E. I. Kompantseva, A. A. Tuganbaev

Multiplication Groups
of Abelian Torsion-Free Groups of Finite Rank

Abstract. For an Abelian group $G$, any homomorphism $\mu\colon G\otimes
G\rightarrow G$ is called a multiplication on $G$. The set $\text{Mult}\,G$ of
all multiplications on an Abelian group $G$ itself is an Abelian group with
respect to addition; the group is called the multiplication group of $G$. Let
$\mathcal{A}_{0}$ be the class of all reduced block-rigid almost completely
decomposable groups of ring type with cyclic regulator quotient. In this
paper, for groups $G\in\mathcal{A}_{0}$, we describe groups $\text{Mult}\,G$.
We prove that for $G\in\mathcal{A}_{0}$, the group $\text{Mult}\,G$ also
belongs to the class $\mathcal{A}_{0}$. For any group $G\in\mathcal{A}_{0}$,
we describe the rank, the regulator, the regulator index, invariants of near-
isomorphism, a main decomposition, and a standard representation of the group
$\text{Mult}\,G$.

Key words. Abelian group, almost completely decomposable Abelian group, ring
on an Abelian group, multiplication group of an Abelian group.

MSC2020 datebase: 20K30, 20K99, 16B99

##  1 Introduction

For an Abelian group $G$, a multiplication on $G$ is a homomorphism $\mu\colon
G\otimes G\rightarrow G$. The set $\text{Mult}\,G$ of all multiplications on
the group $G$ itself is an Abelian group with respect to addition; the group
is called the multiplication group of $G$ or the group of multiplications on
$G$ [9]. An Abelian group $G$ with multiplication on $G$ is called a ring on
the group $G$. The problem of studying the relationship between the structure
of an Abelian group and the properties of ring structures on it is very
multifaceted and has a long history in algebra; see[1], [2], [6], [7], [10],
[11], [13], [14].

In this paper, we consider only additively written Abelian groups and "a
group" means "an Abelian group" in what follows.

In this paper, we study the group $\text{Mult}\,G$ for an almost completely
decomposable Abelian group $G$. A torsion-free group $G$ of finite rank is
called an almost completely decomposable group ($ACD$-group) if $G$ contains
completely decomposable subgroup of finite index. $ACD$-groups were studied in
[3], [4], [5], [12], [18], [19] and other papers. The achieved level of
$ACD$-group theory development is recorded in the book [19].

Any $ACD$-group $G$ contains a special uniquely defined completely
decomposable (see [9]) subgroup $\text{Reg}\,G$ of finite index which is a
fully invariant subgroup of $G$; it is called the regulator of the group $G$.
The regulator of an $ACD$-group can be defined as the intersection of all its
completely decomposable subgroups of lowest index [4]. The factor group
$G/\text{Reg}\,G$ is called the regulator quotient of the group $G$; the index
of the subgroup $\text{Reg}\,G$ in the group $G$ is called a regulator index.
It is denoted by $n(G)$. $ACD$-groups with cyclic regulator quotient are often
called $CRQ$-groups.

Let $G$ be an almost completely decomposable group. Then the group
$\text{Reg}\,G$ can be uniquely, up to isomorphism, represented as a direct
sum of torsion-free groups of rank 1 [8, Proposition 86.1]. For every type
$\tau$, we denote by $\text{Reg}_{\tau}\,G$ the sum of summands of rank 1 and
type $\tau$ in this decomposition of the group $\text{Reg}\,G$. The set of
types

$T(G)=T(\text{Reg}\,G)=\\{\tau\,|\,\text{Reg}_{\tau}\,G\neq 0\\}$

is called the set of critical types of groups $G$ and $\text{Reg}\,G$. If
$T(G)$ consists of pairwise incomparable types, then the groups $G$ and
$\text{Reg}\,G$ are called block-rigid groups. If, in addition, for any
$\tau\in T(G)$, the group $\text{Reg}_{\tau}G$ is of rank $1$, then $G$ and
$\text{Reg}\,G$ are called rigid groups. If all types in $T(G)$ are idempotent
types, then $G$ is called a group of ring type.

We note that a block-rigid $ACD$-group is either divisible or reduced. For a
divisible torsion-free group $G$, the group $\text{Mult}\,G$ is described in
[8, Section 121]; therefore, we consider only reduced groups in what follows.

We denote by $\mathcal{A}_{0}$ the class of all reduced block-rigid
$CRQ$-groups of ring type. In Section 2, we describe the group
$\text{Mult}\,G$ for $G\in\mathcal{A}_{0}$ (Theorem 2.8). The aim of Section 3
is, for groups $G$ in the class $\mathcal{A}_{0}$, to study properties of
$\text{Mult}\,G$. It is proved (Theorem 3.4) that if $G$ is a block-rigid
$CRQ$-group of ring type, then $\text{Mult}\,G$ also is a block-rigid
$CRQ$-group of ring type. We describe the rank, the regulator, the regulator
index, invariants of near-isomorphism, a main decomposition and a standard
representation of the group $\text{Mult}\,G$ for $G\in\mathcal{A}_{0}$.

The multiplication $\mu\colon G\otimes G\rightarrow G$ is often denoted by the
symbol $\times$, i.e.,

$\mu(g_{1}\otimes g_{2})=g_{1}\times g_{2}\text{ for all }g_{1},g_{2}\in G.$

The multiplication $\times$ on the group $G$ induces a ring on this group
which is denoted by $(G,\times)$. Let $G$ be a group and $g\in G$. The
characteristic and the order of the element $g$ are denoted by $\chi(g)$ and
$o(g)$, respectively. The rank and the divisible hull of the group $G$ are
denoted by $r(G)$ and $\tilde{G}$, respectively. If $S\subseteq G$, then $|S|$
is the cardinality of the set $S$ and $\langle S\rangle$ is the subgroup of
the group $G$ generated by the set $S$. We write an element of a group direct
product $\prod_{i\in I}G_{i}$ in the form $(g_{i})_{i\in I}$, where $g_{i}\in
G_{i}$. If $I_{1}\subseteq I$, then for simplicity, we identify the subgroup
$\\{(g_{i})_{i\in I}\in\prod_{i\in I}G_{i}\,|\,g_{i}=0$ for all $i\notin
I_{1}\\}$ of the group $\prod_{i\in I}G_{i}$ with the group $\prod_{i\in
I_{1}}G_{i}$; we write elements of this group in the form $(g_{i})_{i\in
I_{1}}$.

As usual, $\mathbb{N}$ and $\mathbb{P}$ are the sets of positive integers and
all prime integers, respectively, $\mathbb{Z}$ is the group (the ring) of
integers, $\mathbb{Q}$ is the group (the field) of rational numbers. If $R$ is
a unital ring, then $Re$ is the cyclic module over $R$ generated by the
element $e$. If $S$ is a finite subset in $\mathbb{Z}$, then $\text{gcd}(S)$
is the greatest common divisor of all integers in $S$ and $\text{lcm}(S)$ is
the least common multiple of the integers in $S$. If
$P_{1}\subseteq\mathbb{P}$, then $P_{1}$-integer is an integer such that any
prime divisor of it (if it exists) is contained in $P_{1}$. It follows from
the definition that $1$ is a $P_{1}$-integer for any
$P_{1}\subseteq\mathbb{P}$. For any type $\tau$, we set

$P_{\infty}(\tau)=\\{p\in\mathbb{P}\,|\,\tau(p)=\infty\\},\quad
P_{0}(\tau)=\mathbb{P}\setminus P_{\infty}(\tau).$

Unless otherwise stated, we use notation and definitions from [8], [9] and
[17].

##  2 Multiplication Groups of Block-Rigid $CRQ$-groups of Ring Type

All over this section, $G$ is a reduced block-rigid $CRQ$-group of ring type
with regulator $A$, regulator quotient $G/A=\langle d+A\rangle$ where $d\in
G$, regulator index $n$ and set of critical types $T(G)=T(A)$.

By setting $\text{Reg}_{\tau}\,G=A_{\tau}$, we can represent the group $A$ in
the form $A=\oplus_{\tau\in T(G)}A_{\tau}$. According to [19, Proposition
2.4.11], such decomposition of the completely decomposable group $A$ is unique
if and only if $A$ is a block-rigid group. For divisible hulls $\tilde{G}$,
$\tilde{A}$, $\tilde{A}_{\tau}$ of the groups $G$, $A$ and $A_{\tau}$,
respectively, we have relations

$\tilde{G}=\tilde{A}=\oplus_{\tau\in T(G)}\tilde{A}_{\tau}.$

For $\tau\in T(G)$, we denote by $\pi_{\tau}$ the natural projection from the
group $\tilde{G}$ onto $\tilde{A}_{\tau}$.

In [5], positive integers $m_{\tau}=m_{\tau}(G)$ ($\tau\in T(G)$) are defined;
these integers are invariants of near-isomorphism of the group $G$. We can
define integers $m_{\tau}$ ($\tau\in T(G)$ as follows; we take an element
$d\in G/A$ such that $\langle d+A\rangle=G/A$. Let
$d_{\tau}=\pi_{\tau}(d)\in\tilde{A}_{\tau}$ and let $m_{\tau}=o(d_{\tau}+A)$
be the order of the element $d_{\tau}+A$ in the torsion group $\tilde{A}/A$.
In [5], it is shown that integers $m_{\tau}$ ($\tau\in T(G)$) do not depend on
the choice of the element $d$. We note that
$n=o(d+A)=\text{lcm}\\{m_{\tau}\,|\,\tau\in T(G)\\}$.

Remark 2.1.
Let $T$ be a finite set of pair-wise incomparable types and let
$\\{m_{\tau}\,|\,\tau\in T\\}$ be some set of positive integers. We say that
the set $\\{m_{\tau}\,|\,\tau\in T\\}$ satisfies condition $(m)$ if for any
$p\in\mathbb{P}$, $k\in\mathbb{N}$, $\tau\in T$, we have that $p^{k}$ divides
$m_{\sigma}$ for some $\sigma\in T\setminus\\{\tau\\}$ provided $p^{k}$
divides $m_{\tau}$. We note that the set $\\{m_{\tau}\,|\,\tau\in T\\}$
satisfies condition $(m)$ if and only if the set $\\{m_{\tau}\,|\,\tau\in
T,\,m_{\tau}>1\\}$ satisfies condition $(m)$.

According to [19, Theorem 13.1.2], the set $\\{m_{\tau}\,|\,\tau\in T\\}$ is a
system of invariants of a near-isomorphism of some block-rigid $CRQ$-group $G$
with $T(G)=T$ if and only if this set satisfies condition $(m)$ and $m_{\tau}$
are $P_{0}(\tau)$-integers for all $\tau\in T$. $\rhd$

In [3, Theorem 3.5], it is proved that for any of the group
$G\in\mathcal{A}_{0}$, there exists a direct decomposition

$G=G_{1}\oplus C,$ $None$

where $C$ is a completely decomposable group and $G_{1}$ is a rigid
$CRQ$-group which satisfies the following conditions:

$\tau\in T(G_{1})\text{ if and only if }m_{\tau}(G)>1,$ $None$
$m_{\tau}(G_{1})=m_{\tau}(G)\text{ for all }\tau\in T(G_{1}).$ $None$

Decomposition $(1)$, which satisfies conditions $(1^{\prime})$ and
$(1^{\prime\prime})$, is called a main decomposition of the group $G$. In a
main decomposition of the group $G$, the group $G_{1}$ does not contain a
completely decomposable summand; such groups are said to be clipped. We note
that a main decomposition of a $CRQ$-group is not uniquely defined, since it
depends on the choice of the element $d$ participating in the definition of
the group. In what follows, we assume that a main decomposition of the group
$G$ is fixed. We set $T_{0}(G)=\\{\tau\in T(G)\,|\,m_{\tau}>1\\}$. Then
$T_{0}(G)$ is the set of critical types of a clipped direct summand in any
main decomposition of the group $G$.

Let $B$ be the regulator of the group $G_{1}$, then $T(G_{1})=T(B)=T_{0}(G)$
and $\tilde{G}_{1}=\tilde{B}$. There exists a system
$E_{0}=\\{e_{0}^{(\tau)}\in B_{\tau}\,|\,\tau\in T(B)\\}$ such that

$B=\oplus_{\tau\in T(B)}R_{\tau}e_{0}^{(\tau)}.$ $None$

In $(2)$, we assume that $R_{\tau}$ is a unitary subring in $\mathbb{Q}$, the
type of the additive group of $R_{\tau}$ is equal to $\tau$, characteristics
$\chi(e_{0}^{(\tau)})\in\tau$ contain only zeros and symbols $\infty$
($\tau\in T(B)$).

Let $D=\\{d\in G_{1}\,|\,G/A=\langle d+A\rangle\\}$, it is easy to see that
$D\neq\varnothing$. Let $d\in D$. In the group $\tilde{B}$, the element $d$
can be represented in the form $d=\sum_{\tau\in
T(B)}\dfrac{s_{\tau}}{r_{\tau}}e_{0}^{(\tau)}$, where $s_{\tau}\in\mathbb{Z}$,
$r_{\tau}\in\mathbb{N}$, $\text{gcd}(s_{\tau},r_{\tau})=1$. Without loss of
generality, we can assume that $s_{\tau}$, $r_{\tau}$ are
$P_{0}(\tau)$-integers (otherwise, we can replace the system $E_{0}$).

Let $\tau\in T(B)$. By the definition of the integer $m_{\tau}$, the relation
$o\left(\dfrac{s_{\tau}}{r_{\tau}}e_{0}^{(\tau)}+A\right)=m_{\tau}$ holds in
the group $\tilde{A}/A$. Since $r_{\tau}$ is a $P_{0}(\tau)$-integer and
$\text{gcd}(s_{\tau},r_{\tau})=1$, we have $r_{\tau}=m_{\tau}$. Consequently,
the element $d$ of $\tilde{B}$ is of the form

$d=\sum_{\tau\in T(B)}\dfrac{s_{\tau}}{m_{\tau}}e_{0}^{(\tau)},$ $None$

and the integers $n$, $m_{\tau}$ and $s_{\tau}$ satisfy the following
conditions:

$n=\text{lcm}\\{m_{\tau}\,|\,\tau\in T(B)\\}$, $(3^{\prime})$

$\text{gcd}(s_{\tau},m_{\tau})=1$ for all $\tau\in T(B)$, $(3^{\prime\prime})$

$s_{\tau}$ and $m_{\tau}$ are $P_{0}(\tau)$ numbers for any $\tau\in T(B)$.
$(3^{\prime\prime\prime})$

A system $E_{0}=\\{e_{0}^{(\tau)}\in B_{\tau}\,|\,\tau\in T(B)\\}$ which
satisfies conditions $(2)$ and $(3)$, is called a $B$-basis of the group $G$
defined by the element $d$. We note that the pair $(d,E_{0})$ uniquely defines
the numbers $s_{\tau}$ ($\tau\in T(B)$). Relation $(3)$ is called a standard
representation of block-rigid $CRQ$-group $G$ related to the pair $(d,E_{0})$.

Remark 2.2.
We note that a $B$-basis $E_{0}$ can be defined by more than one element $d\in
D$.

$\lhd$ Indeed, let we have a standard representation $(3)$ of the group $G$.
Let $\gamma$ be an integer which is co-prime with the regulator index $n$ and
$d_{1}=\gamma d\in G_{1}$. Then $G/A=\langle d+A\rangle=\langle
d_{1}+A\rangle$, i.e., $d_{1}\in D$. In addition,

$d_{1}=\sum_{\tau\in T(B)}\dfrac{\gamma s_{\tau}}{m_{\tau}}e_{0}^{(\tau)}.$
$None$

If $\gamma$ is a $P_{0}(\tau)$-integer, then the relation $(4)$ is a standard
representation of the group $G$. Consequently, the $B$-basis $E_{0}$ is
defined by each of elements $d$ and $d_{1}$.

We note that if $\gamma$ is not a $P_{0}(\tau)$-integer, then the relation
$(4)$ is not a standard representation of the group $G$. $\rhd$

For a $B$-basis $E_{0}$, we set

$D(E_{0})=\\{d\in D\,|\,B\text{-basis }E_{0}\text{ is determined by the
element }d\\}.$

It follows from the definition of the $B$-basis that $D(E_{0})\neq\varnothing$
and it follows from Remark 2.2 that $D(E_{0})$ can contain more than one
element.

With the right choice of elements $e_{i}^{(\tau)}\in C_{\tau}$
($i=0,1,\ldots,k_{\tau}$), the group $C$ can be written in the form

$C=\oplus_{\tau\in T(C)}C_{\tau}=\oplus_{\tau\in
T(C)}\oplus_{i=1,\ldots,k_{\tau}}R_{\tau}e_{i}^{(\tau)},$

where $R_{\tau}$ is a unitary subring in the field of rational numbers, the
type of the additive group of $R_{\tau}$ is equal to $\tau$, and
characteristics $\chi(e_{i}^{(\tau)})\in\tau$ contain only zeros and symbols
$\infty$.

For $\tau\in T(G)$, we define the following sets:

$I_{\tau}(B)=\begin{cases}\\{0\\},\quad\text{ for }\tau\in T(B)\\\
\quad\emptyset,\quad\text{ for }\tau\notin T(B),\end{cases}$
$I_{\tau}(C)=\begin{cases}\\{1,\dots,k_{\tau}\\},\quad\text{ for }\tau\in
T(C),\,k_{\tau}\in\mathbb{N}\\\ \qquad\emptyset,\qquad\text{ for }\tau\notin
T(C),\end{cases}$ $I_{\tau}=I_{\tau}(B)\cup I_{\tau}(C).$

Then $A_{\tau}=\oplus_{i\in I_{\tau}}R_{\tau}e_{i}^{(\tau)}$ for any $\tau\in
T(G)$ and

$A=\oplus_{\tau\in T(G)}\oplus_{\tau\in I_{\tau}}R_{\tau}e_{i}^{(\tau)}.$
$None$

A system $E=\\{e_{i}^{(\tau)}\in A_{\tau}\,|\,\tau\in T(G),i\in I_{\tau}\\}$
is called an $A$-basis of the group $G$ if $E$ satisfies $(5)$ and the
subsystem $E_{0}=\\{e_{0}^{(\tau)}\in A_{\tau}\,|\,\tau\in T(B)\\}$ of $E$ is
a $B$-basis.

Let $(G,\times)$ be a ring on the group $G\in\mathcal{A}_{0}$. Since $A$ is a
fully invariant subgroup of the group $G$, we have that $A$ is an ideal of the
ring $(G,\times)$ which is a direct sum of ideals $A_{\tau}$ ($\tau\in T(G)$).
Thus, every multiplication on $G$ induces a multiplication on $A$; therefore,
$\text{Mult}\,G\subseteq\text{Mult}\,A$; however, the converse is not true.

Let $E=\\{e_{i}^{(\tau)}\in A_{\tau}\,|\,\tau\in T(G),i\in I_{\tau}\\}$ be an
$A$-basis of the group $G$. Then for any set $\\{u_{ij}^{(\tau)}\in
A_{\tau}\,|\,\tau\in T(G),\;i,j\in I_{\tau}\\}$, there exists a unique ring
$(A,\times)$ such that $e_{i}^{(\tau)}\times e_{j}^{(\tau)}=u_{ij}^{(\tau)}$
for all $\tau\in T(G)$ and $i,j\in I_{\tau}$. The multiplication $\times$ is
uniquely extended to a multiplication on $\tilde{A}=\tilde{G}$, where it is
defined as follows:

$\sum_{i\in I_{\tau}}r_{i}e_{i}^{(\tau)}\times\sum_{i\in
I_{\tau}}r^{\prime}_{i}e_{i}^{(\tau)}=\sum_{i,j\in
I_{\tau}}r_{i}r^{\prime}_{j}(e_{i}^{(\tau)}\times e_{j}^{(\tau)})$ $None$

for all $\tau\in T(G)$, $r_{i},r^{\prime}_{j}\in\mathbb{Q}$; and
$\tilde{A}_{\tau}\times\tilde{A}_{\sigma}=0$ for $\tau\neq\sigma$. However,
$G$ is not necessarily a subring of the ring $(\tilde{A},\times)$. We say that
the set $\\{u_{ij}^{(\tau)}\in A_{\tau}\,|\,\tau\in T(G),\;i,j\in I_{\tau}\\}$
defines a multiplication on $G$ with respect to the $A$-basis $E$ if there
exists a ring $(G,\times)$ such that $e_{i}^{(\tau)}\times
e_{j}^{(\tau)}=u_{ij}^{(\tau)}$ for all $\tau\in T(G)$ and $i,j\in I_{\tau}$.
We note that any set $\\{u_{ij}^{(\tau)}\in A_{\tau}\,|\,\tau\in T(G),\;i,j\in
I_{\tau}\\}$ defines multiplication on group $G$ with respect to the $A$-basis
$E$ at most one way.

Let $G\in\mathcal{A}_{0}$. To describe sets defining multiplications on the
group $G$, we define the following groups.

For any $\tau\in T(G)$, let $n_{\tau}=|I_{\tau}|$ and let
$M_{\tau}^{(0)}=M_{n_{\tau}}(A_{\tau})$ be the additive group of square
matrices of order $n_{\tau}$ with elements in $A_{\tau}$,

$M_{\tau}^{(1)}=\begin{bmatrix}m_{\tau}A_{\tau}&m_{\tau}A_{\tau}&\ldots&m_{\tau}A_{\tau}\\\
m_{\tau}A_{\tau}&A_{\tau}&\ldots&A_{\tau}\\\ \ldots&\ldots&\ldots&\ldots\\\
m_{\tau}A_{\tau}&A_{\tau}&\ldots&A_{\tau}\end{bmatrix},$

where the symbol $[\ldots]$ means the set of matrices of certain form,
$m_{\tau}$ are invariants of near-isomorphism of the group $G$,

$M_{\tau}^{(2)}=\begin{bmatrix}m_{\tau}^{2}A_{\tau}&m_{\tau}A_{\tau}&\ldots&m_{\tau}A_{\tau}\\\
m_{\tau}A_{\tau}&A_{\tau}&\ldots&A_{\tau}\\\ \ldots&\ldots&\ldots&\ldots\\\
m_{\tau}A_{\tau}&A_{\tau}&\ldots&A_{\tau}\end{bmatrix}\subseteq
M_{\tau}^{(1)}.$

We set

$M^{(0)}=\prod_{\tau\in T(G)}M_{\tau}^{(0)},\;M^{(1)}=\prod_{\tau\in
T(G)}M_{\tau}^{(1)},\;M^{(2)}=\prod_{\tau\in T(G)}M_{\tau}^{(2)}.$

Then $M^{(2)}\subseteq M^{(1)}\subseteq M^{(0)}$ and $M^{(2)}\cong
M^{(1)}\cong M^{(0)}\cong\text{Mult}\,A$.

For the standard representation $(3)$ of the group $G$ related to the pair
$(d,E_{0})$ and for every $\tau\in T(G)$, we consider elements

$X^{(\tau)}=X^{(\tau)}(d,E_{0})=\begin{pmatrix}m_{\tau}s_{\tau}^{-1}e_{0}^{(\tau)}&0&\ldots&0\\\
0&0&\ldots&0\\\ \ldots&\ldots&\ldots&\ldots\\\ 0&0&\ldots&0\end{pmatrix}\in
M_{\tau}^{(1)},\;\text{if }\tau\in T(B),$

where $s_{\tau}^{-1}$ is an integer which is inverse to $s_{\tau}$ modulo
$m_{\tau}$,

$X^{(\tau)}=\begin{pmatrix}0&0&\ldots&0\\\ 0&0&\ldots&0\\\
\ldots&\ldots&\ldots&\ldots\\\ 0&0&\ldots&0\end{pmatrix}\in
M_{\tau}^{(1)},\;\text{if }\tau\notin T(B).$

We set

$X=X(d,E_{0})=\left(X^{(\tau)}\right)_{\tau\in
T(G)}=\left(X^{(\tau)}\right)_{\tau\in T(B)}\in M^{(1)}.$ $M(d,E_{0})=\langle
X,M^{(2)}\rangle\subseteq M^{(1)}.$

We note that integral solutions of the congruence $s_{\tau}x\equiv
1\,(\text{mod}\,m_{\tau})$ form the residue class modulo $m_{\tau}$.
Therefore, the set $M(d,E_{0})$ does not depend on the choice of the integers
$s_{\tau}^{-1}$ in the definition of $X$.

We also note that if $\tau\in T(C)\setminus T(B)$, then $m_{\tau}=1$ by
$(1^{\prime})$. Therefore,

$M_{\tau}^{(2)}=M_{\tau}^{(1)}=M_{\tau}^{(0)}$

in this case. For every set $U=\\{u_{ij}^{(\tau)}\in A_{\tau}\,|\,\tau\in
T(G),\;i,j\in I_{\tau}\\}$ and every $\tau\in T(G)$, we consider the matrix

$U^{(\tau)}=\begin{pmatrix}u_{i_{1},i_{1}}^{(\tau)}&u_{i_{1},i_{2}}^{(\tau)}&\ldots&u_{i_{1},i_{n_{\tau}}}^{(\tau)}\\\
u_{i_{2},i_{1}}^{(\tau)}&u_{i_{2},i_{2}}^{(\tau)}&\ldots&u_{i_{2},i_{n_{\tau}}}^{(\tau)}\\\
\ldots&\ldots&\ldots&\ldots\\\
u_{i_{n_{\tau}},i_{1}}^{(\tau)}&u_{i_{n_{\tau}},i_{2}}^{(\tau)}&\ldots&u_{i_{n_{\tau}},i_{n_{\tau}}}^{(\tau)}\end{pmatrix}\;\in
M_{\tau}^{(0)},$

where $i_{k}\in I_{\tau}$, $i_{1}<i_{2}<\ldots<i_{n_{\tau}}$. We set
$\overline{U}=\left(U^{(\tau)}\right)_{\tau\in T(G)}\in M^{(0)}$.

Remark 2.3. Let $G\in\mathcal{A}_{0}$, $G=G_{1}\oplus C$ be a main
decomposition of the group $G$, and let $\text{Reg}\,G=A$,
$\text{Reg}\,G_{1}=B$. In [12], it is proved that for any multiplication
$\times$ on group $G$, we have

$B_{\tau}\times A\subseteq m_{\tau}A_{\tau}\,\text{ and }\,A\times
B_{\tau}\subseteq m_{\tau}A_{\tau}\,\text{ for all }\,\tau\in T(B).\;\rhd$

Theorem 2.4. Let $G$ be a block-rigid $CRQ$-group of ring type with $A$-basis
$E$ containing an $B$-basis $E_{0}$. Let $U=\\{u_{ij}^{(\tau)}\in
A_{\tau}\,|\,\tau\in T(G),\;i,j\in I_{\tau}\\}$. Then the following conditions
are equivalent.

  1. 1)

$U$ defines a multiplication on $G$ with respect to the $A$-basis $E$.

  2. 2)

$\overline{U}\in M(d,E_{0})$ for any $d\in D(E_{0})$.

  3. 3)

$\overline{U}\in M(d,E_{0})$ for some $d\in D(E_{0})$.

$\lhd$ 1) $\Rightarrow$ 2). Let the set $U=\\{u_{ij}^{(\tau)}\in
A_{\tau}\,|\,\tau\in T(G),\;i,j\in I_{\tau}\\}$ induce the multiplication
$\times$ on $G$ with respect to the $A$-basis $E=\\{e_{i}^{(\tau)}\in
A_{\tau}\,|\,\tau\in T(G),\;i\in I_{\tau}\\}$. It follows from Remark 2.3 that
$u_{0i}^{(\tau)},u_{i0}^{(\tau)}\in m_{\tau}A_{\tau}$ and
$u_{00}^{(\tau)}=m_{\tau}v_{00}^{(\tau)}$ ($v_{00}^{(\tau)}\in A_{\tau}$) for
all $\tau\in T(B)$, $i\in I_{\tau}$.

Let $d\in D(E_{0})$ and let a standard representation related to the pair
$(d,E_{0})$ be of the form

$d=\sum_{\tau\in T(B)}\dfrac{s_{\tau}}{m_{\tau}}e_{0}^{(\tau)}.$

Since $d\times d\in G$, we have that $d\times d=\alpha d+a$ for some
$\alpha\in\mathbb{Z}$ and $a\in A$. Then

$d\times d=\sum_{\tau\in T(B)}\dfrac{\alpha
s_{\tau}}{m_{\tau}}e_{0}^{(\tau)}+a.$ $None$

On the other hand, it follows from $(6)$ that

$d\times d=\sum_{\tau\in
T(B)}\dfrac{s_{\tau}}{m_{\tau}}e_{0}^{(\tau)}\times\sum_{\tau\in
T(B)}\dfrac{s_{\tau}}{m_{\tau}}e_{0}^{(\tau)}=\sum_{\tau\in
T(B)}\dfrac{s_{\tau}^{2}}{m_{\tau}^{2}}u_{00}^{(\tau)}=\sum_{\tau\in
T(B)}\dfrac{s_{\tau}^{2}}{m_{\tau}}v_{00}^{(\tau)}.$ $None$

Let $\tau\in T(B)$. It follows from $(7)$ and $(8)$ that

$\pi_{\tau}(d\times
d)=\dfrac{s_{\tau}^{2}}{m_{\tau}}v_{00}^{(\tau)}=\dfrac{\alpha
s_{\tau}}{m_{\tau}}e_{0}^{(\tau)}+a_{\tau},\;\text{where}\;a_{\tau}=\pi_{\tau}(a)\in
A_{\tau}.$

Consequently, $s_{\tau}^{2}v_{00}^{(\tau)}=\alpha
s_{\tau}e_{0}^{(\tau)}+m_{\tau}a_{\tau}$; therefore, $v_{00}^{(\tau)}=\alpha
s_{\tau}^{-1}e_{0}^{(\tau)}+m_{\tau}a_{\tau}^{\prime}$ for some
$a_{\tau}^{\prime}\in A_{\tau}$, where $s_{\tau}^{-1}$ is an integer which is
inverse to $s_{\tau}$ modulo $m_{\tau}$. Therefore,

$u_{00}^{(\tau)}=m_{\tau}v_{00}^{(\tau)}=\alpha
m_{\tau}s_{\tau}^{-1}e_{0}^{(\tau)}+m_{\tau}^{2}a_{\tau}^{\prime}.$

Consequently, $\overline{U}\in\alpha X+M^{(2)}\subseteq M(d,E_{0})$.

2) $\Rightarrow$ 3). The implication is directly verified.

3) $\Rightarrow$ 1). Let $\overline{U}\in M(d,E_{0})$ for some $d\in D(E_{0})$
and let a standard representation related to $(d,E_{0})$ be of the form
$d=\sum_{\tau\in T(B)}\dfrac{s_{\tau}}{m_{\tau}}e_{0}^{(\tau)}$. Then
$\overline{U}=\alpha X+Y$ for some $\alpha\in\mathbb{Z}$, $Y\in M^{(2)}$.
Therefore, for all $\tau\in T(B)$ and $i\in I_{\tau}$, we have

$u_{0i}^{(\tau)}=m_{\tau}v_{0i}^{(\tau)},\quad
u_{i0}^{(\tau)}=m_{\tau}v_{i0}^{(\tau)},\quad u_{00}^{(\tau)}=\alpha
m_{\tau}s_{\tau}^{-1}e_{0}^{(\tau)}+m_{\tau}^{2}a_{\tau},$

where $v_{0i}^{(\tau)},v_{i0}^{(\tau)},a_{\tau}\in A_{\tau}$ and the integers
$s_{\tau}^{-1}$ satisfy conditions $s_{\tau}s_{\tau}^{-1}=1+m_{\tau}x_{\tau}$
for some $x_{\tau}\in\mathbb{Z}$.

There exists a ring $(A,\times)$ such that $e_{i}^{(\tau)}\times
e_{j}^{(\tau)}=u_{ij}^{(\tau)}$ for all $\tau\in T(G)$, $i,j\in I_{\tau}$; and
$A_{\tau}\times A_{\sigma}=0$ for $\tau\neq\sigma$. This multiplication is
extended to a multiplication on divisible hull $\tilde{A}=\tilde{G}$ of the
group $A$. We prove that $G$ is a subring of the ring $(\tilde{A},\times)$.

It follows from $(6)$ that

$d\times d=\sum_{\tau\in
T(B)}\dfrac{s_{\tau}^{2}}{m_{\tau}^{2}}u_{00}^{(\tau)}=\alpha\sum_{\tau\in
T(B)}\dfrac{s_{\tau}^{2}s_{\tau}^{-1}}{m_{\tau}}e_{0}^{(\tau)}+\sum_{\tau\in
T(B)}s_{\tau}^{2}a_{\tau}=$ $=\alpha\sum_{\tau\in
T(B)}\dfrac{s_{\tau}}{m_{\tau}}(1+m_{\tau}x_{\tau})e_{0}^{(\tau)}+\sum_{\tau\in
T(B)}s_{\tau}^{2}a_{\tau}=$ $=\alpha\sum_{\tau\in
T(B)}\dfrac{s_{\tau}}{m_{\tau}}e_{0}^{(\tau)}+\sum_{\tau\in T(B)}(\alpha
s_{\tau}x_{\tau})e_{0}^{(\tau)}+\sum_{\tau\in T(B)}s_{\tau}^{2}a_{\tau}=$
$=\alpha d+\sum_{\tau\in T(B)}(\alpha
s_{\tau}x_{\tau}e_{0}^{(\tau)}+s_{\tau}^{2}a_{\tau})\in G.$

In addition, if $\sigma\in T(B)$ and $i\in I_{\sigma}$, then

$d\times e_{i}^{(\sigma)}=\left(\sum_{\tau\in
T(B)}\dfrac{s_{\tau}}{m_{\tau}}e_{0}^{(\tau)}\right)\times
e_{i}^{(\sigma)}=\dfrac{s_{\sigma}}{m_{\sigma}}u_{0i}^{(\sigma)}=\dfrac{s_{\sigma}}{m_{\sigma}}m_{\sigma}v_{0i}^{(\sigma)}=s_{\sigma}v_{0i}^{(\sigma)}\in
A.$

Similarly, we have $e_{i}^{(\sigma)}\times d\in A$.

If $\sigma\notin T(B)$, then $d\times e_{i}^{(\sigma)}=e_{i}^{(\sigma)}\times
d=0$ for any $i\in I_{\sigma}$. Since $G=\langle d,A\rangle$, we have that $G$
is a subring of the ring $(\tilde{A},\times)$. Therefore, the set $U$ defines
a multiplication on $G$. $\rhd$

It follows from Theorem 2.4 that the group $M(d,E_{0})$ does not depend on the
choice of the element $d\in D(E_{0})$. In the following assertion, we consider
relations between elements of groups $M(d_{1},E_{0})$ and $M(d_{2},E_{0})$ for
$d_{1},d_{2}\in D(E_{0})$. We note that if $d_{1},d_{2}\in D$, then $\langle
d_{1}+A\rangle=\langle d_{2}+A\rangle=G/A$; therefore, $d_{1}=\gamma d_{2}+b$
for some $b\in B$ and some integer $\gamma$ which is co-prime with $n(G)$.

Proposition 2.5. Let $G$ be a group in $\mathcal{A}_{0}$ with main
decomposition $(1)$ and with a $B$-basis $E_{0}$ and regulator index $n$. Then
the following assertions hold.

  1. 1)

$M(d_{1},E_{0})=M(d_{2},E_{0})$ for any $d_{1},d_{2}\in D(E_{0})$.

  2. 2)

If $d_{1},d_{2}\in D(E_{0})$ and $d_{1}=\gamma d_{2}+b$, where
$\gamma\in\mathbb{Z}$, $\text{lcd}(\gamma,n)=1$, $b\in B$,
$X_{1}=X(d_{1},E_{0})\in M(d_{1},E_{0})$, $X_{2}=X(d_{2},E_{0})\in
M(d_{2},E_{0})$, then $X_{1}+M^{(2)}=\gamma^{-1}X_{2}+M^{(2)}$, where
$\gamma^{-1}$ is an integer which is inverse to $\gamma$ modulo $n$.

$\lhd$. Let $d_{1},d_{2}\in D(E_{0})$, $d_{1}=\gamma d_{2}+b$, where
$\gamma\in\mathbb{Z}$, $\text{lcd}\,(\gamma,n)=1$ and $b=\sum_{\tau\in
T(B)}b_{\tau}e_{0}^{(\tau)}$ ($b_{\tau}\in R_{\tau}$). Let

$d_{1}=\sum_{\tau\in T(B)}\dfrac{s_{\tau}}{m_{\tau}}e_{0}^{(\tau)},\quad
d_{2}=\sum_{\tau\in T(B)}\dfrac{t_{\tau}}{m_{\tau}}e_{0}^{(\tau)}$

be standard representations of the group $G$ related to $(d_{1},E_{0})$ and
$(d_{2},E_{0})$, respectively. In the divisible hull $\tilde{G}$ of the group
$G$, we have relations

$\sum_{\tau\in T(B)}\dfrac{s_{\tau}}{m_{\tau}}e_{0}^{(\tau)}=d_{1}=\gamma
d_{2}+b=\sum_{\tau\in T(B)}\dfrac{\gamma
t_{\tau}+m_{\tau}b_{\tau}}{m_{\tau}}e_{0}^{(\tau)}.$

Therefore,

$s_{\tau}=\gamma t_{\tau}+m_{\tau}b_{\tau}\;\text{for all}\;\tau\in T(B).$
$None$

Let $\tau\in T(B)$, $\gamma^{-1}$ be an integer which is inverse to $\gamma$
modulo $n$, and let $t_{\tau}^{-1}$ be an integer which is inverse to
$t_{\tau}$ modulo $m_{\tau}$. Then number $\gamma^{-1}$ is inverse to $\gamma$
modulo $m_{\tau}$ by $(3^{\prime})$; therefore, the integer
$\gamma^{-1}t_{\tau}^{-1}$ is inverse to $s_{\tau}$ modulo $m_{\tau}$ by
$(9)$.

Let $a_{\tau}\in A_{\tau}$. Then

$s_{\tau}^{-1}m_{\tau}e_{0}^{(\tau)}+m_{\tau}^{2}a_{\tau}=\gamma^{-1}t_{\tau}^{-1}m_{\tau}e_{0}^{(\tau)}+m_{\tau}^{2}a_{\tau}^{\prime},\;\text{where}\;a_{\tau}^{\prime}\in
A_{\tau}.$

Consequently,

$X_{1}+M^{(2)}\subseteq\gamma^{-1}X_{2}+M^{(2)}.$ $None$

Since $d_{2}=\gamma^{-1}d_{1}+b_{1}^{\prime}$ (where $b_{1}^{\prime}\in B$),
we have

$\gamma^{-1}X_{2}+M^{(2)}\subseteq\gamma^{-1}(\gamma
X_{1}+M^{(2)})+M^{(2)}=X_{1}+M^{(2)}$

by $(10)$. $\rhd$

It follows from Proposition 2.5 that we can write $M(d,E_{0})=M(E_{0})$,
however we will often write $M(d,E_{0})$ if we want to point which a standard
representation is used in the definition of the group $M(d,E_{0})$.

Remark 2.6.
Let $\overline{U}\in M^{(2)}$ for the set $U=\\{u_{ij}^{(\tau)}\in
A_{\tau}\,|\,\tau\in T(G),\;i,j\in I_{\tau}\\}$. It follows from Theorem 2.4
that, with respect to any $A$-basis $E$ of the group $G$, the set $U$ defines
a multiplication $\times_{U,E}$ on $G$ such that $G\times_{U,E}G\subseteq A$.
Such a multiplication is called a regulator multiplication. $\rhd$

Example 2.7. It is possible that the set $U=\\{u_{ij}^{(\tau)}\,|\,\tau\in
T(G),\;i,j\in I_{\tau}\\}$ defines a multiplication on $G$ with respect to one
$A$-basis and does not define any multiplication on $G$ with respect to
another $A$-basis even for the same main decomposition. Moreover, the
following situation is possible: there exist two $A$-bases $E$ and $F$ such
that any set $U$, which defines non-regulator multiplication with respect to
$E$, does not define any multiplication with respect to $F$. This means that

$M(E_{0})\cap M(F_{0})=M^{(2)}.$

$\lhd$ Let $s_{1}$, $s_{2}$ be two co-prime integers, $s_{1}>1$, $s_{2}>1$,
and let $m$ be a prime integer which does not divide any of the integers
$s_{1}$, $s_{2}$, $s_{1}^{2}-s_{2}^{2}$.

Let $\tau_{i}$ be an idempotent type such that $P_{0}(\tau_{i})$ is the set of
all prime divisors of the integers $m$ and $s_{i}$, respectively ($i=1,2$).
Then types $\tau_{1}$ and $\tau_{2}$ are incomparable.

We consider a group $B=R_{1}e_{1}\oplus R_{2}e_{2}$, where $R_{1}$ and $R_{2}$
are unital subrings in the field $\mathbb{Q}$ whose additive groups are of
types $\tau_{1}$ and $\tau_{2}$, respectively.

It follows from Remark 2.1 that there exists a $CRQ$-group $G=\langle
d,B\rangle$ with regulator $B$ and quasi-isomorphism invariants
$m_{\tau_{1}}=m_{\tau_{2}}=m$. We can choose the group $G$ in such a way that
a standard representation of $G$ is of the form

$d=\dfrac{s_{1}}{m}e_{1}+\dfrac{s_{2}}{m}e_{2}.$

Then the system $E_{0}=\\{e_{1},e_{2}\\}$ is a $B$-basis of the group $G$
defined by the element $d$ (in this case, the $B$-basis coincides with an
$A$-basis).

We set

$B_{1}=B_{\tau_{1}}=R_{1}e_{1},\;B_{2}=B_{\tau_{2}}=R_{2}e_{2}.$

We consider $d_{1}=d+(e_{1}+e_{2})\in G$. Then $d_{1}\in D$; in $\tilde{G}$,
we have

$d_{1}=\dfrac{s_{1}+m}{m}e_{1}+\dfrac{s_{2}+m}{m}e_{2}.$ $None$

Since $s_{i}+m$ is co-prime with each of the integers $s_{i}$ and $m$, we have
that $s_{i}+m$ is a $P_{\infty}(\tau_{i})$-integer for $i=1,2$. Consequently,
$(11)$ is not a standard representation of the group $G$. We set
$f_{1}=(s_{1}+m)e_{1}$ and $f_{2}=(s_{2}+m)e_{2}$. Then the system
$F_{0}=\\{f_{1},f_{2}\\}$ is a $B$-basis defined by the element $d_{1}$. The
standard representation of the group $G$, related to the pair $(d_{1},F_{0})$,
is of the form $d_{1}=\dfrac{1}{m}f_{1}+\dfrac{1}{m}f_{2}$.

Let the set $U=\\{u_{i}\in B_{i}\,|\,i=1,2\\}$ define a non-regulator
multiplication on $G$ with respect to of the $B$-basis
$E_{0}=\\{e_{1},e_{2}\\}$. By Theorem 2.4, we have $(u_{1},u_{2})\in
M(d,E_{0})\setminus M^{(2)}$. Consequently,

$u_{1}\in\alpha ms_{1}^{-1}e_{1}+m^{2}B_{1},\quad u_{2}\in\alpha
ms_{2}^{-1}e_{2}+m^{2}B_{2},$ $None$

for some integer $\alpha$ which is not divided by the prime integer $m$.

We assume that the set $U$ defines multiplication with respect to the
$B$-basis $F_{0}=\\{f_{1},f_{2}\\}$. Then it follows from Theorem 2.4 that for
some $\beta\in\mathbb{Z}$, we have

$u_{1}\in\beta mf_{1}+m^{2}B_{1},\quad u_{2}\in\beta mf_{2}+m^{2}B_{2}.$
$None$

It follows from $(12)$ and $(13)$ that

$\alpha ms_{1}^{-1}e_{1}\in\beta mf_{1}+m^{2}B_{1}=\beta
ms_{1}e_{1}+m^{2}B_{1},$ $\alpha ms_{2}^{-1}e_{2}\in\beta
mf_{2}+m^{2}B_{2}=\beta ms_{2}e_{2}+m^{2}B_{2}.$

Therefore,

$\alpha=\beta s_{1}^{2}+mx_{1},\quad\alpha=\beta s_{2}^{2}+mx_{2}$ $None$

for some $x_{1}\in R_{1}$, $x_{2}\in R_{2}$. Since $x_{i}=\dfrac{\alpha-\beta
s_{i}^{2}}{m}\in R_{i}$ and $m$ is a $P_{0}(\tau_{i})$-integer, we have
$x_{i}\in\mathbb{Z}$ for $i=1,2$. Consequently, it follows from $(14)$ that
$\beta(s_{1}^{2}-s_{2}^{2})=my$ for $y=x_{2}-x_{1}\in\mathbb{Z}$. Since the
prime integer $m$ does not divide $s_{1}^{2}-s_{2}^{2}$, we have that $m$
divides $\beta$; therefore, $m$ divides $\alpha$ by $(14)$. This contradicts
to the property that $(u_{1},u_{2})\notin M^{(2)}$. Consequently, the set
$U=\\{u_{1},u_{2}\\}$ does not define any multiplication with respect to the
$B$-basis $F_{0}$. $\rhd$

Let $\times$ be a multiplication on a group $G\in\mathcal{A}_{0}$. Let
$E=\\{e_{i}^{(\tau)}\in A_{\tau}\,|\,\tau\in T(G),\,i\in I_{\tau}\\}$ be an
$A$-basis of the group $G$ and let

$U_{\times}=U_{\times}(E)=\\{u_{ij}^{(\tau)}=e_{i}^{(\tau)}\times
e_{j}^{(\tau)}\in A_{\tau}\,|\,\tau\in T(G),\;i,j\in I_{\tau}\\}.$

It clearly follows from Theorem 2.4 that the correspondence
$\times\mapsto\overline{U_{\times}}$ defines an isomorphism from the group
$\text{Mult}\,G$ onto $M(E_{0})$.

Theorem 2.8. If $G\in\mathcal{A}_{0}$ and $E_{0}$ is a $B$-basis of the group
$G$, then $\text{Mult}\,G\cong M(E_{0})$. $\rhd$

We note that Theorem 2.8 implies the following property: up to isomorphism,
the group $M(E_{0})$ does not depend on the choice of the $B$-basis $E_{0}$.

Remark 2.9.
Let $G=\langle d,A\rangle\in\mathcal{A}_{0}$ and let $E$ be an $A$-basis of
the group $G$ containing the $B$-basis $E_{0}$.

1. It follows from Theorem 2.8 that the group $\text{Mult}\,G$ can be identified with the group $M(E_{0})=\langle X,M^{(2)}\rangle$ and the multiplication $\times$ can be identified with $\overline{U}_{\times}\in M(E_{0})$.

2. Let $\overline{U}_{\times}=\overline{U}_{\times}(E)\in\text{Mult}\,G=M(d,E_{0})=\langle,M^{(2)}\rangle$, $\alpha\in\mathbb{Z}$. It follows from the proof of Theorem 2.4 that $\overline{U}_{\times}\in\alpha X+M^{(2)}$ if and only if $d\times d\in\alpha d+A$.

3. It follows from 2 that $\overline{U}_{\times}\in M^{(2)}$ if and only if $G\times G\subseteq A$. In the group $\text{Mult}\,G$, this means that the subgroup $\text{Hom}(G\otimes G,A)$ of all regulator multiplications coincides with the group $M^{(2)}$. $\rhd$

##  3 Properties of Multiplication Groups of Block-Rigid $CRQ$-Groups of Ring
Type

The purpose of this section is to show that for any group $G$ in the class
$\mathcal{A}_{0}$, the group $\text{Mult}\,G$ belongs to this class, as well.
We will also describe the rank, the set of critical types, invariants of near-
isomorphism, the regulator, a main decomposition, and a standard
representation of the group $\text{Mult}\,G$, where $G\in\mathcal{A}_{0}$.

Remark 3.1.
Let $A$ be a completely decomposable block-rigid group of finite rank and
$G=\langle d,A\rangle$, where $d\in\tilde{A}$. Let we have
$o(d_{\tau}+A)=m_{\tau}$ in the group $\tilde{A}/A$, where
$d_{\tau}=\pi_{\tau}(d)$ for $\tau\in T(A)$. Then the set
$\\{m_{\tau}\,|\,\tau\in T(A)\\}$ satisfies condition $(m)$ (see Remark 2.1)
if and only if for any $\tau\in T(A)$, the subgroup $A_{\tau}$ is pure in $G$.

$\lhd$ Indeed, let the set $\\{m_{\tau}\,|\,\tau\in T(A)\\}$ satisfy condition
$(m)$, $\sigma\in T(A)$, $a\in A_{\sigma}$ and $a=k(td+x)$ for some
$k,t\in\mathbb{Z}$ and $x\in A$. Let $\tau\neq\sigma$, then

$ktd_{\tau}+kx_{\tau}=0,\;\text{ where }\,x_{\tau}=\pi_{\tau}(x)\in A_{\tau}.$

Therefore, $td_{\tau}\in A_{\tau}$, whence $m_{\tau}$ divides $t$. Since the
set $\\{m_{\tau}\,|\,\tau\in T(A)\\}$ satisfies condition $(m)$, we have that

$\text{lcm}\\{m_{\tau}\,|\,\tau\in
T(A),\tau\neq\sigma\\}=\text{lcm}\\{m_{\tau}\,|\,\tau\in T(A)\\}=n.$

Consequently, $n$ divides $t$; therefore, $td+x\in A$. Since the type of the
element $td+x$ is equal to $\sigma$, we have that $td+x\in A_{\sigma}$.

Conversely, let the subgroup $A_{\tau}$ be pure in $G$ for any $\tau\in T(A)$.
We assume that the set $\\{m_{\tau}\,|\,\tau\in T(A)\\}$ does not satisfy
condition $(m)$. Then there exists a type $\sigma\in T(A)$ such that
$m_{\sigma}$ does not divide $n_{1}=\text{lcm}\\{m_{\tau}\,|\,\tau\in
T(A),\tau\neq\sigma\\}$. Consequently, for
$n=\text{lcm}\\{m_{\tau}\,|\,\tau\in T(A)\\}$, it is true that $n=n_{1}n_{2}$
for some integer $n_{2}>1$. Consequently, $n_{1}d_{\sigma}\notin A_{\sigma}$
and $n_{2}(n_{1}d_{\sigma})\in A_{\sigma}$. Therefore, the subgroup
$A_{\sigma}$ is not pure in $G$. $\rhd$

Remark 3.2.
Let $A$ be a reduced block-rigid (resp., rigid) completely decomposable group
of finite rank and let $G=\langle d,A\rangle$, where $d\in\tilde{A}\setminus
A$. Then $A$ is a subgroup of finite index of the group $G$, and $T(G)=T(A)$.
Therefore, $G$ is a block-rigid (resp., rigid) $ACD$-group by the definition.
We note that if $A$ is a group of ring type, then $G$ also is a group of ring
type. Since $G/A=\langle d+A\rangle$ is a cyclic group, we have that $G$ is a
$CRQ$-group by [3, Section 2]. In addition, $A=\text{Reg}\,G$ if and only if
subgroup $A_{\tau}$ is pure in $G$ [3, Section 2] for any $\tau\in T(G)$.
$\rhd$

Let $G\in\mathcal{A}_{0}$. The following theorem describes properties of the
group $\text{Mult}\,G$. In what follows, $G\in\mathcal{A}_{0}$, $T(G)=T$,
$T_{0}(G)=T_{0}$, $m_{\tau}(G)=m_{\tau}$ $(\tau\in T)$, $n(G)=n$,
$G=G_{1}\oplus C$ is a main decomposition of the group $G$,
$\text{Reg}\,G_{1}=B=\oplus_{\tau\in T_{0}}R_{\tau}e_{0}^{(\tau)}$,
$\text{Reg}\,G=A=B\oplus C$, $G=\langle d,A\rangle$, and a standard
representation of the group $G$ is of the form

$d=\sum_{\tau\in T_{0}}\dfrac{s_{\tau}}{m_{\tau}}e_{0}^{(\tau)}.$

We note that the set of integral solutions of the congruence $s_{\tau}x\equiv
1(\text{mod }m_{\tau})$ always contains a $P_{0}(\tau)$-integer
$s_{\tau}^{\varphi(m_{\tau})-1}$, where $\varphi(x)$ is the Euler function.
Therefore, we always can take this $P_{0}(\tau)$-integer as the integer
$s_{\tau}^{-1}$ inverse to $s_{\tau}$ modulo $m_{\tau}$, .

Theorem 3.3. Let $G\in\mathcal{A}_{0}$. Then the group $\text{Mult}\,G$
satisfies the following conditions.

1. The group $\text{Mult}\,G$ is a block-rigid $CRQ$-group of ring type with regulator $M^{(2)}=\text{Hom}(G\otimes G,A)$.

2. $T(\text{Mult}\,G)=T(G)$ and $T_{0}(\text{Mult}\,G)=T_{0}(G)$, as a corollary.

3. $m_{\tau}(\text{Mult}\,G)=m_{\tau}(G)$ for any $\tau\in T(G)$, $n(\text{Mult}\,G)=n(G)$.

4. $r(\text{Reg}_{\tau}(\text{Mult}\,G))=(r(\text{Reg}_{\tau}G))^{3}$ for any $\tau\in T(G)$.

5. One of main decompositions of the group $\text{Mult}\,G$ is of the form $\text{Mult}\,G=M^{\prime}\oplus M^{\prime\prime}$, where

$M^{\prime}=\langle X,K\rangle,\;K=\prod_{\tau\in T_{0}(G)}K_{\tau},\quad
K_{\tau}=\begin{bmatrix}m_{\tau}^{2}B_{\tau}&0&\ldots&0\\\ 0&0&\ldots&0\\\
\ldots&\ldots&\ldots&\ldots\\\ 0&0&\ldots&0\end{bmatrix}\subseteq
M_{\tau}^{(2)},$ $X=\left(X^{(\tau)}\right)_{\tau\in T_{0}(G)},\qquad
X^{(\tau)}=\begin{pmatrix}m_{\tau}s_{\tau}^{-1}e_{0}^{(\tau)}&0&\ldots&0\\\
0&0&\ldots&0\\\ \ldots&\ldots&\ldots&\ldots\\\ 0&0&\ldots&0\end{pmatrix}\in
M_{\tau}^{(1)}\text{ for }\tau\in T_{0}(G),$ $M^{\prime\prime}=\prod_{\tau\in
T(G)}M_{\tau}^{\prime\prime},\qquad
M_{\tau}^{\prime\prime}=\begin{bmatrix}m_{\tau}^{2}C_{\tau}&m_{\tau}A_{\tau}&\ldots&m_{\tau}A_{\tau}\\\
m_{\tau}A_{\tau}&A_{\tau}&\ldots&A_{\tau}\\\ \ldots&\ldots&\ldots&\ldots\\\
m_{\tau}A_{\tau}&A_{\tau}&\ldots&A_{\tau}\end{bmatrix}\,\subseteq
M_{\tau}^{(2)}.$

In addition, $T(M^{\prime})=T_{0}(G)$, $\text{Reg}\,M^{\prime}=K$.

6. For every $\tau\in T_{0}(G)$, we denote by $s_{\tau}^{-1}$ a $P_{0}(\tau)$-integer which is inverse to $s_{\tau}$ modulo $m_{\tau}$,

$E_{0}^{(\tau)}=\begin{pmatrix}m_{\tau}^{2}e_{0}^{(\tau)}&0&\ldots&0\\\
0&0&\ldots&0\\\ \ldots&\ldots&\ldots&\ldots\\\ 0&0&\ldots&0\end{pmatrix}\in
K_{\tau}.$

Then the system $\\{E_{0}^{(\tau)}\,|\,\tau\in T_{0}(G)\\}$ is one of
$B$-bases of the group $\text{Mult}\,G$. One of standard representations of
the group $\text{Mult}\,G$ is of the form

$X=\left(\dfrac{s_{\tau}^{-1}}{m_{\tau}}E_{0}^{(\tau)}\right)_{\tau\in
T_{0}(G)}.$

$\lhd$ 1. It follows from Theorem 2.8 that $\text{Mult}\,G=\langle
X,M^{(2)}\rangle$, where

$X=\left(X^{(\tau)}\right)_{\tau\in
T_{0}},\;X^{(\tau)}=\begin{pmatrix}m_{\tau}s_{\tau}^{-1}e_{0}^{(\tau)}&0&\ldots&0\\\
0&0&\ldots&0\\\ \ldots&\ldots&\ldots&\ldots\\\ 0&0&\ldots&0\end{pmatrix}\in
M^{(1)}\;\text{ for }\tau\in T_{0}.$

Since $\text{gcd}(s_{\tau}^{-1},m_{\tau})=1$ for $\tau\in T_{0}$, we have
$o(X_{\tau}+M^{(2)})=m_{\tau}$ in the group $\tilde{M}^{(2)}/M^{(2)}$.

By Remark 2.1, the set $\\{m_{\tau}\,|\,\tau\in T\\}$ satisfies condition
$(m)$. Therefore, it follows from Remark 3.1 that the subgroups
$M_{\tau}^{(2)}$ are pure in $\text{Mult}\,G$ for any $\tau\in T$. Since
$M^{(2)}$ is a block-rigid completely decomposable group of ring type,
$\text{Mult}\,G$ is a block-rigid $CRQ$-group of ring type with regulator
$M^{(2)}$ by Remark 3.2. It follows from Remark 2.9(3) that we have
$M^{(2)}=\text{Hom}(G\otimes G,A)$.

2. It follows from 1 and the definition of the group $M^{(2)}$ that

$T(\text{Mult}\,G)=T(M^{(2)})=T(G).$

3. We have $\text{Mult}\,G=\langle X,M^{(2)}\rangle$ and $\text{Reg}(\text{Mult}\,G)=M^{(2)}$ by 1. Let $\tau\in T$, then

$m_{\tau}(\text{Mult}\,G)=o(X_{\tau}+M^{(2)})=m_{\tau}=m_{\tau}(G).$

Therefore, $n(\text{Mult}\,G)=\text{lcm}\\{m_{\tau}\,|\,\tau\in T\\}=n(G)$.

4. Let $\tau\in T$. It follows from 1 that

$\text{Reg}_{\tau}(\text{Mult}\,G)=M_{\tau}^{(2)}=\begin{bmatrix}m_{\tau}^{2}A_{\tau}&m_{\tau}A_{\tau}&\ldots&m_{\tau}A_{\tau}\\\
m_{\tau}A_{\tau}&A_{\tau}&\ldots&A_{\tau}\\\ \ldots&\ldots&\ldots&\ldots\\\
m_{\tau}A_{\tau}&A_{\tau}&\ldots&A_{\tau}\end{bmatrix}\cong
M_{n_{\tau}}(A_{\tau}),$

where $n_{\tau}=r(A_{\tau})$. Consequently,
$r(\text{Reg}_{\tau}(\text{Mult}\,G))=n_{\tau}^{3}=(r(\text{Reg}_{\tau}\,G))^{3}$.

5. In the decomposition $\text{Mult}\,G\cong M^{\prime}\oplus M^{\prime\prime}$, the group $M^{\prime\prime}$ is completely decomposable and $M^{\prime}=\langle X,K\rangle$. By the definition of the group $K$, we have $T(K)=T(M^{\prime})=T_{0}$. It is easy to see that $o(X^{(\tau)}+K)=m_{\tau}$ in the group $\tilde{K}/K$, for any $\tau\in T_{0}$. Since $\\{m_{\tau}\,|\,\tau\in T_{0}\\}$ satisfies condition $(m)$ (by Remark 2.1) and $K$ is a rigid completely decomposable group, we have that $M^{\prime}$ is a rigid group in $\mathcal{A}_{0}$ with $\text{Reg}\,M^{\prime}=K$ by Remark 3.1 and Remark 3.2.

Since $T(M^{\prime})=T_{0}$, we have that $\tau\in T(M^{\prime})$ if and only
if $m_{\tau}(\text{Mult}\,G)=m_{\tau}>1$. In addition,

$m_{\tau}(M^{\prime})=o(X^{(\tau)}+K)=m_{\tau}=m_{\tau}(\text{Mult}\,G),$

by 3. It follows from $(1^{\prime})$, $(1^{\prime\prime})$ that the
decomposition $\text{Mult}\,G=M^{\prime}\oplus M^{\prime\prime}$ is a main
decomposition of the group $\text{Mult}\,G$.

6. Let $\tau\in T_{0}$, $s_{\tau}^{-1}$ be a $P_{0}(\tau)$-integer which is inverse to $s_{\tau}$ modulo $m_{\tau}$, and let

$E_{0}^{(\tau)}=\begin{pmatrix}m_{\tau}^{2}e_{0}^{(\tau)}&0&\ldots&0\\\
0&0&\ldots&0\\\ \ldots&\ldots&\ldots&\ldots\\\ 0&0&\ldots&0\end{pmatrix}\in
K_{\tau}.$

Then $K=\prod_{\tau\in T_{0}}R_{\tau}E_{0}^{(\tau)}$. In addition,

$X=\left(\dfrac{s_{\tau}^{-1}}{m_{\tau}}E_{0}^{(\tau)}\right)_{\tau\in
T_{0}}.$ $None$

Since $s_{\tau}^{-1}$ ($\tau\in T_{0}$) is a $P_{0}(\tau)$-integer, $(15)$ is
a standard representation of the group $\text{Mult}\,G$. Therefore,
$\\{E_{0}^{(\tau)}\,|\,\tau\in T_{0}(G)\\}$ is a $B$-basis of the group
$\text{Mult}\,G$. $\rhd$

##  4 Data Availability Statement

Our manuscript has no associate data.

## References

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