Dgdrxrt

Blender game recording at the wrong time

Was credit for the black hole image misattributed?

If A makes B more likely then B makes A more likely"

What are the performance impacts of 'functional' Rust?

Why does this iterative way of solving of equation work?

Windows 10: How to Lock (not sleep) laptop on lid close?

Cauchy Sequence Characterized only By Directly Neighbouring Sequence Members

What is the largest species of polychaete?

New Order #5: where Fibonacci and Beatty meet at Wythoff

Unexpected result with right shift after bitwise negation

Writing Thesis: Copying from published papers

How to say that you spent the night with someone, you were only sleeping and nothing else?

Determine whether f is a function, an injection, a surjection

What would be Julian Assange's expected punishment, on the current English criminal law?

3 doors, three guards, one stone

Antler Helmet: Can it work?

What computer would be fastest for Mathematica Home Edition?

Why is there no army of Iron-Mans in the MCU?

What do you call a plan that's an alternative plan in case your initial plan fails?

The following signatures were invalid: EXPKEYSIG 1397BC53640DB551

Why use gamma over alpha radiation?

Do working physicists consider Newtonian mechanics to be "falsified"?

What LEGO pieces have "real-world" functionality?

Can the prologue be the backstory of your main character?

Prove that the Galois group is $D_8$.

1

2

$begingroup$

Prove that the Galois group of $x^4-2x^2-2$ over $mathbbQ$ is $D_8$.

Note that the order of $D_8$ is $8$.

How would I find the eight automorphisms of this group? Or can I just use the order of $G(x^4-2x^2-2/mathbbQ)$ to show that its order is $8$?

If I show the order of the Galois group is $8$, is it sufficent enough to show that, since it's nonabelian and order $8$, it's isomorphic to $D_8$?

abstract-algebra galois-theory

share|cite|improve this question

asked Mar 31 at 19:39

numericalorangenumericalorange

1,944314

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  there's also the Quaternion group that has order $8$ and is non-Abelian
  $endgroup$
  – J. W. Tanner
  Mar 31 at 19:47
  
  

  
  
  
  

add a comment | 

1

2

$begingroup$

Prove that the Galois group of $x^4-2x^2-2$ over $mathbbQ$ is $D_8$.

Note that the order of $D_8$ is $8$.

How would I find the eight automorphisms of this group? Or can I just use the order of $G(x^4-2x^2-2/mathbbQ)$ to show that its order is $8$?

If I show the order of the Galois group is $8$, is it sufficent enough to show that, since it's nonabelian and order $8$, it's isomorphic to $D_8$?

abstract-algebra galois-theory

share|cite|improve this question

asked Mar 31 at 19:39

numericalorangenumericalorange

1,944314

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  there's also the Quaternion group that has order $8$ and is non-Abelian
  $endgroup$
  – J. W. Tanner
  Mar 31 at 19:47
  
  

  
  
  
  

add a comment | 

$begingroup$

Prove that the Galois group of $x^4-2x^2-2$ over $mathbbQ$ is $D_8$.

Note that the order of $D_8$ is $8$.

How would I find the eight automorphisms of this group? Or can I just use the order of $G(x^4-2x^2-2/mathbbQ)$ to show that its order is $8$?

If I show the order of the Galois group is $8$, is it sufficent enough to show that, since it's nonabelian and order $8$, it's isomorphic to $D_8$?

abstract-algebra galois-theory

share|cite|improve this question

asked Mar 31 at 19:39

numericalorangenumericalorange

1,944314

$endgroup$

share|cite|improve this question

asked Mar 31 at 19:39

numericalorangenumericalorange

1,944314

share|cite|improve this question

asked Mar 31 at 19:39

numericalorangenumericalorange

1,944314

asked Mar 31 at 19:39

numericalorangenumericalorange

1,944314

asked Mar 31 at 19:39

numericalorangenumericalorange

1,944314

1 Answer
1

active

oldest

votes

1

$begingroup$

First prove that $x^4-2x^2-2$ is irreducible over $BbbQ$. Then its Galois group is isomorphic to a transitive subgroup of $S_4$. That means it is isomorphic to either $BbbZ/4BbbZ$, $(BbbZ/2BbbZ)^2$, $D_8$, $A_4$ or $S_4$.

If you are then able to show that the order of the Galois group is $8$, then it must be isomorphic to $D_8$.

---

Slightly more in line with your approach, you could note that up to isomorphism there are precisely two non-abelian groups of order $8$. They are $D_8$ and $Q_8$, the quaternion group of order $8$. But the latter is not isomorphic to a subgroup of $S_4$, so it cannot be the Galois group of $x^4-2x^2-2$.

share|cite|improve this answer

answered Mar 31 at 19:52

ServaesServaes

30.6k342101

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I totally forgot about $Q_8$, thanks for the heads up on that. I understand now what to do, thanks so much.
  $endgroup$
  – numericalorange
  Mar 31 at 20:13
  

  
  
  
  

add a comment | 

Your Answer

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3169811%2fprove-that-the-galois-group-is-d-8%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

First prove that $x^4-2x^2-2$ is irreducible over $BbbQ$. Then its Galois group is isomorphic to a transitive subgroup of $S_4$. That means it is isomorphic to either $BbbZ/4BbbZ$, $(BbbZ/2BbbZ)^2$, $D_8$, $A_4$ or $S_4$.

If you are then able to show that the order of the Galois group is $8$, then it must be isomorphic to $D_8$.

---

Slightly more in line with your approach, you could note that up to isomorphism there are precisely two non-abelian groups of order $8$. They are $D_8$ and $Q_8$, the quaternion group of order $8$. But the latter is not isomorphic to a subgroup of $S_4$, so it cannot be the Galois group of $x^4-2x^2-2$.

share|cite|improve this answer

answered Mar 31 at 19:52

ServaesServaes

30.6k342101

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I totally forgot about $Q_8$, thanks for the heads up on that. I understand now what to do, thanks so much.
  $endgroup$
  – numericalorange
  Mar 31 at 20:13
  

  
  
  
  

add a comment | 

1

$begingroup$

First prove that $x^4-2x^2-2$ is irreducible over $BbbQ$. Then its Galois group is isomorphic to a transitive subgroup of $S_4$. That means it is isomorphic to either $BbbZ/4BbbZ$, $(BbbZ/2BbbZ)^2$, $D_8$, $A_4$ or $S_4$.

If you are then able to show that the order of the Galois group is $8$, then it must be isomorphic to $D_8$.

---

Slightly more in line with your approach, you could note that up to isomorphism there are precisely two non-abelian groups of order $8$. They are $D_8$ and $Q_8$, the quaternion group of order $8$. But the latter is not isomorphic to a subgroup of $S_4$, so it cannot be the Galois group of $x^4-2x^2-2$.

share|cite|improve this answer

answered Mar 31 at 19:52

ServaesServaes

30.6k342101

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I totally forgot about $Q_8$, thanks for the heads up on that. I understand now what to do, thanks so much.
  $endgroup$
  – numericalorange
  Mar 31 at 20:13
  

  
  
  
  

add a comment | 

$begingroup$

First prove that $x^4-2x^2-2$ is irreducible over $BbbQ$. Then its Galois group is isomorphic to a transitive subgroup of $S_4$. That means it is isomorphic to either $BbbZ/4BbbZ$, $(BbbZ/2BbbZ)^2$, $D_8$, $A_4$ or $S_4$.

If you are then able to show that the order of the Galois group is $8$, then it must be isomorphic to $D_8$.

---

Slightly more in line with your approach, you could note that up to isomorphism there are precisely two non-abelian groups of order $8$. They are $D_8$ and $Q_8$, the quaternion group of order $8$. But the latter is not isomorphic to a subgroup of $S_4$, so it cannot be the Galois group of $x^4-2x^2-2$.

share|cite|improve this answer

answered Mar 31 at 19:52

ServaesServaes

30.6k342101

$endgroup$

share|cite|improve this answer

answered Mar 31 at 19:52

ServaesServaes

30.6k342101

answered Mar 31 at 19:52

ServaesServaes

30.6k342101

answered Mar 31 at 19:52

ServaesServaes

30.6k342101