Groups and normal subgroups












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$begingroup$


Let H be normal subgroup of G, and H∩G'={e} , G'={[x,y]=xyx^(-1)y^(-1) |x,y∈G}. Prove that ∀g ∈ G and ∀h∈ H gh=hg.






abstract-algebra normal-subgroups







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$endgroup$

















    -4












    $begingroup$


    Let H be normal subgroup of G, and H∩G'={e} , G'={[x,y]=xyx^(-1)y^(-1) |x,y∈G}. Prove that ∀g ∈ G and ∀h∈ H gh=hg.






    abstract-algebra normal-subgroups







    share|cite|improve this question











    $endgroup$















      -4












      -4








      -4





      $begingroup$


      Let H be normal subgroup of G, and H∩G'={e} , G'={[x,y]=xyx^(-1)y^(-1) |x,y∈G}. Prove that ∀g ∈ G and ∀h∈ H gh=hg.






      abstract-algebra normal-subgroups







      share|cite|improve this question











      $endgroup$




      Let H be normal subgroup of G, and H∩G'={e} , G'={[x,y]=xyx^(-1)y^(-1) |x,y∈G}. Prove that ∀g ∈ G and ∀h∈ H gh=hg.






      abstract-algebra normal-subgroups




      abstract-algebra normal-subgroups






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 27 at 16:25









      Pierre-Yves Gaillard

      13.5k23184




      13.5k23184










      asked Jan 27 at 13:29









      user638212user638212

      302




      302






















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          $begingroup$

          Note $gh=hgiff ghg^{-1}h^{-1}=e$.



          But, $Htriangleleft Giff ghg^{-1}in H$.



          So, for all $gin G,hin H$, we have $ghg^{-1}h^{-1}in Hcap G'={e}$.






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            Note $gh=hgiff ghg^{-1}h^{-1}=e$.



            But, $Htriangleleft Giff ghg^{-1}in H$.



            So, for all $gin G,hin H$, we have $ghg^{-1}h^{-1}in Hcap G'={e}$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Note $gh=hgiff ghg^{-1}h^{-1}=e$.



              But, $Htriangleleft Giff ghg^{-1}in H$.



              So, for all $gin G,hin H$, we have $ghg^{-1}h^{-1}in Hcap G'={e}$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Note $gh=hgiff ghg^{-1}h^{-1}=e$.



                But, $Htriangleleft Giff ghg^{-1}in H$.



                So, for all $gin G,hin H$, we have $ghg^{-1}h^{-1}in Hcap G'={e}$.






                share|cite|improve this answer









                $endgroup$



                Note $gh=hgiff ghg^{-1}h^{-1}=e$.



                But, $Htriangleleft Giff ghg^{-1}in H$.



                So, for all $gin G,hin H$, we have $ghg^{-1}h^{-1}in Hcap G'={e}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 27 at 13:44









                Chris CusterChris Custer

                14.2k3827




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