11institutetext: IBM Research, Haifa, Israel
11email<EMAIL_ADDRESS>
# Towards a Benchmark for Causal Business Process Reasoning with LLMs

Fabiana Fournier  Lior Limonad  Inna Skarbovsky

###### Abstract

Large Language Models (LLMs) are increasingly used for boosting organizational
efficiency and automating tasks. While not originally designed for complex
cognitive processes, recent efforts have further extended to employ LLMs in
activities such as reasoning, planning, and decision-making. In business
processes, such abilities could be invaluable for leveraging on the massive
corpora LLMs have been trained on for gaining deep understanding of such
processes. In this work, we plant the seeds for the development of a benchmark
to assess the ability of LLMs to reason about causal and process perspectives
of business operations. We refer to this view as Causally-augmented Business
Processes (BPC). The core of the benchmark comprises a set of BPC related
situations, a set of questions about these situations, and a set of deductive
rules employed to systematically resolve the ground truth answers to these
questions. Also with the power of LLMs, the seed is then instantiated into a
larger-scale set of domain-specific situations and questions. Reasoning on BPC
is of crucial importance for process interventions and process improvement.
Our benchmark could be used in one of two possible modalities: testing the
performance of any target LLM and training an LLM to advance its capability to
reason about BPC.

###### Keywords:

Large Language Models Business Processes Causally-augmented Business Processes
Reasoning Benchmark

##  1 Introduction

Large Language Models (LLM) refers to statistical models of natural language
that, based on the large corpora of text data they have been trained on,
predict next plausible tokens (basic units of text) given an input string
[22]. LLMs have been successfully applied to a wide range of applications
including: Chatbots and virtual assistants (e.g., automated customer support);
content generation and automation (e.g., articles and blogs generation);
language translation; text summarization and document analysis; and question
answering.The interaction with LLMs is conventionally attained via a textual
prompt in which the content and the instructions to the LLM are being
constructed, also known as prompt engineering. The process is iterative, with
the model’s output being analyzed and the prompt adjusted accordingly. Prompt
engineering is key to the efficient use of LLMs [18]. However, LLM models can
only answer prompts accurately if they have been fed the right training data
as they lack planning and reasoning capabilities (e.g., [21, 11]). As stated
by Yann
LeCun111https://www.ft.com/content/23fab126-f1d3-4add-a457-207a25730ad9, chief
AI scientist of Facebook and Instagram, LLMs have “very limited understanding
of logic…do not understand the physical world, do not have persistent memory,
cannot reason in any reasonable definition of the term…”. In fact, ChatGPT has
a causal hallucination issue when tackling causal relationships which cannot
be overcome relying solely on prompts [8].

While reasoning may not be inherent in the architecture of LLMs, critics do
argue that the ability to reason does manifest itself as an emergent property
in LLMs [20]. That is, once trained on sufficiently large corpora of examples,
LLMs may become capable of statistically deriving statements that are also
coincidentally sound with the given set of arguments in the input. Similar
emergent abilities may also include planning, decision-making, in-context
learning, and answering in zero-shot settings [15]. Whether reasoning is an
inherent capability or an emergent property of future LLMs, particularly for
interpreting business process models, is likely to remain a heated debate.
This debate emphasizes the importance of establishing benchmarks for
evaluating LLM performance. Our focus is on driving such enablement by using
LLMs to analyze textual descriptions of Causally-augmented Business Processes
(BPCs). This refers to a specific type of business process where causal
execution dependencies among the activities are a first-class citizen of the
process descriptions. Teaching LLMs to reason about BPCs could enhance the
analysis and improvement of such processes.

To standardize and measure the ability of LLMs to facilitate such tasks, a
designated benchmark is being developed. The purpose of the developed novel
benchmark is twofold: as a testing dataset that can be employed to quantify
the ability of an LLM to reason about BPCs, and as a training dataset that can
be used to adapt the ability of an LLM for this task. Our developed prototype
is intended to gradually evolve into a full-scale benchmark. We evaluated two
open-source and three commercial LLMs on a small subset of the situations and
questions. Our results highlight the importance of producing an objective
numeric scale partitioned by different perspectives to compare and assess
various LLM performances. They also indicate that there is room for further
improvement of LLMs in the task of reasoning about BPCs. The evolving
benchmark and corresponding materials are available here:
https://github.com/IBM/SAX/tree/main/NLP4BPM2024.

Given these observations, we are cautious in claiming that LLMs possess a
genuine inherent ability to reason. We approach this by acknowledging that
while LLMs may not inherently reason, they can achieve predictive accuracy
through training on a large set of deductive textual statements. This accuracy
results in syntactic output that reliably aligns with what would be produced
by genuine logical reasoning. Whether reaching such a level of predictive
accuracy is philosophically equivalent to actually having the ability to
reason is something that we leave beyond the scope of our study.

##  2 Background

Business process management is the discipline that combines approaches for the
design, execution, control, measurement, and optimization of business
processes [1]. With the penetration of AI applications into organizations, the
concept of AI-Augmented Business Process Management Systems (ABPMSs) was
coined in [5]. ABPMSs are process-aware information systems that rely on
trustworthy AI technology to continuously adapt and improve a set of business
processes with respect to one or more performance indicators. In such systems,
subsymbolic AI methods are not used to replace human or symbolic reasoning in
crucial tasks, but rather to support human and machine decisions and actions
[12]. A natural way of driving this interaction between humans and AI is
through LLMs. To enable this, it is important for LLMs to support reasoning
about business processes. In analogy to [10], by reasoning we refer here to
the process of thinking about business processes in a logical and systematic
way, using evidence and past experiences to reach a conclusion or make a
decision, for the sake of process improvement. As shown in [7], understanding
the temporal dependencies among the tasks in the process is not enough for
reasoning about the consequences of interventions underlying process
improvement decisions. To this aim, we specify here the Causally-augmented
Business Process (BPC) formalism as the concrete flavor for process
descriptions in which not only the temporal flow but also the causal relations
among the tasks are inherently captured.

A BPC is a business process extended with inter-activity relations that
reflect causal execution dependencies. A causal execution dependence, denoted
as $A\xrightarrow{c}B$, implies that the time task $B$ executes is determined
by the time task $A$ executes in a given process as defined in [7]. From a
process perspective, we consider followed-by as a general form of a temporal
relation among activities, hereafter denoted as $A\rightarrow B$, implying
that according to most process observations (frequently evidenced in many
process execution logs), the execution of activity $B$ occurs either
immediately after the execution of $A$ or sometime after. Considering the
causal process perspective, a BPC may be represented in the basic form of a
graph, with nodes designating activities and edges as causal execution
dependencies among these activities. Three fundamental patterns, or
“junctions” [17], can be composed to characterize any causal network:
$A\xleftarrow{c}B\xrightarrow{c}C$ (confounder),
$A\xrightarrow{c}B\xleftarrow{c}C$ (collider), and
$A\xrightarrow{c}B\xrightarrow{c}C$ (mediator).

The potential of leveraging LLMs in the BPM field has been recently
researched, for example, by analyzing which opportunities and challenges LLMs
pose for the individual stages of the BPM lifecycle [19], in [4], where prompt
engineering techniques are discussed as an alternative to fine-tuning a
specific LLM, and in [12] where the authors introduce the notion of a Large
Process Model (LPM), an envisioned neuro-symbolic software system that
integrates process management knowledge in organizations with LLMs and
statistical and inference methods for the automated inference of insights and
actions. In addition, LLMs have been researched in a wide range of tasks
related to BPM, including process mining [9, 3], automation of portions of
complex tasks 222https://www.infoworld.com/article/3714621/how-llms-can-help-
streamline-business-processes.html, conversational process modelling [13], and
explainability of business process outcomes [6]. To date, neither there is
work on leveraging LLMs for causal reasoning about business processes, nor a
relevant benchmark for testing LLMs. As an early member of this family of
solutions, the PET dataset [2] presents a benchmark for question answering,
featuring an initial corpus of business process descriptions annotated with
activities, gateways, actors, and flow information. The dataset contains 45
documents with narrative descriptions of business processes and their
annotations. In [16] the TORQUE benchmark for temporal ordering has been
investigated. The dataset encompasses 21k user-generated and fully answered
temporal relation questions. Concerning both benchmarks, ours differs in
tackling the task of causal reasoning about business processes.

##  3 Approach

Our goal is to develop a question-and-answer benchmark dataset to test the
capacity of an LLM to produce sound answers to questions about textual
narratives describing BPCs. Regardless of the debate about LLMs’ suitability
for reasoning tasks, we assume that with “sufficient” training and exposure to
massive amounts of data, these models can become proficient. Yet, determining
sufficiency requires some objective performance measurement. Thus, a benchmark
is necessary. We believe the developed instrumentation and methodology, shown
in Figure 1, are suitable for model testing and can also serve as a dataset to
train LLMs to reason about BPC. The benchmark may also be further extended to
more concrete domains and particular aspects of reasoning as the example given
in this paper. We do acknowledge that the realization presented here is the
first step in our longer journey. Hence, in addition to presenting our results
so far, we also elaborate on our future evaluation intentions once the
benchmark is instantiated at a greater scale.

Figure 1: BPC benchmark for testing and training of LLMs

Our approach is to define a core set of template questions and situations for
basic reasoning about BPC textual narratives. These descriptions combine
statements about process activities, time precedence relations, and causal
execution dependencies. At the root of such descriptions, we anticipate
statements of the form “activity $A$ occurs before activity $B$”, and “the
execution of $A$ causes the execution of $B$”. From a process perspective,
textual descriptions typically include manifestations of the basic temporal
relation followed-by. While there is a formal distinction between directly-
follows and eventually-follows [1], natural language often does not strictly
differentiate between them. Hence, our work assumes a variety of
manifestations of the core phrase ‘occurs before/after’ to be expressed in the
populated text. Similarly, from a causal process perspective, the descriptions
in the text are likely to include manifestations of the causal relation of
causal-execution-dependence. Respectively, manifestations of the core phrase
of “causes the execution of” are populated in the text.

Therefore, to account for complete coverage of textual narratives that are
descriptive of any structural form of a BPC, we partition the set of template
situations into three subsets corresponding to each of the three fundamental
causal patterns. Respectively, from a process perspective, each pattern is
associated with either a congruent process structure or a simple execution
sequence of the form $A\rightarrow B\rightarrow C$ (as shown in [7]). The core
set of situations that can be composed to describe any real-world BPC
situation is shown in Table 1. A situation refers to the unified temporal and
causal relations among any subset of three activities $A$, $B$, and $C$. These
conditions provide the core content for generating domain-specific text based
on the respective template phrases listed.

Table 1: The domain of five different BPC situations as spanned by the three
causal patterns and corresponding process structures

| Confounder | Collider | Mediator
---|---|---|---
Situation$\\#$ | 1 | 2 | 3 | 4 | 5
| Causal
---
structure
$A\xleftarrow{C}B\xrightarrow{C}C$ | $A\xrightarrow{C}B\xleftarrow{C}C$ | $A\xrightarrow{C}B\xrightarrow{C}C$
| Process
---
structure
$A\leftarrow B\rightarrow C$ (split) |  | $B\rightarrow A\rightarrow C$
---
(or $B\rightarrow C\rightarrow A$)
$A\rightarrow B\leftarrow C$ (join) |  | $A\rightarrow C\rightarrow B$
---
(or $C\rightarrow A\rightarrow B$)
$A\rightarrow B\rightarrow C$
| Causal phrase
---
template
| B causes the execution of A,
---
B causes the execution of C
| A causes the execution of B,
---
C causes the execution of B
| A causes the execution of B,
---
B causes the execution of C
| Process phrase
---
template
| B occurs before A,
---
B occurs before C
| B occurs before A,
---
A occurs before C
| A occurs before B,
---
C occurs before B
| A occurs before C,
---
C occurs before B
| A occurs before B,
---
B occurs before C

Respectively, we associate each situation with a set of deductive rules to
facilitate the reasoning about each situation. The rules are split between
causal, process, and the combination of the two perspectives, capturing the
meaning of the relations in each situation, unfolding from fundamental
properties of the relations in discrete mathematics, jointly with a closed
world assumption according to which any unknown premise is deduced to be
False. For each situation, we generated a set of “Yes”/“No” template questions
answerable by deductive reasoning that combines the facts stated in the
situation with at least one associated rule. For example, the basic rule of
transitivity on the relation of causal execution dependence, i.e.,
$A\xrightarrow{c}C\Leftarrow(A\xrightarrow{c}B)\land(B\xrightarrow{c}C)$, may
be employed to resolve the answer to the question “Does A cause the execution
of C?” when applied in the context of Mediator situation #5 (Table 1).

For brevity, we include here as an example the set of rules (see Table 2) and
corresponding template questions (see Table 3) created for situation #2\. This
situation addresses a BPC condition in which the confounder pattern in the
causal perspective is associated with a sequence pattern of the same three
activities in the process perspective. Concerning symmetry and reflexivity
rules (Table 2), it is assumed that any cycle in the BPC structure can be
entangled with a method such as k-loop unrolling [14].

Table 2: Deductive rules associated with the confounder situation #2,
considering the manifestation of $\rightarrow$ as “occurs before” and of
$\xrightarrow{c}$ as “causes the execution of”.

| Rules | Comments
---|---|---
Process related | PR1: $A\rightarrow C\Leftarrow(A\rightarrow B)\land(B\rightarrow C)$ | Transitivity of the $\rightarrow$ relation.
PR2: $B\not\rightarrow A\Leftarrow A\rightarrow B$ | Asymmetry of the $\rightarrow$ relation.
PR3: $A\leftarrow B\Leftarrow B\rightarrow A$ |  | Where $\leftarrow$ is an antonym manifestation of $\rightarrow$,
---
e.g., “occurs after”.
PR4: $A\not\rightarrow A$ | No reflexivity.
| PR5: $B\Leftarrow(A\rightarrow B)\land A$ |  | Entailed from the meaning of $\rightarrow$:
---
i.e., if $A$ executes then $B$ will execute at some later time.
Causal related | CR1: $B\not\xrightarrow{c}A\Leftarrow A\xrightarrow{c}B$ | Asymmetry of the $\xrightarrow{c}$ relation.
CR2: $A\xrightarrow{c}C\Leftarrow(A\xrightarrow{c}B)\land(B\xrightarrow{c}C)$ |  | Transitivity of the $\xrightarrow{c}$ relation.
---
Relevant for situation $\\#$5.
CR3: $A\xleftarrow{c}B\Leftarrow B\xrightarrow{c}A$ |  | Where $\xleftarrow{c}$ is an antonym manifestation of $\xrightarrow{c}$,
---
e.g., “because the execution of”.
| CR4: $\lnot B\Leftarrow(A\xrightarrow{c}B)\land\lnot A$ and
---
$A\Leftarrow(A\xrightarrow{c}B)\land B$
| Entailed from the meaning of $\xrightarrow{c}$:
---
i.e., if $A$ doesn’t execute then $B$ doesn’t execute (and vice versa).
Process and causal structure | PCR1: $A\rightarrow B\Leftarrow A\xrightarrow{c}B$ | Causal execution dependence implies time precedence in the process.
PCR2: $B\not\xrightarrow{c}A\Leftarrow A\rightarrow B$ |  | Time precedence in the process implies no causal execution
---
dependence on the opposite direction.

Table 3 shows for each of the populated template questions, which of the rules
in Table 2 were involved in deducing its answer w.r.t the condition
articulated in situation #2\. These answers to the template questions remain
as the “ground truth” for their corresponding instantiated domain-specific
versions.

Table 3: A core set of template questions with answers deduced by the rules
for the confounder situation #2

| Template question | Answer | Related rule(s)
---|---|---|---
Process related | QP1: Does C occur before B? | Yes | PR1
QP2: Does B occur before C? | No | PR2 and QP1
QP3: Does A occur after C? | Yes | PR3
QP4: Does B occur after C? | Yes | PR3 and QP1
QP5: Does C occur after A? | No | PR2
QP6: Does C occur after B? | No | PR2 and QP1
QP7: Does A occur after B? | No | PR3 and PR2
Causal related | QC1: Does A cause the execution of C? | No | CR1
QC2: Does B cause the execution of C? | No | CR1
QC3: Does A execute because of C? | Yes | CR3
QC4: Does B execute because of C? | Yes | CR3
QC5: If C doesn’t execute, will A ever execute? | No | CR4
QC6: If C doesn’t execute, will B ever execute? | No | CR4
Process and causal related | QPC1: Does A cause the execution of B? | No |  | Close world
---
assumption
QPC2: Does B cause the execution of A? | No | PCR2

Similar to the underlying set of situations, domain-specific questions were
methodologically instantiated from these questions. While the core set of
template questions is extensible (see section 3.1), any specific set uniquely
tags the benchmark version and also sets a bound to its expressiveness, that
is, the grammatical richness the benchmark can accommodate. Pragmatically,
this means that any two target LLMs can be compared only when assessed
according to the same benchmark version.

###  3.1 Extending the benchmark

As noted, the core set of questions may be extended with additional template
questions to assess the capacity of an LLM to reason about any additional
aspect of interest. For example, we may define the boolean function
is_shortened($A$) to denote that the execution time of activity $A$ was
expedited to finish its execution earlier. Respectively, the following rule
can be added:

$CR5:is\\_shortened(B)\Leftarrow A\xrightarrow{c}B\land is\\_shortened(A)$

Similarly, other functions can be added to denote other forms of temporal
intervention, such as is_extended($A$), is_halted($A$), and is_delayed($A$).

Respective to adding this rule, the causal set of questions for situation #2
above can be extended with the additional template questions as listed in
Table 4.

Table 4: An extension to the core set of template questions for situation #2.

| Template question | Answer | Related rule(s)
---|---|---|---
Causal related | QC7: If we shorten A, will B be shortened? | No | CR5
QC8: If we shorten B, will A be shortened? | No | CR5
QC9: If we shorten C, will A be shortened? | Yes | CR5
QC10: If we shorten C, will B be shortened? | Yes | CR5

###  3.2 Populating domain-specific questions

We used the open-source Mixtral-
instruct-8x-7b333https://huggingface.co/mistralai/Mixtral-8x7B-Instruct-v0.1#model-
card-for-mixtral-8x7b LLM to instantiate the template questions for situation
#2 to concrete problem domain statements for each of the three perspectives.
For this work, we restricted the instantiation of the 15 questions in Table 3
and the additional 4 in Table 4, to having each template question populated
with one corresponding domain-specific question. The domains were arbitrarily
selected as one of the following: transportation, manufacturing, logistics,
retail, finance, insurance, and medical. An example of a prompt employed for
such instantiation attending to the process perspective in situation #2 (QP1)
is illustrated as prompt #1 in Table 5.

Table 5: Series of LLM prompts used during benchmark development # |  LLM prompt
---|---
1 |  INPUT: Phrase: A, B, and C are activities in some process. C occurs before A. A occurs before B. Question: Does C occur before B? Instruction: Considering the above question, choose any relevant concrete activities A, B, and C in the domain of $\ll$transportation$\gg$ that retain the truth of the phrase statements. Using these terms, instantiate corresponding phrase statements and a question statement in a form that matches the above statements and question. OUTPUT: Concrete Activities: A: Boarding the plane B: Takeoff of the plane C: Check-in at the airport Instantiated Phrase Statements: Check-in at the airport occurs before boarding the plane. Boarding the plane occurs before takeoff of the plane. Instantiated Question Statement: Does check-in at the airport occur before takeoff of the plane?
2 |  INPUT: Template question: If we shorten C, will B be shortened? Instantiated question: If we shorten Damage assessment team inspects the damage and estimates the cost of repair in duration, will Insurance policyholder receives the payout be shortened? Instruction: Considering the instantiated question above a concrete version of the template question where the letters are replaced with process activity descriptions, when replacing these descriptions with their corresponding letters in the template, how would you rate the similarity between the revised instantiated question and the template question (where a 1 rate means they are identical and 0 they are completely different)? In your output, print only the rate value on a 0-1 scale. OUTPUT: 0.9 or 90% similarity.

###  3.3 Filtering out inadequate questions

Our instantiated set of questions so far was relatively small. However, in
full-scale development, covering the complete situation space and more
domains, the number of questions will likely increase significantly. This
process might generate domain-specific questions in a form that becomes
incongruent with the original template questions used as its seed due to the
non-deterministic nature of LLMs. Therefore, we also foresee the use of “LLM-
as-a-judge” to curate the quality of the generated questions. That is,
excluding the ones that do not maintain a faithful linkage to their
corresponding template. To this end, prompt#2 in Table 5 was created to grade
and remove questions scoring below an acceptable threshold.

##  4 Evaluation

We employed the developed benchmark prototype to get a sense of its
applicability against state-of-the-art LLMs, to get an initial measure of
accuracy spanning the three perspectives: process, causal, and the combination
of the two.

At first, we ran the three core sets of template questions from Table 3 with
five different target LLMs:
GPT3.5444https://beta.openai.com/docs/models/gpt-3,
GPT4555https://www.openai.com/research/gpt-4,
GPT4o666https://www.openai.com/research/gpt-4, Mixtral-instruct-8x-7bfootnote
3, and Merlinite-7b777https://huggingface.co/ibm/merlinite-7b. We repeated
each template question ten times with clean-slate prompts, preceding each
question with the corresponding phrase describing the situation and
instruction as illustrated in Table 6. We averaged the proportion of correctly
answered questions for each reasoning perspective: process, causal, and
combined. We then also added the extended set of questions (see section 3.1)
and revised the results accordingly.

As a second step, using the domain-specific questions, we repeated the
benchmark testing for the two open-source LLMs (Mixtral-instruct-8x-7b and
Merlinite-7b) ten times per prompt, measuring the proportion of correct
answers for the three perspectives, both without and with the inclusion of the
extension questions.

Table 6: LLM Prompts used for benchmark testing # |  LLM prompts: process, causal, and both
---|---
1 |  Phrase: A, B, and C are activities in some process. $\ll$process related$\gg$C occurs before A. A occurs before B. $\ll$causal related$\gg$C causes the execution of A. C causes the execution of B. Instruction: Considering the above phrase about activities in a process, answer the following question. Your answer should be limited to either Yes or No and nothing else. Question: $\ll$process related (QP1)$\gg$Does C occur before B? $\ll$causal related (QC1)$\gg$Does A cause the execution of C? $\ll$causal & process related (QPC1)$\gg$Does A cause the execution of B?

###  4.1 Assessing the quality of the Benchmark

We acknowledge that our current evaluation caters strictly to the
applicability of the benchmark to a handful of LLMs. We recognize that such an
assessment lacks characterizing the benchmark quality. For this, we elaborate
here on a set of relevant metrics that we intend to assess once the benchmark
is developed in full scale.

Completeness refers to the range of realistic BPC situations the benchmark
covers. This is ensured by using causal “junctions” as per [17], which
accommodate any causal structure, and by the richness of rules and their
coverage by questions. Our design principle ensures each rule helps resolve at
least one question and that the benchmark is extensible. A metric can capture
the proportion of rules covered by questions. However, completeness is always
limited by the core set of template questions and the finite domains involved.
Therefore, it is crucial to disclose the list of questions and domains for any
benchmark version.

Correctness refers to the degree each question’s answer is adequate to the
targeted process situation. For Yes/No questions, a correct answer indicates
whether the condition holds (or not) in the corresponding situation. We use
rules as a formal mechanism to determine answers and keep the same answers for
the instantiated situations and corresponding questions.

Reliability reflects result consistency across multiple tests. This may also
be influenced by the LLM’s inherent consistency. It can be measured using
conventional metrics like Cronbach’s alpha. To ensure reliability, we reset
the prompt to prevent prior context from affecting interactions.

Validity assesses how well each question captures the specific aspect of its
domain. This can be measured through convergence metrics like factor loading,
and discriminant validity when partitioning the instantiated questions by
different perspectives and domains.

##  5 Results

We report the results for running our benchmark prototype in Table 7. This
table shows in percentage the proportion of questions answered correctly by
each of the LLMs included in the text. The results are split between the
template questions (i.e., domain-independent) and domain-specific questions,
partitioned also by the various perspectives, considering the causal one
without and with the addition of the extending questions. While the core set
of template questions can be seen as exhaustive with respect to domain
coverage, the domain-specific part should be observed with caution, as each
template question was populated only with one corresponding domain-specific
question. As such, we expect that in a full-scale set of domain-specific
questions, the performance is likely to get closer to the accuracy presented
by the template questions.

Table 7: LLM accuracy results using the prototype benchmark for situation #2

|  | Template questions | Domain-specific questions
---|---|---|---
|  | Num of
---
questions
| Merlinite
---
7b
| Mixtral
---
instruct
8x 7b
| GPT
---
4o
| GPT
---
4
| GPT
---
3.5
| Merlinite
---
7b
| Mixtral
---
instruct
8x 7b
Process | 7 | 58% | 71% | 100% | 85% | 65% | 100% | 100%
Causal | 6 | 66% | 100% | 100% | 100% | 56% | 98% | 100%
Process + Causal | 2 | 100% | 100% | 100% | 80% | 15% | 55% | 95%
Extension | 4 | 50% | 50% | 70% | 65% | 55% | 73% | 50%
Causal+Extension | 10 | 60% | 80% | 88% | 86% | 56% | 88% | 80%
Total weighted avg |  | 63% | 79% | 94% | 85% | 55% | 89% | 89%

##  6 Conclusion and Future Work

The benchmark can be used in two ways: testing LLM performance, as
demonstrated here, and training an LLM to improve its reasoning about BPC. For
testing, question answers serve as the “ground truth” to determine model
accuracy, which can be analyzed by perspective and domain. For training, the
questions and answers can be randomly split into training and testing subsets
(e.g., 80/20). The testing subset is used before and after training to measure
improvement.

Setting aside the debate whether LLMs can generally reason, our goal here is
more modest. We would like to equip (or measure) LLMs regarding their capacity
to adequately infer sound conclusions when presented with knowledge about
causal business processes. In this regard, our developed instrumentation can
be employed in two (complementary) manners. It could be used to test such an
ability with respect to a relatively wide variety of process domains. In
addition, the tool can also be used as a model training dataset to augment an
existing LLM with such ability and also to be adopted to additional problem
domains for specific needs. For the former purpose, it could be used “as is”
with only the questions component to facilitate model bench-marking, and with
the ground truth answers for model training. For any newly embarked problem
domains, the core seed of the template questions should be instantiated
methodologically in the same process reported here to derive a corresponding
set of domain-relevant items, to be added. This is also the rationale
underlying our aim to release the model as open source
(https://github.com/IBM/SAX/tree/main/NLP4BPM2024), letting the community
gradually contribute to the dataset to make its domain coverage broader.

The core contribution of this work is twofold. An open-source model that is
developed and a methodology reporting how to construct a benchmark for a
particular task to be facilitated by LLMs, in this case, the one of reasoning
about BPCs. At the time of submitting this paper, we embark on completing the
specification of the template rules and questions corresponding to the other
situations. As next steps, we plan to populate our benchmark with a more
exhaustive set of domains and at a larger scale of instances per domain. Our
vision is that LLM benchmarks may become standardized means to guide the
choice of suitability of LLMs to specific tasks.

## References

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E. I. Kompantseva, A. A. Tuganbaev

Rings on Abelian Torsion-Free Groups of Finite Rank

Abstract. In the class of reduced Abelian torsion-free groups $G$ of finite
rank, we describe $TI$-groups, this means that every associative ring on $G$
is filial. If every associative multiplication on $G$ is the zero
multiplication, then $G$ is called a $nil_{a}$-group. It is proved that a
reduced Abelian torsion-free group $G$ of finite rank is a $TI$-group if and
only if $G$ is a homogeneous Murley group or $G$ is a $nil_{a}$-group. We also
study the interrelations between the class of homogeneous Murley groups and
the class of $nil_{a}$-groups. For any type $t\neq(\infty,\infty,\ldots)$
every integer $n>1$, there exist $2^{\aleph_{0}}$ pairwise non-quasi-
isomorphic homogeneous Murley groups of type $t$ and rank $n$ which are
$nil_{a}$-groups. We describe types $t$ such that there exists a homogeneous
Murley group of type $t$ which is not a $nil_{a}$-group.
This paper will be published in Beiträge zur Algebra und Geometrie /
Contributions to Algebra and Geometry.

Key words. Abelian group, ring on Abelian group, filial ring, $TI$-group.

MSC2020 datebase: 20K30, 20K99, 16B99

##  1 Introduction

For an Abelian group $G$, any homomorphism $\mu\colon G\otimes G\to G$ is
called a multiplication on $G$. An Abelian group $G$, with multiplication
defined on $G$, is called a ring on $G$. On every group $G$, we can define the
multiplication $\mu\colon G\otimes G\to 0$ which is called the zero
multiplication. A ring with such a multiplication is said to be trivial. If a
group $G$ has no multiplication (resp., no associative multiplication) besides
the zero multiplication, then $G$ is called a $nil$-group (resp., a
$nil_{a}$-group). R.A.Beaumont [7] posed the description problem for ring
structures on Abelian groups; he considered rings on direct sums of cyclic
groups.

In the theory of additive groups of rings, one of parts is the study of
Abelian groups on which every ring belongs to a certain class. We consider the
class of filial rings which were defined in [10] and studied in [29],[3],
[12], [13]. An associative ring is called a filial ring if every meta-ideal of
finite index is an ideal. A subring $A$ of an associative ring $R$ is called a
meta-ideal of index $n$ if there exists a chain

$A=A_{0}\subset A_{1}\subset\cdots\subset A_{n}=R$

such that $A_{i}$ is an ideal of $A_{i+1}$ for all $i=0,\cdots,n-1$ [10]. It
is easy to see that an associative ring is filial if and only if the relation
‘is an ideal of’ is transitive. We note that the class of filial rings
contains not only Hamiltonian rings, [28], [2], [23], but (Von Neumann)
regular rings and simple rings; on regular rings, e.g., see[18], [30]. In this
paper, we study Abelian groups, on which every associative ring is a filial
ring; such groups are called $TI$-groups. In [5], the problem of the study of
$TI$-groups is formulated. In the same paper, torsion $TI$-groups are
described. In [21], $TI$-groups are described in the class of algebraically
compact Abelian groups. In [22], $TI$-groups from some classes of Abelian
torsion-free groups are described. In [22], non-reduced $TI$-groups are also
described. This allows us to consider only the reduced case in the studies of
$TI$-groups.

The main aim of this paper is to describe $TI$-groups in the class of Abelian
torsion-free groups of finite rank. In [25] and [26], C.E. Murley
systematically studied reduced Abelian torsion-free groups $G$ of finite rank
such that the $p$-rank of the group $G$ does not exceed $1$, for every prime
$p$; these groups are now called Murley groups. We note that L. Fuchs [17]
decided to break this terminological tradition to pay tribute L. Prochazka, he
calls these groups Prochazka-Murley groups. Indeed, L. Prochazka [2] initiated
the study of groups of this class a few years before the publication of
Murley’s paper [25] and obtained more essential results. About Murley groups,
see also the book [24, Section 44].

Let $\mathcal{MH}$ denote the class of all homogeneous Murley groups and let
$\mathcal{NIL}_{a}$ denote the class of all Abelian torsion-free
$nil_{a}$-groups of finite rank.

The main result of Section 2 is that the class of reduced torsion-free groups
of finite rank coincides with the class
$\mathcal{MH}\,\cup\,\mathcal{NIL}_{a}$ (Theorem 11). In addition, we prove
some properties of Murley groups and rings on them.

Since the class of reduced torsion-free $TI$-groups of finite rank coincides
with the union of the classes $\mathcal{MH}$ and $\mathcal{NIL}_{a}$,
interrelations between these classes are studied in Section 3. It is shown
that the classes $\mathcal{MH}\,\cap\,\mathcal{NIL}_{a}$ and
$\mathcal{MH}\,\setminus\,\mathcal{NIL}_{a}$ are very large and varied.

It is easy to see that every torsion-free group $G$ of rank $1$ belongs to the
class $\mathcal{MH}$. In addition, $G\in\mathcal{MH}\,\cap\,\mathcal{NIL}_{a}$
if the type $t(G)$ is not an idempotent type and
$G\in\mathcal{MH}\,\setminus\,\mathcal{NIL}_{a}$ if the type $t(G)$ is an
idempotent type. But the situation changes significantly if the rank of the
group exceeds $1$. In Theorem 15, it is proved that for any type
$t\neq(\infty,\infty,\ldots)$ and each positive integer $n>1$, the class
$\mathcal{MH}\,\cap\,\mathcal{NIL}_{a}$ contains $2^{\aleph_{0}}$ of pairwise
non-quasi-isomorphic groups for type $t$ and rank $n$.

However, not all types $t\neq(\infty,\infty,\ldots)$ can be realized as types
of some groups in the class $\mathcal{MH}\,\setminus\,\mathcal{NIL}_{a}$. For
example, this class does not contain groups of rank $>1$ and type
$(0,0,\ldots)$. In Theorem 17, a criterion is obtained for a type to be the
type of some group in $\mathcal{MH}\,\setminus\,\mathcal{NIL}_{a}$.

In the paper, we widely use a method of the study of rings on torsion-free
Abelian groups of finite rank. This argument is based on the method of
R.A.Beaumont and R.S.Pierce [8]. For such rings, they proved an analogue of
the main Wedderburn theorem on finite-dimensional algebras. They also offered
a reasonably simple system of invariants for the quasi-isomorphism classes of
quotient divisible torsion-free groups of finite rank.

All groups, considered in the paper, are Abelian, and the word "group" means
"Abelian group" in what follows. A multiplication $\mu\colon G\otimes G\to G$
on the group $G$ is often denoted by $\times$, i.e. $\mu(g_{1}\otimes
g_{2})=g_{1}\times g_{2}$ for all $g_{1},g_{2}\in G$. A group $G$ with
multiplication $\times$, defined on $G$, determines a ring on the group $G$
which is denoted by $(G,\times)$. As usual, $\mathbb{N}$, $\mathbb{N}_{0}$,
and $\mathbb{P}$ are the sets of all positive integers, non-negative integers,
and prime integers, respectively; $\mathbb{Z}$ is the ring of integers,
$\mathbb{Q}$ is the group (or the field) of rational numbers,
$\widehat{\mathbb{Z}}_{p}$ is the group (or the ring) of $p$-adic integers,
and ${\mathbb{Q}}_{p}^{*}$ is the field of $p$-adic numbers. If
$a,b\in\mathbb{Z}$, then $(a,b)$ is the largest common divisor of the integers
$a$ and $b$.

Let $G$ be a group, $g\in G$, and let $(G,\times)$ be a ring on $G$. We denote
by $(g)_{\times}$ the ideal of the ring $(G,\times)$ generated by the element
$g$. We denote by $\chi(g)$ and $t(g)$ the characteristic and the type of the
element $g\in G$, respectively; $t(G)$ is the type of the group $G$ provided
$G$ is homogeneous. We denote by $r(G)$ and $r_{p}(G)$ the rank and the
$p$-rank of the group $G$, respectively; $\bar{G}$ is the divisible hull of
the group $G$. If $S\subseteq G$, then $\langle S\rangle$ and $\langle
S\rangle_{*}$ are the subgroup and the pure subgroup of the group $G$
generated by the set $S$, respectively.

If $R$ is a ring containing $1$, then $Re$ is the cyclic module over $R$
generated by the element $Re$. We write $A\lhd R$ when $A$ is an ideal of the
ring $R$. We denote by $A\oplus B$ and $A\dotplus B$ the direct sum of groups
and the direct sum of rings, respectively. Unless otherwise stated, we use
definitions and designations from [17] and [24].

##  2 Torsion-Free $TI$-Groups of Finite Rank

The aim of this section is to describe $TI$-groups in the class of reduced
torsion-free groups of finite rank.

Lemma 1. Let $n\in\mathbb{N}$, $G$ be a reduced torsion-free group, and let
$A\oplus B$ be a subgroup of index $n$ of $G$ such that $A$ is not a
$nil_{a}$-group and $B\neq 0$. Then $G$ is not a $TI$-group.

Proof. Let $C=A\oplus B$. By [22, Corollary 1], there exists an associative
non-filial ring $(C,\times)$ such that $C\times C\subseteq n^{2}C$. By [22,
Lemma 5], $G$ is not a $TI$-group. $\square$

Remark 2. According to [3, Theorem 1], the ring $(G,\times)$ is filial if and
only if for any $g\in G$, we have

$(g)_{\times}=(g)_{\times}^{2}+\mathbb{Z}g.$

We note that every trivial ring is filial. In the following theorem, it is
proved that there do not exist filial nilpotent rings besides trivial.

Theorem 3. Every filial nilpotent torsion-free ring is trivial.

Proof. Let $(G,\times)$ be a filial ring and let $G^{t}=0$ for some positive
integer $t>2$. We take $x\in G^{t-2}$, $p\in P$, and set $x_{1}=px\in
G^{t-2}$. We have

$(x_{1})_{\times}=x_{1}\times G+G\times x_{1}+\mathbb{Z}x_{1}.$

Since $x_{1}\times G\times G=G\times x_{1}\times G=G\times G\times x_{1}=0,$
we have

$(x_{1})^{2}_{\times}+\mathbb{Z}x_{1}=\mathbb{Z}x_{1}^{2}+\mathbb{Z}x_{1}.$

Since the ring $(G,\times)$ is filial,
$(x_{1})_{\times}=(x_{1})^{2}_{\times}+\mathbb{Z}x_{1}$ by Remark 2.
Consequently,

$(x_{1})_{\times}=\mathbb{Z}x_{1}^{2}+\mathbb{Z}x_{1}.$ $None$

Therefore,

$x_{1}\times x=kx_{1}^{2}+mx_{1}.$ $None$

for some $k,m\in\mathbb{Z}$.

Let $g\in G$.

Case 1. Let $m=0$.

It follows from $(2)$ that $px^{2}=kp^{2}x^{2}$, whence $p(kp-1)x^{2}=0$.
Therefore, $x^{2}=0$, since $G$ is a torsion-free group. Consequently,
$x_{1}^{2}=p^{2}x^{2}=0.$ By $(1)$, we have
$(x_{1})_{\times}=\mathbb{Z}x_{1}$. Then

$x_{1}\times g=sx_{1}$ $None$

for some $s\in\mathbb{Z}$. If $s=0$, then $x_{1}\times g=0$. If $s\neq 0$,
then we multiply the both parts of $(3)$ from the right by $g$ and obtain

$s(x_{1}\times g)=x_{1}\times g\times g=0,\;\text{whence }x_{1}\times g=0.$

Case 2. Let $m\neq 0$.

It follows from $(2)$ that $m(x_{1}\times g)=x_{1}\times x\times
g-kx_{1}^{2}\times g=0$. Therefore, $x_{1}\times g=0$.

Thus, $x\times G=0$ for each $x\in G^{t-2}$. Consequently,
$G^{t-1}=G^{t-2}\times G=0$. If $t-1>2$, then $G^{t-2}=0$, by the above. Thus,
we have $G^{2}=0$, i.e., the multiplication $\times$ is the zero
multiplication. $\square$

Let $G$ be a torsion-free group of finite rank, $\bar{G}=\mathbb{Q}\otimes G$
be the divisible hull of the group $G$, and let $(G,\,\times)$ be an
associative ring on $G$. Then the multiplication $\times$ is uniquely extended
to a multiplication on $\bar{G}$ [17, Theorem 1.3, p. 675].

By the main Wedderburn theorem, the ring $(\bar{G},\times)$ has a
decomposition $\bar{G}=\widetilde{S}\oplus\widetilde{N}$, where
$\widetilde{N}$ is a nilpotent ideal of the ring $(\bar{G},\,\times)$ and
$\widetilde{S}$ is a semisimple subalgebra of the algebra $(\bar{G},\times)$.
By [8], there exists a subring $S$ and a nilpotent ideal $N$ of the ring
$(G,\,\times)$ such that $\overline{S}=\widetilde{S}$,
$\bar{N}=\widetilde{N}$, and the subring $S\oplus N$ has a finite index in
$(G,\,\times)$.

Let $R$ be an associative algebra over $\mathbb{Q}$. According to [8], we call
a ring $(G,\,\times)$ a ring of algebra type $R$ if the induced ring
$(\bar{G},\,\times)$ is isomorphic to $R$. In this case, one says that the
group $G$ admits multiplication of algebra type $R$.

Lemma 4. Every associative ring on a torsion-free group of finite rank either
is nilpotent or contains a subring $C$ of finite index that is of the form
$C=M_{1}\oplus\cdots\oplus M_{k}\oplus N$, where $N$ is a nilpotent ideal,
$k\in\mathbb{N},$ $M_{i}$ is the full matrix ring of order $r_{i}$ over a ring
of rational division algebra type (in particular, $M_{i}$ is a ring of
rational division algebra type for $r_{i}=1$).

Proof. Let $G$ be a torsion-free group of finite rank and let $(G,\times)$ be
an associative ring. Then the induced ring $(\bar{G},\times)$ also is
associative. By the main Wedderburn theorem,
$\bar{G}=\widetilde{S}\oplus\widetilde{N}$, where $\widetilde{S}$ is a
semisimple algebra and $\widetilde{N}$ is a nilpotent ideal. If
$\widetilde{S}=0$, then the ring $(\bar{G},\times)$, whence $(G,\times)$ is
nilpotent.

Let $\widetilde{S}\neq 0$. By [8, Theorem 1.4], the ring $(G,\times)$ has a
subring $S\oplus N$ of finite rank such that $N$ is a nilpotent ideal and $S$
is a ring of semisimple algebra type $\widetilde{S}$. By [8, Corollary 3.5],
the ring $S$ has a subring $A_{1}\dotplus\cdots\dotplus A_{2}$ of finite
index, where $A_{i}$ is a ring of simple algebra type for every
$i=1,\;\ldots,\;k$. By [8, Corollary 3.9], for every $i=1,\;\ldots,\;k$, the
ring $A_{i}$ contains a subring $M_{i}$ of finite index which is a full matrix
ring of order $r_{i}\in\mathbb{N}$ over some ring of rational division algebra
type.

Thus, the subring $C=M_{1}\oplus\cdots\oplus M_{k}\oplus N$ is of finite index
in the ring $(G,\times)$. $\square$

Two Abelian groups $A$ and $B$ are said to be quasi-isomorphic (we write
$A\sim B$ in this case) if there exist subgroups $A^{\prime}\subseteq A$ and
$B^{\prime}\subseteq B$ such that $A^{\prime}\cong B^{\prime}$, $mA\subseteq
A^{\prime}$, and $nB\subseteq B^{\prime}$ for some positive integers $m$ and
$n$. A group $G$ is said to be quasi-decomposable if $G\sim A\oplus B$ for
some non-zero groups $A$ and $B$. Groups, that are not quasi-decomposable, are
called a strongly indecomposable.

Corollary 5. Let $G$ be a strongly indecomposable torsion-free group of finite
rank. Then every associative ring $(G,\times)$ on $G$ is either nilpotent or a
ring of rational division algebra type.

Proof. Let us assume that the ring $(G,\times)$ is not nilpotent. By Lemma 4,
this ring contains a subring $C$ of finite index such that
$C=M_{1}\oplus\cdots\oplus M_{k}\oplus N$, where $N$ is a nilpotent ideal,
$k\in\mathbb{N},$ $M_{i}$ is the full matrix ring of order $r_{i}$ over a ring
$R_{i}$ of rational division algebra type. Then algebra types of the rings
$(G,\times)$ and $C$ coincide. The additive group $C^{+}$ of the ring $C$ is
quasi-isomorphic to the group $G$ and it can be represented in the form
$C^{+}=R_{1}^{+}\oplus B$, where $R_{1}^{+}$ is the additive group of the ring
$R_{1}$, which is a ring of rational division algebra type and $B$ is a group.
If either $N\neq 0$, or $k>1$, or $r>1$, then $B\neq 0$; this contradicts to
the property that $G$ is strongly indecomposable. Therefore, the ring
$C=R_{1}$ is a ring of rational division algebra type. Therefore, the ring
$(G,\times)$ is a ring of rational division algebra type, as well.

Corollary 6. Every associative non-trivial ring on $TI$-group is a ring of
rational division algebra type.

Proof. Let $G$ be a $TI$-group and let $(G,\times)$ be an associative non-
trivial ring. By Theorem 3, this ring is not nilpotent. By Lemma 4, this ring
has a subring of finite index $C=M_{1}\oplus\cdots\oplus M_{k}\oplus N$, where
$k\in\mathbb{N}$ and for every $i=1,\;\ldots,\;k$, the additive group $M_{i}$
is a direct sum of groups on each of which there exists a ring of rational
division algebra type. By Lemma 1, the ring $(G,\times)$ is a ring of rational
division algebra type. $\square$

Further, we prove that the class of reduced torsion-free $TI$-groups of finite
rank coincides with the union of the class of Murley groups and the class of
$nil_{a}$-groups.

We recall that a Murley group is a reduced torsion-free group of finite rank
such that $r_{p}(G)\leq 1$ for any $p\in P$. We prove some properties of
Murley groups and rings on them.

Remark 7. By [25], every indecomposable Murley group is strongly
indecomposable.

Theorem 8.
1. On a Murley group, every multiplication is associative and commutative.

2. On a Murley group, every non-trivial ring is not nilpotent.

3. Every homogeneous Murley group is strongly indecomposable.

Proof. Let $G$ be a Murley group and let $(G,\times)$ be a ring on $G$. The
multiplication $\times$ is uniquely extended to the multiplication on the
pure-injective hull $\widehat{G}$ of the group $G$ [17]. In addition,
$\widehat{G}=\prod\limits_{p\in P_{1}}\widehat{G}_{p}=\prod\limits_{p\in
P_{1}}\widehat{\mathbb{Z}}_{p}e_{p}$, where $P_{1}\subseteq\mathbb{P}$,
$\widehat{G}_{p}$ is the $p$-adic completion of the group $G$,
$e_{p}\in\widehat{G}_{p}$. According to [19], the ring $(\widehat{G},\times)$
is the direct product of ideals $\widehat{\mathbb{Z}}_{p}e_{p}$, $p\in P_{1}$.
In addition, if $e_{p}\times e_{p}=\tau_{p}e_{p}$, where
$\tau_{p}\in\widehat{\mathbb{Z}}_{p},$ then $\alpha e_{p}\times\beta
e_{p}=(\alpha\beta\tau_{p})e_{p}$ for all $p\in P_{1}$ and
$\alpha,\beta\in\widehat{\mathbb{Z}}_{p}.$

1. It is easy to see that the ring $(\widehat{G},\times)$ is associative and commutative. Therefore, the subring $(G,\times)$ of $(\widehat{G},\times)$ is associative and commutative.

2. Let the ring $(G,\times)$ be non-trivial. Then $a\times b\neq 0$ for some $a=(\alpha_{p}e_{p})_{p\in P_{1}}$ and $b=(\beta_{p}e_{p})_{p\in P_{1}}\in G\subseteq\widehat{G},$ where $\alpha_{p},\beta_{p}\in\widehat{\mathbb{Z}}_{p}$, $p\in P_{1}$. In this case, $\alpha_{q}\neq 0$, $\beta_{q}\neq 0$ and $e_{q}\times e_{q}=\tau_{q}e_{q}\neq 0$ for some $q\in P_{1}$. Consequently, $\alpha_{q}^{n}\tau_{q}^{n-1}e_{q}\neq 0$ for every $n\in\mathbb{N}.$ Since $a^{n}=(\alpha_{p}^{n}\tau_{p}^{n-1}e_{p})_{p\in P_{1}}$, we have that $a^{n}\neq 0$ for all $n\in\mathbb{N}.$

3. Let $G$ be a homogeneous Murley group and $G=A\oplus B$. Then there exists a prime integer $p$ with $r_{p}(G)=1$. Since

$1=r_{p}(G)=r_{p}(A)+r_{p}(B),$

exactly one of the numbers $r_{p}(A)$ and $r_{p}(B)$ is equal to $1$. For
definiteness, let $r_{p}(A)=1$ and $r_{p}(B)=0$. This means that
$h_{p}(a)\neq\infty$ and $h_{p}(b)=\infty$ for some $a\in A\setminus\\{0\\}$
and some $b\in B$. Since the group $G$ is homogeneous, $B=0$. Consequently,
$G$ is strongly indecomposable by Remark 7. $\square$

We note that the converse to Theorem 8(3) is not true.

Proposition 9. There exist strongly indecomposable non-homogeneous Murley
groups.

Proof. In the group $V=\mathbb{Q}e_{1}\oplus\mathbb{Q}e_{2}$, the linear
combinations of the form $c=k_{1}e_{1}+k_{2}e_{2}$, where
$k_{1},k_{2}\in\mathbb{Z}$, $k_{1}>0$, $k_{2}\neq 0$ and $(k_{1},k_{2})=1$,
will be ordered in the form of a sequence $\\{c_{j}\\}_{j>2}$. We set
$c_{1}=e_{1}$, $c_{2}=e_{2}$. Let $\\{p_{j}\\}_{j\in\mathbb{N}}$ be the
sequence of all prime integers. In $V$, we consider the subgroup $G=\langle
p_{j}^{-\infty}c_{j}\,|\,j\in\mathbb{N}\rangle$. Then the rank $G$ is equal to
$2$. In addition, $G$ is a Murley group. Indeed, let $a$ and $b$ be two
linearly independent elements of $G$, $j\in\mathbb{N}$. If
$h_{p_{j}}(a)=\infty$ or $h_{p_{j}}(b)=\infty$, then $a$, $b$ are
$p_{j}$-dependent. We assume that $h_{p_{j}}(a)\neq\infty$ and
$h_{p_{j}}(b)\neq\infty$. Then $a$, $b$, $c_{j}$ are linearly dependent, i.e.,
$m_{1}c_{j}=m_{2}a+m_{3}b$ for some $m_{1},m_{2},m_{3}$ which are not equal to
zero simultaneously. It is easy to see that $m_{2}\neq 0$ or $m_{3}\neq 0$.
For definiteness, let $m_{2}\neq 0$. Then $p_{j}^{t}\,\not|\,m_{2}$ for some
positive integer $t$. However, $p_{j}^{t}\,|\,m_{1}c_{j}$. Consequently,
$r_{p_{j}}(G)\leq 1$.
In the group $G$, any two linearly independent elements have incomparable
types. According to [9], the group $G$ is pure-indecomposable in this case,
i.e., every pure subgroup of the group $G$ is indecomposable. By Remark 7, the
group $G$ is strongly indecomposable. $\square$

In the class of Murley groups, it follows from Theorem 8(1) that the notion of
a $nil$-group coincides with the notion of a $nil_{a}$-group.

Theorem 10. Let $G$ be a Murley group which is not a $nil$-group. Then the
following conditions are equivalent.

1) $G$ is a homogeneous group of idempotent type.

2) $G$ is an indecomposable group.

3) $G$ is a strongly indecomposable group.

4) Every non-trivial ring on $G$ is a ring of field type.

5) On the group $G$, there exists a ring of field type.

Proof. The implication 1) $\Rightarrow$ 2) follows from Theorem 8(3).

The implication 2) $\Rightarrow$ 3) follows from Remark 7.

The implication 3) $\Rightarrow$ 4) follows from Corollary 6 and Theorem 8.

The implication 4) $\Rightarrow$ 5) follows from the property that $G$ is not
a $nil$-group, by assumption.

5) $\Rightarrow$ 1). By [8, Lemma 9.7], every group $G$, which admits
multiplication of simple algebra type, is homogeneous. The type of $G$ is an
idempotent type, since otherwise $G$ is a $nil$-group. $\square$

We note that the assertion of Theorem 10 is not true without the condition
that there exists a non-trivial ring on the group $G$. Indeed, the Murley
group $G$, which is constructed in Proposition 9, satisfies condition 3) but
does not satisfy condition 1), since $G$ is a $nil$-group.

In the following theorem, we describe $TI$-groups in the class of reduced
torsion-free groups of finite rank.

Theorem 11. A reduced torsion-free group $G$ of finite rank is a $TI$-group if
and only if $G$ is a homogeneous Murley group or $G$ is a $nil_{a}$-group

Proof. Let $G$ be a $TI$-group such that there exists a non-trivial
associative ring $(G,\times)$. Then $(G,\times)$ is a ring of rational
division algebra type, by Corollary 6. By [8, Lemma 9.7], the group $G$ is
homogeneous.

We assume that $r_{p}(G)\geq 2$ for some $p\in P$. Then $G$ contains
$p$-independent elements $a$ and $b$. Since the induced ring
$(\bar{G},\times)$ on the divisible hull $\bar{G}$ of the group $G$ is a
rational division algebra, $(\bar{G},\times)$ contains the inverse element
$a^{-1}$ of $a$. In addition, $na^{-1}\in G$ for some positive integer $n$.
Let $k$ be a positive integer such that $p^{k}$ does not divide $n$. In the
ring $(G,\times)$, the ideal $(p^{k}a)_{\times}$ satisfies the relation

$(p^{k}a)_{\times}^{2}+\mathbb{Z}p^{k}a\subseteq p^{2k}G+\mathbb{Z}p^{k}a.$

We consider the element $p^{k}nb=p^{k}a\times na^{-1}\times
b\in(p^{k}a)_{\times}$. We assume that
$p^{k}nb\in(p^{k}a)_{\times}^{2}+\mathbb{Z}p^{k}a$. Then
$p^{k}nb=p^{2k}x+sp^{k}a$ for some $x\in G$, $s\in\mathbb{Z}$. Consequently,
$nb-sa=p^{k}x$. It follows from $p$-independence of the elements $a$ and $b$
that $p^{k}$ divides $n$; this contradicts to the choice of the integer $k$.
Therefore, $(p^{k}a)_{\times}\neq(p^{k}a)_{\times}^{2}+\mathbb{Z}p^{k}a$.
Therefore, the ring $(G,\times)$ is not filial. Consequently, $r_{p}(G)\leq 1$
for every $p\in P$.

It is easy to see that every $nil_{a}$-group is a $TI$-group. Let $G$ be a
homogeneous Murley group and there exists a non-trivial ring $(G,\times)$ on
$G$. Then $t(G)$ is an idempotent type. By Theorem 10, the ring $(G,\times)$
is a ring of field type. In this case, it follows from [4, Theorem 7] that the
ring $(G,\times)$ is filial if for any $g\in G$, there exists a positive
integer $m$ such that $mg\in G\times g^{2}$. Let $g\in G$. Since the induced
ring $(\bar{G},\times)$ is a field, it contains the inverse element $g^{-1}$
of $g$. In addition, there exists a positive integer $m$ such that $mg^{-1}\in
G$, whence $mg=mg^{-1}\times g^{2}\in G\times g^{2}.$ Consequently, the ring
$(G,\times)$ is filial. Since the ring $(G,\times)$ is arbitrary, $G$ is a
$TI$-group. $\square$

Corollary 12. A reduced torsion-free group $G$ of finite rank is a $TI$-group
if and only if either $G$ is a $nil_{a}$-group or $G$ is a Murley group which
satisfies one of conditions 1)–5) of Theorem 10.

##  3 Homogeneous Murley Groups

Let $\mathcal{MH}$ be the class of all homogeneous Murley groups and let
$\mathcal{NIL}$ (resp., $\mathcal{NIL}_{a}$) be the class of all torsion-free
$nil$-groups (resp., $nil_{a}$-groups) of finite rank.

According to Theorem 11, the class of all torsion-free $TI$-groups of finite
rank coincides with $\mathcal{MH}\,\cup\,\mathcal{NIL}_{a}$. However, this
theorem does not explain the interrelations between the classes $\mathcal{MH}$
and $\mathcal{NIL}_{a}$. For example, what can we say about classes

$\mathcal{MH}\,\cap\,\mathcal{NIL}_{a}\qquad\text{and}\qquad\mathcal{MH}\setminus\mathcal{NIL}_{a}\;?$

In particular, what types and positive integers can be realized as types and
ranks of groups in these classes provided the classes are not empty?

We note that

$\mathcal{MH}\,\cap\,\mathcal{NIL}_{a}=\mathcal{MH}\,\cap\,\mathcal{NIL},\quad\mathcal{MH}\setminus\mathcal{NIL}_{a}=\mathcal{MH}\setminus\mathcal{NIL}$

by Theorem 8(1).

If $r(G)=1$, then $G\in\mathcal{MH}$; in addition
$G\in\mathcal{MH}\,\cap\,\mathcal{NIL}$ if and only if the type $t(G)$ is not
an idempotent type.

It follows from Theorem 10 that there exists a ring of field type on every
group in the class $\mathcal{MH}\setminus\mathcal{NIL}$. In [8], it is proved
that every group, which admits a multiplication of simple algebra type, is
quotient divisible. According to [8], a torsion-free group $G$ of finite rank
is said to be quotient divisible if $G$ has a free Abelian subgroup $F$ such
that $G/F$ is a divisible torsion group. Now quotient divisible groups are
actively studied by various authors; e.g., see [1], [16], [20] and others.

In [8], there is a quite simple independent invariant system for quasi-
isomorphism classes of quotient divisible groups. These invariants are
constructed similar to Kurosh, Malcev and Derry invariants for isomorphism
classes of arbitrary torsion-free groups of finite rank; they are based on the
following argument.

Let $G$ be a torsion-free group of finite rank. We set $V=\mathbb{Q}\otimes
G=\bar{G}$, $V^{(p)}={\mathbb{Q}}_{p}^{*}\otimes G$. We assume that the group
$V$ is embedded in the group $V^{(p)}$. If $R$ is a subgroup of the additive
group ${\mathbb{Q}}_{p}^{*}$ containing $\mathbb{Z}$ and $A$ is a subgroup of
the group $V^{(p)}$, then we follow [8] and [25] and, instead of $R\otimes A$,
we write

$RA=\\{r_{1}a_{1}+\ldots+r_{k}a_{k}\,|\,r_{i}\in R,\,a_{i}\in
A,\,i=1,\;\ldots,\;k\\}.$

If $S\subseteq V^{(p)}$ and $A=\langle S\rangle_{R}$, then the group $RA$ is
also denoted by $\langle S\rangle_{R}$.

Let $\delta_{p}(G)$ be the maximal divisible subgroup of the group
$\widehat{\mathbb{Z}}_{p}G$. This group is the maximal divisible submodule of
the $\widehat{\mathbb{Z}}_{p}$-module $\widehat{\mathbb{Z}}_{p}G$. Therefore,
$\delta_{p}(G)$ is a ${\mathbb{Q}}_{p}^{*}$-subspace of the space $V^{(p)}$.

For every $p\in P$, we denote by $\mathcal{L}_{p}$ the subspace lattice of the
${\mathbb{Q}}_{p}^{*}$-space $V^{(p)}$. Let
$\mathcal{L}=\prod\limits_{p}\mathcal{L}_{p}$ be the direct product of these
lattices. Then $\delta_{p}(G)\in\mathcal{L}_{p}$ for every $p\in P$. Following
[8], we call $\delta(G)=(\delta_{p}(G))_{p\in P}\in\mathcal{L}$ the quotient
divisible invariant (q.d.-invariant) of the group $G$.

Below we use the following definition everywhere.

Let $P_{1}\subseteq P$ and $n\in\mathbb{N}$. We say that the set
$\\{\alpha_{p}\in\widehat{\mathbb{Z}}_{p}\,|\,p\in P_{1}\\}$ satisfies
condition $(T)$ for $n\in\mathbb{N}$ if for any polynomial
$f(x)\in\mathbb{Z}[x]$ of degree $\leq n$, the following two conditions are
true.

1) $f(\alpha_{p})\neq 0$ for all $p\in P_{1}$;

2) $f(\alpha_{p})\not\equiv 0\text{ (mod }p)$ for almost all $p\in P_{1}$.

For the set $\\{\alpha_{p}\in\widehat{\mathbb{Z}}_{p}\,|\,p\in P_{1}\\}$, the
elements $\alpha_{p}$, $p\in P_{1}\;$, are called $p$-components of this set.
It is easy to see that if the set
$\\{\alpha_{p}\in\widehat{\mathbb{Z}}_{p}\,|\,p\in P_{1}\\}$ satisfies
condition $(T)$ for $n>1$, then it also satisfies condition $(T)$ for any
positive integer $m<n$.

If $t$ is a type, then we set

$P_{\infty}(t)=\\{p\in P\,|\,t(p)=\infty\\},\;P_{0}(t)=\\{p\in
P\,|\,t(p)<\infty\\}.$

Lemma 13. Let $n\in\mathbb{N}$, $n>1$, $t$ be a type which is not equal to
$(\infty,\infty,\ldots)$, and let the set
$\\{\alpha_{p}\in\widehat{\mathbb{Z}}_{p}\,|\,p\in P_{0}(t)\\}$ satisfy
condition $(T)$ for $n-1$. Then there exists a group $G\in\mathcal{MH}$ of
rank $n$ and type $t$ such that for some basis $e_{1},\;\ldots,\;e_{n}$ of the
space $\bar{G}$, the quotient divisible invariant
$\delta(G)=(\delta_{p}(G))_{p\in P}$ has the form

$\delta_{p}(G)=V^{(p)}\text{ for }p\in P_{\infty}(t),\;\delta_{p}(G)=\langle
u_{p,1},\;\ldots,\;u_{p,n-1}\rangle_{\mathbb{{Q}_{p}^{*}}}\text{ for }p\in
P_{0}(t),$ $\text{where
}u_{p,1}=e_{1}+\alpha_{p}e_{2},\;\ldots,\;u_{p,n-1}=e_{n-1}+\alpha_{p}e_{n}.$

In addition, if the type $t$ is an idempotent type, then the group $G$ is
quotient divisible.

Proof. We set $P_{\infty}=P_{\infty}(t)$, $P_{0}=P_{0}(t)$. Let $p\in P_{0}$.
We write the number $\alpha_{p}$ in the form

$\alpha_{p}=a_{p,0}+a_{p,1}p+a_{p,2}p^{2}+\ldots,\text{ where
}a_{p,k}\in\mathbb{Z},\;0\leq a_{p,k}\leq p-1.$

For every $k\in\mathbb{N}$, we set

$\alpha_{p,k}=a_{p,0}+a_{p,1}p+\ldots+a_{p,k-1}p^{k-1}.$

We take a characteristic $\chi=(\chi_{p})_{p\in P}\in t$. For all $p\in
P_{0}$, $s\in\\{1,\;\ldots,\;n-1\\}$ and $k\in\mathbb{N}$, we consider
elements

$u_{p,s}^{(k)}=e_{s}+\alpha_{p,k}e_{s+1},\qquad
x_{p,s}^{(k)}=p^{-(k+\chi_{p})}u_{p,s}^{(k)}$ $None$

of the space $V=\mathbb{Q}e_{1}\oplus\ldots\oplus\mathbb{Q}e_{n}$ Let $R$ be a
subgroup of the group $\mathbb{Q}$ such that $1\in R$ and $\chi_{R}(1)=\chi$.
In the space $V$, we consider the subgroups

$F=\mathbb{Z}e_{1}\oplus\ldots\oplus\mathbb{Z}e_{n},\;A=Re_{1}\oplus\ldots\oplus
Re_{n},\;$ $G=\langle\,\\{x_{p,s}^{(k)}\,|\,p\in
P_{0},\,s\in\\{1,\;\ldots,\;n-1\\},\,k\in\mathbb{N}\\},\,A\rangle_{R},$
$A_{p,k}=\langle
x_{p,1}^{(k)},\;\ldots,\;x_{p,n-1}^{(k)},e_{n}\rangle_{R},\quad p\in
P_{0},\quad k\in\mathbb{N}.$

Let $p\in P_{0}$, $s\in\\{1,\;\ldots,\;n-1\\}$, $k\in\mathbb{N}$. By $(4)$, we
have $e_{s}=p^{k+\chi_{p}}x_{p,s}^{(k)}-\alpha_{p,k}e_{s+1}$. Consequently,

$A\subseteq\langle
x_{p,1}^{(k)},\;\ldots,\;x_{p,n-1}^{(k)},e_{n}\rangle_{R}=A_{p,k}\text{ for
all }p\in P_{0},\,k\in\mathbb{N}.$ $None$

If $m\in\mathbb{N}$ and $m<k$, then

$p^{m+\chi_{p}}x_{p,s}^{(m)}=e_{s}+\alpha_{p,m}e_{s+1},\;p^{k+\chi_{p}}x_{p,s}^{(k)}=e_{s}+\alpha_{p,k}e_{s+1}.$

Therefore,
$p^{m+\chi_{p}}x_{p,s}^{(m)}-\alpha_{p,m}e_{s+1}=p^{k+\chi_{p}}x_{p,s}^{(k)}-\alpha_{p,k}e_{s+1},$

$p^{m+\chi_{p}}x_{p,s}^{(m)}=p^{k+\chi_{p}}x_{p,s}^{(k)}-(\alpha_{p,k}-\alpha_{p,m})e_{s+1}.$

Then
$x_{p,s}^{(m)}=p^{k-m}x_{p,s}^{(k)}-(a_{p,m}+\ldots+a_{p,k-1}p^{k-m-1})p^{-\chi_{p}}e_{s+1}.$
Consequently,

$x_{p,s}^{(m)}\in\langle
x_{p,s}^{(k)},\;\ldots,\;x_{p,n-1}^{(k)},e_{n}\rangle_{R}=A_{p,k}\,\text{for
all }p\in P_{0},\,m<k.$ $None$

Now let $q\in P_{0}$, $\;q\neq p$, $\;k,m\in\mathbb{N}$. Since

$q^{m+\chi_{q}}x_{q,s}^{(m)}=e_{s}+\alpha_{q,m}e_{s+1}\,\text{and
}\,p^{k+\chi_{p}}x_{p,s}^{(k)}=e_{s}+\alpha_{p,k}e_{s+1},$

we have that

$q^{m+\chi_{q}}x_{q,s}^{(m)}=p^{k+\chi_{p}}x_{p,s}^{(k)}-\alpha_{p,k}e_{s+1}+\alpha_{q,m}e_{s+1}=p^{k+\chi_{p}}x_{p,s}^{(k)}+(\alpha_{q,m}-\alpha_{p,k})e_{s+1}.$

Therefore, it follows from $(5)$ that

$q^{m+\chi_{q}}x_{q,s}^{(m)}\in\langle
x_{p,1}^{(k)},\;\ldots,\;x_{p,n-1}^{(k)},e_{n}\rangle_{R}=A_{p,k}\text{ for
all }p,q\in P_{0},\,k,m\in\mathbb{N}.$ $None$

Let $g\in G$. By $(5)$ and $(6)$, the element $g$ can be represented in the
form $g=\sum_{q\in P_{1}}g_{q}$, where $P_{1}$ is some finite subset in
$P_{0}$ and $g_{q}\in A_{q,m_{q}}$ for some $m_{q}\in\mathbb{N}$. We fix $p\in
P_{0}$. It follows from $(7)$ that there exist integer $b,k\in\mathbb{N}$ such
that

$bg\in A_{p,k}\,\text{ and }\,(b,p)=1.$ $None$

We prove that the subgroup $Re_{n}$ is pure in $G$. Let $re_{n}=p^{m}g$ for
some $p\in P_{0}$, $m\in\mathbb{N}$, $r\in R$, and $g\in G$. Then the element
$g$ satisfies condition $(8)$ for some $b\in\mathbb{N}$, $k\in\mathbb{N}$.
Therefore,

$bre_{n}=bp^{m}g=p^{m}(c_{1}x_{p,1}^{(k)}+\ldots+c_{n-1}x_{p,n-1}^{(k)}+c_{n}e_{n}),$

where $c_{1},\ldots,c_{n}\in R$. By substituting
$x_{p,1}^{(k)},\;\ldots,\;x_{p,n-1}^{(k)}$ from $(4)$ to the right part of
this relation, we obtain that the coefficient of $e_{1}$ is equal to
$p^{m}c_{1}$. Therefore, $c_{1}=0$. Similarly, $c_{2}=\ldots c_{n-1}=0$. Thus,
$bre_{n}=p^{m}c_{n}e_{n}$, whence $re_{n}=p^{m}c_{n}^{\prime}e_{n}$, where
$c_{n}^{\prime}\in R$. Consequently, the subgroup $Re_{n}$ is $p$-pure for
every $p\in P_{0}$. Since $Re_{n}$ is $p$-divisible for all $p\in P_{\infty}$,
we have that $Re_{n}$ is a pure subgroup of the group $G$.

We prove that $G$ is a homogeneous group of type $t$. Let
$a=r_{1}e_{1}+\ldots+r_{n}e_{n}\in F$, where $r_{i}\in\mathbb{Z}$. It is easy
to see that

$\chi(a)\geq\chi,$ $None$

since $\chi(e_{i})=\chi$ for all $i\in\\{1,\;\ldots,\;n\\}$. In particular,
$h_{p}(a)=\infty$ for all $p\in P_{\infty}$.

Let $p\in P_{0}$, $k\in\mathbb{N}$. In the relations below, the symbol
$\equiv$ means congruence modulo $p^{k+\chi_{p}}$. First, we note that

$a=r_{1}e_{1}+\ldots+r_{n}e_{n}\equiv
r_{1}e_{1}+\ldots+r_{n}e_{n}-r_{1}p^{k+\chi_{p}}x_{p,1}^{(k)}=$
$=r_{1}e_{1}+\ldots+r_{n}e_{n}-r_{1}(e_{1}+\alpha_{p,k}e_{2})=(r_{2}-r_{1}\alpha_{p,k})e_{2}+r_{3}e_{3}\ldots+r_{n}e_{n},$

and for any $s\in\\{2,\;\ldots,\;n-1\\}$, it follows from $(4)$ that

$(r_{s}-r_{s-1}\alpha_{p,k}+r_{s-2}\alpha_{p,k}^{2}+\ldots+(-1)^{s-1}r_{1}\alpha_{p,k}^{s-1})e_{s}+r_{s+1}e_{s+1}+\ldots+r_{n}e_{n}\equiv$
$\equiv(r_{s}-r_{s-1}\alpha_{p,k}+r_{s-2}\alpha_{p,k}^{2}+\ldots+(-1)^{s-1}r_{1}\alpha_{p,k}^{s-1})e_{s}+r_{s+1}e_{s+1}+\ldots+r_{n}e_{n}-$
$-(r_{s}-r_{s-1}\alpha_{p,k}+r_{s-2}\alpha_{p,k}^{2}+\ldots+(-1)^{s-1}r_{1}\alpha_{p,k}^{s-1})p^{k+\chi_{p}}x_{p,s}^{(k)}=$
$r_{s+1}e_{s+1}+\ldots+r_{n}e_{n}-(r_{s}-r_{s-1}\alpha_{p,k}+\ldots+(-1)^{s-1}r_{1}\alpha_{p,k}^{s-1})\alpha_{p,k}e_{s+1}=$
$=(r_{s+1}-r_{s}\alpha_{p,k}+r_{s-2}\alpha_{p,k}^{2}+\ldots+(-1)^{s}r_{1}\alpha_{p,k}^{s})e_{s+1}+r_{s+2}e_{s+2}+\ldots+r_{n}e_{n}.$

Consequently, $a=r_{1}e_{1}+\ldots+r_{n}e_{n}\equiv$

$\equiv(r_{n}-r_{n-1}\alpha_{p,k}+r_{n-2}\alpha_{p,k}^{2}+\ldots+(-1)^{n-1}r_{1}\alpha_{p,k}^{n-1})e_{n}\,\text{
(mod }p^{k+\chi_{p}}).$ $None$

We set
$f(x)=r_{n}-r_{n-1}x+r_{n-2}x^{2}+\ldots+(-1)^{n-1}r_{1}x^{n-1}\in\mathbb{Z}[x].$
Then the relation $(10)$ can be given in the form

$a\equiv f(\alpha_{p,k})e_{n}\;(\text{mod }p^{k+\chi_{p}}).$ $None$

We assume that $p^{k}$ divides $a$ for every $k\in\mathbb{N}$. Then
$f(\alpha_{p,k})e_{n}\equiv 0\text{ (mod }p^{k+\chi_{p}})$ for every
$k\in\mathbb{N}$ by $(11)$. Since the subgroup $Re_{n}$ is pure in $G$, we
have that $f(\alpha_{p,k})\equiv 0\text{ (mod }p^{k})$ for every
$k\in\mathbb{N}$. Therefore, the sequence of integers
$\\{f(\alpha_{p,k})\\}_{k\in\mathbb{N}}$ is a Cauchy sequence in the $p$-adic
topology on $\widehat{\mathbb{Z}}_{p}$ and the limit
$\lim\\{f(\alpha_{p,k})\\}_{k\in\mathbb{N}}$ of this sequence is equal to
zero. Consequently,

$0=\lim\\{f(\alpha_{p,k})\\}_{k\in\mathbb{N}}=f(\lim\\{\alpha_{p,k}\\}_{k\in\mathbb{N}})=f(\alpha_{p}).$

This contradicts to the property that the set $\\{\alpha_{p}\,|\,p\in
P_{0}\\}$ satisfies condition $(T)$ for $n-1$. Therefore, $h_{p}(a)<\infty$
for all $p\in P_{0}$.

Let $P_{2}=\\{p\in P_{0}\,|\,p^{\chi_{p}+1}|a\\}$. If $p\in P_{2}$, then

$0\equiv a\equiv f(\alpha_{p,\chi_{p}+1})e_{n}\text{ (mod }p^{\chi_{p}+1})$

by $(11)$. Therefore, $f(\alpha_{p,\chi_{p}+1})\equiv 0\text{ (mod }p)$ for
all $p\in P_{2}$. Since the set $\\{\alpha_{p}\,|\,p\in P_{0}\\}$ satisfies
condition $(T)$ for $n-1$, we have that $P_{2}$ is a finite set. Therefore,
$h_{p}(a)=\chi_{p}$ for almost all $p\in P_{0}$. Therefore, $t(a)=t$ by $(9)$.
Since $G/F$ is a torsion group, $G$ also is a homogeneous group of type $t$.

For every $p\in P$, we consider the divisible part $\delta_{p}(G)$ of the
group $\widehat{\mathbb{Z}}_{p}G$. If $p\in P_{\infty}$, then the group
$\widehat{\mathbb{Z}}_{p}G$ is divisible, whence

$\delta_{p}(G)=\mathbb{Q}_{p}^{*}e_{1}\oplus\ldots\oplus\mathbb{Q}_{p}^{*}e_{n}=V^{(p)}.$

Let $p\in P_{0}$. Since the group $\widehat{\mathbb{Z}}_{p}G$ is not
divisible, we have
$\text{dim}_{\mathbb{Q}_{p}^{*}}\delta_{p}(G)\leq n-1$. We consider elements

$u_{p,1}=e_{1}+\alpha_{p}e_{2},\;\ldots,\;u_{p,n-1}=e_{n-1}+\alpha_{p}e_{n}$

of the group $\widehat{\mathbb{Z}}_{p}G$. It is easy to see that
$u_{p,1},\;\ldots,\;u_{p,n-1}$ are linearly independent over
$\mathbb{Q}_{p}^{*}$. In addition, the pure subgroup $\langle
u_{p,1},\;\ldots,\;u_{p,n-1}\rangle_{*}$ of the group
$\widehat{\mathbb{Z}}_{p}G$ is divisible, since

$u_{p,s}=e_{s}+\alpha_{p}e_{s+1}=e_{s}+\alpha_{p,k}e_{s+1}+(\alpha_{p}-\alpha_{p,k})e_{s+1}=$
$=p^{k+\chi_{p}}x_{p,s}^{(k)}+(a_{p,k}p^{k}+a_{p,k+1}p^{k+1}+\ldots)e_{s+1}=$
$=p^{k}(p^{\chi_{p}}x_{p,s}^{(k)}+(a_{p,k}+a_{p,k+1}p+\ldots)e_{s+1})$

for all $s\in\\{1,\;\ldots,\;n-1\\}$ and $k\in\mathbb{N}$. Consequently,
$\delta_{p}(G)=\langle u_{p,1},\,\;\ldots,\;\,u_{p,n-1}\rangle_{Q_{p}^{*}}$.

According to [25], the relation $r_{p}(G)+\text{dim
}_{\mathbb{Q}_{p}^{*}}\delta_{p}(G)=n$ is true for every $p\in P$. Therefore,
$r_{p}(G)=0$ for $p\in P_{\infty}$ and $r_{p}(G)=1$ for $p\in P_{0}$.
Consequently, $G$ is a Murley group.

It is easy to see that

$G/A=\langle x_{p,s}^{(k)}+A\;|\;p\in
P_{0},\,s\in\\{1,\;\ldots,\;n-1\\},\,k\in\mathbb{N}\rangle$

is a divisible torsion group. If $t$ is an idempotent type, then
$R/\mathbb{Z}$ is a divisible torsion group; this implies that $A/F$ is a
divisible torsion group. Since $G/A\cong(G/F)/(A/F)$, we have that $G/F$ also
is a divisible torsion group. Consequently, the group $G$ is quotient
divisible. $\square$

Lemma 14. For every $P_{1}\subseteq P$ and each $n\in\mathbb{N}$, there exist
$2^{\aleph_{0}}$ sets $\\{\alpha_{p}\in\widehat{\mathbb{Z}}_{p}\,|\,p\in
P_{1}\\}$ such that the following properties are true.
The sets $\\{\alpha_{p}\in\widehat{\mathbb{Z}}_{p}\,|\,p\in P_{1}\\}$ satisfy
condition $(T)$ for $n$.
Any two of these sets have algebraically independent $p$-components for at
least one $p\in P_{1}$.

Proof. Let $\\{p_{k}\\}_{k\in\mathbb{N}}$ be the sequence of all prime
integers ordered in ascending order. Let $B_{n}$ be the set of all polynomials
in $\mathbb{Z}[x]$, whose degrees do not exceed $n$ and the coefficients are
coprime in total, and let $\\{f_{k}(x)\\}_{k\in\mathbb{N}}$ be the sequence of
all polynomials in $B_{n}$ which is ordered so that the leading coefficient of
the polynomial $f_{k}(x)$ is less than $p_{k}$, $k\in\mathbb{N}$.

By [15], there exists an integer $c\in\mathbb{N}$ such that $p_{k}>ck\ln k$
for all $k\in\mathbb{N}$. It is easy to see that $c\ln k_{0}>n+1$ for some
$k_{0}\in\mathbb{N}$.

Let $k\geq k_{0}$. The congruence

$f_{1}(x)\cdot\ldots\cdot f_{k}(x)\equiv 0\text{ (mod }p_{k})$ $None$

has at most $\text{deg }(f_{1}(x)\cdot\ldots\cdot f_{k}(x))$ solutions
provided the leading coefficient of the polynomial $f_{1}(x)\cdot\ldots\cdot
f_{k}(x)$ is not divided by $p_{k}$. If $m_{s}$ is the leading coefficient
polynomial $f_{s}(x)$, $s\in\\{1,\;\ldots,\;k\\}$, then $m_{s}<p_{k}$.
Therefore, the leading coefficient of the polynomial $f_{1}(x)\cdot\ldots\cdot
f_{k}(x)$, which is equal to $m_{1}\cdot\ldots\cdot m_{k}$, is not divided by
$p_{k}$. In addition,

$\text{deg }(f_{1}(x)\cdot\ldots\cdot f_{k}(x))\leq kn<k(c\ln k_{0}-1)\leq
kc\ln k-k<p_{k}-k.$

Consequently, there exists a positive integer $a_{k}\in\\{1,\ldots,p_{k}-1\\}$
which is not a solution of comparison $(12)$.

By choosing arbitrary positive integers $a_{k}\in\\{1,\ldots,p_{k}-1\\}$ for
$k<k_{0}$, we obtain a sequence $\\{a_{k}\\}_{k\in\mathbb{N}}$ such that
$a_{k}$ is not a solution of $(12)$ for almost all $k\in\mathbb{N}$. For every
$k\in\mathbb{N}$, we take an element
$\beta_{k}\in\widehat{\mathbb{Z}}_{p_{k}}$ which is transcendental over
$\mathbb{Q}$ and we set $\alpha_{p_{k}}=a_{k}+p_{k}\beta_{k}$. Then the
element $\alpha_{p_{k}}$ is transcendental over $\mathbb{Q}$ for every
$k\in\mathbb{N}$.

We prove that the set $\\{\alpha_{p}\,|\,p\in P_{1}\\}$ satisfies property 2)
of condition $(T)$ for $n$. Let $f(x)\in\mathbb{Z}[x]$, deg$\,f(x)\leq n$.
Then $f(x)=f_{k_{1}}(x)$ for some $k_{1}\in\mathbb{N}$. Let $k_{2}=\text{max
}\\{k_{0},k_{1}\\}$. For for all $k\geq k_{2}$, we have

$f_{1}(\alpha_{p_{k}})\cdot\ldots\cdot f_{k}(\alpha_{p_{k}})\equiv
f_{1}(a_{k})\cdot\ldots\cdot f_{k}(a_{k})\not\equiv 0\,\text{ (mod }p_{k})$

by the choice of the numbers $a_{k}$. Consequently,

$f(\alpha_{p_{k}})=f_{k_{1}}(\alpha_{p_{k}})\not\equiv 0\text{ (mod }p_{k}).$

Therefore, the set $\\{\alpha_{p}\,|\,p\in P\\}$ satisfies condition $(T)$ for
$n$. Therefore, the set $\\{\alpha_{p}\,|\,p\in P_{1}\\}$ also satisfies
condition $(T)$ for $n$.

Let $N_{1}=\\{k\in\mathbb{N}\;|\;p_{k}\in P_{1}\\}$. We assume that for every
$k\in N_{1}$, we have chosen transcendental elements
$\beta_{k}^{\prime}\in\widehat{\mathbb{Z}}_{p_{k}}$ such that $\beta_{m}$ and
$\beta_{m}^{\prime}$ are algebraically independent for some
$m\in\mathbb{N}_{1}$. We set
$\alpha_{p_{k}}^{\prime}=a_{k}+p_{k}\beta_{k}^{\prime}$, $k\in\mathbb{N}_{1}$.
By the above, the set $\\{\alpha_{p}^{\prime}\,|\,p\in P_{1}\\}$ satisfies
condition $(T)$; in addition $\alpha_{p_{m}}$ and $\alpha_{p_{m}}^{\prime}$
are algebraically independent.

For every $p\in P_{1}$, the group $\mathbb{Q}_{p}^{*}$ has $2^{\aleph_{0}}$
transcendental algebraically independent elements. Therefore, there exist
$2^{\aleph_{0}}$ sets $\\{\alpha_{p}\in\widehat{\mathbb{Z}}_{p}\,|\,p\in
P_{1}\\}$ which satisfy condition $(T)$ for $n$ and each two of them have
algebraically independent $p$-components for at least one $p\in P_{1}$.
$\square$

Theorem 15. For every $n\in\mathbb{N}$, $n>1$, and every type
$t\neq(\infty,\infty,\ldots)$, the class $\mathcal{MH}\,\cap\,\mathcal{NIL}$
contains $2^{\aleph_{0}}$ pairwise non-quasi-isomorphic groups from of rank
$n$ and type $t$.

Proof. Let $n\in\mathbb{N}$, $n>1$, $t$ be a type not equal to
$(\infty,\infty,\ldots)$, $P_{\infty}=P_{\infty}(t)$, and let
$P_{0}=P_{0}(t)$. By Lemma 14, there exists a set
$\\{\alpha_{p}\in\widehat{\mathbb{Z}}_{p}\,|\,p\in P_{0}\\}$ which satisfies
condition $(T)$ for $n$.

By Lemma 13, there exists a group $G\in\mathcal{MH}$ of rank $n$ and type $t$.
In addition, there exists a basis $e_{1},\;\ldots,\;e_{n}$ of the space
$V=\bar{G}$ such that

$\delta_{p}(G)=\begin{cases}\widehat{\mathbb{Z}}_{p}G=\mathbb{Q}_{p}^{*}G\text{
for }p\in P_{\infty}\\\ \langle
u_{p,1},\;\ldots,\;u_{p,n-1}\rangle_{\mathbb{Q}_{p}^{*}}\text{ for }p\in
P_{0},\end{cases}$

where $u_{p,s}=e_{s}+\alpha_{p}e_{s+1}$, $s\in\\{1,\;\ldots,\;n-1\\}$.

Let the set $\\{\gamma_{p}\in\widehat{\mathbb{Z}}_{p}\,|\,p\in P_{0}\\}$ also
satisfy condition $(T)$ for $n$. By Lemma 13, there also exists a group
$G^{\prime}\in\mathcal{MH}$ of rank $n$ and type $t$ such that for some basis
$e_{1}^{\prime},\;\ldots,\;e_{n}^{\prime}$ of the space
$\overline{G^{\prime}}$, the spaces $\delta_{p}(G^{\prime})$, $p\in P_{0}$
have the form

$\delta_{p}(G^{\prime})=\begin{cases}\widehat{\mathbb{Z}}_{p}G^{\prime}=\mathbb{Q}_{p}^{*}G^{\prime}\text{
for }p\in P_{\infty}\\\ \langle
v_{p,1},\;\ldots,\;v_{p,n-1}\rangle_{\mathbb{Q}_{p}^{*}}\text{ for }p\in
P_{0},\end{cases}$

where $v_{p,s}=e_{s}^{\prime}+\gamma_{p}e_{s+1}^{\prime}$,
$s\in\\{1,\;\ldots,\;n-1\\}$.

Let $\varphi\colon G\to G^{\prime}$ be a group homomorphism and

$\varphi(e_{s})=c_{s,1}e_{1}^{\prime}+\ldots+c_{s,n}e_{n}^{\prime},\;c_{s,i}\in\mathbb{Q},\;s,i\in\\{1,\;\ldots,\;n-1\\}.$

For any $p\in P$, we have a homomorphism

$\varphi^{(p)}=\text{id}_{\mathbb{Q}_{p}^{*}}\otimes\varphi\colon\widehat{\mathbb{Z}}_{p}G\to\widehat{\mathbb{Z}}_{p}G.$

Let $p\in P_{0}$, $s\in\\{1,\;\ldots,\;n-1\\}$. Since
$\varphi^{(p)}(\delta_{p}(G))\subseteq\delta_{p}(G^{\prime})$, we have

$\varphi^{(p)}(u_{p,s})=m_{1}v_{p,1}+\ldots+m_{n-1}v_{p,n-1}=$
$=m_{1}e_{1}^{\prime}+(m_{1}\gamma_{p}+m_{2})e_{2}^{\prime}+\ldots+(m_{n-2}\gamma_{p}+m_{n-1})e_{n}^{\prime}+m_{n-1}\gamma_{p}e_{n}^{\prime},$

where $m_{i}\in\mathbb{Q}_{p}^{*}$, $i\in\\{1,\;\ldots,\;n-1\\}$. On the other
hand,

$\varphi^{(p)}(u_{p,s})=\varphi^{(p)}(e_{s}+\alpha_{p}e_{s+1})=\varphi(e_{s})+\alpha_{p}\varphi(e_{s+1})=$
$=\sum_{i=1}^{n}c_{s,i}e_{i}^{\prime}+\alpha_{p}\sum_{i=1}^{n}c_{s+1,i}e_{i}^{\prime}=b_{1}e_{1}^{\prime}+\ldots+b_{n}e_{n}^{\prime},$

where $b_{i}=c_{s,i}+\alpha_{p}c_{s+1,i}$.

By equating the coefficients of $e_{1}^{\prime},\;\ldots,\;e_{n}^{\prime}$ in
these relations, we obtain

$m_{1}=b_{1},\;m_{1}\gamma_{p}+m_{2}=b_{2},\;\ldots,\;m_{n-2}\gamma_{p}+m_{n-1}=b_{n-1},\quad
m_{n-1}\gamma_{p}=b_{n}.$ $None$

It follows from $(13)$ that

$0=b_{n}-m_{n-1}\gamma_{p}=b_{n}-(b_{n-1}-m_{n-2}\gamma_{p})\gamma_{p}=b_{n}-b_{n-1}\gamma_{p}+m_{n-2}\gamma_{p}^{2}=$
$=b_{n}-b_{n-1}\gamma_{p}+b_{n-2}\gamma_{p}^{2}+\ldots+(-1)^{k}b_{n-k}\gamma_{p}^{k}+\ldots+(-1)^{n-1}b_{1}\gamma_{p}^{n-1}=$
$=(c_{s,n}+\alpha_{p}c_{s+1,n})-(c_{s,n-1}+\alpha_{p}c_{s+1,n-1})\gamma_{p}+\ldots+(-1)^{n-1}(c_{s,1}+\alpha_{p}c_{s+1,1})\gamma_{p}^{n-1}.$

Therefore, $\alpha_{p}$ and $\gamma_{p}$ satisfy the relation

$c_{s,n}+c_{s+1,n}\alpha_{p}-c_{s,n-1}\gamma_{p}-c_{s+1,n-1}\alpha_{p}\gamma_{p}+$
$+\ldots+(-1)^{n-1}c_{s,1}\gamma_{p}^{n-1}+(-1)^{n-1}c_{s+1,1}\alpha_{p}\gamma_{p}^{n-1}=0$
$None$

for all $p\in P_{0}$ , $s\in\\{1,\ldots,n-1\\}$

We consider two cases.

Case 1. $G^{\prime}=G$.

In this case, we choose the basis $e_{1}^{\prime},\ldots,e_{n}^{\prime}$ of
the space $V$ and bases $v_{p,1},\ldots,v_{p,n-1}$ of the spaces
$\delta_{p}(G)$ for $p\in P$ coinciding with the basis $e_{1},\ldots,e_{n}$
and the basis $u_{p,1},\ldots,u_{p,n-1}$, $p\in P$, respectively.

Let $p\in P_{0}$, $s\in\\{1,\,\ldots\,n-1\\}$. Then $\gamma_{p}=\alpha_{p}$
and relation $(14)$ takes the form

$d_{0}+d_{1}\alpha_{p}+\ldots+d_{n}\alpha_{p}^{n}=0,\quad\text{where}$
$d_{0}=c_{s,n},\quad d_{1}=c_{s+1,n}-c_{s,n-1},\quad
d_{2}=(-1)(c_{s+1,n-1}-c_{s,n-2})$
$\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots$ $None$
$d_{n-1}=(-1)^{n-2}(c_{s+1,2}-c_{s,1}),\quad d_{n}=(-1)^{n-1}c_{s+1,1}.$

Since the set $\\{\alpha_{p}\,|\,p\in P_{0}\\}$ satisfies condition $(T)$ for
$n$ and $d_{i}\in\mathbb{Q}$, $i=0,\;\ldots,\;n$, we have
$d_{0}=\ldots=d_{n}=0$. It follows from $(15)$ that

$c_{s,n}=0,\;c_{s+1,1}=0,\;c_{s,r}=c_{s+1,r+1}\text{ for all
}s,r\in\\{1,\;\ldots,\;n-1\\}.$ $None$

Therefore, if $s<r<n$, then we have
$c_{s,r}=c_{s+1,r+1}=\ldots=c_{n-(r-s),n}=0$, by $(16)$. If $s>r>1$, then
$c_{s,r}=c_{s-1,r-1}=\ldots=c_{s-r+1,1}=0$; in addition
$c_{1,1}=c_{2,2}=\ldots=c_{n,n}$.

Consequently, $\varphi(x)=c_{1,1}x$ for all $x\in G$, whence $\text{End
}G\subseteq\mathbb{Q}$. By [11, Lemma 2.1.8], $G$ is a $nil$-group.

Case 2. $G^{\prime}\neq G$ and the sets $\\{\alpha_{p}\,|\,p\in P_{0}\\}$ and
$\\{\gamma_{p}\,|\,p\in P_{0}\\}$ are chosen so that $\alpha_{p}$,
$\gamma_{p}$ are algebraically independent over $\mathbb{Q}$ for at least one
$p\in P_{0}$ (such sets exist, by Lemma 14).

In this case, it follows from $(14)$ that $c_{s,i}=0$ for all
$s,i\in\\{1,\;\ldots,\;n-1\\}$, whence $\varphi=0$. Consequently $\text{Hom
}(G,G^{\prime})=0$ and the groups $G$ and $G^{\prime}$ are not quasi-
isomorphic to each other.

By Lemma 14, there exist $2^{\aleph_{0}}$ sets
$\\{\alpha_{p}\in\widehat{\mathbb{Z}}_{p}\,|\,p\in P_{0}\\}$ which satisfy
condition $(T)$ for $n$ and each two of which have algebraically independent
$p$-components for at least one $p\in P_{0}$. Therefore, the class
$\mathcal{MH}\,\cap\,NIL$ has $2^{\aleph_{0}}$ pairwise non-quasi-isomorphic
groups of rank $n$ and type $t$. $\square$

For every polynomial $f(x)\in\mathbb{Z}[x]$, we set $P(f)=\\{p\in P\,|\,f(x)$
has a root in $\mathbb{Q}_{p}^{*}\\}$.

Remark 16. It follows from the proof of [8, Theorem 8.4] that if
$f(x)\in\mathbb{Z}[x]$ and $\text{deg }f(x)>1$, then the set $P(f)$ is
infinite. On the other hand, if $f(x)$ is irreducible in $\mathbb{Z}[x]$ and
deg $f(x)>1$, then by [14], the set $P\setminus P(f)$ is also infinite.

Theorem 17. Let $n\in\mathbb{N}$, $n>1$, and let $t$ be a type which is not
equal to $(\infty,\infty,\ldots)$. A group
$G\in\mathcal{MH}\setminus\mathcal{NIL}$ of rank $n$ and type $t$ exists if
and only if $t$ is an idempotent type and $P_{0}(t)\subseteq P(f)$ for some
irreducible polynomial $f(x)\in\mathbb{Z}[x]$ of degree $n$.

Proof. Let $G\in\mathcal{MH}\setminus\mathcal{NIL}$, $r(G)=n>1$, $t(G)=t$. By
Theorem 10, there exists a ring of field type on $G$. By [8, Corollary 4.9],
the type $t$ is an idempotent type. Since $r_{p}(G)+\text{dim
}_{\mathbb{Q}_{p}^{*}}\delta_{p}(G)=n$, $p\in P$, we have that

$\text{dim }_{\mathbb{Q}_{p}^{*}}\delta_{p}(G)=\begin{cases}n\text{ if
}r_{p}(G)=0\\\ n-1\text{ if }r_{p}(G)=1.\end{cases}$

It follows from [8, Corollary 8.5] that the group $G$ is quotient divisible;
in addition, there exist an irreducible polynomial $f(x)\in\mathbb{Z}[x]$ of
degree $n$ and a basis $e_{1},\;\ldots,\;e_{n}$ of the space $\bar{G}$ such
that

$\delta_{p}(G)=\begin{cases}V^{(p)},\text{ if }r_{p}(G)=0\\\ \langle
e_{1}+\alpha_{p}e_{2},\;\ldots,\;e_{n-1}+\alpha_{p}e_{n}\rangle_{Q_{p}^{*}},\text{
if }r_{p}(G)=1,\end{cases}$

where $\alpha_{p}$ is a root of $f(x)$ contained in $\mathbb{Q}_{p}^{*}$.

Let $p\in P_{0}(t)$. Then $\delta_{p}(G)\neq V^{(p)}$. Consequently, $f(x)$
has a root $\alpha_{p}\in\mathbb{Q}_{p}^{*}$, i.e., $p\in P(f)$. Therefore,
$P_{0}(t)\subseteq P(f)$.

Now let $n\in\mathbb{N}$, $n>1$, $t$ be an idempotent type which is not equal
$(\infty,\infty,\ldots)$, and let $P_{0}(t)\subseteq P(f)$ for some
irreducible polynomial $f(x)\in\mathbb{Z}[x]$ of degree $n$. Then for every
$p\in P_{0}(t)$, the polynomial $f(x)$ has a root
$\alpha_{p}\in\mathbb{Q}_{p}^{*}$.

It is easy to see that the set $\\{\alpha_{p}\,|\,p\in P_{0}(t)\\}$ satisfies
condition $(T)$ for $n-1$. Indeed, let $g(x)\in\mathbb{Z}[x]$, $\text{deg
}g(x)\leq n-1$. Then for every $p\in P_{0}(t)$, the element $\alpha_{p}$ is
not a root of $g(x)$, since $f(x)$ is a minimal polynomial for $\alpha_{p}$.
If $h(x)$ is the largest common divisor of the polynomials $f(x)$ and $g(x))$
in the factorial ring $\mathbb{Z}[x]$, then $h(x)=c\in\mathbb{Z}$, since the
polynomial $f(x)$ is irreducible. For every $p\in P_{0}(t)$, the relation
$g(\alpha_{p})\equiv 0\text{ (mod }p)$ holds if and only if
$c=h(\alpha_{p})\equiv 0\text{ (mod }p)$. Consequently,
$g(\alpha_{p})\not\equiv 0\text{ (mod }p)$ for almost all $p\in P_{0}(t)$.

By Lemma 13, there exists a group $G\in\mathcal{MH}$ of rank $n$ and type $t$
such that for some basis $e_{1},\;\ldots,\;e_{n}$ of the space $\bar{G}$, we
have

$\delta_{p}(G)=\begin{cases}V^{(p)}\text{ for }p\in P_{\infty}(t)\\\ \langle
e_{1}+\alpha_{p}e_{2},\;\ldots,\;e_{n-1}+\alpha_{p}e_{n}\rangle_{\mathbb{Q}_{p}^{*}}\text{
for }p\in P_{0}(t).\end{cases}$

By [8, Corollary 8.5], there exists a ring of field type on the group $G$.
Therefore, $G\in\mathcal{MH}\setminus\mathcal{NIL}$. $\square$

It follows from Theorem 17 and Remark 16 that there does not exist a
homogeneous Murley group of zero type $(0;0;,\ldots)$ which admits a non-zero
multiplication. We can formulate a more general assertion.

Corollary 18. If $t$ is a type and the set $P_{\infty}(t)$ is finite, then the
class $\mathcal{MH}\setminus\mathcal{NIL}$ does not contain a group of type
$t$.

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