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Alternative axioms for groups.

Group identities and inversesUniqueness of Inverses in Groups Implies Associativity Holds?Concepts of left group and right group.Prove that (G,*) is a group.Prove an alternative definition of group (which replaces the identity & inverse axioms with another)Is there a non-associative muliplicative closed set, with two-sided inverses and a two-sided identity?

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6

$begingroup$

The usual axioms I've seen for a group are: associativity; existence of two-sided identity; existence of two-sided inverses for all elements.

$$forall a,b,cin G: aleft(bcright)=left(abright)c$$
$$exists ein G, forall ain G: ae=a=ea$$
$$forall ain G exists a'in G: aa'=e=a'a$$

I recently came across a different axiomatisation, and there were no proofs of equivalence. They were: associativity; existence of left-identity; existence of left-inverses.

$$forall a,b,cin G: aleft(bcright)=left(abright)c$$
$$exists ein G, forall ain G: ea=a$$
$$forall ain G, exists a'in G: a'a=e$$

Are these equivalent? I kind of doubt it, since we have associative semigroups with left but not right identities, but maybe the left-inverses part changes things.

There was a proof that given these axioms, a left-inverse is a right-inverse, and hence that the original inverses axiom is proven, but what about right-identity?

Proof:
Let $gin G$ then $g$ has a left inverse, call it $g'in G$ and this too has a left inverse, call it $g''in G$. Then, $g'g=e$, $g''g'=e$ and so
$$gg'=egg'=g''g'gg'=g''g'=e$$
so $g'$ is the right-inverse of $g$ also.

group-theory axioms

share|cite|improve this question

edited 8 hours ago

Joshua Tilley

asked 8 hours ago

Joshua TilleyJoshua Tilley

65333 silver badges1313 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  All semigroups are associative.
  $endgroup$
  – Shaun
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, fair point on language. I meant to emphasise that these satisfy 2/3 of the alternative group axioms.
  $endgroup$
  – Joshua Tilley
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The word "magma" would be more appropriate then.
  $endgroup$
  – Shaun
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I can edit if you like, but do you have any thoughts on the question?
  $endgroup$
  – Joshua Tilley
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  A left identity $e$ satisfies, given the existence of inverses, $ae=aa'a=ea=a$, for $ain G$
  $endgroup$
  – AnotherJohnDoe
  8 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

6

$begingroup$

The usual axioms I've seen for a group are: associativity; existence of two-sided identity; existence of two-sided inverses for all elements.

$$forall a,b,cin G: aleft(bcright)=left(abright)c$$
$$exists ein G, forall ain G: ae=a=ea$$
$$forall ain G exists a'in G: aa'=e=a'a$$

I recently came across a different axiomatisation, and there were no proofs of equivalence. They were: associativity; existence of left-identity; existence of left-inverses.

$$forall a,b,cin G: aleft(bcright)=left(abright)c$$
$$exists ein G, forall ain G: ea=a$$
$$forall ain G, exists a'in G: a'a=e$$

Are these equivalent? I kind of doubt it, since we have associative semigroups with left but not right identities, but maybe the left-inverses part changes things.

There was a proof that given these axioms, a left-inverse is a right-inverse, and hence that the original inverses axiom is proven, but what about right-identity?

Proof:
Let $gin G$ then $g$ has a left inverse, call it $g'in G$ and this too has a left inverse, call it $g''in G$. Then, $g'g=e$, $g''g'=e$ and so
$$gg'=egg'=g''g'gg'=g''g'=e$$
so $g'$ is the right-inverse of $g$ also.

group-theory axioms

share|cite|improve this question

edited 8 hours ago

Joshua Tilley

asked 8 hours ago

Joshua TilleyJoshua Tilley

65333 silver badges1313 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  All semigroups are associative.
  $endgroup$
  – Shaun
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, fair point on language. I meant to emphasise that these satisfy 2/3 of the alternative group axioms.
  $endgroup$
  – Joshua Tilley
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The word "magma" would be more appropriate then.
  $endgroup$
  – Shaun
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I can edit if you like, but do you have any thoughts on the question?
  $endgroup$
  – Joshua Tilley
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  A left identity $e$ satisfies, given the existence of inverses, $ae=aa'a=ea=a$, for $ain G$
  $endgroup$
  – AnotherJohnDoe
  8 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

The usual axioms I've seen for a group are: associativity; existence of two-sided identity; existence of two-sided inverses for all elements.

$$forall a,b,cin G: aleft(bcright)=left(abright)c$$
$$exists ein G, forall ain G: ae=a=ea$$
$$forall ain G exists a'in G: aa'=e=a'a$$

I recently came across a different axiomatisation, and there were no proofs of equivalence. They were: associativity; existence of left-identity; existence of left-inverses.

$$forall a,b,cin G: aleft(bcright)=left(abright)c$$
$$exists ein G, forall ain G: ea=a$$
$$forall ain G, exists a'in G: a'a=e$$

Are these equivalent? I kind of doubt it, since we have associative semigroups with left but not right identities, but maybe the left-inverses part changes things.

There was a proof that given these axioms, a left-inverse is a right-inverse, and hence that the original inverses axiom is proven, but what about right-identity?

Proof:
Let $gin G$ then $g$ has a left inverse, call it $g'in G$ and this too has a left inverse, call it $g''in G$. Then, $g'g=e$, $g''g'=e$ and so
$$gg'=egg'=g''g'gg'=g''g'=e$$
so $g'$ is the right-inverse of $g$ also.

group-theory axioms

share|cite|improve this question

edited 8 hours ago

Joshua Tilley

asked 8 hours ago

Joshua TilleyJoshua Tilley

65333 silver badges1313 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

Joshua Tilley

asked 8 hours ago

Joshua TilleyJoshua Tilley

65333 silver badges1313 bronze badges

share|cite|improve this question

edited 8 hours ago

Joshua Tilley

asked 8 hours ago

Joshua TilleyJoshua Tilley

65333 silver badges1313 bronze badges

asked 8 hours ago

Joshua TilleyJoshua Tilley

65333 silver badges1313 bronze badges

asked 8 hours ago

Joshua TilleyJoshua Tilley

65333 silver badges1313 bronze badges

1 Answer
1

active

oldest

votes

6

$begingroup$

I agree with the proof that the left inverse (wrt the left identity) is also a right inverse.

Now let $e$ be the left inverse and $g in G$. Then

$$ge = g(g'g) = (gg')g = eg= g$$ where we use that $g'$ is both-sided.

Then $e$ is also a right identity.

share|cite|improve this answer

answered 8 hours ago

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges140140 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There we go! Nice, didn't see that myself. Guess I find it a little odd to establish that inverses are two sided before identity is, since what is a right-inverse if the identity isn't two sided?
  $endgroup$
  – Joshua Tilley
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JoshuaTilley you work with the inverse wrt the left-identity (the only one we have at the outset) and $g'$ is two-sided for that and that very fact then enables this double identity, i.e. that $e$ is identity from the right too.
  $endgroup$
  – Henno Brandsma
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand the argument. My point is that on its own, $aa'=e$ for an arbitrary left-identity $e$ does not make $a'$ behave as a right-inverse, that was my point. Of course it does given the other requirements, as per your answer.
  $endgroup$
  – Joshua Tilley
  8 hours ago
  
  

  
  
  
  

add a comment
 | 

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6

$begingroup$

I agree with the proof that the left inverse (wrt the left identity) is also a right inverse.

Now let $e$ be the left inverse and $g in G$. Then

$$ge = g(g'g) = (gg')g = eg= g$$ where we use that $g'$ is both-sided.

Then $e$ is also a right identity.

share|cite|improve this answer

answered 8 hours ago

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges140140 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There we go! Nice, didn't see that myself. Guess I find it a little odd to establish that inverses are two sided before identity is, since what is a right-inverse if the identity isn't two sided?
  $endgroup$
  – Joshua Tilley
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JoshuaTilley you work with the inverse wrt the left-identity (the only one we have at the outset) and $g'$ is two-sided for that and that very fact then enables this double identity, i.e. that $e$ is identity from the right too.
  $endgroup$
  – Henno Brandsma
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand the argument. My point is that on its own, $aa'=e$ for an arbitrary left-identity $e$ does not make $a'$ behave as a right-inverse, that was my point. Of course it does given the other requirements, as per your answer.
  $endgroup$
  – Joshua Tilley
  8 hours ago
  
  

  
  
  
  

add a comment
 | 

6

$begingroup$

I agree with the proof that the left inverse (wrt the left identity) is also a right inverse.

Now let $e$ be the left inverse and $g in G$. Then

$$ge = g(g'g) = (gg')g = eg= g$$ where we use that $g'$ is both-sided.

Then $e$ is also a right identity.

share|cite|improve this answer

answered 8 hours ago

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges140140 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There we go! Nice, didn't see that myself. Guess I find it a little odd to establish that inverses are two sided before identity is, since what is a right-inverse if the identity isn't two sided?
  $endgroup$
  – Joshua Tilley
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JoshuaTilley you work with the inverse wrt the left-identity (the only one we have at the outset) and $g'$ is two-sided for that and that very fact then enables this double identity, i.e. that $e$ is identity from the right too.
  $endgroup$
  – Henno Brandsma
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand the argument. My point is that on its own, $aa'=e$ for an arbitrary left-identity $e$ does not make $a'$ behave as a right-inverse, that was my point. Of course it does given the other requirements, as per your answer.
  $endgroup$
  – Joshua Tilley
  8 hours ago
  
  

  
  
  
  

add a comment
 | 

$begingroup$

I agree with the proof that the left inverse (wrt the left identity) is also a right inverse.

Now let $e$ be the left inverse and $g in G$. Then

$$ge = g(g'g) = (gg')g = eg= g$$ where we use that $g'$ is both-sided.

Then $e$ is also a right identity.

share|cite|improve this answer

answered 8 hours ago

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges140140 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges140140 bronze badges

answered 8 hours ago

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges140140 bronze badges

answered 8 hours ago

Henno BrandsmaHenno Brandsma

133k44 gold badges5353 silver badges140140 bronze badges