Groups, in which almost all subgroups are near to normal

A subgroup H of a group G is said to be nearly normal, if H has a finite index in its normal closure. These subgroups have been introduced by B.H. Neumann. In a present paper is studied the groups whose non polycyclic by finite subgroups are nearly normal. It is not hard to show that under some n...

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Published in:	Algebra and Discrete Mathematics
Date:	2004
Main Authors:	Semko, M.M., Kuchmenko, S.M.
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Language:	English
Published:	Інститут прикладної математики і механіки НАН України 2004
Online Access:	https://nasplib.isofts.kiev.ua/handle/123456789/156421
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Cite this:	Groups, in which almost all subgroups are near to normal / M.M. Semko, S.M. Kuchmenko // Algebra and Discrete Mathematics. — 2004. — Vol. 3, № 2. — С. 92–113. — Бібліогр.: 20 назв. — англ.

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author	Semko, M.M. Kuchmenko, S.M.
author_facet	Semko, M.M. Kuchmenko, S.M.
citation_txt	Groups, in which almost all subgroups are near to normal / M.M. Semko, S.M. Kuchmenko // Algebra and Discrete Mathematics. — 2004. — Vol. 3, № 2. — С. 92–113. — Бібліогр.: 20 назв. — англ.
collection	DSpace DC
container_title	Algebra and Discrete Mathematics
description	A subgroup H of a group G is said to be nearly normal, if H has a finite index in its normal closure. These subgroups have been introduced by B.H. Neumann. In a present paper is studied the groups whose non polycyclic by finite subgroups are nearly normal. It is not hard to show that under some natural restrictions these groups either have a finite derived subgroup or belong to the class S₁F (the class of soluble by finite minimax groups). More precisely, this paper is dedicated of the study of S₁F groups whose non polycyclic by finite subgroups are nearly normal.
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fulltext	Jo u rn al A lg eb ra D is cr et e M at h . Algebra and Discrete Mathematics RESEARCH ARTICLE Number 2. (2004). pp. 92 – 113 c© Journal “Algebra and Discrete Mathematics” Groups, in which almost all subgroups are near to normal M. M. Semko and S. M. Kuchmenko Communicated by L. A. Kurdachenko Abstract. A subgroup H of a group G is said to be nearly normal, if H has a finite index in its normal closure. These sub- groups have been introduced by B.H. Neumann. In a present paper is studied the groups whose non polycyclic by finite subgroups are nearly normal. It is not hard to show that under some natural restrictions these groups either have a finite derived subgroup or belong to the class S1F (the class of soluble by finite minimax groups). More precisely, this paper is dedicated of the study of S1F groups whose non polycyclic by finite subgroups are nearly normal. Let ν be a subgroup-theoretical property. This property can be in- ternal to the group as, for example, in the cases when ν denotes the property of being a normal subgroup, subnormal subgroup, almost nor- mal subgroup, or permutable subgroup, and external to the groups, as in the case when ν refers to belonging to some class of groups X. If G is a group then let Sν(G) = {H\|H is a subgroup of G and H has a property ν}, Snon−ν(G) = {H\|H is a subgroup of G and H does not have a property ν}. One of the important problem of Group Theory is the study of influence of the families Sν(G) and Snon−ν(G) on a structure of group for the most natural properties ν. The starting point of such researches is the paper of R. Dedekind [1], in which finite groups with all subgroups are nor- mal have been described; that is the family Snon−norm(G) is empty. The Jo u rn al A lg eb ra D is cr et e M at h .M. M. Semko, S. M. Kuchmenko 93 next important paper in this direction was the work of G.A. Miller and H. Moreno [2], where finite groups, in which every proper subgroup is abelian, are considered; that is Snon−ab(G) = {G}. The paper of O.Yu. Schmidt [2], where have been described the finite groups with all proper subgroups are nilpotent (Snon−nil(G) = {G}), has played also an signif- icant role in this series. These researches inspired the systematic study of groups, in which the family Snon−ν(G) is "very small" or Sν(G) is "very large" in some sense. This direction became very fruitful. Many prominent mathematicians published significant amount of articles and monographs dedicated to this problematic. In the current paper we will consider groups with the restrictions on the family Snon−nn(G) of all non nearly normal subgroups. A subgroup H of a group G is said to be nearly normal (in G), if H has a finite index in its normal closure HG. Such subgroups without any special title have been introduced and considered first by B.H. Neumann [4]. In [5] they have been named finitely-normal. We think that this is not a best possible name. In [6] these subgroup have been named more successfully as near normal subgroup. Lately the in- terest to such subgroups increased significantly (see, for example, [7, 8]). In the paper [4] B.H. Neumann characterized groups, all subgroups of which are nearly normal (that is the family Snon−nn(G) is empty) as the groups with finite derived subgroup. The subsequent study of groups, in which the family Snon−nn(G) is “significantly small”, has been continued in the papers of other authors ( see, for example, [5, 7, 8]). In our paper we consider the groups, in which the family Snon−nn(G) consists of only polycyclic-by-finite subgroups. In this connection note that the groups, in which the family of all non-normal subgroups consists of polycyclic- by-finite subgroups, have been considered by L.A. Kurdachenko, V.V. Pylaev and G. Cutolo ( see [9, 10, 11]); the groups, in which all non al- most normal subgroups are polycyclic-by-finite, have been studied by S. Franciosi, F. de Giovanni and L.A. Kurdachenko [12] (more precisely, this paper dedicated to the groups, whose non-finitely generated subgroups are almost normal and the description of groups, whose not polycyclic-by- finite subgroups are almost normal, could be derived from these results). As the series of examples constructed by A.Yu. Ol’shanskij [12] show the groups, whose proper subgroups are polycyclic-by-finite have very com- plicated structure and their description is not achievable now. There- fore the study of groups, whose non-(polycyclic-by-finite) subgroups are nearly normal, it is expedient to carry out by some additional conditions, in some way similar to generalized solubility. It is not hard to show that under such conditions these groups either have a finite derived subgroup or belong to the class S1F. Hence the case of S1F-groups is the basic here. This paper is dedicated these case. Jo u rn al A lg eb ra D is cr et e M at h .94Groups, in which almost all subgroups are near to normal Note first some elementary properties of the nearly normal subgroups. Let H, L be the nearly normal subgroups of a group G, then K =< H, L > and H ∩ L are nearly normal. Indeed, we have HG = x1H ∪ . . . ∪ xnH, LG = Ly1 ∪ . . . ∪ Lym. If u ∈ HG, ν ∈ LG, then u ∈ xjH, ν ∈ Lyt for some 1 ≤ j ≤ n , 1 ≤ t ≤ m, so that uν ∈ xjHLyt. Hence HGLG = ⋃ 1≤j≤n, 1≤t≤m xjHLyt. Furthermore HL ≤ K implies xjHLyt ≤ xjKyt = xjyt(y −1 t Kyt), therefore HGLG = ⋃ 1≤j≤n, 1≤t≤m xjyt(y −1 t Kyt). Using 4.1 of [14] we obtain that at least one of the subgroups y−1 t Kyt has a finite index in HGLG, and hence K has a finite index in HGLG. An obviously inclusion KG ≤ HGLG implies that K is nearly normal in G. Consider now an intersection H ∩ L. We have /HG ∩LG/ : H ∩L/ = /HG ∩LG : HG ∩LG ∩L//HG ∩LG ∩L : H ∩L/. Using now the properties of indexes, we obtain that /HG ∩ LG : HG ∩ LG ∩ L/ = /(HG ∩ LG)L : L/ ≤ /LG : L/ and /HG ∩ LG ∩ L : H ∩ L/ = /(HG ∩ L)H : H/ ≤ /HG : H/, what implies that /HG ∩ LG : H ∩ L/ is finite. An obviously inclusion (H ∩L)G ≤ HG ∩LG implies that H ∩L is nearly normal in G. We will use often these properties without special references. Lemma 1. Let G be a group, whose non polycyclic-by-finite subgroups are nearly normal. If H is a normal subgroup of G such that every subgroup including H is non polycyclic-by-finite, then G/H has finite derived subgroup. In fact, every subgroup including H is nearly normal. In this case [G/H, G/H] is finite by a result due to B. Neumann [4]. A group G is called an F-perfect if it does not include the proper sub- groups of finite index. The F-perfect groups are the antipode (in some sense) of the residually finite groups. In every group G the subgroup gen- erated by all its F-perfect subgroups is likewise F-perfect. This subgroup is called the F-perfect part of a group G. Lemma 2. Let G be a group and H be an F-perfect subgroup of G. If H is nearly normal, then it is normal. Indeed, since the index /HG : H/ is finite and H is F-perfect, then H coincides with the F-perfect part of HG. The F-perfect part is a characteristic subgroup, in particular, it is normal in G. Jo u rn al A lg eb ra D is cr et e M at h .M. M. Semko, S. M. Kuchmenko 95 Theorem 1. Let G ∈ S1F. Every non polycyclic-by-finite subgroup of G is nearly normal if and only if G is the groups of one of the following types: (1) G is a group with finite derived subgroup; (2) G is an almost Prüfer group. Proof. Suppose, that G has an infinite derived subgroup. Being periodic G is a Chernikov group. Let D be a divisible (F-perfect ) part of G. Assume that D is not a Prüfer group. Let K be a Prüfer subgroup of D. We have D = K × B where B is a divisible subgroup, in particular, it is F-perfect. By Lemma 2 the both subgroups K and B are normal in G. Lemma 1 yields that [G/K, G/K] and [G/B, G/B] are finite. Since K ∩ B =< 1 >, then by Remak’s theorem G ↪→ G/K × G/B. In turn out this embedding proves that [G, G] is finite. This contradiction proves that D a Prüfer subgroup, that is G has a type (2). Conversely, let G is a group of types (1) – (2). If G has a type (1), then every its subgroup is nearly normal. Consider now the case, when G is a group of a type (2). Let H be a subgroup of G and suppose, that H is not polycyclic-by-finite. Denote again by D a divisible (F-perfect) part of G. Since G/D is finite, an intersection H ∩ D is infinite. Then H ∩D = D, that is D ≤ H and hence H has a finite index in G. Clearly then H is nearly normal in G. A further study of a non-periodic case splits on two following situa- tions: (1) a group includes the infinite periodic subgroups; and (2) all periodic subgroups are finite. Consider these situation in this turn. Lemma 3. Let G ∈ S1F. Suppose that G is not periodic, [G, G] is infinite and ζ(G) does not include the Prüfer subgroups. If every non polycyclic-by-finite subgroup of G is nearly normal, then G includes a normal Prüfer subgroup D such that G/D is a finitely generated group with finite derived subgroup. Proof. Let P be an infinite periodic subgroup of G. Choose a maximal periodic subgroup T including P . Since Let G ∈ S1F, T is Chernikov. Let D be a divisible (F-perfect) part of T . Assume that D is not a Prüfer group. Let K be a Prüfer subgroup of D. We have D = K × B where B is a divisible subgroup, in particular, it is F-perfect. By Lemma 2 Jo u rn al A lg eb ra D is cr et e M at h .96Groups, in which almost all subgroups are near to normal the both subgroups K and B are normal in G. Lemma 1 yields that [G/K, G/K] and [G/B, G/B] are finite. Since K ∩ B =< 1 >, then by Remak’s theorem G ↪→ G/K × G/B. In turn out this embedding proves that [G, G] is finite. This contradiction proves that D a Prüfer subgroup. A factor-group G/D has a finite derived subgroup, in particular, it is an FC-group. Since ζ(G) does not include D, there is an element g ∈ G such that [g, D] 6=< 1 >. If we suppose that [g, D] 6= D, then [g, D] is finite. It follows that a derived subgroup of < D, g > is finite, in particular, < D, g > is an FC-group. However the center of an FC- group includes every divisible subgroup. This contradiction proves an equation [g, D] = D. Put C/D = CG/D(gD) and let x ∈ C. Then gx = gy for some element y ∈ D. By D = [g, D] we have y = [g, u] for some element u ∈ D. Therefore gx = gy = g[g, u] = gu, and this implies xu−1 ∈ CG(g), which proves an equation C = DCG(g). By the selection of g we have D∩CG(g) 6= D, and hence D∩CG(g) is finite. Let E = CG(g) and suppose that E is not polycyclic-by-finite. Then E is nearly normal, i.e. /EG : E/ is finite. A normal closure EG is not polycyclic-by-finite, moreover, every subgroup including EG also is not polycyclic-by-finite and hence is nearly normal. In other words, every subgroup of G/EG is nearly normal. Then G/EG has a finite derived subgroup by a result due to B. Neumann [4]. Since /EG : E/ is finite, F = D ∩ EG is finite. By Remak’s theorem G/F ↪→ G/D × G/EG, what proves that [G/F, G/F ] is finite. Together with a finiteness of F it follows that [G, G] is finite. This contradiction proves that E is polycyclic-by-finite. Since G/D is an FC-group, C has a finite index in G. From C = DE we obtain that C/D is finitely generated, and hence so is G/D. Lemma 4. Let G be a group and A be an abelian torsion-free subgroup having finite 0-rank. If A is nearly normal, then A includes a G-invariant abelian subgroup B such that AG/B is finite. Proof. Since A is nearly normal, /AG : A/ is finite. There exists a positive integer n such that B = (AG)n ≤ A. Clearly B is G-invariant and a factor AG/B is bounded. The finiteness of /AG : A/ implies that AG has a finite special rank. But a bounded abelian-by-finite group of finite special rank is finite. Thus AG/B is finite. Lemma 5. Let G be a non-periodic locally ( soluble-by-finite ) group. Suppose also that [G, G] is infinite and ζ(G) includes the Prüfer sub- groups. If all non-polycyclic-by-finite subgroups of G are nearly normal, then [G/D, G/D] is finite and a section A/(A ∩ D) is finitely generated for every abelian subgroup A. Jo u rn al A lg eb ra D is cr et e M at h .M. M. Semko, S. M. Kuchmenko 97 Proof. Suppose the contrary. Let A/(A ∩ D) is not finitely generated. Since D ≤ ζ(G), AD is abelian. In other words, we can assume that D ≤ A. By the properties of the divisible subgroups of abelian groups (see, for example, [15, Theorem 21.2]) A = D × E for some subgroup E. By our assumption on A/D a subgroup E can not be finitely generated. Then it is nearly normal. Lemma 4 yields that E includes a G-invariant torsion-free subgroup B of finite index. In particular, B is not finitely generated. By Lemma 1 the both factor-groups G/D and G/B have the finite derived subgroups. By D ∩ B ≤ D ∩ E =< 1 > we obtain an embedding G ↪→ G/D × G/B, which proves that [G, G] is finite. This contradiction shows that A/(A ∩ D) is finitely generated. Corollary 1. Let G ∈ S1F. Suppose that G is not periodic, [G, G] is in- finite and ζ(G) includes a Prüfer subgroup D. If every non polycyclic-by- finite subgroup of G is nearly normal and G/D is not finitely generated, then every abelian subgroup of G has an infinite index in G. Proof. Since D is not polycyclic-by-finite, [G/D, G/D] is finite. Suppose that G includes an abelian normal subgroup A of finite index. Being F- perfect D lies in A. If A/D is finitely generated, then so is and G/D. This contradiction shows that A/D can not be finitely generated. However this contradicts to Lemma 5. Corollary 2. Let G ∈ S1F. Suppose that G is not periodic, [G, G] is infinite and ζ(G) includes a Prüfer subgroup D. If all non polycyclic- by-finite subgroups of G are nearly normal, then a section A/(A ∩ D) is finitely generated for every FC-subgroup A. Proof. Suppose the contrary, let A/(A ∩ D) is not finitely generated. Since D ≤ ζ(G), AD is an FC-subgroup. In other words, we can assume that D ≤ A. By Lemma 1 [G/D, G/D] is finite. It follows that the set T of all elements having finite orders is a (characteristic) subgroup of G. Moreover, G/T is abelian. Since Let G ∈ S1F, T is Chernikov. Lemma 5 proves that T/D is finite. Put C/D = CA/D(T/D), then A/C is finite. Since (C/D)∩(T/D) ≤ ζ(C/D) and (C/D)/((C/D)∩(T/D)) ∼= (C/D)(T/D)/(T/D) ∼= CT/T is abelian, then C/D is nilpotent. Finally, an inclusion D ≤ ζ(G) implies that C is nilpotent. A factor-group C/D is not finitely generated, because A/C is finite. Put Z = ζ(C), then C/Z is periodic ( see, for example, [16, Theorem 4.32]). Since Z has a finite special rank, it includes a finitely generated subgroup K such that C/K is periodic. By the nilpotency of C/K we have C/K = P/K × Q/K where P/K is a Sylow p-subgroup, p = Π(D), Q/K is a Sylow p,-subgroup. Since C/D is infinite, either DK/K has an infinite index in Jo u rn al A lg eb ra D is cr et e M at h .98Groups, in which almost all subgroups are near to normal P/K or Q/K is infinite. In every case C/K includes a normal; infinite subgroup U/K such that U/K ∩ DK/K =< 1 >. Then U ∩ D = L is a finite subgroup. Clearly U is not polycyclic, and Lemma 1 shows that [C/U, C/U ] is finite. Using again Remak’s theorem we obtain an embedding C/L ↪→ E/D × E/U , which proves that [C/L, C/L] is finite. Then by finiteness of L a derived subgroup [C, C] is finite. In this case G/[C, C] is abelian-by-finite, what contradicts to Corollary 1 of Lemma 5. This contradiction shows that A/(A ∩ D) is finitely generated. Corollary 3. Let G ∈ S1F. Suppose that G is not periodic, [G, G] is in- finite and ζ(G) includes a Prüfer subgroup D. If every non polycyclic-by- finite subgroup of G is nearly normal and G/D is not finitely generated, then every FC-subgroup of G has an infinite index in G. In particular, G/FC(G) is infinite. Proof. By Lemma 1 [G/D, G/D] is finite. Suppose that G includes a normal FC-subgroup A of finite index. Being F-perfect D lies in A. If we suppose that A/D is finitely generated, then so is and G/D. This contradiction shows that A/D can not be finitely generated. But this contradicts to Corollary 2 of Lemma 5. Lemma 6. Let G ∈ S1F. Suppose that G is not periodic, [G, G] is infinite and ζ(G) includes a Prüfer subgroup D. If every non polycyclic- by-finite subgroup of G is nearly normal, then (i) [G/D, G/D] is finite; (ii) If H is a not finitely generated subgroup of G, then D ≤ H. Proof. By Lemma 1 [G/D, G/D] is finite. Suppose that H does not include D. This means, that an intersection H ≤ D is finite. Since G ∈ S1F, H includes an abelian normal torsion-free subgroup A such that H/A is finite [17, Lemma 3]. Clearly A is not finitely generated. But this contradicts to Lemma 5. Let G be an abelian minimax group. Choose in G a finitely gen- erated torsion-free subgroup H such that G/H is periodic (and hence Chernikov). Let D/H be a divisible part of G/H. Put Sp(G) = Π(D/H). If K is another finitely generated torsion-free subgroup of G such that G/K is periodic, then the both factors H/(H ∩ K) and K/(H ∩ K) are finite. It follows that the divisible parts of G/H and G/K are isomor- phic. In other words, a set Sp(G) is an invariant of a group G. Let p 6∈ Π(G/H), then H/Hp is a finite Sylow p-subgroup of G/Hp and G/Hp = H/Hp × R/Lp ( see, for example, [15, Theorem 27.5]). This shows that G 6= Gp. Jo u rn al A lg eb ra D is cr et e M at h .M. M. Semko, S. M. Kuchmenko 99 Lemma 7. Let G ∈ S1F. Suppose that G is not periodic, [G, G] is infinite and ζ(G) includes a Prüfer subgroup D. If every non polycyclic- by-finite subgroup of G is nearly normal, then (i) [G/D, G/D] is finite; (ii) If H either G/D is finitely generated or G includes a normal periodic subgroup T such that T/D is finite and G/T is an abelian minimax torsion-free group such that Sp(G/T ) = p, where p = Π(D). Proof. By Lemma 1 [G/D, G/D] is finite. It follows that a set T of all elements having finite orders is a subgroup. Since G ∈ S1F, T is Chernikov. By Lemma 5 T/D is finite. Suppose that G/T is not finitely generated. A group G includes a normal subgroup H ≥ D such that H/D is an abelian torsion-free group and G/H is finite [17, Lemma 3]. In particular, H is a nilpotent of class 2 group. For each h ∈ H the mapping Θh : x −→ [h, x], x ∈ H, is an endomorphism, so that [H, h] = ImΘh ∼= H/KerΘh = H/CH(h). Since [H, h] ≤ D, then either H/CH(h) is a finite cyclic p-group or H/CH(h) is a Prüfer p-subgroup where p = Π(D). Since G ∈ S1F, an abelian torsion-free subgroup H/D has a finite 0-rank. Let h1D, . . . , hrD be a maximal Z independent set of H/D and put V = CH(h1, . . . , hr). An obvious equation CH(h1, . . . , hr) = CH(h1) ∩ . . . ∩ CH(hr) and Remak’s theorem give an embedding H/V ↪→ H/CH(h1) × . . . × CH(hr), which proves that H/V is a Chernikov p-group. Suppose that V D/D is not finitely generated. A subgroup W =< h1, . . . , hr > lies in the center of V , in particular, it is normal. By the selection of W a factor V/W is periodic. Since V is nilpotent, V/W = Drq∈Π(V/W )Sq/W where Sq/W is a Sylow q-subgroup of V/W . Since V ∈ S1F, the Sylow subgroups of its periodic factor-group V/W are Chernikov. Remark that the center of a periodic nilpotent group includes every divisible subgroup ( see, for example, [16, Lemma 3.13]), then every Sq/W is central-by -finite, in particular, it is an FC-group. Then V/W is a periodic FC- group. Since W is finitely generated, W ∩ D is finite. An embedding V/(W ∩D) ↪→ V/W × V/D and the fact that V/D is abelian prove that V/(W ∩ D) is an FC-group. It follows that V is an FC-group because W ∩ D is finite. By Corollary 3 of Lemma 5 V/D is finitely generated. Hence H/V is infinite. Then there is an index j, 1 ≤ j ≤ r, such that H/CH(hj) is a Prüfer p-group. We have already noted above that H/V is a Chernikov p-group. Since V/D is finitely generated, H/D is an abelian minimax torsion-free group such that Sp(H/D) = p. Finally, a finiteness of G/H implies that G/T is an abelian minimax torsion-free group such that Sp(G/T ) = p. Jo u rn al A lg eb ra D is cr et e M at h .100Groups, in which almost all subgroups are near to normal Now we may collect the all above results and to finish a consideration of a case when a group includes the infinite periodic subgroups. Theorem 2. Let G ∈ S1F. Suppose that G is not periodic and includes an infinite periodic subgroup. Every non polycyclic-by-finite subgroup of G is nearly normal if and only if G is the groups of one of the following types: (1) G has a finite derived subgroup. (2) [G, G] includes a Prüfer subgroup D such that D is normal in G, [G, G]/D is finite and G/D is finitely generated. (3) G satisfies the following conditions: (3A) [G, G] includes a Prüfer subgroup D such that D ≤ ζ(G); (3B) the set of all elements of G/D having finite orders is a finite sub- group T/D; (3C) G/T is an abelian minimax group; (3D) Sp(G/T ) = {p} = Π(D) where p is a prime; (3E) if A is an FC-subgroup, then A/(A ∩ D) is finitely generated. In particular, every non finitely generated subgroup of G includes D. Proof. Assume that [G, G] is infinite. If ζ(G) does not include the Prüfer subgroups, then by Lemma 3 G is a group of type (2). Suppose now that ζ(G) includes a Prüfer subgroup D. By Lemma 1 [G/D, G/D] is finite. It follows that a set T of all elements having finite orders is a subgroup. Since G ∈ S1F, T is Chernikov. By Lemma 5 T/D is finite. If G/D is finitely generated, then again G is a group of type (2). Suppose that G/D is not finitely generated. By Lemma 7 G satisfies (3C), (3D). Finally, a condition (3E) follows from Corollary 2 of Lemma 5 and Lemma 6. Conversely, let G be a group of types (1) –– (3). If G is a group of type (1), then each its subgroup is nearly normal. Let H be a subgroup of G and suppose that H is not polycyclic-by-finite. First consider a case when G is a group of type (2). Since G/D is finitely generated, H ∩ D can not be finite. Then H ∩D = D, that is D ≤ H. But every subgroup of G/D is nearly normal, so that H is nearly normal in G. Finally, let G be a group of type (3). Suppose that H does not include D. Then H ∩D is finite. The conditions (3B), (3C) and H/(H ∩D) ∼= (HD)/D together with finiteness of [G/D, G/D] imply that [H, H] is finite. In particular, H is an FC-subgroup. Since H ∩ D is finite, H/(H ∩ D) is not finitely Jo u rn al A lg eb ra D is cr et e M at h .M. M. Semko, S. M. Kuchmenko 101 generated, what contradicts (3E). This contradiction proves an inclusion D ≤ H. Now we remark again that every subgroup of G/D is nearly normal, so that H is nearly normal in G. The next result shows that a study of our groups is possible up to the factor-group by finite normal subgroups. Proposition 1. Let G be a group and F be a normal finite subgroup of G. A subgroup H is nearly normal in G if and only if HF/F is nearly normal in G/F . Proof. If H is nearly normal in G, then /HG : H/ is finite. Since F is finite, /HGF : HG/ is finite. It implies that /HGF : H/ is finite. In particular, the index /HGF : HF/ is finite, so that /(HF )G : HF/ is also finite. It proves that HF is nearly normal in G, therefore HF/F is nearly normal in G/F . Conversely, let HF/F is nearly normal in G/F . Then HF is nearly normal in G. Since /HF : H/ is finite, a subgroup H has a finite index in some normal subgroup. In particular, the index /HG : H/ is finite, so that H is nearly normal in G. The next finally phase is a consideration of a case when all periodic subgroups are finite. Proposition 1 shows that for the description of the locally nilpotent groups with finite periodic part it is sufficient to describe only torsion- free locally nilpotent groups. A description of such groups splits on two situation depending of a derived subgroup. Lemma 8. Let G be a locally nilpotent torsion-free group of finite 0- rank. Suppose that G is not finitely generated. If every non polycyclic subgroup of G is nearly normal, then ζ(G) is not finitely generated. In particular, G is nilpotent of class at most 2. Proof. Suppose that [G, G] is infinite. By Maltsev’s theorem (see, for example, [18, Theorem 6.36]) G is nilpotent. Since G is not finitely generated, its center Z can not be finitely generated [19, Lemma 2.6]. By Lemma 1 [G/Z, G/Z] is finite. On the other hand, G/Z is torsion- free (see, for example, [16, Theorem 2.25]). This means that G/Z is abelian. Corollary. Let G be a non-abelian locally nilpotent torsion-free group of finite 0-rank. Suppose that G is not finitely generated but [G, G] is finitely generated. If every non polycyclic subgroup of G is nearly normal, then G/ζ(G) is abelian and finitely generated. Jo u rn al A lg eb ra D is cr et e M at h .102Groups, in which almost all subgroups are near to normal Proof. By Lemma 8 ζ(G) is not finitely generated and G/ζ(G) is abelian. For every element g ∈ G the mapping φg : x −→ [x, g], x ∈ G, is an endomorphism, so that [G, g] = Imφg ∼= A/Kerφg = G/CG(g). Since [G, G] is finitely generated, [G, g] is likewise finitely generated. Then G/CG(g) is finitely generated for each g ∈ G. Let {g1ζ(G), . . . , gtζ(G)} be a maximal Z-independent set of G/ζ(G), H =< g1, . . . , gt >. By Remak’s theorem G/CG(H) ↪→ G/CG(g1) × . . . × G/CG(gt), which proves that G/CG(H) is finitely generated. Put D = Hζ(G), then CG(H) = CG(D), in particular, G/CG(D) is finitely generated. By the selection of D for every element g ∈ G there is an integer m such that gm ∈ D. For each element x ∈ CG(D) we have gmx = xgm. But in nilpotent torsion-free group G a last equation implies gx = xg, which shows that CG(D) ≤ ζ(G), that is CG(D) = ζ(G). In turn out it implies that a factor-group G/ζ(G) is finitely generated. Let A be an abelian torsion-free group. Denote by PMfg(A) the intersection of all pure not finitely generated subgroups of A. A subgroup PMfg(A) is called a pure polycyclic monolith of A. A subgroup PMfg(A) has been introduced in a paper [12]. We remark that it [12] has been used the other name and other designation. Lemma 9. Let A be an abelian torsion-free group. If < 1 > 6= PMfg(A), then r0(A) is finite and A includes a pure finitely generated subgroup B such that r0(A/B) = 1. Proof. Put C = PMfg(A). Then every proper pure subgroup of C is finitely generated. Let M be a maximal Z-independent set of C. We may extend M to a maximal Z-independent set L of a group A. For an element a ∈ M choose in A a maximal subgroup U under < a > ∩U =< 1 > and L {a} ⊆ U . Let T/U be the periodic part of A/U . By the selection of U an element < aU > has infinite order, so that a 6∈ T . It follows that U = T , so that A/U is torsion-free. In other words, U is a pure subgroup of A. Since U does not include C, U is finitely generated. By the selection of U every non-identity cyclic subgroup of A/U has a non- identity intersection with < aU >. In other words, A/U is a locally cyclic group (that is r0(A/U) = 1). Since a finitely generated subgroup of A has finite 0-rank, A has a finite 0-rank. Jo u rn al A lg eb ra D is cr et e M at h .M. M. Semko, S. M. Kuchmenko 103 Theorem 3. Let G be a non-abelian locally nilpotent torsion-free group of finite 0-rank. Suppose that G is not finitely generated but [G, G] is finitely generated. Every non polycyclic subgroup of G is nearly normal if and only if G satisfies the following conditions: (1) G includes a finitely generated subgroup E such that G = Eζ(G) and E ∩ ζ(G) = [G, G]; (2) ζ(G) is not finitely generated; (3) [G, G] PMfg(ζ(G)), in particular, PMfg(ζ(G)) 6=< 1 >, there- fore ζ(G) includes a finitely generated pure subgroup M such that ζ(G)/M is a non-cyclic locally cyclic group. Proof. By Lemma 8 Z = ζ(G) is not finitely generated. Corollary to Lemma 8 yields that G/Z is abelian and finitely generated. In particular, K = [G, G] ≤ ζ(G). An abelian group G/K has a free abelian factor- group (G/K)/(Z/K), therefore G/K = Z/K ×E/K where G/Z ∼= E/K is a finitely generated abelian group (see, for example, [15, Theorem 14.4]). Together with K a subgroup E is finitely generated. Finally, if S is a pure subgroup of Z such that S is not finitely generated, then Lemma 1 yields that G/S has finite derived subgroup. On the other hand, Z/S is torsion-free and also G/Z is torsion-free (see, for example, [16, Theorem 2.25]), so that G/S is torsion-free. Hence this factor-group is abelian, that is K ≤ S. It implies that [G, G] ≤ PMfg(ζ(G)). Finally, an application of Lemma 9 concludes a proof of (3). Conversely, suppose that G satisfies the conditions (1) –– (3). Let H be a not polycyclic subgroup of G. Since G/Z is finitely generated, H∩Z can not be finitely generated. Let L be a pure envelope of a subgroup H ∩ Z in Z, then L is not finitely generated. By (3) [G, G] = K ≤ L. Since [G, G] ≤ Z, K ∩ H = K ∩ H ∩ Z. Using K/(K ∩ H) = K/(K ∩ H ∩ Z) ∼= K(H ∩ Z)/(H ∩ Z) ≤ L/(H ∩ Z), And the fact that L/(H ∩ Z) is periodic, we obtain that K/(K ∩ H) is also periodic. On the other hand, K is finitely generated, so its periodic factor-group K/(K ∩ H) is finite. Thus G/(K ∩ H) has a finite derived subgroup K/(K ∩ H). Then every its subgroup is nearly normal, in particular, so is H/(K ∩H). It follows that H is nearly normal in G. Lemma 10. Let G be a non-abelian locally nilpotent torsion-free group of finite 0-rank. Suppose that [G, G] is not finitely generated. If every non polycyclic subgroup of G is nearly normal, then ζ(G) includes a finitely generated subgroup M such that ζ(G)/M is a Prüfer p-subgroup for some prime p. Jo u rn al A lg eb ra D is cr et e M at h .104Groups, in which almost all subgroups are near to normal Proof. By Lemma 8 Z = ζ(G) is not finitely generated and K = [G, G] ≤ ζ(G). Since G has a finite 0-rank, there exists a finitely generated sub- group E of Z such that Z/E is periodic. If a set Π(Z/E) is finite, then there are two infinite subsets ∆ and Ξ such that ∆ ∩ Ξ = Π(Z/E) and ∆∩Ξ = ®. Let D/E be a Sylow Ξ-subgroup of Z/E and S/E be a Sylow Ξ-subgroup of Z/E, then Z = DS and D∩S = E. Every subgroup D and S are not polycyclic, therefore by Lemma 1 [G/D, G/D] and [G/S, G/S] are finite. From E = D ∩ S we obtain embedding G/E ↪→ G/D × G/S, which proves that [G/E, G/E] is finite. Since E is finitely generated, [G, G] is likewise finitely generated. This contradiction shows that a set Π(Z/E) is finite. In turn, it implies that Z/E is a Chernikov group. Let P/E be a divisible part of Z/E. If it is not a Prüfer subgroup, then P/E = U/E ∩ V/E where the both subgroups U/E are V/E non- identity and divisible, in particular, they are not polycyclic. By Lemma 1 [G/U, G/U ] and [G/V, G/V ] are finite. From E = U ∩ V we obtain embedding G/E ↪→ G/U ×G/V , which proves that [G/E, G/E] is finite. Thus we come again to a contradiction. This contradiction shows that P/E is a Prüfer subgroup. Since P/E is divisible, Z/E = P/E ∩ M/E where a subgroup M/E is finite. Then M is finitely generated and Z/M is a Prüfer p-subgroup. Corollary. Let G be a non-abelian locally nilpotent torsion-free group of finite 0-rank. Suppose that [G, G] is not finitely generated. If every non polycyclic subgroup of G is nearly normal, then G/ζ(G) is an abelian minimax group, moreover, Sp(G/ζ(G)) = Sp(ζ(G)) = {p} where p is a prime. Proof. By Lemma 8 ζ(G) is not finitely generated and G/ζ(G) is abelian. Hence for each element g ∈ G the mapping φg : x −→ [x, g], x ∈ G, is an endomorphism, so that [G, g] = Imφg ∼= A/Kerφg = G/CG(g). An inclusion [G, G] ≤ ζ(G) and lemma 10 show that [G, g] is mini- max and Sp([G, g]) = {p} where p is a prime. Then G/CG(g) is like- wise minimax and Sp(G/CG(g)) = {p} for each element g ∈ G. Let {g1ζ(G), . . . , gtζ(G)} be a maximal Z-independent set of G/ζ(G), H =< g1, . . . , gt >. Using again a Remak’s theorem we obtain an embedding G/CG(H) ↪→ G/CG(g1) × . . . × G/CG(gt), which proves that G/CG(H) is minimax and Sp(G/CG(H)) = {p}. Put D = Hζ(G), then CG(H) = CG(D), in particular, G/CG(D) is minimax and Sp(G/CG(D)) = {p}. By the selection of D for every element g ∈ G there is a positive integer m such that gm ∈ D. We have Jo u rn al A lg eb ra D is cr et e M at h .M. M. Semko, S. M. Kuchmenko 105 for each element x ∈ CG(D) an equation gmx = xgm. The last equation implies in locally nilpotent torsion-free group G an equation gx = xg. In turn it implies that CG(D) ≤ ζ(G), that is CG(D) = ζ(G). Hence G/ζ(G) is minimax and Sp(G/ζ(G)) = {p}. Lemma 11. Let G be a non-abelian locally nilpotent torsion-free group of finite 0-rank. Suppose that [G, G] is not finitely generated. If every non polycyclic subgroup of G is nearly normal, then a section A/(A∩ [G, G]) is finitely generated for each abelian subgroup A. Proof. Suppose the contrary, let A/(A∩ [G, G]) is not finitely generated. By Lemma 8 K = [G, G] ≤ ζ(G), so that AK is abelian. Therefore we may assume that K ≤ A. Let E be a finitely generated subgroup of K such that K/E is a Prüfer p-subgroup. Such subgroup exists by Lemma 10. By the properties of the divisible subgroups in abelian groups (see., for example, [15, Theorem 21.2]) A/E = K/E × L/E for some sub- group L. By our assumption about A/K a subgroup L can not be finitely generated. Then it is nearly normal. Lemma 4 yields that L includes a G- invariant torsion-free subgroup B of finite index. In particular, B is not finitely generated. By Lemma 1 [G/K, G/K] and [G/B, G/B] are finite. We have again G/(K ∩B) ↪→ G/K×G/B, what implies that G/(K ∩B) has a finite derived subgroup. Together with K ∩ B ≤ K ∩ L = E it follows that [G, G] is finitely generated. This contradiction shows that A/K is finitely generated. Lemma 12. Let G be a non-abelian locally nilpotent torsion-free group of finite 0-rank. Suppose that [G, G] is not finitely generated. If every non polycyclic subgroup of G is nearly normal, then an intersection A∩ [G, G] is not finitely generated whenever A is not finitely generated. Proof. Suppose the contrary, let A ∩ [G, G] is finitely generated. By Lemma 4 A includes a G-invariant torsion-free subgroup C of finite index (remark, that A is not polycyclic, so that it is nearly normal). In par- ticular, C can not be polycyclic, and Lemma 1 shows that [G/C, G/C] is finite. Usin an embedding G/(C ∩ [G, G]) ↪→ G/[G, G] × G/C, we can prove that G/(C ∩ [G, G]) has a finite derived subgroup. Since C ∩ [G, G] ≤ A ∩ [G, G], C ∩ [G, G] is finitely generated. It follows that [G, G] is likewise finitely generated. This contradiction shows that A ∩ [G, G] is not finitely generated. Corollary. Let G be a non-abelian locally nilpotent torsion-free group of finite 0-rank. Suppose that [G, G] is not finitely generated. If every non polycyclic subgroup of G is nearly normal, then an intersection A∩ [G, G] has finite index in [G, G] whenever A is not finitely generated. Jo u rn al A lg eb ra D is cr et e M at h .106Groups, in which almost all subgroups are near to normal In fact, by Lemma 12 an intersection A∩ [G, G] can not be polycyclic. By Lemma 1 G/(A ∩ [G, G]) has a finite derived subgroup. This means that A ∩ [G, G] has a finite index in [G, G]. Theorem 4. Let G be a non-abelian locally nilpotent torsion-free group of finite 0-rank. Suppose that [G, G] is not finitely generated. Every non polycyclic subgroup of G is nearly normal if and only if G satisfies the following conditions: (1) ζ(G) includes a finitely generated subgroup M such that ζ(G)/M is a Prüfer p-subgroup for some prime p; (2) G/ζ(G) is an abelian minimax group such that Sp(G/ζ(G)) = Sp(ζ(G)) = {p}; (3) [G, G] ≤ PMfg(ζ(G)), in particular, PMfg(ζ(G)) 6=< 1 >; (4) if A is an abelian subgroup of G then a section A/(A ∩ [G, G]) is finitely generated. Furthermore, if a subgroup H is not finitely generated, then an intersection H∩ [G, G] has finite index in [G, G]. Proof. A condition (1) follows from Lemma 10. A condition (2) follows from Corollary to Lemma 10. Corollary to Lemma 8 proves an inclusion [G, G] ≤ ζ(G), which implies (3). Finally, a condition (4) follows from Lemma 11 and a last statement follows from Corollary to Lemma 12. Conversely, suppose that a group G satisfies all conditions (1) —- (4). Let H be a non polycyclic subgroup of G. Then H includes an abelian subgroup A such that A is not finitely generated. By (4) an intersection A∩ [G, G] is not finitely generated. Let L be a pure envelope of B = A∩ [G, G] in Z = ζ(G). Then L is not polycyclic, and a condition (3) implies an inclusion [G, G] ≤ L. Since L/B is periodic, the conditions (3) and (1) show that L/B is finite. By [G, G]B/B ≤ L/B we obtain now that [G/B, G/B] is finite. Then every subgroup of G/B is nearly normal, in particular, so is and H/B. It follows that H is nearly normal in G. Consequently, now we must consider a case, when G is not locally nilpotent and all periodic subgroups of G are finite. In particular, P(G) is finite. Here P(G) is a maximal normal periodic subgroup of G. Propo- sition 1 make possible the next reduction P(G) =< 1 >. Since G ∈ S1F, G includes the normal subgroups H, A such that H ≤ A, H is nilpotent and torsion-free, A/H is finitely generated free abelian group and G/A is finite [20, Theorem 6]). Jo u rn al A lg eb ra D is cr et e M at h .M. M. Semko, S. M. Kuchmenko 107 Lemma 13. Let G ∈ S1F and suppose that P(G) =< 1 > and [G, G] is infinite. If every non polycyclic-by-finite subgroup of G is nearly normal, then G includes an abelian torsion-free normal subgroup A such that A is not finitely generated and G/A has a finite derived subgroup and finite periodic part. Proof. As we have noted above, G includes the normal subgroups H, L such that H ≤ L, H is nilpotent and torsion-free, L/H is finitely generated free abelian group and G/L is finite. Since G/H is polycyclic- by-finite, H can not be polycyclic. By Lemma 8 A = ζ(H) is not finitely generated. Lemma 1 yields that [G/A, G/A] is finite. Since H/A and L/H are torsion-free and G/L is finite, the periodic part of G/A is finite. Lemma 14. Let G ∈ S1F and suppose that P(G) =< 1 > and [G, G] is infinite but polycyclic-by-finite. If every non polycyclic-by-finite sub- group of G is nearly normal, then G is a nilpotent torsion-free group. Proof. As we have noted above, G includes the normal subgroups H, L such that H∩L, H is nilpotent and torsion-free, L/H is finitely generated free abelian group and G/L is finite. Since [G, G] is polycyclic-by-finite, [H, H] is likewise polycyclic. By Corollary to Lemma 8 A = ζ(H) is not finitely generated but H/A is abelian and finitely generated. Since G/H is polycyclic-by-finite, G/A is polycyclic-by-finite. Lemma 1 yields that G/A has a finite derived subgroup. For each element g ∈ G the mapping φg : a −→ [a, g], a ∈ A, is an endomorphism, so that Kerφg = CA(g) and Imφg = [A, g] are the < g >-invariant subgroups and [A, g] = Imφg ∼= A/Kerφg = A/CA(g). Together with [G, G] a subgroup [A, g] is finitely generated, thus E = CA(g) is not polycyclic, in particular, it is nearly normal.. Let G/A =< g1A, . . . , gnA >. Since E1 = CA(g1) is not poly- cyclic, it is nearly normal. By Lemma 4 E1 includes a G-invariant sub- group B1 of finite index. In particular, B1 is not polycyclic. Repeating the above arguments, we obtain that B2 = B1 ∩ CA(g2) is not poly- cyclic. Using the same arguments after finitely many steps we obtain that C = CA(g1, . . . , gn) is not polycyclic. By Lemma 1 [G/C, G/C] is fi- nite. Hence the set T/C of all elements having finite orders is a subgroup. A subgroup [T, T ] is locally finite (see, for example, [16, Corollary to The- orem 4.12]). In particular, the set P of elements of T having finite orders, is a (characteristic) subgroup of T . Moreover, T/P is abelian torsion-free group. An equation P(G) =< 1 > shows that T is an abelian torsion-free subgroup. This together with inclusion CA(g1, . . . , gn) ≤ ζ(G) imply that T ≤ ζ(G), because T/C is periodic. Since G/T is abelian, G is nilpotent. Finally, an equation P(G) =< 1 > shows that G is torsion-free. Jo u rn al A lg eb ra D is cr et e M at h .108Groups, in which almost all subgroups are near to normal Corollary. Let G be an abelian-by-finite group, whose non polycyclic- by-finite subgroups are nearly normal. Suppose that P(G) =< 1 >. If G is non- abelian, then [G, G] is not finitely generated. Proof. Suppose the contrary, let [G, G] is finitely generated. By Lemma 14 G is nilpotent. In particular, it is torsion-free. Since G is abelian- by-finite, G is abelian. This contradiction proves that [G, G] is not polycyclic-by-finite. Lemma 15. Let G ∈ S1F and suppose that P(G) =< 1 > and [G, G] is not polycyclic-by-finite. If every non polycyclic-by-finite subgroup of G is nearly normal, then every abelian torsion-free subgroup A includes a finitely generated subgroup B such that A/B is a Prüfer p-subgroup for some prime p. Proof. Since G ∈ S1F, A has a finite 0-rank. Then A includes a finitely generated subgroup E such that A/E is periodic. If a set Π(A/E) is infinite, then there are two infinite subsets ∆ and Ξ such that ∆ ∪ Ξ = Π(Z/E) and ∆ ∩ Ξ = ®. Let D/E be a Sylow Ξ-subgroup and S/E be a Sylow Ξ-subgroup of A/E. Then Z = DS and D ∩ S = E. The both subgroups D and S are not polycyclic-by-finite, so that [G/D, G/D] and [G/S, G/S] are finite by Lemma 1. An equation E = D ∩ S gives an embedding G/E ↪→ G/D ×G/S, which shows that [G/E, G/E] is finite. Since E is finitely generated, [G, G] is finitely generated. Thus we obtain a contradiction with Corollary of Lemma 14. This contradiction proves that Π(A/E) is finite. This means that A/E is a Chernikov group. Let P/E be a divisible part of A/E. If we suppose, that it is not a Prüfer subgroup, then P/E = U/E×V/E where U/E and V/E are non-identity and divisible. In particular, U and V are not polycyclic. By Lemma 1 [G/U, G/U ] and [G/V, G/V ] are finite. From E = U ∩ V we obtain an embedding G/E ↪→ G/U×G/V , which shows that G/E has finite derived subgroup. This come again to contradiction, which prove that P/E is a Prüfer subgroup. Since P/E is divisible, Z/E = P/E×B/E where B/E is finite. Then B is finitely generated and A/B is a Prüfer subgroup. Corollary 1. Let G ∈ S1F and suppose that P(G) =< 1 > and [G, G] is not polycyclic-by-finite. If every non polycyclic-by-finite subgroup of G is nearly normal, then [G, G] satisfies the following conditions: (1) [G, G] includes a G-invariant abelian torsion-free subgroup A of finite index; (2) A includes a finitely generated subgroup B such that A/B is a Prüfer p-subgroup for some prime p; Jo u rn al A lg eb ra D is cr et e M at h .M. M. Semko, S. M. Kuchmenko 109 (3) if C is a subgroup of A and r0(C) < r0(A) then C is finitely gener- ated. Proof. Since [G, G] is not polycyclic-by-finite, so is G. Then G includes an abelian subgroup U , which is not finitely generated (see, for example, [16, Theorem 3.31]). By P(G) =< 1 > a periodic part of U is finite. Without loss of generality we can suppose that U is torsion-free. Among the not polycyclic-by-finite subgroups of U choose a subgroup B having the biggest 0-rank. It is possible, because the 0-rank of U is finite. Since B is not polycyclic-by-finite, B is nearly normal. By Lemma 4 B includes a G- invariant subgroup E of finite index. In particular, E is not polycyclic-by- finite. By Lemma 1 [G/E, G/E] is finite. Put A = [G, G]∩E. Then from [G, G]/A ∼= [G, G]/([G, G] ∩ E) ∼= [G, G]E/E we obtain that [G, G]/A is finite. Since [G, G] is not polycyclic-by-finite, A is not finitely generated. Therefore r0(U) = r0(A), and from a selection of U follows (3). Finally, (2) follows from Lemma 15. Corollary 2. Let G ∈ S1F and suppose that P(G) =< 1 > and [G, G] is not polycyclic-by-finite. Assume also that every non polycyclic-by-finite subgroup of G is nearly normal. If A is a G-invariant abelian torsion- free subgroup of [G, G] having finite index, then either CA(g) = A or CA(g) =< 1 > for each element g ∈ G. Proof. Use again an endomorphism φg : a −→ [a, g], a ∈ A. We have Kerφg = CA(g) and Imφg = [A, g], so that [A, g] = Imφg ∼= A/Kerφg = A/CA(g). It is not hard to see, that CA(g) is a pure subgroup of A. If we suppose that A 6= CA(g) 6=< 1 >, then r0(CA(g)) < r0(A) and a condition (3) of Corollary 1 to Lemma 15 shows that CA(g) is finitely generated. Similarly we obtain that r0([A, g]) < r0(A), and using the same arguments we obtain that [A, g] is finitely generated. An isomor- phism [A, g] ∼= A/CA(g) shows that A is finitely generated. Hence [G, G] is polycyclic-by-finite. This contradiction shows that A 6= CA(g) is pos- sible only whenever CA(g) =< 1 >. Lemma 16. Let G ∈ S1F and suppose that P(G) =< 1 > and [G, G] is not polycyclic-by-finite. Assume also that every non polycyclic-by-finite subgroup of G is nearly normal. If A is a G-invariant abelian torsion-free subgroup of [G, G] such that index /[G, G] : A/ is finite and A ≤ ζ(G), then G is nilpotent. Proof. By Lemma 1 [G/A, G/A] is finite. Therefore the set T/A of all elements having finite orders is a subgroup. Then [T, T ] is a locally fi- nite subgroup (see, for example, [16, Corollary to Theorem 4.12]). In Jo u rn al A lg eb ra D is cr et e M at h .110Groups, in which almost all subgroups are near to normal particular, a set P of all elements of T having finite orders is a (charac- teristic) subgroup of T . Moreover, T/P is an abelian torsion-free group. By P(G) =< 1 > we obtain that T is an abelian torsion-free subgroup. Since T/A is periodic, an inclusion A ≤ ζ(G) yields T ≤ ζ(G). Then G is nilpotent because G/T is abelian. Lemma 17. Let G ∈ S1F and suppose that P(G) =< 1 > and [G, G] is not polycyclic-by-finite. Assume also that every non polycyclic-by-finite subgroup of G is nearly normal. If A is a G-invariant abelian torsion-free subgroup of [G, G] such that index /[G, G] : A/ is finite and there is an element g ∈ G with a property CA(g) ∩ A, then G includes a subgroup D = AV of finite index such that A ∩ V =< 1 > and V is an abelian torsion-free subgroup. Proof. By Corollary 2 of Lemma 15 CA(g) =< 1 >. Lemma 1 yields that [G/A, G/A] is finite. Then L/A = CG/A(gA) has finite index in G/A. This together with L/A ∈ S1F, implies that it includes an abelian subgroup M/A of finite index. Since gA ∈ ζ(L/A), < gA, M/A > is abelian. Thus we may assume that L/A is abelian. Using an isomorphism [A, g] ∼= A/CA(g) ∼= A and a condition (2) of Corollary 1 to Lemma 15 we obtain that A/[A, g] is finite. It follows that C = CL(A/[A, g]) has in L a finite index. Let k = /A/[A, g]/ and E = Ck. Since C ∈ S1F, C/E is finite, and therefore L/E is likewise finite. Let c ∈ C. We have [c2, g] = [c, g]c[c, g] = [c, gc][c, g]. Since gc = ga for some element a ∈ A, [c, ga] = [c, g]a[c, a] , thus [c2, g] = [c, g]2[c, a] where [c, a] = v ∈ [A, g]. With the help of similarly arguments we obtain that [ck, g] = [c, g]kw for some element w ∈ [A, g]. From [c, g ∈ A we obtain [c, g]k ∈ Ak ≤ [A, g], therefore [ck, g] ∈ [A, g], that is [ck, g] = [b, g] for some element b ∈ A. It follows that ckb−1 ∈ CL(g) or ck ∈ CL(g)A. In other words, we obtain an inclusion E ≤ CL(g)A. An equation CA(g) =< 1 > implies CL(g)∩A =< 1 >, thus we have CL(g) ∼= CL(g)/(CL(g) ∩ A) ∼= CL(g)A/A ≥ EA/A. As we remarked, EA/A is abelian and has finite index in G/A. Hence V = CL(g) is abelian and D = AE has finite index in G. Corollary. Let G ∈ S1F and suppose that P(G) =< 1 > and the both group [G, G] and G/[G, G] are not polycyclic-by-finite. Assume also that every non polycyclic-by-finite subgroup of G is nearly normal. Then G is nilpotent. Proof. By Corollary 1 of Lemma 15 [G, G] includes a G-invariant abelian torsion-free subgroup A having finite index. If CA(g) = A for each ele- ment g ∈ G, then A ≤ ζ(G). By Lemma 16 G is nilpotent. Consequently we may assume that there is an element g with a property CA(g) 6= A. Jo u rn al A lg eb ra D is cr et e M at h .M. M. Semko, S. M. Kuchmenko 111 By Lemma 17 G includes a subgroup D = AV of finite index such that A ∩ V =< 1 > and V is abelian torsion-free subgroup. In particular, V is not finitely generated and hence is nearly normal. Lemma 4 proves that there exists a G-invariant abelian subgroup B having finite index in V . By Lemma 1 [G/B, G/B] is finite. From A ∩ B ≤ A ∩ V =< 1 > we obtain an embedding G ↪→ G/A × G/B, which proves that [G, G] is finite. This contradiction shows that CA(g) = A for each element g ∈ G. As we see above it implies that G is nilpotent. Now we can obtain a description and for the last case. Theorem 5. Let G ∈ S1F and suppose that G is not polycyclic-by-finite, P(G) =< 1 > and G is not nilpotent. Every non polycyclic-by-finite subgroup of G is nearly normal if and only if G satisfies the following conditions: (1) [G, G] includes a G-invariant abelian torsion-free subgroup of finite index; (2) A includes a finitely generated subgroup B such that A/B is a Prüfer p-subgroup for some prime p; (3) if C is a subgroup of A and r0(C) < r0(A), then C is finitely generated; (4) CG(A) 6= G and CA(g) =< 1 > for each element g ∈ G CG(A); (5) G includes a subgroup D = AV of finite index such that A∩V =< 1 > and V is a finitely generated torsion-free abelian subgroup. Proof. By Lemma 14 [G, G] is not polycyclic-by-finite. Corollary 1 of Lemma 15 shows that G satisfies (1) —- (3). A condition (4) follows from Lemma 16. By Corollary to Lemma 17 G/[G, G] is finitely generated. Finally, using Lemma 17 we obtain that G satisfies (5). Conversely, suppose that G satisfies the conditions (1) —- (5). Let H be an arbitrary non polycyclic-by-finite subgroup of G. Since G/[G, G] is finitely generated, an intersection H ∩ [G, G] can not be polycyclic-by- finite. By (3) we obtain that r0(H ∩ A) = r0(A). It follows that H ∩ A has a finite index s in a subgroup A. Then B = As ≤ H ∩ A. Clearly B is a normal subgroup of G. Since r0(A) is finite, A/B is finite. This implies that G/B has a finite derived subgroup. Then every subgroup of G/B is nearly normal, in particular, so is H/B. Hence and H is nearly normal in G. Jo u rn al A lg eb ra D is cr et e M at h .112Groups, in which almost all subgroups are near to normal References [1] Dedekind R. Über Gruppen, deren sammtliche Teiler Normalteiler sind// Math. Annalen.—-1897.—-48.—- S. 548 –– 561. [2] Miller G.A., Moreno H. C. Non-abelian groups in which every subgroup is abelian // Trans. Amer. Math. Soc.—-1903.—4.—P. 389 — 404. [3] Schmidt O. Yu. Groups with all proper subgroups are nilpotent // Mat. Sb.— 1924.—-31, № 3.—- P. 366 —- 372. [4] Neumann B.H. Groups with finite classes of conjugate subgroups // Math. Z.—- 1955.—- 63, № 1.—- P. 76 —- 96. [5] Kurdachenko L. A., Kuzenny M.F., Semko M.M. Groups with dense systems of infinite subgroups // Dokl. Akad. Nauk Ukr. SSR, Ser. A . —- 1985. —- № 3. —- P. 7 —- 9. [6] Galoppo A. Groups satisfying the maximal condition on non-nearly normal sub- groups // Ricerche Mat.—-2000.—- 49, № 2.—- P. 213 —- 220. [7] Franciosi S., de Giovanni F. Groups satisfying the minimal condition on certain non-normal subgroups// "Groups —- Korea 94".—-1995.—- Berlin: Walter de Gruyter.—-P. 107 —- 118. [8] Musella C. Isomorphisms between lattices of nearly normal subgroups // Note di Matematica.—-2000/2001.—-20, №1.—- P. 43 —- 52. [9] Kurdachenko L.A., Pylaev V.V. Groups dual dedekinds // Dokl. Akad. Nauk Ukr. SSR, Ser. A.—- 1989.—- № 10.—- P. 21 —- 22. [10] Cutolo G. On groups satisfying the maximal condition on non-normal subgroups // Riv. Mat. Pura Appl. —- 1991.—- 9.—- P. 49 —- 59. [11] Cutolo G., Kurdachenko L.A. Groups with a maximality condition for some non- normal subgroups //Geometriae Dedicata. —-1995.—- 55. —- P. 279 —- 292. [12] Franciosi S., de Giovanno F. and Kurdachenko L.A. On groups with many almost normal subgroups // Ann. Mat. Pura Appl. —- 1995.—- 169, № 4.—- P. 35 —- 65. [13] Ol’shanskij A.Yu. Geometry of defining relations in groups. —- M.: Nauka, 1989.— - 447 p. [14] B.H. Neumann. Groups covered by permutable subsets // J. London Math. Soc.— -1954.—- 29, № 114.—-P. 236 —- 248. [15] Fuchs L. Infinite Abelian groups.—-M.: Мir, 1974.—-Vol.1.—- 336 p. [16] Robinson D.J.S. Finiteness conditions and generalized soluble groups, Pt. 1.—- Berlin: Springer, 1972.—- 210 p. [17] Zaitsev D.I.T heory of minimax groups // Ukrain. Mat. J.—-1971.—- 23, № 5.—- P. 652 —- 660. [18] Robinson D.J.S. Finiteness conditions and generalized soluble groups, Part 2.—- Berlin: Springer, 1972.—- 254 p. [19] Zaitsev D.I., Kurdachenko L.A., Tushev A.V. Modules over nilpotent groups of finite rank // Algebra and Logic. —- 1985.—-24, № 6.—- P. 631 —- 666 [20] Charin V.S. On automorphism groups of nilpotent groups // Ukrain. Mat. J.—- 1954.—- 6, №3.—- P. 295 —- 304. Jo u rn al A lg eb ra D is cr et e M at h .M. M. Semko, S. M. Kuchmenko 113 Contact information M. M. Semko, S. M. Kuchmenko Department of Mathematics National State Tax Service Academy of Ukraine, ul. K. Marksa 31, 08200, Irpin, Kyiv obl., Ukraine E-Mail: n_semko@mail.ru Received by the editors: 24.02.2004 and final form in 29.06.2004.
id	nasplib_isofts_kiev_ua-123456789-156421
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
issn	1726-3255
language	English
last_indexed	2025-12-07T13:27:19Z
publishDate	2004
publisher	Інститут прикладної математики і механіки НАН України
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spelling	Semko, M.M. Kuchmenko, S.M. 2019-06-18T13:31:35Z 2019-06-18T13:31:35Z 2004 Groups, in which almost all subgroups are near to normal / M.M. Semko, S.M. Kuchmenko // Algebra and Discrete Mathematics. — 2004. — Vol. 3, № 2. — С. 92–113. — Бібліогр.: 20 назв. — англ. 1726-3255 https://nasplib.isofts.kiev.ua/handle/123456789/156421 A subgroup H of a group G is said to be nearly normal, if H has a finite index in its normal closure. These subgroups have been introduced by B.H. Neumann. In a present paper is studied the groups whose non polycyclic by finite subgroups are nearly normal. It is not hard to show that under some natural restrictions these groups either have a finite derived subgroup or belong to the class S₁F (the class of soluble by finite minimax groups). More precisely, this paper is dedicated of the study of S₁F groups whose non polycyclic by finite subgroups are nearly normal. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics Groups, in which almost all subgroups are near to normal Article published earlier
spellingShingle	Groups, in which almost all subgroups are near to normal Semko, M.M. Kuchmenko, S.M.
title	Groups, in which almost all subgroups are near to normal
title_full	Groups, in which almost all subgroups are near to normal
title_fullStr	Groups, in which almost all subgroups are near to normal
title_full_unstemmed	Groups, in which almost all subgroups are near to normal
title_short	Groups, in which almost all subgroups are near to normal
title_sort	groups, in which almost all subgroups are near to normal
url	https://nasplib.isofts.kiev.ua/handle/123456789/156421
work_keys_str_mv	AT semkomm groupsinwhichalmostallsubgroupsareneartonormal AT kuchmenkosm groupsinwhichalmostallsubgroupsareneartonormal

Groups, in which almost all subgroups are near to normal

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