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algebra permutations

Element of Largest Order in $S_n$Permutation Group NotationSome naive questions about embeddingsStabilizers of permutationsPrimitive Element theorem, permutationsNumber of permutations with given cyclic structureLinear Algebra question on permutationsNumber of Even Permutations $p$ such that $pqp^{-1}=q$Absract Algebra: PermutationsEquivalence relation on $S_n$ involving algebraic permutations

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3

1

$begingroup$

Ok, this is rather a question than a problem. So, if we have $sigma:in S_n$ a permutation, why there is a natural number, say $p$, such that $sigma^p=e$, where $e$ is the identical permutation?

abstract-algebra

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asked 8 hours ago

NumbersNumbers

16166 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It would improve your posts of Questions to work out an example or two, so that you can provide Readers with a better indication of where you got with understanding the underlying mechanism, e.g. in this case taking a power of a permutation of $n$ things.
  $endgroup$
  – hardmath
  8 hours ago
  

  
  
  
  

add a comment
 | 

3

1

$begingroup$

Ok, this is rather a question than a problem. So, if we have $sigma:in S_n$ a permutation, why there is a natural number, say $p$, such that $sigma^p=e$, where $e$ is the identical permutation?

abstract-algebra

share|cite|improve this question

asked 8 hours ago

NumbersNumbers

16166 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It would improve your posts of Questions to work out an example or two, so that you can provide Readers with a better indication of where you got with understanding the underlying mechanism, e.g. in this case taking a power of a permutation of $n$ things.
  $endgroup$
  – hardmath
  8 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Ok, this is rather a question than a problem. So, if we have $sigma:in S_n$ a permutation, why there is a natural number, say $p$, such that $sigma^p=e$, where $e$ is the identical permutation?

abstract-algebra

share|cite|improve this question

asked 8 hours ago

NumbersNumbers

16166 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

NumbersNumbers

16166 bronze badges

share|cite|improve this question

asked 8 hours ago

NumbersNumbers

16166 bronze badges

asked 8 hours ago

NumbersNumbers

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asked 8 hours ago

NumbersNumbers

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3 Answers
3

active

oldest

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4

$begingroup$

In any finite group every element has finite order by Lagrange. Hence for every $sigmain S_n$ there exists a $p$ such that $sigma ^p=id$. However, the largest possible order of an element in $S_n$ is much smaller than the group order, see here:

Element of Largest Order in $S_n$

share|cite|improve this answer

answered 8 hours ago

Dietrich BurdeDietrich Burde

88k66 gold badges5050 silver badges111111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yeah, ok, I forgot to mention that I don t now group theory.
  $endgroup$
  – Numbers
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Lagrange is at the very beginning of group theory, and you don't have to know group theory for it.
  $endgroup$
  – Dietrich Burde
  7 hours ago
  

  
  
  
  

add a comment
 | 

0

$begingroup$

The set $${sigma,sigma^2, sigma^3,...}$$ is finite so at some points we have $sigma ^m=sigma^n$

That is $$sigma ^{m-n}=e$$

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.4k44 gold badges2727 silver badges7474 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

One more methodical idea is the following:

Let $sigma = (a_1a_2dots a_k)$. Consider $$sigma^k = (a_1a_2dots a_k)(a_1a_2dots a_k)cdots (a_1a_2dots a_k).$$ Then
begin{align}
a_1 to a_2 to &cdots to a_k to a_1,\
a_2 to a_3 to &cdots to a_1 to a_2,\
&;; vdots\
a_k to a_1 to &cdots to a_1 to a_k,
end{align}
sending every $a_i$ to itself. Think of this as sort of chaining together $a_i to a_{i+1}, a_{i+1} to a_{i+2},dots$.

An explicit example: If we have $(123)^3 = (123)(123)(123)$, then
begin{align}
1 to 2 to 3 to 1\
2 to 3 to 1 to 2\
3 to 1 to 2 to 3
end{align}

share|cite|improve this answer

answered 8 hours ago

HendrixHendrix

44144 silver badges1515 bronze badges

$endgroup$

add a comment
 | 

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4

$begingroup$

In any finite group every element has finite order by Lagrange. Hence for every $sigmain S_n$ there exists a $p$ such that $sigma ^p=id$. However, the largest possible order of an element in $S_n$ is much smaller than the group order, see here:

Element of Largest Order in $S_n$

share|cite|improve this answer

answered 8 hours ago

Dietrich BurdeDietrich Burde

88k66 gold badges5050 silver badges111111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yeah, ok, I forgot to mention that I don t now group theory.
  $endgroup$
  – Numbers
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Lagrange is at the very beginning of group theory, and you don't have to know group theory for it.
  $endgroup$
  – Dietrich Burde
  7 hours ago
  

  
  
  
  

add a comment
 | 

4

$begingroup$

In any finite group every element has finite order by Lagrange. Hence for every $sigmain S_n$ there exists a $p$ such that $sigma ^p=id$. However, the largest possible order of an element in $S_n$ is much smaller than the group order, see here:

Element of Largest Order in $S_n$

share|cite|improve this answer

answered 8 hours ago

Dietrich BurdeDietrich Burde

88k66 gold badges5050 silver badges111111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yeah, ok, I forgot to mention that I don t now group theory.
  $endgroup$
  – Numbers
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Lagrange is at the very beginning of group theory, and you don't have to know group theory for it.
  $endgroup$
  – Dietrich Burde
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

In any finite group every element has finite order by Lagrange. Hence for every $sigmain S_n$ there exists a $p$ such that $sigma ^p=id$. However, the largest possible order of an element in $S_n$ is much smaller than the group order, see here:

Element of Largest Order in $S_n$

share|cite|improve this answer

answered 8 hours ago

Dietrich BurdeDietrich Burde

88k66 gold badges5050 silver badges111111 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Dietrich BurdeDietrich Burde

88k66 gold badges5050 silver badges111111 bronze badges

answered 8 hours ago

Dietrich BurdeDietrich Burde

88k66 gold badges5050 silver badges111111 bronze badges

answered 8 hours ago

Dietrich BurdeDietrich Burde

88k66 gold badges5050 silver badges111111 bronze badges

0

$begingroup$

The set $${sigma,sigma^2, sigma^3,...}$$ is finite so at some points we have $sigma ^m=sigma^n$

That is $$sigma ^{m-n}=e$$

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.4k44 gold badges2727 silver badges7474 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

The set $${sigma,sigma^2, sigma^3,...}$$ is finite so at some points we have $sigma ^m=sigma^n$

That is $$sigma ^{m-n}=e$$

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.4k44 gold badges2727 silver badges7474 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

The set $${sigma,sigma^2, sigma^3,...}$$ is finite so at some points we have $sigma ^m=sigma^n$

That is $$sigma ^{m-n}=e$$

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.4k44 gold badges2727 silver badges7474 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.4k44 gold badges2727 silver badges7474 bronze badges

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.4k44 gold badges2727 silver badges7474 bronze badges

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.4k44 gold badges2727 silver badges7474 bronze badges

0

$begingroup$

One more methodical idea is the following:

Let $sigma = (a_1a_2dots a_k)$. Consider $$sigma^k = (a_1a_2dots a_k)(a_1a_2dots a_k)cdots (a_1a_2dots a_k).$$ Then
begin{align}
a_1 to a_2 to &cdots to a_k to a_1,\
a_2 to a_3 to &cdots to a_1 to a_2,\
&;; vdots\
a_k to a_1 to &cdots to a_1 to a_k,
end{align}
sending every $a_i$ to itself. Think of this as sort of chaining together $a_i to a_{i+1}, a_{i+1} to a_{i+2},dots$.

An explicit example: If we have $(123)^3 = (123)(123)(123)$, then
begin{align}
1 to 2 to 3 to 1\
2 to 3 to 1 to 2\
3 to 1 to 2 to 3
end{align}

share|cite|improve this answer

answered 8 hours ago

HendrixHendrix

44144 silver badges1515 bronze badges

$endgroup$

add a comment
 | 

0

$begingroup$

One more methodical idea is the following:

Let $sigma = (a_1a_2dots a_k)$. Consider $$sigma^k = (a_1a_2dots a_k)(a_1a_2dots a_k)cdots (a_1a_2dots a_k).$$ Then
begin{align}
a_1 to a_2 to &cdots to a_k to a_1,\
a_2 to a_3 to &cdots to a_1 to a_2,\
&;; vdots\
a_k to a_1 to &cdots to a_1 to a_k,
end{align}
sending every $a_i$ to itself. Think of this as sort of chaining together $a_i to a_{i+1}, a_{i+1} to a_{i+2},dots$.

An explicit example: If we have $(123)^3 = (123)(123)(123)$, then
begin{align}
1 to 2 to 3 to 1\
2 to 3 to 1 to 2\
3 to 1 to 2 to 3
end{align}

share|cite|improve this answer

answered 8 hours ago

HendrixHendrix

44144 silver badges1515 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

One more methodical idea is the following:

Let $sigma = (a_1a_2dots a_k)$. Consider $$sigma^k = (a_1a_2dots a_k)(a_1a_2dots a_k)cdots (a_1a_2dots a_k).$$ Then
begin{align}
a_1 to a_2 to &cdots to a_k to a_1,\
a_2 to a_3 to &cdots to a_1 to a_2,\
&;; vdots\
a_k to a_1 to &cdots to a_1 to a_k,
end{align}
sending every $a_i$ to itself. Think of this as sort of chaining together $a_i to a_{i+1}, a_{i+1} to a_{i+2},dots$.

An explicit example: If we have $(123)^3 = (123)(123)(123)$, then
begin{align}
1 to 2 to 3 to 1\
2 to 3 to 1 to 2\
3 to 1 to 2 to 3
end{align}

share|cite|improve this answer

answered 8 hours ago

HendrixHendrix

44144 silver badges1515 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

HendrixHendrix

44144 silver badges1515 bronze badges

answered 8 hours ago

HendrixHendrix

44144 silver badges1515 bronze badges

answered 8 hours ago

HendrixHendrix

44144 silver badges1515 bronze badges