Is an inverse element of binary operation unique? If yes then how?












2












$begingroup$


I am trying to prove it but not getting any clue how to start it!
$$a*b=b*a=e,$$
$$a*c=c*a=e$$
How to show $b=c$?






abstract-algebra binary-operations nonassociative-algebras







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I am trying to prove it but not getting any clue how to start it!
    $$a*b=b*a=e,$$
    $$a*c=c*a=e$$
    How to show $b=c$?






    abstract-algebra binary-operations nonassociative-algebras







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I am trying to prove it but not getting any clue how to start it!
      $$a*b=b*a=e,$$
      $$a*c=c*a=e$$
      How to show $b=c$?






      abstract-algebra binary-operations nonassociative-algebras







      share|cite|improve this question











      $endgroup$




      I am trying to prove it but not getting any clue how to start it!
      $$a*b=b*a=e,$$
      $$a*c=c*a=e$$
      How to show $b=c$?






      abstract-algebra binary-operations nonassociative-algebras




      abstract-algebra binary-operations nonassociative-algebras






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 14 at 9:38









      Shaun

      9,366113684




      9,366113684










      asked Feb 14 at 9:06









      Vivek DhingraVivek Dhingra

      141




      141






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          If the binary operation is associative you have that for all $ain G$ if $b,cin G$ are elements such that



          $ac=ca=e$ and $ab=ba=e$



          then



          $c=ce=c(ab)=(ca)b=eb =b$



          so



          $c=b$






          share|cite|improve this answer











          $endgroup$









          • 3




            $begingroup$
            All binary operations are no Associative
            $endgroup$
            – Vivek Dhingra
            Feb 14 at 9:19










          • $begingroup$
            Sorry, you’re right, I’ve corrected my answer
            $endgroup$
            – Federico Fallucca
            Feb 14 at 9:54






          • 1




            $begingroup$
            You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
            $endgroup$
            – Dannie
            Feb 14 at 10:00



















          2












          $begingroup$

          Well, I think you are refering to the shortening rule:
          $$a*b=a*c Rightarrow b=c, quad b*a=c*aRightarrow b=c.$$
          If you work in a ring $R$ and $0ne ain R$ is not a zero divisor, then
          $a*b=a*c$ implies that $a*(b-c)=0$. If $a$ is not a zero divisor, then $b-c=0$, i.e., $b=c$. Its not necessary that $ain R$ is invertible!






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            It is not unique. Take $S = {a,b,c,e}$ and set $ab = ba = e$, $ac = ca =e$, $ea = a = ae$, $eb = b= be$, $ec = c = ce$, $ee = e$ and define the missing products $aa, bb, bc, cb, cc$ as you wish. You will get a binary operation for which $e$ is the identity and $a$ has two inverses: $b$ and $c$.






            share|cite|improve this answer









            $endgroup$













              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              If the binary operation is associative you have that for all $ain G$ if $b,cin G$ are elements such that



              $ac=ca=e$ and $ab=ba=e$



              then



              $c=ce=c(ab)=(ca)b=eb =b$



              so



              $c=b$






              share|cite|improve this answer











              $endgroup$









              • 3




                $begingroup$
                All binary operations are no Associative
                $endgroup$
                – Vivek Dhingra
                Feb 14 at 9:19










              • $begingroup$
                Sorry, you’re right, I’ve corrected my answer
                $endgroup$
                – Federico Fallucca
                Feb 14 at 9:54






              • 1




                $begingroup$
                You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
                $endgroup$
                – Dannie
                Feb 14 at 10:00
















              3












              $begingroup$

              If the binary operation is associative you have that for all $ain G$ if $b,cin G$ are elements such that



              $ac=ca=e$ and $ab=ba=e$



              then



              $c=ce=c(ab)=(ca)b=eb =b$



              so



              $c=b$






              share|cite|improve this answer











              $endgroup$









              • 3




                $begingroup$
                All binary operations are no Associative
                $endgroup$
                – Vivek Dhingra
                Feb 14 at 9:19










              • $begingroup$
                Sorry, you’re right, I’ve corrected my answer
                $endgroup$
                – Federico Fallucca
                Feb 14 at 9:54






              • 1




                $begingroup$
                You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
                $endgroup$
                – Dannie
                Feb 14 at 10:00














              3












              3








              3





              $begingroup$

              If the binary operation is associative you have that for all $ain G$ if $b,cin G$ are elements such that



              $ac=ca=e$ and $ab=ba=e$



              then



              $c=ce=c(ab)=(ca)b=eb =b$



              so



              $c=b$






              share|cite|improve this answer











              $endgroup$



              If the binary operation is associative you have that for all $ain G$ if $b,cin G$ are elements such that



              $ac=ca=e$ and $ab=ba=e$



              then



              $c=ce=c(ab)=(ca)b=eb =b$



              so



              $c=b$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Feb 14 at 9:53

























              answered Feb 14 at 9:08









              Federico FalluccaFederico Fallucca

              2,270210




              2,270210








              • 3




                $begingroup$
                All binary operations are no Associative
                $endgroup$
                – Vivek Dhingra
                Feb 14 at 9:19










              • $begingroup$
                Sorry, you’re right, I’ve corrected my answer
                $endgroup$
                – Federico Fallucca
                Feb 14 at 9:54






              • 1




                $begingroup$
                You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
                $endgroup$
                – Dannie
                Feb 14 at 10:00














              • 3




                $begingroup$
                All binary operations are no Associative
                $endgroup$
                – Vivek Dhingra
                Feb 14 at 9:19










              • $begingroup$
                Sorry, you’re right, I’ve corrected my answer
                $endgroup$
                – Federico Fallucca
                Feb 14 at 9:54






              • 1




                $begingroup$
                You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
                $endgroup$
                – Dannie
                Feb 14 at 10:00








              3




              3




              $begingroup$
              All binary operations are no Associative
              $endgroup$
              – Vivek Dhingra
              Feb 14 at 9:19




              $begingroup$
              All binary operations are no Associative
              $endgroup$
              – Vivek Dhingra
              Feb 14 at 9:19












              $begingroup$
              Sorry, you’re right, I’ve corrected my answer
              $endgroup$
              – Federico Fallucca
              Feb 14 at 9:54




              $begingroup$
              Sorry, you’re right, I’ve corrected my answer
              $endgroup$
              – Federico Fallucca
              Feb 14 at 9:54




              1




              1




              $begingroup$
              You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
              $endgroup$
              – Dannie
              Feb 14 at 10:00




              $begingroup$
              You need some extra constraints for your proposition to have a hope of being true VivekDhingra . @Federico 's suggestion that they be a group (and so associative) is a reasonable one. The binary operation x * y = e (for all x,y) satisfies your criteria yet not that b=c.
              $endgroup$
              – Dannie
              Feb 14 at 10:00











              2












              $begingroup$

              Well, I think you are refering to the shortening rule:
              $$a*b=a*c Rightarrow b=c, quad b*a=c*aRightarrow b=c.$$
              If you work in a ring $R$ and $0ne ain R$ is not a zero divisor, then
              $a*b=a*c$ implies that $a*(b-c)=0$. If $a$ is not a zero divisor, then $b-c=0$, i.e., $b=c$. Its not necessary that $ain R$ is invertible!






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Well, I think you are refering to the shortening rule:
                $$a*b=a*c Rightarrow b=c, quad b*a=c*aRightarrow b=c.$$
                If you work in a ring $R$ and $0ne ain R$ is not a zero divisor, then
                $a*b=a*c$ implies that $a*(b-c)=0$. If $a$ is not a zero divisor, then $b-c=0$, i.e., $b=c$. Its not necessary that $ain R$ is invertible!






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Well, I think you are refering to the shortening rule:
                  $$a*b=a*c Rightarrow b=c, quad b*a=c*aRightarrow b=c.$$
                  If you work in a ring $R$ and $0ne ain R$ is not a zero divisor, then
                  $a*b=a*c$ implies that $a*(b-c)=0$. If $a$ is not a zero divisor, then $b-c=0$, i.e., $b=c$. Its not necessary that $ain R$ is invertible!






                  share|cite|improve this answer









                  $endgroup$



                  Well, I think you are refering to the shortening rule:
                  $$a*b=a*c Rightarrow b=c, quad b*a=c*aRightarrow b=c.$$
                  If you work in a ring $R$ and $0ne ain R$ is not a zero divisor, then
                  $a*b=a*c$ implies that $a*(b-c)=0$. If $a$ is not a zero divisor, then $b-c=0$, i.e., $b=c$. Its not necessary that $ain R$ is invertible!







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 14 at 13:42









                  WuestenfuxWuestenfux

                  4,9821513




                  4,9821513























                      1












                      $begingroup$

                      It is not unique. Take $S = {a,b,c,e}$ and set $ab = ba = e$, $ac = ca =e$, $ea = a = ae$, $eb = b= be$, $ec = c = ce$, $ee = e$ and define the missing products $aa, bb, bc, cb, cc$ as you wish. You will get a binary operation for which $e$ is the identity and $a$ has two inverses: $b$ and $c$.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        It is not unique. Take $S = {a,b,c,e}$ and set $ab = ba = e$, $ac = ca =e$, $ea = a = ae$, $eb = b= be$, $ec = c = ce$, $ee = e$ and define the missing products $aa, bb, bc, cb, cc$ as you wish. You will get a binary operation for which $e$ is the identity and $a$ has two inverses: $b$ and $c$.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          It is not unique. Take $S = {a,b,c,e}$ and set $ab = ba = e$, $ac = ca =e$, $ea = a = ae$, $eb = b= be$, $ec = c = ce$, $ee = e$ and define the missing products $aa, bb, bc, cb, cc$ as you wish. You will get a binary operation for which $e$ is the identity and $a$ has two inverses: $b$ and $c$.






                          share|cite|improve this answer









                          $endgroup$



                          It is not unique. Take $S = {a,b,c,e}$ and set $ab = ba = e$, $ac = ca =e$, $ea = a = ae$, $eb = b= be$, $ec = c = ce$, $ee = e$ and define the missing products $aa, bb, bc, cb, cc$ as you wish. You will get a binary operation for which $e$ is the identity and $a$ has two inverses: $b$ and $c$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Feb 14 at 9:40









                          J.-E. PinJ.-E. Pin

                          18.5k21754




                          18.5k21754






























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