A Way to Construct Commutative Hyperstructures ^†

Abstract

This article aims to create commutative hyperstructures, starting with a non-commutative group. Therefore, we consider the starting group to be the dihedral group $D_n$, where n is a natural number, $n>1$, and we determine the HX groups associated with the dihedral group. For a fixed number n, we note ${\mathcal{G}}_n=\left\{{\mathcal{G}}_{p_2}^{p_1}HX-\mathrm{groups},\mathrm{for}\mathrm{any}{p}_1,p_2\in {\mathbb{N}}^{*}\mathrm{such}\mathrm{that}n=p_1{p}_2\right\}$ as the set of all HX groups. This paper analyses this new structure’s properties for particular cases when the dihedral group $D_4$ is the support group.

1. Introduction

The algebraic term hyperstructure denotes an appropriate generalization of structures of classical algebras, such as group, semigroup, and ring. In classical algebraic structures, the composition of two elements is an element, and within the algebraic hyperstructure, the hypercomposition law of two elements represents a set. F. Marty noticed many aspects of a factor group, which became the starting point within the theory of hypergroups. He introduced the concept of the hypergroup in 1934 at the Congress of Mathematicians from Scandinavian Countries. Hypergroups have been studied from the theoretically and for their applications in pure and applied mathematical problems: geometry, topology, cryptography, code theory, graphs, hypergraphs, automata theory, fuzzy degree, probability, etc. [1]. The Chinese mathematician Mi introduced the notion of HX-groups. After this, Honghai and Li Honxing contributed to the theory of HX-groups with Corsini [2,3,4]. Furthermore, Cristea analysed the connection between HX-groups and hypergroups [5]. In the article [6], we studied the form of HX-groups with a dihedral group $D_n$ as a group, where n is a natural number greater than 3. Moreover, we analysed the HX-groups’ commutativity degree and the Chinese hypergroup’s fuzzy grade associated with them. We noticed a connection between the commutativity degree of the HX-groups related to the dihedral group and its commutativity degree. In the article [7] a new concept of neutro HX-groups was defined.

2. Main Results

In this section, we recall the notions of an HX-group, and define the new set forms by HX-groups [8].

Definition 1.

Let $(G,.)$ be a group and $\mathcal{G}\subset$${\mathcal{P}}^{*}\left(G\right)\setminus \{\varnothing \}$, where ${\mathcal{P}}^{*}\left(G\right)$ is the set of non-empty subsets of G. An HX-group is a non-empty subset H of ${\mathcal{P}}^{*}\left(G\right)$ which is a group with respect to the operation “∗”, defined by:

$$\forall A,B\in \mathcal{G},A\ast B=\{a.b|a\in A,b\in B\}.$$

(1)

We assume $\mathcal{G}$ has group G as a support.

Definition 2.

Let $\left(\mathcal{G},\ast \right)$ be an HX-group with the support $\left(G,.\right)$ and E as the identity of group $\mathcal{G}.$ A Chinese hypergroupoid is a hyperstructure $<G^{*},\oplus >$, where

$$G^{*}=\bigcup _{A\in \mathcal{G}}A\mathit{and}\forall \left(x,y\right)\in G^{*}\times G^{*},x\oplus y=\underset{\{A,B\}\subset \mathcal{G}}{\underset{y\in B}{\bigcup _{x\in A}}}A\ast B.$$

Construction of ${\mathcal{G}}_n$

In the following, we consider the set of all HX groups with the dihedral group $(D_4,·)$ as a support. The dihedral group $D_n$, where $n\in \mathbb{N}$, $n\ge 2$, is the group generated by rotation $\rho$ and symmetry $\sigma$, and satisfies the following properties:

$$\begin{array}{ccc}\hfill {\rho }^n& =& {\sigma }^2=e;\hfill \\ \hfill {\rho }^k·\sigma & =& \sigma ·{\rho }^{n-k},\forall k\le n,k\in \mathbb{N}.\hfill \end{array}$$

(2)

It is denoted by $D_n=<\rho ,\sigma >$. Therefore, in the article [6], we determined the form of HX-groups associated with the dihedral group $D_n$, and analysed them in particular cases for $D_4$, $D_5$ and $D_6$. For a fixed number n, we note that

$${\mathcal{G}}_n=\left\{{\mathcal{G}}_{p_2}^{p_1}|HX-\mathrm{groups},\mathrm{for}\mathrm{any}{p}_1,p_2\in {\mathbb{N}}^{*}\mathrm{such}\mathrm{that}n=p_1{p}_2\right\}$$

be the set of all HX-groups. We define the following hyperoperation

$$“\circ ”:{\mathcal{G}}_n\times {\mathcal{G}}_n\to P^{*}\left({\mathcal{G}}_n\right);$$

thus,

$$\begin{array}{ccc}\hfill {\mathcal{G}}_{p_2}^{p_1}\circ {\mathcal{G}}_{p_2^{\prime }}^{p_1^{\prime }}& =& \left\{\bigcup _{0\le s\le 2p_2-1}\left(\bigcup _{0\le t\le 2p_2^{\prime }-1}{C}_{s,t}\right);C_{s,t}=X_s^{p_1}\ast Y_t^{p_1^{\prime }};X_s^{p_1}\subseteq {\mathcal{G}}_{p_2}^{p_1},Y_t^{p_1^{\prime }}\subseteq {\mathcal{G}}_{p_2^{\prime }}^{p_1^{\prime }}\right\},\hfill \\ \hfill n& =& p_1{p}_2=p_1^{\prime }{p}_2^{\prime },p_1,p_2,p_1^{\prime },p_2^{\prime }\in {\mathbb{N}}^{*}.\hfill \end{array}$$

3. Results and Discussion

In the following we present a particular cases for $n=4$, and analyse the hyperstructure $\left({\mathcal{G}}_4,\circ \right).$

Proposition 1.

The hyperstructure $\left({\mathcal{G}}_4,\circ \right)$ is a commutative structure, where

$${\mathcal{G}}_4=\left\{{\mathcal{G}}_{p_2}^{p_1}|HX-\mathit{groups},\mathit{for}\mathit{any}{p}_1,p_2\in {\mathbb{N}}^{*}\mathit{such}\mathit{that}4=p_1{p}_2\right\}.$$

Proof. 

Therefore, $\left({\mathcal{G}}_4,\circ \right)$ is a commutative hyperstructure if and only if ${\mathcal{G}}_{p_2}^{p_1}\circ {\mathcal{G}}_{p_2^{\prime }}^{p_1^{\prime }}={\mathcal{G}}_{p_2^{\prime }}^{p_1^{\prime }}\circ {\mathcal{G}}_{p_2^{\prime }}^{p_1}$, for any $p_1$, $p_2$,$p_1^{\prime }$,$p_2^{\prime }\in {\mathbb{N}}^{*}$, such that $4=p_1{p}_2=p_1^{\prime }{p}_2^{\prime }$. In agreement with our previous work [6], the form of the HX-groups associated with the dihedral group $D_4$ is:

$$\begin{array}{ccc}\hfill {\mathcal{G}}_2^2& =& \{\{e,{\rho }^2\},\{\rho ,{\rho }^3\},\{\sigma ,{\rho }^2\sigma \},\{\rho \sigma ,{\rho }^3\sigma \}\};\hfill \\ \hfill {\mathcal{G}}_1^4& =& \{\{e,\rho ,{\rho }^2,{\rho }^3\},\{\sigma ,\rho \sigma ,{\rho }^2\sigma ,{\rho }^3\sigma \}\};\hfill \\ \hfill {\mathcal{G}}_4^1& =& \{\left\{e\right\},\left\{\rho \right\},\left\{{\rho }^2\right\},\left\{{\rho }^3\right\},\left\{\sigma \right\},\left\{\rho \sigma \right\},\left\{{\rho }^2\sigma \right\},\left\{{\rho }^3\sigma \right\}\}.\hfill \end{array}$$

Therefore,

$${\mathcal{G}}_2^2\circ {\mathcal{G}}_1^4=\{C_{0,0},C_{1,0},C_{2,0},C_{3,0},C_{0,1},C_{1,1},C_{2,1},C_{3,1}\}$$

and the sets $X_s^{p_1}$ and $Y_t^{p_1^{\prime }}$ are:

$$\begin{array}{ccc}\hfill X_0^2& =& \{e,{\rho }^2\},X_1^2=\{\rho ,{\rho }^3\},X_2^2=\{\sigma ,{\rho }^2\sigma \},X_3^2=\{\rho \sigma ,{\rho }^3\sigma \};\hfill \\ \hfill Y_0^4& =& \{e,\rho ,{\rho }^2,{\rho }^3\},Y_1^4=\{\sigma ,\rho \sigma ,{\rho }^2\sigma ,{\rho }^3\sigma \}.\hfill \end{array}$$

In calculating the elements $C_{i,j}$, $i\in \{0,1,2,3\}$ and $j\in \{0,1\}$, we use the rules gives by (2)

$$\begin{array}{ccc}\hfill C_{0,0}& =& X_0^2\ast Y_0^4=\{e,{\rho }^2\}\ast \{e,\rho ,{\rho }^2,{\rho }^3\}=e·e\cup e·\rho \cup e·{\rho }^2\cup \hfill \\ & & \cup e·{\rho }^3\cup {\rho }^2·e\cup {\rho }^2·\rho \cup {\rho }^2·{\rho }^2\cup {\rho }^2·{\rho }^3\hfill \\ & =& \{e,\rho ,{\rho }^2,{\rho }^3\}=Y_0^4;\hfill \\ \hfill C_{1,0}& =& X_1^2\ast Y_0^4=\{\rho ,{\rho }^3\}\ast \{e,\rho ,{\rho }^2,{\rho }^3\}=\rho ·e\cup \rho ·\rho \cup \rho ·{\rho }^2\cup \rho ·{\rho }^3\cup \hfill \\ & & \cup {\rho }^3·e\cup {\rho }^3·\rho \cup {\rho }^3·{\rho }^2\cup {\rho }^3·{\rho }^3.\hfill \\ & =& \{e,\rho ,{\rho }^2,{\rho }^3\}=Y_0^4\hfill \\ \hfill C_{2,0}& =& X_2^2\ast Y_0^4=\{\sigma ,{\rho }^2\sigma \}\ast \{e,\rho ,{\rho }^2,{\rho }^3\}=\{\sigma ,\sigma \rho ,\sigma {\rho }^2,\sigma {\rho }^3,{\rho }^2\sigma ,{\rho }^2\sigma \rho ,{\rho }^2\sigma {\rho }^2,{\rho }^2\sigma {\rho }^3\}\hfill \\ & =& \{\sigma ,{\rho }^3\sigma ,{\rho }^2\sigma ,\rho \sigma \}=\{\sigma ,\rho \sigma ,{\rho }^2\sigma ,{\rho }^3\sigma \}=Y_1^4.\hfill \\ \hfill C_{3,0}& =& X_3^2\ast Y_0^4=\{\rho \sigma ,{\rho }^3\sigma \}\ast \{e,\rho ,{\rho }^2,{\rho }^3\}=Y_1^4.\hfill \end{array}$$

Similarly we calculate the other elements, obtaining

$$C_{0,1}=X_0^2\ast Y_1^4=Y_1^4,C_{1,1}=X_1^2\ast Y_1^4=Y_1^4,C_{2,1}=X_2^2\ast Y_1^4=Y_0^4,C_{3,1}=X_3^2\ast Y_1^4=Y_0^4.$$

Therefore, we have

$${\mathcal{G}}_2^2\circ {\mathcal{G}}_1^4={\mathcal{G}}_1^4.$$

(3)

Now, we analysed the composition

$${\mathcal{G}}_1^4\circ {\mathcal{G}}_2^2=\{C_{0,0}^{\prime },C_{0,1}^{\prime },C_{0,2}^{\prime },C_{0,3}^{\prime },C_{1,0}^{\prime },C_{1,1}^{\prime },C_{1,2}^{\prime },C_{1,3}^{\prime }\}.$$

According to (2) we noticed that ${\rho }^k{\rho }^p={\rho }^{k+p}={\rho }^p{\rho }^k={\rho }^{p+k}$, for any p, $k\in {\mathbb{N}}^{*}$. Therefore, we can conclude that

$$\begin{array}{ccc}\hfill X_0^2\ast Y_0^4& =& Y_0^4\ast X_0^2=Y_0^4=C_{0,0}^{\prime },\hfill \\ \hfill X_1^2\ast Y_0^4& =& Y_0^4\ast X_1^2=C_{0,1}^{\prime }=Y_0^4,\hfill \\ \hfill C_{0,2}^{\prime }& =& Y_0^4\ast X_2^2=C_{0,3}^{\prime }=Y_0^4\ast X_3^2=Y_1^4,\hfill \\ \hfill C_{1,0}^{\prime }& =& Y_1^4\ast X_0^2=C_{1,1}^{\prime }=Y_1^4\ast X_1^2=Y_1^4,\hfill \\ \hfill C_{1,2}^{\prime }& =& Y_1^4\ast X_2^2=C_{1,3}^{\prime }=Y_1^4\ast X_3^2=Y_0^4.\hfill \end{array}$$

Therefore, ${\mathcal{G}}_1^4\circ {\mathcal{G}}_2^2={\mathcal{G}}_2^2\circ {\mathcal{G}}_1^4={\mathcal{G}}_1^4$. Analogously we thus have

$${\mathcal{G}}_1^4\circ {\mathcal{G}}_4^4={\mathcal{G}}_4^1\circ {\mathcal{G}}_1^4={\mathcal{G}}_1^4,{\mathcal{G}}_2^2\circ {\mathcal{G}}_4^1={\mathcal{G}}_4^1\circ {\mathcal{G}}_2^2={\mathcal{G}}_2^2.$$

In conclusion, $\left({\mathcal{G}}_4,\circ \right)$ is a commutative hyperstructure. □

Remark 1.

The elements of hyperstructure $\left({\mathcal{G}}_4,\circ \right)$ satisfy the following equality

$${\mathcal{G}}_{p_2}^{p_1}\circ {\mathcal{G}}_{p_2^{\prime }}^{p_1^{\prime }}={\mathcal{G}}_{p_2^{\prime }}^{p_1^{\prime }}\circ {\mathcal{G}}_{p_2}^{p_1}={\mathcal{G}}_{gcd\{p_2,p_2^{\prime }\}}^{lcm\{p_1,p_1^{\prime }\}},$$

(4)

for any $p_1$, $p_2$,$p_1^{\prime }$,$p_2^{\prime }\in {\mathbb{N}}^{*}$, such that $4=p_1{p}_2=p_1^{\prime }{p}_2^{\prime }.$

Notation 1.

$lcm\{p_1,p_1^{\prime }\}$ represents the least common multiple of numbers $p_1,p_1^{\prime }$, and $gcd\{p_2,p_2^{\prime }\}$ is the greatest common divisor of $p_2,p_2^{\prime }$.

Proposition 2.

The hyperstructure $\left({\mathcal{G}}_4,\circ \right)$ is a semi-hypergroup, but not a quasi-hypergroup.

Proof. 

$\left({\mathcal{G}}_4,\circ \right)$ is a semi-hypergroup if and only if the hyperoperation “∘” is associative, i.e.,

$$\left({\mathcal{G}}_{p_2}^{p_1}\circ {\mathcal{G}}_{p_2^{\prime }}^{p_1^{\prime }}\right)\circ {\mathcal{G}}_{p_2^{″}}^{p_1^{″}}={\mathcal{G}}_{p_2}^{p_1}\circ \left({\mathcal{G}}_{p_2^{\prime }}^{p_1^{\prime }}\circ {\mathcal{G}}_{p_2^{″}}^{p_1^{″}}\right),$$

for any $p_1$, $p_2$,$p_1^{\prime }$,$p_2^{\prime }$,$p_1^{″}$ and $p_2^{″}\in {\mathbb{N}}^{*}$, such that $4=p_1{p}_2=p_1^{\prime }{p}_2^{\prime }=p_1^{″}{p}_2^{″}$. We use the relation (4), and properties of gcd respectively, $lcm$

$$\begin{array}{ccc}\hfill gcd\{gcd\{p_2,p_2^{\prime }\},p_2^{″}\}& =& gcd\{p_2,gcd\{p_2^{\prime },p_2^{″}\}\};\hfill \\ \hfill lcm\{lcm\{p_1,p_1^{\prime }\},p_1^{″}\}& =& lcm\{p_1,lcm\{p_1^{\prime },p_1^{″}\}\}.\hfill \end{array}$$

Therefore,

$$\begin{array}{ccc}\hfill \left({\mathcal{G}}_{p_2}^{p_1}\circ {\mathcal{G}}_{p_2^{\prime }}^{p_1^{\prime }}\right)\circ {\mathcal{G}}_{p_2^{\prime \prime }}^{p_1^{\prime \prime }}& =& {\mathcal{G}}_{gcd\{p_2,p_2^{\prime }\}}^{lcm\{p_1,p_1^{\prime }\}}\circ {\mathcal{G}}_{p_2^{\prime \prime }}^{p_1^{\prime \prime }}={\mathcal{G}}_{gcd\{gcd\{p_2,p_2^{\prime }\},p_2^{\prime \prime }\}}^{lcm\{lcm\{p_1,p_1^{\prime }\},p_1^{\prime \prime }\}}=\hfill \\ & =& {\mathcal{G}}_{gcd\{p_2,gcd\{p_2^{\prime },p_2^{\prime \prime }\}\}}^{lcm\{p_1,lcm\{p_1^{\prime },p_1^{\prime \prime }\}\}}={\mathcal{G}}_{p_2}^{p_1}\circ \left({\mathcal{G}}_{p_2^{\prime }}^{p_1^{\prime }}\circ {\mathcal{G}}_{p_2^{\prime \prime }}^{p_1^{\prime \prime }}\right).\hfill \end{array}$$

To prove that the semi-hypergroup $\left({\mathcal{G}}_4,\circ \right)$ is not a quasi-hypergroup, means that the hyperoperation does not satisfy the reproductive law, which is the following

$${\mathcal{G}}_{p_2}^{p_1}\circ {\mathcal{G}}_4={\mathcal{G}}_4\circ {\mathcal{G}}_{p_2}^{p_1}={\mathcal{G}}_4,$$

for any $p_1$, $p_2\in {\mathbb{N}}^{*}$, such that $4=p_1{p}_2.$ We notice that

$${\mathcal{G}}_1^4\circ {\mathcal{G}}_4={\mathcal{G}}_1^4\circ {\mathcal{G}}_1^4\cup {\mathcal{G}}_1^4\circ {\mathcal{G}}_2^2\cup {\mathcal{G}}_1^4\circ {\mathcal{G}}_4^1={\mathcal{G}}_1^4\ne {\mathcal{G}}_4.$$

Therefore, the semi-hypergroup $\left({\mathcal{G}}_4,\circ \right)$ is not a quasi-hypergroup. □

Remark 2.

The cardinality of the semi-hypergroup $|{\mathcal{G}}_4|$ coincides with the number divisors of four.

4. Conclusions

In this paper, we presented a way to obtain a commutative hyperstructure, starting with a non-abelian group. We construct the hyperstructure $\left({\mathcal{G}}_n,\circ \right)$ as a composition of the HX-groups associated with the dihedral group $D_n$, $n\in {\mathbb{N}}^{*},n>2.$ In Section 2, we analysed this new structure in particular cases, $n=4$. We noticed that we have a commutative semi-hypergroup $\left({\mathcal{G}}_4,\circ \right)$ that is not a quasi-hypergroup. Furthermore, we see that a connection exists between the composition of the HX-groups and the function $lcm,$ $gcd$ associated with them.