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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.

-4 votes
1 answer
79 views

Practical example of differences between associativity and alternativity (and the in-between Bol loop)? [closed]

Associativity is: $$(a * b) * c = a * (b * c)$$ Alternativity is: $$a * (a * b) = (a * a) * b$$ $$(a * b) * b = a * (b * b)$$ Bol loop is: $${\displaystyle a(b(ac))=(a(ba))c}$$ $${\displaystyle ((ca)b)...
Lance's user avatar
  • 3,773
0 votes
1 answer
55 views

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\...
Miles Gould's user avatar
1 vote
0 answers
31 views

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
2 votes
0 answers
47 views

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 ...
Naiim's user avatar
  • 317
2 votes
1 answer
113 views

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
  • 141
0 votes
1 answer
92 views

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
  • 564
3 votes
1 answer
84 views

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 $\ell x=y$...
WillG's user avatar
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1 vote
0 answers
39 views

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
  • 63
2 votes
1 answer
73 views

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
  • 3,054
5 votes
1 answer
160 views

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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2 votes
1 answer
84 views

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
5 votes
0 answers
86 views

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
233 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 ...
Alex C's user avatar
  • 1,120
0 votes
1 answer
48 views

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
2 votes
1 answer
129 views

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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