Commutator group normal in G [duplicate]












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  • How to show that the commutator subgroup is a normal subgroup

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$G$ is a group with $e$ as it's identity element.
The commutator group $[G,G] = <[x,y]|x,y in G>$ such that $[x,y]=x^{-1}y^{-1}xy : : : forall x,y in G$



Show that $[G,G]triangleleft G$



Solution:
If $[G,G]triangleleft G$, then $g^{-1}[x,y]g$ must belong to [G,G].



$$g^{-1}[x,y]g$$
$$g^{-1}x^{-1}y^{-1}xyg$$



I am stuck here. I am not sure how to proceed since we have three elements...



Thanks for the help!






group-theory







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 Dec 8 '18 at 3:54


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    0












    $begingroup$



    This question already has an answer here:




    • How to show that the commutator subgroup is a normal subgroup

      1 answer




    $G$ is a group with $e$ as it's identity element.
    The commutator group $[G,G] = <[x,y]|x,y in G>$ such that $[x,y]=x^{-1}y^{-1}xy : : : forall x,y in G$



    Show that $[G,G]triangleleft G$



    Solution:
    If $[G,G]triangleleft G$, then $g^{-1}[x,y]g$ must belong to [G,G].



    $$g^{-1}[x,y]g$$
    $$g^{-1}x^{-1}y^{-1}xyg$$



    I am stuck here. I am not sure how to proceed since we have three elements...



    Thanks for the help!






    group-theory







    share|cite|improve this question









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    marked as duplicate by Arturo Magidin group-theory
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      0












      0








      0





      $begingroup$



      This question already has an answer here:




      • How to show that the commutator subgroup is a normal subgroup

        1 answer




      $G$ is a group with $e$ as it's identity element.
      The commutator group $[G,G] = <[x,y]|x,y in G>$ such that $[x,y]=x^{-1}y^{-1}xy : : : forall x,y in G$



      Show that $[G,G]triangleleft G$



      Solution:
      If $[G,G]triangleleft G$, then $g^{-1}[x,y]g$ must belong to [G,G].



      $$g^{-1}[x,y]g$$
      $$g^{-1}x^{-1}y^{-1}xyg$$



      I am stuck here. I am not sure how to proceed since we have three elements...



      Thanks for the help!






      group-theory







      share|cite|improve this question









      $endgroup$





      This question already has an answer here:




      • How to show that the commutator subgroup is a normal subgroup

        1 answer




      $G$ is a group with $e$ as it's identity element.
      The commutator group $[G,G] = <[x,y]|x,y in G>$ such that $[x,y]=x^{-1}y^{-1}xy : : : forall x,y in G$



      Show that $[G,G]triangleleft G$



      Solution:
      If $[G,G]triangleleft G$, then $g^{-1}[x,y]g$ must belong to [G,G].



      $$g^{-1}[x,y]g$$
      $$g^{-1}x^{-1}y^{-1}xyg$$



      I am stuck here. I am not sure how to proceed since we have three elements...



      Thanks for the help!





      This question already has an answer here:




      • How to show that the commutator subgroup is a normal subgroup

        1 answer







      group-theory




      group-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 8 '18 at 3:32









      JoeyFJoeyF

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      marked as duplicate by Arturo Magidin group-theory
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       Dec 8 '18 at 3:54


      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 Arturo Magidin group-theory
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          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          HINT :



          $$g^{-1}x^{-1}y^{-1}xyg=g^{-1}x^{-1}gg^{-1}y^{-1}gg^{-1}xgg^{-1}yg$$



          Let $g^{-1}xg=u $ and $g^{-1}yg=v $. Then $$g^{-1}x^{-1}y^{-1}xyg=u^{-1}v^{-1}uv .$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
            $endgroup$
            – Arturo Magidin
            Dec 8 '18 at 3:55


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          HINT :



          $$g^{-1}x^{-1}y^{-1}xyg=g^{-1}x^{-1}gg^{-1}y^{-1}gg^{-1}xgg^{-1}yg$$



          Let $g^{-1}xg=u $ and $g^{-1}yg=v $. Then $$g^{-1}x^{-1}y^{-1}xyg=u^{-1}v^{-1}uv .$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
            $endgroup$
            – Arturo Magidin
            Dec 8 '18 at 3:55
















          1












          $begingroup$

          HINT :



          $$g^{-1}x^{-1}y^{-1}xyg=g^{-1}x^{-1}gg^{-1}y^{-1}gg^{-1}xgg^{-1}yg$$



          Let $g^{-1}xg=u $ and $g^{-1}yg=v $. Then $$g^{-1}x^{-1}y^{-1}xyg=u^{-1}v^{-1}uv .$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
            $endgroup$
            – Arturo Magidin
            Dec 8 '18 at 3:55














          1












          1








          1





          $begingroup$

          HINT :



          $$g^{-1}x^{-1}y^{-1}xyg=g^{-1}x^{-1}gg^{-1}y^{-1}gg^{-1}xgg^{-1}yg$$



          Let $g^{-1}xg=u $ and $g^{-1}yg=v $. Then $$g^{-1}x^{-1}y^{-1}xyg=u^{-1}v^{-1}uv .$$






          share|cite|improve this answer









          $endgroup$



          HINT :



          $$g^{-1}x^{-1}y^{-1}xyg=g^{-1}x^{-1}gg^{-1}y^{-1}gg^{-1}xgg^{-1}yg$$



          Let $g^{-1}xg=u $ and $g^{-1}yg=v $. Then $$g^{-1}x^{-1}y^{-1}xyg=u^{-1}v^{-1}uv .$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 3:44









          Thomas ShelbyThomas Shelby

          2,498221




          2,498221








          • 1




            $begingroup$
            Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
            $endgroup$
            – Arturo Magidin
            Dec 8 '18 at 3:55














          • 1




            $begingroup$
            Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
            $endgroup$
            – Arturo Magidin
            Dec 8 '18 at 3:55








          1




          1




          $begingroup$
          Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
          $endgroup$
          – Arturo Magidin
          Dec 8 '18 at 3:55




          $begingroup$
          Or, $g^{-1}ag = a[a,g]$, so $g^{-1}[x,y]g = [x,y][[x,y],g]$...
          $endgroup$
          – Arturo Magidin
          Dec 8 '18 at 3:55



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