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A group-like structure with multiplicative zero instead of the identity

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

1

$begingroup$

My question is whether the structure in the title is known and has a name.

For clarity: the structure is a finite group like structure $G$ but instead of the identity we have a zero element $0in G$ such that $0 g=0$ for all $gin G$.

The prototypical example of this structure is the subset of $ntimes n$ matrices denoted with $M_ij$ for $i,j=1,...,n$ (with matrix multiplication as the group product) and defined by having $1$ in the $i,j$ position and $0$ everywhere else, and one additional matrix $M_0$ that is the all zeros matrix.

abstract-algebra group-theory

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edited 9 hours ago

oleg

asked 9 hours ago

olegoleg

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$endgroup$

add a comment
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1

$begingroup$

My question is whether the structure in the title is known and has a name.

For clarity: the structure is a finite group like structure $G$ but instead of the identity we have a zero element $0in G$ such that $0 g=0$ for all $gin G$.

The prototypical example of this structure is the subset of $ntimes n$ matrices denoted with $M_ij$ for $i,j=1,...,n$ (with matrix multiplication as the group product) and defined by having $1$ in the $i,j$ position and $0$ everywhere else, and one additional matrix $M_0$ that is the all zeros matrix.

abstract-algebra group-theory

share|cite|improve this question

edited 9 hours ago

oleg

asked 9 hours ago

olegoleg

12355 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

My question is whether the structure in the title is known and has a name.

For clarity: the structure is a finite group like structure $G$ but instead of the identity we have a zero element $0in G$ such that $0 g=0$ for all $gin G$.

The prototypical example of this structure is the subset of $ntimes n$ matrices denoted with $M_ij$ for $i,j=1,...,n$ (with matrix multiplication as the group product) and defined by having $1$ in the $i,j$ position and $0$ everywhere else, and one additional matrix $M_0$ that is the all zeros matrix.

abstract-algebra group-theory

share|cite|improve this question

edited 9 hours ago

oleg

asked 9 hours ago

olegoleg

12355 bronze badges

$endgroup$

share|cite|improve this question

edited 9 hours ago

oleg

asked 9 hours ago

olegoleg

12355 bronze badges

share|cite|improve this question

edited 9 hours ago

oleg

asked 9 hours ago

olegoleg

12355 bronze badges

asked 9 hours ago

olegoleg

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asked 9 hours ago

olegoleg

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2 Answers
2

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6

$begingroup$

A set with a binary associative operation is a semigroup.

A semigroup that has a two-sided identity element is a monoid.

A monoid in which every element has a two-sided inverse is a group.

Given a semigroup $S$, an element $0in S$ such that $0g=0$ for all $gin S$ is called a zero element.

You have "semigroups with zero" and "monoids with zero." Note that a zero element, if it exists, must be unique.

The set of $n times n$ matrices with matrix multiplication are a (non-commutative) monoid with zero. Note that it does not, as you write, "have a zero instead of an identity." Rather, it has a zero in addition to having an identity (which is not a problem, since $0e = 0$).

If your object has a zero but no identity, then it is a semigroup with a zero. An example of a semigroup with zero that is not a monoid could be the even integers under multiplication.

The set of $n times n$ matrices with a single $1$ and zeros elsewhere, plus the zero matrix, would be a semigroup with zero.

share|cite|improve this answer

edited 4 hours ago

red_trumpet

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answered 9 hours ago

Arturo MagidinArturo Magidin

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the answer, "semigroups with zero" sounds like a good name for it. I did not mean all $n times n$ matrices. I edited the post to clarify that better.
  $endgroup$
  – oleg
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @oleg: I've edited the post and deleted the other comment.
  $endgroup$
  – Arturo Magidin
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 Another common phrase for "zero element" is absorbing element
  $endgroup$
  – rschwieb
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

2

$begingroup$

Well, in each commutative ring $R$ (such as the ring of integers or the ring of square matrices), the zero element $0$ is absorbing, i.e., $r0 = 0 = 0r$.

Indeed, $0 = 0 + 0$ and so $r0 = r(0+0) = r0 + r0$. By adding the additive inverse $-r0$ to both sides (i.e., $r0 + (-r0) = 0$), $0 = r0$.

share|cite|improve this answer

answered 9 hours ago

WuestenfuxWuestenfux

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6

$begingroup$

A set with a binary associative operation is a semigroup.

A semigroup that has a two-sided identity element is a monoid.

A monoid in which every element has a two-sided inverse is a group.

Given a semigroup $S$, an element $0in S$ such that $0g=0$ for all $gin S$ is called a zero element.

You have "semigroups with zero" and "monoids with zero." Note that a zero element, if it exists, must be unique.

The set of $n times n$ matrices with matrix multiplication are a (non-commutative) monoid with zero. Note that it does not, as you write, "have a zero instead of an identity." Rather, it has a zero in addition to having an identity (which is not a problem, since $0e = 0$).

If your object has a zero but no identity, then it is a semigroup with a zero. An example of a semigroup with zero that is not a monoid could be the even integers under multiplication.

The set of $n times n$ matrices with a single $1$ and zeros elsewhere, plus the zero matrix, would be a semigroup with zero.

share|cite|improve this answer

edited 4 hours ago

red_trumpet

1,23933 silver badges1919 bronze badges

answered 9 hours ago

Arturo MagidinArturo Magidin

276k3434 gold badges606606 silver badges942942 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the answer, "semigroups with zero" sounds like a good name for it. I did not mean all $n times n$ matrices. I edited the post to clarify that better.
  $endgroup$
  – oleg
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @oleg: I've edited the post and deleted the other comment.
  $endgroup$
  – Arturo Magidin
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 Another common phrase for "zero element" is absorbing element
  $endgroup$
  – rschwieb
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

6

$begingroup$

A set with a binary associative operation is a semigroup.

A semigroup that has a two-sided identity element is a monoid.

A monoid in which every element has a two-sided inverse is a group.

Given a semigroup $S$, an element $0in S$ such that $0g=0$ for all $gin S$ is called a zero element.

You have "semigroups with zero" and "monoids with zero." Note that a zero element, if it exists, must be unique.

The set of $n times n$ matrices with matrix multiplication are a (non-commutative) monoid with zero. Note that it does not, as you write, "have a zero instead of an identity." Rather, it has a zero in addition to having an identity (which is not a problem, since $0e = 0$).

If your object has a zero but no identity, then it is a semigroup with a zero. An example of a semigroup with zero that is not a monoid could be the even integers under multiplication.

The set of $n times n$ matrices with a single $1$ and zeros elsewhere, plus the zero matrix, would be a semigroup with zero.

share|cite|improve this answer

edited 4 hours ago

red_trumpet

1,23933 silver badges1919 bronze badges

answered 9 hours ago

Arturo MagidinArturo Magidin

276k3434 gold badges606606 silver badges942942 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the answer, "semigroups with zero" sounds like a good name for it. I did not mean all $n times n$ matrices. I edited the post to clarify that better.
  $endgroup$
  – oleg
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @oleg: I've edited the post and deleted the other comment.
  $endgroup$
  – Arturo Magidin
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 Another common phrase for "zero element" is absorbing element
  $endgroup$
  – rschwieb
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

$begingroup$

A set with a binary associative operation is a semigroup.

A semigroup that has a two-sided identity element is a monoid.

A monoid in which every element has a two-sided inverse is a group.

Given a semigroup $S$, an element $0in S$ such that $0g=0$ for all $gin S$ is called a zero element.

You have "semigroups with zero" and "monoids with zero." Note that a zero element, if it exists, must be unique.

The set of $n times n$ matrices with matrix multiplication are a (non-commutative) monoid with zero. Note that it does not, as you write, "have a zero instead of an identity." Rather, it has a zero in addition to having an identity (which is not a problem, since $0e = 0$).

If your object has a zero but no identity, then it is a semigroup with a zero. An example of a semigroup with zero that is not a monoid could be the even integers under multiplication.

The set of $n times n$ matrices with a single $1$ and zeros elsewhere, plus the zero matrix, would be a semigroup with zero.

share|cite|improve this answer

edited 4 hours ago

red_trumpet

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answered 9 hours ago

Arturo MagidinArturo Magidin

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$endgroup$

share|cite|improve this answer

edited 4 hours ago

red_trumpet

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answered 9 hours ago

Arturo MagidinArturo Magidin

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edited 4 hours ago

red_trumpet

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edited 4 hours ago

red_trumpet

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answered 9 hours ago

Arturo MagidinArturo Magidin

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answered 9 hours ago

Arturo MagidinArturo Magidin

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2

$begingroup$

Well, in each commutative ring $R$ (such as the ring of integers or the ring of square matrices), the zero element $0$ is absorbing, i.e., $r0 = 0 = 0r$.

Indeed, $0 = 0 + 0$ and so $r0 = r(0+0) = r0 + r0$. By adding the additive inverse $-r0$ to both sides (i.e., $r0 + (-r0) = 0$), $0 = r0$.

share|cite|improve this answer

answered 9 hours ago

WuestenfuxWuestenfux

10.8k22 gold badges66 silver badges1717 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

Well, in each commutative ring $R$ (such as the ring of integers or the ring of square matrices), the zero element $0$ is absorbing, i.e., $r0 = 0 = 0r$.

Indeed, $0 = 0 + 0$ and so $r0 = r(0+0) = r0 + r0$. By adding the additive inverse $-r0$ to both sides (i.e., $r0 + (-r0) = 0$), $0 = r0$.

share|cite|improve this answer

answered 9 hours ago

WuestenfuxWuestenfux

10.8k22 gold badges66 silver badges1717 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Well, in each commutative ring $R$ (such as the ring of integers or the ring of square matrices), the zero element $0$ is absorbing, i.e., $r0 = 0 = 0r$.

Indeed, $0 = 0 + 0$ and so $r0 = r(0+0) = r0 + r0$. By adding the additive inverse $-r0$ to both sides (i.e., $r0 + (-r0) = 0$), $0 = r0$.

share|cite|improve this answer

answered 9 hours ago

WuestenfuxWuestenfux

10.8k22 gold badges66 silver badges1717 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

WuestenfuxWuestenfux

10.8k22 gold badges66 silver badges1717 bronze badges

answered 9 hours ago

WuestenfuxWuestenfux

10.8k22 gold badges66 silver badges1717 bronze badges

answered 9 hours ago

WuestenfuxWuestenfux

10.8k22 gold badges66 silver badges1717 bronze badges