How do I find the order of a bijection?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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:
$
beginBmatrix
1 & 2 & 3 \
1 & 2 & 3 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
1 & 3 & 2 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
2 & 1 & 3 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix
$,$
beginBmatrix
1 & 2 & 3 \
3 & 1 & 2 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
3 & 2 & 1 \
endBmatrix
$
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:
$$
beginBmatrix
1 & 2 & 3 \
1 & 2 & 3 \
endBmatrix
$$
then order $= 0$ and sign = $(-1)^k = (-1)^0 = 1$.
And for: $
beginBmatrix
1 & 2 & 3 \
1 & 3 & 2 \
endBmatrix
$, order $= 1$, sign $= -1$. And for $
beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix
$ order $= 2$, sign $= 1$.
But if my first bijection from $1, 2, 3$ onto $1, 2, 3$ is:
$$
beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix
$$
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 Jan 18 at 20:56
jordan_glen
1
asked Jan 18 at 20:39
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:
$
beginBmatrix
1 & 2 & 3 \
1 & 2 & 3 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
1 & 3 & 2 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
2 & 1 & 3 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix
$,$
beginBmatrix
1 & 2 & 3 \
3 & 1 & 2 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
3 & 2 & 1 \
endBmatrix
$
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:
$$
beginBmatrix
1 & 2 & 3 \
1 & 2 & 3 \
endBmatrix
$$
then order $= 0$ and sign = $(-1)^k = (-1)^0 = 1$.
And for: $
beginBmatrix
1 & 2 & 3 \
1 & 3 & 2 \
endBmatrix
$, order $= 1$, sign $= -1$. And for $
beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix
$ order $= 2$, sign $= 1$.
But if my first bijection from $1, 2, 3$ onto $1, 2, 3$ is:
$$
beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix
$$
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 Jan 18 at 20:56
jordan_glen
1
asked Jan 18 at 20:39
Bn.F76Bn.F76
206
$endgroup$
1
$begingroup$
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
Jan 18 at 20:45
$begingroup$
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
Jan 18 at 20:47
$begingroup$
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
Jan 18 at 20:47
$begingroup$
Summing up all your answers would make a valid answer,I think.
$endgroup$
– J.F
Jan 18 at 20:48
$begingroup$
The identity permutation assigns each element in a set to itself: $id(a) = a$, $id(b) = b$, etc.
$endgroup$
– jordan_glen
Jan 18 at 21:05
|
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:
$
beginBmatrix
1 & 2 & 3 \
1 & 2 & 3 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
1 & 3 & 2 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
2 & 1 & 3 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix
$,$
beginBmatrix
1 & 2 & 3 \
3 & 1 & 2 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
3 & 2 & 1 \
endBmatrix
$
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:
$$
beginBmatrix
1 & 2 & 3 \
1 & 2 & 3 \
endBmatrix
$$
then order $= 0$ and sign = $(-1)^k = (-1)^0 = 1$.
And for: $
beginBmatrix
1 & 2 & 3 \
1 & 3 & 2 \
endBmatrix
$, order $= 1$, sign $= -1$. And for $
beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix
$ order $= 2$, sign $= 1$.
But if my first bijection from $1, 2, 3$ onto $1, 2, 3$ is:
$$
beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix
$$
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 Jan 18 at 20:56
jordan_glen
1
asked Jan 18 at 20:39
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:
$
beginBmatrix
1 & 2 & 3 \
1 & 2 & 3 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
1 & 3 & 2 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
2 & 1 & 3 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix
$,$
beginBmatrix
1 & 2 & 3 \
3 & 1 & 2 \
endBmatrix
$,
$
beginBmatrix
1 & 2 & 3 \
3 & 2 & 1 \
endBmatrix
$
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:
$$
beginBmatrix
1 & 2 & 3 \
1 & 2 & 3 \
endBmatrix
$$
then order $= 0$ and sign = $(-1)^k = (-1)^0 = 1$.
And for: $
beginBmatrix
1 & 2 & 3 \
1 & 3 & 2 \
endBmatrix
$, order $= 1$, sign $= -1$. And for $
beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix
$ order $= 2$, sign $= 1$.
But if my first bijection from $1, 2, 3$ onto $1, 2, 3$ is:
$$
beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix
$$
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 Jan 18 at 20:56
jordan_glen
1
asked Jan 18 at 20:39
Bn.F76Bn.F76
206
edited Jan 18 at 20:56
jordan_glen
1
asked Jan 18 at 20:39
Bn.F76Bn.F76
206
edited Jan 18 at 20:56
jordan_glen
1
edited Jan 18 at 20:56
jordan_glen
1
edited Jan 18 at 20:56
jordan_glen
1
1
asked Jan 18 at 20:39
Bn.F76Bn.F76
206
asked Jan 18 at 20:39
Bn.F76Bn.F76
206
asked Jan 18 at 20:39
Bn.F76Bn.F76
206
206
1
$begingroup$
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
Jan 18 at 20:45
$begingroup$
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
Jan 18 at 20:47
$begingroup$
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
Jan 18 at 20:47
$begingroup$
Summing up all your answers would make a valid answer,I think.
$endgroup$
– J.F
Jan 18 at 20:48
$begingroup$
The identity permutation assigns each element in a set to itself: $id(a) = a$, $id(b) = b$, etc.
$endgroup$
– jordan_glen
Jan 18 at 21:05
|
show 2 more comments
1
$begingroup$
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
Jan 18 at 20:45
$begingroup$
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
Jan 18 at 20:47
$begingroup$
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
Jan 18 at 20:47
$begingroup$
Summing up all your answers would make a valid answer,I think.
$endgroup$
– J.F
Jan 18 at 20:48
$begingroup$
The identity permutation assigns each element in a set to itself: $id(a) = a$, $id(b) = b$, etc.
$endgroup$
– jordan_glen
Jan 18 at 21:05
1
1
$begingroup$
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
Jan 18 at 20:45
$begingroup$
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
Jan 18 at 20:45
$begingroup$
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
Jan 18 at 20:47
$begingroup$
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
Jan 18 at 20:47
$begingroup$
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
Jan 18 at 20:47
$begingroup$
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
Jan 18 at 20:47
$begingroup$
Summing up all your answers would make a valid answer,I think.
$endgroup$
– J.F
Jan 18 at 20:48
$begingroup$
Summing up all your answers would make a valid answer,I think.
$endgroup$
– J.F
Jan 18 at 20:48
$begingroup$
The identity permutation assigns each element in a set to itself: $id(a) = a$, $id(b) = b$, etc.
$endgroup$
– jordan_glen
Jan 18 at 21:05
$begingroup$
The identity permutation assigns each element in a set to itself: $id(a) = a$, $id(b) = b$, etc.
$endgroup$
– jordan_glen
Jan 18 at 21:05
|
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=beginBmatrix
1 & 2 & 3 \
1 & 3 & 2 \
endBmatrix
$ 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 $sigma_2=beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix$ is $3$ because $sigma_2neq epsilon$ and $sigma_2^2neq epsilon$, but $sigma_2^3=epsilon$.
answered Jan 18 at 20:46
Kevin LongKevin Long
3,56121330
$endgroup$
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 Jan 18 at 20:55
BernardBernard
120k740114
$endgroup$
$begingroup$
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: $operatornamelcm(1, 1, 1) = 1$, which is the order of the identity permutation.
$endgroup$
– jordan_glen
Jan 18 at 21:01
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3078725%2fhow-do-i-find-the-order-of-a-bijection%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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=beginBmatrix
1 & 2 & 3 \
1 & 3 & 2 \
endBmatrix
$ 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 $sigma_2=beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix$ is $3$ because $sigma_2neq epsilon$ and $sigma_2^2neq epsilon$, but $sigma_2^3=epsilon$.
answered Jan 18 at 20:46
Kevin LongKevin Long
3,56121330
$endgroup$
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=beginBmatrix
1 & 2 & 3 \
1 & 3 & 2 \
endBmatrix
$ 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 $sigma_2=beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix$ is $3$ because $sigma_2neq epsilon$ and $sigma_2^2neq epsilon$, but $sigma_2^3=epsilon$.
answered Jan 18 at 20:46
Kevin LongKevin Long
3,56121330
$endgroup$
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=beginBmatrix
1 & 2 & 3 \
1 & 3 & 2 \
endBmatrix
$ 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 $sigma_2=beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix$ is $3$ because $sigma_2neq epsilon$ and $sigma_2^2neq epsilon$, but $sigma_2^3=epsilon$.
answered Jan 18 at 20:46
Kevin LongKevin Long
3,56121330
$endgroup$
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=beginBmatrix
1 & 2 & 3 \
1 & 3 & 2 \
endBmatrix
$ 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 $sigma_2=beginBmatrix
1 & 2 & 3 \
2 & 3 & 1 \
endBmatrix$ is $3$ because $sigma_2neq epsilon$ and $sigma_2^2neq epsilon$, but $sigma_2^3=epsilon$.
answered Jan 18 at 20:46
Kevin LongKevin Long
3,56121330
edited Jan 18 at 21:20
answered Jan 18 at 20:46
Kevin LongKevin Long
3,56121330
answered Jan 18 at 20:46
Kevin LongKevin Long
3,56121330
answered Jan 18 at 20:46
Kevin LongKevin Long
3,56121330
3,56121330
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 Jan 18 at 20:55
BernardBernard
120k740114
$endgroup$
$begingroup$
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: $operatornamelcm(1, 1, 1) = 1$, which is the order of the identity permutation.
$endgroup$
– jordan_glen
Jan 18 at 21:01
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 Jan 18 at 20:55
BernardBernard
120k740114
$endgroup$
$begingroup$
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: $operatornamelcm(1, 1, 1) = 1$, which is the order of the identity permutation.
$endgroup$
– jordan_glen
Jan 18 at 21:01
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 Jan 18 at 20:55
BernardBernard
120k740114
$endgroup$
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 Jan 18 at 20:55
BernardBernard
120k740114
answered Jan 18 at 20:55
BernardBernard
120k740114
answered Jan 18 at 20:55
BernardBernard
120k740114
answered Jan 18 at 20:55
BernardBernard
120k740114
120k740114
$begingroup$
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: $operatornamelcm(1, 1, 1) = 1$, which is the order of the identity permutation.
$endgroup$
– jordan_glen
Jan 18 at 21:01
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: $operatornamelcm(1, 1, 1) = 1$, which is the order of the identity permutation.
$endgroup$
– jordan_glen
Jan 18 at 21:01
$begingroup$
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: $operatornamelcm(1, 1, 1) = 1$, which is the order of the identity permutation.
$endgroup$
– jordan_glen
Jan 18 at 21:01
$begingroup$
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: $operatornamelcm(1, 1, 1) = 1$, which is the order of the identity permutation.
$endgroup$
– jordan_glen
Jan 18 at 21:01
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3078725%2fhow-do-i-find-the-order-of-a-bijection%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
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
Jan 18 at 20:45
$begingroup$
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
Jan 18 at 20:47
$begingroup$
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
Jan 18 at 20:47
$begingroup$
Summing up all your answers would make a valid answer,I think.
$endgroup$
– J.F
Jan 18 at 20:48
$begingroup$
The identity permutation assigns each element in a set to itself: $id(a) = a$, $id(b) = b$, etc.
$endgroup$
– jordan_glen
Jan 18 at 21:05