Intersection of two left cosets is empty
$begingroup$
I have $theta,tau, sigmain S_7$, where $theta=(13)(14)(567), tau=(137)(2456),
sigma=(13456)(27)$. I need to find $sigma langlethetaranglecaptaulanglethetarangle$.
I approached this by first computing $langlethetarangle$. I got $langlethetarangle={(13)(24)(567),(576),(13)(24),(675),(13)(24)(576),id}$.
I then worked out the left coset $sigmalanglethetarangle$ by multiplying $(13456)(27)$ by every element in $langlethetarangle$, one by one. That gave me ${(1574632),(134726),(274563),(1346)(257),(263)(457)}$.
Using the same technique, I found $taulanglethetarangle$ to be ${(357)(46),(136247),(37)(456),(1357)(246),(3647),id}$.
But then I get $sigmalanglethetaranglecaptaulanglethetarangle=phi$, since there are no elements occurring in both $sigmalanglethetarangle$ and $taulanglethetarangle$.
Why is the intersection empty? If I have done this right, is there an easier way to tell that the intersection of two cosets will be empty, without having to work out each individual result (since that is tedious)?
abstract-algebra group-theory
asked 7 hours ago
HarmanHarman
122
New contributor
Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have $theta,tau, sigmain S_7$, where $theta=(13)(14)(567), tau=(137)(2456),
sigma=(13456)(27)$. I need to find $sigma langlethetaranglecaptaulanglethetarangle$.
I approached this by first computing $langlethetarangle$. I got $langlethetarangle={(13)(24)(567),(576),(13)(24),(675),(13)(24)(576),id}$.
I then worked out the left coset $sigmalanglethetarangle$ by multiplying $(13456)(27)$ by every element in $langlethetarangle$, one by one. That gave me ${(1574632),(134726),(274563),(1346)(257),(263)(457)}$.
Using the same technique, I found $taulanglethetarangle$ to be ${(357)(46),(136247),(37)(456),(1357)(246),(3647),id}$.
But then I get $sigmalanglethetaranglecaptaulanglethetarangle=phi$, since there are no elements occurring in both $sigmalanglethetarangle$ and $taulanglethetarangle$.
Why is the intersection empty? If I have done this right, is there an easier way to tell that the intersection of two cosets will be empty, without having to work out each individual result (since that is tedious)?
abstract-algebra group-theory
asked 7 hours ago
HarmanHarman
122
New contributor
Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
$endgroup$
– darij grinberg
7 hours ago
add a comment |
$begingroup$
I have $theta,tau, sigmain S_7$, where $theta=(13)(14)(567), tau=(137)(2456),
sigma=(13456)(27)$. I need to find $sigma langlethetaranglecaptaulanglethetarangle$.
I approached this by first computing $langlethetarangle$. I got $langlethetarangle={(13)(24)(567),(576),(13)(24),(675),(13)(24)(576),id}$.
I then worked out the left coset $sigmalanglethetarangle$ by multiplying $(13456)(27)$ by every element in $langlethetarangle$, one by one. That gave me ${(1574632),(134726),(274563),(1346)(257),(263)(457)}$.
Using the same technique, I found $taulanglethetarangle$ to be ${(357)(46),(136247),(37)(456),(1357)(246),(3647),id}$.
But then I get $sigmalanglethetaranglecaptaulanglethetarangle=phi$, since there are no elements occurring in both $sigmalanglethetarangle$ and $taulanglethetarangle$.
Why is the intersection empty? If I have done this right, is there an easier way to tell that the intersection of two cosets will be empty, without having to work out each individual result (since that is tedious)?
abstract-algebra group-theory
asked 7 hours ago
HarmanHarman
122
New contributor
Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have $theta,tau, sigmain S_7$, where $theta=(13)(14)(567), tau=(137)(2456),
sigma=(13456)(27)$. I need to find $sigma langlethetaranglecaptaulanglethetarangle$.
I approached this by first computing $langlethetarangle$. I got $langlethetarangle={(13)(24)(567),(576),(13)(24),(675),(13)(24)(576),id}$.
I then worked out the left coset $sigmalanglethetarangle$ by multiplying $(13456)(27)$ by every element in $langlethetarangle$, one by one. That gave me ${(1574632),(134726),(274563),(1346)(257),(263)(457)}$.
Using the same technique, I found $taulanglethetarangle$ to be ${(357)(46),(136247),(37)(456),(1357)(246),(3647),id}$.
But then I get $sigmalanglethetaranglecaptaulanglethetarangle=phi$, since there are no elements occurring in both $sigmalanglethetarangle$ and $taulanglethetarangle$.
Why is the intersection empty? If I have done this right, is there an easier way to tell that the intersection of two cosets will be empty, without having to work out each individual result (since that is tedious)?
abstract-algebra group-theory
abstract-algebra group-theory
asked 7 hours ago
HarmanHarman
122
New contributor
Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
HarmanHarman
122
New contributor
Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
HarmanHarman
122
New contributor
Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
HarmanHarman
122
asked 7 hours ago
HarmanHarman
122
122
New contributor
Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
$endgroup$
– darij grinberg
7 hours ago
add a comment |
2
$begingroup$
Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
$endgroup$
– darij grinberg
7 hours ago
2
2
$begingroup$
Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
$endgroup$
– darij grinberg
7 hours ago
$begingroup$
Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
$endgroup$
– darij grinberg
7 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Left (or right) cosets of a subgroup $;H;$ in a group $;G;$ are in fact equivalence classes, the equivalence relation $;sim;$ on $;G;$ being:
$$text{For};x,yin G;,;; xsim yiff x^{-1}yin H$$
We usually write $;xH=yH;$ instead of $;xsim y;$ . Check this is in fact an equivalence relation, and then we know two equivalence classes are either identical or else they're disjoint, and this just means $;xH = yH;;text{or};;xHcap yH=emptyset;$
answered 6 hours ago
DonAntonioDonAntonio
177k1492226
$endgroup$
add a comment |
$begingroup$
Yes, this is standard. Given two left cosets of the same subgroup, either they're the same as sets, or they don't intersect at all.
As such, all we need to do to find that two cosets of a subgroup are disjoint is to find a single element of one coset that's not in the other coset. An example that comes up fairly often: if $a$ is not an element of the subgroup $H$, the coset $aH$ doesn't intersect $H$.
answered 7 hours ago
jmerryjmerry
3,737514
$endgroup$
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
});
}
});
Harman is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3075083%2fintersection-of-two-left-cosets-is-empty%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Left (or right) cosets of a subgroup $;H;$ in a group $;G;$ are in fact equivalence classes, the equivalence relation $;sim;$ on $;G;$ being:
$$text{For};x,yin G;,;; xsim yiff x^{-1}yin H$$
We usually write $;xH=yH;$ instead of $;xsim y;$ . Check this is in fact an equivalence relation, and then we know two equivalence classes are either identical or else they're disjoint, and this just means $;xH = yH;;text{or};;xHcap yH=emptyset;$
answered 6 hours ago
DonAntonioDonAntonio
177k1492226
$endgroup$
add a comment |
$begingroup$
Left (or right) cosets of a subgroup $;H;$ in a group $;G;$ are in fact equivalence classes, the equivalence relation $;sim;$ on $;G;$ being:
$$text{For};x,yin G;,;; xsim yiff x^{-1}yin H$$
We usually write $;xH=yH;$ instead of $;xsim y;$ . Check this is in fact an equivalence relation, and then we know two equivalence classes are either identical or else they're disjoint, and this just means $;xH = yH;;text{or};;xHcap yH=emptyset;$
answered 6 hours ago
DonAntonioDonAntonio
177k1492226
$endgroup$
add a comment |
$begingroup$
Left (or right) cosets of a subgroup $;H;$ in a group $;G;$ are in fact equivalence classes, the equivalence relation $;sim;$ on $;G;$ being:
$$text{For};x,yin G;,;; xsim yiff x^{-1}yin H$$
We usually write $;xH=yH;$ instead of $;xsim y;$ . Check this is in fact an equivalence relation, and then we know two equivalence classes are either identical or else they're disjoint, and this just means $;xH = yH;;text{or};;xHcap yH=emptyset;$
answered 6 hours ago
DonAntonioDonAntonio
177k1492226
$endgroup$
Left (or right) cosets of a subgroup $;H;$ in a group $;G;$ are in fact equivalence classes, the equivalence relation $;sim;$ on $;G;$ being:
$$text{For};x,yin G;,;; xsim yiff x^{-1}yin H$$
We usually write $;xH=yH;$ instead of $;xsim y;$ . Check this is in fact an equivalence relation, and then we know two equivalence classes are either identical or else they're disjoint, and this just means $;xH = yH;;text{or};;xHcap yH=emptyset;$
answered 6 hours ago
DonAntonioDonAntonio
177k1492226
answered 6 hours ago
DonAntonioDonAntonio
177k1492226
answered 6 hours ago
DonAntonioDonAntonio
177k1492226
answered 6 hours ago
DonAntonioDonAntonio
177k1492226
177k1492226
add a comment |
add a comment |
$begingroup$
Yes, this is standard. Given two left cosets of the same subgroup, either they're the same as sets, or they don't intersect at all.
As such, all we need to do to find that two cosets of a subgroup are disjoint is to find a single element of one coset that's not in the other coset. An example that comes up fairly often: if $a$ is not an element of the subgroup $H$, the coset $aH$ doesn't intersect $H$.
answered 7 hours ago
jmerryjmerry
3,737514
$endgroup$
add a comment |
$begingroup$
Yes, this is standard. Given two left cosets of the same subgroup, either they're the same as sets, or they don't intersect at all.
As such, all we need to do to find that two cosets of a subgroup are disjoint is to find a single element of one coset that's not in the other coset. An example that comes up fairly often: if $a$ is not an element of the subgroup $H$, the coset $aH$ doesn't intersect $H$.
answered 7 hours ago
jmerryjmerry
3,737514
$endgroup$
add a comment |
$begingroup$
Yes, this is standard. Given two left cosets of the same subgroup, either they're the same as sets, or they don't intersect at all.
As such, all we need to do to find that two cosets of a subgroup are disjoint is to find a single element of one coset that's not in the other coset. An example that comes up fairly often: if $a$ is not an element of the subgroup $H$, the coset $aH$ doesn't intersect $H$.
answered 7 hours ago
jmerryjmerry
3,737514
$endgroup$
Yes, this is standard. Given two left cosets of the same subgroup, either they're the same as sets, or they don't intersect at all.
As such, all we need to do to find that two cosets of a subgroup are disjoint is to find a single element of one coset that's not in the other coset. An example that comes up fairly often: if $a$ is not an element of the subgroup $H$, the coset $aH$ doesn't intersect $H$.
answered 7 hours ago
jmerryjmerry
3,737514
answered 7 hours ago
jmerryjmerry
3,737514
answered 7 hours ago
jmerryjmerry
3,737514
answered 7 hours ago
jmerryjmerry
3,737514
3,737514
add a comment |
add a comment |
Harman is a new contributor. Be nice, and check out our Code of Conduct.
Harman is a new contributor. Be nice, and check out our Code of Conduct.
Harman is a new contributor. Be nice, and check out our Code of Conduct.
Harman is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3075083%2fintersection-of-two-left-cosets-is-empty%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
$begingroup$
Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
$endgroup$
– darij grinberg
7 hours ago