Torsion group and torsion subgroup












0












$begingroup$


I know the exact definition of torsion group.
But the definition of torsion "sub"group is confusing me.




  1. A subgroup which is torsion.

  2. A collcection of all elements in G whose order is finite.


Which of the two is correct definition of torsion subgroup? The difference between two is that 1. need not to contain "all" elements of finite order.



Thanks for helping me.






abstract-algebra







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$endgroup$

















    0












    $begingroup$


    I know the exact definition of torsion group.
    But the definition of torsion "sub"group is confusing me.




    1. A subgroup which is torsion.

    2. A collcection of all elements in G whose order is finite.


    Which of the two is correct definition of torsion subgroup? The difference between two is that 1. need not to contain "all" elements of finite order.



    Thanks for helping me.






    abstract-algebra







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I know the exact definition of torsion group.
      But the definition of torsion "sub"group is confusing me.




      1. A subgroup which is torsion.

      2. A collcection of all elements in G whose order is finite.


      Which of the two is correct definition of torsion subgroup? The difference between two is that 1. need not to contain "all" elements of finite order.



      Thanks for helping me.






      abstract-algebra







      share|cite|improve this question









      $endgroup$




      I know the exact definition of torsion group.
      But the definition of torsion "sub"group is confusing me.




      1. A subgroup which is torsion.

      2. A collcection of all elements in G whose order is finite.


      Which of the two is correct definition of torsion subgroup? The difference between two is that 1. need not to contain "all" elements of finite order.



      Thanks for helping me.






      abstract-algebra




      abstract-algebra






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 6 at 12:52









      SophiaSophia

      4817




      4817






















          1 Answer
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          $begingroup$

          The point is, the set of all torsion elements need not be a subgroup. The possibly simplest example is provided by the infinite dihedral group $D$.



          In fact if you realise $D$ as the following set of bijections over the integers, the group operation being compositon,
          $$
          { x mapsto a x + b : a = pm 1, b in mathbb{Z} },
          $$
          you see that the torsion elements are precisely those with $a = - 1$, which have all period two. But the product of two distinct such elements is an element of infinite order.






          share|cite|improve this answer









          $endgroup$













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            1












            $begingroup$

            The point is, the set of all torsion elements need not be a subgroup. The possibly simplest example is provided by the infinite dihedral group $D$.



            In fact if you realise $D$ as the following set of bijections over the integers, the group operation being compositon,
            $$
            { x mapsto a x + b : a = pm 1, b in mathbb{Z} },
            $$
            you see that the torsion elements are precisely those with $a = - 1$, which have all period two. But the product of two distinct such elements is an element of infinite order.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The point is, the set of all torsion elements need not be a subgroup. The possibly simplest example is provided by the infinite dihedral group $D$.



              In fact if you realise $D$ as the following set of bijections over the integers, the group operation being compositon,
              $$
              { x mapsto a x + b : a = pm 1, b in mathbb{Z} },
              $$
              you see that the torsion elements are precisely those with $a = - 1$, which have all period two. But the product of two distinct such elements is an element of infinite order.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The point is, the set of all torsion elements need not be a subgroup. The possibly simplest example is provided by the infinite dihedral group $D$.



                In fact if you realise $D$ as the following set of bijections over the integers, the group operation being compositon,
                $$
                { x mapsto a x + b : a = pm 1, b in mathbb{Z} },
                $$
                you see that the torsion elements are precisely those with $a = - 1$, which have all period two. But the product of two distinct such elements is an element of infinite order.






                share|cite|improve this answer









                $endgroup$



                The point is, the set of all torsion elements need not be a subgroup. The possibly simplest example is provided by the infinite dihedral group $D$.



                In fact if you realise $D$ as the following set of bijections over the integers, the group operation being compositon,
                $$
                { x mapsto a x + b : a = pm 1, b in mathbb{Z} },
                $$
                you see that the torsion elements are precisely those with $a = - 1$, which have all period two. But the product of two distinct such elements is an element of infinite order.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 13:42









                Andreas CarantiAndreas Caranti

                56.6k34395




                56.6k34395






























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