Function that is homomorphism and bijection












-1












$begingroup$


Find a function from a group
to a permutation group
{1,2,3} that this function be
homomorphism and bijection






abstract-algebra







share|cite|improve this question











$endgroup$

















    -1












    $begingroup$


    Find a function from a group
    to a permutation group
    {1,2,3} that this function be
    homomorphism and bijection






    abstract-algebra







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$


      Find a function from a group
      to a permutation group
      {1,2,3} that this function be
      homomorphism and bijection






      abstract-algebra







      share|cite|improve this question











      $endgroup$




      Find a function from a group
      to a permutation group
      {1,2,3} that this function be
      homomorphism and bijection






      abstract-algebra




      abstract-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 2 at 10:07









      Bernard

      124k741117




      124k741117










      asked Feb 2 at 7:15









      S.gulS.gul

      112




      112






















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          $begingroup$

          One could just take the identity on $S_3$, the group of permutations of ${1,2,3}$.



          For a slightly less trivial example, how about $h:D_3to S_3$ determined by $h(sigma)=(12)$ and $h(rho)=(123)$.



          Here $D_3=langle sigma, rhomidsigma^2,rho^3,(sigmarho)^2rangle $.



          So $D_3cong S_3$.






          share|cite|improve this answer









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            $begingroup$

            One could just take the identity on $S_3$, the group of permutations of ${1,2,3}$.



            For a slightly less trivial example, how about $h:D_3to S_3$ determined by $h(sigma)=(12)$ and $h(rho)=(123)$.



            Here $D_3=langle sigma, rhomidsigma^2,rho^3,(sigmarho)^2rangle $.



            So $D_3cong S_3$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              One could just take the identity on $S_3$, the group of permutations of ${1,2,3}$.



              For a slightly less trivial example, how about $h:D_3to S_3$ determined by $h(sigma)=(12)$ and $h(rho)=(123)$.



              Here $D_3=langle sigma, rhomidsigma^2,rho^3,(sigmarho)^2rangle $.



              So $D_3cong S_3$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                One could just take the identity on $S_3$, the group of permutations of ${1,2,3}$.



                For a slightly less trivial example, how about $h:D_3to S_3$ determined by $h(sigma)=(12)$ and $h(rho)=(123)$.



                Here $D_3=langle sigma, rhomidsigma^2,rho^3,(sigmarho)^2rangle $.



                So $D_3cong S_3$.






                share|cite|improve this answer









                $endgroup$



                One could just take the identity on $S_3$, the group of permutations of ${1,2,3}$.



                For a slightly less trivial example, how about $h:D_3to S_3$ determined by $h(sigma)=(12)$ and $h(rho)=(123)$.



                Here $D_3=langle sigma, rhomidsigma^2,rho^3,(sigmarho)^2rangle $.



                So $D_3cong S_3$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 2 at 8:43









                Chris CusterChris Custer

                14.4k3827




                14.4k3827






























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