# Symmetric skew braces and brace systems

Valeriy G. Bardakov  Sobolev Institute of Mathematics, pr. ak. Koptyuga 4,
Novosibirsk, 630090, Russia and Novosibirsk State University, Novosibirsk,
630090, Russia and Novosibirsk State Agrarian University, Dobrolyubova street,
160, Novosibirsk, 630039, Russia and Regional Scientific and Educational
Mathematical Center of Tomsk State University, 36 Lenin Ave., Tomsk, Russia.
<EMAIL_ADDRESS>,  Mikhail V. Neshchadim  Sobolev Institute of
Mathematics, pr. ak. Koptyuga 4, Novosibirsk, 630090, Russia and Novosibirsk
State University, Novosibirsk, 630090, Russia and Regional Scientific and
Educational Mathematical Center of Tomsk State University, 36 Lenin Ave.,
Tomsk, Russia<EMAIL_ADDRESS>and  Manoj K. Yadav  Harish-Chandra
Research Institute, A CI of Homi Bhabha National Institute, Chhatnag Road,
Jhunsi, Prayagraj-211 019, India<EMAIL_ADDRESS>
###### Abstract.

For a skew left brace $(G,\cdot,\circ)$, the map
$\lambda:(G,\circ)\to\operatorname{Aut}\,(G,\cdot),~{}~{}a\mapsto\lambda_{a},$
where $\lambda_{a}(b)=a^{-1}\cdot(a\circ b)$ for all $a,b\in G$, is a group
homomorphism. Then $\lambda$ can also be viewed as a map from $(G,\cdot)$ to
$\operatorname{Aut}\,(G,\cdot)$, which, in general, may not be a homomorphism.
A skew left brace will be called $\lambda$-anti-homomorphic
($\lambda$-homomorphic) if $\lambda:(G,\cdot)\to\operatorname{Aut}\,(G,\cdot)$
is an anti-homomorphism (a homomorphism). We mainly study such skew left
braces. We device a method for constructing a class of binary operations on a
given set so that the set with any two such operations constitute a
$\lambda$-homomorphic symmetric skew brace. Most of the constructions of
symmetric skew braces dealt with in the literature fall in the framework of
our construction. We then carry out various such constructions on specific
infinite groups.

###### Key words and phrases:

Skew left brace, $\lambda$-anti-homomorphic, $\lambda$-homomorphic, Symmetric
skew brace

###### 2010 Mathematics Subject Classification:

16T25, 81R50

##  1\. Introduction

A triple $(G,\cdot,\circ)$, where $(G,\cdot)$ and $(G,\circ)$ are groups, is
said to be a _skew left brace_ if

(1.1) $a\circ(b\cdot c)=(a\circ b)\cdot a^{-1}\cdot(a\circ c)$

for all $a,b,c\in G$, where $a^{-1}$ denotes the inverse of $a$ in
$(G,\cdot)$. We call $(G,\cdot)$ the _additive group_ and $(G,\circ)$ the
_multiplicative group_ of the skew left brace $(G,\cdot,\circ)$. A skew left
brace $(G,\cdot,\circ)$ is said to be a _left brace_ if $(G,\cdot)$ is an
abelian group. By analogy one can define a skew right brace and right brace. A
triple $(G,\cdot,\circ)$, where $(G,\cdot)$ and $(G,\circ)$ are groups, is
said to be a _skew right brace_ if

(1.2) $(a\cdot b)\circ c=(a\circ c)\cdot c^{-1}\cdot(b\circ c)$

for all $a,b,c\in G$. A skew right brace $(G,\cdot,\circ)$ is said to be a
_right brace_ if $(G,\cdot)$ is an abelian group. A skew left brace which is
also a skew right brace is said to be _two-sided skew brace_.

In this article we mainly consider skew left braces. So, we’ll mostly suppress
the word ‘left’ and only say skew brace(s).

The concept of braces was introduced by Rump [31] in 2007 in connection with
non-degenerate involutive set theoretic solutions of the quantum Yang-Baxter
equation. We remark that, in fact, braces were considered by Kurosh in his
lectures [27, Section 10] of 1969–1970 academic year. The subject received a
tremendous attention of the mathematical community after the work of Rump; see
[2, 13, 21, 26, 32, 34] and the references therein. Interest in the study of
set theoretic solutions of the quantum Yang-Baxter equations was intrigued by
the paper [17] of Drinfeld, published in 1992. Actually, set-theoretic
solutions of the Yang-Baxter equation were studied before Drinfeld formulated
his question. It was Joyce [22] and Matveev [29] who introduced quandles as
invariants of knots and links. Every quandle gives a set-theoretic solution of
the Yang-Baxter equation.

The concept of skew braces was introduced by Guarnieri and Vendramin [19] in
2017 in connection with non-involutive non-degenerate set theoretic solutions
of the quantum Yang-Baxter equation. For a skew left brace $(G,\cdot,\circ)$,
it was proved in [19] that the map

$\lambda:(G,\circ)\to\operatorname{Aut}\,(G,\cdot),~{}~{}a\mapsto\lambda_{a}$

is a group homomorphism, where $\operatorname{Aut}\,(G,\cdot)$ denotes the
automorphism group of $(G,\cdot)$ and $\lambda_{a}$ is given by
$\lambda_{a}(b)=a^{-1}\cdot(a\circ b)$ for all $a,b\in G$. Obviously,
$\lambda$ can also be viewed as a map from $(G,\cdot)$ to
$\operatorname{Aut}\,(G,\cdot)$, which, in general, may not be a homomorphism.
We say that $(G,\cdot,\circ)$ is a $\lambda$-anti-homomorphic skew left brace
($\lambda$-homomorphic skew left brace) if the map
$\lambda:(G,\cdot)\to\operatorname{Aut}\,(G,\cdot)$ is an anti-homomorphism (a
homomorphism). A $\lambda$-homomorphic skew left brace $(G,\cdot,\circ)$ is
said to be a $\lambda$-cyclic skew left brace if the image
$\operatorname{Im}\lambda$ is a cyclic subgroup of
$\operatorname{Aut}\,(G,\cdot)$. $\lambda$-homomorphic skew braces of finite
order were studied in [8] and that of infinite order in [4].

It is a big challenge to describe all skew braces of a given order. For
example, there are 51 groups of order 32, but there are 1223061 skew braces
and 25281 left braces of order 32 (see [3]). It is not even easy to describe
all skew braces with a given additive group. So the construction of certain
specific classes of skew braces is highly desirable. In the present article
our main focus is on the study of the following problem: For a given group
$(G,\cdot)$, define binary operations ‘$\circ_{i}$’, $i\in I$ such that
$(G,\circ_{i},\circ_{j})$, $i,j\in I\cup\\{0\\}$ is a skew brace, where
$\circ_{0}=\cdot$. Various aspects of this problem has already been studied in
[10, 11, 16, 23, 24, 33].

In Section 2 we introduce the notion of brace systems, symmetric brace system,
full symmetric brace system and linear brace system. In Section 3 we observe
that $\lambda$-anti-homomorphic skew braces characterize symmetric skew
braces. A skew brace $(G,\cdot,\circ)$ is said to be symmetric if
$(G,\circ,\cdot)$ is also a skew brace. We then find a procedure of
constructing a symmetric brace system.

In Section 4 we take up $\lambda$-homomorphic skew braces $(G,\cdot,\circ)$
with $\lambda(G)$ abelian. We device recursive procedure for constructing
symmetric (linear) brace systems on a given $\lambda$-homomorphic skew brace.
Section 5 describes a unification method for most of the existing procedures
for constructing symmetric skew braces. Sections 6 contains many interesting
properties of $\lambda$-homomorphic and $\lambda$-anti-homomorphic skew
braces. Specific constructions of skew braces $(G,\cdot,\circ)$ with abelian
$\lambda(G)$ are taken up in Section 7. In the concluding section we discuss
Rota–Baxter groups and skew braces associated with these.

When the additive group of a skew brace is non-abelian, we’ll mostly suppress
‘$\cdot$’ and use $ab$ for $a\cdot b$. For braces, we denote the additive
operation ‘$\cdot$’ by ‘$+$’. For a skew brace $(G,\cdot,\circ)$, the inverse
of $a\in G$ with respect to ‘$\circ$’ is denoted by $\bar{a}$. For an indexing
set $I$, when we deal with many binary operations $\circ_{i}$, $i\in I$, on
$G$ such that $(G,\circ_{i})$ is a group, we shall denote the inverse of $a\in
G$ with respect to the operation $\circ_{i}$ by $a^{\circ_{i}(-1)}$.

##  2\. Brace systems

The concept of a brace block was introduced in [24] and further studied in
[11]. Analogous concept of a multi-brace was introduced in [6]. In this
section we intend to introduce a concept, which we will be calling a brace
system, to generalize the concepts of brace block and multi-brace.

######  Definition 2.1.

Let $G$ and $V$ be non-empty sets, $E\subseteq V\times V$. A brace system on
$G$ associated with the pair $(V,E)$ is an algebraic system

$\mathcal{B}_{V,E}(G):=(G,\\{\circ_{v}\\}_{v\in V}),$

where $\circ_{v}:G\times G\to G$ is a binary operation for all $v\in V$,
satisfying the following conditions:

  1. (1)

$(G,\circ_{v})$ is a group for all $v\in V$,

  2. (2)

$(G,\circ_{u},\circ_{v})$ is a skew left brace for all pairs $(u,v)\in E$.

For any non-empty set $V$ and $E\subseteq V\times V$, we can associate an
oriented graph $\Gamma=(V,E^{\prime})$ with the set of vertices $V$ and there
is an oriented edge $e_{uv}$ between $u,v\in V$ if $(u,v)\in E$. More
precisely $E^{\prime}=\\{e_{uv}=(u,v)\in E\\}$. Because of this association,
the brace system $\mathcal{B}_{V,E}(G)$ can also be denoted as
$\mathcal{B}_{\Gamma}(G)$.

Notice that it is a two way passage. For any brace system
$\mathcal{B}_{V,E}(G)$ one can construct an oriented graph
$\Gamma=(V,E^{\prime})$ with the set of vertices $V$ and there is an edge
between $u,v\in V$ if $(G,\circ_{u},\circ_{v})$ is a skew left brace. The
preceding definition shows that each pair $(u,v)\in E$ is an oriented edge. To
distinguish, we denote $E^{\prime}=\\{e_{uv}=(u,v)\in E\\}.$

######  Definition 2.2.

Let $\mathcal{B}_{\Gamma}(G)$ be a brace system on $G$ associated to the graph
$\Gamma=(V,E)$. Then $\mathcal{B}_{\Gamma}(G)$ is said to be a symmetric brace
system if for each pair $(u,v)\in V\times V$, $e_{uv}\in E$ if and only if
$e_{vu}\in E$. A symmetric brace system is said to be a full symmetric brace
system on $G$ if for any pair $(u,v)\in V\times V$, there exists an edge
$e_{uv}\in E$.

One can see that if $\mathcal{B}_{\Gamma}(G)$ is a symmetric brace system,
then it defines a non-oriented graph since any edge has an opposite edge. If
$\mathcal{B}_{\Gamma}(G)$ is a full symmetric brace system, then it defines a
full (or complete) non-oriented graph.

######  Definition 2.3.

(i) If $V=I$ is a linearly ordered set and $E=\\{(i,j)\in I\times
I~{}|~{}i<j\\},$ then the brace system $\mathcal{B}_{\Gamma}(G)$ is said to be
a linear brace system and is denoted by $\mathcal{B}_{I}(G)$.

(ii) If $\Gamma=(V,E)$ is a rooted tree, meaning, there exists a vertex
$v_{0}\in V$ such that for any $v\in V$ there exists an oriented path from
$v_{0}$ to $v$, then the brace system $\mathcal{B}_{\Gamma}(G)$ is said to be
a rooted brace system.

Examples of linear and rooted brace system will be given in upcoming sections.

##  3\. $\lambda$-anti-homomorphic skew braces

In this section we introduce the notion of $\lambda$-anti-homomorphic skew
braces and see that these objects characterize symmetric skew braces. For any
skew left brace $(G,\cdot,\circ)$, it was proved in [19] that the map

$\lambda:(G,\circ)\to\operatorname{Aut}\,(G,\cdot),~{}~{}a\mapsto\lambda_{a},$

where $\lambda_{a}(b)=a^{-1}(a\circ b)$, is a group homomorphism. Notice that
$\lambda$ can also be viewed as a set map from $(G,\cdot)$ into
$\operatorname{Aut}\,(G,\cdot)$. Not much is known about the properties of
this map. The map $\lambda$ together with its domain, the additive group,
characterizes the skew brace as a linear $q$-cycle set [33]. We here take up
the case when it is an anti-homomorphism, and introduce the following
definition.

######  Definition 3.1.

A skew brace $(G,\cdot,\circ)$ is said to be a $\lambda$-anti-homomorphic skew
brace if the map $\lambda:(G,\cdot)\to\operatorname{Aut}\,(G,\cdot)$, defined
by $\lambda_{a}(b)=a^{-1}(a\circ b)$, $a,b\in(G,\cdot)$, is an anti-
homomorphism.

The following interesting property follows directly from the definition.

######  Lemma 3.2.

Let $(G,\cdot,\circ)$ be a $\lambda$-anti-homomorphic skew brace. Then for
$a\in G$, $\bar{a}=a^{-1}u_{a}$ for some
$u_{a}\in\operatorname{Ker}\,\lambda$, where $\bar{a}$ denotes the inverse of
$a$ in $(G,\circ)$.

###### Proof.

As we know $\lambda:(G,\circ)\to\operatorname{Aut}\,(G,\cdot)$,
$a\mapsto\lambda_{a}$, is a group homomorphism. So by the given hypothesis,
for any $a\in G$, we have

$\lambda_{\bar{a}}=\lambda^{-1}_{a}=\lambda_{a^{-1}},$

which proves the required assertion. ∎

We remark that the skew braces $(G,\cdot,\circ)$ in which the map $\lambda$ is
a homomorphism from $(G,\cdot)$ into $\operatorname{Aut}\,(G,\cdot)$ have been
investigated earlier by some authors [8], [4]. We’ll come back to it little
later. Before we proceed, let us see an example.

######  Example 3.3.

Let $G=(G,\cdot)$ be any group. Define an operation ‘$\circ$’ on the set $G$
by the formula

$a\circ b=a\lambda_{a}(b)=ba$

for all $a,b\in G$. Then it can be readily verified that $(G,\cdot,\circ)$ is
a skew brace. Now the map $\lambda:G\to\operatorname{Aut}\,G$,
$a\mapsto\lambda_{a}$, where

$\lambda_{a}(b)=a^{-1}(a\circ b)=a^{-1}ba\;\;\mbox{for all $b\in G$},$

is an anti-homomorphism. Hence $(G,\cdot,\circ)$ is a $\lambda$-anti-
homomorphic skew brace.

Since the skew braces of the preceding example can be naturally defined for
any group $G$, we would like to give those some name.

######  Definition 3.4.

A skew brace $(G,\cdot,\circ)$ is said to be natural $\lambda$-anti-
homomorphic skew brace if $\circ=\cdot^{op}$.

We’ll show below that $\lambda$-anti-homomorphic skew braces can be
characterised in terms of natural $\lambda$-anti-homomorphic skew brace.

Let $(G,\cdot,\circ)$ is a $\lambda$-anti-homomorphic skew brace. Then for all
$a,b,c\in G$,

$\displaystyle\lambda_{a\circ b}(c)$ $\displaystyle=$ $\displaystyle(a\circ
b)^{-1}(a\circ b\circ c)$ $\displaystyle=$ $\displaystyle(a\circ
b)^{-1}(a\circ(b\lambda_{b}(c)))$ $\displaystyle=$ $\displaystyle
a^{-1}a\lambda_{a}(\lambda_{b}(c))$ $\displaystyle=$
$\displaystyle\lambda_{a}(\lambda_{b}(c))$ $\displaystyle=$
$\displaystyle\lambda_{b\cdot a}(c)\;\;\mbox{ (by $\lambda$-anti-homomorphic
property)}.$

Hence

(3.5) $\lambda_{a\circ b}=\lambda_{b\cdot a}\;\;\mbox{ for all $a,b\in G$}.$

Now we recall the definition of a symmetric skew brace, which is also known by
the name bi-skew brace.

######  Definition 3.6.

A skew brace $(G,\cdot,\circ)$ is said to be a symmetric skew brace if
$(G,\circ,\cdot)$ is also a skew brace.

The necessary and sufficient condition for a skew brace to become symmetric
skew brace is given in the following result [4, Proposition 5.2, Remark 5.5].

######  Proposition 3.7.

A skew brace $(G,\cdot,\circ)$ is a symmetric skew brace if and only if

(3.8) $\displaystyle\lambda_{a\circ b}=\lambda_{b\cdot a}\;\;\mbox{for all
$a,b\in G$}.$

Let $(G,\cdot,\circ)$ be a symmetric skew brace. To differentiate the maps
$\lambda$, we write $\lambda^{\circ}$ for the map
$\lambda:(G,\circ)\to\operatorname{Aut}\,(G,\cdot)$ and $\lambda^{\cdot}$ for
$\lambda:(G,\cdot)\to\operatorname{Aut}\,(G,\circ)$. Rewriting (3.8) with the
new notation, we get

(3.9) $\lambda^{\circ}_{a\circ b}=\lambda^{\circ}_{b\cdot a}\;\;\mbox{for all
$a,b\in G$}.$

Analogously we can also show that

(3.10) $\lambda^{\cdot}_{a\cdot b}=\lambda^{\cdot}_{b\circ a}\;\;\mbox{for all
$a,b\in G$}.$

We can now prove

######  Proposition 3.11.

Every $\lambda$-anti-homomorphic skew brace is symmetric. On the other hand if
$(G,\cdot,\circ)$ is a symmetric skew brace, then both $(G,\cdot,\circ)$ and
$(G,\circ,\cdot)$ are $\lambda$-anti-homomorphic skew braces.

###### Proof.

The first assertion follows from Proposition 3.7 using (3.5). Now assume that
$(G,\cdot,\circ)$ is a symmetric skew brace. Then (3.9) and (3.10) hold true.
So by (3.9) we have

$\lambda^{\circ}_{a\cdot b}=\lambda^{\circ}_{b\circ
a}=\lambda^{\circ}_{b}\lambda^{\circ}_{a}\;\;\mbox{for all $a,b\in G$}.$

This shows that $\lambda^{\circ}:(G,\cdot)\to\operatorname{Aut}\,(G,\cdot)$,
$a\mapsto\lambda^{\circ}_{a}$, is an anti-homomorphism. Hence
$(G,\cdot,\circ)$ is a $\lambda$-anti-homomorphic skew brace. That
$(G,\circ,\cdot)$ is a $\lambda$-anti-homomorphic skew brace can be seen
analogously using (3.10). ∎

######  Remark 3.12.

The preceding proposition characterizes symmetric skew braces in terms of
$\lambda$-anti-homorphic skew braces.

We remark that a similar characterization of symmetric skew braces (under the
name bi-skew braces) in terms of $\gamma$-functions has been recently obtained
in [9, Theorem 3.1].

Let us see another example, obtained through exact factorization ([4, Section
5]). Recall that a group $G$ factorizes through two subgroups $A$ and $B$ if

$G=AB=\\{ab\mid a\in A,b\in B\\}.$

The factorization is said to be exact if $A\cap B=1$. The following
construction was carried out in [35]. Let a group $G$ admit an exact
factorization $AB$. Define the operation ‘$\circ$’ on $G$ by the rule: If
$a_{1}b_{1},a_{2}b_{2}\in G$, where $a_{1},a_{2}\in A$, $b_{1},b_{2}\in B$,
then

(3.13) $(a_{1}b_{1})\circ(a_{2}b_{2})=a_{1}a_{2}b_{2}b_{1}.$

It is proved in [35, Theorem 2.3] that $(G,\cdot,\circ)$ is a skew brace. It
is easy to see that $(G,\circ)$ is isomorphic to the direct product $A\times
B_{op}$, where $B_{op}$ is the opposite group of $B$. If $B$ is abelian, then
obviously $(G,\circ)\cong A\times B$. Notice that ‘$\circ$’ and ‘$\cdot$’
coincide on $A$.

The homomorphism $\lambda:(G,\circ)\to\operatorname{Aut}\,(G,\cdot)$ gives
automorphisms $\lambda_{z}$ of $(G,\cdot)$ for all $z\in(G,\circ)$ defined by

$\lambda_{z}(y)=z^{-1}(z\circ y),\quad y\in G.$

If $z=a_{1}b_{1}$, $y=a_{2}b_{2}$, where $a_{1},a_{2}\in A$, $b_{1},b_{2}\in
B$, then

$\lambda_{z}(y)=(a_{1}b_{1})^{-1}a_{1}a_{2}b_{2}b_{1}=b_{1}^{-1}yb_{1}.$

So, the automorphism $\lambda_{z}$ is the inner automorphism of $G$ induced by
the element $b_{1}$, where $z=a_{1}b_{1}$.

Let $G=AB$ be as above with $A$ normal in $G$. We now consider the map
$\lambda:G\to\operatorname{Aut}G$ given by $z\mapsto\lambda_{z}$, where
$\lambda_{z}$ is as defined in the preceding para. A straightforward
computation shows that $\lambda$ is an anti-homomorphism. Thus, by Proposition
3.11, we have

######  Proposition 3.14.

The skew brace $(G,\cdot,\circ)$, constructed above, is $\lambda$-anti-
homomorphic; hence symmetric.

We now device a method for constructing a $\lambda$-anti-homomorphic skew
brace from a given anti-homomorphism of a group into its automorphism group.
Let $G=(G,\cdot)$ be a group and $\lambda:G\to\operatorname{Aut}\,G$ be an
anti-homomorphism such that
$a\lambda_{a}(b)a^{-1}b^{-1}\in\operatorname{Ker}\,\lambda$ for all $a,b\in
G$. Then we can define a group $(G,\circ)$ such that, for $a,b\in G$,

$a\circ b=a\lambda_{a}(b).$

Notice that $\bar{a}$, the inverse of $a$ with respect to ‘$\circ$’ is
$\lambda^{-1}_{a}(a^{-1})$, where, as usual, $\lambda^{-1}_{a}$ is the inverse
of the automorphism $\lambda_{a}$. An easy verification now shows that
$(G,\cdot,\circ)$ is a $\lambda$-anti-homomorphic skew brace, and therefore a
symmetric skew brace by Proposition 3.11. We have proved

######  Theorem 3.15.

Let $G=(G,\cdot)$ be a group and $\lambda:G\to\operatorname{Aut}\,G$ be an
anti-homomorphism such that
$a\lambda_{a}(b)a^{-1}b^{-1}\in\operatorname{Ker}\,\lambda$ for all $a,b\in
G$. Then $(G,\cdot,\circ)$ is a $\lambda$-anti-homomorphic skew brace, where
$a\circ b=a\lambda_{a}(b)$ for all $a,b\in G$.

It now follows that $\lambda:(G,\circ)\to\operatorname{Aut}\,(G,\cdot)$,
$a\mapsto\lambda_{a}$, is a homomorphism, where
$\lambda_{a}(b)=a^{-1}\cdot(a\circ b)$ for all $a,b\in(G,\circ)$. We first
show that $\lambda(G)$ is a subgroup of $\operatorname{Aut}\,(G,\circ)$. To
establish this, it is sufficient to show that $\lambda_{a}(b\circ
c)=\lambda_{a}(b)\circ\lambda_{a}(c)$ for all $a,b,c\in(G,\circ)$, which can
be easily proved by a direct computation using the fact that $\lambda$ is an
anti-homomorphism and $\lambda_{a^{-1}}=(\lambda_{a})^{-1}$ for all $a\in G$.

It now follows that $\lambda:(G,\circ)\to\operatorname{Aut}\,(G,\circ)$,
$a\mapsto\lambda_{a}$, is also a homomorphism. Notice that, as a map, this
homomorphism coincides with the map
$\lambda:(G,\cdot)\to\operatorname{Aut}\,(G,\cdot)$. Define a map
${\bar{\lambda}}:(G,\circ)\to\operatorname{Aut}\,(G,\circ)$,
$a\mapsto{\bar{\lambda}}_{a}$ by

${\bar{\lambda}}_{a}:=\lambda_{\bar{a}}=\lambda^{-1}_{a}\;\;\mbox{ for all
$a\in(G,\circ)$}.$

Then it easily follows that $\bar{\lambda}$ is an anti-homomorphism and

$\displaystyle{\bar{\lambda}}(a\circ{\bar{\lambda}}_{a}(b)\circ{\bar{a}}\circ{\bar{b}})$
$\displaystyle=$
$\displaystyle\lambda_{\overline{a\circ{\bar{\lambda}}_{a}(b)\circ{\bar{a}}\circ{\bar{b}}}}$
$\displaystyle=$ $\displaystyle\lambda_{b\circ
a\circ\overline{{\bar{\lambda}}_{a}(b)}\circ{\bar{a}}}$ $\displaystyle=$
$\displaystyle\lambda_{a^{-1}\lambda_{a^{-1}}(b^{-1})ab}$ $\displaystyle=$
$\displaystyle\operatorname{Id},$

since $a\lambda_{a}(b)a^{-1}b^{-1}\in\operatorname{Ker}\,\lambda$ for all
$a,b\in G$. So we can now apply Theorem 3.15 to get a symmetric skew brace
$(G,\circ,\circ_{1})$, where $\circ_{1}$ is defined as

$a\circ_{1}b:=a\circ{\bar{\lambda}}_{a}(b)\;\;\mbox{ for all
$a,b\in(G,\circ)$}.$

Notice that for $a,b\in(G,\circ_{1})$, we get

$a\circ_{1}b=a\circ{\bar{\lambda}}_{a}(b)=a\cdot\lambda({\bar{\lambda}}_{a}(b))=a\cdot
b.$

Hence $(G,\circ_{1})=(G,\cdot)$. We remark that further iteration will be a
repetition, and will give no new skew brace.

In the following result we find a criterion to obtain a (symmetric) skew brace
from two $\lambda$-anti-homomorphic skew braces with a common binary
operation. For two subgroups $H$ and $K$ of a group $G$, by $[H,K]$ we denote
the subgroup of $G$ generated by the set $\\{[h,k]\mid h\in H,k\in K\\}$.
Please do not get confused with the notation $[G,\lambda(G)]$ defined in
(4.4).

######  Lemma 3.16.

Let $(G,\cdot,\circ)$ and $(G,\cdot,\star)$ be two $\lambda$-anti-homomorphic
skew braces such that
$[\lambda^{\circ}(G,\circ),\lambda^{\star}(G,\star)]=\operatorname{Id}$ in
$\operatorname{Aut}\,(G,\cdot)$. Then $(G,\circ,\star)$ is a skew brace if and
only if (i)
$[(G,\cdot),\lambda^{\star}(G,\star)]\subseteq\operatorname{Ker}\,\lambda^{\circ}$.
Moreover, $(G,\circ,\star)$ is a symmetric skew brace if and only if in
addition to (i) we also have (ii)
$[(G,\cdot),\lambda^{\circ}(G,\circ)]\subseteq\operatorname{Ker}\,\lambda^{\star}$.

###### Proof.

Both $(G,\circ)$ and $(G,\star)$ are groups. For $a\in G$, let $\bar{a}$
denote the inverse of $a$ in $(G,\circ)$ and $\hat{a}$ that of $a$ in
$(G,\star)$. Obviously $(G,\circ,\star)$ is a skew brace if and only if

$a\star(b\circ c)=(a\star b)\circ{\bar{a}}\circ(a\star c)\;\;\mbox{for all
$a,b,c\in G$}.$

Compute both sides separately in the skew braces $(G,\cdot,\circ)$ and
$(G,\cdot,\star)$ simultaneously. First

$\displaystyle a\star(b\circ c)$ $\displaystyle=$ $\displaystyle
a\star(b\cdot\lambda^{\circ}_{b}(c))$ $\displaystyle=$ $\displaystyle
a\cdot\lambda^{\star}_{a}(b\cdot\lambda^{\circ}_{b}(c))$ $\displaystyle=$
$\displaystyle
a\cdot\lambda^{\star}_{a}(b)\cdot\lambda^{\star}_{a}(\lambda^{\circ}_{b}(c)).$

Next

$\displaystyle(a\star b)\circ{\bar{a}}\circ(a\star c)$ $\displaystyle=$
$\displaystyle(a\cdot\lambda^{\star}_{a}(b))\circ{\bar{a}}\circ(a\cdot\lambda^{\circ}_{a}(c))$
$\displaystyle=$
$\displaystyle(a\cdot\lambda^{\star}_{a}(b))\circ{\bar{a}}^{-1}\cdot(\bar{a}\circ\lambda^{\circ}_{a}(c))$
$\displaystyle=$
$\displaystyle(a\cdot\lambda^{\star}_{a}(b))\circ\lambda^{\circ}_{\bar{a}}(\lambda^{\star}_{a}(c))$
$\displaystyle=$ $\displaystyle
a\cdot\lambda^{\star}_{a}(b)\cdot\lambda^{\circ}_{a\cdot\lambda^{\star}_{a}(b)}(\lambda^{\circ}_{\bar{a}}(\lambda^{\star}_{a}(c)))$
$\displaystyle=$ $\displaystyle
a\cdot\lambda^{\star}_{a}(b)\cdot\lambda^{\circ}_{\lambda^{\star}_{a}(b)}(\lambda^{\star}_{a}(c)).$

Now $(G,\circ,\star)$ is a skew brace if and only if
$\lambda^{\star}_{a}(\lambda^{\circ}_{b}(c))=\lambda^{\circ}_{\lambda^{\star}_{a}(b)}(\lambda^{\star}_{a}(c))$
if and only if
$\lambda^{\circ}_{b^{-1}}(\lambda^{\star}_{a^{-1}}(\lambda^{\circ}_{\lambda^{\star}_{a}(b)}(\lambda^{\star}_{a}(c))))=c$,
for all $a,b,c\in G$. By the given hypothesis this is equivalent to
$\lambda^{\circ}_{\lambda^{\star}_{a}(b)b^{-1}}=\operatorname{Id}$ for all
$a,b\in G$, which is further equivalent to
$[(G,\cdot),\lambda^{\star}(G,\star)]\subseteq\operatorname{Ker}\,\lambda^{\circ}$.
This proves the first assertion. The second assertion holds by interchanging
$\lambda^{\star}_{a}$ and $\lambda^{\circ}_{a}$ for $a\in G$ in the above
procedure. ∎

Let us see an example.

######  Example 3.17.

Let $G=(G,\cdot)$ be any group of nilpotency class $3$. Let $(G,\cdot,\circ)$
be the $\lambda$-anti-homomorphic skew brace of Example 3.3. Obviously
$\operatorname{Ker}\,\lambda^{\circ}=\operatorname{Z}(G)$. Now we construct
another such skew brace. Let $\beta_{x}$ be an inner automorphism induced by a
fixed element $x\in G$, i.e., $\beta_{x}(a)=x^{-1}ax$ for all $a\in G$. Define
$\lambda^{\star}:G\to\operatorname{Aut}\,G$, $a\mapsto\lambda^{\star}_{a}$,
where

$\lambda^{\star}_{a}(b)=(a^{-1}\beta_{x}(a))^{-1}ba^{-1}\beta_{x}(a)\;\;\mbox{for
all $b\in G$}.$

An easy computation shows that $\lambda^{\star}$ is an anti-homomorphism.
Notice that for all $a,b\in G$, we have

$a\lambda^{\star}_{a}(b)a^{-1}b^{-1}=[a,b][b,a^{-1}\beta_{x}(a)]\in\gamma_{2}(G).$

Also notice that $u^{-1}\beta_{x}(u)\in\gamma_{3}(G)\leq\operatorname{Z}(G)$
for all $u\in\gamma_{2}(G)$. It now follows that
$a\lambda^{\star}_{a}(b)a^{-1}b^{-1}\in\operatorname{Ker}\,\lambda^{\star}$.
Hence $(G,\cdot,\star)$ is a $\lambda$-anti-homomorphic skew brace such that
$\gamma_{2}(G)\leq\operatorname{Ker}\,\lambda^{\star}$. It is now not
difficult to see that the hypotheses of Lemma 3.16 are satisfied, and hence
$(G,\circ,\star)$ is a symmetric skew brace.

We now restrict our attention to a special case of $\lambda$-anti-homomorphic
skew braces $(G,\cdot,\circ)$ in which $\lambda(G,\cdot)$ is an abelian
subgroup of $\operatorname{Aut}\,(G,\cdot)$. In this case $(G,\cdot,\circ)$
becomes a $\lambda$-homomorphic skew brace, a concept introduced and studied
in [4]. We take up this study in the next section.

##  4\. Brace systems through $\lambda$-homomorphic skew braces

In this section, we first recall the definition of a $\lambda$-homomorphic
skew brace and its connection with regular subgroups of the holomorph of its
additive group. We’ll then carry out detailed investigation of such skew
braces with $\lambda(G)$ abelian subgroup of $\operatorname{Aut}\,G$.

######  Definition 4.1.

A skew brace $(G,\cdot,\circ)$ is said to be a $\lambda$-homomorphic skew
brace if the map $\lambda:(G,\cdot)\to\operatorname{Aut}\,(G,\cdot)$, defined
by $\lambda_{a}(b)=a^{-1}(a\circ b)$, $a,b\in(G,\cdot)$, is a homomorphism.

Let $G$ be a group. The holomorph of $G$ is the group
$\operatorname{Hol}G:=\operatorname{Aut}G\ltimes G$, in which the product is
given by

$(f,a)(g,b)=(fg,af(b))$

for all $a,b\in G$ and $f,g\in\operatorname{Aut}G$. Any subgroup $H$ of
$\operatorname{Hol}G$ acts on $G$ as follows

$(f,a)\cdot b=af(b),\quad a,b\in G,\,\,f\in\operatorname{Aut}G.$

A subgroup $H$ of $\operatorname{Hol}G$ is said to be regular if the action of
$H$ on $G$ is free and transitive. It is equivalent to the fact that for each
$a\in G$ there exists a unique $(f,x)\in H$ such that $xf(1)=a$. Let
$\pi_{2}:\operatorname{Hol}G\to G$ denote the projection map from
$\operatorname{Hol}G$ onto the second component of $G$. Notice that
$\pi_{2}|_{H}$, the restriction of $\pi_{2}$ to $H$, is a bijection.

The following theorem provides a connection between skew braces and regular
subgroups.

######  Theorem 4.2.

[19, Theorem 4.2] Let $(G,\cdot,\circ)$ be a skew brace. Then
$\left\\{\,(\lambda_{a},a)\mid a\in G\,\right\\}$ is a regular subgroup of
$\operatorname{Hol}G$, where $\lambda_{a}(b)=a^{-1}(a\circ b)$ for all $b\in
G$.

Conversely, if $(G,\cdot)$ is a group and $H$ is a regular subgroup of
$\operatorname{Hol}\,(G,\cdot)$, then $(G,\cdot,\circ)$ is a skew brace such
that $(G,\circ)\cong H$, where $a\circ b=af(b)$ with
$(\pi_{2}|_{H})^{-1}(a)=(f,a)\in H$.

A sufficient condition, in view of Theorem 4.2, on a group $G=(G,\cdot)$ so
that one can get a $\lambda$-homomorphic skew brace $(G,\cdot,\circ)$ is given
in the following result ([4, Theorem 2.3]):

######  Theorem 4.3.

Let $G$ be a group, $\lambda:G\to\mathrm{Aut}\,G$, $a\mapsto\lambda_{a}$, be a
homomorphism of $G$ into the group of its automorphisms. The set

$H_{\lambda}:=\left\\{\,(\lambda_{a},a)\,|\,a\in G\,\right\\}$

is a subgroup of $\mathrm{Hol}\,G=\mathrm{Aut}\,G\ltimes G$ if and only if

(4.4) $[G,\lambda(G)]:=\\{\,b^{-1}\lambda_{a}(b)\mid a,b\in
G\,\\}\subseteq\mathrm{Ker}\,\lambda.$

Moreover, if $H_{\lambda}$ is a subgroup, then it is regular, and therefore by
Theorem 4.2 we get a skew brace $(G,\cdot,\circ)$, where ‘$\circ$’ is defined
by $a\circ b=a\lambda_{a}(b)$.

We remark that this construction can also be carried out directly on the lines
of the construction of $\lambda$-anti-homomorphic skew braces. More precisely,
without getting into the regular subgroup business, we can directly prove

######  Theorem 4.5.

Let $G=(G,\cdot)$ be a group and $\lambda:G\to\operatorname{Aut}\,G$ be a
homomorphism such that $[G,\lambda(G)]\subseteq\operatorname{Ker}\,\lambda$.
Then $(G,\cdot,\circ)$ is a $\lambda$-homomorphic skew brace, where $a\circ
b=a\lambda_{a}(b)$ for all $a,b\in G$.

As a consequence we get

######  Corollary 4.6.

If $\lambda(G)$ is an abelian subgroup of $\operatorname{Aut}\,G$, then the
skew brace $(G,\cdot,\circ)$ constructed in the preceding theorem is
symmetric. More precisely, any $\lambda$-homomorphic skew brace
$(G,\cdot,\circ)$ with $\lambda(G)$ abelian is symmetric.

###### Proof.

Since $\lambda(G)$ is an abelian, $(G,\cdot,\circ)$ is a $\lambda$-anti-
homomorphic skew brace. Assertion now follows from Proposition 3.11. ∎

We take a pause to present examples of a rooted brace system.

######  Example 4.7.

With the preceding setup we can have the following constructions:

  1. (1)

Let $(G,\cdot)$ be a group and $\mathcal{H}$ the set of regular subgroups of
$\operatorname{Hol}\,G$. Then for any regular subgroup $H\in\mathcal{H}$ there
exists the operation $\circ_{H}$ such that $(G,\cdot,\circ_{H})$ is a skew
left brace. Hence we have a rooted brace system $\mathcal{B}_{\Gamma}(G)$ with
the graph $\Gamma=(V,E)$, where
$V=\\{v_{0}\\}\cup\\{\circ_{H}\\}_{H\in\mathcal{H}}$ and
$E=\\{v_{0}\\}\times\\{\circ_{H}\\}_{H\in\mathcal{H}}$.

  2. (2)

By applying the construction as in (1) to the group $(G,\circ_{H})$ for all
$H\in\mathcal{H}$ and continuing this process, we can construct another rooted
brace system.

Coming back, notice that the skew braces constructed through Theorem 4.3 are
all $\lambda$-homomorphic. A variety of examples of such skew braces were
constructed in [4]. Let $(G,\cdot,\circ)$ be a $\lambda$-homomorphic skew
brace. Then $\lambda:(G,\cdot)\to\operatorname{Aut}\,(G,\cdot)$,
$a\mapsto\lambda_{a}$, is also a homomorphism. Hence for all $a,b,c\in G$,

$\displaystyle\lambda_{a\circ b}(c)$ $\displaystyle=$
$\displaystyle\lambda_{a}(\lambda_{b}(c))$ $\displaystyle=$
$\displaystyle\lambda_{a\cdot b}(c)\;\;\mbox{ (by $\lambda$-homomorphic
property)}.$

Hence

(4.8) $\lambda_{a\circ b}=\lambda_{a\cdot b}\;\;\mbox{ for all $a,b\in G$}.$

As a consequence, we immediately get

$[(G,\cdot),\lambda(G,\cdot)]\subseteq\mathrm{Ker}\,\lambda.$

Now onwards we concentrate on the case when $\lambda(G,\cdot)$ is an abelian
subgroup of $\operatorname{Aut}\,(G,\cdot)$. We remark that in this special
situation the concepts of $\lambda$-homomorphic and $\lambda$-anti-homomorphic
skew braces coincide. The following lemma is the base for a recursive process
of constructing skew braces.

######  Lemma 4.9.

Let $(G,\cdot,\circ)$ be a $\lambda$-homomorphic skew brace such that
$\lambda(G,\cdot)$ is an abelian subgroup of $\operatorname{Aut}\,(G,\cdot)$.
Then the homomorphism $\lambda$ is also a homomorphism from $(G,\circ)$ to
$\operatorname{Aut}\,(G,\circ)$. Moreover $\lambda(G,\circ)$ is an abelian
subgroup of $\operatorname{Aut}\,(G,\circ)$ and
$[(G,\circ),\lambda(G,\circ)]\subseteq\mathrm{Ker}\,\lambda$.

###### Proof.

Since $\lambda(G,\cdot)$ is an abelian subgroup of
$\operatorname{Aut}\,(G,\cdot)$, obviously $(G,\cdot,\circ)$ is also a
$\lambda$-anti-homomorphic skew brace. It now follows from the preceding
section (discussion after Theorem 3.15) that $\lambda(G)$ is a subgroup of
$\operatorname{Aut}\,(G,\circ)$. Thus the homomorphism
$\lambda:(G,\circ)\to\operatorname{Aut}\,(G,\cdot)$ is also a homomorphism
from $(G,\circ)$ to $\operatorname{Aut}\,(G,\circ)$ with the image
$\lambda(G,\circ)$.

Obviously $\lambda(G,\circ)$ is abelian. Now using (4.8) and the given
hypothesis, we get

$\lambda({\bar{b}}\circ\lambda_{a}(b))=\lambda_{{\bar{b}}\circ\lambda_{a}(b)}=\lambda^{-1}_{b}\lambda_{\lambda_{a}(b)}=\lambda_{b^{-1}\lambda_{a}(b)}=\operatorname{Id}$

for all $a,b\in(G,\circ)$. Hence
$[(G,\circ),\lambda(G,\circ)]\subseteq\mathrm{Ker}\,\lambda$, which completes
the proof. ∎

We are now going to construct a symmetric linear brace system starting with a
$\lambda$-homomorphic skew brace $(G,\cdot,\circ)$ such that
$\lambda(G,\cdot)$ is an abelian subgroup of $\operatorname{Aut}\,(G,\cdot)$,
and iteratively applying Lemma 4.9. Let us denote $\cdot$ and $\circ$ by
$\circ_{0}$ and $\circ_{1}$ respectively. Notice that
$(G,\circ_{0},\circ_{1})$ is a symmetric skew brace (Proposition 3.11). By
Lemma 4.9 we know that
$\lambda:(G,\circ_{1})\to\operatorname{Aut}\,(G,\circ_{1})$ is a homomorphism,
$\lambda(G,\circ_{1})$ is abelian and
$[(G,\circ_{1}),\lambda(G,\circ_{1})]\subseteq\mathrm{Ker}\,\lambda$. Then by
Theorem 4.3 we get a $\lambda$-homomorphic skew brace
$(G,\circ_{1},\circ_{2})$ such that for $a,b\in G$,

$a\circ_{2}b:=a\circ_{1}\lambda_{a}(b).$

Again using Proposition 3.11, we contend that it is symmetric. By this
iterative process, we can construct, for any integer $i\geq 1$, a symmetric
$\lambda$-homomorphic skew brace $(G,\circ_{i},\circ_{i+1})$ such that for
$a,b\in G$,

$a\circ_{i+1}b:=a\circ_{i}\lambda_{a}(b).$

We now prove

######  Lemma 4.10.

In the above construction $\lambda$ is a homomorphism and
$\operatorname{Ker}\,\lambda$, as a subset of $G$, is the same at all levels.
More precisely, $\operatorname{Ker}\,\lambda$ in $(G,\circ_{i})$ is equal to
$\operatorname{Ker}\,\lambda$ in $(G,\circ_{0})$ for all $i$. Also
$\lambda(G,\circ_{i})=\lambda(G,\circ_{0})$ for all $i$.

###### Proof.

It is pertinent, by the very construction, that the following diagram is
commutative.

${(G,\circ_{0})}$${\operatorname{Aut}\,(G,\circ_{0})}$${(G,\circ_{1})}$${\operatorname{Aut}\,(G,\circ_{1})}$${(G,\circ_{2})}$${\operatorname{Aut}\,(G,\circ_{2})}$${(G,\circ_{3})}$${\operatorname{Aut}\,(G,\circ_{3})}$${\vdots}$${\vdots,}$$\scriptstyle{\lambda}$$\scriptstyle{\operatorname{Id}}$$\scriptstyle{\lambda^{\circ_{1}}}$$\scriptstyle{\lambda}$$\scriptstyle{\operatorname{Id}}$$\scriptstyle{\lambda^{\circ_{2}}}$$\scriptstyle{\lambda}$$\scriptstyle{\operatorname{Id}}$$\scriptstyle{\lambda^{\circ_{3}}}$$\scriptstyle{\lambda}$

where $\operatorname{Id}$ is the identity map on the level of permutations on
$G$ and $\lambda$ is a homomorphism at each level. We start with the fact that
$\lambda$ with respect to $\circ_{0}$ and $\lambda^{\circ_{1}}$ are equal at
element level. Hence
$\operatorname{Ker}\,\lambda^{\circ_{1}}=\operatorname{Ker}\,\lambda$. Now by
the commutativity of the uppermost triangle, we have that the homomorphisms
$\lambda^{\circ_{1}}$ and
$\lambda:(G,\circ_{1})\to\operatorname{Aut}\,(G,\circ_{1})$ are equal. Thus
the kernels of $\lambda:(G,\circ_{0})\to\operatorname{Aut}\,(G,\circ_{0})$ and
$\lambda:(G,\circ_{1})\to\operatorname{Aut}\,(G,\circ_{1})$ are the same as
subsets of $G$. It also follows that
$\lambda(G,\circ_{1})=\lambda(G,\circ_{0})$. A straightforward iterative
argument now completes the proof. ∎

######  Lemma 4.11.

If the exponent of $\lambda(G,\circ_{0})$ is $n$, then $\circ_{n}=\circ_{0}$.

###### Proof.

For $a,b\in G$, we have

$\displaystyle a\circ_{n}b$ $\displaystyle=$ $\displaystyle
a\circ_{n-1}\lambda_{a}(b)$ $\displaystyle=$ $\displaystyle
a\circ_{n-2}\lambda^{2}_{a}(b)$ $\displaystyle\vdots$ $\displaystyle=$
$\displaystyle a\circ_{0}\lambda^{n}_{a}(b)$ $\displaystyle=$ $\displaystyle
a\circ_{0}b.$

Since this is true for all $a,b\in G$, the assertion holds. ∎

######  Lemma 4.12.

Let $0\leq i<j$ be integers. Then $(G,\circ_{i},\circ_{j})$ is a
$\lambda$-homomorphic skew brace with $\lambda(G,\circ_{j})$ is abelian. Hence
$(G,\circ_{i},\circ_{j})$ is a symmetric skew brace.

###### Proof.

For $a,b\in(G,\circ_{j})$, we have

$\displaystyle a\circ_{j}b$ $\displaystyle=$ $\displaystyle
a\circ_{j-1}\lambda_{a}(b)$ $\displaystyle=$ $\displaystyle
a\circ_{j-2}\lambda^{2}_{a}(b)$ $\displaystyle\vdots$ $\displaystyle=$
$\displaystyle a\circ_{i}\lambda^{j-i}_{a}(b).$

As we have seen above
$\lambda:(G,\circ_{i})\to\operatorname{Aut}\,(G,\circ_{i})$ is a homomorphism
which is equal, at the element level, to the homomorphism
$\lambda:(G,\circ_{0})\to\operatorname{Aut}\,(G,\circ_{0})$. Set $j-i=k$. Now
consider the map
$\lambda^{k}:(G,\circ_{i})\to\operatorname{Aut}\,(G,\circ_{i})$,
$a\mapsto\lambda^{k}_{a}$, where $\lambda^{k}_{a}=(\lambda_{a})^{k}$. Since
$\lambda(G,\circ_{i})$ is abelian, it follows that $\lambda^{k}$ is a
homomorphism. We claim that
$[(G,\circ_{i}),\lambda^{k}(G,\circ_{i})]\leq\operatorname{Ker}\,\lambda^{k}$.
We already know that
$[(G,\circ_{i}),\lambda(G,\circ_{i})]\leq\operatorname{Ker}\,\lambda$. So, for
$a,b\in G$, we have

$\displaystyle\lambda_{b^{\circ_{i}(-1)}\circ_{i}\lambda^{k}_{a}(b)}$
$\displaystyle=$
$\displaystyle\lambda_{b^{\circ_{i}(-1)}\circ_{i}\lambda^{k-1}_{a}(b)\circ_{i}\big{(}(\lambda^{k-1}_{a}(b))^{\circ_{i}(-1)}\circ_{i}\lambda_{a}(\lambda^{k-1}_{a}(b))\big{)}}$
$\displaystyle=$
$\displaystyle\lambda_{b^{\circ_{i}(-1)}\circ_{i}\lambda^{k-1}_{a}(b)}$
$\displaystyle\vdots$ $\displaystyle=$
$\displaystyle\lambda_{b^{\circ_{i}(-1)}\circ_{i}\lambda^{1}_{a}(b)}=\lambda_{b^{\circ_{i}(-1)}\circ_{i}\lambda_{a}(b)}$
$\displaystyle=$ $\displaystyle\operatorname{Id}.$

Hence
$\lambda^{k}_{b^{\circ_{i}(-1)}\circ_{i}\lambda^{k}_{a}(b)}=(\lambda_{b^{\circ_{i}(-1)}\circ_{i}\lambda^{k}_{a}(b)})^{k}=\operatorname{Id}$.
The proof now follows by first applying Theorem 4.3 and then Proposition 3.11.
∎

######  Remark 4.13.

Let $(G,\cdot,\circ)$ be a $\lambda$-homomorphic skew brace such that
$\lambda(G,\cdot)$ is an abelian subgroup of $\operatorname{Aut}\,(G,\cdot)$.
Then $(G,\circ,\cdot)$ is a skew brace. Let
$\xi:(G,\cdot)\to\operatorname{Aut}(G,\circ)$, $a\mapsto\xi_{a}$, be the
corresponding map, where $\xi_{a}(b)={\bar{a}}\circ(a\cdot b)$ for all $b\in
G$. Notice that $\xi_{a}=\lambda^{-1}_{a}$ for all $a\in G$. Hence $\xi$ is a
homomorphism and therefore, $(G,\circ,\cdot)$ is a $\xi$-homomorphic skew
brace with $\xi(G,\circ)$ abelian in $\operatorname{Aut}(G,\circ)$. In the
above notation $(G,\circ,\cdot)=(G,\circ_{1},\circ_{0})$. Define a binary
operation $\circ_{-1}$ on $G$ as follows:

$a\circ_{-1}b=a\circ_{0}\xi_{a}(b)\;\;\;(=a\circ_{0}\lambda^{-1}_{a}(b))$

for all $a,b\in G$. Then it follows that $(G,\circ_{0},\circ_{-1})$ is a
symmetric skew brace. For any integer $n\geq 1$, iteratively define

$a\circ_{-n}b=a\circ_{-n-1}\xi_{a}(b)$

for all $a,b\in G$. As above, we see that $(G,\circ_{-i},\circ_{-j})$ is a
symmetric skew brace for all $1\leq i<j$.

We have proved

######  Theorem 4.14.

Let $G$ be a group, $\lambda:G\to\operatorname{Aut}\,G$,
$a\mapsto\lambda_{a}$, be a homomorphism such that $\lambda(G)$ is abelian and
$[G,\lambda(G)]\subseteq\operatorname{Ker}\,\lambda$. Then there exists a
symmetric linear brace system $\mathcal{B}_{I}(G):=(G,\\{\circ_{i}\\}_{i\in
I})$ on $G$, $I=\\{0,\ldots,n-1\\}$, if the exponent of $\lambda(G)$ is $n$.
Otherwise $\mathcal{B}_{I}(G):=(G,\\{\circ_{i}\\}_{i\in I})$, $I=\mathbb{Z}$,
is a symmetric linear brace system on $G$ of infinite size.

Since a $\lambda$-homomorphic skew brace $(G,\cdot,\circ)$ with abelian
$\lambda(G,\cdot)$ is also a $\lambda$-anti-homomorphic skew brace, we can
state Proposition 3.16 as follows.

######  Proposition 4.15.

Let $(G,\cdot,\circ)$ and $(G,\cdot,\star)$ be two $\lambda$-homomorphic skew
braces with abelian $\lambda^{\circ}(G,\circ)$ and $\lambda^{\star}(G,\star)$
such that
$[\lambda^{\circ}(G,\circ),\lambda^{\star}(G,\star)]=\operatorname{Id}$ in
$\operatorname{Aut}\,(G,\cdot)$. Then $(G,\circ,\star)$ is a skew brace if and
only if (i)
$[(G,\cdot),\lambda^{\star}(G,\star)]\subseteq\operatorname{Ker}\,\lambda^{\circ}$.
Moreover, $(G,\circ,\star)$ is a symmetric skew brace if and only if in
addition to (i) we also have (ii)
$[(G,\cdot),\lambda^{\circ}(G,\circ)]\subseteq\operatorname{Ker}\,\lambda^{\star}$.

Let us now see connection between different symmetric brace system constructed
through Theorem 4.14. When the group $G$ and the linearly ordered set $I$ are
clear from the context, for simplicity, we only write $\mathcal{B}^{\circ}$
for the symmetric linear brace system
$\mathcal{B}_{I}(G):=(G,\\{\circ_{i}\\}_{i\in I})$.

######  Theorem 4.16.

Let $\mathcal{B}^{\circ}:=(G,\\{\circ_{i}\\}_{i\in I})$ and
$\mathcal{B}^{\star}:=(G,\\{\star_{i}\\}_{i\in I})$ be symmetric linear brace
systems on $G$ built on homomorphisms
$\lambda^{\circ}:(G,\circ_{0})\to\operatorname{Aut}\,(G,\circ_{0})$ and
$\lambda^{\star}:(G,\star_{0})\to\operatorname{Aut}\,(G,\star_{0})$
respectively through Theorem 4.14, where $G=(G,\circ_{0})=(G,\star_{0})$. Let
the hypotheses of Proposition 4.15 be satisfied for the skew braces
$(G,\circ_{0},\circ_{1})$ and $(G,\star_{0},\star_{1})$. Then
$\mathcal{B}^{\circ}\cup\mathcal{B}^{\star}$ is a symmetric brace system.

###### Proof.

For given $i,j\geq 1$, consider the groups $(G,\circ_{i})$ and
$(G,\star_{j})$. We know that $(G,\circ_{0},\circ_{i})$ and
$(G,\star_{0},\star_{j})$ are symmetric skew braces. Let $\bar{\lambda}_{i}$
and $\tilde{\lambda}_{j}$ denote the homomorphisms
$(G,\circ_{i})\to\operatorname{Aut}\,(G,\circ_{0})$ and
$(G,\star_{j})\to\operatorname{Aut}\,(G,\star_{0})$ respectively. We know that
$\bar{\lambda}_{i}(a)=(\lambda^{\circ}_{a})^{i}$ and
$\tilde{\lambda}_{j}(a)=(\lambda^{\star}_{a})^{j}$ for all $a\in G$. It is now
clear, using the given hypotheses, that
$\bar{\lambda}_{i}:(G,\circ_{0})\to\operatorname{Aut}\,(G,\circ_{0})$,
$\tilde{\lambda}_{j}:(G,\star_{0})\to\operatorname{Aut}\,(G,\star_{0})$ are
homomorphisms and the hypotheses of Proposition 4.15 are satisfied for the
braces $(G,\circ_{0},\circ_{i})$ and $(G,\star_{0},\star_{j})$. Hence
$(G,\circ_{i},\star_{j})$ is a symmetric skew brace. Since $i,j\geq 1$ are
arbitrary, it follows that $\mathcal{B}^{\circ}\cup\mathcal{B}^{\star}$ is a
symmetric brace system. ∎

We now present a generalization of Proposition 4.15. Let $I_{1}$ and $I_{2}$
be two linearly ordered set and $I^{*}_{k}=I_{k}\cup\\{0\\}$, $k=1,2$. Let
$\mathcal{B}_{I^{*}_{1}}(G)=(G,\\{\circ_{i}\\}_{i\in I^{*}_{1}})$ and
$\mathcal{B}_{I^{*}_{2}}(G)=(G,\\{\star_{i}\\}_{i\in I^{*}_{2}})$ be two
linear brace systems on $G$. Let $(G,\cdot,\circ_{i})$ and
$(G,\cdot,\star_{j})$ for some $i\in I_{1}$, $j\in I_{2}$ be two skew braces,
where $\cdot=\circ_{0}=\star_{0}$. Let
$\lambda:(G,\circ_{i})\to\operatorname{Aut}\,(G,\cdot)$ and
$\mu:(G,\star_{j})\to\operatorname{Aut}\,(G,\cdot)$ be the corresponding
homomorphisms. With this setup, we have

######  Lemma 4.17.

The algebraic system $(G,\circ_{i},\circ_{j})$ is a skew brace if

$\mu_{a}(\lambda_{b}(c))=\lambda_{a\mu_{a}(b)}(\lambda_{a}^{-1}(a^{-1}))\left(\lambda_{a\mu_{a}(b)\lambda_{a\mu_{a}(b)}(\lambda_{a}^{-1}(a^{-1}))}(a\mu_{a}(c))\right)$

for any $a,b,c\in G$.

###### Proof.

Obviously $(G,\circ_{i})$ and $(G,\circ_{j})$ are groups. We only weed to
check the compatibility condition of the skew brace:

$a\circ_{j}(b\circ_{i}c)=(a\circ_{j}b)\circ_{i}a^{\circ_{i}(-1)}\circ_{i}(a\circ_{j}c),~{}~{}~{}a,b,c\in
G,$

where $a^{\circ_{i}(-1)}$ is the inverse to $a$ in the group $(G,\circ_{i})$.
Computing the left side, we get

$a\circ_{j}(b\circ_{i}c)=a\mu_{a}(b\lambda_{b}(c))=a\mu_{a}(b)\mu_{a}(\lambda_{b}(c)),$

where we used the fact that $\mu_{a}$ is an automorphism of $(G,\cdot)$.

To compute the right side, notice that the inverse of $a$ in $(G,\circ_{i})$
is $a^{\circ_{i}(-1)}=\lambda_{a}^{-1}(a^{-1})$, where $a^{-1}$ is the inverse
of $a$ in $(G,\cdot)$. Then the right side is equal to

$(a\circ_{j}b)\circ_{i}a^{\circ_{i}(-1)}\circ_{i}(a\circ_{j}c)=a\mu_{a}(b)\lambda_{a\mu_{a}(b)}(\lambda_{a}^{-1}(a^{-1}))\left(\lambda_{a\mu_{a}(b)\lambda_{a\mu_{a}(b)}(\lambda_{a}^{-1}(a^{-1}))}(a\mu_{a}(c))\right).$

Comparing the left and the right sides, we get the required compatibility
condition. ∎

##  5\. A unification for known symmetric brace systems

In this section we present a unification of most of the constructions of
symmetric skew braces available in the literature. The concept was introduced
by L.N. Childs in 2019 ([16]). Alan Koch [23, 24] constructed symmetric brace
systems on a given group $G$ using abelian endomorphisms of $G$. This
construction was generalized by A. Caranti and L. Stefanello [10, 11]. We show
that all theses symmetric skew braces are $\lambda$-homomorphic.

Let $G=(G,\cdot)$ be a group. Let $\bar{A}$ be an abelian subgroup of
$\bar{G}:=G/\operatorname{Z}(G)$. The inverse image of $\bar{A}$ in $G$ is
$A$. Let $\operatorname{Map}(G,A)$ denote the set of all mappings from $G$
into $A$. Define

$\mathcal{A}:=\\{f\in\operatorname{Map}(G,A)\mid f\;\;\mbox{induces an
endomorphism from $\bar{G}$ into $\bar{A}$}\\}.$

We say that $f_{1},f_{2}\in\mathcal{A}$ are related if both of these induce
the same endomorphism $\bar{G}$ into $\bar{A}$. This is an equivalence
relation. Let $\bar{\mathcal{A}}$ denote the set of all equivalence classes of
the elements of $\mathcal{A}$. The equivalence class of an element
$f\in\mathcal{A}$ will be denoted by $[f]$. For a group $G:=(G,\cdot)$, we
define

$\mathcal{A}^{\prime}:=\\{\alpha:G\times
G\to\operatorname{Z}(G)\mid\alpha\mbox{ is bilinear and
}\alpha(\operatorname{Z}(G),G)=\alpha(G,\operatorname{Z}(G))=e\\},$

where $e\in G$ is the identity element. It follows from the definition that
$\alpha(\gamma_{2}(G),G)=\alpha(G,\gamma_{2}(G))=e$.

Let $[f]\in\bar{\mathcal{A}}$ and $\alpha\in\mathcal{A}^{\prime}$. Define a
map $\lambda:G\to\operatorname{Aut}\,G$, $a\mapsto\lambda_{a}$, where

$\lambda_{a}(b)=f(a)^{-\epsilon}bf(a)^{\epsilon}\,\alpha(a,b)$

with $\epsilon=\pm 1$. We claim that $\lambda$ is a homomorphism and
$\lambda(G)$ is an abelian subgroup of $\operatorname{Aut}\,G$. For $a,c\in
G$, we know that $f(ac)=f(a)f(c)$ modulo $\operatorname{Z}(G)$. Thus
$f(ac)=f(a)f(c)z$ for some $z\in\operatorname{Z}(G)$. Let $b\in G$. Since
$f(G)$ is abelian, we have

$\displaystyle\lambda_{ac}(b)$ $\displaystyle=$ $\displaystyle
f(a)^{-\epsilon}f(c)^{-\epsilon}z^{-\epsilon}bz^{\epsilon}f(c)^{\epsilon}f(a)^{\epsilon}\,\alpha(ac,b)$
$\displaystyle=$ $\displaystyle
f(a)^{-\epsilon}f(c)^{-\epsilon}bf(c)^{\epsilon}\,\alpha(c,b)f(a)^{\epsilon}\,\alpha(a,b)$
$\displaystyle=$ $\displaystyle
f(a)^{-\epsilon}\lambda_{c}(b)f(a)^{\epsilon}\,\alpha(a,b)$ $\displaystyle=$
$\displaystyle
f(a)^{-\epsilon}\lambda_{c}(b)f(a)^{\epsilon}\,\alpha(a,\lambda_{c}(b))$
$\displaystyle=$ $\displaystyle\lambda_{a}(\lambda_{c}(b)).$

This proves that $\lambda_{ac}=\lambda_{a}\lambda_{c}$. That $\lambda(G)$ is
abelian follows from the fact that $f(G)$ is abelian and
$\alpha(G,G)\subseteq\operatorname{Z}(G)$. This settles our claim.

Notice that $\gamma_{2}(G)\operatorname{Z}(G)\leq\operatorname{Ker}\,\lambda$
and $[G,\lambda(G)]\subseteq\gamma_{2}(G)\operatorname{Z}(G)$. Corollary 4.6
now gives

######  Proposition 5.1.

$(G,\cdot,\circ)$ is a symmetric skew brace, where $a\circ b=a\lambda_{a}(b)$
for all $a,b\in G$.

For each pair $([f],\alpha)\in(\bar{\mathcal{A}},\mathcal{A}^{\prime})$, by
Theorem 4.14 we get a symmetric linear brace system, which we denote by
$\mathcal{B}_{(f,\alpha)}$. Since any two representatives of the class $[f]$
give rise to the same inner automorphism $\lambda_{a}$ for each $a\in G$, it
follows that $\mathcal{B}_{(f,\alpha)}$ is well defined. We now prove

######  Theorem 5.2.

Let $G$ be a group and $A^{\prime}$ its subgroup such that
$A^{\prime}\operatorname{Z}(G)/\operatorname{Z}(G)$ is abelian. Let
$A:=A^{\prime}\operatorname{Z}(G)$ and the corresponding $\bar{\mathcal{A}}$
be as defined above. Then

$\bigcup_{([f],\alpha)\in(\bar{\mathcal{A}},\mathcal{A}^{\prime})}\mathcal{B}_{(f,\alpha)}$

is a symmetric brace system.

###### Proof.

Let $[f_{1}],[f_{2}]\in\bar{\mathcal{A}}$ be any two equivalence classes,
$\alpha_{1},\alpha_{2}\in\mathcal{A}^{\prime}$ and
$\mathcal{B}_{(f_{i},\alpha_{i})}$, $i=1,2$, be the corresponding symmetric
brace systems. It suffices to show that
$\mathcal{B}_{(f_{1},\alpha_{1})}\cup\mathcal{B}_{(f_{2},\alpha_{2})}$ is a
symmetric brace system. If
$\mathcal{B}_{(f_{1},\alpha_{1})}\subseteq\mathcal{B}_{(f_{2},\alpha_{2})}$ or
$\mathcal{B}_{(f_{2},\alpha_{2})}\subseteq\mathcal{B}_{(f_{1},\alpha_{1})}$,
then there is nothing to prove. So assume that the remaining case holds. To
differentiate, assume that
${}_{(f_{i},\alpha_{i})}\lambda:G\to\operatorname{Aut}\,G$ are the
homomorphisms induced by $(f_{i},\alpha_{i})$, $i=1,2$. Let
$(G,\cdot,\circ_{(f_{1},\alpha_{1})})$ be the first skew brace in
$\mathcal{B}_{(f_{1},\alpha_{1})}$ and $(G,\cdot,\circ_{(f_{2},\alpha_{2})})$
be the first skew brace in $\mathcal{B}_{(f_{2},\alpha_{2})}$. As we know that
$\gamma_{2}(G)\leq\operatorname{Ker}\,_{(f_{i},\alpha_{i})}\lambda$ and
$[G,\,_{(f_{i},\alpha_{i})}\lambda(G)]\subseteq\gamma_{2}(G)$ for $i=1,2$. It
is now easy to see that the hypotheses of Theorem 4.16 are satisfied. Hence
$\mathcal{B}_{(f_{1},\alpha_{1})}\cup\mathcal{B}_{(f_{2},\alpha_{2})}$ is a
symmetric brace system, and the proof is complete. ∎

##  6\. Further properties

In the present section we record various properties on the skew braces
considered in previous sections. Let $(G,\cdot,\circ)$ be a
$\lambda$-homomorphic skew brace such that
$b^{-1}\lambda_{a}(b)\in\operatorname{Z}(G,\cdot)$ for $a,b\in G$. Then it
easily follows that $\lambda_{a}(b)b^{-1}\in\operatorname{Z}(G,\cdot)$.

######  Proposition 6.1.

Let $(G,\cdot,\circ)$ be a $\lambda$-homomorphic skew brace with $\lambda(G)$
abelian. If $[G,\lambda(G)]\subseteq\operatorname{Z}(G)$, then
$(G,\cdot,\circ)$ is a two-sided skew brace.

###### Proof.

Let $(G,\cdot,\circ)$ be a skew braces as given in the hypothesis. We need to
establish the relation

$(a\cdot b)\circ c=(a\circ c)\cdot c^{-1}\circ(b\circ c)$

for all $a,b,c\in G$. Since $G$ is $\lambda$-homomorphic, we have

$\displaystyle(a\cdot b)\circ c$ $\displaystyle=$ $\displaystyle(a\cdot
b)\cdot\lambda_{a\cdot b}(c)$ $\displaystyle=$ $\displaystyle(a\cdot
b)\cdot\lambda_{a}(\lambda_{b}(c))$ $\displaystyle=$ $\displaystyle(a\cdot
b)\cdot\lambda_{a}(c\cdot c^{-1}\cdot\lambda_{b}(c))$ $\displaystyle=$
$\displaystyle(a\cdot
b)\cdot\lambda_{a}(c)\cdot\lambda_{a}(c^{-1}\cdot\lambda_{b}(c))$
$\displaystyle=$ $\displaystyle(a\cdot b)\cdot\lambda_{a}(c)\cdot
c^{-1}\cdot\lambda_{b}(c)$ $\displaystyle=$ $\displaystyle a\cdot
b\cdot(\lambda_{a}(c)\cdot c^{-1})\cdot b^{-1}\cdot(b\cdot\lambda_{b}(c))$
$\displaystyle=$ $\displaystyle(a\cdot\lambda_{a}(c))\cdot
c^{-1}\cdot(b\cdot\lambda_{b}(c))$ $\displaystyle=$ $\displaystyle(a\circ
c)\cdot c^{-1}\cdot(b\circ c).$

The proof is now complete. ∎

Let us see some examples. Let $(G,\cdot)$ be a group of nilpotency class $2$
and $(G,\cdot,\circ)$ be any skew brace considered in Section 5. Then
$[(G,\cdot),\lambda(G,\cdot)]\subseteq\operatorname{Ker}\,\lambda\leq\operatorname{Z}(G)$.
Hence all these skew braces are two-sided. One can then construct radical
rings from these two-sided skew braces.

We now construct symmetric skew braces from $\lambda$-homomorphic skew braces.
For that, we first recall the definition of opposite skew brace introduced in
[33, Section 3] and further studied in [25].

######  Definition 6.2.

Let $(G,\cdot,\circ)$ be a skew left brace. Define $\cdot^{op}:G\times G\to G$
by $a\cdot^{op}b=b\cdot a$ for all $a,b\in G$. Then $(G,\cdot^{op})$ is a
group and $(G,\cdot^{op},\circ)$ is a skew left brace, which is called the
opposite skew left brace of the given skew left brace $(G,\cdot,\circ)$.

If $\circ=\cdot^{op}$, then $(G,\cdot^{op},\circ)$ is a trivial skew brace. So
we are mainly interested in non-trivial opposite skew braces;
$\circ\neq\cdot^{op}$.

######  Proposition 6.3.

Let $(G,\cdot,\circ)$ be a $\lambda$-homomorphic skew left brace. Then the
opposite skew brace $(G,\cdot^{op},\circ)$ is symmetric if and only if
$\operatorname{Inn}\,(G,\cdot)$ centralizes $\lambda\,(G,\cdot)$.

###### Proof.

Since $(G,\cdot^{op},\circ)$ is a skew left brace, there exists a homomorphism
$\lambda^{\prime}:(G,\circ)\to\operatorname{Aut}\,(G,\cdot^{op})$,
$a\mapsto\lambda^{\prime}_{a}$, where
$\lambda^{\prime}_{a}(b)=a^{-1}\cdot^{op}(a\circ b)$. We know that
$\lambda:(G,\cdot)\to\operatorname{Aut}\,(G,\cdot)$, $a\mapsto\lambda_{a}$ is
a homomorphism, where $\lambda_{a}(b)=a^{-1}\cdot(a\circ b)$. Notice that

$\lambda^{\prime}_{a}(b)=a^{-1}\cdot^{op}(a\circ b)=(a\circ b)\cdot
a^{-1}=\iota_{a}(\lambda_{a}(b)),$

where $\iota_{a}(x)=a\cdot x\cdot a^{-1}$ for $x\in G$. Assume that
$\operatorname{Inn}\,(G,\cdot)$ centralizes $\lambda\,(G,\cdot)$. Now for
$a,b,c\in(G,\cdot^{op})$, we have

$\displaystyle\lambda^{\prime}_{a\cdot^{op}b}(c)$ $\displaystyle=$
$\displaystyle\iota_{a\cdot^{op}b}(\lambda_{a\cdot^{op}b}(c))$
$\displaystyle=$ $\displaystyle\iota_{b\cdot a}(\lambda_{b\cdot a}(c))$
$\displaystyle=$
$\displaystyle\iota_{b}(\iota_{a}(\lambda_{b}(\lambda_{a}(c))))$
$\displaystyle=$
$\displaystyle\iota_{b}(\lambda_{b}(\iota_{a}(\lambda_{a}(c))))$
$\displaystyle=$ $\displaystyle\lambda^{\prime}_{b}(\lambda^{\prime}_{a}(c)).$

This implies that
$\lambda^{\prime}:(G,\cdot^{op})\to\operatorname{Aut}\,(G,\cdot^{op})$ is an
anti-homomorphism. Hence $(G,\cdot^{op},\circ)$ is an anti-homomorphic skew
brace, and therefore symmetric by Proposition 3.11.

The converse part is left as an exercise for the reader. ∎

It is now not difficult to prove

######  Corollary 6.4.

Let $(G,\cdot,\circ)$ be a $\lambda$-homomorphic skew left brace such that
$\lambda(G,\cdot)$ is abelian and $\operatorname{Inn}\,(G,\cdot)$ centralizes
$\lambda(G,\cdot)$. Then $(G,\circ,\cdot^{op})$ is a two-sided skew brace.

For a group $G$, let
$\operatorname{Autcent}(G):=\operatorname{C}_{\operatorname{Aut}(G)}(\operatorname{Inn}(G))$
be the group of all central automorphisms of $G$. Then any
$\lambda$-homomorphic skew left brace $(G,\cdot,\circ)$ with
$\lambda(G,\circ)\leq\operatorname{Autcent}(G,\cdot)$ gives rise to a
symmetric skew brace, namely $(G,\cdot^{op},\circ)$. In addition, if
$\lambda(G,\circ)$ is abelian, then $(G,\circ,\cdot^{op})$ is a two-sided skew
brace. _It is tempting to construct examples of such skew braces in which
$\lambda(G,\circ)$ is not contained in $\operatorname{Inn}(G,\cdot)$. _

Let $(G,\\{\circ_{i}\\}_{i\in I})$ be a symmetric linear brace system
constructed from the $\lambda$-homomorphic skew brace
$(G,\circ_{0},\circ_{1})$ with $\lambda(G)$ abelian by the procedure of
Section 4. Then it follows from Lemma 4.10 that each group $(G,\circ_{i})$ is
an extension of $\operatorname{Ker}\lambda$ by the abelian group $\lambda(G)$.
It might very well happen that all $(G,\circ_{i})$ constitute a single
equivalence class of extensions. An example of such a symmetric linear brace
system can be constructed from the $\lambda$-homomorphic brace
$(\mathbb{Z}^{2},+,\circ_{1})$ with $(\mathbb{Z}^{2},+)$ a free abelian group
with free basis $\\{x_{1},x_{2}\\}$ and $a\circ_{1}b=a+\lambda_{a}(b)$ for all
$a,b\in\mathbb{Z}^{2}$, where $\lambda_{x_{i}}$ is the automorphism of
$\mathbb{Z}^{2}$ defined on the basis elements as follows:

$\lambda_{x_{i}}=\left\\{\begin{array}[]{l}x_{1}\mapsto(1+p)x_{1}-px_{2},\\\
x_{2}\mapsto px_{1}+(1-p)x_{2}.\\\ \end{array}\right.$

It follows from [4, Theorem 4.9 (2)] that $(\mathbb{Z}^{2},\circ_{1})$ is
isomorphic to $(\mathbb{Z}^{2},+)$. Using Lemma 4.9 and then Theorem 4.14 we
can construct a symmetric linear brace system
$\mathcal{B}_{I}(\mathbb{Z}^{2})=(\mathbb{Z}^{2},\\{\circ_{i}\\}_{i\in I})$
such that $(\mathbb{Z}^{2},\circ_{i})$ is isomorphic to $(\mathbb{Z}^{2},+)$,
where $I=\\{0\\}\cup\mathbb{N}$ and $\circ_{0}=+$. One more such construction
is carried out in the next section (Theorem 7.2 and its corollary). On the
other hand, as we will see again in the next section (Corollaries 7.5, 7.7),
there do exists a symmetric linear brace system which consists of infinitely
many distinct equivalence classes of the extensions mentioned above.

We recall the definition of isomorphism of skew braces.

######  Definition 6.5.

Let $(G,\cdot,\circ)$ and $(H,\cdot,\circ)$ be skew left braces. We say that
they are isomorphic if there is a map $\varphi:G\rightarrow H$ such that
$\varphi:(G,\cdot)\rightarrow(H,\cdot)$,
$\varphi:(G,\circ)\rightarrow(H,\circ)$ are group isomorphisms.

For any $a,b,c\in G$, the following holds from the definition:

$\varphi(a\circ(b\cdot
c))=(\varphi(a)\circ\varphi(b))\cdot\varphi(a)^{-1}\cdot(\varphi(a)\circ\varphi(c)).$

######  Definition 6.6.

Let $(G,\cdot,\circ)=(H,\cdot,\circ)$, then any brace isomorphism from
$(G,\cdot,\circ)$ to $(H,\cdot,\circ)$ is called a brace automorphism of
$(G,\cdot,\circ)$. Let us denote the group of all brace automorphisms of
$(G,\cdot,\circ)$ by $\operatorname{Autb}\;G$.

From Lemma 4.9 we have

######  Proposition 6.7.

Let $(G,\cdot,\circ)$ be a $\lambda$-homomorphic skew brace with $\lambda(G)$
is abelian. Then $\lambda(G)\leq\operatorname{Autb}\;G$.

In a given symmetric linear brace system $(G,\\{\circ_{i}\\}_{i\in I})$, it is
not easy to specify whether two symmetric skew braces
$(G,\circ_{i},\circ_{j})$ and $(G,\circ_{i^{\prime}},\circ_{j^{\prime}})$ are
isomorphic or not. We now establish a criterion for checking such an
isomorphism. However, applying the same does not seem easy.

Suppose that we have two $\lambda$-homomorphic skew left braces
$(G,\cdot,\circ)$ and $(H,\cdot,\circ)$ which are constructed by homomorphisms
$\lambda:G\rightarrow\mathrm{Aut}\,G$ and $\mu:H\rightarrow\mathrm{Aut}\,H$,
that is, the operations are defined by the rules

$a\circ b=a\lambda_{a}(b),\quad a,b\in G,$ $x\circ y=x\mu_{x}(y),\quad x,y\in
H.$

Let us find conditions under which these braces are isomorphic.

If $\varphi$ is a brace isomorphism, then

$\varphi(ab)=\varphi(a)\varphi(b),\quad\varphi(a\circ
b)=\varphi(a)\circ\varphi(b),\quad a,b\in G.$

The second equality has the form:

$\varphi(a\lambda_{a}(b))=\varphi(a)\mu_{\varphi(a)}(\varphi(b))\Leftrightarrow\varphi(a)\varphi(\lambda_{a}(b))=\varphi(a)\mu_{\varphi(a)}(\varphi(b))\Leftrightarrow\varphi(\lambda_{a}(b))=\mu_{\varphi(a)}(\varphi(b)).$

We can write the last equality in the form

$\varphi\lambda_{a}\varphi^{-1}\varphi(b)=\mu_{\varphi(a)}(\varphi(b)).$

Since $\varphi$ is an isomorphism, we have

$\varphi\lambda_{a}\varphi^{-1}=\mu_{\varphi(a)},\quad a\in G.$

Hence, we have proved

######  Theorem 6.8.

Two $\lambda$-homomorphic skew braces $(G,\cdot,\circ)$ and $(H,\cdot,\circ)$,
which are constructed by homomorphisms $\lambda:G\rightarrow\mathrm{Aut}\,G$
and $\mu:H\rightarrow\mathrm{Aut}\,H$, respectively, are isomorphic if and
only if there is an isomorphism

$\varphi:(G,\cdot)\rightarrow(H,\cdot)$

such that

$\varphi\lambda_{a}\varphi^{-1}=\mu_{\varphi(a)}\quad\mbox{for all }a\in G.$

We conclude this section by generalizing the concept of meta-trivial skew
brace introduced in [4] by defining a poly-trivial skew brace. Let us first
recall the definition of an ideal of a skew brace (see [19]). A non-empty
subset $I$ of a skew left brace $(G,\cdot,\circ)$ is said to be an ideal of
$G$ if

1) $\lambda_{a}(I)\subseteq I$ for all $a\in G$,

2) $I$ is normal in $(G,\cdot)$,

3) $I$ is normal in $(G,\circ)$.

For example, the kernel of any skew left brace homomorphism is an ideal. The
notion of ideal allows us to consider the quotient skew left brace $G/I$. Note
that the condition 1) is equivalent to the equality $a\circ I=aI$ for all
$a\in G$. A skew left brace $(G,\cdot,\circ)$ is said to be meta-trivial if it
admits a ideal $I$ such that both $I$ and the quotient skew skew brace $G/I$
are trivial. This concept can be generalised as follows:

######  Definition 6.9.

A skew left brace $(G,\cdot,\circ)$ is said to be a poly-trivial skew brace if
there is a sequence of ideals of $G$:

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

such that $I_{i+1}/I_{i}$ is a trivial skew brace for all $i=0,1,\ldots,n-1$.
The minimal $n$ with this property is called the step of triviality of $G$ and
is denoted by $st(G)$.

Notice that a meta-trivial skew brace is poly-trivial of step of triviality
$2$. All $\lambda$-homomorphic skew braces are meta-trivial ([4, Theorem
2.12]). We now show that the converse of this statement ([4, Question 2.13])
is not true. For any non-trivial group $G:=(G,\cdot)$ with abelian
$\gamma_{2}(G)$ and trivial $\operatorname{Z}(G)$, let $(G,\cdot,\circ)$,
where $\circ=\cdot^{op}$, be $\lambda$-anti-homomorphic skew brace (see
Example 3.3). It follows from [15, Proposition 2.3] that $(G,\cdot,\circ)$ is
meta-trivial with $G^{2}=\gamma_{2}(G)$, where
$G^{2}=\left<\lambda_{a}(b)b^{-1}\mid a,b\in G\right>$. Since
$\lambda(G)\cong\operatorname{Inn}(G)$ is not abelian, $(G,\cdot,\circ)$ can
not be $\lambda$-homomorphic. The simplest such example is $S_{3}$, the
symmetric group on $3$ letters, which was pointed out by L. Vendramin in the
Zentralblatt and AMS review of [4] and also in [36, Example 3.15].

We now discuss poly-triviality of $\lambda$-anti-homomorphic (symmetric) skew
braces. We start by noting that any $\lambda$-anti-homomorphic skew brace with
$\operatorname{Ker}\,\lambda=1$ is natural. Further we have

######  Proposition 6.10.

Let $(G,\cdot,\circ)$ be a $\lambda$-anti-homomorphic skew brace. Then either
$(G,\cdot,\circ)$ itself or $G/\operatorname{Ker}\,\lambda$ is a natural
$\lambda$-anti-homomorphic skew brace.

###### Proof.

Suppose that $(G,\cdot,\circ)$ is not a natural $\lambda$-anti-homomorphic
skew brace. Notice that $\operatorname{Ker}\,\lambda$ is a normal subgroup of
both $(G,\cdot)$ and $(G,\circ)$. Also

$\operatorname{Ker}\,\lambda=\\{a\in G\mid a\circ b=a\cdot b\mbox{ for all
}b\in G\\}.$

For simplicity, let us denote this set by $G_{0}$. Obviously $G_{0}$ is a
trivial skew brace. Set $\bar{G}=G/G_{0}$. Then $(\bar{G},\cdot)$ and
$\bar{G},\circ)$ are groups such that

$G_{0}a\cdot G_{0}b=G_{0}a\cdot b\,\,\mbox{ and }\,\,G_{0}a\circ
G_{0}b=G_{0}(a\circ b)$

for all $a,b\in G$.

Since $(G,\cdot,\circ)$ is a $\lambda$-anti-homomorphic skew brace, it follows
that

$a\lambda_{a}(b)a^{-1}b^{-1}\in\operatorname{Ker}\,\lambda$

for all $a,b\in G$. Hence

$G_{0}a\circ G_{0}b=G_{0}(a\circ
b)=G_{0}(a\lambda_{a}(b)a^{-1}b^{-1}ba=G_{0}ba=G_{0}b\cdot G_{0}a$

for all $a,b\in G$. This proves that $G/\operatorname{Ker}\,\lambda$ is a
natural $\lambda$-anti-homomorphic skew brace. ∎

As a consequence of the preceding result we conclude

######  Corollary 6.11.

Any $\lambda$-anti-homomorphic skew brace is either natural or trivial-by-
natural.

So, to characterise $\lambda$-anti-homomorphic skew braces, it only needs to
study natural $\lambda$-anti-homomorphic skew braces. Note that the sub-skew
brace $G_{0}=\operatorname{Ker}\,\lambda$ of a $\lambda$-anti-homomorphic skew
brace $(G,\cdot,\circ)$ as defined in the proof of Proposition 6.10 is an
ideal of $(G,\cdot,\circ)$. For, since
$a\lambda_{a}(b)a^{-1}b^{-1}\in\operatorname{Ker}\,\lambda$, for any $a\in G$
and $b\in G_{0}$, $\lambda_{a}(b)\in G_{0}$. So if
$G/\operatorname{Ker}\,\lambda$ is poly-trivial with
$st(G/\operatorname{Ker}\,\lambda)=n$, then $(G,\cdot,\circ)$ is poly-trivial
with $st(G)$ at most $n+1$.

Let us now specialize to natural $\lambda$-anti-homomorphic skew braces.
Suppose $(G,\cdot,\circ)$ is such a skew brace with $\gamma_{2}(G,\cdot)$
abelian. Then as discussed above $(G,\cdot,\circ)$ is meta-trivial. So we have

######  Corollary 6.12.

Let $(G,\cdot,\circ)$ be a $\lambda$-anti-homomorphic skew brace such that
$\gamma_{2}(\bar{G})$ is abelian, where
$\bar{G}=G/\operatorname{Ker}\,\lambda$ is the group with binary operation
induced by $\cdot$. Then $st(G)$ is at most $3$.

Now let $(G,\cdot,\circ)$ be a natural $\lambda$-anti-homomorphic skew brace
such that $(G,\cdot)$ is nilpotent with nilpotency class $n$. Then
$I_{1}:=\operatorname{Ker}\,\lambda$ is precisely $\operatorname{Z}(G,\cdot)$,
the center of $(G,\cdot)$ and is an ideal of $(G,\cdot,\circ)$. It follows
that $(G/I,\cdot,\circ)$ is a natural $\lambda$-anti-homomorphic skew brace.
An easy iterative argument now shows that $(G,\cdot,\circ)$ is poly-trivial
with $st(G)$ at most $n$. We conclude this section with the remark that if
$(G,\cdot)$ is a non-abelian simple group, then the natural $\lambda$-anti-
homomorphic skew brace $(G,\cdot,\circ)$ can not be poly-trivial. It is a
matter of further investigation to get more information on such skew braces.
After this article was written, there followed an article [36] in which
symmetric skew braces, under the name bi-skew braces, are further invested and
some interesting structural results are obtained. Among other things, the
following conjecture of N. Byott [7] (based on the last sentence of Section 1)
is proved for symmetric skew braces: If $(G,\cdot,\circ)$ is a skew brace with
$(G,\cdot)$ solvable, then $(G,\circ)$ is solvable.

##  7\. Some constructions

In previous sections we considered constructions of brace systems
$(G,\\{\circ_{i}\\}_{i\in I})$ for $I=\\{0\\}\cup\mathbb{N}$. In this section
we construct some concrete symmetric brace systems $(F,\\{\circ_{i}\\}_{i\in
I})$, where $F$ is a non-abelian free group and $I$ is a subset of
$\mathbb{Z}$.

We start with the definition of a $\lambda$-cyclic skew brace.

######  Definition 7.1.

A $\lambda$-homomorphic skew brace $(G,\cdot,\circ)$ is said to be
$\lambda$-cyclic if $\lambda(G)$ is cyclic.

Let $F$ be a non-abelian free group with a basis $X$ and
$\theta\in\operatorname{Aut}F$ be an automorphism. We can define a
homomorphism $\lambda:F\to\operatorname{Aut}F$ by the rule

$\lambda_{x}=\lambda(x)=\theta~{}~{}\mbox{for all}~{}x\in X.$

Hence, for $a\in F$ its image $\lambda_{a}=\theta^{l(a)}$, where $l(y)$, for
an element

$y=x_{i_{1}}^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}}\ldots
x_{i_{k}}^{\alpha_{k}},~{}~{}x_{j}\in X,~{}~{}\alpha_{j}\in\mathbb{Z}$

of $F$ is given by $l(y):=\alpha_{1}+\alpha_{2}+\ldots+\alpha_{k}$. It is not
difficult to see that the hypotheses of Theorem 4.14 are satisfied. If
$\theta$ has finite order $n$, then we get symmetric linear brace system
$\mathcal{B}_{I}(F)=(F,\\{\circ_{i}\\}_{i\in I})$, where
$I=\\{0,\ldots,n-1\\}$. , and $\circ_{0}=\cdot$, $\circ_{1}=\circ$. If the
automorphism $\theta$ has infinite order, then we can construct symmetric
linear brace system with $I=\\{0\\}\cup\mathbb{N}$.

In the rest of this section we analyse both of these possibilities. At first,
we consider the case when $F=F_{n}$ has rank $n$ and $\theta$ is the cyclic
permutation of the generators. We prove that in this case $(F_{n},\circ)$ is
isomorphic to $F_{n}$, but the operation $\circ$ is different from $\cdot$ and
$\cdot^{op}$.

######  Theorem 7.2.

Let $F_{n}=\langle x_{1},x_{2},\ldots,x_{n}\rangle$ be the free group of rank
$n$ and $\theta$ be the automorphism which is the permutation of the
generators,

$\theta:x_{1}\mapsto x_{2}\mapsto\ldots\mapsto x_{n}\mapsto x_{1}.$

A homomorphism $\lambda:F_{n}\rightarrow\mathrm{Aut}\,F_{n}$ is defined by the
rule

$\lambda:x_{i}\mapsto\theta,\quad i=1,\ldots,n.$

Then in the corresponding $\lambda$-cyclic skew brace $(F_{n},\cdot,\circ)$
the multiplicative group $(F_{n},\circ)$ is an extension of
$\operatorname{Ker}\,\lambda$ that is a free group of rank $n^{2}-n+1$ with
free generators,

$z_{j,k}=x_{1}^{k}x_{j}x_{1}^{-k-1},\quad k=0,\ldots,n-2,\quad j=2,\ldots,n,$
$y_{i}=x_{1}^{n-1}x_{i},\quad i=1,\ldots,n$

by an infinite cyclic group $\langle s\rangle$, which acts, by conjugation, on
the generators of $\operatorname{Ker}\,\lambda$ as follows:

$\displaystyle s^{-1}y_{1}s$ $\displaystyle=$ $\displaystyle
z_{n0}z_{n1}\cdot\ldots\cdot z_{n,n-2}y_{n},$ $\displaystyle s^{-1}y_{2}s$
$\displaystyle=$ $\displaystyle z_{n0}z_{n1}\cdot\ldots\cdot z_{n,n-3}y_{n},$
$\displaystyle s^{-1}y_{i}s$ $\displaystyle=$ $\displaystyle
z_{n0}z_{n1}\cdot\ldots\cdot z_{n,n-3}z_{i-2,n-2}y_{n},\quad i=3,\ldots,n,$
$\displaystyle s^{-1}z_{j0}s$ $\displaystyle=$ $\displaystyle
y_{n}^{-1}y_{j-1},\quad j=2,\ldots,n,$ $\displaystyle s^{-1}z_{j1}s$
$\displaystyle=$ $\displaystyle z_{j-1,0}z_{n,0}^{-1},\quad j=2,\ldots,n,$
$\displaystyle s^{-1}z_{j,k}s$ $\displaystyle=$
$\displaystyle(z_{n0}z_{n1}\cdot\ldots\cdot
z_{n,k-2})(z_{j-1,k-1}z_{n,k-1}^{-1})(z_{n0}z_{n1}\cdot\ldots\cdot
z_{n,k-2})^{-1},$

where $j=2,\ldots,n$, $k=2,\ldots,n-2$ and we assume $z_{1,k}=1$. Moreover,

$s^{n}=z_{21}z_{32}\cdot\ldots\cdot z_{n-1,n-2}z_{n,n-1}.$

###### Proof.

Take the subgroup

$H_{\lambda}=\left\\{\,(\lambda(a),a)\,|\,a\in F_{n}\,\right\\}$

of the holomorph $\mathrm{Hol}\,F_{n}$.

The kernel of $\lambda$ consists of the words $w=x_{i_{1}}^{\alpha_{1}}\ldots
x_{i_{m}}^{\alpha_{m}}\in F_{n}$ such that $l(w)=0$, where

$l(w)=\alpha_{1}+\cdots+\alpha_{m}.$

Notice that for $1\leq i\leq n-1$, we have

$x_{i}^{-1}\lambda_{x_{j}}(x_{i})=x_{i}^{-1}\theta(x_{i})=x_{i}^{-1}x_{i+1}$

and for $i=n$, we have

$x_{n}^{-1}\lambda_{x_{j}}(x_{n})=x_{n}^{-1}\theta(x_{n})=x_{n}^{-1}x_{1}$

for all $1\leq j\leq n$. Hence, $x_{i}^{-1}\lambda_{x_{j}}(x_{i})$ lies in the
kernel of $\lambda$. It now follows from [4, Proposition 2.14] that
$b^{-1}\lambda_{a}(b)\in Ker\lambda$ for all $a,b\in F_{n}$. It is now not
difficult to see that the hypotheses of [4, Themrem 3.5] are satisfied; hence
$H_{\lambda}$ is a regular subgroup of $\mathrm{Hol}\,F_{n}$. This establishes
the existence of the $\lambda$-cyclic skew brace $(F_{n},\cdot,\circ)$, where
$a\circ b=a\cdot\lambda_{a}(b)$ for all $a,b\in F_{n}$.

Using the Reidemeister-Shraier method let us find the generators of
$\operatorname{Ker}\,\lambda$.

We have

$\bullet$ $\left\\{x_{1},\ldots,x_{n}\right\\}$ is a generating set of
$F_{n}$,

$\bullet$ $\left\\{1,x_{1},x_{1}^{2},\ldots,x_{1}^{n-1}\right\\}$ is a Shraier
set of coset representatives of $F_{n}/\mathrm{Ker}\,\lambda$.

Then the non-trivial elements of the form
$x_{1}^{k}x_{j}(\overline{x_{1}^{k}x_{j}})^{-1}$, $k=0,\ldots,n-1$,
$j=1,\ldots,n$ are free generators of $\mathrm{Ker}\,\lambda$, where $\bar{a}$
denotes the inverse of $a$ in $(F_{n},\circ)$. Note that
$\mathrm{rank}(\mathrm{Ker}\,\lambda)=1+(n-1)n=n^{2}-n+1$. Hence,
$\mathrm{Ker}\,\lambda$ has the basis consisting of the elements

$z_{j,k}:=x_{1}^{k}x_{j}x_{1}^{-k-1},\quad k=0,\ldots,n-2,\quad j=2,\ldots,n,$
$y_{i}:=x_{1}^{n-1}x_{i},\quad i=1,\ldots,n.$

The quotient $H_{\lambda}/\mathrm{Ker}\,\lambda\cong\mathbb{Z}_{n}$ and is
generated, for example, by the image of $s=(\theta,x_{1})$. Let us find the
action of $s$ on the generators of $\mathrm{Ker}\,\lambda$.

At first, we remark that

$s^{-1}=(\theta,x_{1})^{-1}=(\theta^{-1},\theta^{-1}(x_{1}^{-1}))=(\theta^{-1},x_{n}^{-1}).$

Hence,

$\displaystyle s^{-1}z_{j,k}s$ $\displaystyle=$
$\displaystyle(\theta^{-1},x_{n}^{-1})(1,x_{1}^{k}x_{j}x_{1}^{-k-1})(\theta,x_{1})$
$\displaystyle=$
$\displaystyle(\theta^{-1},x_{n}^{-1})(\theta,x_{1}^{k}x_{j}x_{1}^{-k})$
$\displaystyle=$ $\displaystyle(1,x_{n}^{-1}x_{n}^{k}x_{j-1}x_{n}^{-k})$
$\displaystyle=$ $\displaystyle(1,x_{n}^{k-1}x_{j-1}x_{n}^{-k}),$
$\displaystyle s^{-1}y_{1}s$ $\displaystyle=$
$\displaystyle(\theta^{-1},x_{n}^{-1})(1,x_{1}^{n})(\theta,x_{1})$
$\displaystyle=$ $\displaystyle(\theta^{-1},x_{n}^{-1})(\theta,x_{1}^{n+1})$
$\displaystyle=$ $\displaystyle(1,x_{n}^{n})$

and for $i>1$,

$\displaystyle s^{-1}y_{i}s$ $\displaystyle=$
$\displaystyle(\theta^{-1},x_{n}^{-1})(1,x_{1}^{n-1}x_{i})(\theta,x_{1})$
$\displaystyle=$
$\displaystyle(\theta^{-1},x_{n}^{-1})(\theta,x_{1}^{n-1}x_{i}x_{1})$
$\displaystyle=$ $\displaystyle(1,x_{n}^{n-2}x_{i-1}x_{n}).$

So,

$s^{-1}z_{j,k}s=x_{n}^{k-1}x_{j-1}x_{n}^{-k},\quad k=0,\ldots,n-2,\quad
j=2,\ldots,n,$ $s^{-1}y_{1}s=x_{n}^{n},$
$s^{-1}y_{i}s=x_{n}^{n-2}x_{i-1}x_{n},\quad i=2,\ldots,n.$

Further, we need to rewrite the right hand sides in the generators $z_{j,k}$,
$x_{i}$.

We have

$\displaystyle s^{-1}z_{20}s$ $\displaystyle=$ $\displaystyle
x_{n}^{-1}x_{1}=y_{n}^{-1}y_{1},$ $\displaystyle s^{-1}z_{21}s$
$\displaystyle=$ $\displaystyle
x_{1}x_{n}^{-1}=(x_{n}x_{1}^{-1})^{-1}=z_{n0}^{-1},$ $\displaystyle
s^{-1}z_{22}s$ $\displaystyle=$ $\displaystyle
x_{n}x_{1}x_{n}^{-2}=z_{n0}z_{n1}^{-1}z_{n0}^{-1},$ $\displaystyle
s^{-1}z_{23}s$ $\displaystyle=$ $\displaystyle
x_{n}^{2}x_{1}x_{n}^{-3}=z_{n0}z_{n1}z_{n2}^{-1}z_{n1}^{-1}z_{n0}^{-1},$
$\displaystyle\vdots$ $\displaystyle s^{-1}z_{2,n-2}s$ $\displaystyle=$
$\displaystyle x_{n}^{n-3}x_{1}x_{n}^{2-n}$ $\displaystyle=$
$\displaystyle(z_{n0}z_{n1}\cdot\ldots\cdot
z_{n,n-4})z_{n,n-3}^{-1}(z_{n0}z_{n1}\cdot\ldots\cdot z_{n,n-4})^{-1}.$

Further,

$\displaystyle s^{-1}z_{30}s$ $\displaystyle=$ $\displaystyle
x_{n}^{-1}x_{2}=y_{n}^{-1}y_{2},$ $\displaystyle s^{-1}z_{31}s$
$\displaystyle=$ $\displaystyle
x_{2}x_{n}^{-1}=(x_{n}x_{1}^{-1})^{-1}=z_{20}z_{n0}^{-1},$ $\displaystyle
s^{-1}z_{32}s$ $\displaystyle=$ $\displaystyle
x_{n}x_{2}x_{n}^{-2}=z_{n0}z_{21}z_{n1}^{-1}z_{n0}^{-1},$ $\displaystyle
s^{-1}z_{33}s$ $\displaystyle=$ $\displaystyle
x_{n}^{2}x_{2}x_{n}^{-3}=z_{n0}z_{n1}(z_{22}z_{n2}^{-1})z_{n1}^{-1}z_{n0}^{-1},$
$\displaystyle\vdots$ $\displaystyle s^{-1}z_{3,n-2}s$ $\displaystyle=$
$\displaystyle(z_{n0}z_{n1}\cdot\ldots\cdot
z_{n,n-4})(z_{2,n-3}z_{n,n-3}^{-1})(z_{n0}z_{n1}\cdot\ldots\cdot
z_{n,n-4})^{-1}.$

Now for any $3\leq j\leq n$ and $2\leq k\leq n-2$, on the same line, we can
compute

$\displaystyle s^{-1}z_{j0}s$ $\displaystyle=$ $\displaystyle
x_{n}^{-1}x_{j-1}=y_{n}^{-1}y_{j-1},$ $\displaystyle s^{-1}z_{j1}s$
$\displaystyle=$ $\displaystyle x_{j-1}x_{n}^{-1}=z_{j-1,0}z_{n,0}^{-1},$
$\displaystyle\vdots$ $\displaystyle s^{-1}z_{j,k}s$ $\displaystyle=$
$\displaystyle(z_{n0}z_{n1}\cdot\ldots\cdot
z_{n,k-2})(z_{j-1,k-1}z_{n,k-1}^{-1})(z_{n0}z_{n1}\cdot\ldots\cdot
z_{n,k-2})^{-1},$

Since $z_{1k}=1$, the action of $s$ on it is trivial. To summarize

$\displaystyle s^{-1}z_{j0}s$ $\displaystyle=$ $\displaystyle
y_{n}^{-1}y_{j-1},$ $\displaystyle s^{-1}z_{j1}s$ $\displaystyle=$
$\displaystyle z_{j-1,0}z_{n,0}^{-1},$ $\displaystyle\vdots$ $\displaystyle
s^{-1}z_{j,k}s$ $\displaystyle=$ $\displaystyle(z_{n0}z_{n1}\cdot\ldots\cdot
z_{n,k-2})(z_{j-1,k-1}z_{n,k-1}^{-1})(z_{n0}z_{n1}\cdot\ldots\cdot
z_{n,k-2})^{-1},$

where $j=2,\ldots,n$, $k=2,\ldots,n-2$.

Further, we consider the action of $s$ on $y_{i}$ and compute $s^{-1}y_{i}s$,
$i=1,\ldots,n$. First we establish the formulas for

$x_{1}^{-k}z_{n0}x_{1}^{k},\quad k=1,\ldots,n-1.$

We have

$\displaystyle x_{1}^{-1}z_{n0}x_{1}$ $\displaystyle=$ $\displaystyle
x_{1}^{-1}x_{n}=x_{1}^{-n}x_{1}^{n-1}x_{n}=y_{1}^{-1}y_{n},$ $\displaystyle
x_{1}^{-2}z_{n0}x_{1}^{2}$ $\displaystyle=$ $\displaystyle
x_{1}^{-1}y_{1}^{-1}y_{n}x_{1}=y_{1}^{-1}x_{1}^{-1}y_{n}x_{1}=y_{1}^{-1}z_{n,n-2}y_{1},$
$\displaystyle x_{1}^{-3}z_{n0}x_{1}^{3}$ $\displaystyle=$ $\displaystyle
y_{1}^{-1}z_{n,n-3}y_{1},$ $\displaystyle\vdots$ $\displaystyle
x_{1}^{1-n}z_{n0}x_{1}^{n-1}$ $\displaystyle=$ $\displaystyle
y_{1}^{-1}z_{n,1}y_{1}.$

Hence,

$\displaystyle s^{-1}y_{1}s$ $\displaystyle=$ $\displaystyle
x_{n}^{n}=(z_{n0}x_{1})^{n}=(z_{n0}x_{1})(z_{n0}x_{1})\cdot\ldots\cdot(z_{n0}x_{1})$
$\displaystyle=$ $\displaystyle
z_{n0}(x_{1}z_{n0}x_{1}^{-1})(x_{1}^{2}z_{n0}x_{1}^{-2})\cdot\ldots\cdot(x_{1}^{n-1}z_{n0}x_{1}^{1-n})x_{1}^{n}$
$\displaystyle=$ $\displaystyle z_{n0}z_{n1}\cdot\ldots\cdot z_{n,n-2}y_{n},$
$\displaystyle s^{-1}y_{2}s$ $\displaystyle=$ $\displaystyle
x_{n}^{n-2}x_{1}x_{n}=(z_{n0}x_{1})^{n-2}x_{1}x_{n}=(z_{n0}x_{1})\cdot\ldots\cdot(z_{n0}x_{1})x_{1}x_{n}$
$\displaystyle=$ $\displaystyle
z_{n0}(x_{1}z_{n0}x_{1}^{-1})(x_{1}^{2}z_{n0}x_{1}^{-2})\cdot\ldots\cdot(x_{1}^{n-3}z_{n0}x_{1}^{3-n})x_{1}^{n-1}x_{n}$
$\displaystyle=$ $\displaystyle z_{n0}z_{n1}\cdot\ldots\cdot z_{n,n-3}y_{n},$
$\displaystyle s^{-1}y_{3}s$ $\displaystyle=$ $\displaystyle
x_{n}^{n-2}x_{2}x_{n}=(z_{n0}x_{1})^{n-2}x_{2}x_{n}=(z_{n0}x_{1})\cdot\ldots\cdot(z_{n0}x_{1})x_{2}x_{n}$
$\displaystyle=$ $\displaystyle z_{n0}z_{n1}\cdot\ldots\cdot
z_{n,n-3}x_{1}^{n-2}x_{2}x_{n}=z_{n0}z_{n1}\cdot\ldots\cdot
z_{n,n-3}z_{2,n-2}y_{n}.$

For all $i=3,4,\ldots,n$, on the same line, we have

$s^{-1}y_{i}s=x_{n}^{n-2}x_{i-1}x_{n}=z_{n0}z_{n1}\cdot\ldots\cdot
z_{n,n-3}z_{i-2,n-2}y_{n}.$

Finally we compute the power of $s$.

$\displaystyle s^{n}$ $\displaystyle=$
$\displaystyle(\theta,x_{1})^{n}=(\theta,x_{1})(\theta,x_{1})\cdot\ldots\cdot(\theta,x_{1})$
$\displaystyle=$
$\displaystyle(\theta^{n},x_{1}\theta(x_{1})\cdot\ldots\cdot\theta^{n-1}(x_{1}))=(1,x_{1}x_{2}\cdot\ldots\cdot
x_{n})$ $\displaystyle=$
$\displaystyle(1,(x_{1}x_{2}x_{1}^{-2})(x_{1}^{2}x_{3}x_{1}^{-3})\cdot\ldots\cdot(x_{1}^{n-2}x_{n-1}x_{1}^{1-n})(x_{1}^{n-1}x_{n}))$
$\displaystyle=$ $\displaystyle z_{21}z_{32}\cdot\ldots\cdot
z_{n-1,n-2}z_{n,n-1}.$

This completes the proof. ∎

We now get

######  Corollary 7.3.

The skew brace $(F_{n},\cdot,\circ)$ constructed in Theorem 7.2 is non-trivial
and the group $(F_{n},\circ)$ is free group of rank $n$.

###### Proof.

That $(F_{n},\cdot,\circ)$ is a non-trivial skew brace easily follows from the
construction. For the second assertion we use the following theorem of
Stallings [37]: If a finitely generated torsion free group $G$ contains a free
subgroup of finite index, then $G$ is a free group.

The kernel of the homomorphism $\lambda:F_{n}\rightarrow\mathrm{Aut}\,F_{n}$
is a free subgroup of $F_{n}$ and has index $n$. Hence, it suffices to prove
that $(F_{n},\circ)$ does not admit torsion elements. Notice that the group
$(F_{n},\circ)$ is equal to the subgroup

$H_{\lambda}=\left\\{\,(\lambda(a),a)\,|\,a\in F_{n}\,\right\\}$

of $\mathrm{Hol}\,F_{n}$.

Let $(\lambda(a),a)\in H_{\lambda}$ has finite order. Since $\theta^{n}=1$ and
$\mathrm{Ker}\,\lambda$ does not have torsion, we have
$\lambda(a)=\theta^{m}$, $m\neq 0$ and $m$ divides $n$. Hence, $a=x_{1}^{m}w$
for some $w\in\mathrm{Ker}\,\lambda$. We have

$(\theta^{m},x_{1}^{m}w)^{\frac{n}{m}}=(1,(x_{1}^{m}w)\theta^{m}(x_{1}^{m}w)\cdot\ldots\cdot\theta^{m(\frac{n}{m}-1)}(x_{1}^{m}w)).$

Since $\theta(\mathrm{Ker}\,\lambda)\subseteq\mathrm{Ker}\,\lambda$, it
follows that

$x_{1}^{m}\theta^{m}(x_{1}^{m})\cdot\ldots\cdot\theta^{m(\frac{n}{m}-1)}(x_{1}^{m})\in\mathrm{Ker}\,\lambda.$

Again, since $\theta(x_{1})\equiv x_{1}(\mathrm{mod}\,\mathrm{Ker}\,\lambda)$,

$x_{1}^{m(\frac{n}{m}-1)}\in\mathrm{Ker}\,\lambda,$

which means $x_{1}^{n-m}\in\mathrm{Ker}\,\lambda$, a contradiction, because
$\lambda(x_{1}^{n-m})=\theta^{n-m}\neq 1$. The proof is complete. ∎

We now construct a $\lambda$-cyclic skew brace on the free group $F_{n}$ of
rank $n$ such that $\lambda(F_{n})$ is infinite cyclic.

######  Theorem 7.4.

Let $F_{n}=\left<x_{1},\ldots,x_{n}\right>$ be the free group of rank $n$, $w$
be a fixed non-identity element of $F_{n}$ and $\theta:F_{n}\rightarrow F_{n}$
be the inner automorphism

$\theta:a\mapsto waw^{-1},\quad a\in F_{n}.$

The homomorphism $\lambda:F_{n}\rightarrow\mathrm{Aut}\,F_{n}$ is defined by
the rule

$\lambda:x_{i}\mapsto\theta,\quad i=1,\ldots,n.$

Then in the $\lambda$-cyclic skew brace $(F_{n},\cdot,\circ)$ the
multiplicative group $(F_{n},\circ)$ is an extension of the free group
$F_{\infty}$ $(=\operatorname{Ker}\,\lambda)$ of infinite rank with free
generators

$z_{j,k}=x_{1}^{k}x_{j}x_{1}^{-k-1},\quad k\in\mathbb{Z},\quad j=2,\ldots,n,$

by the infinite cyclic group $\langle s\rangle$, which acts by conjugations on
the generators of $F_{\infty}$ by the rule:

$s^{-1}z_{j,k}s=z_{j,k-m-1},\quad k\in\mathbb{Z},\quad j=2,\ldots,n,$

where $m$ is such that $w\equiv
x_{1}^{m}(\mathrm{mod}\,\mathrm{Ker}\,\lambda)$.

In particular, if $m\neq-1$, then we can express the generators $z_{j,k}$ in
terms of generators $z_{j1},\ldots,z_{j,m+1}$, $j=2,\ldots,n$, by the rule
$s^{-1}z_{j,k}s=z_{j,k-m-1}$, and so $\left\langle
F_{n},\circ\right\rangle\cong F_{r}$ is the free group of rank
$r=|(m+1)(n-1)|+1$. If $m=-1$, then $(F_{n},\circ)\cong
F_{\infty}\times\mathbb{Z}$ is the direct product of the free group of
infinite rank and the infinite cyclic group.

###### Proof.

Consider a subgroup

$H_{\lambda}=\left\\{\,(\lambda(a),a)\,|\,a\in F_{n}\,\right\\}$

of $\mathrm{Hol}\,F_{n}$. The kernel of the homomorphism $\lambda$ consists of
the words $w\in F_{n}$ such that $l(w)=0.$ Notice that

$x_{i}^{-1}\lambda_{x_{j}}(x_{i})=x_{i}^{-1}\theta(x_{i})=x_{i}^{-1}wx_{i}w^{-1}.$

Hence, $x_{i}^{-1}\lambda_{x_{j}}(x_{i})$ lies in the kernel of $\lambda$ and
$H_{\lambda}$ is a regular subgroup of $\mathrm{Hol}\,F_{n}$. Existence of the
skew brace $(F_{n},\cdot,\circ)$ now follows.

We find the generators of $\mathrm{Ker}\,\lambda$ using the Reidemeister-
Shraier method as follows. We have

$\bullet$ $\left\\{x_{1},\ldots,x_{n}\right\\}$ is a generating set of
$F_{n}$,

$\bullet$ $\left\\{x_{1}^{k},\,\,k\in\mathbb{Z}\right\\}$ is a Shraier set of
coset representatives of $F_{n}/\mathrm{Ker}\,\lambda$.

Then the non-trivial elements of the form
$x_{1}^{k}x_{j}(\overline{x_{1}^{k}x_{j}})^{-1}$, $k\in\mathbb{Z}$,
$1=2,\ldots,n$ are free generators of $\mathrm{Ker}\,\lambda$. It means that
$\mathrm{Ker}\,\lambda$ has the basis

$z_{j,k}=x_{1}^{k}x_{j}x_{1}^{-k-1},\quad k\in\mathbb{Z},\quad j=2,\ldots,n.$

The quotient $H_{\lambda}/\mathrm{Ker}\,\lambda\cong\mathbb{Z}$ and is
generated, for example, by the images of $s=(\theta,x_{1})$. We now compute
the action of $s$ on the generators of $\mathrm{Ker}\,\lambda$. Notice that

$s^{-1}=(\theta,x_{1})^{-1}=(\theta^{-1},\theta^{-1}(x_{1}^{-1}))=(\theta^{-1},w^{-1}x_{1}^{-1}w).$

Hence,

$\displaystyle s^{-1}z_{j,k}s$ $\displaystyle=$
$\displaystyle(\theta^{-1},w^{-1}x_{1}^{-1}w)(1,z_{j,k})(\theta,x_{1})=(\theta^{-1},w^{-1}x_{1}^{-1}w)(\theta,z_{j,k}x_{1})$
$\displaystyle=$
$\displaystyle(1,w^{-1}x_{1}^{-1}ww^{-1}z_{j,k}x_{1}w)=(1,w^{-1}x_{1}^{-1}z_{j,k}x_{1}w)$
$\displaystyle=$ $\displaystyle(1,w^{-1}z_{j,k-1}x_{1}w).$

So,

$s^{-1}z_{j,k}s=w^{-1}z_{j,k-1}w,\quad k\in\mathbb{Z},\quad j=2,\ldots,n.$

Note that if $w\in\mathrm{Ker}\,\lambda$, then it is a composition of the
shift of the indexes on $k$ and the inner automorphism of
$\mathrm{Ker}\,\lambda$.

Now assume that

$w=x_{1}^{m}w_{0},$

where $m\in\mathbb{Z}$, $w_{0}\in\mathrm{Ker}\,\lambda$. Then

$s^{-1}z_{j,k}s=w_{0}^{-1}z_{j,k-m-1}w_{0},\quad k\in\mathbb{Z},\quad
j=2,\ldots,n.$

Hence, the element $sw_{0}^{-1}$ acts on the generators of
$\mathrm{Ker}\,\lambda$ by the shift of the indexes on $k$:

$(sw_{0}^{-1})^{-1}z_{j,k}(sw_{0}^{-1})=z_{j,k-m-1},\quad k\in\mathbb{Z},\quad
j=2,\ldots,n.$

Thus

$(F_{n},\circ)=H_{\lambda}=\mathrm{Ker}\,\lambda\rtimes\mathbb{Z}\cong
F_{\infty}\rtimes\mathbb{Z},$

where $F_{\infty}=\langle z_{j,k},\quad k\in\mathbb{Z},\quad
j=2,\ldots,n\rangle$ is a free group of infinite rank,
$\mathbb{Z}=\left\langle s\right\rangle$ is the infinite cyclic group and the
action is defined by the rules

$s^{-1}z_{j,k}s=z_{j,k-m-1},\quad k\in\mathbb{Z},\quad j=2,\ldots,n.$

The proof is complete. ∎

As a direct consequence of Theorem 7.4, we get

######  Corollary 7.5.

There exists a symmetric linear brace system
$\mathcal{B}_{I}(F_{n})=(F_{n},\\{\circ_{i}\\}_{i\in I})$, $n\geq 2$, where
$I$ is an infinite linearly ordered set.

######  Corollary 7.6.

For any natural number $n\geq 2$ there exists a homomorphism
$\lambda:F_{2}\rightarrow\mathrm{Aut}\,F_{2}$, such that for the skew left
brace $(F_{2},\cdot,\circ)$ the multiplicative group $(F_{2},\circ)$ is
isomorphic to $F_{n}$.

Let us define a subset of natural numbers consisting of

$r_{0}=2,~{}~{}r_{k+1}=2r_{k}+1,~{}~{}k=0,1,\ldots$

and define the graph $\Gamma$ with the set of vertices
$V=\\{v_{k}~{}|~{}k=0,1,\ldots\\}$, and set of edges

$E=\\{e_{k,k+1}=(k,k+1),e_{k+1,k}=(k+1,k)~{}|~{}k=0,1,\ldots\\}.$

From Theorem 7.4 for $m=1$ it follows that we can construct brace system
$\mathcal{B}_{\Gamma}(F_{2})$ such that

$(F_{2},\circ_{k},\circ_{k+1}),~{}~{}k=0,1,\ldots$

is a symmetric skew brace such that $(F_{2},\circ_{k})\cong F_{r_{k}}$. Let
$I=\\{0\\}\cup\mathbb{N}$. Hence, from this construction and Corollary 7.6, we
have

######  Corollary 7.7.

There is a symmetric linear brace system
$\mathcal{B}_{I}(G)=(G,\\{\circ_{k}\\}_{k\in I})$ such that all groups
$(G,\circ_{k})$ are pairwise non-isomorphic.

##  8\. Rota–Baxter operators and brace systems

L. Guo, H. Lang, Y. Sheng [20] defined the notion of Rota–Baxter operator on a
group. A map $B\colon G\to G$ is called a _Rota–Baxter operator_ (of weight 1)
if

$B(g)B(h)=B(gB(g)hB(g)^{-1})$

for all $g,h\in G$. A group $G$ with a Rota–Baxter operator $B$ is called a
_Rota–Baxter group_.

A new binary operation $\circ\colon G\to G$ on a Rota–Baxter group $(G,B)$ was
defined in [20] .

######  Proposition 8.1 ([20]).

Let $(G,\cdot,B)$ be a Rota–Baxter group.

a) The pair $(G,\circ)$ with the product

(8.2) $g\circ h=gB(g)hB(g)^{-1},$

where $g,h\in G$, is also a group.

b) The operator $B$ is a Rota–Baxter operator on the group $(G,\circ)$.

c) The map $B\colon(G,\circ)\to(G,\cdot)$ is a homomorphism of Rota–Baxter
groups.

As observed in [6, Proposition 3.1], given a Rota–Baxter group $(G,\cdot,B)$,
we get a skew left brace $(G,\cdot,\circ)$, where ‘$\circ$’ is defined (8.2).
Such a skew brace will be called a Rota–Baxter skew brace. The following
result provides a procedure for converting a word written in $(G,\circ)$ to a
word written in $G$.

######  Proposition 8.3 ([5]).

Let $(G,B)$ be a Rota–Baxter group. If $A$ is a subset of $G$ and

$w=a_{i_{1}}^{\circ(k_{1})}\circ a_{i_{2}}^{\circ(k_{2})}\circ\ldots\circ
a_{i_{s}}^{\circ(k_{s})},\quad a_{i_{j}}\in A,\ k_{j}\in\mathbb{Z},$

is presented by a word under the operation $\circ$, then

$w{=}(a_{i_{1}}B(a_{i_{1}}))^{k_{1}}(a_{i_{2}}B(a_{i_{2}}))^{k_{2}}\ldots(a_{i_{s}}B(a_{i_{s}}))^{k_{s}}B(a_{i_{s}})^{-k_{s}}B(a_{i_{s-1}})^{-k_{s-1}}\ldots
B(a_{i_{1}})^{-k_{1}}.$

In particular, $a^{\circ(-1)}=B(a)^{-1}a^{-1}B(a)$.

Suppose that $(G,\cdot,\circ_{B})$ is a Rota–Baxter skew brace. The next
proposition gives condition under which this skew brace is a symmetric skew
brace.

######  Proposition 8.4.

A Rota–Baxter skew brace $(G,\cdot,\circ)$ is symmetric if and only if $B$ is
an anti-homomorphism modulo $Z(G^{(\cdot)})$, that is,

$B(c)^{-1}B(a)^{-1}B(ca)\in Z(G^{(\cdot)})$

for all $a,c\in G$. In particular, if $B$ is an anti-homomorphism, then
$(G,\cdot,\circ)$ is a symmetric skew brace.

###### Proof.

As we know a skew brace is symmetric if and only if

$\lambda_{a\circ c}(b)=\lambda_{c\cdot a}(b),~{}~{}~{}a,b,c\in G.$

Rewrite the left side using part c) of Proposition 8.1, we get

$\lambda_{a\circ c}(b)=B(a\circ c)bB(a\circ
c)^{-1}=B(a)B(c)bB(a)^{-1}B(c)^{-1}.$

Rewriting the right side gives

$\lambda_{c\cdot a}(b)=B(ca)bB(ca)^{-1}.$

Hence,

$[b,B(c)^{-1}B(a)^{-1}B(ca)]=1.$

Since $b$ is arbitrary element of $G$, one gets $B(c)^{-1}B(a)^{-1}B(ca)\in
Z(G^{(\cdot)})$.

Converse part is left as an exercise. ∎

######  Corollary 8.5.

If $B$ is an endomorphism of $G$ onto some abelian subgroup of $G$, then
$(G,\cdot,\circ)$ is a symmetric skew brace.

The next lemma follows from the definition of the Rota–Baxter operator

######  Lemma 8.6.

If a Rota–Baxter operator $B:G\to G$ is an anti-homomorphism, then

$[B(b),B^{2}(a)B(a)]=1$

for all $a,b\in G$.

Connection between Rota–Baxter skew braces and $\lambda$-homomorphic skew left
braces is given by

######  Proposition 8.7 ([6]).

A Rota–Baxter skew brace $(G,\cdot,\circ)$ is $\lambda$-homomorphic if and
only if $B(ac)^{-1}B(a)B(c)\subseteq Z(G^{(\cdot)})$ for all $a,c\in G$.

The following concept of a skew left multibrace was introduced in [6].

######  Definition 8.8.

Let $k$ be a natural number. By a skew left $k$-brace we mean a
$(k+1)$-groupoid $(G,\circ_{0},\circ_{1},\ldots,\circ_{k})$, i. e., a non-
empty set $G$ with $k+1$ binary algebraic operations, such that

1) $(G,\circ_{i})$ is a group for all $i=0,1,\ldots,k$;

2) for $0<i\leq k$

$a\circ_{i}(b\circ_{i-1}c)=(a\circ_{i}b)\circ_{i-1}a^{\circ_{i-1}(-1)}\circ_{i-1}(a\circ_{i}c),$

where $a^{\circ_{i-1}(-1)}$ is the inverse to $a$ in the group
$(G,\circ_{i-1})$.

A skew left 1-brace is just the skew left brace. By a skew left multibrace we
mean a skew left $k$-brace for some $k>1$. In our terminology skew left
multibrace is a brace system.

On a Rota–Baxter group one can construct multibrace using the following
proposition.

######  Proposition 8.9 ([6]).

Let $(G,\cdot,B)$ be an RB-group and $k$ be a natural number. Define on the
set $G$ binary operations $\circ_{1},\circ_{2},\ldots,\circ_{k}$ as follows,

$x\circ_{i+1}y=x\circ_{i}B(x)\circ_{i}y\circ_{i}(B(x))^{\circ_{i}(-1)},$

where $\circ_{0}=\cdot$. Then $(G,\cdot,\circ_{1},\circ_{2},\ldots,\circ_{k})$
is a skew left $k$-brace.

If we take a skew brace $(G,\circ_{i},\circ_{i+1})$, then

$\lambda^{(i)}_{x}(y)=B(x)\circ_{i}y\circ_{i}B(x)^{\circ_{i}(-1)}.$

Just to demonstrate, let us rewrite first two new operations in terms of
original operation $\cdot$. We have

$x\circ y=x\circ_{1}y=x\cdot B(x)\cdot y\cdot(B(x))^{-1},$

and

$\lambda_{x}(y)=\lambda^{(0)}_{x}(y)=B(x)yB(x)^{-1}$

Further,

$x\circ_{2}y=x\circ_{1}B(x)\circ_{1}y\circ_{1}(B(x))^{\circ_{1}(-1)},$

and by Proposition 8.3, using the equality
$(B(x))^{\circ_{1}(-1)}=(B^{2}(x))^{-1}B(x)^{-1}B^{2}(x)$, we get

$x\circ_{2}y=x(B(x))^{2}B^{2}(x)yB(y)(B^{2}(x))^{-1}B(x)^{-1}B^{2}(x)B(y)^{-1}(B^{2}(x))^{-1}B(x)^{-1}.$

In this case

$\displaystyle\lambda^{(1)}_{x}(y)$ $\displaystyle=$ $\displaystyle B(x)\circ
y\circ B(x)^{\circ(-1)}=B(x)\circ
y\circ\left((B^{2}(x))^{-1}B(x)^{-1}B^{2}(x)\right)$ $\displaystyle=$
$\displaystyle
B(x)B^{2}(x)yB(y)(B^{2}(x))^{-1}B(x)^{-1}B^{2}(x)B(y)^{-1}(B^{2}(x))^{-1}.$

Let $(G,\cdot)$ be a group. Then the map $B(g)=g^{-1}$, $g\in G$, is a
Rota–Baxter operator on $G$. We see that $B$ is an anti-homomorphism. The skew
brace constructed from the Rota–Baxter group $(G,\cdot,B)$ is the one as in
Example 3.3. We conclude with another concrete example.

######  Example 8.10.

Let $G=F_{2}(x,y)$ be a free group of rank 2. The homomorphism $B:G\to\langle
x\rangle$, which is defined on the generators by $B(x)=x$, $B(y)=x$ is a
Rota–Baxter operator on $G$. Let us construct multi-brace which it defines.
For $a,b\in G$, we define

$a\circ b=ax^{l(a)}bx^{-l(a)},$

where $l(a)$ is as defined in the preceding section. Inductively, for any
integer $m\geq 1$, we can define binary operations

$a\circ_{m}b=ax^{ml(a)}bx^{-ml(a)}.$

It is not difficult to see that $(G,\cdot,\circ_{1},\circ_{2},\ldots)$ is a
multibrace with infinitely many binary operations.

We close with the comment that connection between Rota–Baxter skew braces and
skew braces constructed from $\gamma$-functions (closely related to
$\lambda$-homomorphic skew braces) in terms of second cohomology group has
been very recently investigated in [12].

Acknowledgements. The first author is supported by the Ministry of Science and
Higher Education of Russia (agreement No. 075-02-2021-1392). The first and
second named authors acknowledge the support of RFBR (project No.
19-01-00569). The third named author acknowledges the support of DST-SERB
Grant MTR/2021/000285. The authors are grateful to the unanimous referee for
making several useful suggestions, which improved the quality of the
manuscript. The authors thank Lorenzo Stefanello for pointed out a gap in an
argument (regarding step of poly-triviality of symmetric skew braces) at the
end of Section 6 (of earlier version) of this article (which we have modified
now).

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