Problem from Artin's Algebra [duplicate]

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  • Proving the kernel is a normal subgroup.

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I want to prove that
let S be the set on which group G operates. Let H = g∈G prove that H is normal subgroup of G.
This group action gives an homomorphism whose kernal is H.
Then it follows directly from statement that H is normal.



How to prove that if the subgroup is the kernel of a homomorphism having the group as its domain then it is normal?



Any help clearing up the statement would be greatly appreciated. Thanks!
Is there any other approch ?






group-theory normal-subgroups







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marked as duplicate by José Carlos Santos, Derek Holt, Bungo, Jendrik Stelzner, Shailesh Sep 6 at 13:18


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    up vote
    0
    down vote

    favorite













    This question already has an answer here:



    • Proving the kernel is a normal subgroup.

      2 answers



    I want to prove that
    let S be the set on which group G operates. Let H = g∈G prove that H is normal subgroup of G.
    This group action gives an homomorphism whose kernal is H.
    Then it follows directly from statement that H is normal.



    How to prove that if the subgroup is the kernel of a homomorphism having the group as its domain then it is normal?



    Any help clearing up the statement would be greatly appreciated. Thanks!
    Is there any other approch ?






    group-theory normal-subgroups







    share|cite|improve this question















    marked as duplicate by José Carlos Santos, Derek Holt, Bungo, Jendrik Stelzner, Shailesh Sep 6 at 13:18


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
















      up vote
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      down vote

      favorite









      up vote
      0
      down vote

      favorite












      This question already has an answer here:



      • Proving the kernel is a normal subgroup.

        2 answers



      I want to prove that
      let S be the set on which group G operates. Let H = g∈G prove that H is normal subgroup of G.
      This group action gives an homomorphism whose kernal is H.
      Then it follows directly from statement that H is normal.



      How to prove that if the subgroup is the kernel of a homomorphism having the group as its domain then it is normal?



      Any help clearing up the statement would be greatly appreciated. Thanks!
      Is there any other approch ?






      group-theory normal-subgroups







      share|cite|improve this question
















      This question already has an answer here:



      • Proving the kernel is a normal subgroup.

        2 answers



      I want to prove that
      let S be the set on which group G operates. Let H = g∈G prove that H is normal subgroup of G.
      This group action gives an homomorphism whose kernal is H.
      Then it follows directly from statement that H is normal.



      How to prove that if the subgroup is the kernel of a homomorphism having the group as its domain then it is normal?



      Any help clearing up the statement would be greatly appreciated. Thanks!
      Is there any other approch ?





      This question already has an answer here:



      • Proving the kernel is a normal subgroup.

        2 answers






      group-theory normal-subgroups




      group-theory normal-subgroups






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Sep 4 at 10:40

























      asked Sep 4 at 10:33









      MathsforSS

      427




      427




      marked as duplicate by José Carlos Santos, Derek Holt, Bungo, Jendrik Stelzner, Shailesh Sep 6 at 13:18


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by José Carlos Santos, Derek Holt, Bungo, Jendrik Stelzner, Shailesh Sep 6 at 13:18


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






















          2 Answers
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          Let $phi$ be a homomorphism. Then for any $kinkerphi$ and $gin G$, then
          $$phi(gkg^-1)=phi(g)phi(k)phi(g^-1)= phi(g)cdot ecdot phi(g)^-1 =e$$
          Therefore $gkg^-1inkerphi$, i.e. $kerphi$ is normal.






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            Directly for the situation that you describe:



            Note that for $gin H$ and $xin G$: $$(xgx^-1).s=(xg).(x^-1cdot s)=x.(g.(x^-1.s))=x.(x^-1.s)=(xx^-1).s=e.s=s$$
            so that we can conclude that $xgx^-1in H$.






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              2 Answers
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              2 Answers
              2






              active

              oldest

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              active

              oldest

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              active

              oldest

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              up vote
              0
              down vote













              Let $phi$ be a homomorphism. Then for any $kinkerphi$ and $gin G$, then
              $$phi(gkg^-1)=phi(g)phi(k)phi(g^-1)= phi(g)cdot ecdot phi(g)^-1 =e$$
              Therefore $gkg^-1inkerphi$, i.e. $kerphi$ is normal.






              share|cite|improve this answer
























                up vote
                0
                down vote













                Let $phi$ be a homomorphism. Then for any $kinkerphi$ and $gin G$, then
                $$phi(gkg^-1)=phi(g)phi(k)phi(g^-1)= phi(g)cdot ecdot phi(g)^-1 =e$$
                Therefore $gkg^-1inkerphi$, i.e. $kerphi$ is normal.






                share|cite|improve this answer






















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                  up vote
                  0
                  down vote









                  Let $phi$ be a homomorphism. Then for any $kinkerphi$ and $gin G$, then
                  $$phi(gkg^-1)=phi(g)phi(k)phi(g^-1)= phi(g)cdot ecdot phi(g)^-1 =e$$
                  Therefore $gkg^-1inkerphi$, i.e. $kerphi$ is normal.






                  share|cite|improve this answer












                  Let $phi$ be a homomorphism. Then for any $kinkerphi$ and $gin G$, then
                  $$phi(gkg^-1)=phi(g)phi(k)phi(g^-1)= phi(g)cdot ecdot phi(g)^-1 =e$$
                  Therefore $gkg^-1inkerphi$, i.e. $kerphi$ is normal.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 4 at 10:44









                  BAI

                  1,725518




                  1,725518




















                      up vote
                      0
                      down vote













                      Directly for the situation that you describe:



                      Note that for $gin H$ and $xin G$: $$(xgx^-1).s=(xg).(x^-1cdot s)=x.(g.(x^-1.s))=x.(x^-1.s)=(xx^-1).s=e.s=s$$
                      so that we can conclude that $xgx^-1in H$.






                      share|cite|improve this answer
























                        up vote
                        0
                        down vote













                        Directly for the situation that you describe:



                        Note that for $gin H$ and $xin G$: $$(xgx^-1).s=(xg).(x^-1cdot s)=x.(g.(x^-1.s))=x.(x^-1.s)=(xx^-1).s=e.s=s$$
                        so that we can conclude that $xgx^-1in H$.






                        share|cite|improve this answer






















                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          Directly for the situation that you describe:



                          Note that for $gin H$ and $xin G$: $$(xgx^-1).s=(xg).(x^-1cdot s)=x.(g.(x^-1.s))=x.(x^-1.s)=(xx^-1).s=e.s=s$$
                          so that we can conclude that $xgx^-1in H$.






                          share|cite|improve this answer












                          Directly for the situation that you describe:



                          Note that for $gin H$ and $xin G$: $$(xgx^-1).s=(xg).(x^-1cdot s)=x.(g.(x^-1.s))=x.(x^-1.s)=(xx^-1).s=e.s=s$$
                          so that we can conclude that $xgx^-1in H$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Sep 4 at 10:54









                          drhab

                          89.1k541122




                          89.1k541122












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