Free Products of digroups

TL;DR

This paper constructs free products of digroups, providing a solution to an open problem and advancing the understanding of algebraic structures called digroups.

Contribution

It introduces a method to construct free products of arbitrary digroups, addressing a previously open problem in the field.

Findings

Successfully constructs free products of digroups

Solves an open problem posed by Zhuchok

Provides a new framework for digroup algebraic structures

Abstract

We construct the free products of arbitrary digroups, and thus we solve an open problem of Zhuchok.

Equations

$$\{\begin{array}{l}(D,⊢)\text{mbox}and(D,⊣)\text{mbox}arebothsemigroups,\\ a⊣(b⊢c)=a⊣(b⊣c),\\ (a⊣b)⊢c=(a⊢b)⊢c,\\ a⊢(b⊣c)=(a⊢b)⊣c,\end{array}$$

$$(\forall a\in D,\exists a^{-1}\in D)a⊢a^{-1}=e=a^{-1}⊣a.$$

$$a=e⊢a=(b⊣c)⊢a=b⊢(c⊢a)=b⊢e=b⊢(c⊢b)=(b⊣c)⊢b=e⊢b=b.$$

$$(a⊢e)⊣a^{-1}$$

$$=(e⊣a)⊣a^{-1}=e⊣(a⊣a^{-1})=e⊣(a⊢a^{-1})=e.$$

$$a^{-1}=e⊢a^{-1}=(a^{-1}⊣a)⊢a^{-1}=a^{-1}⊢(a⊢a^{-1})=a^{-1}⊢e.$$

$$(a⊢e)⊢(b⊢e)$$

$$=(a⊢e)⊣(b⊣e)=(a⊢e)⊣(b⊢e).$$

$$\{d\in D\mid d⊢a=a⊣d=a\text{mbox}foreverya\in D\}$$

$$b=e⊢b=(a^{-1}⊣a)⊢b=a^{-1}⊢(a⊢b)\in J,$$

$$(D_i,e_i)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$D={\cup }_{i\in I}{D}_i^{\prime }\text{mbox}andJ={\cup }_{i\in I}{J}_i.$$

$$\underline{\underline{u}\underline{v}}=\underline{\underline{u}v}=\underline{u\underline{v}}=\underline{uv}\text{mbox}and\underline{eu}=\underline{ue}=\underline{u}.$$

$$a^{♯}=a⊢e.$$

$$u^{♯}=a_1^{♯}{a}_2^{♯}...a_m^{♯}$$

$${\epsilon }^{♯}=e,$$

$$a_1⊢a_2\cdots ⊢a_m⊣\cdots ⊣a_t$$

$$u\dot{a}v=a_1...a_m\dot{a}{b}_1...b_n,$$

$$a_1...a_m\dot{a}{b}_1...b_n=a_1⊢\cdots ⊢a_m⊢a⊣b_1⊣\cdots ⊣b_n.$$

$$a_i⊢a_j=(a_i⊣e)⊢a_j=(a_i⊢e)⊢a_j=a_i^{♯}⊢a_j$$

$$a_i⊣a_j=a_i⊣(a_j⊣e)=a_i⊣(a_j⊢e)=a_i⊣a_j^{♯}.$$

$$CW(D)=\{u\dot{a}v\mid u,v\in J^{\ast },a\in D,uav\in W(D);\text{mbox}ifa\in J,\text{mbox}thenu=\epsilon \}.$$

$$[u\dot{a}v]=\{\begin{array}{ll}u\dot{a}v,& \text{mbox}if{a}_m,a\text{mbox}donotlieinthesameset{D}_i^{\prime }\text{mbox}foranyi\in I,\text{mbox}anda,b_1\\ & \text{mbox}donotlieinthesameset{D}_j^{\prime }\text{mbox}foranyj\in I;\\ a_1...a_{m-1}\dot{c}v,& \text{mbox}if{a}_m,a\in D_i^{\prime }\text{mbox}forsomeindexi\in I,a_m⊢a=c\in D_i^{\prime },b_1\notin D_i^{\prime };\\ u\dot{c}{b}_2...b_n,& \text{mbox}ifa,b_1\in D_i^{\prime }\text{mbox}forsomeindexi\in I,a⊣b_1=c\in D_i^{\prime },a_m\notin D_i^{\prime };\\ a_1...a_{m-1}\dot{c}{b}_2...b_n,& \text{mbox}if{a}_m,a,b_1\in D_i^{\prime }\text{mbox}forsomeindexi\in I,(a_m⊢a)⊣b_1=c\in D_i^{\prime }.\end{array}$$

$$[u]={\dot{a}}_1...a_m.$$

$$c^{♯}=(a_m⊢a)⊢e=(a_m⊢e)⊢(a⊢e)=\underline{a_m^{♯}{a}^{♯}}.$$

$$c^{♯}=(a⊣b_1)⊢e=(a⊢e)⊢(b_1⊢e)=\underline{a^{♯}{b}_1^{♯}}.$$

$$c^{♯}=((a_m⊢a)⊣b_1)⊢e=(a_m⊢e)⊢(a⊢e)⊢(b_1⊢e)=\underline{a_m^{♯}{a}^{♯}{b}_1^{♯}}.$$

$$u_1{\dot{a}}_1{v}_1⊢u_2{\dot{a}}_2{v}_2=\{\begin{array}{ll}[\underline{u_1^{♯}{a}_1^{♯}{v}_1^{♯}{u}_2^{♯}}{\dot{a}}_2{v}_2^{♯}],& \text{mbox}if{a}_2\in D\setminus J;\\ & \\ [\underline{u_1^{♯}{a}_1^{♯}{v}_1^{♯}{u}_2^{♯}{a}_2^{♯}{v}_2^{♯}}],& \text{mbox}if{a}_2\in J,\end{array}$$

$$u_1{\dot{a}}_1{v}_1⊣u_2{\dot{a}}_2{v}_2=\{\begin{array}{ll}[u_1^{♯}{\dot{a}}_1\underline{v_1^{♯}{u}_2^{♯}{a}_2^{♯}{v}_2^{♯}}],& \text{mbox}if{a}_1\in D\setminus J;\\ & \\ [\underline{u_1^{♯}{a}_1^{♯}{v}_1^{♯}{u}_2^{♯}{a}_2^{♯}{v}_2^{♯}}],& \text{mbox}if{a}_1\in J.\end{array}$$

$$u_1{\dot{a}}_1{v}_1⊢u_2{\dot{a}}_2{v}_2=[\underline{u_1^{♯}{a}_1^{♯}{v}_1^{♯}{u}_2^{♯}{a}_2^{♯}{v}_2^{♯}}]=[\underline{\underline{u_1^{♯}{a}_1^{♯}{v}_1^{♯}}{u}_2^{♯}{a}_2^{♯}{v}_2^{♯}}]=[\underline{u_1^{♯}{a}_1^{♯}{v}_1^{♯}}]⊢u_2{\dot{a}}_2{v}_2.$$

$$u_1{\dot{a}}_1{v}_1⊢u_2{\dot{a}}_2{v}_2=[\underline{u_1^{♯}{a}_1^{♯}{v}_1^{♯}{u}_2^{♯}}{\dot{a}}_2{v}_2^{♯}]=[\underline{\underline{u_1^{♯}{a}_1^{♯}{v}_1^{♯}}{u}_2^{♯}}{\dot{a}}_2{v}_2^{♯}]=[\underline{u_1^{♯}{a}_1^{♯}{v}_1^{♯}}]⊢u_2{\dot{a}}_2{v}_2.$$

Full text

∗ Supported by the NNSF of China (11571121, 12071156)

♯ Supported by Young Teacher Research and Cultivation Foundation of South China Normal University 20KJ02

♯ Corresponding author

1. Introduction

A digroup is a generalization of a group, and it first appeared in Loday’s work [6]. Digroups play an important role in the theory of Leibniz algebras in the problem on finding an appropriate generalization of Lie’s third theorem which associates a (local) Lie group to any Lie algebra. Kinyon made a good progress in terms of digroups to build a partial solution to the above open problem on Leibniz algebras [4, Corollary 5.7].

The study of algebraic properties of digroups has attracted considerable attention. A digroup is a set with two binary associative operations satisfying some additional conditions (Definition 2.2). Kinyon [4] showed that every digroup is a product of a group and a trivial digroup [4]. In [8], Phillips offered an equivalent definition for the variety of digroups. Salazar-Díaz, Velásquez and Wills-Toro [9] studied a further generalization of the digroup structure and showed an analogous to the first isomorphism theorem. Ongay, Velásquez and Wills-Toro [7] discussed the notions of normal subdigroups and quotient digroups, and then established the corresponding analogues of the classical Isomorphism Theorems. In [11], A. V. Zhuchok and Y. V. Zhuchok constructed some interesting digroups. For further investigation on digroups, we refer to Zhuchok [10], Felipe [2] and Liu [5]. One of the open problems raised by Zhuchok in [10] is how to construct the free product of digroups. And our aim of this article is to construct the free products of arbitrary digroups.

2. Free products of arbitrary digroups

Recall that a disemigroup $(D,\vdash,\dashv)$ is a set $D$ equipped with two binary operations $\vdash$ and $\dashv$ such that the following conditions are satisfied:

[TABLE]

for all $a,b,c\in D$. An element $e$ in $D$ is called a bar-unit if $e\vdash a=a\dashv e=a$ for every $a\in D$. And a disemigroup $(D,\vdash,\dashv)$ is called a dimonoid if $(D,\vdash,\dashv)$ has a bar-unit. From now on, we shall simply say that $D$ is a disemigroup if $(D,\vdash,\dashv)$ is.

Now we recall an example of a dimonoid, where each element of the dimonoid can be a bar-unit. Moreover, it is in fact a trivial digroup in the sense of [4].

Example 2.1**.**

[6, Example 1.3.b] Let $D=\{e_{i}\mid i\in I\}$, where $I$ is an index set. Define $e_{i}\vdash e_{j}=e_{j}$ and $e_{j}\dashv e_{i}=e_{j}$ for all $i,j\in I$. Then $(D,\vdash,\dashv)$ is a disemigroup and every $e_{i}$ can be a bar-unit of $D$.

Definition 2.2**.**

([4, 2, 1, 5, 8, 11]) A dimonoid $D$ with a fixed bar-unit $e$ is called a digroup with respect to $e$, denoted by $(D,e)$, if $D$ satisfies the following condition:

[TABLE]

In this case, we call $e$ the fixed bar-unit of the digroup $(D,e)$, and call $a^{-1}$ the inverse of $a$ (with respect to $e$).

A more precise notation for the inverse might be $a^{-1}_{e}$, but since the fixed bar-unit is always clear whenever we write a digroup $(D,e)$, there is no harm to use the notation $a^{-1}$ instead of $a^{-1}_{e}$ for the inverse of $a$. We refer to [1] for more interesting properties of digroups. For instance, from [1] we easily deduce that the inverse of $a$ is unique. We shortly repeat the proof for the convenience of the readers. Suppose that $a$ and $b$ are two inverses of $c$, then we have

[TABLE]

Now we recall a useful result which describes the “group part” of a digroup. For the convenience of the readers, we quickly repeat the proof.

Lemma 2.3**.**

([4])* Let $(D,e)$ be a digroup, and let $J=\{a^{-1}\mid a\in D\}$ be the set of all inverses in $D$. Then we have $J=\{a\vdash e\mid a\in D\}$. Moreover, $J$ is a group with unit $e$ such that the operations $\vdash$ and $\dashv$ coincide in $J$.*

Proof.

For every $a\in D$, we have $a^{-1}\vdash(a\vdash e)=(a^{-1}\dashv a)\vdash e=e\vdash e=e$ and

[TABLE]

So we deduce $a\vdash e=(a^{-1})^{-1}\in J$ and $\{a\vdash e\mid a\in D\}\subseteq J$. On the other hand, for every $a^{-1}\in J$, we have

[TABLE]

Thus we obtain $\{a\vdash e\mid a\in D\}=J$.

Now we show that the operations $\vdash$ and $\dashv$ coincide in $J$. For all $a\vdash e,b\vdash e\in J$, we have

[TABLE]

Clearly, $e=e\vdash e\in J$. By the above reasoning, $J$ is a group with the unit $e$. ∎

With the notation of Lemma 2.3, the set $J$ is called the group part of $(D,e)$ and the set

[TABLE]

is called the halo part of the digroup $(D,e)$.

Note that the fixed bar-unit $e$ plays an important role when one describes a digroup. For instance, let $D$ be the disemigroup in Example 2.1. If we let $e_{i}$ be the fixed bar-unit, then $(D,e_{i})$ is a digroup and the inverse of $e_{j}$ is $e_{i}$ for every $j\in I$. In particular, the group part of $(D,e_{i})$ is $\{e_{i}\}$.

Now we show that the product of an element in the group part and an element $b$ not in the group part such that the product “point to” $b$ does not lie in the group part.

Lemma 2.4**.**

Let $(D,e)$ be a digroup with the group part $J$. Then for every $a\in J$ and for every $b\in D\setminus J$, we have $a\vdash b\in D\setminus J$ and $b\dashv a\in D\setminus J$.

Proof.

If $a\vdash b\in J$, then we have $a^{-1}\vdash(a\vdash b)\in J$ by Lemma 2.3. So we obtain

[TABLE]

which is a contradiction. Thus we deduce $a\vdash b\in D\setminus J$. Similarly, we have $b\dashv a\in D\setminus J$. ∎

Our aim in this article is to construct the free products of arbitrary digroups. So we first recall some necessary notions and recall the definition of the free product of digroups. A map $\varphi$ from a disemigroup $D_{1}$ to a disemigroup $D_{2}$ is called a disemigroup homomorphism if we have $\varphi(a\vdash b)=\varphi(a)\vdash\varphi(b)$ and $\varphi(a\dashv b)=\varphi(a)\dashv\varphi(b)$ for all $a,b\in D_{1}$. Let $(D_{1},e_{1})$ and $(D_{2},e_{2})$ be two digroups. A disemigroup homomorphism $\varphi$ from $D_{1}$ to $D_{2}$ is called a digroup homomorphism from $(D_{1},e_{1})$ to $(D_{2},e_{2})$ if $\varphi(e_{1})=e_{2}$.

Definition 2.5**.**

Let $\{(D_{i},e_{i})\mid i\in I\}$ be a family of indexed digroups, and let $(D,e)$ be a digroup such that

(i) there exists an injective digroup homomorphism $\varphi_{i}:(D_{i},e_{i})\longrightarrow(D,e)$ for every $i\in I$;

(ii) if $(G,e^{\prime})$ is a digroup, and if there exists a digroup homomorphism $\psi_{i}:(D_{i},e_{i})\rightarrow(G,e^{\prime})$ for every $i\in I$, then there exists a unique digroup homomorphism $\eta:(D,e)\rightarrow(G,e^{\prime})$ such that the diagram

[TABLE]

commutes for every $i\in I$.

Then $(D,e)$ is called a free product of $\{(D_{i},e_{i})\mid i\in I\}$.

Clearly, if a free product of digroups $\{(D_{i},e_{i})\mid i\in I\}$ exists, then it is unique up to isomorphism.

Remark 2.6**.**

Let $\{(D_{i},e_{i})\mid i\in I\}$ be a family of digroups, and let $\{(D_{i}^{\prime\prime},e_{i}^{\prime\prime})\mid i\in I\}$ be a family of digroups such that $(D_{i}^{\prime\prime},e_{i}^{\prime\prime})$ is isomorphic to $(D_{i},e_{i})$ for every $i\in I$. Then clearly the free product of $\{(D_{i},e_{i})\mid i\in I\}$ is isomorphic to that of $\{(D_{i}^{\prime\prime},e_{i}^{\prime\prime})\mid i\in I\}$. So there is no harm to assume that $\{(D_{i},e_{i})\mid i\in I\}$ is a family of disjoint digroups. From now on, we assume that $\{(D_{i},e_{i})\mid i\in I\}$ is a family of disjoint digroups and we shall identify $(D_{i},e_{i})$ with $((D_{i}\setminus\{e_{i}\})\cup\{e\},e)$ for every $i\in I$ with the obvious multiplication, where $e$ is a symbol that is not in $\cup_{i}D_{i}$. Then $(D_{i},e_{i})$ and $((D_{i}\setminus\{e_{i}\})\cup\{e\},e)$ are isomorphic as digroups. We denote by $(D_{i}^{\prime},e)$ the digroup $((D_{i}\setminus\{e_{i}\})\cup\{e\},e)$ and denote by $(J_{i},e)$ the group part of $(D_{i}^{\prime},e)$. Now we fix the notation $D$ and $J$ defined by

[TABLE]

We shall construct the free product of $\{(D_{i}^{\prime},e)\mid i\in I\}$ in Theorem 2.15.

Let $J^{+}$ (resp. $J^{*}$) be the free semigroup (resp. monoid) generated by $J$, where $J$ is defined in Remark 2.6. Note that $J_{i}$ is a group for every $i\in I$, and for all distinct $i,t\in I$, we have $J_{i}\cap J_{t}=\{e\}$. Denote by $\Pi_{i\in I}^{\ast}J_{i}$ the free product of the groups $\{J_{i}\mid i\in I\}$. Moreover, by the construction of the free products of groups [3, Section 2.9], we can consider $J_{i}$ as a subgroup of $\Pi_{i\in I}^{\ast}J_{i}$ for every $i\in I$.

Now we introduce two special kinds of words over $D$, which resembles the elements in the free product of semigroups or groups.

Definition 2.7**.**

For every $a\in D\setminus\{e\}$, there exists a unique index $i\in I$ such that $a\in D_{i}^{\prime}$, we refer to $i$ as the index of $a$, and denote it by $\sigma(a)$. A word $u$ is called a good word over $D$, if $u=e$, or if $u=c_{1}c_{2}\cdots c_{n}$, $n\geq 1$, each $c_{t}$ lies in $D\setminus\{e\}$ and $\sigma(c_{p})\neq\sigma(c_{p+1})$ for every integer $p\leq n-1$. Denote by $W(D)$ the set of all good words over $D$. A word $u$ (over $D$) is called a reduced word in $J^{+}$ if $u$ lies in $W(D)\cap J^{+}$.

We also note that, the set of all reduced words in $J^{+}$ forms a set of normal forms of elements in $\Pi_{i\in I}^{\ast}J_{i}$. And this is why we use the name “reduced words”. More precisely, let $\theta$ be a semigroup homomorphism from $J^{+}$ to $\Pi_{i\in I}^{\ast}J_{i}$ such that $\theta(a)=a$ for every $a\in J$. Then clearly $\theta$ is a surjective homomorphism. So there exists a congruence $\rho$ on $J^{+}$ such that we have $J^{+}/\rho\cong\Pi_{i\in I}^{\ast}J_{i}$. By the construction of the free products of arbitrary groups, for every word $u\in J^{+}$, there exists a unique reduced word $\underline{u}$ in $J^{+}$ such that we have $u\rho=\underline{u}\rho$ in $J^{+}/\rho$. For instance, let $a\in J_{1}$, $b\in J_{2}\setminus\{e\}$ such that $aa=c\neq e$ in $J_{1}$. Then we have $\underline{aab}=cb$. In particular, if $u$ is a reduced word, then we have $\underline{u}=u$ in $J^{+}$. Moreover, it follows that $\underline{\ }$ is a map from $J^{+}$ to $J^{+}$ such that for all $u,v\in J^{+}$, we have

[TABLE]

Now we introduce a map ♯ from $D^{*}$ to $D^{*}$. For every $a\in D$, we define

[TABLE]

Note that, if $a\in D_{i}^{\prime}$, then $a\vdash e$ means the product in $D_{i}^{\prime}$. Since all the $D_{i}^{\prime}\setminus\{e\}$ and $D_{j}^{\prime}\setminus\{e\}$ are disjoint for all $i\neq j\in I$, the notation $a\vdash e$ makes sense and $a\vdash e$ means an element in $D_{i}^{\prime}$. By Lemma 2.3, we have $a^{\sharp}=a\vdash e\in J\subseteq J^{+}$. Moreover, for every word $u=a_{1}a_{2}...a_{m}\in D^{+}$, where $D^{+}$ is the free semigroup generated by $D$ and every $a_{t}$ lies in $D$, we define

[TABLE]

to be an element in $J^{+}\subseteq D^{*}$. Clearly, $J^{+}$ is a subsemigroup of $D^{*}$, where $D^{*}$ is the free monoid generated by $D$. Then we obtain that $\underline{u^{\sharp}}=\underline{a_{1}^{\sharp}a_{2}^{\sharp}...a_{m}^{\sharp}}$ is a reduced word in $J^{+}$. Now we extend ♯ by defining

[TABLE]

where $\varepsilon$ is the empty word. Then ♯ becomes a map from $D^{*}$ to $D^{*}$ such that the image of $D^{*}$ is contained in $J^{+}$.

The following result is an obvious corollary of Lemma 2.3, and thus the proof is omitted.

Lemma 2.8**.**

The map ♯ is a semigroup homomorphism from $D^{*}$ to $D^{*}$ such that for every $a\in J$, we have $a^{\sharp}=a$. Moreover, for every $u\in D^{*}$, we have $({u}^{\sharp})^{\sharp}=u^{\sharp}\in J^{+}$.

Recall that for every disemigroup $M$, for all $a_{1},...,a_{t}$ in $M$, every parenthesizing of

[TABLE]

gives the same element in $M$ [6]. In light of this, we define a formal expression $u\dot{a}v$ as follows: for all reduced words $u,v\in J^{+}$ and for every $a\in D$, say $u=a_{1}...a_{m}$ and $v=b_{1}...b_{n}$, where all $a_{1},...,a_{m},b_{1},...,b_{n}$ lie in $J$, $m\geq 0$ and $n\geq 0$, we define

[TABLE]

where if $m$ (resp. $n$) is 0, that is, when $u$ (resp. $v$) is the empty word, it means that $u$ (resp. $v$) does not appear. In particular, if $u$ and $v$ are both the empty word, then by $u\dot{a}v$ we mean $\dot{a}$. Therefore, for all words $u,v\in D^{*}$ and for every $a\in D$, the formula $\underline{u^{\sharp}}\dot{a}\underline{v^{\sharp}}$ makes sense.

For the moment, the formula $a_{1}...a_{m}\dot{a}b_{1}...b_{n}$ is just a formal expression that does not lie in any digroup. But we shall construct a set consisting of some of such elements and define certain products on the elements of the set to make it a digroup such that we have

[TABLE]

Suppose that we already have the free product of $\{(D_{i}^{\prime},e)\mid i\in I\}$. Then for all $a_{i}\in D_{i}^{\prime}$ and $a_{j}\in D_{j}^{\prime}$, in the free product, we have

[TABLE]

and

[TABLE]

Therefore, for an arbitrary element $a_{1}\vdash\cdots\vdash a_{m}\vdash a\dashv b_{1}\dashv\cdots\dashv b_{n}$ in the free product of $\{(D_{i}^{\prime},e)\mid i\in I\}$, we may always assume $a_{p}\in J$ and $b_{t}\in J$ for all $p\leq m$ and $t\leq n$. This motivates us to introduce the notion of good center-words as follows. Recall that $W(D)$ is the set of all good words over $D$ (Definition 2.7). We call $u\dot{a}v$ a good center-word over $D$ if we have

(i) $u,v\in J^{*}$, $a\in D$, $uav\in W(D)$;

(ii) if $a\in J$, then $u$ is the empty word.

Denote by $CW(D)$ the set of all good center-words over $D$, that is,

[TABLE]

Now we introduce an extended reduction rule on certain “center-words” as follows. For all reduced words $u=a_{1}...a_{m},v=b_{1}...b_{n}\in J^{+}$, where all $a_{1},...,a_{m},b_{1},...,b_{n}$ lie in $J$, and for every $a\in D\setminus J$, we define

[TABLE]

By Lemma 2.4, it follows that $[u\dot{a}v]$ is a good center-word in $CW(D)$. Finally, for every reduced word $u=a_{1}...a_{m}\in J^{+}$, where $a_{1},...\leavevmode\hbox{}\hskip 0.56905pt,a_{m}$ lie in $J$, we define

[TABLE]

In particular, for every word $u\in D^{*}$, the formula $[\underline{u^{\sharp}}]$ makes sense.

The next lemma shows that the extended reduction rule is compatible with the reduction rule in $\Pi_{i\in I}^{\ast}J_{i}$ in the following sense.

Lemma 2.9**.**

For all reduced words $u=a_{1}...a_{m}$ and $v=b_{1}...b_{n}\in J^{+}$, where all $a_{i},b_{j}$ lie in $J$, $m\geq 0$ and $n\geq 0$, for every $a\in D\setminus J$, if $[u\dot{a}v]=c_{1}...c_{p}\dot{c}d_{1}...d_{q}$, then we have $\underline{a_{1}^{\sharp}...a_{m}^{\sharp}a^{\sharp}b_{1}^{\sharp}...b_{n}^{\sharp}}=\underline{c_{1}^{\sharp}...c_{p}^{\sharp}c^{\sharp}d_{1}^{\sharp}...d_{q}^{\sharp}}$.

Proof.

For every index $i\in I$, we have the following three observations: (i) If $a_{m}\vdash a=c$ in $D_{i}^{\prime}$, then since by Lemma 2.3 the operations $\vdash$ and $\dashv$ coincide in $J_{i}$, we have

[TABLE]

(ii) If $a\dashv b_{1}=c$ in $D_{i}^{\prime}$, then we have

[TABLE]

(iii) If $(a_{m}\vdash a)\dashv b_{1}=c$ in $D_{i}^{\prime}$, then we have

[TABLE]

The proof of the lemma follows immediately by the definition of $[u\dot{a}v]$. ∎

With the above notations, we can define new operations $\vdash$ and $\dashv$ on the set of all good center-words over $D$ in Definition 2.10. Since the operations extend those of $D_{i}^{\prime}$ for every $i\in I$, we use the same symbol $\vdash$ and $\dashv$ but not $\vdash^{\prime},\dashv^{\prime}$. We shall prove that the following definition turns out to be a way of constructing the free products of arbitrary digroups in Theorem 2.15.

Definition 2.10**.**

With the notion of Remark 2.6, let $CW(D)$ be the set of all good center-words over $D$ defined in (2.2). We define two operations $\vdash$ and $\dashv$ on $CW(D)$ as follows. For all $u_{1}\dot{a}_{1}v_{1},u_{2}\dot{a}_{2}v_{2}\in CW(D)$, we define

[TABLE]

and

[TABLE]

Before showing that $(CW(D),\vdash,\dashv)$ is a digroup, we first establish several useful properties of $\vdash$ and $\dashv$. Now we show that, when the product “points to” an element, we may assume that the other element is of certain form.

Lemma 2.11**.**

With the notation of Definition 2.10, for all $u_{1}\dot{a}_{1}v_{1},u_{2}\dot{a}_{2}v_{2}\in CW(D)$, we have $u_{1}\dot{a}_{1}v_{1}\vdash u_{2}\dot{a}_{2}v_{2}=[\underline{u_{1}^{\sharp}a_{1}^{\sharp}v_{1}^{\sharp}}]\vdash u_{2}\dot{a}_{2}v_{2}$ and $u_{1}\dot{a}_{1}v_{1}\dashv u_{2}\dot{a}_{2}v_{2}=u_{1}\dot{a}_{1}v_{1}\dashv[\underline{u_{2}^{\sharp}a_{2}^{\sharp}v_{2}^{\sharp}}]$.

Proof.

We just prove $u_{1}\dot{a}_{1}v_{1}\vdash u_{2}\dot{a}_{2}v_{2}=[\underline{u_{1}^{\sharp}a_{1}^{\sharp}v_{1}^{\sharp}}]\vdash u_{2}\dot{a}_{2}v_{2}$ because the other one can be proved similarly. If $a_{2}\in J$, then we obtain

[TABLE]

If $a_{2}\notin J$, then we have

[TABLE]

The proof is completed. ∎

The next two lemmas show that the extended reduction rule is compatible with the operations $\vdash$ and $\dashv$.

Lemma 2.12**.**

For all reduced words $u,v\in J^{+}$, and for every $a\in D\setminus J$, we have $[\underline{uv}\dot{a}]=[u]\vdash[v\dot{a}]$ and $[\dot{a}\underline{uv}]=[\dot{a}u]\dashv[v]$. Moreover, for every reduced word $w\in J^{+}$, if $aw$ is a good word or if $w=e$, then we have $[\underline{uv}\dot{a}w]=[u]\vdash[v\dot{a}w]$; and if $wa$ is a good word or if $w=e$, then we have $[w\dot{a}\underline{uv}]=[w\dot{a}u]\dashv[v]$.

Proof.

We first show $[\underline{uv}\dot{a}]=[u]\vdash[v\dot{a}]$. Assume that $u=a_{1}...a_{m}$, $v=b_{1}...b_{n}$, where all $a_{p},b_{q}$ lie in $J$ and $m,n$ are positive integers. There are several cases to consider:

Case 1: If $a,b_{n}\notin D_{j}^{\prime}$ for any $j\in I$, then by Definition 2.10, we have

[TABLE]

Case 2: If $a,b_{n}\in D_{j}^{\prime}$ for some $j\in I$ such that $b_{n}\vdash a=b$ in $D_{j}^{\prime}$, and if $\underline{uv}=\underline{ub_{1}...b_{n-1}}b_{n}$ as words in the free semigroup $J^{+}$, then by Definition 2.10, we have

[TABLE]

Case 3: If $a,b_{n}\in D_{j}^{\prime}$ for some $j\in I$ such that $b_{n}\vdash a=b$ in $D_{j}^{\prime}$, and if $\underline{uv}\neq\underline{ub_{1}...b_{n-1}}b_{n}$ as words in the free semigroup $J^{+}$, then we deduce

[TABLE]

as words in the free semigroup $J^{+}$, and we have $a_{m-n+1},b_{n},a\in D_{j}^{\prime}$. Moreover, for every $0\leq k\leq n-2$, we have $b_{k+1}=a_{m-k}^{-1}\in J_{i_{k}}$ for some $i_{k}\in I$. If $b_{n}=a_{m-n+1}^{-1}$, then we have $\underline{uv}=a_{1}...a_{m-n}$ and

[TABLE]

So we deduce

[TABLE]

If $b_{n}\neq a_{m-n+1}^{-1}$, then we have $a_{m-n+1}\vdash b_{n}=c\neq e$. Assume $c\vdash a=d$. Then we have $\underline{uv}=a_{1}...a_{m-n}c$ in the free semigroup $J^{+}$ and

[TABLE]

So we deduce

[TABLE]

Similarly, we can show $[\dot{a}\underline{uv}]=[\dot{a}u]\dashv[v]$.

As for what remained, if $w=e$, then we have $[u\dot{a}w]=[u\dot{a}]$ and $[w\dot{a}u]=[\dot{a}u]$. So the claim follows by the above proof; if $w\neq e$, then by Lemma 2.4, what remained can be proved similarly by noting that $w$ is never involved in the above reasoning on the extended reduction rule. ∎

Lemma 2.13**.**

For all reduced words $u,v\in J^{+}$, and for every $a\in D\setminus J$, we have $[u\dot{a}v]=[u\dot{a}]\dashv[v]=[u]\vdash[\dot{a}v]$.

Proof.

By the definition of $[u\dot{a}v]$ and by Lemma 2.4, we clearly have $[u\dot{a}v]=[u\dot{a}]\dashv[v]$. Similarly, by the fact that $(b\vdash c)\dashv d=b\vdash(c\dashv d)$ for all $b,c,d$ in a disemigroup, we easily deduce $[u\dot{a}v]=[u]\vdash[\dot{a}v]$. ∎

Now we show that $CW(D)$ is a disemigroup with respect to the operations defined in Definition 2.10.

Lemma 2.14**.**

For all $u_{i}\dot{a}_{i}v_{i}\in CW(D)$, $1\leq i\leq 3$, we have

(i)* $(u_{1}\dot{a}_{1}v_{1}\vdash u_{2}\dot{a}_{2}v_{2})\vdash u_{3}\dot{a}_{3}v_{3}=(u_{1}\dot{a}_{1}v_{1}\dashv u_{2}\dot{a}_{2}v_{2})\vdash u_{3}\dot{a}_{3}v_{3}$;*

(ii)* $u_{1}\dot{a}_{1}v_{1}\dashv(u_{2}\dot{a}_{2}v_{2}\dashv u_{3}\dot{a}_{3}v_{3})=u_{1}\dot{a}_{1}v_{1}\dashv(u_{2}\dot{a}_{2}v_{2}\vdash u_{3}\dot{a}_{3}v_{3})$;*

(iii)* $(u_{1}\dot{a}_{1}v_{1}\vdash u_{2}\dot{a}_{2}v_{2})\vdash u_{3}\dot{a}_{3}v_{3}=u_{1}\dot{a}_{1}v_{1}\vdash(u_{2}\dot{a}_{2}v_{2}\vdash u_{3}\dot{a}_{3}v_{3})$;*

(iv)* $(u_{1}\dot{a}_{1}v_{1}\dashv u_{2}\dot{a}_{2}v_{2})\dashv u_{3}\dot{a}_{3}v_{3}=u_{1}\dot{a}_{1}v_{1}\dashv(u_{2}\dot{a}_{2}v_{2}\dashv u_{3}\dot{a}_{3}v_{3})$;*

(v)* $(u_{1}\dot{a}_{1}v_{1}\vdash u_{2}\dot{a}_{2}v_{2})\dashv u_{3}\dot{a}_{3}v_{3}=u_{1}\dot{a}_{1}v_{1}\vdash(u_{2}\dot{a}_{2}v_{2}\dashv u_{3}\dot{a}_{3}v_{3})$.

In particular, $CW(D)$ is a disemigroup.*

Proof.

We first prove Point (i). For every $\delta\in\{\vdash,\dashv\}$, assume $(u_{1}\dot{a}_{1}v_{1})\delta(u_{2}\dot{a}_{2}v_{2})=u\dot{a}v$. Then by Lemma 2.9 and by Definition 2.10, we have

[TABLE]

So by Lemma 2.11, we deduce

[TABLE]

In particular, we have $(u_{1}\dot{a}_{1}v_{1}\vdash u_{2}\dot{a}_{2}v_{2})\vdash u_{3}\dot{a}_{3}v_{3}=(u_{1}\dot{a}_{1}v_{1}\dashv u_{2}\dot{a}_{2}v_{2})\vdash u_{3}\dot{a}_{3}v_{3}$. Point (ii) can be proved similarly.

Now we prove Point (iii). If $a_{3}\notin J$, then by the above reasoning and by Lemmas 2.11 and 2.12, we have

[TABLE]

If $a_{3}\in J$, then by Lemma 2.11 and by the above reasoning, we have

[TABLE]

Point (iv) can be proved similarly.

Finally we prove Point (v). If $a_{2}\in J$, then clearly we have

[TABLE]

If $a_{2}\notin J$, then by Lemma 2.13, we have

[TABLE]

and

[TABLE]

By Points (iii) and (iv), and by Lemma 2.13 again, we deduce

[TABLE]

The proof is completed. ∎

Now we are ready to prove the main result of our article, in which we construct the free products of arbitrary digroups.

Theorem 2.15**.**

With Definition 2.10, $(CW(D),\dot{e})$ is the free product of the family of indexed digroups $\{(D_{i}^{\prime},e)\mid i\in I\}$. Moreover, the group part of $(CW(D),\dot{e})$ is $\{[\underline{u}]\mid u\mbox{ is a reduced word in }J^{+}\}$, which is isomorphic to $\Pi_{i\in I}^{\ast}J_{i}$, and the halo part of $(CW(D),\dot{e})$ is $\{u\dot{a}v\in CW(D)\mid\underline{u^{\sharp}a^{\sharp}v^{\sharp}}=e\}$.

Proof.

We first show that $(CW(D),\dot{e})$ is a digroup. By Lemma 2.14, we know that $CW(D)$ becomes a disemigroup. We proceed to show that $(CW(D),\dot{e})$ is a digroup. Clearly $\dot{e}$ is a bar-unit in the disemigroup $CW(D)$. Moreover, for every element $u\dot{a}v\in CW(D)$, since $\Pi_{i\in I}^{\ast}J_{i}$ is a group, by the construction of the free products of groups, there exists a reduced word $w\in J^{+}$ such that

[TABLE]

Thus we obtain

[TABLE]

Therefore, $(CW(D),\dot{e})$ is a digroup.

Now we show that $(CW(D),\dot{e})$ is the free product of $\{(D_{i}^{\prime},e)\mid i\in I\}$. For every $i\in I$, assume that $\varphi_{i}:(D_{i}^{\prime},e)\rightarrow(CW(D),\dot{e})$ is the injective digroup homomorphism which is defined by $\varphi_{i}(x)=[x]=\dot{x}$ for every $x\in D_{i}^{\prime}$. Let $(G,e^{\prime})$ be an arbitrary digroup such that there exists a digroup homomorphism $\psi_{i}:(D_{i}^{\prime},e)\rightarrow(G,e^{\prime})$ for every $i\in I$. Recall that for every element $d$ in $D\setminus\{e\}$, the notation $\sigma(d)$ means the index $i$ in $I$ such that $d$ lies in $D_{i}^{\prime}$. Then we define a map $\eta$ from $(CW(D),\dot{e})$ to $(G,e^{\prime})$ by the rule $\eta(\dot{e})=e^{\prime}$ and

[TABLE]

where $\dot{e}\neq a_{1}...a_{m}\dot{a}b_{1}...b_{n}\in CW(D)$, $m\geq 0$ and $n\geq 0$. If $m=0$ (resp. $n=0$), then $\psi_{\sigma(a_{1})}(a_{1})\vdash\cdots\vdash\psi_{\sigma(a_{m})}(a_{m})\vdash$ (resp. $\dashv\psi_{\sigma(b_{1})}(b_{1})\dashv\cdots\dashv\psi_{\sigma(b_{n})}(b_{n})$) does not appear. In particular, we have $\eta(\dot{a})=\psi_{\sigma(a)}(a)$ for every $a\in D\setminus\{e\}$.

Before showing that $\eta$ is a digroup homomorphism, we first note that the map $\eta$ is compatible with the extended reduction rule in the following sense: For all $a_{1},\dots,a_{m},b_{1},\dots,b_{n}$ in $J$ and for every $a\in D$, assume that $a_{t}\delta a_{t+1}=c$ in $D_{i}^{\prime}$ for some $i\in I$, where $\delta\in\{\vdash,\dashv\}$. Then clearly we have

[TABLE]

Similarly, if $a_{m}$ and $a$ lie in the same digroup, or if $b_{1},a$ or $b_{t},b_{t+1}$ lie in the same digroup, then a similar result holds. Therefore, if $a_{1}...a_{m}$ and $b_{1}...b_{n}$ are words in $J^{+}$ and if $a\in D\setminus{J}$, then no matter whether $\underline{a_{1}...a_{m}}\dot{a}\underline{b_{1}...b_{n}}$ is a good center-word or not, we have

[TABLE]

In particular, we deduce

[TABLE]

for all $a_{1},...,a_{m}\in J$.

We also note that, for every $c_{1}\in D_{i}^{\prime}$, we have

[TABLE]

By Lemma 2.3, it follows that $\psi_{i}(c_{1}\vdash e)$ lies in the group part of $G^{\prime}$, and in particular, for every $a_{p}\in J$, the element $\eta(\dot{a}_{p})$ lies in the group part of $G^{\prime}$. So by (2.4) and by Lemma 2.3, we obtain

[TABLE]

for every $t$ such that $1\leq t\leq m-1$.

Now we are ready to show that $\eta$ is a digroup homomorphism. For all $u_{1}\dot{c}_{1}v_{1}$ and $u_{2}\dot{c}_{2}v_{2}$ in $CW(D)$, where $u_{1}=a_{1}...a_{m}$, $v_{1}=b_{1}...b_{n}$, $u_{2}=d_{1}...d_{p}$ and $v_{2}=f_{1}...f_{q}$, if $c_{2}\in J$, then we have $p=0$ and

[TABLE]

On the other hand, if $c_{2}\in D\setminus J$, then we have

[TABLE]

Similarly, we have $\eta(u_{1}\dot{c}_{1}v_{1}\dashv u_{2}\dot{c}_{2}v_{2})=\eta(u_{1}\dot{c}_{1}v_{1})\dashv\eta(u_{2}\dot{c}_{2}v_{2})$. It follows that $\eta$ is a digroup homomorphism from $(CW(D),\dot{e})$ to $(G,e^{\prime})$.

Since the set $\{\dot{a}\mid a\in D\}$ generates $(CW(D),\dot{e})$, it is clear that $\eta$ is the unique digroup homomorphism from $(CW(D),\dot{e})$ to $(G,e^{\prime})$ such that $\eta\varphi_{i}=\psi_{i}$ for every $i\in I$. Therefore, $(CW(D),\dot{e})$ is the free product of the family of digroups $\{(D_{i}^{\prime},e)\mid i\in I\}$.

Assume that $u\dot{a}v$ lies in the group part of $(CW(D),\dot{e})$. Then by Lemma 2.3, we have

[TABLE]

which holds if and only if $u$ is the empty word and $\dot{a}v$ lies in $CW(D)$ satisfying $av\in J^{+}$, namely, $u\dot{a}v=[av]$ and $av$ is a reduced word in $J^{+}$. Now the isomorphism is clear by the constructions of $\Pi_{i\in I}^{\ast}J_{i}$ and $(CW(D),\dot{e})$.

Assume that $u\dot{a}v$ lies in the halo part of $(CW(D),\dot{e})$. Then we have

[TABLE]

It follows immediately that we have $\underline{u^{\sharp}a^{\sharp}v^{\sharp}}=e$. On the other hand, by the construction of $(CW(D),\dot{e})$, it is clear that every $u\dot{a}v\in CW(D)$ satisfying $\underline{u^{\sharp}a^{\sharp}v^{\sharp}}=e$ lies in the halo part of $(CW(D),\dot{e})$. ∎

Theorem 2.15 shows that the free product of digroups extends the notion of the free product of groups in the sense that, if every digroup $(D_{i}^{\prime},e)$ is a group, $i\in I$, then $(CW(D),\dot{e})$ is the free product of the groups $\{D_{i}^{\prime}\mid i\in I\}$.

Now we offer two easy examples of the free product of a family of digroups. By Remark 2.6, we may always assume that the considered digroups are disjoint.

Example 2.16**.**

Let $(D_{1},e_{1})$ and $(D_{2},e_{2})$ be two disjoint digroups such that the group part of $(D_{2},e_{2})$ contains exactly one element. Then with the conventions and notations of Remark 2.6, we have $J_{2}=\{e\}$, $J=J_{1}$ and $D=D_{1}^{\prime}\cup D_{2}^{\prime}$. Moreover, the set of all good center-words over $D$ is

[TABLE]

With Definition 2.10, $(CW(D),\dot{e})$ is the free product of $(D_{1}^{\prime},e)$ and $(D_{2}^{\prime},e)$, which is isomorphic to the free product of $(D_{1},e_{1})$ and $(D_{2},e_{2})$.

Example 2.17**.**

Let $\{(D_{i},e_{i})\mid i\in I\}$ be a family of disjoint digroups such that the group part of $(D_{i},e_{i})$ contains exactly one element for every $i\in I$. Then with the conventions and notations of Remark 2.6, we have $J=J_{i}=\{e\}$ for every $i\in I$ and $D=\cup_{i\in I}D_{i}^{\prime}$. Moreover, the set of all good center-words over $D$ is

[TABLE]

With Definition 2.10, $(CW(D),\dot{e})$ is the free product of $\{(D_{i}^{\prime},e)\mid i\in I\}$, which is isomorphic to the free product of $\{(D_{i},e_{i})\mid i\in I\}$.

Acknowledgments

We thank very much the anonymous referee for an extremely careful reading of the original version of this article. The referee’s insightful comments and suggestions help us greatly improve the exposition of the article.