A Survey on the Theory of n-Hypergroups

Abstract

This paper presents a series of important results from the theory of n-hypergroups. Connections with binary relations and with lattices are presented. Special attention is paid to the fundamental relation and to the commutative fundamental relation. In particular, join n-spaces are analyzed.

1. Introduction

The theory of n-ary hypergroups, also called n-hypergroups, contains two generalizations of the notion of a group: n-groups and hypergroups, which are briefly presented in the next paragraph. The two concepts were introduced around the same time.

n-groups, also called polyadic groups, were introduced in 1928 by W. Dörnte [1], and they are a generalization of classical groups. An important role in n-group theory is the paper written by E.L. Post of 143 pages [2]. Such operations are used then in the study of $(m,n)$-rings. Among those who made recently important contributions in the theory of n-groups, we mention W. Dudek and his collaborators; see for instance [3,4,5]. Let $n>2$, and denote the chain $x_i,\dots ,x_j$ by $x_i^j$ (for $j<i,$ the above sequence is the empty symbol). For a nonempty set G with one n-operation, $f:G^n\to G$ is a n-groupoid, which is a n-quasigroup, if, for all $a_1^n,b\in G$, there is exactly one $x_i\in G$ such that $f(a_1^{i-1},x_i,a_{i+1}^n)=b.$ An n-quasigroup with an associative operation is called an n-ary group.

Hypergroup theory is a field of algebra that appeared in 1934 and was introduced by the French mathematician Marty [6]. The theory has known various periods of flourishing: the 1940s, then 1970s, and especially after the 1990s, the theory has been studied on all continents, both theoretically and for a multitude of applications in various fields of knowledge: various chapters of mathematics, computer science, biology, physics, chemistry, and sociology. Several books have been written in this field, which highlight both the theoretical aspects and the applications; for instance, see [7]. Figure 1 suggestively shows the connections between the previously mentioned domains.

This survey is structured as follows: First, basic notions in the field of algebraic hyperstructures are recalled, followed by results, in particular characterizations in the field of n-hypergroups. Special attention is given to the connections with binary relations and fundamental relations. Finally, join n-spaces with connections to lattice theory are presented.

2. Hypergroups

An algebraic hyperstructure is a nonempty set H together with one or some functions from $H\times H$ to the set ${\mathcal{P}}^{*}(H)$ of nonempty subsets of H. For all $(x,y)\in H^2$, one denotes by $x\circ y$ the image $f(x,y),$ where f is the function $f:H\times H\to {\mathcal{P}}^{*}(H).$ Then, $(H,\circ )$ is called a hypergroupoid.

If $S,T\in {\mathcal{P}}^{*}(H),$$S\circ T$ denotes the set ${\bigcup }_{s\in S,t\in T}s\circ t.$

Definition 1.

The pair $(H,\circ )$ is called a semihypergroup if

$$\forall (r,s,t)\in H^3,(r\circ s)\circ t=r\circ (s\circ t),$$

where $(r\circ s)\circ t$ denotes the union

$$\bigcup _{a\in r\circ s}a\circ t.$$

Analogously,

$$r\circ (s\circ t)=\bigcup _{b\in s\circ t}r\circ b.$$

Definition 2.

A hypergroup $(H,\circ )$ is a semihypergroup such that

$$\forall (a,b)\in H^2,\exists (x,y)\in H^2\mathrm{such}\mathrm{that}$$

$$a\in b\circ x\mathrm{and}a\in y\circ b$$

Several types of hypergroup homomorphisms are analyzed. We refer to [8]. Furthermore, several classes of subhypergroups are introduced and studied, such as canonical hypergroups, join spaces, and complete hypergroups. Join Spaces were introduced by Prenowitz.

Definition 3.

Let $(H,\circ )$ be a commutative hypergroup. Then, $(H,\circ )$ is a join space if the following implication is satisfied:

$$\forall (r,s,t,w)\in H^4,$$

$$r/s\cap t/w\ne \varnothing \Rightarrow r\circ w\cap s\circ t\ne \varnothing ,$$

where $r/s$ denotes the set

$$\{a\in H\mid r\in a\circ s\}.$$

Example 1.

Suppose that $(L,\vee ,\wedge )$ is a lattice. Then, L is a distributive lattice if and only if $(L,\star )$ is a join space, where $a\star b=\{x\in L|a\wedge b\le x\le a\vee b\}.$

Example 2.

Suppose that $(L,\vee ,\wedge )$ is a lattice. Then, L is a modular lattice if and only if $(L,\circ )$ is a join space, where $a\circ b=\{x\in L|a\vee b=b\vee x=a\vee x\}.$ Clearly, $a\vee b\in a\circ b.$

Canonical hypergroups have a structure close to that of a commutative group: they are commutative, have a scalar identity e (that is, $\forall x\in H,x\circ e=e\circ x=x)$, every element has a unique inverse, and they are reversible (that is, if $x\in y\circ z,$ then $z\in y^{-1}\circ x,y\in x\circ z^{-1}).$

An important result is the next one:

Theorem 1.

Let $(H,\circ )$ be a commutative hypergroup. Then, it is a canonical hypergroup iff it is a join space with a scalar identity.

One of the most-investigated hypergroups associated with binary relations is that introduced by Rosenberg [9] in 1998. It represents a theme of research of numerous papers. Rosenberg associated a partial hypergroupoid $H_{\rho }=(H,\circ )$ with a binary relation $\rho$ defined on a set H, where, for any $x,y\in H,$ we have $x\circ x=\{z\in H|(x,z)\in \rho \}$ and $x\circ y=x\circ x\cup y\circ y.$

Definition 4.

An element b in H is an outer element of ρ if there exists $a\in H$ such that $(a,b)\notin {\rho }^2$.

Theorem 2.

$(H,\circ )$ is a hypergroup iff:

(1)

ρ has full domain;

(2)

ρ has full range;

(3)

$\rho \subseteq {\rho }^2$;

(4)

If $(a,b)\in {\rho }^2$, then $(a,b)\in \rho$, where b is an outer element of $\rho .$

Special attention is paid to the fundamental $\beta$ relation, which leads to a group quotient structure.

Definition 5.

Suppose that $(H,\circ )$ is a semihypergroup and n is a natural number greater than 1. We can consider the relation ${\beta }_n$ on H as follows: $x{\beta }_{n}y$ if there exist $a_1,a_2,\dots ,a_n$ in H, such that $\{x,y\}\subseteq {\displaystyle \prod _{i=1}^n}{a}_i$, and assume that $\beta ={\displaystyle \bigcup _{n\ge 1}}{\beta }_n$, where ${\beta }_1=\left\{(r,r)\right|r\in H\}$.

In [10], Freni showed that, in every hypergroup, the relation $\beta$ is transitive, so the following result holds:

Theorem 3.

If $(H,\circ )$ is a hypergroup, then $(H/\beta ,·)$ is a group, where $\overline{x}·\overline{y}=\overline{z}$, where z is an arbitrary element of $x\circ y.$ Moreover, the canonical projection $\phi :H\to H/\beta$ is a good homomorphism.

3. n-Hypergroups

Davvaz and Vougiouklis [11] defined the notion of n-hypergroups for the first time. This concept is a generalization of n-groups, as well as hypergroups in the sense of Marty. Some properties of such hyperstructures were investigated in [12,13,14,15,16,17,18]. Moreover, some researchers have pointed out the relation between n-hypergroup and fuzzy sets.

Suppose that H is a nonempty set. A function $f:\underset{n\mathrm{times}}{\underbrace{H\times \dots \times H}}\to {\mathcal{P}}^{*}(H)$ is called an n-hyperoperation. As usual, we may write $H^n=H\times \dots \times H$, where H appears n times. An element of $H^n$ is denoted by $(x_1,\dots ,x_n)$, where $x_i\in H$ for any i with $1\le i\le n$. Let $P_1,\dots ,P_n$ be nonempty subsets of H. We define

$$f(P_1,\dots ,P_n)=\bigcup \left\{f(p_1,\dots ,p_n)\right|p_i\in P_i,i=1,\dots ,n\}$$

The pair $(H,f)$ is called an n-hypergroupoid. An n-hypergroupoid $(H,f)$ is called an n-semihypergroup iff

$$f(h_1^{i-1},f(h_i^{n+i-1}),h_{n+i}^{2n-1})=f(h_1^{j-1},f(h_j^{n+j-1}),h_{n+j}^{2n-1}),$$

for all $1\le i,j\le n$ and $h_1,h_2,\dots ,h_{2n-1}\in H$. An n-semihypergroup $(H,f)$ in which the equation:

$$b\in f(h_1^{i-1},x_i,h_{i+1}^n)$$

(1)

has the solution $x_i\in H$ for every $h_1,\dots ,h_{i-1},h_{i+1},\dots ,h_n,b\in H$, and $1\le i\le n$ is called an n-hypergroup. If the value of $f(h_1,\dots ,h_n)$ is independent of the permutation of elements $h_1,\dots ,h_n$, then we have a commutative n-hypergroup.

Example 3.

If $(H,\star )$ is a hypergroup, then obtain an n-hypergroup by defining $f(h_1,\dots ,h_n)=h_1\star \dots \star h_n$, for all $h_1,\dots ,h_n\in H$.

Example 4.

Let $\mathbb{Z}$ be the set of integer numbers. If we define

$$f(h_1,\dots ,h_n)=\{m_1{h}_1+\dots +m_n{h}_n|m_1,\dots ,m_n\in \mathbb{Z}\},$$

then $(\mathbb{Z},f)$ is an n-hypergroup.

Example 5.

Assume $(L,\vee ,\wedge )$ is a modular lattice. For every $h_1,\dots ,h_n\in L$ and $i\in \{1,\dots ,n\}$, we define

$$\begin{array}{c}{A}_n^{(i)}=h_1\vee \dots \vee h_{i-1}\vee h_{i+1}\vee \dots \vee h_n,\hfill \\ \hfill \\ A_n=h_1\vee \dots \vee h_n.\hfill \end{array}$$

If we define:

$$f(h_1,\dots ,h_n)=\{x\in L|x\vee A_n^{(i)}=A_n,\mathrm{for}\mathrm{all}1\le i\le n\},$$

then $(L,f)$ is a commutative n-hypergroup.

Theorem 4.

Suppose that $(H,f)$ is an n-semihypergroup. Then, $(H,f)$ is an n-hypergroup iff Equation (1) is solvable at the first place and at the last place or at least one place $1<i<n$.

Proof. 

If Equation (1) is solvable at the place $i=1$ and $i=n$, then, for every $h_1,\dots ,h_n,b\in H$, there are $x_0,z_0\in H$ such that

$$b\in f(x_0,h_2^n)\mathrm{and}{x}_0\in f(h_1^{n-1},z_0).$$

If $j\in \{1,\dots ,n\}$ is arbitrary, then we have

$$b\in f(f(h_1^{n-1},z_0),h_2^n)=f(h_1^{j-1},f(h_j^{n-1},z_0,h_2^j),h_{j+1}^n).$$

Hence, there is $x\in f(h_j^{n-1},z_0,h_2^j)$ such that $b\in f(h_1^{j-1},x,h_{j+1}^n)$.

Now, assume that Equation (1) is solvable at place $1<i<n$. Assume that $j<i$, then for every $a_1,\dots ,a_n,b\in H$, there is $y_1\in H$ such that

$$b\in f(h_1^{i-1},y_1,f(\underset{n-(i-j+1)}{\underbrace{h_1,\dots ,h_1}},h_{j+1}^{i+1}),h_{i+2}^n).$$

This implies that

$$b\in f(h_1^{j-1},f(h_j^{i-1},y_1,\underset{n-(i-j+1)}{\underbrace{h_1,\dots ,h_1}}),h_{j+1}^n).$$

Hence, there is $x\in f(h_j^{i-1},y_1,h_1,\dots h_1)$ such that $b\in f(h_1^{j-1},x,h_{j+1}^n)$. If we consider $i<j$, then in a similar, way we can prove that Equation (1) is solvable. □

An n-hyperoperation f is called weakly $(i,j)$-associative if

$$f(x_1^{i-1},f(x_i^{n+i-1}),x_{n+i}^{2n-1})\cap f(x_1^{j-1},f(x_j^{n+j-1}),x_{n+j}^{2n-1})\ne \varnothing ,$$

and $(i,j)$-associative if

$$f(x_1^{i-1},f(x_i^{n+i-1}),x_{n+i}^{2n-1})=f(x_1^{j-1},f(x_j^{n+j-1}),x_{n+j}^{2n-1}),$$

holds for fixed $1\le i<j\le n$ and all $x_1,x_2,\dots ,x_{2n-1}\in H$.

We say that the element $a\in H$ is in the center of an n-hypergroupoid $(G,f)$, if

$$f(a,x_2^n)=f(x_2,a,x_3^n)=f(x_2^3,a,x_4^n)=\dots =f(x_2^n,a),$$

for all $x_2,\dots ,x_n\in H$. An $(i,i+k)$-associative n-hypergroupoid $(G,f)$ containing cancelable elements in the center (cancelable elements belong to the center) is $(1,n)$-associative [12].

Theorem 5

([12]). An n-hypergroupoid containing cancellative elements in the center is an n-semihypergroup iff it is $(i,j)$-associative for some $1\le i<j\le n$.

An n-hypergroupoid $(H,f)$ is called a b-derived from a binary hypergroupoid $(G,\star )$ [12], and denote this fact by $(H,f)=de{r}_b(H,\star )$ if the hyperoperation f has the form

$$f(x_1^n)=(x_1\star x_2\star \dots \star x_n)\star b.$$

Theorem 6

([12]). An n-semihypergroup has a neutral element iff it is derived from a binary semihypergroup with the identity.

Theorem 7

([12]). An n-semihypergroup derived from a binary semihypergroup has a neutral polyad iff it has a neutral element.

Consequently, if an n-semihypergroup without neutral elements is derived from a binary semihypergroup, then it does not possess any neutral polyad.

Theorem 8

([12]). If an n-semihypergroup $(H,f)$ does not contain any neutral elements, then to $(H,f)$, we can adjoint the neutral element if and only if $(H,f)$ is derived from a binary semihypergroup.

Theorem 9

([12]). To an n-semihypergroup $(H,f)$ we can adjoint the neutral element iff $(H,f)$ is derived from a binary semihypergroup.

Theorem 10

([12]). For any n-semihypergroup $(H,f)$ with a right neutral polyad, there is a semihypergroup $(H,\star )$ with a right identity and an endomorphism φ of $(H,\star )$ such that

$$f(x_1^n)=x_1\star \phi (x_2)\star {\phi }^2(x_3)\star \dots \star {\phi }^{n-1}(x_n)\star b,$$

for some $b\in H$.

Theorem 11

([12]). For any n-semihypergroup $(H,f)$ with a left neutral polyad, there is a semihypergroup $(H,\star )$ with a left identity and an endomorphism ψ of $(H,\star )$ such that

$$f(x_1^n)=b\star {\psi }^{n-1}(x_1)\star {\psi }^{n-2}(x_2)\star \dots \star {\varphi }^2(x_{n-2})\star \varphi (x_{n-1})\star x_n$$

for some $b\in H$.

4. Binary Relations and Fundamental Relations

Suppose that R is a binary relation on a nonempty set H. We define a partial n-hypergroupoid $(H,f_R)$, as follows:

$$f_R(\underset{n}{\underbrace{w,\dots ,w}})=\{y\mid (w,y)\in R\},$$

for all w in H and

$$f_R(w_1,w_2,\dots ,w_n)=f_R(\underset{n}{\underbrace{w_1,\dots ,w_1}})\cup f_R(\underset{n}{\underbrace{w_2,\dots ,w_2}})\cup \dots \cup f_R(\underset{n}{\underbrace{w_n,\dots ,w_n}}),$$

for every $w_1,w_2,\dots ,w_n\in H$. It is clear that $(H,f_R)$ is commutative. The partial n-hypergroupoid $(H,f_R)$ is a generalization of the Rosenberg partial hypergroupoid. We denote $f_R(w_1,w_2,\dots ,w_n)$ by $f_R(w_1^n)$. The relation R is transitive iff, for any w in H, we have

$$f_R(f_R(\underset{n}{\underbrace{w,\dots ,w}}),\underset{n-1}{\underbrace{w,\dots ,w}})=f_R(\underset{n}{\underbrace{w,\dots ,w}}).$$

Moreover, $(H,f_R)$ is an n-hypergroupoid if the domain of R is H.

Theorem 12

([17]). Suppose that R is a binary relation on H, with full domain. Then, $(H,f_R)$ is an n-semihypergroup iff $R\subset R^2$ and for each outer element y of R, if $(x,y)\in R^2$ implies $(x,y)\in R.$

It follows that:

Corollary 1.

Suppose that R is a binary relation with full domain. Then, $(H,f_R)$ is an n-hypergroup iff the following hold:

(1)

R has a full range;

(2)

$R\subset R^2$;

(3)

$(x,y)\in R^2$ implies $(x,y)\in R$ for every outer element $y\in R$.

Note that if R is a subset of $R^2$, then x is an outer element of R iff $x\notin f_R(f_R(\underset{n}{\underbrace{w,\dots ,w}}),$$\underset{n-1}{\underbrace{w,\dots ,w}})$ for some $w\in H$.

If R is a subset of $R^2$, then there are no outer elements of R iff, for each $w\in H$, we have

$$f_R(f_R(\underset{n}{\underbrace{w,\dots ,w}}),\underset{n-1}{\underbrace{w,\dots ,w}})=H.$$

Theorem 13

([17]). Suppose that the relation R is reflexive and symmetric. Then, $(H,f_R)$ is an n-hypergroup iff, for every $u,w\in H$, we have

$$f_R(f_R(\underset{n}{\underbrace{u,\dots ,u}}),\underset{n-1}{\underbrace{u,\dots ,u}})-f_R(\underset{n}{\underbrace{u,\dots ,u}})\subset f_R(f_R(\underset{n}{\underbrace{w,\dots ,w}}),\underset{n-1}{\underbrace{w,\dots ,w}}).$$

Corollary 2.

Suppose that the relation R is reflexive and symmetric, but not transitive. Then, $(H,f_R)$ is an n-hypergroup iff $R^2=H^2.$

The concept of mutually associative hypergroupoids was introduced by Corsini [19]. We generalize this concept to n-hypergroupoids. Two partial n-hypergroupoids $(H,f_1)$ and $(H,f_2)$ are mutually associative if, for every $w_1,\dots ,w_{2n-1}\in H$, we have:

(i${}_1$)

$f_2(f_1(w_1^n),w_{n+1}^{2n-1})=f_1(w_1^{n-1},f_2(w_n^{2n-1}));$

(i${}_2$)

$f_2(w_1,f_1(w_2^{n+1}),w_{n+2}^{2n-1})=f_1(w_1^{n-2},f_2(w_{n-1}^{2n-2}),w_{2n-1});$

(i${}_3$)

$f_2(w_1,w_2,f_1(w_3^{n+2}),w_{n+3}^{2n-1})=f_1(w_1^{n-3},f_2(w_{n-2}^{2n-3}),w_{2n-2},w_{2n-1});$

…

(i${}_{n-1}$)

$f_2(w_1^{n-2},f_1(w_{n-1}^{2n-2}),w_{2n-1})=f_1(w_1,f_2(w_2^{n+1}),w_{n+2}^{2n-1});$

(i${}_n$)

$f_2(w_1^{n-1},f_1(w_n^{2n-1}))=f_1(f_2(w_1^n),w_{n+1}^{2n-1}).$

Let $f_1$ and $f_2$ be two ordinary hyperoperations. Then, we obtain two mutually associative partial hypergroupoids. If R is a binary relation on H and $A\subset H$, we denote

$$R(A)=\{b\mid (a,b)\in R,\mathrm{for}\mathrm{some}a\in A\}.$$

If $A=\{w_1,w_2,\dots ,w_k\}$, we write $R(w_1^k)$ for $R(A)$. If R and S are binary relations on H, then we denote by $SR$ the relation $\{(a,c)\in H^2\mid (a,b)\in R$ and $(b,c)\in S$, for some $b\in H\}$.

Theorem 14

([17]). Let R and S be two relations on H with full domains. Then, $(H,f_R)$ and $(H,f_S)$ are mutually associative iff, for every $w_1,w_2,\dots ,w_{2n-1}\in H$, we have

$$SR(w_1^n)\cup S(w_{n+1}^{2n-1})=RS(w_n^{2n-1})\cup R(w_1^{n-1}).$$

Theorem 15

([17]). If $(H,f_R)$ and $(H,f_S)$ are mutually associative n-hypergroups, then $(H,f_{R\cup S})$ is also an n-hypergroup.

Theorem 16.

Let R and S be relations on H, such that $R\subset SR$. If $(H,f_R)$ is an n-hypergroup, $(H,f_R)$ and $(H,f_S)$ are mutually associative and one of the following two conditions holds:

(1)

$RS\cap \{(w,w)\mid w\in H\}=\varnothing ;$

(2)

The domain ${\displaystyle (RS)}$ of $RS$ is different from H.

Then, $(H,f_{SR})$ is an n-hypergroup, as well.

Now, suppose that $(H,f)$ is an n-semihypergroup. We denote

$$\begin{array}{c}{f}_{(1)}=\{f(w_1^n)\mid w_i\in H,1\le i\le n\}\},\hfill \\ f_{(2)}=\{f(f(u_1^n),w_2^n)\mid u_i\in H,w_j\in H,1\le i\le n,\hfill \\ \forall 2\le j\le n\},\\ f_{(3)}=\{f(f(f(v_1^n),u_2^n),w_2^n)\mid v_i\in H,u_j\in H,w_j\in H,\hfill \\ \forall 1\le i\le n,\forall 2\le j\le n\},\end{array}$$

and so on. Denote ${\displaystyle \mathcal{U}=\bigcup _{k\in {\mathrm{I}\mathrm{N}}^{*}}{f}_{(k)}.}$ We define ${\displaystyle \beta =\bigcup _{k\ge 1}{\beta }_k}$, where, for all $x,y$ of H,

$$a{\beta }_{k}y\iff \exists u\in f_{(k)},\mathrm{such}\mathrm{that}\{x,y\}\subseteq u.$$

Denote ${\displaystyle \bigcup _{{}_{u\in \mathcal{U}}^{a\in u}}u}$ by ${\mathcal{C}}_1(a)$, which means

$${\mathcal{C}}_1(w)=\{a\mid \mathrm{there}\mathrm{exists}u\in \mathcal{U}\mathrm{such}\mathrm{that}w\in u,a\in u\}.$$

For every $n\in {\mathrm{I}\mathrm{N}}^{*},$ denote

$${\mathcal{C}}_{n+1}(w)=\{a\mid \mathrm{there}\mathrm{exists}u\in \mathcal{U}\mathrm{such}\mathrm{that}{\mathcal{C}}_n(w)\cap u\ne \varnothing ,a\in u\}.$$

A subsets B is a complete part of $(H,f)$ if, for every $u\in \mathcal{U},$

$$B\cap u=\varnothing ⟹u\subset B.$$

Suppose that $\mathcal{C}(w)$ is the complete closure of w. We have ${\displaystyle \mathcal{C}(w)=\bigcup _{i\in {\mathrm{I}\mathrm{N}}^{*}}{\mathcal{C}}_i(w),}$ for all $w\in H$.

Theorem 17

([17]). Suppose that $(H,f)$ is an n-semihypergroup. The relation β is transitive iff $\mathcal{C}(w)={\mathcal{C}}_1(w)$, for all $w\in H.$

Theorem 18

([17]). If $(H,f)$ is an n-hypergroup, then β is transitive.

Suppose that $(H_1,f)$ and $(H_2,g)$ are n-hypergroups. We define $(f,g):{(H_1\times H_2)}^n⟶{\mathcal{P}}^{*}(H_1\times H_2)$ by $(f,g)((u_1,v_1),\dots ,(u_n,v_n))=\{(u,v)|u\in f(u_1,\dots ,u_n),v\in g(v_1,\dots ,$$v_n)\}$. Clearly, $(H_1\times H_2,(f,g))$ is an n-hypergroup, and it is the direct hyperproduct of $H_1$ and $H_2$.

Theorem 19

([11]). Let $(H_1,f)$ and $(H_2,g)$ be two n-hypergroups, and let ${\beta }_1^{*}$, ${\beta }_2^{*}$, and ${\beta }^{*}$ be fundamental equivalence relations on $H_1$, $H_2$, and $H_1\times H_2$, respectively. Then,

$$(H_1\times H_2)/{\beta }^{*}\cong H_1/{\beta }_1^{*}\times H_2/{\beta }_2^{*}.$$

Let $(H,f)$ be an n-semihypergroup and $\rho$ be an equivalence relation on H; we define

$$X\overline{\overline{\rho }}Y⟺x\rho y\mathrm{for}\mathrm{all}x\in X,y\in Y.$$

The relation $\rho$ is a strongly regular relation if $x_i\rho y_i$ for all $1\le i\le n$, then,

$$f(x_1,\dots ,x_n)\overline{\overline{\rho }}f(y_1,\dots ,y_n).$$

If $\rho$ is a strongly regular relation on an n-semihypergroup $(H,f)$, then the quotient $(H/\rho ,f/\rho )$ is an n-semigroup such that

$$f/\rho (\rho (x_1),\dots ,\rho (x_n))=\rho (z)\mathrm{for}\mathrm{all}z\in f(x_1,\dots ,x_n)$$

where $x_1,\dots ,x_n\in H$.

Similar to the relation defined by Freni [20,21] on semihypergroups, Davvaz et al. [13] introduced the following relation on an n-semihypergroup so that the quotient is a commutative n-semigroup. Let $(H,f)$ be an n-semihypergroup. Then, $\widehat{\gamma }$ denotes the transitive closure of the relation ${\displaystyle \gamma =\bigcup _{k\ge 1}{\gamma }_k}$, where ${\gamma }_1=\left\{(w,w)\right|w\in H\}$, and for every integer $k>1$, we define

$$x{\gamma }_{k}y⟺x\in u_{(k)}andy\in u_{(k)}^{\sigma }.$$

When $m=k(n-1)+1$, there are $a_1^m\in H^m$ and $\sigma \in {\mathbb{S}}_m$ such that $u_{(k)}=f_{(k)}(a_1^m)$ and $u_{(k)}^{\sigma }=f_{(k)}(a_{\sigma (1)}^{\sigma (m)}).$$x{\gamma }_{1}y$ (i.e., $x=y$), then we write $x\in u_{(0)}$ and $y\in u_{(0)}^{\sigma }=u_{(0)}$. We define ${\gamma }^{*}$ as the smallest equivalence relation such that the quotient $(H/{\gamma }^{*},f/{\gamma }^{*})$ is a commutative n-semigroup.

Theorem 20

([13]). The fundamental relation ${\gamma }^{*}$ is the transitive closure of the relation γ.

Proof. 

The n-operation $f/\widehat{\gamma }$ in $H/\widehat{\gamma }$ is defined in the usual manner:

$$f/\widehat{\gamma }(\widehat{\gamma }(x_1),\dots ,\widehat{\gamma }(x_n))=\left\{\widehat{\gamma }(y)\right|y\in f(\widehat{\gamma }(x_1),\dots ,\widehat{\gamma }(x_n))\}$$

for all $x_1,\dots ,x_n\in H$. Let $a_1\in \widehat{\gamma }(x_1),\dots ,a_n\in \widehat{\gamma }(x_n)$. Then, we have:

• $a_1\widehat{\gamma }{x}_1$ iff there exist $x_{11},\dots x_{1m_1+1}$ with $x_{11}=a_1$, $x_{1m_1+1}=x_1$ such that

  $$\begin{array}{c}{x}_{1i_1}\in u_{(k_1)}(1\le i_1\le m_1-1),\hfill \\ x_{1i_1+1}\in u_{(k_1)}^{{\sigma }_1}(2\le i_1\le m_1).\hfill \end{array}$$
• …
• $a_n\widehat{\gamma }{x}_n$ iff there exist $x_{n1},\dots x_{n{m}_n+1}$ with $x_{n1}=a_n$, $x_{n{m}_n+1}=x_n$ such that

  $$\begin{array}{c}{x}_{n{i}_n}\in u_{(k_n)}(1\le i_n\le m_n-1),\hfill \\ x_{n{i}_n+1}\in u_{(k_n)}^{{\sigma }_n}(2\le i_n\le m_n).\hfill \end{array}$$

Therefore, we obtain

$$\begin{array}{cc}f(x_{1i_1},x_{21},\dots ,x_{n1})\subseteq u_{(k_1)}\hfill & 1\le i_1\le m_1-1,\hfill \\ f(x_{1i_1+1},x_{21},\dots ,x_{n1})\subseteq u_{(k_1)}^{{\sigma }_1}\hfill & 2\le i_1\le m_1,\hfill \\ f(x_{1m_1+1},x_{2i_2},\dots ,x_{n1})\subseteq u_{(k_2)}\hfill & 1\le i_2\le m_2-1,\hfill \\ f(x_{1m_1+1},,x_{2i_2+1},\dots ,x_{n1})\subseteq u_{(k_2)}^{{\sigma }_2}\hfill & 2\le i_2\le m_2,\hfill \\ \dots \hfill & \dots \hfill \\ f(x_{1m_1+1},x_{2m_2+1},\dots ,x_{n{i}_n})\subseteq u_{(k_n)}\hfill & 1\le i_n\le m_n-1,\hfill \\ f(x_{1m_1+1},x_{2m_2+1},\dots ,x_{n{i}_n+1})\subseteq u_{(k_n)}^{{\sigma }_n}\hfill & 2\le i_n\le m_n.\hfill \end{array}$$

This yields that $f/\widehat{\gamma }(\widehat{\gamma }(x_1),\dots ,\widehat{\gamma }(x_n))$ is singleton. Therefore, we can write

$$f/\widehat{\gamma }(\widehat{\gamma }(x_1),\dots ,\widehat{\gamma }(x_n))=\widehat{\gamma }(z)\mathrm{for}\mathrm{all}z\in f(\widehat{\gamma }(x_1),\dots ,\widehat{\gamma }(x_n)).$$

Moreover, since f is associative, we obtain that $f/\widehat{\gamma }$ is associative, and consequently, $H/\widehat{\gamma }$ is an n-semigroup.

$(H/\widehat{\gamma },f/\widehat{\gamma })$ is commutative because, if $\sigma \in {\mathbb{S}}_n$ and $a\in f(x_1^n)$ and $b\in f(x_{\sigma (1)}^{\sigma (n)})$, then $a\gamma b$, and so, $\widehat{\gamma }(a)=\widehat{\gamma }(b)$. Therefore, $f/\widehat{\gamma }(\widehat{\gamma }(x_1),\dots ,\widehat{\gamma }(x_n))=f/\widehat{\gamma }(\widehat{\gamma }(x_{\sigma (1)}),\dots ,\widehat{\gamma }(x_{\sigma (n)}));$ thus $(H/\widehat{\gamma },f/\widehat{\gamma })$ is commutative.

Now, assume that $\theta$ is an equivalence relation on H such that $H/\theta$ is a commutative n-semigroup. Then, for all $w_1,\dots ,w_n\in H$,

$$f/\theta (\theta (w_1),\dots ,\theta (w_n))=\theta (z)\mathrm{for}\mathrm{all}z\in f(\theta (w_1),\dots ,\theta (w_n)).$$

However, for any $\sigma \in {\mathbb{S}}_n$ and $w_1,\dots ,w_n\in H$ and $X_i\subseteq \theta (w_i)(i=1,\dots ,n)$, we have

$$f/\theta (\theta (w_1),\dots ,\theta (w_n))=\theta (f(w_{\sigma (1)},\dots ,w_{\sigma (n)}))=\theta (f(X_{\sigma (1)},\dots ,X_{\sigma (n)})).$$

Therefore,

$$\theta (w)=\theta (u_{(k)}^{\sigma })\mathrm{for}\mathrm{all}k\ge 0\mathrm{and}\mathrm{for}\mathrm{all}w\in u_{(k)}.$$

This gives that, for all $y\in H$,

$$w\in \gamma (y)\mathrm{implies}w\in \theta (y).$$

Since $\theta$ is transitively closed, it follows that

$$w\in \widehat{\gamma }(y)\mathrm{implies}w\in \theta (y).$$

Consequently, we obtain $\widehat{\gamma }={\gamma }^{*}$. □

Relation $\gamma$ is a strongly regular relation.

Now, we present some necessary and sufficient conditions such that the relation $\gamma$ is transitive. These conditions are analogous to those determined in [20] for the transitivity of relation $\gamma$ in hypergroups. Let M be a nonempty subset of n-semihypergroup $(H,f).$ We say that M is a $\gamma$-$part$ if, for any $k\in \mathbb{N},$ $i=1,2,\dots ,m=k(n-1)+1,$$\forall (w_1,w_2,\dots ,w_m)\in H^m$, $\forall \sigma \in {\mathbb{S}}_m,$ we have

$$f_{(k)}(w_1^m)\cap M\ne \varnothing ⟹f_{(k)}(w_{\sigma (1)}^{\sigma (m)})\subseteq M.$$

Theorem 21.

Suppose that M is a nonempty subset of an n-semihypergroup $H.$ Then, the following statements are equivalent:

(1)

M is a γ-part of H;

(2)

$x\in M,x\gamma y$ implies that $y\in M;$

(3)

$x\in M,x{\gamma }^{*}y$ implies that $y\in M.$

Proof. 

$(1\Rightarrow 2):$ If $(x,y)\in H^2$ is a pair such that $x\in M$ and $x\gamma y,$ then $\exists k\in \mathbb{N}$ for $i=1,\dots ,m=k(n-1)+1,\exists \sigma \in {\mathbb{S}}_m$ and $\exists (z_1,\dots ,z_m)\in H^m,$ such that $x\in f_{(k)}(z_1^m)\bigcap M$ and $y\in f_{(k)}(z_{\sigma (1)}^{\sigma (m)}).$ Since M is a $\gamma$-part of $H,$ we have $f_{(k)}(z_{\sigma (1)}^{\sigma (m)})\subseteq M$ and $y\in M.$

$(2\Rightarrow 3):$ Assume that $(x,y)\in H^2$ such that $x\in M$ and $x{\gamma }^{*}y.$ Then, there exist $p\in \mathbb{N}$ and $(x=w_0,w_1,\dots ,w_{p-1},w_p=y)\in H^{p+1}$ such that $x=w_0\gamma w_1\gamma \dots \gamma w_{p-1}\gamma w_p=y.$ Since $x\in M,$ applying $(2)$p times, it follows that $y\in M.$

$(3\Rightarrow 1):$ Suppose that $f_{(k)}(z_1^m)\bigcap M\ne \varnothing ,$ and $x\in f_{(k)}(z_1^m)\bigcap M.$ For any $\sigma \in {\mathbb{S}}_m$ and $y\in f_{(k)}(z_{\sigma (1)}^{\sigma (m)}),$ we have $x\gamma y.$ This yields that $x\in M$ and $x{\gamma }^{*}y.$ Finally, by $(3),$ we obtain $y\in M.$ This means that $f_{(k)}(z_{\sigma (1)}^{\sigma (m)})\subseteq M$. □

For every element x of an n-semihypergroup $(H,f),$ set:

$T_k(w)=\{(w_1,\dots ,w_m)\in H^m|m=k(n-1)+1,w\in f_{(k)}(x_1^m)\}$

${\displaystyle P_k(w)=\bigcup \left\{f_{(k)}(w_{\sigma (1)}^{\sigma (m)})\right|\sigma \in S_m,(w_1,\dots ,w_m)\in T_k(w),m=k(n-1)+1\}}$

${\displaystyle P_{\sigma }(w)=\bigcup _{k\ge 1}{P}_k(w).}$

From the preceding notations and definitions, it follows that

Corollary 3

([13]). For every $x\in H,P_{\sigma }(x)=\{y\in H|x\gamma y\}.$

Theorem 22

([13]). Let $(H,f)$ be an n-semihypergroup. The following statements are equivalent:

(1)

γ is transitive;

(2)

For any $w\in H,{\gamma }^{*}(w)=P_{\sigma }(w);$

(3)

For any $w\in H,$$P_{\sigma }(w)$ is a γ-part of $H.$

Let $(H,f)$ be an n-hypergroup; we consider the canonical projection $\phi :H\to H/{\gamma }^{*}$ with $\phi (x)={\gamma }^{*}(x).$

Corollary 4

([13]). Let $(H,f)$ be an n-hypergroup and $\delta \in H/{\gamma }^{*},$ then ${\phi }^{-1}(\delta )$ is a γ-part of $H.$

Corollary 5

([13]). If $(H,f)$ is a commutative n-semihypergroup, then $\gamma =\beta .$

Theorem 23

([13]). For every nonempty subset M of an n-hypergroup $(H,f)$, we have:

(1)

If $H/{\gamma }^{*}$ has a neutral element ε and $D={\phi }^{-1}(\epsilon ),$ then for every $i=1,\dots ,n,$

$$f(D^{i-1},M,D^{n-i})\subseteq {\phi }^{-1}(\phi (M));$$

(2)

Moreover if $H/{\gamma }^{*}$ is one-cancellative, then $f(D^{i-1},M,D^n)={\phi }^{-1}(\phi (M));$

(3)

If M is a γ-part of $H,$ then ${\phi }^{-1}(\phi (M))=M.$

Proof. 

(1) For any $x\in f(D^{i-1},M,D^{n-i}),$ there exist $d_2,\dots ,d_n\in D$ and $b\in M$ such that $x\in f(d_2^{i-1},b,d_{i+1}^n),$ so $\phi (x)=f/{\gamma }^{*}({\epsilon }^{i-1},\phi (b),{\epsilon }^{n-i})=\phi (b);$ therefore, $x\in {\phi }^{-1}(\phi (x))\subseteq {\phi }^{-1}(\phi (M)).$

(2) For any $x\in {\phi }^{-1}(\phi (M)),$ an element $b\in M$ exists such that $\phi (x)=\phi (b)$. Let $d\in D$. Then, there exists $a\in H$ such that $x\in f(a,b,d^{n-2}).$ Therefore,

$$\phi (b)=\phi (x)=f/{\gamma }^{*}(\phi (a),\phi {(d)}^{i-2},\phi (b),\phi {(d)}^{n-i})=f/{\gamma }^{*}(\phi (a),{\epsilon }^{i-2},\phi (b),{\epsilon }^{n-2}).$$

However, $f({\epsilon }^{i-1},\phi (b),{\epsilon }^{n-i})=\phi (b)$, and since $(H,f)$ is one-cancellative, thus $\phi (a)=\epsilon$ and $a\in {\phi }^{-1}(\epsilon )=D.$ Therefore, $x\in f(a,b,d^{n-2})=f(D^{i-1},M,D^{n-i}).$ This and (1) prove that ${\phi }^{-1}(\phi (M))=f(D^{i-1},M,D^{n-i}).$

(3) Clearly, we have $M\subseteq {\phi }^{-1}(\phi (M)).$ Furthermore, if $x\in {\phi }^{-1}(\phi (M)),$ then there exists $b\in M$ such that $\phi (x)=\phi (b).$ This yields that $x\in {\gamma }^{*}(x)={\gamma }^{*}(b)\subseteq M$ and ${\phi }^{-1}(\phi (M))\subseteq M.$ □

Theorem 24

([13]). If $(H,f)$ is an n-hypergroup with neutral (identity) $e,$ such that $H/\gamma *$ is j-cancellative, then we have:

(1)

If $x\in P_{\sigma }(e)$ and $x\gamma y$, then $y\in P_{\sigma }(e);$

(2)

γ is transitive.

Proof. 

(1) If $x\in P_{\sigma }(e)$ and $x\gamma y,$ then $\exists (k,k^{\prime })\in \mathbb{N}\times \mathbb{N},m=k(n-1)+1,m^{\prime }=k^{\prime }(n-1)+1,$ $\exists (x_1,\dots ,x_m)\in H^m$, $\exists (y_1,\dots ,y_{m^{\prime }})\in H^{m^{\prime }},$$\exists \sigma \in {\mathbb{S}}_m$ and $\exists {\sigma }^{\prime }\in {\mathbb{S}}_{m^{\prime }},$ such that $e\in f_{(k)}(x_1^m),x\in f_{(k)}(x_{\sigma (1)}^{\sigma (m)}),x\in f_{(k^{\prime })}(y_1^{m^{\prime }}),y\in f_{(k^{\prime })}(y_{{\sigma }^{\prime }(1)}^{{\sigma }^{\prime }(m^{\prime })}).$ Therefore, if $x^{\prime }$ is an element of H such that

$$\begin{array}{cc}e\in f(e^{n-2},x,x^{\prime })\hfill & \subseteq f(e^{n-2},f(e^{n-2},x,e),x^{\prime })\hfill \\ & \subseteq f(e^{n-2},f(e^{n-2},f_{(k^{\prime })}(y_1^{m^{\prime }}),f_{(k)}(x_1^m)),x^{\prime }).\hfill \end{array}$$

Moreover, we have

$$\begin{array}{cc}y\in f(e^{n-2},y,e)\hfill & \subseteq f(e^{n-2},y,f(e^{n-2},x,x^{\prime }))\hfill \\ & \subseteq f(e^{n-2},f_{(k^{\prime })}(y_{{\sigma }^{\prime }(1)}^{{\sigma }^{\prime }(m^{\prime })}),f(e^{n-2},f_{(k)}(x_{\sigma (1)}^{\sigma (m)}),x^{\prime })).\hfill \end{array}$$

Thus, $y\in P_{\sigma }(e).$

(2) By (1), we have $P_{\sigma }(e)={\gamma }^{*}(e)=D.$ Moreover, if $x{\gamma }^{*}y,$ then $x\in {\gamma }^{*}(y)$, so $x\in {\phi }^{-1}(\phi (y))=f(D^{i-1},y,D^{n-i}).$ Therefore, there exist $(a_2^n)\in D^n$ such that $x\in f(a_2^{i-1},y,a_{i+1}^n).$ Thus, there exist $k_i\in \mathbb{N}$, and there are $(x_{i1},\dots ,x_{i{m}_i})\in H^{m_i}$, where $m_i=k_i(n-1)+1,$ and ${\sigma }_i\in {\mathbb{S}}_{m_i}$ such that $e\in f_{(k_i)}(x_{i1}^{i{m}_i})=A_i$ and $a_i\in f_{(k)}(x_{i{\sigma }_i(1)}^{i{\sigma }_i(m_i)})=A_{\sigma (i)},$ where $i=2,\dots ,n.$ If $j\in \{1,\dots ,n\}$, it follows that

$$x\in f(a_2^{j-1},y,a_{j+1}^n)\subseteq f(A_{\sigma (2)}^{\sigma (j-1)},y,A_{\sigma (j+1)}^{\sigma (n)})\mathrm{and}y\in f(e^{j-1},y,e^{j-n)})\subseteq f(A_2^{j-1},y,A_{j+1}^n).$$

Whence $x\gamma y$ and ${\gamma }^{*}=\gamma .$ □

5. Join $\mathit{n}$-Spaces

Let $(L,\le ,\vee )$ be a join semi-lattice and $a_1^n$ be elements of L. We denote

$$\begin{array}{cc}{A}_n=a_1\vee a_2\vee \dots \vee a_n,\hfill & A_n^{(1)}=a_2\vee \dots \vee a_n,\hfill \\ A_n^{(n)}=a_1\vee \dots \vee a_{n-1},\hfill & A_n^{(i)}=a_1\vee \dots \vee a_{i-1}\vee a_{i+1}\vee \dots \vee a_n,\hfill \end{array}$$

for any $2\le i\le n-1.$ For any $a_1^n$ of L, we define the following n-hyperoperation:

$$f(a_1^n)=\{x\mid x\vee A_n^{(i)}=A_n,foranyi\in \{1,2,\dots ,n\}\}.$$

Notice that $A_n\in f(a_1^n)$. Notice also that the n-hyperoperation f is commutative.

If $(L,\le ,\vee )$ is a join semi-lattice, then the following statements hold:

(1)

For any $b,a_1^{n-1}$ of L, there is $x=b\vee A_{n-1}$ such that $b\in f(x,a_1^{n-1})$.

(2)

If L has a 0, then 0 is a scalar identity for $(L,f)$.

(3)

If $n\ge 3$, then any $x\in L$ is an identity for $(L,f)$.

(4)

For any $a,x,b_1^{n-1}$ of L, we have the equivalence:

$$a\in f(x,b_1^{n-1})\mathrm{iff}x\in f(a,b_1^{n-1}).$$

(5)

For any $a,b_1^{n-1}$ of L, we have $a/b_1^{n-1}=f(a,b_1^{n-1}).$

Theorem 25

([18]). If $(L,f)$ is an n-semihypergroup, then for any $a,c,b_1^{n-1},d_1^{n-1}$ of L, we have

$$f(a,b_1^{n-1})\cap f(c,d_1^{n-1})\ne \varnothing ⟹f(a,d_1^{n-1})\cap f(c,b_1^{n-1})\ne \varnothing .$$

Theorem 26

([18]). For any $a_1^{2n-1}$ of L, if we denote

$$S=\{y\mid A_{2n-1}=A_{2n-1}^{(i)}\vee y,foranyi\in \{1,2,\dots ,2n-1\}\},$$

then $f(a_1^{n-1},f(a_n^{2n-1}))\subset S.$

Theorem 27

([18]). If $(L,\vee ,\wedge )$ is a modular lattice, then $S\subset f(a_1^{n-1},f(a_n^{2n-1})).$

Proof. 

Let $y\in S$. Set $z\in (y\vee A_{n-1})\wedge (a_n\vee \dots \vee a_{2n-1}).$ We check $z\in f(a_n^{2n-1})$ and $y\in f(a_1^{n-1},z)$. Indeed, for any $i\in \{1,2,\dots ,n-2\}$, we have

$$\begin{array}{c}{a}_n\vee \dots \vee a_{n+i-1}\vee z\vee a_{n+i+1}\vee \dots \vee a_{2n-1}=\hfill \\ =(a_n\vee \dots \vee a_{n+i-1}\vee a_{n+i-1}\vee \dots \vee a_{2n-1})\vee [(y\vee A_{n-1})\wedge (a_n\vee \dots \vee a_{2n-1})]=\hfill \\ =(A_{2n-1}^{(n+i)}\vee y)\wedge (a_n\vee \dots \vee a_{2n-1})=A_{2n-1}\wedge (a_n\vee \dots \vee a_{2n-1})=a_n\vee \dots \vee a_{2n-1}.\hfill \end{array}$$

Similarly, we have

$$z\vee a_{n+1}\vee \dots \vee a_{2n-1}=a_n\vee \dots \vee a_{2n-2}\vee z=a_n\vee \dots \vee a_{2n-1}.$$

Hence, $z\in f(a_n^{2n-1}).$ On the other hand,

$$\begin{array}{c}{A}_{n-1}\vee z=A_{n-1}\vee [(y\vee A_{n-1})\wedge (a_n\vee \dots \vee a_{2n-1})]=\hfill \\ =(y\vee A_{n-1})\wedge A_{2n-1}=(y\vee A_{n-1})\wedge (A_{2n-2}\vee y)=y\vee A_{n-1}\hfill \end{array}$$

and for any $i\in \{1,2,\dots ,n-1\}$, we have

$$\begin{array}{c}{A}_n^{(i)}\vee y\vee z=(A_{n-1}^{(i)}\vee y)\vee [(y\vee A_{n-1})\wedge (a_n\vee \dots \vee a_{2n-1})=\hfill \\ =(y\vee A_{n-1})\wedge (A_{n-1}^{(i)}\vee y\vee a_n\vee \dots \vee a_{2n-1})=\hfill \\ =(y\vee A_{n-1})\wedge (A_{2n-1}^{(i)}\vee y)=\hfill \\ =(y\vee A_{n-1})\wedge (A_{2n-2}\vee y)=y\vee A_{n-1}.\hfill \end{array}$$

Therefore, $y\in f(a_1^{n-1},z)\subset f(a_1^{n-1},f(a_n^{2n-1})).$ □

Corollary 6

([18]). If $(L,\vee ,\wedge )$ is a modular lattice, then $(L,f)$ is an n-semihypergroup.

Theorem 28

([18]). If $(L,\vee ,\wedge )$ is a lattice and $(L,f)$ is an n-semihypergroup, then the lattice $(L,\vee ,\wedge )$ is modular.

Proof. 

Assume that L is not modular. Hence, L contains a five-element sublattice, isomorphic to this one: $\{m,a,b,c,M\}$, where $m<b<a<M$, $m<c<M$, $a,c,$ and $b,c$, respectively, are not comparable. We have $c\in f(a,\underset{n-2}{\underbrace{b,\dots ,b}},M)$ and $M\in f(b,\underset{n-1}{\underbrace{c,\dots ,c}})$, since $a\vee c=b\vee c=M.$ Hence,

$$c\in f(a,\underset{n-2}{\underbrace{b,\dots ,b}},f(b,\underset{n-1}{\underbrace{c,\dots ,c}}))=f(f(a,\underset{n-1}{\underbrace{b,\dots ,b}}),\underset{n-1}{\underbrace{c,\dots ,c}}).$$

Therefore, there exists $x\in f(a,\underset{n-1}{\underbrace{b,\dots ,b}})$, such that $c\in f(x,\underset{n-1}{\underbrace{c,\dots ,c}})$. We have $a=a\vee b=b\vee x=a\vee x\vee b=a\vee x$ and $c\vee x=c$, whence $x\le a$ and $x\le c$, that is $x\le a\wedge c=m.$ Hence, $x<b$, which contradicts $a=b\vee x.$ Therefore, $(L,\vee ,\wedge )$ is modular. □

Corollary 7

([18]). A lattice $(L,\vee ,\wedge )$ is modular iff $(L,f)$ is an n-semihypergroup.

Corollary 8

([18]). The lattice $(L,\vee ,\wedge )$ is modular iff the n-hypergroupoid $(L,f)$ is a join n-space.

Now, we can consider the following dual-n-hyperoperation $f^{\circ }$ on a meet semilattice $(L,\le ,\wedge )$, defined by: for any $a_1^n$ of L, we have:

$$f^{\circ }(a_1^n)=\{x\in L\mid x\wedge B_n^{(i)}=B_n,\mathrm{for}\mathrm{any}i\in \{1,2,\dots ,n\}\},$$

where $B_n=a_1\wedge a_2\wedge \dots \wedge a_n$, $B_n^{(1)}=a_2\wedge \dots \wedge a_n,$$B_n^{(n)}=a_1\wedge \dots \wedge a_{n-1}$ and for any $i\in \{2,\dots ,n-1\}$, $B_n^{(i)}=a_1\wedge \dots \wedge a_{i-1}\wedge a_{i+1}\wedge \dots \wedge a_n.$ By duality, the following result holds:

Theorem 29

([18]). A lattice $(L,\vee ,\wedge )$ is modular iff the n-hypergroupoid $(L,f^{\circ })$ is a join n-space:

• If L has the greatest element 1, then 1 is a scalar identity for $(L,f^{\circ })$.
• If $n\ge 3$, then any $x\in L$ is an identity for $(L,f^{\circ })$.

Theorem 30

([18]). Let $(L,\vee ,\wedge )$ be a modular lattice:

(1)

A subset I of L is an n-subhypergroup of $(L,f)$ iff I is an ideal of L.

(2)

A subset I of L is an n-subhypergroup of $(L,f^{\circ })$ iff I is a filter of L.

Proof. 

(1) Let $(I,f)$ be an n-subhypergroupoid of $(L,f)$. Then, for any $a_1,a_2\in I$, we have

$$a_1\vee a_2\in f(a_1,\underset{n-1}{\underbrace{a_2,\dots ,a_2}})\subset I.$$

If $a\in I$ and $x\le a$, then $x\in f(\underset{n}{\underbrace{a,\dots ,a}})\subset I.$

“⟸” Let $a_1^n$ be elements of I. If $z\in f(a_1^n)$, then $A_n=z\vee A_n^{(i)}$, for any $i\in \{1,2,\dots ,n\}$, whence $z\le A_n$. Since $A_n\in I$, it follows that $z\in I$. On the other hand, for any $a,a_1^{i-1},a_{i+1}^n$ of I and $1\le i\le n$, there is $x_i=a\vee A_n^{(i)}$ such that $a\in f(a_1^{i-1},x_i,a_{i+1}^n).$ Hence, I is an n-subhypergroup of $(L,f)$.

(2) It follows by duality. □

Theorem 31

([18]). Let $(L,\vee ,\wedge )$ be a lattice and $\phi :L\to L$ a bijective map. The following conditions are equivalent:

(1)

For any $a_1^n$ of L, we have $\phi (A_n)=\phi (a_1)\wedge \dots \wedge \phi (a_n).$

(2)

φ is a morphism from $(L,f)$ to $(L,f^{\circ })$.

Proof. 

(1⟹2): For any $a_1^n$ of L, we have $\phi (f(a_1^n))=\{\phi (z)\mid z\in f(a_1^n)\}=\{\phi (z)\mid A_n=z\vee A_n^{(i)}$, for any $i\in \{1,2,\dots ,n\}\},$ whence $\phi (a_1)\wedge \dots \wedge \phi (a_n)=\phi (A_n)=\phi (z\vee A_n^{(i)})=\phi (z)\wedge \phi (a_1)\wedge \dots \wedge \phi (a_{i-1})\wedge \phi (a_{i+1})\wedge \dots \wedge \phi (a_n),$ that is

$$\phi (z)\in f^{\circ }(\phi (a_1),\dots ,\phi (a_n)).$$

Now, let $t\in f^{\circ }(\phi (a_1),\dots ,\phi (a_n)).$ Since there is x such that $t=\phi (x)$, it follows that

$$\phi (x)\wedge [\phi (a_1)\wedge \dots \wedge \phi (a_{i-1})\wedge \phi (a_{i+1})\wedge \dots \wedge \phi (a_n)]=\phi (a_1)\wedge \dots \wedge \phi (a_n),$$

for any $i\in \{1,2,\dots ,n\},$ and according to $(1)$, we obtain $\phi (x\vee A_n^{(i)})=\phi (A_n),$ for any $i\in \{1,2,\dots ,n\}.$ Since $\phi$ is bijective, it follows that $x\vee A_n^{(i)}=A_n$, for any $i\in \{1,2,\dots ,n\}$, that is $x\in f(a_1^n).$ Hence,

$$t=\phi (x)\in \phi (f(a_1^n)).$$

(2⟹1): Let $a_1^n$ be elements of L. If $z\in f(a_1^n),$ then

$$\phi (z)\in f^{\circ }(\phi (a_1),\dots ,\phi (a_n))$$

that is

$$\phi (z)\wedge \phi (a_1)\wedge \dots \wedge \phi (a_{i-1})\wedge \phi (a_{i+1})\wedge \dots \wedge \phi (a_n)=\phi (a_1)\wedge \dots \wedge \phi (a_n),$$

for any $i\in \{1,2,\dots ,n\}.$ Hence,

$$\phi (a_1)\wedge \dots \wedge \phi (a_n)\le \phi (z).$$

For $z=A_n\in f(a_1^n),$ we obtain $\phi (a_1)\wedge \dots \wedge \phi (a_n)\le \phi (A_n).$ On the other hand, for any $i\in \{1,2,\dots ,n\}$, $A_n\in f(a_i,\underset{n-1}{\underbrace{A_n,\dots ,A_n}}),$ so

$$\phi (A_n)\in \phi (f(a_i,\underset{n-1}{\underbrace{A_n,\dots ,A_n}}))=f^{\circ }(\phi (a_i),\underset{n-1}{\underbrace{\phi (A_n),\dots ,\phi (A_n)}})$$

whence $\phi (A_n)=\phi (a_i)\wedge \phi (A_n),$ that is $\phi (A_n)\le \phi (a_i).$ It follows that

$$\phi (A_n)\le \phi (a_1)\wedge \dots \wedge \phi (a_n).$$

Therefore, the condition (1) holds. □

By duality, we obtain the following.

Theorem 32

([18]). Let $(L,\vee ,\wedge )$ be a lattice and $\phi :L\to L$ a bijective map. The following conditions are equivalent:

(1)

For any $a_1^n$ of L, we have

$$\phi (B_n)=\phi (a_1)\vee \dots \vee \phi (a_n).$$

(2)

φ is a morphism from $(L,f^{\circ })$ to $(L,f)$.

Let $(L,\vee ,\wedge )$ be an arbitrary lattice. We define on L the following n-hyperoperation: for any $a_1^n$ of L, we have

$$\begin{array}{ccc}\hfill g(a_1^n)& =& \{x\in L\mid B_n\le x\le A_n\},\mathrm{where}\hfill \\ \hfill B_n& =& a_1\wedge a_2\wedge \dots \wedge a_n\mathrm{and}{A}_n=a_1\vee a_2\vee \dots \vee a_n.\hfill \end{array}$$

The n-hypergroupoid $(L,g)$ has the following properties:

(1)

g is commutative;

(2)

For any $a\in L,$ we have $g(\underset{n}{\underbrace{a,\dots ,a}})=a;$

(3)

for any $a_1^n$ of L, we have $\left\{a_i^n\right\}\subset g(a_i^n);$

(4)

For any $a_1^{n-1}$ of $L,$ we have $b\in b/a_1^{n-1};$

(5)

For any $a\in L,$ we have $a/\{\underset{n-1}{\underbrace{a,\dots ,a}}\}=L;$

(6)

For any $a,b\in L,$ we have $x\in a/\{\underset{n-1}{\underbrace{b,\dots ,b}}\}\cap b/\{\underset{n-1}{\underbrace{a,\dots ,a}}\}$ iff $a\wedge x=b\wedge x$ and $a\vee x=b\vee x.$

Theorem 33

([18]). If the lattice $(L,\vee ,\wedge )$ is distributive, then for any $a_1^{2n-1}$ of L, we have

$$g(g(a_1^n),a_{n+1}^{2n-1})=[B_{2n-1},A_{2n-1}].$$

Proof. 

Indeed, for any $a_1^{2n-1}$ of L, we have

$$g(g(a_1^n),a_{n+1}^{2n-1})\subset [B_{2n-1},A_{2n-1}].$$

Conversely, let $z\in [B_{2n-1},A_{2n-1}].$ If $x=(z\wedge A_n)\vee B_n$, then $B_n\le x\le A_n,$ that is $x\in g(a_1^n).$ On the other hand,

$$z\in g(x,a_{n+1}^{2n-1}).$$

Indeed, by distributivity, we have

$$\begin{array}{c}{a}_{n+1}\wedge \dots \wedge a_{2n-1}\wedge x=a_{n+1}\wedge \dots \wedge a_{2n-1}\wedge [(z\wedge A_n)\vee B_n]=\hfill \\ =(z\wedge A_n\wedge a_{n+1}\wedge \dots \wedge a_{2n-1})\vee B_{2n-1}\le z\hfill \end{array}$$

and

$$\begin{array}{c}{a}_{n+1}\vee \dots \vee a_{2n-1}\vee x=a_{n+1}\vee \dots \vee a_{2n-1}\vee (z\wedge A_n)\vee B_n=\hfill \\ =(a_{n+1}\vee \dots \vee a_{2n-1}\vee B_n\vee z)\wedge (a_{n+1}\vee \dots \vee a_{2n-1}\vee B_n\vee A_n)=\hfill \\ =A_{2n-1}\wedge (a_{n+1}\vee \dots \vee a_{2n-1}\vee B_n\vee z)\ge z.\hfill \end{array}$$

Hence $z\in g(x,a_{n+1}^{2n-1}),$ whence $z\in g(g(a_1^n),a_{n+1}^{2n-1}).$ We obtain

$$g(g(a_1^n),a_{n+1}^{2n-1})=[B_{2n-1},A_{2n-1}].$$

□

Corollary 9

([18]). If $(L,\vee ,\wedge )$ is a distributive lattice, then $(L,g)$ is an n-hypergroup.

Proof. 

Since the subset $[B_{2n-1},A_{2n-1}]$ is invariant to any permutation $(a_{i_1},\dots ,a_{i_{2n-1}})$ of $(a_1,\dots ,a_{2n-1})$, it follows that

$$[B_{2n-1},A_{2n-1}]=g(g(a_{i_1},\dots ,a_{i_n}),a_{i_{n+1}},\dots ,a_{i_{2n-1}}).$$

Moreover, g is commutative, so it follows that g is associative. Therefore, we obtain that $(L,g)$ is an n-hypergroup. □

Theorem 34

([18]). If $(L,\vee ,\wedge )$ is a distributive lattice, then $(L,g)$ is a join n-space.

Proof. 

We still have to check the join n-space condition. Let $x\in a/b_1^{n-1}\cap c/d_1^{n-1},$ that is

$$\begin{array}{cc}x\wedge b_1\wedge \dots \wedge b_{n-1}\le a\le x\vee b_1\vee \dots \vee b_{n-1}\hfill & \mathrm{and}\hfill \\ x\wedge d_1\wedge \dots \wedge d_{n-1}\le c\le x\vee d_1\vee \dots \vee d_{n-1}.\hfill \end{array}$$

We have to prove that there is $z\in g(a,d_1^{n-1})\cap g(c,b_1^{n-1}),$ that is

$$\begin{array}{c}(a\wedge d_1\wedge \dots \wedge d_{n-1})\vee (c\wedge b_1\wedge \dots \wedge b_{n-1})\le z\le \hfill \\ \le (a\vee d_1\vee \dots \vee d_{n-1})\wedge (c\vee b_1\vee \dots \vee b_{n-1}).\hfill \end{array}$$

We have $a\wedge d_1\wedge \dots \wedge d_{n-1}\le (x\vee b_1\vee \dots \vee b_{n-1})\wedge (d_1\wedge \dots \wedge d_{n-1})=(x\wedge d_1\wedge \dots \wedge d_{n-1})\vee [(b_1\vee \dots \vee b_{n-1})\wedge d_1\wedge \dots \wedge d_{n-1}]\le c\vee b_1\vee \dots \vee b_{n-1}.$ Hence, $(a\wedge d_1\wedge \dots \wedge d_{n-1})\vee (c\wedge b_1\wedge$$\dots \wedge b_{n-1})\le c\vee b_1\vee \dots \vee b_{n-1}.$ Similarly, we have $(a\wedge d_1\wedge \dots \wedge d_{n-1})\vee (c\wedge b_1\wedge \dots \wedge b_{n-1})\le a\vee d_1\vee \dots \vee d_{n-1}.$ Therefore,

$$(a\wedge d_1\wedge \dots \wedge d_{n-1})\vee (c\wedge b_1\wedge \dots \wedge b_{n-1})\le (a\vee d_1\vee \dots \vee d_{n-1})\wedge (c\vee b_1\vee \dots \vee b_{n-1}),$$

that is

$$g(a,d_1^{n-1})\cap g(c,b_1^{n-1})\ne \varnothing .$$

□

Theorem 35

([18]). If $(L,\vee ,\wedge )$ is a join n-space, then the lattice $(L,\vee ,\wedge )$ is distributive.

Proof. 

Suppose that L is not distributive. Then, L contains a five-element sublattice $\{m,a,b,c,M\}$, where $a\vee c=b\vee c=M$, $a\wedge c=b\wedge c=m$, and either $a>b$ or $a,b,c$ are mutually non-comparable. We have $c\in a/\{\underset{n-1}{\underbrace{b,\dots ,b}}\}\cap b/\{\underset{n-1}{\underbrace{a,\dots ,a}}\}$, and since $(L,g)$ is a join n-space, we obtain

$$g(\underset{n}{\underbrace{a,\dots a}})\cap g(\underset{n}{\underbrace{b,\dots ,b}})\ne \varnothing ,$$

that is $a=b$, which is a contradiction.

Therefore, $(L,\vee ,\wedge )$ is distributive. □

Corollary 10

([18]). The n-hypergroupoid $(L,g)$ is a join n-space iff the lattice $(L,\vee ,\wedge )$ is distributive.

Theorem 36

([18]). Let $(L,\vee ,\wedge )$ be a distributive lattice. If I is an ideal and F is a filter of L, then $(I,g)$ and $(F,g)$ are n-subhypergroups of $(L,g)$.

Proof. 

Let I be an ideal of L. For any $a_1^n$ of I, we have $g(a_1^n)=\{z\mid B_n\le z\le A_n\}.$ Since $A_n=a_1\vee \dots \vee a_n\in I$ and $z\le A_n$, it follows $z\in I.$ Hence, $g(a_1^n)\subset I.$ On the other hand, we have $a\in g(a,a_1^{n-1})$ for any $a,a_1^{n-1}$ of I. Therefore, $(I,g)$ is an n-subhypergroup of $(L,g)$. Similarly, it follows that $(F,g)$ is an n-subhypergroup of $(L,g)$. □

The converse fails, as can be seen from the following example:

Example 6.

Let us consider the distributive lattice $(\mathcal{P}(M),\cup ,\cap )$, where M is a set with at least three elements. Let $a,b\in M$, $a\ne b$ and $S=\{M-\{a\},M-\{a,b\left\}\right\}$. Then, $(S,g)$ is an n-subhypergroup of $(\mathcal{P}(M),g)$, but S is neither an ideal, nor a filter of $\mathcal{P}(M)$, since $\varnothing \notin S$ and $M\notin S$, respectively.