$G_{x_1x_2+x_3x_4}cong D_8$
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 at 15:19
J. Doe
936
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 at 15:19
J. Doe
936
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 at 15:19
J. Doe
936
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 at 15:19
J. Doe
936
asked Nov 29 at 15:19
J. Doe
936
edited Nov 29 at 15:30
asked Nov 29 at 15:19
J. Doe
936
asked Nov 29 at 15:19
J. Doe
936
asked Nov 29 at 15:19
J. Doe
936
936
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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 at 15:33
Pedro Tamaroff♦
95.9k10150295
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018761%2fg-x-1x-2x-3x-4-cong-d-8%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 at 15:33
Pedro Tamaroff♦
95.9k10150295
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 at 15:33
Pedro Tamaroff♦
95.9k10150295
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 at 15:33
Pedro Tamaroff♦
95.9k10150295
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 at 15:33
Pedro Tamaroff♦
95.9k10150295
answered Nov 29 at 15:33
Pedro Tamaroff♦
95.9k10150295
answered Nov 29 at 15:33
Pedro Tamaroff♦
95.9k10150295
answered Nov 29 at 15:33
Pedro Tamaroff♦
95.9k10150295
95.9k10150295
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018761%2fg-x-1x-2x-3x-4-cong-d-8%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
