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 office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK Communications in Algebra Publication details, including instructions for authors and subscription information: http://www.tandfonline.com/loi/lagb20 Commutativity Pattern of Finite Non-Abelian p-groups Determine Their Orders A. Abdollahi a, S. Akbari b, H. Dorbidi b & H. Shahverdi a a Department of Mathematics, University of Isfahan, Isfahan, Iranand School of Mathematics, Institute for Research in Fundamental Sciences (IPM), Tehran, Iran b Department of Mathematical Sciences, Sharif University of Technology, Tehran, Iranand School of Mathematics, Institute for Research in Fundamental Sciences (IPM), Tehran, Iran To cite this article: A. Abdollahi, S. Akbari, H. Dorbidi & H. Shahverdi (2013): Commutativity Pattern of Finite Non-Abelian p-groups Determine Their Orders, Communications in Algebra, 41:2, 451-461 To link to this article: http://dx.doi.org/10.1080/00927872.2011.627075 PLEASE SCROLL DOWN FOR ARTICLE Full terms and conditions of use: http://www.tandfonline.com/page/terms-and-conditions This article may be used for research, teaching, and private study purposes. Any substantial or systematic reproduction, redistribution, reselling, loan, sub-licensing, systematic supply, or distribution in any form to anyone is expressly forbidden. The publisher does not give any warranty express or implied or make any representation that the contents will be complete or accurate or up to date. The accuracy of any instructions, formulae, and drug doses should be independently verified with primary sources. The publisher shall not be liable for any loss, actions, claims, proceedings, demand, or costs or damages whatsoever or howsoever caused arising directly or indirectly in connection with or arising out of the use of this material.
Communications in Algebra, 41: 451 461, 2013 Copyright Taylor & Francis Group, LLC ISSN: 0092-7872 print/1532-4125 online DOI: 10.1080/00927872.2011.627075 COMMUTATIVITY PATTERN OF FINITE NON-ABELIAN p-groups DETERMINE THEIR ORDERS Key Words: A. Abdollahi 1, S. Akbari 2, H. Dorbidi 3, and H. Shahverdi 4 1 Department of Mathematics, University of Isfahan, Isfahan, Iran; and School of Mathematics, Institute for Research in Fundamental Sciences (IPM), Tehran, Iran 2 Department of Mathematical Sciences, Sharif University of Technology, Tehran, Iran; and School of Mathematics, Institute for Research in Fundamental Sciences (IPM), Tehran, Iran 3 Department of Mathematical Sciences, Sharif University of Technology, Tehran, Iran 4 Department of Mathematics, University of Isfahan, Isfahan, Iran Let G be a non-abelian group and Z G be the center of G. Associate a graph G (called noncommuting graph of G) with G as follows: Take G\Z G as the vertices of G, and join two distinct vertices x and y, whenever xy yx. Here, we prove that the commutativity pattern of a finite non-abelian p-group determine its order among the class of groups"; this means that if P is a finite non-abelian p-group such that P H for some group H, then P = H. Graph isomorphism; Groups with abelian centralizers; Non-commuting graph; p-group. 2010 Mathematics Subject Classification: 20D15; 20D45. 1. INTRODUCTION AND RESULTS Given a finite non-abelian group G, one can associate in many different ways a graph to G (e.g., [3, 11]). Here we consider the noncommuting graph G of G: the set of vertices of G is G\Z G, and two vertices x and y are adjacent if and only if xy yx. The noncommuting graph was first considered by Paul Erdös in 1975 [8]. The noncommuting graph of finite groups has been studied by many people (e.g., [1, 7]). The noncommuting graph of a group is a discrete way to reflect the commutativity pattern of the group. In [1] the following conjecture was formulated. Conjecture 1.1 (Conjecture 1.1 of [1]). Let G and H be two finite non-abelian groups such that G H. Then G = H. Received June 25, 2011; Revised September 16, 2011. Communicated by A. Turull. Address correspondence to Dr. Alireza Abdollahi, Department of Mathematics, University of Isfahan, Isfahan 81746-73441, Iran; E-mail: a.abdollahi@math.ui.ac.ir 451
452 ABDOLLAHI ET AL. Conjecture 1.1 was refuted in [7] by exhibiting two groups G and H of orders G =2 10 5 3 2 3 5 6 = H with isomorphic noncommuting graphs. In [1], it is proved that Conjecture 1.1 holds whenever one of the groups in question is a symmetric group, dihedral group, alternative group, or a nonsolvable AC-group (where by an AC-group we mean a group in which the centralizer of every noncentral element is abelian). Recently Darafsheh [5] has proved the validity of Conjecture 1.1 whenever one of the groups G or H is a non-abelian finite simple group. The main result of the present article shows that any pair of groups consisting a counterexample for Conjecture 1.1 cannot contain a group of prime power order. Theorem 1.2. If P is a finite non-abelian p-group such that P G for some group G, then P = G. This is a curious general phenomenon for non-abelian groups of prime power order: The order of a prime power order group can be determined among all finite groups by a proper model of its commutativity behavior, i.e, the noncommuting graph. 2. PRELIMINARY RESULTS It is not hard to prove that the finiteness or the being non-abelian of a group can be transferred under graph isomorphism whenever two groups have the same noncommuting graph. Throughout, P denotes a fixed but arbitrary finite nonabelian p-group of order p n whose center Z P is of order p r and 1 <p a 1 <p a 2 < <p a k are all distinct conjugacy class sizes of P, where p a i is the size of conjugacy class g i G of the element g i. Throughout, we also denote by u the greatest common divisor gcd a 1 a k n r of a 1 a k n r. Lemma 2.1. Let G be a finite non-abelian group and H be a group such that G H is a graph isomorphism. Then the following statements hold: (1) C H h divides g G 1 Z H Z G, where h = g ; (2) If Z G Z H and G contains a non-central element g such that C G g 2 G Z G, then G = H. Proof. (1) Since G H, we have G Z G = H Z H H = G Z G + Z H (a) and C G g Z G = C H h Z H C H h = C G g + Z H Z G (b)
COMMUTATIVITY PATTERN OF FINITE p-groups 453 As C H h divides H, C H h divides G C G g + Z H Z G C G g (c) it follows from a b c that C H h divides g G 1 Z H Z G (2) Let h = g. By part (1), we have C H h = C G g + Z H Z G divides g G 1 Z H Z G. Now, the inequality C G g 2 G Z G implies that 0 C H h g G 1 Z G Z H < C G g + Z H Z G = C H h and this yields g G 1 Z G Z H = 0. Hence Z G = Z H. Lemma 2.2. Suppose that H = P 1 A is a finite group, where P 1 is a p-group, A is a finite abelian group such that gcd p A = 1. If P H, then P = H. Proof. Let be a graph isomorphism from P to H. Suppose h = g t for some 1 t k and P 1 =p Z P 1 =p A =a, and C H h =ap. Since P H, we have P Z P =p r p n r 1 = ap p 1 = H Z H P C P g t =p n a t p a t 1 = ap p 1 = H C H h since gcd a p = 1, it follows that r = and n a t =. Therefore, C P g t Z P =p r p n a t r 1 = ap p 1 = C H h Z H Therefore, a = 1. Since r =, Z P = Z H. Hence, P = H. Lemma 2.3. Suppose H = Q A, where Q is a q-group for some prime q, A is an abelian group and gcd A q = 1. If P H, then H = P. Proof. If p = q, then Lemma 2.2 completes the proof. Suppose, for a contradiction, that p q. Note that g P 1 =pa 1. Let be a graph isomorphism from P to H, and let g 1 = h A =a Q =q C H h =aq Z H =aq It is clear that > >. Since P H, we have the following equation: C H h Z H =aq q 1 = p r p n a1 r 1 = C P g 1 Z P (1) H C H h =aq q 1 = p n a 1 p a 1 1 = P CP g 1 (2)
454 ABDOLLAHI ET AL. Since g P 1 gp for all g P\Z P, h H has the minimum size among all conjugacy classes of noncentral elements of H. By considering the conjugacy class equation of H, we have where aq = aq + q + s x H i i=1 x i H i = 1 s = g H g H\Z H \ h H Since gcd a q = 1 and q s i=1 x i H, it follows that Equation (1) implies that the largest p-power number possibly dividing a is p r. Now it follows from Eqs. (1), (2) and the inequality that p n a 1 r q 1 q 1 p n a 1 r 2 which is a contradiction. This completes the proof. Lemma 2.4. Let H be a group such that P H. Then Z H divides p r p u 1, where u = gcd a 1 a k n r. Proof. (1) Since P H, P Z P = H Z H and P C P g i = H C H h i, for every i 1 k and h i = g 1, where G H. Therefore, we have the following equalities: ( ) H p r p n r 1 = Z H Z H 1 ( ) p n a i p a i H 1 = C H h i C H h i 1 for each i 1 k. Thus Z H divides the great common divisors of the lefthand side of two latter equalities which is p r p u 1. A class of groups arising in the proof of our main result is the class of ACgroups; as we mentioned, a group G is called an AC-group whenever the centralizer of every non-central element is abelian. AC-groups was studied by many people (e.g., [10]). It is easy to see that C G x C G y = Z G for any two noncentral elements x y G with distinct centralizers. This implies that G = { C G x /Z G x G\Z G } ( ) is a partition of G/Z G, where by a partition for a group H we mean a collection of proper subgroups of H such that H = S S and S T = 1 for any two distinct S T. Each element of is called a component of the partition. If each component is abelian, we call an abelian partition. Thus G is an abelian
COMMUTATIVITY PATTERN OF FINITE p-groups 455 partition for G/Z G. The size of G is an invariant of the noncommuting graph G, called the clique number, where by definition the clique number of a finite graph is the maximum number of vertices which are pairwise adjacent. The clique number of the noncommuting graph H of a non-abelian group H will be denoted by H. Thus H is simply the maximum number of pairwise noncommuting elements in the group. Lemma 2.5. Suppose that G is a finite non-abelian AC-group such that G/Z G is a p-group. Then G 1 mod p. Proof. Since G is an AC-group, = G = G, where G = { C G x x G\Z G } On the other hand, C G x C G y = Z G for any two noncentral elements x y G such that C G x C G y. Therefore, This completes the proof. G = 1 Z G + S G Lemma 2.6 (Mann [2]: Lemma 39.8, p. 354). Suppose C is a subgroup of group G and let a G, be such that CC a = C a C. Then CC a = C C a. Proof. We have S CC a = Cc a = Cc 1 c a = C c a C C a c C c C c C Thus CC a C C a. Since all the generators c a = c 1 c a (c C) of C a belong to CC a, we have C C a CC a, because by hypothesis CC a is a group. This completes the proof. In the following proposition we will use this property of any AC-groups G; for any two commuting noncentral elements x and y of G, we have C G x = C G y. Proposition 2.7. Let G be a nilpotent AC-group of nilpotency class greater than 2, then the set of all centralizers of noncentral elements of G has exactly one normal member T in G. In particular, T is a characteristic subgroup of G. Moreover, the latter normal subgroup T has the maximum order among all members of. Proof. Let x be any element of Z 2 G \Z G. Then C G x is a normal subgroup of G containing G : For the map defined on G by g = x g for all g G is a group homomorphism and its image is contained in Z G and its kernel is C G x. Since G is of nilpotency class greater than 2, there exists an element g G \Z G. Since Z 2 G G = 1, the remark preceding the proposition implies that C G x = C G g for all x Z 2 G \Z G ( )
456 ABDOLLAHI ET AL. Now suppose that N = C G y is a normal centralizer of G for some noncentral element y. Then there exists an element t ( N Z 2 G ) \Z G, since Z G N. Since yt = ty, it follows from that C G t = C G y = C G g. Hence, we have so far proved that has exactly one normal member in G. This implies that C G x is a characteristic subgroup of G. Now, we prove C G x has the maximum order among all members of. Suppose that C = C G h for some h G\Z G. We may assume that C is not normal in G. Thus there exists an element a N G N G C \N G C, since G is nilpotent. Then C a C, and C a is a subgroup of N G C. Let A = CC a. By Lemma 2.6, we have CC G x CG Z G C C a Z G = CC a Z G = CC a = A It follows that C C G x Z G = CC G x A = CC a = Thus C G x C. This completes the proof. C 2 Z G The proof of existence of unique normal centralizer is due to Rocke [9, Lemma 3.8]; the argument to prove the existence of a normal centralizer of maximal order is due to Mann [2, Theorem 39.7, p. 354]. He has proved among all abelian subgroups of maximal order in a metabelian p-group, there exists a normal subgroup. The latter was first proved by Gillam [4]. Lemma 2.8. Let P be of nilpotency class 2. Then a i r for every i. Proof. Since P is of nilpotency class 2, for every x P\Z P with class size p a i, the conjugacy class of x is contained in xp xz P. Hence p a i p r. This completes the proof. Now we will need the following two well-known results about Frobenius groups. Proposition 2.9. (1) (See e.g., Theorem 6.7 of [6]) Let N be a normal subgroup of a finite group G, and suppose that C G n N for every non-identity element n N. Then N is complemented in G, and if 1 <N<G, then G is a Frobenius group with kernel N. (2) (see e.g., Lemma 6.1 of [6]) Let H be a Frobenius group with the kernel F and a complement K. Then K divides F 1. Lemma 2.10. Let H = KF be a Frobenius group with the kernel F and a complement K. Suppose 1 F 1 F is a normal subgroup of H. Then H 1 = KF 1 is a Frobenius group with the kernel F 1 and a complement K. Proof. For every non-identity element f 1 of F 1, C H1 f 1 = C H f 1 H 1 F H 1 = F F 1 K = F 1
COMMUTATIVITY PATTERN OF FINITE p-groups 457 by the Dedekind modular law. Therefore H 1 is a Frobenius group with the kernel F 1. It is clear that K is a complement for F 1 in H 1. 3. PROOF OF THE MAIN RESULT In this section we prove our main result, Theorem 1.2. We argue by induction on the order of P. If P =p 3, then P = G by Proposition 3.20 of [1]. If P is not an AC-group, there exists a non-central element x P such that C P x is non-abelian. If y = x, then CP x CG y. Now induction hypothesis implies that C P x = C G y and since P C P x = G C G y, we have P = G. Thus, we may assume that P is an AC-group and so G is also an AC-group. By Proposition 3.14 of [1], we may assume that G is solvable. Therefore, by the classification of non-abelian solvable AC-groups in [10], G is isomorphic to one the following groups H i (i = 1 5): (1) H 1 is non-nilpotent and it has an abelian normal subgroup N of prime index and H 1 = N Z H 1 +1. (2) H 2 /Z H 2 is a Frobenius group with the Frobenius kernel and complement F/Z H 2 and K/Z H 2, respectively, and F and K are abelian subgroups of H 2 and H 2 = F Z H 2 +1. (3) H 3 /Z H 3 S 4 and V is a non-abelian subgroup of H 3 such that V/Z H 3 is the Klein 4-group of H 3 /Z H 3 and H 3 = 13, where S 4 is the symmetric group of on 4 letters. (4) H 4 = A Q, where A is an abelian subgroup and Q is an AC-group of prime power order. (5) H 5 /Z H 5 is a Frobenius group with the Frobenius kernel and complement F/Z H 5 and K/Z H 5, respectively, and K is an abelian subgroup of H. Z F = Z H 5, and F/Z H 5 is of prime power order and H 5 = F Z H 5 + F. By Lemmas 3.11 and 3.12 of [1] and Lemma 2.3, we may assume that G is isomorphic to either H 1 or H 5. Suppose that G H 1. Then, obviously P H1. Since N is abelian, there exists h H 1 \Z H 1 such that C H1 h = N.AsP is an ACp-group, it follows from Lemma 2.5 that Since P H1, we have P 1 mod p H 1 = C H1 h Z H 1 +1 1 mod p and so p C H1 h Z H 1. On the other hand Lemma 2.1(1) implies that, C H1 h divides p a t 1 p r Z H 1, where g t maps to h under a graph isomorphism from P to H1. Thus p divides Z H 1 and so p 2 C H1 h. This follows that p 2 divides Z H 1. By continuing this latter process, one obtains that p r divides Z H 1 and so Z H 1 Z P. Now, let y H 1 \C H1 h so that H 1 = C H1 h C H1 y and H 1 Z H 1 = C H1 h C H1 y max C H1 h 2 C H1 y 2 Now, Lemma 2.1(2) implies that P = H 1.
458 ABDOLLAHI ET AL. Thus, it remains to deal with the case G H 5. Let H = H 5 and note that P H. We need to introduce some new notation for the group H. Since F/Z F is a q-group for some prime q, we set F =bq, for some positive integer b such that gcd b q = 1 and therefore Z H =bq and C F f i = C H f i =bq i for some f i F\Z F. (Recall that Z F = Z H in this case) Since F is nilpotent and non-abelian, we have 1 < i <. Since gcd K/Z H F/Z H = 1, we have C H h = K =aq for some h H\F and for some positive integer a. It is clear that b a and gcd a q = 1. Therefore, H =aq. Suppose that under a graph isomorphism from H to P, h maps to g t for some integer 1 t k and f i maps to g i, where 1 i k and i t. Here note that f t is not defined. Suppose further that = a t. We need to prove the following (a), (b), (c), and (d). (a) p q. (b) If p l divides a, for some integer l, then p l divides b and p r+1 does not divide a. This simply means that the largest p-power part of a and b are the same and p r is the largest p-power possibly dividing a. (c) F is a regular graph so that there exists integers and such that i = and a i = for all 1 i k and i t. (d) 2 and 3. Proof of (a). Suppose p = q. Since P H, ap p 1 = p n p 1 and bp p i 1 = p r p n ai r 1. Therefore, n = = r, a contradiction. Proof of (b). Since P H, we have a b q = p r p n r 1 (3) Thus p r a b. This proves the first part of (b) for all l 1 r. Now, suppose t>rand p t divides a and p t b. Equation (3) shows p r+1 b. Now let i 1 k such that i t. Then by the graph isomorphism, we have p n a t p n a i = aq bq i ( ) Since r + 1 n a t and r + 1 n a i, it follows from that p r+1 divides b, a contradiction. Now Eq. (3) implies that p r+1 a, and since b divides a, the proof of part (b) follows. Proof of (c). Suppose F is not regular. Therefore, F has two centralizers C H f i1 and C H f i2 of order bq i 1 and bq i 2, respectively, where i1 i2. We may assume that the conjugacy class of f i1 in F is of minimum size among all conjugacy classes of noncentral elements of F. We distinguish two cases to reach a contradiction. (I) Suppose that i1 i2. p n a i 2 p r = bq i 2 bq (4) p n a i 1 p n a i 2 = bq i 1 bq i 2 (5)
COMMUTATIVITY PATTERN OF FINITE p-groups 459 Now it follows from Eqs. (4), (5) and part (b) that p n a i 2 r q i 1 i2 1 q 1 p n a i 2 r 2 a contradiction. (II) Suppose that i1 i2 >. We claim that the nilpotency class of F is greater than 2. If not, then Lemma 2.8 implies that i2 ( ) Since i1 i2 >, ( ) is a contradiction. Therefore, the nilpotency class of F is greater than 2. Since H is an AC-group, F is also an AC-group. Therefore, every maximal abelian subgroup of F is centralizer of noncentral element of F. By Proposition 2.7, F has a characteristic centralizer C F f j of order bq j = bq i 1 having the maximum order among the proper centralizers. Thus i1 = j and so a i1 = a j. Since F is normal subgroup of H, C F f j is normal in H. Since H/Z H is Frobenius group, by Lemma 2.10 K/Z H C F f j /Z H is a Frobenius group with the kernel C F f j /Z H and a complement K/Z H. Thus a b q i 1 1 By the graph isomorphism, we have ( ) bq q i 1 1 = p r p n a i r 1 1 (6) ( bq i a ) 1 b q i 1 1 = p n a i 1 p a i 1 1 (7) Since gcd a p = 1, Eqs. ( ) and (6) imply that a b b q p n a i r 1 1. Eq. (7) implies that p n a i r 1 a b q i 1 1, and by the conjugacy class equation, i1. Therefore, a b q < a b q, a contradiction. Proof of (d). Since F is regular and F is an AC-group, we have F = q 1 q 1. Therefore, divides. Now, by considering the conjugacy class equation of F, we find that and 3. Now we have two different possibilities on the centralizer orders of H. (I) bq >aq. Since P H, we have p n p n = bq aq where = a t. It follows from the latter equation, Lemma 2.4, and parts (b),(d) that p n r q a b <q p u 1 <p n r a contradiction. (II) aq >bq.
460 ABDOLLAHI ET AL. Since P H, we have We consider two cases: aq bq = p n p n (8) (i) u<n r. Since u n r, 2u n r. Since H/Z H is a Frobenius group, K/Z H =a/b divides F/Z F 1. Now it follows from parts (b) and (d), Lemma 2.4(1), and Eq. (8), we have p n r a b q q 1 q <q 2 p u 1 2 <p 2u a contradiction. (ii) u = n r. Since u n r, n r 2u. By the graph isomorphism p n p n = aq q 1 This latter equation, Lemma 2.4 and parts (b) and (d) imply that a contradiction. This completes the proof. ACKNOWLEDGMENT p n r q 1 <q 2 p u 1 2 <p 2u The first author s research was in part supported by a grant from IPM (No. 90050219) as well as by the Center of Excellence for Mathematics, University of Isfahan. The second author s research was in part supported by a grant from IPM (No. 90050212). REFERENCES [1] Abdollahi, A., Akbari, S., Maimani, H. R. (2006). Non-commuting graph of a group. J. Algebra 298:468 492. [2] Berkovich, Y. (2008). Groups of prime power order. Vol. 1. de Gruyter Expositions in Mathematics. Vol. 46. Berlin: Walter de Gruyter GmbH & Co. KG. [3] Bertram, B. A., Herzog, M., Mann, A. (1990). On a graph related to conjugacy classes of groups. Bull. London Math. Soc. 22(6):569 575. [4] Gillam, J. D. (1970). A note on finite metabelian p-groups. Proc. Amer. Math. Soc. 25:189 190. [5] Darafsheh, M. R. (2009). Groups with the same non-commuting graph. Discrete Appl. Math. 157(4):833 837. [6] Isaacs, I. M. (2008). Finite group theory. Graduate Studies in Mathematics. Vol. 92. Providence, RI: Amer. Math. Soc. [7] Moghaddamfar, A. R. (2005). About noncommuting graphs. Siberian Math. J. 47(5):1112 1116.
COMMUTATIVITY PATTERN OF FINITE p-groups 461 [8] Neumann, B. H. (1976). A problem of Paul Erdös on groups. J. Aust. Math. Soc. Ser. A21:467 472. [9] Rocke, D. M. (1975). p-groups with abelian centralizers. Proc. London Math. Soc. 30(3):55 75. [10] Schmidt, R. (1975). Zenralisatorverbände endlicher Gruppen. Rend. Sem. Mat. Univ. Padova 44:55 75. [11] Williams, J. S. (1981). Prime graph components of finite groups. J. Algebra 69:487 513.