A Zassenhaus Lemma for Digroups

Abstract

In this paper, we construct a quotient structure on digroups. This construction yields a new functor from the category of digroups to the category of groups. We obtain a modular property for digroups and use it to prove an analogue of the Zassenhaus lemma in this framework.

1. Introduction

Digroups are a generalization of groups which were independently introduced by J. L. Loday et al. in their study of dialgebras [1], by K. Liu [2] and R. Felipe [3], and were further studied in [4]. A formal axiomatic definition was provided by M. Kinyon in [5], in which Kinyon used digroups to construct Lie racks in an attempt to solve the Coquecigrue problem, which consists of finding an appropriate generalization of Lie’s third theorem in the category of Leibniz algebras [6]. This problem was formulated by J. L. Loday. He used this terminology to suggest that this structure is elusive or perhaps fictional.

The notion of normal subgroups plays a fundamental role in defining quotient groups and obtaining classical isomorphism theorems, which are fundamental tools in the development of Group Theory (see [7]). Recently, Ongay, Velasquez and Wills-Toro defined normal subdigroups [8] and studied a construction of quotient digroups and the corresponding classical isomorphism theorems. In [9], the concept of trigroups is defined as a generalization of digroups, essentially following Loday’s axiomatic definition of associative trioids [1]. The authors in [9] also constructed quotient trigroups and proved classical isomorphism theorems in the category of trigroups [10]. In this paper, we obtain the same results on digroups by considering that digroups have a trivial trigroup structure. This study produces a new functor from the category of digroups to the category of groups. More precisely, we use the notion of the conjugation of digoups provided in [5] to define a congruence for which the quotient set has a group structure, i.e., a trivial digroup structure. In Section 2, we study the concept quotient digroups and state classical isomorphism theorems yielding from this concept. These theorems are independent of the theorems obtained in [8]. In Section 3, we establish an analogue of the Zassenhaus lemma in the category of digroups.

2. Preliminaries

2.1. Normal Subdigroups

In this section, we provide a few results on normal subdigroups. Recall from [5] (Definition 4.1) that a digroup $(D,⊢,⊣)$ is a set D equipped with two binary associative operations ⊢ and ⊣, respectively, called left and right, satisfying the following conditions:

$$\left\{\begin{array}{cc}x⊢(y⊢z)=(x⊣y)⊢z\hfill & \left(p_1\right)\hfill \\ x⊢(y⊣z)=(x⊢y)⊣z\hfill & \left(p_2\right)\hfill \\ x⊣(y⊣z)=x⊣(y⊢z)\hfill & \left(p_3\right)\hfill \end{array}\right.$$

for all $x,y,z\in D,$ and there exists an element $1\in D$ satisfying

$$1⊢x=x=x⊣1\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x\in D$$

(1)

and for all $x\in D,$ there exists $x^{-1}\in D$ (called inverse of x) such that

$$x⊢x^{-1}=1=x^{-1}⊣x.$$

(2)

A subset S of a digroup D is said to be a subdigroup of D if $(S,⊢,⊣)$ is a digroup with a distinguished bar unit 1.

Note that the set ${\mathfrak{U}}_D:=\{e\in D:e⊢x=x=x⊣e\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x\in D\}$ of the bar units of D is a subdigroup of $D.$

Also, recall that a morphism between two digroups is a map that preserves the two binary operations and is compatible with bar units and inverses.

Remark 1.

Let $(D,⊢,⊣)$ be a digroup. Thus,

(a) 

The set $J_D=\{x^{-1}:x\in D\}$ is a group in which $⊢=⊣.$

(b) 

The mapping $\varphi :D\to J_D$ defined by $x\mapsto {\left(x^{-1}\right)}^{-1}$ is an epimorphism of digroups that fixes $J_D$, and $\mathsf{Ker}\varphi ={\mathfrak{U}}_D.$

(c) 

${\left(x^{-1}\right)}^{-1}=x⊢1=1⊣x$ for all $x\in D.$

(d) 

${(x⊢y)}^{-1}=y^{-1}⊢x^{-1}=y^{-1}⊣x^{-1}={(x⊣y)}^{-1}$ for all $x,y\in D.$ Consequently, ${\left({\left(x^{-1}\right)}^{-1}\right)}^{-1}=x^{-1}.$

(e) 

$x^{-1}⊢1=x^{-1}=1⊣x^{-1}$ for all $x\in D.$

(f) 

$x⊢y={\left(x^{-1}\right)}^{-1}⊢y$ for all $x,y\in D.$

(g) 

$x⊣y=x⊣{\left(y^{-1}\right)}^{-1}$ for all $x,y\in D.$

Proof. 

The proofs of items (a), (b), (d) and (e) are given in [5] (Lemma 4.5), and item (c) in [5] (Lemma 4.3). Items (f) and (g) follow from (c) and (1). □

Remark 2.

Let $(D,⊢,⊣)$ be a digroup. Then,

$$a^{-1}⊣b=a^{-1}⊣{\left(b^{-1}\right)}^{-1}=a^{-1}⊢{\left(b^{-1}\right)}^{-1}$$

for all $a,b\in S.$

Proof. 

This is a consequence of Remark 1(d) and Remark 1(g). □

Definition 3.

Let S be a subdigroup of a digroup $(D,⊢,⊣).$ We say that S is closed under conjugation by $x\in D$ if $x⊢S⊣x^{-1}\subseteq S.$

Definition 4

([8] (Definition 4)). A subdigroup S of a digroup $(D,⊢,⊣)$ is said to be normal if S is closed under a conjugation by all elements in $D.$

Remark 5.

By Assertions (f) and (g) of Remark 1, it follows that if S is normal in D then $x^{-1}⊢S⊣x\subseteq S$ for all $x\in D.$

For any subdigroup S of a digroup $D,$ we denote

$$\widehat{S}=\{x\in D|x^{-1}\in S\}.$$

The following lemma is the modular property for groups.

Lemma 6.

Let $(D,⊢,⊣)$ be a digroup, let S and $S^{\prime }$ be two subdigroups of D and let R be a subdigroup of $S.$ Then,

$$\widehat{S}\cap S^{\prime }⊢R=S\cap (S^{\prime }⊢R)andR⊣\widehat{S}\cap S^{\prime }=S\cap (R⊣S^{\prime }).$$

Proof. 

Let $x\in \widehat{S}\cap S^{\prime }$ and $r\in R.$ Clearly, $x⊢r\in S^{\prime }⊢R$ and $x⊢r={\left(x^{-1}\right)}^{-1}⊢r\in S$ since $x^{-1}\in S$ and $R\subseteq S.$ So, $\widehat{S}\cap S^{\prime }⊢R\subseteq S\cap (S^{\prime }⊢R).$ For the other inclusion, let $x⊢r\in S\cap (S^{\prime }⊢R).$ It is enough to show that $x\in \widehat{S}$, i.e., $x^{-1}\in S.$ Indeed, ${\left(x^{-1}\right)}^{-1}=x⊢1=x⊢(r⊢r^{-1})=(x⊢r)⊢r^{-1}\in S$ since $x⊢r\in S$ and $r^{-1}\in R\subseteq S,$ and thus $x^{-1}={\left({\left(x^{-1}\right)}^{-1}\right)}^{-1}\in S,$ thanks to Remark 1(d). This proves the first identity. The proof of the second identity is similar. □

Lemma 7.

Let $(D,⊢,⊣)$ be a digroup. If S and R are two normal subdigroups of $D,$ then $S⊢R$ is also a normal subdigroup of $D.$

Proof. 

First, we show that $S⊢R$ is closed under the digroup operations ⊢ and $⊣.$ Indeed, for all $x,y\in S$ and $r_1,r_2\in R,$ R is normal in D and by Remark 5, $y^{-1}⊢r_1⊣y\in R.$ Similarly, $(y^{-1}⊢r_2^{-1})⊢(r_2⊢r_1)⊣(y⊢r_2)\in R.$ So,

$$\begin{array}{cc}\hfill (x⊢r_1)⊢(y⊢r_2)& =(x⊢((y⊢y^{-1})⊢r_1))⊢(y⊢r_2)\hfill \\ & =((x⊢y)⊢y^{-1})⊢r_1⊢(y⊢r_2)\hfill \\ & =(x⊢y)⊢(y^{-1}⊢r_1)⊢(y⊢r_2)\hfill \\ & \stackrel{p_1}{=}(x⊢y)⊢((y^{-1}⊢r_1)⊣y)⊢r_2)\hfill \\ & =(x⊢y)⊢((y^{-1}⊢r_1⊣y)⊢r_2)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill (x⊢r_1)⊣(y⊢r_2)& =(x⊣((y⊢y^{-1})⊢(r_2^{-1}⊢r_2)⊢r_1))⊣(y⊢r_2)\hfill \\ & \stackrel{p_3}{=}((x⊣y)⊣y^{-1})⊢(r_2^{-1}⊢(r_2⊢r_1)⊣(y⊢r_2))\hfill \\ & \stackrel{p_1}{=}((x⊣y)⊢(y^{-1}⊢r_2^{-1})⊢(r_2⊢r_1)⊣(y⊢r_2)).\hfill \end{array}$$

So, $(x⊢r_1)⊢(y⊢r_2),(x⊢r_1)⊣(y⊢r_2)\in S⊢R.$

Now for $x\in S$ and $r\in R,$

$$\begin{array}{cc}\hfill {(x⊢r)}^{-1}& =r^{-1}⊢x^{-1}=(r^{-1}⊢x^{-1})⊣(r⊢r^{-1})\stackrel{p_3}{=}(r^{-1}⊢x^{-1}⊣r)⊣r^{-1}\hfill \\ & =(r^{-1}⊢x^{-1}⊣{\left(r^{-1}\right)}^{-1})⊣r^{-1}={(r^{-1}⊢x⊣r)}^{-1}⊣r^{-1}\hfill \\ & ={(r^{-1}⊢x⊣r)}^{-1}⊢r^{-1}\in S⊢R.\hfill \end{array}$$

Since $1=1⊢1,$ we conclude that $S⊢R$ is a subdigroup of $D.$ To show that $S⊢R$ is normal, let $s\in S,r\in R$ and $x\in D.$ Then,

$$\begin{array}{cc}\hfill s⊢r& =(s⊣1)⊢r=(s⊣(x^{-1}⊣x))⊢r=((s⊣x^{-1})⊣x)⊢r\hfill \\ & \stackrel{p_1}{=}(s⊣x^{-1})⊢(x⊢r).\hfill \end{array}$$

It follows that

$$\begin{array}{cc}\hfill x⊢(s⊢r)⊣x^{-1}& =x⊢((s⊣x^{-1})⊢(x⊢r))⊣x^{-1}\hfill \\ & =(x⊢s⊣x^{-1})⊢(x⊢r⊣x^{-1})\in S⊢R.\hfill \end{array}$$

□

Lemma 8.

Let D be a digroup, let $I_1,I_2$ and J be three subdigroups of D such that $I_1$ is a normal subdigroup of J and let $I_2$ be closed under a conjugation by all elements in $J.$ Then, $I_1⊢I_2$ is a normal subdigroup of $J⊢I_2.$

Proof. 

Note that with the given hypotheses, it follows from the first part of the proof of Lemma 7 that $I_1⊢I_2$ is a subdigroup of $J⊢I_2.$ We still need to show that $I_1⊢I_2$ is normal in $J⊢I_2.$ Indeed, let $r_1\in I_1,$$s,r_2\in I_2$ and $r\in J.$ We need to show that $(r⊢s)⊢(r_1⊢r_2)⊣{(r⊢s)}^{-1}\in I_1⊢I_2.$ Set $c_1:=s⊢r_1⊣s^{-1}$ and $c_2:=s⊢r_2⊣s^{-1}.$ Clearly $c_1\in I_1$ by the normality of $I_1$ in J and $c_2\in I_2$ via closure under a conjugation by all elements in $J.$ So, $b_1:=r⊢c_1⊣r^{-1}\in I_1$ and $b_2:=r⊢c_2⊣r^{-1}\in I_2$ for the same reason. We claim that $(r⊢s)⊢(r_1⊢r_2)⊣{(r⊢s)}^{-1}=b_1⊢b_2.$ Indeed,

$$\begin{array}{cc}\hfill b_1⊢b_2& =(r⊢c_1⊣r^{-1})⊢(r⊢c_2⊣r^{-1})\stackrel{p_1}{=}(r⊢c_1⊣(r^{-1}⊣r))⊢(c_2⊣r^{-1})\hfill \\ & =(r⊢c_1⊣1)⊢(c_2⊣r^{-1})=r⊢c_1⊢c_2⊣r^{-1}\hfill \\ & =r⊢(s⊢r_1⊣s^{-1})⊢(s⊢r_2⊣s^{-1})⊣r^{-1}\hfill \\ & \stackrel{p_1}{=}r⊢(s⊢r_1⊣(s^{-1}⊣s))⊢r_2⊣s^{-1}⊣r^{-1}\hfill \\ & =r⊢(s⊢r_1⊣1)⊢r_2⊣s^{-1}⊣r^{-1}=r⊢(s⊢r_1)⊢r_2⊣s^{-1}⊣r^{-1}\hfill \\ & =r⊢s⊢(r_1⊢r_2)⊣s^{-1}⊣r^{-1}=(r⊢s)⊢(r_1⊢r_2)⊣{(r⊢s)}^{-1}.\hfill \end{array}$$

□

Lemma 9.

Let $(D,⊢,⊣)$ be a digroup. If $S,$$S^{\prime }$, R and $R^{\prime }$ are subdigroups of D such that R is a normal subdigroup of S and $R^{\prime }$ is a normal subdigroup of $S^{\prime },$ then

(a) 

$(S\cap R^{\prime })⊢R$ is a normal subdigroup of $(S\cap S^{\prime })⊢R.$

(b) 

$(S^{\prime }\cap R)⊢R^{\prime }$ is a normal subdigroup of $(S^{\prime }\cap S)⊢R^{\prime }.$

Proof. 

Since R and $R^{\prime }$ are, respectively, normal subdigroups of S and $S^{\prime },$ one can easily verify that they are, along with $S\cap R^{\prime }$ and $S^{\prime }\cap R,$ normal subgroups of $S\cap S^{\prime }.$ The results of (a) and (b) now follow from Lemma 8 since R and $R^{\prime }$ are closed under a conjugation by elements in $S\cap S^{\prime }$. □

2.2. Quotient Digroups

This section proposes a new notion of a quotient of a given digroup by a normal subdigroup. We construct an equivalence relation for which the equivalence classes are the co-sets of the normal subdigroup, and the equivalence class of the identity element is the normal subdigroup. This construction is identical to the work presented in [9] on trigroups by considering their underlying digroup structure. Consequently, the proofs of all results in this section follow their corresponding results in [9].

Lemma 10

([9] (Lemma 4.1)). Let $(D,⊢,⊣)$ be a digroup, and let S be a subdigroup of $D.$ Then, the following assertions are true:

(a)

$g⊢S=S\iff g^{-1}\in S\iff S⊣g=S$ for all $g\in D.$

(b)

$g⊢S=h⊢S\iff g^{-1}⊣h\in S.$

(c)

$S⊣g=S⊣h,\iff g⊢h^{-1}\in S.$

Proposition 11

([9] (Proposition 4.2)). Let $(D,⊢,⊣)$ be a digroup and let S be a subdigroup of $D.$ Define the following relation: For $x,y\in D,$

$$x\sim y\iff x^{-1}⊣y\in S.$$

Then, ∼ is an equivalence relation, and the equivalence classes are the left co-sets $x⊢S,$ $x\in D$ (orbits of the action of S on D).

By the fundamental theorem of equivalence relations, the relation ∼ partitions D into the left co-sets $x⊢S,$ $x\in D$. Let $D/S$ be the set of left co-sets. Define the following binary operations $⊩,⫣:D/S\times D/S\to D/S$ by

$$(g⊢S)⊩(h⊢S)=(g⊢h)⊢S$$

$$(g⊢S)⫣(h⊢S)=(g⊣h)⊢S.$$

The following proposition provides a functor from the category of digroups to the category of groups.

Proposition 12

([9] (Proposition 4.4)). Let $(D,⊢,⊣)$ be a digroup and let S be a normal subdigroup of D. Then, the binary operations $⊩,⫣$ are well defined and equip $D/S$ with a structure of a group with identity $\left\{S\right\}$, and the inverse of the class $g⊢S$ is the class $g^{-1}⊢S.$

Example 13.

Let $Z(\mathbb{K},n)$ be the center of $GL(n,\mathbb{K}),$ the general linear group of degree n with coefficients in $\mathbb{K}.$ Define the following binary operations on $G={\mathbb{K}}^n\times Z(n,\mathbb{K})$:

(i)

$(x,M)⊢(y,N)=(My,MN).$

(ii)

$(x,M)⊣(y,N)=(x,MN).$

For all $x,y\in {\mathbb{K}}^n$ and $M,N\in Z(n,\mathbb{K}).$ Thus, $(G,⊢,⊣)$ is a digroup with a distinguished bar unit $I_G:=(0,I_n)$ in which $(0,M^{-1})$ is the inverse of $(x,M),$ where $I_n$ is the identity matrix. It is easy to verify on $G/I_G$ that

$$((x,M)⊢I_G)⊩((y,N)⊢I_G)=\left\{(0,MN)\right\}=((x,M)⊢I_G)⫣((y,N)⊢I_G).$$

So, the operations ⊩ and ⫣ are equal, and thus the quotient $G/I_G$ is a group.

The following results are obtained from the theorems proven in [9] on trigroups by using the trivial trigroup structure of digroups.

Proposition 14

([9] (Proposition 4.8)). Let D and $D^{\prime }$ be two digroups and let S be a normal subdigroup of $D.$ Let $\varphi :D\to D^{\prime }$ be a morphism of digroups such that $\mathsf{Ker}\left(\varphi \right)\subseteq S.$ Then, there is an isomorphism of groups $\widehat{\varphi }:D/S\cong Im\varphi /\varphi \left(S\right).$ In particular, if $S=ker\left(\varphi \right)$, then this isomorphism becomes $\widehat{\varphi }:D/ker\left(\varphi \right)\cong Im\varphi /\left\{1\right\}.$

Proposition 15

([9] (Corollary 4.3)). Let D be a digroup, and let S and R be two subdigroups of D such that $s⊢R=R⊣s$ for all $s\in S.$ Then, there is a group isomorphism

$$(S⊢R)/R\cong S/(S\cap \widehat{R}).$$

Proposition 16

([9] (Proposition 4.17)). Let D be a digroup, and let S and R be two normal subdigroups of D such that S is a normal subgroup of $R.$ Then, there is a group isomorphism

$$(D/S)/(\widehat{R}/S)\cong D/R.$$

3. The Zassenhaus Lemma for Digroups

In this section, we prove the Zassenhaus lemma for digroups. We use the following lemma.

Lemma 17.

Let $(D,⊢,⊣)$ be a digroup. If $S_1,$$S_2$ are normal subdigroups of $D,$ then there is a group isomorphism

$${\displaystyle \frac{D}{S_1⊢S_2}}\cong {\displaystyle \frac{D}{S_2⊢S_1}}.$$

Proof. 

Consider the map

$$\mathrm{\Gamma }:{\displaystyle \frac{D}{S_1⊢S_2}}⟶{\displaystyle \frac{D}{S_2⊢S_1}}\mathrm{d}\mathrm{e}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{d} \mathrm{b}\mathrm{y}a⊢(S_1⊢S_2)⟼a⊢(S_2⊢S_1).$$

Notice that for all $a,b\in D,$ we have the following via Remark 2:

$$a⊢(S_1⊢S_2)=b⊢(S_1⊢S_2)\iff a^{-1}⊢{\left(b^{-1}\right)}^{-1}\in (S_1⊢S_2)$$

which implies $a^{-1}⊢{\left(b^{-1}\right)}^{-1}=s_1⊢s_{2}forsome{s}_1\in S_{1}and{s}_2\in S_2,$ and thus $b^{-1}⊢{\left(a^{-1}\right)}^{-1}=s_2^{-1}⊢s_1^{-1}\in S_2⊢S_1$; hence, $a⊢(S_2⊢S_1)=b⊢(S_2⊢S_1).$ So, $\mathrm{\Gamma }$ is well defined. $\mathrm{\Gamma }$ is clearly a digroup homomorphism by the definition of the operations $⊩,$ on left co-sets. By interchanging the positions of $S_1$ and $S_2$ in the proof of the well-defined $\mathrm{\Gamma },$ one proves injection and that $\mathrm{\Gamma }$ is surjective and trivial. □

The following is an analogue of the Zassenhaus lemma (also known as the Butterfly lemma on groups) [7] for digroups.

Lemma 18.

Let $(D,⊢,⊣)$ be a digroup. If $S,$$S^{\prime }$, R and $R^{\prime }$ are subdigroups of D such that R is a normal subdigroup of S and $R^{\prime }$ is a normal subdigroup of $S^{\prime },$ then

$${\displaystyle \frac{S\cap (S^{\prime }⊢R)}{S\cap (R^{\prime }⊢R)}}\cong {\displaystyle \frac{S\cap (S^{\prime }⊢R^{\prime })}{S^{\prime }\cap (R⊢R^{\prime })}}$$

Proof. 

By the modular property of digroups (Lemma 6), it suffices to show that

$${\displaystyle \frac{(\widehat{S}\cap S^{\prime })⊢R}{(\widehat{S}\cap R^{\prime })⊢R}}\cong {\displaystyle \frac{(\widehat{S^{\prime }}\cap S)⊢R^{\prime }}{(\widehat{S^{\prime }}\cap R)⊢R^{\prime }}}.$$

Set $\mathfrak{N}:=(\widehat{S}\cap R^{\prime })⊢(\widehat{S^{\prime }}\cap R).$ Clearly, $\widehat{S}\cap R^{\prime }$ and $\widehat{S^{\prime }}\cap R$ are normal subdigroups of $\widehat{S}\cap \widehat{S^{\prime }}.$ It follows from Lemma 7 that $\mathfrak{N}$ is a normal subdigroup of $\widehat{S}\cap \widehat{S^{\prime }}.$ Now consider the map

$$\mathrm{\Phi }:(\widehat{S}\cap S^{\prime })⊢R⟶{\displaystyle \frac{\widehat{S}\cap \widehat{S^{\prime }}}{\mathfrak{N}}}\mathrm{d}\mathrm{e}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{d} \mathrm{b}\mathrm{y}x⊢r⟼x⊢\mathfrak{N}.$$

$\mathrm{\Phi }$ is well defined since for all $x,y\in \widehat{S}\cap S^{\prime }$ and $r_1,r_2\in R$ such that $x⊢r_1=y⊢r_2,$ we have $r_1⊢r_1^{-1}=1$ and $y^{-1}⊢y\in {\mathfrak{U}}_D,$

$$\begin{array}{cc}\hfill y^{-1}⊣x& =y^{-1}⊣{\left(x^{-1}\right)}^{-1}=y^{-1}⊢{\left(x^{-1}\right)}^{-1}=y^{-1}⊢(x⊢1)\hfill \\ & =y^{-1}⊢(x⊢(r_1⊢r_1^{-1}))=y^{-1}⊢(x⊢r_1)⊢r_1^{-1}\hfill \\ & =y^{-1}⊢(y⊢r_2)⊢r_1^{-1}=(y^{-1}⊢y)⊢(r_2⊢r_1^{-1})=r_2⊢r_1^{-1}.\hfill \end{array}$$

$y^{-1}⊣x=r_2⊢r_1^{-1}\in S^{\prime }\cap R=1⊢S^{\prime }\cap R\subseteq 1⊢\widehat{S^{\prime }}\cap R\subseteq \mathfrak{N}.$ So, $x\sim y$, i.e., $x⊢\mathfrak{N}=y⊢\mathfrak{N}.$ To show that $\mathrm{\Phi }$ is a digroup homomorphism, notice that as R is normal in $S,$ for all $x,y\in S\cap S^{\prime }$ and $r_1,r_2\in R,$ we have $y^{-1}⊢r_1⊣y\in R$ and $(y^{-1}⊢r_2^{-1})⊢(r_2⊢r_1)⊣(y⊢r_2)\in R,$ thanks to Remark 5. So, we have the following from the proof of Lemma 7:

$$(x⊢r_1)⊢(y⊢r_2)=(x⊢y)⊢\left((y^{-1}⊢r_1⊣y)⊢r_2\right)\in (S\cap S^{\prime })⊢R,$$

and

$$(x⊢r_1)⊣(y⊢r_2)=(x⊣y)⊢\left((y^{-1}⊢r_2^{-1})⊢(r_2⊢r_1)⊣(y⊢r_2)\right)\in (S\cap S^{\prime })⊢R.$$

Therefore,

$$\mathrm{\Phi }((x⊢r_1)⊢(y⊢r_2))=(x⊢y)⊢\mathfrak{N}=(x⊢\mathfrak{N})⊩(y⊢\mathfrak{N})=\mathrm{\Phi }(x⊢r_1)⊩\mathrm{\Phi }(y⊢r_2)$$

and

$$\mathrm{\Phi }((x⊢r_1)⊣(y⊢r_2))=(x⊢y)⊢\mathfrak{N}=(x⊢\mathfrak{N})⫣(y⊢\mathfrak{N})=\mathrm{\Phi }(x⊢r_1)⫣\mathrm{\Phi }(y⊢r_2).$$

$\mathrm{\Phi }$ is surjective since if $x\in \widehat{S}\cap \widehat{S^{\prime }},$ then ${\left(x^{-1}\right)}^{-1}\in \widehat{S}\cap S^{\prime },$ and we have by Remark 1(f) that $\mathrm{\Phi }({\left(x^{-1}\right)}^{-1}⊢R)={\left(x^{-1}\right)}^{-1}⊢\mathfrak{N}=x⊢\mathfrak{N}.$ It remains to show that $\mathrm{\Phi }$ is injective. Indeed, $x⊢a\in \mathsf{Ker}\left(\mathrm{\Phi }\right)$⇔$x⊢\mathfrak{N}=\mathfrak{N}$⇔$x^{-1}\in \mathfrak{N}$⇔${\left(x^{-1}\right)}^{-1}\in \mathfrak{N},$ thanks to Lemma 10(a). So, ${\left(x^{-1}\right)}^{-1}=y_1⊢y_2$ for some $y_1\in \widehat{S}\cap R^{\prime }$ and $y_2\in \widehat{S^{\prime }}\cap R.$ It follows that

$$x⊢a={\left(x^{-1}\right)}^{-1}⊢a=(y_1⊢y_2)⊢a=y_1⊢(y_2⊢a)\in \widehat{S}\cap R^{\prime }⊢R.$$

So, $\mathsf{Ker}\left(\mathrm{\Phi }\right)\subseteq \widehat{S}\cap R^{\prime }⊢R.$ However, $\widehat{S}\cap R^{\prime }⊢R\subseteq \mathsf{Ker}\left(\mathrm{\Phi }\right)$ since for all $x\in \widehat{S}\cap R^{\prime }$ and $a\in R,$${\left(x^{-1}\right)}^{-1}=x⊢1\in \mathfrak{N}$ implying that $\mathrm{\Phi }(x⊢a)=\mathrm{\Phi }({\left(x^{-1}\right)}^{-1}⊢a)={\left(x^{-1}\right)}^{-1}⊢\mathfrak{N}=\mathfrak{N}.$ Hence, $\mathsf{Ker}\left(\mathrm{\Phi }\right)=(\widehat{S}\cap R^{\prime })⊢R.$ Therefore, via Proposition 14, we have

$${\displaystyle \frac{(\widehat{S}\cap S^{\prime })⊢R}{(\widehat{S}\cap R^{\prime })⊢R}}\cong {\displaystyle \frac{\widehat{S}\cap \widehat{S^{\prime }}}{\mathfrak{N}}}/\left\{\mathfrak{N}\right\}\cong {\displaystyle \frac{\widehat{S}\cap \widehat{S^{\prime }}}{\mathfrak{N}}}.$$

The last isomorphism holds since $\left\{\mathfrak{N}\right\}$ is the identity element in the group ${\displaystyle \frac{\widehat{S}\cap \widehat{S^{\prime }}}{\mathfrak{N}}}.$

Similarly, by setting ${\mathfrak{N}}^{\prime }:=(\widehat{S^{\prime }}\cap R)⊢(\widehat{S}\cap R^{\prime })$ and considering the map

$$\mathrm{\Phi }:(\widehat{S^{\prime }}\cap S)⊢R^{\prime }⟶{\displaystyle \frac{\widehat{S}\cap \widehat{S^{\prime }}}{{\mathfrak{N}}^{\prime }}}\mathrm{d}\mathrm{e}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{d} \mathrm{b}\mathrm{y}x⊢r⟼x⊢{\mathfrak{N}}^{\prime }.$$

we show by that

$${\displaystyle \frac{(\widehat{S^{\prime }}\cap S)⊢R^{\prime }}{(\widehat{S^{\prime }}\cap R)⊢R^{\prime }}}\cong {\displaystyle \frac{\widehat{S}\cap \widehat{S^{\prime }}}{{\mathfrak{N}}^{\prime }}}.$$

Since by Lemma 17 we have ${\displaystyle \frac{\widehat{S}\cap \widehat{S^{\prime }}}{\mathfrak{N}}}\cong {\displaystyle \frac{\widehat{S}\cap \widehat{S^{\prime }}}{{\mathfrak{N}}^{\prime }}},$ it follows that

$${\displaystyle \frac{(\widehat{S}\cap S^{\prime })⊢R}{(\widehat{S}\cap R^{\prime })⊢R}}\cong {\displaystyle \frac{(\widehat{S^{\prime }}\cap S)⊢R^{\prime }}{(\widehat{S^{\prime }}\cap R)⊢R^{\prime }}}.$$

This completes the proof. □

4. Conclusions

In this paper, we constructed a quotient structure on digroups and proved an analogue of the Zassenhaus lemma, which may be useful in analyzing the structure of a digroup and its normal subdigroups by visualizing certain subdigroup relationships through a Hasse diagram. This construction allowed us to obtain some isomorphism theorems that are independent of the results obtained in [8], with a different quotient structure. It would be interesting to investigate the relationship between these two structures.