Does every subgroup of an abelian group have to be abelian?












3












$begingroup$


My original problem is to show that E/L is an abelian extension over L and L/F is an abelian extension over F, given that E/F is an abelian extension over F and that L is a normal extension of F such that $Fsubseteq L subseteq E$.



So far I have proved that E is a normal extension of F, E is a normal extension of L, and L is a normal extension of F. I know that to prove abelian extension I must also prove that Gal(E/L) is an abelian group. I have shown that Gal(E/L) $subseteq$ Gal (E/F). In my mind it makes sense that I cannot lose commutativity therefore my subgroup must be Abelian too. How do I show this in a proof? Is it enough to show two elements in the subgroup must also exist in the larger group and that they must be commutative in the larger group? I feel like I know what needs to be done, just not how to phrase it.






abstract-algebra group-theory galois-theory abelian-groups







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$endgroup$

















    3












    $begingroup$


    My original problem is to show that E/L is an abelian extension over L and L/F is an abelian extension over F, given that E/F is an abelian extension over F and that L is a normal extension of F such that $Fsubseteq L subseteq E$.



    So far I have proved that E is a normal extension of F, E is a normal extension of L, and L is a normal extension of F. I know that to prove abelian extension I must also prove that Gal(E/L) is an abelian group. I have shown that Gal(E/L) $subseteq$ Gal (E/F). In my mind it makes sense that I cannot lose commutativity therefore my subgroup must be Abelian too. How do I show this in a proof? Is it enough to show two elements in the subgroup must also exist in the larger group and that they must be commutative in the larger group? I feel like I know what needs to be done, just not how to phrase it.






    abstract-algebra group-theory galois-theory abelian-groups







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      My original problem is to show that E/L is an abelian extension over L and L/F is an abelian extension over F, given that E/F is an abelian extension over F and that L is a normal extension of F such that $Fsubseteq L subseteq E$.



      So far I have proved that E is a normal extension of F, E is a normal extension of L, and L is a normal extension of F. I know that to prove abelian extension I must also prove that Gal(E/L) is an abelian group. I have shown that Gal(E/L) $subseteq$ Gal (E/F). In my mind it makes sense that I cannot lose commutativity therefore my subgroup must be Abelian too. How do I show this in a proof? Is it enough to show two elements in the subgroup must also exist in the larger group and that they must be commutative in the larger group? I feel like I know what needs to be done, just not how to phrase it.






      abstract-algebra group-theory galois-theory abelian-groups







      share|cite|improve this question











      $endgroup$




      My original problem is to show that E/L is an abelian extension over L and L/F is an abelian extension over F, given that E/F is an abelian extension over F and that L is a normal extension of F such that $Fsubseteq L subseteq E$.



      So far I have proved that E is a normal extension of F, E is a normal extension of L, and L is a normal extension of F. I know that to prove abelian extension I must also prove that Gal(E/L) is an abelian group. I have shown that Gal(E/L) $subseteq$ Gal (E/F). In my mind it makes sense that I cannot lose commutativity therefore my subgroup must be Abelian too. How do I show this in a proof? Is it enough to show two elements in the subgroup must also exist in the larger group and that they must be commutative in the larger group? I feel like I know what needs to be done, just not how to phrase it.






      abstract-algebra group-theory galois-theory abelian-groups




      abstract-algebra group-theory galois-theory abelian-groups






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 4 hours ago









      J. W. Tanner

      5,1651520




      5,1651520










      asked 6 hours ago









      MT mathMT math

      253




      253






















          2 Answers
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          active

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          3












          $begingroup$

          Huh, funny, we just went over this today in my algebra class.



          Yes, subgroups of abelian groups are indeed abelian, and your thought process has the right idea.



          Showing this is pretty easy. Take an abelian group $G$ with subgroup $H$. Then we know that, for all $a,bin H$, $ab=ba$ since it must also hold in $G$ (as $a,b in G ge H$ and $G$ is given to be abelian).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            can I use essentially the same reasoning to prove that L/F is an abelian extension as well?
            $endgroup$
            – MT math
            5 hours ago










          • $begingroup$
            I believe so, yes.
            $endgroup$
            – Eevee Trainer
            4 hours ago



















          2












          $begingroup$

          If $G$ is an abelian group and $H$ is a subgroup, suppose $x, yin H$. Then in particular $x, yin G$, so $xy=yx$. Since $x, y$ were arbitrary, $H$ is abelian.






          share|cite|improve this answer









          $endgroup$














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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

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            active

            oldest

            votes









            3












            $begingroup$

            Huh, funny, we just went over this today in my algebra class.



            Yes, subgroups of abelian groups are indeed abelian, and your thought process has the right idea.



            Showing this is pretty easy. Take an abelian group $G$ with subgroup $H$. Then we know that, for all $a,bin H$, $ab=ba$ since it must also hold in $G$ (as $a,b in G ge H$ and $G$ is given to be abelian).






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              can I use essentially the same reasoning to prove that L/F is an abelian extension as well?
              $endgroup$
              – MT math
              5 hours ago










            • $begingroup$
              I believe so, yes.
              $endgroup$
              – Eevee Trainer
              4 hours ago
















            3












            $begingroup$

            Huh, funny, we just went over this today in my algebra class.



            Yes, subgroups of abelian groups are indeed abelian, and your thought process has the right idea.



            Showing this is pretty easy. Take an abelian group $G$ with subgroup $H$. Then we know that, for all $a,bin H$, $ab=ba$ since it must also hold in $G$ (as $a,b in G ge H$ and $G$ is given to be abelian).






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              can I use essentially the same reasoning to prove that L/F is an abelian extension as well?
              $endgroup$
              – MT math
              5 hours ago










            • $begingroup$
              I believe so, yes.
              $endgroup$
              – Eevee Trainer
              4 hours ago














            3












            3








            3





            $begingroup$

            Huh, funny, we just went over this today in my algebra class.



            Yes, subgroups of abelian groups are indeed abelian, and your thought process has the right idea.



            Showing this is pretty easy. Take an abelian group $G$ with subgroup $H$. Then we know that, for all $a,bin H$, $ab=ba$ since it must also hold in $G$ (as $a,b in G ge H$ and $G$ is given to be abelian).






            share|cite|improve this answer









            $endgroup$



            Huh, funny, we just went over this today in my algebra class.



            Yes, subgroups of abelian groups are indeed abelian, and your thought process has the right idea.



            Showing this is pretty easy. Take an abelian group $G$ with subgroup $H$. Then we know that, for all $a,bin H$, $ab=ba$ since it must also hold in $G$ (as $a,b in G ge H$ and $G$ is given to be abelian).







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 6 hours ago









            Eevee TrainerEevee Trainer

            10.8k31843




            10.8k31843












            • $begingroup$
              can I use essentially the same reasoning to prove that L/F is an abelian extension as well?
              $endgroup$
              – MT math
              5 hours ago










            • $begingroup$
              I believe so, yes.
              $endgroup$
              – Eevee Trainer
              4 hours ago


















            • $begingroup$
              can I use essentially the same reasoning to prove that L/F is an abelian extension as well?
              $endgroup$
              – MT math
              5 hours ago










            • $begingroup$
              I believe so, yes.
              $endgroup$
              – Eevee Trainer
              4 hours ago
















            $begingroup$
            can I use essentially the same reasoning to prove that L/F is an abelian extension as well?
            $endgroup$
            – MT math
            5 hours ago




            $begingroup$
            can I use essentially the same reasoning to prove that L/F is an abelian extension as well?
            $endgroup$
            – MT math
            5 hours ago












            $begingroup$
            I believe so, yes.
            $endgroup$
            – Eevee Trainer
            4 hours ago




            $begingroup$
            I believe so, yes.
            $endgroup$
            – Eevee Trainer
            4 hours ago











            2












            $begingroup$

            If $G$ is an abelian group and $H$ is a subgroup, suppose $x, yin H$. Then in particular $x, yin G$, so $xy=yx$. Since $x, y$ were arbitrary, $H$ is abelian.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              If $G$ is an abelian group and $H$ is a subgroup, suppose $x, yin H$. Then in particular $x, yin G$, so $xy=yx$. Since $x, y$ were arbitrary, $H$ is abelian.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                If $G$ is an abelian group and $H$ is a subgroup, suppose $x, yin H$. Then in particular $x, yin G$, so $xy=yx$. Since $x, y$ were arbitrary, $H$ is abelian.






                share|cite|improve this answer









                $endgroup$



                If $G$ is an abelian group and $H$ is a subgroup, suppose $x, yin H$. Then in particular $x, yin G$, so $xy=yx$. Since $x, y$ were arbitrary, $H$ is abelian.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 6 hours ago









                Matt SamuelMatt Samuel

                39.5k63870




                39.5k63870






























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