EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 14, No. 1, 2021, 314-326 ISSN 1307-5543 – ejpam.com Published by New York Business Global On γ-Sets in Rings Eva Jenny C. Sigasig1, Cristoper John S. Rosero2, Michael P. Baldado Jr.3,∗ 1 Lourdes Ledesma Del Prado Memorial National High School, Tanjay City, Philippines 2 Mathematics and ICT Department, Cebu Normal University, Cebu City, Philippines 3 Mathematics Department, Negros Oriental State University, Dumaguete City, Philippines Abstract. Let R be a ring with identity 1R. A subset J of R is called a γ-set if for every a ∈ R\J , there exist b, c ∈ J such that a+ b = 0 and ac = 1R = ca. A γ-set of minimum cardinality is called a minimum γ-set. In this study, we identified some elements of R that are necessarily in a γ-sets, and we presented a method of constructing a new γ-set. Moreover, we gave: necessary and sufficient conditions for rings to have a unique γ-set; an upper bound for the total number of minimum γ-sets in a division ring; a lower bound for the total number of minimum γ-sets in a division ring; necessary and sufficient conditions for T (x) and T to be equal; necessary and sufficient conditions for a ring to have a trivial γ-set; necessary and sufficient conditions for an image of a γ-set to be a γ-set also; necessary and sufficient conditions for a ring to have a trivial γ-set; and, necessary and sufficient conditions for the families of γ-sets of two division rings to be isomorphic. 2020 Mathematics Subject Classifications: 16 Key Words and Phrases: D-set, γ-set, minimum γ-set, separating γ-set, ring 1. Introduction Let G be a group with identity e. A subset D of G is called a D-set of G if for every x in G\D, there exists y ∈ D such that xy = e = yx. In other words, a subset of a group G is a D-set only if every element not in D has its inverse in D. A smallest D-set of G is called a minimum D-set of G. The number of minimum D-set of G is called the index minimum. If G is a finite group and S = {s ∈ G : s2 = e} (the elements of S will be called involutions), then the c-number of G is given by |(G\S)| /2. Let R be a ring with identity 1R. A subset J of R is called a γ-set of R if for every a ∈ R\J , there exist b, c ∈ J such that a+ b = 0 and ac = 1R = ca. For example, consider the field Z5. Then the γ-sets of Z5 are {0, 1, 4, 2}, {0, 1, 4, 3}, and Z5. A γ-set of a finite ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v14i1.3873 Email addresses: evajenny@yahoo.com (E.J. Sigasig), crisrose 18@yahoo.com (C.J. Rosero), michaelpbaldadojr@yahoo.com (M. Baldado Jr.) http://www.ejpam.com 314 c© 2021 EJPAM All rights reserved. E.J. Sigasig, C.J. Rosero, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (1) (2021), 314-326 315 ring having minimum cardinality is called a minimum γ-set. For example, {0, 1, 4, 2} and {0, 1, 4, 3} are minimum γ-sets Z5. Here after please refer to [4], [5], [6], [7], [8], [9], [10] for the other concepts. Motivated by the concept dominating sets in graphs, Buloron et al. [3] introduced the concept D-set in a group. The concept D-set uses the idea of dominating sets in some sense. For example, a D-set E in a group requires that every element not in E must have its inverse in E in the same way that a dominating set D in a graph requires every element not in D must be a neighbor of some element in D. Buloron et al. [3] gave some fundamental properties of D-sets and some characteriza- tions. Ontolan et al. [12] gave the number of minimum D-sets in a group. Corcino et al. [2] presented some isomorphism results for some families of D-sets. Rosero and Baldado [1] continued the study of D-sets by investigating the D-sets that are generated by a set. Moreover, they introduced and investigated a parallel concept for rings, called γ-sets [11]. In this study, we continued the investigation of γ-sets. 2. Preliminary Results This section presents some elementary properties of a γ-set. We denote by TR the set of all γ-sets of R. Note that TR 6= ∅ since R is a γ-set. The next theorem, Theorem 1, is taken from [3]. It shows that the set of all γ-sets in a ring is a semi-group under the set operation union, and the set TC R = {JC : J is a γ-set} is a semi-group under the set operation intersection. Theorem 1. [3] Let R be a ring with identity 1R. Let TR be the set of all γ-sets of R and TC R = {JC : J is a γ-set}. Then a.) The set TR is a semi-group under the set operation union; b.) The set TC R is a semi-group under the set operation intersection. Remark 1 (b) is found in [4], while (c) is an exercise in [6] (Prob 24E, Chapter 5.1). Remark 1 (d) is a contrapositive of (c). Remark 1. Let R be a ring with identity 1R and a ∈ R. a.) If a is a unit, then −(a−1) = (−a)−1. b.) If a is a unit, then so is −a. c.) If a is a unit, then a is not a zero divisor. d.) If a is a zero divisor, then a is not a unit. Remark 2 is clear, and sometimes are given in the exercises of some books. E.J. Sigasig, C.J. Rosero, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (1) (2021), 314-326 316 Remark 2. Let R be a ring with identity 1R 6= 0 and a be a unit of R. a.) −a = a−1 if and only if a = (−a)−1. b.) −a 6= a−1 if and only if a 6= (−a)−1 c.) a2 = 1R if and only if −a = (−a)−1. d.) a2 6= 1R if and only if −a 6= (−a)−1. e.) 2a = 0 if and only if (a)−1 = (−a)−1. f.) 2a 6= 0 if and only if (a)−1 6= (−a)−1. Theorem 2, identified the elements of a ring that are necessarily in a γ-sets. Theorem 2. Let R be a ring with identity 1R 6= 0 and J be a γ-set of R. a.) If 2a = 0, then a ∈ J . b.) If a2 = 1R, then a ∈ J . c.) If a is not a unit, then a ∈ J . d.) If a is a zero-divisor, then a ∈ J . Proof. Let R be a ring with identity 1R 6= 0 and J be a γ-set of R. (a) Assume that 2a = 0 and a /∈ J . Since J is a γ-set, there exists b ∈ J such that a + b = 0. Hence, a = a + 0 = a + (a + b) = (a + a) + b = 2a + b = 0 + b = b, that is a = b. This is a contradiction. (b) Assume that a2 = 1R and a /∈ J . Since J is a γ-set, there exists c ∈ J such that ac = 1R = ca. Hence, a = a1R = a(ac) = (aa)c = 1Rc = c, that is a = c. This is a contradiction. (c) If a is not a unit, then a has no multiplicative inverse. Clearly, a is necessarily in J . (d) If a is a zero-divisor, then by Remark 2 (b), a is not a unit. Hence, by (c) a must be in J . 3. Constructing a γ-Set In this section, we presented a method of constructing a γ-set from a γ-set. The next theorem, Theorem 3, says that a unit a with a2 6= 1R and 2a 6= 0 determines a γ-set. Theorem 3. Let R be a ring with identity 1R 6= 0, and J is a γ-set of R. If a is a unit with a2 6= 1R and 2a 6= 0, then ( J\{a, (−a)−1} ) ∪ {a−1,−a} and ( J\{a−1,−a} ) ∪ {a, (−a)−1} are γ-sets of R. E.J. Sigasig, C.J. Rosero, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (1) (2021), 314-326 317 Proof. Let R be a ring with identity 1R 6= 0, and J is a γ-set of R. Let a be a unit of R with a2 6= 1R and 2a 6= 0. Then by Remark 2 (d) and Remark 2 (f), −a 6= (−a)−1 and a−1 6= (−a)−1. Consider J1 = (J\{a, (−a)−1}) ∪ {a−1,−a} and J2 = (J\{a−1,−a}) ∪ {a, (−a)−1}. Claim 1. J1 = (J\{a, (−a)−1}) ∪ {a−1,−a} is a γ-set To show Claim 1 consider the following cases: Case 1. a /∈ J If a /∈ J , then J = J1. Hence J1 is a γ-set. Case 2. a ∈ J If a ∈ J , then let b ∈ R\J1 and consider the following subcases: Subcase 1. b 6= a and b 6= (−a)−1 If b 6= a and b 6= (−a)−1, then b ∈ R\J ∪{a−1,−a}. Since J is a γ-set, there exist c, d ∈ J1 such that b+ c = 0 = c+ b and bd = 1R = db. Subcase 2. b = a If b = a, then a+ (−a) = 0 = (−a) + a and aa−1 = 1R = a−1a. Subcase 3. b = (−a)−1 If b = (−a)−1, then by Remark 1 (a) (−a)−1 + a−1 = −a−1 + a−1 = 0 = a−1 + −a−1 = a−1 + (−a)−1 and (−a)−1(−a) = 1R = (−a)(−a)−1. This shows the claim. Claim 2. J2 = (J\{a−1,−a}) ∪ {a, (−a)−1} is a γ-set Proved similarly. Let R be a ring with identity 1R 6= 0, and J is a γ-set of R. A unit a with a2 6= 1R and 2a 6= 0 is called a super-couple. Theorem 5 suggests that every super-couple determines a minimum γ-set, in the same way as in [3] that every non-involution determines a D-set. Theorem 4 give some of the conditions wherein a ring R has a unique γ-set, that is, |TR| = 1. Theorem 4. Let R be a ring with identity 1R 6= 0 and J be a γ-set of R. Then |TR| > 1 if and only if there exists a unit u ∈ R such that u2 6= 1R and 2u 6= 0. Proof. Let R be a ring with identity 1R 6= 0 and J be a γ-set of R. Suppose that |TR| > 1. Then there exists a γ-set J in R with J 6= R. Let x ∈ R\J . Since J is a γ-set, there exists y, z ∈ J such that x + y = 0 = y + x and xz = 1R = zx. Thus, x is a unit. Moreover, since y, z ∈ J and x ∈ R\J , x 6= y and x 6= z. Hence, by Remark 2 (d) and Remark 2 (f), x2 6= 1R and 2x 6= 0, respectively. Conversely, assume that there exists a unit x ∈ R such that x2 6= 1R and 2x 6= 0. Then by Theorem 5, ( R\{x−1,−x} ) ∪ {x, (−x)−1} is a nontrivial γ-sets of R. Therefore, |TR| > 1. Theorem 5. Let R be a ring with identity 1R 6= 0 and J be a γ-set of R. Then |TR| = 1 if and only if for all a ∈ R either a2 = 1R or 2a = 0 or a is a zero-divisor. Proof. Proved similarly. E.J. Sigasig, C.J. Rosero, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (1) (2021), 314-326 318 4. An Equivalence Relation in R\S In this section, we presented an equivalence relation in R\S which will be useful in the next section. Lemma 1. Let R be a division ring and S = {x ∈ R : x2 = 1R or 2x = 0}. The relation ∼ on R\S given by x ∼ y if and only if x = y or x = y−1 or x = y or x = (−y)−1 is an equivalence relation. Proof. Let R be a division ring and S = {x ∈ R : x2 = 1R or 2x = 0}. Define a relation ∼ on R\S as follows: x ∼ y if and only if x = y or x = y−1 or x = (−y)−1. Since x = x for all x ∈ R, we have x ∼ x for all x ∈ R\S. Hence, ∼ is reflexive. It can easily be shown that ∼ is symmetric and transitive. Thus, ∼ is an equivalence relation. Remark 3. Let R be a division ring, and let S = {x ∈ R : x2 = 1R or 2x = 0}. The equivalence relation ∼ in R\S of Lemma 1 partitions R\S into equivalence classes [a] = {x ∈ R\S : x ∼ a} = {x ∈ R\S : x = a, or x = a−1, or x = −a, or x = (−a)−1}. 5. Some Bounds on the Number of Minimum γ-Set In this section, we established a sharp upperbound and a sharp lowerbound for the number of minimum γ-set in a finite division ring. If R is a finite division ring, then we denote the partition of R\S in Remark 3 by C = {[a1] , [a2] , . . . , [ac]}. In this case, we call c the c-number of R. Lemma 2. Let R be a finite division ring, and let S = {x ∈ R : x2 = 1R or 2x = 0}. If C = {[a1] , [a2] , . . . , [ac]} is the partition of R\S in the sense of Remark 3, then 2 ≤ |[ai]| ≤ 4. Proof. Let R be a finite division ring and let S = {x ∈ R : x2 = 1R or 2x = 0}. If C = {[a1] , [a2] , . . . , [ac]} is the partition of R\S in the sense of Remark 3, then [ai] = {ai,−ai, a−1i ,−a−1i , } for all i = 1, 2, . . . , c. Since a2i 6= 1R and 2ai 6= 0 for all i, Remark 2 implies that {ai,−a−1i } ∩ {−ai, a −1 i } = ∅ for all i. If ai = −a−1i , then by Remark 2, −ai = a−1i . Hence, in this case |[ai]| = 2. On the hand, if ai 6= −a−1i , then by Remark 2, −ai 6= a−1i . Hence, in this case |[ai]| = 4. Accordingly, 2 ≤ |[ai]| ≤ 4. By Theorem 3, each equivalence class [ai] determines two minimum γ-sets. Lemma 3. Let R be a finite division ring, and let S = {x ∈ R : x2 = 1R or 2x = 0}. Then, c ≥ (|R| − |S|)/4. Proof. Let R be a finite division ring and let S = {x ∈ R : x2 = 1R or 2x = 0}. By Lemma 2, |[ai]| ≤ 4. Therefore, c ≥ (|R| − |S|)/4. E.J. Sigasig, C.J. Rosero, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (1) (2021), 314-326 319 Lemma 4. Let R be a finite division ring, and let S = {x ∈ R : x2 = 1R or 2x = 0}. Then, c ≤ (|R| − |S|)/2. Proof. Let R be a finite division ring and let S = {x ∈ R : x2 = 1R or 2x = 0}. By Lemma 2, 2 ≤ |[ai]|. Therefore, c ≤ (|R| − |S|)/2. Theorem 6 give a necessary and sufficient condition for a γ-set to be minimum. Theorem 6. Let R be a finite division ring. Then, E is a minimum γ-set of R if and only if E = S ∪ {−x1, x−11 ,−x2, x−12 , . . . ,−xc, x−1c } where xi ∈ [ai] for i = 1, 2, . . . , c, and {[a1] , [a2] , . . . , [ac]} is the partition of R\S in the sense of Remark 3. Proof. Suppose that E is a minimum γ-set of R and E is not of the form S ∪ {−x1, x−11 ,−x2, x−12 , . . . ,−xc, x−1c }. If E is not of the form S ∪ {−x1, x−11 , −x2, x−12 , . . . ,−xc, x−1c }, then there exists i ∈ {1, 2, . . . , j} such that −xi, x−1i , (−xi)−1 ∈ E or xi,−xi, x−1i , (−xi)−1 ∈ E. Thus, S ∪ {−x1, x−11 ,−x2, x−12 , . . . , −xc, x−1c } is a γ-set smaller than E. This is a contradiction. Conversely, suppose that E is of the form S ∪ {−x1, x−11 ,−x2, x−12 , . . . , −xc, x−1c } and E is not a minimum γ-set of R. If E is not a minimum γ-set of R, then then there exists i ∈ {1, 2, . . . , j} such that xi,−xi, x−1i /∈ E. Since E is a γ-set and x−1i , xi = (x−1i )−1 ∈ E. This is a contradiction. The index minimum of a finite ring R is the number of minimum γ-sets of R and is denoted by ind(R). Corollary 1 gives an upper bound on the number of minimum γ-set of a finite division ring, while Corollary 2 gives a lower bound on the number of minimum γ-set of a finite division ring. Corollary 1. Let R be a finite division ring. Then ind(R) ≤ 2(|R|−|S|)/2. Proof. LetR be a finite division ring and let S = {x ∈ R : x2 = 1R or 2x = 0}. Then by Lemma 4, c ≤ (|R|−|S|)/2. Moreover, by Theorem 3, each equivalence class [ai] determines two minimum γ-sets. Therefore, by the multiplication principle, ind(R) ≤ 2(|R|−|S|)/2. The bound in Corollary 1 is sharp. Equality holds for some fields. For example, if R is Z5, then the equality holds. To see this, we note that the minimum γ-sets of Z5 are J1 = {0, 1, 4, 3} and J1 = {0, 1, 4, 3}. Hence, ind(Z5) = 2. This is equal to 2(|R|−|S|)/2 = 2(|Z5|−|{0,1,4}|)/2 = 2(5−3)/2 = 2. Corollary 2. Let R be a finite division ring. Then ind(R) ≥ 2(|R|−|S|)/4. Proof. LetR be a finite division ring and let S = {x ∈ R : x2 = 1R or 2x = 0}. Then by Lemma 3, c ≥ (|R|−|S|)/4. Moreover, by Theorem 3, each equivalence class [ai] determines two minimum γ-sets. Therefore, by multiplication principle, ind(R) ≤ 2(|R|−|S|)/2. The bound in Corollary 2 is also sharp. Equality holds for some fields. For example, if R is Z7, then the equality holds. To see this, we note that the minimum γ-sets of Z7 are J1 = {0, 1, 2, 3, 6} and J2 = {0, 1, 4, 5, 6}. Hence, ind(Z7) = 2. This is equal to 2(|R|−|S|)/2 = 2(|Z7|−|{0,1,6}|)/2 = 2(7−3)/4 = 2. E.J. Sigasig, C.J. Rosero, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (1) (2021), 314-326 320 6. γ-Sets and Homomorphism of Rings In this section, we gave some properties of γ-sets in relation to its homomorphic image. We say that a set A precedes a set B if there exists an injective map from A to B. In this case, we write A≺B. Theorem 7. Let J be a γ-set of a ring. Then R\J≺J . Proof. Let f : R\J → J be a given by f(x) = x−1, and let x, y ∈ J with x = y. Since J is a γ-set and each unit of a ring has a unique multiplicative inverse, x = y implies that f−1(x) = f−1(y). This means that f is injective. Hence, R\J≺J . Theorem 7 says that in a finite ring, a γ-set has more elements than its complement. Lemma 5. Let R be a ring, and x ∈ R. Then T (x) = {J ⊆ R : J is a γ-set and x ∈ J} is a semigroup under the operation union. Proof. It suffices to show that T (x) is closed under the operation union. Let J1 and J2 be element of T (x). Then by Theorem 1, J1 ∪ J2 is a γ-set. Since clearly x ∈ J1 ∪ J2, J1 ∪ J2 ∈ T (x). An element x of a ring R is called an involution if x2 = 1R, or 2x = 0, or x 6= −x−1. Theorem 8. Let x be a non-identity element of a ring. Then x is an involution if and only if T (x) = T . Proof. Let R be a ring, and suppose that x is an involution of R. If x is an involution, then by Theorem 2 x is contained in every γ-set of R. Thus, if J is a γ-set, then J ∈ T (x), that is T ⊆ T (x). Since clearly T ⊇ T (x), we must have T = T (x). Conversely, assume that T (x) = T and x is not an involution. If x is not an involution and x 6= 1R, then x 6= x−1. By Theorem 3, J ′ = ( J\{x, (−x)−1} ) ∪ {x−1,−x} is also a γ-set. Note that J ′ ∈ T , but J ′ /∈ T (x), that is T (x) 6= T . This is a contradiction. Theorem 9. Let R be a ring with identity and x be a non-zero element of R with 2x = 0. Then every γ-set of R contains a non-trivial subring. Proof. Let J be a γ-set of a ring R and x be a non-zero element of R with 2x = 0. Theorem 2 implies that 0 and x are elements of J . Thus, {0, x} is a subring of R contained in J . Theorem 10. Let R be a ring with identity. R has a trivial γ-set if and only if R is trivial. Proof. Let R = {0, 1}. Then clearly {0, 1} is a γ-set of R. Conversely, suppose that J = {0, 1} is a γ-set of R and R is non-trivial. Let x ∈ R with x 6= 0 and x 6= 1R. Since J is a γ-set of R, there exists y, z ∈ J such that x+ y = 0 and xz = 1R. Since the elements of J are 0 and 1 only, this implies that x + 1 = 0, that is x = −1. Hence, x2 = (−1)2 = 1R. By Theorem 2, x ∈ J . This is a contradiction. E.J. Sigasig, C.J. Rosero, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (1) (2021), 314-326 321 Theorem 11. Let T be the set of all γ-sets of a division ring R, and S = {x ∈ R : x2 = 1R} ∪ {0}. Then |T | = 1 if and only if R = S. Proof. Assume that |T | = 1, and R 6= S. If R 6= S, then there exists x ∈ R\S such that x2 = 1R. Let J ∈ T and consider the following cases: Case 1. x /∈ J If x ∈ J , then by Theorem 3, J ′ = ( J\{x−1,−x} ) ∪{x, (−x)−1} is another γ-set. This is a contradiction. Case 2. x ∈ J If x ∈ J , then by Theorem 3, J ′ = ( J\{x, (−x)−1} ) ∪{x−1,−x} is another γ-set. This is a contradiction. Conversely, suppose that R = S. Since R is a division ring, every non-zero element is an involution. Hence, by Theorem 10 if J is a γ-set of R, we must have J = R, that is R is the only γ-set R. Thus, |T | = 1. Theorem 12. Let Q be a subring of R, and J be a γ-set of R. Then J is a γ-set of Q if and only if Q = R. Proof. Assume that J is a γ-set of Q, and Q 6= R. If Q 6= R, then there exists x ∈ R\Q. Since x /∈ Q, x /∈ J . Since J , x−1 ∈ J . Since J is also a γ-set of Q, x = xx−1 ∈ T . This is a contradiction. The converse is clear. Theorem 13. Let R1 and R2 be rings, and φ : R1 → R2 be an epimorphism of rings. If J is a γ-set of R1, then φ(J) is a γ-set of R2. Proof. Let J be a γ-set of R1, and y ∈ R2\φ(J). If y ∈ R2\φ(J) and φ is an epimor- phism, then there exists x ∈ R1 such that φ(x) = y. Note that x ∈ R1\J , otherwise y = φ(x) ∈ φ(J). Since J is a γ-set, there exists u, v ∈ J such that xu = 1R1 and x+ v = 0R1 . Clearly, φ(u), φ(v) ∈ φ(J). Since φ is a homomorphism, φ(x)φ(u) = φ(xu) = φ(1R1) = 1R2 and φ(x) + φ(v) = φ(x + v) = φ(0R1) = 0R2 . Hence, there exists φ(u), φ(v) ∈ φ(J) such that φ(x)φ(u) = 1R2 and φ(x) + φ(v) = 0R2 . This shows that φ(J) is a γ-set of R2. Theorem 14. Let R1 and R2 be rings, and φ : R1 → R2 be an isomorphism of rings. Then, J is a γ-set of R1 if and only if φ(J) is a γ-set of R2. Proof. Let J be a γ-set of R1. Then by Theorem 7, φ(J) is a γ-set of R2. Conversely, let J be a subset of R1, and suppose that φ(J) is a γ-set of R2. Let x ∈ R1\J . Then φ(x) ∈ R2\φ(J), otherwise x = φ−1φ(x) ∈ J . Since φ(J) is a γ-set of R2, there exists u, v ∈ φ(J) such that φ(x)u = 1R2 and φ(x) + v = 0R2 . Note that φ−1(u), φ−1(v) ∈ J , otherwise u = φ(φ−1(u)) ∈ R2\φ(J) and v = φ(φ−1(v)) ∈ R2\φ(J). Since φ is an isomorphism, xφ−1(u) = φ−1(φ(x))φ−1(u) = φ−1(φ(x)u) = φ−1(1R2) = 1R1 and x + φ−1(v) = φ−1(φ(x)) + φ−1(v) = φ−1(φ(x) + v) = φ−1(0R2) = 0R1 . Hence, there exists φ−1(u), φ−1(v) ∈ J such that xφ−1(u) = 1R1 and x + φ−1(v) = 0R1 . This shows that J is a γ-set of R1. E.J. Sigasig, C.J. Rosero, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (1) (2021), 314-326 322 Theorem 15. Let R be a ring, T = {J ⊆ R : J is a γ-set of R}, and T ′ = {J ′ : J ∈ T} where J ′ is the complement of J . Then, T is isomorphic to T ′. Proof. Let R be a ring, T = {J ⊆ R : J is a γ-set of R}, and T ′ = {J ′ : J ∈ T}. Define φ : T → T ′ by J 7→ J ′ where J ′ is the complement of J . Then clearly φ is bijective. Now, let J1, J2 ∈ T . Then φ(J1 ∪ J2) = (J1 ∪ J2)′ = J ′1 ∩ J ′2 = φ(J1) ∩ φ(J2). This shows that φ is an isomorphism, that is T is isomorphic to T ′ as a semigroup. 7. Separating γ-Sets Our objective in this section is to show the statement: Let R and S be rings. Then TR is isomorphic to TS if and only if |G\SR| = |H\SS |. We borrowed here some ideas presented by Joris N. Buloron in [2] to show the results. We denote the set of all involutions of a division ring D by SD, that is, SD = {x ∈ R : x2 = 1D, or 2x = 0, or a 6= −a−1}. Let J be a γ-set of a division ring D. Then J is called a separating γ-set of D if for every x ∈ J\SD, x−1 /∈ D. Note that for a finite division ring D, the separating γ-sets are just the minimum γ-sets. Also note that if J is not a separating γ-set, then there exists x ∈ D\SD such that x, x−1 ∈ J . Lemma 6. Let D be a division ring and J be a γ-set. If J is not a separating γ-set, then J can be expressed as a union of two distinct separating γ-sets. Proof. Let J be a γ-set that is not separating. Define a relation ∼ on J\SD as follows: x ∼ y if and only if x = y or y = x−1. Then ∼ is an equivalence relation, that is, ∼ partitions J\SD into equivalence classes. For each x ∈ J\SD, the equivalence class con- taining x is x̄ = {x, x−1}. By the Axiom of Choice, there exists a set ∆ such that ∆ ∩ x̄ is a singleton set for all x ∈ J\SD. It is easy to see that ∆ ∪ SD and J\∆ is a separating γ-set, and J = (∆ ∪ SD) ∪ (J\∆). A careful observation would suggest that a separating γ-set cannot be expressed as a union of two distinct γ-sets. The next lemma is anchored on this idea. Lemma 7. Let D be a division ring and J be a γ-set of D. J is not a separating D-set if and only if it is a union of two or more distinct γ-sets. E.J. Sigasig, C.J. Rosero, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (1) (2021), 314-326 323 Proof. Let J be a γ-set of D and assume that J is not a separating γ-set. Then by Lemma 6, J = E ∪ F for some separating γ-sets E and F . Note that E and F must be distinct, otherwise, J = E ∪F = E (which is a contradiction since E is a separating γ-set while J is not). Conversely, assume that J = E ∪F for some γ-sets E and F , with E 6= F . If one of E and F is not a separating γ-set, then clearly, J = E ∪ F is not a separating γ-set. So we assume that E and F are both separating γ-sets. Since E 6= F , E\F 6= ∅. Let x ∈ E\F . Since F is a γ-set, x−1 ∈ F . Hence, x, x−1 ∈ D. This implies that D is not a separating γ-set. At this point, we will now state some consequence of the above lemma. The following definitions are helpful in the succeeding statements. Let x be an element of a division ring D. We denote by TD(x) the family of all γ-sets containing x, that is, TD(x) = {D ∈ TD : x ∈ D}. Similarly, we denote by Tsep(D)(x) the family of all separating γ-sets containing x, that is, Tsep(D)(x) = {D ∈ Tsep(D) : x ∈ D}. Lemma 8. Let D be a non-trivial division ring and x be an element of D. Then the following statements are equivalent. (i) x is an involution. (ii) TD(x) = TD. (iii) Tsep(D)(x) = Tsep(D). Proof. (1) ⇒ (2) Suppose that x is an involution and TD(x) 6= TD. If TD(x) 6= TD, then there exists a γ-set B such that x /∈ B. Since x is an involution and B is a γ-set, x = x−1 ∈ B. This is a contradiction. Hence, TD(x) = TD. (2) ⇒ (1) Suppose that TD(x) = TD and x is not an involution. Let J be a γ-set of D and x be a non-involution. Consider the following cases: Case 1. x2 6= 1D If x2 6= 1D, that is, x 6= x−1, then we note that J\{x} is a γ-set that do not contain x. This is a contradiction since TD(x) = TD. Therefore, x must be an involution. Case 2. 2x 6= 0 If 2x 6= 0, that is, x 6= −x, then we note that J\{x} is a γ-set that do not contain x. This is a contradiction since TD(x) = TD. Therefore, x must be an involution. Case 3. x = −x−1 If x = −x−1, then we note that J\{x} is a γ-set that do not contain x. This is a contradiction since TD(x) = TD. Therefore, x must be an involution. (1) ⇒ (3) Suppose that x is an involution and Tsep(D)(x) 6= Tsep(D). If Tsep(D)(x) 6= Tsep(D), then there exists a separating γ-set B such that x /∈ B. Since x is an involution and B is a γ-set, x = x−1 ∈ B. This is a contradiction. Hence, Tsep(D)(x) = Tsep(D). (3) ⇒ (1) Suppose that Tsep(D)(x) = Tsep(D) and x is not an involution. If x is not an involution, then consider the following cases: Case 1. x2 6= 1D E.J. Sigasig, C.J. Rosero, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (1) (2021), 314-326 324 If x2 6= 1D, then x 6= x−1. Let H be a separating γ-set. Then (H\{x}) ∪ {x−1} is a separating γ-set that do not contain x. This is a contradiction since Tsep(D)(x) = Tsep(D). Therefore, x must be an involution. Case 2. 2x 6= 0 If 2x 6= 0, then x 6= −x. Let H be a separating γ-set. Then (H\{x})∪{−x} is a separating γ-set that do not contain x. This is a contradiction since Tsep(D)(x) = Tsep(D). Therefore, x must be an involution. Case 3. x = −x−1 If x = −x−1, then x = (−x)−1. Let H be a separating γ-set. Then (H\{x}) ∪ {−x} is a separating γ-set that do not contain x. This is a contradiction since Tsep(D)(x) = Tsep(D). Therefore, x must be an involution. The next proposition shows that an isomorphism preserves the state of being separating in the same way as it preserves other properties. Lemma 9. Let D1 and D2 be division rings, and ϕ : TD1 → TD2 be an isomorphism. Then J is a separating γ-set of D1 if and only if ϕ(J) is a separating γ-set of D2. Proof. Let D1 and D2 be division rings, and ϕ : TD1 → TD2 be an isomorphism. Suppose that J is a separating γ-set of D1 and ϕ(J) is not a separating γ-set of D2. If ϕ(J) is not a separating γ-set of D2, then by Lemma 6, ϕ(J) = E ∪ F for some distinct separating γ-sets E and F in D2. It is easy to see that there exist distinct γ-sets E′ and F ′ such that ϕ(E′) = E and ϕ(F ′) = F . Thus, ϕ(J) = E ∪ F = ϕ(E′)∪ ϕ(F ′) = ϕ(E′ ∪ F ′). Since ϕ is injective, we have J = E′∪F ′. This is a contradiction (by Lemma 6). Therefore, ϕ(J) must be a separating γ-set of D2. Conversely, assume that ϕ(J) is a separating γ-set of H and J is not a separating γ-set of D1. If J is not a separating γ-set of D1, then by Lemma 6, J = E ∪ F for some γ-sets E and F with E 6= F . Thus, ϕ(J) = ϕ(E ∪F ) = ϕ(E)∪ϕ(F ). Since ϕ is injective, ϕ(E) 6= ϕ(F ). This is a contradiction (by Lemma 6). Therefore, J must be a separating γ-set of D1. Lemma 10. Let D1 and D2 be division rings and ϕ : TD1 → TD2 be an isomorphism. Let J be a separating γ-set of D1 and x ∈ D1\J . Then there exists a unique y ∈ D2\ϕ(J) such that ϕ(J ∪ {x}) = ϕ(J) ∪ {y}. Proof. Let D1 and D2 be division rings and ϕ : TD1 → TD2 be an isomorphism. Let J be a separating γ-set of D1 and x ∈ D1\J . If x ∈ D1\J , then x /∈ J . Note that ϕ(J) ∪ {x} 6= ϕ(J) since J is a separating γ-set and ϕ(J) ∪ {x} is not (by Lemma 9). Hence, (ϕ(J) ∪ {x})\ϕ(J) 6= ∅. Now, we claim that (ϕ(J) ∪ {x})\ϕ(J) is singleton. Suppose it is not. Without loss of generality, assume that {u, v} = (ϕ(J) ∪ {x})\ϕ(J). If {u, v} = (ϕ(J)∪{x})\ϕ(J), then u, v /∈ ϕ(J). Since ϕ(J) is a γ-set u−1, u−1 ∈ ϕ(J). Thus, A = ϕ(J), B = (ϕ(J)\{u−1})∪ {u}, and C = (ϕ(J)\{v−1})∪ {v} are three distinct sepa- rating γ-sets. Note that ϕ(J ∪ {x}) = A∪B ∪C. Hence, J ∪ {x} = J ∪ϕ−1(B)∪ϕ−1(C) where J , ϕ−1(B), ϕ−1(C) are three distinct separating γ-sets. This is a contradiction. E.J. Sigasig, C.J. Rosero, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (1) (2021), 314-326 325 Therefore, (ϕ(J) ∪ {x})\ϕ(J) must be singleton. Let y ∈ (ϕ(J) ∪ {x})\ϕ(J). Then there exists y ∈ D2\ϕ(J) such that ϕ(J ∪ {x}) = ϕ(J) ∪ {y}. The next result give necessary and sufficient conditions for two division rings to have isomorphic families of γ-set. Theorem 16. Let D1 and D2 be division rings. Then TD1 is isomorphic to TD2 if and only if there exists a bijection σ : D1\SD1 → D2\SD2. Proof. Let ϕ : TD1 → TD2 be an isomorphism. Define σ : D1\SD1 → D2\SD2 as follows. Let J be a separating γ-set of D1 and x ∈ D1\SD1 . Without loss of generality, choose x /∈ J . If x /∈ J , then x ∈ D1\J . By Lemma 10, there exists y ∈ D2\ϕ(J) with ϕ(J ∪ {x}) = ϕ(J) ∪ {y}. Now, we define σ(x) = y and σ(x−1) = y−1. We first show that σ is injective. Let a, b ∈ D1\SD1 with a 6= b. Let Jj = (J\{a}) ∪ {a−1} and Jk = (J\{b}) ∪ {b−1}. Then a ∈ D1\Jj and b ∈ D1\Jk. By Lemma 10, there exist u ∈ D2\ϕ(Jj), and v ∈ D2\ϕ(Jk) such that ϕ(Jj ∪ {a}) = ϕ(Jj) ∪ {u} and ϕ(Jk ∪ {b}) = ϕ(Jk) ∪ {v}. Without loss of generality, assume that a /∈ J and b /∈ J . If a /∈ J and b /∈ J , then σ(a) = u and σ(b) = v. In the sense of the proof of Lemma 10, ϕ(J)\ϕ(Jj) and ϕ(J)\ϕ(Jj) are singleton sets. Thus, if u = v, then ϕ(J∪{a}) = ϕ(J∪{b}). Since ϕ is an isomorphism, J ∪ {a} = J ∪ {b}. Thus, if a, b /∈ J , then a = b. This is a contradiction.This shows that σ is injective. Next, we show that σ is surjective. Let y ∈ D2\SD2 and J be a separating γ-set of D1. Without loss of generality, assume that y /∈ J . If y /∈ J , then y ∈ D2\ϕ(J). Since ϕ−1 is also an isomorphism, by Lemma 10, there exists x ∈ D1\J such that ϕ−1(ϕ(J) ∪ {y}) = J ∪ {x}, that is ϕ(J)∪ {y} = ϕ(J ∪ {x}). This implies that there exists x ∈ D1\SD1 such that σ(x) = y. This shows that σ is surjective. Accordingly, σ is bijective. For the converse, consider the bijective function σ : D1\SD1 → D2\SD2 given by σ(x) = y and σ(x−1) = y−1 where x /∈ J , and y and J are in the same sense as in the above arguments. Define ϕ : TD1 → TD2 as follows. Let J be in TD1 , then J = SD1 ∪ A for some subset A of D1\SD1 . Let ϕ(J) = SD2 ∪ σ(A). Then it is easy to show that ϕ is an isomorphism. Corollary 3. Let D1 and D2 be division rings. Then, TD1 is isomorphic to TD2 if and only if |D1\SD1 | = |D2\SD2 |. Proof. The given statement follows from Theorem 16. 8. 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