Relation between odd order theorem and the fact that every polynomial of odd order has at least one root
$begingroup$
It is known that there is connection between solvability of a group and expressing the roots of a polynomial by radicals, which is something I will study in this semester; however, by the odd order theorem
Every finite group of odd order is solvable.
However, I was wondering is there any relation between the fact that every polynomial of odd order has at least one root, with the fact that every finite group of odd order is solvable ?
Edit:
Since I haven't studied the main subject in detail (we just started in Ring theory in the graduate Algebra course that I'm taking)
group-theory polynomials field-theory finite-groups solvable-groups
$endgroup$
|
show 1 more comment
$begingroup$
It is known that there is connection between solvability of a group and expressing the roots of a polynomial by radicals, which is something I will study in this semester; however, by the odd order theorem
Every finite group of odd order is solvable.
However, I was wondering is there any relation between the fact that every polynomial of odd order has at least one root, with the fact that every finite group of odd order is solvable ?
Edit:
Since I haven't studied the main subject in detail (we just started in Ring theory in the graduate Algebra course that I'm taking)
group-theory polynomials field-theory finite-groups solvable-groups
$endgroup$
2
$begingroup$
Just a coincidence. "Odd order" for groups is the number of elements. For polynomials it's the degree. The assertion for polynomials doesn't help find algebraic solutions - it depends on real analysis.
$endgroup$
– Ethan Bolker
Dec 3 '18 at 17:19
2
$begingroup$
Nothing apparent. The statement regarding polynomials is topological, not algebraic. It isn't true over $mathbb Q$, say. $x^3-2$ has no rational root. It is true over $mathbb R$ because it goes to $pm infty$ as $x$ gets very large or very small.
$endgroup$
– lulu
Dec 3 '18 at 17:19
3
$begingroup$
Not really - equations of odd order having a root is an elementary result in real analysis, whereas the solvability of groups of odd order is a very deep and fiendishly difficult result in abstract algebra.
$endgroup$
– Derek Holt
Dec 3 '18 at 17:20
1
$begingroup$
I'll also add that the Galois groups of odd-order polynomials can very often be even-order.
$endgroup$
– Hempelicious
Dec 4 '18 at 0:59
1
$begingroup$
It's a consequence of continuity. Any continuous function from $mathbb R to mathbb R$ that takes positive and negative values must have a real root. Don't need it to be a polynomial. As I say, the parallel claim is false over $mathbb Q$.
$endgroup$
– lulu
Dec 5 '18 at 18:32
|
show 1 more comment
$begingroup$
It is known that there is connection between solvability of a group and expressing the roots of a polynomial by radicals, which is something I will study in this semester; however, by the odd order theorem
Every finite group of odd order is solvable.
However, I was wondering is there any relation between the fact that every polynomial of odd order has at least one root, with the fact that every finite group of odd order is solvable ?
Edit:
Since I haven't studied the main subject in detail (we just started in Ring theory in the graduate Algebra course that I'm taking)
group-theory polynomials field-theory finite-groups solvable-groups
$endgroup$
It is known that there is connection between solvability of a group and expressing the roots of a polynomial by radicals, which is something I will study in this semester; however, by the odd order theorem
Every finite group of odd order is solvable.
However, I was wondering is there any relation between the fact that every polynomial of odd order has at least one root, with the fact that every finite group of odd order is solvable ?
Edit:
Since I haven't studied the main subject in detail (we just started in Ring theory in the graduate Algebra course that I'm taking)
group-theory polynomials field-theory finite-groups solvable-groups
group-theory polynomials field-theory finite-groups solvable-groups
edited Dec 3 '18 at 17:19
onurcanbektas
asked Dec 3 '18 at 17:17
onurcanbektasonurcanbektas
3,37611036
3,37611036
2
$begingroup$
Just a coincidence. "Odd order" for groups is the number of elements. For polynomials it's the degree. The assertion for polynomials doesn't help find algebraic solutions - it depends on real analysis.
$endgroup$
– Ethan Bolker
Dec 3 '18 at 17:19
2
$begingroup$
Nothing apparent. The statement regarding polynomials is topological, not algebraic. It isn't true over $mathbb Q$, say. $x^3-2$ has no rational root. It is true over $mathbb R$ because it goes to $pm infty$ as $x$ gets very large or very small.
$endgroup$
– lulu
Dec 3 '18 at 17:19
3
$begingroup$
Not really - equations of odd order having a root is an elementary result in real analysis, whereas the solvability of groups of odd order is a very deep and fiendishly difficult result in abstract algebra.
$endgroup$
– Derek Holt
Dec 3 '18 at 17:20
1
$begingroup$
I'll also add that the Galois groups of odd-order polynomials can very often be even-order.
$endgroup$
– Hempelicious
Dec 4 '18 at 0:59
1
$begingroup$
It's a consequence of continuity. Any continuous function from $mathbb R to mathbb R$ that takes positive and negative values must have a real root. Don't need it to be a polynomial. As I say, the parallel claim is false over $mathbb Q$.
$endgroup$
– lulu
Dec 5 '18 at 18:32
|
show 1 more comment
2
$begingroup$
Just a coincidence. "Odd order" for groups is the number of elements. For polynomials it's the degree. The assertion for polynomials doesn't help find algebraic solutions - it depends on real analysis.
$endgroup$
– Ethan Bolker
Dec 3 '18 at 17:19
2
$begingroup$
Nothing apparent. The statement regarding polynomials is topological, not algebraic. It isn't true over $mathbb Q$, say. $x^3-2$ has no rational root. It is true over $mathbb R$ because it goes to $pm infty$ as $x$ gets very large or very small.
$endgroup$
– lulu
Dec 3 '18 at 17:19
3
$begingroup$
Not really - equations of odd order having a root is an elementary result in real analysis, whereas the solvability of groups of odd order is a very deep and fiendishly difficult result in abstract algebra.
$endgroup$
– Derek Holt
Dec 3 '18 at 17:20
1
$begingroup$
I'll also add that the Galois groups of odd-order polynomials can very often be even-order.
$endgroup$
– Hempelicious
Dec 4 '18 at 0:59
1
$begingroup$
It's a consequence of continuity. Any continuous function from $mathbb R to mathbb R$ that takes positive and negative values must have a real root. Don't need it to be a polynomial. As I say, the parallel claim is false over $mathbb Q$.
$endgroup$
– lulu
Dec 5 '18 at 18:32
2
2
$begingroup$
Just a coincidence. "Odd order" for groups is the number of elements. For polynomials it's the degree. The assertion for polynomials doesn't help find algebraic solutions - it depends on real analysis.
$endgroup$
– Ethan Bolker
Dec 3 '18 at 17:19
$begingroup$
Just a coincidence. "Odd order" for groups is the number of elements. For polynomials it's the degree. The assertion for polynomials doesn't help find algebraic solutions - it depends on real analysis.
$endgroup$
– Ethan Bolker
Dec 3 '18 at 17:19
2
2
$begingroup$
Nothing apparent. The statement regarding polynomials is topological, not algebraic. It isn't true over $mathbb Q$, say. $x^3-2$ has no rational root. It is true over $mathbb R$ because it goes to $pm infty$ as $x$ gets very large or very small.
$endgroup$
– lulu
Dec 3 '18 at 17:19
$begingroup$
Nothing apparent. The statement regarding polynomials is topological, not algebraic. It isn't true over $mathbb Q$, say. $x^3-2$ has no rational root. It is true over $mathbb R$ because it goes to $pm infty$ as $x$ gets very large or very small.
$endgroup$
– lulu
Dec 3 '18 at 17:19
3
3
$begingroup$
Not really - equations of odd order having a root is an elementary result in real analysis, whereas the solvability of groups of odd order is a very deep and fiendishly difficult result in abstract algebra.
$endgroup$
– Derek Holt
Dec 3 '18 at 17:20
$begingroup$
Not really - equations of odd order having a root is an elementary result in real analysis, whereas the solvability of groups of odd order is a very deep and fiendishly difficult result in abstract algebra.
$endgroup$
– Derek Holt
Dec 3 '18 at 17:20
1
1
$begingroup$
I'll also add that the Galois groups of odd-order polynomials can very often be even-order.
$endgroup$
– Hempelicious
Dec 4 '18 at 0:59
$begingroup$
I'll also add that the Galois groups of odd-order polynomials can very often be even-order.
$endgroup$
– Hempelicious
Dec 4 '18 at 0:59
1
1
$begingroup$
It's a consequence of continuity. Any continuous function from $mathbb R to mathbb R$ that takes positive and negative values must have a real root. Don't need it to be a polynomial. As I say, the parallel claim is false over $mathbb Q$.
$endgroup$
– lulu
Dec 5 '18 at 18:32
$begingroup$
It's a consequence of continuity. Any continuous function from $mathbb R to mathbb R$ that takes positive and negative values must have a real root. Don't need it to be a polynomial. As I say, the parallel claim is false over $mathbb Q$.
$endgroup$
– lulu
Dec 5 '18 at 18:32
|
show 1 more comment
0
active
oldest
votes
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%2f3024363%2frelation-between-odd-order-theorem-and-the-fact-that-every-polynomial-of-odd-ord%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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%2f3024363%2frelation-between-odd-order-theorem-and-the-fact-that-every-polynomial-of-odd-ord%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
2
$begingroup$
Just a coincidence. "Odd order" for groups is the number of elements. For polynomials it's the degree. The assertion for polynomials doesn't help find algebraic solutions - it depends on real analysis.
$endgroup$
– Ethan Bolker
Dec 3 '18 at 17:19
2
$begingroup$
Nothing apparent. The statement regarding polynomials is topological, not algebraic. It isn't true over $mathbb Q$, say. $x^3-2$ has no rational root. It is true over $mathbb R$ because it goes to $pm infty$ as $x$ gets very large or very small.
$endgroup$
– lulu
Dec 3 '18 at 17:19
3
$begingroup$
Not really - equations of odd order having a root is an elementary result in real analysis, whereas the solvability of groups of odd order is a very deep and fiendishly difficult result in abstract algebra.
$endgroup$
– Derek Holt
Dec 3 '18 at 17:20
1
$begingroup$
I'll also add that the Galois groups of odd-order polynomials can very often be even-order.
$endgroup$
– Hempelicious
Dec 4 '18 at 0:59
1
$begingroup$
It's a consequence of continuity. Any continuous function from $mathbb R to mathbb R$ that takes positive and negative values must have a real root. Don't need it to be a polynomial. As I say, the parallel claim is false over $mathbb Q$.
$endgroup$
– lulu
Dec 5 '18 at 18:32