What is an example of a groupoid which is not a semigroup?
up vote
3
down vote
favorite
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?
abstract-algebra semigroups magma
edited Nov 21 at 14:37
Arnaud D.
15.6k52343
add a comment |
up vote
3
down vote
favorite
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?
abstract-algebra semigroups magma
edited Nov 21 at 14:37
Arnaud D.
15.6k52343
2
You should be careful with the word "groupoid". In the 60s some people used this word to refer to "sets with a binary operation" (possibly non-associative), but these are now called magmas. Groupoids, on the other hand, are sets (or classes) with a partially defined binary operation satisfying group-like properties.
– Luiz Cordeiro
May 19 at 17:50
@LuizCordeiro Thanks. I didn't know that. However, I guess in the context of this question: "Give an example of a groupoid which is not a semigroup." I think they mean "magma" by "groupoid"?
– user563280
May 19 at 17:57
Yes, since you're looking for an example of a non-associative binary operation, you're looking for a magma which isn't a semigroup.
– Luiz Cordeiro
May 19 at 18:01
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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?
abstract-algebra semigroups magma
edited Nov 21 at 14:37
Arnaud D.
15.6k52343
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?
abstract-algebra semigroups magma
abstract-algebra semigroups magma
edited Nov 21 at 14:37
Arnaud D.
15.6k52343
edited Nov 21 at 14:37
Arnaud D.
15.6k52343
edited Nov 21 at 14:37
Arnaud D.
15.6k52343
edited Nov 21 at 14:37
Arnaud D.
15.6k52343
edited Nov 21 at 14:37
Arnaud D.
15.6k52343
15.6k52343
asked May 19 at 17:42
user563280
2
You should be careful with the word "groupoid". In the 60s some people used this word to refer to "sets with a binary operation" (possibly non-associative), but these are now called magmas. Groupoids, on the other hand, are sets (or classes) with a partially defined binary operation satisfying group-like properties.
– Luiz Cordeiro
May 19 at 17:50
@LuizCordeiro Thanks. I didn't know that. However, I guess in the context of this question: "Give an example of a groupoid which is not a semigroup." I think they mean "magma" by "groupoid"?
– user563280
May 19 at 17:57
Yes, since you're looking for an example of a non-associative binary operation, you're looking for a magma which isn't a semigroup.
– Luiz Cordeiro
May 19 at 18:01
add a comment |
2
You should be careful with the word "groupoid". In the 60s some people used this word to refer to "sets with a binary operation" (possibly non-associative), but these are now called magmas. Groupoids, on the other hand, are sets (or classes) with a partially defined binary operation satisfying group-like properties.
– Luiz Cordeiro
May 19 at 17:50
@LuizCordeiro Thanks. I didn't know that. However, I guess in the context of this question: "Give an example of a groupoid which is not a semigroup." I think they mean "magma" by "groupoid"?
– user563280
May 19 at 17:57
Yes, since you're looking for an example of a non-associative binary operation, you're looking for a magma which isn't a semigroup.
– Luiz Cordeiro
May 19 at 18:01
2
2
You should be careful with the word "groupoid". In the 60s some people used this word to refer to "sets with a binary operation" (possibly non-associative), but these are now called magmas. Groupoids, on the other hand, are sets (or classes) with a partially defined binary operation satisfying group-like properties.
– Luiz Cordeiro
May 19 at 17:50
You should be careful with the word "groupoid". In the 60s some people used this word to refer to "sets with a binary operation" (possibly non-associative), but these are now called magmas. Groupoids, on the other hand, are sets (or classes) with a partially defined binary operation satisfying group-like properties.
– Luiz Cordeiro
May 19 at 17:50
@LuizCordeiro Thanks. I didn't know that. However, I guess in the context of this question: "Give an example of a groupoid which is not a semigroup." I think they mean "magma" by "groupoid"?
– user563280
May 19 at 17:57
@LuizCordeiro Thanks. I didn't know that. However, I guess in the context of this question: "Give an example of a groupoid which is not a semigroup." I think they mean "magma" by "groupoid"?
– user563280
May 19 at 17:57
Yes, since you're looking for an example of a non-associative binary operation, you're looking for a magma which isn't a semigroup.
– Luiz Cordeiro
May 19 at 18:01
Yes, since you're looking for an example of a non-associative binary operation, you're looking for a magma which isn't a semigroup.
– Luiz Cordeiro
May 19 at 18:01
add a comment |
5 Answers
5
active
oldest
votes
up vote
5
down vote
These are called magmas, not groupoids.
The ``midpoint'' operation $sast t=frac{s+t}{2}$ on $mathbb{R}$ makes it a magma which is not a semigroup.
answered May 19 at 17:55
Luiz Cordeiro
12.5k1143
s*t = s-t => Is also a groupoid but not a semi-group. I learned that these are being called magmas now. Thanks!
– router
yesterday
add a comment |
up vote
2
down vote
Here's three different examples.
Take an abelian group $(A,+)$ and define a new binary operation $circ$ on $A$ by $xcirc y=x+(-y)$. This is an example of a quasigroup.
Take a group $(G,cdot)$ and define a new binary operation $triangleleft$ on $G$ by $xtriangleleft y=xcdot y cdot x^{-1}$. This is an example of a quandle.
Take a digraph $(V,A)$ with the property that for any two distinct vertices $v,win V$, exactly one of the arcs $vw$ or $wv$ is in $A$. Define a commutative binary operation $cdot$ on $V$ by $vcdot v=v$ and $vcdot w=w$ if and only if $vwin A$. 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).
answered May 20 at 1:32
Eran
1,178818
add a comment |
up vote
0
down vote
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.
answered May 19 at 17:54
rschwieb
104k1299241
add a comment |
up vote
0
down vote
Let ${a,b}$ be a set with two distinct elements.
Define a partial multiplication by $atimes a=a$ and $btimes b=b$ and nothing more (so $atimes b$ and $btimes a$ 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.
answered May 19 at 17:56
Vera
2,467617
add a comment |
up vote
0
down vote
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.
answered May 19 at 17:58
Berci
59.4k23672
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
These are called magmas, not groupoids.
The ``midpoint'' operation $sast t=frac{s+t}{2}$ on $mathbb{R}$ makes it a magma which is not a semigroup.
answered May 19 at 17:55
Luiz Cordeiro
12.5k1143
s*t = s-t => Is also a groupoid but not a semi-group. I learned that these are being called magmas now. Thanks!
– router
yesterday
add a comment |
up vote
5
down vote
These are called magmas, not groupoids.
The ``midpoint'' operation $sast t=frac{s+t}{2}$ on $mathbb{R}$ makes it a magma which is not a semigroup.
answered May 19 at 17:55
Luiz Cordeiro
12.5k1143
s*t = s-t => Is also a groupoid but not a semi-group. I learned that these are being called magmas now. Thanks!
– router
yesterday
add a comment |
up vote
5
down vote
up vote
5
down vote
These are called magmas, not groupoids.
The ``midpoint'' operation $sast t=frac{s+t}{2}$ on $mathbb{R}$ makes it a magma which is not a semigroup.
answered May 19 at 17:55
Luiz Cordeiro
12.5k1143
These are called magmas, not groupoids.
The ``midpoint'' operation $sast t=frac{s+t}{2}$ on $mathbb{R}$ makes it a magma which is not a semigroup.
answered May 19 at 17:55
Luiz Cordeiro
12.5k1143
answered May 19 at 17:55
Luiz Cordeiro
12.5k1143
answered May 19 at 17:55
Luiz Cordeiro
12.5k1143
answered May 19 at 17:55
Luiz Cordeiro
12.5k1143
12.5k1143
s*t = s-t => Is also a groupoid but not a semi-group. I learned that these are being called magmas now. Thanks!
– router
yesterday
add a comment |
s*t = s-t => Is also a groupoid but not a semi-group. I learned that these are being called magmas now. Thanks!
– router
yesterday
s*t = s-t => Is also a groupoid but not a semi-group. I learned that these are being called magmas now. Thanks!
– router
yesterday
s*t = s-t => Is also a groupoid but not a semi-group. I learned that these are being called magmas now. Thanks!
– router
yesterday
add a comment |
up vote
2
down vote
Here's three different examples.
Take an abelian group $(A,+)$ and define a new binary operation $circ$ on $A$ by $xcirc y=x+(-y)$. This is an example of a quasigroup.
Take a group $(G,cdot)$ and define a new binary operation $triangleleft$ on $G$ by $xtriangleleft y=xcdot y cdot x^{-1}$. This is an example of a quandle.
Take a digraph $(V,A)$ with the property that for any two distinct vertices $v,win V$, exactly one of the arcs $vw$ or $wv$ is in $A$. Define a commutative binary operation $cdot$ on $V$ by $vcdot v=v$ and $vcdot w=w$ if and only if $vwin A$. 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).
answered May 20 at 1:32
Eran
1,178818
add a comment |
up vote
2
down vote
Here's three different examples.
Take an abelian group $(A,+)$ and define a new binary operation $circ$ on $A$ by $xcirc y=x+(-y)$. This is an example of a quasigroup.
Take a group $(G,cdot)$ and define a new binary operation $triangleleft$ on $G$ by $xtriangleleft y=xcdot y cdot x^{-1}$. This is an example of a quandle.
Take a digraph $(V,A)$ with the property that for any two distinct vertices $v,win V$, exactly one of the arcs $vw$ or $wv$ is in $A$. Define a commutative binary operation $cdot$ on $V$ by $vcdot v=v$ and $vcdot w=w$ if and only if $vwin A$. 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).
answered May 20 at 1:32
Eran
1,178818
add a comment |
up vote
2
down vote
up vote
2
down vote
Here's three different examples.
Take an abelian group $(A,+)$ and define a new binary operation $circ$ on $A$ by $xcirc y=x+(-y)$. This is an example of a quasigroup.
Take a group $(G,cdot)$ and define a new binary operation $triangleleft$ on $G$ by $xtriangleleft y=xcdot y cdot x^{-1}$. This is an example of a quandle.
Take a digraph $(V,A)$ with the property that for any two distinct vertices $v,win V$, exactly one of the arcs $vw$ or $wv$ is in $A$. Define a commutative binary operation $cdot$ on $V$ by $vcdot v=v$ and $vcdot w=w$ if and only if $vwin A$. 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).
answered May 20 at 1:32
Eran
1,178818
Here's three different examples.
Take an abelian group $(A,+)$ and define a new binary operation $circ$ on $A$ by $xcirc y=x+(-y)$. This is an example of a quasigroup.
Take a group $(G,cdot)$ and define a new binary operation $triangleleft$ on $G$ by $xtriangleleft y=xcdot y cdot x^{-1}$. This is an example of a quandle.
Take a digraph $(V,A)$ with the property that for any two distinct vertices $v,win V$, exactly one of the arcs $vw$ or $wv$ is in $A$. Define a commutative binary operation $cdot$ on $V$ by $vcdot v=v$ and $vcdot w=w$ if and only if $vwin A$. 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).
answered May 20 at 1:32
Eran
1,178818
answered May 20 at 1:32
Eran
1,178818
answered May 20 at 1:32
Eran
1,178818
answered May 20 at 1:32
Eran
1,178818
1,178818
add a comment |
add a comment |
up vote
0
down vote
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.
answered May 19 at 17:54
rschwieb
104k1299241
add a comment |
up vote
0
down vote
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.
answered May 19 at 17:54
rschwieb
104k1299241
add a comment |
up vote
0
down vote
up vote
0
down vote
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.
answered May 19 at 17:54
rschwieb
104k1299241
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.
answered May 19 at 17:54
rschwieb
104k1299241
answered May 19 at 17:54
rschwieb
104k1299241
answered May 19 at 17:54
rschwieb
104k1299241
answered May 19 at 17:54
rschwieb
104k1299241
104k1299241
add a comment |
add a comment |
up vote
0
down vote
Let ${a,b}$ be a set with two distinct elements.
Define a partial multiplication by $atimes a=a$ and $btimes b=b$ and nothing more (so $atimes b$ and $btimes a$ 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.
answered May 19 at 17:56
Vera
2,467617
add a comment |
up vote
0
down vote
Let ${a,b}$ be a set with two distinct elements.
Define a partial multiplication by $atimes a=a$ and $btimes b=b$ and nothing more (so $atimes b$ and $btimes a$ 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.
answered May 19 at 17:56
Vera
2,467617
add a comment |
up vote
0
down vote
up vote
0
down vote
Let ${a,b}$ be a set with two distinct elements.
Define a partial multiplication by $atimes a=a$ and $btimes b=b$ and nothing more (so $atimes b$ and $btimes a$ 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.
answered May 19 at 17:56
Vera
2,467617
Let ${a,b}$ be a set with two distinct elements.
Define a partial multiplication by $atimes a=a$ and $btimes b=b$ and nothing more (so $atimes b$ and $btimes a$ 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.
answered May 19 at 17:56
Vera
2,467617
answered May 19 at 17:56
Vera
2,467617
answered May 19 at 17:56
Vera
2,467617
answered May 19 at 17:56
Vera
2,467617
2,467617
add a comment |
add a comment |
up vote
0
down vote
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.
answered May 19 at 17:58
Berci
59.4k23672
add a comment |
up vote
0
down vote
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.
answered May 19 at 17:58
Berci
59.4k23672
add a comment |
up vote
0
down vote
up vote
0
down vote
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.
answered May 19 at 17:58
Berci
59.4k23672
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.
answered May 19 at 17:58
Berci
59.4k23672
answered May 19 at 17:58
Berci
59.4k23672
answered May 19 at 17:58
Berci
59.4k23672
answered May 19 at 17:58
Berci
59.4k23672
59.4k23672
add a comment |
add a comment |
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2
You should be careful with the word "groupoid". In the 60s some people used this word to refer to "sets with a binary operation" (possibly non-associative), but these are now called magmas. Groupoids, on the other hand, are sets (or classes) with a partially defined binary operation satisfying group-like properties.
– Luiz Cordeiro
May 19 at 17:50
@LuizCordeiro Thanks. I didn't know that. However, I guess in the context of this question: "Give an example of a groupoid which is not a semigroup." I think they mean "magma" by "groupoid"?
– user563280
May 19 at 17:57
Yes, since you're looking for an example of a non-associative binary operation, you're looking for a magma which isn't a semigroup.
– Luiz Cordeiro
May 19 at 18:01