Finding Homomorphisms from dihedral groups to cyclical groups












1












$begingroup$


Ok so there was another question very similar to this on here however it leaves me a little confused.
$bf{Question}$



Let G = $D_{14}$ the Dihedral group order 14 and A = $c_7$ be the cyclical group order 7. Find every homomorphism.



I started out by finding the trivial homomorphism when Im($varphi$)=1 then using the first isomorphism theorem the non trivial homomorphism we get $frac{|G|}{|K|}=7$ so we get |K|=2, however i understand that the answer to this is that there are no homomorphisms as the other subgroups aren't normal. But how do i work that out. And why does that mean there are no homomorphisms.






abstract-algebra group-theory group-isomorphism group-homomorphism dihedral-groups







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Ok so there was another question very similar to this on here however it leaves me a little confused.
    $bf{Question}$



    Let G = $D_{14}$ the Dihedral group order 14 and A = $c_7$ be the cyclical group order 7. Find every homomorphism.



    I started out by finding the trivial homomorphism when Im($varphi$)=1 then using the first isomorphism theorem the non trivial homomorphism we get $frac{|G|}{|K|}=7$ so we get |K|=2, however i understand that the answer to this is that there are no homomorphisms as the other subgroups aren't normal. But how do i work that out. And why does that mean there are no homomorphisms.






    abstract-algebra group-theory group-isomorphism group-homomorphism dihedral-groups







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Ok so there was another question very similar to this on here however it leaves me a little confused.
      $bf{Question}$



      Let G = $D_{14}$ the Dihedral group order 14 and A = $c_7$ be the cyclical group order 7. Find every homomorphism.



      I started out by finding the trivial homomorphism when Im($varphi$)=1 then using the first isomorphism theorem the non trivial homomorphism we get $frac{|G|}{|K|}=7$ so we get |K|=2, however i understand that the answer to this is that there are no homomorphisms as the other subgroups aren't normal. But how do i work that out. And why does that mean there are no homomorphisms.






      abstract-algebra group-theory group-isomorphism group-homomorphism dihedral-groups







      share|cite|improve this question









      $endgroup$




      Ok so there was another question very similar to this on here however it leaves me a little confused.
      $bf{Question}$



      Let G = $D_{14}$ the Dihedral group order 14 and A = $c_7$ be the cyclical group order 7. Find every homomorphism.



      I started out by finding the trivial homomorphism when Im($varphi$)=1 then using the first isomorphism theorem the non trivial homomorphism we get $frac{|G|}{|K|}=7$ so we get |K|=2, however i understand that the answer to this is that there are no homomorphisms as the other subgroups aren't normal. But how do i work that out. And why does that mean there are no homomorphisms.






      abstract-algebra group-theory group-isomorphism group-homomorphism dihedral-groups




      abstract-algebra group-theory group-isomorphism group-homomorphism dihedral-groups






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 4 '18 at 19:31









      L GL G

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          2 Answers
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          1












          $begingroup$

          Here's a productive way to go about this question.



          The isomorphism theorems tell us that the image of a homomorphism is determined, up to isomorphism, by the kernel: the image is isomorphic to the quotient modulo the kernel.



          Also, the kernel is a normal subgroup.



          So, let's start by listing all normal subgroups of $D_{14}$:




          • the whole group $D_{14}$;

          • its cyclic subgroup of order 7;

          • the trivial subgroup.


          Next, let's compute the quotient (up to isomorphism) of $D_{14}$ by each of these normal subgroups:




          • the quotient by the whole group is the trivial group;

          • the quotient by the cyclic subgroup of order 7 is the order 2 cyclic group $c_2$;

          • the quotient by the trivial subgroup is the group $D_{14}$.


          Finally, observe that the image of any homomorphism $D_{14} to c_7$ cannot be isomorphic to $c_2$ or to $D_{14}$, so the only possibility is that the image is trivial subgroup of $c_7$.



          Thus, the only homomorphism $D_{14} to c_7$ is the trivial homomorphism.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            If you found that there were only two possible subgroups, then they would have to be G itself and {e}, meaning there are no homomorphisms.



            I'm really confused by some of the stuff you wrote, though. I would think that any of the rotations of the heptagon (1/7, 2/7, ..., 6/7) could be generators.



            I might just be completely misunderstanding though, apologies.






            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









              1












              $begingroup$

              Here's a productive way to go about this question.



              The isomorphism theorems tell us that the image of a homomorphism is determined, up to isomorphism, by the kernel: the image is isomorphic to the quotient modulo the kernel.



              Also, the kernel is a normal subgroup.



              So, let's start by listing all normal subgroups of $D_{14}$:




              • the whole group $D_{14}$;

              • its cyclic subgroup of order 7;

              • the trivial subgroup.


              Next, let's compute the quotient (up to isomorphism) of $D_{14}$ by each of these normal subgroups:




              • the quotient by the whole group is the trivial group;

              • the quotient by the cyclic subgroup of order 7 is the order 2 cyclic group $c_2$;

              • the quotient by the trivial subgroup is the group $D_{14}$.


              Finally, observe that the image of any homomorphism $D_{14} to c_7$ cannot be isomorphic to $c_2$ or to $D_{14}$, so the only possibility is that the image is trivial subgroup of $c_7$.



              Thus, the only homomorphism $D_{14} to c_7$ is the trivial homomorphism.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Here's a productive way to go about this question.



                The isomorphism theorems tell us that the image of a homomorphism is determined, up to isomorphism, by the kernel: the image is isomorphic to the quotient modulo the kernel.



                Also, the kernel is a normal subgroup.



                So, let's start by listing all normal subgroups of $D_{14}$:




                • the whole group $D_{14}$;

                • its cyclic subgroup of order 7;

                • the trivial subgroup.


                Next, let's compute the quotient (up to isomorphism) of $D_{14}$ by each of these normal subgroups:




                • the quotient by the whole group is the trivial group;

                • the quotient by the cyclic subgroup of order 7 is the order 2 cyclic group $c_2$;

                • the quotient by the trivial subgroup is the group $D_{14}$.


                Finally, observe that the image of any homomorphism $D_{14} to c_7$ cannot be isomorphic to $c_2$ or to $D_{14}$, so the only possibility is that the image is trivial subgroup of $c_7$.



                Thus, the only homomorphism $D_{14} to c_7$ is the trivial homomorphism.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Here's a productive way to go about this question.



                  The isomorphism theorems tell us that the image of a homomorphism is determined, up to isomorphism, by the kernel: the image is isomorphic to the quotient modulo the kernel.



                  Also, the kernel is a normal subgroup.



                  So, let's start by listing all normal subgroups of $D_{14}$:




                  • the whole group $D_{14}$;

                  • its cyclic subgroup of order 7;

                  • the trivial subgroup.


                  Next, let's compute the quotient (up to isomorphism) of $D_{14}$ by each of these normal subgroups:




                  • the quotient by the whole group is the trivial group;

                  • the quotient by the cyclic subgroup of order 7 is the order 2 cyclic group $c_2$;

                  • the quotient by the trivial subgroup is the group $D_{14}$.


                  Finally, observe that the image of any homomorphism $D_{14} to c_7$ cannot be isomorphic to $c_2$ or to $D_{14}$, so the only possibility is that the image is trivial subgroup of $c_7$.



                  Thus, the only homomorphism $D_{14} to c_7$ is the trivial homomorphism.






                  share|cite|improve this answer









                  $endgroup$



                  Here's a productive way to go about this question.



                  The isomorphism theorems tell us that the image of a homomorphism is determined, up to isomorphism, by the kernel: the image is isomorphic to the quotient modulo the kernel.



                  Also, the kernel is a normal subgroup.



                  So, let's start by listing all normal subgroups of $D_{14}$:




                  • the whole group $D_{14}$;

                  • its cyclic subgroup of order 7;

                  • the trivial subgroup.


                  Next, let's compute the quotient (up to isomorphism) of $D_{14}$ by each of these normal subgroups:




                  • the quotient by the whole group is the trivial group;

                  • the quotient by the cyclic subgroup of order 7 is the order 2 cyclic group $c_2$;

                  • the quotient by the trivial subgroup is the group $D_{14}$.


                  Finally, observe that the image of any homomorphism $D_{14} to c_7$ cannot be isomorphic to $c_2$ or to $D_{14}$, so the only possibility is that the image is trivial subgroup of $c_7$.



                  Thus, the only homomorphism $D_{14} to c_7$ is the trivial homomorphism.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 5 '18 at 18:20









                  Lee MosherLee Mosher

                  48.5k33681




                  48.5k33681























                      0












                      $begingroup$

                      If you found that there were only two possible subgroups, then they would have to be G itself and {e}, meaning there are no homomorphisms.



                      I'm really confused by some of the stuff you wrote, though. I would think that any of the rotations of the heptagon (1/7, 2/7, ..., 6/7) could be generators.



                      I might just be completely misunderstanding though, apologies.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        If you found that there were only two possible subgroups, then they would have to be G itself and {e}, meaning there are no homomorphisms.



                        I'm really confused by some of the stuff you wrote, though. I would think that any of the rotations of the heptagon (1/7, 2/7, ..., 6/7) could be generators.



                        I might just be completely misunderstanding though, apologies.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          If you found that there were only two possible subgroups, then they would have to be G itself and {e}, meaning there are no homomorphisms.



                          I'm really confused by some of the stuff you wrote, though. I would think that any of the rotations of the heptagon (1/7, 2/7, ..., 6/7) could be generators.



                          I might just be completely misunderstanding though, apologies.






                          share|cite|improve this answer









                          $endgroup$



                          If you found that there were only two possible subgroups, then they would have to be G itself and {e}, meaning there are no homomorphisms.



                          I'm really confused by some of the stuff you wrote, though. I would think that any of the rotations of the heptagon (1/7, 2/7, ..., 6/7) could be generators.



                          I might just be completely misunderstanding though, apologies.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 5 '18 at 18:04









                          nessness

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