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.











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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







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    up vote
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    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






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 18 hours ago









      david D

      835




      835






















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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.)






          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





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          active

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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
            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


















           

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