Intersection of two left cosets is empty
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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 Jan 15 at 22:54
HarmanHarman
302
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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 Jan 15 at 22:54
HarmanHarman
302
$endgroup$
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
Jan 15 at 22:56
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 Jan 15 at 22:54
HarmanHarman
302
$endgroup$
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 Jan 15 at 22:54
HarmanHarman
302
asked Jan 15 at 22:54
HarmanHarman
302
asked Jan 15 at 22:54
HarmanHarman
302
asked Jan 15 at 22:54
HarmanHarman
302
asked Jan 15 at 22:54
HarmanHarman
302
302
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
Jan 15 at 22:56
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
Jan 15 at 22:56
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
Jan 15 at 22:56
$begingroup$
Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
$endgroup$
– darij grinberg
Jan 15 at 22:56
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:
$$textFor;x,yin G;,;; xsim yiff x^-1yin H$$
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 $;xH = yH;;textor;;xHcap yH=emptyset;$
answered Jan 15 at 22:59
DonAntonioDonAntonio
178k1493228
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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 Jan 15 at 22:57
jmerryjmerry
6,457718
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add a comment |
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2 Answers
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2 Answers
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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:
$$textFor;x,yin G;,;; xsim yiff x^-1yin H$$
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 $;xH = yH;;textor;;xHcap yH=emptyset;$
answered Jan 15 at 22:59
DonAntonioDonAntonio
178k1493228
$endgroup$
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:
$$textFor;x,yin G;,;; xsim yiff x^-1yin H$$
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 $;xH = yH;;textor;;xHcap yH=emptyset;$
answered Jan 15 at 22:59
DonAntonioDonAntonio
178k1493228
$endgroup$
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:
$$textFor;x,yin G;,;; xsim yiff x^-1yin H$$
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 $;xH = yH;;textor;;xHcap yH=emptyset;$
answered Jan 15 at 22:59
DonAntonioDonAntonio
178k1493228
$endgroup$
Left (or right) cosets of a subgroup $;H;$ in a group $;G;$ are in fact equivalence classes, the equivalence relation $;sim;$ on $;G;$ being:
$$textFor;x,yin G;,;; xsim yiff x^-1yin H$$
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 $;xH = yH;;textor;;xHcap yH=emptyset;$
answered Jan 15 at 22:59
DonAntonioDonAntonio
178k1493228
answered Jan 15 at 22:59
DonAntonioDonAntonio
178k1493228
answered Jan 15 at 22:59
DonAntonioDonAntonio
178k1493228
answered Jan 15 at 22:59
DonAntonioDonAntonio
178k1493228
178k1493228
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 Jan 15 at 22:57
jmerryjmerry
6,457718
$endgroup$
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 Jan 15 at 22:57
jmerryjmerry
6,457718
$endgroup$
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 Jan 15 at 22:57
jmerryjmerry
6,457718
$endgroup$
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 Jan 15 at 22:57
jmerryjmerry
6,457718
answered Jan 15 at 22:57
jmerryjmerry
6,457718
answered Jan 15 at 22:57
jmerryjmerry
6,457718
answered Jan 15 at 22:57
jmerryjmerry
6,457718
6,457718
add a comment |
add a comment |
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$begingroup$
Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
$endgroup$
– darij grinberg
Jan 15 at 22:56