Any Subgroup containing commutator subgroup is normal.

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I can prove that commutator is minimal subgroup such that factor group of it is abelian. I had encountered one statement as




If $H$ is a subgroup containing commutator subgroup then $H$ is
normal.




I.e. we have to show that $forall gin G$ such that $gHg^-1=H$ with fact that $G'subset H$



It is for elements in $G'$ to show condition for normality.



But how to do for elements not in $G'$ but in $H$ that in $H/G'$?




abstract-algebra group-theory normal-subgroups




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    See this question.
    – Dietrich Burde
    2 days ago














up vote
6
down vote

favorite












I can prove that commutator is minimal subgroup such that factor group of it is abelian. I had encountered one statement as




If $H$ is a subgroup containing commutator subgroup then $H$ is
normal.




I.e. we have to show that $forall gin G$ such that $gHg^-1=H$ with fact that $G'subset H$



It is for elements in $G'$ to show condition for normality.



But how to do for elements not in $G'$ but in $H$ that in $H/G'$?




abstract-algebra group-theory normal-subgroups




share|cite|improve this question

















  • 1




    See this question.
    – Dietrich Burde
    2 days ago












up vote
6
down vote

favorite









up vote
6
down vote

favorite











I can prove that commutator is minimal subgroup such that factor group of it is abelian. I had encountered one statement as




If $H$ is a subgroup containing commutator subgroup then $H$ is
normal.




I.e. we have to show that $forall gin G$ such that $gHg^-1=H$ with fact that $G'subset H$



It is for elements in $G'$ to show condition for normality.



But how to do for elements not in $G'$ but in $H$ that in $H/G'$?




abstract-algebra group-theory normal-subgroups




share|cite|improve this question













I can prove that commutator is minimal subgroup such that factor group of it is abelian. I had encountered one statement as




If $H$ is a subgroup containing commutator subgroup then $H$ is
normal.




I.e. we have to show that $forall gin G$ such that $gHg^-1=H$ with fact that $G'subset H$



It is for elements in $G'$ to show condition for normality.



But how to do for elements not in $G'$ but in $H$ that in $H/G'$?





abstract-algebra group-theory normal-subgroups





share|cite|improve this question












share|cite|improve this question




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edited 2 days ago









Marcus Müller

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asked 2 days ago









SRJ

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984216







  • 1




    See this question.
    – Dietrich Burde
    2 days ago












  • 1




    See this question.
    – Dietrich Burde
    2 days ago







1




1




See this question.
– Dietrich Burde
2 days ago




See this question.
– Dietrich Burde
2 days ago










2 Answers
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If $gin G$ and $hin H$, then $ghg^-1h^-1=h'$, for some $h'in H$ (since $H$ contains the commutator subgroup). But then $ghg^-1=h'hin H$. Therefore, $gHg^-1subset H$.






share|cite|improve this answer






























    up vote
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    down vote













    $G'$ is certainly normal in $G$, and $G/G'$ is Abelian. Every
    subgroup of an Abelian group is normal. But $H/G'$ is a subgroup
    of $G/G'$ so $H/G'$ is normal in $G/G'$. Therefore, by the third
    isomorphism theorem for groups, $H$ is normal in $G$.






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      2 Answers
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      2 Answers
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      up vote
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      down vote



      accepted










      If $gin G$ and $hin H$, then $ghg^-1h^-1=h'$, for some $h'in H$ (since $H$ contains the commutator subgroup). But then $ghg^-1=h'hin H$. Therefore, $gHg^-1subset H$.






      share|cite|improve this answer



























        up vote
        11
        down vote



        accepted










        If $gin G$ and $hin H$, then $ghg^-1h^-1=h'$, for some $h'in H$ (since $H$ contains the commutator subgroup). But then $ghg^-1=h'hin H$. Therefore, $gHg^-1subset H$.






        share|cite|improve this answer

























          up vote
          11
          down vote



          accepted







          up vote
          11
          down vote



          accepted






          If $gin G$ and $hin H$, then $ghg^-1h^-1=h'$, for some $h'in H$ (since $H$ contains the commutator subgroup). But then $ghg^-1=h'hin H$. Therefore, $gHg^-1subset H$.






          share|cite|improve this answer















          If $gin G$ and $hin H$, then $ghg^-1h^-1=h'$, for some $h'in H$ (since $H$ contains the commutator subgroup). But then $ghg^-1=h'hin H$. Therefore, $gHg^-1subset H$.







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago


























          answered 2 days ago









          José Carlos Santos

          112k1696172




          112k1696172




















              up vote
              9
              down vote













              $G'$ is certainly normal in $G$, and $G/G'$ is Abelian. Every
              subgroup of an Abelian group is normal. But $H/G'$ is a subgroup
              of $G/G'$ so $H/G'$ is normal in $G/G'$. Therefore, by the third
              isomorphism theorem for groups, $H$ is normal in $G$.






              share|cite|improve this answer

























                up vote
                9
                down vote













                $G'$ is certainly normal in $G$, and $G/G'$ is Abelian. Every
                subgroup of an Abelian group is normal. But $H/G'$ is a subgroup
                of $G/G'$ so $H/G'$ is normal in $G/G'$. Therefore, by the third
                isomorphism theorem for groups, $H$ is normal in $G$.






                share|cite|improve this answer























                  up vote
                  9
                  down vote










                  up vote
                  9
                  down vote









                  $G'$ is certainly normal in $G$, and $G/G'$ is Abelian. Every
                  subgroup of an Abelian group is normal. But $H/G'$ is a subgroup
                  of $G/G'$ so $H/G'$ is normal in $G/G'$. Therefore, by the third
                  isomorphism theorem for groups, $H$ is normal in $G$.






                  share|cite|improve this answer













                  $G'$ is certainly normal in $G$, and $G/G'$ is Abelian. Every
                  subgroup of an Abelian group is normal. But $H/G'$ is a subgroup
                  of $G/G'$ so $H/G'$ is normal in $G/G'$. Therefore, by the third
                  isomorphism theorem for groups, $H$ is normal in $G$.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered 2 days ago









                  Lord Shark the Unknown

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