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












2












$begingroup$


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
    4 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
    4 hours ago
















2












$begingroup$


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
    4 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
    4 hours ago














2












2








2


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 4 hours ago









sepehr78sepehr78

675




675












  • $begingroup$
    Composition isn't invertible either.
    $endgroup$
    – jgon
    4 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
    4 hours ago


















  • $begingroup$
    Composition isn't invertible either.
    $endgroup$
    – jgon
    4 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
    4 hours ago
















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




$begingroup$
Composition isn't invertible either.
$endgroup$
– jgon
4 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
4 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
4 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

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









$endgroup$





















    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











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      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









      $endgroup$


















        3












        $begingroup$

        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









        $endgroup$
















          3












          3








          3





          $begingroup$

          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 3 hours ago









          Mark FischlerMark Fischler

          33.4k12452




          33.4k12452























              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 3 hours ago

























                  answered 3 hours ago









                  cspruncsprun

                  2,00829




                  2,00829






























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                      濃尾地震

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                      濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸...
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                      シテ島

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                      シテ島 Île de la Cité シテ島（トゥルネル橋からの眺望） 所在地 フランス･パリ 座標 北緯48度51分17秒 東経2度20分45秒 座標: 北緯48度51分17秒 東経2度20分45秒 面積 0.225 km² 最高標高 48.854722 m  プロジェクト 地形 テンプレートを表示 シテ島 （シテとう、仏 : Île de la Cité）は、サン＝ルイ島と並ぶ、パリの中心部を流れるセーヌ川の中州である。パリ市内で最も古い歴史をもつパリ1区及び4区に属し、“パリ発祥の地”とも称される。ユリウス・カエサルの『ガリア戦記』では、すでに紀元前1世紀にパリシイ族 (Parisii) が住んでいたことが述べられている。 目次 1 概要 2 歴史ギャラリー 3 名所 4 シテ島の交通 4.1 橋 4.2 鉄道 5 ギャラリー 6 関連項目 7 脚注 8 外部リンク 概要 東半分が4区に、西半分は1区に属する。 ノートルダム大聖堂とパリ警視庁とパリ市立病院（ オテル・デュー・ド・パリ （フランス語版） 、直訳では「パリの神の館」）がある。病院の名前は、旅の病人を宿泊させ、病になれば看病した教会、巡礼教会からの発展の名残である。近代外科医学の開拓者で、床屋医者の出身でもある外科医アンブロワーズ・パレが勤務していた病院としてもその名が知られている。位置関係は大聖堂が警視庁庁舎と向き合い、警視庁の隣に市立病院。 また、この島はとても狭いので、ここの住民たちは夜も一定の時間を過ぎるとトイレの水は流さない等、互いに気配りしつつ一緒に住んでいくことの不便さを味わってきた。ここから「シテ島に住む者」を表す古仏語：citeain （現代仏語：シトワイヤン citoyen 、現代英語：シティズン citizen）の意味が転じ、「 市民 」という言葉が成立したという（仏語：cité は英語：city に相当する）。 中世以来からの狭く入り組んだ路地は、19世紀に入りセーヌ川左岸と共に貧民窟化していた。そうした狭く不衛生で入り組んだ空間は、2月革命に見るようなバリケード戦といった民衆騒...
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