$X$ has a normal group $Y$ so $[X: Y]=4$.












0












$begingroup$


Consider the following set: $$X = {sigma in S_6 | sigma(k) is odd if and only if k is odd}.$$



I would like to show that $X$ has a normal group $Y$ so $[X: Y]=4$.



How should I prove this theorem?










share|cite|improve this question











$endgroup$












  • $begingroup$
    You should verify that it is a subgroup and that it is normal. Finally, you should figure out how big it is, so you can calculate its index.
    $endgroup$
    – Cheerful Parsnip
    Jan 12 at 18:56
















0












$begingroup$


Consider the following set: $$X = {sigma in S_6 | sigma(k) is odd if and only if k is odd}.$$



I would like to show that $X$ has a normal group $Y$ so $[X: Y]=4$.



How should I prove this theorem?










share|cite|improve this question











$endgroup$












  • $begingroup$
    You should verify that it is a subgroup and that it is normal. Finally, you should figure out how big it is, so you can calculate its index.
    $endgroup$
    – Cheerful Parsnip
    Jan 12 at 18:56














0












0








0


1



$begingroup$


Consider the following set: $$X = {sigma in S_6 | sigma(k) is odd if and only if k is odd}.$$



I would like to show that $X$ has a normal group $Y$ so $[X: Y]=4$.



How should I prove this theorem?










share|cite|improve this question











$endgroup$




Consider the following set: $$X = {sigma in S_6 | sigma(k) is odd if and only if k is odd}.$$



I would like to show that $X$ has a normal group $Y$ so $[X: Y]=4$.



How should I prove this theorem?







abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 18:46









Shaun

9,083113683




9,083113683










asked Jan 12 at 17:53









BadukBaduk

171




171












  • $begingroup$
    You should verify that it is a subgroup and that it is normal. Finally, you should figure out how big it is, so you can calculate its index.
    $endgroup$
    – Cheerful Parsnip
    Jan 12 at 18:56


















  • $begingroup$
    You should verify that it is a subgroup and that it is normal. Finally, you should figure out how big it is, so you can calculate its index.
    $endgroup$
    – Cheerful Parsnip
    Jan 12 at 18:56
















$begingroup$
You should verify that it is a subgroup and that it is normal. Finally, you should figure out how big it is, so you can calculate its index.
$endgroup$
– Cheerful Parsnip
Jan 12 at 18:56




$begingroup$
You should verify that it is a subgroup and that it is normal. Finally, you should figure out how big it is, so you can calculate its index.
$endgroup$
– Cheerful Parsnip
Jan 12 at 18:56










1 Answer
1






active

oldest

votes


















0












$begingroup$

Solution (sketch):



(a) How can we relate X to some group we are familiar with? Prove that $Xcong S_3times S_3$.



Let $S={1,2,3,4,5,6}$ be the set of elements in $S_6$ and let $S_1$ and $S_2$ denote the set of all even resp. odd numbers in $S$.
Let $f:Xrightarrow S_3times S_3$, $sigmamapsto (sigma_1,sigma_2)$, where $sigma_1=sigma|_{S_1}$, $sigma_2=sigma|_{S_2}$. You can show that $f$ is indeed an isomorphism.



(b) Now, we can calculate the size of Y: $$[X:Y]=frac{|X|}{|Y|}=frac{|S_3times S_3|}{Y}=frac{6cdot 6}{|Y|}=4.$$



(c) Take a look at the subgroups of $S_3$ (or, $S_3times S_3$). Which subgroup would be applicable? According to the required size, $Y$ should be isomorphic to $A_3times A_3$.






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Solution (sketch):



    (a) How can we relate X to some group we are familiar with? Prove that $Xcong S_3times S_3$.



    Let $S={1,2,3,4,5,6}$ be the set of elements in $S_6$ and let $S_1$ and $S_2$ denote the set of all even resp. odd numbers in $S$.
    Let $f:Xrightarrow S_3times S_3$, $sigmamapsto (sigma_1,sigma_2)$, where $sigma_1=sigma|_{S_1}$, $sigma_2=sigma|_{S_2}$. You can show that $f$ is indeed an isomorphism.



    (b) Now, we can calculate the size of Y: $$[X:Y]=frac{|X|}{|Y|}=frac{|S_3times S_3|}{Y}=frac{6cdot 6}{|Y|}=4.$$



    (c) Take a look at the subgroups of $S_3$ (or, $S_3times S_3$). Which subgroup would be applicable? According to the required size, $Y$ should be isomorphic to $A_3times A_3$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Solution (sketch):



      (a) How can we relate X to some group we are familiar with? Prove that $Xcong S_3times S_3$.



      Let $S={1,2,3,4,5,6}$ be the set of elements in $S_6$ and let $S_1$ and $S_2$ denote the set of all even resp. odd numbers in $S$.
      Let $f:Xrightarrow S_3times S_3$, $sigmamapsto (sigma_1,sigma_2)$, where $sigma_1=sigma|_{S_1}$, $sigma_2=sigma|_{S_2}$. You can show that $f$ is indeed an isomorphism.



      (b) Now, we can calculate the size of Y: $$[X:Y]=frac{|X|}{|Y|}=frac{|S_3times S_3|}{Y}=frac{6cdot 6}{|Y|}=4.$$



      (c) Take a look at the subgroups of $S_3$ (or, $S_3times S_3$). Which subgroup would be applicable? According to the required size, $Y$ should be isomorphic to $A_3times A_3$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Solution (sketch):



        (a) How can we relate X to some group we are familiar with? Prove that $Xcong S_3times S_3$.



        Let $S={1,2,3,4,5,6}$ be the set of elements in $S_6$ and let $S_1$ and $S_2$ denote the set of all even resp. odd numbers in $S$.
        Let $f:Xrightarrow S_3times S_3$, $sigmamapsto (sigma_1,sigma_2)$, where $sigma_1=sigma|_{S_1}$, $sigma_2=sigma|_{S_2}$. You can show that $f$ is indeed an isomorphism.



        (b) Now, we can calculate the size of Y: $$[X:Y]=frac{|X|}{|Y|}=frac{|S_3times S_3|}{Y}=frac{6cdot 6}{|Y|}=4.$$



        (c) Take a look at the subgroups of $S_3$ (or, $S_3times S_3$). Which subgroup would be applicable? According to the required size, $Y$ should be isomorphic to $A_3times A_3$.






        share|cite|improve this answer









        $endgroup$



        Solution (sketch):



        (a) How can we relate X to some group we are familiar with? Prove that $Xcong S_3times S_3$.



        Let $S={1,2,3,4,5,6}$ be the set of elements in $S_6$ and let $S_1$ and $S_2$ denote the set of all even resp. odd numbers in $S$.
        Let $f:Xrightarrow S_3times S_3$, $sigmamapsto (sigma_1,sigma_2)$, where $sigma_1=sigma|_{S_1}$, $sigma_2=sigma|_{S_2}$. You can show that $f$ is indeed an isomorphism.



        (b) Now, we can calculate the size of Y: $$[X:Y]=frac{|X|}{|Y|}=frac{|S_3times S_3|}{Y}=frac{6cdot 6}{|Y|}=4.$$



        (c) Take a look at the subgroups of $S_3$ (or, $S_3times S_3$). Which subgroup would be applicable? According to the required size, $Y$ should be isomorphic to $A_3times A_3$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 at 20:28









        fiatluxfiatlux

        132




        132






























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