ON FINITE GROUPS WITH EXACTLY SEVEN ELEMENT ...

J. Appl. Math. & Computing Vol. 22(2006), No. 1 - 2, pp. 403 - 410

Website: http://jamc.net

ON FINITE GROUPS WITH EXACTLY SEVEN ELEMENT CENTRALIZERS ALI REZA ASHRAFI∗ AND BIJAN TAERI

Abstract. For a finite group G, #Cent(G) denotes the number of centralizers of its elements. A group G is called n-centralizer
if #Cent(G) = n, G and primitive n-centralizer if #Cent(G) = #Cent Z(G) = n. The first author in [1], characterized the primitive 6-centralizer finite groups. In this paper we continue this problem and characterize the primitive 7-centralizer finite groups. We prove that a finite group G is primitive G ∼ 7-centralizer if and only if Z(G) = D10 or R, where R is the semidirect product of a cyclic group of order 5 by a cyclic group of order 4 acting faithfully. Also, we compute #Cent(G) for some finite groups, using the structure of G modulu its center. AMS Mathematics Subject Classification : 20D99, 20E07. Key words and phrases : Finite group, n-centralizer group, primitive ncentralizer group.

1. Notation and preliminaries In this section we describe some notations which will be kept throughout. Following B. H. Neumann [9] we shall say that a group G is covered by a family of cosets or subgroups if G is simply the set-theoretic union of the family. Throughout this paper we consider covering by subgroups. Suppose Γ = {Xi }ni=1 is a covering for G. We shall say that S Γ is irredundant if none of the subgroups Xi can be omitted; that is, Xi 6⊆ j6=i Xj , for each i. Finally, a group G is called H . capable if there exists a group H such that G ∼ = Z(H) Tn The maximum value of |G : i=1 Xi | in a group G with an irredundant covering by n subgroups will be denoted by f2 (n). In [12] Tomkinson has proved that f2 (3) = 4 and f2 (4) = 9. He also has announced that by a detailed investigation Received March 13, 2005. ∗ Corresponding author. This research was in part supported by a grant from MSRT. c 2006 Korean Society for Computational & Applied Mathematics and Korean SIGCAM.

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Ali Reza Ashrafi and Bijan Taeri

of the different situations which can arise when n = 5 he has been able to show that f2 (5) = 16. The unpublished proof of that result was a lengthy calculation using the known structure of groups covered by three or four subgroups but gave no structural information about the groups covered by five subgroups. In [4] Bryce, Fedri and Serena presented a short proof for the mentioned result. Following Belcastro and Sherman [3] we shall use the notation #Cent(G) to denote the number of distinct centralizers in the group G. That is, if we define Cent(G) to be the set of all element centralizers of the group G, then #Cent(G) = |Cent(G)|. Definition 1. A group G is called n-centralizer if #Cent(G)
= n. Also, we say G that G is primitive n-centralizer if moreover #Cent Z(G) = n. It is obvious that G is 1-centralizer if and only if G is abelian. In [3] Belcastro and Sherman proved that there is no n-centralizer group, for n = 2, 3, and that G ∼ G is 4-centralizer if and only if Z(G) = Z2 × Z2 . Furthermore, they proved that G ∼ G is 5-centralizer if and only if Z(G) = Z3 × Z3 or S3 , the symmetric group on three letters. By these results one can see that there is no primitive 4-centralizer G ∼ group and that G is a primitive 5-centralizer group if and only if Z(G) = S3 . The first author in [1] and [2] studied this problem and prove that if G is G ∼ 6-centralizer then Z(G) = D8 , A4 , Z2 × Z2 × Z2 or Z2 × Z2 × Z2 × Z2 . In this connection one might ask about the structure of 7-centralizer and primitive 7centralizer groups. In this paper we study this problem and prove the following theorem: G ∼ Theorem. A finite group G is primitive 7-centralizer if and only if Z(G) = D10 or R, the semidirect product of a cyclic group of order 5 by a cyclic group of order 4 acting faithfully.

Throughout this paper all groups mentioned are assumed to be finite. Z(G) denotes the center of G and Dn denotes the dihedral group of order n. Most notation is standard and it is taken mainly from [8] and [10]. 2. Main results In this section we obtain the number of distinct centralizers of a finite group G according to a knowledge of the structure of G modulu its center. We generalize a result of Belcastro and Sherman [3] and apply the result to obtain the structure of primitive 7-centralizer group. G ∼ In [2] Ashrafi proved that if G is a 6-centralizer group, then Z(G) = A4 , D8 , Z2 × G ∼ Z2 × Z2 or Z2 × Z2 × Z2 × Z2 . Furthermore, he proved that if Z(G) = D8 , then G G ∼ is 6-centralizer and if Z(G) = A4 , then G is 6- or 8-centralizer. Now it is natural G ∼ to ask about #Cent(G) when Z(G) = Z2 × Z2 × Z2 .

On finite groups with exactly seven element centralizers

Lemma 1. Let G be a finite group and or 8.

G Z(G)

405

∼ = Z2 ×Z2 ×Z2 . Then #Cent(G) = 6

Proof. It is clear that #Cent(G) ≤ |G : Z(G)| = 8. Suppose Z = Z(G), G Z(G) = Z ∪ x1 Z ∪ x2 Z ∪ · · · ∪ x7 Z and #Cent(G) < 8. Then there are i and j such that 1 ≤ i 6= j ≤ 7 and CG (xi ) = CG (xj ). It is an easy fact CG (xj ) CG (xi ) CG (xi xj ) = = {Z, xi Z, xj Z, xi xj Z} = . This shows that that Z Z Z #Cent(G) ≤ 6 and since G is not abelian, #Cent(G) ≥ 4. Now by Theorems 2 and 4 of [3], #Cent(G) = 6, proving the lemma.  G ∼ Belcastro and Sherman, in [3, Theorem 5], proved that if Z(G) = D2p , p is odd prime, then #Cent(G) = p + 2. In the following lemma we generalize this result to the case where p is an arbitrary positive integer.

Lemma 2. Let G be a finite group and n + 2.

G Z(G)

∼ = D2n , n ≥ 2. Then #Cent(G) =

G Proof. Set Z = Z(G) and suppose that Z(G) has the following presentation: D E xZ, yZ | xn Z = y 2 Z = Z , yZxZyZ = x−1 Z .

Then it is obvious that {xi y j | 0 ≤ i ≤ n − 1; 0 ≤ j ≤ 1} is a left transversal to Z = Z(G) in G. It is enough to investigate the element centralizers CG (xi y j ) for 0 ≤ i ≤ n − 1; 0 ≤ j ≤ 1. Suppose aZ ∈ hxZi Z . Then there exists i, 0 ≤ i ≤ n − 1, such that a ∈ xi Z. Therefore, CG (a) = CG (xi ) = hxi or G. If n is odd, then CG (xi y) = Z ∪ xi yZ, 0 ≤ i ≤ n − 1 and #Cent(G) = n + 2. We now assume i y) that n = 2k. Using a simple calculation we can see that CG (x ⊆ C G (xi yZ) = Z i

Z

y) = C G (xi yZ) for some i, 0 ≤ i ≤ n − 1. {Z, xi yZ, xk Z, xi+k yZ}. Suppose CG (x Z Z Then CG (xi+k y) CG (xi y) ⊆ C G (xi+k yZ) = C G (xi yZ) = . Z Z Z Z Thus, CG (xi+k y) ⊆ CG (xi y). So xk y = yxk . This means that xk ∈ Z(G), which i y) is a contradiction. Hence, | CG (x | = 2 and CG (xi y) = Z ∪ xi yZ for every Z i, 0 ≤ i ≤ n − 1. Therefore,

Cent(G) = {G, CG (x), CG (y), CG (xy), · · · , CG (xn−1 y)}. This completes the proof.  It is an elementary fact that D12 , T and A4 , are the only non-abelian group of order 12, in which the group T presented by: ha, b | a6 = 1, a3 = b2 , ba = a−1 bi.

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Ali Reza Ashrafi and Bijan Taeri

G ∼ If Z(G) = A4 , then by Theorem 3.6 of [1], #Cent(G) = 6 or 8. Also, by the G ∼ previous lemma if Z(G) = D12 , then #Cent(G) = 8. In the following lemma we G ∼ compute #Cent(G) when Z(G) = T.

Lemma 3. Let G be a finite group and

G Z(G)

∼ = T . Then #Cent(G) = 8.

G Proof. First of all, we can assume that Z(G) has the following presentation: E D aZ, bZ | a6 Z = Z, a3 Z = b2 Z, bZaZ = a−1 ZbZ .

It is obvious that haZi ⊆ i

CG (a ) Z

CG (ai ) , Z

0 ≤ i ≤ 5. If for some i, 0 ≤ i ≤ 5, haZi = 6

i

then a ∈ Z(G), a contradiction. Thus for every i, 0 ≤ i ≤ 5 we have: CG (ai ) = Z ∪ aZ ∪ a2 Z ∪ a3 Z ∪ a4 Z ∪ a5 Z.

Using similar argument as in Lemma 2, we can see that the element centralizers CG (ai b), 0 ≤ i ≤ 5, are different and so #Cent(G) = 8, proving the lemma.  Suppose r(G) denotes the maximal size of a subset of pairwise non-commuting elements of G. In the following lemma, under certain condition on r(G), we prove that every element centralizer of G is abelian. In fact, we prove that: Lemma 4. Let G be a finite n−centralizer group and r(G) = n − 1. Then every proper element centralizer of G is abelian. Moreover, for every non-central elements x and y of G, CG (x) = CG (y) or CG (x) ∩ CG (y) = Z(G). Proof. Let {x1 , x2 , · · · , xn−1 } be a set of pairwise non-commuting elements of G having maximal size. Then we have, Cent(G) = {G, CG (x1 ), · · · , CG (xn−1 )}. Suppose for some i, 1 ≤ i ≤ n − 1, X = CG (xi ) is non-abelian. Choose elements a, b ∈ X such that ab 6= ba. It is an easy fact that CG (a) 6= G, CG (b) 6= G and CG (a) 6= CG (b). Without loss of generality, we can assume that CG (a) = CG (xj ) for some j 6= i. So xi ∈ CG (a) = CG (xj ), which is a contradiction. We now assume that x and y are non-central and CG (x) 6= CG (y). Set T = CG (x) ∩ CG (y). Choose an element a ∈ T − Z(G). It is obvious that CG (a) 6= G and CG (x) ∪ CG (y) ⊆ CG (a), which contradicts by our assumption. This proves the lemma.  Let R be a group of order 20 presented by D E R = a, b | a5 = b4 = 1, b−1 ab = a2 . It is easy to see that R is centerless and it is an extension of a cyclic group of order 5 by a cyclic group of order 4 acting faithfully. Let R1 be the unique Sylow 5-subgroup and R2 , R3 , · · · , R6 be the Sylow 2-subgroups of R. We can

On finite groups with exactly seven element centralizers

407

see that Cent(R) = {R, R1 , · · · , R6 } and so R is 7-centralizer. In the following lemma, we investigate the number of element centralizers of the group G, when G ∼ Z(G) = R. Lemma 5. Let G be a finite group and

G Z(G)

∼ = R. Then G is 7-centralizer.

Proof. Suppose Z = Z(G). By assumption, we have E G D = aZ, bZ | a5 Z = b4 Z = Z, b−1 abZ = a2 Z . Z It is clear that {ai bj | 0 ≤ i ≤ 4; 0 ≤ j ≤ 3} is a left transversal to Z = Z(G) in G. So, it is enough to investigate the element centralizers CG (ai bj ) for 0 ≤ i ≤ 4; 0 ≤ j ≤ 3. Since haZi is the unique Sylow 5-subgroup of G Z , C G (aZ) = haZi. Z

On the other hand,

CG (a) Z

⊆ C G (aZ) = haZi and so Z

2

CG (a) = CG (a ) = CG (a3 ) = CG (a4 ) = Z ∪ aZ ∪ a2 Z ∪ a3 Z ∪ a4 Z. We now compute the element centralizers CG (ab), CG (a2 b), CG (a2 b3 ), CG (b) and CG (a3 b). To do this, we note that habZi ⊆ CGZ(ab) ⊆ C G (abZ) = habZi. On the Z other hand, Y1 = habZi = {Z, abZ, a4 b2 Z, a3 b3 Z} Y2 = ha2 bZi = {Z, a2 bZ, a3 b2 Z, ab3 Z} Y3 = ha2 b3 Zi = {Z, a2 b3 Z, ab2 Z, a4 bZ} Y4 = hbZi = {Z, bZ, b2Z, b3 Z} Y5 = ha3 bZi = {Z, a3 bZ, a2 b2 Z, a4 b3 Z}. Thus, CG (ab) = Z ∪ abZ ∪ a4 b2 Z ∪ a3 b3 Z. We now prove that CG (ab) = CG (a4 b2 ) = CG (a3 b3 ). Since CG (a3 b3 ) ⊆ C G (a3 b3 Z) = ha3 b3 Zi = habZi, Z Z 3 3 4 2 CG (a b ) = CG (ab). Suppose CG (a b ) = Z ∪ a4 b2 Z. Then CG (a4 b2 ) is abelian and CG (a4 b2 ) ⊆ CG (ab). On the other hand, CGZ(ab) is cyclic and so CG (ab) is abelian, a contradiction. Thus CG (a4 b2 ) = CG (ab). This shows that CG (ab) = CG (a4 b2 ) = CG (a3 b3 ). Using a similar argument as in above, we can prove that for every non-identity elements uZ, vZ ∈ Yi , 1 ≤ i ≤ 5, CG (u) = CG (v). This completes the proof.  habZi = ha3 b3 Zi ⊆

Now we are ready to state the main result of the paper. Theorem. A group G is primitive 7-centralizer if and only if

G Z(G)

∼ = D10 or R.

G ∼ Proof. Suppose Z(G) = D10 or R. Then by Lemmas 2 and 5, G is 7-centralizer. On the other hand, by Lemma 2.4 of [2], D10 is 7-centralizer, and by the paragraph before Lemma 5, R is 7-centralizer. Hence, G is primitive 7-centralizer.

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We now assume that G is a primitive 7-centralizer finite group. We also suppose that {x1 , · · · , xr } is a set of pairwise non-commuting elements of G having maximal size. Set Xi = CG (xi ), 1 ≤ i ≤ r. By Theorem 5.2 of [12], 3 ≤ r ≤ 6. Thus, it is enough to investigate four cases that which we treat separately. G Case 1: r = 3. In this case by Corollary 5.2 of Tomkinson [12], | Z(G) | ≤ G ∼ f2 (3) = 4. So Z(G) = Z2 × Z2 , a contradiction. G Case 2: r = 4. Using Corollary 5.2 of Tomkinson [12], we can see that | Z(G) |≤ G f2 (4) = 9. So Z(G) is isomorphic to S3 , D8 or Q8 . But, Q8 is not capable and if G ∼ Z(G) = S3 then by Theorem 4 of [3], G is 5-centralizer, a contradiction. Finally, G ∼ if Z(G) = D8 then by Lemma 2, G is 6-centralizer, which is a contradiction. G Case 3: r = 5. In this case | Z(G) | ≤ f2 (5) and by a result of Bryce, Fedri and G Serena [4], f2 (5) = 16. So | Z(G) | ≤ 16 and we can assume that |G : Z(G)| = 10, 12, 14, 16. Our computation together with the computational group theory system GAP [11] in investigating groups of order 16 shows that there is no 7G G ∼ centralizer group of order 16. Suppose | Z(G) | = 14. Then Z(G) = D14 and by G Lemma 2, G is 9-centralizer, a contradiction. We now assume that | Z(G) | = 12. G ∼ G ∼ Then Z(G) = D12 , A4 or T . If Z(G) = D12 or T , then by Lemmas 2 and 3, G is G ∼ 8-centralizer. On the other hand, if Z(G) = A4 then by Theorem 3.6 of Ashrafi G G ∼ [1], #Cent(G) = 6 or 8. So | Z(G) | = 10 and Z(G) = D10 . Case 4: r = 6. Set βi = |G : Xi |, β1 ≤ β2 ≤ · · · ≤ β6 . By Lemma 3.3 of Tomkinson [12], β2 ≤ 5 and by Lemma 4, Xi ∩Xj = Z(G) for every i, j such that G ∼ 1 ≤ i 6= j ≤ 6. If β2 = 2, then β1 = 2 and and Z(G) = Z2 × Z2 , a contradiction. Suppose β2 = 3. If β1 = 3 and one of element centralizers of index 3 is normal in G, then G is 5-centralizer, which is impossible. If β1 = 2 or β1 = 3 and G does not have a normal element centralizer then |G : Z(G)| = 6, which is a contradiction. We now assume that β2 = 4. β1 = 2 implies that |G : Z(G)| = 8 and β1 = 3 implies that |G : Z(G)| = 12. Thus β1 = 4 and |G : Z(G)| ≤ 16. But these cases cannot occur. Therefore β2 = 5 and by Lemma 3.3 of Tomkinson [12], β3 = · · · = β6 = 5. If β1 = 5, then by Ito’s Theorem [7], G ∼ = A × P in which A is an abelian G P G ∼ group and P is a 5−group. Since Z(G) = Z(P ) and #Cent( Z(G) ) = 7, by G ∼ Lemma 2.7 of [1], Z(G) = Z5 × Z5 , a contradiction. Thus β1 < 5. Now by a result of Haber and Rosenfeld, since G is an irredundant union of X1 , · · · , X6 , 6 6 \ \ Xi = Xi = Z(G). If X2  G, then Xi ’s, 2 ≤ i ≤ 6 are normal in G and i=1

i=2

G ∼ so Z(G) = Z5 × Z5 which is impossible. So X2 is a non-normal subgroup of G G and Z(G) = CoreG X2 . This shows that Z(G) is isomorphic to a subgroup of S5 . Using a simple GAP program, we can see that #Cent(S5 ) > 7 and #Cent(A5 ) > G 7. Thus Z(G) is a solvable subgroup of S5 . On the other hand, S5 does not have a

On finite groups with exactly seven element centralizers

409

G G G ∼ subgroup of index 3 or 4. Hence, | Z(G) | = 10 or 20. If | Z(G) | = 10, then Z(G) = G D10 , as desired. Suppose | Z(G) | = 20. But there is three non-abelian groups of order 20, SmallGroup(20, 1), SmallGroup(20, 3) and SmallGroup(20, 4), in GAP notation. Using a GAP program, we can see that these groups are 7centralizer. But, SmallGroup(20, 1) and SmallGroup(20, 4) have an abelian element centralizer of order 10 and S5 does not have an element of order 10, G ∼ which is a contradiction. Therefore, Z(G) = SmallGroup(20, 3) ∼ = R. This completes the proof. 

References 1. A. R. Ashrafi, On finite groups with a given number of centralizers, Algebra Coloquium, 7(2) (2000), 139-146. 2. A. R. Ashrafi, Counting the centralizers of some finite groups, J. Appl. Math. & Computing(old:KJCAM) 7(1) (2000), 115-124. 3. S. M. Belcastro and G. J. Sherman, Counting Centralizers in Finite Groups, Math. Mag. 5 (1994), 366-374. 4. R. A. Bryce, V. Fedri and L. Serena, Covering groups with subgroups, Bull. Austral. Math. Soc. 55(3) (1997), 469-476. 5. J. H. E. Cohn, On n-sum groups, Math. Scand. 75 (1994), 44-58. 6. M. R. Darafsheh and Z. Mostaghim, Computation of the complex characters of the group AU T (GL7 (2)), J. Appl. Math. & Computing(old:KJCAM) 4 (1997), 193-210. 7. M. R. Darafsheh and F. Nowroozi Larki, The character table of the group GL2 (q) when extended by a certain group of order two, J. Appl. Math. & Computing(old:KJCAM) 7 (2000), 643-654. 8. S. Haber and A. Rosenfeld, Groups as unions of proper subgroups, Amer. Math. Monthly 66 (1959), 491-494. 9. N. Ito, On the degrees of irreducible representations of a finite group, Nagoya Math. J. 3 (1951) 5-6. 10. G. Karpilovsky, Group Representations, Volume 2, North-Holland Mathematical Studies, 177, Amesterdam-New York-Oxford-Tokyo. 11. B. H. Neumann, Groups covered by permutable subsets, J. London Math. Soc. 29 (1954), 236-248. 12. D. J. S. Robinson, A course in the theory of groups, Springer-Verlag New York, 1980. 13. M. Schonert et al., GAP: Groups, algorithms, and programming, Lehrstuhl D fur Mathematik, RWTH Aachen, 1994. 14. M. J. Tomkinson, Groups covered by finitely many cosets or subgroups, Comm. Alg. 15 (1987), 845-859. Ali Reza Ashrafi received his BSc from Teacher Training University of Tehran, MSc from Shahid Beheshti University and Ph. D at the University of Tehran under the direction of Mohammad Reza Darafsheh. Since 1996 he has been at the University of Kashan. Now he is an associate professor of mathematics and supervised twenty MSc students in the field of computational group theory, hyperstructures and mathematical chemistry. His research interests focus on computational group theory, finite lattice, hyperstructures and mathematical chemistry. Department of Mathematics, Faculty of Science, University of Kashan, Kashan, Iran e-mail: [email protected]

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Ali Reza Ashrafi and Bijan Taeri

Bijan Taeri received his BSc from Isfahan University of Technology, MSc from Isfahan University and Ph. D at the Isfahan University under the direction of Aliakbar Mohammadi Hassanabadi. Since 1990 he has been at the Isfahan University of Technology. Now he is an associate professor of mathematics. His main intrests include finite and infinite group theory. Department of Mathematics, Isfahan University of Technology, Isfahan, Iran e-mail: [email protected]