$G_{x_1x_2+x_3x_4}cong D_8$












2














Let $R=F[x_1,x_2,x_3,x_4]$ be the set of polynoms in 4 variables over a field $F$. Let a map $varphi:S_4to operatorname{Sym}(R)$ by
$$
\ (f(x_1,x_2,x_3,x_4))varphi(sigma)=f(x_{1sigma},x_{2sigma},x_{3sigma},x_{4sigma}).
$$

Prove that
$$
G_{x_1x_2+x_3x_4}cong D_8
$$

($D_8$ is the dihedral group).





I've shown that $varphi$ is a group action of $S_4$ on $R$. My problem is that I know that
$$
\ G_{x_1 x_2+x_3 x_4 }={σ∈S_4:(x_1 x_2+x_3 x_4 )σ=x_1 x_2+x_3 x_4 }
={id,(1 2),(3 4),(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}
$$

then $|G_{x_1 x_2+x_3 x_4 }|=6ne 8=|D_8|$ so I don't see how it is possible that $G_{x_1x_2+x_3x_4}cong D_8$.










share|cite|improve this question





























    2














    Let $R=F[x_1,x_2,x_3,x_4]$ be the set of polynoms in 4 variables over a field $F$. Let a map $varphi:S_4to operatorname{Sym}(R)$ by
    $$
    \ (f(x_1,x_2,x_3,x_4))varphi(sigma)=f(x_{1sigma},x_{2sigma},x_{3sigma},x_{4sigma}).
    $$

    Prove that
    $$
    G_{x_1x_2+x_3x_4}cong D_8
    $$

    ($D_8$ is the dihedral group).





    I've shown that $varphi$ is a group action of $S_4$ on $R$. My problem is that I know that
    $$
    \ G_{x_1 x_2+x_3 x_4 }={σ∈S_4:(x_1 x_2+x_3 x_4 )σ=x_1 x_2+x_3 x_4 }
    ={id,(1 2),(3 4),(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}
    $$

    then $|G_{x_1 x_2+x_3 x_4 }|=6ne 8=|D_8|$ so I don't see how it is possible that $G_{x_1x_2+x_3x_4}cong D_8$.










    share|cite|improve this question



























      2












      2








      2


      1





      Let $R=F[x_1,x_2,x_3,x_4]$ be the set of polynoms in 4 variables over a field $F$. Let a map $varphi:S_4to operatorname{Sym}(R)$ by
      $$
      \ (f(x_1,x_2,x_3,x_4))varphi(sigma)=f(x_{1sigma},x_{2sigma},x_{3sigma},x_{4sigma}).
      $$

      Prove that
      $$
      G_{x_1x_2+x_3x_4}cong D_8
      $$

      ($D_8$ is the dihedral group).





      I've shown that $varphi$ is a group action of $S_4$ on $R$. My problem is that I know that
      $$
      \ G_{x_1 x_2+x_3 x_4 }={σ∈S_4:(x_1 x_2+x_3 x_4 )σ=x_1 x_2+x_3 x_4 }
      ={id,(1 2),(3 4),(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}
      $$

      then $|G_{x_1 x_2+x_3 x_4 }|=6ne 8=|D_8|$ so I don't see how it is possible that $G_{x_1x_2+x_3x_4}cong D_8$.










      share|cite|improve this question















      Let $R=F[x_1,x_2,x_3,x_4]$ be the set of polynoms in 4 variables over a field $F$. Let a map $varphi:S_4to operatorname{Sym}(R)$ by
      $$
      \ (f(x_1,x_2,x_3,x_4))varphi(sigma)=f(x_{1sigma},x_{2sigma},x_{3sigma},x_{4sigma}).
      $$

      Prove that
      $$
      G_{x_1x_2+x_3x_4}cong D_8
      $$

      ($D_8$ is the dihedral group).





      I've shown that $varphi$ is a group action of $S_4$ on $R$. My problem is that I know that
      $$
      \ G_{x_1 x_2+x_3 x_4 }={σ∈S_4:(x_1 x_2+x_3 x_4 )σ=x_1 x_2+x_3 x_4 }
      ={id,(1 2),(3 4),(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}
      $$

      then $|G_{x_1 x_2+x_3 x_4 }|=6ne 8=|D_8|$ so I don't see how it is possible that $G_{x_1x_2+x_3x_4}cong D_8$.







      abstract-algebra group-theory permutations group-actions dihedral-groups






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 29 '18 at 15:30







      J. Doe

















      asked Nov 29 '18 at 15:19









      J. DoeJ. Doe

      1286




      1286






















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          Your stabilizer is not complete, it is not a subgroup, for example!



          To compute the size of the stabilizer $G$ of $p = x_1x_2+x_3x_4$ you can begin by computing its orbit: it consists of three elements: $x_1x_2+x_3x_4,x_1x_3+x_2x_4,x_1x_4+x_2x_3$. By the orbit--stabilizer theorem, the stabilizer must consist of eight elements, since $24/3=8$.



          You already noted that $G$ contains ${1,(12),(34),(13)(24),(14)(23),(12)(13)}$. But this is not a subgroup. It is missing $(1423)$, which is $(13)(24)(12)$, and it is missing $(1324)$, its inverse. This makes $8$ elements, so $G = {1,(12),(34),(13)(24),(14)(23),(12)(13),(1423), (1324)}$ is your stabilizer.






          share|cite|improve this answer





















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            active

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            2














            Your stabilizer is not complete, it is not a subgroup, for example!



            To compute the size of the stabilizer $G$ of $p = x_1x_2+x_3x_4$ you can begin by computing its orbit: it consists of three elements: $x_1x_2+x_3x_4,x_1x_3+x_2x_4,x_1x_4+x_2x_3$. By the orbit--stabilizer theorem, the stabilizer must consist of eight elements, since $24/3=8$.



            You already noted that $G$ contains ${1,(12),(34),(13)(24),(14)(23),(12)(13)}$. But this is not a subgroup. It is missing $(1423)$, which is $(13)(24)(12)$, and it is missing $(1324)$, its inverse. This makes $8$ elements, so $G = {1,(12),(34),(13)(24),(14)(23),(12)(13),(1423), (1324)}$ is your stabilizer.






            share|cite|improve this answer


























              2














              Your stabilizer is not complete, it is not a subgroup, for example!



              To compute the size of the stabilizer $G$ of $p = x_1x_2+x_3x_4$ you can begin by computing its orbit: it consists of three elements: $x_1x_2+x_3x_4,x_1x_3+x_2x_4,x_1x_4+x_2x_3$. By the orbit--stabilizer theorem, the stabilizer must consist of eight elements, since $24/3=8$.



              You already noted that $G$ contains ${1,(12),(34),(13)(24),(14)(23),(12)(13)}$. But this is not a subgroup. It is missing $(1423)$, which is $(13)(24)(12)$, and it is missing $(1324)$, its inverse. This makes $8$ elements, so $G = {1,(12),(34),(13)(24),(14)(23),(12)(13),(1423), (1324)}$ is your stabilizer.






              share|cite|improve this answer
























                2












                2








                2






                Your stabilizer is not complete, it is not a subgroup, for example!



                To compute the size of the stabilizer $G$ of $p = x_1x_2+x_3x_4$ you can begin by computing its orbit: it consists of three elements: $x_1x_2+x_3x_4,x_1x_3+x_2x_4,x_1x_4+x_2x_3$. By the orbit--stabilizer theorem, the stabilizer must consist of eight elements, since $24/3=8$.



                You already noted that $G$ contains ${1,(12),(34),(13)(24),(14)(23),(12)(13)}$. But this is not a subgroup. It is missing $(1423)$, which is $(13)(24)(12)$, and it is missing $(1324)$, its inverse. This makes $8$ elements, so $G = {1,(12),(34),(13)(24),(14)(23),(12)(13),(1423), (1324)}$ is your stabilizer.






                share|cite|improve this answer












                Your stabilizer is not complete, it is not a subgroup, for example!



                To compute the size of the stabilizer $G$ of $p = x_1x_2+x_3x_4$ you can begin by computing its orbit: it consists of three elements: $x_1x_2+x_3x_4,x_1x_3+x_2x_4,x_1x_4+x_2x_3$. By the orbit--stabilizer theorem, the stabilizer must consist of eight elements, since $24/3=8$.



                You already noted that $G$ contains ${1,(12),(34),(13)(24),(14)(23),(12)(13)}$. But this is not a subgroup. It is missing $(1423)$, which is $(13)(24)(12)$, and it is missing $(1324)$, its inverse. This makes $8$ elements, so $G = {1,(12),(34),(13)(24),(14)(23),(12)(13),(1423), (1324)}$ is your stabilizer.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 29 '18 at 15:33









                Pedro TamaroffPedro Tamaroff

                96.4k10151296




                96.4k10151296






























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