## 1 Introduction

Throughout this paper, all groups are finite, and *G* always denotes a finite group; *G*.
The group *G* is said to be *strongly supersoluble* [6] if *G* is supersoluble and *G* induces on any of its chief factor

In what follows, *N* of *G* with *formation* if every homomorphic image of *G*.
The formation *saturated* if *hereditary* (Mal’cev [9]) if

Recall that

If *G* and

We say that a chief factor *G* is in

*G*(see [12]) if

We say that a subgroup *A* of *G* is

- (i)
if every chief factor -normal in$\U0001d509p$ *G* of$H/K$ *G*between and${A}_{G}$ with${A}^{G}$ is$p\in \pi \left(H/K\right)$ -central in$\U0001d509$ *G*, - (ii)
provided -normal in$\U0001d509$ *G**A*is -normal in$\U0001d509p$ *G*for all primes*p*, - (iii)
in -abnormal$\U0001d509$ *G*if*V*is not -normal in$\U0001d509$ *W*for every two subgroups of$V<W$ *G*such that .$A\le V$

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

Before continuing, consider some well-known examples.

- (i)A subgroup
*A*of*G*is said to be*quasinormal*or*permutable*if*A*permutes with all subgroups*H*of*G*, that is, . In view of [8] (see also [1, Corollary 1.5.6] or [10, Theorem 5.2.3]), every quasinormal subgroup of$AH=HA$ *G*is -normal in$\U0001d511$ *G*. - (ii)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 of all subgroups of$\mathcal{L}\left(G\right)$ *G*, that is,- (a)
for all$\langle X,M\cap Z\rangle =\langle X,M\rangle \cap Z$ ,$X\le G$ such that$Z\le G$ ,$X\le Z$ - (b)
for all$\langle M,Y\cap Z\rangle =\langle M,Y\rangle \cap Z$ ,$Y\le G$ such that$Z\le G$ .$M\le Z$

*G*is -normal in${\U0001d518}_{s}$ *G*. We say that a subgroup*A*of*G*is*abmodular in*if*G**V*is not modular in*W*for every two subgroups of$V<W$ *G*such that . In view of [10, Lemma 5.1.2], a subgroup$A\le V$ *A*of*G*is abmodular in*G*if and only if it is -abnormal in${\U0001d518}_{s}$ *G*. - (a)
- (iii)Let
, where$G=\left({C}_{7}\u22ca\left({C}_{2}\times {C}_{3}\right)\right)\times P$ and${C}_{2}\times {C}_{3}=\mathrm{Aut}\left({C}_{7}\right)$ *P*is a non-abelian group of order of exponent${p}^{3}$ *p*for some prime . Then the subgroup$p>2$ is not${C}_{2}$ -normal in${\U0001d518}_{s}$ *G*. Now let*L*be a subgroup of*P*of order*p*with . Then$L\nleqq Z\left(P\right)$ *L*is neither quasinormal nor modular in*G*, but*L*evidently is -normal and so$\U0001d511$ -normal in${\U0001d518}_{s}$ *G*.

We say that *G* is a * DM-group*
if

- (1)
is abelian,$D={G}^{\prime}\ne 1$ - (2)
is a cyclic abnormal Sylow$M=\langle x\rangle $ *p*-subgroup of*G*, where*p*is the smallest prime dividing ,$\left|G\right|$ - (3)
,${M}_{G}=\langle {x}^{p}\rangle =Z\left(G\right)$ - (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.

*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:*

- (I)
.$G\in \U0001d509$ - (II)
$G=D\u22caM$ *is a**DM**-group, where*$D={G}^{\U0001d509}$ *, and**M**is an*$\U0001d509$ *-abnormal subgroup of**G**with* .${M}_{G}={Z}_{\U0001d509}\left(G\right)$

*Conversely, in a group G of type *(I)

*or*(II)

*, every subgroup is either*$F$ -normal or $F$ -abnormal.

In this theorem, , that is, the product of all normal subgroups

*N*of

*G*such that either

*G*below

*N*is

*G*.

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

*Every subgroup of G is either *

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.

*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.

*The following statements are equivalent:*

- (i)
*G**is strongly supersoluble.* - (ii)
*Every Sylow subgroup of**G**is*${\U0001d518}_{s}$ *-normal in**G*. - (iii)
*Each subgroup of**G**is either*${\U0001d518}_{s}$ *-normal or*${\U0001d518}_{s}$ *-abnormal in**G*.

In fact, *G* is an * M-group* [10], that is, the lattice

*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.

*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 $L\left(G\right)$.*

## 2 Proof of Proposition 1.8

The first two lemmas can be proved by direct checking.

*Let N, M and *

- (1)
*If*$N\le K$ *, then*$\begin{array}{cc}& \left(H/K\right)\u22ca\left(G/{C}_{G}\left(H/K\right)\right)\\ & \simeq \left(\left(H/N\right)/\left(K/N\right)\right)\u22ca\left(\left(G/N\right)/{C}_{G/N}\left(\left(H/N\right)/\left(K/N\right)\right)\right).\end{array}$ - (2)
*If*$T/L$ *is a chief factor of**G**and*$H/K$ *and*$T/L$ *are**G**-isomorphic, then*${C}_{G}\left(H/K\right)={C}_{G}\left(T/L\right)$ *and*$\left(H/K\right)\u22ca\left(G/{C}_{G}\left(H/K\right)\right)\simeq \left(T/L\right)\u22ca\left(G/{C}_{G}\left(T/L\right)\right).$ - (3)
.$\left(MN/N\right)\u22ca\left(G/{C}_{G}\left(MN/N\right)\right)\simeq \left(M/M\cap N\right)\u22ca\left(G/{C}_{G}\left(M/M\cap N\right)\right)$

Recall that *G* is called a * pd-group* if

*Let *

*G*. Suppose that every chief

*pd*-factor of

*G*between

*K*and

*H*is

*G*.

- (1)
*If every chief**pd**-factor of**G**between**K**and**KN**is*$\U0001d509$ *-central in**G**, then every chief**pd**-factor of**G**between*$K\cap N$ *and**N**is*$\U0001d509$ *-central in**G*. - (2)
*If every chief**pd**-factor of**G**between**W**and**V**is*$\U0001d509$ *-central in**G**, then every chief**pd**-factor of**G**between*$K\cap W$ *and*$H\cap V$ *is*$\U0001d509$ *-central in**G*. - (3)
*If every chief**pd**-factor of**G**between**K**and**V**is*$\U0001d509$ *-central in**G**, then every chief**pd**-factor of**G**between**K**and**HV**is*$\U0001d509$ *-central in**G*.

Let *G* such that *G*.
Then every chief *pd*-factor of *G* between *G*.

First we show that *G*-isomorphism

we get that every chief *pd*-factor of *G* between *G* by Lemma 2.2 (1).
Similarly, we get that every chief *pd*-factor of *G* between *G*.
But then we get that every chief *pd*-factor of *G* between *G* by Lemma 2.2 (2).
It is clear also that *pd*-factor of *G* between *G*.
Therefore,

Now we show that *G*-isomorphisms

we get that every chief *pd*-factor of *G* between *G*.
Similarly, every chief *pd*-factor of *G* between *G*.
Moreover,

and so every chief *pd*-factor of *G* between *G* by Lemma 2.2 (3).

Next note that *pd*-factor of *G* between *G*.
Hence

Therefore, *G*, we have

and so

## 3 Proof of Theorem 1.4

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

*Let *

*Let *

*N*and

*E*be subgroups of

*G*, where

*N*is normal in

*G*.

- (1)
*If*$N\le Z$ *, then* .$Z/N={Z}_{\U0001d509}\left(G/N\right)$ - (2)
.$Z\cap E\le {Z}_{\U0001d509}\left(E\right)$ - (3)
*If* .$NZ/N\le {Z}_{\U0001d509}\left(G/N\right)$

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

*Let $F$ be a hereditary saturated formation, and let A and $N\le E$ be subgroups of G, where N is normal and A is $F$-normal in G.
Then*

- (1)
$AN/N$ *is*$\U0001d509$ *-normal in* .$G/N$ - (2)
*If*$E/N$ *is*$\U0001d509$ *-normal in*$G/N$ *, then**E**is*$\U0001d509$ *-normal in**G*. - (3)
$A\cap E$ *is*$\U0001d509$ *-normal in**E*.

(1) From the *G*-isomorphisms

and Lemma 2.1, we get that every chief factor of *G* between

Hence every chief factor of

(2) This follows from the *G*-isomorphism

(3) First note that

by Lemma 3.2 (2) since, by hypothesis, we have

where

where

and so

by Lemma 3.2 (2) and (3).
Hence *E*.
∎

*Let *

- (i)
.$G\in \U0001d509$ - (ii)
*Every chief factor of**G**is*$\U0001d509$ *-central in**G*. - (iii)
*Every Sylow subgroup of**G**is*$\U0001d509$ *-normal in**G*.

(i)

(ii)

(ii)

(iii) *P* be a Sylow *p*-subgroup of *G*, where *p* is any prime dividing *p*-subgroup of *P* is *G* by hypothesis.
Therefore, the hypothesis holds for *G* has a minimal normal subgroup *G*, then, by the Jordan–Hölder theorem for the chief series, every chief factor of *G* is *G*, and so

Now assume that *G* and that *G*.
Let *p*-subgroup of *G*.
Then *V* of *G*, we have *G*, a contradiction.
Hence *p*-group for some prime *p*.
Let *M* be a maximal subgroup of *G* such that *G* is a *p*-group, then *Q* be a Sylow *q*-subgroup of *G*, a contradiction.
The implication is proved.
∎

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

*Let P be a normal p-subgroup of G.
If every chief factor of G between *

*Necessity.* Assume that this is false, and let *G* be a counterexample of minimal order.
Then *G* is not nilpotent since *R* be a minimal normal subgroup of *G*.

(1) *Every proper subgroup *: Let

*A*be any subgroup of

*E*. If

*A*is

*G*, then

*A*is evidently

*E*. On the other hand, if

*A*is

*G*, then

*A*is

*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* is not *G*, and so, by hypothesis and Lemma 3.3 (1), *G*.

(2) : Assume

*P*be a Sylow

*p*-subgroup of

*G*, where

*p*is the smallest prime dividing

*P*is not cyclic since otherwise

*G*has a normal

*p*-complement

*E*by [7, Chapter IV, Satz 2.8] and

*V*of

*P*, we have

*V*is not

*G*, so

*V*is

*G*. Assume

(3) * D is nilpotent, and every element of G induces a power automorphism in *: Let

*V*be a maximal subgroup of

*D*. Then

*V*is not

*G*by claim (2), so

*V*is

*G*. Assume that

*V*is not normal in

*G*. Then we have

*V*is normal in

*G*, so

*D*is nilpotent, and every element of

*G*induces a power automorphism in

Now assume that *p* divides *p* divides *V* of *D*, we have *V* is normal in *G*.
But then *p* is the smallest prime dividing

(4) * D is a Hall subgroup of G.
Hence D has a complement M in G, and p divides *: Suppose that this is false, and let

*P*be a Sylow

*r*-subgroup of

*D*such that

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 *D* is nilpotent by claim (3), *N* is a *q*-group for some prime *q*.
Moreover, *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,

Let *V* be a maximal subgroup of *N*.
Then *V* is *G* by claim (2).
Assume

Now let *S* be a maximal subgroup of *G* such that *G*.
Assume *L* be a minimal normal subgroup of *G* contained in *r* of *G*.
Then *G*-isomorphism *G*.
Since *G* is supersoluble.
Then a Sylow *q*-subgroup *Q* of *G*, where *q* is the largest prime dividing *G*.
Therefore, *Q* is an elementary abelian *r*-group, which implies *D* is a Hall subgroup of *G*.
Therefore, *D* has a complement *M* in *G* by the Schur–Zassenhaus theorem.
Moreover, *p* divides

(5) : By claim (4),

*p*-subgroup of

*G*, and every proper subgroup of

*M*is

*G*

*p*divides

*V*of

*M*is not

*G*, so

*V*is

*G*by hypothesis. Let

*p*-subgroup of

*M*. Assume

*G*. Moreover, in this case, each Sylow subgroup

*P*of

*M*is

*G*, and

*P*is a Sylow subgroup of

*G*by claim (4). It follows that all Sylow subgroups of

*G*are

*G*by claims (3) and (4), and therefore

*G*and every proper subgroup of

*M*is

*G*. Therefore,

*M*is cyclic since the set of all

*G*forms a sublattice of the lattice of subgroups of

*G*by Proposition 1.8. Hence we have (5).

(6) : Let

*M*is not

*G*, contrary to claim (5). Therefore,

(7) : Let

*V*be a maximal subgroup of

*D*such that

*V*is normal in

*G*by claim (3). Let

*M*are not

*G*, contrary to claim (5). Furthermore, claim (1) implies

(8) *Every proper subgroup H of D is normal in G,
and hence D is abelian*: Claim (7) implies

*M*is

*G*by claim (5), and so

*D*is a Dedekind group, and

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 *G* is *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 *A* is *G* since the subgroups *M* and *G*.
Now assume that *p*-subgroup *A* and for some *G* by hypothesis.
Now note that *G*, and so *A* is *G* by Proposition 1.8.
∎

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

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