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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 either

  1. as a magma with identity and invertibility, or
  2. as a magma with identity and divisibility.

It is easy to see that from 1. follows 2. But I have trouble proving the converse. Here is a proof for the case that the loop is finite:

By divisibility $xa=xb\implies a=b$ for all $x,a,b\in L$. Thus, the map $a\mapsto xa$ is injective. If the loop is finite, the map is also surjective. Thus, there is a $x'\in L$ with $xx'=e$. This $x'$ is therefore the inverse of $x$.

But what if $L$ is not finite?

M. Winter
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    Where did you find this diagram? Can you give precise definitions of "divisibility" and "invertibility"? – Christoph Feb 01 '21 at 14:36
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    Note that $(\mathbb Z_{\ge 0}, +)$ has the identity element $0$ and allows cancellation $x+a=x+b\ \Rightarrow\ a=b$ (divisibility?) but does not contain inverses of its non-identity elements ($-x\notin\mathbb Z_{\ge 0}$). – Christoph Feb 01 '21 at 14:37
  • @Christoph Actually, I do not have definitions but only the words on the figure. But I too assumed "divisibility = cancellation", and this seems to be where the problem lies. Wikipedia defines loop as a quasigroup with identity. But checking the definition of quasigroup on the same page seems to use "divisibility" in a different sense. Hm – M. Winter Feb 01 '21 at 14:58

2 Answers2

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It should be pointed out that the accepted answer's definition of "has division" is not how the term is used in quasigroup theory or universal algebra. A magma $(M,\cdot)$ (or binar or groupoid in the old noncategorical sense) is said to be a division magma if for every $a,b\in M$, there exist $x,y\in M$ such that $a\cdot x = b$ and $y\cdot a = b$. Uniqueness is not assumed. A quasigroup is a cancellative, division magma.

A better way to think about it is in terms of the translation maps $L_a : M\to M; x\mapsto a\cdot x$ and $R_a :M\to M; x\mapsto x\cdot a$ for all $a\in M$. In a division magma, these are all surjective. In a cancellative magma, they are injective. In a quasigroup, they are bijective.

Michael Kinyon
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Having division is a stronger property than having cancellation. A magma $M$ (that is, a set with a binary operation, which we write multiplicatively) has division (and is called a quasigroup) if for all $a, b \in M$, there exist unique $x, y \in M$ such that $ax = b$ and $ya = b$. If $M$ has an identity element $e \in M$, then this trivially implies the existence of inverses: the left (resp. right) inverse of $a \in M$ is the unique $x \in M$ (resp. $y \in M$) such that $ax = e$ (resp. $ya = e$), which is exactly the definition of an inverse in a unital magma.

On the other hand, as Christoph mentions in a comment, the nonnegative integers with addition give an example of a commutative monoid that has cancellation but not inverses, and hence also doesn't have division (which we would think of as "having subtraction" in this case since it's written additively).

By the way, you've used the word "divisible" for what I've called "has division", but I've avoided the word "divisible" in this context because it has a different meaning: a commutative group $G$ is called a divisible group if for all $x \in G$ and all positive integers $n$, there exists $y \in G$ such that $y^n = x$. (If the group operation is written additively, this means that every element is "divisible by integers", hence the name.) So make sure not to mix up the notion of a magma having division (the defining property of a quasigroup) with the notion of a commutative group being divisible.

Daniel Hast
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