Handling a subgroup of $S_{6}$ [closed]












0












$begingroup$


I came across with the following question:



Let $H={sigma in S_6 | sigma(x) textrm{ is odd if and only if } x textrm{ is odd}}$. Prove that $H$ has a normal subgroup $Y$ so $[H : Y ]=4$.



How should I prove this theorem?










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$endgroup$



closed as off-topic by Derek Holt, Alexander Gruber Jan 8 at 22:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Try to understand $H$ a little. For example, what is its order?
    $endgroup$
    – verret
    Jan 8 at 20:28
















0












$begingroup$


I came across with the following question:



Let $H={sigma in S_6 | sigma(x) textrm{ is odd if and only if } x textrm{ is odd}}$. Prove that $H$ has a normal subgroup $Y$ so $[H : Y ]=4$.



How should I prove this theorem?










share|cite|improve this question











$endgroup$



closed as off-topic by Derek Holt, Alexander Gruber Jan 8 at 22:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Try to understand $H$ a little. For example, what is its order?
    $endgroup$
    – verret
    Jan 8 at 20:28














0












0








0





$begingroup$


I came across with the following question:



Let $H={sigma in S_6 | sigma(x) textrm{ is odd if and only if } x textrm{ is odd}}$. Prove that $H$ has a normal subgroup $Y$ so $[H : Y ]=4$.



How should I prove this theorem?










share|cite|improve this question











$endgroup$




I came across with the following question:



Let $H={sigma in S_6 | sigma(x) textrm{ is odd if and only if } x textrm{ is odd}}$. Prove that $H$ has a normal subgroup $Y$ so $[H : Y ]=4$.



How should I prove this theorem?







group-theory






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 20:34









Sam Hughes

526112




526112










asked Jan 8 at 0:47









abuka123abuka123

344




344




closed as off-topic by Derek Holt, Alexander Gruber Jan 8 at 22:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Derek Holt, Alexander Gruber Jan 8 at 22:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Derek Holt, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Try to understand $H$ a little. For example, what is its order?
    $endgroup$
    – verret
    Jan 8 at 20:28


















  • $begingroup$
    Try to understand $H$ a little. For example, what is its order?
    $endgroup$
    – verret
    Jan 8 at 20:28
















$begingroup$
Try to understand $H$ a little. For example, what is its order?
$endgroup$
– verret
Jan 8 at 20:28




$begingroup$
Try to understand $H$ a little. For example, what is its order?
$endgroup$
– verret
Jan 8 at 20:28










2 Answers
2






active

oldest

votes


















0












$begingroup$

I'll separate the steps into spoilers so you can use them as hints



The set $H$:




First we need to understand the set $H$, we know that a permutation in $Sym(6)$ moves points in the set $Omega:={1,2,3,4,5,6}$. Now, $H$ is defined to be the set of points which fix the subset $Lambda:={1,3,5}$ of odd integers setwise.




The group $H$:




The elements of $Sym(6)$ which fix $Lambda$ setwise are exactly the permutations that permute elements of $Lambda$, that is, $Sym(Lambda)=Sym({1,3,5})$ and the elements which permute $Gamma:={2,4,6}$, these are exactly the elements in $Sym(Gamma)$. Since no element of $Gamma$ can be sent to an element of $Lambda$ we conclude $H$ is the direct product $Sym(Lambda)times Sym(Gamma)cong Sym(3)times Sym(3)$.




The normal subgroup:




First we recall that $Sym(3)$ has a normal subgroup isomorphic to $C_3:=mathbb{Z}_3$ so we deduce that $N_1:=langle(135)rangle$ is normal in $Sym(Lambda)$ and $N_2:=langle(246)rangle$ is normal in $Sym(Gamma)$. As these are normal subgroups in different factors of the direct product we conclude there is a normal subgroup $N:=N_1times N_2$ in $H$.




The order:




A quick calculation gives $|H|=6*6=36$ and $|N|=3*3=9$. Therefore $|H:N|=4$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great, thank you! How to compute $[S_6 : G cap A_6]$?
    $endgroup$
    – abuka123
    Jan 12 at 18:39



















2












$begingroup$

Hint: for permutations of $H$, you can separate their action on odd and even integers. Thus $H cong S_3 times S_3$.






share|cite|improve this answer









$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    I'll separate the steps into spoilers so you can use them as hints



    The set $H$:




    First we need to understand the set $H$, we know that a permutation in $Sym(6)$ moves points in the set $Omega:={1,2,3,4,5,6}$. Now, $H$ is defined to be the set of points which fix the subset $Lambda:={1,3,5}$ of odd integers setwise.




    The group $H$:




    The elements of $Sym(6)$ which fix $Lambda$ setwise are exactly the permutations that permute elements of $Lambda$, that is, $Sym(Lambda)=Sym({1,3,5})$ and the elements which permute $Gamma:={2,4,6}$, these are exactly the elements in $Sym(Gamma)$. Since no element of $Gamma$ can be sent to an element of $Lambda$ we conclude $H$ is the direct product $Sym(Lambda)times Sym(Gamma)cong Sym(3)times Sym(3)$.




    The normal subgroup:




    First we recall that $Sym(3)$ has a normal subgroup isomorphic to $C_3:=mathbb{Z}_3$ so we deduce that $N_1:=langle(135)rangle$ is normal in $Sym(Lambda)$ and $N_2:=langle(246)rangle$ is normal in $Sym(Gamma)$. As these are normal subgroups in different factors of the direct product we conclude there is a normal subgroup $N:=N_1times N_2$ in $H$.




    The order:




    A quick calculation gives $|H|=6*6=36$ and $|N|=3*3=9$. Therefore $|H:N|=4$.







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Great, thank you! How to compute $[S_6 : G cap A_6]$?
      $endgroup$
      – abuka123
      Jan 12 at 18:39
















    0












    $begingroup$

    I'll separate the steps into spoilers so you can use them as hints



    The set $H$:




    First we need to understand the set $H$, we know that a permutation in $Sym(6)$ moves points in the set $Omega:={1,2,3,4,5,6}$. Now, $H$ is defined to be the set of points which fix the subset $Lambda:={1,3,5}$ of odd integers setwise.




    The group $H$:




    The elements of $Sym(6)$ which fix $Lambda$ setwise are exactly the permutations that permute elements of $Lambda$, that is, $Sym(Lambda)=Sym({1,3,5})$ and the elements which permute $Gamma:={2,4,6}$, these are exactly the elements in $Sym(Gamma)$. Since no element of $Gamma$ can be sent to an element of $Lambda$ we conclude $H$ is the direct product $Sym(Lambda)times Sym(Gamma)cong Sym(3)times Sym(3)$.




    The normal subgroup:




    First we recall that $Sym(3)$ has a normal subgroup isomorphic to $C_3:=mathbb{Z}_3$ so we deduce that $N_1:=langle(135)rangle$ is normal in $Sym(Lambda)$ and $N_2:=langle(246)rangle$ is normal in $Sym(Gamma)$. As these are normal subgroups in different factors of the direct product we conclude there is a normal subgroup $N:=N_1times N_2$ in $H$.




    The order:




    A quick calculation gives $|H|=6*6=36$ and $|N|=3*3=9$. Therefore $|H:N|=4$.







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Great, thank you! How to compute $[S_6 : G cap A_6]$?
      $endgroup$
      – abuka123
      Jan 12 at 18:39














    0












    0








    0





    $begingroup$

    I'll separate the steps into spoilers so you can use them as hints



    The set $H$:




    First we need to understand the set $H$, we know that a permutation in $Sym(6)$ moves points in the set $Omega:={1,2,3,4,5,6}$. Now, $H$ is defined to be the set of points which fix the subset $Lambda:={1,3,5}$ of odd integers setwise.




    The group $H$:




    The elements of $Sym(6)$ which fix $Lambda$ setwise are exactly the permutations that permute elements of $Lambda$, that is, $Sym(Lambda)=Sym({1,3,5})$ and the elements which permute $Gamma:={2,4,6}$, these are exactly the elements in $Sym(Gamma)$. Since no element of $Gamma$ can be sent to an element of $Lambda$ we conclude $H$ is the direct product $Sym(Lambda)times Sym(Gamma)cong Sym(3)times Sym(3)$.




    The normal subgroup:




    First we recall that $Sym(3)$ has a normal subgroup isomorphic to $C_3:=mathbb{Z}_3$ so we deduce that $N_1:=langle(135)rangle$ is normal in $Sym(Lambda)$ and $N_2:=langle(246)rangle$ is normal in $Sym(Gamma)$. As these are normal subgroups in different factors of the direct product we conclude there is a normal subgroup $N:=N_1times N_2$ in $H$.




    The order:




    A quick calculation gives $|H|=6*6=36$ and $|N|=3*3=9$. Therefore $|H:N|=4$.







    share|cite|improve this answer











    $endgroup$



    I'll separate the steps into spoilers so you can use them as hints



    The set $H$:




    First we need to understand the set $H$, we know that a permutation in $Sym(6)$ moves points in the set $Omega:={1,2,3,4,5,6}$. Now, $H$ is defined to be the set of points which fix the subset $Lambda:={1,3,5}$ of odd integers setwise.




    The group $H$:




    The elements of $Sym(6)$ which fix $Lambda$ setwise are exactly the permutations that permute elements of $Lambda$, that is, $Sym(Lambda)=Sym({1,3,5})$ and the elements which permute $Gamma:={2,4,6}$, these are exactly the elements in $Sym(Gamma)$. Since no element of $Gamma$ can be sent to an element of $Lambda$ we conclude $H$ is the direct product $Sym(Lambda)times Sym(Gamma)cong Sym(3)times Sym(3)$.




    The normal subgroup:




    First we recall that $Sym(3)$ has a normal subgroup isomorphic to $C_3:=mathbb{Z}_3$ so we deduce that $N_1:=langle(135)rangle$ is normal in $Sym(Lambda)$ and $N_2:=langle(246)rangle$ is normal in $Sym(Gamma)$. As these are normal subgroups in different factors of the direct product we conclude there is a normal subgroup $N:=N_1times N_2$ in $H$.




    The order:




    A quick calculation gives $|H|=6*6=36$ and $|N|=3*3=9$. Therefore $|H:N|=4$.








    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 13 at 21:11

























    answered Jan 8 at 16:03









    Sam HughesSam Hughes

    526112




    526112












    • $begingroup$
      Great, thank you! How to compute $[S_6 : G cap A_6]$?
      $endgroup$
      – abuka123
      Jan 12 at 18:39


















    • $begingroup$
      Great, thank you! How to compute $[S_6 : G cap A_6]$?
      $endgroup$
      – abuka123
      Jan 12 at 18:39
















    $begingroup$
    Great, thank you! How to compute $[S_6 : G cap A_6]$?
    $endgroup$
    – abuka123
    Jan 12 at 18:39




    $begingroup$
    Great, thank you! How to compute $[S_6 : G cap A_6]$?
    $endgroup$
    – abuka123
    Jan 12 at 18:39











    2












    $begingroup$

    Hint: for permutations of $H$, you can separate their action on odd and even integers. Thus $H cong S_3 times S_3$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint: for permutations of $H$, you can separate their action on odd and even integers. Thus $H cong S_3 times S_3$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint: for permutations of $H$, you can separate their action on odd and even integers. Thus $H cong S_3 times S_3$.






        share|cite|improve this answer









        $endgroup$



        Hint: for permutations of $H$, you can separate their action on odd and even integers. Thus $H cong S_3 times S_3$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 0:50









        MindlackMindlack

        4,585210




        4,585210















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