On Quasi S-Propermutable Subgroups of Finite Groups

School of Information Science and Engineering, Chengdu University, Chengdu 610106, China Department of Mathematics, Division of Science and Technology, University of Education Lahore, Lahore, Pakistan Department of Mathematics, COMSAT University, Vehari Campus, Vehari, Pakistan Department of Mathematics and Statistics, *e University of Lahore, Lahore, Pakistan Department of Basic Science and Humanities, University of Engineering and Technology, Lahore, Narowal Campus, Pakistan


Introduction
A finite group is a group, of which the underlying set contains a finite number of elements. roughout this paper, all groups are finite and G always denotes a finite group. Moreover Sylow subgroups are denoted by Syl(G) and the set of primes is denoted by π(G), if order of G is divisible by some prime. For any q ∈ π(G) implies G q is a Syl q (G). Furthermore, supersolvable groups are denoted by U 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. e term S-propermutable was introduced by Yi and Skiba in [3]. Recall that a subgroup H is S-quasinormal if H ≤ G and is S-permutable, if it commutes with all Sylow subgroups syl(G) of G [4]. For interesting properties of Spermutable, we refer the readers to [5,6]. e c-normal subgroups were introduced by Wang [7] as follows: a subgroup S of G is called c-normal if G � ST with S ∩ T ≤ S G , where S contains the largest normal subgroup S G with T ⊴ G. [3]. e 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 G, and weakly s-permutable subgroups of G were studied by Skiba in 2007. Beidleman and Ragland in [9] studied some properties of subnormal, permutable, and embedded subgroups in G. In 2012, Zhang and Wang studied the influence of s-semipermutable subgroups of G. Some generalizations of permutability and S-permutability are given in [10]. For details, we refer the readers to [11][12][13]. In this paper, we aim to study some interesting properties of quasi S-propermutable subgroups of G.
where H qsG is the subgroup formed by all subgroups of H which are S-propermutable in G.
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).
In the theory of propermutable subgroups, our contributions are the following theorems. Theorem 1. Let G be a Sylow q-group, where q is a prime and divides |G| and (|G|, q − 1) � 1. en, any Q 1 ≤ Q, which is quasi S-propermutable in G, does not have a q-nilpotent supplement in G and is hence solvable.
To prove our main contribution, somehow, we used same methodology as we used in [14].

Preliminaries
In this section, we present some lemmas that will be helpful to prove eorems 1-3.
In the following lemma, the sufficient conditions for S-subgroup to be S-permutable are given.
Lemma 1 (see [4,15]). Let X be Ssubgroup of G; then, the following statements hold: In the following lemma, some interesting properties of S-propermutable and normal subgroups are given.
Lemma 2 (see [3], Lemma 2.3). Let X and M be S-propermutable and normal in G, respectively. en, the following statements hold: (2) For a prime divisor q of |G|, X commutes with some Sylow q-subgroup of G. (3) If G is π-solvable, then X commutes with some Hall π-subgroup of G.
In the following lemma, we give equivalent statements for a q-subgroup of a group G. Lemma 3. Let K be a q-subgroup of a group G. en, the following statements are equivalent: erefore, O q (G) is a subgroup of N G (K), and hence K is S-permutable [16].
Some properties of S-propermutable subgroups are given in the following lemma. □ Lemma 4. Let X be S-propermutable and suppose that T⊴G and Y ≤ G. en, we have the following statements: By using Dedekind identity, we have So, D 1 be the supplement of X in Y.

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Furthermore, there exists S ∈ Syl(D) for any A ∈ Syl(D 1 ) such that A ≤ S. So, we have XS � SX, and hence Furthermore, T is contained in Y. us, Hence, Y is S-propermutable in G. e following two lemmas are about the basic properties of subgroups of group G. □ Lemma 5. Suppose X ≤ Y ≤ G. en, we have the following statements: and X⊴G.
Proof. ese results can be easily proved by using Lemmas 3 and 4. □ Lemma 6. Suppose that X ≤ G. en, we have the following statements: Using Lemma 5 (2), we have are S-permutable in (G/C). Now by Lemma 1 (2) and (|XC: X|, |XC: Using Lemma 5 (3), we have Journal of Mathematics Now if S � B∩Y, then S will be normal in G, and is S-permutable. Now, using Lemma 1 (5), we have XS ≤ Y and Hence, the desired result is proved. e relation between q-supersoluble, q-nilpotent, cyclic Sylow q-subgroup, and normal subgroups is given in the following lemma. □ Lemma 7. Let q be a prime divisor of |G| such that (|G|, q − 1) � 1. en, Proof. One can prove (1) by using the approach of [14]. Proofs of (2)-(4) are obvious and can be seen in ( [17], eorem 2.8). Now, we give some known lemmas that are very important to prove our main theorems. □ Lemma 8 (see [18] Lemma 9 (see [16], eorem A). If Q is an S-permutable q-subgroup of G, then N G (Q) ≥ O q (G).
Lemma 10 (see [2], VI, 4.10). Let C, D ≤ G such that G ≠ C D. en, a nontrivial normal subgroup of G contains either C or D satisfying CD g � D g C, for any g ∈ G.
Lemma 11 (see [19], Lemma 2.12). Let q be a prime divisor of |G| such that (|G|, q − 1) � 1 and Q be a Sylow q-subgroup of G. en, G is q-nilpotent if every largest subgroup of Q has a q-nilpotent supplement in G.
en, we have A⊴G where A is the largest subgroup of X which is also quasi S-propermutable.
Proof. If order of X is q, then the result holds.
If Y ≤ X is a normal q-subgroup and X ≠ Y, then by using Lemma 6 (2), we can easily obtain the required result.
If the subgroup (L/Y) of (X/Y) ⊲ (G/Y), then obviously L ≤ X and L ⊴ G and the result holds.
If X � Y and L is any largest subgroup of X, then there will be E⊴G such that LE is S-permutable and L ∩ E ≤ L qsG . Let L ≠ L qs G. en, LE ≠ L and Y ≠ 1. If X ≤ LE, then Hence, X ≤ E, which shows that L � L ∩ E � L qsG . us, it is a contradiction. Now, if X ≰ LE, then So using Lemma 1 (5), LE ∩ X is S-permutable, which is again a contradiction. us, L � L qsG . So, using Lemma 3, L is S-permutable in G. Consequently, we have largest subgroup X such that X ⊲ G, and by using Lemma 12, the result is proved.

Proofs of Main Theorems
In this section, we prove our main theorems.
Proof of eorem 1. We divide our proofs into 6 steps.
Step 1. First we prove that O q (G) � 1. Let and take en obviously, (Q/C) is a Syl q (G/C). Suppose (Q 1 /C) is the largest subgroup of (Q/C). Clearly Q 1 will be the largest subgroup of Q and (Q 1 /C) has a q-nilpotent supplement (AC/C) in (G/C) provided Q 1 has supplement of A in G which is q-nilpotent. If Q 1 is quasi S-propermutable, then by using Lemma 6 (2), (Q 1 /C) is quasi S-propermutable in (G/C). Now, as G is smallest, so (G/C) is solvable. Hence, our supposition is wrong, and thus O q (G) � 1.
Step 2. In this step, we prove that O q ′ (G) � 1. Suppose on contrary that If then clearly (QV/V) is a Sylow q-subgroup of (G/V). Consider (J/V) to be the largest subgroup of (QV/V). So, there will be a largest subgroup Q 1 of Q such that J � Q 1 V. en, (J/V) has a q-nilpotent supplement (BV/V) in (G/V). If Q 1 has a q-nilpotent supplement B in G, then by using Lemma 6 (3), we obtain that (J/V) is quasi S-propermutable in (G/V) provided Q 1 is quasi S-propermutable in G. Since G has smallest order, (G/V) is solvable and by using the Feit-ompson theorem, V is solvable. It follows that G is solvable, which contradicts our supposition, and hence O q ′ (G) � 1.
Step 3. Here, we prove that Q is not cyclic. Let Q be a cyclic group; then, by Lemma 7, G is q-nilpotent. So, G is solvable, which is a contradiction to our supposition, and hence Q is not cyclic.
which is a contradiction to step (1) or (2). So our supposition is wrong and Y is not solvable. Now, we will prove the later part. For this, let en, by Lemma 6 (1), every largest subgroup Q 1 of Q is quasi S-propermutable in QY. As Q 1 does not have a q-nilpotent supplement in QY, QY fulfills all the conditions of our theorem. Since G is of smallest order, this implies QY and Y are also solvable, which is a contradiction, and thus QY � G.
Step 5. Here, we prove that Y is unique and smallest subgroup of G, such that Y ⊲ G. Since by step 4, QY � G for every Y ⊴ G, (G/Y) is solvable. Hence, Y is smallest and unique, and Y ⊲ G.
By Lemma 11, G is q-nilpotent if every largest subgroup of Q has a q-nilpotent supplement in G, which shows that G is solvable, a contradiction. So, we can assume a largest subgroup Q 1 of Q such that Q 1 is quasi S-propermutable, so as Q 1 N is S-permutable and then Q 1 is S-permutable in G. Now by Lemma 9, we have In view of step (5), By step (4), Y is not solvable. is implies Hence, Q is cyclic, which is a contradiction to step (3). So, Consequently, Now, for any Sylow p-subgroup Y p of Y with p ≠ q, we may write by using step (2) So, that is, Hence the desired result is proved.

□
Proof of eorem 2. Here, we use the contradiction method to prove this theorem. ere are seven steps.
Step 1. Firstly, we will prove that C is q-nilpotent.

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Suppose that Q 1 is the largest subgroup of Q and Q 1 has a q-supersolvable supplement X∩C in C provided Q 1 has a q-supersolvable supplement X in G. Because this implies X∩C is q-nilpotent by Lemma 7 (1). If Q 1 is quasi S-propermutable in G, then Q 1 is also quasi Spropermutable in C by Lemma 6 (1). Also, Q 1 does not have any q-nilpotent supplement in C. So, by eorem 1, C is q-nilpotent.
Step 2. In this step, we show that Q � C.
We can check it easily that our theorem is true for (G/O q ′ (C, (C/O q ′ (C)))). Using induction, we can see that every chief factor of (G/O q ′ (C)), between 1 and (C/O q ′ (C)), is cyclic, which implies that each factor between C and O q ′ (C) is cyclic, so O q ′ (C) � 1, and hence Q � C.
Step 4. Here, we prove that every largest subgroup of Q is quasi S-propermutable in G. Consider Q 1 is the largest subgroup of Q such that J is q-supersolvable supplement of Q 1 in G. us, with Q ∩ J ≠ 1. Because we suppose that Q ∩ J contains a smallest normal subgroup Y of J. Here, obviously |Y| � q.
Since Q is elementary abelian and G � QJ, this implies Here, we can check that our theorem holds for ((G/Y), (Q/Y)). By our selection of (G, C), we can see that every chief factor of (G/Y) under (Q/Y) is cyclic. As a consequence, every chief factor of G under Q is cyclic, which is a contradiction, and hence (4) holds.
Step 5. Now, we prove that G does not have a smallest normal subgroup Q. Let Q ⊲ G, so by Lemma 13, G contains some largest normal subgroup of Q, which cannot be true because Q is of smallest order.
If |Y| � q, then Y is a cyclic group, which contradiction of our supposition. Now if Q contains two smallest normal subgroups S and Y of G, then and from the isomorphism it follows that a contradiction again. us, step (6) is true.
Step 7. Finally, to prove our theorem, we need the following contradiction.
Suppose that y ⊲ Q of G and Y 1 is a largest subgroup of Y. To show Y 1 is S-permutable, we may suppose that B is a complement of Y in Q, as Q is an elementary abelian q-group.
Also, take W � Y 1 B. Clearly, W is a largest subgroup of Q, so by step (4), W is quasi S-propermutable in G, and by Lemma 6 (4), there will be S ⊴ G satisfying the condition and WS is S-permutable in G. So by virtue of Lemma 3, W qsG is an S-permutable subgroup in G. Now, if S � Q, then W � W qsG is S-permutable; by Lemma 1 (5), is S-permutable. If S � 1, this gives W � WS is S-permutable. As a result, Y 1 is S-permutable. Consider 1 < S < Q; then, Y ≤ S by step (6). So, by Lemma 1 (5), is S-permutable. is implies |Y| � q, which contradicts step (6). is completes the proof of our eorem 2.
□ Proof of eorem 3. Consider q-nilpotent group G, so G contains a normal Hall q′-subgroup G q ′ . Suppose that the largest Q 1 ≤ Q; then, Using Lemma 7 (3), we obtain Journal of Mathematics Clearly, us, Q 1 is quasi S-propermutable in G. 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 G is solvable, which can be proved easily by eorem 1.
Step 2. Here, we show that (G/Y) is q-nilpotent provided Y is the smallest unique normal subgroup. Let Y ⊲ G, which is smallest. By step (1), G is solvable; this implies that Y is an elementary abelian. Hence, in light of Lemma 6, (G/Y) satisfies our theorem. Following this, (G/Y) is q-nilpotent as G is of smallest order, which is the required result.
Step 3. Here, we need to show that Φ(G) � 1, which is clear from step (2).
Step 4. Now, we show that Q is not cyclic. Let Q be cyclic; then, by Lemma 7 (2), G will be q-nilpotent, which is against our supposition. us, Q is not cyclic.
Step 6. In this step, we prove that G contained the q-nilpotent supplement of every largest subgroup of Q. Obviously, So by step (3), we can select a largest K of G satisfying Let Q 1 be the largest subgroup of Q. So, we need to show that G contains a q-nilpotent supplement of Q 1 . As Y has the q-nilpotent supplement K, we will show Y ≤ Q 1 , where Q 1 is quasi S-propermutable in G. For this, suppose that L ⊴ G, and Q 1 L is S-permutable in G. ere are two possibilities.
It follows that Q 1 is S-permutable. Also, by Lemma 9, In view of step (3) and Lemma 7 (2), we have So by step (2), we have is implies that By using step (4), we obtain where G p is any Syl p (G)(q ≠ p). en, Obviously, at is why Since Y is smallest subgroup, it follows If then |Y| � q, because largest subgroup As a result, G is q-nilpotent by Lemma 7 (4) and step (2). Hence, Step 7. Finally, we prove the contradiction.
By step (6), G contained the q-nilpotent supplement of every largest subgroup of Q, so by Lemma 11, G is q-nilpotent, hence a contradiction.
is completes the proof of eorem 3.

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

Theorem 4.
Suppose that a saturated formation is denoted by F, having all the supersolvable groups and Y⊴G such that (G/Y) ∈ F. en, G ∈ F provided every noncyclic Sylow subgroup Q of F * (Y) is quasi S-propermutable in G such that every largest subgroup of Q does not have any supersolvable supplement in G.

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