When do permutations commute?
6
$begingroup$
When do two permutations commute?
How do you find out something like how many permutations in $S_7$ commute with $(12)(345)$?
abstract-algebra permutations
asked Nov 13 '17 at 23:45
ZazZaz
5581826
$endgroup$
|
show 1 more comment
6
$begingroup$
When do two permutations commute?
How do you find out something like how many permutations in $S_7$ commute with $(12)(345)$?
abstract-algebra permutations
asked Nov 13 '17 at 23:45
ZazZaz
5581826
$endgroup$
6
$begingroup$
Conjugate permutations hardly ever commute.
$endgroup$
– user228113
Nov 13 '17 at 23:50
$begingroup$
$ ab = ba iff b = a^{-1}ba $ ?
$endgroup$
– Zaz
Nov 13 '17 at 23:52
$begingroup$
I second what you just wrote: $b$ commutes with $a$ if and only if the conjugate of $b$ with respect to $a$ is $b$ itself.
$endgroup$
– user228113
Nov 13 '17 at 23:54
$begingroup$
@G.Sassatelli: Aah, I see I've misunderstood the meaning of conjugate. What is the criteria for when permutations commute then?
$endgroup$
– Zaz
Nov 13 '17 at 23:58
3
$begingroup$
The question is a duplicate of this.
$endgroup$
– Alex Ravsky
Nov 23 '17 at 3:45
|
show 1 more comment
6
6
6
3
$begingroup$
When do two permutations commute?
How do you find out something like how many permutations in $S_7$ commute with $(12)(345)$?
abstract-algebra permutations
asked Nov 13 '17 at 23:45
ZazZaz
5581826
$endgroup$
When do two permutations commute?
How do you find out something like how many permutations in $S_7$ commute with $(12)(345)$?
abstract-algebra permutations
abstract-algebra permutations
asked Nov 13 '17 at 23:45
ZazZaz
5581826
asked Nov 13 '17 at 23:45
ZazZaz
5581826
edited Nov 22 '17 at 16:55
Zaz
asked Nov 13 '17 at 23:45
ZazZaz
5581826
asked Nov 13 '17 at 23:45
ZazZaz
5581826
asked Nov 13 '17 at 23:45
ZazZaz
5581826
5581826
6
$begingroup$
Conjugate permutations hardly ever commute.
$endgroup$
– user228113
Nov 13 '17 at 23:50
$begingroup$
$ ab = ba iff b = a^{-1}ba $ ?
$endgroup$
– Zaz
Nov 13 '17 at 23:52
$begingroup$
I second what you just wrote: $b$ commutes with $a$ if and only if the conjugate of $b$ with respect to $a$ is $b$ itself.
$endgroup$
– user228113
Nov 13 '17 at 23:54
$begingroup$
@G.Sassatelli: Aah, I see I've misunderstood the meaning of conjugate. What is the criteria for when permutations commute then?
$endgroup$
– Zaz
Nov 13 '17 at 23:58
3
$begingroup$
The question is a duplicate of this.
$endgroup$
– Alex Ravsky
Nov 23 '17 at 3:45
|
show 1 more comment
6
$begingroup$
Conjugate permutations hardly ever commute.
$endgroup$
– user228113
Nov 13 '17 at 23:50
$begingroup$
$ ab = ba iff b = a^{-1}ba $ ?
$endgroup$
– Zaz
Nov 13 '17 at 23:52
$begingroup$
I second what you just wrote: $b$ commutes with $a$ if and only if the conjugate of $b$ with respect to $a$ is $b$ itself.
$endgroup$
– user228113
Nov 13 '17 at 23:54
$begingroup$
@G.Sassatelli: Aah, I see I've misunderstood the meaning of conjugate. What is the criteria for when permutations commute then?
$endgroup$
– Zaz
Nov 13 '17 at 23:58
3
$begingroup$
The question is a duplicate of this.
$endgroup$
– Alex Ravsky
Nov 23 '17 at 3:45
6
6
$begingroup$
Conjugate permutations hardly ever commute.
$endgroup$
– user228113
Nov 13 '17 at 23:50
$begingroup$
Conjugate permutations hardly ever commute.
$endgroup$
– user228113
Nov 13 '17 at 23:50
$begingroup$
$ ab = ba iff b = a^{-1}ba $ ?
$endgroup$
– Zaz
Nov 13 '17 at 23:52
$begingroup$
$ ab = ba iff b = a^{-1}ba $ ?
$endgroup$
– Zaz
Nov 13 '17 at 23:52
$begingroup$
I second what you just wrote: $b$ commutes with $a$ if and only if the conjugate of $b$ with respect to $a$ is $b$ itself.
$endgroup$
– user228113
Nov 13 '17 at 23:54
$begingroup$
I second what you just wrote: $b$ commutes with $a$ if and only if the conjugate of $b$ with respect to $a$ is $b$ itself.
$endgroup$
– user228113
Nov 13 '17 at 23:54
$begingroup$
@G.Sassatelli: Aah, I see I've misunderstood the meaning of conjugate. What is the criteria for when permutations commute then?
$endgroup$
– Zaz
Nov 13 '17 at 23:58
$begingroup$
@G.Sassatelli: Aah, I see I've misunderstood the meaning of conjugate. What is the criteria for when permutations commute then?
$endgroup$
– Zaz
Nov 13 '17 at 23:58
3
3
$begingroup$
The question is a duplicate of this.
$endgroup$
– Alex Ravsky
Nov 23 '17 at 3:45
$begingroup$
The question is a duplicate of this.
$endgroup$
– Alex Ravsky
Nov 23 '17 at 3:45
|
show 1 more comment
2 Answers
2
active
oldest
votes
2
$begingroup$
Not true at all. For example, the cycles $(1,2,3)$ and $(2,3,4)$ have the same cycle structure but do not commute.
answered Nov 13 '17 at 23:50
Robert IsraelRobert Israel
330k23219473
$endgroup$
$begingroup$
In support of this: $P_{12}P_{23}ne P_{23}P_{12}$ although both elements have the same cycle structure.
$endgroup$
– user160660
Nov 13 '17 at 23:57
add a comment |
2
+50
$begingroup$
The answer below needs an edit, please see discussion below.
In your case (and in this earlier question), all cycles have different lengths, but in general, a permutation may have some equal length cycles in its disjoint cycle decomposition. Permutations $sigma$ and $pi$ commute when $pi$
- permutes elements within disjoint cycles of $sigma$, and/or
- permutes the sets of elements in equal length disjoint cycles of $sigma$.
Klein 4-group is the smallest nontrivial example of the second action. E.g. $(12)(34)$ commutes with $(13)(24)$ because $(13)(24)$ maps the set ${1,2}$ onto set ${3,4}$ and vice versa.
answered Nov 29 '17 at 3:09
Alexander BursteinAlexander Burstein
1,254218
$endgroup$
$begingroup$
is this a necessary and sufficient condition for two permutation to commute? Thank you.
$endgroup$
– GA316
Dec 17 '18 at 10:03
1
$begingroup$
It’s easier to think of the conditions above as those for $pisigmapi^{-1}=sigma$. Also $sigma: imapsto j$ if and only if $pisigmapi^{-1}: pi(i)mapstopi(j)$. So, yes, the above conditions are necessary and sufficient for $pisigmapi^{-1}=sigma$.
$endgroup$
– Alexander Burstein
Dec 17 '18 at 15:25
$begingroup$
Thank a lot for the clarification :)
$endgroup$
– GA316
Dec 17 '18 at 16:19
$begingroup$
I have a doubt. please clarify. a = (124)(356) permutes elements within the cycle b = (123456). ie. It sends 1,2,3,4,5,6 to 1,2,3,4,5,6 . but these two cycles are not commuting. So can you please explain me your term "permutes elements within disjoint cycles of $sigma$"?. Thank a lot.
$endgroup$
– GA316
Dec 19 '18 at 10:35
$begingroup$
You’re right, this doesn’t quite work as desired. I think I meant it to be in a way that preserves the cycles, i.e. for $(123456)$, that would be e.g. $(135)(246)$ or $(14)(25)(36)$. Those are both powers of $(123456)$, so I think this part can be made even more precise.
$endgroup$
– Alexander Burstein
Dec 19 '18 at 15:30
|
show 1 more comment
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Not true at all. For example, the cycles $(1,2,3)$ and $(2,3,4)$ have the same cycle structure but do not commute.
answered Nov 13 '17 at 23:50
Robert IsraelRobert Israel
330k23219473
$endgroup$
$begingroup$
In support of this: $P_{12}P_{23}ne P_{23}P_{12}$ although both elements have the same cycle structure.
$endgroup$
– user160660
Nov 13 '17 at 23:57
add a comment |
2
$begingroup$
Not true at all. For example, the cycles $(1,2,3)$ and $(2,3,4)$ have the same cycle structure but do not commute.
answered Nov 13 '17 at 23:50
Robert IsraelRobert Israel
330k23219473
$endgroup$
$begingroup$
In support of this: $P_{12}P_{23}ne P_{23}P_{12}$ although both elements have the same cycle structure.
$endgroup$
– user160660
Nov 13 '17 at 23:57
add a comment |
2
2
2
$begingroup$
Not true at all. For example, the cycles $(1,2,3)$ and $(2,3,4)$ have the same cycle structure but do not commute.
answered Nov 13 '17 at 23:50
Robert IsraelRobert Israel
330k23219473
$endgroup$
Not true at all. For example, the cycles $(1,2,3)$ and $(2,3,4)$ have the same cycle structure but do not commute.
answered Nov 13 '17 at 23:50
Robert IsraelRobert Israel
330k23219473
answered Nov 13 '17 at 23:50
Robert IsraelRobert Israel
330k23219473
answered Nov 13 '17 at 23:50
Robert IsraelRobert Israel
330k23219473
answered Nov 13 '17 at 23:50
Robert IsraelRobert Israel
330k23219473
330k23219473
$begingroup$
In support of this: $P_{12}P_{23}ne P_{23}P_{12}$ although both elements have the same cycle structure.
$endgroup$
– user160660
Nov 13 '17 at 23:57
add a comment |
$begingroup$
In support of this: $P_{12}P_{23}ne P_{23}P_{12}$ although both elements have the same cycle structure.
$endgroup$
– user160660
Nov 13 '17 at 23:57
$begingroup$
In support of this: $P_{12}P_{23}ne P_{23}P_{12}$ although both elements have the same cycle structure.
$endgroup$
– user160660
Nov 13 '17 at 23:57
$begingroup$
In support of this: $P_{12}P_{23}ne P_{23}P_{12}$ although both elements have the same cycle structure.
$endgroup$
– user160660
Nov 13 '17 at 23:57
add a comment |
2
+50
$begingroup$
The answer below needs an edit, please see discussion below.
In your case (and in this earlier question), all cycles have different lengths, but in general, a permutation may have some equal length cycles in its disjoint cycle decomposition. Permutations $sigma$ and $pi$ commute when $pi$
- permutes elements within disjoint cycles of $sigma$, and/or
- permutes the sets of elements in equal length disjoint cycles of $sigma$.
Klein 4-group is the smallest nontrivial example of the second action. E.g. $(12)(34)$ commutes with $(13)(24)$ because $(13)(24)$ maps the set ${1,2}$ onto set ${3,4}$ and vice versa.
answered Nov 29 '17 at 3:09
Alexander BursteinAlexander Burstein
1,254218
$endgroup$
$begingroup$
is this a necessary and sufficient condition for two permutation to commute? Thank you.
$endgroup$
– GA316
Dec 17 '18 at 10:03
1
$begingroup$
It’s easier to think of the conditions above as those for $pisigmapi^{-1}=sigma$. Also $sigma: imapsto j$ if and only if $pisigmapi^{-1}: pi(i)mapstopi(j)$. So, yes, the above conditions are necessary and sufficient for $pisigmapi^{-1}=sigma$.
$endgroup$
– Alexander Burstein
Dec 17 '18 at 15:25
$begingroup$
Thank a lot for the clarification :)
$endgroup$
– GA316
Dec 17 '18 at 16:19
$begingroup$
I have a doubt. please clarify. a = (124)(356) permutes elements within the cycle b = (123456). ie. It sends 1,2,3,4,5,6 to 1,2,3,4,5,6 . but these two cycles are not commuting. So can you please explain me your term "permutes elements within disjoint cycles of $sigma$"?. Thank a lot.
$endgroup$
– GA316
Dec 19 '18 at 10:35
$begingroup$
You’re right, this doesn’t quite work as desired. I think I meant it to be in a way that preserves the cycles, i.e. for $(123456)$, that would be e.g. $(135)(246)$ or $(14)(25)(36)$. Those are both powers of $(123456)$, so I think this part can be made even more precise.
$endgroup$
– Alexander Burstein
Dec 19 '18 at 15:30
|
show 1 more comment
2
+50
$begingroup$
The answer below needs an edit, please see discussion below.
In your case (and in this earlier question), all cycles have different lengths, but in general, a permutation may have some equal length cycles in its disjoint cycle decomposition. Permutations $sigma$ and $pi$ commute when $pi$
- permutes elements within disjoint cycles of $sigma$, and/or
- permutes the sets of elements in equal length disjoint cycles of $sigma$.
Klein 4-group is the smallest nontrivial example of the second action. E.g. $(12)(34)$ commutes with $(13)(24)$ because $(13)(24)$ maps the set ${1,2}$ onto set ${3,4}$ and vice versa.
answered Nov 29 '17 at 3:09
Alexander BursteinAlexander Burstein
1,254218
$endgroup$
$begingroup$
is this a necessary and sufficient condition for two permutation to commute? Thank you.
$endgroup$
– GA316
Dec 17 '18 at 10:03
1
$begingroup$
It’s easier to think of the conditions above as those for $pisigmapi^{-1}=sigma$. Also $sigma: imapsto j$ if and only if $pisigmapi^{-1}: pi(i)mapstopi(j)$. So, yes, the above conditions are necessary and sufficient for $pisigmapi^{-1}=sigma$.
$endgroup$
– Alexander Burstein
Dec 17 '18 at 15:25
$begingroup$
Thank a lot for the clarification :)
$endgroup$
– GA316
Dec 17 '18 at 16:19
$begingroup$
I have a doubt. please clarify. a = (124)(356) permutes elements within the cycle b = (123456). ie. It sends 1,2,3,4,5,6 to 1,2,3,4,5,6 . but these two cycles are not commuting. So can you please explain me your term "permutes elements within disjoint cycles of $sigma$"?. Thank a lot.
$endgroup$
– GA316
Dec 19 '18 at 10:35
$begingroup$
You’re right, this doesn’t quite work as desired. I think I meant it to be in a way that preserves the cycles, i.e. for $(123456)$, that would be e.g. $(135)(246)$ or $(14)(25)(36)$. Those are both powers of $(123456)$, so I think this part can be made even more precise.
$endgroup$
– Alexander Burstein
Dec 19 '18 at 15:30
|
show 1 more comment
2
+50
2
+50
2
+50
$begingroup$
The answer below needs an edit, please see discussion below.
In your case (and in this earlier question), all cycles have different lengths, but in general, a permutation may have some equal length cycles in its disjoint cycle decomposition. Permutations $sigma$ and $pi$ commute when $pi$
- permutes elements within disjoint cycles of $sigma$, and/or
- permutes the sets of elements in equal length disjoint cycles of $sigma$.
Klein 4-group is the smallest nontrivial example of the second action. E.g. $(12)(34)$ commutes with $(13)(24)$ because $(13)(24)$ maps the set ${1,2}$ onto set ${3,4}$ and vice versa.
answered Nov 29 '17 at 3:09
Alexander BursteinAlexander Burstein
1,254218
$endgroup$
The answer below needs an edit, please see discussion below.
In your case (and in this earlier question), all cycles have different lengths, but in general, a permutation may have some equal length cycles in its disjoint cycle decomposition. Permutations $sigma$ and $pi$ commute when $pi$
- permutes elements within disjoint cycles of $sigma$, and/or
- permutes the sets of elements in equal length disjoint cycles of $sigma$.
Klein 4-group is the smallest nontrivial example of the second action. E.g. $(12)(34)$ commutes with $(13)(24)$ because $(13)(24)$ maps the set ${1,2}$ onto set ${3,4}$ and vice versa.
answered Nov 29 '17 at 3:09
Alexander BursteinAlexander Burstein
1,254218
edited Dec 20 '18 at 3:27
answered Nov 29 '17 at 3:09
Alexander BursteinAlexander Burstein
1,254218
answered Nov 29 '17 at 3:09
Alexander BursteinAlexander Burstein
1,254218
answered Nov 29 '17 at 3:09
Alexander BursteinAlexander Burstein
1,254218
1,254218
$begingroup$
is this a necessary and sufficient condition for two permutation to commute? Thank you.
$endgroup$
– GA316
Dec 17 '18 at 10:03
1
$begingroup$
It’s easier to think of the conditions above as those for $pisigmapi^{-1}=sigma$. Also $sigma: imapsto j$ if and only if $pisigmapi^{-1}: pi(i)mapstopi(j)$. So, yes, the above conditions are necessary and sufficient for $pisigmapi^{-1}=sigma$.
$endgroup$
– Alexander Burstein
Dec 17 '18 at 15:25
$begingroup$
Thank a lot for the clarification :)
$endgroup$
– GA316
Dec 17 '18 at 16:19
$begingroup$
I have a doubt. please clarify. a = (124)(356) permutes elements within the cycle b = (123456). ie. It sends 1,2,3,4,5,6 to 1,2,3,4,5,6 . but these two cycles are not commuting. So can you please explain me your term "permutes elements within disjoint cycles of $sigma$"?. Thank a lot.
$endgroup$
– GA316
Dec 19 '18 at 10:35
$begingroup$
You’re right, this doesn’t quite work as desired. I think I meant it to be in a way that preserves the cycles, i.e. for $(123456)$, that would be e.g. $(135)(246)$ or $(14)(25)(36)$. Those are both powers of $(123456)$, so I think this part can be made even more precise.
$endgroup$
– Alexander Burstein
Dec 19 '18 at 15:30
|
show 1 more comment
$begingroup$
is this a necessary and sufficient condition for two permutation to commute? Thank you.
$endgroup$
– GA316
Dec 17 '18 at 10:03
1
$begingroup$
It’s easier to think of the conditions above as those for $pisigmapi^{-1}=sigma$. Also $sigma: imapsto j$ if and only if $pisigmapi^{-1}: pi(i)mapstopi(j)$. So, yes, the above conditions are necessary and sufficient for $pisigmapi^{-1}=sigma$.
$endgroup$
– Alexander Burstein
Dec 17 '18 at 15:25
$begingroup$
Thank a lot for the clarification :)
$endgroup$
– GA316
Dec 17 '18 at 16:19
$begingroup$
I have a doubt. please clarify. a = (124)(356) permutes elements within the cycle b = (123456). ie. It sends 1,2,3,4,5,6 to 1,2,3,4,5,6 . but these two cycles are not commuting. So can you please explain me your term "permutes elements within disjoint cycles of $sigma$"?. Thank a lot.
$endgroup$
– GA316
Dec 19 '18 at 10:35
$begingroup$
You’re right, this doesn’t quite work as desired. I think I meant it to be in a way that preserves the cycles, i.e. for $(123456)$, that would be e.g. $(135)(246)$ or $(14)(25)(36)$. Those are both powers of $(123456)$, so I think this part can be made even more precise.
$endgroup$
– Alexander Burstein
Dec 19 '18 at 15:30
$begingroup$
is this a necessary and sufficient condition for two permutation to commute? Thank you.
$endgroup$
– GA316
Dec 17 '18 at 10:03
$begingroup$
is this a necessary and sufficient condition for two permutation to commute? Thank you.
$endgroup$
– GA316
Dec 17 '18 at 10:03
1
1
$begingroup$
It’s easier to think of the conditions above as those for $pisigmapi^{-1}=sigma$. Also $sigma: imapsto j$ if and only if $pisigmapi^{-1}: pi(i)mapstopi(j)$. So, yes, the above conditions are necessary and sufficient for $pisigmapi^{-1}=sigma$.
$endgroup$
– Alexander Burstein
Dec 17 '18 at 15:25
$begingroup$
It’s easier to think of the conditions above as those for $pisigmapi^{-1}=sigma$. Also $sigma: imapsto j$ if and only if $pisigmapi^{-1}: pi(i)mapstopi(j)$. So, yes, the above conditions are necessary and sufficient for $pisigmapi^{-1}=sigma$.
$endgroup$
– Alexander Burstein
Dec 17 '18 at 15:25
$begingroup$
Thank a lot for the clarification :)
$endgroup$
– GA316
Dec 17 '18 at 16:19
$begingroup$
Thank a lot for the clarification :)
$endgroup$
– GA316
Dec 17 '18 at 16:19
$begingroup$
I have a doubt. please clarify. a = (124)(356) permutes elements within the cycle b = (123456). ie. It sends 1,2,3,4,5,6 to 1,2,3,4,5,6 . but these two cycles are not commuting. So can you please explain me your term "permutes elements within disjoint cycles of $sigma$"?. Thank a lot.
$endgroup$
– GA316
Dec 19 '18 at 10:35
$begingroup$
I have a doubt. please clarify. a = (124)(356) permutes elements within the cycle b = (123456). ie. It sends 1,2,3,4,5,6 to 1,2,3,4,5,6 . but these two cycles are not commuting. So can you please explain me your term "permutes elements within disjoint cycles of $sigma$"?. Thank a lot.
$endgroup$
– GA316
Dec 19 '18 at 10:35
$begingroup$
You’re right, this doesn’t quite work as desired. I think I meant it to be in a way that preserves the cycles, i.e. for $(123456)$, that would be e.g. $(135)(246)$ or $(14)(25)(36)$. Those are both powers of $(123456)$, so I think this part can be made even more precise.
$endgroup$
– Alexander Burstein
Dec 19 '18 at 15:30
$begingroup$
You’re right, this doesn’t quite work as desired. I think I meant it to be in a way that preserves the cycles, i.e. for $(123456)$, that would be e.g. $(135)(246)$ or $(14)(25)(36)$. Those are both powers of $(123456)$, so I think this part can be made even more precise.
$endgroup$
– Alexander Burstein
Dec 19 '18 at 15:30
|
show 1 more comment
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6
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Conjugate permutations hardly ever commute.
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– user228113
Nov 13 '17 at 23:50
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$ ab = ba iff b = a^{-1}ba $ ?
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– Zaz
Nov 13 '17 at 23:52
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I second what you just wrote: $b$ commutes with $a$ if and only if the conjugate of $b$ with respect to $a$ is $b$ itself.
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– user228113
Nov 13 '17 at 23:54
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@G.Sassatelli: Aah, I see I've misunderstood the meaning of conjugate. What is the criteria for when permutations commute then?
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– Zaz
Nov 13 '17 at 23:58
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The question is a duplicate of this.
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– Alex Ravsky
Nov 23 '17 at 3:45