What is an example of a groupoid which is not a semigroup?

Question

I know that groupoid refers to an algebraic structure with a binary operation. The only necessary condition is closure.

However, I couldn't find any easy-to-understand example of a groupoid which is not a semigroup. I did come across some examples of (certain type of) matrices but then matrix multiplication is always associative (thus making it a semi-group).

So, could someone please provide me an example of a groupoid which isn't a semigroup?

Answer 1

Let $a,b,c$ be distinct members of a three element set and $ab=c=cc $, $bc=a=aa$ and define $ac, bb $ however you like (but in $\{a,b,c\}$.) You have a nonassociative binary operation.

Answer 2

These are called magmas, not groupoids.

The ``midpoint'' operation $s\ast t=\frac{s+t}{2}$ on $\mathbb{R}$ makes it a magma which is not a semigroup.

Answer 3

Let $\{a,b\}$ be a set with two distinct elements.

Define a partial multiplication by $a\times a=a$ and $b\times b=b$ and nothing more (so $a\times b$ and $b\times a$ are not defined).

Then $\{a,b\}$ equipped with this multiplication is a groupoid, but not a semigroup.

I suspect you use another definition of groupoid. If it is what I would call a magma then see the other answers.

Answer 4

First, the term 'groupoid' recently rather means primarily a category with invertible arrows, and the term 'magma' is arising for an algebraic structure with a binary operation.

For a familiar example, consider $\Bbb Z$ (or almost any Abelian group) with the subtraction.

Or, define $a*b:=a+b+1$ (or whatever..)

Other examples arise e.g. from finite quasigroups whose multiplication table is a latin square: having each element once in every row and in every column.

Answer 5

Here's three different examples.

1. Take an abelian group $(A,+)$ and define a new binary operation $\circ$ on $A$ by $x\circ y=x+(-y)$. This is an example of a quasigroup.

2. Take a group $(G,\cdot)$ and define a new binary operation $\triangleleft$ on $G$ by $x\triangleleft y=x\cdot y \cdot x^{-1}$. This is an example of a quandle.

3. Take a digraph $(V,A)$ with the property that for any two distinct vertices $v,w\in V$, exactly one of the arcs $vw$ or $wv$ is in $A$. Define a commutative binary operation $\cdot$ on $V$ by $v\cdot v=v$ and $v\cdot w=w$ if and only if $vw\in A$. This is an example of a tournament.

A quasigroup is associative if and only if it is a group, a quandle is associative if and only if it is trivial, and a tournament is associative if and only if it is a commutative idempotent semigroup (aka a semilattice).

Answer 6

$( \Bbb Z , -)$

This is a groupoid and not a semi-group.

Answer 7

What about the set $\mathbb{R}^+$ of positive real numbers with exponentiation?

It is not associative: $$2^{(1^3)}=2\ne 8=(2^1)^3$$ and not commutative: $$2^3=8\ne 9=3^2.$$ It neither has identity element. Actually, $1$ is a left-absorbing and right identity element: $${1^n=1\ \forall\,n}\quad\text{and}\quad{n^1=n\ \forall\,n}.$$

Answer 8

A non-numerical example of non-associative magma is the Jankenpon game. Let's assume the dyadic operation "*" determines the winner of a match:

(rock * paper) * scissors = paper * scissors = scissors

rock * (paper * scissors) = rock * scissors = rock

Therefore,

(rock * paper) * scissors $\neq$ rock * (paper * scissors).

Incidentally, this magma happens to be commutative and idempotent.