Does invertability and closure imply identity?












2














Sorry if this is a basic question or I'm overthinking it, but if an algebraic structure has inverse elements (or at least for a member $a$), that means $a^{-1}a=e$, and if there's closure then e is an element of the set. So in the case of defining a group, for instance, why do we need to include identity? Is it already implied?






abstract-algebra binary-operations







share|cite|improve this question
























  • What alternate axiom(s) are you proposing, precisely?
    – lulu
    3 hours ago










  • @lulu none, I was wondering why we can't remove this axiom because it appears to already be implied. I think I get it now though.
    – Benjamin Thoburn
    3 hours ago












  • Because you need that $, ae=ea=a,$ for all $,a.$
    – Somos
    2 hours ago
















2














Sorry if this is a basic question or I'm overthinking it, but if an algebraic structure has inverse elements (or at least for a member $a$), that means $a^{-1}a=e$, and if there's closure then e is an element of the set. So in the case of defining a group, for instance, why do we need to include identity? Is it already implied?






abstract-algebra binary-operations







share|cite|improve this question
























  • What alternate axiom(s) are you proposing, precisely?
    – lulu
    3 hours ago










  • @lulu none, I was wondering why we can't remove this axiom because it appears to already be implied. I think I get it now though.
    – Benjamin Thoburn
    3 hours ago












  • Because you need that $, ae=ea=a,$ for all $,a.$
    – Somos
    2 hours ago














2












2








2







Sorry if this is a basic question or I'm overthinking it, but if an algebraic structure has inverse elements (or at least for a member $a$), that means $a^{-1}a=e$, and if there's closure then e is an element of the set. So in the case of defining a group, for instance, why do we need to include identity? Is it already implied?






abstract-algebra binary-operations







share|cite|improve this question















Sorry if this is a basic question or I'm overthinking it, but if an algebraic structure has inverse elements (or at least for a member $a$), that means $a^{-1}a=e$, and if there's closure then e is an element of the set. So in the case of defining a group, for instance, why do we need to include identity? Is it already implied?






abstract-algebra binary-operations




abstract-algebra binary-operations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









José Carlos Santos

150k22120221




150k22120221










asked 3 hours ago









Benjamin Thoburn

15210




15210












  • What alternate axiom(s) are you proposing, precisely?
    – lulu
    3 hours ago










  • @lulu none, I was wondering why we can't remove this axiom because it appears to already be implied. I think I get it now though.
    – Benjamin Thoburn
    3 hours ago












  • Because you need that $, ae=ea=a,$ for all $,a.$
    – Somos
    2 hours ago


















  • What alternate axiom(s) are you proposing, precisely?
    – lulu
    3 hours ago










  • @lulu none, I was wondering why we can't remove this axiom because it appears to already be implied. I think I get it now though.
    – Benjamin Thoburn
    3 hours ago












  • Because you need that $, ae=ea=a,$ for all $,a.$
    – Somos
    2 hours ago
















What alternate axiom(s) are you proposing, precisely?
– lulu
3 hours ago




What alternate axiom(s) are you proposing, precisely?
– lulu
3 hours ago












@lulu none, I was wondering why we can't remove this axiom because it appears to already be implied. I think I get it now though.
– Benjamin Thoburn
3 hours ago






@lulu none, I was wondering why we can't remove this axiom because it appears to already be implied. I think I get it now though.
– Benjamin Thoburn
3 hours ago














Because you need that $, ae=ea=a,$ for all $,a.$
– Somos
2 hours ago




Because you need that $, ae=ea=a,$ for all $,a.$
– Somos
2 hours ago










2 Answers
2






active

oldest

votes


















5














Because if we were not essuming the existence of the identity element $e$, we would not be able to express the idea that $a^{-1}a=e$.






share|cite|improve this answer

















  • 1




    +1 One can express the idea that $xa=b$ has a solution for all $a$ and $b$, though. Together with identities (and associativity) this is equivalent to having left inverses, but it makes sense to speak of even in structures without identities,
    – Henning Makholm
    3 hours ago








  • 1




    (And if both $xa=b$ and $ax=b$ always have solutions, then you automatically have identities and inverses).
    – Henning Makholm
    3 hours ago










  • To check my understanding, while invertability does require identity it adds something extra, so it's worth specifying. But indentity is a basis for invertability though so it too has to be specified... I suppose I'm still a bit confused though since while you do need identity for invertability I don't see why you have to say it explicitly as an axiom because it's not adding anything new.
    – Benjamin Thoburn
    2 hours ago










  • If you assert that it adds nothing new, please explan how to express the existence of an inverse of each element without it. Unless that the existence of an inverse becomes$$(forall ain G)(exists a'in G)(forall bin G):a'ab=ba'a=b,$$which is a bit heavy in my opinion.
    – José Carlos Santos
    2 hours ago










  • What I'm trying to get at isn't identity existing for invertability, but rather that it seems that inverse is a sort of "compound" statement, asserting both that inverse elements exist but also with this statement you can conclude there has to be an identity, because without it, as you said, you couldn't even define invertability. In your answer you said WHY identity has to be there but what I'm asking about is if it has to be there in one way WHY add it in a second way? It feels unnessesary to be redundant, introducing this as its own new and seperate axiom.I hope I'm explaining myself well?
    – Benjamin Thoburn
    34 mins ago



















2














It's a reasonable question whether identity needs to be its own axiom. The inverse axiom requires that $a^{-1}$ exist for each $a$ such that $a^{-1} a=a a^{-1} = e$, which of course uses the identity in the definition. Instead, you might extend the inverse axiom as follows:




For each $ain G$, there exists an element $a^{-1}in G$, such that $a^{-1}a=aa^{-1}$ (call the result $e_a$), and such that for all $bin G$, $e_a b=be_a =b$.




This seems to be weaker, allowing multiple distinct identity-ish elements, but it's not hard to show that they must actually all be the same element. So extending the inverse axiom in this way gets you back to the same definition of a group, except for one thing: without the identity axiom, the empty set would vacuously constitute a group.






share|cite|improve this answer

















  • 1




    $e_a e_b=e_b=e_a$
    – Zachary Selk
    3 hours ago











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2 Answers
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active

oldest

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5














Because if we were not essuming the existence of the identity element $e$, we would not be able to express the idea that $a^{-1}a=e$.






share|cite|improve this answer

















  • 1




    +1 One can express the idea that $xa=b$ has a solution for all $a$ and $b$, though. Together with identities (and associativity) this is equivalent to having left inverses, but it makes sense to speak of even in structures without identities,
    – Henning Makholm
    3 hours ago








  • 1




    (And if both $xa=b$ and $ax=b$ always have solutions, then you automatically have identities and inverses).
    – Henning Makholm
    3 hours ago










  • To check my understanding, while invertability does require identity it adds something extra, so it's worth specifying. But indentity is a basis for invertability though so it too has to be specified... I suppose I'm still a bit confused though since while you do need identity for invertability I don't see why you have to say it explicitly as an axiom because it's not adding anything new.
    – Benjamin Thoburn
    2 hours ago










  • If you assert that it adds nothing new, please explan how to express the existence of an inverse of each element without it. Unless that the existence of an inverse becomes$$(forall ain G)(exists a'in G)(forall bin G):a'ab=ba'a=b,$$which is a bit heavy in my opinion.
    – José Carlos Santos
    2 hours ago










  • What I'm trying to get at isn't identity existing for invertability, but rather that it seems that inverse is a sort of "compound" statement, asserting both that inverse elements exist but also with this statement you can conclude there has to be an identity, because without it, as you said, you couldn't even define invertability. In your answer you said WHY identity has to be there but what I'm asking about is if it has to be there in one way WHY add it in a second way? It feels unnessesary to be redundant, introducing this as its own new and seperate axiom.I hope I'm explaining myself well?
    – Benjamin Thoburn
    34 mins ago
















5














Because if we were not essuming the existence of the identity element $e$, we would not be able to express the idea that $a^{-1}a=e$.






share|cite|improve this answer

















  • 1




    +1 One can express the idea that $xa=b$ has a solution for all $a$ and $b$, though. Together with identities (and associativity) this is equivalent to having left inverses, but it makes sense to speak of even in structures without identities,
    – Henning Makholm
    3 hours ago








  • 1




    (And if both $xa=b$ and $ax=b$ always have solutions, then you automatically have identities and inverses).
    – Henning Makholm
    3 hours ago










  • To check my understanding, while invertability does require identity it adds something extra, so it's worth specifying. But indentity is a basis for invertability though so it too has to be specified... I suppose I'm still a bit confused though since while you do need identity for invertability I don't see why you have to say it explicitly as an axiom because it's not adding anything new.
    – Benjamin Thoburn
    2 hours ago










  • If you assert that it adds nothing new, please explan how to express the existence of an inverse of each element without it. Unless that the existence of an inverse becomes$$(forall ain G)(exists a'in G)(forall bin G):a'ab=ba'a=b,$$which is a bit heavy in my opinion.
    – José Carlos Santos
    2 hours ago










  • What I'm trying to get at isn't identity existing for invertability, but rather that it seems that inverse is a sort of "compound" statement, asserting both that inverse elements exist but also with this statement you can conclude there has to be an identity, because without it, as you said, you couldn't even define invertability. In your answer you said WHY identity has to be there but what I'm asking about is if it has to be there in one way WHY add it in a second way? It feels unnessesary to be redundant, introducing this as its own new and seperate axiom.I hope I'm explaining myself well?
    – Benjamin Thoburn
    34 mins ago














5












5








5






Because if we were not essuming the existence of the identity element $e$, we would not be able to express the idea that $a^{-1}a=e$.






share|cite|improve this answer












Because if we were not essuming the existence of the identity element $e$, we would not be able to express the idea that $a^{-1}a=e$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









José Carlos Santos

150k22120221




150k22120221








  • 1




    +1 One can express the idea that $xa=b$ has a solution for all $a$ and $b$, though. Together with identities (and associativity) this is equivalent to having left inverses, but it makes sense to speak of even in structures without identities,
    – Henning Makholm
    3 hours ago








  • 1




    (And if both $xa=b$ and $ax=b$ always have solutions, then you automatically have identities and inverses).
    – Henning Makholm
    3 hours ago










  • To check my understanding, while invertability does require identity it adds something extra, so it's worth specifying. But indentity is a basis for invertability though so it too has to be specified... I suppose I'm still a bit confused though since while you do need identity for invertability I don't see why you have to say it explicitly as an axiom because it's not adding anything new.
    – Benjamin Thoburn
    2 hours ago










  • If you assert that it adds nothing new, please explan how to express the existence of an inverse of each element without it. Unless that the existence of an inverse becomes$$(forall ain G)(exists a'in G)(forall bin G):a'ab=ba'a=b,$$which is a bit heavy in my opinion.
    – José Carlos Santos
    2 hours ago










  • What I'm trying to get at isn't identity existing for invertability, but rather that it seems that inverse is a sort of "compound" statement, asserting both that inverse elements exist but also with this statement you can conclude there has to be an identity, because without it, as you said, you couldn't even define invertability. In your answer you said WHY identity has to be there but what I'm asking about is if it has to be there in one way WHY add it in a second way? It feels unnessesary to be redundant, introducing this as its own new and seperate axiom.I hope I'm explaining myself well?
    – Benjamin Thoburn
    34 mins ago














  • 1




    +1 One can express the idea that $xa=b$ has a solution for all $a$ and $b$, though. Together with identities (and associativity) this is equivalent to having left inverses, but it makes sense to speak of even in structures without identities,
    – Henning Makholm
    3 hours ago








  • 1




    (And if both $xa=b$ and $ax=b$ always have solutions, then you automatically have identities and inverses).
    – Henning Makholm
    3 hours ago










  • To check my understanding, while invertability does require identity it adds something extra, so it's worth specifying. But indentity is a basis for invertability though so it too has to be specified... I suppose I'm still a bit confused though since while you do need identity for invertability I don't see why you have to say it explicitly as an axiom because it's not adding anything new.
    – Benjamin Thoburn
    2 hours ago










  • If you assert that it adds nothing new, please explan how to express the existence of an inverse of each element without it. Unless that the existence of an inverse becomes$$(forall ain G)(exists a'in G)(forall bin G):a'ab=ba'a=b,$$which is a bit heavy in my opinion.
    – José Carlos Santos
    2 hours ago










  • What I'm trying to get at isn't identity existing for invertability, but rather that it seems that inverse is a sort of "compound" statement, asserting both that inverse elements exist but also with this statement you can conclude there has to be an identity, because without it, as you said, you couldn't even define invertability. In your answer you said WHY identity has to be there but what I'm asking about is if it has to be there in one way WHY add it in a second way? It feels unnessesary to be redundant, introducing this as its own new and seperate axiom.I hope I'm explaining myself well?
    – Benjamin Thoburn
    34 mins ago








1




1




+1 One can express the idea that $xa=b$ has a solution for all $a$ and $b$, though. Together with identities (and associativity) this is equivalent to having left inverses, but it makes sense to speak of even in structures without identities,
– Henning Makholm
3 hours ago






+1 One can express the idea that $xa=b$ has a solution for all $a$ and $b$, though. Together with identities (and associativity) this is equivalent to having left inverses, but it makes sense to speak of even in structures without identities,
– Henning Makholm
3 hours ago






1




1




(And if both $xa=b$ and $ax=b$ always have solutions, then you automatically have identities and inverses).
– Henning Makholm
3 hours ago




(And if both $xa=b$ and $ax=b$ always have solutions, then you automatically have identities and inverses).
– Henning Makholm
3 hours ago












To check my understanding, while invertability does require identity it adds something extra, so it's worth specifying. But indentity is a basis for invertability though so it too has to be specified... I suppose I'm still a bit confused though since while you do need identity for invertability I don't see why you have to say it explicitly as an axiom because it's not adding anything new.
– Benjamin Thoburn
2 hours ago




To check my understanding, while invertability does require identity it adds something extra, so it's worth specifying. But indentity is a basis for invertability though so it too has to be specified... I suppose I'm still a bit confused though since while you do need identity for invertability I don't see why you have to say it explicitly as an axiom because it's not adding anything new.
– Benjamin Thoburn
2 hours ago












If you assert that it adds nothing new, please explan how to express the existence of an inverse of each element without it. Unless that the existence of an inverse becomes$$(forall ain G)(exists a'in G)(forall bin G):a'ab=ba'a=b,$$which is a bit heavy in my opinion.
– José Carlos Santos
2 hours ago




If you assert that it adds nothing new, please explan how to express the existence of an inverse of each element without it. Unless that the existence of an inverse becomes$$(forall ain G)(exists a'in G)(forall bin G):a'ab=ba'a=b,$$which is a bit heavy in my opinion.
– José Carlos Santos
2 hours ago












What I'm trying to get at isn't identity existing for invertability, but rather that it seems that inverse is a sort of "compound" statement, asserting both that inverse elements exist but also with this statement you can conclude there has to be an identity, because without it, as you said, you couldn't even define invertability. In your answer you said WHY identity has to be there but what I'm asking about is if it has to be there in one way WHY add it in a second way? It feels unnessesary to be redundant, introducing this as its own new and seperate axiom.I hope I'm explaining myself well?
– Benjamin Thoburn
34 mins ago




What I'm trying to get at isn't identity existing for invertability, but rather that it seems that inverse is a sort of "compound" statement, asserting both that inverse elements exist but also with this statement you can conclude there has to be an identity, because without it, as you said, you couldn't even define invertability. In your answer you said WHY identity has to be there but what I'm asking about is if it has to be there in one way WHY add it in a second way? It feels unnessesary to be redundant, introducing this as its own new and seperate axiom.I hope I'm explaining myself well?
– Benjamin Thoburn
34 mins ago











2














It's a reasonable question whether identity needs to be its own axiom. The inverse axiom requires that $a^{-1}$ exist for each $a$ such that $a^{-1} a=a a^{-1} = e$, which of course uses the identity in the definition. Instead, you might extend the inverse axiom as follows:




For each $ain G$, there exists an element $a^{-1}in G$, such that $a^{-1}a=aa^{-1}$ (call the result $e_a$), and such that for all $bin G$, $e_a b=be_a =b$.




This seems to be weaker, allowing multiple distinct identity-ish elements, but it's not hard to show that they must actually all be the same element. So extending the inverse axiom in this way gets you back to the same definition of a group, except for one thing: without the identity axiom, the empty set would vacuously constitute a group.






share|cite|improve this answer

















  • 1




    $e_a e_b=e_b=e_a$
    – Zachary Selk
    3 hours ago
















2














It's a reasonable question whether identity needs to be its own axiom. The inverse axiom requires that $a^{-1}$ exist for each $a$ such that $a^{-1} a=a a^{-1} = e$, which of course uses the identity in the definition. Instead, you might extend the inverse axiom as follows:




For each $ain G$, there exists an element $a^{-1}in G$, such that $a^{-1}a=aa^{-1}$ (call the result $e_a$), and such that for all $bin G$, $e_a b=be_a =b$.




This seems to be weaker, allowing multiple distinct identity-ish elements, but it's not hard to show that they must actually all be the same element. So extending the inverse axiom in this way gets you back to the same definition of a group, except for one thing: without the identity axiom, the empty set would vacuously constitute a group.






share|cite|improve this answer

















  • 1




    $e_a e_b=e_b=e_a$
    – Zachary Selk
    3 hours ago














2












2








2






It's a reasonable question whether identity needs to be its own axiom. The inverse axiom requires that $a^{-1}$ exist for each $a$ such that $a^{-1} a=a a^{-1} = e$, which of course uses the identity in the definition. Instead, you might extend the inverse axiom as follows:




For each $ain G$, there exists an element $a^{-1}in G$, such that $a^{-1}a=aa^{-1}$ (call the result $e_a$), and such that for all $bin G$, $e_a b=be_a =b$.




This seems to be weaker, allowing multiple distinct identity-ish elements, but it's not hard to show that they must actually all be the same element. So extending the inverse axiom in this way gets you back to the same definition of a group, except for one thing: without the identity axiom, the empty set would vacuously constitute a group.






share|cite|improve this answer












It's a reasonable question whether identity needs to be its own axiom. The inverse axiom requires that $a^{-1}$ exist for each $a$ such that $a^{-1} a=a a^{-1} = e$, which of course uses the identity in the definition. Instead, you might extend the inverse axiom as follows:




For each $ain G$, there exists an element $a^{-1}in G$, such that $a^{-1}a=aa^{-1}$ (call the result $e_a$), and such that for all $bin G$, $e_a b=be_a =b$.




This seems to be weaker, allowing multiple distinct identity-ish elements, but it's not hard to show that they must actually all be the same element. So extending the inverse axiom in this way gets you back to the same definition of a group, except for one thing: without the identity axiom, the empty set would vacuously constitute a group.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









mjqxxxx

31.1k23985




31.1k23985








  • 1




    $e_a e_b=e_b=e_a$
    – Zachary Selk
    3 hours ago














  • 1




    $e_a e_b=e_b=e_a$
    – Zachary Selk
    3 hours ago








1




1




$e_a e_b=e_b=e_a$
– Zachary Selk
3 hours ago




$e_a e_b=e_b=e_a$
– Zachary Selk
3 hours ago


















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Souastre

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Souastre — miejscowość i gmina — Herb Państwo   Francja Region Hauts-de-France Departament Pas-de-Calais Okręg Arras Kod INSEE 62800 Powierzchnia 7,18 km² Populacja  (1990) • liczba ludności 352 • gęstość 49 os./km² Kod pocztowy 62111 Położenie na mapie Francji Souastre 50°09′N   2°34′E / 50,150000   2,566667 Portal Francja Souastre – miejscowość i gmina we Francji, w regionie Hauts-de-France, w departamencie Pas-de-Calais. Według danych na rok 1990 gminę zamieszkiwały 352 osoby, a gęstość zaludnienia wynosiła 49 osób/km² (wśród 1549 gmin regionu Nord-Pas-de-Calais Souastre plasuje się na 887. miejscu pod względem liczby ludności, natomiast pod względem powierzchni na miejscu 507.). Bibliografia | Francuski urząd statystyczny ( fr. ) . p   •   d   •   e Gminy okręgu Arras Ablain-Saint-Nazaire Ablainzevelle Acheville Achicourt Achiet-le-Grand Achiet-le-Pe...
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