Let $H$ be a subgroup of $G$. Show that if $G/H$ is abelian, then $ghg^{-1}h^{-1}$ is in $H$ for all $g, h$...











up vote
1
down vote

favorite












Let $H$ be a subgroup of $G$. Show that if $G/H$ is abelian, then $ghg^{-1}h^{-1}$ is in $H$ for all $g, h$ in $G$..



I know if $G/H$ is abelian, $(g_1H)(g_2H)=(g_2H)(g_1H)$. But I don't know how to approach from here.










share|cite|improve this question




























    up vote
    1
    down vote

    favorite












    Let $H$ be a subgroup of $G$. Show that if $G/H$ is abelian, then $ghg^{-1}h^{-1}$ is in $H$ for all $g, h$ in $G$..



    I know if $G/H$ is abelian, $(g_1H)(g_2H)=(g_2H)(g_1H)$. But I don't know how to approach from here.










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $H$ be a subgroup of $G$. Show that if $G/H$ is abelian, then $ghg^{-1}h^{-1}$ is in $H$ for all $g, h$ in $G$..



      I know if $G/H$ is abelian, $(g_1H)(g_2H)=(g_2H)(g_1H)$. But I don't know how to approach from here.










      share|cite|improve this question















      Let $H$ be a subgroup of $G$. Show that if $G/H$ is abelian, then $ghg^{-1}h^{-1}$ is in $H$ for all $g, h$ in $G$..



      I know if $G/H$ is abelian, $(g_1H)(g_2H)=(g_2H)(g_1H)$. But I don't know how to approach from here.







      abstract-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 20 at 11:24









      1ENİGMA1

      960316




      960316










      asked Nov 19 at 4:34









      david D

      875




      875






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          2
          down vote



          accepted










          You have it. So what is $(g_1g_2g_1^{-1}g_2^{-1})H$? (P.S. Probably it's better not to use the letter $h$ for an element of $G$; it suggests an element of $H$. So I think your switching to $g_1$ and $g_2$ was preferable.)






          share|cite|improve this answer





















          • Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
            – david D
            Nov 19 at 4:56






          • 1




            If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
            – Ted Shifrin
            Nov 19 at 4:59











          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',
          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%2f3004526%2flet-h-be-a-subgroup-of-g-show-that-if-g-h-is-abelian-then-ghg-1h-1%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          You have it. So what is $(g_1g_2g_1^{-1}g_2^{-1})H$? (P.S. Probably it's better not to use the letter $h$ for an element of $G$; it suggests an element of $H$. So I think your switching to $g_1$ and $g_2$ was preferable.)






          share|cite|improve this answer





















          • Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
            – david D
            Nov 19 at 4:56






          • 1




            If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
            – Ted Shifrin
            Nov 19 at 4:59















          up vote
          2
          down vote



          accepted










          You have it. So what is $(g_1g_2g_1^{-1}g_2^{-1})H$? (P.S. Probably it's better not to use the letter $h$ for an element of $G$; it suggests an element of $H$. So I think your switching to $g_1$ and $g_2$ was preferable.)






          share|cite|improve this answer





















          • Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
            – david D
            Nov 19 at 4:56






          • 1




            If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
            – Ted Shifrin
            Nov 19 at 4:59













          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          You have it. So what is $(g_1g_2g_1^{-1}g_2^{-1})H$? (P.S. Probably it's better not to use the letter $h$ for an element of $G$; it suggests an element of $H$. So I think your switching to $g_1$ and $g_2$ was preferable.)






          share|cite|improve this answer












          You have it. So what is $(g_1g_2g_1^{-1}g_2^{-1})H$? (P.S. Probably it's better not to use the letter $h$ for an element of $G$; it suggests an element of $H$. So I think your switching to $g_1$ and $g_2$ was preferable.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 19 at 4:44









          Ted Shifrin

          62.5k44489




          62.5k44489












          • Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
            – david D
            Nov 19 at 4:56






          • 1




            If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
            – Ted Shifrin
            Nov 19 at 4:59


















          • Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
            – david D
            Nov 19 at 4:56






          • 1




            If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
            – Ted Shifrin
            Nov 19 at 4:59
















          Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
          – david D
          Nov 19 at 4:56




          Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
          – david D
          Nov 19 at 4:56




          1




          1




          If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
          – Ted Shifrin
          Nov 19 at 4:59




          If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
          – Ted Shifrin
          Nov 19 at 4:59


















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3004526%2flet-h-be-a-subgroup-of-g-show-that-if-g-h-is-abelian-then-ghg-1h-1%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

          How to change which sound is reproduced for terminal bell?

          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

          Can I use Tabulator js library in my java Spring + Thymeleaf project?