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











up vote
1
down vote

favorite












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



I know if G/H is Abelian, (g1H)(g2H)=(g2H)(g1H). But I don't know how to approach from here.






abstract-algebra







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^-1h^-1 is in H for all g, h in G.



    I know if G/H is Abelian, (g1H)(g2H)=(g2H)(g1H). But I don't know how to approach from here.






    abstract-algebra







    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^-1h^-1 is in H for all g, h in G.



      I know if G/H is Abelian, (g1H)(g2H)=(g2H)(g1H). But I don't know how to approach from here.






      abstract-algebra







      share|cite|improve this question













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



      I know if G/H is Abelian, (g1H)(g2H)=(g2H)(g1H). But I don't know how to approach from here.






      abstract-algebra




      abstract-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 18 hours ago









      david D

      835




      835






















          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
            17 hours ago






          • 1




            If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
            – Ted Shifrin
            17 hours ago











          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-is-in-h-f%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
            17 hours ago






          • 1




            If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
            – Ted Shifrin
            17 hours ago















          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
            17 hours ago






          • 1




            If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
            – Ted Shifrin
            17 hours ago













          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 17 hours ago









          Ted Shifrin

          62k44489




          62k44489












          • Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
            – david D
            17 hours ago






          • 1




            If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
            – Ted Shifrin
            17 hours ago


















          • Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
            – david D
            17 hours ago






          • 1




            If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
            – Ted Shifrin
            17 hours ago
















          Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
          – david D
          17 hours ago




          Sorry I am still confused, why (g1g2g1−1g2−1)H can prove the statement?
          – david D
          17 hours ago




          1




          1




          If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
          – Ted Shifrin
          17 hours ago




          If that element of $G/H$ equals $eH$, then $g_1g_2g_1^{-1}g_2^{-1}in H$.
          – Ted Shifrin
          17 hours ago


















           

          draft saved


          draft discarded



















































           


          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-is-in-h-f%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

          'app-layout' is not a known element: how to share Component with different Modules

          Image
          0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
          Read more

          android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
          Read more

          WPF add header to Image with URL pettitions [duplicate]

          Image
          up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
          Read more