#### 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 [11–13]. 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 1–3.

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*(2)

*S*-propermutable in .*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*(2)

*S*-permutable in .*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*(2)

*S*-propermutable in .*If is any*(3)

*S*-propermutable subgroup in , then is*S*-propermutable in .*If is -propermutable in , then is*(4)

*S*-propermutable in .*If and is*

*S*-propermutable in , then is*S*-propermutable in .*Proof. *(1)Suppose is a supplement of such that By using Dedekind identity, we have So, be the supplement of in . Furthermore, there exists for any such that . So, we have , and hence Therefore, 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*(3)

*S*-propermutable, where is a prime, then is quasi*S*-propermutable in .*If is a normal -subgroup of and is a*(4)

*q*-subgroup and quasi*S*-propermutable in , then is quasi*S*-propermutable in .*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 and is *S*-permutable in . Using Lemma 1 (3), we have Hence, 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 have Thus, is quasi *S*-propermutable in .(3)Let be quasi *S*-propermutable in , so we have such that is *S*-permutable and . Obviously, and are *S*-permutable in . Now by Lemma 1 (2) and , we have Using Lemma 5 (3), we have Hence, 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 . Let and take Then 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 that If then 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, either As so Thus, 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, let Then, 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 . Because this 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 . Because we suppose that contains a smallest normal subgroup of . Here, obviously . Since is elementary abelian and , this implies Here, 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 , then and from the isomorphism it follows that a 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 satisfying Let 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 that and 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 have So by step (2), we have(ii)If , then This implies that By using step (4), we obtain where is any . Then, Obviously, That is why Since is smallest subgroup, it follows If then because largest subgroup As 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.