ISRN Algebra

Volume 2014, Article ID 607984, 4 pages

http://dx.doi.org/10.1155/2014/607984

## On 10-Centralizer Groups of Odd Order

School of Mathematics, Iran University of Science and Technology, Narmak, Tehran 1684613114, Iran

Received 17 February 2014; Accepted 12 March 2014; Published 1 April 2014

Academic Editors: J. Rada and S. Yang

Copyright © 2014 Z. Foruzanfar and Z. Mostaghim. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

Let be a group, and let denote the number of distinct centralizers of its elements. A group is called -centralizer if . In this paper, we investigate the structure of finite groups of odd order with and prove that there is no finite nonabelian group of odd order with .

#### 1. Introduction

Throughout this paper all groups mentioned are assumed to be finite, and we will use usual notation; for example, denotes the cyclic group of order , and denotes the semidirect product of and with normal subgroup , where is a positive integer and is a prime. For a group , denotes the center of , and , where is the centralizer of the element in ; that is, . A group is a -group if is abelian for every . Many authors have studied the influence of on finite group (see [1–9]). It is clear that a group is 1-centralizer if and only if it is abelian. In [6] Belcastro and Sherman proved that there is no -centralizer group for . They also proved that is 4-centralizer if and only if , and is 5-centralizer if and only if or . In [2] Ashrafi proved that if is 6-centralizer, then , , × × , or × × × . In [1] Abdollahi et al. proved that is 7-centralizer if and only if , or , . They also proved that if is 8-centralizer, then , or .

Our main result is as follows.

Theorem 1. *There is no finite nonabelian group of odd order with .*

#### 2. Preliminary Results

By [1], a cover for a group is a collection of proper subgroups whose union is the whole . We use the term -cover for a cover with members. A cover is called irredundant if no proper subcollection is also a cover. A cover is called a partition with kernel if the intersection of pairwise members of the cover is . Neumann in [10] obtained a uniform bound for the index of the intersection of an irredundant -cover in terms of , and Tomkinson [11] improved this bound. For a natural number , let denote the largest index , where is a group with an irredundant -cover whose intersection of all of them is . We know that , , , and (see [12–15], resp.). Now we present some lemmas and propositions that will be used in the proof of Theorem 1.

Lemma 2 (Lemma 3.3 of [11]). *Let be a proper subgroup of the finite group , and let , be subgroups of with and . If , then . Furthermore, if , then and for any two distinct and .*

*Definition 3 (Definition 2.1 of [1]). *A nonempty subset of a finite group is called a set of pairwise noncommuting elements if for all distinct . A set of pairwise noncommuting elements of is said to have maximal size if its cardinality is the largest one among all such sets.

*Remark 4. *Let be a finite group, and let be a set of pairwise noncommuting elements of having maximal size. Then (1) is an irredundant -cover with the intersection (see Theorem 5.1 of [11]).(2) (see Corollary 5.2 of [11]).(3), , , , and (see [12–16], resp.).(4)Let be a group such that every proper centralizer in is abelian. Then for all either or .

If , then contains both and , since and are abelian. Since is not in , and . Thus, . Hence, in such a group , forms a partition with kernel . It follows that forms a partition whose kernel is the trivial subgroup (see also Proposition 1.2 of [17]).

Lemma 5 (Lemma 2.4 of [1]). *Let be a finite nonabelian group, and let be a set of pairwise noncommuting elements of with maximal size. Then*(1)*;*(2)*;*(3)* if and only if ;*(4)* if and only if .*

*Proposition 6 (Proposition 2.5 of [1]). Let be a finite group, and let be a set of pairwise noncommuting elements of having maximal size.(a)If , then(1)for each element , is abelian if and only if for some ;(2)if is a maximal subgroup of for some , then for all . In particular, if , then , and if , then .(b)If , then there exists a proper nonabelian centralizer which contains , , and for three distinct .(c)If , then there exists a proper nonabelian centralizer which contains and for two distinct .*

*Lemma 7 (Lemma 2.6 of [1]). Let be a finite nonabelian group. Then every proper centralizer of is abelian if and only if , where is the maximal size of a set of pairwise noncommuting elements of .*

*Theorem 8 (Theorem 4.2 of [11]). Suppose that is covered by abelian subgroups ; then(i)if , then ;(ii)if , then .*

*Definition 9 (see [18]). *A group is said to be an -sum group if it can be written as the sum of of its proper subgroups and of no smaller number. One then writes , where, for each , is a proper subgroup, which can be assumed to be maximal where convenient.

*Theorem 10 (Theorem 1 of [18]). If , then , with equality if and only if ; and ; .*

*Remark 11. *If and are subgroups of finite index in a group , and and are relatively prime, then .

*Proposition 12 (Proposition 2.2 of [19]). Let be the smallest prime that is dividing . If , then or .*

*Proposition 13 (Proposition 2.9 of [19]). Let be nonabelian, an integer, and a prime. If , then .*

*Now we are ready to state the main result of this paper.*

*3. The Proof of Theorem 1*

*3. The Proof of Theorem 1*

*In this section, we give the proof of the main theorem.*

*Suppose, on the contrary, that is a finite nonabelian group of odd order with . Let be a set of pairwise noncommuting elements of finite nonabelian group having maximal size. Then , , is an irredundant -cover with intersection . Assume that , where . Since is a 10-centralizer group, therefore, by Lemma 5, we have .*

*Suppose . By Remark 4, and ; therefore by Lemma 2, . Since is odd, then . On the other hand, 3 is a divisor of ; therefore, we have . If , then by [6, Theorem 6], which is a contradiction. Now if , then is abelian which is not possible. It implies that .*

*Suppose . By Remark 4, and ; therefore by Lemma 2, . If , since 5 is a divisor of , then . If , then is abelian, which is not possible. If , then by [6, Theorem 6], , a contradiction. Now suppose ; since 3 is a divisor of , therefore . If , then is abelian which is not possible. If , then , and by Proposition 13, . If , then . If , then by Proposition 12, or 14. Hence, and so .*

*Now suppose . By Remark 4, , and by Lemma 2, . Now if , then or 5. If , then by Remark 11, , and by Proposition 6, . Hence, which is not possible, since is nonabelian. If , then , and since 5 is a divisor of , then , which is again a contradiction. Hence, .*

*Now suppose . By Remark 4, , and by Lemma 2, . Now if , by Proposition 6, , a contradiction. If , then or 5. If , then by Remark 11, , and by Proposition 6, . Hence, which is not possible, since is nonabelian. If , then . Now since 5 is a divisor of , , which is again a contradiction. Finally, suppose that . By Lemma 2, . Therefore, , and by Theorem 10, . Again by Proposition 6, . Since is odd, , or 7. If , then . So , and by Proposition 13, , which is a contradiction. If , then . Therefore, is abelian, which is not possible. If , then . Now by [6, Theorem 6], , which is a contradiction. Hence, .*

*Thus, . By Lemma 7, is a -group, and by Remark 4, for all distinct . Now , and by Lemma 2, . If , then , a contradiction. If , then or 5. If , then by Remark 11, . Since , , which is a contradiction. If , then , and since 5 is a divisor of , , which is again a contradiction. If , then , or 7. If , then by Remark 11, , and since , therefore and so , and by Proposition 13, , which is a contradiction. Similarly, If , then , a contradiction. Finally, if , then . Therefore by Theorem 8, . Since is odd and is nonabelian, . If , then , and by Proposition 13, , which is a contradiction. If , then is abelian which is not possible. If , then by [6, Theorem 6], , which is a contradiction. Now suppose that . If or is normal in , then . Since , , and by [6, Theorem 6], , a contradiction. It is easy to see that for any . Therefore, , or 21, for . Let be an element of order 7 in . Then is the normal Sylow 7-subgroup of . Since , . Hence by [19, Lemma 3.1], has exactly one centralizer of index 9. Suppose . Then and , which is a contradiction by Theorem 10. Therefore, . Similarly, we can show that . So we have , , and . By and the property for all distinct , it is easy to see that . So we have , a contradiction.*

*
Now the proof of Theorem 1 is complete.*

*Conflict of Interests*

*Conflict of Interests*

*The authors declare that there is no conflict of interests regarding the publication of this paper.*

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