What is $D_{16}/ Z(D_{16})$?












1












$begingroup$


I was asked the following: Let $D_{16}$ be the dihedral group of order $16$. What is $D_{16} / Z(D_{16})$?



I know that the center of $D_{16}$ har order $2$. So therefore, the quotient has order $16/2 = 8$. I know that there are $5$ groups of order $8$, but I am not sure which of these $D_{16} / Z(D_{16})$ is isomorphic to.



I am pretty sure that it is not $mathbb{Z}_{8}$ since that would imply that $D_{16}$ is abelian (which it is not). So what is it?



EDIT: I see Find $G/Z(G)$ given the following information about the group? but I am not sure that being generated by two elements mean.






abstract-algebra group-theory dihedral-groups quotient-group







share|cite|improve this question











$endgroup$












  • $begingroup$
    Possible duplicate of Find $G/Z(G)$ given the following information about the group?
    $endgroup$
    – user10354138
    Dec 8 '18 at 17:43






  • 2




    $begingroup$
    I think duplicate is a bit of an exaggeration here.
    $endgroup$
    – A. Pongrácz
    Dec 8 '18 at 17:51


















1












$begingroup$


I was asked the following: Let $D_{16}$ be the dihedral group of order $16$. What is $D_{16} / Z(D_{16})$?



I know that the center of $D_{16}$ har order $2$. So therefore, the quotient has order $16/2 = 8$. I know that there are $5$ groups of order $8$, but I am not sure which of these $D_{16} / Z(D_{16})$ is isomorphic to.



I am pretty sure that it is not $mathbb{Z}_{8}$ since that would imply that $D_{16}$ is abelian (which it is not). So what is it?



EDIT: I see Find $G/Z(G)$ given the following information about the group? but I am not sure that being generated by two elements mean.






abstract-algebra group-theory dihedral-groups quotient-group







share|cite|improve this question











$endgroup$












  • $begingroup$
    Possible duplicate of Find $G/Z(G)$ given the following information about the group?
    $endgroup$
    – user10354138
    Dec 8 '18 at 17:43






  • 2




    $begingroup$
    I think duplicate is a bit of an exaggeration here.
    $endgroup$
    – A. Pongrácz
    Dec 8 '18 at 17:51
















1












1








1





$begingroup$


I was asked the following: Let $D_{16}$ be the dihedral group of order $16$. What is $D_{16} / Z(D_{16})$?



I know that the center of $D_{16}$ har order $2$. So therefore, the quotient has order $16/2 = 8$. I know that there are $5$ groups of order $8$, but I am not sure which of these $D_{16} / Z(D_{16})$ is isomorphic to.



I am pretty sure that it is not $mathbb{Z}_{8}$ since that would imply that $D_{16}$ is abelian (which it is not). So what is it?



EDIT: I see Find $G/Z(G)$ given the following information about the group? but I am not sure that being generated by two elements mean.






abstract-algebra group-theory dihedral-groups quotient-group







share|cite|improve this question











$endgroup$




I was asked the following: Let $D_{16}$ be the dihedral group of order $16$. What is $D_{16} / Z(D_{16})$?



I know that the center of $D_{16}$ har order $2$. So therefore, the quotient has order $16/2 = 8$. I know that there are $5$ groups of order $8$, but I am not sure which of these $D_{16} / Z(D_{16})$ is isomorphic to.



I am pretty sure that it is not $mathbb{Z}_{8}$ since that would imply that $D_{16}$ is abelian (which it is not). So what is it?



EDIT: I see Find $G/Z(G)$ given the following information about the group? but I am not sure that being generated by two elements mean.






abstract-algebra group-theory dihedral-groups quotient-group




abstract-algebra group-theory dihedral-groups quotient-group






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 17:45







John Doe

















asked Dec 8 '18 at 17:39









John DoeJohn Doe

25621346




25621346












  • $begingroup$
    Possible duplicate of Find $G/Z(G)$ given the following information about the group?
    $endgroup$
    – user10354138
    Dec 8 '18 at 17:43






  • 2




    $begingroup$
    I think duplicate is a bit of an exaggeration here.
    $endgroup$
    – A. Pongrácz
    Dec 8 '18 at 17:51




















  • $begingroup$
    Possible duplicate of Find $G/Z(G)$ given the following information about the group?
    $endgroup$
    – user10354138
    Dec 8 '18 at 17:43






  • 2




    $begingroup$
    I think duplicate is a bit of an exaggeration here.
    $endgroup$
    – A. Pongrácz
    Dec 8 '18 at 17:51


















$begingroup$
Possible duplicate of Find $G/Z(G)$ given the following information about the group?
$endgroup$
– user10354138
Dec 8 '18 at 17:43




$begingroup$
Possible duplicate of Find $G/Z(G)$ given the following information about the group?
$endgroup$
– user10354138
Dec 8 '18 at 17:43




2




2




$begingroup$
I think duplicate is a bit of an exaggeration here.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:51






$begingroup$
I think duplicate is a bit of an exaggeration here.
$endgroup$
– A. Pongrácz
Dec 8 '18 at 17:51












2 Answers
2






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$begingroup$

It cannot be the cyclic group of order $8$ because the factor with respect to the center cannot be a nontrivial cyclic group. (I mean in general. This is a well-known exercise problem.)



Hint: try to compute the order of elements in the factor group. (The center is the $2$-element group generated by the rotation with $pi$.)
Also note that it cannot be commmutative: pick two reflections such that the angle between the two axes is $pi/8$. These elements do not commute in the factor group.
So you just need to choose between the quaternion group and $D_8$: the order of elements will seal the deal.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    $D_{16} / Z(D_{16})cong D_8$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      It cannot be the cyclic group of order $8$ because the factor with respect to the center cannot be a nontrivial cyclic group. (I mean in general. This is a well-known exercise problem.)



      Hint: try to compute the order of elements in the factor group. (The center is the $2$-element group generated by the rotation with $pi$.)
      Also note that it cannot be commmutative: pick two reflections such that the angle between the two axes is $pi/8$. These elements do not commute in the factor group.
      So you just need to choose between the quaternion group and $D_8$: the order of elements will seal the deal.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        It cannot be the cyclic group of order $8$ because the factor with respect to the center cannot be a nontrivial cyclic group. (I mean in general. This is a well-known exercise problem.)



        Hint: try to compute the order of elements in the factor group. (The center is the $2$-element group generated by the rotation with $pi$.)
        Also note that it cannot be commmutative: pick two reflections such that the angle between the two axes is $pi/8$. These elements do not commute in the factor group.
        So you just need to choose between the quaternion group and $D_8$: the order of elements will seal the deal.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          It cannot be the cyclic group of order $8$ because the factor with respect to the center cannot be a nontrivial cyclic group. (I mean in general. This is a well-known exercise problem.)



          Hint: try to compute the order of elements in the factor group. (The center is the $2$-element group generated by the rotation with $pi$.)
          Also note that it cannot be commmutative: pick two reflections such that the angle between the two axes is $pi/8$. These elements do not commute in the factor group.
          So you just need to choose between the quaternion group and $D_8$: the order of elements will seal the deal.






          share|cite|improve this answer











          $endgroup$



          It cannot be the cyclic group of order $8$ because the factor with respect to the center cannot be a nontrivial cyclic group. (I mean in general. This is a well-known exercise problem.)



          Hint: try to compute the order of elements in the factor group. (The center is the $2$-element group generated by the rotation with $pi$.)
          Also note that it cannot be commmutative: pick two reflections such that the angle between the two axes is $pi/8$. These elements do not commute in the factor group.
          So you just need to choose between the quaternion group and $D_8$: the order of elements will seal the deal.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 8 '18 at 17:51

























          answered Dec 8 '18 at 17:45









          A. PongráczA. Pongrácz

          5,9931929




          5,9931929























              2












              $begingroup$

              $D_{16} / Z(D_{16})cong D_8$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                $D_{16} / Z(D_{16})cong D_8$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  $D_{16} / Z(D_{16})cong D_8$.






                  share|cite|improve this answer









                  $endgroup$



                  $D_{16} / Z(D_{16})cong D_8$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 8 '18 at 17:54









                  yavaryavar

                  843




                  843






























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