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Animation: customize bounce interpolation

Writing a Permutation as a product of Disjoint Cycles

How to Write Permutation as the Product of Transpositions?Writing a Permutation as a product of Disjoint CyclesHow to Write Permutation as the Product of Transpositions?Permutation of cycles as transpositionExpressing a permuation as a product of disjoint cycles.Abstract product of disjoint cyclesPermutation written as product of transpositionsIf a permutation is a cycle, can I factorize it as a product of disjoint cycles?Permutation and Disjoint cycles questionProve that every permutation is the product of disjoint cyclesPermutation as disjoint cycles

1

0

$begingroup$

How do i write a permutation as a product of disjoint cycles ?

I know that in order to determine a cycle we need to start with the smallest element and move on till the mapping points to itself.Then start with the next non repeating smallest element..But how to write this as a product of disjoint cycles?

group-theory permutations

share|cite|improve this question

edited Oct 8 '15 at 11:46

Andrés E. Caicedo

65.7k8160251

asked Oct 8 '15 at 7:07

technotechno

3222826

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I suppose it is worth mentioning that disjoint cycles commute, that is, may be multiplied in any order. Hence (4)(1532) is also a valid answer.
  $endgroup$
  – A.Sh
  Oct 8 '15 at 7:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @A.Sh From the method i have described i got (1532)(4).Is this itself the way to get Disjoint cycles?
  $endgroup$
  – techno
  Oct 8 '15 at 7:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, yes, you do (always) end up with disjoint cycles by performing your algorithm. Also, if you do exactly as described, you will end up with (1532)(4), so you have done correctly. Since disjoint cycles are permutations that act on entirely different sets of elements, they can be applied in any order. Just mentioning it, in case you would encounter a similar situation in the future : )
  $endgroup$
  – A.Sh
  Oct 8 '15 at 7:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @A.Sh okay.thanks
  $endgroup$
  – techno
  Oct 8 '15 at 9:38
  

  
  
  
  

add a comment | 

1

0

$begingroup$

How do i write a permutation as a product of disjoint cycles ?

I know that in order to determine a cycle we need to start with the smallest element and move on till the mapping points to itself.Then start with the next non repeating smallest element..But how to write this as a product of disjoint cycles?

group-theory permutations

share|cite|improve this question

edited Oct 8 '15 at 11:46

Andrés E. Caicedo

65.7k8160251

asked Oct 8 '15 at 7:07

technotechno

3222826

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I suppose it is worth mentioning that disjoint cycles commute, that is, may be multiplied in any order. Hence (4)(1532) is also a valid answer.
  $endgroup$
  – A.Sh
  Oct 8 '15 at 7:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @A.Sh From the method i have described i got (1532)(4).Is this itself the way to get Disjoint cycles?
  $endgroup$
  – techno
  Oct 8 '15 at 7:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, yes, you do (always) end up with disjoint cycles by performing your algorithm. Also, if you do exactly as described, you will end up with (1532)(4), so you have done correctly. Since disjoint cycles are permutations that act on entirely different sets of elements, they can be applied in any order. Just mentioning it, in case you would encounter a similar situation in the future : )
  $endgroup$
  – A.Sh
  Oct 8 '15 at 7:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @A.Sh okay.thanks
  $endgroup$
  – techno
  Oct 8 '15 at 9:38
  

  
  
  
  

add a comment | 

$begingroup$

How do i write a permutation as a product of disjoint cycles ?

I know that in order to determine a cycle we need to start with the smallest element and move on till the mapping points to itself.Then start with the next non repeating smallest element..But how to write this as a product of disjoint cycles?

group-theory permutations

share|cite|improve this question

edited Oct 8 '15 at 11:46

Andrés E. Caicedo

65.7k8160251

asked Oct 8 '15 at 7:07

technotechno

3222826

$endgroup$

share|cite|improve this question

edited Oct 8 '15 at 11:46

Andrés E. Caicedo

65.7k8160251

asked Oct 8 '15 at 7:07

technotechno

3222826

share|cite|improve this question

edited Oct 8 '15 at 11:46

Andrés E. Caicedo

65.7k8160251

asked Oct 8 '15 at 7:07

technotechno

3222826

edited Oct 8 '15 at 11:46

Andrés E. Caicedo

65.7k8160251

edited Oct 8 '15 at 11:46

Andrés E. Caicedo

65.7k8160251

asked Oct 8 '15 at 7:07

technotechno

3222826

asked Oct 8 '15 at 7:07

technotechno

3222826

3 Answers
3

active

oldest

votes

5

$begingroup$

First, we note that writing it as a product of disjoint cycles means that each number appears only once throughout all of the cycles.

We see that $1mapsto 5$, $5mapsto 3$, $3mapsto 2$, $2mapsto 1$. So, we can express this in cycle notation as
$$(1532).$$
Now, we see what is left over... well, that is just $4$, which is fixed by the permutation in question. So, the permutation can be written as

$$(1532)(4),mbox or equivalently, just (1532).$$

share|cite|improve this answer

edited Oct 5 '18 at 8:06

answered Oct 8 '15 at 7:14

eloiprimeeloiprime

2,1651721

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks.As i have described in the question what i get by applying is (1532)(4) =>(1532).Is this itself the method to get disjoint cycles?
  $endgroup$
  – techno
  Oct 8 '15 at 7:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's all you need to do :)
  $endgroup$
  – eloiprime
  Oct 8 '15 at 7:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  okay.. so if i use the method i described in the question ie: finding the smallest etc... i will get the product of disjoint cycles.
  $endgroup$
  – techno
  Oct 8 '15 at 7:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I mean, that's all I did above. Wasn't it?
  $endgroup$
  – eloiprime
  Oct 8 '15 at 8:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah...got it now
  $endgroup$
  – techno
  Oct 8 '15 at 9:39
  

  
  
  
  

 | 
show 5 more comments

0

$begingroup$

$4$ is the only one invariant, while the others mingle among them, thus there are only two cycles
$$(1,5,3,2)$$
It's customary not to write any singleton.

share|cite|improve this answer

answered Oct 8 '15 at 7:15

MASLMASL

708313

$endgroup$

add a comment | 

0

$begingroup$

You start with the first number than you go to the number with is under this number and so on until you get back to the start number...... In your case you start with 1 then go to 5 then to 3 to 2 and are back at one,.... So the first cycle is (1;5;2;3)

Then you look at the numbers which aren't in the first cycle e.g. 4 and do the same again.... ( you can, but you do not have to mention cycles with length 1 if it is known in which Sn the permutation is or it is otherwise clear how long the permutation is.)

share|cite|improve this answer

answered Oct 8 '15 at 7:19

BörgeBörge

1,037415

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  okay.. so if i use the method i described in the question ie: finding the smallest etc... i will get the product of disjoint cycles?
  $endgroup$
  – techno
  Oct 8 '15 at 7:23
  
  

  
  
  
  

add a comment | 

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5

$begingroup$

First, we note that writing it as a product of disjoint cycles means that each number appears only once throughout all of the cycles.

We see that $1mapsto 5$, $5mapsto 3$, $3mapsto 2$, $2mapsto 1$. So, we can express this in cycle notation as
$$(1532).$$
Now, we see what is left over... well, that is just $4$, which is fixed by the permutation in question. So, the permutation can be written as

$$(1532)(4),mbox or equivalently, just (1532).$$

share|cite|improve this answer

edited Oct 5 '18 at 8:06

answered Oct 8 '15 at 7:14

eloiprimeeloiprime

2,1651721

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks.As i have described in the question what i get by applying is (1532)(4) =>(1532).Is this itself the method to get disjoint cycles?
  $endgroup$
  – techno
  Oct 8 '15 at 7:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's all you need to do :)
  $endgroup$
  – eloiprime
  Oct 8 '15 at 7:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  okay.. so if i use the method i described in the question ie: finding the smallest etc... i will get the product of disjoint cycles.
  $endgroup$
  – techno
  Oct 8 '15 at 7:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I mean, that's all I did above. Wasn't it?
  $endgroup$
  – eloiprime
  Oct 8 '15 at 8:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah...got it now
  $endgroup$
  – techno
  Oct 8 '15 at 9:39
  

  
  
  
  

 | 
show 5 more comments

5

$begingroup$

First, we note that writing it as a product of disjoint cycles means that each number appears only once throughout all of the cycles.

We see that $1mapsto 5$, $5mapsto 3$, $3mapsto 2$, $2mapsto 1$. So, we can express this in cycle notation as
$$(1532).$$
Now, we see what is left over... well, that is just $4$, which is fixed by the permutation in question. So, the permutation can be written as

$$(1532)(4),mbox or equivalently, just (1532).$$

share|cite|improve this answer

edited Oct 5 '18 at 8:06

answered Oct 8 '15 at 7:14

eloiprimeeloiprime

2,1651721

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks.As i have described in the question what i get by applying is (1532)(4) =>(1532).Is this itself the method to get disjoint cycles?
  $endgroup$
  – techno
  Oct 8 '15 at 7:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's all you need to do :)
  $endgroup$
  – eloiprime
  Oct 8 '15 at 7:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  okay.. so if i use the method i described in the question ie: finding the smallest etc... i will get the product of disjoint cycles.
  $endgroup$
  – techno
  Oct 8 '15 at 7:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I mean, that's all I did above. Wasn't it?
  $endgroup$
  – eloiprime
  Oct 8 '15 at 8:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah...got it now
  $endgroup$
  – techno
  Oct 8 '15 at 9:39
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

First, we note that writing it as a product of disjoint cycles means that each number appears only once throughout all of the cycles.

We see that $1mapsto 5$, $5mapsto 3$, $3mapsto 2$, $2mapsto 1$. So, we can express this in cycle notation as
$$(1532).$$
Now, we see what is left over... well, that is just $4$, which is fixed by the permutation in question. So, the permutation can be written as

$$(1532)(4),mbox or equivalently, just (1532).$$

share|cite|improve this answer

edited Oct 5 '18 at 8:06

answered Oct 8 '15 at 7:14

eloiprimeeloiprime

2,1651721

$endgroup$

share|cite|improve this answer

edited Oct 5 '18 at 8:06

answered Oct 8 '15 at 7:14

eloiprimeeloiprime

2,1651721

answered Oct 8 '15 at 7:14

eloiprimeeloiprime

2,1651721

answered Oct 8 '15 at 7:14

eloiprimeeloiprime

2,1651721

0

$begingroup$

$4$ is the only one invariant, while the others mingle among them, thus there are only two cycles
$$(1,5,3,2)$$
It's customary not to write any singleton.

share|cite|improve this answer

answered Oct 8 '15 at 7:15

MASLMASL

708313

$endgroup$

add a comment | 

0

$begingroup$

$4$ is the only one invariant, while the others mingle among them, thus there are only two cycles
$$(1,5,3,2)$$
It's customary not to write any singleton.

share|cite|improve this answer

answered Oct 8 '15 at 7:15

MASLMASL

708313

$endgroup$

add a comment | 

$begingroup$

$4$ is the only one invariant, while the others mingle among them, thus there are only two cycles
$$(1,5,3,2)$$
It's customary not to write any singleton.

share|cite|improve this answer

answered Oct 8 '15 at 7:15

MASLMASL

708313

$endgroup$

share|cite|improve this answer

answered Oct 8 '15 at 7:15

MASLMASL

708313

answered Oct 8 '15 at 7:15

MASLMASL

708313

answered Oct 8 '15 at 7:15

MASLMASL

708313

0

$begingroup$

You start with the first number than you go to the number with is under this number and so on until you get back to the start number...... In your case you start with 1 then go to 5 then to 3 to 2 and are back at one,.... So the first cycle is (1;5;2;3)

Then you look at the numbers which aren't in the first cycle e.g. 4 and do the same again.... ( you can, but you do not have to mention cycles with length 1 if it is known in which Sn the permutation is or it is otherwise clear how long the permutation is.)

share|cite|improve this answer

answered Oct 8 '15 at 7:19

BörgeBörge

1,037415

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  okay.. so if i use the method i described in the question ie: finding the smallest etc... i will get the product of disjoint cycles?
  $endgroup$
  – techno
  Oct 8 '15 at 7:23
  
  

  
  
  
  

add a comment | 

0

$begingroup$

You start with the first number than you go to the number with is under this number and so on until you get back to the start number...... In your case you start with 1 then go to 5 then to 3 to 2 and are back at one,.... So the first cycle is (1;5;2;3)

Then you look at the numbers which aren't in the first cycle e.g. 4 and do the same again.... ( you can, but you do not have to mention cycles with length 1 if it is known in which Sn the permutation is or it is otherwise clear how long the permutation is.)

share|cite|improve this answer

answered Oct 8 '15 at 7:19

BörgeBörge

1,037415

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  okay.. so if i use the method i described in the question ie: finding the smallest etc... i will get the product of disjoint cycles?
  $endgroup$
  – techno
  Oct 8 '15 at 7:23
  
  

  
  
  
  

add a comment | 

$begingroup$

You start with the first number than you go to the number with is under this number and so on until you get back to the start number...... In your case you start with 1 then go to 5 then to 3 to 2 and are back at one,.... So the first cycle is (1;5;2;3)

Then you look at the numbers which aren't in the first cycle e.g. 4 and do the same again.... ( you can, but you do not have to mention cycles with length 1 if it is known in which Sn the permutation is or it is otherwise clear how long the permutation is.)

share|cite|improve this answer

answered Oct 8 '15 at 7:19

BörgeBörge

1,037415

$endgroup$

share|cite|improve this answer

answered Oct 8 '15 at 7:19

BörgeBörge

1,037415

answered Oct 8 '15 at 7:19

BörgeBörge

1,037415

answered Oct 8 '15 at 7:19

BörgeBörge

1,037415