Publicly Available Published by De Gruyter March 21, 2019

# Finite groups with only 𝔉-normal and 𝔉-abnormal subgroups

Bin Hu , Jianhong Huang and Alexander N. Skiba
From the journal Journal of Group Theory

## Abstract

Let G be a finite group, and let 𝔉 be a class of groups. A chief factor H/K of G is said to be 𝔉-central (in G) if the semidirect product (H/K)(G/CG(H/K))𝔉. We say that a subgroup A of G is 𝔉-normal in G if every chief factor H/K of G between AG and AG is 𝔉-central in G and 𝔉-abnormal in G if V is not 𝔉-normal in W for every two subgroups V<W of G such that AV. We give a description of finite groups in which every subgroup is either 𝔉-normal or 𝔉-abnormal.

## 1 Introduction

Throughout this paper, all groups are finite, and G always denotes a finite group; (G) denotes the lattice of all subgroups of G. The group G is said to be strongly supersoluble [6] if G is supersoluble and G induces on any of its chief factor H/K an automorphism group of square free order.

In what follows, 𝔉 is a class of groups containing all nilpotent groups; G𝔉 denotes the intersection of all normal subgroups N of G with G/N𝔉. The class 𝔉 is said to be a formation if every homomorphic image of G/G𝔉 belongs to 𝔉 for every group G. The formation 𝔉 is said to be saturated if G𝔉 whenever G𝔉Φ(G) and hereditary (Mal’cev [9]) if H𝔉 whenever HG𝔉.

Recall that 𝔑, 𝔘 and 𝔘s are the classes of all nilpotent, all supersoluble and all strongly supersoluble groups, respectively. It is well known that 𝔑 and 𝔘 are hereditary saturated formations. In the paper [13], it is proved that the class 𝔘s is also a hereditary saturated formation.

If KH are normal subgroups of G and CCG(H/K), then we can form the semidirect product (H/K)(G/C) putting

(hK)gC=g-1hgKfor allhKH/KandgCG/C.

We say that a chief factor H/K of G is 𝔉-central in G (see [12]) if

(H/K)(G/CG(H/K))𝔉.

## Definition 1.1.

We say that a subgroup A of G is

1. 𝔉p-normal in G if every chief factor H/K of G between AG and AG with pπ(H/K) is 𝔉-central in G,

2. 𝔉-normal in G provided A is 𝔉p-normal in G for all primes p,

3. 𝔉-abnormal in G if V is not 𝔉-normal in W for every two subgroups V<W of G such that AV.

By definition, all normal subgroups of G are 𝔉-normal in G. Moreover, G is the unique subgroup of G which simultaneously is 𝔉-normal and 𝔉-abnormal in G. Every maximal subgroup of G is either 𝔉-normal or 𝔉-abnormal in G. In this paper, we study those groups in which each subgroup is either 𝔉-normal or 𝔉-abnormal.

## Example 1.2.

Before continuing, consider some well-known examples.

1. A subgroup A of G is said to be quasinormal or permutable if A permutes with all subgroups H of G, that is, AH=HA. In view of [8] (see also [1, Corollary 1.5.6] or [10, Theorem 5.2.3]), every quasinormal subgroup of G is 𝔑-normal in G.

2. A subgroup M of G is called modular if M is a modular element (in the sense of Kurosh [10, p. 43]) of the lattice (G) of all subgroups of G, that is,

1. X,MZ=X,MZ for all XG, ZG such that XZ,

2. M,YZ=M,YZ for all YG, ZG such that MZ.

In view of [10, Theorem 5.1.14], every modular subgroup of G is 𝔘s-normal in G. We say that a subgroup A of G is abmodular in G if V is not modular in W for every two subgroups V<W of G such that AV. In view of [10, Lemma 5.1.2], a subgroup A of G is abmodular in G if and only if it is 𝔘s-abnormal in G.

3. Let G=(C7(C2×C3))×P, where C2×C3=Aut(C7) and P is a non-abelian group of order p3 of exponent p for some prime p>2. Then the subgroup C2 is not 𝔘s-normal in G. Now let L be a subgroup of P of order p with LZ(P). Then L is neither quasinormal nor modular in G, but L evidently is 𝔑-normal and so 𝔘s-normal in G.

## Definition 1.3.

We say that G is a DM-group if G=DM, where

1. D=G1 is abelian,

2. M=x is a cyclic abnormal Sylow p-subgroup of G, where p is the smallest prime dividing |G|,

3. MG=xp=Z(G),

4. The element x induces a fixed-point-free power automorphism on D.

In [4], Fattahi defined B-groups to be groups in which every subgroup is either normal or abnormal, and he showed that a non-nilpotent group G is a B-group if and only if G is a DM-group. Later the results in [4] were generalized in several directions. In particular, Qinhai Zhang proved in [14] that every non-nilpotent group in which each subgroup is either quasinormal or abnormal is a B-group, and he posed a general problem of finding other conditions under which G is a B-group.

In this paper, we prove the following result in this research line.

## Theorem 1.4.

Let F be a hereditary saturated formation containing all nilpotent groups. If every subgroup of G is either F-normal or F-abnormal in G, then G is of either of the following types:

1. G𝔉.

2. G=DM is a DM-group, where D=G𝔉, and M is an 𝔉-abnormal subgroup of G with MG=Z𝔉(G).

Conversely, in a group G of type (I) or (II), every subgroup is either F-normal or F-abnormal.

In this theorem, Z𝔉(G) denotes the 𝔉-hypercenter of G, that is, the product of all normal subgroups N of G such that either N=1 or every chief factor of G below N is 𝔉-central in G.

In view of [11, Chapter IV, Theorem 17.1], A is an 𝔑-abnormal subgroup of a soluble group G if and only if A is abnormal in G. Therefore, we get from Theorem 1.4 the following:

## Corollary 1.5.

Every subgroup of G is either N-normal or N-abnormal in G if and only if G is either nilpotent or a B-group.

Corollary 1.5 covers the main result in [4]. Moreover, in view of Example 1.2 (i), we get from Corollary 1.5 the following result.

## Corollary 1.6 (Zhang [14]).

Let G be a non-nilpotent group. Then the following statements are equivalent:

1. G is a B-group.

2. Every subgroup of G is either quasinormal or abnormal in G.

Strongly supersoluble groups have found applications in many works (see, for example, [6, 13, 15]). Since every DM-group is evidently strongly supersoluble and the class of all strongly supersoluble groups is a hereditary saturated formation, we get from Theorem 1.4 and Proposition 3.4 the following characterizations of such groups.

## Corollary 1.7.

The following statements are equivalent:

1. G is strongly supersoluble.

2. Every Sylow subgroup of G is 𝔘s-normal in G.

3. Each subgroup of G is either 𝔘s-normal or 𝔘s-abnormal in G.

In fact, G is an M-group [10], that is, the lattice (G) is modular if and only if every subgroup of G is modular in G. From Corollary 1.7 and Example 1.2 (ii), we get that G is an M-group also in the case when every non-abmodular subgroup of G is modular in G.

In conclusion of this section, note that one of the main tools in the proof of Theorem 1.4 is the following useful fact.

## Proposition 1.8.

The class of all F-normal subgroups and, for any prime p, the class of all Fp-normal subgroups of G are sublattices of the lattice L(G).

## 2 Proof of Proposition 1.8

The first two lemmas can be proved by direct checking.

## Lemma 2.1.

Let N, M and K<HG be normal subgroups of G, where H/K is a chief factor of G.

1. If NK, then

(H/K)(G/CG(H/K))((H/N)/(K/N))((G/N)/CG/N((H/N)/(K/N))).
2. If T/L is a chief factor of G and H/K and T/L are G-isomorphic, then CG(H/K)=CG(T/L) and

(H/K)(G/CG(H/K))(T/L)(G/CG(T/L)).
3. (MN/N)(G/CG(MN/N))(M/MN)(G/CG(M/MN)).

Recall that G is called a pd-group if pπ(G).

## Lemma 2.2.

Let KH, KV, WV and NH be normal subgroups of G. Suppose that every chief pd-factor of G between K and H is F-central in G.

1. If every chief pd-factor of G between K and KN is 𝔉-central in G, then every chief pd-factor of G between KN and N is 𝔉-central in G.

2. If every chief pd-factor of G between W and V is 𝔉-central in G, then every chief pd-factor of G between KW and HV is 𝔉-central in G.

3. If every chief pd-factor of G between K and V is 𝔉-central in G, then every chief pd-factor of G between K and HV is 𝔉-central in G.

## Proof of Proposition 1.8.

Let A,B be subgroups of G such that A,Bpc𝔉(G), where pc𝔉(G) is the class of all 𝔉p-normal subgroups of G. Then every chief pd-factor of G between AG and AG is 𝔉-central in G.

First we show that ABpc𝔉(G). Note that (AB)G=AGBG. On the other hand, from the G-isomorphism

(AGBG)/(AGBG)=(AGBG)/(AGBGAG)AG(BGAG)/AGAG/AG,

we get that every chief pd-factor of G between AGBG and AGBG is 𝔉-central in G by Lemma 2.2 (1). Similarly, we get that every chief pd-factor of G between BGAG and BGAG is 𝔉-central in G. But then we get that every chief pd-factor of G between (AGBG)(BGAG)=AGBG and AGBG is 𝔉-central in G by Lemma 2.2 (2). It is clear also that (AB)GAGBG. Thus every chief pd-factor of G between (AB)G=AGBG and (AB)G is 𝔉-central in G. Therefore, ABpc𝔉(G).

Now we show that A,Bpc𝔉(G). In view of the G-isomorphisms

AG(AGBG)/AGBGAG/(AGAGBG)=AG/AG(AGBG)(AG/AG)/(AG(AGBG)/AG),

we get that every chief pd-factor of G between AGBG and AG(AGBG) is 𝔉-central in G. Similarly, every chief pd-factor of G between AGBG and BG(AGBG) is 𝔉-central in G. Moreover,

AGBG/AGBG=(AG(AGBG)/AGBG)(BG(AGBG)/AGBG),

and so every chief pd-factor of G between AGBG and AGBG is 𝔉-central in G by Lemma 2.2 (3).

Next note that A,BG=AGBG and AGBGA,BG. Therefore, every chief pd-factor of G between A,BG and A,BG=AGBG is 𝔉-central in G. Hence A,Bpc𝔉(G).

Therefore, pc𝔉(G) is a sublattice of the lattice (G). Finally, for the class c𝔉(G), of all 𝔉-normal subgroups of G, we have

c𝔉(G)=pπ(G)pc𝔉(G),

and so c𝔉(G) is also a sublattice of (G). ∎

## 3 Proof of Theorem 1.4

The following lemma is well known (see, for example, [12, Lemma 3.29]).

## Lemma 3.1.

Let H/K be an abelian chief factor of G, and let V be a maximal subgroup of G with KM and HM=G. Then

G/VG(H/K)(G/CG(H/K)).

## Lemma 3.2 ([5, Chapter 1, Theorem 2.7]).

Let F be a hereditary saturated formation, and let Z=ZF(G). Let N and E be subgroups of G, where N is normal in G.

1. If NZ, then Z/N=Z𝔉(G/N).

2. ZEZ𝔉(E).

3. If NZ/NZ𝔉(G/N).

In fact, the following lemma is a corollary of [3, Chapter IV, Theorem 6.7].

## Lemma 3.3.

Let F be a hereditary saturated formation, and let A and NE be subgroups of G, where N is normal and A is F-normal in G. Then

1. AN/N is 𝔉-normal in G/N.

2. If E/N is 𝔉-normal in G/N, then E is 𝔉-normal in G.

3. AE is 𝔉-normal in E.

## Proof.

(1) From the G-isomorphisms

(AGN/N)/(AGN/N)AGN/AGNAG/(AGAGN)=AG/AG(AGN)(AG/AG)/(AG(AGN)/AG)

and Lemma 2.1, we get that every chief factor of G/N between AGN/N and AGN/N is 𝔉-central in G/N since every chief factor of G between AG and AG is 𝔉-central in G/N by hypothesis. On the other hand, we have

(AN/N)G/N=(AN)G/N=AGN/NandAGN/N(AN/N)G/N.

Hence every chief factor of G/N between (AN/N)G/N and (AN/N)G/N is 𝔉-central in G/N, so AN/N is 𝔉-normal in G/N.

(2) This follows from the G-isomorphism

EG/EG(EG/N)/(EG/N)=(E/N)G/N/(E/N)G/N.

(3) First note that

(AG/AG)(EAG/AG)=AG(AGE)/AGZ𝔉(EAG/AG)

by Lemma 3.2 (2) since, by hypothesis, we have AG/AGZ𝔉(G/AG). On the other hand, we have

f(Z𝔉(EAG/AG))=Z𝔉(E/AGE),

where f:EAG/AGE/EAG is the canonical isomorphism from EAG/AG onto E/EAG. Hence

f(AG(AGE)/AG)=(AGE)/(AGE)Z𝔉(E/AGE),

where

AGE(EA)EAE(AE)EAGE,

and so

(AE)E/(AE)EZ𝔉(E/(AE)E)

by Lemma 3.2 (2) and (3). Hence AE is 𝔉-normal in E. ∎

## Proposition 3.4.

Let F be a saturated formation containing all nilpotent groups. Then the following statements are equivalent:

1. G𝔉.

2. Every chief factor of G is 𝔉-central in G.

3. Every Sylow subgroup of G is 𝔉-normal in G.

## Proof.

(i) (ii) This directly follows from the Barnes–Kegel result [3, Chapter IV, Proposition 1.5].

(ii) (iii) This implication is evident.

(ii) (i) In fact, this application is well known, and it can be easily proved by using Lemma 3.1 and induction on |G|.

(iii) (i) Let P be a Sylow p-subgroup of G, where p is any prime dividing |G|. Then PR/R is a Sylow p-subgroup of G/R and PR/R is 𝔉-normal in G/R by Lemma 3.3 (1) since P is 𝔉-normal in G by hypothesis. Therefore, the hypothesis holds for G/R, so G/R𝔉 by induction. Therefore, if either RΦ(G) or G has a minimal normal subgroup NR, then G𝔉. Moreover, if R/1 is 𝔉-central in G, then, by the Jordan–Hölder theorem for the chief series, every chief factor of G is 𝔉-central in G, and so G𝔉 by the implication (ii) (i).

Now assume that RΦ(G) is the unique minimal normal subgroup of G and that R/1 is not 𝔉-central in G. Let pπ(R), and let Gp be a Sylow p-subgroup of G. Then Rp=RGp is normal in Gp. Suppose that RpR; then, for some maximal subgroup V of G, we have GpNG(Rp)V, and so G=RV by the Frattini argument. Then (Gp)GVG=1, so 1<GpZ𝔉(G). But then RZ𝔉(G), and so R/1 is 𝔉-central in G, a contradiction. Hence R=Rp is an abelian p-group for some prime p. Let M be a maximal subgroup of G such that G=RM=RM. If G is a p-group, then G𝔉 since 𝔉 contains all nilpotent groups by hypothesis. Now assume |π(G)|>1, and let Q be a Sylow q-subgroup of M, where qπ(G){p}. Then QG=1, and so 1<QGZ𝔉(G), which again implies that R/1 is 𝔉-central in G, a contradiction. The implication is proved. ∎

In fact, the following lemma is a corollary of [3, Chapter IV, Theorem 6.7].

## Lemma 3.5.

Let P be a normal p-subgroup of G. If every chief factor of G between Φ(P) and P is cyclic, then every chief factor of G below P is cyclic.

## Proof of Theorem 1.4.

Necessity. Assume that this is false, and let G be a counterexample of minimal order. Then G𝔉, so D=G𝔉1, and also G is not nilpotent since 𝔉 contains all nilpotent groups by hypothesis. Let R be a minimal normal subgroup of G.

(1) Every proper subgroup E𝔉 is of type (II) and if G/R𝔉, then G/R is of type (II): Let A be any subgroup of E. If A is 𝔉-abnormal in G, then A is evidently 𝔉-abnormal in E. On the other hand, if A is 𝔉-normal in G, then A is 𝔉-normal in E by Lemma 3.3 (3). Hence the hypothesis holds for E, so the choice of G implies that E is of type (II).

Finally, if A/R is a non-𝔉-abnormal subgroup of G/R, then A is not 𝔉-abnormal in G, and so, by hypothesis and Lemma 3.3 (1), A/R is 𝔉-normal in G/R. Therefore, G/R is of type (II) by the choice of G.

(2) D<G: Assume D=G, and let P be a Sylow p-subgroup of G, where p is the smallest prime dividing |G|. Then P is not cyclic since otherwise G has a normal p-complement E by [7, Chapter IV, Satz 2.8] and G/E𝔉, which implies DE<G. Then, for a maximal subgroup V of P, we have V1, and V is not 𝔉-abnormal in G, so V is 𝔉-normal in G. Assume VG1 and RVG. If G/R𝔉, then D=R<G, which is impossible by our assumption D=G. Hence G/R𝔉, so statement (II) holds for G/R by claim (1), which implies G/R=D/R=(G/R)𝔉=(G/R)<G/R by [2, Proposition 2.2.8]. Then D=G<G. This contradiction shows that VG=1, so 1<VGZ𝔉(G), and hence we can assume without loss of generality RZ𝔉(G). Then G/R𝔉 by Lemma 3.2 (1), and so, as above, we get D=G<G. This contradiction completes the proof of Claim (2).

(3) D is nilpotent, and every element of G induces a power automorphism in D/Φ(D). Moreover, if p is the smallest prime dividing |G|, then p does not divide |D|: Let V be a maximal subgroup of D. Then V is not 𝔉-abnormal in G by claim (2), so V is 𝔉-normal in G. Assume that V is not normal in G. Then we have VG=D and VG/VGZ𝔉(G/VG), which implies G/VG𝔉 by Lemma 3.2 (1). But then DVGD. This contradiction shows that V is normal in G, so D is nilpotent, and every element of G induces a power automorphism in D/Φ(D) since every subgroup of D/Φ(D) can be written as the intersection of some maximal subgroup of D/Φ(D).

Now assume that p divides |D|. Then p divides |D/Φ(D)|, so, from the previous paragraph, we know that, for some maximal subgroup V of D, we have |D:V|=p, and V is normal in G. But then G/CG(D/V) is a cyclic group of order dividing p-1. Since p is the smallest prime dividing |G|, it follows that CG(D/V)=G, and so D/VZ(G/V). Hence G/V𝔉 by Lemma 3.2 (1) since 𝔉 contains all nilpotent groups by hypothesis, and so DV<D, a contradiction. Thus |D| is a p-number. Hence we have (3).

(4) D is a Hall subgroup of G. Hence D has a complement M in G, and p divides |M|: Suppose that this is false, and let P be a Sylow r-subgroup of D such that 1<P<Gr, where GrSylr(G).

First we show that D is a minimal normal subgroup of G. Assume that this is false. Then, for a minimal normal subgroup N of G contained in D, we have G/N𝔉. Since D is nilpotent by claim (3), N is a q-group for some prime q. Moreover, D/N=(G/N)𝔉 is a Hall subgroup of G/N by claim (1) and [2, Proposition 2.2.8]. Suppose that PN/N1. Then we have PN/NSylr(G/N). If qr, then PSylr(G). This contradicts the fact that P<Gr. Hence q=r, so NP, and therefore P/NSylr(G/N). It follows that PSylr(G). This contradiction shows that PN/N=1, which implies that N=P is a Sylow r-subgroup of D.

Therefore, N is the unique minimal normal subgroup of G contained in D. It is also clear that a p-complement U of D is a Hall subgroup of G. Claim (3) implies that U is characteristic in D, so it is normal in G. Therefore, U=1, and hence D=N. So NΦ(G) since the formation 𝔉 is saturated by hypothesis.

Let V be a maximal subgroup of N. Then V is 𝔉-normal in G by claim (2). Assume V1. Then VG<V<VG, where N/1=VG/VGZ𝔉(G/1), which implies G𝔉 by Lemma 3.2 (1). This contradiction shows that V=1, so we have |N|=r.

Now let S be a maximal subgroup of G such that NS, and let C=CG(N). Then G=NS, and so C=N(CS), where CS is normal in G. Assume CS1, and let L be a minimal normal subgroup of G contained in CS. Then NL/L=DL/L=(G/L)𝔉 is a Hall subgroup of G/L, and so NL/L=GrL/L, which implies NL=GrL. Hence Gr=N×(GrL). Let N=a, and let A=b be a subgroup of order r of GrL. Let L0=ab. Then |L0|=r, and L0N=1=L0GrL. First assume that L0 is not normal in G. Then L0L0GZ𝔉(G). On the other hand, from the G-isomorphism NL/NL, it follows that LZ𝔉(G) by Lemma 2.1 and Proposition 3.4. Moreover, Gr=L0(GrL) since |N|=r and L0(GrL)=1. Hence we have NGrZ𝔉(G), and so G𝔉 by Lemma 3.2 (1). This contradiction shows that CS=1, and so C=N is the unique minimal normal subgroup of G. Since |N|=r, it follows that G/N=G/C is cyclic, so G is supersoluble. Then a Sylow q-subgroup Q of G, where q is the largest prime dividing |G|, is normal in G. Therefore, NQ and NΦ(Q)Φ(G). Hence q=r, and Q is an elementary abelian r-group, which implies D=N=Q=Gr. This contradiction completes the proof of the fact that D is a Hall subgroup of G. Therefore, D has a complement M in G by the Schur–Zassenhaus theorem. Moreover, p divides |M| by claim (3). Hence we have (4).

(5) M=x is a cyclic 𝔉-abnormal Sylow p-subgroup of G, and every proper subgroup of M is 𝔉-normal in G: By claim (4), MG/D𝔉, and p divides |M|. Therefore, every proper subgroup V of M is not 𝔉-abnormal in G, so V is 𝔉-normal in G by hypothesis. Let Mp be a Sylow p-subgroup of M. Assume MpM. Then Mp is 𝔉-normal in G. Moreover, in this case, each Sylow subgroup P of M is 𝔉-normal in G, and P is a Sylow subgroup of G by claim (4). It follows that all Sylow subgroups of G are 𝔉-normal in G by claims (3) and (4), and therefore G𝔉 by Proposition 3.4. This contradiction shows that M=Mp is 𝔉-abnormal in G and every proper subgroup of M is 𝔉-normal in G. Therefore, M is cyclic since the set of all 𝔉-normal subgroups of G forms a sublattice of the lattice of subgroups of G by Proposition 1.8. Hence we have (5).

(6) Z𝔉(G)=MG=xp: Let Z=Z𝔉(G), V=xp. Assume ZD1. Let U=(ZD)M. Then ZDZUZ𝔉(U) by Lemma 3.2 (2), so U/Z𝔉(U)M/MZ𝔉(U)𝔉, and hence U𝔉 by Lemma 3.2 (1). But then M is not 𝔉-abnormal in G, contrary to claim (5). Therefore, ZD=1, so ZM, and hence, in fact, ZVG. It is clear that DCG(VG), so VGZ(G) by claim (5). Therefore, VVGZV by Lemma 3.2 (1). Hence we have MG=VG=V=Z(G)=Z.

(7) MyNG(H) for every proper subgroup H of D and every yG: Let V be a maximal subgroup of D such that HV. Then V is normal in G by claim (3). Let E=VMy and D0=E𝔉. It is clear that D0V. Moreover, D01 since otherwise E𝔉, which implies that the subgroups My and M are not 𝔉-abnormal in G, contrary to claim (5). Furthermore, claim (1) implies E=D0My=VMy, and every subgroup of D0=V is normalized by My. Hence MyNG(H).

(8) Every proper subgroup H of D is normal in G, and hence D is abelian: Claim (7) implies MGNG(H), where M is 𝔉-abnormal in G by claim (5), and so G=MGNG(H). Therefore, D is a Dedekind group, and |D| is odd by claim (3). Hence we have (8).

From claims (4), (5), (6) and (8), it follows that condition (II) holds for G, which is impossible by the choice of G. This contradiction completes the proof of the necessity condition of the theorem.

Sufficiency. If G𝔉, then every subgroup of G is 𝔉-normal in G by Proposition 3.4. Now assume that G is a group of type (II), and let A be any subgroup of G. First suppose that |M| divides |A|. Then, for some aG, we have MaA, so A is 𝔉-abnormal in G since the subgroups M and Ma are 𝔉-abnormal in G. Now assume that |M| does not divide |A|. Then, for a Sylow p-subgroup Ap of A and for some bG, we have (Ap)b<M, so the subgroups (Ap)b and Ap are 𝔉-normal in G by hypothesis. Now note that A=(AD)Ap, where AD is normal in G, and so A is 𝔉-normal in G by Proposition 1.8. ∎

Communicated by Evgenii I. Khukhro

Award Identifier / Grant number: 11401264

Award Identifier / Grant number: PPZY 2015A013

Funding statement: Research is supported by an NNSF grant of China (Grant No. 11401264) and a TAPP of Jiangsu Higher Education Institutions (PPZY 2015A013).

## Acknowledgements

The authors are very grateful to the helpful suggestions of the referee.

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Revised: 2019-01-18
Published Online: 2019-03-21
Published in Print: 2019-09-01

© 2020 Walter de Gruyter GmbH, Berlin/Boston