1

**Figure 2.** SBG of *G*.

*γ is transitive and the SBG of G is not connected, because vertices 0 and 1 are not adjacent. The SBG of G is not complete, which results in the invalidity of the reverse Theorem 6.*

**Corollary 1.** *Let G* = &*H*, *E*' *be an SBG, and let H be a semihypergroup. If the SBG of G is complete, then H*/*γ*<sup>∗</sup> *is a singleton, and diam*(*G*) = 1.

**Proof.** By applying Theorems 4 and 6, the relation *γ* is transitive and *H*/*γ*<sup>∗</sup> is a singleton. Since the SBG of *G* is complete, then every path from vertex *x* to vertex *y* has a maximum length of 1, which means *diam*(*G*) = 1.

**Proposition 4.** *Suppose that H is a hypergroup on the SBG of G* = &*H*, *E*'. *Then, the degree of vertex x in SBG of G is equal to* |*γ*∗(*x*)|*.*

**Proof.** Let *H* be a hypergroup. By employing Theorem 4 and *γ*∗(*x*) as an equivalence class of x, the results show that the number of edges incident with vertex *x* is equal to |*γ*∗(*x*)|.

**Corollary 2.** *Let G* = &*H*, *E*' *be an SBG, and let H be a hypergroup. Assume that* |*γ*∗(*x*)| = *k for all x* ∈ *H*. *Then, the SBG of G is a k-regular graph.*

**Theorem 7.** *Let H*<sup>1</sup> *be a hypergroup on SBG of G*<sup>1</sup> = &*H*1, *E*1'. *Let H*<sup>2</sup> *be a subhypergroup of H*<sup>1</sup> *on SBG of G*<sup>2</sup> = &*H*2, *E*2'. *Then, the SBG of G*<sup>2</sup> *is a sub-SBG of G*1.

**Proof.** Assume that *H*<sup>2</sup> is a subhypergroup of *H*1, then *H*<sup>2</sup> ⊂ *H*1. Therefore, the vertices SBG of *G*<sup>2</sup> is contained in the vertices SBG of *G*<sup>1</sup> and the edges *G*<sup>2</sup> is included in the edges of *G*1. Then, the SBG of *G*<sup>2</sup> is a sub-SBG of *G*1.

**Theorem 8.** *Let H be a hypergroup. The SBG of G* = &*H*, *E*' *is Eulerian if and only if* |*γ*∗(*z*)| = 2*k for all z* ∈ *H*, *k* ∈ N.

**Proof.** Let *H* be a hypergroup. Then, the relation *γ* is transitive [9]. By applying Theorem 4, the SBG of *G* is a connected graph. Additionally, with Proposition 4, *d*(*z*) = |*γ*∗(*z*)|, for all *z* ∈ *H* and by Theorem 3, the proof is completed.

**Example 2.** *Let* (*H*, ◦) *be a hypergroup in [34] (Example 28 (3)).*

*The corresponding SBG of G is shown in Figure 3, which is a connected and complete graph. Moreover, H*/*γ*<sup>∗</sup> = {{*a*, *b*, *c*}} *and* |*H*/*γ*∗| = 1. *Additionally,* |*γ*∗(*a*)| = |*γ*∗(*b*)| = |*γ*∗(*c*)| = 2, *that means d*(*a*) = *d*(*b*) = *d*(*c*) = 2. *Furthermore, diam*(*G*) = 1 *and the SBG of G is a* 2*-regular and Eulerian graph.*

**Figure 3.** SBG of *G*.

**Definition 6.** *The SBG of G is isomorphic to the SBG of G* , *if there exists a bijection φ from the set vertices of G to the set vertices of G* , *such that xγGy* ⇐⇒ *<sup>φ</sup>*(*x*)*γGφ*(*y*), *written by G* <sup>∼</sup><sup>=</sup> *<sup>G</sup>* .

**Theorem 9.** *Let* (*H*1, ◦1) *and* (*H*2, ◦2) *be two isomorphic hypergroups and let G*<sup>1</sup> *and G*<sup>2</sup> *be two SBGs associated with H*<sup>1</sup> *and H*2, *respectively. Then, the SBG of G*<sup>1</sup> *and the SBG of G*<sup>2</sup> *are isomorphisms.*

**Proof.** Assume *H*<sup>1</sup> and *H*<sup>2</sup> are isomorphisms. Then, |*H*1| = |*H*2| and we have |*G*1| = |*G*2|. Furthermore, if vertex *x* is connected to vertex *y*, then *xγ*∗*y* and we have ∃(*a*1, ... , *an*) ∈ *Hn*, ∃*σ* ∈ *Sn*; *x* ∈ ∏*<sup>n</sup> <sup>i</sup>*=<sup>1</sup> *ai*, *<sup>y</sup>* ∈ <sup>∏</sup>*<sup>n</sup> <sup>i</sup>*=<sup>1</sup> *aσ*(*i*). Let *φ* : *H*<sup>1</sup> −→ *H*<sup>2</sup> be an isomorphism and let *φ*(*x*) = *x* , *φ*(*y*) = *y* and *φ*(*ai*) = *a i* . Furthermore, *φ*(∏*<sup>n</sup> <sup>i</sup>*=<sup>1</sup> *ai*) = <sup>∏</sup>*<sup>n</sup> <sup>i</sup>*=<sup>1</sup> *<sup>φ</sup>*(*ai*) = <sup>∏</sup>*<sup>n</sup> <sup>i</sup>*=<sup>1</sup> *<sup>a</sup> i* , which yields *x* that is connected to *y* . Hence, *<sup>G</sup>*<sup>1</sup> ∼= *<sup>G</sup>*2.

**Example 3.** *To show that the reverse of Theorem 9 is not satisfied, consider two hypergroups* (*H*, ◦) *and* (*H* , ◦ ) *in [34] (Example 16 (3)).*

*Let f* : *H* −→ *H with f*(*a*) = 1, *f*(*b*) = 1, *f*(*c*) = 2. *Since f*(*a* ◦ *b*) = *f*(*H*) = {1, 2} *and f*(*a*) ◦ *f*(*b*) = 1 ◦ 1 = 1, *means that f is not an isomorphism* (*i*.*e*., *f*(*a* ◦ *b*) = *f*(*a*) ◦ *f*(*b*)). *The two SBGs are isomorphisms, as depicted in Figure 4.*

**Figure 4.** SBGs of *G* and *G* associated with *H* and *H* .

**Definition 7.** *Let G* = &*H*, *E*' *and G* = &*H* , *E* ' *be two SBGs, where H and H are two hypergroups and E* = {*E*1, ... , *Em*} *and E* = {*E* <sup>1</sup>, ... , *<sup>E</sup> <sup>n</sup>*}. *Define the Cartesian product G G with the vertices set H* × *H and edges set El* × *E <sup>k</sup> for* 1 ≤ *l* ≤ *m*, 1 ≤ *k* ≤ *n*.

**Example 4.** *Consider two SBGs in Example 3. By considering G* = &*H*, *E*' *and G* = &*H* , *E* ', *the Cartesian product of two SBGs G and G is depicted in Figure 5. The vertices of G G are H* × *H* = {(*a*, 0),(*a*, 1),(*a*, 2),(*b*, 0),(*b*, 1),(*b*, 2),(*c*, 0),(*c*, 1),(*c*, 2)} *and the corresponding edges are E* × *E* = {[(*a*, 0),(*b*, 0)], [(*a*, 0),(*a*, 1)],..., [(*a*, 2),(*c*, 2)]}.

**Figure 5.** Cartesian product of SBGs *G* and *G* .

**Proposition 5.** *Let G* = &*H*, *E*' *and G* = &*H* , *E* ' *be two SBGs and let* (*a*1, *b*1),(*a*2, *b*2) ∈ *H* × *H* . *Then,*

$$(a\_1, b\_1) \gamma\_{G \times G'}(a\_2, b\_2) \iff a\_1 \gamma\_G a\_2. \ b\_1 \gamma\_{G'} b\_2.$$

## *Geometric Concept of SBG*

A *geometric space* is a couple (*S*, *V*) where *S* is a nonvoid set and *V* is the family of a nonvoid subset of *S*. The elements of *S* are considered points and the elements of *V* are represented as blocks. If *V* covers *S*, then a *polygonal* of (*S*, *V*) is an *n*-tuple of blocks (*V*1, *V*2, ... , *Vn*) so that *Vi* ∩ *Vi*+<sup>1</sup> = ∅, for every *i* ∈ {1, 2, ... , *n* − 1}. Introduce the relation ≈ on *S* as follows:

$$\forall x \approx y \iff \exists (V\_1, V\_2, \dots, V\_n); \ x \in V\_1, \ y \in V\_n.$$

If *V* covers *S*, then the relation is an equivalence relation. The equivalence class [*x*] is determined as a *connected component* of *x* in *S* [10,11].

According to the SBG of *G*, we consider a pair &*H*, *E*' as a *geometric space of SBG*, where *H* is a semihypergroup (set of vertices) and *E* is the set of relations *γ<sup>n</sup>* (set of edges) for *n* ∈ N on *H*. For every *x*, *y* ∈ *H*, we have *xEiy* ⇐⇒ *xγny* with the given relation *γ<sup>n</sup>* as follows:

$$\text{x}\gamma\_n y \Longleftrightarrow \exists (a\_1, \dots, a\_n) \in H^n, \exists \sigma \in S\_n : \ x \in \prod\_{i=1}^n a\_{i\prime} \ y \in \prod\_{i=1}^n a\_{\sigma(i)}$$

Take a polygonal SBG of *G* = &*H*, *E*' as (*E*1, *E*2, ... , *En*), so that *Ei* ∩ *Ei*+<sup>1</sup> = ∅ (i.e., (*x*, *x* ) ∈ *Ei*,(*x* , *x*) ∈ *Ei*+1) for 1 ≤ *i* ≤ *n* − 1. By applying the polygonal concept of SBG, the relation ≈ is defined as follows:

$$x \approx y \iff \exists E\_i, 1 \le i \le n; \ (x, z) \in E\_1, \ (z, y) \in E\_n$$

The relation ≈ is an equivalence relation. The SBG of *G* is connected and the equivalence class [*x*] = {*y* |*xγ*∗*y*} =|*γ*∗(*x*) |, where [*x*] is a connected component by Theorem 4. Indeed, the connected components SBG of *G* = &*H*, *E*' are equivalence classes modulo *γ*∗. The geometric space *G* = &*H*, *E*' is connected if it includes only one connected component, i.e., *H* = [*x*], for *x* ∈ *H*. Clearly, the relation ≈ is the transitive closure of the relation *γ* = *n*∈*N γn*. The blocks of the geometric space SBG of *G* = &*H*, *E*' using relation *γ<sup>n</sup>* are the constructed sets with permuting finite hyperproducts of distinct finite points (vertices).
