isomorphism of dihedral group with these elements












1












$begingroup$


So I have a group of order $2m$ with these elements:
$$(overline{0},overline{0}),(overline{1},overline{0})...(overline{m-1},overline{0})$$
$$(overline{0},overline{1})(overline{-1},overline{1})...(overline{-(m-1)},overline{1})$$



Dihedral group $D_m$ has these elements:
$$(r^0s^0)(r^1s^0)...(r^{m-1}s^0)$$
$$(r^0s^1)(r^1s^1)...(r^{m-1}s^1)$$



I want write that it is isomorphic to the dihedral group of order m. It is so obvious to me that I'm not sure how to prove it.



Thanks for the help






group-theory dihedral-groups







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the group operation on your $2m$ elements?
    $endgroup$
    – Servaes
    Dec 2 '18 at 20:40






  • 1




    $begingroup$
    Group G of order 2m is: $$mathbb{Z}_m rtimes_{varphi} mathbb{Z}_2$$ and I am given that $varphi_{1+2mathbb{Z}}(1+mmathbb{Z}) = m-1+mmathbb{Z}$
    $endgroup$
    – JoeyF
    Dec 2 '18 at 20:40


















1












$begingroup$


So I have a group of order $2m$ with these elements:
$$(overline{0},overline{0}),(overline{1},overline{0})...(overline{m-1},overline{0})$$
$$(overline{0},overline{1})(overline{-1},overline{1})...(overline{-(m-1)},overline{1})$$



Dihedral group $D_m$ has these elements:
$$(r^0s^0)(r^1s^0)...(r^{m-1}s^0)$$
$$(r^0s^1)(r^1s^1)...(r^{m-1}s^1)$$



I want write that it is isomorphic to the dihedral group of order m. It is so obvious to me that I'm not sure how to prove it.



Thanks for the help






group-theory dihedral-groups







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the group operation on your $2m$ elements?
    $endgroup$
    – Servaes
    Dec 2 '18 at 20:40






  • 1




    $begingroup$
    Group G of order 2m is: $$mathbb{Z}_m rtimes_{varphi} mathbb{Z}_2$$ and I am given that $varphi_{1+2mathbb{Z}}(1+mmathbb{Z}) = m-1+mmathbb{Z}$
    $endgroup$
    – JoeyF
    Dec 2 '18 at 20:40
















1












1








1





$begingroup$


So I have a group of order $2m$ with these elements:
$$(overline{0},overline{0}),(overline{1},overline{0})...(overline{m-1},overline{0})$$
$$(overline{0},overline{1})(overline{-1},overline{1})...(overline{-(m-1)},overline{1})$$



Dihedral group $D_m$ has these elements:
$$(r^0s^0)(r^1s^0)...(r^{m-1}s^0)$$
$$(r^0s^1)(r^1s^1)...(r^{m-1}s^1)$$



I want write that it is isomorphic to the dihedral group of order m. It is so obvious to me that I'm not sure how to prove it.



Thanks for the help






group-theory dihedral-groups







share|cite|improve this question









$endgroup$




So I have a group of order $2m$ with these elements:
$$(overline{0},overline{0}),(overline{1},overline{0})...(overline{m-1},overline{0})$$
$$(overline{0},overline{1})(overline{-1},overline{1})...(overline{-(m-1)},overline{1})$$



Dihedral group $D_m$ has these elements:
$$(r^0s^0)(r^1s^0)...(r^{m-1}s^0)$$
$$(r^0s^1)(r^1s^1)...(r^{m-1}s^1)$$



I want write that it is isomorphic to the dihedral group of order m. It is so obvious to me that I'm not sure how to prove it.



Thanks for the help






group-theory dihedral-groups




group-theory dihedral-groups






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 20:15









JoeyFJoeyF

62




62












  • $begingroup$
    What is the group operation on your $2m$ elements?
    $endgroup$
    – Servaes
    Dec 2 '18 at 20:40






  • 1




    $begingroup$
    Group G of order 2m is: $$mathbb{Z}_m rtimes_{varphi} mathbb{Z}_2$$ and I am given that $varphi_{1+2mathbb{Z}}(1+mmathbb{Z}) = m-1+mmathbb{Z}$
    $endgroup$
    – JoeyF
    Dec 2 '18 at 20:40




















  • $begingroup$
    What is the group operation on your $2m$ elements?
    $endgroup$
    – Servaes
    Dec 2 '18 at 20:40






  • 1




    $begingroup$
    Group G of order 2m is: $$mathbb{Z}_m rtimes_{varphi} mathbb{Z}_2$$ and I am given that $varphi_{1+2mathbb{Z}}(1+mmathbb{Z}) = m-1+mmathbb{Z}$
    $endgroup$
    – JoeyF
    Dec 2 '18 at 20:40


















$begingroup$
What is the group operation on your $2m$ elements?
$endgroup$
– Servaes
Dec 2 '18 at 20:40




$begingroup$
What is the group operation on your $2m$ elements?
$endgroup$
– Servaes
Dec 2 '18 at 20:40




1




1




$begingroup$
Group G of order 2m is: $$mathbb{Z}_m rtimes_{varphi} mathbb{Z}_2$$ and I am given that $varphi_{1+2mathbb{Z}}(1+mmathbb{Z}) = m-1+mmathbb{Z}$
$endgroup$
– JoeyF
Dec 2 '18 at 20:40






$begingroup$
Group G of order 2m is: $$mathbb{Z}_m rtimes_{varphi} mathbb{Z}_2$$ and I am given that $varphi_{1+2mathbb{Z}}(1+mmathbb{Z}) = m-1+mmathbb{Z}$
$endgroup$
– JoeyF
Dec 2 '18 at 20:40












2 Answers
2






active

oldest

votes


















0












$begingroup$

As you want to show an isomorphism with a dihedral group, how about constructing a faithful action of your group on the regular $m$-gon? This yields an injection into $D_m$, and because they are both groups of order $2m$ this must be an isomorphism.



Given that the group is constructed as a semidirect product, another approach is to take subgroups $T,Ssubset D_m$ such that $TcongBbb{Z}_m$ and $ScongBbb{Z}_2$. And show that $D_m=Trtimes_{psi}S$ where $psi: S longrightarrow operatorname{Aut}T$ is the map obtained from $varphi$ through the isomorphisms $TcongBbb{Z}_m$ and $ScongBbb{Z}_2$. Of course you want $psi$ to turn out to be the conjugation action of $S$ on $T$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I haven't seen group actions yet in my class. But thanks for the help!
    $endgroup$
    – JoeyF
    Dec 2 '18 at 20:52



















0












$begingroup$

Probably the easiest way to see is to show you have 2 generators, $r$ and $s$ with orders n and 2 respectively. Furthermore, $sr^{-1}=rs$. Hence, it is the dihedral group with 2n elements.






share|cite|improve this answer









$endgroup$













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    2 Answers
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    2 Answers
    2






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    active

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    active

    oldest

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    0












    $begingroup$

    As you want to show an isomorphism with a dihedral group, how about constructing a faithful action of your group on the regular $m$-gon? This yields an injection into $D_m$, and because they are both groups of order $2m$ this must be an isomorphism.



    Given that the group is constructed as a semidirect product, another approach is to take subgroups $T,Ssubset D_m$ such that $TcongBbb{Z}_m$ and $ScongBbb{Z}_2$. And show that $D_m=Trtimes_{psi}S$ where $psi: S longrightarrow operatorname{Aut}T$ is the map obtained from $varphi$ through the isomorphisms $TcongBbb{Z}_m$ and $ScongBbb{Z}_2$. Of course you want $psi$ to turn out to be the conjugation action of $S$ on $T$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I haven't seen group actions yet in my class. But thanks for the help!
      $endgroup$
      – JoeyF
      Dec 2 '18 at 20:52
















    0












    $begingroup$

    As you want to show an isomorphism with a dihedral group, how about constructing a faithful action of your group on the regular $m$-gon? This yields an injection into $D_m$, and because they are both groups of order $2m$ this must be an isomorphism.



    Given that the group is constructed as a semidirect product, another approach is to take subgroups $T,Ssubset D_m$ such that $TcongBbb{Z}_m$ and $ScongBbb{Z}_2$. And show that $D_m=Trtimes_{psi}S$ where $psi: S longrightarrow operatorname{Aut}T$ is the map obtained from $varphi$ through the isomorphisms $TcongBbb{Z}_m$ and $ScongBbb{Z}_2$. Of course you want $psi$ to turn out to be the conjugation action of $S$ on $T$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I haven't seen group actions yet in my class. But thanks for the help!
      $endgroup$
      – JoeyF
      Dec 2 '18 at 20:52














    0












    0








    0





    $begingroup$

    As you want to show an isomorphism with a dihedral group, how about constructing a faithful action of your group on the regular $m$-gon? This yields an injection into $D_m$, and because they are both groups of order $2m$ this must be an isomorphism.



    Given that the group is constructed as a semidirect product, another approach is to take subgroups $T,Ssubset D_m$ such that $TcongBbb{Z}_m$ and $ScongBbb{Z}_2$. And show that $D_m=Trtimes_{psi}S$ where $psi: S longrightarrow operatorname{Aut}T$ is the map obtained from $varphi$ through the isomorphisms $TcongBbb{Z}_m$ and $ScongBbb{Z}_2$. Of course you want $psi$ to turn out to be the conjugation action of $S$ on $T$.






    share|cite|improve this answer









    $endgroup$



    As you want to show an isomorphism with a dihedral group, how about constructing a faithful action of your group on the regular $m$-gon? This yields an injection into $D_m$, and because they are both groups of order $2m$ this must be an isomorphism.



    Given that the group is constructed as a semidirect product, another approach is to take subgroups $T,Ssubset D_m$ such that $TcongBbb{Z}_m$ and $ScongBbb{Z}_2$. And show that $D_m=Trtimes_{psi}S$ where $psi: S longrightarrow operatorname{Aut}T$ is the map obtained from $varphi$ through the isomorphisms $TcongBbb{Z}_m$ and $ScongBbb{Z}_2$. Of course you want $psi$ to turn out to be the conjugation action of $S$ on $T$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 2 '18 at 20:49









    ServaesServaes

    25.6k33996




    25.6k33996












    • $begingroup$
      I haven't seen group actions yet in my class. But thanks for the help!
      $endgroup$
      – JoeyF
      Dec 2 '18 at 20:52


















    • $begingroup$
      I haven't seen group actions yet in my class. But thanks for the help!
      $endgroup$
      – JoeyF
      Dec 2 '18 at 20:52
















    $begingroup$
    I haven't seen group actions yet in my class. But thanks for the help!
    $endgroup$
    – JoeyF
    Dec 2 '18 at 20:52




    $begingroup$
    I haven't seen group actions yet in my class. But thanks for the help!
    $endgroup$
    – JoeyF
    Dec 2 '18 at 20:52











    0












    $begingroup$

    Probably the easiest way to see is to show you have 2 generators, $r$ and $s$ with orders n and 2 respectively. Furthermore, $sr^{-1}=rs$. Hence, it is the dihedral group with 2n elements.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Probably the easiest way to see is to show you have 2 generators, $r$ and $s$ with orders n and 2 respectively. Furthermore, $sr^{-1}=rs$. Hence, it is the dihedral group with 2n elements.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Probably the easiest way to see is to show you have 2 generators, $r$ and $s$ with orders n and 2 respectively. Furthermore, $sr^{-1}=rs$. Hence, it is the dihedral group with 2n elements.






        share|cite|improve this answer









        $endgroup$



        Probably the easiest way to see is to show you have 2 generators, $r$ and $s$ with orders n and 2 respectively. Furthermore, $sr^{-1}=rs$. Hence, it is the dihedral group with 2n elements.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 20:49









        Baran ZadeogluBaran Zadeoglu

        296




        296






























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