Elements of prime order in symmetric groups
I have been assigned the following task:
“Explain briefly why there can be no odd elements of prime order except for order $2$ in any symmetric group $S_n$.”
Unless I am largely mistaken in my understanding of the notions of order and the sign of a permutation, I have come up with the following:
$[3,3,3,1]$ is an element of $S_{10}$ with order $3$, and is even.
$[7, 1, 1]$ is an element of $S_9$ with order $7$, and is odd.
Am I mistaken in thinking that this shows that not only do there exist odd elements of prime order $p≠2$, but also that there exist even elements of order $2$ (the former contradicting the statement in the question)?
abstract-algebra group-theory
asked Nov 29 at 18:50
TomHanney
174
add a comment |
I have been assigned the following task:
“Explain briefly why there can be no odd elements of prime order except for order $2$ in any symmetric group $S_n$.”
Unless I am largely mistaken in my understanding of the notions of order and the sign of a permutation, I have come up with the following:
$[3,3,3,1]$ is an element of $S_{10}$ with order $3$, and is even.
$[7, 1, 1]$ is an element of $S_9$ with order $7$, and is odd.
Am I mistaken in thinking that this shows that not only do there exist odd elements of prime order $p≠2$, but also that there exist even elements of order $2$ (the former contradicting the statement in the question)?
abstract-algebra group-theory
asked Nov 29 at 18:50
TomHanney
174
What is this notation: $[3,3,3,1]$?
– Berci
Nov 29 at 19:04
Anyway, it's odd, but a cycle of odd length is an even permutation..
– Berci
Nov 29 at 19:07
1
@Berci It's the lengths of the disjoint cycles.
– verret
Nov 29 at 20:11
add a comment |
I have been assigned the following task:
“Explain briefly why there can be no odd elements of prime order except for order $2$ in any symmetric group $S_n$.”
Unless I am largely mistaken in my understanding of the notions of order and the sign of a permutation, I have come up with the following:
$[3,3,3,1]$ is an element of $S_{10}$ with order $3$, and is even.
$[7, 1, 1]$ is an element of $S_9$ with order $7$, and is odd.
Am I mistaken in thinking that this shows that not only do there exist odd elements of prime order $p≠2$, but also that there exist even elements of order $2$ (the former contradicting the statement in the question)?
abstract-algebra group-theory
asked Nov 29 at 18:50
TomHanney
174
I have been assigned the following task:
“Explain briefly why there can be no odd elements of prime order except for order $2$ in any symmetric group $S_n$.”
Unless I am largely mistaken in my understanding of the notions of order and the sign of a permutation, I have come up with the following:
$[3,3,3,1]$ is an element of $S_{10}$ with order $3$, and is even.
$[7, 1, 1]$ is an element of $S_9$ with order $7$, and is odd.
Am I mistaken in thinking that this shows that not only do there exist odd elements of prime order $p≠2$, but also that there exist even elements of order $2$ (the former contradicting the statement in the question)?
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 29 at 18:50
TomHanney
174
asked Nov 29 at 18:50
TomHanney
174
asked Nov 29 at 18:50
TomHanney
174
asked Nov 29 at 18:50
TomHanney
174
asked Nov 29 at 18:50
TomHanney
174
174
What is this notation: $[3,3,3,1]$?
– Berci
Nov 29 at 19:04
Anyway, it's odd, but a cycle of odd length is an even permutation..
– Berci
Nov 29 at 19:07
1
@Berci It's the lengths of the disjoint cycles.
– verret
Nov 29 at 20:11
add a comment |
What is this notation: $[3,3,3,1]$?
– Berci
Nov 29 at 19:04
Anyway, it's odd, but a cycle of odd length is an even permutation..
– Berci
Nov 29 at 19:07
1
@Berci It's the lengths of the disjoint cycles.
– verret
Nov 29 at 20:11
What is this notation: $[3,3,3,1]$?
– Berci
Nov 29 at 19:04
What is this notation: $[3,3,3,1]$?
– Berci
Nov 29 at 19:04
Anyway, it's odd, but a cycle of odd length is an even permutation..
– Berci
Nov 29 at 19:07
Anyway, it's odd, but a cycle of odd length is an even permutation..
– Berci
Nov 29 at 19:07
1
1
@Berci It's the lengths of the disjoint cycles.
– verret
Nov 29 at 20:11
@Berci It's the lengths of the disjoint cycles.
– verret
Nov 29 at 20:11
add a comment |
1 Answer
1
active
oldest
votes
A $7$-cycle is even because we can break it into transpositions $$(a_1,a_7)(a_1,a_6)(a_1,a_5)(a_1,a_4)(a_1,a_3)(a_1,a_2)$$
And as you can see there are six transpositions, so this permutation is even.
answered Nov 29 at 20:39
CyclotomicField
2,1821313
Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
– TomHanney
Nov 29 at 22:09
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%2f3019025%2felements-of-prime-order-in-symmetric-groups%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
A $7$-cycle is even because we can break it into transpositions $$(a_1,a_7)(a_1,a_6)(a_1,a_5)(a_1,a_4)(a_1,a_3)(a_1,a_2)$$
And as you can see there are six transpositions, so this permutation is even.
answered Nov 29 at 20:39
CyclotomicField
2,1821313
Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
– TomHanney
Nov 29 at 22:09
add a comment |
A $7$-cycle is even because we can break it into transpositions $$(a_1,a_7)(a_1,a_6)(a_1,a_5)(a_1,a_4)(a_1,a_3)(a_1,a_2)$$
And as you can see there are six transpositions, so this permutation is even.
answered Nov 29 at 20:39
CyclotomicField
2,1821313
Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
– TomHanney
Nov 29 at 22:09
add a comment |
A $7$-cycle is even because we can break it into transpositions $$(a_1,a_7)(a_1,a_6)(a_1,a_5)(a_1,a_4)(a_1,a_3)(a_1,a_2)$$
And as you can see there are six transpositions, so this permutation is even.
answered Nov 29 at 20:39
CyclotomicField
2,1821313
A $7$-cycle is even because we can break it into transpositions $$(a_1,a_7)(a_1,a_6)(a_1,a_5)(a_1,a_4)(a_1,a_3)(a_1,a_2)$$
And as you can see there are six transpositions, so this permutation is even.
answered Nov 29 at 20:39
CyclotomicField
2,1821313
answered Nov 29 at 20:39
CyclotomicField
2,1821313
answered Nov 29 at 20:39
CyclotomicField
2,1821313
answered Nov 29 at 20:39
CyclotomicField
2,1821313
2,1821313
Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
– TomHanney
Nov 29 at 22:09
add a comment |
Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
– TomHanney
Nov 29 at 22:09
Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
– TomHanney
Nov 29 at 22:09
Thank you, it was an oversight on my part that I thought that a cycle of length $7$ consisted of $7$ transpositions.
– TomHanney
Nov 29 at 22:09
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%2f3019025%2felements-of-prime-order-in-symmetric-groups%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
What is this notation: $[3,3,3,1]$?
– Berci
Nov 29 at 19:04
Anyway, it's odd, but a cycle of odd length is an even permutation..
– Berci
Nov 29 at 19:07
1
@Berci It's the lengths of the disjoint cycles.
– verret
Nov 29 at 20:11