Journal of Mathematics

Journal of Mathematics / 2020 / Article

Research Article | Open Access

Volume 2020 |Article ID 9292171 | https://doi.org/10.1155/2020/9292171

Hong Yang, Abid Mahboob, Asim Zafer, Iftikhar Ali, Zafar Ullah, Absar Ul Haq, "On Quasi S-Propermutable Subgroups of Finite Groups", Journal of Mathematics, vol. 2020, Article ID 9292171, 8 pages, 2020. https://doi.org/10.1155/2020/9292171

On Quasi S-Propermutable Subgroups of Finite Groups

Academic Editor: Ji Gao
Received26 Feb 2020
Accepted22 Jun 2020
Published01 Aug 2020

Abstract

A subgroup of a finite group is said to be quasi S-propermutable in if such that is S-permutable in and , where is the subgroup formed by all those subgroups of which are -propermutable in . In this paper, we give some generalizations of finite group by using the properties and effects of quasi S-propermutable subgroups.

1. Introduction

A finite group is a group, of which the underlying set contains a finite number of elements. Throughout this paper, all groups are finite and always denotes a finite group. Moreover Sylow subgroups are denoted by and the set of primes is denoted by , if order of is divisible by some prime. For any implies is a . Furthermore, supersolvable groups are denoted by here. Other notions that are used and not defined in this paper are taken from [1, 2]. A solvable group (also called as soluble group) can be constructed from the abelian groups by using extensions.

The term S-propermutable was introduced by Yi and Skiba in [3]. Recall that a subgroup is S-quasinormal if and is S-permutable, if it commutes with all Sylow subgroups of [4]. For interesting properties of S-permutable, we refer the readers to [5, 6]. The c-normal subgroups were introduced by Wang [7] as follows: a subgroup of is called c-normal if with , where contains the largest normal subgroup with . A subgroup of , where , is called S-propermutable in if is -permutable in [3]. The structure of finite groups in which permutability is a transitive relation is discussed in [8] by Robinson in 2001. In 2002, Ballester-Bolinches and Esteban-Romero discussed the Sylow permutable subnormal subgroups of , and weakly s-permutable subgroups of were studied by Skiba in 2007. Beidleman and Ragland in [9] studied some properties of subnormal, permutable, and embedded subgroups in . In 2012, Zhang and Wang studied the influence of s-semipermutable subgroups of . Some generalizations of permutability and S-permutability are given in [10]. For details, we refer the readers to [1113]. In this paper, we aim to study some interesting properties of quasi S-propermutable subgroups of .

Definition 1. (quasi S-propermutable subgroup). A subgroup of a finite group is said to be quasi S-propermutable in if , such that is S-permutable in and , where is the subgroup formed by all subgroups of which are S-propermutable in .
In it clear from the definition of quasi S-propermutable subgroup that both the ideas of c-normal subgroups and S-propermutable subgroups are covered by quasi S-propermutable subgroups. But converse is not true (see Examples 1 and 2).

Example 1. Suppose and . If and , then is quasi S-propermutable in , but does not commute with all Sylow subgroups of , so is not S-propermutable in .

Example 2. Suppose and is normal subgroup of of order four. If is , then and . This implies that is not c-normal but is quasi S-propermutable.
In the theory of propermutable subgroups, our contributions are the following theorems.

Theorem 1. Let G be a Sylow q‐group, where is a prime and divides and . Then, any , which is quasi S-propermutable in , does not have a -nilpotent supplement in and is hence solvable.

Theorem 2. Let us consider a Sylow -subgroup of , where and is a prime divisor of satisfying . If every largest subgroup of is quasi S-propermutable in such that does not have a -supersolvable supplement in , then each chief factor of between and is cyclic.

Theorem 3. Let be contained in , where is a prime division of and . Then, is -nilpotent if every largest subgroup of is quasi S-propermutable in such that does not have a -nilpotent supplement in .
To prove our main contribution, somehow, we used same methodology as we used in [14].

2. Preliminaries

In this section, we present some lemmas that will be helpful to prove Theorems 13.

In the following lemma, the sufficient conditions for -subgroup to be -permutable are given.

Lemma 1. (see [4, 15]). Let be - subgroup of ; then, the following statements hold:(1)If , then is -permutable in .(2)If , then is S-permutable in and is S-permutable in .(3)If , then is -permutable in .(4) is subnormal in .(5)If is -permutable subgroup in , then is S-permutable in .In the following lemma, some interesting properties of S-propermutable and normal subgroups are given.

Lemma 2. (see [3], Lemma 2.3). Let and be S-propermutable and normal in , respectively. Then, the following statements hold:(1) is S-propermutable in .(2)For a prime divisor of , commutes with some Sylow -subgroup of .(3)If is -solvable, then commutes with some Hall -subgroup of .In the following lemma, we give equivalent statements for a q-subgroup of a group .

Lemma 3. Let be a q-subgroup of a group . Then, the following statements are equivalent:(1) is S-permutable in .(2) is S-propermutable in and .

Proof. (1). Suppose is S-permutable in . Since is S-permutable in , is also subnormal in [4], that is, . As an S-permutable subgroup is also S-propermutable, (2) holds.(2) Suppose is S-propermutable in . By definition, there will be such thatParticularly, if , , then andTherefore, is a subgroup of , and hence is S-permutable [16].
Some properties of S-propermutable subgroups are given in the following lemma.

Lemma 4. Let be S-propermutable and suppose that and . Then, we have the following statements:(1)If , then is S-propermutable in .(2)If is any S-propermutable subgroup in , then is S-propermutable in .(3)If is -propermutable in , then is S-propermutable in .(4)If and is S-propermutable in , then is S-propermutable in .

Proof. (1)Suppose is a supplement of such thatBy using Dedekind identity, we haveSo, be the supplement of in .Furthermore, there exists for any such that . So, we have , and henceTherefore,Hence, is S-propermutable in .(2)This follows immediately from Lemma 2 (1).(3)Suppose , such that and , . Since is S-propermutable in , and is a supplement of in . Thus,Hence, is S-propermutable in .(4)Let be S-propermutable in . Then, by definition, there exists , which is supplement of to . As is also a supplement of to , is a , for any . This impliesSo,Furthermore, is contained in . Thus,Hence, is S-propermutable in .
The following two lemmas are about the basic properties of subgroups of group .

Lemma 5. Suppose . Then, we have the following statements:(1).(2)Let be p-group and . Then, .(3), where and .

Proof. These results can be easily proved by using Lemmas 3 and 4.

Lemma 6. Suppose that . Then, we have the following statements:(1)If is quasi S-propermutable and , then is quasi S-propermutable in .(2)If such that and is -group and quasi S-propermutable, where is a prime, then is quasi S-propermutable in .(3)If is a normal -subgroup of and is a q-subgroup and quasi S-propermutable in , then is quasi S-propermutable in .(4)If is quasi S-propermutable in , such that , then such that is S-permutable in with and .

Proof. (1)Let and such that is S-permutable in and . Then, is normal andis S-permutable in . Using Lemma 1 (3), we haveHence, is quasi S-propermutable in .(2)Let be a quasi S-propermutable in , so we have such that is S-permutable in and .By Lemma 1 (2), we have and is S-permutable in .Using Lemma 5 (2), we haveThus, is quasi S-propermutable in .(3)Let be quasi S-propermutable in , so we have such that is S-permutable and .Obviously, andare S-permutable in . Now by Lemma 1 (2) and , we haveUsing Lemma 5 (3), we haveHence, is quasi S-propermutable in .(4)Let be quasi S-propermutable in . So, we have such that is S-permutable in and .Now if , then will be normal in , andis S-permutable. Now, using Lemma 1 (5), we have andHence, the desired result is proved.
The relation between -supersoluble, -nilpotent, cyclic Sylow -subgroup, and normal subgroups is given in the following lemma.

Lemma 7. Let be a prime divisor of such that (, q 1) = 1. Then,(1)If is -supersoluble, then is -nilpotent.(2)If has cyclic Sylow -subgroup, then is -nilpotent.(3)If and , then is normal in .(4)If and , then lies in .

Proof. One can prove (1) by using the approach of [14]. Proofs of (2)–(4) are obvious and can be seen in ([17], Theorem 2.8).
Now, we give some known lemmas that are very important to prove our main theorems.

Lemma 8. (see [18]). Let . Then, if and only if .

Lemma 9. (see [16], Theorem A). If is an S-permutable -subgroup of , then .

Lemma 10. (see [2], VI, 4.10). Let such that . Then, a nontrivial normal subgroup of contains either or satisfying , for any .

Lemma 11. (see [19], Lemma 2.12). Let be a prime divisor of such that and be a Sylow -subgroup of . Then, is -nilpotent if every largest subgroup of has a -nilpotent supplement in .

Lemma 12. (see [20], Lemma 2.11). Suppose is elementary abelian normal subgroup of . Let satisfying and such that is S-permutable in . Then, contains largest normal subgroup of .

Lemma 13. Suppose is q-subgroup, where is a prime. Then, we have where is the largest subgroup of which is also quasi S-propermutable.

Proof. If order of is , then the result holds.
If is a normal q-subgroup and , then by using Lemma 6 (2), we can easily obtain the required result.
If the subgroup of , then obviously and and the result holds.
If and is any largest subgroup of , then there will be such that LE is S-permutable and . Let . Then, and . If , thenHence, , which shows that . Thus, it is a contradiction.
Now, if , thenSo using Lemma 1 (5), is S-permutable, which is again a contradiction. Thus, . So, using Lemma 3, is S-permutable in . Consequently, we have largest subgroup such that , and by using Lemma 12, the result is proved.

3. Proofs of Main Theorems

In this section, we prove our main theorems.

Proof of Theorem 1. We divide our proofs into 6 steps.Step 1. First we prove that .Letand takeThen obviously, is a . Suppose is the largest subgroup of . Clearly will be the largest subgroup of and has a q-nilpotent supplement in provided has supplement of A in which is -nilpotent.If is quasi S-propermutable, then by using Lemma 6 (2), is quasi S-propermutable in . Now, as is smallest, so is solvable. Hence, our supposition is wrong, and thus .Step 2. In this step, we prove that .Suppose on contrary thatIfthen clearly is a Sylow -subgroup of . Consider to be the largest subgroup of . So, there will be a largest subgroup of such that . Then, has a q-nilpotent supplement in . If has a q-nilpotent supplement in , then by using Lemma 6 (3), we obtain that is quasi S-propermutable in provided is quasi S-propermutable in . Since has smallest order, is solvable and by using the Feit–Thompson theorem, is solvable. It follows that is solvable, which contradicts our supposition, and hence .Step 3. Here, we prove that is not cyclic.Let be a cyclic group; then, by Lemma 7, is -nilpotent. So, is solvable, which is a contradiction to our supposition, and hence is not cyclic.Step 4. Here, we prove that is not solvable if and .Let be q-soluble; then, eitherAssoThus,which is a contradiction to step (1) or (2). So our supposition is wrong and is not solvable.Now, we will prove the later part. For this, letThen, by Lemma 6 (1), every largest subgroup of is quasi S-propermutable in .As does not have a q-nilpotent supplement in , fulfills all the conditions of our theorem. Since is of smallest order, this implies and are also solvable, which is a contradiction, and thus .Step 5. Here, we prove that is unique and smallest subgroup of , such that .Since by step 4, for every , is solvable. Hence, is smallest and unique, and .Step 6. .By Lemma 11, is -nilpotent if every largest subgroup of has a -nilpotent supplement in , which shows that is solvable, a contradiction. So, we can assume a largest subgroup of such that is quasi S-propermutable, soas is S-permutable andNow ifthen is S-permutable in .
Now by Lemma 9, we haveIn view of step (5),By step (4), is not solvable. This impliesHence, is cyclic, which is a contradiction to step (3). So,Consequently,Now, for any Sylow p-subgroup of with , we may write by using step (2)So,that is,is S-permutable in . LetBy Lemma 2 (1), is S-permutable in . Hence,for any , where is a with . As () , so by Lemma 10, is not simple, which is a contradiction.
Hence the desired result is proved.

Proof of Theorem 2. Here, we use the contradiction method to prove this theorem. There are seven steps.Step 1. Firstly, we will prove that is -nilpotent.Suppose that is the largest subgroup of and has a -supersolvable supplement in provided has a -supersolvable supplement in . Becausethis implies is q-nilpotent by Lemma 7 (1). If is quasi S-propermutable in , then is also quasi S-propermutable in by Lemma 6 (1). Also, does not have any -nilpotent supplement in . So, by Theorem 1, is -nilpotent.Step 2. In this step, we show that .Using step (1), is the normal Hall -subgroup of .Let . We can check it easily that our theorem is true for . Using induction, we can see that every chief factor of , between 1 and , is cyclic, which implies that each factor between C and is cyclic, so  = 1, and hence Q = C.Step 3. Here, we prove that (Q) = 1.First, we let (Q)  1; then, by Lemma 3 (2), our theorem holds for . Every chief factor of under is cyclic by our selection of by Lemma 8, which is a contradiction.Step 4. Here, we prove that every largest subgroup of is quasi S-propermutable in G.Consider is the largest subgroup of such that J is -supersolvable supplement of in . Thus,with . Becausewe suppose that contains a smallest normal subgroup of . Here, obviously .Since is elementary abelian and , this impliesHere, we can check that our theorem holds for . By our selection of , we can see that every chief factor of under is cyclic. As a consequence, every chief factor of under is cyclic, which is a contradiction, and hence (4) holds.Step 5. Now, we prove that does not have a smallest normal subgroup .Let , so by Lemma 13, contains some largest normal subgroup of , which cannot be true because is of smallest order.Step 6. Let of ; then,Moreover, using Lemma 3 (2), our theorem is satisfied . Thus, from our selection of , every chief factor of under is cyclic.If , then Y is a cyclic group, which contradiction of our supposition. Now if contains two smallest normal subgroups and of , thenand from the isomorphismit follows thata contradiction again. Thus, step (6) is true.Step 7. Finally, to prove our theorem, we need the following contradiction.Suppose that of and is a largest subgroup of . To show is S-permutable, we may suppose that is a complement of in , as is an elementary abelian -group.
Also, take . Clearly, is a largest subgroup of , so by step (4), is quasi S-propermutable in , and by Lemma 6 (4), there will be satisfying the conditionand is S-permutable in . So by virtue of Lemma 3, is an S-permutable subgroup in .
Now, if , then is S-permutable; by Lemma 1 (5),is S-permutable.
If , this gives is S-permutable. As a result, is S-permutable. Consider ; then, by step (6). So, by Lemma 1 (5),is S-permutable. This implies , which contradicts step (6).
This completes the proof of our Theorem 2.

Proof of Theorem 3. Consider -nilpotent group , so contains a normal Hall -subgroup . Suppose that the largest ; then,Using Lemma 7 (3), we obtainClearly,Thus, is quasi S-propermutable in .
For sufficient condition, we suppose that hypothesis is wrong. So, our proof consists of the following seven steps.Step 1. Firstly, we need to prove that is solvable, which can be proved easily by Theorem 1.Step 2. Here, we show that is -nilpotent provided is the smallest unique normal subgroup.Let , which is smallest. By step (1), is solvable; this implies that is an elementary abelian. Hence, in light of Lemma 6, satisfies our theorem. Following this, is -nilpotent as is of smallest order, which is the required result.Step 3. Here, we need to show that , which is clear from step (2).Step 4. Now, we show that is not cyclic.Let be cyclic; then, by Lemma 7 (2), will be -nilpotent, which is against our supposition. Thus, is not cyclic.Step 5. Now, it is obvious that .Step 6. In this step, we prove that contained the -nilpotent supplement of every largest subgroup of .Obviously,So by step (3), we can select a largest of satisfyingLet be the largest subgroup of . So, we need to show that contains a -nilpotent supplement of . As has the -nilpotent supplement , we will show , where is quasi S-propermutable in . For this, suppose thatand is S-permutable in . There are two possibilities.(i)If L = 1.It follows that is S-permutable. Also, by Lemma 9,In view of step (3) and Lemma 7 (2), we haveSo by step (2), we have(ii)If , thenThis implies thatBy using step (4), we obtainwhere is any .Then,Obviously,That is whySince is smallest subgroup, it followsIfthenbecause largest subgroupAs a result, is q-nilpotent by Lemma 7 (4) and step (2). Hence,Step 7. Finally, we prove the contradiction.By step (6), contained the -nilpotent supplement of every largest subgroup of , so by Lemma 11, is -nilpotent, hence a contradiction.
This completes the proof of Theorem 3.

4. Concluding Remarks

In this paper, we gave some properties of quasi S-propermutable subgroups of a finite group. We relate quasi S-propermutable subgroups with solvable subgroups and cyclic subgroups. Lastly, we gave necessary and sufficient condition for quasi S-propermutable subgroups. The following theorem can be obtained immediately from our results.

Theorem 4. Suppose that a saturated formation is denoted by , having all the supersolvable groups and such that . Then, provided every noncyclic Sylow subgroup of is quasi S-propermutable in such that every largest subgroup of does not have any supersolvable supplement in .

Data Availability

All data required for this study are included within the article.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

All authors contributed equally to this study.

Acknowledgments

This research was supported by HEC Pakistan.

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