# On QuasiS-Propermutable Subgroups of Finite Groups

Authors: Hong Yang; Abid Mahboob; Asim Zafer; Iftikhar Ali; Zafar Ullah; Absar Ul Haq
Journal: Journal of Mathematics
 (2020)

---

## Abstract

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


---

## Body

## 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 andG always denotes a finite group. Moreover Sylow subgroups are denoted by SylG and the set of primes is denoted by πG, if order of G is divisible by some prime. For any q∈πG implies Gq is a SylqG. 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.The termS-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 sylG of G [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 S of G is called c-normal if G=ST with S∩T≤SG, where S contains the largest normal subgroup SG with T⊴G. A subgroup K of G, where G=NGKA, is called S-propermutable in G if K is S-permutable in A [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 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 [11–13]. In this paper, we aim to study some interesting properties of quasi S-propermutable subgroups of G.Definition 1.
(quasiS-propermutable subgroup). A subgroup H of a finite group G is said to be quasi S-propermutable in G if K⊲¯G, such that HK is S-permutable in G and H∩K≤HqsG, where HqsG is the subgroup formed by all subgroups of H which are S-propermutable in G.
In it clear from the definition of quasiS-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.
SupposeG=S4 and K=14. If G=A4K and K∩A4=1, then K is quasi S-propermutable in G, but K does not commute with all Sylow subgroups of A4, so K is not S-propermutable in G.Example 2.
SupposeG=S4 and M is normal subgroup of G of order four. If K is Syl3G, then KM=A4⊴G and K∩M=1. This implies that K is not c-normal but K is quasi S-propermutable.
In the theory of propermutable subgroups, our contributions are the following theorems.Theorem 1.
LetG be a Sylow q‐group, where q is a prime and divides G and G,q−1=1. Then, any Q1≤Q, which is quasi S-propermutable in G, does not have a q-nilpotent supplement in G and is hence solvable.Theorem 2.
Let us consider a Sylowq-subgroup Q of M, where M⊴G and q is a prime divisor of M satisfying M,q−1=1. If every largest subgroup Q1 of Q is quasi S-propermutable in G such that Q1 does not have a q-supersolvable supplement in G, then each chief factor of G between M and Oq′M is cyclic.Theorem 3.
LetSylqG be contained in G , where q is a prime division of G and G,q−1=1. Then, G is q-nilpotent if every largest subgroup Q1 of Q is quasi S-propermutable in G such that Q1 does not have a q-nilpotent supplement in G.
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 Theorems1–3.In the following lemma, the sufficient conditions forS-subgroup to be S-permutable are given.Lemma 1.
(see [4, 15]). Let X be S- subgroup of G; then, the following statements hold:(1)
IfX≤Y≤G, then X is S-permutable in Y.(2)
IfM⊴G, then XM is S-permutable in G and XM/M is S-permutable in G/M.(3)
IfY≤G, then X∩Y is S-permutable in Y.(4)
X is subnormal in G.(5)
IfY is S-permutable subgroup in G, then X∩Y is S-permutable in G.
In the following lemma, some interesting properties ofS-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. Then, the following statements hold:(1)
XM/M is S-propermutable in G/M.(2)
For a prime divisorq of G, X commutes with some Sylow q-subgroup of G.(3)
IfG is π-solvable, then X commutes with some Hall π-subgroup of G.
In the following lemma, we give equivalent statements for aq-subgroup of a group G.Lemma 3.
LetK be a q-subgroup of a group G. Then, the following statements are equivalent:(1)
K is S-permutable in G.(2)
K is S-propermutable in G and K≤OqG.Proof.
(1)
⟶2. Suppose K is S-permutable in G. Since K is S-permutable in G, K is also subnormal in G [4], that is, K≤OqG. As an S-permutable subgroup is also S-propermutable, (2) holds.(2)
⟶1 Suppose K is S-propermutable in G. By definition, there will be D≤G such that(1)G=NGKD,KY=YK,∀Y∈SylD.
Particularly, ifY=S∈SylpD, p≠q, then KS=SK and(2)K=OqG∩KS⊲KS.
Therefore,OqG is a subgroup of NGK, and hence K is S-permutable [16].
Some properties ofS-propermutable subgroups are given in the following lemma.Lemma 4.
LetX be S-propermutable and suppose that T⊴G and Y≤G. Then, we have the following statements:(1)
IfX≤Y, then X is S-propermutable in Y.(2)
IfA is any S-propermutable subgroup in G, then AT/T is S-propermutable in G/T.(3)
IfY is S-propermutable in G, then XY is S-propermutable in G.(4)
IfT≤Y and Y/T is S-propermutable in G/T, then Y is S-propermutable in G.Proof.
(1)
SupposeD is a supplement of X such that(3)G=NGXD,XY=YX,∀Y∈SylD.By using Dedekind identity, we have(4)Y=NGXD∩Y=NGXD∩Y=NGXD1.So,D1 be the supplement of X in Y.Furthermore, there existsS∈SylD for any A∈SylD1 such that A≤S. So, we have XS=SX, and hence(5)XS∩Y=XS∩Y=XA,Y∩SX=Y∩SX=AX.Therefore,(6)XA=AX,∀A∈SylD1.Hence,X is S-propermutable in Y.(2)
This follows immediately from Lemma2 (1).(3)
SupposeD≤G, such that G=NGXD and XV=VX, ∀V∈SylD. Since Y is S-propermutable in G, XY≤G and D is a supplement of NGXY in G. Thus,(7)XYV=XYV=XVY,XVY=XVY=VXY=VXY.Hence,XY is S-propermutable in G.(4)
LetY/T be S-propermutable in G/T. Then, by definition, there exists D/T, which is supplement of NGY/T to G/T. As D is also a supplement of NGY to G, RT/T is a SylD/T, for any R∈SylD. This implies(8)YT⋅RTT=RTT⋅YT.
So,(9)YRT=RTY.
Furthermore,T is contained in Y. Thus,(10)YR=RY,∀R∈SylD.
Hence,Y is S-propermutable in G.
The following two lemmas are about the basic properties of subgroups of groupG.Lemma 5.
SupposeX≤Y≤G. Then, we have the following statements:(1)
XqsG≤XqsK.(2)
LetY be p-group and X⊴G. Then, YqsG/X≤Y/XqsG/H.(3)
DqsGX/X≤DX/XqsG/X, where X,D=1 and X⊴G.Proof.
These results can be easily proved by using Lemmas3 and 4.Lemma 6.
Suppose thatX≤G. Then, we have the following statements:(1)
IfX is quasi S-propermutable and X≤Y≤G, then X is quasi S-propermutable in Y.(2)
IfC⊴G such that B≤X and X is q-group and quasi S-propermutable, where q is a prime, then X/C is quasi S-propermutable in G/C.(3)
IfC is a normal q′-subgroup of G and X is a q-subgroup and quasi S-propermutable in G, then XC/C is quasi S-propermutable in G/C.(4)
IfX is quasi S-propermutable in G, such that X≤Y⊴G, then B⊴G such that XB is S-permutable in G with X∩B≤XqsG and XB≤Y.Proof.
(1)
LetX≤Y≤G and B⊴G such that XB is S-permutable in G and X∩B≤XqsG. Then, Y∩B is normal and(11)XY∩B=XB∩Y,isS-permutable in Y. Using Lemma 1 (3), we have(12)X∩Y∩B=X∩B≤XqsG≤XqsK.Hence,X is quasi S-propermutable in Y.(2)
LetX be a quasi S-propermutable in G, so we have Y⊴G such that XY is S-permutable in G and X∩Y≤XqsG.By Lemma1 (2), we have YC/C⊴G/C and X/CYC/C=XY/C is S-permutable in G/C.Using Lemma5 (2), we have(13)XC∩YCC=X∩YCC≤XqsGCC=XqsGC≤XCqsG/C.Thus,X/C is quasi S-propermutable in G/C.(3)
LetX be quasi S-propermutable in G, so we have Y⊴G such that XY is S-permutable and X∩Y≤XqsG.Obviously,YC/C⊴G and(14)XCYCC=XYCC,areS-permutable in G/C. Now by Lemma 1 (2) and XC:X,XC:C=1, we have(15)XCC∩YCC=XC∩YCC=X∩YC∩YCC=X∩YCC≤XqsGCC.Using Lemma5 (3), we have(16)XqsGCC≤XCCqsG/C.Hence,XC/C is quasi S-propermutable in G/C.(4)
LetX be quasi S-propermutable in G. So, we have Y⊴G such that XY is S-permutable in G and X∩Y≤XqsG.
Now ifS=B∩Y, then S will be normal in G, and(17)BS=XB∩Y=XB∩Y,is S-permutable. Now, using Lemma 1 (5), we have XS≤Y and(18)X∩S=X∩B∩Y=X∩B≤XqsG.
Hence, the desired result is proved.
The relation betweenq-supersoluble, q-nilpotent, cyclic Sylow q-subgroup, and normal subgroups is given in the following lemma.Lemma 7.
Letq be a prime divisor of G such that (G, q − 1) = 1. Then,(1)
IfG is q-supersoluble, then G is q-nilpotent.(2)
IfG has cyclic Sylow q-subgroup, then G is q-nilpotent.(3)
IfG:X=q and X≤G, then X is normal in G.(4)
IfM=q and M⊲¯G, then N lies in ZG.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 Y⊴G. Then, Y/ΦY≤ZUG/ΦY if and only if Y≤ZUG.Lemma 9.
(see [16], Theorem A). If Q is an S-permutable q-subgroup of G, then NGQ≥OqG.Lemma 10.
(see [2], VI, 4.10). Let C,D≤G such that G≠CD. Then, a nontrivial normal subgroup of G contains either C or D satisfying CDg=DgC, 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. Then, G is q-nilpotent if every largest subgroup of Q has a q-nilpotent supplement in G.Lemma 12.
(see [20], Lemma 2.11). SupposeM is elementary abelian normal subgroup of G. Let A≤M satisfying 1<A<M and K≤M such that K=A is S-permutable in G. Then, G contains largest normal subgroup of M.Lemma 13.
SupposeX⊲¯G is q-subgroup, where q is a prime. Then, we have A⊴G where A is the largest subgroup of X which is also quasi S-propermutable.Proof.
If order ofX is q, then the result holds.
IfY≤X is a normal q-subgroup and X≠Y, then by using Lemma 6 (2), we can easily obtain the required result.
If the subgroupL/Y of X/Y⊲¯G/Y, then obviously L≤X and L⊴G and the result holds.
IfX=Y and L is any largest subgroup of X, then there will be E⊴G such that LE is S-permutable and L∩E≤LqsG. Let L≠LqsG. Then, LE≠L and Y≠1. If X≤LE, then(19)X=X∩LE=LX∩E.
Hence,X≤E, which shows that L=L∩E=LqsG. Thus, it is a contradiction.
Now, ifX≰LE, then(20)L=LE∩X.
So using Lemma1 (5), LE∩X is S-permutable, which is again a contradiction. Thus, L=LqsG. 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.


## 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 OqG=1.Let(21)OqG≠1,and take(22)OqG=C.Then obviously,Q/C is a SylqG/C. Suppose Q1/C is the largest subgroup of Q/C. Clearly Q1 will be the largest subgroup of Q and Q1/C has a q-nilpotent supplement AC/C in G/C provided Q1 has supplement of A in G which is q-nilpotent.IfQ1 is quasi S-propermutable, then by using Lemma 6 (2), Q1/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 OqG=1.Step 2. In this step, we prove that Oq′G=1.Suppose on contrary that(23)Oq′G≠1.If(24)Oq′G=V,then clearlyQV/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 Q1 of Q such that J=Q1V. Then, J/V has a q-nilpotent supplement BV/V in G/V. If Q1 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 Q1 is quasi S-propermutable in G. Since G has smallest order, G/V is solvable and by using the Feit–Thompson theorem, V is solvable. It follows that G is solvable, which contradicts our supposition, and hence Oq′G=1.Step 3. Here, we prove that Q is not cyclic.LetQ 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.Step 4. Here, we prove that Y is not solvable if Y⊲¯G and QY=G.LetY be q-soluble; then, either(25)OqY≠1,orOq′Y≠1.As(26)OqY≤OqG,so(27)Oq′Y≤Oq′G.Thus,(28)OqG≠1,orOq′G≠1,which is a contradiction to step (1) or (2). So our supposition is wrong andY is not solvable.Now, we will prove the later part. For this, let(29)QY<G.Then, by Lemma6 (1), every largest subgroup Q1 of Q is quasi S-propermutable in QY.AsQ1 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.Step 6. Q1∩Y=Q1qsG∩Y.
By Lemma11, 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 Q1 of Q such that Q1 is quasi S-propermutable, so(30)N⊴G,as Q1N is S-permutable and(31)Q1∩N≤Q1qsG.
Now if(32)N=1,then Q1 is S-permutable in G.
Now by Lemma9, we have(33)Q1⊴QOqG=G.
In view of step (5),(34)Q1=1orY≤Q1.
By step (4),Y is not solvable. This implies(35)Q1=1.
Hence,Q is cyclic, which is a contradiction to step (3). So,(36)N≠1,Y≤N.
Consequently,(37)Q1∩Y=Q1qsG∩Y.
Now, for any Sylow p-subgroupYp of Y with p≠q, we may write by using step (2)(38)Q1qsGYp=YpQ1qsG.
So,(39)Q1qsGYp∩Y=YpQ1qsG∩Y=YpQ1∩Y,that is,(40)Q1∩Y,is S-permutable in Y. Let(41)Y≅Y1×⋯×Yk.
By Lemma2 (1), Q1∩Y is S-permutable in Q1∩YY1. Hence,(42)Q1∩YY1pm1∩Y1=Y1pm1Q1∩Y∩Y1=Y1pm1Q1∩Y1,for any m1∈Y1, where Y1p is a SylpY1 with p≠q. As (Y1p)m1Q1∩Y1≠Y1, so by Lemma 10, Y1 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 C is q-nilpotent.Suppose thatQ1 is the largest subgroup of Q and Q1 has a q-supersolvable supplement X∩C in C provided Q1 has a q-supersolvable supplement X in G. Because(43)C,q−1=1,this impliesX∩C is q-nilpotent by Lemma 7 (1). If Q1 is quasi S-propermutable in G, then Q1 is also quasi S-propermutable in C by Lemma 6 (1). Also, Q1 does not have any q-nilpotent supplement in C. So, by Theorem 1, C is q-nilpotent.Step 2. In this step, we show that Q=C.Using step (1),Oq′C is the normal Hall q′-subgroup of C.LetOq′C≠1. We can check it easily that our theorem is true for G/Oq′C,C/Oq′C. Using induction, we can see that every chief factor of G/Oq′C, between 1 and C/Oq′C, is cyclic, which implies that each factor between C and Oq′C is cyclic, so Oq′C = 1, and hence Q = C.Step 3. Here, we prove that Φ(Q) = 1.First, we letΦ(Q) ≠ 1; then, by Lemma 3 (2), our theorem holds for G/ΦQ,Q/ΦQ. Every chief factor of G/ΦQ under Q/ΦQ is cyclic by our selection of G,C by Lemma 8, which is a contradiction.Step 4. Here, we prove that every largest subgroup of Q is quasi S-propermutable in G.ConsiderQ1 is the largest subgroup of Q such that J is q-supersolvable supplement of Q1 in G. Thus,(44)QJ=G,withQ∩J≠1. Because(45)Q∩J⊴J,we suppose thatQ∩J contains a smallest normal subgroup Y of J. Here, obviously Y=q.SinceQ is elementary abelian and G=QJ, this implies(46)Y⊴G.Here, we can check that our theorem holds forG/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.LetQ⊲¯G, so by Lemma 13, G contains some largest normal subgroup of Q, which cannot be true because Q is of smallest order.Step 6. Let Y⊲¯Q of G; then,(47)QY≤ZUGY,Y>q.Moreover, using Lemma3 (2), our theorem is satisfied G/Y,Q/Y. Thus, from our selection of G,C=G,Q, every chief factor of G/Y under Q/Y is cyclic.IfY=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(48)YSS≤QS,and from the isomorphism(49)YSS≅Y,it follows that(50)Y=q,a contradiction again. Thus, step (6) is true.Step 7. Finally, to prove our theorem, we need the following contradiction.
Suppose thaty⊲¯Q of G and Y1 is a largest subgroup of Y. To show Y1 is S-permutable, we may suppose that B is a complement of Y in Q, as Q is an elementary abelian q-group.
Also, takeW=Y1B. 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(51)W∩S≤WqsG,WS≤Q,and WS is S-permutable in G. So by virtue of Lemma 3, WqsG is an S-permutable subgroup in G.
Now, ifS=Q, then W=WqsG is S-permutable; by Lemma 1 (5),(52)W∩Y=Y1C∩Y=Y1C∩Y=Y1,is S-permutable.
IfS=1, this gives W=WS is S-permutable. As a result, Y1 is S-permutable. Consider 1<S<Q; then, Y≤S by step (6). So, by Lemma 1 (5),(53)Y1=W∩Y=WqsG∩Y,is S-permutable. This implies Y=q, which contradicts step (6).
This completes the proof of our Theorem2.Proof of Theorem 3.
Considerq-nilpotent group G, so G contains a normal Hall q′-subgroup Gq′. Suppose that the largest Q1≤Q; then,(54)G:Q1Gq′=q.
Using Lemma7 (3), we obtain(55)Q1Gq′⊴G.
Clearly,(56)Q1∩Gq′=1.
Thus,Q1 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 Theorem 1.Step 2. Here, we show that G/Y is q-nilpotent provided Y is the smallest unique normal subgroup.LetY⊲¯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 ΦG=1, which is clear from step (2).Step 4. Now, we show that Q is not cyclic.LetQ be cyclic; then, by Lemma 7 (2), G will be q-nilpotent, which is against our supposition. Thus, Q is not cyclic.Step 5. Now, it is obvious that Oq′G=1.Step 6. In this step, we prove that G contained the q-nilpotent supplement of every largest subgroup of Q.Obviously,(57)Y≤OqG.So by step (3), we can select a largestK of G satisfying(58)G=YK,GY≅K.LetQ1 be the largest subgroup of Q. So, we need to show that G contains a q-nilpotent supplement of Q1. As Y has the q-nilpotent supplement K, we will show Y≤Q1, where Q1 is quasi S-propermutable in G. For this, suppose that(59)L⊴G,Q1∩L≤Q1qsG,andQ1L is S-permutable in G. There are two possibilities.(i)
IfL = 1.It follows thatQ1 is S-permutable. Also, by Lemma 9,(60)Q1⊴QOqG=G.In view of step (3) and Lemma7 (2), we have(61)Q1≠1.So by step (2), we have(62)Y≤Q1.(ii)
IfL≠1, then(63)Y≤L.This implies that(64)Q1∩Y=Q1qsG∩Y.By using step (4), we obtain(65)Q1qsGGp=GpQ1qsG,whereGp is any SylpGq≠p.Then,(66)Q1qsG∩Y=Q1qsGGp∩Y⊴Q1qsGGp.Obviously,(67)Q1∩Y⊴Q.That is why(68)Q1∩Y⊴G.SinceY is smallest subgroup, it follows(69)Q1∩Y=1,Q1∩Y=Y.If(70)Q1∩Y=1,then(71)Y=q,because largest subgroup(72)Q1∩Y≤Y.As a result,G is q-nilpotent by Lemma 7 (4) and step (2). Hence,(73)Q1∩Y=Y,Y≤Q1.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.
This completes the proof of Theorem3.


## 4. Concluding Remarks

In this paper, we gave some properties of quasiS-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 byℱ, having all the supersolvable groups and Y⊴G such that G/Y∈ℱ. Then, 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.



---

*Source: 9292171-2020-08-01.xml*