#### 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 [19]). 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 [1215], 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 [1216], 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

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

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