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











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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.










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    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






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      edited Nov 20 at 11:24









      1ENİGMA1

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      asked Nov 19 at 4:34









      david D

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      875






















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          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.)






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          • 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











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          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


















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