Classification of groups with toroidal coprime graphs

AUSTRALASIAN JOURNAL OF COMBINATORICS Volume 69(2) (217), Pages 17 183 Classification of groups with toroidal coprime graphs K. Selvakumar M. Subajini Department of Mathematics Manonmaniam Sundaranar University Tirunelveli 627, Tamil Nadu India selva 158@yahoo.co.in subashini.m5@gmail.com Abstract The coprime graph of a group G, denoted by Γ G, is a graph whose vertices are elements of G and two distinct vertices x and y are adjacent if and only if ( x, y ) = 1. In this paper we discuss some basic properties of Γ G and try to classify all finite groups whose coprime graphs are toroidal and projective. 1 Introduction In order to get a better understanding of a given algebraic structure A, onecan associate to it a graph G and study an interplay of algebraic properties of A and combinatorial properties of G. In particular there are many ways to associate a graph to a group. For example, commuting graph of groups, non-commuting graph of groups, non-cyclic graph of groups and generating graph of groups have attracted many researchers towards this dimension. One can refer to [1, 2, 3, 6, 1, 15] for such studies. Let G be a finite group. One can associate a graph to G in many different ways. Since the order of an element is one of the most basic concepts of group theory, X. Ma et al. [13] defined the coprime graph of a group G, denoted by Γ G, as follows: take G as the vertices of Γ G and two distinct vertices x and y are adjacent if and only if ( x, y ) = 1. In this paper, we discuss some basic properties of Γ G and try to classify all finite groups whose coprime graphs are toroidal and projective. Now recall some definitions of graph theory which are necessary in this paper. AgraphGin which each pair of distinct vertices is joined by an edge is called a complete graph. We use K n to denote the complete graph with n vertices. A subset ΩofV (G) is called a clique if the induced subgraph of Ω is complete. The order of the largest clique in G is its clique number, which is denoted by ω(g). The chromatic number of G, χ(g), is the minimum k for which there is an assignment of k colors, 1,...k, to the vertices of G such that adjacent vertices have different colors. An

K. SELVAKUMAR ET AL. / AUSTRALAS. J. COMBIN. 69 (2) (217), 17 183 175 r-partite graph is one whose vertex set can be partitioned into r subsets so that no edge has both ends in any one subset. A complete r-partite graph is one in which each vertex is joined to every vertex that is not in the same subset. A complete bipartite graph (2-partite graph) with part sizes m and n is denoted by K m,n. If G = K 1,n where n 1, then G is a star graph. A split graph is a simple graph in which the vertices can be partitioned into a clique and an independent set. A graph G is said to be unicycle graph if it contains a unique cycle. A tree is a connected acyclic graph. A graph is outerplanar if it can be embedded into the plane so that all its vertices lie on the same face. The genus of a graph is the minimal integer t such that the graph can be drawn without crossing itself on a sphere with t handles (that is an oriented surface of genus t). Thus a planar graph has genus zero, because it can be drawn on a sphere without self-crossing. A genus one graph is called a toroidal graph. In other words, a graph G is toroidal if it can be embedded on a torus, this means that, the graph s vertices can be placed on a torus such that no edges cross. Usually it is assumed that G is also non-planar. A subdivision of a graph is any graph that can be obtained from the original graph by replacing edges by paths. A remarkable characterization of planar graphs was given by Kuratowski in 193. Kuratowski s Theorem says that a graph G is planar if and only if it contains no subdivision of K 5 or K 3,3. Let G be a group and N and H be subgroups of G with N normal (but H is not necessarily normal in G) andn H = {e} and every g G can be written uniquely as g = nh, wheren N,h H. Define φ : H G,byφ(h) =h = hn, which N is an isomorphism. Then one can construct G = N φ H a new group called the semidirect product of N and H. Let π(n) be the set of prime divisors of a natural number n, and denote π( G ) byπ(g). Throughout this paper, we assume that G is a finite group. We denote the group of integers addition modulo n by Z n.forbasic definitions on groups, one may refer to [9]. In this paper, we will prove the following main results. Theorem 1.1. Let G be a finite group. Then γ(γ G )=1if and only if G is isomorphic to one of the following groups: S 3, Z 2 Z 6, Z, Z 15, Z 2, Z 2 Z 1, or Z 21. Theorem 1.2. Let G be a finite group. Then γ(γ G )=1if and only if G is isomorphic to S 3, Z, Z 6 Z 2, or Z 15. 2 Some basic properties of Γ G In this section, we study some fundamental properties of the coprime graph. Remark 2.1. Let G be a p-group, where p is prime. Then for any two non-identity elements x, y G, ( x, y ) > 1 and so the subgraph induced by G {e} in Γ G is isomorphic to K G 1. Hence Γ G = K1, G 1. The characterization for split graphs was given by S. Földers et al. [11]. Using this characterization, we charcterize all finite groups G whose Γ G is a split graph.

K. SELVAKUMAR ET AL. / AUSTRALAS. J. COMBIN. 69 (2) (217), 17 183 176 Theorem 2.2. [11, Theorem 6.3] Let G be a connected graph. Then G is a split graph if and only if G contains no induced subgraph isomorphic to 2K 2, C, C 5. Theorem 2.3. Let G be a finite group. Then Γ G is a split graph if and only if G is a p-group or G = Z 2 Q where Q is a q-group with q odd. Proof. Assume that Γ G is a split graph. If G is a p-group, then clearly Γ G is a split graph. Suppose G is not a p-group. Then G = p α 1 1 p α 2 2 p α k k where p i s are distinct primes, p 1 <p 2 < <p k, k 2andα i 1 for all i. LetS pi be the union of Sylow p i -subgroups of G. Ifk 3, then every vertex in S p2 {e} is adjacent to every vertex in S p3 {e} and so Γ G contains C as a subgraph, a contradiction. Hence k =2. If S p1 1 2, then every vertex in S p1 {e} is adjacent to every vertex in S p2 {e} so that C as a induced subgraph of Γ G.Thusp α 1 1 1 S p1 1 < 2, p α 1 1 =2and so G =2p α 2 2. Thus the Sylow 2-subgroup and Sylow p 2 -subgroup are normal and hence G = Z 2 Q, where Q is a p 2 -group. The converse is clear. The following theorem is used in the subsequent theorem. Theorem 2.. [1] χ(γ G )=ω(γ G )=π(g)+1. Theorem 2.5. Let G be a finite group. Then Γ G is not a unicycle graph. Proof. Assume that Γ G is unicycle. Clearly G is not a p-group. Therefore G = p α 1 1 p α 2 2 p αn n,wherep 1,p 2,...,p n are distinct prime integers, n 2, p i <p j for i<jand α i 1fori {1, 2,...,n}. Let a i be an element of order p i. Then Γ G contains two cycles e a 1 a 2 e and e a 1 a 1 2 e, which is a contradiction. Theorem 2.6. Let G be a finite group. Then Γ G is a tree if and only if G is isomorphic to a p-group. Proof. Assume that Γ G is a tree. Suppose G is not a p-group. Then at least two prime integers divides G. By Theorem 2., ω(γ G ) 3andsoΓ G contains a cycle, a contradiction. Thus G is isomorphic to a p-group. Conversely, if G is a p-group, then Γ G = K1, G 1 and hence Γ G is tree. The following characterization of outerplanar graphs was given by Chartrand and Harary [8]. Using this characterization, we charcterize all finite groups G whose Γ G is outerplanar. Theorem 2.7. [8] A graph G is outerplanar if and only if it contains no subdivision of K or K 2,3. Theorem 2.8. Let G be a finite group. Then Γ G is outerplanar if and only if G is a p-group or G = Z 6.

K. SELVAKUMAR ET AL. / AUSTRALAS. J. COMBIN. 69 (2) (217), 17 183 177 Proof. Assume that Γ G is outerplanar. Suppose G is not a p-group. Let G = p α 1 1 p α 2 2 p α k k,wherep i s are distinct primes with p 1 <p 2 < <p k, k 2and α i 1. If k 3, then K 2,3 is a subgraph of Γ G, a contradiction. Therefore k =2. Suppose α i > 1forsomei, let it be α 1, then Γ G contains K 2,3 as a subgraph, a contradiction. Hence α 1 = α 2 =1. Ifp i 5forsomei, thenk 2,3 is a subgraph of Γ G, which is again a contradiction. Thus G =6andsoG = Z 6 or S 3. But Γ S3 = K1,2,3, which is not possible. Thus G = Z 6. 2 1 5 3 Fig. 1: Γ Z6 Conversely, if G is a p-group or Z 6,thenΓ G is isomorphic to either K 1, G 1 or the graph as given in Fig. 1. Lemma 2.9. Let G be a group of order p α 1 1 pα 2 2 pα k k integers for i {1,...,k} and k 2, then Γ G K 1,p α 1 1 1,pα 2 2 1,...,pα k k 1. where p i s are distinct prime has a subgraph isomorphic to Proof. Note that G has a Sylow p i -subgroup of order p α i i for every i {1, 2,...,k}. Also every element of Sylow p i -subgroup is adjacent to every element of Sylow p j - subgroup for all i j, which completes the proof. Lemma 2.1. Let G be a finite cyclic group and not a p-group. Then Γ G is the union of K φ(d1 ),φ(d 2 ),...,φ(d k ) where d i s are divisors of G and (d i,d j )=1for all i j. Proof. It is straight forward. A subset S of a graph G is said to be an independent set if no two vertices in S are adjacent. The independent number α(g) is the number of vertices in the largest independent set in G. Theorem 2.11. Let G be a group of order p α 1 1 p α 2 2 p αn n, where p i s are distinct primes, n 1. Thenα(Γ G ) max{ n pi : i =1,...,n} where n pi = {x G : p i x }. Moreover if G is cyclic, then α(γ G )=max{ n pi : i =1,...,n}. Proof. Consider the set n pi = {x G : p i x }, 1 i n. Since ( x, y ) 1for every x, y n pi, x and y are not adjacent in Γ G. Therefore n pi s are independent sets of Γ G and by defintion of independent number, α(γ G ) max{ n pi : i =1,...,n}.

K. SELVAKUMAR ET AL. / AUSTRALAS. J. COMBIN. 69 (2) (217), 17 183 178 Remark 2.. Let G = Z 2 Z 2 Z 3 Z 5. Let S be the collection of set of all x G such that x =6, 1, 15 or 3. In Γ G, n p5 = 8, which is maximum. But S =5> n p5. 3 Proof of Theorem 1.1 The main goal of this section is to determine all finite groups G whose coprime graph has genus one. Dorbidi [1] determine the finite groups G for which Γ G is planar. The following observation proved by Dorbidi [1] is used frequently in this article and hence given below. Theorem 3.1. [1, Theorem 3.6] Let G be a finite group. Then Γ G is a planar graph if and only if G is a p-group or G = Z 2 Q where Q is a q-group. It is well known that any compact surface is either homeomorphic to a sphere, or to a connected sum of g tori, or to a connected sum of k projective planes (see [, Theorem 5.1]). We denote by S g the surface formed by a connected sum of g tori, and by N k the one formed by a connected sum of k projective planes. The number g is called the genus of the surface S g and k is called the crosscap of N k. When considering the orientability, the surfaces S g and sphere are among the orientable class and the surfaces N k are among the non-orientable one. A simple graph which can be embedded in S g but not in S g 1 is called a graph of genus g. Similarly, if it can be embedded in N k but not in N k 1,thenwecall it a graph of crosscap k. The notations γ(g) andγ(g) are denoted for the genus and crosscap of a graph G, respectively. It is easy to see that γ(h) γ(g) and γ(h) γ(g) for all subgraph H of G. AlsoagraphG is called planar if γ(g) =, and it is called toroidal if γ(g) =1. For a rational number q, q is the first integer number greater or equal than q. In the following lemma we bring some well-known formulas for genus of a graph (see [5]). Lemma 3.2. The following statements hold: (i) γ(k n )= 1 (n 3)(n ) if n 3; (ii) γ(k m,n )= 1 (m 2)(n 2) if m, n 2. If G is a graph and V (G) ={x V (G) :deg(x) =1}, then we use G for the subgraph G V and call it the reduction of G. Then we can easily observe that γ(g) =γ(g ). Proof of Theorem 1.1. Assume that γ(γ G ) = 1. Then by Theorem 3.1, G is not a p-group. Let G = p α 1 1 pα 2 2 pα k k,wherep i s are distinct primes with p 1 <p 2 < < p k, k 2andα i 1. If k, then by Lemma 2.9, K 1,1,2,,6 is a subgraph of Γ G and so by Lemma 3.2, γ(γ G ) > 2, which is a contradiction. Therefore k 3.

K. SELVAKUMAR ET AL. / AUSTRALAS. J. COMBIN. 69 (2) (217), 17 183 179 Case 1. k =2. Then G = p α 1 1 p α 2 2. Suppose G is an odd integer. If α i > 1forsomei, letitbeα 1, then by Lemma 2.9, K 1,8, is a subgraph of Γ G and hence γ(γ G ) 3. Thus α i =1fori =1, 2. If p i > 7forsomei, then by Lemma 2.9, Γ G contains a subgraph isomorphic to K 1,2,1 and hence γ(γ G ) 2, a contradiction. Therefore p i 7andso G =15, 21 or 35. If G =35, then by Lemma 2.9, K 1,,6 is a subgraph of Γ G, which is a contradiction. If G =15, then G is isomorphic to Z 15. If G =21, then G is isomorphic to Z 21 or Z 7 Z 3. Suppose G = Z 7 Z 3 = x, y x 3 = y 7 =1,x 1 yx = y 2. Consider the vertex sets W 1 = {xi y j :1 i<3, j<7} and W 2 = y {1}. Since the order of every element in W 1 is 3 and the order of every element in W 2 is 7, every element in W 1 is adjacent to every element in W 2 and so Γ G contains a subgraph isomorphic to K 1,6.Thusγ(Γ Z7 Z 3 ), which is a contradiction. Hence G = Z 21. Suppose G is an even integer. Then G =2 α 1 p α 2 2. If α 1 =1, then the Sylow 2-subgroup is not normal because if it is normal then Γ G is planar. Therefore in this case, consider the Sylow 2-subgroup is not normal. If α i 3forsomei, then K 3,7 is a subgraph of Γ G and so γ(γ G ) 2, which is a contradiction and hence α i 2fori =1, 2. Suppose α 1 = α 2 =2. Then by Lemma 2.9, Γ G contains K 1,3,7 as a subgraph, a contradiction. (i) Consider α 1 =2andα 2 = 1. Suppose that p 2 7. Then by Lemma 2.9, K 1,3,6 is a subgraph of Γ G, a contradiction. Therefore p 2 < 7andso G = or 2. If G =, then G is isomorphic to one of the following groups: Z 3 Z, Z,A,D, Z 6 Z 2. Suppose G = Z 3 Z = x, y x = y 3 =1,x 1 yx = y 1. Consider the vertex sets V 1 = {x i y j :1 i 3, j 2} {x 2 y, x 2 y 2 } and V 2 = y. Since every element of V 1 is adjacent to every element of V 2, K 3,7 is a subgraph of Γ G and so γ(γ G ) 2. Consider G = A.SinceA contains3elementsoforder2and8elementsoforder 3, Γ A contains an induced subgraph isomorphic to K 3,8 induced by these elements and hence γ(γ A ) 2. Suppose G = D = r, s r 6 = s 2 =1,sr = r 1 s. Let us consider the vertex sets S 1 = {r 2,r, 1} and S 2 = {r 3 } {sr i : i 5}. Since every vertex of S 1 is adjacent to every vertex of S 2,Γ G contains a subgraph isomorphic to K 3,7 so that γ(γ D ) γ(k 3,7 )=2. HenceG = Z or Z 6 Z 2. If G =2. Then G is isomorphic to one of the following groups: Z 5 1 Z, Z 2, Z 5 2 Z,D 2,or Z 1 Z 2. If G = Z 5 1 Z = x, y x = y 5 =1,x 1 yx = y 1, then consider the vertex sets A 1 =( x {1}) {xy, xy 2 } and A 2 = y {1}. It is easily seen that A 1 and A 2 induces a subgraph isomorphic to K,5 and so γ(γ Z5 1 Z ) 2.

K. SELVAKUMAR ET AL. / AUSTRALAS. J. COMBIN. 69 (2) (217), 17 183 18 If G = Z 5 2 Z = x, y x = y 5 =1,x 1 yx = y 2, then Γ Z5 2 Z contains a subgraph isomorphic to K,5 induced by the vertex sets A 1 and A 2 where A 1 = {x 2 y i : i } and A 2 = y {1}. Therefore γ(γ Z 5 2 Z ) 2. Suppose G = D 2 = r, s r 1 = s 2 =1,sr= r 1 s. It is easily seen that Γ D is a subgraph of Γ D2. Therefore γ(γ G ), a contradiction. Hence G = Z 2 or Z 2 Z 1. (ii) Suppose α 1 = 1 and α 2 = 2. Then clearly K 3,8 is a subgraph of Γ G, a contradiction. (iii) Suppose α 1 = α 2 =1. Ifp i 11, then Γ G contains a copy of K 3,1. Hence γ(γ G ) 2, a contradiction and so p i < 11 for i =1, 2. In this case the possible orders of G are 6, 1 and 1. If G =1, then G = Z 1 or D 1. By Theorem 3.1, G Z 1. If G = D 1 = r, s r 5 = s 2 =1,sr= r 1 s, thenk,5 is a subgraph of Γ D1 induced by the vertex sets S 1 and S 2 where S 1 = r {1} and S 2 = {sr i i }. Thusγ(Γ G ) 2, a contradiction. If G =1, then G = Z 1 or D 1. By Theorem 3.1, Γ Z1 is planar. Suppose G = D 1 = r, s r 7 = s 2 =1,sr= r 1 s. It is clear that Γ D1 is a subgraph of Γ D1. Therefore γ(γ D1 ) 2, which is a contradiction. Hence G = S 3. Case 2. Suppose k =3. Ifp i 7forsomei, then by Lemma 2.9, Γ G contains a induced subgraph isomorphic to K 1,1,2,6. So γ(γ G ) 2 and hence p i 5 for all i. Since p i <p j for i<j, G =2 α 1 3 α 2 5 α 3. If α i 2forsomei, let it be α 1, then Γ G contains K 1,3,2, as a subgraph, a contradiction. Thus α i =1foralliand so G = 3. Hence G is isomorphic to one of the following groups: Z 3, Z 3 D 1,D 3, or Z 5 S 3. Consider G = Z 3. Since G is cyclic and by Lemma 2.1, 2K 3, as a subgraph of Γ G formed by the vertex sets {V 1,V 2 } and {U 1,U 2 } where V 1 contains elements of order 5, V 2 contains elements of order 2 and 6 and U 1 contains elements of order 1, U 2 contains elements of order 3 and identity. Hence γ(γ G ) 2. Consider the graph Γ Z3 D 1. Since Γ D1 is a subgraph of Γ Z3 D 1, γ(γ Z3 D 1 ) γ(γ D1 ) 2. Suppose G = D 3. It is clear that Γ D1 is a subgraph of Γ D3. Therefore γ(γ G ) 2, a contradiction. Consider the group G = Z 5 S 3. Let us consider the vertex sets W 1 = {a i G : a i =15} and W 2 = {b i G : b i =2}. Then it is easily seen that W 1 =8and W 2 = 3 and every vertex in W 1 is adjacent to every vertex in W 2.Thus{W 1,W 2 } induced K 3,8 as a subgraph of Γ G and hence γ(γ Z5 S 3 ) 2, a contradiction. Conversely, suppose G is isomorphic to one of the following groups: S 3, Z 2 Z 6, Z, Z 15, Z 2, Z 2 Z 1 and Z 21. If G = Z 15, then by Lemma 2.1, K 3, is a subgraph of the graph induced by the vertex sets V 1 whose elements of order 3 and identity and V 2 whose elements of

K. SELVAKUMAR ET AL. / AUSTRALAS. J. COMBIN. 69 (2) (217), 17 183 181 order 5 and hence γ(γ Z15 ) 1. Consider Γ Z 15 =Γ Z15 {x Z 15 : x =15}. Then the embedding in Fig. 2 explicitly shows that γ(γ Z15 )=1. 6 9 3 3 5 3 3 a 6 9 18 7 a 1 15 1 15 3 6 9 3 3 3 6 9 18 Fig. 2: Embedding of Γ Z 15 Fig. 3: Embedding of Γ Z 21 9 3 6 9 16 8 a 5 1 16 8 8 b 15 b 3 6 8 a Fig. : Embedding of Γ Z Fig. 5: Embedding of Γ Z 2 Suppose G = Z 21, then by Lemma 2.9, K 1,2,6 is a subgraph of Γ G and hence γ(γ Z21 ) 1. Let Γ Z 21 =Γ Z21 {x Z 21 : x =21} and the embedding in Fig. 3 explicitly shows that γ(γ Z15 )=1. Consider the graph Γ Z. Since Z is cyclic, by Lemma 2.1, Γ Z contains a subgraph isomorphic to K 3,3 and hence γ(γ Z ) 1. Here Γ Z =Γ Z {x Z : x =or6}. Then the embedding in Fig. explicitly shows that γ(γ Z )=1. If G = Z 2 Z 6, then it is easily seen that Γ Z = ΓZ2 Z 6. Therefore γ(γ Z2 Z 6 )=1. Consider G = Z 2.SinceGis cyclic, by Lemma 2.1, K, as a subgraph induced by the vertex sets Ω 1 and Ω 2 where Ω 1 contains elements of order 5 and Ω 2 contains elements of order 2, and identity. Hence γ(γ G ) 1. Consider Γ G =Γ G {x G : x =2or1}. Then the embedding in Fig. 5 explicitly shows that γ(γ G )=1. Suppose G = Z 2 Z 1, then it is easily seen that Γ Z2 = ΓZ2 Z 1. Hence γ(γ Z2 Z 1 )=1. If G = S 3, then Γ S3 = K1,2,3.AlsoΓ S3 is a subgraph of Γ Z,thenγ(Γ S3 )=1.

K. SELVAKUMAR ET AL. / AUSTRALAS. J. COMBIN. 69 (2) (217), 17 183 182 Proof of Theorem 1.2. The main goal of this section is to determine all finite groups G whose coprime graph has crosscap one. The following two results about the crosscap formulae of a complete graph and a complete bipartite graph are very useful in the proof of Theorem 1.2. Lemma.1. [16] The following statements hold: { 1 6 (i) γ(k n )= (n 3)(n ) if n 3 and n 7; 3 if n = 7 (ii) γ(k m,n )= 1 (m 2)(n 2),wherem, n 2. 2 By slight modifications in the proof of Theorem 1.1 with Lemma.1, and using Figs. 6 and 7, one can prove Theorem 1.2. A C A C 5 6 3 B 3 1 B B 9 B 9 8 6 C A C A Fig. 6: Embedding of Γ Z 15 in N 1 Fig. 7: Embedding of Γ Z in N 1 Acknowledgments The authors are deeply grateful to the referees for careful reading of the manuscript and helpful suggestions. The work reported here is supported by the UGC Major Research Project F. No. 2-8/213(SR) awarded to the first author by the University Grants Commission, Government of India. Also the work is supported by the INSPIRE programme (IF 17) of Department of Science and Technology, Government of India for the second author. References [1] A. Abdollahi, S. Akbari and H. R. Maimani, Noncommuting graph of a group, J. Algebra 298 (26), 68 92.

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