Let $G$ be a non-abelian finite group. In this paper, we prove that $Gamma(G)$ is $K_4$-free if and only if $G cong A times P$, where $A$ is an abelian group, $P$ is a $2$-group and $G/Z(G) cong mathbb{ Z}_2 times mathbb{Z}_2$. Also, we show that $Gamma(G)$ is $K_{1,3}$-free if and only if $G cong {mathbb{S}}_3,~D_8$ or $Q_8$

Hajar Mousavi

Neda Ahanjideh

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Journal of Algebraic SystemsVol. 2, No. 2, (2014), pp 147-151ON THE GROUPS WITH THE PARTICULARNON-COMMUTING GRAPHSN. AHANJIDEH∗ AND H. MOUSAVIAbstract. Let G be a non-abelian finite group. In this paper,we prove that Γ(G) is K4-free, if and only if G ∼= A× P , where Ais an abelian group, P is a 2-group and G/Z(G) ∼= Z2 × Z2. Also,we show that Γ(G) is K1,3-free if and only if G ∼= S3, D8 or Q8.1. IntroductionFor an integer z > 1, we denote by π(z) the set of all prime divisorsof z. If G is a finite group, then π(|G|) is denoted by π(G). Let G bea non-abelian finite group and Z(G) be its center. For x ∈ G, supposethat clG(x) denotes the conjugacy class in G containing x and CG(x)denotes the centralizer of x in G. We will associate a graph Γ(G)to G which is called the non-commuting graph of G. The vertex setV (Γ(G)) is G − Z(G) and the edge set E(Γ(G)) consists of (x, y) (wewrite x ∼ y), where x and y are distinct non-central elements of G suchthat xy ̸= yx. Here, we are considering simple graphs, i.e., graphs withno loops or directed or repeated edges. The non-commuting graphsof the non-abelian finite groups have been studied in some literatures.For example in [1], the authors classified non-abelian finite groups withHamiltonian non-commuting graphs, regular non-commuting graphsand planner non-commuting graphs. Also, it has been shown in [1] thatthe non-commuting graph of a non-abelian group is connected. Notethat for a graph H, the H-free graph L is a graph that does not have anMSC(2010): Primary: 20D60; Secondary: 05C25Keywords: Non-commuting graph, K4-free graph, K1,3-free graph.Received: 14 September 2014, Revised: 8 February 2015.∗Corresponding author .147148 AHANJIDEH AND MOUSAVIinduced subgraph isomorphic to L. Because of the special propertiesof K1,3-free graphs and K4-free graphs, they have been studied in somepapers. In this paper, we are going to study non-abelian finite groupswhich their non-commuting graphs are K1,3-free and non-abelian finitegroups which their non-commuting graphs are K4-free. Throughoutthis paper, we will use the following notation: let G be a finite non-abelian group and M(G) denote a set of the orders of maximal abeliansubgroups G. A set of vertices of a graph Γ is called an independentset, if its elements are pairwise nonadjacent. The independent numberof a graph Γ, which is denoted by α(Γ), is the cardinality of the largestits independent set.2. Some LemmasIn this section, we bring some lemmas which will be used in the proofof the main theorem:Lemma 2.1. If G is a finite group and H,K and L are distinct propersubgroups of G such that G = H ∪ K ∪ L, then [G : H] = [G : K] =[G : L] = 2 and H ∩ L = H ∩K = K ∩ L = H ∩K ∩ L.Proof. It follows immediately by considering the order of G. □Lemma 2.2. [3] If for every x ∈ G−Z(G), |clG(x)| = m, then m is apower of the prime p and G = P × A, where P is a p-Sylow subgroupof G and A is abelian.Lemma 2.3. For every x ∈ G − Z(G), there is a triangular in Γ(G)containing the vertex x.Proof. Since x ̸∈ Z(G), there exists y ∈ G− Z(G) such that xy ̸= yx.Thus x ∼ y in Γ(G). Since CG(x) ∪ CG(y) ̸= G, we deduce that x, y, zform a triangular, where z ∈ G− (CG(x) ∪ CG(y)). □It follows from Lemma 2.3 that:Corollary 2.4. Γ(G) contains a triangular.Lemma 2.5. Let p, q ∈ π(G).(i) IfM(G) ⊆ {p, q}, thenM(G) = {p, q} and G is the non-abeliangroup of order pq.(ii) If M(G) ⊆ {p, p2}, then G is a p-group, |Z(G)| = p andM(G) = {p2}.Proof. Since every maximal abelian subgroup of a finite non-abelian p-group has order at least p2, we get that G is not a p-group and hence,G is a {p, q}-group such that every Sylow subgroup of G has primeON THE GROUPS WITH THE PARTICULAR 149order and its center is trivial. Thus M(G) = {p, q}, as claimed in (i).The same reasoning completes the proof of (ii). □Lemma 2.6. If M(G) ⊆ {2, 4}, then G ∼= D8 or Q8.Proof. By Lemma 2.5,M(G) = {4}, |Z(G)| = 2 and G is a non-abelian2-group. Thus there exists x ∈ G−Z(G) such that O(x) = 4. Also, wecan see that G/Z(G) is a 2-elementary abelian group and CG(x) = ⟨x⟩.So ⟨x⟩/Z(G) is a normal subgroup of G/Z(G) and hence, ⟨x⟩ is normalin G. Therefore, G/⟨x⟩ = G/CG(x) ↪→ Aut(⟨x⟩) ∼= Z2. This forces|G| = 8 and hence, lemma follows. □Lemma 2.7. If α(Γ(G)) ≤ 2, then G ∼= S3, D8 or Q8.Proof. By [1, Remak 2.5], we can see that M is a maximal abeliansubgroup of G if and only if M − Z(G) is a maximal independent setof Γ(G). Thus “α(Γ(G)) ≤ 2” implies that for every maximal abeliansubgroup M of G, |M | − |Z(G)| ≤ 2. Thus |Z(G)|(|M |/|Z(G)| −1) ∈ {1, 2}. This forces (|M |, |Z(G)|) ∈ {(2, 1), (3, 1), (4, 2)}. Thus byLemmas 2.5 and 2.6, the result follows. □2.1. Main results.Theorem 2.8. Let G be a non-abelian finite group.(i) Γ(G) is K4-free if and only if G ∼= A×P , where A is an abeliangroup, P is a 2-group and G/Z(G) ∼= Z2 × Z2.= S3, D8 or Q8.(ii) Γ(G) is K1,3-free if and only if G ∼Proof. (i) “=⇒ ” Let x be an arbitrary element of G − Z(G) andy ∈ G − CG(x). Then for every element z ∈ G − (CG(x) ∪ CG(y)),G = CG(x) ∪ CG(y) ∪ CG(z). Fix K = CG(x) ∩ CG(y) ∩ CG(z). ByLemma 2.1, CG(x)∩CG(y) = K is a normal subgroup of G, |G/K| = 4and G/K contains different elements xK, yK and zK of order 2. ThusG/K ∼= Z2 × Z2. Now for every a, b ∈ CG(x) − (CG(x) ∩ CG(y)) =CG(x) − K, we have a, b ∈ G − (CG(y) ∪ CG(z)) and hence, G =CG(a) ∪ CG(y) ∪ CG(z) = CG(b) ∪ CG(y) ∪ CG(z). Thus CG(a)−K =CG(b) − K, so a ∈ CG(b). Consequently, CG(x) = ⟨CG(x) − K⟩ isabelian. Similarly, we can see that CG(y) and CG(z) are abelian andhence, we get that K ≤ Z(G). Since G is non-abelian and |G/K| = 4,we get K = Z(G) and G/Z(G) ∼= Z2×Z2. Now Lemma 2.2 completesthe proof.“ ⇐= ” Let G/Z(G) = ⟨aZ(G)⟩ × ⟨bZ(G)⟩ ∼= Z2 × Z2. Then CG(a),CG(b) and CG(ab) are abelian and G = CG(a) ∪ CG(b) ∪ CG(ab). Forevery subset T of G−Z(G) with four elements, at least one of the setsT ∩CG(a), T ∩CG(b) and T ∩CG(ab) contains more than two elements.150 AHANJIDEH AND MOUSAVIThis forces T not to form K4, as desired.= S3, D8 or Q8, then it is obvious that Γ(G) is K1,3-free. Now(ii) If G ∼let Γ(G) be K1,3-free. Let M = {x1, ..., xt} be a maximal independentset of Γ(G) such that |M | = α(Γ(G)). We continue the proof in thefollowing cases:(a) Let α(Γ(G)) ≥ 3 and xi1 , xi2 , xi3 be three arbitrary elementsof M . Since Γ(G) is K1,3-free, we deduce that for every y ∈G − Z(G), y ∈ CG(xi1), CG(xi2) or CG(xi3). This shows thatG = CG(xi1) ∪ CG(xi2) ∪ CG(xi2). Thus Lemma 2.1 shows that∩3j=1 CG(xij) = CG(xi1) ∩ CG(xi2). (2.1)Now let z ∈ G − (M ∪ Z(G)). If there exists 1 ≤ i, j ≤ tsuch that z ̸∼ xi and z ̸∼ xj, then by (2.1), for every u ∈{1, .., t} − {i, j}, z ̸∼ xu. Thus M ∪ {z} is an independent set,which is a contradiction. Therefore, there exists at most onei ∈ {1, ..., t} such that z ̸∼ xi and hence, z is adjacent to everyxj ∈M − {xi}. This means that z has at least t− 1 neighborsin M . Since Γ(G) is K1,3-free, t − 1 ≤ 2. But α(Γ(G)) ≥3 and hence, α(Γ(G)) = 3. So [2, Lemma 2.4] shows thatthere exists a maximal abelian subgroup M ′ of G such that|M ′ − Z(G)| = 3 and for every maximal abelian subgroup M ′′of G, we have |M ′′ − Z(G)| ≤ 3. If |Z(G)| = 3, then |M ′| = 6,which is an abelian subgroup. Assume that M ′ = ⟨a⟩. Thena3 ̸∈ Z(G). Consequently for b ∈ G−CG(a3), {b}∪(M ′−Z(G))is the vertex set of K1,3, a contradiction. Thus |Z(G)| = 1 and|M ′| = 4. If G has a maximal abelian subgroup M ′′ of order 3,then for some x ∈ M ′′ of order 3, CG(x) = M ′′. Consequently(M ′ − Z(G)) ∪ {x} is the vertex set of K1,3, a contradiction.ThusM(G) = {2, 4} and hence, by Lemma 2.6, G ∼= D8 or Q8.So α(G) = 2, which is a contradiction.(b) If α(Γ(G)) ≤ 2, then Lemma 2.7 completes the proof.□AcknowledgmentsThe authors express their gratitude to the referee for carefully read-ing and several useful comments, which improved the manuscript, inparticular, Theorem 2.8(ii). The first author was partially supportedby the center of Excellence for Mathematics, University of Shahrekord,Iran.ON THE GROUPS WITH THE PARTICULAR 151References1. A. Abdollahi, S. Akbari and H. R. Maimani, Non-commuting graph of a group,J. Algebra 298 (2006), 468-492.2. N. Ahanjideh and A. Iranmanesh, On the relation between the non-commutinggraph and the prime graph, Int. J. Group Theory 1 (2012), 25-28.3. N. Ito, On finite groups with given conjugate type I, Nagoya Math. J. 6 (1953),17-28.Neda AhanjidehDepartment of pure Mathematics, Shahrekord University, P.O.Box 115, Shahrekord,Iran.Email: ahanjideh.neda@sci.sku.acHajar MousaviDepartment of pure Mathematics, Shahrekord University, P.O.Box 115, Shahrekord,Iran.Email: h.sadat68@yahoo.com

ON THE GROUPS WITH THE PARTICULAR NON-COMMUTING GRAPHS

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ON THE GROUPS WITH THE PARTICULAR NON-COMMUTING GRAPHS

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