Generalization of the non-commuting graph of a

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ScienceAsia 42 (2016): 231–235

doi: 10.2306/scienceasia1513-1874.2016.42.231

Generalization of the non-commuting graph of a group via a normal subgroup Fereshteh Kakeria , Ahmad Erfanianb,∗ , Farzaneh Mansooria a b

∗

Ferdowsi University of Mashhad, International Campus, Mashhad, Iran Department of Pure Mathematics and Centre of Excellence in Analysis on Algebraic Structures, Ferdowsi University of Mashhad, Mashhad, Iran Corresponding author, e-mail: [email protected] Received 18 Feb 2015 Accepted 23 May 2016

ABSTRACT: Let G be a finite group and N be a normal subgroup of G. We define an undirected simple graph ΓN ,G to be a graph whose vertex set is all elements in G \ Z N (G) and two vertices x and y are adjacent iff [x, y] ∈ / N , where Z N (G) = {g ∈ G : [x, g] ∈ N for all x ∈ G}. If N = 1, then we obtain the known non-commuting graph of G. We give some basic results about connectivity, regularity, planarity, 1-planarity and some numerical invariants of the graph which are mostly improvements of the results given for non-commuting graphs. Also, a probability related to this graph is defined and a formula for the number of edges of the graph in terms of this probability is given. KEYWORDS: commutativity degree, graph diameter, girth, strongly regular, planar graph MSC2010: 05C25 20P05

INTRODUCTION There are many graphs which are associated with groups: for instance, prime graphs 1 , non-cyclic graphs 2 , and conjugate graphs 3 . An important one is the non-commuting graph. Let G be a group. Then the non-commuting graph of G, denoted by ΓG , is a graph whose vertices are elements of G \ Z(G) and two distinct vertices x and y are adjacent if x y 6= y x. Neumann 4 obtained the first remarkable result on the non-commuting graphs by answering a problem of Erd˝ os. From that time on, noncommuting graphs have been studied extensively in the literature 5, 6 . The aim of this paper is to give a generalization of the non-commuting graph of a group G with respect to a given normal subgroup N of G. In the next section, we introduce the generalized noncommuting graph G through a normal subgroup N of G denoted by ΓN ,G and state some of the basic graph theoretical properties of this graph which are mostly new or a generalization of some results in Ref. 5. We also give a connection between ΓN ,G and the probability that the commutator of two arbitrary elements of G belongs to the normal subgroup N of G. We also state some conditions under which the graph is regular or strongly regular.

GENERALIZED NON-COMMUTING GRAPH Definition 1 Let G be a finite group and N be a normal subgroup of G. The non-commuting graph of G can be generalized using the subgroup N is such a way that its vertices are G \ Z N (G) where Z N (G) = {g ∈ G : gN ∈ Z(G/N )} and two distinct vertices are adjacent when [x, y] ∈ / N . This graph is denoted by ΓN ,G . Notice that if N = 1 then the graph ΓN ,G is the ordinary non-commuting graph of G. Clearly, N ⊆ Z N (G) and so the elements of N do not belong to the vertex set of ΓN ,G . One can easily see that the graph ΓN ,G is null whenever G is abelian or N = G. Hence, throughout this paper, it is always assumed that G is a non-abelian group and N is a proper normal subgroup of G. We also note that if [G : N ] = k and {x 1 , . . . , x k } is a left transversal of N in G such that [x i , x j ] ∈ N for all 1 ¶ i, j ¶ k, then ΓN ,G is a null graph. Example 1 Let D12 = 〈a, b | a6 = b2 = 1; bab = a−1 〉 be the dihedral group of order 12 and put N = Z(D12 ) = {e, a3 }. Then Γ D12,N is drawn as in Fig. 1. Lemma 1 Let G be a group and N be a normal www.scienceasia.org

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We also note that if [G : N ] = k and {x1 , . . . , xk } is a left transversal to N in G such that [xi , xj ] ∈ N for all 1 ⩽ i, j ⩽ k, then ΓN,G is a null graph. Example 1 Let D12 = ⟨a, b a6 = b2 = 1; bab = a−1 ⟩ be a dihedral group of order 12 and put N = Z(D12 ) = {e, a3 }. Then ΓD12,N is drawn as Figure 1. Fig. 1 aaaaaaaaaaaaaaaaaaaaaa

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|CG (h)|. Thus |Z (G)| = 1, which contradicts the choice of N . 2 Now, we obtain diameter and girth of the graph ΓN,G .

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By the above lemma, it can be easily′ checked that 69 [x, y] ∈ / N ⇔ [xn, yn ] ∈ /N the graph ΓN ,G does not have any isolated vertex and is 70 neverfor anall empty n, n′ graph. ∈ N.

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determine some conditions for N andgroup G when Proposition 1 Let G be a non-abelian and Nthe be aΓnormal subgroup of G. Then Γ is Hamiltograph is Eulerian. N,G N ,G nian.

Proposition 2 If |N | is even, then ΓN ,G is Eulerian.

71 ByLet the G above can N be be easily check that Proof Proof OneVcan see that deg(x) ⩾ |V (ΓN,G )|/2 Definition 2 be alemma, groupitand a normal : :Let (ΓNeasily ,G ) = x 1 N ∪ · · · ∪ x r N . Let x be 72 theofgraph ΓN,G for doseevery not have all x ∈ V (Γ ). the proof is similar to subgroup G. Then x ∈any G, isolated vertex and anyforvertex N,G of ΓN ,G and xNow, i1 , . . . , x i t be those vertices 73 is never an empty graph. Proposition 2.2 in 1 . 2 § ª among x 1 , . . . , x r adjacent to x. Hence x ∼ nx i j An Eulerian tour is a walk which passes every N 74 Definition 2 Let G be a group and N be a normal for all n ∈ N and 1 ¶ j ¶ t, which implies that C G (x) = g ∈ G : g N ∈ CG/N (x N ) . edge exactly one times. A graph is Eulerian if it con75 subgroup of G. Then for every x ∈ G, deg(x) = Eulerian t|N | is even. by Theorem 4.1 in tains an tour. InHence the following propositions, N Ref. 7 the result follows. N we determine some conditions for N and G when the 2 It is clear that C (x) is a subgroup of G and 76

deg(x) =

CG (x) G = {g ∈ G : gN ∈ C G (xN )}

|G| − |CGN (x)|

for all x ∈ V (ΓN ,G ). N

graph ΓN,G is Eulerian.

Proposition 3 If |G| is odd, then ΓN ,G is Eulerian.

N It is clear that CG (x) is a subgroup of G and deg(x) = Proposition 2 If |N | is even, then ΓN,G is Eulerian. N 78 |G|1−Let |CG G (x)|beforaallfinite x ∈ Vgroup (ΓN,G ).and N be a Proof : Let x ∈ V (ΓN ,G ) be an arbitrary vertex. We Theorem N |G|x1−N|C∪GN· (x)|. Since (x) nontrivial subgroup of G. Then ΓN ,G is not a complete know Proofthat : Letdeg(x) V (ΓN,G=) = · · ∪ xr N . Let C xGbe 79 Theorem 1 Let G be a finite group and N be a non- is any a subgroup odd and hence vertex of Γof anditsxi1order , . . . , xiist be those vertices graph. N,GG, 80 trivial subgroup of G. Then ΓN,G is not a complete among x , . . . , x adjacent to x. Hence x ∼ nx deg(x) is even. The result follows by Theorem i4.1 1 r j 81 graph. for all8. n ∈ N and 1 ⩽ j ⩽ t, which implies that 2 Proof : Suppose on the contrary that ΓN ,G is com- in Ref. t|N | is even. Therefore, by aTheorem 4.1 plete. Then In the= following theorem, we give lower bound 82 Proof : Suppose on the contrary that ΓN,G is complete. deg(x) 11 in , the result follows. 83 Then for the chromatic number χ(Γ ) of the graph Γ2 . 77

deg(h) = |G| − |CGN (h)| = |G| − |Z N (G)| − 1

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N deg(h) = |G| − |CG (h)| = |G| − |Z N (G)| − 1

for all vertices h of ΓN ,G . Then |CGN (h)| = |Z N (G)|+1. On the www.scienceasia.org other hand, |Z N (G)| divides |CGN (h)|. Thus N |Z (G)| = 1 which contradicts the choice of N . 2 Theorem 2 diam(ΓN ,G ) = 2 and girth(ΓN ,G ) = 3.

Proof : Let a, b be two non-adjacent vertices. Then there exist x, y such that [a, x] ∈ / N and [b, y] ∈ / N . Now if a is adjacent to y or b is adjacent to x, then d(x, y) = 2. Suppose [a, y] ∈ N and [b, x] ∈ N . It is obvious that x y is a vertex and therefore x y is adjacent to both x and y. Hence diam(ΓN ,G ) = 2. 2 Theorem 3 A lower bound for the minimum degree of vertices of ΓN ,G , which we denote by δ(ΓN ,G ), satisfies δ(ΓN ,G ) ¾ 3|N |. www.scienceasia.org

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Proof : Let a, b be two non-adjacent vertices. Then there exist x, y such that [a, x] ̸∈ N and [b, y] ̸∈ N . ScienceAsia 42 (2016) Now if a join y or b join x, then d(x, y) = 2. Suppose [a, y] ∈ N and [b, x] ∈ N . It is obvious that xy is a vertex and therefore xy is adjacent to both x and y. Proof : Let x be an arbitrary vertex of ΓN ,G . Since Hence diam(ΓN,G ) = 2. ΓN ,G does not have any isolated vertex, there exists 2 a vertex to theorem, x. Hencewex ∼ n y for all n ∈ In ytheadjacent following obtain a lower N . bound Sinceforx ytheisminimum a vertexdegree of ΓN ,Gof adjacent x, we vertices oftoΓN,G , have x ∼isnx y for by allδ(Γ n∈ N . As x y = 6 y x, x ∼ nyx ). which denoted N,G

a vertex y adjacent to x. Hence, x ∼ ny for all n ∈ 63 N . : Since xy iseasily a vertex ΓN,G adjacent we Proof One can see of that deg(x) ¾ |Vto(Γx, 64 N ,G )|/2 Fig. 1 Generalized non-commuting graph Γ D12,N . ∼ Vnxy allThe n ∈proof N . Asisxy ̸= yx, to x∼ nyxfor forhave all xx ∈ (ΓNfor ). similar that ,G 65 Now, we state the following simple lemma, which for all n ∈ N , from which it follows that δ(ΓN,G ) ⩾ Proposition 2.2 in Ref. 5. 2 66 will be used frequently in the proof of next results. 3|N |. 2 An Eulerian tour is a walk which passes every subgroup of G. If x, y ∈ G, then [x, y] ∈ / N ⇔ Utilizing Definition 2, we show that Γ is a 0 a group and N be a normal edge exactly once. A graph is Eulerian if itN,G 1 Let G nbe contains [x67n, y n0Lemma ]∈ / N for all n, ∈ N. Hamiltonian graph. 68 subgroup of G. If x, y ∈ G, then an Eulerian tour. In the following propositions, we a5 b

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Proof : Let 1x be vertex of Γgroup N,G . Since Proposition LetanG arbitrary be a non-abelian and N ΓN,G does not have any isolated vertex, there exists be a normal subgroup of G. Then ΓN ,G is Hamiltonian.

.

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Theorem 2 diam(ΓN,G ) = 2 and girth(ΓN,G ) = 3.

for all n ∈ N , from which it follows that δ(ΓN ,G ) ¾ 3 We have δ(ΓN,G ) ⩾ 3|N |. 3|NTheorem |. 2

4

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Proposition 3 If |G| [G is odd, is Eulerian. Theorem 4 Suppose : N ]then = kΓ+N,G 1 and G = x0 N ∪ x 1 N ∪ · · · ∪ x k N . Then

χ(ΓN ,G ) ¾ 1 + max{|A| : A ⊆ {x 0 , . . . , x k }, x i , x j ∈ A, [x i , x j ] ∈ / N , }. Proof : Put t 1 = |{x i : [x i , x 1 ] ∈ / N }|. Then it is clear that χ(ΓN ,G ) ¾ t 1 + 1. Similarly, if t j = |{x i : [x i , x j ] ∈ / N }|, then we need at least t j + 1 colours in order to colour ΓN ,G and again χ(ΓN ,G ) ¾ t j + 1. Now, if t = max{t j : 1 ¶ j ¶ k}, then χ(ΓN ,G ) ¾ t +1, as required. 2 We now may state some results for the dominating number of the graph ΓN ,G . A subset of the graph is a dominating set if every vertex which is not in the subset is adjacent to at least one member of the subset. The size of the minimum dominating set is called the dominating number.

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Theorem 5 Let G be a non-abelian group and N be a normal subgroup of G such that x N is a dominating set for ΓN ,G for every vertex x of ΓN ,G . Then Z N (G) = N , x 2 = 1, and CG/N = 〈x N 〉. Proof : To prove Z N (G) = N , assume on the contrary that Z N (G) 6= N . Then there exists an element a ∈ Z N (G) − N and so [a, g] ∈ N for all g ∈ G. Since x is a vertex, x ∈ / Z N (G) which implies that [x, b] ∈ /N for some b ∈ G. Thus [x a, b] ∈ / N . Consequently, x a is a vertex. We now have [x a, x] = [a, x] ∈ N and so x a is not adjacent to x which is a contradiction. If x 2 6= 1, then x −1 6= x and again x −1 is not adjacent to x is a contradiction. The last part directly follows from the point that x is a vertex in the dominating set and x 2 = 1. 2 One can easily see that a subset S of V (ΓN ,G ) is a dominating set if and only if CGN (S) ⊆ Z N (G) ∪ S. Also, if X is a generating set for G, then X N\Z N (G)N is a dominating set for ΓN ,G . A cut-set of ΓN ,G is a set of edges of the graph which, if removed, disconnects the graph. The vertex connectivity of ΓN ,G , denoted by κ(ΓN ,G ), is the minimum size of all cut sets. In the following proposition we determine the vertex connectivity of ΓN ,G . Proposition 4 Let G be a non-abelian group and S be a cut set of ΓN ,G . If x, y ∈ V (ΓN ,G )\S belong to distinct components of ΓN ,G \S then S can be written as a union of double cosets of the subgroup CGN (x)∩CGN ( y). Also, if G is finite then κ(ΓN ,G ) = t|Z N (G)|, in which t is an integer greater that 1. Proof : It is similar to the proof of Proposition 2.4 in Ref. 5. Put H = CGN (x) ∩ CGN ( y). Since x and y are vertices, x, y ∈ / Z N (G) S and consequently, H 6= G. We now prove that S = a∈G H aH. Firstly, we show that for every a ∈ G, if H aH ∩S 6= ∅, then H aH ⊆ S. On the contrary, assume that H aH * S. Then there exist elements h1 , h2 ∈ H such that h1 ah2 ∈ / S. It can be easily seen that xS∼ h1 ah2 ∼ y which is a contradiction. Hence a∈G H aH ⊆ SSand S ⊆ S G = a∈G H aH which imply that S = a∈G H aH whenever H aH ∩ S 6= ∅. Secondly, assume that k(ΓN ,G ) = |S|, where S is a minimum cut set of ΓN ,G . Then Z N (G) ¶ H allows as to write S as the union of cosets of Z N (G). Thus there exists a positive integer t such that k(ΓN ,G ) = t|Z N (G)|. We claim that t > 1. If t = 1 then S = bZ N (G) for some b∈ / Z N (G). If x 1 and y1 and two vertices belong to different connected components of ΓN ,G \ S then [x, y] ∈ N . Since diam(ΓN ,G ) = 2, we should have

a path between x 1 and y1 which is a contradiction. Hence t > 1. 2 The commutativity degree d(G) of a finite group G is the probability that two randomly chosen elements of G commute. We may extend it to the commutativity degree of G with respect to a normal subgroup N of G, denoted by d N (G): d N (G) =

|{(x, y) ∈ G × G : [x, y] ∈ N }| . |G|2

It is obvious that if N = 1, then d(G) = d N (G). Also, one can see that d N (G) = d(G/N ). Utilizing the above definition, we may find a formula for the number of edges of ΓN ,G . Lemma 2 The number of edges of ΓN ,G is 12 |G|2 (1 − d N (G)). Proof : Let A = {(x, y) ∈ G × G | [x, y] ∈ N } and B = {(x, y) ∈ G × G | [x, y] ∈ / N }. It is easy to see that d N (G) = |A|/|G|2 . On the other hand, |G|2 = |{(x, y) ∈ G × G : [x, y] ∈ N }| + |{(x, y) ∈ G × G : [x, y] ∈ / N }| = |A| + |B|. Also, as |B| = 2|E(ΓN ,G )|, we have 2|E(ΓN ,G )| = |B| = |G|2 − |A|2 = |G|2 (1 − d N (G)) and the result follows. 2 Using the above lemma, we obtain the following inequalities. Proposition 5 Let G be a finite group and N be a normal subgroup of G. Then (i) d N (G) ¾ 2|Z N (G)|/|G|+1/|G|−|Z N (G)|2 /|G|2 − |Z N (G)|/|G|2 ; 3 (ii) |E(ΓN ,G )| ¾ 16 |G|2 , if G/N is not abelian. Proof : They follow from Lemma 2 and an upper bound 58 for d(G) by a result in Ref. 9. 2 A graph is k-regular if every vertex has degree k. Also, a k-regular graph with n vertices is called strongly regular with parameter (n, k, r, s) if it is neither empty nor complete such that any two adjacent vertices are adjacent to r common vertices and any two non-adjacent vertices are adjacent to s common vertices. Theorem 6 If ΓN ,G is a k-regular graph, then |N | divides k. Proof : Let x ∈ V (ΓN ,G ) and y be a vertex adjacent to x. Then x ∼ y n for all n ∈ N so that |N | divides deg(x). Since deg(x) = k, it follows that |N | divides k, as required. 2 www.scienceasia.org

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Theorem 7 If ΓN ,G is strongly regular with parameters (n, k, r, s), then s = k. Proof : Let x ∈ V (ΓN ,G ) and z ∈ Z N (G) \ N . Clearly, xz ∈ V (ΓN ,G ). Since x and xz are not adjacent, s = |G \ (CGN (x) ∪ CGN (xz))|. Since CGN (x) = CGN (xz) and ΓN ,G is k-regular, the result follows. 2 To prove the next theorem, we need the following lemma concerning AC-groups. A group is an ACgroup if its nontrivial centralizers are abelian. Lemma 3 The following statements are equivalent: (i) G is an AC-group; (ii) if x, y ∈ G \ Z(G) such that [x, y] = 1, then CG (x) = CG ( y); (iii) if x, y, z ∈ G\Z(G) such that [x, y] = [ y, z] = 1, then [x, z] = 1; (iv) if A and B are subgroups of G such that Z(G) ⊂ CG (A) ⊆ CG (B) ⊂ G, then CG (A) = CG (B). Proof : See Lemma 3.2 in Ref. 10.

2

Corollary 1 If ΓN ,G is a strongly regular graph with parameters (n, k, r, s), then G/N is an AC-group. Proof : Let x, y be two non-adjacent vertices. Then CG/N (x N ) = CG/N ( y N ) since deg(x) = |G \ CGN (x)| and deg( y) = |G \ CGN ( y)| and hence |CGN (x)| = |CGN ( y)|. As k = |G| − |CGN (x)| = |N (x) ∩ N ( y)| = |G \ CGN (x) ∪ CGN ( y)|, we conclude that |CGN (x)| = |CGN (x) ∪ CGN ( y)|. Hence CGN ( y) ⊆ CGN (x) and N consequently CG (x) = CGN ( y). Thus CGN (x)/N = CGN ( y)/N . Hence CG/N (N ) = CG/N ( y N ) and, by Lemma 3, G/N is an AC-group. Let x, y be two non-adjacent vertices. Then [x, y] ∈ N . Since deg(x) = |G \ CGN (x)| = |G \ CGN ( y)| and ΓN ,G is regular, we have |CGN (x)| = |CGN ( y)|. Assume that t ∈ G \ (CGN (x) ∪ CGN ( y)). Then t is adjacent to x and y. The number of common neighbours of x and y are equal and is k, so we should have k = |G \ (CGN (x) ∪ CGN ( y))| = |G \ CGN (x)|. Consequently, |CGN (x)| = |CGN (x) ∪ CGN ( y)| and so CGN (x) = CGN ( y). NG N Hence CG\N (x N ) = CG\N ( y N ) and by Lemma 3, G/N is an AC-group as required. 2 Theorem 8 If ΓN ,G is a strongly regular graph with parameters (n, k, r, s) then r = 2k − n. Proof : Let x, y be two adjacent vertices. Then [x, y] ∈ / N . Hence, CGN (x) ∩ CGN ( y) = Z N (G), by Corollary 1 and a consequence of Lemmas 4.2 and www.scienceasia.org

4.3 in Ref. 5. We now have r = |N (x) ∩ N ( y)| = |G \ CGN (x) ∪ CGN ( y)| = |G| − |CGN (x)| − |CGN ( y)| + |CGN (x) ∩ CGN ( y)| = |G| − |CGN (x)| + |G| − |CGN ( y)| + |Z N (G)| − |G| = 2k − n, where N (x) is the set of neighbours of x. 2 In the following two propositions, we give a link between this graph and the nilpotency property in group theory. Proposition 6 Suppose G = x 1 N ∪ x 2 N ∪ . . . ∪ x k N such that x i is a vertex in ΓN ,G , for each 2 ¶ i ¶ k. Then G is not nilpotent. Proof : Since all x i for 2 ¶ i ¶ k are vertices, Z N (G) = N and therefore |Z(G/N )| = 1. Hence G cannot be nilpotent. 2

Proposition 7 Let N be a normal subgroup of G and G/N be non-abelian. If ΓN ,G is regular, then G/N is a nilpotent group.

Proof : For every vertex x in ΓN ,G we have deg(x) = |G| − |CGN (x)|. Moreover, if y ∈ / Z N (G), then deg(x) = deg( y) which implies that |CGN (x)| = |CGN ( y)|. On the other hand, if zN ∈ Z(G/N ), then CG/N (zN ) = G/N and consequently, the size of all conjugacy classes of G/N has only two values: 1 and a positive integer n 6= 1. Hence G/N is a nilpotent by a theorem of Ito 11 . 2 Recall that a graph is planar if it can be drawn in the plane such that edges intersect only at vertices. The planarity of ΓN ,G is investigated in Ref. 5 when N = 1 is the trivial subgroup. In the following theorem, we show that ΓN ,G is not planar when N 6= 1. Theorem 9 The graph ΓN ,G is never planar for every nontrivial normal subgroup N of G. Proof : Let N 6= 1 and ΓN ,G be planar. Then δ(ΓN ,G ) ¶ 5 (see Corollary 3.5.9 in Ref. 8). On the other hand, δ(ΓN ,G ) ¾ 3|N | by Theorem 3. Hence 3|N | ¶ 5, which is a contradiction. 2 A graph is called 1-planar if it can be drawn in the plane in such a way that each edge is crossed by no more than one other edge. Note that graphs K7 , K3,7 and K4,5 are not 1-planar in Ref. 12. Lemma 4 There is a vertex in ΓN ,G whose order in G is greater than 2.

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235 5 (x) = 311 xyn3 then 312 x)| 313 yxn1 yn1 Hence it is not a 1-planar graph. Also, if G ∼ = T , then are=vertices, so 305 348 then 314 Γ has a subgraph isomorphic to K as drawn in = 1. Hence G 306 N ,G 4,5 Figure 2. 349 yn2 ugacy 315 2 the right of Fig. 3. Hence it is not 1-planar. 307 Hence 316 Finally, if G ∼ = D14 , then ΓN ,G has a subgraph isoyn3 xn1 If |N | = 2 then, by the previous lemma, the graph 350 bgroup G and 308 2 of 317 morphic to K7 so that it is not a 1-planar graph. 2 G a satisfying b 1be a 351 N,G xyn xn2 |a| > 2. Let 309 is has a a vertex nlar,thethen ΓN 318 A graph is called outer planar if it can be drawn vertex adjacent to a. If N = {n , n }, then Γ has 352 1 2 N,G 310 Recall 319 in the plane without crossing edges in such a way xyn2 ??, subgraph isomorphic to Kxn 353 4,5 3as drawn in Figure plane a 320 that all vertices belong to the unbounded face of the hence Γ is not 1-planar. e. The have deg(x) 321 =N,G 311 xyn3 drawing. N ∈ / =Z1 (G),322then 312 N thatwe|CG (x)| 313 yxn1 m, 323 = Theorem 11 The graph ΓN ,G is never an outer-planar 348 GFig. 3 Induced subgraph K4,5 of ΓN,G . ), N ∈ Z( N 314 324then graph. Figure 2. 349 Fig. 2 Induced subgraph of Γ isomorphic to K . N ,G 3,7 6 Submitted to ScienceAsia e of all conjugacy 315 354 every 325 Proof : If |N | ¾ 2, then clearly ΓN ,G is not planar integer n. Hence 316 326 x If |N | = 2 then, by the previous lemma, the graph 350Theorem 9 and ? 11. Tolue B, Erfanian A (2013) by consequently is not outerRelative planar. non commuting bn 2 317 1 y Γ318 has 351 graphthat of a finite group. J. that Algebra Appl. N,Gan 1 a vertex a satisfying |a| > 2. Let b be a Thus we may assume N = 1 and ΓN ,G is 12(2), 1250157 327 the weThen study on 3 adjacent to a. bn If N 352 (11w(Γ pages).) ¶ 3. As in the proof of 2 = {n1 , n2 }, then ΓN,Gxhas outer planar. Hence hand, ph ΓN,G . 328 Recall vertex 319 an N ,G y 2 JS (1981) Primethat graph|G| components of finite subgraph isomorphic to K4,5 as drawn in Figurexy ??, 353 Proposition 2.312. in Williams Ref. 5, one can show ¶ ⩽ 5,in the329 awn plane a 320 abn1 groups. J. Algebra 69 (2), 487-513. 2 hence Γ is not 1-planar. x y 2 330The 8. The only non-abelian groups of order less or nd vertices. 321a2 n1N,G ? the x3 y abn2 in 331= 1 equal to 8 are S3 , D8 and Q 8 for which the graph when N 322 x2 y 2 a2 n2 e than 332 we ΓN ,G has a subgraph isomorphic to K2,3 , which is a wing theorem, 323 2 xy abn Fig. 3 Induced subgraph K14,5 of ΓN,G . 355 333 contradiction. 371 Hence ΓN ,G is not an outer planar NK ̸=4,5 1. 324 Figure 3. 356 334 Figure 4. 372 Fig. 3 Induced subgraphs of ΓN ,G isomorphic to K4,5 . graph. 2 354 r planar for every 325 der in 335 Finally, if GΓ ∼ , then 357 ΓN,G has REFERENCES a subgraph 373 = Dis14not Finally, if |N326 | = 1, then we show that N,G bn 1 2 336 isomorphic to K so that it is not a 1-planar graph. 2 374 7 that x = 1 Supposeon onthe the contrary a 1-planar Proof graph.: Suppose 358for all 1. Williams JS (1981) Prime graph components of finite an 1 contrary that ΓN,G planar. Then 327 A graph is called outer-planar if it can be drawn in 375 ∈ V (Γthe ). Let x and be two adjacent vertices. is337 1-planar.x Since graphy with 7 bn vertices 359 groups. J Algebra 69, 487–513. N ,Gcomplete 2 or all other ). In hand, 328 2 the plane without crossing edges in such a way that all 376 an 2 Then [x, y] ∈ / N and hence 1 = (x y) ∈ / N which is is not a 1-planar graph, we must have w(Γ ) ⩽ 6. 360 2. Abdollahi A, Mohammadi Hassanabadi A (2007) N,G rtices. 3|N338 Hence | ⩽ 5, 329 ? abn vertices belong to the1that unbounded face of the drawing. 377 The same as Proposition 2.3 in , one can show 361 a contradiction. 2 Non-cyclic graph of a group. Comm Algebra 35, ich is 339 2 330 a2 n1 2057–81. 378 |G| 362 2 340 ⩽ 14. Hence, it remains to consider non-abelian abn2 n be drawn in the Theorem 331 The214, graph are ΓN ,GS3is, D 1-planar if and 3. Erfanian A, Tolue B (2012) Conjugate graphs of finite groups of order less or10 equal 363 only 8 , Q8 , Γ a 4n2 that Theorem 11 The graph 379 N,G is never an outerby no more than 332 3G ∼ S x −1, and groups. Discrete Math Algorithm Appl 4, 1250035. if it is planar and hence , D , Q N = 1. D , A , D , T = ⟨x, y : x = y = 1, y = y ⟩ 364 = 3 abn d only 341 10 4 12 8 1 8 planar graph. 380 355 , K and K 333 7 3,7 and D4,5 . If G ∼ S , D or Q , then Γ 4. Neumann BH (1976) A problem of Paul Erd˝ os on 365 = 3 8 = 1. 342 14 8 N,G is planar so Figure 3. 356 334 groups. J Aust Math Soc Ser A 21, 467–72. Proof : Now, If |N |if¾GProof 4∼ Γ, NA,G has a subgraph isomorthat it is 1-planar. D:10If or D , 366 |N | ⩾ 2, then clearly Γ is not planar 381 =then 4 12 N,G 5. Abdollahi A,382 Akbari S, Maimani HR (2006) Nonh iso- Γ343 to K4,5 and it4,5 is?? 1-planar graph. subgraph isomorphic to K , not hence it is 367 by hence Theorem anda consequently is not outer-planar. N,G has aphic 335 G whose order in Finally, if Also, |N{n | Thus, = 1,2 , ∼ then we show that ΓN,G is 1not 357 commuting graph of a group. J Algebra 298, 468–92. graph. not 344a 1-planar graph. if G T , then Γ has 368 Also, if N = , n n } and x, y are two adjacent = we may assume that N = and that Γ is 383 N,G 1 3 N,G 336 6. same Solomon Woldar AJ (2013) Simple groups a 1-planar graph. Suppose on the contrary that Γ⩽ 358 jacent a 345 N,G subgraphvertices, isomorphic to ΓKNouter-planar. as drawn inHence, Figure ??, 369 then a subgraph isomorphic K3,7 4,5 w(ΓN,G )to 3. The as RM, 384 ,G has 2 are characterized by their non-commuting graphs. is 1-planar. Since the complete graph with 7 vertices 359 ? otKx3,7 346 it is not 1-planar. 370 = 1hence for all as337 drawn in Fig.Proposition 2. Hence Γ2.3 is not 1-planar. , one can show that |G| ⩽ 8. The 385 N ,G in J Group Theor 16, 793–824. is not a 1-planar graph, we must have w(Γ ) ⩽ 6. 360 ar. 347 N,G adjacent vertices. 338If |N | = 2 then, Lemma 4, the graph a or equal 8 are onlyby non-abelian groups of Γorder less 386 N ,G has ? 7. Tolue B, Erfanian A (2013) Relative non commuting The as Proposition , one canthe show thatΓ 361 )2 ∈ / N , which is vertex 339 same a satisfying 2.2.3 Let be a vertex adjacent S3 , |a| D8>and Qin which graph a sub-of a finite 387 8 bfor N,G hasgraph group. J Algebra Appl 12, 1250157. Fig. 4 2 Induced subgraph |G| Hence, 362 4,5 of Γ 340 to a.⩽ 14. If NK = {n1 ,itN,T nremains },. thentoΓconsider a, which subgraph graph to has K2,3non-abelian is a contradiction. 388 2isomorphic N ,G 8. Bondy JA, Murty JSR (1997) Graph Theory with groups of order or equal 14,inis that are S3 ,ofDFig. ,graph. 363 8 , Q83. isomorphic to Therefore, Kless the ΓN,G not aleft outer-planar 2 389 4,5 as shown Applications, Elsevier. 4 3 x −1 D341 , T 1-planar. = ⟨x, y : x = y = 1, y = y ⟩ 364 lanar if and only Hence 10 , AΓ4 , Dis 12not N ,G 9. Gustafson WH (1973) What is the probability that REFERENCES S , D or Q , then Γ is planar so 365 = , Q8 and N = 1. and 342D14 . If G ∼ 3 8 8 N,G Finally, if |N | = 1, thenwww.scienceasia.org we show that ΓN ,G is not two group elements commute? Amer Math Mon 80, ∼ that it is 1-planar. Now, if Gon DAkbari , A4 S, or D 366 =A,the 10contrary 12 , then Abdollahi Maimani Non390 a 1-planar graph.1.Suppose that ΓN ,GHR (2006) 1031–304. a subgraph iso- Γ343 isomorphic to of Ka4,5group. , hence it is 298, 367 468–492. N,G has a subgraph commuting graph J. Algebra 391 p-groups with abelian centralizers. is 1-planar. Since the complete graph with 7 vertices 10. Rocke DM (1975) ∼ a 1-planar graph. not 344a 1-planar graph. Also, ifA,GMohammadi 368 = T , then Γ N,G has 2. Abdollahi Hassanabadi A (2007) Non392 Proc Lond Math Soc 30, 55–75. is not a 1-planar graph, we must have w(ΓN ,G ) ¶ 6. are two adjacent a 345 subgraph isomorphic to K in Figure 369 cyclic graph ofdrawn a group. Comm.??, Algebra11. 35,Ito2057– 393 4,5 as N (1953) On finite groups with given conjugate As in the proof of Proposition 2.3 in Ref. 5, one can omorphic to K3,7 hence 346 it is not 1-planar. 370 2081. 394 types I. Nagoya Math J 6, 17–28. show that |G| ¶ 14. Hence we only need to consider s not 1-planar. 347 3. Bondy JA, Murty JSR (1997) Graph Theory with Ap395 12. Czap J, Hudák D (2012) 1-planarity of complete non-abelian groups plications. of order less than or equal to 14, Elsevier. 396 4 3 multipartite graphs. Discrete Appl Math 160, 505–12. i.e., D8 , Q 8subgraph ,D , AK D12of , ΓTTolue = 〈x, y : x = y = 4 , 4,5 Fig. S 43 ,Induced 4.10Erfanian A, 397 N,T . B (2012) Conjugate graphs of fix −1 ∼ 1, y = y 〉 and D14 . Ifgroups. G = S3Discrete , D8 or QMath. ΓN ,G 8 , then nite Algorithms Appl. (4) 2, 398 is planar so that it is 1-planar. Now, if G ∼ = D10 , A4 1250035 (8 pages). 399 or D12 , then ΓN ,G 5.has a subgraph isomorphic tothe K4,5 . Gustafson WH (1973) What is probability that two 400 ScienceAsia 42 (2016)

groups elements commute?. Amer. Math. Monthly 80, www.scienceasia.org 1031-1304. 6. J´ulius, Hud´ak, D´avid (2012) 1-planarity of complete multipartite graphs. Discrete Appl. Math. 160, (4-5)

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