aa r X i v : . [ m a t h . G R ] S e p Intersection graph of cyclic subgroups of groups
R. Rajkumar ∗ , P. Devi † Department of Mathematics, The Gandhigram Rural Institute – Deemed University,Gandhigram – 624 302, Tamil Nadu, India.
Abstract
Let G be a group. The intersection graph of cyclic subgroups of G , denoted by I c ( G ), isa graph having all the proper cyclic subgroups of G as its vertices and two distinct verticesin I c ( G ) are adjacent if and only if their intersection is non-trivial. In this paper, we classifythe finite groups whose intersection graph of cyclic subgroups is one of totally disconnected,complete, star, path, cycle. We show that for a given finite group G , girth ( I c ( G )) ∈ { , ∞} .Moreover, we classify all finite non-cyclic abelian groups whose intersection graph of cyclicsubgroups is planar. Also for any group G , we determine the independence number, cliquecover number of I c ( G ) and show that I c ( G ) is weakly α -perfect. Among the other results, wedetermine the values of n for which I c ( Z n ) is regular and estimate its domination number. Keywords:
Intersection graph, cyclic subgroups, girth, weakly α -perfect, planar. Associating graphs to algebraic structures and studying their properties using the methods ofgraph theory has been an interesting topic for mathematicians in the last decade. For a givenfamily F = { S i | i ∈ I } of sets, the intersection graph of F is a graph with the members of F asits vertices and two vertices S i and S j are adjacent if and only if i = j and S i ∩ S j = {∅} . For theproperties of these graphs and several special class of intersection graphs, we refer the reader to [7].In the past fifty years, it has been a growing interest among mathematicians, when the members ∗ e-mail: [email protected] † e-mail: [email protected] F have some specific algebraic structures. In 1964 Bosak [2] defined the intersection graphs ofsemigroups. Motivated by this, Cs´ a k´ a ny and Poll´ a k [4] defined the intersection graph of subgroupsof a finite group.Let G be a group. The intersection graph of subgroups of G , denoted by I ( G ), is a graphhaving all the proper subgroups of G as its vertices and two distinct vertices in I ( G ) are adjacentif and only if the intersection of the corresponding subgroups is non-trivial.Some properties of the intersection graphs of subgroups of finite abelian groups were studied byZelinka in [13]. Inspired by these, there are several papers appeared in the literature which havestudied the intersecting graphs on algebraic structures, viz., rings and modules. See, for instance[1, 3, 9, 10, 12] and the references therein.In this paper, for a given group G , we define the intersection graph of cyclic subgroups of G ,denoted by I c ( G ), is a graph having all the proper cyclic subgroups of G as its vertices andtwo distinct vertices in I c ( G ) are adjacent if and only if their intersection is non-trivial. Clearly I c ( G ) is a subgraph of I ( G ) induced by all the proper cyclic subgroups of G . When G is cyclic,then I c ( G ) and I ( G ) are the same. In [3], Chakrabarty et al have studied several properties ofintersection graphs of subgroups of cyclic groups.Now we recall some basic definitions and notations of graph theory. We use the standardterminology of graphs (e.g., see [5]). Let G be a graph. We denote the degree of a vertex v in G by deg ( v ). A graph whose edge set is empty is called a null graph or totally disconnected graph. K n denotes the complete graph on n vertices. K m,n denotes the complete bipartite graph withone partition consists of m vertices and other partition consists of n vertices. In particular K ,n iscalled a star . P n and C n respectively denotes the path and cycle with n edges. A graph is regular if all the vertices have the same degree. A graph is planar if it can be drawn in a plane such thatno two edges intersect except (possibly) at their end vertices. The girth of G , denoted by girth ( G ),is the length of its shortest cycle, if it exist; other wise girth ( G ) = ∞ . An independent set of G is a subset of V ( G ) having no two vertices are adjacent. The independence number of G , denotedby α ( G ), is the cardinality of the largest independent set. A clique of G is a complete subgraph of G . The clique cover number of G , denoted by θ ( G ), is the minimum number of cliques in G whichcover all the vertices of G . G said to be weakly α -perfect if α ( G ) = θ ( G ). A dominating set of G isa subset D of V ( G ) such that every vertex not in D is adjacent to at least one member of D . The domination number γ ( G ) is the number of vertices in a smallest dominating set for G .In this paper, we classify the finite groups whose intersection graph of cyclic subgroups is oneof totally disconnected, complete, star graph, cycle, path and we give a characterization for groupswhose intersection graph of cyclic subgroups is one of acyclic, bipartite, C -free. We show that fora given finite group G , girth ( I c ( G )) ∈ { , ∞} . Moreover, we classify all finite non-cyclic abeliangroups whose intersection graph of cyclic subgroups is planar. Also we determine the independencenumber, clique cover number of I c ( G ) for any group G . As a consequence, we show that I c ( G ) isweakly α -perfect. Finally, we determine the values of n for which I c ( Z n ) is regular and estimateits domination number.Throughout this paper p , q , r denotes the distinct prime numbers. Theorem 2.1.
Let G and G be two groups. If G ∼ = G , then I c ( G ) ∼ = I c ( G ) .Proof. Let f : G → G be a group isomorphism. Define a map ψ : V ( I c ( G )) → V ( I c ( G )) by ψ ( H ) = f ( H ), for every H ∈ V ( I c ( G )). It is easy to see that ψ is a graph isomorphism. Remark 2.1.
The converse of Theorem 2.1 is not true. For example, let G ∼ = Z p and G ∼ = Q = h a, b | a = b = 1 , b = a , ab = ba − i . Here H i , i = 1 , 2, 3, 4 are subgroups of G of orders p i respectively; H is a subgroup of H i , i = 2 , 3, 4. h a i , h b i , h ab i , h a i are the subgroups of G ; h a i is a subgroup of h a i , h b i , h ab i . It follows that I c ( G ) = K = I c ( G ) , but G ≇ G . Theorem 2.2.
Let G be a group. Then I c ( G ) is totally disconnected if and only if the order ofevery element of G is prime.Proof. If the order of every element of G is prime, then obviously proof follows. Suppose G hasan element of composite order, then the subgroup generated by that element is adjacent with itsproper cyclic subgroups in I c ( G ) and so I c ( G ) is not totally disconnected. This completes theproof. Corollary 2.1.
Let G be a finite group. Then I c ( G ) is totally disconnected if and only if G is oneof p -group with exponent p or nilpotent group of order p α q or A .Proof. It is proved in [6] that in a finite group, every element of it is of prime order if and only ifit is one of p -group with exponent p or nilpotent group of order p α q or A . This fact together withTheorem 2.2 completes the proof. Remark 2.2.
Tarski Monster’s group is an infinite non-abelian group in which every proper sub-group is of order a fixed prime p . The existence of such a group was given by Olshanski in [8]. Thisshows the validity of Theorem 2.2 for infinite groups. Theorem 2.3.
Let G be a group. Then I c ( G ) is complete if and only if G has a unique subgroupof prime order.Proof. Suppose G has two subgroups of prime order, then they are not adjacent in I c ( G ). It followsthat if I c ( G ) is complete then G has a unique subgroup of prime order. Conversely, suppose G has a unique subgroup of prime order, then every subgroup of G contains that subgroup. It followsthat the intersection of any two subgroups of G is non-trivial and so I c ( G ) is complete. Hence theproof. Corollary 2.2.
Let G be a finite group. Then I c ( G ) is complete if and only if G is isomorphic toeither Z p α or Q α ( α ≥ . In this case, I c ( Z p α ) ∼ = K α − and I c ( Q α ) ∼ = K α − + α − .Proof. By Theorem 2.3, G must be a p -group. Since G has a unique subgroup of prime order, soby [11, Proposition 1.3], G ∼ = Z p α or Q α . If G ∼ = Z p α , then G has α − I c ( G ) ∼ = K α − . If G ∼ = Q α , then it has2 α − + α − G contains a unique subgroupof order 2 and so I c ( Q α ) ∼ = K α − + α − . Remark 2.3. Z p ∞ is an example of an infinite group such that I c ( Z p ∞ ) is complete. This showsthe validity of Theorem 2.3 for infinite groups. In [3], Chakarabarthy et al classified all the cyclic groups whose intersection graph of subgroupsare planar. In the next result, we classify all non-cyclic finite abelian groups whose intersectiongraph of cyclic subgroups are planar. We mainly use the Kuratowski’s Theorem to check theplanarity of a graph.
Theorem 2.4.
Let G be a finite non-cyclic abelian group. Then I c ( G ) is planar if and only if G is isomorphic to one of Z np , Z × Z , Z × Z , Z q × Z or Z × Z .Proof. We divide the proof into several cases.
Case 1: If G ∼ = Z np , n ≥
1, then every element of G is prime and so by Theorem 2.1, Γ Ic ( G ) istotally disconnected. Case 2: If G ∼ = Z p × Z p . Here h (1 , i , h (1 , i , . . . , h (1 , p − i , h ( p, i , h ( p, i , . . . , h ( p, p − i , h (0 , i are the only proper cyclic subgroups of G . Note that h ( p, i is a subgroup of h (1 , i , h (1 , i , . . . , h (1 , p − i ; also no two remaining subgroups intersect non-trivially. Therefore, I c ( G ) ∼ = K p +1 ∪ K p . Thus I c ( G ) is planar if and only if p = 2, 3. Case 3: If G ∼ = Z pq × Z p . Here G has p + 1 subgroups of order p , let them be H i , i = 1, 2, . . . , p + 1; unique subgroup of order q , say H ; for each i = 1, 2, . . . , p + 1, HH i is a cyclic subgroups of G of order pq . So H is a subgroup of HH i , i = 1, 2, . . . , p + 1; for each i = 1, 2, . . . , p + 1, H i is asubgroup of HH i ; no two remaining cyclic subgroups intersect non-trivially. So I c ( G ) is planar ifand only if p = 2. Case 4: If G ∼ = Z p × Z p . If p ≥
5, then Z p × Z p is a subgroup of G and so by case 2, I c ( G ) isnon-planar. If p = 2, then h (1 , i , h (1 , i , h (2 , i , h (0 , i , h (2 , i , h (0 , i , h (1 , i , h (2 , i are theonly proper cyclic subgroups of G . Also h (2 , i is a proper subgroup of h (1 , i , h (1 , i ; h (0 , i is asubgroup of h (0 , i , h (2 , i ; h (2 , i is a subgroup of h (1 , i ; no two remaining subgroups intersect.Therefore, I c ( G ) ∼ = 2 K ∪ K , which is planar. If p = 3, then G ∼ = h a, b | a = b = 1 , ab = ba i .Here h a i is a subgroup of h a i , h ab i , h ab i , h a b i . Therefore, these five subgroups adjacent witheach other and so they form K as a subgraph of I c ( G ). It follows that I c ( G ) is non-planar. Case 5: If G ∼ = Z p q × Z p , then G has the unique subgroup of order q , let it be H ; G has at leastthree subgroups of order p q and at least three subgroups of order pq . Therefore, G has at leastfour cyclic subgroups containing H as a subgroup. Hence I c ( G ) contains K as a subgraph and so I c ( G ) is non-planar. Case 6: If G ∼ = Z p × Z p , then G has at least four cyclic subgroups of order p , p , p , p respectivelyand they have a unique subgroup of order p in common. Therefore, these five subgroups are adjacentwith each other and they form K as a subgraph of I c ( G ), so I c ( G ) is non-planar. Case 7: If G ∼ = Z p × Z p , then G has a subgroup isomorphic to Z p × Z p . So by Case 6, I c ( G )contains K as a subgraph and hence I c ( G ) is non-planar. Case 8: If G ∼ = Z p α × Z p α × . . . × Z p αkk , where p i ’s are primes with at least two p i ’s are equaland α i ≥
1. If k ≥
2, then G has one of the following groups as its subgroup: Z p × Z p , Z p q × Z p or Z p × Z p and so by Cases 5, 6, 7, I c ( G ) contains K as a subgraph. Therefore, I c ( G ) isnon-planar.Combining all the cases together the proof follows. Z ∞ p , Q and Tarski Monster groups shows the existence of an infinite abelian group, finitenon-abelian group and infinite non-abelian group respectively, whose intersection graph of cyclicsubgroups is planar. Now we pose the following: Problem 2.1.
Classify the non-abelian groups and infinite abelian groups whose intersection graphof cyclic subgroups are planar.
Theorem 2.5.
Let G be a finite group. Then I c ( G ) is a star if and only if G ∼ = Z p .Proof. If G ∼ = Z p , then I c ( G ) ∼ = K , which is a star.Conversely let I c ( G ) be a star. Let H be the center vertex and H , H , . . . , H n be the pendentvertices of I c ( G ). Now we consider the following cases: Case 1:
Let H be a subgroup of H i , for all i = 1, . . . , n . Then H , H , H forms C as a subgraphof I c ( G ), which is not possible. So the only possibility is n = 1 and H is a subgroup of H . Case 2:
Let H i be a subgroup of H , for all i = 1, 2, . . . , n . Then we show that | H i | = p , for all i . For otherwise, | H i | = q or m for some i , where m is a composite number. If | H i | = m , then H , H i and a proper subgroup of H i forms C as a subgraph of I c ( G ), which is a contradiction to ourassumption. If | H i | = q , then | H | = pq and so | G | = p α q β , α + β ≥ pqk , k >
1. But | G | = p α q β is not possible, since it has a subgroup of order p . | G | = pqk is also not possible, since it has asubgroup of order other than p and q .Thus we have | H i | = p , for all i and so | H | = p . It follows that | G | = p α . Since G has a uniquesubgroup of order p , so by [11, Proposition 1.3], G ∼ = Z p α or Q α . But G can not be isomorphicto Q α , since Q α contains more than one subgroup of order 4. So by Corollary 2.2, we must have G ∼ = Z p . This completes the proof. Theorem 2.6.
Let G be a finite group. Then I c ( G ) ∼ = P n if and only if n = 1 and G ∼ = Z p .Proof. If G ∼ = Z p , then I c ( G ) ∼ = K , which is a path.Conversely, let I c ( G ) be a path H − H − . . . − H r . Let r ≥
3. Firstly note that for distinct i , j , k , H i ∩ H j = H k is not possible, since these subgroups forms C as a subgraph of I c ( G ). Sofor each i = 2, 3, . . . , r −
1, the subgroups H i − , H i , H i +1 must satisfy one of the following: (i) H i − ⊆ H i ⊆ H i +1 , (ii) H i − ⊇ H i ⊇ H i +1 , (iii) H i − ⊆ H i ⊇ H i +1 . But first two cases are notpossible, since these three subgroups forms C as a subgraph of I c ( G ). So the only possibility isthe case (iii). If r ≥
4, then H ⊆ H ⊇ H ⊆ H , so that H , H , H forms C as a subgraph of I c ( G ), which is a contradiction. Thus r ≤
3. If r = 3, then I c ( G ) ∼ = K , is a star, which is notpossible by Theorem 2.5. So we must have r = 2 and it follows that G ∼ = Z p . Theorem 2.7.
Let G be a finite group. Then I c ( G ) ∼ = C n if and only if n = 3 and G ∼ = Z p .Proof. If G ∼ = Z p , then I c ( G ) ∼ = C .Conversely, Suppose that I c ( G ) is a cycle H − H − · · · − H r − H . Let 2 < r ≥
4. Thenby the same argument as in the proof of Theorem 2.6, we can show that r ≤
3. If r = 3, then byTheorem 2.2, I c ( G ) ∼ = K if and only if G ∼ = Z p . Corollary 2.3.
Let G be a finite group. Then girth ( I c ( G )) ∈ { , ∞} .Proof. In the proof of Theorem 2.7, we observed that if I c ( G ) contains a cycle of length at leastfour, then it must contains C as a subgraph, so its girth is 3. In the remaining cases, I c ( G ) isacyclic and so its girth is infinity. Hence the proof. Theorem 2.8.
Let G be a finite group. Then the following are equivalent: (1) G has a maximal cyclic subgroup of order one of p , p or pq and no two cyclic subgroups oforder p or pq have a non-trivial intersection; (2) I c ( G ) is acyclic; (3) I c ( G ) is bipartite; (4) I c ( G ) is C -free.Proof. Clearly (1) ⇒ (2) ⇒ (3) ⇒ (4).Now assume that I c ( G ) is acyclic. Suppose either G has a maximal cyclic subgroup of order m > p, p , pq and two maximal cyclic subgroups intersect non-trivially, then I c ( G ) contains C ,which is a contradiction to our assumption. This prove the part (2) ⇒ (1).Suppose I c ( G ) is C -free, then by Corollary 2.3, girth ( I c ( G )) = ∞ and so I c ( G ) is acyclic. Thisprove the part (4) ⇒ (2). Hence the proof. Theorem 2.9.
Let G be a group. Then α ( I c ( G )) = m , where m is the number of prime ordersubgroups of G .Proof. Let A be the set of all prime order subgroups of G . Clearly A is a maximal independent setin I c ( G ). We show that α ( I c ( G )) = | A | . Let B be a another one independent set I c ( G ), whichis different from A . Then B has at least one cyclic subgroup of G of composite order, say H . If | H | = p , then H has a unique subgroup of order p . If | H | 6 = p , then H has at least two subgroupsof prime orders. In either case, these prime order subgroups are adjacent with H in I c ( G ). So foreach vertex in B , there corresponds at least one vertex in A . It follows that | B | ≤ | A | . Hence theproof. Theorem 2.10.
Let G be a group. Then Θ( I c ( G )) = m , where m is the number of prime ordersubgroups of G .Proof. Let | G | = p α p α . . . p α k k , where p i ’s are distinct primes, α i ≥
1. For each i = 1, 2, . . . , k , let t i be the number of subgroups of order p i and let H ( j, p i ), j = 1, 2, . . . , t i be a subgroup of G oforder p i . For each j = 1, 2, . . . , t i , let S ( j, p i ) be the set of all proper cyclic subgroups of G having H ( j, p i ) in common. Clearly S ( j, p i ) forms a clique in I c ( G ) and S := { S ( j, p i ) | i = 1, 2, . . . , k and j = 1, 2, . . . , t i } forms a clique cover of I c ( G ) with | S | = t + t + · · · + t k . Therefore,Θ( I c ( G )) ≤ | S | . Let T be a clique cover of I c ( G ) such that Θ( I c ( G )) = | T | . If | T | < | S | ,then by pigeonhole principle, T has a clique which contains atleast two subgroups, say H ( j, p i ), H ( l, p r ), for some i = r , j ∈ {
1, 2, . . . , t i } , l ∈ {
1, 2, . . . , t r } , which is not possible, since H ( j, p i )and H ( l, p r ) are not adjacent in I c ( G ). So | T | = | S | = the number of prime order subgroups of G . A similar argument also works when G is infinite. Remark 2.4.
In [13], Zelinka proved that for any group G , α ( I ( G )) = m , where m is the numberof prime order proper subgroups of G . In [9], the authors showed that Θ( I ( G )) = α ( I ( G )) . It isinteresting to note that by Theorems 2.9 and 2.10, we have α ( I ( G )) = α ( I c ( G )) and Θ( I ( G )) =Θ( I c ( G )) . Corollary 2.4.
Let G be a group. Then I c ( G ) is weakly α -perfect.Proof. Proof follows from Theorems 2.9 and 2.10.
Theorem 2.11. I c ( Z n ) is regular if and only if n = p α , α ≥ .Proof. Let n = p α p α . . . p α k k , where p i ′ s are distinct primes, α i ≥
1. We need to consider thefollowing cases.
Case 1:
Let k ≥
2. It is shown in [3, p. 5387], that for any subgroup H of Z n , if | H | = p t i i p t i i . . . p t ir i r , then deg( H ) = [ τ ( n ) − n Y j =1 j / ∈{ i , i , . . . , i r } ( α i + 1)] −
2; if | H | = p α i p α i . . . p α ik k , thendeg( H ) = τ ( n ) − K be the subgroup of Z n of order p α r r . If some α i >
1, let A be a subgroup of Z n of order p α i p α i . . . p α ik k , where 0 ≤ α i t ≤ α t . Then deg( K ) = τ ( n ) − − k Y i =1 i = r ( α i + 1) and deg( A ) = τ ( n ) − A ) > deg( K ). Therefore, I c ( Z n ) is not regular. If α i = 1 for every i = 1, 2, . . . , k ,then let B be the subgroup of Z n of order p p . . . p i − p i +1 . . . p k . Then deg( B ) = 2 k − K ) = 2 k − −
2. But deg( B ) = 2 k − k − − > k − − K ). Therefore, I c ( G ) isnot regular. Case 2: k = 1, then by Theorem 2.2, I c ( Z n ) ∼ = K α − , which is regular.Combining the above two cases we get the result. Theorem 2.12.
Let n = p α p α . . . p α k k , where p i ′ s are distinct primes, α i ≥ . Then γ ( I c ( Z n )) = , if α i > for some i ;2 , if α i = 1 for every i. Proof. If α i >
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