On the Weak Norm of Up-Residuals of All Subgroups of a Finite Group

Home , Fitting length

arXiv:2105.11637v1 [math.GR] 25 May 2021 Keywords: p Introduction 1 rvne hn GatNme:21KDM5)adteSinea Science 201804010088). the number: and (Grant 2017KZDXM058) China Municipality, Number: Guangzhou (Grant China Province, [email protected] undn nvriyo dcto,unzo,501,China, 510310, Education,Guangzhou, of University Guangdong n oain si 1.Frteseic edenote we specific, the For [1]. in as notations and ntewa omof norm weak the On -length; ∗ † § ‡ hogotti ae,algop r nt.W s h standard the use We finite. are groups all paper, this Throughout S 2020: MSC • upre ytemjrpoeto ai n ple eerh(Na Research Applied and Basic of project major the by Supported colo ahmtclSine,GihuNra nvriy Guiy University, Normal Guizhou Sciences, Mathematical of School olg fMteais uzo omlUiest,Giag 55000 Guiyang, University, Normal Guizhou Mathematics, of College author Corresponding oeo u anrslsmyrgr sacniuto fmn n many of continuation a as regard may results work. main our of Some U N emnltr ftesre.I hsppr o h case the for paper, this In series. the of term terminal N ytepoete fwa omo oesbrusin subgroups some of norm weak of properties so of the structure by the characterize we groups,respectively), in G p F F nt group. finite a : 0 G ( Let ( ( U p G G ihrsetto respect with Ftiglength. -Fitting eknr;Nraie;sprovberesidual; supersolvable Normalizer; norm; Weak rsetvl)i h ls falfinite all of class the is ,respectively) 1, = ) = ) F eafrainand formation a be 01,20D25. 20D10, N F N ( F i ,G G, +1 ( G uoLv Yubo .Let ). F ) /N sdfie by defined is F i fafiiegroup finite a of ( G N = ) G F i † ( nt ru.Tewa omo subgroup a of norm weak The group. finite a G U N Abstract ), F i p N ( ≥ G/N rsdaso l subgroups all of -residuals 1 F ,b pe eisof series upper a be 1, ( agigLi Yangming ,H G, p F sprovbegroups(supersolvable -supersolvable i ( G = ) )addntdby denoted and )) T T ≤ ∗ H [email protected] N egvnfiiegroups finite given me G G F ‡§ ( p ua cec)i Guangdong in Science) tural ihrsetto respect with T sprovberesidual; -supersolvable dTcnlg rga of Program Technology nd { ∈ F ,Cia et fMath., of Dept. China; 1, .I particular, In ). U n,500,China, 550001, ang, p G , c previous ice N U ysetting by F ∞ } ( where , G terminology the ) F H . • |G|: the order of G.

• π(G): the set of prime divisors of |G|.

• p: a prime.

• Gp: the Sylow p-subgroup of G.

• Fp(G): the p-Fitting subgroup of G. • G: the class of all ﬁnite groups.

• F : a formation of groups, that is a class of ﬁnite groups satisfying the following: (1) if G ∈ F and N is a normal subgroup of G, then G/N ∈ F , and

(2) if N1,N2 are normal subgroups of G such that G/Ni ∈ F (i = 1, 2), then

G/(N1 ∩ N2) ∈ F . • GF : the F -residual of G, that is the intersection of all those normal subgroups N of G such that G/N ∈ F .

• F1F2: the formation product or Gasch¨etz product of F1 and F2, that is the F2 2 class {G ∈ G | G ∈ F1}. In particular, denote F = F F .

• A : the class of all Abelian groups.

• N : the class of all nilpotent groups.

• Np: the class of all p-nilpotent groups. • U : the class of all supersolvable groups.

• Up: the class of all p-supersolvable groups.

• lp(G): the p-length of a p-solvable group G, that is the number of the p-factor groups in the upper p-series of G:

1= P0(G) E M0(G) E P1(G) E M1(G) E ··· E Pn(G) E Mn(G)= G

such that Mi(G)/Pi(G)= Op′ (G/Pi(G)) and Pi(G)/Mi−1(G)= Op(G/Mi−1(G)).

• hp(G): the p-Fitting length of a p-solvable group G, that is the smallest positive integer n such that

0 1 n−1 n 1= Fp (G) ≤ Fp (G) ≤···≤ Fp (G) ≤ Fp (G)= G,

1 i+1 i i where Fp (G)= Fp(G) and Fp (G)/Fp(G)= Fp(G/Fp(G)) for i =1, 2, ··· , n− 1.

2 • h(G): the Fitting length of a solvable group G, that is the positive integer n such that 1 = F 0(G) ≤ F 1(G) ≤ ··· ≤ F n−1(G) ≤ F n(G) = G, where F 1(G)= F (G) and F i+1(G)/F i(G)= F (G/F i(G)) for i =1, 2, ··· , n − 1.

• N(G): the intersection of the normalizers of all subgroups of G.

• S(G)(or NN (G)): the intersection of the normalizers of the nilpotent residuals of all subgroups of G.

N • N p (G): the intersection of the normalizers of the p-nilpotent residuals of all subgroups of G.

It is interesting to characterize the structure of a given group by using of some special subgroups. For example, Gashëtz and N. Itô[1, Satz 5.7,p.436] proved that G is solvable with Fitting length at most 3 if all minimal subgroups of G are normal. Also, it is well know that G is nilpotent if G′ normalizers each subgroup of G(see Baer’s theorem in [2]). Further more, R. Baer in [3] defined the subgroup N(G), the norm of a group G. Obviously, a group G is a Dedeking group if and only if G = N(G). The norm of a group has many other good properties and has studied further by many scholars. In recent years, some weaker versions of the concepts of norm of groups have been introduced. Let F be a non-empty formation. Recently, Su and Wang in [4, 5] introduced the subgroup NF (G), the norm of F -residual of a group G as follows

F NF (G)= \ NG(H ). (⋆) H≤G 0 i i+1 i As in [4], we set NF (G) = 1 and if NF (G) is deﬁned, set NF (G)/NF (G) = i ∞ NF (G/NF (G)). The subgroup NF (G) is the terminal term of the ascending series. In ∞ k k k+1 k+2 fact, NF (G)= NF (G) for some integer k such that NF (G)= NF (G)= NF (G)= ··· .

Many scholars also call NF (G)(F = A , N , Np, respectively) the generalized norm of group G. Obviously, NF (G) is a characteristic subgroup and every element of NF (G) normalize the F -residual of each subgroup of G. The so called norm of F -residual has many other nice properties and also closely related to the global properties of a given group. For some given formation F , there are many papers devoted to study the p-length, Fitting length, solvability, (p-)nilpotency and so on. For example, for the case F = A , Li and Shen in [6] denoted NA (G) by D(G). They fingered out that G is solvable with Fitting length at most 3 if all elements of G of prime order are in D(G)(see [6, Theorem 4.1]). It is a dual problem of Gashëtz and N. Itô[1, Satz 5.7,p.436]. Li and Shen also defined the D-group, i.e., G = D(G),

3 they characterized the relationship between D(G) and G. Shen, Shi and Qian in [7] considered the case F = N , they denoted NN (G) by S(G). Shen et al., deeply studied the dual problem of Gashëtz and N. Itôand characterized the Fnn-groups by ∞ ∞ ∞ means of the subgroup S (G), where S (G)= NN (G) and Fnn-groups are class of groups belong to N N . They also introduced the S-group, i.e., G = S(G) and given some sufficient and necessary conditions involved S-groups. Meanwhile, Gong and Guo in [8] also consider the case F = N and given some meaningful conclusions. In the case F = Np, Guo and Li in [9] introduced the norm of Np-residual of a group Np G, they denoted NNp (G) by N (G). As a local version of Gong’s results, Li and Np Np Guo characterized the relationship between CG(G ) and N (G). In particular, Li N and Guo in [9] also investigated the relationship between NN (G) and N p (G). For more detail and other relevant conclusions about the norm of F -residual, please see [4–13]. We wonder whether the above conclusions hold for general formations. A natural idea is to replace F (F ∈{Np, N , A }) by Up or U in (⋆).

Remark 1.1. (1) In general, for a group G and some p ∈ π(G), NUp (G) =6 NF (G) is possible, where F ∈{Np, N , A }. For example, let G = S4, the symmetric group of degree 4. Obviously, G is a 3-supersolvable non-3-nilpotent group, so G = NU3 (G). ∼ N3 N3 Pick a subgroup H = S3, the symmetric group of degree 3, but NG(H )= NG(S3 )=

NG(C3)= S3

(3) The case that NU (G) = NUp (G)=1 is possible for a solvable group G and 3 3 ∼ p ∈ π(G). Let H = ha, b | a = b = 1, [a, b]=1i = C3 × C3 and Q8 = hc,d | 4 2 2 d −1 c = 1,c = d = e, c = c i. Considering the irreducible action of Q8 on H by c −1 c d −1 d ∼ a = a b, b = ab, a = b , b = a, denote T = H⋊Q8 = (C3×C3)⋊Q8. Let C = hfi be a cyclic group of order 3 and let C act on T by af = b−1, bf = ab−1,cf = d3,df = cd. ∼ Then G = ha, b, c, d, e, fi = ((C3 × C3) ⋊ Q8) ⋊ C3 is solvable(IdGroup=[216,153]). U2 ∼ U2 It is easy to see that G = ha, b, c, d, ei = (C3 × C3) ⋊ Q8 and CG(G )=1. By

Theorem 3.4, NU2 (G)=1. Further more, by Theorem 3.14, NU (G)=1. So we have

NU (G)= NU2 (G)=1.

∗ Recall that the weak centralizer of H in G, CG(H), introduced in [14, P.33,Definitions], is defined by ∗ CG(H)= \{NG(K): K ≤ H}. In the above investigation, we introduce the following more general and interesting definition.

4 Deﬁnition 1.2. Let G be a group and F a formation. We deﬁne NF (G,H), the weak norm of H in G with respect to F as follows:

F NF (G,H)= \ NG(T ). T ≤H

U U U U p In particular, N (G,H)= T NG(T ) and N p (G,H)= T NG(T ). T ≤H T ≤H

∼ 4 2 3 e Example 1.3. Let G = ha,b,c,di ⋊ he, fi = C2 ⋊ C6, where e = f = 1 and a = ac, be = bd,ce = c,de = d, af = b, bf = ab, cf = d,df = cd (IdGroup=[96,70] in GAP ∼ 4 [15]). Let H = ha, b, fi, then NU2 (G)= NU2 (G,H)= ha, b, c, d, fi = C2 ⋊ C3

Obviously, NF (G) = NF (G,G) ≤ NF (G,H) ≤ G. Without causing confusion, we call NF (G) = NF (G,G) the norm of G with respect to F . Moreover, for a subgroup H of G, we call G is an N1-group with respect to H and F if G = NF (G,H). In this paper, we mainly investigate properties of NF (G,H) and the inﬂuence of NF (G,H) on the structure of group G. Actually, we mainly consider the case F ∈ {U , Up}. Our main work may be regard as the continuation of some conclusions in [6–9].

2 Preliminaries

In this section, we always assume that F is a non-empty formation and G is a group. We ﬁrst give some important lemmas.

Lemma 2.1. Let H,K,N be subgroups of G and N E G. Then (1) If H ≤ K, then NF (G,K) ≤ NF (G,H); (2) K ∩ NF (G,H) ≤ NF (K,K ∩ H), in particular, if H ≤ K, then K ∩ NF (G,H) ≤ NF (K,H). (3) If N ≤ H, then NF (G,H)N/N ≤ NF (G/N, H/N).

F F Proof. (1) By deﬁnition, NF (G,K)= T NG(T ) ≤ T NG(T )= NF (G,H). T ≤K T ≤H F F (2) Obviously, K∩NF (G,H)= K∩( T NG(T )) ≤ T NK(T )= NF (K,H∩ T ≤H T ≤H∩K K). In particular, if H ≤ K, then K ∩ NF (G,H) ≤ NF (K,H). F (3) Let x ∈ NF (G,H), then x normalize T for every T ≤ H, so xN normalize F F F T N/N =(T N/N) . Thus every element of NF (G,H)N/N normalize (T/N) for all subgroups T/N of H/N, so NF (G,H)N/N ≤ NF (G/N, H/N).

As a corollary, we have

5 Lemma 2.2. Let G be a group, let K be a subgroup of G and N a normal subgroup of G, then (1) K ∩ NF (G) ≤ NF (K). (2) NF (G)N/N ≤ NF (G/N). Lemma 2.3. [4, Lemmas 2.2,2.3] Let G be a group, let K be a subgroup of G and N a normal subgroup of G, then ∞ ∞ (1) K ∩ NF (G) ≤ NF (K). ∞ ∞ (2) NF (G)N/N ≤ NF (G/N). ∞ ∞ ∞ (3) If N ≤ NF (G), then NF (G)/N = NF (G/N). Lemma 2.4. Let G be a p-solvable group.

(1) If N E G, then lp(G/N) ≤ lp(G). (2) If U ≤ G, then lp(U) ≤ lp(G).

(3) Let N1 and N2 be two normal subgroups of G, then lp(G/(N1 ∩ N2)) ≤ max{lp(G/N1),lp(G/N2)}.

(4) lp(G/Φ(G)) = lp(G). ′ (5) If N is a normal p -group of G, then lp(G/N)= lp(G).

Proof. The proof of (1)-(4) follows from [1, VI,6.4] and (5) is obviously. Lemma 2.5. Let G be a group and N a normal subgroup of G. Then (1) if N/N ∩ Φ(G) is p-nilpotent, then N is p-nilpotent; (2) Let H be a subgroup of G and N be a p′-group, if HN/N is p-nilpotent, then H is p-nilpotent.

Proof. The statement (1) is directly form [16, Lemma 2.5], and (2) is easy. Lemma 2.6. Let G be a p-solvable group and H a subgroup of G, let H,N,A,B be subgroups of G and N, A, B are normal in G, then

(1) hp(H) ≤ hp(G).

(2) hp(G/N) ≤ hp(G).

(3) If G = A × B, then hp(G) = max{hp(A), hp(B)}.

(4) If hp(G/A) ≤ k and hp(G/N) ≤ k, then hp(G/(A ∩ B)) ≤ k.

(5) hp(G/Φ(G))) = hp(G). ′ (6) If N is a p -group, then hp(G/N)= hp(G).

Proof. Let 1 = N0 ≤ N1 ≤ N2 ≤···≤ Nr = G be the shortest normal chain of G with p-nilpotent factors Ni/Ni−1 for all i =1, 2, ··· ,r.

(1) If H ≤ G, then 1 = N0 ∩ H ≤ N1 ∩ H ≤ N2 ∩ H ≤···≤ Nr ∩ H = G ∩ H = H ∼ is a normal chain of H with Ni ∩ H/Ni−1 ∩ H = (Ni ∩ H)Ni−1/Ni−1 a p-nilpotent factors for all i =1, 2, ··· ,r, so hp(H) ≤ hp(G).

6 (2) If N E G, obviously, 1¯ = N0N/N ≤ N1N/N ≤ N2N/N ≤ ··· ≤ Nr/N = ∼ G/N is a normal chain of G/N. Note that NiN/N/Ni−1N/N = NiN/Ni−1N =

Ni/Ni−1(Ni ∩ N) ≤ Ni/Ni−1 is p-nilpotent, so hp(G/N) ≤ hp(G).

(3) Let 1 = A0 ≤ A1 ≤ A2 ≤ ··· ≤ Ar = A and 1 = B0 ≤ B1 ≤ B2 ≤

··· ≤ Bt = B be the shortest normal chain of A and B with p-nilpotent factors respectively. Without loss of generality, assume r ≤ t, then 1 = A0B0 ≤ A1B1 ≤

A2B2 ≤ ··· ≤ ArBr ≤ ArBr+1 ≤ ArBr+2 ≤ ArBt = AB is a normal chain. ∼ Since AiBi/Ai−1Bi−1 = Ai/Ai−1(Ai ∩ Bi−1) · Bi/Bi−1(Bi ∩ Ai−1) whenever i ≤ r ∼ and ArBj/ArBj−1 = Bj/Bj−1(Bj ∩ Ai) whenever r < j ≤ t are p-nilpotent. So hp(G) = max{hp(A), hp(B)}. (4) Since G/(A ∩ B) ∼= G/A × G/B, the result follows from (1) and (3).

(5) Assume that 1¯ = Φ(G)/Φ(G) = T0/Φ(G) ≤ T1/Φ(G) ≤ T2/Φ(G) ≤ ··· ≤

Ts/Φ(G) = G/Φ(G) is the normal chain of G with p-nilpotent factors, so s ≤ r and ∼ Ti/Ti−1 = Ti/Φ(G)/Ti−1/Φ(G) is p-nilpotent. Since T1/Φ(G) is p-nilpotent, then T1 is p-nilpotent by Lemma 2.5, so 1 ≤ T1 ≤ T2 ≤···≤ Ts = G is a normal chain with p-nilpotent factors. Now we have r ≤ s, hence s = r.

(6) Let 1=¯ N0/N ≤ N1/N ≤ N2/N ≤···≤ Nr/N = G/N be a normal chain of ∼ G/N with p-nilpotent factors Ni/N/Ni−1/N = Ni/Ni−1 for all i =1, 2, ··· ,r. Now by ′ Lemma 2.5, N1 is p-nilpotent since N is a p -group, so 1 ≤ N1 ≤ N2 ≤···≤ Nr = G is a normal chain of G with p-nilpotent factors Ni/Ni−1 for all i =1, 2, ··· ,r, which implies that hp(G) ≤ hp(G/N), now by (1), we have hp(G/N)= hp(G).

As a local vision of [17, Lemma 3.2], we have

Lemma 2.7. Let G be a solvable group, then lp(G) ≤ 1 if h(G) ≤ 2 for every p ∈ π(G). Lemma 2.8. [18, Theorem 1 and Proposition 1] Let F be a saturated formation. (1) Assume that G is a group such that G does not belong to F , but all its proper subgroups belong to F . Then F ′(G)/Φ(G) is the unique minimal normal subgroup of G/Φ(G), where F ′(G)= Soc(G mod Φ(G)), and F ′(G)= GF Φ(G). In addition, if the derived subgroup of GF is a proper subgroup of GF , then GF is a soluble group. Furthermore, if GF is soluble, then F ′(G)= F (G), the Fitting subgroup of G. F ′ F Moreover (G ) = T ∩ G for every maximal subgroup T of G such that G/TG 6∈ F and F ′(G)T = G. (2) Assume that G is a group such that G does not belong to F and there exists a maximal subgroup M of G such that M ∈ F and G = MF (G). Then GF /(GF )′ is a chief factor of G, GF is a p-group for some prime p, GF has exponent p if p > 2 and exponent at most 4 if p = 2. Moreover, either GF is elementary abelian

7 or (GF )′ = Z(GF )=Φ(GF ) is an elementary abelian group. U Lemma 2.9. Let G be a p-solvable minimal non-p-supersolvable group, then G p is a p-group.

Proof. By Lemma 2.8(1), G/Φ(G) has the unique minimal normal subgroup F ′(G)/Φ(G), U where F ′(G)= G p Φ(G), so F ′(G)/Φ(G) is an elementary abelian p-group since G is a p-solvable group. Otherwise, F ′(G)/Φ(G) is a p′-group and the p-supersolvability of G/Φ(G)/F ′(G)/Φ(G) implies that G/Φ(G) is p-supersolvable, thus G is p-supersolvable, U a contradiction. Furthermore, Φ(G) = 1. Now we have G p is solvable and F ′(G)= U F (G), G = F (G)T for some maximal subgroup T of G. Now by Lemma 2.8(2), G p is a p-group.

U Up1 Up2 Up Lemma 2.10. Let G be a group and π(G)= {p1,p2, ··· ,pn}, then G = G G ··· G n .

U Up U Proof. It follows from the supersolvability of G/G that G i ≤ G for all pi ∈ π(G), U U U U U U U p1 p2 pn p1 p2 pn U so G G ··· G ≤ G . Conversely, obviously, G/(G G ··· G ) ∈ pi for U Up1 Up2 Upn any pi ∈ π(G), so G ≤ G G ··· G , as desired.

3 Basic properties

Now we give some elementary properties. Firstly, we have

Proposition 3.1. Let G = A × B with (|A|, |B|)=1 and H ≤ G, then NUp (G,H)=

NUp (A, A ∩ H) × B whenever p ∈ π(A) or A × NUp (B, B ∩ H) whenever p ∈ π(B).

Proof. By hypothesis, we may assume that p ∈ π(A) \ π(B). Note that H = (H ∩ U U U U A)×(H ∩B) and T =(T ∩A)×(T ∩B), so T p =(T ∩A) p ×(T ∩B) p =(T ∩A) p , Up Up Up where T ≤ H. Now NG(T )= NA×B(T )= NA((T ∩ A) ) × B. We have

Up NUp (G,H)= \ NG(T ) T ≤H

Up =( \ NA((T ∩ A) )) × B T ≤H

Up =( \ NA(T )) × B T ≤H∩A

= NUp (A, A ∩ H) × B.

Similarly, if p ∈ π(B)\π(A), we have NUp (G,H)= A × NUp (B, B ∩ H). U Proposition 3.2. Let G be a p-solvable group, then the p-residual of NUp (G) is a p-group.

8 Proof. Denote X = NUp (G). We ﬁrst prove this result in the case that X = G. Up Assume G ∈ Up, obviously G = 1 is a p-group. Assume now that G 6∈ Up, if

Op(G) > 1, then by Lemma 2.2(2), G/Op(G) = NUp (G/Op(G)). By induction on U Up p Up |G|, (G/Op(G)) = NUp (G/Op(G)) is a p-group. Obviously G is a p-group. If

Op(G) = 1, then for any K

Assume now that X

Denote Z∞(G) be the terminal term of the ascending central series of G. As we E know, Z∞(G)= T{N G | Z(G/N)=1}, we have the following similar result. ∞ Theorem 3.3. Let G be a group, then NU (G)= {N E G | NU (G/N)=1}. p T p ∞ E U Proof. Let N G such that N p (G/N) = 1, then NUp (G/N) = 1. By Lemma ∞ ∞ ∞ ∞ 2.3(2), NU (G)N/N ≤ NU (G/N)=1, so NU (G) ≤ N, thus NU (G) ≤ {N E G | p p p p T

NUp (G/N)=1}. ∞ n Conversely, choice the least positive integer n such that NUp (G) = NUp (G). By n n+1 n n n U deﬁnition, N p (G/NUp (G)) = NUp (G)/NUp (G) = NUp (G)/NUp (G) = 1. Obviously, ∞ {N E G | NU (G/N)=1} ≤ NU (G). T p p

Up Now we give some characterizations between NUp (G,H) and CG(H ). Theorem 3.4. Let G be a p-solvable group and H a normal subgroup of G, then Up NUp (G,H)=1 if and only if CG(H )=1.

Up Proof. As we know, CG(H ) ≤ NUp (G,H), the necessary is obviously. Assume Up now CG(H ) = 1, we prove that NUp (G,H) = 1. If not, let N be a minimal Up normal subgroup of G contained in NUp (G,H), if N H, then N normalize H U U by deﬁnition, we have [N,H p ] ≤ N ∩ H p = 1 by the minimal normality of N, so Up N ≤ CG(H ) = 1, a contradiction. If H

By induction on |G||H| and Lemma 2.1(2), N ≤ H ∩ NUp (G,H) ≤ NUp (H,H) = 1, a contradiction, thus H = G. Now by the p-solvability of G, N is a p′-group or an ′ elementary abelian p-group. If N is a p -group, then G/CG(N) ∈ Up, otherwise, there is a minimal non-p-supersolvable subgroup T/CG(N) of G/CG(N). Let T = CG(N)L ∼ such that CG(N) ∩ L ≤ Φ(L). Since L/Φ(L) ≤ T/CG(N) = L/L ∩ CG(N) is a U U minimal non-p-supersolvable group, we have (L/Φ(L)) p = L p Φ(L)/Φ(L) is a p- Up Up Up group by Lemma 2.9. Now let P ∈ Sylp(L ), then L = P (L ∩Φ(L)). By Frattini Up Up Up argument, L = NL(P )L = NL(P )P (L ∩ Φ(L)) = NL(P ), we have P E L and

9 Up N normalize P , so P ≤ CG(N) ∩ L ≤ Φ(L), thus (L/Φ(L)) = 1, L ∈ Up, contrary Up Up to our choice of T/CG(N). Consequently, G ≤ CG(N), hence N ≤ CG(G ) = 1, a contradiction. So N is an elementary abelian p-group and N ≤ CG(N). Note that

Φ(G) ≤ F (G) ≤ CG(N), we consider the following two cases:

Case 1. Φ(G)= CG(N).

In this case, F (G)= CG(N)=Φ(G) ≤ Fp(G), so N ≤ Φ(G). By the p-solvability of G, Fp(G/Φ(G)) = Fp(G)/Φ(G) =6 1, so Φ(G)= F (G) < Fp(G). Now follows from Fp(G)/Op′ (G) = Op′p(G)/Op′ (G) = Op(G/Op′ (G)) that Op′ (G) > 1, so Op′ (G) ≤

CG(N)=Φ(G). On the other hand, Let P ∈ Sylp(Fp(G)), then Fp(G) = [Op′ (G)]P .

By the Frattini argument, we have G = Fp(G)NG(P ) = Op′ (G)NG(P ) = NG(P ), so

Fp(G)= P × Op′ (G)= F (G), a contradiction.

Case 2. Φ(G)

As corollaries of Theorem 3.4, we have Corollary 3.5. Let G be a solvable group and H a normal subgroup of G, then U NU (G,H)=1 if and only if CG(H )=1.

Corollary 3.6. Let G be a p-solvable group. Then NUp (G) =1 if and only if Up CG(G )=1.

Up Up Corollary 3.7. Let G be a p-solvable group, if Z(G )=1, then CG(G )= NUp (G).

Up Proof. By Corollary 3.6, we may assume that CG(G ) > 1. Moreover, obviously,

10 Up Up CG(G ) ≤ NUp (G). Now consider G/CG(G ), let

Up Up Up Up Up Up Up Up gCG(G ) ∈ CG/CG(G )((G/CG(G ) )= CG/CG(G )(G CG(G )/CG(G )),

Up Up Up Up Up Up Up Up then [g,G ] ≤ CG(G )∩G = Z(G ) = 1, thus g ∈ CG(G ) and CG/CG(G )(G/CG(G ) )= U ¯ p ¯ 1. By Corollary 3.6, we have NUp (G/CG(G )) = 1, so by Lemma 2.2(2), NUp (G) ≤ Up CG(G ), as desired. U U Corollary 3.8. Let G be a p-solvable group, then Z(G p )=1 if and only if G p ∩

NUp (G)=1.

Up Up Proof. Since Z(G ) ≤ G ∩NUp (G), so the “only if” part is obviously. Now assume Up Up Up that Z(G ) = 1, it follows form Corollary 3.7 that CG(G ) = NUp (G), so G ∩ Up Up Up NUp (G)= G ∩ CG(G )= Z(G ) = 1, as desired.

Up ∞ Theorem 3.9. Let G be a group, then Z∞(G ) ≤ NUp (G). U U Proof. If Z(G p ) = 1, obviously the result holds. So we assume Z(G p ) > 1. Up Up Up Now, Z(G ) ≤ NUp (G) by deﬁnition. By induction on |G|, Z∞(G /Z(G )) = Up Up ∞ Up Up ∞ Z∞((G/Z(G )) ) ≤ NUp (G/Z(G )), so Z∞(G ) ≤ NUp (G) by Lemma 2.3(3).

As a similar argument above, we have

U ∞ Corollary 3.10. Let G be a group, then Z∞(G ) ≤ NUp (G) for every p ∈ π(G).

At the end of this section, we investigate the relationship between NUp (G,H) and NU (G,H).

Lemma 3.11. U Let G be a group and H a subgroup of G, then T N p (G,H) ≤ p∈π(H) NU (G,H).

U Proof. Denote T = T N p (G,H), then for any subgroup K of H, by deﬁnition, T p∈π(G) U U normalize K p for every p ∈ π(K). Now by Lemma 2.10, T normalize K for any K ≤ H, so T ≤ NU (G,H), as desired. Lemma 3.12. Let G be a group and H a subgroup of G. If HU is a p-group for some p ∈ π(H), where π(H)= {p1,p2, ··· ,pn}, then NU (G,H)= NU (G,H). T pi pi∈π(H)

Proof. By Lemma 3.11, we only need to prove NU (G,H) ≤ NU (G,H). With- T pi pi∈π(H) Up out loss of generality , we assume p = p1. By Lemma 2.10, H i is a p-group for every U pj U pi ∈ π(H). Now if j =6 1, the pj-supersolvability of H/H implies that H ∈ pj . U U p U Thus by Lemma 2.10, K = K for every subgroup K of H and G = N pj (G,H)

11 for every p =6 pj ∈ π(H), so NU (G,H) = NU (G,H) = NU (G,H), as de- T pi p pi∈π(H) sired. Theorem 3.13. Let G be a solvable group and H a normal subgroup of G, then U U T N p (G,H)= G if and only if N (G,H)= G. p∈π(H)

Proof. Denote π(H) = {p1,p2, ··· ,pn}, the “⇒” part of this result follows from U Lemma 3.11. Now we prove the “⇐” part, this is equivalent to prove that N pi (G,H)= G for arbitrary pi ∈ π(H). Assume false, then there is at least one prime, say Up p in π(H) such that NUp (G,H) < G. If p 6∈ π(H ), then the p-supersolvability U U p U p of H/H implies that H ∈ p, then H = 1, so G = NUp (G,H), a contradic- U tion. By Lemma 2.2(1), H = NU (H). So H is nilpotent by [5, Theorem B]. Note that p ∈ π(HU ), so there is a minimal normal subgroup N of G contained in U Op(H ). By Lemma 2.1, G/N = NU (G/N, H/N). By induction on |G||H|, we have U U pi G/N = N pi (G/N, H/N) for any pi ∈ π(H). Now by Proposition 3.2, (H/N) is Up a pi-group for any pi ∈ π(H). In particular, H is a p-group. If there exists some Up p =6 pi ∈ π(H) such that H i =6 1, then as a similar proof above, there is a minimal U pi normal subgroup T of G contained in Opi (H ) such that G/T = NUp (G/T,H/T ). U U For any subgroup K of H, we have (KT/T ) p E G/T , so K p T E G. Note that Up Up Up Up Up [K , T ] ≤ [H , T ] ≤ H ∩ T = 1 since p =6 pi, hence K char K T E G, which U U p pi implies that K E G. This is also contrary to that G = NUp (G,H). So H = 1 U Up1 Up2 Upn Up for any p =6 pi ∈ π(H), thus by Lemma 2.10, H = H H ··· H = H is a p-group, now by Lemma 3.12, we have NU (G,H)= NU (G,H), a contradic- T pi pi∈π(H) tion. Theorem 3.14. Let G be a solvable group and H a normal subgroup of G, then U U T N p (G,H)=1 if and only if N (G,H)=1. p∈π(H)

U U Proof. Assume that N (G,H) = 1, then by Lemma 3.11, T N p (G,H) = 1. p∈π(H) U U U p U Now assume that T N p (G,H) = 1, since CG(H ) ≤ CG(H ) ≤ N p (G,H), so p∈π(H) U U U 1= CG(H ) ≤ T N p (G,H). By Corollary 3.5, N (G,H) = 1. p∈π(H)

4 N1-group

Let G be a group and H a subgroup of G. We now consider the case that N U G = NUp (G,H), i.e., G is an 1-group with respect to H and p. Obviously, G is an

12 Up N1-group with respect to H and Up if and only if K E G for every K ≤ H. There are many groups which are N1-groups, here we list some of them.

Proposition 4.1. The following groups are N1-groups with respect to some subgroup

H and Up: (1) All (p-)supersolvable groups. (2) Groups with H a (p-)supersolvable subgroup. (3) Groups all of whose non-(p-)supersolvable subgroups are normal. U (4) Groups with G p a cyclic subgroup. (5) Groups with a normal minimal non-p-supersolvable subgroup H.

Proposition 4.2. Let G be an N1-group with respect to H and Up, then

(1) For any K ≤ G, K is an N1-group with respect to H ∩ K and Up; (2) Let N E G and N ≤ H, then G/N is an N1-group with respect to H/N and

Up;

Proof. The proof of (1) and (2) follows from Lemmas 2.1(2)(3) respectively.

In general, if both of a normal subgroup N and corresponding quotient group G/N are N1-groups with respect to themselves and Up, then G may be not an N1-group. ∼ 4 2 3 Example 4.3. Let G = ha,b,c,di ⋊ he, fi = C2 ⋊ C6, where e = f = 1 and ae = ac, be = bd,ce = c,de = d, af = b, bf = ab, cf = d,df = cd (IdGroup=[96,70] in

GAP [15]). Let N = ha, b, c, d, ei, then N = NU2 (N) and G/N = NU2 (G/N). Now ∼ U2 let H = ha, b, fi = A4, the alternating group of degree 4, then H = ha, bi is not normal in G, so G is not an N1-group with respect to itself and U2.

In the case F ∈ {A , N }, the solvability of NF (G) has been investigated by many scholars. For example, see [6, 7]. For the case F = U , we have

Theorem 4.4. Let G be an N1-group with respect to itself and U . Then G is ∞ solvable. In particular, NU (G) and NU are solvable.

Proof. Obviously, every subgroup and every homomorphic image of G satisﬁes our hypothesis. If G is not solvable, then G is a non-abelian simple group. Let H be a proper subgroup of G, then HU E G, so HU = 1 and H is supersolvable. This implies that G is a minimal non-supersolvable group. Now by Doerk’s result [19], G is solvable, a contradiction.

F U Let H be a subgroup of G, for the case of = p, the solvability of NUp (G,H) does not always holds. Here, we have

13 Theorem 4.5. Let G be a group and H a subgroup of G, if G is an N1-group with respect to H and Up, where p is the smallest prime in π(H), then H is p-solvable.

Proof. Assume that the result is false and let the pair (G,H) be a counterexample with |G||H| minimal. By Feit-Thompson’s theorem, we may assume that p = 2. If

G = NU2 (G). If G is a non-abelian simple group, we prove the contradiction that G is a solvable simple group. Let K be any proper subgroup of G, then KU2 E G, so KU2 = 1 and K is 2-supersolvable. Which implies that G is a non-2-supersolvable group all of whose proper subgroups are 2-supersolvable. Now by [20, 1.1], G is solvable, a contradiction. Let N be a minimal normal subgroup of G, then Lemma

2.2 implies that G/N = NU2 (G/N) and N = NU2 (N), thus G/N and N are 2-solvable by our choice of (G,H), hence G is 2-solvable, the ﬁnial contradiction complement our proof. Corollary 4.6. Let G be a group and p is the smallest prime in π(G). If G is an

N1-group with respect to itself and Up, then G is p-solvable. In Theorem 4.5, the restriction that p is smallest prime in π(H) is necessary. Furthermore, in general, we can’t obtain the p-solvability of G even though p = 2. Example 4.7. Let G = SL(2, 5)(IdGroup=[120,5], see [15]), then π(G) = {2, 3, 5} and G is not 2-solvable and not 3-solvable. But G = NU3 (G). ∼ Furthermore, G has a normal subgroup, say H = C2, obviously, G = NU2 (G,H). But we have the following obvious conclusions:

Theorem 4.8. Let G be a group and H a subgroup of G . Assume G is an N1-group with respect to H and Up, where p is the smallest prime divisor of |H|. Then G is p-solvable if one of the statement holds: (1) H E G and G/H is p-solvable; U (2) G p ≤ H; Theorem 4.9. Let G be a group and P a Sylow p-subgroup of G. Assume that P EG and G is an N1-group with itself and Up, then G is solvable.

Theorem 4.10. Let G be a p-solvable N1-group with respect to itself and Up, then U (1) G p is a p-group.

(2) NUp (G/N) > 1 for any proper normal subgroup N of G. (3) hp(G) ≤ 3;

14 Proof. (1) It follows from Proposition 3.2.

(2) Let N be a proper normal subgroup of G, by Lemma 2.2(2), G/N = NUp (G/N), so NUp (G/N) > 1 by our choice of N. Up Up (3) By (1), G is a p-group, then G ≤ Fp(G). Consider G¯ = G/Fp(G), then G¯ ′ ′ is p-supersolvable, thus G¯ = G Fp(G)/Fp(G) is p-nilpotent. So hp(G) ≤ hp(G¯)+1 ≤ ′ ′ hp(G/¯ G¯ )+ hp(G¯ ) + 1 = 3. As a corollary of Theorem 4.10, we have

Corollary 4.11. Let G be a N1-group with respect to itself and Up for every p ∈ π(G), then (1) G is solvable. (2) G = NU (G). (3) GU is nilpotent. (4) G/N ∈ N U for every normal subgroup N of G. (5) h(G) ≤ 3.

(6) lr(G) ≤ 2 for every r ∈ π(G).

Proof. (1) Let p ∈ π(G) be the smallest prime, if p> 2, obviously G is solvable by the well know Feit-Thompson Theorem. If p = 2, then by Theorem 4.5, G is 2-solvable, so G is solvable, as desired.

(2) By hypothesis, G = NUp (G) for every p ∈ π(G). Now by Theorem 3.13, G = NU (G). Up (3) Let p ∈ π(G), if G ∈ Up, then G = 1, if G 6∈ Up, by Theorem 4.10(1), Up U G is a p-group. Now assume π(G)= {p1,p2, ··· ,pn}, then by Lemma 2.10, G = U U U U G p1 G p2 ··· G pn , so G is nilpotent. (4) By (2), GU is nilpotent. Let N be a proper normal subgroup of G, then (G/N)U = GU N/N ∼= GU /GU ∩ N is nilpotent, as desired. (5) It follows from Theorem 4.10(3).

′ ′ ′ (6) If Or (G) > 1, then NUp (G/Or (G)) = G/Or (G) for every p ∈ π(G) by Lemma 2.2(2), so G/Or′ (G) satisﬁes our hypothesis. By induction, lr(G/Or′ (G)) ≤ 2, which implies that lr(G) ≤ 2. As a similarly argument above and by Lemma 2.4, we have lr(G) = lr(G/Φ(G)) ≤ 2, hence we may assume that Or′ (G) = Φ(G) = 1, so

F (G) = Fr(G) = Or(G). Now consider G/F (G), by (5), h(G/F (G)) ≤ 2. Now by

Lemma 2.7, lr(G/F (G)) = lr(G/Or(G)) ≤ 1, so lr(G) ≤ lr(G/Or(G))+1 ≤ 2, as desired.

Recall that the well-know results of P.Hall and D.J.S. Robinson as follows: Theorem 4.12. Let G be a group and N a nilpotent normal subgroup of G. Then (1) (P. Hall) If G/N ′ is nilpotent, then G is nilpotent.

15 (2) (D.J.S. Robinson). If G/N ′ is supersolvable, then G is supersolvable. Here we have

Theorem 4.13. Let G be a p-solvable N1-group with respect to itself and Up. Then N U 2 G ∈ N if G ≤ G p .

Theorem 4.14. Let G be an N1-group with respect to itself and U . Then G ∈ U if 2 GU ≤ GU .

5 Applications

Theorem 5.1. Let G be a p-solvable group, then the following statements are equivalent:

(1) G ∈ NpUp. N U (2) G/NUp (G) ∈ p p.

Proof. (1)⇒ (2). It is obviously. (2) ⇒ (1). Assume the result is false and let G be a counterexample of minimal order. If NUp (G) = 1, there is nothing to prove. Now we assume that NUp (G) >

1. Let N be a minimal normal subgroup of G contained in NUp (G). Then either ′ N is an elementary abelian p-group or a p -group. Note that G/N/NUp (G/N) ≤ ∼ N U G/N/NUp (G)/N = G/NUp (G) by Lemma 2.2(2), then G/N ∈ p p by the choice of U G. So we may assume that N is an elementary abelian p-group, otherwise, G p N/N ∼= Up Up G /(G ∩ N) ∈ Np, so G ∈ NpUp by Lemma 2.5(2), a contradiction. Furthermore, Up if N ≤ Φ(G), then G/Φ(G) ∈ NpUp, so G Φ(G)/Φ(G) ∈ Np, hence G ∈ NpUp by Lemma 2.5(1), a contradiction again. Now there exist some maximal subgroup M Up Up of G such that G = MN and M ∩ N = 1. Also we have M = (G/N) ∈ Np. Up Up Up Note that N ≤ NUp (G), then N normalize M , thus [N, M ] = 1 and NM = U U U U U U N ×M p EG. Since G/NM p = MN/NM p = MNM p /NM p ∼= M/M ∩NM p = Up Up Up M/M ∈ Up, then G ≤ NM ∈ Np, a contradiction. ∞ Corollary 5.2. Let G be a p-solvable group, if G = NUp (G), then (1) G ∈ NpUp. ∞ (2) NUp (G/N) > 1 for any proper normal subgroup N of G. (3) hp(G) ≤ 3;

n Proof. (1) Set G = NUp (G) for some positive integer n. If n = 1, the result follows from Theorem 4.10(1). If n ≥ 2, then NUp (G) < G. By Lemma 2.3, G/NUp (G) = ∞ U U U N U NUp (G/N p (G)), so G/N p (G) satisﬁes our hypothesis, thus G/N p (G) ∈ p p by induction on |G|. Now by Theorem 5.1, G ∈ NpUp, as desired.

16 ∞ (2) If there exists some proper normal subgroup N of G such that NUp (G/N) = 1, ∞ then G = NUp (G) ≤ N by Lemma 2.3(3), that is impossible. The proof of (3) is similar to Theorem 4.10(3).

∞ We have proved in Theorem 4.4 that G is solvable if G = NU (G). Now as similar argument above, we have ∞ Corollary 5.3. Let G be a group, if G = NU (G), then (1) G ∈ N U . (2) h(G) ≤ 3.

(3) lp(G) ≤ 2 for every p ∈ π(G). Theorem 5.4. Let G be a group and p ∈ π(G). ∞ (1) Assume G is p-solvable, then hp(G) ≤ k if and only if hp(G/NUp (G)) ≤ k, where k ≥ 3. ∞ (2) Assume G is solvable, then lp(G) ≤ k if and only if lp(G/NU (G)) ≤ k, where k ≥ 2.

Proof. We only prove (1) since the proof of (2) is similar and based on Corollary 5.3(4). The “⇒” part of this theorem is obviously, so we prove the“⇐” part. Setting ∞ n n NUp (G)= NUp (G) for some positive integer n. If G = NUp (G), then the result follows n form Corollary 5.2(3). So we assume that NUp (G)

N be a minimal normal subgroup of G, then by Lemma 2.2(2), G/N/NUp (G/N) ≤ ∼ ∼ G/N/NUp (G)N/N = G/NUp (G)N = G/NUp (G)/NUp (G)N/NUp (G), so hp(G/N/NUp (G/N)) ≤ k by Lemma 2.6(2), thus hp(G/N) ≤ k by the choice of (G, 1). Suppose that N is not unique, that is G has at least two minimal normal subgroups, say N1, N2 and

N1 =6 N2, then as above, hp(G/N1), hp(G/N2) ≤ k, so hp(G) ≤ k by Lemma 2.6(4), a contradiction. With similar argument and by using of Lemma 2.6(5), we have

Φ(G) = 1. Furthermore, if Op′ (G) > 1, then it follows from Lemma 2.6(6) that hp(G/Op′ (G)) = hp(G) ≤ k, a contradiction. So Fp(G) = F (G) = CG(N) = N ≤

NUp (G). Therefore, G = MN, M ∩ N = 1 for some maximal subgroup M of G. U ∞ p U If M = 1, that is G/N is p-supersolvable, so G/N = N p (G/N) = NUp (G/N) = ∞ ∞ Up Up NUp (G)/N, thus G = NUp (G), a contradiction. If M > 1, then N normalize M , Up Up Up so [N, M ] = 1 and hence M ≤ CG(N)= N, M ≤ N ∩ M = 1, a contradiction n ∼ n−1 n n−1 again. Now assume that n> 1, since G/NUp (G) = G/NUp (G)/NUp (G)/NUp (G)= n−1 n−1 U G/NUp (G)/N p (G/NUp (G)), so by induction and combining the proof of the case n−1 n = 1, we have that hp(G/NUp (G)) ≤ k , hence hp(G) ≤ k by our choice of (G, n), a contradiction.

17 Theorem 5.5. Let G be a p-solvable group. If all elements of order p and order 4(if ∞ p =2) of G are in NUp (G), then hp(G) ≤ 3. Proof. Assume that the result is false and let G be a counterexample with minimal order, we prove this result by several steps:

Step 1. Let H

Step 2. Op′ (G) = 1, CG(Op(G)) ≤ Op(G)= F (G). ′ ∞ Let T = Op (G), if T =6 1, for any xT ∈ NUp (G)T/T with |x| = p or |x| = 4(if ∞ p = 2), xT ≤ NUp (G/T ) by Lemma 2.3(3), so G/T satisﬁes our hypothesis, thus hp(G/T ) ≤ 3 by the choice of G. It is easy to see that hp(G) ≤ 3, a contradiction. Further more, we have CG(Op(G)) ≤ Op(G) = F (G) since G is p-solvable and

Op′ (G) = 1.

Step 3. Φ(G)

By (2), Φ(G) ≤ Op(G), if Op(G)=Φ(G), then Opp′ (G)/Φ(G)= Op′ (G/Φ(G)) > 1 since G is p-solvable and Op(G/Φ(G)) = 1. Let Opp′ (G) = Φ(G)T , where T > 1 is ′ a p -group, then G = NG(T )Opp′(G) = NG(T ) by Frattini argument, it follows that T E G and then T ≤ Op′ (G) = 1 by Step 2, a contradiction. Step 4. Finial contradiction.

Denote by Fp the class of all p-solvable groups whose p-Fitting length are less than 3, then by Lemma 2.6, Fp is a saturated formation. Now by Step 1, G is an

Fp-critical group, i.e, G is not belongs to Fp, but all proper subgroups of G belong to Fp. Now by Steps 1,3, there exists some maximal subgroup M of G such that Fp G = Op(G)M = F (G)M and M ∈ Fp, so by Lemma 2.8(2), G is a p-group and Fp Fp G has exponent p if p > 2 and exponent at most 4 if p = 2, hence G ≤ Op(G) Fp ∞ ∞ and G ≤ NUp (G), it follows that hp(G/NUp (G)) ≤ 3. Now by Theorem 5.4(2), hp(G) ≤ 3, a contradiction. Corollary 5.6. Let G be a p-solvable group, if every cyclic subgroup of order prime p and order 4(if p =2) of G is contained in NUp (G), then hp(G) ≤ 3. Combining Theorem 5.4, Lemma 2.4 and Lemma 2.8, we have the following The- orem 5.7. It’s proof is similar to Theorem 5.5. Theorem 5.7. Let G be a solvable group, if every cyclic subgroup of order prime p ∞ and order 4(if p =2) of G is contained in NU (G), then lp(G) ≤ 2. As a local vision of Theorem 5.7, we have the following question.

18 Question 5.8. Let G be a p-solvable group, if every cyclic subgroup of order prime ∞ p and order 4(if p =2) of G is contained in NUp (G), does lp(G) ≤ 2? Remark 5.9. (1) In Theorem 5.5, Corollary 5.6 and Question 5.8, the Theorem 4.5 implies the hypothesises that G is p-solvable are necessary. For example, consider

G = A5, the alternating group of degree 5, then G = NU3 (G)= NU5 (G), so G satisﬁes our condition, but G is simple. (2) In Theorem 5.7, the solvability of G is necessary. For example, choice the non-solvable group G = C7 × A5(IdGroup=[420,13]) and let p = 7. Then NU (G) = ∞ NU (G)= C7 and every cyclic subgroup of order 7 is in C7. (3) In Question 5.8, the integer 2 is the minimum upper bound. For example, let G = C5 × S4(IdGroup=[120,37] in GAP). Obviously, G is 2-solvable and the proper non-2-supersolvable groups of G isomorphism to one of A4,S4,C5 × A4. As

A4,S4,C5 × A4 are normal in G, so G = NU2 (G). Note that 1 < C5 < C10 × C2 <

C5 × A4

References

[1] B. Huppert.(1967). Endliche Gruppen, Springer, Berlin-Heidelberg-New York.

[2] R. Baer.(1956).Norm and hypernorm, Publ. Math. Debrecen 4:347-356.

[3] R. Baer, Der Kern.(1935). eine charkteristishe Untergruppe, Compos. Math. 1:254-283.

[4] Su Ning, Wang Yanming.(2013). On the normalizers of F -residuals of all subgroups of a ﬁnite group, Journal of Algebra, 185-198.

[5] Su, Ning, Wang, Yanming.(2014). On the Intersection of the Normalizers of the F -Residuals of Subgroups of a Finite Group. Algebras and Representation Theory. 17:507-518.

[6] Li, Shirong, Shen, Zhencai.(2010). On the intersection of the normalizers of derived subgroups of all subgroups of a ﬁnite group. Journal of Algebra. 323:1349-1357.

[7] Shen Zhencai, Shi Wujie, Qian Guohua.(2012). On the norm of the nilpotent residuals of all subgroups of a ﬁnite group, Journal of Algebra. 352:290-298.

[8] Gong, L¨u, Guo, Xiuyun.(2013). On the Intersection of the Normalizers of the Nilpotent Residuals of All Subgroups of a Finite Group. Algebra Colloquium. 20(2):349-360.

[9] Li, Xuan, Guo, Xiuyun.(2015). On the normalizers of p-nilpotency-residuals of all subgroups in a ﬁnite group. Journal of Algebra and Its Applications. 14(10).1550146.

[10] Shen, Zhencai, Li, Shirong, Shi, Wujie.(2014). On the norm of the derived subgroups of all subgroups of a ﬁnite group. Bulletin of the Iranian Mathematical Society. 40:281- 291.

19 [11] Li, Xuan, Guo, Xiuyun.(2016).On Generalized Norms of Finite Groups. Communica- tions in Algebra. 44:1088-1095.

[12] Wang, Junxin, Guo, Xiuyun.(2007). On the Norm of Finite Groups. Algebra Collo- quium. 14(4):605-612.

[13] Hu, Bin, Huang, Jianhong, Skiba, Alexander.(2020). On the σ-nilpotent norm and the σ-nilpotnet length of a ﬁnite group. Glasgow Mathematical Journal. 63:1-12.

[14] Weinstein, M., Between Nilpotent and Solvable, Passaic/New York/Jersey:Polygonal Publishing House, 1982.

[15] The GAP Group: GAPgroups, algorithms, and programming.(2020). Version 4.11.0. http://www.gap-system.org.

[16] A. N. Skiba.(2015). On σ-subnormal and σ-permutable subgroups of ﬁnite groups, Journal of Algebra. 436:1-16.

[17] E. Jabara.(2018). The ﬁtting length of a product of mutually permutable ﬁnite groups. Acta Math. Hungar.

[18] A. Ballester-Bolinches, M.C.Pedraza-Aguilera.(1996). On minimal subgroups of ﬁnite groups, Acta Math. Hungar.73(4):335-342.

[19] Doerk, K.(1966). Minimal nicht überauflösbare, endliche Gruppen. Math. Z. 91, 198C205.

[20] Tuccillo F.(1992). On ﬁnite minimal non-p-supersoluble groups. Colloquium Mathematicum,63:119-131.