Locally soluble groups with the restrictions on the generalized norms
The author studies groups with given restrictions on norms of decomposable and Abelian non-cyclic subgroups. The properties of non-periodic locally soluble groups, in which such norms are nonidentity and have the identity intersection, are described.
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irk-123456789-1885042023-03-04T01:27:14Z Locally soluble groups with the restrictions on the generalized norms Lukashova, T. The author studies groups with given restrictions on norms of decomposable and Abelian non-cyclic subgroups. The properties of non-periodic locally soluble groups, in which such norms are nonidentity and have the identity intersection, are described. 2020 Article Locally soluble groups with the restrictions on the generalized norms / T. Lukashova // Algebra and Discrete Mathematics. — 2020. — Vol. 29, № 1. — С. 85–98. — Бібліогр.: 17 назв. — англ. 1726-3255 DOI:10.12958/adm1527 2010 MSC: 20D25, 20E28. http://dspace.nbuv.gov.ua/handle/123456789/188504 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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The author studies groups with given restrictions on norms of decomposable and Abelian non-cyclic subgroups. The properties of non-periodic locally soluble groups, in which such norms are nonidentity and have the identity intersection, are described. |
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Lukashova, T. Locally soluble groups with the restrictions on the generalized norms Algebra and Discrete Mathematics |
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Lukashova, T. |
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Lukashova, T. |
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Locally soluble groups with the restrictions on the generalized norms |
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Locally soluble groups with the restrictions on the generalized norms |
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Locally soluble groups with the restrictions on the generalized norms |
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Locally soluble groups with the restrictions on the generalized norms |
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Locally soluble groups with the restrictions on the generalized norms |
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locally soluble groups with the restrictions on the generalized norms |
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Інститут прикладної математики і механіки НАН України |
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2020 |
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http://dspace.nbuv.gov.ua/handle/123456789/188504 |
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Locally soluble groups with the restrictions on the generalized norms / T. Lukashova // Algebra and Discrete Mathematics. — 2020. — Vol. 29, № 1. — С. 85–98. — Бібліогр.: 17 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT lukashovat locallysolublegroupswiththerestrictionsonthegeneralizednorms |
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2025-07-16T10:35:42Z |
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2025-07-16T10:35:42Z |
_version_ |
1837799459476471808 |
fulltext |
“adm-n1” — 2020/5/14 — 19:35 — page 85 — #93
© Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 29 (2020). Number 1, pp. 85–98
DOI:10.12958/adm1527
Locally soluble groups with the restrictions on
the generalized norms
T. Lukashova∗
Communicated by I. Ya. Subbotin
Abstract. The author studies groups with given restrictions
on norms of decomposable and Abelian non-cyclic subgroups. The
properties of non-periodic locally soluble groups, in which such norms
are nonidentity and have the identity intersection, are described.
Introduction
In group theory findings related to the study of the impact of properties
of the different systems of the subgroups on the group are in focus. This
direction includes findings when the restrictions are imposed on the
different Σ-norms.
Let Σ be the system of all subgroups of G with a certain theoretical
group property. The maximal subgroup of G which normalizes every
subgroup of Σ is called Σ-norm of a group G. It is clear that the Σ-norm
of a group G coincides with the intersection of the normalizers of all
subgroups included in Σ, and contains the center of a group.
In the case when Σ-norm of a group coincides with the group, all
subgroups of Σ are normal in the group (assuming that the system Σ is
non-empty). For the first time non-Abelian groups with such property
∗The author expresses his deep gratitude to F. M. Lyman for his helpful suggestions
during the preparation of this work.
2010 MSC: 20D25, 20E28.
Key words and phrases: locally soluble groups, non-periodic group, decompos-
able group, generalized norms of groups, norm of decomposable subgroups, norm of
Abelian non-cyclic subgroups, non-Dedekind group.
https://doi.org/10.12958/adm1527
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86 Generalized norms of groups
were considered in the second part of the XIX century by R. Dedekind,
who characterized groups, all subgroups of which are normal (nowadays
these groups are called Dedekind groups). However, the systematic study
of groups with different systems of normal subgroups were continued only
in the second part of the XX century, that slowed down the study of
Σ-norms. So, the question on the study of the properties of groups, in
which the Σ-norm is a proper subgroup, arises naturally.
For the first time such problem was formulated by R. Baer in 30s of
the previous century for the system Σ of all subgroups of a group [1].
Such Σ-norm was called the norm N(G) of a group G and denoted as the
intersection of normalizers of all subgroups of a group G. Later the findings
of R. Baer on the norm of a group were extended on the different systems of
subgroups Σ and on the different restrictions, which the Σ-norms satisfies
(see e.g. [2]–[7]). It is clear that the norm N(G) is contained in the other
Σ-norms, which, in turn, can be regarded as its generalizations.
In this paper, we consider the relations between the norm of decompos-
able subgroups and the norm of Abelian non-cyclic subgroups of a group.
The norm Nd
G of decomposable subgroups of a group G is the intersection
of the normalizers of all decomposable subgroups of a group or group
itself, if the system of such subgroups is empty [7]. Recall that a subgroup
of a group G is called decomposable if it can be representable in the form
of the direct product of two non-trivial factors [8].
It is clear, that in the case when Nd
G = G, all decomposable subgroups
are normal in a group G or the system of such subgroups is empty. Non-
Abelian groups with such property were studied in [8] and called di-groups.
Obviously, the presence of decomposable subgroups in a group is
directly related to the existence of decomposable Abelian subgroups,
which in most cases are non-cyclic. So, the norm Nd
G of decomposable
subgroups of group G is closely related to the norm NA
G of Abelian non-
cyclic subgroups.
The intersection of normalizers of all non-cyclic Abelian subgroups of
a group G (provided that the system of these subgroups is non-empty) is
called the norm of non-cyclic Abelian subgroups of a group G and denoted
by NA
G (see e.g. [6, 9]). If the norm NA
G contains at least one Abelian non-
cyclic subgroup, then each such a subgroup is normal in NA
G . Non-Abelian
groups with this property were studied by F. Lyman in [10] and called
HA–group. So, the norm of Abelian non-cyclic subgroups is Dedekind or
non-Hamiltonian HA-group.
The relations between these norms has been investigated in [7, 11, 12]
for quite broad classes of groups. In [7] it was proved that in locally finite
“adm-n1” — 2020/5/14 — 19:35 — page 87 — #95
T. Lukashova 87
groups, that contain an Abelian non-cyclic subgroup, one of the ratios
holds: NA
G ⊆ Nd
G, N
A
G ⊇ Nd
G.
In particular, it was found that a periodic locally nilpotent group has
the non-Dedekind norm of decomposable subgroups if and only if it is
a locally finite p-group and Nd
G = NA
G . The same relations between norms
can be traced in an arbitrarily locally finite group with the non-Dedekind
locally nilpotent norm Nd
G of decomposable subgroups (see [11]).
In [12] the study of the relations between the norm of decomposable
subgroups and the norm of Abelian non-cyclic subgroups was continued
in the class of non-periodic locally soluble groups. It was proved, if at
least one of the norms NA
G or Nd
G is non-Dedekind and the subgroup Nd
G
is infinite, then one of the following inclusions takes place: NA
G ⊆ Nd
G or
Nd
G ⊆ NA
G .
The purpose of the article is to study the properties of locally soluble
groups in which the norm of decomposable subgroups and the norm
of Abelian non-cyclic subgroups are nonidentity and have the identity
intersection Nd
G ∩NA
G = E.
1. Preliminary results
The next statements are actively used in the further research.
Lemma 1. ([7]) If a group G contains a nonidentity Nd
G-admissible
subgroup H such that Nd
G
⋂
H = E, where Nd
G is the norm of decomposable
subgroups of G, then Nd
G is Dedekind.
Lemma 2. ([6]) If a group G contains an Abelian non-cyclic subgroup
H such that NA
G
⋂
H = E, where NA
G is the norm of Abelian non-cyclic
subgroups of G, then the norm NA
G is Dedekind (Abelian, if a group is
non-periodic).
Lemma 3. ([6]) Let G be a non-periodic group, NA
G be the norm of Abelian
non-cyclic subgroups and a group G contain a nonidentity NA
G -admissible
subgroup H such that NA
G
⋂
H = E. If the norm NA
G is non-periodic, then
all infinite cyclic subgroups are normal in it.
The following statement reduces the study of groups, in which the
norms NA
G of Abelian non-cyclic subgroups and the norm Nd
G of decom-
posable subgroups are nonidentity and Nd
G ∩NA
G = E, to the study of
non-periodic groups.
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88 Generalized norms of groups
Theorem 1. If a locally soluble group G contains an Abelian non-cyclic
subgroup, the norm NA
G of Abelian non-cyclic subgroups and the norm Nd
G
of decomposable subgroups are nonidentity and Nd
G ∩NA
G = E, then G is
a non-periodic group.
Proof. Suppose that G is a periodic group. Then it is locally finite and
either NA
G ⊇ Nd
G or NA
G ⊆ Nd
G by Theorem 1.4 [7], which contradicts the
condition of the Theorem. Thus, G is a non-periodic locally soluble group
and the Theorem is proved.
Further we will consider only non-periodic groups in which the norms
Nd
G and NA
G are nonidentity and their intersection is identity. The existence
of non-periodic groups with given restrictions on the norm of Abelian
non-cyclic and the norm of decomposable subgroups is confirmed by the
following examples.
Example 1. ([12], Example 3.5). Let G = (〈a〉⋋B)⋋ 〈c〉, where |a| = p
(p is prime, p 6= 2), B be a group, isomorphic to an additive group of
p-adic fractions, B = B1〈x〉, x
2 ∈ B1, x
−1ax = a−1, [B1, 〈a〉] = E, |c| =
2, [c, a] = 1, c−1bc = b−1 for any element b ∈ B.
In this group the norm of decomposable subgroups Nd
G = 〈a〉 is a cyclic
subgroup of prime order. At the same time, the norm of non-cyclic Abelian
subgroups is non-Dedekind, NA
G = B1 ⋋ 〈c〉 and Nd
G ∩NA
G = E.
2. The main results
The aim of this section is to study the properties of non-periodic locally
soluble groups and the structure of the norms NA
G and Nd
G, provided that
Nd
G ∩ NA
G = E. The first of the following theorems characterizes the
groups with the non-Dedekind norm NA
G , respectively, the second theorem
describes groups in which the norm NA
G is the Dedekind.
Theorem 2. If a non-periodic locally soluble group G has an Abelian
non-cyclic subgroup, the norm NA
G of Abelian non-cyclic subgroups is non-
Dedekind, the norm Nd
G of decomposable subgroups is nonidentity and
Nd
G ∩NA
G = E, then the following conditions take place:
1) Z (G) = N (G) = E, where N (G) is the norm of G;
2) the norm of decomposable subgroups Nd
G = 〈c〉 is a cyclic group of
a prime odd order p;
3) the norm NA
G of Abelian non-cyclic subgroups is a group of the type
NA
G = A⋋ 〈b〉, where A is a group isomorphic to an additive group
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T. Lukashova 89
of p-adic fractions (p is prime, (p, 2) = 1), |b| = 2 and b−1ab = a−1
for any element a ∈ A;
4) any infinite cyclic subgroup has a nonidentity intersection with the
norm NA
G ;
5) a group G does not contain free Abelian subgroups of rank 2;
6) a group G does not contain finite non-cyclic Abelian subgroups;
7) a group G does not contain periodic non-cyclic locally cyclic sub-
groups;
8) the factor-group G/NA
G is periodic.
Proof. Let group G satisfy the conditions of the theorem. Then the first
statement of the theorem follows from the inclusions:
Z (G) ⊆ N (G) ⊆ Nd
G ∩NA
G = E.
Let’s show that the norm Nd
G of decomposable subgroups is a cyclic
group. Indeed, since Nd
G ∩ NA
G = E, the subgroup Nd
G is Dedekind by
Lemma 1. If Nd
G contains non-cyclic Abelian subgroups, then the norm
NA
G of Abelian non-cyclic subgroups is also Dedekind by Lemma 2, which
contradicts the condition. Thus, Nd
G does not contain non-cyclic Abelian
subgroups.
Since Nd
G is a Dedekind group by the proved above, Nd
G is a finite group
and its Sylow p-subgroups are cyclic for p 6= 2 and the Sylow 2-subgroup
is either a cyclic group or the quaternion group. Taking into account the
condition Z (G) = E, we make a conclusion that the order of the norm Nd
G
is not divided to 2, because otherwise Nd
G contains a central involution.
Therefore, Nd
G is a cyclic group, Nd
G = 〈c〉 and (|c| , 2) = 1.
Considering that
∣
∣Nd
G
∣
∣ < ∞, we obtain
[
G : CG(N
d
G)
]
< ∞ and xm ∈
CG
(
Nd
G
)
,m ∈ N for an arbitrary element x ∈ G, |x| = ∞. Then the
subgroup 〈xm, c〉 is Abelian non-cyclic and NA
G -admissible. If 〈x〉∩NA
G = E,
then 〈xm, c〉∩NA
G = E, and the norm NA
G is Dedekind by Lemma 2, which
is impossible. Therefore, NA
G is a non-periodic group, and any infinite
cyclic subgroup 〈x〉 of a group G has a nonidentity intersection with the
norm NA
G .
Since the subgroup 〈c〉 is NA
G -admissible and 〈c〉∩NA
G = E, all infinite
cyclic subgroups are normal in NA
G by Lemma 3. By the description of
such groups (see [5]) NA
G is a group of the type NA
G = A〈b〉, where A is
a non-periodic Abelian group, |b| ∈ {2, 4}, b2 ∈ A and b−1ab = a−1 for
any element a ∈ A.
Let’s prove that the norm Nd
G = 〈c〉 of decomposable subgroups is
a group of prime odd order p. Suppose that 〈c〉 ⊇ 〈c1〉 × 〈c2〉, where
“adm-n1” — 2020/5/14 — 19:35 — page 90 — #98
90 Generalized norms of groups
|c1| = p, |c2| = q, (p, q) = 1, p and q are odd prime. Then by the condition
[Nd
G, N
A
G ] ⊆ NA
G ∩Nd
G = E,
it follows that |ac1| = ∞ for an arbitrary element a ∈ A, |a| = ∞. Since
the subgroup 〈ac1, c2〉 is Abelian non-cyclic, it is NA
G -admissible. Hence,
the subgroup 〈(ac1)
q〉 is also NA
G -admissible. It is clear that the element
b ∈ NA
G cannot be permutable with the element (ac1)
q , because in this
case [(ac1)
q , b] = [aq, b] = 1, which is impossible. So,
b−1(ac1)
qb = (ac1)
−q = a−qcq1 = a−qc−q
1
and c2q1 = 1. We have a contradiction. Therefore, |c| = pk, where p is
prime, k ∈ N.
If k > 1, then the subgroups 〈ac〉 × 〈cp
k−1
〉 and 〈(ac)p〉 = 〈apcp〉 are
NA
G -admissible. Therefore, considering that |ac| = ∞ and b−1ab = a−1,
where b ∈ NA
G , we have
b−1apcpb = (apcp)−1 = a−pc−p = a−pcp.
Then c−p = cp and c2p = 1, which is impossible. Therefore, Nd
G = 〈c〉,
where |c| = p, p 6= 2.
Let’s specify the structure of the subgroup A ⊆ NA
G . Assume that
A is a mixed group and T (A) is its periodic part. Since Nd
G ∩NA
G = E,
the group G contains an indecomposable non-cyclic Abelian subgroup
H which is not Nd
G-admissible. Clearly, H cannot be a complete group,
because otherwise H ⊆ CG
(
Nd
G
)
, which contradicts its choice. Therefore,
H is the incomplete non-cyclic Abelian torsion-free group of rank 1.
Since [〈a〉, H] ⊆ T (A) ∩H = E for an arbitrary non-identity element
a ∈ T (A), the subgroup 〈a〉 ×H is Abelian decomposable and, therefore,
Nd
G-admissible. Then by the condition
[Nd
G, H] ⊆ Nd
G ∩ (〈a〉 ×H) = E,
the subgroup H is also Nd
G-admissible, which is impossible. Thus, A is an
Abelian torsion-free group. Since b2 ∈ A, |b| = 2 and NA
G = A⋋ 〈b〉, where
b−1ab = a−1 for an arbitrary element a ∈ A.
Suppose that the group A contains a free abelian subgroup 〈a1〉×〈a2〉,
where |a1| = |a2| = ∞. Since H is a non-periodic Abelian torsion-free
group of rank 1 and NA
G ∩ H 6= E by the proved, at least one of the
subgroups 〈a1〉 or 〈a2〉 has an identity intersection with H . Let 〈a1〉∩H =
E. Considering that H is a NA
G -admissible subgroup, let a−1
1 h1a1 = h2,
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T. Lukashova 91
where h1, h2 ∈ H. Then by the condition A ∩ H 6= E we have hk1 ∈ A
for some positive integer k. Thus, a−1
1 hk1a1 = hk1 = hk2 h1 = h2 and
[〈a1〉, H] = E.
It is clear that the subgroup 〈am1 〉×H is Nd
G-admissible for an arbitrary
positive integer number m. Therefore, H =
⋂
∞
m=1(〈a
m
1 〉 ×H) is also Nd
G-
admissible subgroup, which contradicts the choice of H . So, the subgroup
A does not contain free Abelian subgroups of rank 2 and is an Abelian
torsion-free group of rank 1.
Let us consider the group
G1 = NA
G ×Nd
G = A⋋ 〈b〉 × 〈c〉,
where A is a torsion-free group of rank 1, |b| = 2, b−1ab = a−1 for any
element a ∈ A and |c| = p, p 6= 2. Since NA
G is a subgroup of the norm
NA
G1
of Abelian non-cyclic subgroups of the group G1 and c ∈ Z(G1), we
have G1 = NA
G1
and G1 is a HA-group. By the description of such groups
(see, e.g. [10]) A is an infinite cyclic group or a group isomorphic to an
additive group of p-adic fractions.
Assume that A = 〈a〉 is an infinite cyclic group. Then
(
NA
G
)
′
= 〈a2〉 ⊳
G, C = CG(〈a
2〉) ⊳ G and [G : C] 6 2. Since b /∈ C, we have G = C⋋ 〈b〉,
where |b| = 2. By the results of [13] the centralizer C contains all elements
of infinite order of a group G and all its Abelian non-cyclic subgroups.
Moreover, the periodic part T (C) of the subgroup C is normal in G and
C/T (C) is an Abelian torsion-free group of rank 1. Therefore, in this case
the commutant C ′ is periodic and C ′ ⊆ T (C).
As [Nd
G, N
A
G ] = E, it follows that Nd
G = 〈c〉 ⊆ T (C). Taking into
account the non-Dedekindness of the norm NA
G and Lemma 2, we conclude
that T (C) does not contain Abelian non-cyclic subgroups, so |T (C)| < ∞.
Let u be a non-identity element of T (C). Since the group 〈u, a2〉 =
〈u〉 × 〈a2〉 and its characteristic subgroup 〈u〉 are NA
G -admissible,
[〈u〉, NA
G ] ⊆ NA
G ∩ 〈u〉 = E.
Hence, [T (C), NA
G ] = E and [u, b] = 1 for an arbitrary element u ∈ T (C).
Suppose that T (C) contains an involution z. Then subgroup 〈b〉 × 〈z〉
is Abelian non-cyclic and, therefore, NA
G -admissible. Thus,
〈b〉 = NA
G ∩ (〈b〉 × 〈z〉) ⊳ NA
G ,
which is impossible. So, 2 /∈ π(T (C).
Let us prove that T (C) = Nd
G. Suppose for a contradiction that there
exists an element u ∈ T (C)\〈c〉. Since T (C) does not contain non-cyclic
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92 Generalized norms of groups
Abelian subgroups, then |u| 6= p and the element c is contained in each
cyclic p-subgroup of the composite order. Therefore, if |u| = pk > p, k ∈ N,
then 〈c〉 ⊆ 〈u〉 and [u, c] = 1. Now let (|u|, p) = 1. Thus,
[u, c] ∈ (〈c〉 ∩ (〈u〉 × 〈a2〉)) = E
and again [u, c] = 1. Since the subgroup 〈ua2〉 × 〈c〉 is Abelian non-cyclic
and therefore NA
G -admissible, we conclude that the subgroup 〈ua2〉p =
〈upa2p〉 is also NA
G -admissible. Then by the condition [b, a] 6= 1 we have
that [b, upa2p] 6= 1 and
b−1upa2pb = (upa2p)−1 = u−pa−2p.
On the other hand, b−1upa2pb = upa−2p, because [u, b] = 1. Therefore,
u−p = up and u2p = 1, which is impossible. So, T (C) = Nd
G = 〈c〉, where
|c| = p, p 6= 2.
Let C1 = CC(N
d
G) be the centralizer of the subgroup Nd
G = 〈c〉 in C.
Since C ′ ⊆ 〈c〉, the group C1 is Abelian with the complementary subgroup
〈c〉, e.g., C1 = 〈c〉×Y , where Y is an Abelian torsion-free group of rank 1.
By the proved above C contains all elements of infinite order of a group.
Therefore, H ⊆ C, where H is a non-cyclic Abelian torsion-free subgroup
that is not Nd
G-admissible, and C1 6= C.
By the cyclicity of the factor group C/C1 and the previous consid-
erations, we obtaine that the subgroup Y is non-cyclic. Suppose that it
contains an infinite sequence of subgroups
〈y1〉 ⊂ 〈y2〉 ⊂ ... ⊂ 〈yn〉 ⊂ ...,
where yn = y
kn+1
n+1 , (kn+1, p) = 1 for all n ∈ N. Then the isolator I
(see [14]) of the subgroup 〈cy1〉 is non-cyclic and hence, I is a NA
G -admissible
subgroup. Therefore,
b−1(cy1)b ∈ (I ∩ (〈c〉 × 〈cy1〉)) = 〈cy1〉.
Since 〈y1〉 ∩ 〈a〉 6= E and b−1y1b = y−1
1 , then b−1cy1b = (cy1)
−1 =
c−1y−1
1 = cy−1
1 and c2 = 1, which is impossible. So, Y does not contain
such chains and hence is a group isomorphic to an additive group of p-adic
numbers.
Let’s prove that the subgroup 〈c〉 is complemented in C. By the proved
above we have 1 6= [C : C1] = k, where k|(p− 1). Since we can uniquely
find the root of k degree for each element of the subgroup 〈c〉 and 〈c〉 is
complemented in its centralizer, it is also complemented in C (Theorem
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T. Lukashova 93
1, [15]), e.g. C = 〈c〉⋋D, where D is an incomplete Abelian group of rank
1.
It is obvious, that the group G = (〈c〉 ⋋ D) ⋋ 〈b〉 does not contain
periodic Abelian non-cyclic subgroups, all mixed Abelian subgroups belong
to the group 〈c〉⋋D, contain 〈c〉 and are normal in G. Moreover, all tortion-
free Abelian non-cyclic subgroups are contained either in the subgroup
D or in subgroups g−1Dg, g ∈ G, conjugated to this subgroup. Then
the normalizer of each Abelian non-cyclic subgroup of rank 1 contains
a subgroup Y ⊳ G and, as a consequence, the norm NA
G contains this
subgroup, which contradicts the assumption of its structure. Therefore, A
cannot be an infinite cyclic group. So, NA
G = A⋋ 〈b〉, where A is a group
isomorphic to an additive group of p-adic fractions, (p, 2) = 1, |b| = 2 and
b−1ab = a−1 for any element a ∈ A.
By the proved above, every infinite cyclic subgroup has a nonidentity
intersection with the norm NA
G . On the other hand, the norm NA
G does not
contain free Abelian subgroups of rank 2. So, the group G also does not
contain such subgroups. A similar statement holds for non-cyclic Abelian
subgroups of finite order. Indeed, if a group G contains finite Abelian
non-cyclic subgroups, then their intersection with NA
G , is a finite subgroup
normal in NA
G , which is impossible, or is an identity subgroup, which
contradicts Lemma 2.
Suppose that G contains a periodic non-cyclic locally cyclic subgroup
P . If P contains an infinite subgroup which has the identity intersection
with the norm NA
G , then the norm NA
G is Dedekind by Lemma 2, which is
impossible. So, P is a quasicyclic subgroup and P ∩NA
G 6= E. But in this
case (P ∩NA
G ) ⊳ NA
G , which contradicts the structure of the norm NA
G .
Therefore, the group G does not contain periodic non-cyclic locally cyclic
subgroups.
Finally, since the intersection 〈x〉 ∩NA
G is nonidentity for an arbitrary
element x ∈ G, |x| = ∞, the factor-group G/NA
G is periodic. The Theorem
is proved.
Theorem 3. If a non-periodic locally soluble group G has an Abelian non-
cyclic subgroup, the norm NA
G of Abelian non-cyclic subgroups is Dedekind,
NA
G 6= E, Nd
G 6= E and Nd
G ∩NA
G = E, then:
1) Z (G) = N (G) = E;
2) the norm NA
G of Abelian non-cyclic subgroups is an Abelian torsion-
free group of rank 1;
3) the norm Nd
G of decomposable subgroups is a cyclic group, Nd
G =
〈c〉, (|c|, 2) = 1.
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94 Generalized norms of groups
Proof. The first statement is proved in the same way as in Theorem 2.
By the condition Nd
G ∩ NA
G = E and Lemma 1 we have that the norm
Nd
G is Dedekind. Moreover, the group G contains a non-primary, not NA
G -
admissible cyclic subgroup 〈g〉 and an indecomposable Abelian non-cyclic
subgroup H, which is not Nd
G-admissible.
Suppose that the norm Nd
G is non-periodic and an element c ∈ Nd
G such
that |c| = ∞ exists. Since the subgroup 〈g〉 is Nd
G-admissible, the subgroup
〈g, ck〉 = 〈g〉 × 〈ck〉 is Abelian non-cyclic for some positive integer k, and
therefore NA
G -admissible. So, the subgroup 〈g〉 is also NA
G -admissible, which
contradicts its choice. Therefore, the norm Nd
G is a periodic Dedekind
group.
Assume that Nd
G does not satisfy the minimal condition for Abelian
subgroups. Then the intersection CG(g)∩Nd
G contains non-cyclic Abelian
subgroups A1 and A2 such that (A1 ∪A2)∩ 〈g〉 = E. Since the subgroups
A1×〈g〉 and A2×〈g〉 are non-cyclic Abelian, they are NA
G -admissible. So,
the group
〈g〉 = (A1 × 〈g〉) ∩ (A2 × 〈g〉)
is also NA
G -admissible, which is impossible. This contradiction shows
that Nd
G is a group with the minimal condition for Abelian subgroups.
Moreover, since the subgroup Nd
G is Dedekind it follows from Corollary
4.2 [16] that Nd
G is a finite extension of the direct product of a finite
number of quasicyclic subgroups.
Let denote the subgroup generated by elements of the prime order
of the norm Nd
G by ω(Nd
G). By the proved above
∣
∣ω(Nd
G)
∣
∣ < ∞, so [G :
CG(ω(N
d
G))] < ∞. If an indecomposable non-cyclic Abelian subgroup
H, which is not Nd
G-admissible, is complete, then H ⊆ CG(ω(N
d
G)) and
the group B = H · ω(Nd
G) is Abelian. If B is decomposable, then it is
Nd
G-admissible. But in this case the subgroup
B|ω(Nd
G
)| = H|ω(Nd
G
)| = H
is also Nd
G-admissible, which contradicts its choice. Thus, B is a non-
decomposable Abelian group and as a consequence, H is a quasicyclic
p-group. So, ω(Nd
G) ⊆ H and
∣
∣ω(Nd
G)
∣
∣ = p. Since Z(G) = E and the
norm Nd
G is Dedekind and contains an only one subgroup of prime order
by the proved above, it is either a cyclic or a quasicyclic p-group. In
both cases we conclude that H ⊆ CG(N
d
G). Therefore, the subgroup H is
Nd
G-admissible, which is impossible. Hence, H is an incomplete non-cyclic
Abelian torsion-free group of rank 1.
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T. Lukashova 95
Let’s prove that the norm NA
G of Abelian non-cyclic subgroups is
a torsion-free Abelian group. Indeed, otherwise, there exists a nonidentity
element x ∈ NA
G , |x| < ∞. Then, taking into account that the norm NA
G
is Dedekind and the subgroup H is NA
G -admissible, we have
[〈x〉, H] ⊆ T (NA
G ) ∩H = E,
where T (NA
G ) is the periodic part of the subgroup NA
G . Therefore, the
subgroup 〈x〉 ×H is decomposable Abelian and, as a consequence, Nd
G-
admissible. But in this case
[Nd
G, H] ⊆ Nd
G ∩ (〈x〉 ×H) = E.
Hence, H is Nd
G-admissible subgroup, which contradicts its choice. So,
NA
G is an Abelian torsion-free group.
IfNA
G∩H = E, then [NA
G , H] = E and for any element a ∈ NA
G , |a| = ∞
the subgroup 〈a〉×H is Abelian decomposable and, hence, Nd
G-admissible.
But then [Nd
G, H] ⊆ Nd
G ∩ (〈a〉 ×H) = E, which is impossible, because
in this case H will be Nd
G-admissible subgroup. Therefore, NA
G ∩H 6= E,
and for an arbitrary element h ∈ H there exists a non-zero integer k such
that hk ∈ NA
G .
Suppose that the norm NA
G contains a free Abelian subgroup 〈a1〉×〈a2〉,
where |a1| = |a2| = ∞. Then by the proved at least one of the subgroups
〈a1〉 or 〈a2〉 has the identity intersection with H . Let 〈a1〉 ∩H = E. Since
H is a NA
G -admissible subgroup, then a−1
1 h1a1 = h2, where h1, h2 ∈ H.
Moreover, by the condition NA
G ∩H 6= E we have hk1 ∈ NA
G for some integer
k 6= 0. Hence, a−1
1 hk1a1 = hk1 = hk2, and h1 = h2. Therefore, [〈a1〉, H] = E
and the subgroup 〈am1 〉 ×H is Nd
G-admissible for an arbitrary natural m.
Thus, the subgroup H =
⋂
∞
m=1(〈a
m
1 〉 ×H) is also Nd
G-admissible, which
contradicts its choice. So, the norm NA
G does not contain free abelian
subgroups of rank 2 and is an Abelian torsion-free group of rank 1.
Let 〈g〉 be a non-primary subgroup, which is not NA
G -admissible. It is
clear that at least one of its Sylow p-subgroups is also not NA
G -admissible.
Let it be a subgroup 〈g〉p, where p is prime. Since the factor-group
G/CG(N
A
G ) is isomorphic to a subgroup of automorphisms of an Abelian
torsion-free group of rank 1 with the periodic part of order 2 ([17], p. 294],
we conclude that 〈g〉p = 〈g〉2 = 〈ḡ〉 is a 2-group, |ḡ| = 2n, n ∈ N.
Let’s prove that all Sylow p-subgroups of Nd
G are cyclic. Suppose that
Nd
G contains an elementary Abelian subgroup N of order p2, p 6= 2. Since
[N, 〈ḡ〉] ⊆ (Nd
G)p ∩ 〈ḡ〉 = E,
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96 Generalized norms of groups
where
(
Nd
G
)
p
is a Sylow p-subgroup of the norm Nd
G, the subgroup N×〈ḡ〉
is an Abelian non-cyclic and 〈ḡ〉 is NA
G -admissible as its characteristic
subgroup, which is impossible. Therefore, any Sylow p-subgroup of the
norm Nd
G for p 6= 2 contains a unique subgroup of prime order, so, it is
a cyclic or a quasicyclic p-group.
Suppose that the norm Nd
G contains quasicyclic p–subgroup P for some
prime p 6= 2. Then P×〈ḡ〉 is Abelian non-cyclic, and hence, NA
G -admissible
group. Thus, the subgroup 〈ḡ〉 is NA
G -admissible, which contradicts its
choice. So, any Sylow p-subgroup of the norm Nd
G is cyclic for p 6= 2.
Let us consider the Sylow 2-subgroup
(
Nd
G
)
2
of the norm Nd
G. If
(
Nd
G
)
2
∩ 〈ḡ〉 = E, then for an arbitrary element c ∈
(
Nd
G
)
2
the subgroup
〈c, ḡ〉 = 〈c〉 × 〈ḡ〉 is Abelian non-cyclic, and therefore, is NA
G -admissible.
Then [〈ḡ〉, NA
G ] ⊆ (〈c〉 × 〈ḡ〉) ∩ NA
G = E, which is impossible. Thus,
(
Nd
G
)
2
∩ 〈ḡ〉 6= E.
Let’s denote the lower layer of the Sylow 2-subgroup
(
Nd
G
)
2
by M
and consider the group G2 = 〈ḡ〉M . Let M be a non-cyclic group. Then
by the condition 〈ḡ〉 ⊳ G2 we have [〈ḡ〉,M ] ⊆ M ∩ 〈ḡ〉 = 〈c1〉, where
c1 ∈ M, |c1| = 2. As 〈ḡ〉 = CG2
(〈ḡ〉), it follows M = 〈c1〉 × 〈c2〉 and
[〈ḡ〉, 〈c2〉] = 〈c1〉. Then by M ⊳ G, [G : CG(M)] = 2 and ḡ /∈ CG(M), we
conclude that G = CG (M) 〈ḡ〉. However, in this case c1 ∈ Z(G), which is
impossible. Therefore, the lower layer of
(
Nd
G
)
2
contains one involution,
which again contradicts the condition Z(G) = E. So, 2 /∈ π(Nd
G) and
Nd
G = 〈c〉, (|c|, 2) = 1. The Theorem is proved.
The following example confirms the existence of groups satisfying the
conditions of Theorem 3 and generalizes Example 3.4 of [12]. Let’s note
that the order of the norm of decomposable subgroups in this case can
be a composite number (unlike the groups satisfying the conditions of
Theorem 2).
Example 2. Let G = (〈a〉 ⋋ B) ⋋ 〈c〉, where |a| = m > 1, (m, 2) = 1,
B is a group isomorphic to an additive group of q-adic fractions, q is
prime, (q, 2m) = 1, B = B1〈x〉, x
2 ∈ B1, x
−1ax = a−1, [B1, 〈a〉] = E,
|c| = 2, [c, a] = 1 and c−1bc = b−1 for any element b ∈ B.
In this group, all periodic decomposable subgroups have the order 2d,
d|m, d > 1 and are groups of the form 〈a
m
d
scbk1〉, where b1 ∈ B1, (s, d) =
1, k ∈ {0, 1}. Thus, all nonperiodic decomposable subgroups are mixed
and contained in the group B1 × 〈a〉 and, hence, they are normal in G.
Since NG(〈a
m
d
scbk1〉) = 〈acbk1〉, we conclude that Nd
G = 〈a〉.
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T. Lukashova 97
Let’s determine the norm NA
G of non-cyclic Abelian subgroups of
the group G. It is obvious that G does not contain periodic non-cyclic
Abelian subgroups but all mixed Abelian subgroups contain 〈a
m
d
s〉, and
are subgroups of the group B1 × 〈a〉. It is easy to prove that all these
subgroups are normal in G. Further, all non-cyclic Abelian subgroups
of rank 1 are contained either in the subgroup B or in the subgroups
g−1Bg, g ∈ G, conjugate to this subgroup, or in the group B1 × 〈a〉. Let’s
consider an infinite sequence of subgroups in B1:
〈y1〉 ⊂ 〈y2〉 ⊂ ... ⊂ 〈yn〉 ⊂ ...,
where yn = y
kn+1
n+1 , (kn+1,m) = 1 for all n ∈ N.
It is easy to prove that the isolator I of the subgroup 〈ay1〉 is non-cyclic
because the root of the element a of any power mutually prime with m
exists. Moreover, NG(I) = B1 × 〈a〉. Since NG(B) = B⋋ 〈c〉, we conclude
that NA
G = B1 is a torsion-free Abelian group of rank 1 and Nd
G∩NA
G = E.
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Contact information
Tetiana Lukashova Taras Shevchenko National University of Kyiv,
Volodymyrska 60, Kyiv, Ukraine, 01033
E-Mail(s): tanya.lukashova2015@gmail.com
Received by the editors: 15.01.2020.
T. Lukashova
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