What is a groupoid?

  • Thread starter AdrianZ
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What is a groupoid? I'm reading a book about gyrovectors and in the first chapter it starts defining something which it calls it a groupoid but doesn't inform the reader about its axioms. so I felt uncomfortable to face a new mathematical definition without knowing the axioms that it should satisfy.
Here is what the book says:

Definition 2.1 (Binary Operations, Groupoids, and Groupoid
Automorphisms). A binary operation + in a set S is a function + :
S x S -> S. We use the notation a + b to denote +(a, b) for any a, b in S.
A groupoid (S, +) is a nonempty set, S, with a binary operation, +. An
automorphism phi of a groupoid (S, +) is a bijective (that is, one-to-one)
self-map of S, phi : S -> S, which preserves its groupoid operation, that is,
phi(a + b) = phi(a) + phi(b) for all a, b in S.

What axioms should a groupoid satisfy? and what's the difference between a groupoid and group? because the name apparently has been derived from the word group.
 

Answers and Replies

  • #2
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Hi AdrianZ! :smile:

A groupoid doesn't have any axioms. It's just any set with any binary operation.
So [itex](\mathbb{N},+)[/itex], [itex](\mathbb{N},\cdot)[/itex], [itex](\mathbb{Z},\cdot)[/itex], etc are all groupoids.

The only requirement is that the operation must really be an operation. For example [itex](\mathbb{R},/)[/itex] isn't a groupoid as 1/0 doesn't exist.

A more conventional name for a groupoid is a magma: http://en.wikipedia.org/wiki/Magma_(algebra)
The name magma is preferred, because groupoid often has another meaning in category theory and Lie algebras.
 
  • #3
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So you mean the binary operation must be well-defined for any two elements in a groupoid? right?
 
  • #4
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So you mean the binary operation must be well-defined for any two elements in a groupoid? right?

Yes, the operation must be well-defined and the set must be closed under the operation. (for example [itex](\mathbb{N},-)[/itex] is not a groupoid since 2-3 is not a natural number. That are the only requirements.

In other words, the operation

[tex]*:M\times M\rightarrow M[/tex]


must be a function.
 
  • #5
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Okay. Thanks.
 

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