EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 1, 2023, 243-252 ISSN 1307-5543 – ejpam.com Published by New York Business Global Cliques and Supercliques in a Graph Sergio R. Canoy, Jr.1,∗, Ramil H. Dela Cerna1, Armalene Abragan1 1 Department of Mathematics and Statistics, College of Science and Mathematics, Center for Graph Theory, Algebra and Analysis-PRISM, MSU-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. A set S ⊆ V (G) of an undirected graph G is a clique if every two distinct vertices in S are adjacent. A clique is a superclique if for every pair of distinct vertices v, w ∈ S, there exists u ∈ V (G) \S such that u ∈ NG(v) \NG(w) or u ∈ NG(w) \NG(v). The maximum cardinality of a clique (resp. superclique) in G is called the clique (resp. superclique) number of G. In this paper, we determine the clique and superclique numbers of some graphs. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Clique, clique number, superclique, superclique number 1. Introduction Recently, Dela Cerna and Canoy (see [3]) initiated the study of the concept of super- clique in a graph. It is known that the superclique number of a graph is at most equal to the clique number of the graph. Moreover, it was shown that any two positive integers a and b with 2 ≤ a ≤ b are, respectively, realizable as the superclique number and clique number of a connected graph. This result also implies that the difference of the clique number and the superclique number can be made arbitrarily large. As pointed out in an earlier study, superclique and superclique number were introduced and first used in the study of Acal, Monsanto, Sumaoy, and Rara in [1], [14], and [21] when they investigated some variations of resolving domination for graphs under some binary operations. Their study was motivated by the concepts of strong resolving set, strong metric dimension, and resolving domination which were introduced and studied in [2], [16], and [19]. These lat- ter studies, in turn, came after Slater in [20] introduced the concepts of resolving set and metric dimension. The same concepts were also independently investigated by Harary and Melter in [9]. Chartrand et al. (see [4]) also studied resolving set and metric dimension of a graph. Domination and some variations of domination are found in [10]. Other studies on domination are in [11], [12], [17], and [22]. Some studies involving cliques can be found in [5], [6], [7], [8], [13], [15], [18], and [23]. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i1.4651 Email addresses: , sergio.canoy@g.msuiit.edu.ph (S. Canoy, Jr.), ramil.delacerna@g.msuiit.edu.ph (R. Dela Cerna), armalene.abragan@g.msuiit.edu.ph (A. Abragan) https://www.ejpam.com 243 © 2023 EJPAM All rights reserved. S. Canoy, Jr., R. Dela Cerna, A. Abragan / Eur. J. Pure Appl. Math, 16 (1) (2023), 243-252 244 2. Terminologies and Notations Let G = (V (G), E(G)) be a simple undirected graph. The distance between two vertices u and v of G, denoted by dG(u, v), is equal to the length of a shortest path connecting u and v. Any path connecting u and v of length dG(u, v) is called a u-v geodesic. The open neighbourhood of a vertex v of G is the set NG(v) = {u ∈ V (G) : uv ∈ E(G)} and its closed neighbourhood is the set NG[v] = NG(v) ∪ {v}. The open neighbourhood of a subset S of V (G) is the set NG(S) = ∪v∈SNG(v) and its closed neighbourhood is the set NG[S] = NG(S) ∪ S. The degree of v, denoted by degG(v), is equal to |NG(v)|. A set S ⊆ V (G) is a dominating set of G if NG[S] = V (G). The smallest cardinality of a dominating set of G, denoted by γ(G), is called the domination number of G. A dominating set of G with with cardinality γ(G) is called a γ-set of G. A set S ⊆ V (G) is a clique in a graph G if the graph G[S] = ⟨S⟩ induced by S is a complete subgraph of G. A clique C in G is called a superclique if for every pair of distinct vertices u, v ∈ C, there exists w ∈ V (G) \ C such that w ∈ NG(u) \ NG(v) or w ∈ NG(v)\NG(u). The clique number (resp. superclique number) of G, denoted by ω(G) (resp. ωs(G)), is the largest cardinality of a clique (resp. superclique) in G. Any clique (resp. superclique) in G with cardinality ω(G) (resp. ωs(G)) is called a maximum clique or ω-set (resp. maximum superclique or ωs-set). LetG andH be graphs. The edge corona G⋄H of graphsG andH is the graph obtained by taking one copy of G and |E(G)| copies of H and joining each of the end vertices u and v of every edge uv in G to every vertex of the copy Huv of H. ( that is forming the join ⟨{u, v}⟩+Huv for each uv ∈ E(G)). The Tensor product G⊠H of graphs G and H is the graph with vertex set V (G)×V (H) and (u, v) is adjacent with (u′, v′) whenever uu′ ∈ E(G) and uv′ ∈ E(H). The strong product G⊗H of graphs G and H is the graph with vertex set V (G)× V (H) and (u, v) is adjacent with (u′, v′) whenever [uu′ ∈ E(G) and v = v′] or [vv′ ∈ E(H) and u = u′] or [uu′ ∈ E(G) and vv′ ∈ E(H)]. We note that every non-empty subset C of V (G)× V (H) can be expressed as C = ∪x∈S [{x} × Tx], where S ⊆ V (G) and Tx = {a ∈ V (H) : (x, a) ∈ C} for each x ∈ S. 3. Results The first result is found in [3]. Recall that two adjacent vertices v and w of a graph G are true twins if NG[v] = NG[w]. Theorem 1. Let G be any graph. Then each of the following statements holds: (i) G admits a superclique and 1 ≤ ωs(G) ≤ ω(G). (ii) ωs(G) = 1 if and only if every component of G is complete. (iii) ωs(G) = ω(G) if and only if G has a maximum clique containing no true twin vertices. Theorem 2. Let G be a nontrivial connected graph and H be any graph. Then S is clique in G ⋄H if and only if one of the following holds: S. Canoy, Jr., R. Dela Cerna, A. Abragan / Eur. J. Pure Appl. Math, 16 (1) (2023), 243-252 245 (i) S is a clique in G. (ii) S is clique in Huv for some uv ∈ E(G). (iii) S = Suv∪D, where Suv is a clique in Huv and ∅ ̸= D ⊆ {u, v} for some uv ∈ E(G). Proof. Suppose S is a clique in G ⋄ H. If S ⊆ V (G), then S is a clique in G and (i) holds. Suppose that S ⊆ V (Huv) for some uv ∈ E(G). Since S is a clique in G ⋄H, Suv is a clique in Huv. Hence, (ii) holds. Suppose now that D = S ∩ {u, v} ≠ ∅ and Suv = S ∩ V (Huv) ̸= ∅ for some uv ∈ E(G). Then clearly, Suv is a clique in Huv and S = Suv ∪D. Thus, (iii) holds. The converse is clear. The next result is immediate from Theorem 2. Corollary 1. Let G be a nontrivial connected graph and let H be any graph. Then ω(G ⋄H) = max{ω(G), ω(H) + 2}. Theorem 3. Let G be a nontrivial connected graph such that G ̸= K2 and let H be any graph. Then S is superclique in G ⋄H if and only if one of the following holds: (i) S is a clique in G. (ii) S is superclique in Huv for some uv ∈ E(G). (iii) S = Suv ∪ D for some uv ∈ E(G), where Suv is a superclique in Huv and D is a nonempty subset of {u, v} such that D = {u} if degG(v) = 1 and D = {v} if degG(u) = 1. Proof. Suppose S is a superclique in G ⋄H. Then S is a clique in G ⋄H. If S ⊆ V (G) or S ⊆ V (Huv) for some uv ∈ E(G), then S is a clique in G or Huv, respectively, by (i) and (ii) of Theorem 2. Suppose S = Suv for some uv ∈ E(G) and let a, b ∈ Suv. Since S is a superclique in G ⋄H, there exists c ∈ V (G ⋄H) \ S such that c ∈ NG⋄H(a) \NG⋄H(b) or c ∈ NG⋄H(b) \NG⋄H(a). This implies that c ∈ V (Huv) \Suv and c ∈ NHuv(a) \NHuv(b) or c ∈ NHuv(b) \ NHuv(a). Hence, Suv is a superclique in Huv, showing that (i) or (ii) holds. Next, suppose that S ∩ {u, v} ̸= ∅ and S ∩ V (Huv) ̸= ∅ for some uv ∈ V (G). Then S = Suv ∪D, where Suv is a clique in Huv and ∅ ̸= D ⊆ {u, v} for some uv ∈ E(G), by Theorem 2(iii). Again, since S is a superclique in G ⋄ H, Suv is a superclique in Huv. Suppose now that degG(u) = 1 or degG(v) = 1, say degG(v) = 1. Pick any x ∈ Suv. Then NG⋄H(x)∩ [V (G ⋄H) \ S] = NG⋄H(u)∩ [V (G ⋄H) \ S]. Thus, u /∈ D. Therefore, |D| = 1. In particular, D = {u} showing that (iii) holds. For the converse, suppose first that (i) holds. Let u, v ∈ S with u ̸= v. Since G is connected and G ̸= K2, degG(u) ≥ 2 or degG(v) ≥ 2. Assume that degG(u) ≥ 2. Let w ∈ NG(u) \ {v} and pick any q ∈ V (Huw). Then q ∈ V (G ⋄ H) \ S and q ∈ NG⋄H(u)\NG⋄H(v). Therefore, S is a superclique in G⋄H. Next, suppose that (ii) holds. S. Canoy, Jr., R. Dela Cerna, A. Abragan / Eur. J. Pure Appl. Math, 16 (1) (2023), 243-252 246 Since S is a superclique in Huv, it is a superclique in G ⋄H. Finally, suppose that (iii) holds. By Theorem 2(iii), S is a clique in G ⋄H. Let x, y ∈ S with x ̸= y. If x, y ∈ Suv, then there exists z ∈ V (Huv \ Suv) ⊆ V (G ⋄H) \ S such that z ∈ NHuv(x) \NHuv(y) or z ∈ NHuv(y) \NHuv(x) because Suv is a superclique in Huv. Therefore, z ∈ V (G ⋄H) \ S and z ∈ NG⋄H(x) \ NG⋄H(y) or z ∈ NG⋄H(y) \ NG⋄H(x). Suppose x ∈ D and y ∈ Suv. Assume, without lost of generality, that x = u. Then degG(u) ≥ 2. Let w ∈ NG(u) \ {v}. Then w ∈ V (G ⋄H) \ S and w ∈ NG⋄H(x) \NG⋄H(y). Lastly, suppose that x, y ∈ D . In particular, let x = u and y = v. Then, by assumption, degG(u) ≥ 2 and degG(v) ≥ 2. Let z ∈ NG(u) \ {v} and choose any p ∈ V (Huz). Then p ∈ V (G ⋄ H) \ S and p ∈ NG⋄H(x) \NG⋄H(y). Therefore, in either case, S is a superclique in G ⋄H. Corollary 2. Let G be a nontrivial connected graph such that G is not a star and let H be any graph. Then ωs(G ⋄H) = max{ω(G), ωs(H) + 2}. Proof. Let S be a maximum clique in G and let u, v ∈ V (G) with degG(u) ≥ 2 and degG(v) ≥ 2 (these vertices exist because G is not a star). Let Suv be a maximum superclique in Huv. Then S and S∗ = Suv ∪ {u, v} are supercliques in G ⋄H by Theorem 3. This implies that ωs(G ⋄H) ≥ max{|S|, |S∗|} = max{ω(G), ωs(H) + 2}. On the other hand, if S0 is a maximum superclique in G ⋄H, then S0 is a clique in G or S0 = Suv ∪D for some uv ∈ E(G) satisfying the conditions in Theorem 3(iii). Hence, ωs(G ⋄H) = |S0| ≤ max{ω(G), ωs(H) + 2}, establishing the desired equality. Theorem 4. Let G = K1,m = ⟨v0⟩+Km, where m ≥ 2, and let H be any graph. Then S is a superclique in G ⋄H if and only if one of the following holds: (i) S is a clique in G. (ii) S is a superclique in Huv0 for some u ∈ V (G) \ {v0}. (iii) S = Suv0 ∪ {v0} for some u ∈ V (G) \ {v0}, where Suv0 is a superclique in Huv0. Proof. Suppose S is a superclique in G ⋄H. Then (i), (ii), or (iii) holds by Theorem 3. The converse also follows from Theorem 3. Corollary 3. Let G = K1,m, where m ≥ 2, and let H be any graph. Then ωs(G ⋄H) = ωs(H) + 1. S. Canoy, Jr., R. Dela Cerna, A. Abragan / Eur. J. Pure Appl. Math, 16 (1) (2023), 243-252 247 Proof. Clearly, ω(G) = 2. Since ωs(H) ≥ 1, it follows that ω(G) ≤ ωs(H) + 1. The desired equality now follows from Theorem 4. Theorem 5. Let G and H be nontrivial connected graphs. Then C = ∪x∈S({x} × Tx), where S ⊆ V (G) and Tx ⊆ V (H) for each x ∈ S, is a clique in G ⊠H if and only if the following statements hold: (i) S is a clique in G. (ii) |Tx| = 1 for each x ∈ S, Tx ̸= Ty if x ̸= y, and ∪x∈STx is a clique in H. Proof. Suppose C is a clique in G⊠H. Let x, y ∈ S such that x ̸= y and let a ∈ Tx and b ∈ Ty. Then (x, a)(y, b) ∈ C and (x, a) ̸= (y, b). It follows that (x, a)(y, b) ∈ E(G ⊠H). Hence, xy ∈ E(G), showing that S is a clique in G. Thus, (i) holds. Next, let x ∈ S and suppose that |Tx| ≥ 2. Let p, q ∈ Tx such that p ̸= q. Then (x, p)(x, q) ∈ C and (x, p) ̸= (x, q). Since C is a clique in G⊠H, (x, p)(x, q) ∈ E(G⊠H) which is not possible. It follows that |Tx| = 1 for each x ∈ S. Now, let s, t ∈ ∪x∈STx such that s ̸= t. Then s ∈ Tv and t ∈ Tw for some v, w ∈ S with v ̸= w. Since (v, s), (w, t) ∈ C and C is a clique in G ⊠ H, (v, s)(w, t) ∈ E(G ⊠ H). This implies that st ∈ E(H). Therefore, ∪x∈STx is a clique in H, showing that (ii) holds. Conversely, suppose that C satisfies (i) and (ii). Let (v, p), (w, q) ∈ C such that (v, p) ̸= (w, q). If v = w, then p ̸= q and p, q ∈ Tv contrary to the assumption that |Tx| = 1 for all x ∈ S. Thus, v ̸= w and p ̸= q. Since S and ∪x∈STx are cliques in G and H, respectively, vw ∈ E(G) and pq ∈ E(H). Thus, (v, p), (w, q) ∈ E(G ⊠H). Therefore, C is a clique in (G⊠H). Corollary 4. Let G and H be nontrivial connected graphs. Then ω(G⊠H) = min{ω(G), ω(H)}. Proof. Let S and D be maximum cliques in G and H. Suppose first that ω(G) = |S| ≤ |D| = ω(H). Let D′ ⊆ D such that |S| = |D′|. Let S = {v1, v2, . . . , vk} and D′ = {a1, a2, . . . , ak}. Set Tvj = {aj} for each j ∈ [k]. Then C = ∪k j=1({vj} × Tvj ) is a clique in G ⊠H by Theorem 5. Consequently, ω(G ⊠H) ≥ |C| = |S| = ω(G). A similar argument can be used to show that ω(G⊠H) ≥ |D| = ω(G) if |D| ≤ |S|. Suppose now that C0 = ∪x∈S0({x} ×Rx) is a maximum clique in G⊠H. Then S is a clique in G, |Rx| = 1 for each x ∈ S, Rx ̸= Ry if x ̸= y, and ∪x∈SRx is a clique in H by Theorem 5. It follows that ω(G⊠H) = |C0| = ∑ x∈S |Rx| = |S| = | ∪x∈S Rx| ≤ min{ω(G), ω(H)}. This establishes the desired equality. S. Canoy, Jr., R. Dela Cerna, A. Abragan / Eur. J. Pure Appl. Math, 16 (1) (2023), 243-252 248 Theorem 6. Let G and H be nontrivial connected graphs such that G ̸= K2 or H ̸= K2. Then C = ∪x∈S({x} × Tx), where S ⊆ V (G) and Tx ⊆ V (H) for each x ∈ S, is a superclique in G⊠H if and only if it is a clique. Proof. Since every superclique is a clique, it remains to show that the converse is true. To this end, suppose that C is a clique in G⊠H. Then C satisfies (i) and (ii) of Theorem 5. Let (v, p), (w, q) ∈ C such that (v, p) ̸= (w, q). Then (v, p), (w, q) ∈ C. Hence, vw ∈ E(G) and pq ∈ E(H). Suppose G ̸= K2. Since G is connected, degG(v) ≥ 2 or degG(w) ≥ 2. Assume that degG(v) ≥ 2 and let z ∈ NG(v) \ {w}. Since C is a clique and (z, q)(w, q) /∈ E(G⊠H), it follows that (z, q) /∈ C. Hence, (z, q) ∈ NG⊠H((v, p)) \NG⊠H((w, q)). Next, suppose that H ̸= K2. Then degH(p) ≥ 2 or degH(q) ≥ 2. Assume that degH(p) ≥ 2 and let t ∈ NH(p) \ {q}. Since C is a clique and (w, t)(w, q) /∈ E(G ⊠H), it follows that (w, t) /∈ C. Hence, (w, t) ∈ NG⊠H((v, p))\NG⊠H((w, q)). In either case, C is a superclique in G⊠H. It is clear that ωs(K2 ⊠K2) = ωs(K2 ∪K2) = 1. Corollary 5. Let G and H be nontrivial connected graphs such that G ̸= K2 or H ̸= K2. Then ωs(G⊠H) = ω(G⊠H) = min{ω(G), ω(H)}. Theorem 7. Let G and H be nontrivial connected graphs. Then C = ∪x∈S({x} × Tx), where S ⊆ V (G) and Tx ⊆ V (H) for each x ∈ S, is a clique in G ⊗H if and only if the following statements hold: (i) S is a clique in G. (ii) Tx is a clique in H for each x ∈ S. (iii) ∪x∈STx is a clique in H. Proof. Suppose C is a clique in G⊗H. Let x, y ∈ S such that x ̸= y and let a ∈ Tx and b ∈ Ty. Then (x, a)(y, b) ∈ C and (x, a) ̸= (y, b). By assumption, (x, a)(y, b) ∈ E(G⊗H). This implies that xy ∈ E(G), showing that (i) is true. Let x ∈ S and let p, q ∈ Tx such that p ̸= q. Then (x, p)(x, q) ∈ C and (x, p) ̸= (x, q). Since C is a clique in G⊗H, (x, p)(x, q) ∈ E(G⊗H). Adjacency in G⊗H would imply that pq ∈ E(H). This shows that Tx is a clique in H as asserted in (ii). Next, let s, t ∈ ∪x∈STx such that s ̸= t. If s, t ∈ Tx for x ∈ S, then st ∈ E(H) because Tx is a clique. Suppose s ∈ Tv and t ∈ Tw for some v, w ∈ S with v ̸= w. Since S is a clique in G, vw ∈ E(G). Also, since C is a clique in G⊗H, (v, s)(w, t) ∈ E(G⊗H). Consequently, st ∈ E(H) by the definition of the adjacency in G⊗H. This proves (iii). For the converse, suppose that C satisfies (i), (ii), and (iii). Let (x, p), (y, q) ∈ C such that (x, p) ̸= (y, q). If x = y, then p ̸= q and p, q ∈ Tx. It follows from (ii) that pq ∈ E(H). Hence, (x, p)(y, q) ∈ E(G ⊗ H). Suppose now that x ̸= y. Then xy ∈ E(G) by (i). If p = q, then (x, p)(y, q) ∈ E(G ⊗ H) by the definition of G ⊗ H. Suppose p ̸= q. The assumption that (iii) holds would imply that Tx ∪ Ty is a clique in H. Thus, pq ∈ E(G) and (x, p)(y, q) ∈ E(G⊗H). This proves that C is a clique in G⊗H. S. Canoy, Jr., R. Dela Cerna, A. Abragan / Eur. J. Pure Appl. Math, 16 (1) (2023), 243-252 249 Corollary 6. Let G and H be nontrivial connected graphs. Then ω(G⊗H) = ω(G)ω(H). Proof. Let S and D be maximum cliques in G and H, respectively. Set Tx = D for each x ∈ S. Then C = ∪x∈S({x} × Tx) = S × D is a clique in G ⊗ H by Theorem 7. Therefore, ω(G⊗H) ≥ |C| = |S||D| = ω(G)ω(H). Next, suppose that C0 = ∪x∈S0({x} × Rx) is a maximum clique in G ⊗H. Then C0 satisfies properties (i), (ii), and (iii) of Theorem 7. It follows that ω(G⊗H) = |C0| = ∑ x∈S0 |Rx| ≤ |S0|ω(H) ≤ ω(G)ω(H). This establishes the desired equality. Theorem 8. Let G and H be non-trivial connected graphs. Then C = ∪x∈S({x} × Tx), where S ⊆ V (G) and Tx ⊆ V (H) for each x ∈ S, is a superclique in G⊗H if and only if it satisfies the following conditions: (i) S is a clique in G. (ii) Tx is a superclique in H for each x ∈ S. (iii) ∪x∈STx is a clique in H. (iv) For each distinct pairs of vertices v, w ∈ S such that Tw ∩ Tv ̸= ∅, there exists u ∈ V (G) \ S such that u ∈ NG(v) \NG(w) or u ∈ NG(w) \NG(v). (v) For each pair of distinct vertices v, w ∈ S such that NG[v] = NG[w], and for each distinct vertices a and b such that a ∈ Tv and b ∈ Tw, there exists c ∈ V (H) such that [c /∈ Tv and c ∈ NH(a) \NH(b)] or [c /∈ Tw and c ∈ NH(b) \NH(a)]. Proof. Suppose C is superclique in G⊗H. Since C is a clique, each Tx is clique in H, and (i) and (iii) hold by Theorem 7. Let x ∈ S and p, q ∈ Tx such that p ̸= q. Since Tx is a clique, pq ∈ E(H). Also, since C is a superclique in ⊗H, we may assume that there exists (u, t) ∈ V (G⊗H) \C such that (u, t) ∈ NG⊗H((x, p)) \NG⊗H((x, q)). Suppose first that u ̸= x. Then ux ∈ E(G). If p = t, then (u, t)(x, q) ∈ E(G ⊗ H), a contradiction. Thus, pt ∈ E(H) because (u, t) ∈ NG⊗H((x, p)). Since (u, t) /∈ NG⊗H((x, q)), t ̸= q and tq /∈ E(H). This implies that t ∈ V (H) \ Tx and t ∈ NH(p) \ NH(q). Next, suppose that u = x. Then pt ∈ E(H) and qt /∈ E(H). It follows that t ∈ V (H) \ Tx and t ∈ NH(p)\NH(q). In either case, Tx is a superclique in H, showing that (ii) holds. Next, suppose that v, w ∈ S such that v ̸= w and Tv∩Tw ̸= ∅, say p ∈ Tv∩Tw. Again, since C is a superclique and (v, p), (w, p) ∈ C, we may assume that there exists (z, s) ∈ V (G⊗H)\C such that (z, s) ∈ NG⊗H((v, p)) \NG⊗H((w, p)). Suppose z = v. Then ps ∈ E(H) because (z, s) ∈ NG⊗H((v, p)). Further, since vw ∈ E(G), (z, s) ∈ NG⊗H((w, p)), a contradiction. Thus, z ̸= v. This implies that zv ∈ E(G). Suppose zw ∈ E(G). If p = s, then S. Canoy, Jr., R. Dela Cerna, A. Abragan / Eur. J. Pure Appl. Math, 16 (1) (2023), 243-252 250 (z, s) ∈ NG⊗H((w, p)). If p ̸= s, then ps ∈ E(H) since (z, s) ∈ NG⊗H((v, p)). Hence, (z, s) ∈ NG⊗H((w, p)). In either case, we get a contradiction. Therefore, zw /∈ E(G), that is, z ∈ V (G) \ S and z ∈ NG(v) \NG(w), showing that (iv) holds. For the converse, suppose that C satisfies (i), (ii), (iii), and (iv). Then C is a clique by Theorem 7. Let (x, a), (z, b) ∈ C such that (x, a) ̸= (z, b). Consider the following cases: Case 1. x = z. Then a, b ∈ Tx and a ̸= b Since Tx is a clique, ab ∈ E(H). Moreover, since Tx is a su- perclique, we may assume that there exists p ∈ V (H) \ Tx such that p ∈ NH(a) \NH(b). Hence, (x, p) ∈ V (G⊗H) \ C and (x, p) ∈ NG⊗H((x, a)) \NG⊗H((x, b)). Case 2. x ̸= z. Then xz ∈ E(G) because S is a clique in G. Suppose a = b. Then a ∈ Tx ∩ Tz. By (iv), we may assume that there exists y ∈ V (G) \ S such that y ∈ NG(x) \ NG(z). It follows that (y, a) ∈ V (G ⊗H) \ C and (y, a) ∈ NG⊗H(x, a) \NG⊗H(z, b). Finally, suppose that a ̸= b. Then ab ∈ E(G) by (iii). Suppose NG[x] ̸= NG[z]. We may assume that there exists v ∈ NG(x) \ NG(z). Clearly, v ∈ V (G) \ S. Hence, (v, a) ∈ V (G ⊗ H) \ C and (v, a) ∈ NG⊗H(x, a) \NG⊗H(z, b). If NG[x] = NG[z], then there exists c ∈ V (H) such that [c ∈ V (H)\Tx and c ∈ NH(a)\NH(b)] or [c ∈ V (H)\Tz and c ∈ NH(b)\NH(a)] by property (v). Assume that c ∈ V (H) \ Tx and c ∈ NH(a) \NH(b). Then (x, c) ∈ V (G⊗H) \C and (x, c) ∈ NG⊗H(x, a) \NG⊗H(z, b). Accordingly, C is a superclique in G⊗H. Corollary 7. Let G and H be non-trivial connected graphs. Then ωs(G)ωs(H) ≤ ωs(G⊗H) ≤ ω(G)ωs(H). Moreover, if ωs(G) = ω(G), then ωs(G⊗H) = ωs(G)ωs(H). Proof. Let S and D be ωs-sets in G and H, repectively. Set Tx = D for each x ∈ S. Then C = ∪x∈S({x} × Tx) = S ×D is a superclique in G⊗H by Theorem 8. Therefore, ωs(G⊗H) ≥ |C| = |S||D| = ωs(G)ωs(H). On the other hand, suppose that C ′ = ∪x∈S′({x} ×Dx) is an ωs-set in G⊗H. Then S′ is a clique in G and each Dx is a superclique in H by Theorem 8. It follows that ωs(G⊗H) = |C ′| ≤ ω(G)ωs(H). This proves the assertion. Example 1. For any two positive integers m ≥ 3 and n ≥ 3, ωs(Pm ⊗ Pn) = 4 = ωs(Pm)ωs(Pn) = ω(Pm)ωs(Pn). Note that ωs(P2 ⊗ P3) = ωs(P2)ωs(P3) = 2 ̸= 4 = ω(P2)ωs(P3). REFERENCES 251 4. Conclusion Cliques and supercliques in the edge corona, Tensor product, and strong product of two graphs have been characterized and the corresponding clique and supercliques numbers have been described. These concepts can be studied further for other graphs. Moreover, it may be interesting to investigate the complexity of the superclique problem. Acknowledgment: The authors would like to thank the referees for the suggestions and comments that helped improve the paper. 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