How to show that $Kleq S_4$ is a normal subgroup?












1












$begingroup$


Let $K={(1),(1,2),(3,4),(2,4),(1,4),(2,3)}$ is a subgrope of $S_4$, we want to show that it is a normal subgroup.



I know that i have to examine the condition of normality in $S_4$ i.e. for all $gin S_4, g^{-1}Kg=K$. I think this takes times or I am missing something near. Please give me the right hint. Thank you.



Edit: $K = {(1),(12)(34),(13)(24),(14)(23)}.$






group-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    It is not a subgroup! $(1,2)(2,4)=(1,2,4)notin K$.
    $endgroup$
    – Berci
    Feb 23 '13 at 15:53










  • $begingroup$
    It is printed as above as in my text note.
    $endgroup$
    – Nancy Rutkowskie
    Feb 23 '13 at 15:57






  • 5




    $begingroup$
    What is your text book? Most probably it is a typo or you simply miscopied... It could probably be $,{(1), (12)(34), (13)(24), (14)(23)}triangleleft S_4,$
    $endgroup$
    – DonAntonio
    Feb 23 '13 at 16:56
















1












$begingroup$


Let $K={(1),(1,2),(3,4),(2,4),(1,4),(2,3)}$ is a subgrope of $S_4$, we want to show that it is a normal subgroup.



I know that i have to examine the condition of normality in $S_4$ i.e. for all $gin S_4, g^{-1}Kg=K$. I think this takes times or I am missing something near. Please give me the right hint. Thank you.



Edit: $K = {(1),(12)(34),(13)(24),(14)(23)}.$






group-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    It is not a subgroup! $(1,2)(2,4)=(1,2,4)notin K$.
    $endgroup$
    – Berci
    Feb 23 '13 at 15:53










  • $begingroup$
    It is printed as above as in my text note.
    $endgroup$
    – Nancy Rutkowskie
    Feb 23 '13 at 15:57






  • 5




    $begingroup$
    What is your text book? Most probably it is a typo or you simply miscopied... It could probably be $,{(1), (12)(34), (13)(24), (14)(23)}triangleleft S_4,$
    $endgroup$
    – DonAntonio
    Feb 23 '13 at 16:56














1












1








1





$begingroup$


Let $K={(1),(1,2),(3,4),(2,4),(1,4),(2,3)}$ is a subgrope of $S_4$, we want to show that it is a normal subgroup.



I know that i have to examine the condition of normality in $S_4$ i.e. for all $gin S_4, g^{-1}Kg=K$. I think this takes times or I am missing something near. Please give me the right hint. Thank you.



Edit: $K = {(1),(12)(34),(13)(24),(14)(23)}.$






group-theory







share|cite|improve this question











$endgroup$




Let $K={(1),(1,2),(3,4),(2,4),(1,4),(2,3)}$ is a subgrope of $S_4$, we want to show that it is a normal subgroup.



I know that i have to examine the condition of normality in $S_4$ i.e. for all $gin S_4, g^{-1}Kg=K$. I think this takes times or I am missing something near. Please give me the right hint. Thank you.



Edit: $K = {(1),(12)(34),(13)(24),(14)(23)}.$






group-theory




group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 16:41









Dietrich Burde

80.1k647104




80.1k647104










asked Feb 23 '13 at 15:51









Nancy RutkowskieNancy Rutkowskie

7051718




7051718








  • 3




    $begingroup$
    It is not a subgroup! $(1,2)(2,4)=(1,2,4)notin K$.
    $endgroup$
    – Berci
    Feb 23 '13 at 15:53










  • $begingroup$
    It is printed as above as in my text note.
    $endgroup$
    – Nancy Rutkowskie
    Feb 23 '13 at 15:57






  • 5




    $begingroup$
    What is your text book? Most probably it is a typo or you simply miscopied... It could probably be $,{(1), (12)(34), (13)(24), (14)(23)}triangleleft S_4,$
    $endgroup$
    – DonAntonio
    Feb 23 '13 at 16:56














  • 3




    $begingroup$
    It is not a subgroup! $(1,2)(2,4)=(1,2,4)notin K$.
    $endgroup$
    – Berci
    Feb 23 '13 at 15:53










  • $begingroup$
    It is printed as above as in my text note.
    $endgroup$
    – Nancy Rutkowskie
    Feb 23 '13 at 15:57






  • 5




    $begingroup$
    What is your text book? Most probably it is a typo or you simply miscopied... It could probably be $,{(1), (12)(34), (13)(24), (14)(23)}triangleleft S_4,$
    $endgroup$
    – DonAntonio
    Feb 23 '13 at 16:56








3




3




$begingroup$
It is not a subgroup! $(1,2)(2,4)=(1,2,4)notin K$.
$endgroup$
– Berci
Feb 23 '13 at 15:53




$begingroup$
It is not a subgroup! $(1,2)(2,4)=(1,2,4)notin K$.
$endgroup$
– Berci
Feb 23 '13 at 15:53












$begingroup$
It is printed as above as in my text note.
$endgroup$
– Nancy Rutkowskie
Feb 23 '13 at 15:57




$begingroup$
It is printed as above as in my text note.
$endgroup$
– Nancy Rutkowskie
Feb 23 '13 at 15:57




5




5




$begingroup$
What is your text book? Most probably it is a typo or you simply miscopied... It could probably be $,{(1), (12)(34), (13)(24), (14)(23)}triangleleft S_4,$
$endgroup$
– DonAntonio
Feb 23 '13 at 16:56




$begingroup$
What is your text book? Most probably it is a typo or you simply miscopied... It could probably be $,{(1), (12)(34), (13)(24), (14)(23)}triangleleft S_4,$
$endgroup$
– DonAntonio
Feb 23 '13 at 16:56










2 Answers
2






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oldest

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3












$begingroup$

This isn't going to work on two fronts:



First your subset $K$ is not a subgroup: $(1 2)(2 3) = (1 2 3)$ so $K$ is not closed under the group operation.



Second your subset $K$ is not closed under conjugation: $(2 3)^{-1}(1 2)(2 3) = (1 3)$.



Maybe you meant the subgroup generated by $K$? But $K$ generates all of $S_4$ so in that case it is trivially normal.






share|cite|improve this answer









$endgroup$





















    6












    $begingroup$

    Maybe your course instructor intended $K = {(1),(12)(34),(13)(24),(14)(23)}.$ That is, $K$ consists of the identity, together with all products of two disjoint $2$-cycles. This is a subgroup. It is also normal because the identity is conjugate only to the identity, and because the set of products of two disjoint $2$-cycles is closed under conjugacy, so that $sigma Ksigma^{-1} = K$ for all $sigma in S_{4}.$






    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









      3












      $begingroup$

      This isn't going to work on two fronts:



      First your subset $K$ is not a subgroup: $(1 2)(2 3) = (1 2 3)$ so $K$ is not closed under the group operation.



      Second your subset $K$ is not closed under conjugation: $(2 3)^{-1}(1 2)(2 3) = (1 3)$.



      Maybe you meant the subgroup generated by $K$? But $K$ generates all of $S_4$ so in that case it is trivially normal.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        This isn't going to work on two fronts:



        First your subset $K$ is not a subgroup: $(1 2)(2 3) = (1 2 3)$ so $K$ is not closed under the group operation.



        Second your subset $K$ is not closed under conjugation: $(2 3)^{-1}(1 2)(2 3) = (1 3)$.



        Maybe you meant the subgroup generated by $K$? But $K$ generates all of $S_4$ so in that case it is trivially normal.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          This isn't going to work on two fronts:



          First your subset $K$ is not a subgroup: $(1 2)(2 3) = (1 2 3)$ so $K$ is not closed under the group operation.



          Second your subset $K$ is not closed under conjugation: $(2 3)^{-1}(1 2)(2 3) = (1 3)$.



          Maybe you meant the subgroup generated by $K$? But $K$ generates all of $S_4$ so in that case it is trivially normal.






          share|cite|improve this answer









          $endgroup$



          This isn't going to work on two fronts:



          First your subset $K$ is not a subgroup: $(1 2)(2 3) = (1 2 3)$ so $K$ is not closed under the group operation.



          Second your subset $K$ is not closed under conjugation: $(2 3)^{-1}(1 2)(2 3) = (1 3)$.



          Maybe you meant the subgroup generated by $K$? But $K$ generates all of $S_4$ so in that case it is trivially normal.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 23 '13 at 15:58









          JimJim

          24.4k23370




          24.4k23370























              6












              $begingroup$

              Maybe your course instructor intended $K = {(1),(12)(34),(13)(24),(14)(23)}.$ That is, $K$ consists of the identity, together with all products of two disjoint $2$-cycles. This is a subgroup. It is also normal because the identity is conjugate only to the identity, and because the set of products of two disjoint $2$-cycles is closed under conjugacy, so that $sigma Ksigma^{-1} = K$ for all $sigma in S_{4}.$






              share|cite|improve this answer









              $endgroup$


















                6












                $begingroup$

                Maybe your course instructor intended $K = {(1),(12)(34),(13)(24),(14)(23)}.$ That is, $K$ consists of the identity, together with all products of two disjoint $2$-cycles. This is a subgroup. It is also normal because the identity is conjugate only to the identity, and because the set of products of two disjoint $2$-cycles is closed under conjugacy, so that $sigma Ksigma^{-1} = K$ for all $sigma in S_{4}.$






                share|cite|improve this answer









                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  Maybe your course instructor intended $K = {(1),(12)(34),(13)(24),(14)(23)}.$ That is, $K$ consists of the identity, together with all products of two disjoint $2$-cycles. This is a subgroup. It is also normal because the identity is conjugate only to the identity, and because the set of products of two disjoint $2$-cycles is closed under conjugacy, so that $sigma Ksigma^{-1} = K$ for all $sigma in S_{4}.$






                  share|cite|improve this answer









                  $endgroup$



                  Maybe your course instructor intended $K = {(1),(12)(34),(13)(24),(14)(23)}.$ That is, $K$ consists of the identity, together with all products of two disjoint $2$-cycles. This is a subgroup. It is also normal because the identity is conjugate only to the identity, and because the set of products of two disjoint $2$-cycles is closed under conjugacy, so that $sigma Ksigma^{-1} = K$ for all $sigma in S_{4}.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 23 '13 at 16:54









                  Geoff RobinsonGeoff Robinson

                  20.6k13043




                  20.6k13043






























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