understanding the commutator of dihedral group [duplicate]












0












$begingroup$



This question already has an answer here:




  • Commutator subgroup of a dihedral group.

    2 answers




Let $G=D2n=⟨x,y|x^2=y^n=e, $ $yx=xy^{n-1}⟩$



i need to find G' [ the commutattor of G]



now i understand the G' is the subgroup that is generated from $ U=xyx^{-1}y^{-1} , $ $forall x,y in G$



so $<u> ={u^1,u^2...e}$



whats the strategy here ?






abstract-algebra group-theory dihedral-groups







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$endgroup$



marked as duplicate by Dietrich Burde, Iuʇǝƃɹɐʇoɹ, Anastasiya-Romanova 秀, Stefan Hamcke, Cheerful Parsnip Dec 12 '14 at 15:41


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    I'd suggest trying with small $n$ first. Also you're missing generators, you need the commutators of all elements while here you've only written $[x,y]$.
    $endgroup$
    – GPerez
    Dec 12 '14 at 14:49










  • $begingroup$
    @GPerez how can i find the commutattor of group in general..i'm not getting it
    $endgroup$
    – Nizar Halloun
    Dec 12 '14 at 14:57










  • $begingroup$
    @NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators.
    $endgroup$
    – Cheerful Parsnip
    Dec 12 '14 at 15:40
















0












$begingroup$



This question already has an answer here:




  • Commutator subgroup of a dihedral group.

    2 answers




Let $G=D2n=⟨x,y|x^2=y^n=e, $ $yx=xy^{n-1}⟩$



i need to find G' [ the commutattor of G]



now i understand the G' is the subgroup that is generated from $ U=xyx^{-1}y^{-1} , $ $forall x,y in G$



so $<u> ={u^1,u^2...e}$



whats the strategy here ?






abstract-algebra group-theory dihedral-groups







share|cite|improve this question











$endgroup$



marked as duplicate by Dietrich Burde, Iuʇǝƃɹɐʇoɹ, Anastasiya-Romanova 秀, Stefan Hamcke, Cheerful Parsnip Dec 12 '14 at 15:41


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    I'd suggest trying with small $n$ first. Also you're missing generators, you need the commutators of all elements while here you've only written $[x,y]$.
    $endgroup$
    – GPerez
    Dec 12 '14 at 14:49










  • $begingroup$
    @GPerez how can i find the commutattor of group in general..i'm not getting it
    $endgroup$
    – Nizar Halloun
    Dec 12 '14 at 14:57










  • $begingroup$
    @NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators.
    $endgroup$
    – Cheerful Parsnip
    Dec 12 '14 at 15:40














0












0








0


1



$begingroup$



This question already has an answer here:




  • Commutator subgroup of a dihedral group.

    2 answers




Let $G=D2n=⟨x,y|x^2=y^n=e, $ $yx=xy^{n-1}⟩$



i need to find G' [ the commutattor of G]



now i understand the G' is the subgroup that is generated from $ U=xyx^{-1}y^{-1} , $ $forall x,y in G$



so $<u> ={u^1,u^2...e}$



whats the strategy here ?






abstract-algebra group-theory dihedral-groups







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Commutator subgroup of a dihedral group.

    2 answers




Let $G=D2n=⟨x,y|x^2=y^n=e, $ $yx=xy^{n-1}⟩$



i need to find G' [ the commutattor of G]



now i understand the G' is the subgroup that is generated from $ U=xyx^{-1}y^{-1} , $ $forall x,y in G$



so $<u> ={u^1,u^2...e}$



whats the strategy here ?





This question already has an answer here:




  • Commutator subgroup of a dihedral group.

    2 answers







abstract-algebra group-theory dihedral-groups




abstract-algebra group-theory dihedral-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '14 at 14:53







Nizar Halloun

















asked Dec 12 '14 at 14:44









Nizar HallounNizar Halloun

739




739




marked as duplicate by Dietrich Burde, Iuʇǝƃɹɐʇoɹ, Anastasiya-Romanova 秀, Stefan Hamcke, Cheerful Parsnip Dec 12 '14 at 15:41


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Dietrich Burde, Iuʇǝƃɹɐʇoɹ, Anastasiya-Romanova 秀, Stefan Hamcke, Cheerful Parsnip Dec 12 '14 at 15:41


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    I'd suggest trying with small $n$ first. Also you're missing generators, you need the commutators of all elements while here you've only written $[x,y]$.
    $endgroup$
    – GPerez
    Dec 12 '14 at 14:49










  • $begingroup$
    @GPerez how can i find the commutattor of group in general..i'm not getting it
    $endgroup$
    – Nizar Halloun
    Dec 12 '14 at 14:57










  • $begingroup$
    @NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators.
    $endgroup$
    – Cheerful Parsnip
    Dec 12 '14 at 15:40














  • 1




    $begingroup$
    I'd suggest trying with small $n$ first. Also you're missing generators, you need the commutators of all elements while here you've only written $[x,y]$.
    $endgroup$
    – GPerez
    Dec 12 '14 at 14:49










  • $begingroup$
    @GPerez how can i find the commutattor of group in general..i'm not getting it
    $endgroup$
    – Nizar Halloun
    Dec 12 '14 at 14:57










  • $begingroup$
    @NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators.
    $endgroup$
    – Cheerful Parsnip
    Dec 12 '14 at 15:40








1




1




$begingroup$
I'd suggest trying with small $n$ first. Also you're missing generators, you need the commutators of all elements while here you've only written $[x,y]$.
$endgroup$
– GPerez
Dec 12 '14 at 14:49




$begingroup$
I'd suggest trying with small $n$ first. Also you're missing generators, you need the commutators of all elements while here you've only written $[x,y]$.
$endgroup$
– GPerez
Dec 12 '14 at 14:49












$begingroup$
@GPerez how can i find the commutattor of group in general..i'm not getting it
$endgroup$
– Nizar Halloun
Dec 12 '14 at 14:57




$begingroup$
@GPerez how can i find the commutattor of group in general..i'm not getting it
$endgroup$
– Nizar Halloun
Dec 12 '14 at 14:57












$begingroup$
@NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators.
$endgroup$
– Cheerful Parsnip
Dec 12 '14 at 15:40




$begingroup$
@NizarHalloun: Terminology issue: A "commutator" is an element of a group. You are talking about the "commutator subgroup," which is the subgroup generated by commutators.
$endgroup$
– Cheerful Parsnip
Dec 12 '14 at 15:40










1 Answer
1






active

oldest

votes


















4












$begingroup$

Some ideas: if we write



$$D_{2n}=langle;x,y;:;;x^2=y^n=1;,;;xyx=y^{n-1};rangle$$



then we get that



$$forall,kinBbb N;,;;[x,y^k]=x^{-1}y^{-k}xy^k=left(xy^{-k}xright)y^k=left(y^{n-1}right)^{-k};y^k=y^{2k}$$



and this means every element in $;langle;y^2;rangle;$ is a commutator.



OTOH, we have that



$$xy(yx)^{-1}=xyxy^{-1}=y^{n-1}y^{-1}=y^{-2}inlangle;y^2;rangleimplies G/langle;y^2;rangle;;text{is abelian}iff G'lelangle;y^2;rangle$$



The above yields $;G'=[G:G]=langle;y^2;rangle;$ .






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Some ideas: if we write



    $$D_{2n}=langle;x,y;:;;x^2=y^n=1;,;;xyx=y^{n-1};rangle$$



    then we get that



    $$forall,kinBbb N;,;;[x,y^k]=x^{-1}y^{-k}xy^k=left(xy^{-k}xright)y^k=left(y^{n-1}right)^{-k};y^k=y^{2k}$$



    and this means every element in $;langle;y^2;rangle;$ is a commutator.



    OTOH, we have that



    $$xy(yx)^{-1}=xyxy^{-1}=y^{n-1}y^{-1}=y^{-2}inlangle;y^2;rangleimplies G/langle;y^2;rangle;;text{is abelian}iff G'lelangle;y^2;rangle$$



    The above yields $;G'=[G:G]=langle;y^2;rangle;$ .






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Some ideas: if we write



      $$D_{2n}=langle;x,y;:;;x^2=y^n=1;,;;xyx=y^{n-1};rangle$$



      then we get that



      $$forall,kinBbb N;,;;[x,y^k]=x^{-1}y^{-k}xy^k=left(xy^{-k}xright)y^k=left(y^{n-1}right)^{-k};y^k=y^{2k}$$



      and this means every element in $;langle;y^2;rangle;$ is a commutator.



      OTOH, we have that



      $$xy(yx)^{-1}=xyxy^{-1}=y^{n-1}y^{-1}=y^{-2}inlangle;y^2;rangleimplies G/langle;y^2;rangle;;text{is abelian}iff G'lelangle;y^2;rangle$$



      The above yields $;G'=[G:G]=langle;y^2;rangle;$ .






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Some ideas: if we write



        $$D_{2n}=langle;x,y;:;;x^2=y^n=1;,;;xyx=y^{n-1};rangle$$



        then we get that



        $$forall,kinBbb N;,;;[x,y^k]=x^{-1}y^{-k}xy^k=left(xy^{-k}xright)y^k=left(y^{n-1}right)^{-k};y^k=y^{2k}$$



        and this means every element in $;langle;y^2;rangle;$ is a commutator.



        OTOH, we have that



        $$xy(yx)^{-1}=xyxy^{-1}=y^{n-1}y^{-1}=y^{-2}inlangle;y^2;rangleimplies G/langle;y^2;rangle;;text{is abelian}iff G'lelangle;y^2;rangle$$



        The above yields $;G'=[G:G]=langle;y^2;rangle;$ .






        share|cite|improve this answer









        $endgroup$



        Some ideas: if we write



        $$D_{2n}=langle;x,y;:;;x^2=y^n=1;,;;xyx=y^{n-1};rangle$$



        then we get that



        $$forall,kinBbb N;,;;[x,y^k]=x^{-1}y^{-k}xy^k=left(xy^{-k}xright)y^k=left(y^{n-1}right)^{-k};y^k=y^{2k}$$



        and this means every element in $;langle;y^2;rangle;$ is a commutator.



        OTOH, we have that



        $$xy(yx)^{-1}=xyxy^{-1}=y^{n-1}y^{-1}=y^{-2}inlangle;y^2;rangleimplies G/langle;y^2;rangle;;text{is abelian}iff G'lelangle;y^2;rangle$$



        The above yields $;G'=[G:G]=langle;y^2;rangle;$ .







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '14 at 15:26









        TimbucTimbuc

        31k22145




        31k22145















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