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Orbit of a Permutation

Centralizer of a given element in $S_n$?Help with Conrad's “Recognizing Galois Groups”Group Action on $S_n$Calculation related to the number of conjugacy classes of the symmetric groupFinding the centralizer of a permutationThe sum of orbit size of some element over the image of group “polynomial”Exercise 3A.7 of “Finite group theory”, M. IsaacsFinding Subgroups Of $S_15$Why is this the right answer to this orbit question?Show that the cycle decomposition of a permutation can be recovered by considering the orbits of the action of its cyclic group on $1, 2, …, n$

2

$begingroup$

On page 66 of these notes is proposition 4.26:

Every permutation can be written (in essentially one way) as a product of disjoint cycles.

The proof begins as follows:

Let $sigma in S_n$, and let $O subseteq 1,...,n$ be an orbit for $langle sigma rangle$....

What does it mean for $O$ to be an orbit for $langle sigma rangle$? I am unfamiliar with this terminology. From what I gather, the implicit action is of $S_n$ on $1,...,n$ by functional evaluation. So, $O$ will be the orbit of some element in $1,...,n$. How can it be an orbit for $langle sigma rangle$?

EDIT

Also, the author writes $O = i,sigma (i),..., sigma^r-1(i)$. How do we know this equality holds? What if $sigma$ has order smaller than $r-1$?

group-theory terminology group-actions

share|cite|improve this question

asked Mar 14 at 15:22

user193319user193319

2,4292927

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  How did they define $r$?
  $endgroup$
  – Mike Earnest
  Mar 14 at 15:32
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $O$ is the orbit of $i$ under the action of the cyclic subgroup $langle sigma rangle$ [this action naturally being the restriction of the action of $S_n$].
  $endgroup$
  – M. Vinay
  Mar 14 at 15:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See also Exercise 7 on UMN Fall 2017 Math 4990 homework set #7 for a version of this proof with all details filled in. It is one of the most painful to formalize proofs in basic abstract algebra. (Note that my $sim$-equivalence classes are exactly the orbits of $sigma$, although I define them a bit differently.)
  $endgroup$
  – darij grinberg
  Mar 14 at 15:37
  
  

  
  
  
  

add a comment | 

2

$begingroup$

On page 66 of these notes is proposition 4.26:

Every permutation can be written (in essentially one way) as a product of disjoint cycles.

The proof begins as follows:

Let $sigma in S_n$, and let $O subseteq 1,...,n$ be an orbit for $langle sigma rangle$....

What does it mean for $O$ to be an orbit for $langle sigma rangle$? I am unfamiliar with this terminology. From what I gather, the implicit action is of $S_n$ on $1,...,n$ by functional evaluation. So, $O$ will be the orbit of some element in $1,...,n$. How can it be an orbit for $langle sigma rangle$?

EDIT

Also, the author writes $O = i,sigma (i),..., sigma^r-1(i)$. How do we know this equality holds? What if $sigma$ has order smaller than $r-1$?

group-theory terminology group-actions

share|cite|improve this question

asked Mar 14 at 15:22

user193319user193319

2,4292927

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  How did they define $r$?
  $endgroup$
  – Mike Earnest
  Mar 14 at 15:32
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $O$ is the orbit of $i$ under the action of the cyclic subgroup $langle sigma rangle$ [this action naturally being the restriction of the action of $S_n$].
  $endgroup$
  – M. Vinay
  Mar 14 at 15:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See also Exercise 7 on UMN Fall 2017 Math 4990 homework set #7 for a version of this proof with all details filled in. It is one of the most painful to formalize proofs in basic abstract algebra. (Note that my $sim$-equivalence classes are exactly the orbits of $sigma$, although I define them a bit differently.)
  $endgroup$
  – darij grinberg
  Mar 14 at 15:37
  
  

  
  
  
  

add a comment | 

$begingroup$

On page 66 of these notes is proposition 4.26:

Every permutation can be written (in essentially one way) as a product of disjoint cycles.

The proof begins as follows:

Let $sigma in S_n$, and let $O subseteq 1,...,n$ be an orbit for $langle sigma rangle$....

What does it mean for $O$ to be an orbit for $langle sigma rangle$? I am unfamiliar with this terminology. From what I gather, the implicit action is of $S_n$ on $1,...,n$ by functional evaluation. So, $O$ will be the orbit of some element in $1,...,n$. How can it be an orbit for $langle sigma rangle$?

EDIT

Also, the author writes $O = i,sigma (i),..., sigma^r-1(i)$. How do we know this equality holds? What if $sigma$ has order smaller than $r-1$?

group-theory terminology group-actions

share|cite|improve this question

asked Mar 14 at 15:22

user193319user193319

2,4292927

$endgroup$

share|cite|improve this question

asked Mar 14 at 15:22

user193319user193319

2,4292927

share|cite|improve this question

asked Mar 14 at 15:22

user193319user193319

2,4292927

asked Mar 14 at 15:22

user193319user193319

2,4292927

asked Mar 14 at 15:22

user193319user193319

2,4292927