Mixing
Orbit of permutaton and orbit of group action
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
1) Suppose $G$ is a group and $S$ is a set and we consider the action of the group $G$ on a set $S$. Then the orbit of $xin S$ is the following subset of $S$, $$textOrb_x=gin G.$$
2) But let's take some set $S$ and it's symmetric group $A(S)$ and some $thetain A(S)$. We say that $asim b$ iff $theta^i(a)=b$ for some integer $i$. We can easily check that this is an equivalence relation. We call the equivalence class of an element $sin S$ the orbit of $s$ under $theta$; thus the orbit of $s$ under $theta$ consists of all the elements $theta^i(s): iin mathbbZ$.
I would like to ask the following question: Are the orbit notions in 1) and 2) are the same? If yes, can anyone explain what is the group action in 2)?
P.S. I have found a couple of topics in MSE which contain the same question but did not find an answer to my question. So please do not duplicate. Moreover, it could be interested to people who have the same question.
abstract-algebra group-theory permutations
asked 45 mins ago
RFZ
4,78631337
add a comment |Â
up vote
2
down vote
favorite
1) Suppose $G$ is a group and $S$ is a set and we consider the action of the group $G$ on a set $S$. Then the orbit of $xin S$ is the following subset of $S$, $$textOrb_x=gin G.$$
2) But let's take some set $S$ and it's symmetric group $A(S)$ and some $thetain A(S)$. We say that $asim b$ iff $theta^i(a)=b$ for some integer $i$. We can easily check that this is an equivalence relation. We call the equivalence class of an element $sin S$ the orbit of $s$ under $theta$; thus the orbit of $s$ under $theta$ consists of all the elements $theta^i(s): iin mathbbZ$.
I would like to ask the following question: Are the orbit notions in 1) and 2) are the same? If yes, can anyone explain what is the group action in 2)?
P.S. I have found a couple of topics in MSE which contain the same question but did not find an answer to my question. So please do not duplicate. Moreover, it could be interested to people who have the same question.
abstract-algebra group-theory permutations
asked 45 mins ago
RFZ
4,78631337
1
The equivalence relation in $2$ is on some set $S$ : is it $S_n$, or is it $1,2,...,n$? Because both give equivalence classes : one is the orbits given by the cyclic group generated by $theta$, and the other is the cycle decomposition of $theta$. That is, are $a,b in S_n$ or $a,b$ are between $1$ and $n$?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
40 mins ago
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó, I have edited please take a look
– RFZ
35 mins ago
Thank you for the edit.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
1) Suppose $G$ is a group and $S$ is a set and we consider the action of the group $G$ on a set $S$. Then the orbit of $xin S$ is the following subset of $S$, $$textOrb_x=gin G.$$
2) But let's take some set $S$ and it's symmetric group $A(S)$ and some $thetain A(S)$. We say that $asim b$ iff $theta^i(a)=b$ for some integer $i$. We can easily check that this is an equivalence relation. We call the equivalence class of an element $sin S$ the orbit of $s$ under $theta$; thus the orbit of $s$ under $theta$ consists of all the elements $theta^i(s): iin mathbbZ$.
I would like to ask the following question: Are the orbit notions in 1) and 2) are the same? If yes, can anyone explain what is the group action in 2)?
P.S. I have found a couple of topics in MSE which contain the same question but did not find an answer to my question. So please do not duplicate. Moreover, it could be interested to people who have the same question.
abstract-algebra group-theory permutations
asked 45 mins ago
RFZ
4,78631337
1) Suppose $G$ is a group and $S$ is a set and we consider the action of the group $G$ on a set $S$. Then the orbit of $xin S$ is the following subset of $S$, $$textOrb_x=gin G.$$
2) But let's take some set $S$ and it's symmetric group $A(S)$ and some $thetain A(S)$. We say that $asim b$ iff $theta^i(a)=b$ for some integer $i$. We can easily check that this is an equivalence relation. We call the equivalence class of an element $sin S$ the orbit of $s$ under $theta$; thus the orbit of $s$ under $theta$ consists of all the elements $theta^i(s): iin mathbbZ$.
I would like to ask the following question: Are the orbit notions in 1) and 2) are the same? If yes, can anyone explain what is the group action in 2)?
P.S. I have found a couple of topics in MSE which contain the same question but did not find an answer to my question. So please do not duplicate. Moreover, it could be interested to people who have the same question.
abstract-algebra group-theory permutations
abstract-algebra group-theory permutations
asked 45 mins ago
RFZ
4,78631337
asked 45 mins ago
RFZ
4,78631337
edited 36 mins ago
asked 45 mins ago
RFZ
4,78631337
asked 45 mins ago
RFZ
4,78631337
asked 45 mins ago
RFZ
4,78631337
4,78631337
1
The equivalence relation in $2$ is on some set $S$ : is it $S_n$, or is it $1,2,...,n$? Because both give equivalence classes : one is the orbits given by the cyclic group generated by $theta$, and the other is the cycle decomposition of $theta$. That is, are $a,b in S_n$ or $a,b$ are between $1$ and $n$?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
40 mins ago
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó, I have edited please take a look
– RFZ
35 mins ago
Thank you for the edit.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35 mins ago
add a comment |Â
1
The equivalence relation in $2$ is on some set $S$ : is it $S_n$, or is it $1,2,...,n$? Because both give equivalence classes : one is the orbits given by the cyclic group generated by $theta$, and the other is the cycle decomposition of $theta$. That is, are $a,b in S_n$ or $a,b$ are between $1$ and $n$?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
40 mins ago
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó, I have edited please take a look
– RFZ
35 mins ago
Thank you for the edit.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35 mins ago
1
1
The equivalence relation in $2$ is on some set $S$ : is it $S_n$, or is it $1,2,...,n$? Because both give equivalence classes : one is the orbits given by the cyclic group generated by $theta$, and the other is the cycle decomposition of $theta$. That is, are $a,b in S_n$ or $a,b$ are between $1$ and $n$?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
40 mins ago
The equivalence relation in $2$ is on some set $S$ : is it $S_n$, or is it $1,2,...,n$? Because both give equivalence classes : one is the orbits given by the cyclic group generated by $theta$, and the other is the cycle decomposition of $theta$. That is, are $a,b in S_n$ or $a,b$ are between $1$ and $n$?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
40 mins ago
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó, I have edited please take a look
– RFZ
35 mins ago
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó, I have edited please take a look
– RFZ
35 mins ago
Thank you for the edit.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35 mins ago
Thank you for the edit.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
Yes, they are the same, in that 2) is a special case of 1).
How so ? Well consider the set $S=1,...,n$ and $G=langle theta rangle$, the subgroup of $S_n$ generated by $theta$. Then $G$ acts on $S$, and the orbit of $a$ under this action is the equivalence class of $a$ under your equivalence relation
answered 35 mins ago
Max
11.3k11037
Indeed, this makes sense!
– RFZ
34 mins ago
As far as I understand, we just need to consider the action of subgroup on the set $S$, where the subgroup is the cyclic group generated by our $theta$, right?
– RFZ
32 mins ago
add a comment |Â
up vote
3
down vote
Consider the natural action of $mathfrak S_n$ on the set $1,ldots ,n$ defined by $$forall sigma in mathfrak S_n, forall x in 1,ldots ,n, quadsigma cdot x := sigma (x)$$
The orbit of an element $x$ under this action is the set $ sigma in mathfrak S_n$.
Now, fix $theta in mathfrak S_n$ and consider the restriction of this action to the subgroup $langle theta rangle$ generated by $theta$. More precisely, if $rho: mathfrak S_n rightarrow mathfrak S_n$ is the morphism of groups describing the action, consider its restriction $left.rhoright|_langle theta rangle$.
Then, the orbit of $x$ under this restricted action is given by $textOrb_x = theta^i(x): iin mathbbZ$, as $langle theta rangle = theta^i: iin mathbbZ$.
answered 37 mins ago
Suzet
2,533527
add a comment |Â
up vote
1
down vote
For a fixed permutation $theta$, the equivalence relation $asim b$ iff $theta^i(a)=b$, means that $a$ and $b$ are on the same cycle of the permutation $theta$. In the last defintion you apply the same permutation several times to a
Notice that when you define the orbit of an element $x$ the permutations in $G=S_n$ are variable (this is your $g$ running in $S_n$) while your element $x$ is fixed.
answered 39 mins ago
Hector Blandin
1,819816
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Yes, they are the same, in that 2) is a special case of 1).
How so ? Well consider the set $S=1,...,n$ and $G=langle theta rangle$, the subgroup of $S_n$ generated by $theta$. Then $G$ acts on $S$, and the orbit of $a$ under this action is the equivalence class of $a$ under your equivalence relation
answered 35 mins ago
Max
11.3k11037
Indeed, this makes sense!
– RFZ
34 mins ago
As far as I understand, we just need to consider the action of subgroup on the set $S$, where the subgroup is the cyclic group generated by our $theta$, right?
– RFZ
32 mins ago
add a comment |Â
up vote
4
down vote
Yes, they are the same, in that 2) is a special case of 1).
How so ? Well consider the set $S=1,...,n$ and $G=langle theta rangle$, the subgroup of $S_n$ generated by $theta$. Then $G$ acts on $S$, and the orbit of $a$ under this action is the equivalence class of $a$ under your equivalence relation
answered 35 mins ago
Max
11.3k11037
Indeed, this makes sense!
– RFZ
34 mins ago
As far as I understand, we just need to consider the action of subgroup on the set $S$, where the subgroup is the cyclic group generated by our $theta$, right?
– RFZ
32 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Yes, they are the same, in that 2) is a special case of 1).
How so ? Well consider the set $S=1,...,n$ and $G=langle theta rangle$, the subgroup of $S_n$ generated by $theta$. Then $G$ acts on $S$, and the orbit of $a$ under this action is the equivalence class of $a$ under your equivalence relation
answered 35 mins ago
Max
11.3k11037
Yes, they are the same, in that 2) is a special case of 1).
How so ? Well consider the set $S=1,...,n$ and $G=langle theta rangle$, the subgroup of $S_n$ generated by $theta$. Then $G$ acts on $S$, and the orbit of $a$ under this action is the equivalence class of $a$ under your equivalence relation
answered 35 mins ago
Max
11.3k11037
answered 35 mins ago
Max
11.3k11037
answered 35 mins ago
Max
11.3k11037
answered 35 mins ago
Max
11.3k11037
11.3k11037
Indeed, this makes sense!
– RFZ
34 mins ago
As far as I understand, we just need to consider the action of subgroup on the set $S$, where the subgroup is the cyclic group generated by our $theta$, right?
– RFZ
32 mins ago
add a comment |Â
Indeed, this makes sense!
– RFZ
34 mins ago
As far as I understand, we just need to consider the action of subgroup on the set $S$, where the subgroup is the cyclic group generated by our $theta$, right?
– RFZ
32 mins ago
Indeed, this makes sense!
– RFZ
34 mins ago
Indeed, this makes sense!
– RFZ
34 mins ago
As far as I understand, we just need to consider the action of subgroup on the set $S$, where the subgroup is the cyclic group generated by our $theta$, right?
– RFZ
32 mins ago
As far as I understand, we just need to consider the action of subgroup on the set $S$, where the subgroup is the cyclic group generated by our $theta$, right?
– RFZ
32 mins ago
add a comment |Â
up vote
3
down vote
Consider the natural action of $mathfrak S_n$ on the set $1,ldots ,n$ defined by $$forall sigma in mathfrak S_n, forall x in 1,ldots ,n, quadsigma cdot x := sigma (x)$$
The orbit of an element $x$ under this action is the set $ sigma in mathfrak S_n$.
Now, fix $theta in mathfrak S_n$ and consider the restriction of this action to the subgroup $langle theta rangle$ generated by $theta$. More precisely, if $rho: mathfrak S_n rightarrow mathfrak S_n$ is the morphism of groups describing the action, consider its restriction $left.rhoright|_langle theta rangle$.
Then, the orbit of $x$ under this restricted action is given by $textOrb_x = theta^i(x): iin mathbbZ$, as $langle theta rangle = theta^i: iin mathbbZ$.
answered 37 mins ago
Suzet
2,533527
add a comment |Â
up vote
3
down vote
Consider the natural action of $mathfrak S_n$ on the set $1,ldots ,n$ defined by $$forall sigma in mathfrak S_n, forall x in 1,ldots ,n, quadsigma cdot x := sigma (x)$$
The orbit of an element $x$ under this action is the set $ sigma in mathfrak S_n$.
Now, fix $theta in mathfrak S_n$ and consider the restriction of this action to the subgroup $langle theta rangle$ generated by $theta$. More precisely, if $rho: mathfrak S_n rightarrow mathfrak S_n$ is the morphism of groups describing the action, consider its restriction $left.rhoright|_langle theta rangle$.
Then, the orbit of $x$ under this restricted action is given by $textOrb_x = theta^i(x): iin mathbbZ$, as $langle theta rangle = theta^i: iin mathbbZ$.
answered 37 mins ago
Suzet
2,533527
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Consider the natural action of $mathfrak S_n$ on the set $1,ldots ,n$ defined by $$forall sigma in mathfrak S_n, forall x in 1,ldots ,n, quadsigma cdot x := sigma (x)$$
The orbit of an element $x$ under this action is the set $ sigma in mathfrak S_n$.
Now, fix $theta in mathfrak S_n$ and consider the restriction of this action to the subgroup $langle theta rangle$ generated by $theta$. More precisely, if $rho: mathfrak S_n rightarrow mathfrak S_n$ is the morphism of groups describing the action, consider its restriction $left.rhoright|_langle theta rangle$.
Then, the orbit of $x$ under this restricted action is given by $textOrb_x = theta^i(x): iin mathbbZ$, as $langle theta rangle = theta^i: iin mathbbZ$.
answered 37 mins ago
Suzet
2,533527
Consider the natural action of $mathfrak S_n$ on the set $1,ldots ,n$ defined by $$forall sigma in mathfrak S_n, forall x in 1,ldots ,n, quadsigma cdot x := sigma (x)$$
The orbit of an element $x$ under this action is the set $ sigma in mathfrak S_n$.
Now, fix $theta in mathfrak S_n$ and consider the restriction of this action to the subgroup $langle theta rangle$ generated by $theta$. More precisely, if $rho: mathfrak S_n rightarrow mathfrak S_n$ is the morphism of groups describing the action, consider its restriction $left.rhoright|_langle theta rangle$.
Then, the orbit of $x$ under this restricted action is given by $textOrb_x = theta^i(x): iin mathbbZ$, as $langle theta rangle = theta^i: iin mathbbZ$.
answered 37 mins ago
Suzet
2,533527
answered 37 mins ago
Suzet
2,533527
answered 37 mins ago
Suzet
2,533527
answered 37 mins ago
Suzet
2,533527
2,533527
add a comment |Â
add a comment |Â
up vote
1
down vote
For a fixed permutation $theta$, the equivalence relation $asim b$ iff $theta^i(a)=b$, means that $a$ and $b$ are on the same cycle of the permutation $theta$. In the last defintion you apply the same permutation several times to a
Notice that when you define the orbit of an element $x$ the permutations in $G=S_n$ are variable (this is your $g$ running in $S_n$) while your element $x$ is fixed.
answered 39 mins ago
Hector Blandin
1,819816
add a comment |Â
up vote
1
down vote
For a fixed permutation $theta$, the equivalence relation $asim b$ iff $theta^i(a)=b$, means that $a$ and $b$ are on the same cycle of the permutation $theta$. In the last defintion you apply the same permutation several times to a
Notice that when you define the orbit of an element $x$ the permutations in $G=S_n$ are variable (this is your $g$ running in $S_n$) while your element $x$ is fixed.
answered 39 mins ago
Hector Blandin
1,819816
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For a fixed permutation $theta$, the equivalence relation $asim b$ iff $theta^i(a)=b$, means that $a$ and $b$ are on the same cycle of the permutation $theta$. In the last defintion you apply the same permutation several times to a
Notice that when you define the orbit of an element $x$ the permutations in $G=S_n$ are variable (this is your $g$ running in $S_n$) while your element $x$ is fixed.
answered 39 mins ago
Hector Blandin
1,819816
For a fixed permutation $theta$, the equivalence relation $asim b$ iff $theta^i(a)=b$, means that $a$ and $b$ are on the same cycle of the permutation $theta$. In the last defintion you apply the same permutation several times to a
Notice that when you define the orbit of an element $x$ the permutations in $G=S_n$ are variable (this is your $g$ running in $S_n$) while your element $x$ is fixed.
answered 39 mins ago
Hector Blandin
1,819816
answered 39 mins ago
Hector Blandin
1,819816
answered 39 mins ago
Hector Blandin
1,819816
answered 39 mins ago
Hector Blandin
1,819816
1,819816
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2972373%2forbit-of-permutaton-and-orbit-of-group-action%23new-answer', 'question_page');
);
Post as a guest
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
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
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
1
The equivalence relation in $2$ is on some set $S$ : is it $S_n$, or is it $1,2,...,n$? Because both give equivalence classes : one is the orbits given by the cyclic group generated by $theta$, and the other is the cycle decomposition of $theta$. That is, are $a,b in S_n$ or $a,b$ are between $1$ and $n$?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
40 mins ago
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó, I have edited please take a look
– RFZ
35 mins ago
Thank you for the edit.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
35 mins ago