p}(B,\cdot)=\\{B_{p}\\}$,
$B_{p}$ is locally centrally-nilpotent and $B$ is the direct product of the
$B_{p}$’s.

###### Proof.

Only one direction is in doubt. Since $B$ is locally finite, we may assume $B$
is finite and centrally nilpotent. Let $p$ be a prime. Since both $(B,+)$ and
$(B,\cdot)$ are nilpotent groups, there exist Sylow $p$-subgroups
$B_{p}\unlhd(B,+)$ and $\bar{B}_{p}\unlhd(B,\cdot)$. Observe that $B_{p}$ is
also $\lambda$-invariant, as it is a characteristic subgroup of $(B,+)$.
Therefore, $B_{p}=\bar{B}_{p}$ is an ideal of $B$. ∎

With respect to the definition of order of an element, we note the following
interesting fact.

######  Proposition 4.14.

Let $B$ be a brace whose additive and multiplicative groups are cyclic. Then
there is $x\in B$ which is a generator of both $(B,+)$ and $(B,\cdot)$.

###### Proof.

By Theorem 4.6 of [19], we may assume $B$ is finite. If
$\operatorname{Ker}(\lambda)=0$, then $(B,\cdot)$ embeds into
$\operatorname{Aut}(B,+)$, a contradiction. Thus, $\zeta(B)\neq 0$. Iterating
this argument, we see that $B$ is centrally nilpotent, so $B$ factorizes into
the direct product of its Sylow $p$-subgroups. It is therefore possible to
assume that $B$ has prime power order $p^{n}$.

Let $I$ be a subbrace of $\zeta(B)$ of order $p$. By induction there is an
element $x\in B$ such that $x+I$ is a both a generator of $(B/I,+)$ and
$(B/I,\cdot)$. If $\langle x^{p^{n-1}}\rangle_{\cdot}\cap I=0$, then
$(B,\cdot)$ is not cyclic, a contradiction. Thus, $\langle
x^{p^{n-1}}\rangle_{\cdot}=I$ and $x$ is a generator of $(B,\cdot)$.
Similarly, $x$ is a generator of $(B,+)$. ∎

Our next result is a huge generalisation of [17], Lemma 4.1. In order to state
it, we need the following definition.

######  Definition 4.15.

Let $B$ be a brace, and let $\pi$ be a set of prime numbers. We say that $B$
is $\pi$-free if it does not contain $\pi$-elements.

Obviously, a trivial brace $B$ is $\pi$-free if and only if $(B,+)$ and/or
$(B,\cdot)$ are $\pi$-free as groups.

######  Theorem 4.16.

Let $B$ be a brace and let $\pi$ be a set of primes. If $\zeta(B)$ is
$\pi$-free, then each factor of the upper central series of $B$, and therefore
the hypercentre of $B$, is $\pi$-free.

###### Proof.

Suppose the theorem is false and let $\alpha$ be the first ordinal such that
$\zeta_{\alpha+1}(B)/\zeta_{\alpha}(B)$ is not $\pi$-free; in particular,
there is $x\in\zeta_{\alpha+1}(B)\setminus\zeta_{\alpha}(B)$ such that
$x^{m}\in\zeta_{\alpha}(B)$ for some positive $\pi$-number $m$. We divide the
proof in two parts according to $\alpha$ being limit or not.

Suppose first $\alpha$ is limit. Then
$x^{m}\in\zeta_{\beta+1}(B)\setminus\zeta_{\beta}(B)$ for some $\beta<\alpha$.
Since $x\notin\zeta_{\alpha}(B)$, there is $b\in B$ and $\gamma\geq\beta$ such
that one of the elements $x\ast b$, $[x,b]_{+}$, $[x,b]_{\cdot}$ belongs to
$\zeta_{\gamma+1}(B)\setminus\zeta_{\gamma}(B)$; call $c$ such an element.
Assume $c=x\ast b$. Then

$(x\ast b)^{m}\equiv x^{m}\ast b\quad\textnormal{(mod $\zeta_{\gamma}(B)$)}$

and so $x^{m}\ast b\in\zeta_{\beta}(B)\leq\zeta_{\gamma}(B)$. Therefore
$(x\ast b)^{m}\in\zeta_{\gamma}(B)$. But $x\ast b\in\zeta_{\gamma+1}(B)$ and
$\zeta_{\gamma+1}(B)/\zeta_{\gamma}(B)$ is $\pi$-free, so $c=x\ast
b\in\zeta_{\gamma}(B)$, a contradiction. Similarly, we deal with the cases in
which $c=[b,x]_{\cdot}$ and $c=[b,x]_{+}$.

Suppose now that $\alpha$ is successor; in this case, we may assume
$\alpha=1$, so $x\in\zeta_{2}(B)\setminus\zeta(B)$, $\zeta(B)$ is $\pi$-free,
and $x^{m}\in\zeta(B)$. Put $C=\langle x\rangle+\zeta(B)$. By Theorem 3.5 of
[9], we have that $|C\ast C|$ is a $\pi$-number. On the other hand, $C\ast
C\leq\zeta(B)$, and so $C\ast C=0$. Thus, $x^{m}=mx$.

Let $b$ be any element of $B$. Then $m[x,b]_{+}=[mx,b]_{+}=0$, so
$[x,b]_{+}=0$. Similarly, $[x,b]_{\cdot}=x\ast b=0$. Therefore $x$ belongs to
$\zeta(B)$, the final contradiction. ∎

######  Corollary 4.17.

Let $B$ be a brace. If $\zeta(B)$ is torsion-free, then
$\zeta_{\alpha+1}(B)/\zeta_{\alpha}(B)$ is torsion-free for every ordinal
$\alpha$.

Conversely, if we have information on the exponent of $\zeta(B)$, then we can
obtain information on the exponent of the factors of the upper central series.

######  Theorem 4.18.

Let $B$ be a brace. If $\zeta(B)$ has exponent $n$, then
$\zeta_{i+1}(B)/\zeta_{i}(B)$ has exponent dividing $n^{2^{i}}$ for each
positive integer $i$.

###### Proof.

It is enough to show that $\zeta_{2}(B)/\zeta(B)$ has exponent dividing
$n^{2}$. Let $b\in\zeta_{2}(B)$ and $a\in B$. Then $b\ast a\in\zeta(B)$, so

$b^{n}\ast a=n(b\ast
a)=0\quad\textnormal{and}\quad[a,b^{n}]_{\cdot}=[a,b]^{n}_{\cdot}=0.$

Thus, if we put $c=nb^{n}=b^{n^{2}}$, then
$c\in\operatorname{Ker}(\lambda)\cap Z(B,\cdot)$. But also

$[a,nb^{n}]_{+}=n[a,b^{n}]_{+}=0$

and so $c\in\zeta(B)$. ∎

######  Corollary 4.19.

Let $B$ be a brace such that $\zeta(B)$ has exponent $n$. If $B$ is centrally
nilpotent of class $c$, then $B$ has exponent at most $n^{2^{c}-1}$.

###  4.2 The index of a subbrace

One of the ways in which we cope with infinite groups is by using finite-index
subgroups. For braces, things are much more complicated, since we could deal
at the same time with two distinct unrelated indices. The aim of this section
is to study the “index” of a subbrace with particular emphasis to case of
locally centrally-nilpotent braces. The following definition provides us with
an invaluable tool in this study.

######  Definition 4.20.

Let $C$ be a subbrace of the brace $B$. We say that $C$ is serial in $B$ if
there is a chain of subbraces $\mathcal{C}$ connecting $C$ to $B$ such that if
$D<E$ are consecutive elements of $\mathcal{C}$, then $D\trianglelefteq E$ —
as in the case of $c$-series, we usually assume that these chains of subbraces
are complete, meaning that they contain arbitrary unions and intersections of
their members.

Now, $C$ is ascendant (resp. descendant) if $\mathcal{C}$ can be well-ordered
(resp. inversely well-ordered) with respect to the inclusion and its order
type is $\lambda$ (resp. the inverse of $\lambda$) for some ordinal number
$\lambda$. If $C$ is ascendant, then $\mathcal{C}$ can be written as

$\begin{array}[]{c}C=C_{0}\trianglelefteq C_{1}\trianglelefteq\ldots
C_{\alpha}\trianglelefteq C_{\alpha+1}\trianglelefteq\ldots
C_{\lambda}=B,\end{array}$

where $\alpha<\lambda$ are ordinal numbers; while, if $C$ is descendant, then
$\mathcal{C}$ takes the form

$\begin{array}[]{c}C=C_{\lambda}\ldots\trianglelefteq
C_{\alpha+1}\trianglelefteq C_{\alpha}\trianglelefteq\ldots\trianglelefteq
C_{1}\trianglelefteq C_{0}=B,\end{array}$

where $\alpha<\lambda$ are ordinal numbers. If $\mathcal{I}$ is finite, we say
that $C$ is subideal.

If $C$ is ascendant (resp. descendant) in $B$, then the smallest ordinal
number $\lambda$ for which there is a chain of subbraces of type (4.20) (resp.
of type (4.20)) is the ascendant length (resp. descendant length) of $C$ in
$B$. In case $C$ is subideal, then the ascendant length of $C$ in $B$ is
finite and is also called the subideal defect of $C$ in $B$.

Let $C$ be a subbrace of a brace $B$. Put $C^{B,0}:=C$ and recursively define
$C^{B,\alpha+1}=C^{C^{B,\alpha}}$ for every ordinal $\alpha$, and
$C^{B,\lambda}=\bigcap_{\alpha<\lambda}C^{B,\alpha}$ for every limit ordinal
$\lambda$. The family $\\{C^{B,\alpha}\\}_{\alpha\in\operatorname{Ord}}$ is
the ideal closure series of $C$ in $B$. It is easy to show that $C$ is
descendant (resp. subideal) in $B$ if and only if $C=C^{B,\mu}$ for some
ordinal $\mu$ (resp. for some finite ordinal $\mu$). If $C$ is descendant
(resp. subideal), then the descendant length (resp. the subideal defect) of
$C$ is then the smallest ordinal number $\lambda$ for which $C=C^{B,\lambda}$.

The following easy lemma contains all the basic properties of subideal,
ascendant, descendant and serial subbraces.

######  Lemma 4.21.

Let $B$ be a brace.

  * •

Every subideal of $B$ is ascendant, descendant and serial.

  * •

Ascendant (resp. descendant) subbraces are serial.

  * •

If $C$ is subideal (resp. ascendant, descendant, serial) in $B$, and $D\leq
B$, then $C\cap D$ is subideal (resp. ascendant, descendant, serial) in $D$.

  * •

If $C$ is subideal (resp. ascendant), then $CI/I$ is subideal (resp.
ascendant) in $B/I$ for every ideal $I$ of $B$.

  * •

If $C$ is subideal in $B$ of defect $n$, then $C$ is subideal in $C^{B}$ of
defect $n-1$.

Serial (resp. ascendant) subbraces play a relevant role in the theory of
locally centrally-nilpotent braces (resp. hypercentral braces), as the
following result shows.

######  Lemma 4.22.

Let $B$ be a brace.

  * (1)

If $B$ is hypercentral, then every subbrace $C$ of $B$ is ascendant.

  * (2)

If $B$ is centrally-nilpotent, then every subbrace $C$ of $B$ is subideal.

  * (3)

If $B$ is locally centrally-nilpotent, then every subbrace $C$ of $B$ is
serial.

###### Proof.

(1) Let $\lambda$ be the hypercentral length of $B$. Since $C$ is an ideal of
$C+\zeta(B)$, we see that

$C\trianglelefteq C+\zeta(B)\trianglelefteq\ldots
C+\zeta_{\alpha}(B)\trianglelefteq C+\zeta_{\alpha+1}(B)\trianglelefteq\ldots
C+\zeta_{\lambda}(B)=B$

is an ascending chain of subbraces of $B$ connecting $C$ to $B$.

(2) The proof is the same of (1).

(3) Zorn’s lemma implies that there is a maximal chain of subbraces between
$C$ and $B$. By Lemma 4.6 (2), if $D<E$ are consecutive terms of this chain,
then $D$ is an ideal of $E$. Therefore $C$ is serial in $B$. ∎

######  Remark 4.23.

Let $B$ be a brace and let $C$ be a (strong) left ideal of $B$. The proof of
Lemma 4.22 can actually be employed to prove that if we have an ascending
$c$-series of $B$, then there is an ascending chain of (strong) left ideals
connecting $C$ to $B$.

Lemma 4.22 has some rather interesting consequences concerning the “index” of
a subbrace.

######  Definition 4.24.

Let $B$ be a brace. A subbrace $C$ of $B$ is said to have finite index in $B$
if both $n_{+}=|(B,+):(C,+)|$ and $n_{\cdot}=|(B,\cdot):(C,\cdot)|$ are
finite; if $n_{+}=n_{\cdot}=n$, we define the index $|B:C|$ of $C$ in $B$ as
$n$. If $C$ has not finite index, we say that $C$ has infinite index.

The following easy result comprises some of the basic statements about
subbraces of finite index.

######  Lemma 4.25.

Let $B$ be a brace, $C,D\leq B$ and $I\trianglelefteq B$. Then:

  * (1)

If $C$ and $D$ have finite index in $B$, then $C\cap D$ has finite index in
$B$.

  * (2)

Suppose $C\leq D$. If $C$ has finite index in $D$, and $D$ has finite index in
$B$, then $C$ has finite index in $B$. Moreover, if $|D:C|$ and $|B:D|$ exist,
then also $|B:C|=|B:D|\cdot|D:C|$ exists.

  * (3)

If $I$ has finite index, then $|B:I|$ exists and is equal to $|B/I|$.

The following result shows that serial subbraces of finite index have a well-
defined index.

######  Lemma 4.26.

Let $C$ be a serial subbrace of the brace $B$. The following conditions are
equivalent:

  * (1)

$|(B,+):(C,+)|$ is finite.

  * (2)

$|(B,\cdot):(C,\cdot)|$ is finite.

  * (3)

$C$ has finite index in $B$.

  * (4)

$|B:C|$ exists.

In particular, if any of the above equivalent statements hold, then all the
indices are equal.

###### Proof.

Clearly, (4) $\implies$ (1), (2) and (3). Assume (1). Since $C$ is serial in
$B$, there is a chain $\mathcal{C}$ of subbraces connecting $C$ to $B$, and in
which $E\trianglelefteq F$ whenever $E<\\!\\!\\!\cdot\,F$. Looking at the
corresponding additive parts of the members of $\mathcal{C}$, we see that
$\mathcal{C}$ is actually finite, so $C$ is subideal in $B$. We prove the
result by induction on the subideal defect $n$ of $C$ in $B$. If $n\leq 1$,
then $C$ is an ideal of $B$ such that $(B,+)/(C,+)$ is finite, so $B/C$ is
finite and we are done. Assume $n>1$ and let $D=C^{B}$. Then the subideal
defect of $C$ in $D$ is strictly less than $n$ and so induction yields that
$|D:C|$ exists. Since $|B:D|$ trivially exists, we have that $|B:C|$ exists by
Lemma 4.25. Thus, (4) is proved. Similarly, we can prove that (2) implies (4),
and that (3) implies (4). ∎

A combination of Lemma 4.26 and Lemma 4.22 shows that every finite-index
subbrace of a locally centrally-nilpotent brace has a well-defined index. Our
next result is a considerable extension of this fact.

######  Theorem 4.27.

Let $B$ be a brace having an ascendant chain of ideals

$0=I_{0}\leq I_{1}\leq\ldots I_{\alpha}\leq I_{\alpha+1}\leq\ldots
I_{\lambda}=B$

such that $I_{\beta+1}/I_{\beta}$ is either finite or locally centrally-
nilpotent for all ordinal numbers $\beta<\lambda$. If $C$ is a subbrace of
$B$, then the following are equivalent:

  * (1)

$|(B,+):(C,+)|$ is finite.

  * (2)

$|(B,\cdot):(C,\cdot)|$ is finite.

  * (3)

$C$ has finite index in $B$.

  * (4)

$|B:C|$ exists.

###### Proof.

We prove the result by induction on $\lambda$. To this aim it is sufficient to
show that (1) implies (4).

If $\lambda\leq 1$, then $B$ is either finite or locally centrally-nilpotent.
The former case is obvious, while the latter is a consequence of Lemma 4.22
(3) and Lemma 4.26. Assume $\lambda>1$.

Suppose $\lambda$ is successor. Since $(C,+)$ has finite index in $(B,+)$, it
follows that $(C\cap I_{\lambda-1},+)$ has finite index $n$ in
$(I_{\lambda-1},+)$. By induction,

$n=\big{|}(I_{\lambda-1},\cdot):(C\cap
I_{\lambda-1},\cdot)\big{|}=\big{|}(CI_{\lambda-1},\cdot):(C,\cdot)\big{|}=\big{|}(C+I_{\lambda-1},+):(C,+)\big{|}.$

Thus the index $|C+I_{\lambda-1}:C|$ exists. Since also the index
$|B:C+I_{\lambda-1}|$ exists, it follows that the index $|B:C|$ exists.

Now, assume $\lambda$ is limit. Let $F_{+}$ be a transversal for $(C,+)$ in
$(B,+)$; in particular, $F$ is finite. Also, let $F_{\cdot}$ be a transversal
for $(C,\cdot)$ in $(B,\cdot)$. Suppose $|F_{\cdot}|>|F_{+}|$, and let
$E_{\cdot}$ be a finite subset of $F_{\cdot}$ such that $|E_{\cdot}|>|F_{+}|$.
Then there is an ordinal number $\mu<\lambda$ such that $F_{+}\cup
E_{\cdot}\subseteq I_{\mu}$. By induction,

$\big{|}(B,+):(C,+)\big{|}=\big{|}(I_{\mu},+):(C\cap
I_{\mu},+)\big{|}=\big{|}(I_{\mu},\cdot):(C\cap I_{\mu},\cdot)\big{|},$

a contradiction. Thus $|F_{\cdot}|\leq|F_{+}|$. By a symmetric argument,
$|F_{+}|\leq|F_{\cdot}|$ and hence the index $|B:C|$ exists. ∎

The range of applicability of Theorem 4.27 is not restricted to local
centrally-nilpotent brace. It follows in fact from [9], Theorems 3.14, that
Theorem 4.27 applies even to any good brace with property (S) (see [9] for the
definitions).

We end this discussion by showing that subbraces of finite index can sometimes
be employed to prove the existence of large proper ideals.

######  Theorem 4.28.

Let $B$ be a brace such that $B/\zeta_{2}(B)$ is finite. If $C$ is any finite-
index subbrace of $B$, then $B/C_{B}$ is finite.

###### Proof.

Without loss of generality we may assume $C_{B}=0$, so in particular,
$C\cap\zeta(B)=0$ and $\zeta(B)$ is finite. Moreover, we may replace $C$ by
$C\cap\zeta_{2}(B)$, so assuming $C\leq\zeta_{2}(B)$. Then $C+\zeta(B)$ is an
ideal of $B$.

Since $C\cong C+\zeta(B)/\zeta(B)$, we have that $C$ is an abelian brace. Let
$n=|\zeta(B)|$. Then Theorem 4.18 shows that

$C^{n^{2},\cdot}\leq\zeta(B)\cap C=0,$

so $C$ is periodic. Thus, as a group, $C$ can be described as a direct product
of infinitely many cyclic subgroups $\langle b_{i}\rangle$, $i\in I$, of order
dividing $n$.

Let $F$ be a finitely generated subbrace of $B$ such that $C+\zeta(B)+F=B$.
Since $B$ is periodic, then it is locally finite by [21], Lemma 3.1, so $F$ is
finite.

For every $b\in F$, $b_{i}^{b,+}=b_{i}+u_{b,i,+}$ for some
$u_{b,i,+}\in\zeta(B)$. On the other hand, $\zeta(B)$ is finite, so there is
an infinite subset $J_{1}$ of $I$ such that

$b_{i}^{b,+}=b_{i}+u_{+,b}$

for all $i\in J_{1}$, and for a fixed $u_{+,b}\in\zeta(B)$. Repeating this
argument for all $b\in F$, we may assume

$b_{i}^{b,+}=b_{i}+u_{+,b}$

for some $u_{+,b}\in\zeta(B)$ and for all $i\in J_{1}$, $b\in F$. Similarly,
there is an infinite subset $J_{2}$ of $J_{1}$ such that

$b_{i}^{b,\cdot}=b_{i}+u_{\cdot,b}$

for some $u_{\cdot,b}\in\zeta(B)$ and for all $i\in J_{2}$, $b\in F$. Finally,
there is an infinite subset $J_{3}$ of $J_{2}$ such that

$b_{i}\ast b=b_{j}\ast b$

for all $i,j\in J_{3}$ and $b\in F$.

Now, for each $i,j\in J_{3}$ with $i\neq j$, we have that
$d_{i}=b_{i}-b_{j}=b_{i}\cdot b_{j}^{-1}$ is additively and multiplicatively
centralised by $F$, and that $d_{i}\ast F=0$. Since $B=C+\zeta(B)+F$, it
follows that $d_{i}\in\zeta(B)$ for all $i\in J_{3}$, a contradiction. ∎

##  5 Central nilpotency for ideals

A celebrated result of Fitting states that a product of nilpotent normal
subgroups of a group is nilpotent. Example B in Section 6 shows that the
product of two centrally nilpotent ideals, regarded as independent braces, is
not centrally nilpotent in general. In this section, we aim to define a
nilpotency concept for ideals that allows us to define a suitable Fitting
ideal. It turns out that, for such a Fitting ideal, it is possible to
generalise remarkable results of group theory concerning with the Fitting
subgroup. For example, we give a characterization of Fitting ideal in terms of
chief factors (see Theorem 5.7), we prove that Fitting ideals are self-
centralising in soluble braces (see Theorem 5.8), and we give an analogue of a
well-known result of Gaschütz stating that the Fitting modulo the Frattini
subgroup of a finite group is a product of all its minimal abelian normal
subgroups (see Theorem 5.12); note that to prove the latter result to prove we
need a suitable Frattini-like ideal for braces. In the final part of the
section we discuss hypercentral (resp. locally nilpotent) concepts for ideals.

Let $B$ be a brace. We start by defining $B$-centrally nilpotent braces. Let
$I$ be an ideal of $B$. We can define the _lower central series of $I$ with
respect to $B$_, or simply the _lower $B$-central series of $I$_, as follows:
take $\Gamma_{1}(I)^{B}=I$ and $\Gamma_{n+1}(I)^{B}=[\Gamma_{n}(I),I]^{B}$,
for every $n\geq 1$. Therefore,

$I=\Gamma_{1}(I)^{B}\geq\Gamma_{2}(I)^{B}\geq\ldots\geq\Gamma_{n}(I)^{B}\geq\ldots$

is a descending chain of ideals of $B$ such that, for every $n\in\mathbb{N}$,

$\Gamma_{n}(I)^{B}/\Gamma_{n+1}(I)^{B}\leq\zeta\big{(}I/\Gamma_{n+1}(I)^{B}\big{)}.$

Similarly, we may define the upper central series of $I$ with respect to $B$
(or simply the upper $B$-central series of $I$), as follows: take
$\zeta_{0}(B)=0$ and let $\zeta_{n+1}(I)^{B}$ satisfy

$\zeta_{n+1}(I)^{B}/\zeta_{n}(I)^{B}=\zeta\big{(}I/\zeta_{n}(I)^{B}\big{)}_{B/\zeta_{n}(I)^{B}}.$

Then

$\zeta_{0}(I)^{B}\leq\zeta_{1}(I)^{B}\leq\ldots\leq\zeta_{n}(I)^{B}\leq\ldots$

is an ascending chain of ideals of $B$.

######  Definition 5.1.

An ideal $I$ of a brace $B$ is defined to be _centrally nilpotent with respect
of $B$_, or simply a _$B$ -centrally nilpotent ideal_, if there exists
$n\in\mathbb{N}$ such that $\Gamma_{n+1}(I)^{B}=0$, or, equivalently,
$\zeta_{n}(I)^{B}=0$. For practical purposes, we often use the following
equivalent definition: $I$ is $B$-centrally nilpotent if there exists a chain

$0=J_{0}\leq J_{1}\leq\ldots\leq J_{n}=I$

of ideals of $I$ such that $J_{i}/J_{i-1}\leq\zeta(I/J_{i-1})$, for every
$1\leq i\leq n$.

To simplify notation, if $J$ is an ideal of $B$ contained in $I$ and such that
$I/J$ is $B/I$-centrally nilpotent, we just say that $I/J$ is _centrally
nilpotent with respect of $B$_, or _$B$ -centrally nilpotent_. If $I/J$ is
$B$-centrally nilpotent, then the smallest $n$ such that
$\Gamma_{n+1}(I/J)^{B/J}=0$ is referred to as its class.

Clearly, a brace $B$ is centrally nilpotent if and only if it is $B$-centrally
nilpotent; in this trivial case, the upper central series (resp. lower central
series) and the upper $B$-central series (resp. lower $B$-central series)
coincide. Moreover, if $I$ is a $B$-centrally nilpotent ideal of a brace $B$,
and $J$ is any ideal of $B$, then $(I+J)/J$ is $B$-centrally nilpotent (of
class less than or equal to that of $I$), and $I\cap C$ is $C$-centrally
nilpotent for any subbrace $C$ of $B$ (also in this case the class of $I\cap
C$ is less than or equal to that of $I$). Our next result shows that an analog
of Fitting theorem holds for $B$-centrally nilpotent ideals, but first, we
need the following property of commutators of ideals in braces.

######  Lemma 5.2.

Let $B$ be a brace and let $I,J,K$ be ideals of $B$. Then

$[I,JK]_{B}=[I,J+K]_{B}=[I,J]_{B}+[I,K]_{B}=[I,J]_{B}[I,K]_{B}.$

###### Proof.

We prove the equality for the sum. Observe that only one inclusion is in doubt
so, by Theorem 3.6, it suffices to show that

$[I,J+K]_{+},\,I\ast(J+K),\,(J+K)\ast I\subseteq[I,J]_{B}+[I,K]_{B}.$

Since $(I,+)$, $(J,+)$ and $(K,+)$ are normal subgroups of $(B,+)$, we have
that $[I,J+K]_{+}=[I,J]_{+}+[I,K]_{+}$ is contained in $[I,J]_{B}+[I,K]_{B}$.
Then, applying Eq. (3), we have that for every $i\in I$, $j\in J$ and $k\in
K$,

$i\ast(j+k)=i\ast j+j+i\ast k-j\in[I,J]_{B}+[I,K]_{B}$

Finally, note that $(J+K)\ast I=(JK)\ast I$, so applying Eq. (1) we see that

$(jk)\ast i=j\ast(k\ast i)+k\ast i+j\ast i\in J\ast I+K\ast I+J\ast
I\subseteq[I,J]_{B}+[I,K]_{B}$

for every $j\in J$, $k\in K$, $i\in I$. ∎

We also need the following notation in the proof: if $I_{1},\ldots,I_{n}$ are
ideals of a brace $B$, we put $[I_{1}]^{B}=I_{1}$, and then, recursively,

$[I_{1},\ldots,I_{k}]^{B}:=\big{[}[I_{1},\ldots,I_{k-1}]^{B},I_{k}\big{]}^{B}\quad\text{for
every $2\leq k\leq n$.}$

######  Theorem 5.3.

Let $I,J$ be $B$-centrally nilpotent ideals of a brace $B$. If $I$ and $J$
have classes $n_{0}$ and $m_{0}$, respectively, then $I+J$ is $B$-centrally
nilpotent of class at most $n_{0}+m_{0}$.

###### Proof.

Set $K=I+J$. First, we show by induction that for every $n\in\mathbb{N}$,
$\Gamma_{n}(K)^{B}$ is the sum of all commutators of the form
$[L_{1},\ldots,L_{n}]_{B}$ with either $L_{i}=I$ or $L_{i}=J$, for every
$1\leq i\leq n$. The base case is clear. Assume the assertion is true for some
$1\leq n\in\mathbb{N}$. Then,

$\Gamma_{n+1}(K)^{B}=[\Gamma_{n}(K),K]^{B}=[\Gamma_{n}(K),I]^{B}+[\Gamma_{n}(K),J]^{B}$

by Lemma 5.2. Using iteratively Lemma 5.2, the assertion also holds for $n+1$.

In particular, for $r=n_{0}+m_{0}+1$, $\Gamma_{r}(K)^{B}$ is the sum of all
commutators of the form $[L_{1},\ldots,L_{r}]^{B}$, where either $I$ occurs
$n_{0}+1$ times or $J$ occurs $m_{0}+1$ times. Thus, it follows that each
$[L_{1},\ldots,L_{r}]^{B}$ is contained in either $\Gamma_{n_{0}+1}(I)^{B}=0$
or $\Gamma_{m_{0}+1}(J)^{B}=0$. Hence, $\Gamma_{r}(K)^{B}=0$ and so $K$ is
$B$-centrally nilpotent. ∎

We will later generalise Theorem 5.3 to Theorem 5.16.

######  Definition 5.4.

Let $B$ be a brace. The Fitting ideal $\operatorname{Fit}(B)$ of $B$ is the
ideal generated by all $B$-centrally nilpotent ideals of $B$.

It follows from Theorem 5.3 that in a finite brace $B$,
$\operatorname{Fit}(B)$ is $B$-centrally nilpotent. More in general, the same
result shows that this is true for a broader class of braces.

######  Corollary 5.5.

Let $B$ be a noetherian brace. Then $\operatorname{Fit}(B)$ is a $B$-centrally
nilpotent ideal.

Now, in order to obtain a characterisation of the Fitting ideal in terms of
chief factors, we need the following definition (recall Proposition 4.2). In
[5, Propostion 4.19], the _centraliser of an ideal $I$_ of a brace $B$,
$\operatorname{C}_{B}(I)$, is defined as the largest ideal that centralises
$I$, i.e. $[\operatorname{C}_{B}(I),I]^{B}=0$.

Moreover, if $I/J$ is a chief factor of $B$, we define the _centraliser in $B$
of $I/J$_ as the ideal $\operatorname{C}_{B}(I/J)$ of $B$ satisfying
$\operatorname{C}_{B/J}(I/J)=\operatorname{C}_{B}(I/J)/J$. Equivalently,
$\operatorname{C}_{B}(I/J)$ is the largest ideal of $B$ such that
$[\operatorname{C}_{B}(I/J),I]^{B}\leq J$.

######  Lemma 5.6.

Let $I$ be a $B$-centrally nilpotent ideal of a brace $B$. If $J$ is a minimal
ideal of $B$, then $[J,I]^{B}=0$.

###### Proof.

Since $J$ is a minimal ideal of $B$, then either $[J,I]^{B}=0$ or
$[J,I]^{B}=J$. However, in the latter case we contradict Definition 5.1. ∎

######  Theorem 5.7.

Let $B$ be a brace with a finite chief series $\mathcal{S}$. Then
$\operatorname{Fit}(B)$ is the intersection of the centralisers in $B$ of the
factors of $\mathcal{S}$.

###### Proof.

Let

$0=I_{0}\leq I_{1}\leq\ldots\leq I_{n}=B$

be a finite chief series of $B$ and set
$C:=\bigcap\\{\operatorname{C}_{B}(I_{k}/I_{k-1}):1\leq k\leq n\\}$. Then $C$
is an ideal of $B$ and

$0=C\cap I_{0}\leq C\cap I_{1}\leq\ldots\leq C\cap I_{n}=C$

is a finite chain of ideals of $B$ such that $(I_{i}\cap C)/(I_{i-1}\cap
C)\leq\zeta\big{(}C/I_{i-1}\cap C\big{)}$ for all $1\leq i\leq n$. Thus, $C$
is $B$-centrally nilpotent and hence $C\leq\operatorname{Fit}(B)$.

Conversely, $B$ is noetherian (see [2]) and so $F:=\operatorname{Fit}(B)$ is
$B$-centrally nilpotent by Corollary 5.5. If $I/J$ is any chief factor of $B$,
then $I/J$ is centralised by $(F+J)/J$ by Lemma 5.6. ∎

######  Theorem 5.8.

Let $B$ be a brace and put $F=\operatorname{Fit}(B)$. Then,
$(\operatorname{C}_{B}(F)+F)/F$ does not contain any non-zero soluble ideal
with respect of $B/F$. In particular, if $B$ is a soluble brace, then
$\operatorname{C}_{B}(F)=\zeta(F)$.

###### Proof.

Assume that $(\operatorname{C}_{B}(F)+F)/F$ contains a non-zero soluble ideal
$I/F$ with respect of $B/F$. Then it also contains a non-zero ideal $J/F$
which is an abelian brace. Let $C=\operatorname{C}_{B}(F)$. Then $J\cap C\cap
F\leq\zeta(J\cap C)$ and $(J\cap C)/(J\cap C\cap F)$ is an abelian brace.
Thus, $J\cap C$ is $B$-centrally nilpotent and so $J\cap C\leq F$. Finally,

$J=J\cap(C+F)=(J\cap C)+F=F,$

a contradiction.

If $B$ is a soluble brace, then $B/F$ is soluble and therefore,
$(\operatorname{C}_{B}(F)+F)/F$ must be zero. Hence,
$\operatorname{C}_{B}(F)\leq F$ which yields
$\operatorname{C}_{B}(F)=\zeta(F)$. ∎

It is well-known that the Fitting subgroup of a finite group modulo its
Frattini subgroup is the product of all its abelian minimal normal subgroups.
The following Frattini-like ideal leads to a brace-theoretic analogue of this
result.

######  Definition 5.9.

Let $B$ be a finite brace. We define Frattini ideal of $F$ as

$\operatorname{Frat}(B):=\bigcap\\{\,L\,|\,\text{$L$ is a maximal left ideal
of $B$}\,\\}\cap\operatorname{Fit}(B).$

Clearly, the Frattini ideal of a finite brace is a left ideal, but the
following result shows that it is actually an ideal (hence providing a
justification for its name).

######  Lemma 5.10.

Let $B$ be a finite brace. If $L$ is any maximal left ideal of $B$, then
$L\cap\operatorname{Fit}(B)$ is an ideal of $B$.

###### Proof.

Let $F=\operatorname{Fit}(B)$ and assume $F\not\leq L$, so in particular
$B=FL$.

Since $F\cap L\unlhd(L,+)$, it follows that
$L\leq\operatorname{N}_{(B,+)}(F\cap L)$. Moreover, $F\cap L$ is properly
contained in $\operatorname{N}_{(F,+)}(F\cap L)$, as $(F,+)$ is nilpotent.
Therefore, $L$ is properly contained in $\operatorname{N}_{(B,+)}(F\cap L)$.

Because $F\cap L$ is $\lambda$-invariant, it holds that
$\operatorname{N}_{(B,+)}(F\cap L)$ is also $\lambda$-invariant. Thus,
$\operatorname{N}_{(B,+)}(F\cap L)=B$ and so, $F\cap L$ is a strong left
ideal. Then, by Lemma 4.22 and Remark 4.23, we can find a strong left ideal
$T$ of $B$ contained in $F$ and such that $F\cap L$ is a proper ideal of $T$.
Therefore, it follows that $F\cap L$ is an ideal of $T+L=TL$, with $T+L$ being
a left ideal of $B$. Hence, $TL=B$ by the maximality of $L$ and the result
follows. ∎

######  Corollary 5.11.

Let $B$ be a finite brace. Then $\operatorname{Frat}(B)$ is an ideal of $B$.

######  Theorem 5.12.

Let $B$ be a finite brace with $\operatorname{Frat}(B)=0$. Then
$\operatorname{Fit}(B)$ is the product of all the abelian minimal ideals of
$B$.

###### Proof.

Let $F=\operatorname{Fit}(B)$. We claim that $\partial_{B}(F)=0$. Indeed,
suppose that $L$ is a maximal left ideal of $B$ such that $\partial_{B}(F)$ is
not included in $L$. Then, $B=\partial_{B}(F)L$, so

$F=F\cap\partial_{B}(F)L=(F\cap L)\partial_{B}(F).$

By Lemma 5.10, $F\cap L$ is an ideal of $B$. Since $F/(F\cap L)$ is non-zero
and $B$-centrally nilpotent, we have that

$\partial_{B}(F)(F\cap L)/(F\cap L)\leq\partial_{B/(F\cap L)}\big{(}F/(F\cap
L)\big{)}<F/(F\cap L),$

a contradiction. Thus, $F$ is an abelian brace.

Let $N$ be the product of all abelian minimal ideals of $B$. Then $N\leq F$.
For the other inclusion, take $S$ a minimal subbrace subject to $B=SN$.
Consider

$X=\bigcap\\{L\,|\,\text{$L$ is a maximal left ideal of $S$}\\},$

a left ideal of $S$. If $S\cap N\not\leq X$, then there exists a maximal left
ideal $L$ of $S$ such that $S\cap N$ is not included in $L$. Thus, $(S\cap
N)L=S$ and then $B=SN=(S\cap N)LN=LN$, which contradicts the minimality of
$S$. Therefore, $S\cap N\leq X$.

Now, $S\cap N$ is an ideal of $B$, as $N$ is abelian and $B=SN$ (see [3, Lemma
27]). Assume that there exists a maximal left ideal $L$ of $B$ such that
$S\cap N\not\leq L$. Thus, $B=(S\cap N)L$ and then, $S=S\cap(S\cap N)L=(S\cap
L)(S\cap N)$. Take $L^{\prime}$ a left ideal of $S$ maximal subject to $S\cap
L\leq L^{\prime}$ and $S\cap N$ not included in $L^{\prime}$. Then,
$L^{\prime}$ is indeed a maximal left ideal of $S$, because the existence of a
left ideal $L^{\prime\prime}$ of $S$ such that
$L^{\prime}<L^{\prime\prime}\leq S$ yields $S=(S\cap L)(S\cap N)\leq
L^{\prime\prime}$. Therefore, $S\cap N\leq X\leq L^{\prime}$, a contradiction.
Thus, $S\cap N=0$.

Finally, $S\cap F$ is an ideal of $B$, as $F$ is abelian and $SF=B$ (see again
[3, Lemma 27]), and consequently $S\cap F$ contains an abelian minimal ideal
of $B$, contradicting $S\cap N=0$. ∎

In a centrally nilpotent brace, the Frattini ideal behaves pretty well. For
example, it is possible to prove that it coincides with the set of non-
generators.

######  Definition 5.13.

Let $B$ be a brace. We say that an element $b\in B$ is a non-generator of $B$
if for all $S\leq B$ such that $B=\langle b,S\rangle$, we have $B=S$.

######  Theorem 5.14.

Let $B$ be a centrally nilpotent finite brace. Then $\operatorname{Frat}(B)$
coincides with the set of all non-generators of $B$.

###### Proof.

Since in a centrally nilpotent brace the maximal left ideals coincide with the
maximal ideals and with the maximal subbraces, the usual group theoretical
proof adapts to prove the following result. ∎

However, we note that there exist a brace $B$ of order $6$ in which
$\operatorname{Fit}(B)=\operatorname{Frat}(B)$ is the only non-zero proper
left ideal of $B$ and has order $3$. This shows that there is no possible
analogue of two well-known group theoretical theorems concerning the Frattini
subgroup of a group $G$: (1) $G/\operatorname{Frat}(G)$ nilpotent implies $G$
nilpotent; (2) if $p$ is a prime dividing $|G|$, then $p$ divides
$|G/\operatorname{Frat}(G)|$ too.

In the final part of this section we discuss further aspects of $B$-centrally
nilpotence and hypercentral/locally nilpotent concepts for (sub)ideals.

The definition of upper $B$-central series (and lower $B$-central series) for
an ideal $I$ of a brace $B$ can be extended by using transfinite numbers (just
how we did in Section 4), and this allows us to define $B$-hypercentral ideals
of braces. However, for our purposes, the following equivalent definition is
more convenient.

######  Definition 5.15.

Let $B$ be a brace. An ideal $I$ of $B$ is said to be $B$-hypercentral if
there is an ascending chain of ideals of $B$

$0=:I_{0}\leq I_{1}\leq\ldots I_{\alpha}\leq I_{\alpha+1}\leq\ldots
I_{\lambda}=I$

such that $I_{\alpha+1}/I_{\alpha}\leq\zeta(B/I_{\alpha})$ for all ordinals
$\alpha<\lambda$. The smallest $\lambda$ for which such a chain exists is the
length of $I$.

Clearly, if $B$ is a brace, then $B$ is hypercentral if and only if $B$ is
$B$-hypercentral, and every $B$-centrally nilpotent ideal is $B$-hypercentral.
The following result generalises Theorem 5.3.

######  Theorem 5.16.

Let $B$ be a brace.

  * (1)

If $C$ and $D$ are ideals of $B$ which are $B$-hypercentral of lengths
$\alpha$ and $\beta$, respectively, then $C+D$ is a $B$-hypercentral ideal of
length at most $\beta\alpha+\operatorname{max}\\{\alpha,\beta\\}$.

  * (2)

Suppose $C$ is subideal of defect $n$ and centrally nilpotent of class $c$,
and $D$ is a $B$-centrally nilpotent ideal of class $d$. Then $C+D$ is
centrally nilpotent of class at most $(c+n)d+c$.

  * (3)

Suppose $C$ is ascendant of length $\mu$ and hypercentral of length $\gamma$,
and $D$ is a $B$-hypercentral ideal of length $\delta$. Then $C+D$ is
hypercentral of length at most $(\gamma+\mu)\delta+\gamma$.

###### Proof.

(1) Let $E=C+D$. Then $E$ is an ideal of $B$ and to show that $E$ is
$B$-hypercentral, it suffices to prove that $\zeta(E)$ contains a non-zero
ideal $I$ of $B$. To this aim we may certainly assume that $C$ and $D$ are
non-zero.

Suppose first $C\cap D=0$. By hypothesis $\zeta(C)$ contains a non-zero ideal
$I$ of $B$. On the other hand, $\zeta(C)\leq\zeta(E)$ and so we are done.

Assume $C\cap D\neq 0$. Then $\zeta(C)\cap D$ contains a non-zero ideal $I$ of
$B$, and $I\cap\zeta(D)$ contains a non-zero ideal $J$ of $B$. Thus

$J\leq\zeta(C)\cap\zeta(D)\leq\zeta(E)$

and we are done. The bound on the hypercentral length can be easily deduced
from the proof.

(3) Since $D$ is $B$-hypercentral of length $\delta$, there is an ascending
chain of ideals of $B$

$0=D_{0}<D_{1}<\ldots D_{\alpha}<D_{\alpha+1}<\ldots D_{\delta}=D$

such that $D_{\beta+1}/D_{\beta}\leq\zeta(D/D_{\beta})$ for all ordinals
$\beta<\delta$.

Let $E=C+D$. Since $C$ is hypercentral of length $\gamma$, it follows that
$C\cap D_{1}\leq\zeta_{\gamma}(E)$. Thus, we may factor out $C\cap D_{1}$ and
assume $C\cap D_{1}=0$. Let $F=\langle C,D_{1}\rangle=CD_{1}$. Now, since $C$
is ascendant of length $\mu$, there is an ascending chain

$C=C_{0}\trianglelefteq C_{1}\trianglelefteq\ldots C_{\alpha}\trianglelefteq
C_{\alpha+1}\trianglelefteq\ldots C_{\mu}=F$

connecting $C$ to $F$. It is easy to see that

$(C_{\beta+1}\cap D_{1})/(C_{\beta}\cap D_{1})\leq\zeta\big{(}E/(C_{\beta}\cap
D_{1})\big{)}$

for all $\beta<\mu$. Therefore $D_{1}\leq\zeta_{\mu}(E)$.

We factor $D_{1}$ out and we repeat the above argument. This shows that
$D\leq\zeta_{\rho}(E)$, where $\rho=(\gamma+\mu)\delta$, so $E$ is
hypercentral of length at most $\rho+\gamma$.

(2) The proof is essentially the same as that of (3), but easier. ∎

On some occasions, the product of all $B$-hypercentral ideals of a brace is
still $B$-hypercentral. This is clearly the case for instance if $B$ satisfies
the maximal condition on ideals, but it is also the case if $B$ satisfies the
minimal condition on ideals, as the following result shows.

######  Theorem 5.17.

Let $B$ be a brace admitting an ascending chief series $\mathcal{S}$. Then the
maximal ideal centralising all factors of $\mathcal{S}$ is precisely the
unique maximal $B$-hypercentral ideal of $B$.

###### Proof.

The proof runs along the same lines as that of Theorem 5.7. ∎

The following result shows that in the universe of locally centrally-nilpotent
braces, the class of $B$-hypercentral braces is closed with respect to forming
extensions by finitely generated hypercentral braces.

######  Theorem 5.18.

Let $N$ be an ideal of the locally centrally-nilpotent brace $B$. If $N$ is
$B$-hypercentral and $B/N$ is finitely generated, then $B$ is hypercentral.

###### Proof.

Let $S$ be a finite subset of $B$ that generates $B$ modulo $N$, and let
$Z=\overline{\zeta}(B)$ be the hypercentre of $B$. Assume by contradiction
$Z\neq B$. Now, $B/N$ is centrally nilpotent, so $N\not\leq Z$ and hence the
ideal $K:=Z\cap N$ of $G$ is strictly contained in $N$. Since $N/K$ is
$B$-hypercentral, there is a non-zero ideal $A/K$ of $B/K$ such that

$A/K\leq\zeta(N/K).$

Let $a\in A\setminus K$ and $U=\langle a,S,K\rangle$; in particular, $U/K$ is
centrally nilpotent. Since $(A\cap U)/K$ is a non-zero ideal of $U/K$, we have
that

$V/K:=\zeta(U/K)\cap\big{(}(A\cap U)/K\big{)}\neq 0.$

Now, the fact that $V\leq A$ implies that $[V,N]_{+}$, $[V,N]_{\cdot}$ and
$V\ast N$ are all contained in $K$. Similarly, the fact that
$V/K\leq\zeta(U/K)$ shows that $[V,T]_{+}$, $[V,T]_{\cdot}$ and $V\ast T$ are
all contained in $K$, where $T=\langle S\rangle$. Since $B=N+T=NT$, we easily
see that $[V,B]_{+}$ and $[V,B]_{\cdot}$ are contained in $K$. Moreover, if
$u\in N$, $v\in T$, and $a\in V$, then

$a\ast(u+v)=a\ast u+u+a\ast v-u\in K.$

This shows that $V\ast B\leq K$ and proves that $V/K\leq\zeta(B/K)$. Since
$K\leq Z$, it follows that $V\leq Z$, so $V\leq Z\cap N$, a contradiction. ∎

Also $B$-central nilpotency (resp. hypercentrality) can be locally detected,
and our next result is in fact a generalisation of Theorem 4.4.

######  Theorem 5.19.

Let $B$ be a brace and let $I\trianglelefteq B$. Then:

  * (1)

$I$ is $B$-centrally nilpotent of class at most $c$ if and only if $I\cap F$
is $F$-centrally nilpotent of class at most $c$ for every finitely generated
subbrace $F$ of $B$.

  * (2)

$I$ is $B$-centrally nilpotent if and only if $I\cap C$ is $C$-centrally
nilpotent for every countable subbrace $C$ of $B$.

  * (3)

$I$ is $B$-hypercentral if and only if $I\cap C$ is $C$-hypercentral for every
countable subbrace $C$ of $B$.

###### Proof.

We only deal with the proof of (1), since (2) and (3) then follow in a similar
fashion using ideas from Theorem 4.4.

For each $u,v\in B$, we write $u\circ v$ to denote one (but we do not know
which one) of the following operations $[u,v]_{\cdot},[u,v]_{+},u\ast v$. Then
(1) is a direct consequence of the fact that $\zeta_{c}(I)^{B}$ can be easily
characterised as the set of all elements $b\in I$ such that

$\langle(\ldots((b\circ b_{1})\circ\ldots)\circ
b_{c-i})\rangle^{B}\leq\zeta_{i}(I)$

for all $i=0,1,\ldots,c-1$ and for all $b_{1},\ldots,b_{c-i}\in B$. ∎

To provide a good definition of “locally $B$-nilpotent ideal” is not an
obvious task. We now concern ourselves with a couple of possible definitions,
mostly sketching proofs and results. The first idea that comes in mind is that
of using central chains of ideals, just as we did for $B$-hypercentral and
$B$-centrally nilpotent ideals. In fact, it follows from Theorem 4.6 (and
Zorn’s lemma) that in every chief series of a locally centrally-nilpotent
brace $B$, two consecutive ideals $K\leq H$ satisfy $H/K\leq\zeta(B/K)$.

######  Definition 5.20.

Let $B$ be a brace. An ideal $I$ of $B$ is said to be $\zeta_{B}$-nilpotent if
every quotient $I/J$ of $I$ by an ideal $J$ of $B$ admits a maximal chain
$\mathcal{S}$ of ideals of $B$ in which $H/K\leq\zeta(I/K)$ for all
consecutive terms $K/J\leq H/J$ of the chain $\mathcal{S}$.

Using ideas from the proof of Theorem 5.16, it is not difficult to see that
the product of arbitrarily many $\zeta_{B}$-nilpotent ideals is
$\zeta_{B}$-nilpotent. Thus, any brace $B$ has a unique maximal
$\zeta_{B}$-nilpotent ideal, we call it the $\zeta_{B}$-radical of $B$: it
turns out that the largest ideal of $B$ centralising all quotients of a chief
series of $B$ is precisely the $\zeta_{B}$-radical of $B$. The following
result shows that an analogue of Theorem 5.8 is possible for the
$\zeta_{B}$-radical.

######  Theorem 5.21.

Let $B$ be a brace admitting an ascending chain of ideals

$0=B_{0}\leq B_{1}\leq\ldots\ldots B_{\alpha}\leq B_{\alpha+1}\leq\ldots
B_{\lambda}=B$

such that $B_{\beta+1}/B_{\beta}$ is a $\zeta_{B}$-nilpotent ideal of
$B/B_{\beta}$ for all $\beta<\lambda$. Then $C_{B}(H)\leq H$, where $H$ is the
$\zeta_{B}$-radical of $B$.

###### Proof.

Suppose $C=C_{B}(H)\not\leq H$. Then $(C+H)/H$ contains a non-zero
$\zeta_{B}$-nilpotent ideal $I/H$ of $B/H$. Since $I\cap C\cap
H\leq\zeta(I\cap C)$, it follows that $I\cap C$ is a $\zeta_{B}$-nilpotent
ideal of $B$. Thus, $I\cap C\leq H$ and

$I=I\cap(C+H)=(I+C)\cap H=H,$

a contradiction. ∎

However, there is one reason for which this is not a really convincing good
definition of “locally $B$-nilpotent ideal”: a $\zeta_{B}$-ideal could not be
locally centrally-nilpotent (there are examples even among groups) — although
if $B$ is locally finite, then a $\zeta_{B}$-ideal is locally centrally-
nilpotent.

A more fruitful approach could deal with finitely generated subbraces and the
way the ideal embeds into them. There are several ways in which this case be
achieved, but the most reasonable one seems to be the following.

######  Definition 5.22.

Let $B$ be a brace. An ideal $I$ of $B$ is locally $B$-nilpotent if the
following property holds:

  * •

For every finitely generated subbrace $F$ of $B$, the finitely generated
subbraces of $I\cap F$ are contained in $F$-centrally nilpotent ideals of $F$.

Trivially, every locally $B$-nilpotent ideal is locally centrally-nilpotent,
so this solves the previous issue for $\zeta_{B}$-nilpotency.

######  Theorem 5.23.

Let $B$ be a brace. The sum of arbitrarily many locally $B$-nilpotent ideals
of $B$ is locally $B$-nilpotent.

###### Proof.

It is clearly enough to prove the statement for two locally $B$-nilpotent
ideals $I$ and $J$. Let $F$ be a finitely generated subbrace of $B$. Choose a
finitely generated subbrace $E$ of $F\cap(I+J)$. In order to prove that $E$ is
contained in an $F$-centrally nilpotent ideal of $F$, we may assume
$E=E_{1}\cup E_{2}$, where $E_{1}\subseteq I$ and $E_{2}\subseteq J$, by
suitably replacing $F$. Now, $E_{1}$ and $E_{2}$ are respectively contained in
$F$-centrally nilpotent ideals $I_{1}$ and $I_{2}$ of $F$. Since $I_{1}+I_{2}$
is $F$-centrally nilpotent, we are done. ∎

By Theorem 5.23, every brace admits a unique maximal locally $B$-nilpotent
ideal, we call it the Hirsch-Plotkin radical of $B$, and we denote it by
$\operatorname{HP}(B)$. Using ideas from the proof of Theorem 4.6 (1), we see
that every locally $B$-nilpotent ideal is actually $\zeta_{B}$-nilpotent.
Finally, we note that for a locally finite brace $B$, the concepts of locally
$B$-nilpotent ideal and $\zeta_{B}$-ideal coincide.

##  6 Worked examples

In this section we describe the main examples of the paper. These examples are
all constructed in a similar fashion, which we now explain, and all the
computations can be done with the computer algebra system GAP [13] and the
functions of its package YangBaxter [22].

Braces can be defined from bijective $1$-cocycles associated with actions of
groups. Let $(C,\cdot)$ and $(B,+)$ be groups such that $C$ acts on $B$ by
means of a group homomorphsim $\lambda\colon
C\rightarrow\operatorname{Aut}(B,+)$, $c\mapsto\lambda_{c}$. A bijective map
$\delta\colon C\rightarrow A$ is said to be a _bijective $1$-cocycle_
associated with $\lambda$, if $\delta(cd)=\delta(c)+\lambda_{c}(\delta(d))$,
for every $c,d\in C$. Following [1], bijective $1$-cocycles can be constructed
by means of trifactorised groups. In the previous situation, take $G=[B]C$ the
semidirect product associated with $\lambda$, written in multiplicative
notation for the sake of uniformity. If $D$ is a subgroup of $G$ such that
$G=BC=BD=DC$ and $B\cap D=D\cap C=1$, then $G$ is said to be a _trifactorised
group_ , and there exists a bijective $1$-cocycle $\delta\colon
C\rightarrow(B,+)$, given by $D=\\{\delta(c)c:c\in C\\}$. At this point,
observe that $\delta(c)$ must be translated in additive notation. Then,
$(B,+)$ admits a brace structure by means of
$ab:=\delta(\delta^{-1}(a)\delta^{-1}(b))$, for every $a,b\in B$ (see [16,
Proposition 1.11]), for example).

We are now ready to delve into our examples. The first of them shows that the
idealiser of a subbrace (as introduced in [17]) does not exist in general
(even for braces of abelian type).

######  Example A.

Let $B=\langle a_{1}\rangle\times\langle a_{2}\rangle\times\langle
a_{3}\rangle\cong C_{4}\times C_{4}\times C_{2}$, whose operation will be
written additively, and let

$\begin{array}[]{c}C=\langle m_{1},m_{2},m_{3},m_{4},m_{5}\mid
m_{1}^{2}=m_{4},m_{2}^{2}=1,m_{3}^{2}=m_{4}^{2}=m_{5},m_{5}^{2}=1,\\\\[2.84544pt]
m_{1}m_{2}{m_{1}^{-1}}=m_{3}m_{2},m_{1}m_{3}{m_{1}^{-1}}=m_{5}m_{3},m_{2}m_{3}m_{2}^{-1}=m_{5}m_{3},\\\\[2.84544pt]
m_{1}m_{4}{m_{1}^{-1}}=m_{2}m_{4}{m_{2}^{-1}}=m_{3}m_{4}{m_{3}^{-1}}=m_{4},\\\\[2.84544pt]
m_{1}m_{5}{m_{1}^{-1}}=m_{2}m_{5}{m_{2}^{-1}}=m_{3}m_{5}{m_{3}^{-1}}=m_{4}m_{5}{m_{4}^{-1}}=m_{5}\rangle.\end{array}$

We note that $B$ and $C$ are groups of order $32$. Since
$m_{3}=m_{1}m_{2}m_{1}^{-1}m_{2}^{-1}$, $m_{4}=m_{1}^{2}$,
$m_{5}=m_{1}m_{3}m_{1}^{-1}m_{3}=m_{3}^{2}$, we have that $\langle
m_{3},m_{4},m_{5}\rangle$ is contained in $\operatorname{Frat}(C)$ (in fact,
they coincide) and so $C=\langle m_{1},m_{2}\rangle$. We have that $C$ acts on
$B$ by means of an action $\lambda$ defined by

$\displaystyle\lambda_{m_{1}}(a_{1})$ $\displaystyle=3a_{1}+a_{3},$
$\displaystyle\lambda_{m_{2}}(a_{1})$ $\displaystyle=3a_{1},$
$\displaystyle\lambda_{m_{1}}(a_{2})$ $\displaystyle=a_{1}+a_{2}+a_{3},$
$\displaystyle\lambda_{m_{2}}(a_{2})$ $\displaystyle=a_{1}+a_{2}+a_{3},$
$\displaystyle\lambda_{m_{1}}(a_{3})$ $\displaystyle=a_{3},$
$\displaystyle\lambda_{m_{2}}(a_{3})$ $\displaystyle=a_{3}.$

We note that

$\displaystyle\lambda_{m_{3}}(a_{1})$
$\displaystyle=\lambda_{m_{1}m_{2}m_{1}^{-1}m_{2}^{-1}}(a_{1})=a_{1},$
$\displaystyle\lambda_{m_{4}}(a_{1})$
$\displaystyle=\lambda_{m_{1}^{2}}(a_{1})=a_{1}$
$\displaystyle\lambda_{m_{3}}(a_{2})$
$\displaystyle=\lambda_{m_{1}m_{2}m_{1}^{-1}m_{2}^{-1}}(a_{2})=a_{2}+a_{3},$
$\displaystyle\lambda_{m_{4}}(a_{2})$
$\displaystyle=\lambda_{m_{1}^{2}}(a_{2})=a_{2}+a_{3},$
$\displaystyle\lambda_{m_{3}}(a_{3})$
$\displaystyle=\lambda_{m_{1}m_{2}m_{1}^{-1}m_{2}^{-1}}(a_{3})=a_{3},$
$\displaystyle\lambda_{m_{4}}(a_{3})$
$\displaystyle=\lambda_{m_{1}^{2}}(a_{3})=a_{3},$

and $\lambda_{m_{5}}$ is the identity map on $B$.

We can consider the semidirect product $G=[B]C$ with respect to this action.
Then $G$ turns out to be a trifactorised group as it possesses a subgroup
$D=\langle a_{1}a_{2}^{3}m_{1},a_{1}m_{2}\rangle$ such that $D\cap C=D\cap
B=1$, $DC=BD=G$. Thus, there exists a bijective $1$-cocycle $\delta\colon
C\longrightarrow B$ with respect to $\lambda$ given by Table 2). This yields a
product $\cdot$ in $B$ and we get a brace of abelian type $(B,+,\cdot)$ of
order $32$. This brace corresponds to SmallBrace(32, 14649) in the Yang–Baxter
library for GAP.

$\begin{array}[]{llll}c&\delta(c)&c&\delta(c)\\\ \hline\cr
1&0&m_{1}&a_{1}+3a_{2}\\\ m_{5}&2a_{1}&m_{1}m_{5}&3a_{1}+3a_{2}\\\
m_{4}&3a_{1}+2a_{2}+a_{3}&m_{1}m_{4}&a_{2}\\\
m_{4}m_{5}&a_{1}+2a_{2}+a_{3}&m_{1}m_{4}m_{5}&2a_{1}+a_{2}\\\
m_{3}&3a_{1}+a_{3}&m_{1}m_{3}&2a_{1}+3a_{2}\\\
m_{3}m_{5}&a_{1}+a_{3}&m_{1}m_{3}m_{5}&3a_{2}\\\
m_{3}m_{4}&2a_{1}+2a_{2}&m_{1}m_{3}m_{4}&a_{1}+a_{2}\\\
m_{3}m_{4}m_{5}&2a_{2}&m_{1}m_{3}m_{4}m_{5}&3a_{1}+a_{2}\\\
m_{2}&a_{1}&m_{1}m_{2}&3a_{2}+a_{3}\\\
m_{2}m_{5}&3a_{1}&m_{1}m_{2}m_{5}&2a_{1}+3a_{2}+a_{3}\\\
m_{2}m_{4}&2a_{2}+a_{3}&m_{1}m_{2}m_{4}&3a_{1}+a_{2}+a_{3}\\\
m_{2}m_{4}m_{5}&2a_{1}+2a_{2}+a_{3}&m_{1}m_{2}m_{4}m_{5}&a_{1}+a_{2}+a_{3}\\\
m_{2}m_{3}&2a_{1}+a_{3}&m_{1}m_{2}m_{3}&3a_{1}+3a_{2}+a_{3}\\\
m_{2}m_{3}m_{5}&a_{3}&m_{1}m_{2}m_{3}m_{5}&a_{1}+3a_{2}+a_{3}\\\
m_{2}m_{3}m_{4}&a_{1}+2a_{2}&m_{1}m_{2}m_{3}m_{4}&2a_{1}+a_{2}+a_{3}\\\
m_{2}m_{3}m_{4}m_{5}&3a_{1}+2a_{2}&m_{1}m_{2}m_{3}m_{4}m_{5}&a_{2}+a_{3}\\\
\hline\cr\end{array}$ Table 2: Associated bijective $1$-cocycle

We have that $\langle 2a_{1}+2a_{2}\rangle_{+}\leq(B,+)$, corresponding to
$\langle m_{3}m_{4}\rangle\leq C$ (through $\delta$), defines a subbrace $S$
of $B$ of order $2$. Also, $\langle
2a_{1},2a_{2},a_{1}+a_{2}+a_{3}\rangle_{+}\leq(B,+)$, corresponding to
$\langle m_{5},m_{3}m_{4}m_{5},m_{1}m_{2}m_{4}m_{5}\rangle\leq C$, defines a
subbrace $T$ of $B$ of order $8$. Furthermore $\langle
2a_{1},2a_{2},a_{1}+a_{3}\rangle_{+}\leq(B,+)$, corresponding to $\langle
m_{5},m_{3}m_{4}m_{5},m_{3}m_{5}\rangle\leq C$, defines another subbrace $U$
of $B$ of order $8$.

We note that $S$ is not a left ideal of $B$, because
$\lambda_{m_{1}}(2a_{1}+2a_{2})=2(3a_{1}+a_{3})+2(a_{1}+a_{2}+a_{3})=2a_{2}\notin
S$. On the other hand, $S$ is a left ideal of $T$, since
$\lambda_{m_{5}}(2a_{1}+2a_{2})=\lambda_{m_{3}m_{4}m_{5}}(2a_{1}+2a_{2})=\lambda_{m_{1}m_{2}m_{4}m_{5}}(2a_{1}+2a_{2})=2a_{1}+2a_{2}$.
Furthermore, $\langle m_{3}m_{4}\rangle$ is a normal subgroup of $\langle
m_{5},m_{3}m_{4}m_{5},m_{1}m_{2}m_{4}m_{5}\rangle$. Therefore, $S$ is an ideal
of $T$. We also have that $S$ is a left ideal of $U$, since
$\lambda_{m_{5}}(2a_{1}+2a_{2})=\lambda_{m_{3}m_{4}m_{5}}(2a_{1}+2a_{2})=\lambda_{m_{3}m_{5}}(2a_{1}+2a_{2})=2a_{1}+2a_{2}$.
Moreover, $\langle m_{3}m_{4}\rangle$ is a normal subgroup of $\langle
m_{5},m_{3}m_{4}m_{5},m_{3}m_{5}\rangle$. Therefore, $S$ is an ideal of $U$.

We prove now that the subbrace $D=\langle T,U\rangle$ of $B$ generated by $T$
and $U$ is $B$. Let $H$ be the additive group of $D$. Then $H\geq\langle
2a_{1},a_{2},a_{1}+a_{3}\rangle_{+}$. Thus, if $R=\delta^{-1}(H)$ is the
corresponding multiplicative group, then $\delta^{-1}(2a_{1})=m_{5}\in R$,
$\delta^{-1}(a_{2})=m_{1}m_{4}\in R$, $\delta^{-1}(2a_{2})=m_{3}m_{4}m_{5}\in
R$, $\delta^{-1}(a_{1}+a_{3})=m_{3}m_{5}\in R$,
$\delta^{-1}(a_{1}+a_{2}+a_{3})=m_{1}m_{2}m_{4}m_{5}\in R$, which implies that
$C=\langle m_{1},m_{2},m_{3},m_{4},m_{5}\rangle=R$. Thus, $H=(B,+)$ and
$\langle T,U\rangle=B$.

Finally, suppose that $S$ possesses an idealiser in $B$. Since it must contain
every subbrace of $B$ in which $S$ is an ideal, it must contain $T$ and $U$.
It follows that the idealiser of $S$ in $B$ must be $B$, but $S$ is not an
ideal of $B$.

Our second example shows that there is no analogue of Fitting’s theorem for
central nilpotency, even for braces of abelian type

######  Example B.

Let $B=\langle a\rangle\times\langle c\rangle\times\langle
d\rangle\times\langle e\rangle\cong C_{4}\times C_{2}\times C_{2}\times
C_{2}$, written additively. Let us consider the automorphisms $g_{1}$,
$g_{2}$, $g_{3}$ of $B$ given by

$\displaystyle g_{1}\colon a$ $\displaystyle\longmapsto 3a+d$ $\displaystyle
g_{2}\colon a$ $\displaystyle\longmapsto a$ $\displaystyle g_{3}\colon a$
$\displaystyle\longmapsto a$ $\displaystyle c$ $\displaystyle\longmapsto c$
$\displaystyle c$ $\displaystyle\longmapsto c$ $\displaystyle c$
$\displaystyle\longmapsto 2a+c$ $\displaystyle d$ $\displaystyle\longmapsto d$
$\displaystyle d$ $\displaystyle\longmapsto 2a+d$ $\displaystyle d$
$\displaystyle\longmapsto d$ $\displaystyle e$ $\displaystyle\longmapsto c+e$
$\displaystyle e$ $\displaystyle\longmapsto 2a+e$ $\displaystyle e$
$\displaystyle\longmapsto 2a+e$

If $g_{4}=g_{1}g_{2}g_{1}^{-1}g_{2}^{-1}$ and
$g_{5}=g_{1}g_{3}g_{1}^{-1}g_{3}^{-1}$, then their action on $(B,+)$ is

$\displaystyle g_{4}\colon a$ $\displaystyle\longmapsto 3a$ $\displaystyle
g_{5}\colon a$ $\displaystyle\longmapsto a$ $\displaystyle c$
$\displaystyle\longmapsto c$ $\displaystyle c$ $\displaystyle\longmapsto c$
$\displaystyle d$ $\displaystyle\longmapsto d$ $\displaystyle d$
$\displaystyle\longmapsto d$ $\displaystyle e$ $\displaystyle\longmapsto e$
$\displaystyle e$ $\displaystyle\longmapsto 2a+e$

and we have that $C=\langle g_{1},g_{2},g_{3}\rangle=\langle
g_{1},g_{2},g_{3},g_{4},g_{5}\rangle$ satisfies the following relations:

$\displaystyle g_{1}^{2}$ $\displaystyle=1,$ $\displaystyle g_{2}^{2}$
$\displaystyle=1,$ $\displaystyle g_{1}g_{2}g_{1}^{-1}$
$\displaystyle=g_{4}g_{2},$ $\displaystyle g_{3}^{2}$ $\displaystyle=1,$
$\displaystyle g_{1}g_{3}g_{1}^{-1}$ $\displaystyle=g_{5}g_{3},$
$\displaystyle g_{2}g_{3}g_{2}^{-1}$ $\displaystyle=g_{3},$ $\displaystyle
g_{4}^{2}$ $\displaystyle=1,$ $\displaystyle g_{1}g_{4}g_{1}^{-1}$
$\displaystyle=g_{4},$ $\displaystyle g_{2}g_{4}g_{2}^{-1}$
$\displaystyle=g_{4},$ $\displaystyle g_{3}g_{4}g_{3}^{-1}$
$\displaystyle=g_{4},$ $\displaystyle g_{5}^{2}$ $\displaystyle=1,$
$\displaystyle g_{1}g_{5}g_{1}^{-1}$ $\displaystyle=g_{5},$ $\displaystyle
g_{2}g_{5}g_{2}^{-1}$ $\displaystyle=g_{5},$ $\displaystyle
g_{3}g_{5}g_{3}^{-1}$ $\displaystyle=g_{5},$ $\displaystyle
g_{4}g_{5}g_{4}^{-1}$ $\displaystyle=g_{5}.$

It follows that $C$ is a group of order $32$. We can consider the semidirect
product $G=[B]C$ with respect to this action $\lambda\colon
C\longrightarrow\operatorname{Aut}(B)$. Let

$D=\langle a^{2}cg_{1},cdeg_{2},a^{3}cg_{3}\rangle=\langle
a^{2}cg_{1},cdeg_{2},a^{3}cg_{3},cg_{4},dg_{5}\rangle;$

then $D$ is a group of order $32$, satisfying the same relations as $C$, and,
since $\langle a^{2}c,cde,a^{3}c,c,d\rangle=B$, we obtain that $G=DC=BD$ and
$D\cap C=D\cap B=1$. This leads to the bijective $1$-cocycle $\delta\colon
C\rightarrow B$ given in Table 3, and we can define a structure of a brace of
abelian type on $(B,+,\cdot)$ of order $32$. This brace corresponds to
SmallBrace(32, 23060) in the Yang–Baxter library for GAP.

$\begin{array}[]{rlrl}x&\delta(x)&x&\delta(x)\\\ \hline\cr
1&0&g_{5}g_{3}g_{2}&a+e\\\ g_{1}&2a+c&g_{3}g_{2}g_{1}&3a+e\\\
g_{2}&c+d+e&g_{4}g_{2}g_{1}&2a+d+e\\\ g_{3}&3a+c&g_{5}g_{2}g_{1}&c+e\\\
g_{4}&c&g_{4}g_{3}g_{1}&a+c+d\\\ g_{5}&d&g_{5}g_{3}g_{1}&3a\\\
g_{2}g_{1}&2a+c+d+e&g_{5}g_{4}g_{1}&2a+d\\\
g_{3}g_{1}&3a+d&g_{4}g_{3}g_{2}&a+c+d+e\\\
g_{4}g_{1}&2a&g_{5}g_{4}g_{2}&2a+e\\\ g_{5}g_{1}&2a+c+d&g_{5}g_{4}g_{3}&a+d\\\
g_{3}g_{2}&3a+d+e&g_{4}g_{3}g_{2}g_{1}&a+c+e\\\
g_{4}g_{2}&d+e&g_{5}g_{3}g_{2}g_{1}&a+d+e\\\
g_{5}g_{2}&2a+c+d&g_{5}g_{4}g_{2}g_{1}&e\\\
g_{4}g_{3}&a&g_{5}g_{4}g_{3}g_{1}&a+c\\\
g_{5}g_{3}&3a+c+d&g_{5}g_{4}g_{3}g_{2}&3a+c+e\\\
g_{5}g_{4}&c+d&g_{5}g_{4}g_{3}g_{2}g_{1}&3a+c+d+e\\\ \hline\cr\end{array}$
Table 3: Associated bijective $1$-cocycle

Since $C$ has order $32$, we have that $\operatorname{Ker}\lambda=1$. In
particular, we have that $\zeta(B)=0$, and $B$ is not centrally nilpotent.

Now, let us compute the ideals of $B$. Suppose that $I$ is a non-zero ideal of
$B$ with additive group $L$ and multiplicative group $E$. Since $E$ is a
normal subgroup of $C$, it must contain a minimal normal subgroup of $E$. All
minimal normal subgroups of $C$ are contained in $\operatorname{Z}(C)=\langle
g_{4},g_{5}\rangle$. Hence $E$ must contain $\langle g_{4}\rangle$, $\langle
g_{5}\rangle$ or $\langle g_{4}g_{5}\rangle$. In the first case, $L$ must
contain $\delta(g_{4})=c$. Since $L$ must be invariant under the action of
$C$, it should contain $g_{3}(c)=2a+c$. Consequently, $\langle
2a,c\rangle_{+}\leq L$. In particular, $\delta^{-1}(2a)=g_{4}g_{1}\in E$ and
$\langle g_{1},g_{4}\rangle\leq E$. Since $E\trianglelefteq C$, we have that
$g_{3}g_{1}g_{3}^{-1}=g_{5}g_{1}\in E$, so $\langle
g_{1},g_{4},g_{5}\rangle\leq E$.

Similarly, if $g_{5}\in E$, then $\delta(g_{5})=d\in L$. Thus,
$g_{2}(d)=2a+d\in L$ and hence $\langle 2a,d\rangle\leq L$. Now,
$g_{1}g_{4}\in E$, so $g_{3}^{-1}g_{1}g_{4}g_{3}=g_{1}g_{4}g_{5}\in E$ and
also $g_{5}\in E$. Therefore $\langle g_{1},g_{4},g_{5}\rangle\leq E$.

Finally, if $g_{4}g_{5}\in E$, then $\delta(g_{4}g_{5})=c+d\in L$, so
$g_{2}(c+d)=2a+c+d\in L$ and hence $\langle 2a,c+d\rangle_{+}\leq L$. Again,
$g_{1}g_{4}\in E$, so $g_{3}^{-1}g_{1}g_{4}g_{3}=g_{1}g_{4}g_{5}\in E$ and
also $g_{5}\in E$. Thus, $\langle g_{1},g_{4},g_{5}\rangle\leq E$.

In all cases, we found out that $\langle g_{1},g_{4},g_{5}\rangle\leq E$.
Since $\delta(\langle g_{1},g_{4},g_{5}\rangle)=\langle
2a,c,d\rangle_{+}\leq(B,+)$ is a $\delta$-invariant subgroup and $\langle
g_{1},g_{4},g_{5}\rangle\trianglelefteq C$, we have that $J=\langle
2a,c,d\rangle_{+}$ is the unique ideal of $B$ of order $8$. We observe that
$B/J$ is abelian. Therefore, the only three ideals of order $16$ of $B$ are
$I_{1}=\langle 2a,c,d,e\rangle_{+}$, $I_{2}=\langle a+e,2a,c,d\rangle_{+}$,
$I_{3}=\langle a,c,d\rangle_{+}$.

It can be easily checked that

$0\leq\langle c\rangle\leq\langle 2a,c\rangle_{+}\leq I_{1},\quad 0\leq\langle
c+d\rangle_{+}\leq\langle 2a,c+d\rangle_{+}\leq I_{2}$

and

$0\leq\langle d\rangle_{+}\leq\langle 2a,d\rangle_{+}\leq I_{3}$

are $c$-series of $I_{1}$, $I_{2}$ and $I_{3}$, respectively. In particular,
$I_{1}$, $I_{2}$ and $I_{3}$ are centrally nilpotent braces. However,
$B=I_{1}+I_{2}=I_{1}+I_{3}=I_{2}+I_{3}$, but, as we have mentioned, $B$ is not
centrally nilpotent.

Our third example shows that another classical property of nilpotency of
finite groups fails for central nilpotency in finite braces: there may be
abelian subideals that are not contained in any centrally nilpotent ideals.

######  Example C.

Let

$(B,+)=\langle a\rangle\times\langle
b\rangle\cong\operatorname{C}_{2}\times\operatorname{C}_{12}\quad\textnormal{
and
}\quad(C,\cdot)=[\langle\sigma\rangle]\langle\tau\rangle\cong\operatorname{Dih}_{24}.$

We have that $C$ acts on $B$ by means of the action $\lambda$ defined by

$\displaystyle\lambda_{\sigma}(a)$ $\displaystyle=a+6b,$
$\displaystyle\lambda_{\tau}(a)$ $\displaystyle=a,$
$\displaystyle\lambda_{\sigma}(b)$ $\displaystyle=a+b,$
$\displaystyle\lambda_{\tau}(b)$ $\displaystyle=a-b,$

We can consider the semidirect product $G$ of $B$ and $C$ with respect to this
action. Then $G$ turns out to be a trifactorised group as it possesses a
subgroup $D=\langle ab^{7}\sigma,b^{6}\tau\rangle$ such that $D\cap C=D\cap
B=\\{1\\}$, $DC=BD=G$. Thus, there is a bijective $1$-cocycle $\delta\colon
C\longrightarrow B$ with respect to $\lambda$ given by Table 4. This yields a
product in $B$ and we get a brace of abelian type $(B,+,\cdot)$ of order $24$.
This brace corresponds to SmallBrace(24, 57) in the Yang–Baxter library for
GAP.

$\begin{array}[]{llllllll}c&\delta(c)&c&\delta(c)&c&\delta(c)&c&\delta(c)\\\
\hline\cr 1&0&\sigma^{6}&a&\tau&6b&\sigma^{6}\tau&a+6b\\\
\sigma&a+7b&\sigma^{7}&b&\sigma\tau&a+b&\sigma^{7}\tau&7b\\\
\sigma^{2}&a+8b&\sigma^{8}&8b&\sigma^{2}\tau&a+2b&\sigma^{8}\tau&2b\\\
\sigma^{3}&9b&\sigma^{9}&a+3b&\sigma^{3}\tau&3b&\sigma^{9}\tau&a+9b\\\
\sigma^{4}&4b&\sigma^{10}&a+4b&\sigma^{4}\tau&10b&\sigma^{10}\tau&a+10b\\\
\sigma^{5}&11b&\sigma^{11}&5b&\sigma^{5}\tau&a+5b&\sigma^{11}\tau&11b\\\
\hline\cr\end{array}$ Table 4: Associated bijective $1$-cocycle

Let $I$ be any ideal of $B$ of order $12$ put $E=\delta^{-1}(I,+)$. Since
$(I,+)$ is a maximal subgroup of $(B,+)$, it must contain its Frattini
subgroup, which is $\langle 6b\rangle$. As $\delta^{-1}(6b)=\tau$ and
$E\trianglelefteq C$, it follows that $\sigma\tau\sigma^{-1}=\sigma^{2}\tau\in
E$. Therefore, $\delta(\sigma^{2}\tau)=a+2b\in I$ and then $(I,+)=\langle
a,2b\rangle_{+}$. Since $I$ is $\lambda$-invariant, we get that $I=\langle
a,2b\rangle_{+}$ is the only ideal of order $12$ of $B$.

Observe that $I$ is not abelian as $\operatorname{Soc}(I)=\langle
a+4b\rangle_{+}$. Therefore, $(I,\cdot)$ is isomorphic to
$\operatorname{Dih}_{12}$ and so $I$ can not be centrally nilpotent. Hence,
$\operatorname{Soc}(I)$ is an abelian subideal of $B$ of order $6$ such that
it is not contained in any centrally nilpotent ideal of $B$.

Our last example shows that another sufficient condition for finitely
generated nilpotent groups also falls in braces: there may be non-centrally
nilpotent braces such that all subbraces are subideals.

######  Example D.

Let $(B,+,\cdot)$ be the brace of abelian type of order $32$ studied in [3,
Example 37]. It corresponds to SmallBrace(32,24003) in the Yang–Baxter library
for GAP, so that

$\begin{array}[]{c}(B,+)=\langle a\rangle\times\langle b\rangle\times\langle
c\rangle\times\langle d\rangle\cong C_{4}\times C_{2}\times C_{2}\times
C_{2},\quad\textnormal{and}\\\\[2.84544pt] (B,\cdot)\cong\langle
e,f,h\rangle\cong[C_{2}\times\operatorname{Q}_{8}]C_{2}\end{array}$

with bijective $1$-cocycle given by Table 5

$\begin{array}[]{llllllll}x&\delta(x)&x&\delta(x)&x&\delta(x)&x&\delta(x)\\\
\hline\cr 1&0&h&c&f&a&fh&a+c\\\ e&3a+b&eh&3a&ef&b+c+d&efh&c+d\\\
e^{2}&b+c&e^{2}h&2a+b+c+d&e^{2}f&3a+d&e^{2}fh&a+c+d\\\
e^{3}&3a+b+d&e^{3}h&a+d&e^{3}f&b&e^{3}fh&2a\\\
e^{4}&2a+b+c&e^{4}h&2a+b&e^{4}f&a+b+c&e^{4}fh&a+b\\\
e^{5}&3a+c&e^{5}h&3a+b+c&e^{5}f&2a+d&e^{5}fh&2a+b+d\\\
e^{6}&2a+c+d&e^{6}h&d&e^{6}h&3a+b+c+d&e^{6}fh&a+b+d\\\
e^{7}&3a+c+d&e^{7}h&a+b+c+d&e^{7}f&2a+c&e^{7}fh&b+d\\\ \hline\cr\end{array}$
Table 5: Associated bijective $1$-cocycle

and associated action given by

$\displaystyle e\colon a$ $\displaystyle\longmapsto a+c+d$ $\displaystyle
f\colon a$ $\displaystyle\longmapsto a+b+c$ $\displaystyle h\colon a$
$\displaystyle\longmapsto a$ $\displaystyle b$ $\displaystyle\longmapsto 2a+c$
$\displaystyle b$ $\displaystyle\longmapsto 2a+b$ $\displaystyle b$
$\displaystyle\longmapsto b$ $\displaystyle c$ $\displaystyle\longmapsto b$
$\displaystyle c$ $\displaystyle\longmapsto c$ $\displaystyle c$
$\displaystyle\longmapsto c$ $\displaystyle d$ $\displaystyle\longmapsto
2a+c+d$ $\displaystyle d$ $\displaystyle\longmapsto c+d$ $\displaystyle d$
$\displaystyle\longmapsto 2a+d.$

We start by providing all subbraces of order $2$. These are generated by those
elements $x\in(B,+)$ of order $2$ such that $\lambda_{x}(x)=x$. We have:

$\begin{array}[]{lll}S_{1}=\\{1,2a+b+d\\},&S_{2}=\\{1,c\\},&S_{3}=\\{1,b+c\\},\\\
S_{4}=\\{1,c+d\\},&S_{5}=\\{1,2a\\},&S_{6}=\\{1,2a+b\\},\\\
S_{7}=\\{1,2a+b+c\\}.&&\end{array}$

(here, $S_{5}$ is the only left ideal). For subbraces of order $4$, we need
those subgroups $H\leq(B,+)$ of order $4$ such that $\delta^{-1}(H)$ is also a
subgroup of $\langle e,f,h\rangle$. We find

$\begin{array}[]{llll}H&\delta^{-1}(H)&H&\delta^{-1}(H)\\\ \hline\cr
S_{8}=\langle 2a,b+c\rangle&\langle e^{3}fh,e^{7}fh\rangle&S_{9}=\langle
2a+b,c\rangle&\langle e^{4},h\rangle\\\ S_{10}=\langle b,2a+c\rangle&\langle
e^{3}f\rangle&S_{11}=\langle c+d,2a+b+d\rangle&\langle efh,e^{5}fh\rangle\\\
S_{12}=\langle 2a+d,b+c+d\rangle&\langle e^{5}f\rangle&S_{13}=\langle
d,2a+b+c\rangle&\langle e^{6}h\rangle\\\ S_{14}=\langle
2a+b+c,b+d\rangle&\langle e^{2}\rangle&\\\ \hline\cr\end{array}$

(here, $S_{8}$ is the only left ideal). For subbraces of order $8$, we need
those subgroups $H\leq(B,+)$ of order $8$ such that $\delta^{-1}(H)\leq\langle
e,f,h\rangle$. Thus,

$\begin{array}[]{ll}H&\delta^{-1}(H)\\\ \hline\cr S_{15}=\langle
2a,b,c\rangle&\langle e^{3}fh,e^{3}f\rangle\\\ S_{16}=\langle
2a,b+c,b+d\rangle&\langle e^{3}fh,e^{2}\rangle\\\ S_{17}=\langle
2a,b+c,d\rangle&\langle e^{3}fh,e^{6}h\rangle\\\ \hline\cr\end{array}$

(here, $S_{15}$ is the only left ideal). The only subbrace of order $16$ is
the only non-zero proper ideal of $B$, that is, $S_{18}=\langle
2a,b,c,d\rangle$.

The following relations can be easily checked to hold:

$\begin{array}[]{r}S_{1},S_{4},S_{7}\unlhd S_{11}\unlhd S_{15}\unlhd
S_{18}\unlhd B\\\ S_{3},S_{5},S_{7}\unlhd S_{8}\unlhd S_{16}\unlhd
S_{18}\unlhd B\\\ S_{2},S_{6},S_{7}\unlhd S_{9}\unlhd S_{17}\unlhd
S_{18}\unlhd B\\\ S_{14}\unlhd S_{15}\unlhd S_{18}\unlhd B\\\
S_{12},S_{13}\unlhd S_{17}\unlhd S_{18}\unlhd B\\\ S_{10}\unlhd S_{17}\unlhd
S_{18}\unlhd B\end{array}$

Therefore, all subraces are subideals but $B$ is not soluble.

##  7 Acknowledgments

All authors are members of the non-profit association “Advances in Group
Theory and Applications” (www.advgrouptheory.com). The third and fifth authors
are supported by GNSAGA (INdAM) and wish to express their gratitude to the
Department of Mathematics of the University of Valencia for its support during
their stay. The third author has been supported by a research visiting grant
issued by the Istituto Nazionale di Alta Matematica (INdAM).

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