algebra permutationsElement 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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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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$begingroup$
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
asked 8 hours ago
NumbersNumbers
1616 bronze badges
$endgroup$
add a comment
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$begingroup$
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
asked 8 hours ago
NumbersNumbers
1616 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
asked 8 hours ago
NumbersNumbers
1616 bronze badges
$endgroup$
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
abstract-algebra
asked 8 hours ago
NumbersNumbers
1616 bronze badges
asked 8 hours ago
NumbersNumbers
1616 bronze badges
asked 8 hours ago
NumbersNumbers
1616 bronze badges
asked 8 hours ago
NumbersNumbers
1616 bronze badges
asked 8 hours ago
NumbersNumbers
1616 bronze badges
1616 bronze badges
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
|
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
1
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
$begingroup$
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 Answers
3
active
oldest
votes
$begingroup$
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 $S_n$
answered 8 hours ago
Dietrich BurdeDietrich Burde
88k6 gold badges50 silver badges111 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
8 hours ago
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 $$sigma ^m-n=e$$
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
54.4k4 gold badges27 silver badges74 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
beginalign
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,
endalign
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
beginalign
1 to 2 to 3 to 1\
2 to 3 to 1 to 2\
3 to 1 to 2 to 3
endalign
answered 8 hours ago
HendrixHendrix
4414 silver badges15 bronze badges
$endgroup$
add a comment
|
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $S_n$
answered 8 hours ago
Dietrich BurdeDietrich Burde
88k6 gold badges50 silver badges111 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
8 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 $S_n$
answered 8 hours ago
Dietrich BurdeDietrich Burde
88k6 gold badges50 silver badges111 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
8 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 $S_n$
answered 8 hours ago
Dietrich BurdeDietrich Burde
88k6 gold badges50 silver badges111 bronze badges
$endgroup$
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 $S_n$
answered 8 hours ago
Dietrich BurdeDietrich Burde
88k6 gold badges50 silver badges111 bronze badges
answered 8 hours ago
Dietrich BurdeDietrich Burde
88k6 gold badges50 silver badges111 bronze badges
answered 8 hours ago
Dietrich BurdeDietrich Burde
88k6 gold badges50 silver badges111 bronze badges
answered 8 hours ago
Dietrich BurdeDietrich Burde
88k6 gold badges50 silver badges111 bronze badges
88k6 gold badges50 silver badges111 bronze badges
$begingroup$
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
8 hours ago
add a comment
|
$begingroup$
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
8 hours ago
$begingroup$
yeah, ok, I forgot to mention that I don t now group theory.
$endgroup$
– Numbers
8 hours ago
$begingroup$
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
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
8 hours ago
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 $$sigma ^m-n=e$$
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
54.4k4 gold badges27 silver badges74 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 $$sigma ^m-n=e$$
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
54.4k4 gold badges27 silver badges74 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 $$sigma ^m-n=e$$
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
54.4k4 gold badges27 silver badges74 bronze badges
$endgroup$
The set $$sigma,sigma^2, sigma^3,...$$ is finite so at some points we have $sigma ^m=sigma^n$
That is $$sigma ^m-n=e$$
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
54.4k4 gold badges27 silver badges74 bronze badges
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
54.4k4 gold badges27 silver badges74 bronze badges
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
54.4k4 gold badges27 silver badges74 bronze badges
answered 8 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
54.4k4 gold badges27 silver badges74 bronze badges
54.4k4 gold badges27 silver badges74 bronze badges
add a comment
|
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
beginalign
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,
endalign
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
beginalign
1 to 2 to 3 to 1\
2 to 3 to 1 to 2\
3 to 1 to 2 to 3
endalign
answered 8 hours ago
HendrixHendrix
4414 silver badges15 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
beginalign
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,
endalign
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
beginalign
1 to 2 to 3 to 1\
2 to 3 to 1 to 2\
3 to 1 to 2 to 3
endalign
answered 8 hours ago
HendrixHendrix
4414 silver badges15 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
beginalign
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,
endalign
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
beginalign
1 to 2 to 3 to 1\
2 to 3 to 1 to 2\
3 to 1 to 2 to 3
endalign
answered 8 hours ago
HendrixHendrix
4414 silver badges15 bronze badges
$endgroup$
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
beginalign
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,
endalign
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
beginalign
1 to 2 to 3 to 1\
2 to 3 to 1 to 2\
3 to 1 to 2 to 3
endalign
answered 8 hours ago
HendrixHendrix
4414 silver badges15 bronze badges
answered 8 hours ago
HendrixHendrix
4414 silver badges15 bronze badges
answered 8 hours ago
HendrixHendrix
4414 silver badges15 bronze badges
answered 8 hours ago
HendrixHendrix
4414 silver badges15 bronze badges
4414 silver badges15 bronze badges
add a comment
|
add a comment
|
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$begingroup$
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