On some non-periodic groups whose cyclic subgroups are \(GNA\)-subgroups

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DOI:10.12958/adm548

On some non-periodic groups whose cyclic

subgroups are

GNA

_-subgroups

A. A. Pypka

Communicated by L. A. Kurdachenko

To Professor L. A. Kurdachenko on the occasion of his 70th birthday

A b s t r a c t . _{In this paper we obtain the description of} non-periodic locally generalized radical groups whose cyclic subgroups areGNA-subgroups.

Introduction

LetGbe a group. Recall that a subgroupH ofGis calledabnormal in Gif g∈ hH, Hg_i _{for every element}_g_∈_{G. Recall also that a subgroup} _H

ofGis self-normalizing inG ifNG(H) =H. It is well known that every

abnormal subgroup ofG is self-normalizing in G. Clearly abnormal and self-normalizing subgroups are antipodes of normal subgroups. On the one hand, a subgroupH of Gis both normal and abnormal in Giff H=G. On the other hand, ifH is a normal subgroup of G, thenNG(H) = G.

This remarks shows that the properties of normal subgroups and abnormal (respectively, self-normalizing) subgroups are diametrically opposite.

In the same time, there are subgroups that combine the concepts of normality and abnormality. Recall that a subgroupHof a groupGis called pronormal inGif for every elementg∈Gthe subgroups H and Hg are conjugate inhH, Hg_i_{. Thus, every normal and abnormal subgroup of}_G_is

2010 MSC:20F16, 20F18, 20F19, 20F22.

Key words and phrases: normal subgroup, abnormal subgroup, pronormal subgroup, self-normalizing subgroup,GNA_{-subgroup, locally nilpotent radical, locally}

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pronormal in G. Note that the normalizer NG(H)of pronormal subgroup

H is abnormal in G (see, for example, [1]), and hence self-normalizing inG.

In the paper [6] the authors introduced the following generalization of normal and abnormal subgroups.

Definition 1. A subgroup H of a group Gis called aGNA-subgroup of G if for every elementg ∈G either Hg ₌_H _or_N

K(NK(H)) =NK(H),

whereK =hH, gi.

Clearly every pronormal subgroup is a GNA-subgroup. Moreover, example from [6] shows that there are GNA-subgroups, which are not pronormal.

In the paper [6], the authors obtained the description of locally finite groups whose all subgroups are GNA-subgroups. Later, in the paper [5], it has been obtained the description of locally finite groups whose cyclic subgroups areGNA-subgroups.

In this article, we continue to study the influence of GNA-subgroups on the group structure. More precisely, we investigate the structure of some non-periodic groups whose cyclic subgroups areGNA-subgroups.

First, we recall some definitions. A locally nilpotent radical of a group Gis a subgroup generated by all normal locally nilpotent subgroups of G. We will denote this subgroup by Lnr(G). We recall also that alocally finite radical of a group Gis a subgroup generated by all normal locally finite subgroups ofG. We will denote this subgroup byLfr(G).

A group G is calledradical if Ghas an ascending series whose factors are locally nilpotent. A groupG is calledgeneralized radical ifGhas an ascending series whose factors are locally nilpotent or locally finite. Hence a generalized radical groupG either has an ascendant locally nilpotent subgroup or an ascendant locally finite subgroup. In the first case, the locally nilpotent radical Lnr(G) ofGis non-identity. In the second case, it is not hard to see thatGcontains a non-identity normal locally finite subgroup. Clearly, in every group Gthe subgroup Lfr(G) is the largest normal locally finite subgroup. Thus, every generalized radical group has an ascending series of normal subgroups with locally nilpotent or locally finite factors.

Observe also that a periodic generalized radical group is locally finite, and hence periodic locally generalized radical group is also locally finite.

The main result of this paper is the following

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of G is a GNA-subgroup, then eitherG is abelian or G=Rhbi, whereR is abelian, b2

∈R and ab₌_a−1 for each element a∈R. Moreover, in the second case, the following conditions hold:

(i) if b2

= 1, then the Sylow 2-subgroup D of R is elementary abelian;

(ii) ifb2

6

= 1, then eitherDis elementary abelian orD=E× hvi, where E is elementary abelian and hb, vi is a quaternion group.

Conversely, if a group Gsatisfies the above conditions, then every cyclic subgroup ofG is a GNA-subgroup.

1. Preliminary results

Lemma 1. LetG be a group whose cyclic subgroups are GNA-subgroups.

(i) If H is a subgroup ofG, then every cyclic subgroup of H is a GNA -subgroup.

(ii) If H is a normal subgroup of G, then every cyclic subgroup of G/H is a GNA-subgroup.

Proof. It follows from the definition of GNA-subgroups.

In the paper [3], B.H. Neumann proved the following classical result: if the factor-group G/ζ(G) is finite, then the derived subgroup [G, G] is also finite. As a corollary, we can come to the following generalization: if the factor-group G/ζ(G) is locally finite, then the derived subgroup [G, G]

is also locally finite.

Lemma 2. Let Gbe a generalized radical group. If every cyclic subgroup of G is a GNA-subgroup, thenG is soluble of class at most 3.

Proof. Suppose that the locally finite radical Lfr(G) = F of G is non-identity. Then [F, F] is abelian [5, Corollary 14]. It follows that in any case the locally nilpotent radicalLnr(G) =RofGis non-identity. We will prove thatG is a radical group. Suppose the contrary. ThenGincludes the normal subgroupsT andS such thatR6T 6S, T is radical,S/T is locally finite andLnr(S/T) =h1i. By [5, Corollary 4],R is a Dedekind group. Corollary 1 from [5] shows that every subgroup ofRisG-invariant. ThenS/CS(R)is abelian (see, for example [7, Theorem 1.5.1]). We observe

thatCS(R)∩T 6R (see [4, Lemma 4]). Suppose first thatR is periodic.

Then

CS(R)/(CS(R)∩R) =CS(R)/(CS(R)∩T)∼=CS(R)T /T 6S/T.

In particular,CS(R)/(CS(R)∩R) is locally finite. SinceR is periodic and

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metabelian by [5, Corollary 14]. Since S/T does not include non-identity normal abelian subgroups,CS(R)6T. We have now

S/T ∼= (S/CS(R))/(T /CS(R)).

We have remarked above that the factor-groupS/CS(R) is abelian, and

thereforeS/T is abelian. Contradiction.

Suppose now thatR is not periodic. Corollary 4 from [5] shows thatR is abelian. LetV be the periodic part ofRand putC =CS(R). By proved

above, C/R∼=C/(C∩R) is locally finite. Also, the inclusion R 6ζ(C)

implies that [C, C] is a locally finite subgroup. Using [5, Corollary 14], we obtain that C is soluble. It follows that CS(R) 6T, and using the

arguments from above, we again obtain a contradiction. This contradiction shows thatGis a radical group.

Then CG(R)6R [4, Lemma 4]. By [5, Corollary 4],R is a Dedekind

group, in particular, R is metabelian. Corollary 1 from [5] shows that every subgroup ofR is G-invariant. ThenG/CG(R) is abelian (see, e.g.,

Theorem 1.5.1 in [7]). The inclusion CG(R) 6 R implies that G/R is

abelian, so thatG is soluble andscl(G)63.

Corollary 1. Let G be a locally generalized radical group. If every cyclic subgroup of G is aGNA-subgroup, then Gis soluble of class at most 3.

Lemma 3. Let Gbe a group and A be a normal abelian subgroup of G. Suppose thatG=Ahbiwhereb2

∈Aandab ₌_a−1 for each elementa∈A. If the subgroup hbi is a GNA-subgroup, then

(i) if b2

= 1, then the Sylow 2-subgroupD of A is elementary abelian;

(ii) ifb2

6

= 1, then eitherD is elementary abelian orD=E× hvi where E is elementary abelian and hb, vi is a quaternion group.

Proof. Suppose thata∈CA(b), then ab =a. On the other hand, by our

conditions, ab = a−1, that is a−1 = a and 1 = a2. Thus C

A(b) is an

elementary abelian 2-subgroup. If c = b2

6

= 1, then c ∈ CA(b), and by

proved above, 1 =c2

=b4

. Conversely, if|a|= 2, thena∈CA(b).

Note that if a∈ hbi, thenhbia ₌_h_b_i_{. Let}_a _{be an arbitrary element}

of A. Then b−1a−1ba = aa = a2, and ba = a−1ba = ba2. Furthermore,

b−1ab=a−1 andab=ba−1. Then we have

(ba)(ba) =b(ab)a=b(ba−1₎_a₌_b2_.

Since this is valid for arbitrary elementa, we obtain(ba2

)2

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Sincehbiis aGNA-subgroup, we have two possibilities: eitherhbia=hbi orNK(hbi) = NK(NK(hbi)), where K =hhbi, ai = hb, ai, a∈ A. In the

first case, we obtain that a subgroup

hbi=hbia=hba2i=hb, a2i

is a 2-subgroup, in particular, a2

(and hence a) is a 2-element. In the second case, we again obtain that a subgroup

hbi=NK(hbi) =NK(NK(hbi))

is a 2-subgroup.

Suppose first |b| = 2. Then hbi ∩A = h1i. Assume that A has an element u of order 4. By proved above u−1bu = bu2. Since |u2| = 2,

u2

∈ CA(b). It follows that hb, u2i is abelian. On the one hand, hbi 6=

hbiu_{. On the other hand} _N

K(hbi) = hb, u2i 6= hbi, K =hhbi, ui = hb, ui.

So that NK(hbi) 6= NK(NK(hbi)), and we obtain a contradiction. This

contradiction shows that a Sylow 2-subgroup ofAis elementary abelian. Suppose now that c = b2

6

= 1. Let D be a Sylow 2-subgroup of A. Since the subgrouphciis normal inG, its image in the factor-groupG/hci is a GNA-subgroup. As proved above, D/hci is an elementary abelian 2-subgroup. Then eitherD is elementary abelian or Dhas an elementv of order 4 such thatv2

=c=b2

. Consider the last situation. Since v has a maximal order among all the elements ofD,D=E× hvi. Sincehvi is hbi-invariant, we have

|hbihvi|= (|hbi||hvi|)/|hbi ∩ hvi|= 8.

Furthermore, as proved above,v−1bv =bv2 = bb2 = b3. Hence hb, vi is

a product of two normal cyclic subgroups of order 4. It follows thathb, vi is a quaternion group.

Corollary 2. Let G be a group andA be a normal abelian non-periodic subgroup of G. Suppose that G=Ahbi where b2

∈ A, and ab ₌_a−1 _for each elementa∈A. ThenGhas a subgroup, which is not aGNA-subgroup.

Proof. Indeed, let h be an element ofA of infinite order. PutH =hh4 i. ThenHis normal inG, the elementhH has order 4, andhhHi∩hbHi=H. Lemma3shows that the subgrouphb, h4

ican not be aGNA-subgroup. Lemma 4. LetG be a non-periodic finitely generated soluble group. Sup-pose thatR is a locally nilpotent radical of G. If every cyclic subgroup of Gis a GNA-subgroup, then either G is abelian or G=Rhbi where R is abelian, b2

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Proof. By [5, Corollary 4],R is a Dedekind group. Corollary 1 from [5] shows that every subgroup of R isG-invariant. Then G/CG(R)is abelian

(see, for example [7, Theorem 1.5.1]). The inclusionCG(R)6R[4, Lemma

4] implies thatG/Ris abelian. Being abelian and finitely generatedG/Ris finitely presented. It follows that R has the elementsx1, . . . , xk such that

R=hx1iG_{. . .}_h_x

kiG (see, for example, [2, p. 421]). Since every subgroup

of R isG-invariant,hxjiG=hxji,16j6k. It follows thatR is finitely

generated. If we suppose thatRis periodic, thenRis finite. The inclusion CG(R)6R [4, Lemma 4] implies thatG/R is also finite, and hence G is

finite. This contradiction proves that R is non-periodic.

Then Corollary 2 and 3 from [5] shows thatR is abelian. Suppose that the centerζ(G)contains every element ofR of infinite order. Clearly,R is generated by elements of infinite order, so thatR 6ζ(G). Then the fact thatG/R is abelian implies thatG is nilpotent. Using again Corollary 2 and 3 from [5] we obtain that G is abelian. Therefore, we consider the case when a subgroup R contains an element of infinite order, which is not central. Since R is abelian and finitely generated,

R=hu1i ×. . .× hu_ni × hv1i ×. . .× hv_ti,

where the elements u1, . . . , u_n have infinite orders and v1, . . . , v_t have finite orders. Suppose thatuj ∈ζ(G)for allj,16j6n. Sinceζ(G)does

not include R, there exists an indexm such thatvm 6∈ζ(G). Then there

exists an element gsuch that vmg =vrm6=vm where r is a certain positive

integer. Consider the elementu1vm. We have

(u1vm)g=ug1vmg =u1vmr 6=u1vm.

We remark that u1v_m has infinite order. By [5, Corollary 1], a subgroup hu1vmiisG-invariant. Then the fact thatg6∈CG(u1vm)implies(u1vm)g =

(u1vm)−1 =u−

1 1 v−

1

m . On the other hand, we have(u1vm)g =u1vmr, which

implies that u1 =u− 1

1 . Contradiction. So, there exists an index j such thatuj 6∈ζ(G). Without loss of generality we can suppose thatj= 1. Let

b be an element of G such that G =hbiCG(hu1i). Thenub1 = u− 1 1 , and b2

∈CG(hu1i). Suppose now that there exists an index s, 1< s6n, such that[b, us] = 1. Then

(u1us)b =ub1ubs=u−

1

1 us6=u1us.

On the other hand, an infinite cyclic subgrouphu1usi is G-invariant by

[5, Corollary 1]. Then it follows that

(u1us)b = (u1us)−1 =u−11u− 1

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Henceus =u−s1, and we obtain a contradiction. This contradiction shows

that ub j = u−

1

j for all j,1 6 j 6n. Using the same arguments we can

prove thatvb j =v−

1

j for all j, 16j6t. It follows that ab =a−

1 for all elementsa∈R.

With the help of similar arguments we can prove that CG(hu1i) =C_G(R) =R.

HenceG=Rhbiand b2 ∈R.

Corollary 3. Let G be a non-periodic locally generalized radical group. Suppose thatR is a locally nilpotent radical ofG. If every cyclic subgroup of Gis a GNA-subgroup, then either G is abelian or G=Rhbi whereR is abelian, b2

∈R, and ab ₌_a−1 _{for each element}a∈R_.

Proof. By Corollary1, G is soluble. Suppose thatG is not abelian. Then Gincludes a non-periodic finitely generated non-abelian subgroup K. By Lemma4,K = Lnr(K)hbi, whereLnr(K)is abelian,b2

∈Lnr(K),b4

= 1, andab ₌_a−1 for each elementa∈Lnr(K).

Choose inGa local familyL_{of finitely generated subgroups containing} K, and letL∈L_{. Using again Lemma}₄_{we obtain that} _L_{= Lnr(}_L₎_h_b1_i_, where Lnr(L) is abelian, b2

1 ∈ Lnr(L), b 4

1 = 1, and ab1 = a− 1

for each elementa∈Lnr(L). Since K is not locally nilpotent, Lnr(L)∩K 6= K. On the other hand,

|K : Lnr(L)∩K|6|L: Lnr(L)|= 2,

so thatLnr(K) = Lnr(L)∩K. In particular,b6∈Lnr(L). It follows that b=b1u for some element u ∈Lnr(L). As in the proof of Lemma 3, we can show thatb2

= (b1u)2

=b2

1. So, instead ofb1 we can putb. In other words, ifL is an arbitrary subgroup of the familyL_{, then} _L_{= Lnr(}_L₎_h_b_i_, where Lnr(L) is abelian, b2

∈ Lnr(L), b4

= 1, and ab ₌ _a−1 for each

element a ∈ Lnr(L). Since L _{is a local family,} _G _{= Lnr(}_G₎_h_b_i_{, where}

Lnr(G) is abelian, b2

∈Lnr(G), b4

= 1 and ab ₌ _a−1 for each element

a∈Lnr(G).

2. Proof of the main result, Theorem 1

The necessity follows from Lemma3 and Corollary3.

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elementu∈R. In this case,G=Rhxi. As in the proof of Lemma3, we can show thatx2

= (bu)2

=b2

. SinceRis abelian,ax₌_ab ₌_a−1 for each

elementa∈R.

Let g be an arbitrary element ofG, then g =xka for some element a∈R. It follows thatg−1xg =a−1xa. We havex−1a−1xa=aa=a2, and

a−1xa=xa2. Furthermore,x−1ax=a−1, and ax=xa−1. Then we have (xa)(xa) =x(ax)a=x(xa−1)a=x2.

Consider hxia_{. We have} _h_x_ia₌_h_xa2

i=hx, a2

i. In particular, it shows that hxia _{is a 2-subgroup. In turn, it follows that} _a2

is a 2-element, so that ais also a 2-element. Then a=vcwhere c2

= 1. A subgrouphb, vi is a quaternion group, so that hbi is hvi-invariant. It follows that hxi is hvi-invariant. Since c2

= 1,[c, x] = 1. It follows that hxia ₌_h_x_i_{, which}

shows thathxi is aGNA-subgroup.

The following result follows directly from Theorem 1 and Corollary 2. Corollary 4. Let G be a non-periodic locally generalized radical group. Then every subgroup of G is a GNA-subgroup if and only ifG is abelian.

References

[1] M.S. Ba, Z.I. Borevich,On arrangement of intermediate subgroups, Rings and Linear Groups, Kubansk. Univ., Krasnodar (1988), 14–41.

[2] P. Hall,Finiteness conditions for soluble groups, Proc. London Math. Soc.4 (1954), 419–436.

[3] B.H. Neumann,Groups with finite classes of conjugate elements, Proc. London Math. Soc.1(1951), 178–187.

[4] B.I. Plotkin,Radical groups, Math. Sb.37(1955), 507–526.

[5] A.A. Pypka,On locally finite groups whose cyclic subgroups are GNA-subgroups, Algebra Discrete Math.24(2017), no. 2, 308–319.

[6] A.A. Pypka, N.A. Turbay,OnGNA-subgroups in locally finite groups, Proc. of Francisk Scorina Gomel state university93(2015), no. 6, 97–100.

[7] R. Schmidt,Subgroup lattices of groups, Walter de Gruyter, Berlin, 1994.

C o n ta c t i n f o r m at i o n

Aleksandr A. Pypka Department of Geometry and Algebra, Faculty of Mechanics and Mathematics, Oles Honchar Dnipro National University, Gagarin ave., 72, Dnipro, 49010, Ukraine

E-Mail(s):[email protected]