FINITE GROUPS WITH GIVENσ-EMBEDDED ANDσ-n-EMBEDDED SUBGROUPS1
Zhenfeng Wu∗, Chi Zhang∗and Jianhong Huang∗∗
∗Department of Mathematics, University of Science and Technology of China
Hefei, 230026, P. R. China
∗∗School of Mathematics and Statistics, Jiangsu Normal University
Xuzhou, 221116, P. R. China
e-mails: [email protected], [email protected]; [email protected]
(Received 22 December 2016; accepted 27 March 2017)
LetGbe a finite group andσ ={σi|i∈I}be a partition of the set of all primesP. A setHof
subgroups ofGis said to be a complete Hallσ-set ofGif every non-identity member ofHis a Hallσi-subgroup ofGandHcontains exactly one Hallσi-subgroup ofGfor everyσi ∈σ(G).
A subgroup H is said to be σ-permutable if Gpossesses a complete Hallσ-setHsuch that HAx=AxHfor allA∈ Hand allx∈G. LetHbe a subgroup ofG. Then we say that: (1)H
isσ-embedded inGif there exists aσ-permutable subgroupT ofGsuch thatHT =HσGand H∩T ≤HσG, whereHσGis the subgroup ofH generated by all those subgroups ofHwhich
areσ-permutable inG, andHσGis theσ-permutable closure ofH, that is, the intersection of all σ-permutable subgroups ofGcontainingH. (2)Hisσ-n-embedded inGif there exists a normal subgroupTofGsuch thatHT =HGandH∩T ≤H
σG. In this paper, we study the properties
of the new embedding subgroups and use them to determine the structure of finite groups.
Key words : Finite group;σ-embedded subgroup;σ-n-embedded subgroup;σ-permutable
sub-group; supersoluble.
1. INTRODUCTION
Throughout this paper, all groups are finite andGalways denotes a group. Moreover,nis an integer,P is the set of all primes. The symbolπ(n)denotes the set of all primes dividingnandπ(G) =π(|G|), the set of all primes dividing the order ofG.
1
Research was supported by the NNSF of China (11371335 and 11401264) and Wu Wen-Tsun Key Laboratory of
In what follows,σ = {σi|i ∈ I} is some partition ofP, that is,P =
S
i∈Iσi andσi∩σj = ∅
for alli6=j. Πis always supposed to be a non-empty subset of the setσandΠ0 =σ\Π. We write
σ(n) ={σi|σi∩π(n)6=∅}andσ(G) =σ(|G|).
Following [1] and [2], G is said to be σ-primary if G = 1or |σ(G)| = 1; n is a Π-number if π(n) ⊆ Sσi∈Πσi; a subgroup H of G is called a Π-subgroup of G if |H| is a Π-number; a
subgroupH ofG is called a HallΠ-subgroup of Gif H is a Π-subgroup of Gand |G : H|is a
Π0-number. We use |G|σi to denote the order of the Hall σi-subgroup of G, OΠ(G) denotes the
subgroup ofGgenerated by all its Π0-subgroups andOΠ(G) denotes the subgroup ofGgenerated by all its normalΠ-subgroups. Instead ofO{σi}(G)we writeOσi(G). A setHof subgroups ofG is said to be a complete HallΠ-set ofGif every non-identity member ofHis a Hallσi-subgroup of Gfor someσi ∈ Π andHcontains exactly one Hallσi-subgroup ofGfor everyσi ∈ Π∩σ(G).
In particular, whenΠ = σ, we call Ha complete Hallσ-set ofG. IfGhas a complete Hallσ-set
H={H1,· · ·, Ht}such thatHi 6= 1andHiHj =HjHifor alli, j, thenHis said to be aσ-basis of G. Following [1] and [3],Gis said to beΠ-full (resp.σ-full or aσ-group) ifGpossesses a complete
HallΠ-set (resp. Hallσ-set);a Π-full (resp. σ-full) group of Sylow type if every subgroup ofGis aDσi-group for allσi ∈Π∩σ(G)(resp. σi ∈σ(G)); a subgroupH ofGis calledσ-subnormal in Gif there is a subgroup chainH =H0 ≤ H1 ≤ · · · ≤ Ht =Gsuch that eitherHi−1is normal in
Hi orHi/(Hi−1)Hi isσ-primary for alli= 1,2,· · ·, t; a groupGis calledσ-soluble if every chief
factor ofGisσ-primary.
It is well known that embedded subgroups and supplemented subgroups play an important role in
the theory of finite group. For example, a subgroupH ofGis said to bec-normal [4] inGifGhas a normal subgroupT ofGsuch thatG = HT andH∩T ≤HG, whereHGis the normal core of H. A subgroupHofGis calleds-embedded [5] inGifGhas ans-permutable subgroupTsuch that
HsG =HT andH∩T ≤H
sG, whereHsGis the subgroup ofH generated by all those subgroups
ofH which ares-permutable inG(note that a subgroupAofGis said to bes-permutable inGif
AP =P Afor any Sylow subgroupP ofG) andHsGthe intersection of alls-permutable subgroups
ofGcontainingH. A subgroupH ofGis calledn-embedded [5] inGifGhas a normal subgroup
Tsuch thatHG=HT andH∩T ≤HsG. A subgroupHofGis calledσ-permutable [1] inGifG
possesses a complete Hallσ-setHsuch thatHAx =AxHfor allA∈ Hand allx∈G. Note that in the case whenσ={{2},{3},· · · }, then aσ-permutable subgroup is just ans-permutable subgroup. By using the above embedded subgroups and supplemented subgroups, people have obtained a series
of interesting results (see, for example [1-8]). Based on this fact, we consider the following new
Definition 1.1 — LetHbe a subgroup of a groupG. We say that:
(1)Hisσ-embedded inGif there exists aσ-permutable subgroupT ofGsuch thatHT =HσG
andH∩T ≤HσG.
(2)Hisσ-n-embedded inGif there exists a normal subgroupT ofGsuch thatHT =HGand
H∩T ≤HσG.
HereHσGis theσ-core ofH, that is, the subgroup ofH generated by all those subgroups ofH
which areσ-permutable inGandHσGis theσ-permutable closure ofH, that is, the intersection of allσ-permutable subgroups ofGcontainingH.
It is clear that everyc-normal subgroup, everyn-embedded subgroup and everyσ-permutable sub-group areσ-n-embedded inG. Moreover, ifGis aσ-full group of Sylow type, then everys-embedded subgroup, everyn-embedded subgroup, everyσ-n-embedded subgroup and everyσ-permutable sub-group ofGare allσ-embedded subgroup ofG. But, the following example shows that the converse is not true.
Example 1.2 : LetG= (C29oC7)×A5, whereC29oC7is a non-abelian group of order 203 and
A5is the alternating group of degree 5. LetBbe a subgroup ofA5of order 12. Letσ={σ1, σ2, σ3},
whereσ1 = {2,3,5}, σ2 = {29} andσ3 = {2,3,5,29}
0
. It is clear thatGisσ-soluble and soG
is aσ-full group of Sylow type by [9, Theorem B]. We claim thatB isσ-n-embedded in Gand so
σ-embedded inG, butBis neithers-embedded inG, norn-embedded inGand norc-normal inG. In
fact, it is easy to see thatH={A5, C29, C7}is a complete Hallσ-set ofG. LetH =C29oC7. Then
H£GandC7x ≤H ≤CG(A5)for allx ∈ G. HenceBC7x =C7xB for allx ∈G, which implies
thatBisσ-permutable inG, and soBisσ-n-embedded inGand alsoσ-embedded inG(see Lemma 2.7(1) and Lemma 2.8(1) below). Clearly,BsGis subnormal inGby [10, Lemmas 2.6 and 2.8], and
so is subnormal inA5 by [11, Chapter A, 14.1]. It follows thatBsG= 1forA5is a simple group. If
Biss-embedded inG, then there exists ans-permutable subgroupT ofGsuch thatBT =BsGand
B∩T ≤BsG = 1. Since16=BsGiss-permutable inGby [5, Lemma 2.5], and soBsG =A5by
the same discussion as above. HenceBT =A5andB∩T = 1. But asT iss-permutable inGand
T ≤A5, we have thatT is subnormal inGand soT is subnormal inA5. This implies thatT = 1
orT = A5. IfT = 1, thenB = A5, a contradiction. If T = A5, thenB = B ∩T = 1, also a contradiction. HenceB is nots-embedded in G. ConsequentlyB is neithern-embedded inGand norc-normal inG.
CG(Oσ2(G)) = CG(C29), whereΠ ={σ1, σ3}, which contradicts the fact thatC29oC7 is a
non-abelian group. ThusAis notσ-permutable andAσG=B. It is clear thatT =C29A5£G,G=AT andA∩T = B. ThenAG = A(AG∩T) andA∩(AG∩T) = B = AσG. This shows that Ais σ-n-embedded inG.
Note that ifσ is the smallest partition ofP, that is,σi is an one-element set for anyi∈I, then it
is clear thatH iss-embedded (n-embedded) inGif and only ifH isσ-embedded (σ-n-embedded) inG. It is also clear that ifGis aσ-full group of Sylow type andHisσ-n-embedded inG, thenH isσ-embedded inG. But the converse is not true (see [5, Example 1.2]).
Our main goal here is to prove the following theorem.
Main Theorem — LetF be a saturated formation containing all supersoluble groups andGa group with a normal subgroupEsuch thatG/E ∈ F. Suppose thatGis aσ-full group of Sylow type andH={W1, W2,· · ·, Wt}is a complete Hallσ-set ofGsuch thatWi is a nilpotentσi-subgroup for alli = 1,· · · , t. If for every non-cyclic Hallσi-subgroupWi ∩E ofE either every maximal subgroup ofWi∩Eor every cyclic subgroupHofWi∩Ewith prime order and order 4 (if the Sylow 2-subgroupP ofEis non-abelian andH Z∞(G)) isσ-n-embedded inGfor alli = 1,2,· · ·, t,
thenG∈ F.
In section 3, we give the proof of the main Theorem. In section 4, we will give some application
of our results.
All unexplained terminologies and notations are standard, as in [11] and [12].
2. PRELIMINARIES
Lemma 2.1 — (See [1, Lemma 2.6] and [3, Lemma 2.1]). LetA, K and N be subgroups of G. Suppose thatAisσ-subnormal inGandN is normal inG. Then:
(1)A∩K isσ-subnormal inK.
(2)AN/N isσ-subnormal inG/N.
(3)IfN ≤KandK/Nisσ-subnormal inG/N, thenKisσ-subnormal inG.
(4)If H 6= 1is a Hall Π-subgroup of GandA is not aΠ0-group, then A∩H 6= 1is a Hall
Π-subgroup ofA.
(5)IfGisΠ-full andAis aΠ-group, thenA≤OΠ(G).
it byRσ(G)andRσ(G)isσ-soluble (see [9]). Moreover, any extension of theσ-soluble group by a σ-soluble group is aσ-soluble group (see also [9, Lemma 2.1]).
Lemma 2.2 — IfGhas a complete Hallσ-setH ={W1,· · · , Wt}such thatWiis nilpotent for
alli= 1,· · · , tandHis aσ-subnormalσ-soluble subgroup ofG, thenH ≤Rσ(G).
PROOF: Assume that this Lemma is false and let(G, H)be a counterexample with|G|+|H|as
small as possible.
By hypothesis, there exists a subgroup chain H = H0 ≤ H1 ≤ · · · ≤ Hl−1 ≤ Hl = G
such that eitherHi−1 is normal inHi orHi/(Hi−1)Hi isσ-primary for alli = 1,· · · , l. LetM = Hl−1. ThenM is normal inGorG/MGisσ-primary. Without loss of generality, We can assume
thatM 6= G. By Lemma 2.1(1), H is σ-subnormal in M. Hence H ≤ Rσ(M) by the choice
of (G, H). If M is normal in G, then H ≤ Rσ(M) ≤ Rσ(G) for Rσ(M) is characteristic in M, a contradiction. ThereforeG/MG isσ-primary, which means thatG/MG is aσi-group and so G/MG=WiMG/MGis nilpotent. It follows thatHMGis subnormal inG. SinceHMG≤M < G
andH isσ-subnormal inHMG by Lemma 2.1(1),H ≤ Rσ(HMG)by the choice of(G, H). But
asRσ(HMG) is subnormal inG, we have thatRσ(HMG) ≤ Rσ(G) by [9, Lemma 2.3]. Hence
H≤Rσ(HMG)≤Rσ(G). This contradiction completes the proof. 2
Let L be some non-empty set of subgroups of G and E a subgroup of G. Following [1], a subgroupAofGis called L-permutable ifAH = HA for allH ∈ L;LE-permutable ifAHx =
HxAfor allH ∈ Land allx ∈E. In particular, a subgroupHofGisσ-permutable ifGpossesses
a complete Hallσ-setHsuch thatHisHG-permutable.
Lemma 2.3 — (See [1, Lemma 2.8] and [3, Lemma 2.2]). Let H, K andN be subgroups of a
σ-groupG. LetH ={H1, H2,· · ·, Ht}be a complete Hallσ-set ofGandL =HK. Suppose that HisL-permutable andN is normal inG.
(1)IfH≤E ≤G, thenHisL∗-permutable, whereL∗ ={H1∩E, H2∩E,· · ·, Ht∩E}K∩E.
In particular, ifGis aσ-full group of Sylow type andHisσ-permutable inG, thenHisσ-permutable inE.
(2)The subgroupHN/N isL∗∗-permutable, whereL∗∗={H1N/N,· · · , HtN/N}KN/N.
(3)IfGis aσ-full group of Sylow type andE/N is aσ-permutable subgroup ofG/N, thenEis
σ-permutable inG.
(4) IfK isL-permutable, then hH, Ki is L-permutable (see also [11, Chapter A, 1.6(a)]). In
Lemma 2.4 — (See [1, Lemma 3.1]). Let H be a σ1-subgroup of a σ-group G. Then H is
σ-permutable inGif and only ifOσ1(G)≤N
G(H).
Lemma 2.5 — (See [1, Lemma 3.2]). LetHandK be subgroups ofG. IfGis aσ-full group of Sylow type andHisσ-permutable inG, thenH∩K isσ-permutable inK.
The following Lemma directly follows from Lemma 2.3, Lemma 2.5 and [1, Theorem C].
Lemma 2.6 — LetGbe aσ-full group of Sylow type andH≤K ≤G. Then:
(1)HσG≤HσK.
(2)HσGisσ-permutable inGandHσK ≤HσG≤HG.
(3)IfHis normal inG, then(K/H)σ(G/H)=KσG/Hand(K/H)σ(G/H)=KσG/H.
(4)IfHisσ-permutable inG, thenHxisσ-permutable inGfor allx∈G.
Lemma 2.7 — LetGbe aσ-full group of Sylow type andH≤K ≤G.
(1)IfHisσ-permutable inG, thenHisσ-embedded inG.
(2)IfHisσ-embedded inG, thenHisσ-embedded inK.
(3) Suppose that H is normal in G. ThenK/H isσ-embedded in G/H if and only if K is
σ-embedded inG.
(4) Suppose that H is normal in G. Then for every σ-embedded subgroup E of G with
(|H|,|E|) = 1,HE/Hisσ-embedded inG/H.
(5)IfHisσ-embedded inG, thenHxisσ-embedded inGfor allx∈G.
PROOF:(1)This is evident.
(2)IfHisσ-embedded inG, then there exists aσ-permutable subgroupT ofGsuch thatHT =
HσGandH∩T ≤H
σG. LetT0=HσK ∩T. ThenT0 =K∩T ∩HσK. By Lemma 2.5,K∩T is
σ-permutable inK and so by [1, Theorem C]T0isσ-permutable inK. By Lemma 2.6(2), we have
thatHσK ≤HσGand soHσK =H(HσK∩T) =HT0. Moreover,H∩T0=H∩T ≤HσG≤HσK
by Lemma 2.6(1). This shows thatHisσ-embedded inK.
(3)First assume thatK/H isσ-embedded inG/Hand letT /H be aσ-permutable subgroup of
G/Hsuch that (K/H)(T /H) = (K/H)σ(G/H)and(K/H)∩(T /H) ≤ (K/H)σ(G/H). Then by
Lemma 2.3(3),T is σ-permutable inG, and by Lemma 2.6(3), KT = KσG andK ∩T ≤ KσG.
σ-permutable subgroupT such thatKT = KσGandK ∩T ≤ KσG. By Lemma 2.3(2), T H/H
isσ-permutable in G/H. By Lemma 2.6(3), (K/H)(T H/H) = KσG/H = (K/H)σ(G/H) and
(K/H)∩(T H/H) = (K∩T)H/H ≤KσG/H = (K/H)σ(G/H). HenceK/H isσ-embedded in
G/H.
(4)Suppose that E isσ-embedded inG and letT be aσ-permutable subgroup of Gsuch that
ET = EσGandE∩T ≤ E
σG. Clearly, HET = HEσG = (HE)σG. Since(|H|,|E|) = 1, we
have that(|HE∩T :H∩T|,|HE∩T :E∩T|) = (|(HE∩T)H:H|,|(HE∩T)E :E|) = 1. HenceHE∩T = (H∩T)(E∩T)by [11, Chapter A, 1.6(b)]. ThenHE∩T = (H∩T)(E∩T)≤
HEσG≤(HE)σGby Lemma 2.3(4). This shows thatHE isσ-embedded inG. It follows from (3)
thatHE/H isσ-embedded inG/H.
(5)This follows from Lemma 2.6(4). 2
Lemma 2.8 — LetGbe aσ-full group of Sylow type andH≤K≤G.
(1)IfHisσ-permutable inG, thenHisσ-n-embedded inG.
(2)IfHisσ-n-embedded inG, thenHisσ-n-embedded inK.
(3)Suppose that H is normal inG. Then K/H isσ-n-embedded inG/H if and only ifK is
σ-n-embedded inG.
(4) Suppose that H is normal in G. Then for every σ-n-embedded subgroup E of G with
(|H|,|E|) = 1,HE/Hisσ-n-embedded inG/H.
(5)IfHisσ-n-embedded inG, thenHxisσ-n-embedded inGfor allx∈G.
PROOFSee the proof of Lemma 2.7. 2
LetP be ap-group. IfP is not a non-abelian2-group, then we useΩ(P)to denote the subgroup
Ω1(P). Otherwise,Ω(P) = Ω2(P).
Lemma 2.9 — ( See [13, Lemma 4.3]). LetCbe a Thompson critical subgroup (see [14, p. 185]) of a nontrivialp-group ofP.
(1)Ifpis odd, then the exponent ofΩ(C)isp.
(2)IfP is a non-abelian 2-group, then the exponent ofΩ(C)is4.
Recall that a class of groups F is called a formation if it is closed under taking homomorphic
Gwith quotient inF. In this paper, we useU to denote the class of all supersoluble groups;ZU(G)
denotes theU-hypercenter of a groupG, that is, the product of all such normal subgroupsH ofG whoseG-chief factors have prime order.
Lemma 2.10 — (See [13, Lemma 4.4] and [15, Lemma 2.12]). LetP be a normalp-subgroup of a group G andC a Thompson critical subgroup of P. If either P/Φ(P) ≤ ZU(G/Φ(P)) or
Ω(C)≤ZU(G), thenP ≤ZU(G).
Lemma 2.11 — (See [10, Lemma 2.16]). LetF be a saturated formation containingU andGbe a group with a normal subgroupEsuch thatG/E ∈ F. IfEis cyclic, thenG∈ F.
Lemma 2.12 — (See [16, Theorem 1.8.17]). LetP be a nilpotent normal subgroup of a groupG. IfP∩Φ(G) = 1, thenP is a direct product of some minimal normal subgroups ofG.
Recall that a classFis called Fitting formation ifF is both Fitting class and formation. A group
Gis said to bep-closed ifGhas a normal Sylowp-subgroup.
Lemma 2.13 — (See [17, p. 34]). Let pbe a prime. Then the class of allp-closed groups is a saturated fitting formation.
3. PROOF OFMAINTHEOREM
Lemma 3.1 — LetGbe aσ-full group of Sylow type andH = {W1, W2,· · ·, Wt} be a complete
Hallσ-set ofGsuch that the Hallσi-subgroupWi ofGis nilpotent for alli= 1,· · · , t. Letp∈σ1, wherepis the smallest prime dividing|G|. If every maximal subgroup ofW1or every cyclic subgroup
HofW1 of prime order and order 4 (if a Sylowp-subgroupP ofW1 is a non-abelian2-group and
HZ∞(G)) isσ-embedded inG, thenGis soluble.
PROOF: First note that if Gisσ-soluble, thenGis soluble. In fact, if Gisσ-soluble, then for every chief factorH/K ofG,H/K isσ-primary, and soH/K is aσi-group, for someσi ∈ σ(G).
HenceH/K ≤WiK/K sinceWiK/K is the Hallσi-subgroup ofG/K. But sinceWiis nilpotent, H/Kis elementary abelianp-group. This shows thatGis soluble. Therefore we only prove thatGis
σ-soluble.
Suppose that it is false and let Gbe a counterexample of minimal order. Then clearly, t > 1. By the well-known Feit-Thompson’s theorem, we havep = 2. Without loss of generality, we may assume thatP is a Sylow2-subgroup ofW1. We now proceed by the following steps.
(1)Gis not2-nilpotent.
(2)W1is not cyclic (This follows from (1) and [18, IV, Theorem 2.8]).
(3)Every maximal subgroup ofW1isσ-embedded inG.
Suppose that some maximal subgroup ofW1 is notσ-embedded inG. Then by hypothesis every cyclic subgroupHofW1of prime order and order 4 (ifP is a non-abelian2-group andH Z∞(G)) isσ-embedded inG. SinceGis not2-nilpotent by (1), by [18, IV, Theorem 5.4] and [16, Theorem 3.4.11],Ghas a2-closed Schmidt subgroupA=A2oAq, whereA2is a Sylow2-subgroup ofAof exponent 2 or 4 (ifA2is non-abelian),Aqis a Sylowq-subgroup ofA,A2/Φ(A2)is aA-chief factor,
Z∞(A) = Φ(A)andΦ(A)∩A2 = Φ(A2).
We claim that|A2/Φ(A2)|= 2. Without loss of generality, we can assume thatA2 is contained inW1. By Lemma 2.7(2), every cyclic subgroupHofA2with prime order and order 4 (ifA2is non-abelian andHZ∞(G)) isσ-embedded inA. Ifq∈π(W1), thenAis aσ1-group. SoA≤W1xfor
Gis aσ-full group of Sylow type, which means thatAis nilpotent, a contradiction. Henceq /∈σ1.
Assume that there exists a minimal subgroupX/Φ(A2)ofA2/Φ(A2)such thatX/Φ(A2)is notσ -permutable inA/Φ(A2). Letx ∈ X\Φ(A2) andL =hxi. ThenX = LΦ(A2)and|L| = 2or4. IfL = LσA, then Lis σ-permutable inA by Lemma 2.3(4). So X/Φ(A2) = LΦ(A2)/Φ(A2) is
σ-permutable inA/Φ(A2)by Lemma 2.3(2), a contradiction. Hence we may assume thatLσA < L.
IfL ≤ Z∞(G), thenL ≤ Z∞(A)∩A2 = Φ(A)∩A2 = Φ(A2), a contradiction. Hence by the hypothesis, there exists aσ-permutable subgroupT ofAsuch thatLT =LσA ≤ A2 andL∩T ≤
LσA < L. Since TΦ(A2)/Φ(A2) isσ-permutable inA/Φ(A2) by Lemma 2.3(2), it follows that
AqΦ(A2)/Φ(A2)≤NA/Φ(A2)(TΦ(A2)/Φ(A2))by Lemma 2.4. ThenTΦ(A2)/Φ(A2)£A/Φ(A2).
HenceTΦ(A2)/Φ(A2) = 1orA2/Φ(A2). IfTΦ(A2)/Φ(A2) =A2/Φ(A2), thenT = A2, and so
L=L∩T ≤LσA< L, a contradiction. HenceT ≤Φ(A2). SinceLT =LσAisσ-permutable inA
by Lemma 2.6(2), we have thatX/Φ(A2) = LΦ(A2)/Φ(A2) = LTΦ(A2)/Φ(A2)isσ-permutable inA/Φ(A2), a contradiction. The contradiction shows that every minimal subgroup ofA2/Φ(A2) isσ-permutable inA/Φ(A2). It follows that every minimal subgroup ofA2/Φ(A2)iss-permutable inA/Φ(A2). Hence|A2/Φ(A2)| = 2 by [10, Lemma 2.11]. This implies that A is nilpotent, a contradiction. Thus we have (3).
(4)Oσ1(G) = 1.
If Oσ1(G) 6= 1, thenH = {W1/Oσ1(G), W2Oσ1(G)/Oσ1(G),· · ·, WtOσ1(G)/Oσ1(G)} is a
complete Hall σ-set of G/Oσ1(G) and WiOσ1(G)/Oσ1(G) ' Wi/Wi ∩Oσ1(G) is nilpotent. If
Oσ1(G) =W1, then|G/W1|is an odd number and soG/W1is soluble. HenceGis soluble, a
ofW1. By (3) and Lemma 2.7(3),M/Oσ1(G)isσ-embedded inG/Oσ1(G). Therefore the
hypothe-sis holds forG/Oσ1(G). The choice ofGimplies thatG/Oσ1(G)isσ-soluble and soGisσ-soluble,
a contradiction. HenceOσ1(G) = 1.
(5)Final contradiction.
For any maximal subgroup V of W1 we haveVσG = 1. Indeed, by Lemma 2.3(4)VσG isσ
-permutable in G, so it is σ-subnormal in G by [1, Theorem B]. Hence VσG ≤ Oσ1(G) = 1 by
(4) and Lemma 2.1(5). By (2), W1 = V1V2 for some maximal subgroupsV1 and V2 of W1. By (3) Vi is σ-embedded in G. So G has a σ-permutable subgroup Ti such that ViσG = ViTi and Vi∩Ti ≤ (Vi)σG = 1fori= 1,2. By [1, Theorem B],Ti isσ-subnormal inG. Hence by Lemma
2.1(4),W1∩Tiis a Hallσ1-subgroup ofTi. But sinceVi∩Ti = 1, we have that|Ti|σ1 =|W1∩Ti|=
|W1∩Ti:Vi∩Ti| ≤ |W1:Vi|=q, whereq∈σ1. This implies thatTiis soluble and soTi ≤Rσ(G)
by Lemma 2.2. It follows thatViσG= ViTi = Vi(Rσ(G)∩ViσG), and soViσGisσ-soluble. By [1,
Theorem B and Theorem C]ViσGis σ-subnormal inG. SoViσG ≤ Rσ(G)by Lemma 2.2. Since W1 =V1V2 ≤ hV1σG, V2σGi ≤Rσ(G). |G/Rσ(G)|is aσ
0
1-number and so it is an odd number. This implies thatGisσ-soluble. The final contradiction completes the proof. 2
Theorem 3.2 — LetGbe aσ-full group of Sylow type andH={W1,· · ·, Wt}be a complete Hall σ-set ofGsuch thatWiis a nilpotentσi-subgroup for alli= 1,· · · , t. If every maximal subgroup of
every non-cyclic subgroupWiisσ-n-embedded inGfori= 1,· · ·, t, thenGis supersoluble.
PROOF : Suppose that this assertion is false and letG be a counterexample of minimal order. Then:
(1)Gis soluble and soGis not a simple group.
Letp ∈ σ1 andP be a Sylow p-subgroup ofG, where pis the smallest prime dividing|G|. If
W1 is cyclic, then P is cyclic. SoG isp-nilpotent by [18, IV, Theorem 2.8] and soG is soluble.
Now assume thatW1 is non-cyclic. Since aσ-n-embedded subgroup ofGisσ-embedded inG, by hypothesis and Lemma 3.1 we have thatGis soluble. The choice ofGimplies thatGis not a simple group.
(2)G/Ris supersoluble for every non-identity minimal normal subgroup ofG.
Let R be a minimal normal subgroup of G. Then R is a p-group by (1). It is clear that the hypothesis holds onG/R. HenceG/Ris supersoluble by the choice ofG.
(3)Ris the unique minimal normal subgroup ofG,|R|> pandRΦ(G)(This directly follows
(4)Final contradiction.
Without loss of generality, we may assume that R ≤ W1. Then by (3) and [18, III, Lemma 3.3(a)],RΦ(W1). Hence there exists a maximal subgroupV ofW1such thatW1 =RV,VG= 1
and|W1 :V|= p = |R : R∩V|. By (3), W1 is not cyclic. Then by hypothesis, Ghas a normal subgroupT ofGsuch thatV T =VGandV ∩T ≤VσG. Then by [1, Theorem C],V∩T =VσG∩T
isσ-permutable inG. It follows from Lemma 2.4 thatOσ1(G) ≤ N
G(V ∩T). On the other hand, V ∩T is a normal subgroup ofW1. HenceV ∩T is normal inG, and soV ∩T = 1forVG = 1.
By (3),T = 1orR ≤T. IfT = 1, thenV =VG£G, and soV = 1. It follows that|R| =p, a
contradiction. HenceR ≤T andR∩V ≤T ∩V = 1, which also means that|R|=p, contrary to
(3). The final contradiction completes the proof. 2
Lemma 3.3 — LetGbe aσ-full group of Sylow type andH={W1,· · · , Wt}be a complete Hall σ-set ofGsuch thatWiis a nilpotentσi-subgroup for alli= 1,· · · , t. LetP be a normalp-subgroup
ofGandP ≤Wi for somei. If every cyclic subgroupH ofP with prime order and order 4 (ifP is
a non-abelian2-group andH Z∞(G)) isσ-n-embedded inG, thenP ≤ZU(G).
PROOF: Assume that this assertion is false and let(G, P) be a counterexample with|G|+|P|
minimal. Then:
(1)Ghas a unique normal subgroupR such thatP/Ris a chief factor ofG,R ≤ ZU(G)and
|P/R|> p.
Let P/R be a chief factor of G. Then clearly, (G, R) satisfies the hypothesis of the Lemma. The choice of (G, P) implies that R ≤ ZU(G). If |P/R| = p, then P/R ≤ ZU(G/R) and so
P ≤ ZU(G), a contradiction. Hence |P/R| > p. Assume that P/N is a chief factor ofG with
P/N 6= P/R. A same discussion as above, we have thatN ≤ ZU(G). Then P/R = N R/R ≤
RZU(G)/R≤ZU(G/R). It follows fromR≤ZU(G)thatP ≤ZU(G), a contradiction. Hence we
have (1).
(2)The exponent ofP ispor4(whenP is a non-abelian2-group).
LetCbe a Thompson critical subgroup ofP (see [14, p. 185]). IfΩ(C)< P, thenΩ(C)≤R≤
ZU(G)by (1). SoP ≤ZU(G) by Lemma 2.10, a contradiction. HenceP = C = Ω(C). Then by
Lemma 2.9, the exponent ofP ispor4(whenP is a non-abelian2-group).
(3)Final contradiction.
non-abelian2-group) by (2). IfH = HσG, then by Lemma 2.3(2)(4),HR/Risσ-permutable inG/R.
Obviously, HR/R£Wi/R and by Lemma 2.4 Oσi(G/R) ≤ NG/R(HR/R). This implies that HR/R£G/R. But asP/Ris a chief factor ofGandH R, we have thatP =HR. It follows that
P/R=HR/R=L/Ris cyclic, which contradicts (1). HenceHσG< H. IfH ≤Z∞(G), then16=
L/R=HR/R≤Z∞(G)R/R≤Z∞(G/R). SoP/R∩Z∞(G/R)6= 1. HenceP/R≤Z∞(G/R).
But asR ≤ ZU(G), it follows thatP ≤ ZU(G), a contradiction. SoH isσ-n-embedded inGby hypothesis, there exists a normal subgroupT ofGsuch thatHT =HG≤P andH∩T ≤HσG< H.
By (1) we have thatT ≤RorT =P. IfT =P, thenH =H∩T ≤HσG< H, a contradiction.
SoT ≤ R. Similarly, we have thatHG ≤ R orHG = P. IfHG ≤ R, thenH ≤ HG ≤ R, a contradiction. SoHG = P. HenceP = HG = HT = HR, and so P/R = HR/R = L/R is cyclic, which contradicts (1). This final contradiction completes the Lemma. 2
PROOF OFMAINTHEOREM: Suppose that this theorem is false and let(G, E)be a
counterex-ample with|G|+|E|minimal. LetP be a Sylow p-subgroup of E wherep is the smallest prime dividing|E|. Without loss of generality we may assume thatP ≤W1∩E. We now proceed by the following steps.
(1)W1∩Eis non-cyclic.
IfW1∩Eis cyclic, thenP is cyclic. It follows from [18, IV, Theorem 2.8] thatE isp-nilpotent. LetEp0 be the normal Hallp0-subgroup ofE. ThenEp0 6= 1by Lemma 2.11. ClearlyEp0 £Gand E/Ep0 is a cyclicp-group. Since(G/E
p0)/(E/Ep0) ' G/E ∈ F, we have thatG/Ep0 ∈ F by
Lemma 2.11. Clearly,Wi∩Ep0 =Wi∩Efori6= 1, andW1∩Ep0 is cyclic. Hence the hypothesis
holds for(G, Ep0)and soG∈ F by the choice of(G, E), a contradiction. SoW1∩Eis non-cyclic.
(2)IfE = P and every cyclic subgroupH ofP with prime order and order 4 (ifP is a
non-abelian2-group andH Z∞(G)) isσ-n-embedded inG, thenE ≤ZU(G). (This directly follows from Lemma 3.3).
(3)If every cyclic subgroupH ofW1∩E with prime order and order 4 (ifP is a non-abelian
2-group andH Z∞(G)) isσ-n-embedded inG, thenEis notp-nilpotent.
Assume thatEisp-nilpotent. ThenEhas a normal Hallp0-subgroupEp0 andEp0£G. IfEp0 = 1,
then by (2) we have that E ≤ ZU(G), and so G ∈ F. This contradiction shows that Ep0 6= 1.
ThenH = {W1Ep0/E
p0,· · ·, WtEp0/Ep0}is a complete Hallσ-set ofG/Ep0 andWiEp0/Ep0 ' Wi/Wi∩Ep0 is nilpotent. ClearlyWiEp0/Ep0 ∩E/Ep0 = 1 fori = 2,· · ·, t, and W1Ep0/Ep0 ∩ E/Ep0 =E/E
cyclic subgroupH/Ep0 ofE/Ep0 with prime orderpor order 4, we have thatH =Ep0 oL, where L = hxi is of order p or 4 by Schur-Zassenhaus Theorem. So by the hypothesis, Lemma 2.8(4)
and [12, Chapter 1, Theorem 2.6(d)], we have thatH/Ep0 =Ep0L/Ep0 isσ-n-embedded inG/Ep0
orH/Ep0 ≤ Z∞(G/Ep0). This shows that the hypothesis holds for (G/Ep0, E/Ep0). Therefore G/Ep0 ∈ F by the choice of(G, E)andE
p0 6= 1. It is clear thatWi∩E =Wi∩Ep0 fori= 2,· · · , t
andW1 ∩Ep0 ≤ W1 ∩E. So the hypothesis holds for (G, Ep0), and so G ∈ F by the choice of (G, E), a contradiction. HenceEis notp-nilpotent.
(4)Every maximal subgroup ofW1∩Eisσ-n-embedded inG.
By hypothesis and (1) either every maximal subgroup ofW1∩Eisσ-n-embedded inGor every cyclic subgroup H of W1 ∩E with prime order and order 4 (if P is a non-abelian 2-group and
HZ∞(G)) isσ-n-embedded inG. Suppose that we have the second case. Then by (3),Eis notp
-nilpotent. So by [18, IV, Theorem 5.4] and [16, Theorem 3.4.11],Ehas ap-closed Schmidt subgroup
A=ApoAr, whereAp is a Sylowp-subgroup ofAof exponentpor4(whenApis a non-abelian 2-group) andAp/Φ(Ap)is an eccentric chief factor ofA. By the same discussion as Lemma 3.1(3),
we have that|Ap/Φ(Ap)|=p. HenceAp/Φ(Ap)is central inAforpis the smallest prime dividing |E|. This contradiction shows that every maximal subgroup ofW1∩E isσ-n-embedded inG.
(5)Eisq-closed, whereqis the largest prime dividing|E|.
Letqbe the largest prime dividing|E|. Without loss of generality, we may assume thatq∈σi,Q
is a Sylowq-subgroup ofEandQ≤Wi∩E. We useMto denote the class of allq-closed groups.
Suppose thatEis not inM. Then:
(a)E=G.
Indeed, ifE 6=G, then the hypothesis holds for(E, E)by Lemma 2.8(2). The choice of(G, E) implies thatEis supersoluble and so it isq-closed, which contradicts to our assumption onE.
(b)Gis soluble (This follows from (4) and Lemma 3.1).
(c)t= 2, that is,G=W1W2.
Assume thatt > 2. By(b)and [9, Theorem A],Ghas aσ-basis{H1, H2,· · ·, Ht}andWj = Hjxjby [9, Theorem B]. SoWx
−1
i i W
x−1
j
j =HiHj is a subgroup ofGand soWiW x−1
j xi
j is a subgroup
ofGfor allj 6=i. By Lemma 2.8(2)(5) and(a), the hypothesis holds for(WiWx −1
j xi j , WiW
x−j1xi j ).
Sincet > 2, WiW x−j1xi
j < G. HenceWiW x−j1xi
j isq-closed. SoW x−j1xi
j ≤ NG(Q)forj 6= i. It
(d)Every minimal subgroup ofW2isσ-n-embedded inG.
IfW2 is cyclic or every maximal subgroup of W2 is σ-n-embedded inG, then by (4), (c) and Theorem 3.2,Gis supersoluble, and soG∈ F, a contradiction. Hence we have(d).
(e)W2is ar-group for some primerand soG=W1RwhereRis a Sylowr-subgroup ofG.
If not, then by (b) and [9, Theorem B]W1 isG-permutable with every Sylow subgroup of G. This means for any Sylowr-subgroupRofW2, we have thatW1Rx=RxW1for somex∈G. Then
W1Rx < G. By (4) and(d),W
1Rx isq-closed. If Q ≤ W1, thenRx ≤ NG(Q). ThereforeQis
normal inGforW1is nilpotent, a contradiction. SoQ≤W2. ThenW1Qxisq-closed. This means thatW1x−1 ≤NG(Q). HenceQis normal inGforW2is nilpotent. This contradiction shows thatW2 is ar-group.
(f)r=q, that is,W2 =Qis aq-group.
If not, then by (e) Q ≤ W1. Now we claim that Oσ1(G) = 1. In fact, ifOσ1(G) 6= 1, then
by (4) and Lemma 2.8(3)(4), the hypothesis holds onG/Oσ1(G), and soG/Oσ1(G)isq-closed by
induction onG=E. ThenQOσ1(G)£G. But sinceQOσ1(G)≤W1andW1is nilpotent, we have
thatQcharQOσ1(G), and soQ£G, a contradiction. HenceOσ1(G) = 1. Sincep, q∈σ1. LetV be
a maximal subgroup ofW1such thatP V. Then by (4) and(a),Ghas a normal subgroupT such thatVG =V T andV ∩T ≤V
σG. By Lemma 2.3(4) and [1, Theorem B]VσGisσ-subnormal inG.
SoV∩T ≤VσG≤Oσ1(G) = 1by Lemma 2.1(5). Clearly,|T|σ1 ≤p. If|T|σ1 =p, thenW1 ≤VG
by (b). HenceVG =W1(VG∩R)by(e). IfVG < G, thenVG satisfies the hypothesis, soVG is
q-closed and soQ£G, a contradiction. HenceG=VG =T oV, soR ≤T. Since|T|σ1 =p, so
T isp-nilpotent by [18, IV, Theorem 2.8]. HenceR£T and soR£G. ThereforeG/R ' W1 is
nilpotent. By Lemma 3.3 and(d),R ≤ZU(G). This implies thatGis supersoluble and soG∈ F, a contradiction. Hence|T|σ1 = 1, soT ≤R. By(d)and Lemma 2.8(2), every minimal subgroup ofT
isσ-n-embedded inVG. Hence by Lemma 3.3 we have thatT ≤ZU(VG). ButVG/T 'V ≤W1 is nilpotent, it follows thatVGis supersoluble. ThenQcharVG£G, soQ£G. This contradiction shows thatr=qandW2=Qis aq-group.
(g)Oq(G) =Gor|Q|=q.
Assume thatOq(G) 6= Gand letM be a maximal subgroup ofGcontainingOq(G). ThenM
is normal inGand by(e) and(f) M = W1(M ∩Q). By Lemma 2.8(2), the hypothesis holds on
(M, M). HenceM∩Qis normal inM. Suppose that|Q|> q. ThenM ∩Q6= 1. It is clear that the
Q/M ∩Qis normal inG/M ∩Q. But thenGisq-closed, a contradiction. Hence|Q| =q. So we
have(g).
(h)IfV is a maximal subgroup ofW1andVG= 1, thenVGisq-closed.
Indeed, by (1), (4),V 6= 1andGhas a normal subgroupTsuch thatVG =V TandV∩T ≤V σG.
ThenV ∩T =VσG∩T is normal inGby the same discussion as Theorem 3.2(4). It follow from VG = 1thatV ∩T = 1. SoVG =T oV. Hence|T|σ1 ≤tfor some primetsinceW1is nilpotent.
ThereforeW1∩T = 1orW1 ∩T is cyclic. ThenT satisfies the hypothesis by Lemma 2.8(2) and
(d), soT isq-closed. LetQ0 be the Sylowq-subgroup ofT. ThenQ0 is normal inVG. This means
thatVGisq-closed.
(i)IfW1 has two maximal subgroupsV1 andV2 such thatW1 =V1V2and(V1)G= (V2)G = 1,
thenW1Gisq-closed.
Since(V1)G = (V2)G = 1,V1GandV2Gareq-closed by(h). Hence by Lemma 2.13,D=V1GV2G isq-closed. ButW1 ≤DandW1G≤D. Hence we have(i).
(j)Oσ1(G)6= 1.
Suppose thatOσ1(G) = 1. Since G = E = W1Qby(c)and(f)andGis soluble by(b), we
have that1 6= Oq(G) < Q. Then by(g) Oq(G) = G. SinceW1 is not cyclic by (1) and(a), for some maximal subgroupsV1andV2ofW1we have thatW1 =V1V2and(V1)G= (V2)G= 1. By(i) W1G isq-closed. HenceG6=W1Gand soOq(G)≤W1G < Gby(f), which contradicts(g). Hence
Oσ1(G)6= 1.
(k)N =Oσ1(G)is the only minimal normal subgroup ofGcontained inW1,N is ar-group for
some primer∈σ1andN =GM Φ(G).
By (4),(d)and Lemma 2.8(3)(4), the hypothesis is still holds onG/N for every minimal normal subgroupN ofGcontained inW1. By(j)N 6= 1. HenceG/N isq-closed. But as the class of all
q-closed groups is a saturated formation by Lemma 2.13, we have thatN Φ(G)and soN =GMis
the only minimal normal subgroup ofGcontained inW1. SinceGis soluble by(b),Nis ar-group for some primer. HenceN ≤Or(G) ≤Oσ1(G). SinceOσ1(G)≤W1is nilpotent,Or(G) =Oσ1(G).
LetMbe a maximal subgroup ofGsuch thatG=N M. ThenN ∩M£G, soN ∩M = 1. Hence
Or(G) = N(Or(G)∩M). But Or(G) ∩M is normal in G by [11, Chapter A, Lemma 8.4], so Or(G)∩M = 1. HenceN =Or(G) =Oσ1(G).
Assume thatOq(G) = 1. Then by(k), F(G) = Oσ1(G) = N. SinceGis soluble by(b), we
have thatCG(N) = CG(F(G)) ≤ F(G) = N by [16, Theorem 1.8.18]. Since N Φ(G) by (k), there exists a maximal subgroupV ofW1 such thatW1 = N V, andVG = 1by (k). Clearly, |W1:V|=r=|N :N∩V|. By (4), there exists a normal subgroupTofGsuch thatV T =VGand
V ∩T ≤VσG. By the same discussion in Theorem 3.2(4), we have thatV ∩T =VσG∩T is normal
inG, soV ∩T = 1. SinceN is a minimal normal subgroup ofG,N ∩T = 1orN. IfN∩T = 1, thenT ≤CG(N)≤N, soT =T ∩N = 1. HenceVG =V T =V is normal inG, and soV = 1.
It follows thatW1 = N and|W1|=r, which contradicts (1). HenceN ∩T =N, and soN ≤ T. Consequently,N∩V ≤T ∩V = 1. SoW1 =N V =N×V. This implies thatV ≤CG(N)≤N.
HenceV = 1. This shows thatW1=N and|W1|=r, a contradiction also. HenceOq(G)6= 1.
Final contradiction for (5).
By (k) there exists a maximal subgroup V ofW1 such thatW1 = N V, VG = 1and |W1 :
V| = r = |N : N ∩V|. By (4), there exists a normal subgroupT ofGsuch thatV T = VG and
V ∩T ≤VσG. The same discussion as Theorem 3.2(4) we have thatV ∩T = VσG∩T is normal
inG, soV ∩T = 1. Clearly,|T|σ1 ≤ r. If|T|σ1 = 1, then T is aq-group and soT ≤ Q. Hence
W1∩VG =W1∩V T =V(W1∩T) =V. AsN is a minimal normal subgroup ofG, we have that
N ∩VG = 1orN. IfN ∩VG = 1, thenN ∩V = 1and so|N|= r. SinceG/N isq-closed by (k),QN is normal inG. FromQN/N ' Q, it follows thatQN/N is supersoluble. HenceQN is
supersoluble. ThenQ£QN. This implies thatQ£G, a contradiction. HenceN∩VG=N and so
N ≤W1∩VG =V. This contradiction shows that|T|σ1 =r. Thus|VG|σ1 =|V||T|σ1 =|W1|and
W1 ≤VG. Obviously,W1G ≤VG. SinceVGisq-closed by(h),W1Gisq-closed. HenceW1G6=G.
By(e)and(f),G/W1Gis aq-group. It follows thatOq(G) ≤W1G< G. Now by(g)|Q|=q. But asOq(G)6= 1by(l), we have thatQ=Oq(G)£G. This contradiction completes the proof of (5).
(6)Letqbe the largest prime dividing|E|andQbe a Sylowq-subgroup ofE. ThenQ£Gand
G/Q∈ F. ConsequentlyQis not cyclic.
Indeed, by (5),E isq-closed, so Q£G. By Lemma 2.8(3)(4), the hypothesis is still true for
(G/Q, E/Q). HenceG/Q∈ F by the choice of(G, E). Thus we have (6).
(7)Suppose thatQ < Eand without loss of generality we may assume thatq∈σi. ThenWi∩E
is non-cyclic and every maximal subgroup ofWi∩Eisσ-n-embedded inG.
IfWi∩E is cyclic, thenQ is cyclic, which contradicts (6). If there is a maximal subgroup of Wi∩E is notσ-n-embedded inG, then by hypothesis, every minimal subgroup ofWi∩Eisσ-n
satisfies the hypothesis. The choice of(G, E)implies thatG∈ F, a contradiction. Hence (7) holds.
(8)IfQ < E, thenQ=Oσi(E), and soQ=Oσi(G)∩E =Oσi(E).
Assume thatQ < Oσi(E). Then sinceWiis nilpotent, there exists a minimal normal subgroup RofGsuch thatR ≤ Or(E) ≤ Oσi(E), wherer ∈ σi andr 6= q. By (7) and Lemma 2.8(3)(4), (G/R, E/R) satisfies the hypothesis, soG/R ∈ F. Then G ∈ F by (6) and R∩Q = 1. This
contradiction shows that (8) holds.
(9)IfQ < E, thenQis a minimal normal subgroup ofG.
We first claim thatQ∩Φ(G) = 1. In fact, if Q∩Φ(G) 6= 1and letN be a minimal normal subgroup ofGcontained inQ∩Φ(G), then by (7),(G/N, E/N)satisfies the hypothesis, soG/N ∈
F. Consequently G ∈ F, a contradiction. Therefore Q ∩Φ(G) = 1. By Lemma 2.12, Q = N1 × · · · ×Ns, whereNi is a minimal normal subgroup ofG. It is also clear that(G/Ni, E/Ni)
satisfies the hypothesis. HenceG/Ni ∈ F. This implies that s = 1 andQ is a minimal normal
subgroup ofG.
(10)Q=E =P.
IfQ < E, thenQ ≤ Wi∩E. Assume that Q= Wi ∩E. LetU be a maximal subgroup ofQ
such thatU£Wi. ThenU 6= 1by (6) andUG= 1by (9). Now by (7) there exists a normal subgroup T ofGsuch thatU T =UGandU ∩T ≤UσG. By the same discussion as Theorem 3.2(4), we have
thatU∩T =UσG∩T is normal inG, soU∩T = 1. By (9)UG=Q. SinceQis a minimal normal
subgroup ofG,T = 1orQ. IfT = 1, thenU =UG =Q, a contradiction. HenceQ =T and so U = U ∩T = 1. This contradiction shows thatQ < Wi ∩E. Since Wi ∩E is nilpotent and Q
is a Sylowq-subgroup ofWi∩E, we have that Wi∩E = Q×S, whereS is a non-identity Hall
subgroup ofWi∩E. HenceS£Wi. LetW be a maximal subgroup ofQsuch thatW £Wi. Let V = W S. ThenV £Wi,V is a maximal subgroup of Wi∩E, andVG = 1by (8) and (9). Then Wi∩E=QV,|Wi∩E:V|=q =|Q:Q∩V|andQ∩V =W 6= 1. By (7) there exists a normal
subgroupT ofGsuch thatV T =VGandV ∩T ≤VσG. By a same discussion as above, we have
thatV ∩T = 1. SinceQis a minimal normal subgroup ofG,Q∩VG = 1orQ. IfQ∩VG = 1, thenQ∩V = 1. This contradiction shows thatQ ≤ VG. Then Wi ∩E = QV ≤ VG ≤ E, so Wi∩E =Wi∩VG=V(Wi∩T). Henceq =|Wi∩E :V|=|V(Wi∩T) :V|=|Wi∩T|, and so Wi∩T is aq-group. But asT ≤VG≤E,Wi∩T ≤E. HenceWi∩T ≤Q, soWi∩T ≤Q∩T.
(11)P =GF is a minimal normal subgroup ofGandPis not cyclic.
LetN be a minimal normal subgroup ofGcontained inP. By (10)W1∩E =E =P. Then by (4) and Lemma 2.8(3),(G/N, E/N)satisfies the hypothesis. HenceG/N ∈ F. This implies thatN is the only minimal normal subgroup ofGcontained inP andN Φ(G). ClearlyP is not cyclic. LetM be a maximal subgroup ofGnot containingN. ThenG=N M =P M. By [11, Chapter A, Lemma 8.4]P∩M£G. HenceP ∩M = 1, soP =N =GF is a minimal normal subgroup ofG.
(12)Final contradiction.
LetV be a maximal subgroup of P such thatV £W1. Then by (4) Ghas a normal subgroup
T such thatV T = VG andV ∩T ≤ VσG. Similarly as above, we have thatV ∩T = VσG∩T is
normal inGand so V ∩T = 1 by (11). SinceP is non-cyclic, V 6= 1. HenceP = VG = V T. But sinceP is a minimal normal subgroup ofGby (11),T = 1orT =P. IfT = 1, thenP =V, a contradiction. HenceT = P, soV = V ∩T = 1, a contradiction also. The final contradiction completes the theorem.
4. SOMEAPPLICATIONS OFTHERESULTS
The main theorem has many corollaries, we here cite some of them.
Corollary 4.1 — (Buckley [7]). LetGbe a group of odd order. If all subgroups ofGof prime order are normal inG, thenGis supersoluble.
Corollary 4.2 — (Srinivasan [19]). If every maximal subgroups of every Sylow subgroups ofG
is normal ors-permutable inG, thenGis supersoluble.
Corollary 4.3 — (Asaad [20]). If every subgroup of prime order and every cyclic subgroup of
order 4 is permutable inG, thenGis supersoluble.
Corollary 4.4 — (Wang [4]). If all cyclic subgroups of G with prime order and order 4 are
c-normal inG, thenGis supersoluble.
Corollary 4.5 — (Wang [4]). If all maximal subgroups of all Sylow subgroups ofGarec-normal inG, thenGis supersoluble.
Corollary 4.6 — (Wei [21]). LetF be a saturated formation containing U andGa group with a normal subgroup E such that G/E ∈ F. If all maximal subgroups of every non-cyclic Sylow subgroup ofEarec-normal inG, thenG∈ F.
U. If all minimal subgroups and all cyclic subgroups with order 4 ofGF arec-normal inG, then
G∈ F.
Corollary 4.8 — (Ballester-Bolinches and Pedraza-Aguilera [22]). LetFbe a saturated formation
containingU andGa group with a normal subgroupE such thatG/E ∈ F. Assume that a Sylow
2-subgroup ofGis abelian. If all minimal subgroups ofEare permutable inG, thenG∈ F.
Corollary 4.9 — (Ballester-Bolinches and Pedraza-Aguilera [22]). LetFbe a saturated formation
containingUandGa group with a normal subgroupEsuch thatG/E ∈ F. If all minimal subgroups and all cyclic subgroups with order 4 ofEare permutable inG, thenG∈ F.
Corollary 4.10 — (Asaad [23]). LetFbe a saturated formation containingU andGa group with a normal subgroupEsuch thatG/E∈ F. If every maximal subgroup of every Sylow subgroup ofE
iss-permutable inG, thenG∈ F.
Corollary 4.11 — (Guo and Skiba [5]). Let F be a saturated formation containingU andG a group with a normal subgroupEsuch thatG/E ∈ F. If for every non-cyclic Sylow subgroupP of
Eevery maximal subgroup ofP or every every cyclic subgroupsHofP with prime order and order
4 (ifPis a non-abelian 2-group andH*Z∞(G))isn-embedded inG, thenG∈ F.
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