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Questions tagged [quasigroups]

A quasigroup is a grouplike structure $(Q, \ast)$, that satisfies the Latin square property but need not have an identity element, nor need it be associative. It coincides with the notion of a divisible magma.

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Term for a Set Equipped With a Binary Operation Which Contains Inverses

Let $A$ be a set and let $\circ:A\times A\rightarrow B,$ $A\subseteq B$ be a binary operation ($A$ is not necessarily closed under $\circ$). If there exists some unique $e\in A$ such that $e\circ a=a\...
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Salzmann's topological loop example

I've been looking for loop examples and stumbled upon this post. The answer describes a specific operation on $\mathbb{R} $, namely this: $$x\ast t = \begin{cases} \ x+\frac{1}{2}t, &\text{ if }\ ...
Dmitry Ivanov's user avatar
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Can we characterize the “associate classes” of a unipotent quasi-commutative quasigroup as some combinatorial design?

$I_n$ is the $n \times n$ or order $n$ identity matrix, $J_n$ is the order $n$ matrix of all ones, and $n \in \mathbb{Z}^+$. We define a Latin square $\mathcal{L_n}$ to be a set of $n$ permutation ...
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How can a quasigroup have a division operation if a group has only one operation?

I'm teaching myself abstract algebra. We can define a magma $\left(M,\cdot\right)$ as $a,b\in M\implies a\cdot b\in M$. I get confused when we talk about quasigroups. A group must satisfy all ...
Oliver's user avatar
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References to for Quasigroup Theory

Can anyone suggest some references (books or articles) to understand Quasigroup Theory. I need very easy to understand reference if there is one. Thank you very much. Edit: Note: I have checked ...
Jins's user avatar
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Uniqueness of quotients in semigroup with divisibility?

Let $G$ be a semigroup, e.g., a set with an associative binary operation. Suppose further that $G$ has the divisibility property, e.g., for all $x,y\in G$ there exist $\ell,r\in G$ such that $\ell x=y$...
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Matrix representation of Quasigroups

This paper says that, each quasigroup of order 4 can be represented in matrix form using the following equation, \begin{equation} x \ast y \equiv m^T +Ax^T +By^T +CA\cdot x^T \circ CB\cdot y^T \end{...
Anisha's user avatar
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Can $S^2$ be given a (semi-)topological quasigroup structure?

It is known from this question that $S^2$ cannot be made into a topological group. Indeed, $S^2$ cannot even be made into an $H$-Space, a much looser requirement than a topological group. However, an $...
volcanrb's user avatar
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Do the Moufang identities *themselves* imply diassociativity / Moufang's theorem / Artin's theorem?

A Moufang loop is a loop satisfying the Moufang identities. Famously, these are diassociative -- the subloop generated by any two elements is associative (is a group) -- and more generally, they ...
Harry Altman's user avatar
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Why is the formal definition of Latin square equivalent with the informal?

Informally, a latin square is a table where each element appears exactly once in each row and each column. I know that this is probrably not an official definition of, however, it should somehow match ...
Tereza Tizkova's user avatar
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Is there a proof that the total number of idempotent elements over all quasigroups of order n equals the number of quasigroups of that order?

To clarify, as requested by Community I am looking for a proof that the total number of idempotent elements over all quasigroups of order n equals the number of quasigroups of that order. Could be ...
John Palmer's user avatar
1 vote
1 answer
216 views

Inverse element of a magma

It is accepted that two elements are inverse to each other if their product is equal to the identity element: Inverse element in a magma https://en.wikipedia.org/wiki/Inverse_element The definition ...
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Example of a pair of non-isomorphic quasi-groups with parastrophic Latin squares?

A Latin square $\Lambda$ over an alphabet $A$ is a set of triples of elements of $A$ such that for every $\alpha,\beta\in A$, there is exactly one $\gamma\in A$ for which $(\alpha,\beta,\gamma)\in \...
Emil Sinclair's user avatar
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An example of an algebraic loop which has different L and R inverses?

Can anyone point me toward a simple example of a non-associative algebraic loop (i.e. a quasigroup with an identity) for which at least one element has a left inverse which is not equal to its right ...
KesterKester's user avatar
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Suspicious diagrams on wiki about group-like structures

It seems to me that the diagrams on wiki about group-like structures are not quite right. For example, the following https://en.wikipedia.org/wiki/Monoid#/media/File:Algebraic_structures_-...
ALife's user avatar
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Defining loops: why is divisibility and identitiy implying invertibility?

Wikipedia contains the following figure (to be found, e.g. here) in order to visualize the relations between several algebraic structures. I highlighted a part that I find especially interesting. It ...
M. Winter's user avatar
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Is there a general theory of linear equations?

Let's define a linear equation as a tuple $\left\langle a, b, x, /, \cdot, =, \sim \right\rangle$, expressing an equation $x \cdot a = b$ (with respect to the unknown $x$) and a solution $x \sim b / a$...
Charlie's user avatar
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1 answer
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Guessing colored hats without repetition

This entire question is inspired by Problem $12082$ of the Problems and Solutions section of the American Mathematical Monthly (see the May $2020$ issue for the solution to said problem). First, I ...
Nikhil Sahoo's user avatar
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Are all non-associative (not necessarily associative) finite division rings finite fields?

According to the Artin–Zorn theorem, any finite alternative division ring is a finite field, but I'm interested in the general non-associative (i.e. not necessarily associative) case. Are there any ...
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1 answer
171 views

Does this almost-group uniquely define a group?

Consider a quasigroup $(S,+)$ such that for every $a,b,c,d\in S$, $$(a+(b+c))+d=a+(b+(c+d)).$$ This is almost a group, but not quite. For instance, $(\mathbb Z,-)$ satisfies those axioms. You can ...
Michał Zapała's user avatar
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Is there a generalized definition of symmetry which results in a symmetry quasigroup?

I've been looking at generalized definitions of symmetry. Symmetries, the way we usually define them, form a group. However, there have been efforts to generalize this, creating algebras which have ...
Cort Ammon's user avatar
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What is the difference between quasigroup with associativity and semigroup with inverse?

Or is there no actual difference? Is the difference just in nomenclature. Literature gives no clear answer. What I understood is that quasigroup with associativity makes it to become group. Howevere ...
Josef Hlava's user avatar
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1 answer
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Is there any name for this property of quasigroups: $x \times (y \times (z \times t)) = (x \times y) \times (t \times z)?$

Is there a name for the property $x \times (y \times (z \times t)) = (x \times y) \times (t \times z)$? Some basic facts about it I was able to figure out: It is shared by all four basic arithmetic ...
Michał Zapała's user avatar
1 vote
1 answer
362 views

Prove that there are no idempotent, commutative quasigroups of even order.

I'm trying to prove this using a counting argument - Meaning of counting argument? I understand one can proved this for Latin Squares, as done in this post -Idempotent and commutative Latin squares ...
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Property of normal subloops

Let $w(x_1, . . . , x_n)$ be a loop term, and let $A$ be a normal subloop of $Q$. Prove that $w(a_1A, · · · , a_nA) = w(a_1, . . . , a_n)A$ for every $a_1, . . . , a_n \in A$. A subloop $A$ of $Q$ ...
wasatar's user avatar
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1 answer
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All loops of order less than $5$ are groups

I am studying loop theory and found that there are no non-trivial loops of order less than $5$. Can anybody help me in this, with some reference of a book or a research article? A set $Q$ with a ...
Ratan Lal's user avatar
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Imbedding in the quasi-group

Theorem: Every finite quasi-group $G$ can be imbedded in a quasi-group $H$ generated by a single element. Let $\varphi$ is the imbedding. Let $\langle h \rangle.$ Let $g_1 \in G$ and $\varphi(g_1) = ...
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1 vote
3 answers
107 views

What are some examples of infinite strict quasigroups?

By strict quasigroup I mean a quasigroup with no identity. I've come across one so far in the answer to this question, but I can't seem to find any others. I am particularly interested in finding ...
Nika's user avatar
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2 answers
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How can one construct a strict quasigroup with neither a left nor a right identity on $n$ elements?

When I say $Q$ is a strict quasigroup, I only mean that $Q$ is a quasigroup that is neither a loop nor is it associative. In general, I can find a quasigroup on $n$ elements with neither a left nor ...
Nika's user avatar
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Why quasi-random sequences are generated in the interval [0,1]? Is it a normalized sequence generation?

The quasi-random sequences are generated using low discrepancy sequences and Koksma-Hlawka inequality explains the quasi sequence clearly. However, it is observed that these sequences are generated in ...
Saurabh Kumar Pandey's user avatar
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1 answer
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Multiplication table of a commutative quasigroup is a symmetric latin square. Is the converse also true? [closed]

Or can we find an example where the Latin square is symmetric but its corresponding quasi group is not commutative?
user637862's user avatar
1 vote
1 answer
191 views

Definition of Additive Loop

I have thought of loops as having the operations of multiplication and left and right division. I read the D. R. Hughes article on Additive and Multiplicative Loops of Planar Ternary Rings and it ...
bblohowiak's user avatar
4 votes
1 answer
390 views

Commutative subtraction

It is well known that subtraction is not commutative in general. However, it is commutative in some groups: $\mathbb I$, $\mathbb C_2$, $\mathbb K_4$. I am trying to understand the logic. ...
Alex C's user avatar
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Invertibility as Criteria for a Loop

I try to understand the correct criteria for a Loop. I see in Wikipedia https://en.wikipedia.org/wiki/Inverse_element#In_a_unital_magma that “A unital magma in which all elements are invertible is ...
Avichai's user avatar
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1 answer
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Up to what level can associativity be guaranteed?

My question is generated from the following question: It turns out that the inverse of product with an assumption of inverse existence is a necessary condition of associative. Then is there any set ...
Hamio Jiang's user avatar
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In a finite monoid (M, $\circ$) if the identity element $e$ is the only idempotent element, prove that each element of the monoid is invertible.

In a finite monoid (M, $\circ$) if the identity element $e$ is the only idempotent element, prove that each element of the monoid is invertible. As the set $M$ is finite, $\exists$ $y \in M$, s.t $y \...
Subhasis Biswas's user avatar
6 votes
3 answers
159 views

Congruence lattice of direct power of Steiner seven element quasigroup

The Steiner seven element quasigroup is the algebra $\mathbf{S}_7 = \langle S_7, \cdot \rangle$, where $S_7 = \{ 1,2,3,4,5,6,7 \}$, and, up to isomorphism, its multiplication table is the one given ...
amrsa's user avatar
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9 votes
1 answer
272 views

Complexity of testing if a binary operation is a group

Given a binary operation specified as an $n \times n$ Cayley table, what is the complexity of the best deterministic algorithm for testing if the binary operation is a group? There's a fairly simple ...
Qudit's user avatar
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Conflict with finite Moufang loop solvability propositions

I really got stuck with the following contradiction. Say we have a Moufang loop $Q$, $|Q| < \infty$. To put it briefly, Moufang loops are groups that not necessary be associative, with extra ...
Evgeny's user avatar
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1 answer
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Examples of proper loops in $\mathbb{R}$

A loop $(L, \cdot)$ is a binary structure that satisfies every group axiom except for the associative property. A loop which is not a group is called a proper loop. A topological loop $(L,\cdot)$ is a ...
tree detective's user avatar
1 vote
1 answer
98 views

Composition Law to make a Quasigroup

Here is the problem I am attempting to solve: Let $S$ be a set with a composition law $\cdot$ possessing the following properties: $$(i)~x\cdot y = y\cdot x~\forall~x,y\in S$$ $$(ii)~x\cdot (x\cdot y)...
Makenzie's user avatar
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2 votes
2 answers
111 views

Prove that a map is not quasi-isometric embedding

I am not sure that I’m using the word metric correctly though: Is that OK? Does anyone has a different way of proving it?
matan's user avatar
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1 vote
1 answer
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When a loop with inverse property is commutative

Question How to prove that a loop $L$ with inverse property and $x^3=e$ for all $x$ is commutative iff $(x y)^2=x^2 y^2$ for all $x,y$? Definitions: A loop is a quasigroup with identity $e$. $...
Mark's user avatar
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1 vote
0 answers
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Quasi-Group represented by a graph which is not a Triangle-Free Graph locally

Can each of all quasi-groups be represented by a graph (latin square graph), which is not locally triangle free graph ? Quasi Group can be represented by Latin Square matrix, thus by a Latin ...
Michael's user avatar
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1 answer
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In a loop, if $(xy)x^r = y$ then $x(yx^r)=y$

Consider a loop $L$, that is, a quasigroup with an identity, and recall that a quasigroup $L$ is a set together with a binary operation such that, for every $a$ and $b$ in $L$, the equations $ax=b$ ...
Mark's user avatar
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2 votes
1 answer
131 views

If $H$ is a subloop of a finte loop $L$ and $N$ is a normal subloop of $L$, then $HN$ is a subloop of $L$.

To prove this is a subloop, I have to show that for $x, y \in HN$, the following are also in $HN$: (a) $xy$, (b) $L^{-1}_x(y)$ and (c) $R^{-1}_{x}(y)$. Here $L_x(y) = xy$ and $R_x(y)=yx$. We have to ...
Mark's user avatar
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2 votes
1 answer
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Does every non-empty quasigroup have a left or right identity?

I know that some quasigroups are not loops, meaning they don't have a two-sided identity. But are there non-empty quasigroups that don't even have one-sided identities?
user107952's user avatar
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A simple question about rational numbers without a simple proof?

As in this question, study the quasigroup $(Q_+,/)$ of positive rational numbers under division. There are two obvious identities: $a/(b/c)=c/(b/a)$, for all $a,b,c\in Q_+$ $(a/b)/c=(a/c)/b$, for all ...
Lehs's user avatar
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9 votes
2 answers
148 views

A familiar quasigroup - about independent axioms

A quasigroup is a pair $(Q,/)$, where $/$ is a binary operation on $Q$, such that (1) for each $a,b\in Q$ there exists unique solutions to the equations $a/x=b$ and $y/a=b$. Now I want to extract a ...
Lehs's user avatar
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1 vote
2 answers
844 views

Looking for examples of finite loops and monoids

I am looking for examples of (small) finite loops and monoids that are not groups for demonstrating what happens if you omit some of the group axioms. Does anyone know some ressources for this? I ...
flawr's user avatar
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