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1 This is a sample cover image for this issue. The actual cover is not yet available at this time.) This article appeared in a journal published by Elsevier. The attached copy is furnished to the author for internal non-commercial research and education use, including for instruction at the authors institution and sharing with colleagues. Other uses, including reproduction and distribution, or selling or licensing copies, or posting to personal, institutional or third party websites are prohibited. In most cases authors are permitted to post their version of the article e.g. in Word or Tex form) to their personal website or institutional repository. Authors requiring further information regarding Elsevier s archiving and manuscript policies are encouraged to visit:

Journal of Algebra 357 01) 03 07 Contents lists available at SciVerse ScienceDirect Journal of Algebra www.elsevier.

81746-73441, Iran b School of Mathematics, Institute for Research in Fundamental Sciences IPM), P.O.

2 Journal of Algebra ) Contents lists available at SciVerse ScienceDirect Journal of Algebra Characterization of the alternating group by its non-commuting graph Alireza Abdollahi a,b,, Hamid Shahverdi a a Department of Mathematics, University of Isfahan, Isfahan, , Iran b School of Mathematics, Institute for Research in Fundamental Sciences IPM), P.O. Box , Tehran, Iran article info abstract Article history: Received 18 January 01 Available online xxxx Communicated by Ronald Solomon MSC: 0D06 0D60 Keywords: Non-commuting graph Finite simple groups Alternating groups Let G be any non-abelian group and ZG) be its center. The noncommuting graph Γ G of G is the simple graph whose vertex set is G \ ZG), with two vertices x and y adjacent whenever xy xy. We prove that if Γ G is isomorphic to the non-commuting graph of the alternating group A n n 4), then G = An. This result together with a recent one due to Solomon and Woldar gives a complete positive answer to a conjecture proposed in [A. Abdollahi, S. Akbari, H.R. Maimani, Non-commuting graph of a group, J. Algebra ) ]: If S is any finite non-abelian simple group such that Γ S = ΓG for some group G, theng = S. 01 Elsevier Inc. All rights reserved. 1. Introduction and results Let G be a non-abelian group and let ZG) denote its center. The non-commuting graph Γ G of G is the simple graph whose vertex set is G \ ZG) and two vertices x and y are adjacent whenever xy yx. The non-commuting graph of a group was introduced by P. Erdős in 1975 see [10]), when he posed a question on the size of the cliques of the graph. The non-commuting graph of a group has been studied by many people e.g., [1,,6,8,9]). One of the problems about non-commuting graphs of groups which has drawn the attention of several authors e.g., [11,1]), is the following conjecture proposed in [1]. Conjecture 1.1. See Conjecture 1.3 of [1].) Let S be a finite non-abelian simple group and G a group such that Γ G = ΓS.ThenG = S. This research was in part supported by a grant from IPM No ). * Corresponding author at: Department of Mathematics, University of Isfahan, Isfahan , Iran. addresses: a.abdollahi@math.ui.ac.ir A. Abdollahi), hamidshahverdi@gmail.com H. Shahverdi) /$ see front matter 01 Elsevier Inc. All rights reserved. doi: /j.jalgebra

3 04 A. Abdollahi, H. Shahverdi / Journal of Algebra ) Our main result is to prove the validity of Conjecture 1.1 for the alternating groups A n. Theorem 1.. Let n 4 be an integer and let G be a group such that Γ G = ΓAn.ThenG = An. By a recent result of Solomon and Woldar, Conjecture 1.1 is valid for all finite simple groups of Lie type. This together with the main result of [7] and Theorem 1. shows that Conjecture 1.1 is valid for all non-abelian finite simple groups S.. Proofs Throughout this section, n 4 is an integer and G is a group such that Γ G = ΓAn. Lemma.1. 1) G = A n. ) NG) = NA n ),whereforanyfinitegrouph, 3) If n 15, theng = An. NH) = { n N H has a conjugacy class C such that C =n }. Proof. 1) This is part ) of Theorem 3.16 of [1]. ) Since the set of vertex degrees of two isomorphic graphs are the same, part 1) implies that the sets of centralizer sizes of groups G and A n are the same. This completes the proof of part ). See also the proof of Theorem 3.4 of [1] to see that the multisets of conjugacy class sizes are even thesameforgroupsg and A n. 3) Suppose first n = 4. By part 1), G =1. Since there are only three non-abelian groups of order 1 and only one of them, i.e. A 4 has trivial center, G = A 4. Now suppose n 5. A conjecture of Thompson states that if M is a finite non-abelian simple group and G is a centerless group such that NG) = NM), theng = M. Chen [5] has shown that Thompson s conjecture is valid for finite non-abelian simple groups M whose prime graph [13] is disconnected. By [13], the prime graph of A n is disconnected whenever n {p, p + 1, p + } for some prime number p. Thusifn 15 and n 10, then G = A n and by the main result of [11] the latter is also valid for n = 10. We use the following well-known result in the proof of Lemma.3. Lemma.. Let g A n and suppose the cycle decomposition of g contains exactly c i = c i g) cycles of length i for each i {1,...,n} so that n = n i=1 ic i.letz= n! k i=1 ic i k i=1 c i!) 1. Then for the size of the conjugacy class g A n of g in A n we have: 1) If for all even i, c i = 0 and for all odd i, c i {0, 1},then g A n =z/. ) In all other cases, g A n =z. In particular, g A n z/. Lemma.3. Let n 16, g A n and c 1 = c 1 g) 4.Then g A n nn 1)n )n 3). Proof. We use the following inequality: i h h! ih ih!, where i, h are integers such that i and h 0excepti, h) = 3, 1).

4 A. Abdollahi, H. Shahverdi / Journal of Algebra ) So we can account for the exception and bound g A n n! n i=1 i c i 1 n c i!) i=1 To complete the proof, it is enough to show that n 4)! > 3 c 1! n c 1 n!. 3 c 1! n c 1 n c 1!) n c1 )!. ) We prove the latter inequality by induction on n. Sincec 1 4, for n = 16 the inequality ) holds. Suppose n > 16.We prove the inequality ) for n + 1 by supposing inductively that ) holds for n. Therefore Since n 3 > n+1 c 1,wehave n 3)! > 3 c 1! n c 1 > 3 c 1! n+1 c 1 n c1 n c1 n 3)! > 3 c 1! n+1 c 1 ) n 3!! ) n 3. n + 1 c1 )!. This completes the proof. Lemma.4. Let A be the pointwise stabilizer of a subset X of {1,...,n} in A n such that X n 4. Then C An A) is equal to the pointwise stabilizer of {1,...,n}\XinA n. Proof. We see that Y ={1,...,n}\X is exactly the set of points moved by A, so any element a in A n that centralizes A leaves this set invariant. It follows that a = uv, where u fixes all points in X and v fixes all points in Y. Thusv centralizes A, and since by assumption uv also centralizes A, it follows that u centralizes A. Butu is in A, sou belongs to the center ZA) of A, which is trivial because Y 4. Thus a = v, and thus a fixes all the points in Y.Conversely,ifa fixes all of the points in Y,it is obvious that a centralizes A. Proof of Theorem 1.. We argue by induction on n. By Lemma.1, the result holds for n {4, 5, 6,...,15}. We may assume that n 16 and suppose inductively that if 4 m < n and Γ H = ΓAm for some group H, thenh = A m.letφ : Γ An Γ G be a graph isomorphism. By Lemma.1, ZG) = 1 and so φ is a bijective map from A n \{1} to G \{1} with the property that xy = yx if and only if φx)φy) = φy)φx) x, y A n ). We extend φ to A n by defining φ1) = 1. By [3,4] and part 1) of Lemma.1, it is enough to prove that G is a simple group. Suppose N is a non-trivial normal subgroup of G. We shall prove that N = G. Weproveourclaim in four steps. 1) Let k n 4 be a positive integer. If A = Stab An X) is the pointwise stabilizer of a subset X {1,...,n} of size n k, then φa) = A k. If k = 1 or, there is nothing to prove. If k = 3, by Lemma.4 C An E) = A, where E = Stab An Y ) and Y ={1,...,n} \X. Therefore C G φe)) = φa). Thus φa) is a subgroup of order 3. Hence φa) = A 3.

5 06 A. Abdollahi, H. Shahverdi / Journal of Algebra ) If 4 k n 4, then by Lemma.4, we have C An A) = D = A n k and C An D) = A. SinceΓ An = ΓG, we have Γ CAn D) = Γ CG φd)). Now, by induction C G φd)) = Ak. Therefore φa) = φc An D)) = C G φd)) = Ak. ) If L = Stab An X) is the pointwise stabilizer of any subset X {1,...,n} of size 4, thenφl) N. Suppose G 1 = φl). By part 1), G 1 = An 4 is simple and so G 1 N ={1} or G 1 N = G 1.We prove the latter is true. Suppose, by way of contradiction, that G 1 N ={1}. Then G 1 N = G 1 N divides G. Therefore n 4)! N divides n!.thus N divides nn 1)n )n 3). SinceN is normal, it is a union of G-classes. By Lemma.3, N has a non-trivial element x such that φ 1 x) A n fixes pointwise a subset S of {1,,...,n} of size at least 4. Fix Y S with Y =4, and let D 1 be the pointwise stabilizer of {1,,...,n}\Y.ThenD 1 = A4. Therefore [φ 1 x), D 1 ]=1 and so φ 1 x) C An D 1 ). By Lemma.4, C An D 1 ) = A n 4. Now it follows by induction that A n 4 = CAn D 1 ) = C G φd 1 )) and x C G φd 1 )). SinceC G φd 1 )) is simple and x C G φd 1 )) N, C G φd1 ) ) N = C G φd1 ) ). Therefore C G φd 1 )) N and so N has a subgroup isomorphic to A n 4. In particular, N n 4)!. Thus n 4)! nn 1)n )n 3), ) since N divides nn 1)n )n 3). Itiseasytoseethatforn 16, the inequality ) does not hold, a contradiction. Hence G 1 N. 3) C G φg)) N for all cycles g = i, j, k) of length 3 in A n. We may assume that g = 1,, 3). ThenC An g) = A3 A n 3.Bypart)φ g ) and φh i ) are contained in N, where H i = Stab An {1,, 3, i}) for each i {4,...,n}. By part 1), both φ g ) and φh i ) are subgroups of G. Nowsince g H i ={1} and [ g, H i ]=1, it follows from the graph isomorphism that φ g ) φh i ) = φ g ),φh i ). It is clear that φ g ) φh i ) C G φg)) N. Let Y = n i=4 φ g ) φh i)). Then ) n 3 Y =3 A n 4 A n 5 1 n 3)! n 3)! n 3)! = n 3 ) + A n 6 ). n 3 3 ) ) It follows that Y > 3 n 3)! 4 = C G φg)) and so Y =C G φg)). Therefore C G φg)) N. 4) N = G. Let h = n,n 1,n) and g = 1,, 3). By part 3), C G φg))c G φh)) N. Since C An g) C An h) = 1,, 3) A n 6 n,n 1,n), ) ) CG φg) CG φh) = 3 n 3)! 3 n 3)! 9 n 6)!.

6 A. Abdollahi, H. Shahverdi / Journal of Algebra ) Now as n 16 it follows that G : N G : CG φg) ), CG φh) ) < and so G = N. This completes the proof. Acknowledgments The authors are grateful to the referee for his/her helpful comments. The authors are indebted to Professor R. Solomon for his encouragement and invaluable comments. They also wish to thank Professor A.J. Woldar for reading a first version of the paper and pointing out some errors. This research was financially supported by the Center of Excellence for Mathematics, University of Isfahan. The authors are grateful to the Graduate Studies of University of Isfahan for moral and financial support. References [1] A. Abdollahi, S. Akbari, H.R. Maimani, Non-commuting graph of a group, J. Algebra ) [] N. Ahanjideh, A. Iranmanesh, On the relation between the non-commuting graph and the prime graph, Int. J. Group Theory 1 1) 01) 5 8. [3] E. Artin, The orders of the linear groups, Comm. Pure Appl. Math ) [4] E. Artin, The orders of the classical simple groups, Appl. Math ) [5] G.Y. Chen, On Thompson s conjecture, J. Algebra 18 1) 1996) [6] M.R. Darafsheh, Groups with the same non-commuting graph, Discrete Appl. Math ) 009) [7] Z. Han, G. Chen, X. Guo, A characterization theorem for sporadic simple groups, Sib. Math. J. 49 6) 008) [8] A.R. Moghaddamfar, About noncommuting graphs, Sib. Math. J. 47 5) 005) [9] A.R. Moghaddamfar, W.J. Shi, W. Zhou, A.R. Zokayi, On the noncommuting graph associated with a finite group, Sib. Math. J. 46 ) 005) [10] B.H. Neumann, A problem of Paul Erdős on groups, J. Austral. Math. Soc. Ser. A ) [11] L. Wang, W. Shi, A new characterization of A 10 by its non-commuting graph, Comm. Algebra ) [1] L. Wang, W. Shi, New characterization of S 4 q) by its noncommuting graph, Sib. Math. J. 50 3) 009) [13] J.S. Williams, Prime graph components of finite groups, J. Algebra )

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Communications in Algebra Publication details, including instructions for authors and subscription information: This article was downloaded by: [Professor Alireza Abdollahi] On: 04 January 2013, At: 19:35 Publisher: Taylor & Francis Informa Ltd Registered in England and Wales Registered Number: 1072954 Registered