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










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










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










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






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

81.5k648106




81.5k648106










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






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









$endgroup$





















    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









    $endgroup$














      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f312071%2fhow-to-show-that-k-leq-s-4-is-a-normal-subgroup%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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

          24.5k23370




          24.5k23370























              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

                  20.7k13144




                  20.7k13144






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f312071%2fhow-to-show-that-k-leq-s-4-is-a-normal-subgroup%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Probability when a professor distributes a quiz and homework assignment to a class of n students.

                      Aardman Animations

                      Are they similar matrix