**Theorem 6.** *Let* A∈O(*G*)*.*

(1) A ∈ *OG*1(*G*) ⇐⇒ **A** ∈ *IVG*(*G*), *A* ∈ *IFG*(*G*), *λ* ∈ *FG*(*G*). (2) A ∈ *OG*2(*G*) ⇐⇒ **A** ∈ *IVG*(*G*), *A* ∈ *IFG*(*G*) *and λ satisfies the following conditions: for every x*, *y* ∈ *G,*

$$
\lambda(\mathfrak{x}y) \le \lambda(\mathfrak{x}) \lor \lambda(y), \; \lambda(\mathfrak{x}^{-1}) \le \lambda(\mathfrak{x}).
$$

(3) A ∈ *OG*3(*G*) ⇐⇒ **A** ∈ *IVG*(*G*), *λ* ∈ *FG*(*G*) *and A satisfies the following conditions: for every x*, *y* ∈ *G,*

(i) *A*∈(*xy*) ≤ *A*∈(*x*) ∨ *A*∈(*y*), *A*∈(*xy*) ≥ *A*∈(*x*) ∧ *A*∈(*y*), (ii) *<sup>A</sup>*∈(*x*−1) <sup>≤</sup> *<sup>A</sup>*∈(*x*), *<sup>A</sup>*∈(*x*−1) <sup>≥</sup> *<sup>A</sup>*∈(*x*). (4) A ∈ *OG*4(*G*) ⇐⇒ **A** ∈ *IVG*(*G*)*, A and λ satisfies the following conditions: for every x*, *y* ∈ *G,* (i) *<sup>λ</sup>*(*xy*) <sup>≤</sup> *<sup>λ</sup>*(*x*) <sup>∨</sup> *<sup>λ</sup>*(*y*), *<sup>λ</sup>*(*x*−1) <sup>≤</sup> *<sup>λ</sup>*(*x*), (ii) *A*∈(*xy*) ≤ *A*∈(*x*) ∨ *A*∈(*y*), *A*∈(*xy*) ≥ *A*∈(*x*) ∧ *A*∈(*y*),

(iii) *<sup>A</sup>*∈(*x*−1) <sup>≤</sup> *<sup>A</sup>*∈(*x*), *<sup>A</sup>*∈(*x*−1) <sup>≥</sup> *<sup>A</sup>*∈(*x*).

**Example 5.** *(1) Consider the additive group* (Z, +)*. We define five mappings* **<sup>A</sup>** = [*A*−, *<sup>A</sup>*+] : <sup>Z</sup> <sup>→</sup> [*I*]*, <sup>A</sup>* = (*A*∈, *<sup>A</sup>*∈), *<sup>B</sup>* = (*B*∈, *<sup>B</sup>*∈) : <sup>Z</sup> <sup>→</sup> *<sup>I</sup>* <sup>⊕</sup> *I and <sup>λ</sup>*, *<sup>μ</sup>* : <sup>Z</sup> <sup>→</sup> *I, respectively as follows: for each* <sup>0</sup> <sup>=</sup> *<sup>n</sup>* <sup>∈</sup> <sup>Z</sup>*,*

$$\mathbf{A}(0) = [1, 1], \ A(0) = (1, 0), \ B(0) = (0, 1), \ \lambda(0) = 1, \ \mu(0) = \frac{1}{6},\tag{1}$$

$$\mathbf{A}(n) = \begin{cases} \begin{bmatrix} \frac{1}{2}\sqrt{\frac{2}{3}} \end{bmatrix} & \text{if } n \text{ is odd} \\\\ \begin{bmatrix} \frac{1}{3}\sqrt{\frac{4}{5}} \end{bmatrix} & \text{if } n \text{ is even} \end{cases} \tag{2}$$

$$A(n) = \begin{cases} \begin{pmatrix} \frac{1}{2}, \frac{1}{3} \end{pmatrix} & \text{if } n \text{ is odd} \\\\ \begin{pmatrix} \frac{2}{3}, \frac{1}{5} \end{pmatrix} & \text{if } n \text{ is even,} \end{cases} \tag{3}$$

$$B(n) = \begin{cases} \begin{pmatrix} \frac{2}{5}, \frac{1}{5} \end{pmatrix} & \text{if } n \text{ is odd} \\\\ \begin{pmatrix} \frac{1}{2}, \frac{1}{5} \end{pmatrix} & \text{if } n \text{ is even} \end{cases} \tag{4}$$

$$\lambda(n) = \begin{cases} \frac{3}{2} & \text{if } n \text{ is odd} \\\\ \frac{3}{5} & \text{if } n \text{ is even,} \end{cases} \tag{5}$$

$$\mu(n) = \begin{cases} \frac{3}{5} & \text{if } n \text{ is odd} \\\\ \frac{1}{2} & \text{if } n \text{ is even.} \end{cases} \tag{6}$$

*Then we can easily check that <sup>λ</sup>* <sup>∈</sup> *FG*(Z)*. Moreover,* **<sup>A</sup>** <sup>∈</sup> *IVG*(Z) *and A* <sup>∈</sup> *IFG*(Z) *from Example 4.1 in [18] and Example 2.1 in [20]. Thus,* **A**, *<sup>A</sup>*, *<sup>λ</sup> is an octahedron subgroup of* <sup>Z</sup>*.*

*On the other hand, we can easily check that μ and B satisfy the conditions of Theorem 6 (2) and (i) and (ii) of Theorem 6 (3), respectively. Hence by Theorem 6 (2) and (3),* **A**, *A*, *μ* ∈ *OG*2(*X*) *and* **A**, *B*, *λ* ∈ *OG*3(*X*)*. Thus, by Theorem 6 (2),* A = **A**, *A*, *μ* ∈ *OG*2(*X*)*. Furthermore, from Theorem 6 (4), we can easily see that* **A**, *B*, *μ* ∈ *OG*4(*X*)*.*

*(2) If* **A** ∈ *IVGP*(*G*)*, then* O**<sup>A</sup>** ∈ *OG*1(*G*)*. Also if A* ∈ *IFGP*(*G*)*, then* O*<sup>A</sup>* ∈ *OG*1(*G*)*. (3) If* A ∈ *OGi*(*X*)*, then clearly,* [ ]A, A∈ *OGi*(*X*) *(i* = 1, 2, 3, 4*).*

**Remark 8.** *(1) If λ* ∈ *FG*(*G*)*, then we have*

$$
\langle [\lambda, \lambda], (\lambda, \lambda^c), \lambda \rangle \in OG\_1(G), \ \langle [\lambda, \lambda], (\lambda, \lambda^c), \lambda^c \rangle \in OG\_2(G),
$$

$$
\left\langle [\lambda, \lambda], (\lambda^C, \lambda), \lambda \right\rangle \in OG\_3(G), \ \left\langle [\lambda, \lambda], (\lambda^C, \lambda), \lambda^c \right\rangle \in OG\_4(G).
$$

*(2) If* **A** ∈ *IVG*(*G*)*, then we have*


*(3) If A* ∈ *IFG*(*G*)*, then we have*

$$
\left\langle [A^{\in}, (A^{\notin})^{c}], A, A^{\in} \right\rangle \in \mathrm{OG}\_{1}(\mathrm{G}), \\
\left\langle [A^{\in}, (A^{\notin})^{c}], A, A^{\notin} \right\rangle \in \mathrm{OG}\_{2}(\mathrm{G}).
$$

$$
\left\langle [A^{\in}, (A^{\notin})^{c}], A^{\in}, A^{\in} \right\rangle \in \mathrm{OG}\_{3}(\mathrm{G}), \\
\left\langle [A^{\in}, (A^{\notin})^{c}], A^{\in}, A^{\notin} \right\rangle \in \mathrm{OG}\_{4}(\mathrm{G}).
$$

The following is an immediate result of Theorem 6 (1) and Remark 3.

**Proposition 15.** *For every H* ⊂ *G, H is a subgroup of G if and only if <sup>χ</sup>*<sup>H</sup> ∈ *OG*1(*G*).

The following is an immediate result of Theorem 6 (1) and Proposition 8.

**Proposition 16.** *If* (A*j*)*j*∈*<sup>J</sup>* = (- **A***j*, *Aj*, *λ<sup>j</sup>* . )*j*∈*<sup>J</sup>* <sup>⊂</sup> *OG*1(*G*)*, then* /<sup>1</sup> *<sup>j</sup>*∈*<sup>J</sup>* <sup>A</sup>*<sup>j</sup>* <sup>∈</sup> *OG*1(*G*)*, where <sup>J</sup> denotes an index set.*

The following is an immediate result of Proposition 16 and Corollary 1.

**Corollary 2.** *Let* A∈O(*G*) *and let*

$$(\mathcal{A}) = \bigcap^1 \{ \mathcal{B} \in \mathrm{OG}\_1(G) : \mathcal{A} \subset\_1 \mathcal{B} \}.$$

*Then* (A) ∈ *OG*1(*G*).

In this case, (A) is called the octahedron subgroup of *G* generated by A.

The following is an immediate result of Proposition 9 and Corollary 2.

**Corollary 3.** *For each <sup>A</sup>* <sup>∈</sup> <sup>2</sup>*X, let* (*A*) *be the subgroup generated by <sup>A</sup> and let χ*(*A*) = + [*χ*(*A*), *χ*(*A*)],(*χ*(*A*), *χ*(*A<sup>c</sup>* )), *χ*(*A*) , *. Then*

$$(\chi\_A) = \chi\_{(A)}.$$

**Proposition 17.** *Let* A ∈ *OGi*(*G*) *(i* = 1, 2, 3, 4*). Then for each x* ∈ *G*,

$$\mathcal{A}(\mathfrak{x}^{-1}) = \mathcal{A}(\mathfrak{x}), \; \mathcal{A}(\mathfrak{e}) \ge\_{\mathfrak{i}} \mathcal{A}(\mathfrak{x}).$$

**Proof.** Case 1: Suppose A ∈ *OG*1(*G*). Then by Theorem 6 (1), we have

$$\mathbf{A} \in IVG(\mathbf{G}), \ A \in IFG(\mathbf{G}), \ \lambda \in FG(\mathbf{G}).$$

Thus, by Propositions 3.1 in [21], 2.6 in [20] and 5.4 in [19], we have

$$\mathbf{A}(\mathbf{x}^{-1}) = \mathbf{A}(\mathbf{x}) \; \; \; A(\mathbf{x}^{-1}) = A(\mathbf{x}) \; \; \; \lambda(\mathbf{x}^{-1}) = \lambda(\mathbf{x}),$$

$$\mathbf{A}(\boldsymbol{\varepsilon}) \ge \mathbf{A}(\mathbf{x}) \; \; \; \; A(\boldsymbol{\varepsilon}) \ge A(\mathbf{x}) \; \; \; \; \lambda(\boldsymbol{\varepsilon}) \ge \lambda(\mathbf{x}).$$

So <sup>A</sup>(*x*−1) = <sup>A</sup>(*x*) and <sup>A</sup>(*e*) <sup>≥</sup><sup>1</sup> <sup>A</sup>(*x*). Case 2: Suppose A ∈ *OG*2(*G*). Then by Theorem 6 (2), we have

$$\mathbf{A} \in IVG(\mathbf{G}) , \ A \in IFG(\mathbf{G})$$

and

$$
\lambda(xy) \le \lambda(x) \lor \lambda(y), \\
\lambda(x^{-1}) \le \lambda(x), \text{ for every } x, \ y \in G.
$$

Thus, *<sup>λ</sup>*(*x*) = *<sup>λ</sup>*((*x*−1)−1) <sup>≤</sup> *<sup>λ</sup>*(*x*−1) <sup>≤</sup> *<sup>λ</sup>*(*x*), i.e., *<sup>λ</sup>*(*x*−1) = *<sup>λ</sup>*(*x*). On the other hand, *<sup>λ</sup>*(*e*) = *<sup>λ</sup>*(*xx*−1) <sup>≤</sup> *<sup>λ</sup>*(*x*) <sup>∨</sup> *<sup>λ</sup>*(*x*−1) = *<sup>λ</sup>*(*x*). By Case 1, we have

**<sup>A</sup>**(*x*−1) = **<sup>A</sup>**(*x*), *<sup>A</sup>*(*x*−1) = *<sup>A</sup>*(*x*) and **<sup>A</sup>**(*e*) <sup>≥</sup> **<sup>A</sup>**(*x*), *<sup>A</sup>*(*e*) <sup>≥</sup> *<sup>A</sup>*(*x*). So <sup>A</sup>(*x*−1) = <sup>A</sup>(*x*) and <sup>A</sup>(*e*) <sup>≥</sup><sup>2</sup> <sup>A</sup>(*x*).

Case 3: Suppose A ∈ *OG*2(*G*). Then by Case 1 and Theorem 6 (3), we have

$$\mathbf{A}(\mathbf{x}^{-1}) = \mathbf{A}(\mathbf{x}), \; \lambda(\mathbf{x}^{-1}) = \lambda(\mathbf{x}), \; \mathbf{A}(\boldsymbol{\varepsilon}) \ge \mathbf{A}(\mathbf{x}), \; \lambda(\boldsymbol{\varepsilon}) \ge \lambda(\mathbf{x})$$

and *A* satisfies the conditions (i) and (ii). By (ii),

$$A^{\in}(\mathbf{x}) = A^{\in}((\mathbf{x}^{-1})^{-1}) \le A^{\in}(\mathbf{x}^{-1}) \le A^{\in}(\mathbf{x}), \text{ i.e., } A^{\in}(\mathbf{x}) = A^{\in}(\mathbf{x}^{-1}).$$

Similarly, we have *A*∈(*x*) = *A*∈(*x*−1). Thus, *A*(*x*) = *A*(*x*−1). By (i), we can easily prove that *<sup>A</sup>*(*e*) <sup>≤</sup> *<sup>A</sup>*(*x*).. So <sup>A</sup>(*x*−1) = <sup>A</sup>(*x*) and <sup>A</sup>(*e*) <sup>≥</sup><sup>3</sup> <sup>A</sup>(*x*).

Case 4: Suppose A ∈ *OG*2(*G*). Then by Cases 1 and 2, we have <sup>A</sup>(*x*−1) = <sup>A</sup>(*x*) and <sup>A</sup>(*e*) <sup>≥</sup><sup>4</sup> A(*x*). This completes the proof.

**Theorem 7.** *Let* A∈O(*G*) *and let <sup>i</sup>* <sup>=</sup> 1, 2, 3, 4*. Then* A ∈ *OGi*(*G*) *if and only if* <sup>A</sup>(*xy*−1) <sup>≥</sup>*<sup>i</sup>* A(*x*) ∧*<sup>i</sup>* A(*x*)(*y*), for every *x*, *y* ∈ *G*.

**Proof.** We prove only the necessity of the condition.

Case 1: Suppose A ∈ *OG*1(*G*) and let *x*, *y* ∈ *G*. Then by Theorem 6 (1), we have

$$\mathbf{A} \in IVG(\mathbf{G}), \ A \in IFG(\mathbf{G}), \ \lambda \in FG(\mathbf{G}).$$

Thus, by Propositions 3.2 in [21] and 5.6 in [19], we have

$$\mathbf{A}(xy^{-1}) = \mathbf{A}(x) \wedge \mathbf{A}(y), \; \lambda(xy^{-1}) = \lambda(x) \wedge \lambda(y).$$

On the other hand, by Proposition 17,

$$A^{\in}(xy^{-1}) \ge A^{\in}(x) \land A^{\in}(y^{-1}) = A^{\in}(x) \land A^{\in}(y),$$

$$A^{\notin}(xy^{-1}) \le A^{\notin}(x) \lor A^{\notin}(y^{-1}) = A^{\notin}(x) \lor A^{\notin}(y).$$

So <sup>A</sup>(*xy*−1) <sup>≥</sup><sup>1</sup> <sup>A</sup>(*x*) <sup>∧</sup><sup>1</sup> <sup>A</sup>(*y*).

Case 2: Suppose A ∈ *OG*2(*G*) and let *x*, *y* ∈ *G*. Then by Theorem 6 (2) and Case 1, we have

$$\mathbf{A}(xy^{-1}) = \mathbf{A}(x) \wedge \mathbf{A}(y), \ A(xy^{-1}) = A(x) \wedge A(y)$$

and

$$
\lambda(xy) \le \lambda(\mathfrak{x}) \lor \lambda(y), \; \lambda(\mathfrak{x}^{-1}) \le \lambda(\mathfrak{x}).
$$

Thus, by Proposition 17,

$$
\lambda(xy^{-1}) \le \lambda(x) \lor \lambda(y^{-1}) = \lambda(x) \lor \lambda(y).
$$

So <sup>A</sup>(*xy*−1) <sup>≥</sup><sup>2</sup> <sup>A</sup>(*x*) <sup>∧</sup><sup>2</sup> <sup>A</sup>(*y*). Case 3: Suppose A ∈ *OG*2(*G*) and let *x*, *y* ∈ *G*. Then by Case 1 and Theorem 6 (3), we have

$$\mathbf{A}(xy^{-1}) \ge \mathbf{A}(x) \land \mathbf{A}(x), \ \lambda(xy^{-1}) \ge \lambda(x) \land \lambda(y)$$

and *A* satisfies the conditions (i) and (ii). Thus, by (i) and Proposition 17,

$$A^{\in}(xy^{-1}) \le A^{\in}(x) \lor A^{\in}(y^{-1}) = A^{\in}(x) \lor A^{\in}(y),$$

$$A^{\notin}(xy^{-1}) \ge A^{\notin}(x) \land A^{\in}(y^{-1}) = A^{\in}(x) \lor A^{\in}(y).$$
 $(1)$ 

So <sup>A</sup>(*xy*−1) <sup>≥</sup><sup>3</sup> <sup>A</sup>(*x*) <sup>∧</sup><sup>3</sup> <sup>A</sup>(*y*).

Case 4: Suppose A ∈ *OG*2(*G*). Then by Cases 1 and 2, we can easily prove that <sup>A</sup>(*xy*−1) <sup>≥</sup><sup>4</sup> <sup>A</sup>(*x*) <sup>∧</sup><sup>4</sup> <sup>A</sup>(*y*). This completes the proof.

The following is an immediate consequence of Corollary in [19], Propositions 4.6 in [18] and 2.7 in [20].

**Proposition 18.** *If* A ∈ *OG*1(*G*). *Then G*<sup>A</sup> = {*<sup>x</sup>* ∈ *<sup>G</sup>* : A(*x*) = A(*e*)} *is a subgroup of G.*

**Proposition 19.** *Let* A ∈ *OG*1(*G*) *(i* <sup>=</sup> 1, 2, 3, 4*) and let <sup>x</sup>*, *<sup>y</sup>* <sup>∈</sup> *G. If* <sup>A</sup>(*xy*−1) = <sup>A</sup>(*e*), *then* <sup>A</sup>(*x*) = A(*y*).

**Proof.** (1) Case 1: Suppose A ∈ *OG*1(*G*) and let *x*, *y* ∈ *G*. Then by Theorem 6 (1), we have

**A** ∈ *IVG*(*G*), *A* ∈ *IFG*(*G*), *λ* ∈ *FG*(*G*).

Thus, by Propositions 4.7 in [18], 2.8 in [20] and 5.5 in [19], we have

$$\mathbf{A}(\mathbf{x}) = \mathbf{A}(y), \ A(\mathbf{x}) = A(y)\ \lambda(\mathbf{x}) = \lambda(y).$$

So A(*x*) = A(*y*).

Case 2: Suppose A ∈ *OG*2(*G*) and let *x*, *y* ∈ *G*. Then by Theorem 6 (2) and Case 1, we have

$$\mathbf{A}(\mathbf{x}) = \mathbf{A}(y), \ A(\mathbf{x}) = A(y).$$

and

*<sup>λ</sup>*(*xy*) <sup>≤</sup> *<sup>λ</sup>*(*x*) <sup>∨</sup> *<sup>λ</sup>*(*y*), *<sup>λ</sup>*(*x*−1) <sup>≤</sup> *<sup>λ</sup>*(*x*).

By Proposition 17,

*<sup>λ</sup>*(*x*) = *<sup>λ</sup>*((*xy*−1)*y*) <sup>≤</sup> *<sup>λ</sup>*(*xy*−1) <sup>∨</sup> *<sup>λ</sup>*(*y*) = *<sup>λ</sup>*(*e*) <sup>∨</sup> *<sup>λ</sup>*(*y*) = *<sup>λ</sup>*(*y*) <sup>=</sup> *<sup>λ</sup>*((*yx*−1)*x*) <sup>≤</sup> *<sup>λ</sup>*(*e*) <sup>∨</sup> *<sup>λ</sup>*(*x*) = *<sup>λ</sup>*(*x*). Thus, *λ*(*x*) = *λ*(*y*). So A(*x*) = A(*y*).

Case 3: Suppose A ∈ *OG*2(*G*) and let *x*, *y* ∈ *G*. Then by Case 1 and Theorem 6 (3), we have **A**(*x*) = **A**(*y*), *λ*(*x*) = *λ*(*y*) and *A* satisfies the conditions (i) and (ii). By (i) and Proposition 17,

*<sup>A</sup>*∈(*x*) = *<sup>A</sup>*∈((*xy*−1)*y*) <sup>≤</sup> *<sup>A</sup>*∈(*e*) <sup>∨</sup> *<sup>A</sup>*∈(*y*) = *<sup>A</sup>*∈(*y*) <sup>=</sup> *<sup>A</sup>*∈((*yx*−1)*x*) <sup>≤</sup> *<sup>A</sup>*∈(*e*) <sup>∨</sup> *<sup>A</sup>*∈(*x*) = *<sup>A</sup>*∈(*x*). Thus, *A*∈(*x*) = *A*∈(*y*). Similarly, we have *A*∈(*x*) = *A*∈(*y*).

$$A^\pounds(xy^{-1}) \ge A^\pounds(x) \land A^\in(y^{-1}) = A^\in(x) \lor A^\in(y).$$

So A(*x*) = A(*y*).

Case 4: Suppose A ∈ *OG*2(*G*). Then by Cases 1 and 2, we can easily prove that A(*x*) = A(*y*). This completes the proof.

The following is an immediate consequence of Corollary in [19], Corollaries 4.7-1 in [18] and 2.8-1 in [20].

**Corollary 4.** *Let* A ∈ *OG*1(*G*)*. If G*<sup>A</sup> *is a normal subgroup of G, then* A *is constant on each coset of G*A*.*

The following is an immediate result of Corollary in [19], Corollaries 4.7-2 in [18] and 2.8-2 in [20].

**Corollary 5.** *Let* A ∈ *OG*1(*G*) *and let <sup>G</sup>*<sup>A</sup> *be a normal subgroup of G. If <sup>G</sup>*<sup>A</sup> *has a finite index, then* A *has the sup property.*

The following is an immediate result of Propositions 5.7 in [19], 4.8 in [18] and 2.10 in [20].

**Proposition 20.** *A group G cannot be the* 1*-union of two proper* 1*-octahedron subgroups of G.*

**Theorem 8.** A ∈ *OLIi*(*G*) *[resp.,* A ∈ *ORIi*(*G*) *and* A ∈ *OIi*(*G*)*] if and only if* A *is a constant mapping (i* = 1, 2, 3, 4*).*

**Proof.** We prove only that A ∈ *OLIi*(*G*) if and only if A is a constant mapping. Suppose A is a constant mapping. Then we can easily show that A ∈ *OLIi*(*G*). Thus, it is sufficient to prove only that the necessary condition holds.

Case 1: Suppose A ∈ *OLI*1(*G*). Then by Propositions 4.14 in [18], 2.16 in [20] and 5.9 in [19], **A**, *A* and *λ* are constant mappings. Thus, A is a constant mapping.

Case 2: Suppose A ∈ *OLI*2(*G*). Since **A** and *A* are constant mappings by Case 1, it is enough to show that *λ* is a constant mapping. Let *x*, *y* ∈ *G*. Since A ∈ *OLI*2(*G*), by Remark 4 (1), we have *λ*(*xy*) ≤ *λ*(*x*) ∨ *λ*(*y*). Let *y* = *e*. Then by Proposition 17, we have

$$
\lambda(\mathfrak{x}) \le \lambda(\mathfrak{x}) \lor \lambda(\mathfrak{e}) = \lambda(\mathfrak{e}), \text{ for each } \mathfrak{x} \in G.
$$

Now let *x* = *y*−1. Then by Proposition 17, we have

$$
\lambda(e) \le \lambda(y^{-1}) \lor \lambda(y) = \lambda(y), \text{ for each } y \in G.
$$

Thus, *λ*(*x*) = *λ*(*y*) = *λ*(*e*). So *λ* is a constant mapping. Hence A is a constant mapping.

Case 3: Suppose A ∈ *OLI*3(*G*). Then by Remark 4 (1) and Proposition 17, we can easily see that *A* is a constant mapping. Thus, A is a constant mapping.

Case 4: Suppose A ∈ *OLI*4(*G*). Then by Cases (2) and (3), we can easily prove that A is a constant mapping. This completes the proof.

**Proposition 21.** *Let <sup>f</sup>* : *<sup>G</sup>* <sup>→</sup> *<sup>G</sup> be a group homomorphism, let* A ∈ *OGi*(*G*) *and let* B ∈ *OGi*(*<sup>G</sup>* ) *(i* = 1, 2, 3, 4*).*

(1) *If* <sup>A</sup> *has the i-sup-property, f*(A) <sup>∈</sup> *OGi*(*<sup>G</sup>* ). (2) *<sup>f</sup>* <sup>−</sup>1(B) <sup>∈</sup> *OGi*(*G*)*.*

**Proof.** (1) Since *<sup>f</sup>*(A) <sup>∈</sup> *OGPi*(*<sup>G</sup>* ) by Proposition <sup>12</sup> (1), It is sufficient to show that *<sup>f</sup>*(A)(*y*−1) <sup>≥</sup>*<sup>i</sup> f*(A)(*y*) for each *y* ∈ *f*(*G*).

Case 1: Suppose A ∈ *OG*1(*G*). Then by Propositions 4.11 in [18], 2.13 in [20] and 5.8 in [19], *<sup>f</sup>*(**A**) <sup>∈</sup> *IVG*(*<sup>G</sup>* ), *<sup>f</sup>*(*A*) <sup>∈</sup> *IFG*(*<sup>G</sup>* ), *<sup>f</sup>*(*λ*) <sup>∈</sup> *FG*(*<sup>G</sup>* ). Thus, by Theorem <sup>6</sup> (1), *<sup>f</sup>*(A) <sup>∈</sup> *OG*1(*<sup>G</sup>* ).

Case 2: Suppose A ∈ *OG*2(*G*) and let *<sup>y</sup>* <sup>∈</sup> *<sup>f</sup>*(*G*). Since *<sup>f</sup>*(**A**) <sup>∈</sup> *IVG*(*<sup>G</sup>* ) and *<sup>f</sup>*(*A*) <sup>∈</sup> *IFG*(*<sup>G</sup>* ) by Case 1, it is enough to prove that *<sup>f</sup>*(*λ*)(*y*−1) <sup>≤</sup> *<sup>f</sup>*(*λ*)(*y*). Since <sup>A</sup> has the 2-sup-property, there is *<sup>x</sup>*<sup>0</sup> <sup>∈</sup> *<sup>f</sup>* <sup>−</sup>1(*y*) such that

$$
\lambda(\mathfrak{x}\_0) = \bigwedge\_{t \in f^{-1}(y)} \lambda(t).
$$

Then *f*(*λ*)(*y*−1) = *<sup>t</sup>*∈*<sup>f</sup>* <sup>−</sup>1(*y*−1) *<sup>λ</sup>*(*t*) <sup>≤</sup> *<sup>λ</sup>*(*x*−<sup>1</sup> <sup>0</sup> ) ≤ *λ*(*x*0) = *f*(*λ*)(*y*). Thus, by Theorem 6 (2), *<sup>f</sup>*(A) <sup>∈</sup> *OG*2(*<sup>G</sup>* ).

Case 3: Suppose A ∈ *OG*3(*G*) and let *<sup>y</sup>* <sup>∈</sup> *<sup>f</sup>*(*G*). Since *<sup>f</sup>*(**A**) <sup>∈</sup> *IVG*(*<sup>G</sup>* ) and *<sup>f</sup>*(*λ*) <sup>∈</sup> *FG*(*<sup>G</sup>* ) by Case 1, it is sufficient to show that [ *<sup>f</sup>*(*A*)]∈(*y*−1) <sup>≤</sup> [ *<sup>f</sup>*(*A*)]∈(*y*) and [ *<sup>f</sup>*(*A*)]∈/(*y*−1) <sup>≥</sup> [ *<sup>f</sup>*(*A*)]∈(*y*). Since <sup>A</sup> has the 3-sup-property, there is *<sup>x</sup>*<sup>0</sup> <sup>∈</sup> *<sup>f</sup>* <sup>−</sup>1(*y*) such that

$$A^{\in}(\mathbf{x}\_0) = \bigwedge\_{t \in f^{-1}(y)} A^{\in}(t), \ A^{\notin}(\mathbf{x}\_0) = \bigvee\_{t \in f^{-1}(y)} A^{\notin}(t).$$

Then [ *f*(*A*)]∈(*y*−1) = *<sup>t</sup>*∈*<sup>f</sup>* <sup>−</sup>1(*y*−1) *<sup>A</sup>*∈(*t*) <sup>≤</sup> *<sup>A</sup>*∈(*x*−<sup>1</sup> <sup>0</sup> ) ≤ *A*∈(*x*0)=[ *f*(*A*)]∈(*y*). Similarly, we have [ *<sup>f</sup>*(*A*)]∈(*y*−1) <sup>≥</sup> [ *<sup>f</sup>*(*A*)]∈(*y*−1). Thus, by Theorem <sup>6</sup> (3), *<sup>f</sup>*(A) <sup>∈</sup> *OG*3(*<sup>G</sup>* ).

Case 4: Suppose A ∈ *OG*4(*G*). Then from Cases (2), (3) and Theorem 6 (4), we can easily prove that *<sup>f</sup>*(A) <sup>∈</sup> *OG*4(*<sup>G</sup>* ).

(2) Case 1: Suppose B ∈ *OG*1(*<sup>G</sup>* ). Then Propositions 4.11 in [18], 2.13 in [20] and 5.8 in [19], *<sup>f</sup>* <sup>−</sup>1(**B**) <sup>∈</sup> *IVG*(*G*), *<sup>f</sup>* <sup>−</sup>1(*B*) <sup>∈</sup> *IFG*(*G*), *<sup>f</sup>* <sup>−</sup>1(*μ*) <sup>∈</sup> *FG*(*G*). Thus, *<sup>f</sup>* <sup>−</sup>1(B) <sup>∈</sup> *OG*1(*G*).

Case 2: Suppose B ∈ *OG*2(*<sup>G</sup>* ) and let *<sup>x</sup>* <sup>∈</sup> *<sup>G</sup>*. Since *<sup>f</sup>* <sup>−</sup>1(**B**) <sup>∈</sup> *IVG*(*G*), *<sup>f</sup>* <sup>−</sup>1(*B*) <sup>∈</sup> *IFG*(*G*) by Case 1, it is sufficient to prove that *<sup>f</sup>* <sup>−</sup>1(*μ*)(*x*−1) <sup>≤</sup> *<sup>f</sup>* <sup>−</sup>1(*μ*)(*x*). Then

$$f^{-1}(\mu)(\mathbf{x}^{-1}) = \mu(f(\mathbf{x}^{-1})) = \mu(f(\mathbf{x})^{-1}) \le \mu(f(\mathbf{x})) = f^{-1}(\mu)(\mathbf{x}).$$

Thus, by Theorem <sup>6</sup> (2), *<sup>f</sup>*(A) <sup>∈</sup> *OG*2(*<sup>G</sup>* ).

Case 3: Suppose B ∈ *OG*3(*<sup>G</sup>* ) and let *<sup>x</sup>* <sup>∈</sup> *<sup>G</sup>*. Since *<sup>f</sup>* <sup>−</sup>1(**B**) <sup>∈</sup> *IVG*(*G*), *<sup>f</sup>* <sup>−</sup>1(*μ*) <sup>∈</sup> *FG*(*G*) by Case 1, it is enough to show that [ *<sup>f</sup>* <sup>−</sup>1(*A*)]∈(*x*−1) <sup>≤</sup> [ *<sup>f</sup>* <sup>−</sup>1(*A*)]∈(*x*) and [ *<sup>f</sup>* <sup>−</sup>1(*A*)]∈(*x*−1) <sup>≥</sup> [ *<sup>f</sup>* <sup>−</sup>1(*A*)]∈(*x*). Then

$$[f^{-1}(A)]^\in(\mathbf{x}^{-1}) = A^\in(f(\mathbf{x}^{-1})) = A^\in(f(\mathbf{x})^{-1}) \le A^\in(f(\mathbf{x})) = [f^{-1}(A)]^\in(\mathbf{x}).$$

Similarly, we have [ *<sup>f</sup>* <sup>−</sup>1(*A*)]∈(*x*−1) <sup>≥</sup> [ *<sup>f</sup>* <sup>−</sup>1(*A*)]∈(*x*). Thus, by Theorem <sup>6</sup> (3), *<sup>f</sup>*(A) <sup>∈</sup> *OG*3(*<sup>G</sup>* ).

Case 4: Suppose B ∈ *OG*3(*<sup>G</sup>* ). Then from Cases (2), (3) and Theorem 6 (4), we can easily prove that *<sup>f</sup>*(A) <sup>∈</sup> *OG*4(*<sup>G</sup>* ). This completes the proof.

From Propositions 4.16 and 4.17 in [18], 2.18 and 2.19 in [20], and Theorems 2.1 and 2.2 in [22], we have the following result.

**Theorem 9.** *If* A ∈ *OG*1(*G*)*, then* [A]*<sup>a</sup>*¯ *is a subgroup of G, for each <sup>a</sup>*¯ <sup>∈</sup> [*I*] <sup>×</sup> (*<sup>I</sup>* <sup>⊕</sup> *<sup>I</sup>*) <sup>×</sup> *<sup>I</sup> such that <sup>a</sup>*¯ <sup>≤</sup><sup>1</sup> <sup>A</sup>(*e*)*. Conversely, if* A∈O(*G*) *such that* [A]*<sup>a</sup>*¯ *is a subgroup of G, for each <sup>a</sup>*¯ <sup>∈</sup> [*I*] <sup>×</sup> (*<sup>I</sup>* <sup>⊕</sup> *<sup>I</sup>*) <sup>×</sup> *<sup>I</sup> such that <sup>a</sup>*¯ <sup>≤</sup><sup>1</sup> <sup>A</sup>(*e*)*, then* A ∈ *OG*1(*G*)*.*

**Theorem 10.** *Let Gp be the cyclic group of prime order p. Then* A ∈ *OGi*(*Gp*) *if and only if* A(*x*) = A(1) ≤*<sup>i</sup>* A(0), *for i* = 1, 2, 3, 4*.*

**Proof.** We prove only for *i* = 1, 2 and the proofs are omitted for *i* = 3, 4.

For *i* = 1, from Propositions 4.12 in [18], 2.14 in [20] and 5.10 in [19], we have for each 0 = *x* ∈ *Gp*,

$$\mathbf{A} \in IVG(\mathbb{G}\_p) \text{ iff } \mathbf{A}^-(\mathbf{x}) = \mathbf{A}^-(1) \le \mathbf{A}^-(0) \text{ and } \mathbf{A}^+(\mathbf{x}) = \mathbf{A}^+(1) \le \mathbf{A}^+(0), \tag{7}$$

$$A \in IFG(G\_p) \text{ iff } A^{\in}(\mathbf{x}) = A^{\in}(1) \le A^{\in}(0) \\ \text{and } A^{\notin}(\mathbf{x}) = A^{\notin}(1) \ge A^{\notin}(0), \tag{8}$$

$$
\lambda \in FG(G\_{\mathcal{P}}) \text{ iff } \lambda(\mathbf{x}) = \lambda(1) \le \lambda(0). \tag{9}
$$

Thus, by Theorem 6 (1), A ∈ *OG*1(*Gp*) iff A(*x*) = A(1) ≤*<sup>i</sup>* A(0).

For *i* = 2, suppose A ∈ *OG*2(*Gp*) and let *y* ∈ *Gp*. Then by Theorem 6 (2), *λ*(*xy*) ≤ *λ*(*x*) ∨ *λ*(*y*). Since *Gp* is the cyclic group of prime order *p*, *Gp* = {0, 1, 2, ..., *p* − 1}. Since *x* is the sum of *i s* and *i* is the sum of *x s*, *λ*(*x*) ≤ *λ*(1) ≤ *λ*(*x*). Thus, *λ*(*x*) = *λ*(1). Since 0 is the identity of *Gp*, *λ*(0) ≤ *λ*(*x*). Thus, *λ*(*x*) = *λ*(1) ≥ *λ*(0). So A(*x*) = A(1) ≤<sup>2</sup> A(0).

Conversely, suppose A(*x*) = A(1) ≤<sup>2</sup> A(0). Then by Theorem 7, A ∈ *OG*2(*Gp*). This completes the proof.

**Definition 23.** *Let* A∈O(*G*) *and let i* = 1, 2, 3, 4*. Then* A *is called an i-octahedron normal subgroup (briefly, an i-ONG) of G, if it satisfies the following conditions:*

$$\mathcal{A} \in OG\_i(G) \text{ and } \mathcal{A}(\mathfrak{x}y) = \mathcal{A}(y\mathfrak{x}), \text{ for every } \mathfrak{x}, \text{ } y \in G.$$

*We will denote the set of all i-ONGs of G as ONGi*(*G*)*. It is obvious that if G is abelian, then* A ∈ *ONGi*(*G*)*, for each* A ∈ *OGi*(*G*)*. Furthermore,*

$$\mathcal{A} \in OG\_1(G) \iff \mathbf{A} \in IVNG(G), \; A \in IFNG(G), \; \lambda \in FNG(G), \; \lambda$$

*where IVNG*(*G*) *[resp., IFNG*(*G*) *and FNG*(*G*)*] is denoted by the set of all interval-valued fuzzy normal subgroups [resp., intuitionistic fuzzy normal subgroups and fuzzy normal subgroups or fuzzy invariant subgroups] of G (See [18] [resp., See [16,20]]).*

**Example 6.** *Let GL*(*n*, *R*) *be the general linear group of degree n and let In be the unit matrix of GL*(*n*, *R*)*. Then clearly, GL*(*n*, *R*) *is a non abelian group. We define the interval-valued fuzzy set* **A***, two intuitionistic fuzzy sets A*, *B and two fuzzy sets λ*, *μ in GL*(*n*, *R*) *as follows: for each In* = *M* ∈ *GL*(*n*, *R*)*,*

$$\mathbf{A}(I\_{\mathbb{H}}) = [1, 1], \; A(I\_{\mathbb{H}}) = (1, 0), \; B(I\_{\mathbb{H}}) = (0, 1), \; \lambda) = 1, \; \mu(I\_{\mathbb{H}}) = 0,$$

**A**(*M*) = ⎧ ⎪⎪⎨ ⎪⎪⎩ 1 5 , 2 3 *if M* is not a triangular matrix 1 3 , 1 2 *if M* is a triangular matrix, *A*(*M*) = ⎧ ⎪⎪⎨ ⎪⎪⎩ 2 3 , 1 5 *if M* is not a triangular matrix 1 2 , 1 3 *if M* is a triangular matrix, *A*(*M*) = ⎧ ⎪⎪⎨ ⎪⎪⎩ 1 5 , 2 3 *if M* is not a triangular matrix 1 3 , 1 2 *if M* is a triangular matrix, *λ*(*M*) = ⎧ ⎪⎨ ⎪⎩ 2 <sup>3</sup> *if M* is not a triangular matrix 1 <sup>3</sup> *if M* is a triangular matrix, *μ*(*M*) = ⎧ ⎪⎨ ⎪⎩ 1 <sup>3</sup> *if M* is not a triangular matrix 2 <sup>3</sup> *if M* is a triangular matrix.

*Then we can easily check that that the followings hold:*

$$
\langle \mathbf{A}, A, \lambda \rangle \in \operatorname{ONG}\_1(GL(n, \mathbb{R})), \quad \langle \mathbf{A}, A, \mu \rangle \in \operatorname{ONG}\_2(GL(n, \mathbb{R})).
$$

$$
\langle \mathbf{A}, B, \lambda \rangle \in \operatorname{ONG}\_3(GL(n, \mathbb{R})), \quad \langle \mathbf{A}, B, \mu \rangle \in \operatorname{ONG}\_4(GL(n, \mathbb{R})).
$$

From Propositions 5.2 in [18], 3.2 in [20] and 2.1 (i) in [16], and Remark 1 (1), we have the following.

**Proposition 22.** *Let* A∈O(*G*) *and let* B ∈ *ONG*1(*G*)*. Then* A ◦<sup>1</sup> B = B ◦<sup>1</sup> A.

Also from Propositions 5.3 in [18], 3.3 in [20] and 2.1 (ii) in [16], and Remark 1 (1), we have the following.

**Proposition 23.** *Let* A ∈ *ONG*1(*G*)*. if* B ∈ *OG*1(*G*)*, then* B ◦<sup>1</sup> A ∈ *OG*1(*G*)*.*

**Proposition 24.** *If* A ∈ *ONG*1(*G*)*, then G*<sup>A</sup> *is a normal subgroup of G.*

**Proof.** From Propositions 5.4 in [18], 3.5 in [20], 2.2 (ii) in [16] and Proposition 18, the proof is clear.

It is obvious that *<sup>A</sup>* is a normal subgroup of *<sup>G</sup>*, then *<sup>χ</sup>*<sup>A</sup> ∈ *ONG*1(*G*) and *<sup>G</sup>χ*<sup>A</sup> = *<sup>A</sup>*.

**Definition 24.** *Let* A ∈ *ONG*1(*G*)*. Then the quotient group <sup>G</sup>*/*G*<sup>A</sup> *is called the octahedron quotient group (briefly, OQG) of G with respect to* A*.*

Now let *π* : *G* → *G*/*G*<sup>A</sup> be the natural projection.

**Proposition 25.** *If* A ∈ *ONG*1(*G*) *and* B∈O(*G*)*, then <sup>π</sup>*−1(*π*(B)) = *<sup>G</sup>*/*G*<sup>A</sup> ◦<sup>1</sup> <sup>B</sup>*.*

**Proof.** From Propositions 5.6 in [18], 3.7 in [20] and 2.3 (ii) in [16], the proof is obvious.

#### **5. Octahedron Ideals**

**Definition 25.** *Let* (*R*, +, ·) *be a ring and let* A∈O(*R*)*. Then*

*(i)* 0¨ <sup>=</sup> <sup>A</sup> *is called a* <sup>1</sup>*-octahedron subring of R, if it satisfies the following conditions: for every <sup>x</sup>*, *<sup>y</sup>* <sup>∈</sup> *R, (a)* <sup>A</sup>(*<sup>x</sup>* <sup>+</sup> *<sup>y</sup>*) <sup>≥</sup><sup>1</sup> <sup>A</sup>(*x*) <sup>∧</sup><sup>1</sup> <sup>A</sup>(*y*)*,*

$$(b)\ \mathcal{A}(x^{-1}) \ge\_1 \mathcal{A}(x)\_{\prime\prime}$$

$$(\mathfrak{c})\underset{\longleftarrow}{\mathcal{A}}(\mathfrak{x}y) \ge\_1 \mathcal{A}(\mathfrak{x}) \land^1 \mathcal{A}(y).$$

*(ii)* + 0, **<sup>0</sup>**¯, 1, = A *is called a* 2*-octahedron subring of R, if it satisfies the following conditions: for every x*, *y* ∈ *R,*

*(a)* <sup>A</sup>(*<sup>x</sup>* <sup>+</sup> *<sup>y</sup>*) <sup>≥</sup><sup>2</sup> <sup>A</sup>(*x*) <sup>∧</sup><sup>2</sup> <sup>A</sup>(*y*)*,*

*(b)* <sup>A</sup>(*x*−1) <sup>≥</sup><sup>2</sup> <sup>A</sup>(*x*)*,*

*(c)* <sup>A</sup>(*xy*) <sup>≥</sup><sup>2</sup> <sup>A</sup>(*x*) <sup>∧</sup><sup>2</sup> <sup>A</sup>(*y*)*,*

*(iii)* + 0, **<sup>1</sup>**¯, 0, = A *is called a* 3*-octahedron subring of R, if it satisfies the following conditions: for every x*, *y* ∈ *R,*

*(a)* <sup>A</sup>(*<sup>x</sup>* <sup>+</sup> *<sup>y</sup>*) <sup>≥</sup><sup>3</sup> <sup>A</sup>(*x*) <sup>∧</sup><sup>3</sup> <sup>A</sup>(*y*)*, (b)* <sup>A</sup>(*x*−1) <sup>≥</sup><sup>3</sup> <sup>A</sup>(*x*)*,*

$$\mathcal{A}(c)\,\mathcal{A}(xy)\_{\,\,\,\,'} \geq \mathcal{A}(x)\,\wedge^3\,\mathcal{A}(y)\_{\,\,,\,'}$$

*(iv)* + 0, **<sup>1</sup>**¯, 1, = A *is called a* 4*-octahedron subring of R, if it satisfies the following conditions: for every x*, *y* ∈ *R,*

*(a)* <sup>A</sup>(*<sup>x</sup>* <sup>+</sup> *<sup>y</sup>*) <sup>≥</sup><sup>4</sup> <sup>A</sup>(*x*) <sup>∧</sup><sup>4</sup> <sup>A</sup>(*y*)*, (b)* <sup>A</sup>(*x*−1) <sup>≥</sup><sup>4</sup> <sup>A</sup>(*x*)*,*

*(c)* <sup>A</sup>(*xy*) <sup>≥</sup><sup>4</sup> <sup>A</sup>(*x*) <sup>∧</sup><sup>4</sup> <sup>A</sup>(*y*)*.*

*We will denote the set of all i-octahedron subrings of R as ORi*(*R*) *(i* = 1, 2, 3, 4*). It is clear that if A is a subring of R, then <sup>χ</sup>*<sup>A</sup> ∈ *OR*1(*R*)*.*

**Example 7.** *Consider the ring* (Z2, <sup>+</sup>, ·)*, where* <sup>Z</sup><sup>2</sup> <sup>=</sup> {0, 1}*. Let us define the interval-valued fuzzy set* **<sup>A</sup>***, two intuitionistic fuzzy sets A*, *B and two fuzzy sets λ*, *μ in* Z<sup>2</sup> *as follows:*

$$\mathbf{A}(0) = [0.5, 0.8], \; \mathbf{A}(1) = [0.4, 0.6], \; \mathbf{A}(2) = [0.2, 0.8]$$

$$A(0) = (0.7, 0.2), \; A(1) = (0.5, 0.3), \; B(0) = (0.6, 0.3), \; B(1) = (0.8, 0.2),$$

$$\lambda(0) = 0.8, \; \lambda(1) = 0.5, \; \mu(0) = 0.6, \; \mu(1) = 0.7.$$

*Then we can easily check that the followings hold:*

$$
\langle \mathbf{A}, A, \lambda \rangle \in \operatorname{OR}\_1(\mathbb{Z}\_2), \quad \langle \mathbf{A}, A, \mu \rangle \in \operatorname{OR}\_2(\mathbb{Z}\_2),
$$

$$
\langle \mathbf{A}, B, \lambda \rangle \in \operatorname{OR}\_3(\mathbb{Z}\_2), \ \langle \mathbf{A}, B, \mu \rangle \in \operatorname{OR}\_4(\mathbb{Z}\_2).
$$

From the definitions of orders of two octahedron numbers and Definition 11, and Theorems 1–4 and 6, we have the following.

**Theorem 11.** *Let* (*R*, +, ·) *be a ring and let* A∈O(*R*)*. Then*

(1) 0¨ <sup>=</sup> A ∈ *OR*1(*R*) ⇐⇒ A ∈ *OG*1((*R*, +)), A ∈ *OGP*1((*R*, ·)), (2) +

0, **<sup>0</sup>**¯, 1, = A ∈ *OR*2(*R*) ⇐⇒ A ∈ *OG*2((*R*, +)), A ∈ *OGP*2((*R*, ·)),

(3) + 0, **<sup>0</sup>**¯, 1, = A ∈ *OR*3(*R*) ⇐⇒ A ∈ *OG*3((*R*, +)), A ∈ *OGP*3((*R*, ·)),

(4) + 0, **<sup>0</sup>**¯, 1, = A ∈ *OR*4(*R*) ⇐⇒ A ∈ *OG*4((*R*, +)), A ∈ *OGP*4((*R*, ·)).

*The following is an immediate result of Theorems 7 and 11.*

**Corollary 6.** *Let R be a ring and let* A∈O(*R*)*. Then* A ∈ *ORi*(*R*) *if and only if it satisfies the following conditions: for every x*, *y* ∈ *R and for every i* = 1, 2, 3, 4*,*

(i) <sup>A</sup>(*<sup>x</sup>* <sup>−</sup> *<sup>y</sup>*) <sup>≥</sup>*<sup>i</sup>* <sup>A</sup>(*x*) <sup>∧</sup>*<sup>i</sup>* <sup>A</sup>(*y*)*,* (ii) <sup>A</sup>(*xy*) <sup>≥</sup>*<sup>i</sup>* <sup>A</sup>(*x*) <sup>∧</sup>*<sup>i</sup>* <sup>A</sup>(*y*)*. The following is an immediate result of Remark 3 and Proposition 15.*

**Theorem 12.** *Let R be a ring. Then A be a subring of R if and only if <sup>χ</sup>*<sup>A</sup> ∈ *OR*1(*R*)*.*

**Definition 26.** *Let R be a ring and let* A ∈ *ORi*(*R*) *(I* = 1, 2, 3, 4*). Then (i)* A *is called an i-octahedron left ideal (briefly, i-OLI) if for every x*, *y* ∈ *R,*

$$\mathcal{A}(xy) \ge\_i \mathcal{A}(y)\_{\prime\prime}$$

*(ii)* A *is called an i-octahedron right ideal (briefly, i-ORI), if for every x*, *y* ∈ *R,*

$$\mathcal{A}(xy) \ge\_i \mathcal{A}(x),$$

*(iii)* A *is called an i-octahedron ideal (briefly, i-OI), if it is an i-octahedron left ideal and an i-octahedron right ideal of R.*

*we will denote the set of all i-OLIs [resp., i-ORIs and i-OIs] in R as OLIi*(*R*) *[resp., ORIi*(*R*) *and OIi*(*R*)*].*

**Example 8.** *Consider the ring* (Z4, <sup>+</sup>, ·)*, where* <sup>Z</sup><sup>4</sup> <sup>=</sup> {0, 1, 2, 3}*. Let us define an interval-valued fuzzy set* **A**, **B***, two intuitionistic fuzzy sets A*, *B and two fuzzy sets λ*, *μ*, *η*, *δ in* Z<sup>2</sup> *as follows:*

$$\mathbf{A}(0) = [0.6, 0.8], \; \mathbf{A}(1) = \mathbf{A}(3) = [0.4, 0.6], \; \mathbf{A}(2) = [0.5, 0.7],$$

$$A(0) = (0.7, 0.2), \; A(1) = A(3) = (0.5, 0.4), \; A(2) = (0.6, 0.3),$$

$$B(0) = (0.5, 0.4), \; B(1) = B(3) = (0.7, 0.2), \; B(2) = (0.6, 0.3),$$

$$\lambda(0) = 0.9, \; \lambda(1) = \lambda(3) = 0.6, \; \lambda(2) = 0.7,$$

$$\mu(0) = 0.5, \; \mu(1) = \mu(3) = 0.8, \; \mu(2) = 0.6.$$

*Then we can easily check that the followings hold:*

**A**, *<sup>A</sup>*, *<sup>λ</sup>* <sup>∈</sup> *OLI*1(Z4), **A**, *<sup>A</sup>*, *<sup>μ</sup>* <sup>∈</sup> *OLI*2(Z4), **A**, *<sup>B</sup>*, *<sup>λ</sup>* <sup>∈</sup> *OLI*3(Z4), **A**, *<sup>B</sup>*, *<sup>μ</sup>* <sup>∈</sup> *OLI*4(Z4).

**Remark 9.** *(1) Let R be a ring. If λ* ∈ *FLI*(*R*) *[resp., FRI*(*R*) *and F I*(*R*)*], then* [*λ*, *<sup>λ</sup>*],(*λ*, *<sup>λ</sup>c*), *<sup>λ</sup>* <sup>∈</sup> *OLI*1(*R*) *[resp., ORI*1(*R*) *and OI*1(*R*)*],* [*λ*, *<sup>λ</sup>*],(*λ*, *<sup>λ</sup>c*), *<sup>λ</sup><sup>c</sup>* <sup>∈</sup> *OLI*2(*R*) *[resp., ORI*2(*R*) *and OI*2(*R*)*],*

[*λ*, *<sup>λ</sup>*],(*λc*, *<sup>λ</sup>*), *<sup>λ</sup>* <sup>∈</sup> *OLI*3(*R*) *[resp., ORI*3(*R*) *and OI*3(*R*)*],* [*λ*, *<sup>λ</sup>*],(*λc*, *<sup>λ</sup>*), *<sup>λ</sup><sup>c</sup>* <sup>∈</sup> *OLI*4(*R*) *[resp., ORI*4(*R*) *and OI*4(*R*)*]. (2) Let R be a ring. If A* ∈ *IFLI*(*R*) *[resp., IFRI*(*R*) *and IF I*(*R*)*], then* - [*A*∈,(*A*∈)*c*], *A*, *A*∈. ∈ *OLI*1(*R*) *[resp., ORI*1(*R*) *and OI*1(*R*)*],* - [*A*∈,(*A*∈)*c*], *A*,(*A*∈)*<sup>c</sup>* . ∈ *OLI*2(*R*) *[resp., ORI*2(*R*) *and OI*2(*R*)*],* - [*A*∈,(*A*∈)*c*], *Ac*, *A*∈. ∈ *OLI*3(*R*) *[resp., ORI*3(*R*) *and OI*3(*R*)*],* - [*A*∈,(*A*∈)*c*], *Ac*,(*A*∈)*<sup>c</sup>* . ∈ *OLI*4(*R*) *[resp., ORI*4(*R*) *and OI*4(*R*)*].*

The following is an immediate result of Propositions 11, 12 and 21.

**Proposition 26.** *Let f* : *<sup>R</sup>* <sup>→</sup> *<sup>R</sup> be a ring homomorphism and let i* = 1, 2, 3, 4*.* (1) *If* A ∈ *ORi*(*R*) *or* A ∈ *OLIi*(*R*) *[resp., ORIi*(*R*) *and OIi*(*R*)*], then so is f*(A)*.* (2) *If* B ∈ *ORi*(*<sup>R</sup>* ) *or* B ∈ *OLIi*(*<sup>R</sup>* ) *[resp., ORIi*(*R* ) *and OIi*(*R* )*], then so is f* <sup>−</sup>1(B)*.*

The following is an immediate result of Corollary 6 and Definition 26.

**Theorem 13.** *Let R be a ring,* A∈O(*R*) *and let i* = 1, 2, 3, 4*. Then* A ∈ *OIi*(*R*) *[resp.,* A ∈ *OLIi*(*R*) *and* A ∈ *ORIi*(*R*)*] if and only if it satisfies the following conditions: for every x*, *y* ∈ *R,* (i) <sup>A</sup>(*<sup>x</sup>* <sup>−</sup> *<sup>y</sup>*) <sup>≥</sup>*<sup>i</sup>* <sup>A</sup>(*x*) <sup>∧</sup>*<sup>i</sup>* <sup>A</sup>(*y*)*,* (ii) <sup>A</sup>(*xy*) <sup>≥</sup>*<sup>i</sup>* <sup>A</sup>(*x*) <sup>∨</sup>*<sup>i</sup>* <sup>A</sup>(*y*) *[resp.,* <sup>A</sup>(*xy*) <sup>≥</sup>*<sup>i</sup>* <sup>A</sup>(*y*) *and* <sup>A</sup>(*xy*) <sup>≥</sup>*<sup>i</sup>* <sup>A</sup>(*x*)*].*

The following is an immediate result of Theorems 12 and 13.

**Theorem 14.** *Let <sup>R</sup> be a ring. Then <sup>A</sup> an ideal [resp., a left ideal and a right ideal] of <sup>R</sup> if and only <sup>χ</sup>*<sup>A</sup> ∈ *OI*1(*R*) *[resp., OLI*1(*R*) *and ORI*1(*R*)*].*

**Theorem 15.** *Let R be a skew field (also division ring) and let* A∈O(*R*)*, where* 0 *and e denote the identity for "*+*" and "*·*". Then* A ∈ *OIi*(*R*) *[resp., OLIi*(*R*) *and ORIi*(*R*)*] if and only if for each* 0 = *x* ∈ *R,* A(*x*) = A(*e*) ≤*<sup>i</sup>* A(0) *for i* = 1, 2, 3, 4*.*

**Proof.** We show only that A ∈ *OLIi*(*R*) iff for each 0 = *x* ∈ *R*, A(*x*) = A(*e*) ≤*<sup>i</sup>* A(0) for *i* = 1, 2. The remainder's proofs are omitted.

Case 1: Let *i* = 1. Then from Propositions 6.6 in [18], 4.7 in [20] and 3.3 in [16],

A ∈ *OLI*1(*R*) iff for each 0 = *x* ∈ *R*, A(*x*) = A(*e*) ≤<sup>1</sup> A(0), i.e.,

$$\mathbf{A}(\mathbf{x}) = \mathbf{A}(\varepsilon) \le \mathbf{A}(0), \ A(\mathbf{x}) = A(\varepsilon) \le A(0), \ \lambda(\mathbf{x}) = \lambda(\varepsilon) \le \lambda(0). \tag{10}$$

Case 2: Let *i* = 2. Suppose A ∈ *OLI*2(*R*) and let 0 = *x* ∈ *R*. Then by Definition 26, *λ*(*x*) = *<sup>λ</sup>*(*xe*) <sup>≤</sup> *<sup>λ</sup>*(*e*) and *<sup>λ</sup>*(*e*) = *<sup>λ</sup>*(*x*−1*x*) <sup>≤</sup> *<sup>λ</sup>*(*x*). Thus, *<sup>λ</sup>*(*x*) = *<sup>λ</sup>*(*e*). Since A ∈ *OLI*2(*R*), A ∈ *OR*2(*R*). By Corollary 6 (i) and Definition 2,

$$
\lambda(0) = \lambda(e - e) \le \lambda(e) \lor \lambda(e) = \lambda(e).
$$

So *λ*(*x*) = *λ*(*e*) ≥ *λ*(0). Hence by the first and the second parts of (5.1),

$$\mathcal{A}(\mathfrak{x}) = \mathcal{A}(e) \le\_2 \mathcal{A}(0).$$

Conversely, suppose the necessary condition holds and let *x*, *y* ∈ *R*. Then we have four cases: (i) *x* = 0, *y* = 0, *x* = *y*, (ii) *x* = 0, *y* = 0, *x* = *y*, (iii) *x* = 0, *y* = 0, (iv) *x* = 0, *y* = 0.

223

Case (i). Suppose *x* = 0, *y* = 0, *x* = *y*. Then by the hypothesis, we have

$$\mathbf{A}(\mathbf{x}) = \mathbf{A}(y) = \mathbf{A}(\boldsymbol{\varepsilon}) = \mathbf{A}(\mathbf{x} - \boldsymbol{y}) = \mathbf{A}(\mathbf{x}\mathbf{y}),$$

$$A(\mathbf{x}) = A(\boldsymbol{y}) = A(\boldsymbol{\varepsilon}) = A(\mathbf{x} - \boldsymbol{y}) = A(\mathbf{x}\mathbf{y}),$$

$$\lambda(\mathbf{x}) = \lambda(\mathbf{y}) = \lambda(\boldsymbol{\varepsilon}) = \lambda(\mathbf{x} - \mathbf{y}) = \lambda(\mathbf{x}\mathbf{y}).$$

Thus,

$$\mathbf{A}(\mathbf{x} - \mathbf{y}) \ge \mathbf{A}(\mathbf{x}) \land \mathbf{A}(\mathbf{y}), \; \mathbf{A}(\mathbf{x}\mathbf{y}) \ge \mathbf{A}(\mathbf{y}),$$

$$A(\mathbf{x} - \mathbf{y}) \ge A(\mathbf{x}) \land A(\mathbf{y}), \; A(\mathbf{x}\mathbf{y}) \ge A(\mathbf{y}),$$

$$\lambda(\mathbf{x} - \mathbf{y}) \le \lambda(\mathbf{x}) \lor \lambda(\mathbf{y}), \; \lambda(\mathbf{x}\mathbf{y}) \le \lambda(\mathbf{y}).$$

So <sup>A</sup>(*<sup>x</sup>* <sup>−</sup> *<sup>y</sup>*) <sup>≥</sup><sup>2</sup> <sup>A</sup>(*x*) <sup>∧</sup><sup>2</sup> <sup>A</sup>(*y*) and <sup>A</sup>(*xy*) <sup>≥</sup><sup>2</sup> <sup>A</sup>(*y*).

Case (ii). Suppose *x* = 0, *y* = 0, *x* = *y*. Then by the hypothesis, we have

$$\mathbf{A}(\mathbf{x}) = \mathbf{A}(y) = \mathbf{A}(\boldsymbol{\varepsilon}) = \mathbf{A}(\mathbf{x}y) = \mathbf{A}(\mathbf{x} - y) = \mathbf{A}(0),$$

$$A(\mathbf{x}) = A(y) = A(\boldsymbol{\varepsilon}) = A(\mathbf{x}y) = A(\mathbf{x} - y) = A(0),$$

$$\boldsymbol{\lambda}(\mathbf{x}) = \boldsymbol{\lambda}(y) = \boldsymbol{\lambda}(\boldsymbol{\varepsilon}) \boldsymbol{) = \boldsymbol{\lambda}(\mathbf{x} - y) = \boldsymbol{\lambda}(\mathbf{x}y) = \boldsymbol{\lambda}(0).$$

Thus, we have the same result in Case (i):

$$\mathcal{A}(\mathfrak{x} - \mathfrak{y}) \ge\_2 \mathcal{A}(\mathfrak{x}) \land^2 \mathcal{A}(\mathfrak{y}) \text{ and } \mathcal{A}(\mathfrak{x}\mathfrak{y}) \ge\_2 \mathcal{A}(\mathfrak{y}).$$

Case (iii). Suppose *x* = 0, *y* = 0. Then by the hypothesis, we have

$$\mathbf{A}(\mathbf{x} - \mathbf{y}) = \mathbf{A}(\mathbf{x}) = \mathbf{A}(\mathbf{e}) \ge \mathbf{A}(\mathbf{x}) \land \mathbf{A}(\mathbf{y}),\\\mathbf{A}(\mathbf{x}\mathbf{y}) = \mathbf{A}(0) \ge \mathbf{A}(\mathbf{y})$$

$$A(\mathbf{x} - \mathbf{y}) = A(\mathbf{x}) = A(\mathbf{e}) \ge A(\mathbf{x}) \land A(\mathbf{y}),\\\mathbf{A}(\mathbf{x}\mathbf{y}) = A(0) \ge A(\mathbf{y}).$$

$$\lambda(\mathbf{x} - \mathbf{y}) = \lambda(\mathbf{x}) = \lambda(\mathbf{e}) \le \lambda(\mathbf{x}) \lor \lambda(\mathbf{y}),\\\lambda(\mathbf{x}\mathbf{y}) = \lambda(0) \le \lambda(\mathbf{y}).$$

Thus, we have the same result in Case (i):

$$\mathcal{A}(\mathbf{x} - \mathbf{y}) \ge\_2 \mathcal{A}(\mathbf{x}) \land^2 \mathcal{A}(\mathbf{y}) \text{ and } \mathcal{A}(\mathbf{x}\mathbf{y}) \ge\_2 \mathcal{A}(\mathbf{y}).$$

Case (iv). Suppose *x* = 0, *y* = 0. Then by the similar proof to Case (iii), we have

$$\mathcal{A}(\mathbf{x} - \mathbf{y}) \ge\_2 \mathcal{A}(\mathbf{x}) \land^2 \mathcal{A}(\mathbf{y}) \text{ and } \mathcal{A}(\mathbf{x}\mathbf{y}) \ge\_2 \mathcal{A}(\mathbf{y}).$$

Hence in either cases, by Theorem 13, A ∈ *OLI*2(*R*).

**Remark 10.** *Theorem 15 shows that an i-OLI (ORI) is an i-OI in a skew field.*

The following gives a characteristic of a (usual) field by a 1-OI.

**Proposition 27.** *Let R be a commutative ring with a unity e. Suppose for each* A ∈ *OI*1(*R*),

$$\mathcal{A}(\mathfrak{x}) = \mathcal{A}(\mathfrak{e}) \le\_1 \mathcal{A}(\mathfrak{0}) \text{ for each } \mathfrak{0} \ne \mathfrak{x} \in \mathbb{R}.$$

*Then R is a field.*

**Proof.** Let *<sup>A</sup>* be an ideal of *<sup>R</sup>* such that *<sup>A</sup>* = *<sup>R</sup>*. Then clearly by Theorem 14, *<sup>χ</sup>*<sup>A</sup> ∈ *OI*1(*R*) such that *<sup>A</sup>* <sup>=</sup> 1¨. Thus, there is *<sup>y</sup>* <sup>∈</sup> *<sup>R</sup>* such that *<sup>y</sup>* ∈ *<sup>A</sup>*. So *<sup>χ</sup>*A(*y*) = 0¨. By the hypothesis, *<sup>χ</sup>*A(*x*) = *<sup>χ</sup>*A(*e*) <sup>≤</sup><sup>1</sup> *<sup>χ</sup>*A(0). Hence *<sup>χ</sup>*A(0) = 1, i.e., ¨ *<sup>A</sup>* <sup>=</sup> {0}. Therefore *<sup>R</sup>* is a field.

### **6. Conclusions**

By using the *i*-product of two octahedron sets, we introduce the concept of *i*-octahedron subgroupoids of a groupoid. In particular, we obtain four characterizations of *i*-octahedron groupoids. Also, we defined an *i*-OLI [resp., *i*-ORI and *i*-OI]] of a groupoid and investigated some of their properties. Moreover, we obtain some properties for the image and preimage of an *i*-octahedron subgroupoid [resp., *i*-OLI, *i*-ORI and *i*-OI] under groupoid homomorphism. Next, we define *i*-octahedron subgroups of a group and study some of their properties. In particular, we obtain two characterizations of *i*-octahedron subgroups and *i*-OLI [resp., *i*-ORI and *i*-OI] of a group. We introduce the concepts of *i*-octahedron subrings [resp., *i*-OLIs, *i*-ORIs and *i*-OIs] of a ring and obtain their characterizations. Furthermore, we found a sufficient condition for which a commutative ring with a unity *e* is a field.

In the future, we expect that one applies octahedron sets to *BCI*/*BCK*-algebras, topologies, decision-making, measures and entropy measures, etc.

**Author Contributions:** Created and conceptualized ideas, J.-G.L. and K.H.; writing—original draft preparation, J.-G.L. and K.H.; writing—review and editing, Y.B.J.; funding acquisition, J.-G.L. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2018R1D1A1B07049321).

**Acknowledgments:** We are very grateful to the reviewers for their careful reading and their meaningful suggestions.

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **Abbreviations**

The following abbreviations are used in this manuscript:




#### **References**


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