Is it possible to have an Abelian group under two different binary operations but the binary operations are...












3












$begingroup$


I am trying to show that if $(R, +)$ is an Abelian group and $(R - {0_R}, cdot)$ is an Abelian group, then $(R, +, cdot)$ is not necessarily a field. Note that $0_R$ is the identity element of $(R, +)$. I know that a field is a commutative division ring and one of a ring's properties is that $forall a,b in R, ~ acdot (b + c) = a cdot b + acdot c$. Therefore, I am trying to come up with a set and two binary operations that satisfy the first property, but together do not form a field.



So far, I have come up with a group over polynomials with $+$ being normal addition and $cdot$ being composition, but then $(R - {0_R})$ is not commutative. I would appreciate any help/guidance.



Thanks.






group-theory ring-theory field-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Composition isn't invertible either.
    $endgroup$
    – jgon
    5 hours ago






  • 5




    $begingroup$
    Let $R$ be any six-element set, and put any Abelian group structures you like on $R$ and $R-{0}$.
    $endgroup$
    – Lord Shark the Unknown
    5 hours ago










  • $begingroup$
    You're working too hard. Just literally take any random abelian group structures at all and they almost certainly will not be distributive.
    $endgroup$
    – Eric Wofsey
    1 hour ago
















3












$begingroup$


I am trying to show that if $(R, +)$ is an Abelian group and $(R - {0_R}, cdot)$ is an Abelian group, then $(R, +, cdot)$ is not necessarily a field. Note that $0_R$ is the identity element of $(R, +)$. I know that a field is a commutative division ring and one of a ring's properties is that $forall a,b in R, ~ acdot (b + c) = a cdot b + acdot c$. Therefore, I am trying to come up with a set and two binary operations that satisfy the first property, but together do not form a field.



So far, I have come up with a group over polynomials with $+$ being normal addition and $cdot$ being composition, but then $(R - {0_R})$ is not commutative. I would appreciate any help/guidance.



Thanks.






group-theory ring-theory field-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Composition isn't invertible either.
    $endgroup$
    – jgon
    5 hours ago






  • 5




    $begingroup$
    Let $R$ be any six-element set, and put any Abelian group structures you like on $R$ and $R-{0}$.
    $endgroup$
    – Lord Shark the Unknown
    5 hours ago










  • $begingroup$
    You're working too hard. Just literally take any random abelian group structures at all and they almost certainly will not be distributive.
    $endgroup$
    – Eric Wofsey
    1 hour ago














3












3








3


1



$begingroup$


I am trying to show that if $(R, +)$ is an Abelian group and $(R - {0_R}, cdot)$ is an Abelian group, then $(R, +, cdot)$ is not necessarily a field. Note that $0_R$ is the identity element of $(R, +)$. I know that a field is a commutative division ring and one of a ring's properties is that $forall a,b in R, ~ acdot (b + c) = a cdot b + acdot c$. Therefore, I am trying to come up with a set and two binary operations that satisfy the first property, but together do not form a field.



So far, I have come up with a group over polynomials with $+$ being normal addition and $cdot$ being composition, but then $(R - {0_R})$ is not commutative. I would appreciate any help/guidance.



Thanks.






group-theory ring-theory field-theory







share|cite|improve this question









$endgroup$




I am trying to show that if $(R, +)$ is an Abelian group and $(R - {0_R}, cdot)$ is an Abelian group, then $(R, +, cdot)$ is not necessarily a field. Note that $0_R$ is the identity element of $(R, +)$. I know that a field is a commutative division ring and one of a ring's properties is that $forall a,b in R, ~ acdot (b + c) = a cdot b + acdot c$. Therefore, I am trying to come up with a set and two binary operations that satisfy the first property, but together do not form a field.



So far, I have come up with a group over polynomials with $+$ being normal addition and $cdot$ being composition, but then $(R - {0_R})$ is not commutative. I would appreciate any help/guidance.



Thanks.






group-theory ring-theory field-theory




group-theory ring-theory field-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









sepehr78sepehr78

725




725












  • $begingroup$
    Composition isn't invertible either.
    $endgroup$
    – jgon
    5 hours ago






  • 5




    $begingroup$
    Let $R$ be any six-element set, and put any Abelian group structures you like on $R$ and $R-{0}$.
    $endgroup$
    – Lord Shark the Unknown
    5 hours ago










  • $begingroup$
    You're working too hard. Just literally take any random abelian group structures at all and they almost certainly will not be distributive.
    $endgroup$
    – Eric Wofsey
    1 hour ago


















  • $begingroup$
    Composition isn't invertible either.
    $endgroup$
    – jgon
    5 hours ago






  • 5




    $begingroup$
    Let $R$ be any six-element set, and put any Abelian group structures you like on $R$ and $R-{0}$.
    $endgroup$
    – Lord Shark the Unknown
    5 hours ago










  • $begingroup$
    You're working too hard. Just literally take any random abelian group structures at all and they almost certainly will not be distributive.
    $endgroup$
    – Eric Wofsey
    1 hour ago
















$begingroup$
Composition isn't invertible either.
$endgroup$
– jgon
5 hours ago




$begingroup$
Composition isn't invertible either.
$endgroup$
– jgon
5 hours ago




5




5




$begingroup$
Let $R$ be any six-element set, and put any Abelian group structures you like on $R$ and $R-{0}$.
$endgroup$
– Lord Shark the Unknown
5 hours ago




$begingroup$
Let $R$ be any six-element set, and put any Abelian group structures you like on $R$ and $R-{0}$.
$endgroup$
– Lord Shark the Unknown
5 hours ago












$begingroup$
You're working too hard. Just literally take any random abelian group structures at all and they almost certainly will not be distributive.
$endgroup$
– Eric Wofsey
1 hour ago




$begingroup$
You're working too hard. Just literally take any random abelian group structures at all and they almost certainly will not be distributive.
$endgroup$
– Eric Wofsey
1 hour ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Here is a concrete example, inspired by LStU:



The set is ${0,1,2,3,4,5}$. Addition is just addition mod $6$.



Multiplication is defined by
$$
acdot b = left{ begin{array}{cl} 0& a=0 \ 0 & b=0 \
1 & a = b= 5 \
5 & a=5 wedge b in [1,4]\
5 & b=5 wedge a in [1,4]\
ab pmod{5}& mbox{otherwise}end{array} right.
$$
or as a table
$$
begin{array}{c|cccccc} cdot&0&1&2&3&4&5 \ hline
0 & 0&0&0&0&0&0 \
1 & 0&1&2&3&4&5 \
2 & 0&2&4&1&3&5 \
3 & 0&3&1&4&2&5 \
4 & 0&4&3&2&1&5 \
5 & 5&5&5&5&5&1
end{array}
$$
The group properties, as well as commutativity, are easily checked.



Now consider $$ (1+4)cdot 5 = 5cdot 5 = 1 \
1cdot 5 + 4 cdot 5 = 5+5 = 4 neq 1
$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Um, is the "group" (ignoring 0) actually a group under the operation .? I mean, first year group theory is a long, long time ago now, but I remember one of the features being that each row and column of the Cayley table featurwa each member exactly once. Plus is it associative?
    $endgroup$
    – SamBC
    59 mins ago








  • 1




    $begingroup$
    Yeah, that's not a group. Not associative. 2(5.5) = 2.1 = 2, while (2.5)5 = 5.5 = 1.
    $endgroup$
    – SamBC
    41 mins ago










  • $begingroup$
    The questioner specified that it be a group under both operations. I'm just going with the questioner's actual request. Plus a monoid's operation is still associative.
    $endgroup$
    – SamBC
    39 mins ago












  • $begingroup$
    Ah, it's not even associative, so even if it did just have to be a monoid (for a ring), it's not.
    $endgroup$
    – Joseph Sible
    38 mins ago



















4












$begingroup$

As @LordSharktheUnknown implicitly points out, if you just take a finite set with non-prime-power order (six is the first such integer $ge 2$) and put any group structures you want, it will have to work, because finite fields have prime-power order.



But just to be clear, you can also do it with infinite sets. Pretty much anything you try will work, provided you let loose a bit. Take $R = mathbb{Z}$, with $+$ being regular addition. Let $S = mathbb{Z}setminus {0}$, let $phi:S to R$ be the bijection which shifts negative numbers up by one and is constant on positive numbers. Now define $acdot b = phi^{-1}(phi(a)+phi(b))$. We're just relabeling $S$ to be $mathbb{Z}$ again and then doing regular addition. Now, doing addition first, we have
$$-2cdot(1+1) = -2cdot 2 = phi^{-1}(-1+2) = 1,$$
but distributing first, we have
$$-2cdot(1+1) = -2cdot 1 + -2cdot 1 = phi^{-1}(-1+1) + phi^{-1}(-1+1) = -1+(-1) = -2.$$



In terms of guidance, you should expect that you'll need to do something perverse like this, because most of the examples you'll think of where two binary operations already exist are rings, where distributivity necessarily holds.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Here is a concrete example, inspired by LStU:



    The set is ${0,1,2,3,4,5}$. Addition is just addition mod $6$.



    Multiplication is defined by
    $$
    acdot b = left{ begin{array}{cl} 0& a=0 \ 0 & b=0 \
    1 & a = b= 5 \
    5 & a=5 wedge b in [1,4]\
    5 & b=5 wedge a in [1,4]\
    ab pmod{5}& mbox{otherwise}end{array} right.
    $$
    or as a table
    $$
    begin{array}{c|cccccc} cdot&0&1&2&3&4&5 \ hline
    0 & 0&0&0&0&0&0 \
    1 & 0&1&2&3&4&5 \
    2 & 0&2&4&1&3&5 \
    3 & 0&3&1&4&2&5 \
    4 & 0&4&3&2&1&5 \
    5 & 5&5&5&5&5&1
    end{array}
    $$
    The group properties, as well as commutativity, are easily checked.



    Now consider $$ (1+4)cdot 5 = 5cdot 5 = 1 \
    1cdot 5 + 4 cdot 5 = 5+5 = 4 neq 1
    $$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Um, is the "group" (ignoring 0) actually a group under the operation .? I mean, first year group theory is a long, long time ago now, but I remember one of the features being that each row and column of the Cayley table featurwa each member exactly once. Plus is it associative?
      $endgroup$
      – SamBC
      59 mins ago








    • 1




      $begingroup$
      Yeah, that's not a group. Not associative. 2(5.5) = 2.1 = 2, while (2.5)5 = 5.5 = 1.
      $endgroup$
      – SamBC
      41 mins ago










    • $begingroup$
      The questioner specified that it be a group under both operations. I'm just going with the questioner's actual request. Plus a monoid's operation is still associative.
      $endgroup$
      – SamBC
      39 mins ago












    • $begingroup$
      Ah, it's not even associative, so even if it did just have to be a monoid (for a ring), it's not.
      $endgroup$
      – Joseph Sible
      38 mins ago
















    2












    $begingroup$

    Here is a concrete example, inspired by LStU:



    The set is ${0,1,2,3,4,5}$. Addition is just addition mod $6$.



    Multiplication is defined by
    $$
    acdot b = left{ begin{array}{cl} 0& a=0 \ 0 & b=0 \
    1 & a = b= 5 \
    5 & a=5 wedge b in [1,4]\
    5 & b=5 wedge a in [1,4]\
    ab pmod{5}& mbox{otherwise}end{array} right.
    $$
    or as a table
    $$
    begin{array}{c|cccccc} cdot&0&1&2&3&4&5 \ hline
    0 & 0&0&0&0&0&0 \
    1 & 0&1&2&3&4&5 \
    2 & 0&2&4&1&3&5 \
    3 & 0&3&1&4&2&5 \
    4 & 0&4&3&2&1&5 \
    5 & 5&5&5&5&5&1
    end{array}
    $$
    The group properties, as well as commutativity, are easily checked.



    Now consider $$ (1+4)cdot 5 = 5cdot 5 = 1 \
    1cdot 5 + 4 cdot 5 = 5+5 = 4 neq 1
    $$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Um, is the "group" (ignoring 0) actually a group under the operation .? I mean, first year group theory is a long, long time ago now, but I remember one of the features being that each row and column of the Cayley table featurwa each member exactly once. Plus is it associative?
      $endgroup$
      – SamBC
      59 mins ago








    • 1




      $begingroup$
      Yeah, that's not a group. Not associative. 2(5.5) = 2.1 = 2, while (2.5)5 = 5.5 = 1.
      $endgroup$
      – SamBC
      41 mins ago










    • $begingroup$
      The questioner specified that it be a group under both operations. I'm just going with the questioner's actual request. Plus a monoid's operation is still associative.
      $endgroup$
      – SamBC
      39 mins ago












    • $begingroup$
      Ah, it's not even associative, so even if it did just have to be a monoid (for a ring), it's not.
      $endgroup$
      – Joseph Sible
      38 mins ago














    2












    2








    2





    $begingroup$

    Here is a concrete example, inspired by LStU:



    The set is ${0,1,2,3,4,5}$. Addition is just addition mod $6$.



    Multiplication is defined by
    $$
    acdot b = left{ begin{array}{cl} 0& a=0 \ 0 & b=0 \
    1 & a = b= 5 \
    5 & a=5 wedge b in [1,4]\
    5 & b=5 wedge a in [1,4]\
    ab pmod{5}& mbox{otherwise}end{array} right.
    $$
    or as a table
    $$
    begin{array}{c|cccccc} cdot&0&1&2&3&4&5 \ hline
    0 & 0&0&0&0&0&0 \
    1 & 0&1&2&3&4&5 \
    2 & 0&2&4&1&3&5 \
    3 & 0&3&1&4&2&5 \
    4 & 0&4&3&2&1&5 \
    5 & 5&5&5&5&5&1
    end{array}
    $$
    The group properties, as well as commutativity, are easily checked.



    Now consider $$ (1+4)cdot 5 = 5cdot 5 = 1 \
    1cdot 5 + 4 cdot 5 = 5+5 = 4 neq 1
    $$






    share|cite|improve this answer









    $endgroup$



    Here is a concrete example, inspired by LStU:



    The set is ${0,1,2,3,4,5}$. Addition is just addition mod $6$.



    Multiplication is defined by
    $$
    acdot b = left{ begin{array}{cl} 0& a=0 \ 0 & b=0 \
    1 & a = b= 5 \
    5 & a=5 wedge b in [1,4]\
    5 & b=5 wedge a in [1,4]\
    ab pmod{5}& mbox{otherwise}end{array} right.
    $$
    or as a table
    $$
    begin{array}{c|cccccc} cdot&0&1&2&3&4&5 \ hline
    0 & 0&0&0&0&0&0 \
    1 & 0&1&2&3&4&5 \
    2 & 0&2&4&1&3&5 \
    3 & 0&3&1&4&2&5 \
    4 & 0&4&3&2&1&5 \
    5 & 5&5&5&5&5&1
    end{array}
    $$
    The group properties, as well as commutativity, are easily checked.



    Now consider $$ (1+4)cdot 5 = 5cdot 5 = 1 \
    1cdot 5 + 4 cdot 5 = 5+5 = 4 neq 1
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 5 hours ago









    Mark FischlerMark Fischler

    33.4k12452




    33.4k12452








    • 1




      $begingroup$
      Um, is the "group" (ignoring 0) actually a group under the operation .? I mean, first year group theory is a long, long time ago now, but I remember one of the features being that each row and column of the Cayley table featurwa each member exactly once. Plus is it associative?
      $endgroup$
      – SamBC
      59 mins ago








    • 1




      $begingroup$
      Yeah, that's not a group. Not associative. 2(5.5) = 2.1 = 2, while (2.5)5 = 5.5 = 1.
      $endgroup$
      – SamBC
      41 mins ago










    • $begingroup$
      The questioner specified that it be a group under both operations. I'm just going with the questioner's actual request. Plus a monoid's operation is still associative.
      $endgroup$
      – SamBC
      39 mins ago












    • $begingroup$
      Ah, it's not even associative, so even if it did just have to be a monoid (for a ring), it's not.
      $endgroup$
      – Joseph Sible
      38 mins ago














    • 1




      $begingroup$
      Um, is the "group" (ignoring 0) actually a group under the operation .? I mean, first year group theory is a long, long time ago now, but I remember one of the features being that each row and column of the Cayley table featurwa each member exactly once. Plus is it associative?
      $endgroup$
      – SamBC
      59 mins ago








    • 1




      $begingroup$
      Yeah, that's not a group. Not associative. 2(5.5) = 2.1 = 2, while (2.5)5 = 5.5 = 1.
      $endgroup$
      – SamBC
      41 mins ago










    • $begingroup$
      The questioner specified that it be a group under both operations. I'm just going with the questioner's actual request. Plus a monoid's operation is still associative.
      $endgroup$
      – SamBC
      39 mins ago












    • $begingroup$
      Ah, it's not even associative, so even if it did just have to be a monoid (for a ring), it's not.
      $endgroup$
      – Joseph Sible
      38 mins ago








    1




    1




    $begingroup$
    Um, is the "group" (ignoring 0) actually a group under the operation .? I mean, first year group theory is a long, long time ago now, but I remember one of the features being that each row and column of the Cayley table featurwa each member exactly once. Plus is it associative?
    $endgroup$
    – SamBC
    59 mins ago






    $begingroup$
    Um, is the "group" (ignoring 0) actually a group under the operation .? I mean, first year group theory is a long, long time ago now, but I remember one of the features being that each row and column of the Cayley table featurwa each member exactly once. Plus is it associative?
    $endgroup$
    – SamBC
    59 mins ago






    1




    1




    $begingroup$
    Yeah, that's not a group. Not associative. 2(5.5) = 2.1 = 2, while (2.5)5 = 5.5 = 1.
    $endgroup$
    – SamBC
    41 mins ago




    $begingroup$
    Yeah, that's not a group. Not associative. 2(5.5) = 2.1 = 2, while (2.5)5 = 5.5 = 1.
    $endgroup$
    – SamBC
    41 mins ago












    $begingroup$
    The questioner specified that it be a group under both operations. I'm just going with the questioner's actual request. Plus a monoid's operation is still associative.
    $endgroup$
    – SamBC
    39 mins ago






    $begingroup$
    The questioner specified that it be a group under both operations. I'm just going with the questioner's actual request. Plus a monoid's operation is still associative.
    $endgroup$
    – SamBC
    39 mins ago














    $begingroup$
    Ah, it's not even associative, so even if it did just have to be a monoid (for a ring), it's not.
    $endgroup$
    – Joseph Sible
    38 mins ago




    $begingroup$
    Ah, it's not even associative, so even if it did just have to be a monoid (for a ring), it's not.
    $endgroup$
    – Joseph Sible
    38 mins ago











    4












    $begingroup$

    As @LordSharktheUnknown implicitly points out, if you just take a finite set with non-prime-power order (six is the first such integer $ge 2$) and put any group structures you want, it will have to work, because finite fields have prime-power order.



    But just to be clear, you can also do it with infinite sets. Pretty much anything you try will work, provided you let loose a bit. Take $R = mathbb{Z}$, with $+$ being regular addition. Let $S = mathbb{Z}setminus {0}$, let $phi:S to R$ be the bijection which shifts negative numbers up by one and is constant on positive numbers. Now define $acdot b = phi^{-1}(phi(a)+phi(b))$. We're just relabeling $S$ to be $mathbb{Z}$ again and then doing regular addition. Now, doing addition first, we have
    $$-2cdot(1+1) = -2cdot 2 = phi^{-1}(-1+2) = 1,$$
    but distributing first, we have
    $$-2cdot(1+1) = -2cdot 1 + -2cdot 1 = phi^{-1}(-1+1) + phi^{-1}(-1+1) = -1+(-1) = -2.$$



    In terms of guidance, you should expect that you'll need to do something perverse like this, because most of the examples you'll think of where two binary operations already exist are rings, where distributivity necessarily holds.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      As @LordSharktheUnknown implicitly points out, if you just take a finite set with non-prime-power order (six is the first such integer $ge 2$) and put any group structures you want, it will have to work, because finite fields have prime-power order.



      But just to be clear, you can also do it with infinite sets. Pretty much anything you try will work, provided you let loose a bit. Take $R = mathbb{Z}$, with $+$ being regular addition. Let $S = mathbb{Z}setminus {0}$, let $phi:S to R$ be the bijection which shifts negative numbers up by one and is constant on positive numbers. Now define $acdot b = phi^{-1}(phi(a)+phi(b))$. We're just relabeling $S$ to be $mathbb{Z}$ again and then doing regular addition. Now, doing addition first, we have
      $$-2cdot(1+1) = -2cdot 2 = phi^{-1}(-1+2) = 1,$$
      but distributing first, we have
      $$-2cdot(1+1) = -2cdot 1 + -2cdot 1 = phi^{-1}(-1+1) + phi^{-1}(-1+1) = -1+(-1) = -2.$$



      In terms of guidance, you should expect that you'll need to do something perverse like this, because most of the examples you'll think of where two binary operations already exist are rings, where distributivity necessarily holds.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        As @LordSharktheUnknown implicitly points out, if you just take a finite set with non-prime-power order (six is the first such integer $ge 2$) and put any group structures you want, it will have to work, because finite fields have prime-power order.



        But just to be clear, you can also do it with infinite sets. Pretty much anything you try will work, provided you let loose a bit. Take $R = mathbb{Z}$, with $+$ being regular addition. Let $S = mathbb{Z}setminus {0}$, let $phi:S to R$ be the bijection which shifts negative numbers up by one and is constant on positive numbers. Now define $acdot b = phi^{-1}(phi(a)+phi(b))$. We're just relabeling $S$ to be $mathbb{Z}$ again and then doing regular addition. Now, doing addition first, we have
        $$-2cdot(1+1) = -2cdot 2 = phi^{-1}(-1+2) = 1,$$
        but distributing first, we have
        $$-2cdot(1+1) = -2cdot 1 + -2cdot 1 = phi^{-1}(-1+1) + phi^{-1}(-1+1) = -1+(-1) = -2.$$



        In terms of guidance, you should expect that you'll need to do something perverse like this, because most of the examples you'll think of where two binary operations already exist are rings, where distributivity necessarily holds.






        share|cite|improve this answer











        $endgroup$



        As @LordSharktheUnknown implicitly points out, if you just take a finite set with non-prime-power order (six is the first such integer $ge 2$) and put any group structures you want, it will have to work, because finite fields have prime-power order.



        But just to be clear, you can also do it with infinite sets. Pretty much anything you try will work, provided you let loose a bit. Take $R = mathbb{Z}$, with $+$ being regular addition. Let $S = mathbb{Z}setminus {0}$, let $phi:S to R$ be the bijection which shifts negative numbers up by one and is constant on positive numbers. Now define $acdot b = phi^{-1}(phi(a)+phi(b))$. We're just relabeling $S$ to be $mathbb{Z}$ again and then doing regular addition. Now, doing addition first, we have
        $$-2cdot(1+1) = -2cdot 2 = phi^{-1}(-1+2) = 1,$$
        but distributing first, we have
        $$-2cdot(1+1) = -2cdot 1 + -2cdot 1 = phi^{-1}(-1+1) + phi^{-1}(-1+1) = -1+(-1) = -2.$$



        In terms of guidance, you should expect that you'll need to do something perverse like this, because most of the examples you'll think of where two binary operations already exist are rings, where distributivity necessarily holds.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 4 hours ago

























        answered 5 hours ago









        cspruncsprun

        2,03229




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