# Supper Bassian and Nearly Generalized Bassian
Abelian Groups

Peter Danchev  Institute of Mathematics and Informatics, Bulgarian Academy of
Sciences, 1113 Sofia, Bulgaria<EMAIL_ADDRESS><EMAIL_ADDRESS>and
Brendan Goldsmith  Technological University, Dublin, Dublin 7, Ireland
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In connection to two recent publications of ours in Arch. Math. Basel (2021)
and Acta Math. Hung. (2022), respectively, and in regard to the results
obtained in Arch. Math. Basel (2012), we have the motivation to study the near
property of both Bassian and generalized Bassian groups. Concretely, we prove
that if an arbitrary reduced group has the property that all of its proper
subgroups are generalized Bassian, then it is also generalized Bassian itself,
that is, if $G$ is an arbitrary nearly generalized Bassian group, then either
$G$ is a quasi-cyclic group or $G$ is generalized Bassian of finite torsion-
free rank. Moreover, we show that there is a complete description of the so-
called super Bassian groups, that are those groups whose epimorphic images are
all Basian, and give their full characterization. Also, we establish that the
hereditary property of Bassian groups gives nothing new as it coincides with
the ordinary Bassian property.

###### Key words and phrases:

Abelian groups, Bassian groups, supper Bassian groups, generalized Bassian
groups, (hereditarily, nearly, super) generalized Bassian groups, Hopfian and
co-Hopfian groups

###### 2010 Mathematics Subject Classification:

20K10, 20K20, 20K21

##  1\. Introduction and Main Facts

Throughout the current paper, the term group will mean an additively written
Abelian group. Our notation and terminology are at all standard and may be
found in the books [4, 5, 6]. We will make extensive use of the notions of
torsion-free rank and $p$-rank by using the corresponding notations $r_{0}(G)$
and $r_{p}(G)$, respectively as full details of these notions are given in [4,
Sections 16,35] or [6, Sections 4.4, 5.6]. If $G$ is a mixed group, we will
denote its torsion subgroup by $T(G)$ (and often we will simplify this
abbreviation to $T$ if there is no possibility of ambiguity). The $p$-primary
components of such a mixed group $G$ will be denote by $T_{p}(G)$ or $T_{p}$
when this is more convenient. We also adopt one slightly non-standard piece of
terminology thus: a group $G$ is said to be a genuine mixed group if $G$ has
both non-trivial torsion and torsion-free elements and $G$ is not splitting,
i.e., in other words, the torsion subgroup of $G$ is not a direct summand of
$G$. Thus, the groups that are not genuinely mixed fall into one of the three
categories as follows: they are either torsion, torsion-free or splitting
mixed. We also use the concept an elementary $p$-group, or sometimes a
$p$-elementary subgroup, to denote a (possibly infinite) direct sum of cyclic
groups of order $p$ and the name elementary group is used to denote a torsion
group such that all its $p$-primary components are elementary $p$-groups.

Any other specific concepts will be explained as needed in what follows. For
instance, let us recall that if $(\mathcal{P})$ is a property of groups, then
a group $G$ is said to be hereditarily $(\mathcal{P})$ if $G$ and all its
subgroups have together the property $(\mathcal{P})$. In this section we wish
to investigate a weak variant of this concept. For simplicity of notation, we
say that a group $G$ is nearly $(\mathcal{P})$ if every proper subgroup of $G$
has property $(\mathcal{P})$; it is certainly not assumed a priory that $G$,
itself, has the property $(\mathcal{P})$. Specifically, we intend to
investigate which groups are nearly $(\mathcal{P})$ when $(\mathcal{P})$ is
the property of being Bassian or generalized Bassian, respectively.
Additionally, we shall examine the ”super” property $(\mathcal{P})$ of Bassian
groups in the sense that each epimorphic image of the group is Bassian; this
is certainly a stronger requirement than that of being Bassian. Thus, by what
we have noted, our basic motivating tool in writing up the present article is
to expand somewhat the corresponding results for Hopfian and co-Hopfian groups
from [7]. We achieve that in Corollary 2.2, Proposition 2.6, Theorem 2.5 and
Theorems 3.3, 3.11, respectively.

The following two basic notions, that are in the focus of our present
interest, appeared in [1] and [2], respectively.

An Abelian group $G$ is called Bassian if the existence of an injective
homomorphism $G\to G/H$ for some subgroup $H$ of $G$ implies that $H=\\{0\\}$
and, more generally, that $G$ is generalized Bassian if the existence of an
injective homomorphism $G\to G/H$ for some subgroup $H$ of $G$ implies that
$H$ is a direct summand of $G$.

Unfortunately, as already commented above, there is no a full description of
the generalized Bassian groups so far. However, the following complete
characterization of Bassian groups was obtained in [1, Main Theorem]:

Theorem (Bassian). (i) A reduced Abelian group $G$ is Bassian if, and only if,
all of the ranks $r_{0}(G),r_{p}(G)$ (for $p$ a prime) are finite;

(ii) A non-reduced Abelian group $G$ is Bassian if, and only if, it has the
form $G=D\oplus R$, where $D$ is a finite dimensional $\mathbb{Q}$-vector
space and $R$ is a reduced Bassian group.

It follows at once from these two necessary and sufficient conditions that a
Bassian group is necessarily countable and that the finite direct sum of
Bassian groups is again a Bassian group. Note also that a consequence of this
classification is that the Bassian groups are hereditarily (see, e.g., [7] for
both Hopfian and co-Hopfian groups): a subgroup of a Bassian group is
necessarily Bassian. The converse is definitely not true since the quasi-
cyclic groups $\mathbb{Z}(p^{\infty})$ are examples of non-Bassian groups all
of whose proper subgroups are Bassian.

Furthermore, a detailed analysis of the fundamental properties of generalized
Bassian groups was given in [2] and [3], respectively, although a full
characterization result was not obtained there. In fact, it is not clarified
yet whether a subgroup of a generalized Bassian groups is again generalized
Bassian, and even finding a suitable approach how to attack this question is
still problematic. However, some further advantage in this matter was made in
the latter paper. Concretely, it was shown in [3] that any generalized Bassian
group must have finite torsion-free rank (thus answering in the negative of a
recent question posed in [2]), that any generalized Bassian group is a direct
sum of a Bassian group and an elementary group and that every subgroup of a
generalized Bassian group is also a direct sum of a Bassian group and an
elementary group, as well as it was conjectured there that the direct sum of a
Bassian group and an elementary group has to be a generalized Bassian group
(compare also with our final Problem 3).

##  2\. The Hereditary, Near and Super Bassian Properties

As mentioned in the Introduction, a group with a property $(\mathcal{P})$ is
said to have the hereditary $(\mathcal{P})$ property, provided the whole group
along with all of its proper subgroups possess the same property
$(\mathcal{P})$; a slight variation of this concept, which we have named
nearly is the following: group is said to have the near $(\mathcal{P})$
property, provided each of its proper subgroup possesses property
$(\mathcal{P})$. We are not assuming that a group which is nearly
$(\mathcal{P})$ necessarily has the property $(\mathcal{P})$.

###  2.1. Hereditary Bassian group

First of all, we show that Bassian groups are, in fact, a familiar class of
groups involving a hereditary condition.

######  Proposition 2.1.

A group is Bassian if, and only if, it is a hereditarily Hopfian group.

###### Proof.

If $G$ is a Bassian group, then, as observed in [1], every subgroup of $G$ is
also Bassian. Furthermore, since Bassian groups are also Hopfian, $G$ is then
hereditarily Hopfian.

Conversely, if $G$ is a hereditarily Hopfian group, then, by the
classification given in [7], $G$ is an extension of a direct sum of finite
primary groups by a finite rank torsion-free group. It thus follows from the
characterization in [1] that $G$ is Bassian, as asserted. ∎

As an immediate consequence, we have the following surprising claim.

######  Corollary 2.2.

A reduced group is hereditarily Bassian if, and only if, it is Bassian.

###  2.2. Nearly Bassian groups

The next result gives us a complete characterization of nearly Bassian groups.

######  Proposition 2.3.

An arbitrary group $G$ is nearly Bassian if, and only if, either $G$ is the
quasi-cyclic group $\mathbb{Z}(p^{\infty})$ for some prime $p$, or $G$ is
Bassian.

###### Proof.

The sufficiency is straightforward: indeed, a proper subgroup of a quasi-
cyclic group is finite and hence Bassian, while a subgroup of a Bassian group
is again Bassian – see, e.g., the discussion in the Introduction of [2].

For the necessity, let $T$ denote the torsion subgroup of $G$ and $d(T)$ the
maximal divisible subgroup of $T$. Clearly, $d(T)$ is a summand of $G$, so
that $G=d(T)\oplus X$ for some subgroup $X$ of $G$. We consider three possible
cases:

Case (i): $X=\\{0\\}$. Then $G$ is of the form $G=\bigoplus\limits_{p\in
I}D_{p}$, where each $D_{p}$ is a divisible $p$-group and $I$ is a set of
primes. If $\mid I\mid~{}>1$, then $G$ has a proper subgroup $D_{q}$ for some
prime $q$, which is, by hypothesis, Bassian – contrary to [1, Proposition
3.1]. We conclude that $G$ must be a divisible $p$-group for some prime $p$,
and so an identical argument shows that it must have rank $1$. Thus, in Case
(i), we obtain that $G\cong\mathbb{Z}(p^{\infty})$ for some prime $p$, as
asserted.

Case (ii): $X\neq\\{0\\}\neq d(T)$. In this case, $d(T)$ is a proper subgroup
of $G$ and hence Bassian - it is, however, impossible as observed in Case (i).
So, this case cannot occur.

Case (iii): $d(T)=\\{0\\}$, that is, $G=X$. We claim that $G$ must be Bassian;
for if not, then it follows from the classification of Bassian groups in [1]
that at least one of the ranks $r_{0}(G),r_{p}(G)$ (for a prime $p$) is
infinite. If any $r_{p}(G)$ is infinite, then the $p$-primary component of
$G$, and hence $G$ itself, has a proper subgroup, $H$ say, with $r_{p}(H)$
infinite; but this is nonsense since $H$ would then also be Bassian. A similar
argument shows that, if $r_{0}(G)$ is infinite, then again there is a proper
subgroup $H$ of $G$ with $r_{0}(H)$ infinite, which is also untrue because $H$
would also be Bassian. Thus, in Case (iii), $G$ is necessarily Bassian, as
required. ∎

###  2.3. Super Bassian Groups

In this subsection we investigate a property that is stronger than the
property of being Bassian. In fact, our objective is to determine those groups
having the property that every epimorphic image of the group is Bassian; such
groups are, of course, Bassian but the converse manifestly fails: a non-
reduced Bassian group will always have an epimorphic image which is a quasi-
cyclic group, $\mathbb{Z}(p^{\infty})$, and the latter is not Bassian - see
the Main Theorem of [1] quoted in the Introduction. However, a group which is
reduced may have a divisible epimorphic image: indeed, Corner has constructed,
for each positive integer $n\geq 2$, a torsion-free group of rank $n$ such
that all its subgroups of rank $n-1$ are cyclic while the torsion-free
quotient groups of rank $1$ are all divisible – see, for example, [6, Exercise
5, Section 4, Chapter 12] and, of course, every reduced unbounded $p$-group
$G$ has a divisible epimorphic image of the form $G/B$, where $B$ is a basic
subgroup of $G$.

Thus, motivated by these facts, we will make use of the following, somewhat ad
hoc definition: A group $G$ is said to have property $(\mathfrak{P})$ if no
quasi-cyclic group is an epimorphic image of $G$.

In keeping with standard terminology used in relation to Hopfian groups (see,
e.g., [7]), we make the following definition.

######  Definition 2.4.

A group $G$ is said to be super Bassian if every epimorphic image of $G$ is
Bassian.

Evidently, a super Bassian group is Bassian and it follows from our discussion
above that a super Bassian group is necessarily reduced and also that $G$ has
property $(\mathfrak{P})$.

The classification we are seeking is given by our next result.

######  Theorem 2.5.

Let $G$ be an arbitrary group. The following three statements hold:

(i) if $G$ is torsion, then $G$ is super Bassian if, and only if, $G$ is
reduced Bassian;

(ii) if $G$ is torsion-free, then $G$ is super Bassian if, and only if, $G$ is
Bassian and has property $(\mathfrak{P})$;

(iii) if $G$ is mixed, then $G$ is super Bassian if, and only if, $G$ is
Bassian and has property $(\mathfrak{P})$.

###### Proof.

Observe that in each of the three cases to be considered necessity holds
obviously, so we need only give the arguments guaranteeing sufficiency.

Case (i): As $G$ is reduced torsion and Bassian, then it has a primary
decomposition of the form $G=\bigoplus\limits_{p\ {\rm prime}}G_{p}$ and each
$p$-primary component is a finite $p$-group. But then every epimorphic image
of $G$ must have a similar primary decomposition and thus is Bassian by the
classification of Bassian groups. So, $G$ is super-Bassian.

Case (ii): Let $G$ be a torsion-free Bassian group satisfying property
$(\mathfrak{P})$. Then $G$ is reduced and of finite torsion-free rank. Let
$\phi:G\twoheadrightarrow X$ be an arbitrary epimorphism from $G$ onto $X$. We
consider the three possibilities for $X$: (a) $X$ is torsion; (b) $X$ is
torsion-free; (c) $X$ is mixed.

Subcase (a): As $X$ is torsion, $X$ must be reduced since otherwise it would
have a quasi-cyclic summand, $Y$ say, which is then an epimorphic image of $G$
contrary to our hypothesis that $G$ has property $\mathfrak{P}$. Let $X$ have
a primary decomposition $X=\bigoplus_{p}X_{p}$ and consider a surjection
$\psi_{p}:G\twoheadrightarrow X_{p}$, where $\psi_{p}$ is the composition of
$\phi$ with the canonical projection of $X$ onto $X_{p}$. If each $X_{p}$ is
bounded then $X_{p}$ is finite since it is then a bounded image of a torsion-
free group of finite rank. We claim all the components $X_{p}$ must be
bounded. Suppose, for a contradiction, that some component $X_{p}$ is
unbounded and let $B_{p}$ be a basic subgroup of $X_{P}$. Now the composition
of $\psi_{p}$ with the canonical projection of $X_{p}$ onto $X_{p}/pX_{p}$ is
a surjection and, as $X_{p}/pX_{p}$ is bounded, we have, as observed above,
that $X_{p}/pX_{p}$ is finite. This is impossible since $X_{p}/pX_{p}\cong
B_{p}/pB_{p}$ would then force $B_{p}/pB_{p}$ to be finite, contrary to our
assumption that $X_{p}$ is unbounded. Hence, $X$ has the property that each of
its primary components is finite and so $X$ is Bassian and reduced.

Subcase (b): If $X$ is torsion-free, then clearly $X$ is also torsion-free of
finite rank and so Bassian.

Subcase (c): If $X$ is mixed then it must be reduced and $r_{0}(X)$ is finite.
So, $X$ is an extension of the torsion group $T(X)$ by a torsion-free group of
finite rank; in particular, $X$ is an extension of a torsion group by a
countable torsion-free group. Let $T(X)=\bigoplus_{p}X_{p}$ be the primary
decomposition of the torsion subgroup $T(X)$ of $X$ and let $B_{p}$ be a basic
subgroup of $X_{p}$ for each prime $p$ in the decomposition. Then, by an
unpublished result of Corner, $B_{p}$ is an epimorphic image of $X_{p}$, say
$\eta_{p}:X_{p}\twoheadrightarrow B_{p}$. (The result of Corner has been used
previously in [7] and [8, Theorem 4.1]; some details of his arguments may be
found in Section 4 of the latter.) Then, the composition of $\eta_{p}$ with
the surjections of $G$ onto $X_{p}$ yields an epimorphism of $G$ onto the
torsion reduced group $B_{p}$. The argument in part (i) tells us that each
$B_{p}$ is finite, whence each $X_{p}$ is also finite. Thus, for each prime
$p$, we have that $r_{p}(X)$ is finite and, as $r_{0}(X)$ is also finite, we
conclude that $X$ is Bassian.

Combining the three subcases we see that in case (ii), each epimorphic image
$X$ of $G$ is Bassian, so $G$ is super Bassian, as claimed.

Case (iii): Here we are assuming that $G$ is mixed Bassian and has property
$(\mathfrak{P})$. Let $G\twoheadrightarrow X$ be an arbitrary epimorphism. We
again consider the three subcases (d) $X$ is torsion; (e) $X$ is torsion-free;
(f) $X$ is mixed.

Subcase (d): As before, $X$ is necessarily reduced. Set $Z=\phi(T(G))\leq X$.
Since each $p$-primary component of $T(G)$ is finite, the same holds for $Z$
and hence $Z$ is Bassian. Furthermore, $X/Z=\phi(G)/\phi(T(G))$ is an
epimorphic image of $G/T(G)$, a torsion-free group of finite rank. Therefore,
$G/T(G)$ satisfies property $(\mathfrak{P})$ since, by assumption $G$ does.
Thus, by what we established in part (ii) above, the group $X/Z$ is Bassian.
So, we have an exact sequence

$0\to Z\to X\to X/Z\to 0,$

where $Z$ is torsion Bassian and $X/Z$ is also Bassian. It now follows from
Lemma 2.6 below that $X$ is Bassian, as asserted.

Subcase (e): If $G$ is mixed Bassian, then $r_{0}(G)$ is finite and thus $X$,
as being a torsion-free epimorphic image of $G$, is also of finite rank and
Bassian.

Subcase (f): Again we have that $r_{0}(X)$ is finite and if $Y=\phi(T(G))$,
then $Y$ is torsion Bassian. Exactly as in subcase (d), we have that
$X/Y=\phi(G)/\phi(T(G))$ is an epimorphic image of the finite rank torsion-
free group $G/T(G)$. Observe that $G/T(G)$ has property $(\mathfrak{P})$ since
$G$ has this property. It now follows from case (ii) that $X/Y$ is Bassian, so
that $X$ is an extension of the torsion Bassian group $Y$ by the Bassian group
$X/Y$. Applying Lemma 2.6 stated below, we see that $X$ is also Bassian, as
wanted.

Combining the subcases (d),(e) and (f), we have established that if $G$ is
mixed Bassian and has property $(\mathfrak{P})$, then $G$ is super Bassian, as
desired. ∎

It remains only to establish the cited above Lemma 2.6, which is a simple
extension of [1, Proposition 2.3].

######  Lemma 2.6.

If $0\to B\to G\to C\to 0$ is an extension of a torsion Bassian group $B$ by a
Bassian group $C$, then $G$ is Bassian.

###### Proof.

Choose $H\leq G$ such that $H/B=T(C)$, the torsion subgroup of $C$. Then, we
have an exact sequence

$0\to H\to G\to G/H\cong C/T(C)\to 0.$

Since $C$ is Bassian, the factor-group $C/T(C)$ is torsion-free of finite rank
and so, in view of [1, Proposition 2.3], it suffices to show that $H$ is
Bassian since it is certainly torsion. To that goal, let $H=\bigoplus H_{p}$
be the primary decomposition of $H$ and note that, if $B_{p}$ is the
corresponding $p$-primary component of $B$, then $B_{p}\leq H_{p}$ and
$H_{p}/B_{p}\cong C_{p}$, where the latter is the $p$-primary component of
$T(C)$. Since $B,T(C)$ are both Bassian, the subgroups $B_{p},C_{p}$ are
finite giving that each primary component of $H$ is also finite and so $H$ is
Bassian, as expected. ∎

The requirement in Theorem 2.2 above that $G$ have property ($\mathfrak{P}$)
is a little unsatisfactory, because it is not immediate how to determine this
from the internal structure of $G$. Fortunately, when a group is Bassian, we
have the following helpful equivalence.

######  Proposition 2.7.

Let $G$ be a Bassian group, then $G$ has property ($\mathfrak{P}$) if, and
only if, the torsion-free quotient $G/T(G)$ has property ($\mathfrak{P}$).

###### Proof.

If $G$ has property ($\mathfrak{P}$), then it is clear that $G/T(G)$ has the
same property even when $G$ is not Bassian.

Conversely, assume the contrary that $G$ does not have property
($\mathfrak{P}$), and let $\phi:G\twoheadrightarrow\mathbb{Z}(p^{\infty})$ be
a surjection onto some quasi-cyclic group $\mathbb{Z}(p^{\infty})$. If $T_{p}$
denotes the $p$-primary component of $G$, then $\phi(T(G))=\phi(T_{p})$ is
finite since $G$ is Bassian. Thus, $\phi(T(G))=C$, where $C$ is a finite
cyclic subgroup of $\mathbb{Z}(p^{\infty})$. Hence,

$\phi(G)/\phi(T(G))=\mathbb{Z}(p^{\infty})/C\cong\mathbb{Z}(p^{\infty}).$

But $\phi(G)/\phi(T(G))$ is an epic image of $G/T(G)$, so that there is an
epimorphism $G/T(G)\twoheadrightarrow\mathbb{Z}(p^{\infty})$ and,
consequently, $G/T(G)$ does not have property ($\mathfrak{P}$), as required. ∎

It is also worthwhile noticing that, for a finite rank torsion-free group $G$,
the following are equivalent (see [3] as well):

(a) $G$ has no homomorphic images that are quasi-cyclic, i.e. $G$ has property
$(\mathfrak{P})$;

(b) $F$ is a free subgroup of $G$ such that $G/F$ has finite $p$-torsion for
all primes $p$;

(c) For every free subgroup $F$ of $G$ such that $G/F$ is torsion, $G/F$ has
finite $p$-torsion for all primes $p$;

(d) For every prime $p$, the localization $G_{(p)}$ is a free
$\mathbb{Z}_{(p)}$-module (i.e., it is locally free).

We end our work in this section with the following interesting question:

Problem 1. Describe the structure of hereditary generalized Bassian groups and
super generalized Bassian groups by finding a complete structural
characterization of them.

##  3\. Nearly Generalized Bassian Groups

In this section, we consider the problem analogous to that addressed in
Proposition 2.6 in Section 2, where Bassian groups are replaced by generalized
Bassian groups. Thus, we shall say that a group $G$ is nearly generalized
Bassian if every proper subgroup of $G$ is generalized Bassian. Before
proceeding to consideration of this problem, we present a brief review of
known material about generalized Bassian groups. Fuller details and proofs may
be found in [2] and [3], respectively.

$\bullet$ A generalized Bassian group has finite torsion-free rank.

Since this fact is very important for our further presentation, we shall give
an independent and more conceptual verification by modifying some of the
arguments used in [2].

To that purpose, assume that $G$ is generalized Bassian and is not torsion. If
$G$ is torsion-free, then it necessarily has finite rank by [2, Proposition
2.8]. If $G$ is splitting mixed, then again it is of finite torsion-free rank
by [2, Proposition 2.11].

So, consider the case where $G$ is genuine mixed and assume, for a
contradiction, that $r_{0}(G)=\kappa$ is infinite. However, we know from [2,
Proposition 3.3] that if $T$ is the torsion subgroup of $G$, then $\kappa<\mid
T\mid$. Furthermore, $T=\bigoplus_{p}T_{p}$, where each $T_{p}$ has the form
$T_{p}E_{p}\oplus F_{p}$, with $E_{p}$ an elementary $p$-group and $F_{p}$ a
finite $p$-group. It follows from Remark 3.4 in [2] that $\mid
T_{p}\mid>r_{0}(G)$ for infinitely many primes $p$, so that infinitely many of
the $E_{p}$ have $p$-rank $>\kappa$.

Since $G/T$ is torsion-free of infinite rank $\kappa$, there is a subgroup $H$
of $G$ with $G=H+T$ and $\mid H\mid=\kappa$. Set $B=H+\bigoplus+pF_{p}$, so
that $\mid B\mid$ is also $\kappa$. Now, $\bigoplus_{p}F_{p}\leq T\cap B\leq
T=\bigoplus_{p}F_{p}\oplus\bigoplus E_{p}$, so there is a subgroup
$D\leq\bigoplus_{p}E_{p}$ with $T=(T\cap B)\oplus D$. But $B\cap D\leq T\cap
B\cap D=\\{0\\}$ and so

$G=H+T\leq B+T=B+(T\cap B)+D\leq G,$

hence $G=B+D=B\oplus D$.

Since $G$ is generalized Bassian and $B$ is a summand of $G$, then $B$ is also
generalized Bassian utilizing [2, Lemma 2.3]. Note that, as $D$ is torsion,
$r_{0}(G)=r_{0}(B)=\kappa$. However, $B=H+\bigoplus_{p}F_{p}$, so our
assumption that $\kappa$ is infinite yields $\mid B\mid=\mid H\mid=\kappa$,
and thus it follows that the torsion subgroup $t(B)$ of $B$ must have
cardinality $\leq\kappa$. This means that $\mid t(B)\mid\leq\kappa=r_{0}(B)$
contrary to [2, Proposition 3.3]. This contradiction forces at once $r_{0}(G)$
to be finite when $G$ is genuine mixed, thereby establishing the result after
all, as asked for.

$\bullet$ If $G$ is generalized Bassian, then the torsion subgroup $T$ of $G$
is of the form $T=\bigoplus_{p}T_{p}$ where each $T_{p}$ is a $p$-primary
group of the form $T_{p}=F_{p}\oplus E_{p}$ where $F_{p}$ is a finite
$p$-group and $E_{p}$ is an elementary $p$-group.

We note now a simple result that gives an initial insight into the structure
of generalized Bassian groups.

######  Proposition 3.1.

If $G$ is a genuinely mixed group which is generalized Bassian, then $G$ is an
extension of an elementary group by a Bassian group.

###### Proof.

As noted above, each $p$-primary component of $G$ is of the form
$T_{p}(G)=F_{p}\oplus E_{p}$, where $F_{p}$ is a finite $p$-group and $E_{p}$
is an elementary $p$-group. Furthermore, $r_{0}(G)$ is finite. Let $E$ be the
direct sum over the various primes $p$ of the elementary groups $E_{p}$, and
let $F$ be the corresponding direct sum of the finite $p$-groups $F_{p}$; thus
the torsion subgroup $T$ of $G$ has the form $T=E\oplus F$.

Consider now the group $G/E$: we have an exact sequence

$0\to T/E\to G/E\to G/T\to 0.$

Since $T/E\cong\bigoplus_{p}F_{p}$ and each $F_{p}$ is a finite $p$-group, we
get that $T/E$ is a torsion Bassian group. Furthermore, $G/T$ is torsion-free
of finite rank, and hence is Bassian. It follows from [2, Proposition 2.3]
that $G/E$ is Bassian. Thus, $G$ is an extension of the elementary group $E$
by the Bassian group $G/E$, as stated. ∎

In fact, Danchev and Keef have shown in [3] the more general fact that a
generalized Bassian group is actually a split extension of an elementary group
by a Bassian group. Their arguments are developed in the context of their deep
investigations into mixed Abelian groups with bounded $p$-torsion. We present
here another version of this result based on the results established in [2].

Recall that in [2], the class $\mathcal{P}$ of groups $G$ was considered,
where $G\in\mathcal{P}$ if there is an infinite set of primes $\Pi$ with
$T\leq G\leq\bar{T}$, where $T$ is the torsion subgroup of $G$ and $T$ is a
direct sum of $p$-primary bounded components $T_{p}$,
$T=\bigoplus_{p\in\Pi}T_{p}$ and $\bar{T}=\prod_{p\in\Pi}T_{p}$. In [2,
Corollary 3.14] a decomposition of a generalized Bassian group in
$\mathcal{P}$ was given: $G=A\oplus H$, where $A$ is torsion and each
$p$-primary component of $A$ is a direct sum of an elementary $p$-group and a
finite $p$-group, while $H$ is a Bassian group. An examination of the proof
shows that the generalized Bassian property od $G$ was used only for two
purposes (i) to show that $r_{0}(G)$ was finite and (ii) that $A$ had the
desired decomposition with primary components of the form “elementary plus
finite”. Thus, in fact, an extended version of [2, Corollary 3.14] holds as
follows.

######  Proposition 3.2.

Suppose that $G\in\mathcal{P}$ with $r_{0}(G)$ finite, and each $p$-primary
component $T_{p}$ of $G$ has the form $T_{p}=E_{p}\oplus S_{p}$, where $E_{p}$
is an elementary $p$-group and $S_{p}$ is a finite $p$-group. Then $G$ has a
decomposition $G=A\oplus H$ with $A$ torsion and having primary components of
the form $E^{\prime}_{p}\oplus F_{p}$ with $E^{\prime}_{p}$ elementary and
$F_{p}$ finite, and $H$ Bassian.

Now, consider an arbitrary genuinely mixed group $G$ which is generalized
Bassian. Then, certainly each primary component of $G$ has the form
“elementary plus finite” , so that the torsion subgroup of $G$ has the form
$T=E\oplus S$, with $E$ elementary and $S$ a direct sum over primes $p$ of
finite $p$-groups. Therefore, we have an exact sequence

$0\to T\to G\to G/T\leq D\to 0,$

where $D$ is a finite dimensional $\mathbb{Q}$-space. Applying Proposition
24.6 in [4] (see also Exercise (6) in [6]), we obtain a group $G^{\prime}$
with

$0\to T\to G^{\prime}\to D\to 0;$

note that $G^{\prime}\in\mathcal{P}$ and has the same torsion subgroup as $G$.
Thus, Proposition 3.2 enables us to get a decomposition $G^{\prime}=A\oplus H$
with $A$ torsion and $H$ Bassian. Furthermore, $A$ has the form $A=E\oplus F$
with $E$ elementary and $F$ is a direct sum over primes $p$ of finite
$p$-groups. Absorbing $F$ into $H$, we write that $G^{\prime}=E\oplus
H^{\prime}$, where $H^{\prime}=F\oplus H$. Since $H$ is Bassian and each
primary component of $F$ is finite, it now follows easily that $H^{\prime}$ is
Bassian, as required.

Finally, observe that as $E\leq T$, modularity gives us that

$G=G\cap(E\oplus H^{\prime})=E\oplus(G\cap H^{\prime});$

since Bassian groups are hereditarily Bassian - see, for example, the comment
in the Introduction to [2] \- it follows that $H_{0}=G\cap H^{\prime}$ is
Bassian, so that $G=E\oplus H_{0}$ with $E$ elementary and $H_{0}$ Bassian.

Thus, we have established the following statement from [3].

######  Theorem 3.3.

If $G$ is a genuine mixed group which is generalized Bassian, then $G$ has a
decomposition of the form $G=E\oplus H$ with $E$ elementary and $H$ Bassian.

It follows from the classification of torsion, torsion-free and mixed
generalized Bassian groups - see, for instance, Proposition 2.8, Theorem 2.10
and Proposition 2.11 in [2] \- that an “elementary plus Bassian” decomposition
also holds in each of these three cases. So, we derive the following
consequence.

######  Corollary 3.4.

If $G$ is an arbitrary generalized Bassian group, then $G$ has a decomposition
$G=E\oplus H$, where $E$ is elementary and $H$ is Bassian.

We return now to consideration of groups which are nearly generalized Bassian.
Such groups are, of course, of finite torsion-free rank. (We remark that this
does not require the full force of the result of Danchev and Keef mentioned in
the first bullet point above: if a nearly generalized Bassian group is not of
finite torsion-free rank then it will have a free subgroup of infinite rank,
but such a group is definitely not generalized Bassian by [2, Example 2.7].)
In addition, if $G$ is a nearly generalized Bassian group, then we have:

$\bullet$ If $X$ is a proper torsion subgroup of $G$, then $X$ is generalized
Bassian and so $X$ is of the form $X=\bigoplus X_{p}$, where each $X_{p}$ is a
$p$-group of the form $X_{p}=E_{p}\oplus F_{p}$ with $E_{p}$ elementary and
$F_{p}$ finite.

Our first step in classifying groups which are nearly generalized Bassian is
to deal with divisible groups.

######  Proposition 3.5.

Let $G$ be a divisible group which is nearly generalized Bassian. Then, the
following three points are true:

(i) $G$ is either torsion or torsion-free;

(ii) if $G$ is torsion, then $G\cong\mathbb{Z}(p^{\infty})$ for some fixed but
arbitrary prime $p$;

(iii) if $G$ is torsion-free, then $G$ is a finite dimensional
$\mathbb{Q}$-vector space and $G$ is generalized Bassian (and even Bassian).

###### Proof.

(i) If $G$ is mixed and divisible, then it must have a proper subgroup which
is isomorphic to $\mathbb{Z}(p^{\infty})$ for some prime $p$. Since
$\mathbb{Z}(p^{\infty})$ is not generalized Bassian, no such group exists.

(ii) If $G$ is torsion divisible of rank greater than one, then $G$ has a
proper quasi-cyclic subgroup which would then be generalized Bassian. As
observed above, this is impossible, so $G$ has rank one and thus is isomorphic
to $\mathbb{Z}(p^{\infty})$ for some prime $p$.

(iii) If $G$ is torsion-free divisible, then, as observed above, $G$ must be
of finite rank, as required. ∎

It follows from an easy application of the above Proposition 3.5 that if $G$
is a nearly generalized Bassian group which is neither divisible nor reduced,
then $G$ has the form $G=\bigoplus_{n}\mathbb{Q}\oplus G^{\prime}$, where
$0\neq n$ is finite and $G^{\prime}$ is a reduced nearly generalized Bassian
group.

Our next result clarifies the situation for reduced nearly generalized Bassian
groups which are either torsion, torsion-free or splitting mixed,
respectively.

######  Proposition 3.6.

Let $G$ be a reduced group which is nearly Bassian. Then if $G$ is either
torsion, torsion-free or splitting mixed, then $G$ is generalized Bassian.

###### Proof.

Since finite groups are always generalized Bassian (and even Bassian), we may
assume that $G$ is infinite. Furthermore, we have observed above that
$r_{0}(G)$ is finite, so if $G$ is torsion-free it is a reduced group of
finite rank which is then certainly generalized Bassian.

If $G$ is a $p$-group, then we see that $G$ has a direct decomposition
$G=\mathbb{Z}(p^{n})\oplus H$ for some finite $n$. But then, by hypothesis,
$H$ is a generalized Bassian $p$-group and hence has the form $H=E\oplus F$,
where $E$ is an elementary $p$-group and $F$ is finite. But then $G$, itself,
has a similar decomposition and hence, with the aid of [2, Theorem 2.10], $G$
is generalized Bassian. If, however, $G$ is torsion, then it follows easily
using the primary decomposition of $G$ and the result just established for
$p$-groups that $G$ is generalized Bassian, as wanted.

Finally, if $G$ is splitting mixed, both its torsion and torsion-free parts
are reduced generalized Bassian and so $G$, itself, is then generalized
Bassian by virtue of [2, Corollary 2.12], as desired. ∎

Combining the results obtained above and using the classification results in
[2], it follows readily that we obtain:

######  Proposition 3.7.

Let $G$ be a nearly generalized Bassian group which is either torsion,
torsion-free or splitting mixed, then either $G\cong\mathbb{Z}(p^{\infty})$
for some prime $p$, or $G$ is generalized Bassian.

To complete the classification of nearly generalized Bassian groups, it
remains to consider the situation in which $G$ is nearly generalized Bassian
and genuinely mixed. For the remainder of this section we shall assume $G$ is
of this type.

The following observations will be useful in dealing with the genuinely mixed
nearly generalized Bassian groups.

$\bullet$ If $\varphi:G\to G/N$ is an injection then, since $r_{0}(G)$ is
finite, we have $r_{0}(G)\leq r_{0}(G/N)=r_{0}(G)-r_{0}(N)$, so that
$r_{0}(N)=0$ or, equivalently that $N\leq T$.

$\bullet$ If $\varphi$ embeds $G\to G/N$, then we have that $\varphi(T)\leq
T(G/N)=T/N$ since $N\leq T$. Furthermore, as we are assuming $G$ is genuinely
mixed, $T$ itself is generalized Bassian, and it follows that $N$ is a direct
summand of $T$, so that if $N=\bigoplus N_{p}$, then each primary component
$N_{p}$ of $N$ is a direct summand of the corresponding $T_{p}$.

$\bullet$ Since $\varphi(T)\leq T/N$, the full invariance of the primary
components gives that $\varphi(T_{p})\leq T_{p}/N_{p}$ and so
$\varphi(pT_{p})\leq p(T_{p}/N_{p})=(pT_{p}+N_{p})/N_{p}$. Hence, as
$pT_{p}=pF_{p}$, we have that

$pF_{p}=\varphi(pT_{p})\leq(pF_{p}+N_{p})/N_{p}\cong pF_{p}/(N_{p}\cap pF).$

However, $pF_{p}$ is finite so that $N_{p}\cap pF_{p}=\\{0\\}$, and it now
follows from Lemma 3.8 below that $N_{p}$ is necessarily elementary.

We are now ready to prove the promised lemma applied above.

######  Lemma 3.8.

Suppose that $A,B,C$ are $p$-groups with $A\leq B\oplus C$. If $B$ is
elementary and $A\cap pC=\\{0\\}$, then $A$ is elementary.

###### Proof.

Let $x\in A$, so that $x=b+c$ with $b\in B,c\in C$. Then $px=pb+pc=pc\in A\cap
pC=\\{0\\}$. ∎

######  Remark 3.9.

We can actually say a bit more. When we work with a single prime $p$, then we
can modify the $p$-primary component so that in the decomposition $E_{p}\oplus
F_{p}$ that $F_{p}$ has no summands of order $p$. The resulting enlarged
elementary group will continue to be a direct summand of $G$. In this
situation, we can conclude that $N_{p}\leq E_{p}$: if $N_{p}$ is not contained
in $E_{p}$, then $N_{p}\cap F_{p}$ is a non-zero summand of the elementary
group $N_{p}$ and hence a summand of $T_{p}$. This, however, is impossible
since if $\langle x\rangle$ is a summand of $N_{p}$, then the height
$ht_{T_{p}}(x)$ is necessarily $0$, but any $x\in N_{p}\cap F_{p}$ must be of
the form $x=py$, because $F_{p}$ has no summands of order $p$. In fact, it is
easy to see that this procedure will work for any finite set of primes. Thus,
in the genuinely mixed situation, we may assume that $N_{p}$ is actually a
summand of $T_{p}$.

We, thus, now come to the following key statement.

######  Proposition 3.10.

Let $G$ be a genuinely mixed group which is nearly generalized Bassian. Then,
$G$ is generalized Bassian.

###### Proof.

Let us notice the crucial fact that, if $\varphi:G\to G/N$ is an embedding for
some $N\neq 0$, then from our observations above we can find a decomposition
$G/N=(T_{p}/N_{p})\oplus(G^{\prime}/N^{\prime})$, where
$N^{\prime}=\bigoplus\limits_{q\neq p}N_{q}$ and $G=T_{p}\oplus G^{\prime}$.

So, we are now in the following situation: $\varphi$ is an embedding of $G$
into $G/N$ and we can decompose $G=T_{p}\oplus G^{\prime}$ with
$T_{p}\neq\\{0\\}$ and $N=N_{p}\oplus N^{\prime}$, where
$N^{\prime}=\bigoplus\limits_{q\neq p}N_{q}$. Then,
$G/N=(T_{p}/N_{p})\oplus(G^{\prime}/N^{\prime})$. Now, let $\pi$ be the
canonical projection of $G/N$ onto $G^{\prime}/N^{\prime}$ and set
$\psi=\pi\varphi$. Consider the restriction $\psi^{\prime}=\psi\upharpoonright
G^{\prime}$, a mapping from $G^{\prime}\to G^{\prime}/N^{\prime}$. We claim
that $\psi^{\prime}$ is monic; if not, then there is a nonzero $x\in
G^{\prime}$ such that $\psi^{\prime}(x)=0$, so that $\varphi(x)\in{\rm
Ker}\pi=T_{p}/N_{p}$, which quotient is a $p$-group. However, if $x$ is a
torsion element of $G^{\prime}$, then its order is relatively prime to $p$.
Since $\varphi$ is monic, it preserves orders, so that $x=0$, a contradiction.
Thus, $x$ must be an element of infinite order, which maps to an element of
order $p^{n}$ for some finite $n$. This, too, is impossible, so
$\psi^{\prime}$ is monic, as claimed.

The mapping $\pi\varphi:G^{\prime}\to G^{\prime}/N^{\prime}$ is an embedding
and, by the hypothesis posed on $G$, we know that $G^{\prime}$ is generalized
Bassian, whence $N^{\prime}$ is a direct summand of $G$. It now follows
immediately that $N$ is then a direct summand of $G$, since, as observed
above, $N_{p}$ is a direct summand of $T_{p}$. Consequently, $G$ is also
generalized Bassian, as asserted. ∎

Summarizing the results obtained above, we have established the following
curious assertion that is our main motivating result.

######  Theorem 3.11.

If $G$ is an arbitrary nearly generalized Bassian group, then either $G$ is a
quasi-cyclic group or $G$ is generalized Bassian.

We close our work in this section with the following two difficult questions
(compare also with the queries from [3]).

Problem 2. If the Bassian group $B$ has no direct summands isomorphic to
$\mathbb{Z}(p)$ for each prime $p$, is then the direct sum $E\oplus B$ a
generalized Bassian group, whenever $E$ is an elementary group?

We would like to notice that in [3] was established that the direct sum of an
elementary group and a Bassian group such that every its elementary summand is
finite is a generalized Bassian group.

Problem 3. Does it follow that the direct sum of a Bassian group and a
generalized Bassian group remains a generalized Bassian group?

##  4\. Concluding Remarks

Two obvious questions arise: what can be said about the hereditarily and super
generalized Bassian groups (see Problem 1 quoted above)? A partial answer is
easily given in relation to the first question.

If $G$ is generalized Bassian and either torsion, torsion-free or splitting
mixed, then every subgroup of $G$ is likewise generalized Bassian. By
utilizing the ‘elementary plus Bassian’ decomposition of Danchev-Keef given in
Theorem 3.3 in Section 3, it is clear that any subgroup of a generalized
Bassian group is again the direct sum of an elementary group and a Bassian
group. However, it is not clear, and it seems to be an extremely difficult
question, as to how to show that such a group is generalized Bassian. However,
several examples are given in [3] of sufficient conditions to ensure the
generalized Bassian property holds, but no a complete result is known yet; see
Conjecture 1.3 in [3].

Funding: The scientific work of the first-named author (P.V. Danchev) was
supported in part by the Bulgarian National Science Fund under Grant KP-06 No.
32/1 of December 07, 2019, as well as by the Junta de Andalucía under Grant
FQM 264, and by the BIDEB 2221 of TÜBÍTAK.

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