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 6 hours ago
HarmanHarman
122
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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 6 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
6 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 6 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 6 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 6 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 6 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 6 hours ago
HarmanHarman
122
asked 6 hours ago
HarmanHarman
122
122
New contributor
Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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.
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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
6 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
6 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
6 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
6 hours ago
add a comment |
2 Answers
2
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$begingroup$
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 6 hours ago
jmerryjmerry
3,737514
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add a comment |
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$begingroup$
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 6 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 6 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 6 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 6 hours ago
jmerryjmerry
3,737514
answered 6 hours ago
jmerryjmerry
3,737514
answered 6 hours ago
jmerryjmerry
3,737514
answered 6 hours ago
jmerryjmerry
3,737514
3,737514
add a comment |
add a comment |
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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
6 hours ago