$G_{x_1x_2+x_3x_4}cong D_8$
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
asked Nov 29 '18 at 15:19
J. DoeJ. Doe
1286
add a comment |
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
asked Nov 29 '18 at 15:19
J. DoeJ. Doe
1286
add a comment |
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
asked Nov 29 '18 at 15:19
J. DoeJ. Doe
1286
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
abstract-algebra group-theory permutations group-actions dihedral-groups
asked Nov 29 '18 at 15:19
J. DoeJ. Doe
1286
asked Nov 29 '18 at 15:19
J. DoeJ. Doe
1286
edited Nov 29 '18 at 15:30
J. Doe
asked Nov 29 '18 at 15:19
J. DoeJ. Doe
1286
asked Nov 29 '18 at 15:19
J. DoeJ. Doe
1286
asked Nov 29 '18 at 15:19
J. DoeJ. Doe
1286
1286
add a comment |
add a comment |
1 Answer
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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.
answered Nov 29 '18 at 15:33
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
add a comment |
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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.
answered Nov 29 '18 at 15:33
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
add a comment |
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.
answered Nov 29 '18 at 15:33
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
add a comment |
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.
answered Nov 29 '18 at 15:33
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
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.
answered Nov 29 '18 at 15:33
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
answered Nov 29 '18 at 15:33
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
answered Nov 29 '18 at 15:33
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
answered Nov 29 '18 at 15:33
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
96.4k10151296
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add a comment |
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