# Finite Groups Whose Certain Subgroups of Prime Power Order AreS-Semipermutable

Authors: Mustafa Obaid
Journal: ISRN Algebra
 (2011)

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## Abstract

LetG be a finite group. A subgroup H of G is said to be S-semipermutable in G if H permutes with every Sylow p-subgroup of G with (p,|H|)=1. In this paper, we study the influence of S-permutability property of certain abelian subgroups of prime power order of a finite group on its structure.


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## Body

## 1. Introduction

All groups considered in this paper will be finite. Two subgroupsH and K of a group G are said to permute if HK=KH. It is easily seen that H and K permute if and only if HK is a subgroup of G. We say, following Kegel [1], that a subgroup of G is S-quasinormal in G if it permutes with every Sylow subgroup of G. Chen [2] introduced the following concept: a subgroup H of G is said to be S-semipermutable in G if H permutes with every Sylow p-subgroup of G with (p,|H|)=1. Obviously, every S-quasinormal subgroup of G is an S-semipermutable subgroup of G. In contrast to the fact that every S-quasinormal sub-group of G is a subnormal subgroup of G (see [1]), it does not hold in general that every S-semipermutable subgroup of G is a subnormal subgroup of G. It suffices to consider the alternating group of degree 4.Several authors have investigated the structure of a finite group when some information is known about some subgroups of prime power order in the group. Huppert [3] proved that a finite group G is solvable provided that all subgroups of prime order are normal in G. Buckley [4], proved that a group G of odd order is supersolvable provided that all subgroups of prime order are normal in G. Srinivasan [5], and proved that a finite group G is supersolvable if the maximal subgroups of every Sylow subgroup of G are normal in G. Developing the result of Srinivasan, Ramadan [6] proved that if G is a solvable group and the maximal subgroups of every Sylow subgroup of the Fitting subgroup F(G) of G are normal in G, then G is supersolvable.For a finitep-group P, we denote
(1.1)Ω(P)=Ω1(P)ifp>2,Ω(P)=〈Ω1(P),Ω2(P)〉ifp=2,
where Ωi(P)=〈x∈P:|x|=pi〉.Of late there has been a considerable interest to investigate the influence of the abelian subgroups of largest possible exponent of prime power order (we call such subgroups ALPE-subgroups) on the structure of the group. Asaad et al. [7] proved that if G is a group such that for every prime p and every Sylow p-subgroup Gp of G, the ALPE-subgroups of Gp (resp.,   Ω(Gp)) are normal in G, then G is supersolvable. Ramadan [8] proved the following two results. (1) Let G be a group such that for every prime p and every Sylow p-subgroup Gp of G, the ALPE-subgroups of Gp (resp., Ω(Gp)) are S-quasinormal in G, then G is supersolvable. (2) Let K be a normal subgroup of G such that G/K is supersolvable. If for every prime p and every Sylow p-subgroup Kp of K, the ALPE-subgroups of Kp (resp., Ω(Kp)) are S-quasinormal in G, then G is supersolvable.In this paper, we study the structure of a finite group under the assumption that certain subgroups of prime power order areS-semipermutable in the group. We focus our attention on S-semipermutability property of the ALPE-subgroups of a fixed ALPE-subgroup having maximal order of the Sylow subgroups of a finite group. Furthermore, we improve and extend the above-mentioned results by using the concept of S-semipermutability.


## 2. Preliminaries

In this section, we give some results which will be useful in the sequal.Lemma 2.1 (see [2, Lemmas 1 and 2]).
LetG be a group.(i)
IfH is a S-semipermutable subgroup of G and K is a subgroup of G such that H≤K≤G, then H is S-semipermutable in K.(ii)
Letπ be a set of primes, N a normal π′-subgroup of G, and H a π-subgroup of G. If H is S-semipermutable in G, then HN/N is S-semipermutable in G/N.Lemma 2.2 (see [9, Lemma A]).
LetH be a p-subgroup of G; for some prime p. Then H is S-quasinormal in G if and only if Op(G)≤NG(H), where Op(G) is the normal subgroup of G generated by all p′-elements of G.Lemma 2.3.
LetH be a p-subgroup of G, p is a prime. Then the following statements are equivalent:(i)
H is S-quasinormal in G;(ii)
H≤Op(G) and H is S-semipermutable in G.Proof.
(i)⇒(ii): Suppose that H is S-quasinormal in G. So it follows by [1, Satz 1, page 209] that H is subnormal in G and then by [10, Lemma 8.6(a), page 28] that H≤Op(G). Since H is S-quasinormal in G, obviously, it is S-semipermutable in G. Thus (ii) holds.
(ii)⇒(i): Since H is S-semipermutable in G, then HGq=GqH for every Sylow q-subgroup Gq of G with (q,|H|)=1. Clearly, H=Op(G)∩HGq is normal in HGq and so Gq≤NG(H). Thus Op(G)≤NG(H). Applying Lemma 2.2, we have that H is S-quasinormal in G. Thus (i) holdes.Lemma 2.4 (see [7, Theorem 4, page 253]).
LetP be a normal p-subgroup of G. If the ALPE-subgroups of P are normal in G, then P is supersolvably embedded in G.Lemma 2.5 (see [11, Lemma 3.8, page 2245]).
Letp be the smallest prime dividing the order of a group G, and let Gp be a Sylow p-subgroup of G. If Ω(Gp)≤genz∞(G), then G is p-nilpotent.Lemma 2.6 (see [12, Lemma 2.6]).
LetN be a nontrivial normal subgroup of a group G. If N∩Φ(G)=1, then the Fitting subgroup F(N) of N is the direct product of minimal normal subgroups of G which are contained in F(N).Lemma 2.7 (see [13, Lemma 3.3.1, page 23]).
Suppose thatGp is a normal Sylow p-subgroup of G and that Ω(Gp)K is supersolvable, where K is a p′-Hall subgroup of G. Then G is supersolvable.


## 3. Main Results

Theorem 3.1.
Letp be the smallest prime dividing the order of a group G, and let Gp be a Sylow p-subgroup of G. Fix an ALPE-subgroup P of Gp having maximal order. If the ALPE-subgroups of P are S-semipermutable in G, then G is p-nilpotent.Proof.
Suppose that the theorem is false, and letG be a counterexample of minimal order. We prove the following steps.
(1) IfP≤M<G, Then M Is p-Nilpotent
It is clear to see by Lemma2.1 that the ALPE-subgroups of P are S-semipermutale in M, so that M satisfies the hypothesis of the theorem. Thus, the minimality of G yields that M is p-nilpotent.
(2)NG(P) Is p-Nilpotent
Suppose thatP is normal in G. Let H be an ALPE-subgroup of P (in particular, we may take H=P). By hypothesis, H is S-semipermutable in G and so by Lemma 2.3, we have that H is S-quasinormal in G. Hence HGq is a subgroup of G, where Gq is a Sylow q-subgroup of G with q≠p. Clearly, H is a subnormal Hall subgroup of HGq. Thus H is normal in HGq and hence H is normal in PGq as P is abelian. Thus P is supersolvably embedded in PGq by Lemma 2.4 and so P≤Q∞(PGq). Since Q∞(PGq)≤genz∞(PGq) by [14, page 34], it follows by Lemma 2.5 that PGq is p-nilpotent. Thus PGq=P×Gq. Hence Gq≤CG(P), so that Op(G)≤CG(P).IfCG(P)<G, then CG(P) is p-nilpotent by (1). Thus Op(G) is p-nilpotent and so G is p-nilpotent: a contradiction. Thus we may assume that CG(P)=G. Then P≤Z(G), in particular, P≤Z(Gp). So, P=Gp by the maximality of P and we have by [15, Theorem 4.3, page 252] that G is p-nilpotent: a contradiction. Thus we may assume that NG(P)<G. According to (1), we have that NG(P) is p-nilpotent.
(3)Op′(G)=1
IfOp′(G)≠1, we consider the quotient group G/Op′(G). Clearly, GpOp′(G)/Op′(G) is a Sylow p-subgroup of G/Op′(G) and POp′(G)/Op′(G) is an ALPE-Subgroup of GpOp′(G)/Op′(G) having maximal order. By Lemma 2.1, the hypotheses are inherited over G/Op′(G). Thus, the minimality of G implies that G/Op′(G) is p-nilpotent, hence G is p-nilpotent, which is a contradiction.
(4)G=GpGq, Where Gq Is a Sylow q-Subgroup of G with q≠p
SinceG is not p-nilpotent by [15, Theorem 4.5, page 253], there exists a subgroup H of Gp such that NG(H) is not p-nilpotent. But NG(Gp) is p-nilpotent by a similar argument of the proof of the step (2). Thus we may choose a subgroup H of Gp such that NG(H) is not p-nilpotent but NG(K) is p-nilpotent for every subgroup K of Gp with H<K≤Gp. It is easy to see that NG(Gp)≤NG(H)≤G. If NG(H)<G, it follows by (1) that NG(H) is p-nilpotent: a contradiction. Thus NG(H)=G. This leads to Op(G)≠1 and NG(K) is p-nilpotent for every subgroup K of Gp with Op(G)<K≤Gp. Now, by [15, Theorem 4.5, page 253] again, we see that G/Op(G) is p-nilpotent and therefore that G is p-solvable. Since G is p-solvable, for any q∈π(G) with q≠p, there exists a Sylow q-subgroup Gq of G such that GpGq≤G by [15, Theorem 3.5, page 229]. If GpGq<G, then GpGq is p-nilpotent by (1) and hence Op(G)Gq is p-nilpotent. Thus Op(G)Gq=Op(G)×Gq. This leads to Gq≤CG(Op(G))≤Op(G) by [15, Theorem 3.2, page 228] as Op′(G)=1 by (3), which is a contradiction. Thus G=GpGq.
(5) The Final Contradiction
LetN be a minimal normal subgroup of G such that N≤Op(G). Clearly, N∩Z(Gp)≠1 and so Z(Gp)≤P by the maximality of P. Hence 1≠N∩Z(Gp)≤N∩P. By hypothesis, PGq≤G for any Sylow q-subgroup Gq of G with (q,|P|)=1. It is easy to see that N∩P=N∩PGq⊲PGq. Thus Op(G)≤NG(N∩P). If NG(N∩P)<G, then by (1), NG(N∩P) is p-nilpotent. Hence Op(G) is p-nilpotent and so also does G: a contradiction. Thus we may assume that NG(N∩P)=G. By the minimality of N and since N∩P≠1, we have that N∩P=N and so N≤P. If PGq<G, then PGq is p-nilpotent by (1) and hence NGq is p-nilpotent. Thus NGq=N×Gq and so Gq≤CG(N). Thus by (4), G/CG(N) is a p-group and so by [14, Theorem 6.3, page 221], N≤Z∞(G). Since Z∞(G)≤Q∞(G), we have that N≤Q∞(G) which implies that N is supersolvably embedded in G and so clearly that |N|=p. Thus, it is easy to see that the quotient group G/N satisfies the hypothesis of the theorem by Lemma 2.1. Now, by the minimality of G, we see that G/N is p-nilpotent. Since the class of all p-nilpotent groups is a saturated formation, we have that N is the unique minimal normal subgroup of G and NΦ(G). Thus Φ(G)=1 and hence N=Op(G) by Lemma 2.6 and so F(G)=Op(G)=N by (3). Hence Gq≤CG(F(G)). Since G is solvable, it follows by [15, Theorem 2.6, page 216] that CG(F(G))≤F(G)=Op(G): a contradiction. Thus we must have G=PGq. Let Gq* be a Sylow q-subgroup of NG(P). By (2), we have that Gq*⊲NG(P). Hence NG(P)=PGq*=P×Gq*. Thus P≤Z(NG(P)), and, therefore, G is p-nilpotent by [15, Theorem 4.3, page 252]: a final contradiction.We need the following result.Theorem 3.2.
Letℱ be a saturated formation containing the class of supersolvable groups 𝒰. Let Gp be a normal Sylow p-subgroup of a group G such that G/Gp∈ℱ. Fix an ALPE-subgroup P of Gp having maximal order. If the ALPE-subgroups of P are S-semipermutable in G, then G∈ℱ.Proof.
We treat the following two cases.
Case 1.
Op′(G)≠1.
Clearly,GpOp′(G)/Op′(G) is a normal Sylow p-subgroup of G/Op′(G) and POp′(G)/Op′(G) is an ALPE-subgroup of GpOp′(G)/Op′(G) having maximal order. By hypothesis and Lemma 2.1, the ALPE-subgroups of POp′(G)/Op′(G) are S-semipermutable in G/Op′(G). Clearly,
(3.1)(G/Gp)(GpOp′(G)/Gp)≅GGpOp′(G)≅(G/Op′(G))(GpOp′(G)/Op′(G))∈F.
Thus, our hypothesis carries over toG/Op′(G) and so G/Op′(G)∈ℱ by induction on the order of G. Therefore, G/(Op′(G)∩Gp)≅G∈ℱ.
Case 2.
Op′(G)=1.
LetH be an ALPE-subgroup of P. Then H is S-quasinormal in G by Lemma 2.3 and hence Op(G)≤NG(H) by Lemma 2.2. Let  T=POp(G). Then H is normal in T. Thus Lemma 2.4 implies that P is supersolvably embedded in T. Then, T/CT(P) is supersolvable by [14, Lemma 7.15, page 35]. Clearly, Tp=Gp∩T⊲T, where Tp is a Sylow p-subgroup of T. Let Q be a p′-subgroup of CT(P). Then QP=Q×P is a group of automorphisms of Tp=Op(T). But CTp(P)=P, and consequently, Q acts trivially on CTp(P). Then Q acts trivially on Tp by [15, Theorem 3.4, page 179], that is, Q≤CT(Tp). It is easy to see that T is subnormal in G and so Op′(T)≤Op′(G)=1.   Hence F(T)=Tp.  Since T is solvable, it follows by [15, Theorem 2.6, page 216] that Q≤CT(F(T))≤F(T)=Tp: a contradiction. Hence CT(P) must be a p-group and so CT(P)=P. Thus, T/CT(P)=T/P is supersolvable which implies that T is supersolvable by [16, Theorem 4]. Thus Op(G) is supersolvable and therefore, G=GpOp(G) is supersolvable by [17, Exercise 7.2.23, page 159]. Hence, G∈𝒰⊆ℱ.As an immediate consequence of Theorem3.2, we have the following theorem.Corollary 3.3.
LetGp be a normal Sylow p-subgroup of a group G such that G/Gp is supersolvable. Fix an ALPE-subgroup P of Gp having maximal order. If the ALPE-subgroups of P are S-semipermutable in G, then G is supersolvable.We now prove the following theorem.Theorem 3.4.
LetG be a group. For every prime p and every Sylow p-subgroup Gp of G, fix an ALPE-subgroup P of Gp having maximal order. If the ALPE-subgroups of P are S-semipermutable in G, then G is supersolvable.Proof.
By repeated applications of Theorem3.1, the group G has a Sylow tower of supersolvable type. Hence G has a normal Sylow p-subgroup Gp, where p is the largest prime dividing the order of G. By Lemma 2.1, our hypothesis carries over to G/Gp. Thus G/Gp is supersolvable by induction on the order of G. Now, it follows from Corollary 3.3 that G is supersolvable.As an immediate consequence of Theorem3.4, we have the following corollary.Corollary 3.5 (Asaad et al. [7]).
IfG is a group such that the ALPE-subgroups of every Sylow subgroup of G are normal in G, then G is supersolvable.Corollary 3.6 (Ramadan [8]).
IfG is a group such that the ALPE-subgroups of every Sylow subgroup of G are S-quasinormal in G, then G is supersolvable.We need the following Lemma.Lemma 3.7.
LetK be a normal p-subgroup of a group G such that G/K is supersolvable. Fix an ALPE-subgroup P of K having maximal order. If the ALPE-subgroups of P are S-semipermutable in G, then G is supersolvable.Proof.
LetGp be a Sylow p-subgroup of G. We treat the following two cases.
Case 1.
K=Gp.
Then by Corollary3.3, G is supersolvable.
Case 2.
K<Gp.
Putπ(G)={p1,p2,…,pn}, where p1>p2>⋯>pn. Since G/K is supersolvable, it follows by [18, Theorem 5, page 5] that G/K possesses supersolvable subgroups M/K and L/K such that |G/K:M/K|=p1 and |G/K:L/K|=pn. Since M/K and L/K are supersolvable, it follows that M and L are supersolvable by induction on the order of G. Since |G:M|=|G/K:M/K|=p1 and |G:L|=|G/K:L/K|=pn, it follows again by [18, Theorem 5, page 5] that G is supersolvable.Now, we can prove the following theorem.Theorem 3.8.
LetK be a normal subgroup of G such that G/K is supersolvable. For every prime p dividing the order of K and every Sylow p-subgroup Kp of K, fix an ALPE-subgroup P of Kp having maximal order. If the ALPE-subgroups of P are S-semipermutable in G, then G is supersolvable.Proof.
By Lemma2.1, the ALPE-subgroups of P are S-semipermutable in K. Hence K is supersolvable by Theorem 3.4. Thus K has a normal Sylow p-subgroup Kp, where p is the largest prime dividing the order of K. Since Kp is characteristic in K and K⊲G, we have that Kp⊲G. Clearly, (G/Kp)/(K/Kp)≅G/K is supersolvable. By Lemma 2.1, our hypothesis carries over to G/Kp and hence G/Kp is supersolvable by induction on the order of G. Now, it follows from Lemma 3.7 that G is supersolvable.As an immediate consequence of Theorem3.8, we have the following corollary.Corollary 3.9 (Ramadan [8]).
LetK be a normal subgroup of a group G such that G/K is supersolvable. If the ALPE-subgroups of every Sylow subgroup of K are S-quasinormal in G, then G is supersolvable.


## 4. Similar Results

Following similar arguments to those of Theorem3.1, it is possible to prove the following result.Theorem 4.1.
Letp be the smallest prime dividing the order of a group G and let Gp be a Sylow p-subgroup of G. Fix an ALPE-subgroup P of Ω(Gp) having maximal order. If the ALPE-subgroups of P are S-semipermutable in G, then G is p-nilpotent.We prove the following lemma.Lemma 4.2.
LetK be a normal p-subgroup of a group G such that G/K is supersolvable. Fix an ALPE-subgroup P of Ω(K) having maximal order. If the ALPE-subgroups of P are S-semipermutable in G, then G is supersolvable.Proof.
LetGp be a Sylow p-subgroup of G. We treat the following two cases.
Case 1K=Gp.
By [15, Theorem 2.1, page 221], there exists a p′-Hall subgroup T, which is a complement to Gp in G. Hence G/Gp≅T is supersolvable. Since Ω(Gp) is characteristic in Gp and Gp⊲G, we have that Ω(Gp)⊲G. Clearly, Ω(Gp)T/Ω(Gp)≅T is supersolvable. Thus, our hypothesis and Corollary 3.3 imply that Ω(Gp)T is supersolvable. Therefore, G is supersolvable by Lemma 2.7.
Case 2K<G.
Putπ(G)={p1,p2,…,pn}, where p1>p2>⋯>pn. Since G/K is supersolvable, it follows by [18, Theorem 5, page 5] that G/K possesses supersolvable subgroups M/K and L/K such that |G/K:M/K|=p1 and |G/K:L/K|=pn. Since M/K and L/K are supersolvable, it follows that M and L are supersolvable by induction on the order of G. Since |G:M|=|G/K:M/K|=p1 and |G:L|=|G/K:L/K|=pn, it follows again by [18, Theorem 5, page 5] that G is supersolvable.By a similar proof to the proof of Theorem3.4, we can prove the following theorem.Theorem 4.3.
LetG be a group. For every prime p and every Sylow p-subgroup Gp of G,fix an ALPE-subgroup P of Ω(Gp) having maximal order. If the ALPE-subgroups of P are S-semipermutable in G, then G is supersolvable.As an immediate consequence of Theorem4.3, we have the following corollary.Corollary 4.4 (Asaad et al. [7]).
IfG is a group such that for every prime p and every Sylow p-subgroup Gp, the ALPE-subgroups of Ω(Gp) are normal in G, then G is supersolvable.Corollary 4.5 (Ramadan [8]).
IfG is a group such that for every prime p and every Sylow p-subgroup Gp, the ALPE-subgroups of Ω(Gp) are S-quasinormal in G, then G is supersolvable.We can now prove the following corollary.Corollary 4.6.
LetK be a normal subgroup of G such that G/K is supersolvable. For every prime p dividing the order of K and every Sylow p-subgroup Kp of K, fix an ALPE-subgroup P of Ω(Kp) having maximal order. If the ALPE-subgroups of P are S-semipermutable in G, then G is supersolvable.Proof.
By Lemma2.1, the ALPE-subgroups of P are S-semipermutable in K. Hence K is supersolvable by Theorem 4.3. Thus K has a normal Sylow p-subgroup Kp, where p is the largest prime dividing the order of K. Since Kp is characteristic in K and K⊲G, we have that Kp⊲G. Clearly, (G/Kp)/(K/Kp)≅G/K is supersolvable. By Lemma 2.1, the hypothesis of our theorem carries over to G/Kp. Thus G/Kp is supersolvable by induction on the order of G and it follows that G is supersolvable by Lemma 4.2.Remarks 4.7.
(a) The converse of Theorem3.4 is not true. For example, set G=S3×Z3, where S3=〈x,y∣x3=y2=1,yx=x2y〉 and Z3=〈z∣z3=1〉. Clearly, G is supersolvable and G has an abelian Sylow 3-subgroup of exponent 3. It is easy to check that G contains a subgroup 〈xz〉 of order 3 which fails to be S-semipermutable in G.
(b) Theorem4.3 is not true when the smallest prime dividing the order of G is even and Ω(Gp)=Ω1(Gp), where Gp is a Sylow p-subgroup of G. For example, if Q is the quaternion group 〈a,b∣a4=1,b2=a2,b-1ab=a-1〉,  C9 is a cyclic group of order 9 with generator c, and the action of C9 on Q is given by ac=b,bc=ab, then the semidirect product of Q by C9 is a group of even order in which every subgroup of prime order is S-semipermutable. Clearly, the semidirect product of Q by C9 is not supersolvable (see Buckley [4, Examples (ii)]).



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