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Discrete &Continuous Models
&Applied Computational Science
2021, 29 (4) 387–398
http://journals.rudn.ru/miph
Research article
On involutive division on monoids
Oleg K. Kroytor, Mikhail D. Malykh,
Peoples’ Friendship University of Russia (RUDN University)
6, Miklukho-Maklaya St., Moscow, 117198, Russian Federation
Meshcheryakov Laboratory of Information Technologies
Joint Institute for Nuclear Research, Dubna, Russia
6, Joliot-Curie St., Dubna, Moscow Region, 141980, Russian Federation
Key words and phrases:
1. Introduction
http://creativecommons.org/licenses/by/4.0/
&
2. Divisions on monoids
Denition 1.
Denition 2.
∗ ∗
′ ′
′ ′′
′
′
Remark 1.
ℒ(𝑈)
Example 1.
∗ ∗
Denition 3.
𝑑
Theorem 1.
If
1𝑠
are multiplicative elements of
for
with respect
to , then 𝑗1
1𝑗𝑠
𝑠.
Proof.
𝑑
Theorem 2.
Suppose that a nite set
generates a monoid
and some
mapping is given 𝑑 .
Let us dene the function
as follows:
if and only if
and there exists a product of
∗
elements from
𝑑
such that
∗
. The function
denes division by
if and only if the
embedding ′implies
𝑑𝑑′ ′
Remark 2.
𝑥∈𝑋𝑗𝑥
𝑥∈𝑋𝑗𝑥
Proof.
∗
∗
′
′
𝑑
′′
′ ′
′
𝑑
′
&
′
′
′
∗
𝑑
∗
𝑑 𝑑′
Example 2.
1𝑛 𝑖𝑗1
1𝑗𝑛
𝑛𝑖
𝑖𝑑𝑖
𝑣∈𝑈 𝑖
′
𝑖𝑑
𝑖 ′ ′
𝑣∈𝑈′𝑖
𝑣∈𝑈 𝑖
3. Involutive divisions on monoids
Denition 4.
′
′ ′
Remark 3.
′
′
′
′
′′
′
′
′ ′′
Denition 5.
Example 3.
𝑛
𝑖
𝑤∈𝑈 𝑖
𝑑 𝑖𝑖𝑖
𝑖𝑑 𝑖 𝑖𝑖 𝑖𝑖𝑖
𝑖𝑑 𝑖 𝑖𝑖 𝑖𝑖𝑖
𝑖
𝑖𝑖𝑖𝑖
𝑖𝑖
𝑖𝑖
𝑖𝑖𝑖
𝑖 𝑖𝑖𝑖𝑖
𝑖𝑖𝑖
4. Complete sets and completely involutive divisions
Denition 6.
Denition 7.
𝑑
𝑑
Denition 8.
𝑑
Remark 4.
Example 4.
𝑛
𝑖𝑖
&
𝑖𝑖𝑖
𝑖𝑖𝑖
1𝑛
𝑖𝑖𝑖
𝑖𝑖𝑖𝑖𝑖
Denition 9.
∗
∗
∗
𝑑∗∗
∗
Denition 10.
Remark 5.
Example 5.
∗ 𝑛
𝑖𝑖𝑖
∗
∗
∗
∗
∗ ∗
𝑛
∗
𝑖𝑖∗
∗
𝑖𝑖
∗
∗
𝑖∗𝑖𝑖
∗
𝑖𝑖𝑖𝑖
∗𝑇∗
∗
Remark 6.
5. Necessary and sucient conditions
for the completeness of a set
Theorem 3 (necessary completeness condition).
Let the monoid
be
generated by elements of a nite set
. For the set
to be complete with
respect to the division of , it is necessary that
Denition 11.
01
𝑖
𝑖𝑖𝑖
𝑖+1 𝑖𝑖
Denition 12.
Remark 7.
Remark 8.
Theorem 4. If is a nite involutive division on the monoid generated
by elements of the nite set
, then for the set
to be complete it is necessary
and sucient that condition be satised.
Proof.
0
0
0
0
000
1 100
1
1 111
2
211
01
&
Example 6.
𝑖𝑘𝑖
𝑖+1 𝑗𝑖+1 𝑗𝑖𝑘𝑖𝑗
𝑖𝑗𝑖+1 𝑗𝑖𝑗𝑗𝑖
𝑖 𝑖
𝑘𝑖𝑖+1 𝑘𝑖𝑖𝑘𝑖𝑘𝑖𝑖
𝑘𝑖𝑖 𝑘𝑖
𝑖
𝑖+1 𝑖𝑘𝑖
6. Set completion
Problem.
∗
1
∗
∗
∗
𝑛
𝑛
𝑛∗
∗
∗
∗
1∗
Theorem 5. If , then is not empty.
Proof.
∗
1
1
11
1
1
1
111
1
1
1
11
111
1 111
11
2
12
2 112′
1122 21′
1
1
1
1
12
1122
22
2
𝑛
∗
Example 7.
∗
𝑖𝑖𝑖
𝑖𝑖𝑖
∗
𝑖𝑖
7. Discussion
&
Acknowledgments
References
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Programming and Computer Software
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Saratovskogo universiteta
https://events.rudn.ru/event/102
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For citation:
Information about the authors
Kroytor, Oleg K.
kroytor_ok@pfur.ru
Malykh, Mikhail D.
malykh_md@pfur.ru
&
Об инволютивном делении на моноидах
О. К. Кройтор, М. Д. Малых,
Российский университет дружбы народов
ул. Миклухо-Маклая, д. 6, Москва, 117198, Россия
Лаборатория информационных технологий им. М. Г. Мещерякова
Объединённый институт ядерных исследований
ул. Жолио-Кюри, д. 6, Дубна, Московская область, 141980, Россия
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