finite abelian groups tensor product.












3












$begingroup$


Is the following question obvious ?



Let $G$ be an abelian group, such that for any finite abelian group $A$, we have
$Gotimes_{mathbf{Z}}A=0$, does it mean that $G$ is a $mathbf{Q}$-vector space ?






abstract-algebra group-theory finite-groups tensor-products abelian-groups







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$endgroup$

















    3












    $begingroup$


    Is the following question obvious ?



    Let $G$ be an abelian group, such that for any finite abelian group $A$, we have
    $Gotimes_{mathbf{Z}}A=0$, does it mean that $G$ is a $mathbf{Q}$-vector space ?






    abstract-algebra group-theory finite-groups tensor-products abelian-groups







    share|cite|improve this question







    New contributor




    lab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Is the following question obvious ?



      Let $G$ be an abelian group, such that for any finite abelian group $A$, we have
      $Gotimes_{mathbf{Z}}A=0$, does it mean that $G$ is a $mathbf{Q}$-vector space ?






      abstract-algebra group-theory finite-groups tensor-products abelian-groups







      share|cite|improve this question







      New contributor




      lab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Is the following question obvious ?



      Let $G$ be an abelian group, such that for any finite abelian group $A$, we have
      $Gotimes_{mathbf{Z}}A=0$, does it mean that $G$ is a $mathbf{Q}$-vector space ?






      abstract-algebra group-theory finite-groups tensor-products abelian-groups




      abstract-algebra group-theory finite-groups tensor-products abelian-groups






      share|cite|improve this question







      New contributor




      lab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      lab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






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      asked 7 hours ago









      lablab

      183




      183




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      lab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          $begingroup$

          You have $Gotimes(Bbb Z/nBbb Z)=0$ for all
          $nin Bbb N$. That means $G/nG=0$, so $G=nG$, that is all elements of $G$ are divisible by $n$. Then $G$ is a divisible Abelian group. Conversely if $G$ is a divisible Abelian group, then
          $Gotimes(Bbb Z/nBbb Z)=0$ and so $Gotimes A=0$ for all finitely generated
          Abelian groups.



          But not all divisible Abelian groups are $Bbb Q$-modules: they may have torsion.
          As an example, let $G=Bbb Q/Bbb Z$.






          share|cite|improve this answer









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            $begingroup$

            You have $Gotimes(Bbb Z/nBbb Z)=0$ for all
            $nin Bbb N$. That means $G/nG=0$, so $G=nG$, that is all elements of $G$ are divisible by $n$. Then $G$ is a divisible Abelian group. Conversely if $G$ is a divisible Abelian group, then
            $Gotimes(Bbb Z/nBbb Z)=0$ and so $Gotimes A=0$ for all finitely generated
            Abelian groups.



            But not all divisible Abelian groups are $Bbb Q$-modules: they may have torsion.
            As an example, let $G=Bbb Q/Bbb Z$.






            share|cite|improve this answer









            $endgroup$


















              6












              $begingroup$

              You have $Gotimes(Bbb Z/nBbb Z)=0$ for all
              $nin Bbb N$. That means $G/nG=0$, so $G=nG$, that is all elements of $G$ are divisible by $n$. Then $G$ is a divisible Abelian group. Conversely if $G$ is a divisible Abelian group, then
              $Gotimes(Bbb Z/nBbb Z)=0$ and so $Gotimes A=0$ for all finitely generated
              Abelian groups.



              But not all divisible Abelian groups are $Bbb Q$-modules: they may have torsion.
              As an example, let $G=Bbb Q/Bbb Z$.






              share|cite|improve this answer









              $endgroup$
















                6












                6








                6





                $begingroup$

                You have $Gotimes(Bbb Z/nBbb Z)=0$ for all
                $nin Bbb N$. That means $G/nG=0$, so $G=nG$, that is all elements of $G$ are divisible by $n$. Then $G$ is a divisible Abelian group. Conversely if $G$ is a divisible Abelian group, then
                $Gotimes(Bbb Z/nBbb Z)=0$ and so $Gotimes A=0$ for all finitely generated
                Abelian groups.



                But not all divisible Abelian groups are $Bbb Q$-modules: they may have torsion.
                As an example, let $G=Bbb Q/Bbb Z$.






                share|cite|improve this answer









                $endgroup$



                You have $Gotimes(Bbb Z/nBbb Z)=0$ for all
                $nin Bbb N$. That means $G/nG=0$, so $G=nG$, that is all elements of $G$ are divisible by $n$. Then $G$ is a divisible Abelian group. Conversely if $G$ is a divisible Abelian group, then
                $Gotimes(Bbb Z/nBbb Z)=0$ and so $Gotimes A=0$ for all finitely generated
                Abelian groups.



                But not all divisible Abelian groups are $Bbb Q$-modules: they may have torsion.
                As an example, let $G=Bbb Q/Bbb Z$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 7 hours ago









                Lord Shark the UnknownLord Shark the Unknown

                106k1161133




                106k1161133






















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