How do I find the order of a bijection?
$begingroup$
Question: List all bijections (permutations) from ${1, 2, 3}$ onto ${1, 2, 3}$. Find their order and sign.
I understand there will be n! permutations, namely:
$
begin{Bmatrix}
1 & 2 & 3 \
1 & 2 & 3 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
2 & 1 & 3 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}
$,$
begin{Bmatrix}
1 & 2 & 3 \
3 & 1 & 2 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
3 & 2 & 1 \
end{Bmatrix}
$
I understand that order of a permutation $sigma$ is the smallest possible integer $k$ such that $sigma^k = epsilon$, where $epsilon$ is the identity permutation.
But I am confused by the definition of "identity permutation". If my identity is:
$$
begin{Bmatrix}
1 & 2 & 3 \
1 & 2 & 3 \
end{Bmatrix}
$$
then order $= 0$ and sign = $(-1)^k = (-1)^0 = 1$.
And for: $
begin{Bmatrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{Bmatrix}
$, order $= 1$, sign $= -1$. And for $
begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}
$ order $= 2$, sign $= 1$.
But if my first bijection from ${1, 2, 3}$ onto ${1, 2, 3}$ is:
$$
begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}
$$
then order $= 0$ and sign $= 1$. So depending on which projection I chose as an identity, the order differs. Can someone clarify?
group-theory elementary-set-theory permutations symmetric-groups
edited 7 hours ago
jordan_glen
1548
asked 7 hours ago
Bn.F76Bn.F76
206
$endgroup$
|
show 2 more comments
$begingroup$
Question: List all bijections (permutations) from ${1, 2, 3}$ onto ${1, 2, 3}$. Find their order and sign.
I understand there will be n! permutations, namely:
$
begin{Bmatrix}
1 & 2 & 3 \
1 & 2 & 3 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
2 & 1 & 3 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}
$,$
begin{Bmatrix}
1 & 2 & 3 \
3 & 1 & 2 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
3 & 2 & 1 \
end{Bmatrix}
$
I understand that order of a permutation $sigma$ is the smallest possible integer $k$ such that $sigma^k = epsilon$, where $epsilon$ is the identity permutation.
But I am confused by the definition of "identity permutation". If my identity is:
$$
begin{Bmatrix}
1 & 2 & 3 \
1 & 2 & 3 \
end{Bmatrix}
$$
then order $= 0$ and sign = $(-1)^k = (-1)^0 = 1$.
And for: $
begin{Bmatrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{Bmatrix}
$, order $= 1$, sign $= -1$. And for $
begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}
$ order $= 2$, sign $= 1$.
But if my first bijection from ${1, 2, 3}$ onto ${1, 2, 3}$ is:
$$
begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}
$$
then order $= 0$ and sign $= 1$. So depending on which projection I chose as an identity, the order differs. Can someone clarify?
group-theory elementary-set-theory permutations symmetric-groups
edited 7 hours ago
jordan_glen
1548
asked 7 hours ago
Bn.F76Bn.F76
206
$endgroup$
$begingroup$
The identity function $id_X~:~ Xto X$ is very specifically the function such that $id_X(x) = x$ for all $xin X$. You do not get to "decide" which is the identity. You have as a result, the identity function satisfies $id_Xcirc f = fcirc id_X = f$ for all functions $f~:~Xto X$. This is just like how in a group you have the identity $e$ satisfies $e*x = x*e = x$ for all group members $x$.
$endgroup$
– JMoravitz
7 hours ago
$begingroup$
Every permutation $sigma$ satisfies $sigma^0=epsilon$. That's not of interest. The identity permutation satisfies $epsilon^1=epsilon$, so it has order $1$. Also, sign and order aren't directly related.
$endgroup$
– Arthur
7 hours ago
$begingroup$
The order of $sigma$ is the smallest positive integer $k$ such that $sigma^k = e$. By convention, $sigma^0 = e$ for any permutation, so it is meaningless to say the order is 0.
$endgroup$
– Rob Arthan
7 hours ago
$begingroup$
Summing up all your answers would make a valid answer,I think.
$endgroup$
– G.F
7 hours ago
$begingroup$
The identity permutation assigns each element in a set to itself: $id(a) = a$, $id(b) = b$, etc.
$endgroup$
– jordan_glen
6 hours ago
|
show 2 more comments
$begingroup$
Question: List all bijections (permutations) from ${1, 2, 3}$ onto ${1, 2, 3}$. Find their order and sign.
I understand there will be n! permutations, namely:
$
begin{Bmatrix}
1 & 2 & 3 \
1 & 2 & 3 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
2 & 1 & 3 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}
$,$
begin{Bmatrix}
1 & 2 & 3 \
3 & 1 & 2 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
3 & 2 & 1 \
end{Bmatrix}
$
I understand that order of a permutation $sigma$ is the smallest possible integer $k$ such that $sigma^k = epsilon$, where $epsilon$ is the identity permutation.
But I am confused by the definition of "identity permutation". If my identity is:
$$
begin{Bmatrix}
1 & 2 & 3 \
1 & 2 & 3 \
end{Bmatrix}
$$
then order $= 0$ and sign = $(-1)^k = (-1)^0 = 1$.
And for: $
begin{Bmatrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{Bmatrix}
$, order $= 1$, sign $= -1$. And for $
begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}
$ order $= 2$, sign $= 1$.
But if my first bijection from ${1, 2, 3}$ onto ${1, 2, 3}$ is:
$$
begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}
$$
then order $= 0$ and sign $= 1$. So depending on which projection I chose as an identity, the order differs. Can someone clarify?
group-theory elementary-set-theory permutations symmetric-groups
edited 7 hours ago
jordan_glen
1548
asked 7 hours ago
Bn.F76Bn.F76
206
$endgroup$
Question: List all bijections (permutations) from ${1, 2, 3}$ onto ${1, 2, 3}$. Find their order and sign.
I understand there will be n! permutations, namely:
$
begin{Bmatrix}
1 & 2 & 3 \
1 & 2 & 3 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
2 & 1 & 3 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}
$,$
begin{Bmatrix}
1 & 2 & 3 \
3 & 1 & 2 \
end{Bmatrix}
$,
$
begin{Bmatrix}
1 & 2 & 3 \
3 & 2 & 1 \
end{Bmatrix}
$
I understand that order of a permutation $sigma$ is the smallest possible integer $k$ such that $sigma^k = epsilon$, where $epsilon$ is the identity permutation.
But I am confused by the definition of "identity permutation". If my identity is:
$$
begin{Bmatrix}
1 & 2 & 3 \
1 & 2 & 3 \
end{Bmatrix}
$$
then order $= 0$ and sign = $(-1)^k = (-1)^0 = 1$.
And for: $
begin{Bmatrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{Bmatrix}
$, order $= 1$, sign $= -1$. And for $
begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}
$ order $= 2$, sign $= 1$.
But if my first bijection from ${1, 2, 3}$ onto ${1, 2, 3}$ is:
$$
begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}
$$
then order $= 0$ and sign $= 1$. So depending on which projection I chose as an identity, the order differs. Can someone clarify?
group-theory elementary-set-theory permutations symmetric-groups
group-theory elementary-set-theory permutations symmetric-groups
edited 7 hours ago
jordan_glen
1548
asked 7 hours ago
Bn.F76Bn.F76
206
edited 7 hours ago
jordan_glen
1548
asked 7 hours ago
Bn.F76Bn.F76
206
edited 7 hours ago
jordan_glen
1548
edited 7 hours ago
jordan_glen
1548
edited 7 hours ago
jordan_glen
1548
1548
asked 7 hours ago
Bn.F76Bn.F76
206
asked 7 hours ago
Bn.F76Bn.F76
206
asked 7 hours ago
Bn.F76Bn.F76
206
206
$begingroup$
The identity function $id_X~:~ Xto X$ is very specifically the function such that $id_X(x) = x$ for all $xin X$. You do not get to "decide" which is the identity. You have as a result, the identity function satisfies $id_Xcirc f = fcirc id_X = f$ for all functions $f~:~Xto X$. This is just like how in a group you have the identity $e$ satisfies $e*x = x*e = x$ for all group members $x$.
$endgroup$
– JMoravitz
7 hours ago
$begingroup$
Every permutation $sigma$ satisfies $sigma^0=epsilon$. That's not of interest. The identity permutation satisfies $epsilon^1=epsilon$, so it has order $1$. Also, sign and order aren't directly related.
$endgroup$
– Arthur
7 hours ago
$begingroup$
The order of $sigma$ is the smallest positive integer $k$ such that $sigma^k = e$. By convention, $sigma^0 = e$ for any permutation, so it is meaningless to say the order is 0.
$endgroup$
– Rob Arthan
7 hours ago
$begingroup$
Summing up all your answers would make a valid answer,I think.
$endgroup$
– G.F
7 hours ago
$begingroup$
The identity permutation assigns each element in a set to itself: $id(a) = a$, $id(b) = b$, etc.
$endgroup$
– jordan_glen
6 hours ago
|
show 2 more comments
$begingroup$
The identity function $id_X~:~ Xto X$ is very specifically the function such that $id_X(x) = x$ for all $xin X$. You do not get to "decide" which is the identity. You have as a result, the identity function satisfies $id_Xcirc f = fcirc id_X = f$ for all functions $f~:~Xto X$. This is just like how in a group you have the identity $e$ satisfies $e*x = x*e = x$ for all group members $x$.
$endgroup$
– JMoravitz
7 hours ago
$begingroup$
Every permutation $sigma$ satisfies $sigma^0=epsilon$. That's not of interest. The identity permutation satisfies $epsilon^1=epsilon$, so it has order $1$. Also, sign and order aren't directly related.
$endgroup$
– Arthur
7 hours ago
$begingroup$
The order of $sigma$ is the smallest positive integer $k$ such that $sigma^k = e$. By convention, $sigma^0 = e$ for any permutation, so it is meaningless to say the order is 0.
$endgroup$
– Rob Arthan
7 hours ago
$begingroup$
Summing up all your answers would make a valid answer,I think.
$endgroup$
– G.F
7 hours ago
$begingroup$
The identity permutation assigns each element in a set to itself: $id(a) = a$, $id(b) = b$, etc.
$endgroup$
– jordan_glen
6 hours ago
$begingroup$
The identity function $id_X~:~ Xto X$ is very specifically the function such that $id_X(x) = x$ for all $xin X$. You do not get to "decide" which is the identity. You have as a result, the identity function satisfies $id_Xcirc f = fcirc id_X = f$ for all functions $f~:~Xto X$. This is just like how in a group you have the identity $e$ satisfies $e*x = x*e = x$ for all group members $x$.
$endgroup$
– JMoravitz
7 hours ago
$begingroup$
The identity function $id_X~:~ Xto X$ is very specifically the function such that $id_X(x) = x$ for all $xin X$. You do not get to "decide" which is the identity. You have as a result, the identity function satisfies $id_Xcirc f = fcirc id_X = f$ for all functions $f~:~Xto X$. This is just like how in a group you have the identity $e$ satisfies $e*x = x*e = x$ for all group members $x$.
$endgroup$
– JMoravitz
7 hours ago
$begingroup$
Every permutation $sigma$ satisfies $sigma^0=epsilon$. That's not of interest. The identity permutation satisfies $epsilon^1=epsilon$, so it has order $1$. Also, sign and order aren't directly related.
$endgroup$
– Arthur
7 hours ago
$begingroup$
Every permutation $sigma$ satisfies $sigma^0=epsilon$. That's not of interest. The identity permutation satisfies $epsilon^1=epsilon$, so it has order $1$. Also, sign and order aren't directly related.
$endgroup$
– Arthur
7 hours ago
$begingroup$
The order of $sigma$ is the smallest positive integer $k$ such that $sigma^k = e$. By convention, $sigma^0 = e$ for any permutation, so it is meaningless to say the order is 0.
$endgroup$
– Rob Arthan
7 hours ago
$begingroup$
The order of $sigma$ is the smallest positive integer $k$ such that $sigma^k = e$. By convention, $sigma^0 = e$ for any permutation, so it is meaningless to say the order is 0.
$endgroup$
– Rob Arthan
7 hours ago
$begingroup$
Summing up all your answers would make a valid answer,I think.
$endgroup$
– G.F
7 hours ago
$begingroup$
Summing up all your answers would make a valid answer,I think.
$endgroup$
– G.F
7 hours ago
$begingroup$
The identity permutation assigns each element in a set to itself: $id(a) = a$, $id(b) = b$, etc.
$endgroup$
– jordan_glen
6 hours ago
$begingroup$
The identity permutation assigns each element in a set to itself: $id(a) = a$, $id(b) = b$, etc.
$endgroup$
– jordan_glen
6 hours ago
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The identity is not something you choose; the identity means the permutation that fixes each element, so the first one you considered is the identity permutation. Secondly, the order of the identity permutation is $1$, because $epsilon^1=epsilon$ (the order must be positive). The order of $
sigma_1=begin{Bmatrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{Bmatrix}
$ is $2$ because $sigma_1$ is not the identity permutation, but if you compose it with itself, you get the identity, so $sigma_1^2=epsilon$. The order of $sigma_2=begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}$ is $3$ because $sigma_2neq epsilon$ and $sigma_2^2neq epsilon$, but $sigma_2^3=epsilon$.
answered 7 hours ago
Kevin LongKevin Long
3,45621330
$endgroup$
add a comment |
$begingroup$
You obtain a direct answer to all questions decomposing permutations into a product of disjoint cycles. Concerning permutations of ${1,2,3}$, a permutation can be:
- a single $3$-cycle: $(1,2,3)$ or $(1,3,2)$
- a transposition: $(1,2)$, $(1,3)$ or $(2,3)$
- the empty cycle $(,)$ (corresponding to the identity)
Now, a cycle of length $k$ has order $k$ and signature $(-1)^{k-1}$.
answered 7 hours ago
BernardBernard
119k740113
$endgroup$
$begingroup$
The identity permutation represented as a product of disjoint cycles is most often depicted by $(1)$, short for $(1)(2)(3)$ in this particular case, which is considered a cycle of length $1$, because the order of two or more cycles is the least common multiple of the lengths of its disjoint cycles: $operatorname{lcm}(1, 1, 1) = 1$, which is the order of the identity permutation.
$endgroup$
– jordan_glen
6 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The identity is not something you choose; the identity means the permutation that fixes each element, so the first one you considered is the identity permutation. Secondly, the order of the identity permutation is $1$, because $epsilon^1=epsilon$ (the order must be positive). The order of $
sigma_1=begin{Bmatrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{Bmatrix}
$ is $2$ because $sigma_1$ is not the identity permutation, but if you compose it with itself, you get the identity, so $sigma_1^2=epsilon$. The order of $sigma_2=begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}$ is $3$ because $sigma_2neq epsilon$ and $sigma_2^2neq epsilon$, but $sigma_2^3=epsilon$.
answered 7 hours ago
Kevin LongKevin Long
3,45621330
$endgroup$
add a comment |
$begingroup$
The identity is not something you choose; the identity means the permutation that fixes each element, so the first one you considered is the identity permutation. Secondly, the order of the identity permutation is $1$, because $epsilon^1=epsilon$ (the order must be positive). The order of $
sigma_1=begin{Bmatrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{Bmatrix}
$ is $2$ because $sigma_1$ is not the identity permutation, but if you compose it with itself, you get the identity, so $sigma_1^2=epsilon$. The order of $sigma_2=begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}$ is $3$ because $sigma_2neq epsilon$ and $sigma_2^2neq epsilon$, but $sigma_2^3=epsilon$.
answered 7 hours ago
Kevin LongKevin Long
3,45621330
$endgroup$
add a comment |
$begingroup$
The identity is not something you choose; the identity means the permutation that fixes each element, so the first one you considered is the identity permutation. Secondly, the order of the identity permutation is $1$, because $epsilon^1=epsilon$ (the order must be positive). The order of $
sigma_1=begin{Bmatrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{Bmatrix}
$ is $2$ because $sigma_1$ is not the identity permutation, but if you compose it with itself, you get the identity, so $sigma_1^2=epsilon$. The order of $sigma_2=begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}$ is $3$ because $sigma_2neq epsilon$ and $sigma_2^2neq epsilon$, but $sigma_2^3=epsilon$.
answered 7 hours ago
Kevin LongKevin Long
3,45621330
$endgroup$
The identity is not something you choose; the identity means the permutation that fixes each element, so the first one you considered is the identity permutation. Secondly, the order of the identity permutation is $1$, because $epsilon^1=epsilon$ (the order must be positive). The order of $
sigma_1=begin{Bmatrix}
1 & 2 & 3 \
1 & 3 & 2 \
end{Bmatrix}
$ is $2$ because $sigma_1$ is not the identity permutation, but if you compose it with itself, you get the identity, so $sigma_1^2=epsilon$. The order of $sigma_2=begin{Bmatrix}
1 & 2 & 3 \
2 & 3 & 1 \
end{Bmatrix}$ is $3$ because $sigma_2neq epsilon$ and $sigma_2^2neq epsilon$, but $sigma_2^3=epsilon$.
answered 7 hours ago
Kevin LongKevin Long
3,45621330
edited 6 hours ago
answered 7 hours ago
Kevin LongKevin Long
3,45621330
answered 7 hours ago
Kevin LongKevin Long
3,45621330
answered 7 hours ago
Kevin LongKevin Long
3,45621330
3,45621330
add a comment |
add a comment |
$begingroup$
You obtain a direct answer to all questions decomposing permutations into a product of disjoint cycles. Concerning permutations of ${1,2,3}$, a permutation can be:
- a single $3$-cycle: $(1,2,3)$ or $(1,3,2)$
- a transposition: $(1,2)$, $(1,3)$ or $(2,3)$
- the empty cycle $(,)$ (corresponding to the identity)
Now, a cycle of length $k$ has order $k$ and signature $(-1)^{k-1}$.
answered 7 hours ago
BernardBernard
119k740113
$endgroup$
$begingroup$
The identity permutation represented as a product of disjoint cycles is most often depicted by $(1)$, short for $(1)(2)(3)$ in this particular case, which is considered a cycle of length $1$, because the order of two or more cycles is the least common multiple of the lengths of its disjoint cycles: $operatorname{lcm}(1, 1, 1) = 1$, which is the order of the identity permutation.
$endgroup$
– jordan_glen
6 hours ago
add a comment |
$begingroup$
You obtain a direct answer to all questions decomposing permutations into a product of disjoint cycles. Concerning permutations of ${1,2,3}$, a permutation can be:
- a single $3$-cycle: $(1,2,3)$ or $(1,3,2)$
- a transposition: $(1,2)$, $(1,3)$ or $(2,3)$
- the empty cycle $(,)$ (corresponding to the identity)
Now, a cycle of length $k$ has order $k$ and signature $(-1)^{k-1}$.
answered 7 hours ago
BernardBernard
119k740113
$endgroup$
$begingroup$
The identity permutation represented as a product of disjoint cycles is most often depicted by $(1)$, short for $(1)(2)(3)$ in this particular case, which is considered a cycle of length $1$, because the order of two or more cycles is the least common multiple of the lengths of its disjoint cycles: $operatorname{lcm}(1, 1, 1) = 1$, which is the order of the identity permutation.
$endgroup$
– jordan_glen
6 hours ago
add a comment |
$begingroup$
You obtain a direct answer to all questions decomposing permutations into a product of disjoint cycles. Concerning permutations of ${1,2,3}$, a permutation can be:
- a single $3$-cycle: $(1,2,3)$ or $(1,3,2)$
- a transposition: $(1,2)$, $(1,3)$ or $(2,3)$
- the empty cycle $(,)$ (corresponding to the identity)
Now, a cycle of length $k$ has order $k$ and signature $(-1)^{k-1}$.
answered 7 hours ago
BernardBernard
119k740113
$endgroup$
You obtain a direct answer to all questions decomposing permutations into a product of disjoint cycles. Concerning permutations of ${1,2,3}$, a permutation can be:
- a single $3$-cycle: $(1,2,3)$ or $(1,3,2)$
- a transposition: $(1,2)$, $(1,3)$ or $(2,3)$
- the empty cycle $(,)$ (corresponding to the identity)
Now, a cycle of length $k$ has order $k$ and signature $(-1)^{k-1}$.
answered 7 hours ago
BernardBernard
119k740113
answered 7 hours ago
BernardBernard
119k740113
answered 7 hours ago
BernardBernard
119k740113
answered 7 hours ago
BernardBernard
119k740113
119k740113
$begingroup$
The identity permutation represented as a product of disjoint cycles is most often depicted by $(1)$, short for $(1)(2)(3)$ in this particular case, which is considered a cycle of length $1$, because the order of two or more cycles is the least common multiple of the lengths of its disjoint cycles: $operatorname{lcm}(1, 1, 1) = 1$, which is the order of the identity permutation.
$endgroup$
– jordan_glen
6 hours ago
add a comment |
$begingroup$
The identity permutation represented as a product of disjoint cycles is most often depicted by $(1)$, short for $(1)(2)(3)$ in this particular case, which is considered a cycle of length $1$, because the order of two or more cycles is the least common multiple of the lengths of its disjoint cycles: $operatorname{lcm}(1, 1, 1) = 1$, which is the order of the identity permutation.
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– jordan_glen
6 hours ago
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The identity permutation represented as a product of disjoint cycles is most often depicted by $(1)$, short for $(1)(2)(3)$ in this particular case, which is considered a cycle of length $1$, because the order of two or more cycles is the least common multiple of the lengths of its disjoint cycles: $operatorname{lcm}(1, 1, 1) = 1$, which is the order of the identity permutation.
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– jordan_glen
6 hours ago
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The identity permutation represented as a product of disjoint cycles is most often depicted by $(1)$, short for $(1)(2)(3)$ in this particular case, which is considered a cycle of length $1$, because the order of two or more cycles is the least common multiple of the lengths of its disjoint cycles: $operatorname{lcm}(1, 1, 1) = 1$, which is the order of the identity permutation.
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– jordan_glen
6 hours ago
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The identity function $id_X~:~ Xto X$ is very specifically the function such that $id_X(x) = x$ for all $xin X$. You do not get to "decide" which is the identity. You have as a result, the identity function satisfies $id_Xcirc f = fcirc id_X = f$ for all functions $f~:~Xto X$. This is just like how in a group you have the identity $e$ satisfies $e*x = x*e = x$ for all group members $x$.
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– JMoravitz
7 hours ago
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Every permutation $sigma$ satisfies $sigma^0=epsilon$. That's not of interest. The identity permutation satisfies $epsilon^1=epsilon$, so it has order $1$. Also, sign and order aren't directly related.
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– Arthur
7 hours ago
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The order of $sigma$ is the smallest positive integer $k$ such that $sigma^k = e$. By convention, $sigma^0 = e$ for any permutation, so it is meaningless to say the order is 0.
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– Rob Arthan
7 hours ago
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Summing up all your answers would make a valid answer,I think.
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– G.F
7 hours ago
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The identity permutation assigns each element in a set to itself: $id(a) = a$, $id(b) = b$, etc.
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– jordan_glen
6 hours ago