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.

A quasigroup is a grouplike structure $(Q, \ast)$, where $Q$ is a set and $$\ast: Q \times Q \to Q$$ is a binary product that satisfies the satisfying the Latin square property (or divisibility property), namely that for all $q, r \in Q$ there is a unique $a \in Q$ such that $r = a q$ and a unique $b \in Q$ such that $r = q b$, or equivalently, the multiplication table of $\ast$ is a Latin square.

A quasigroup $(Q, \ast)$ with an identity element (an element $1 \in Q$ such that $1q = q = q1$ for all $q \in Q$) is called a loop, and an associative loop is the same thing as a group.

Apart from groups, important examples of quasigroups include the integers $\mathbb{Z}$ endowed with subtraction $-$, $\mathbb{Q}$ endowed with division $\div$, vector spaces over fields of characteristic not $2$ endowed with the map $x \ast y := \frac{1}{2} (x + y)$. The nonzero elements of any division algebra always form a quasigroup; in the case of the octonions, the quasigroup of nonzero elements is a Moufang loop (a loop satisfying an additional algebraic identity) but not a group.

Quasi-group (Encyclopedia of Mathematics)

Quasigroup (Wikipedia)

J. Dénes, A.D. Keedwell, Latin squares and their applications, English Univ. Press (1974)

R. Moufang, Zur Struktur von Alternativkörpern Math. Ann., 110 (1935) pp. 416–430 (in German)

62 questions
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What are the commutative quasigroups satisfying $a/b=b/a$?

There's a harder question lurking behind this question that was just asked. The context is quasigroup theory. A commutative quasigroup can be defined as a set $Q$ together with commutative binary operation $*$ such that for all $a,b \in Q$, there is…
goblin GONE
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Existence of finite non-associative fields

Do finite non-associative fields exist? By “non-associative field”, I mean a field with associativity of multiplication dropped but all other properties remain the same. This other question asks something similar but does not require commutativity…
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The unique loop (quasigroup with unit) $L$ of order $5$ satisfying $x^2 = 1$ for all $x \in L$

Recall that a quasigroup is a pair $(Q, \ast)$ , where $Q$ is a set and $\ast$ is a binary product $$\ast: Q \times Q \to Q$$ satisfying the Latin square property , namely that for all $x, y \in Q$ there is a unique $a \in Q$ such that $y = ax$ and…
Travis Willse
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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 easily prove that for any element $x$ , the…
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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 class of quasigroups that captures characteristics…
Lehs
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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 non-associative finite division rings different…
user735141
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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 seems to suggest that a loop can be defined…
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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 topological space which is also a loop such…
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How does this algorithm "rank" amongst other in the literature?

By merging together the contributions from: a) this answer , b) the comments under this answer , we come up to the following: Claim . For $n\in\mathbb N$ , let $Q=(\{1,\dots,n\},*)$ be a quasigroup. Then, there is a deterministic $O(n^2\log n)$…
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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 operations (ie. addition, multiplication,…
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Is there a standard category-theoretic way to express a loop or quasigroup?

The standard way to encode a group as a category is as a "category with one object and all arrows invertible". All of the arrows are group elements, and composition of arrows is the group operation. A loop obeys similar axioms to a group, but does…
wnoise
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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 here. The defining identities for the variety of…
amrsa
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Examples of quasigroups with no identity elements

If you scroll to the bottom of this page , there is a table claiming quasigroups have divisibility but not identity (in general). What would be some examples of quasigroups without an identity element?
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The name of an algebraic structure satisfying $y*(y*x)=x=(x*y)*y$

This is an abstract algebra exercise. See also here. Let $(X,*)$ be an algebraic structure with $*$ a binary operation. If $$ (x*y)*y=x=y*(y*x),\qquad \forall x,y\in X$$ show that $*$ is commutative. Proof Set $a:=x*y$ , $b:=y*x$ , then $$…
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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 satisfy Moufang's theorem, that if any three elements…
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