Help me understand how action is applied on the example given:
$begingroup$
Let $operatorname{Sym}(x)$ be the symetric group and then $operatorname{Sym}(2)={id,(1,2)}$. Let $X={1,2,3}$. (This might be wrong as i think this set must only contain $1$ and $2$ as elements.)
Define the function $f:operatorname{Sym}(2)times Xto X$, where $f(x,y)=xcirc y$ and $circ$ is the composition of maps.
So therefore $f:{(id,1),(id,2),(id,3),((1,2),1),((1,2),2),((1,2),3)}to{1,2,3}$.
The question now it is how can I apply an action on an element of the domain. So for example $f((1,2),2)=(1,2)circ 2=$?
This has been taken from lecture notes as being an example written as: $S_n$ acts on ${1, 2, . . . , n}$ via application of each map.
group-theory group-actions
edited Dec 22 '18 at 17:16
Shaun
10.6k113687
asked Dec 22 '18 at 15:27
ValVal
557
$endgroup$
add a comment |
$begingroup$
Let $operatorname{Sym}(x)$ be the symetric group and then $operatorname{Sym}(2)={id,(1,2)}$. Let $X={1,2,3}$. (This might be wrong as i think this set must only contain $1$ and $2$ as elements.)
Define the function $f:operatorname{Sym}(2)times Xto X$, where $f(x,y)=xcirc y$ and $circ$ is the composition of maps.
So therefore $f:{(id,1),(id,2),(id,3),((1,2),1),((1,2),2),((1,2),3)}to{1,2,3}$.
The question now it is how can I apply an action on an element of the domain. So for example $f((1,2),2)=(1,2)circ 2=$?
This has been taken from lecture notes as being an example written as: $S_n$ acts on ${1, 2, . . . , n}$ via application of each map.
group-theory group-actions
edited Dec 22 '18 at 17:16
Shaun
10.6k113687
asked Dec 22 '18 at 15:27
ValVal
557
$endgroup$
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Dec 22 '18 at 17:56
add a comment |
$begingroup$
Let $operatorname{Sym}(x)$ be the symetric group and then $operatorname{Sym}(2)={id,(1,2)}$. Let $X={1,2,3}$. (This might be wrong as i think this set must only contain $1$ and $2$ as elements.)
Define the function $f:operatorname{Sym}(2)times Xto X$, where $f(x,y)=xcirc y$ and $circ$ is the composition of maps.
So therefore $f:{(id,1),(id,2),(id,3),((1,2),1),((1,2),2),((1,2),3)}to{1,2,3}$.
The question now it is how can I apply an action on an element of the domain. So for example $f((1,2),2)=(1,2)circ 2=$?
This has been taken from lecture notes as being an example written as: $S_n$ acts on ${1, 2, . . . , n}$ via application of each map.
group-theory group-actions
edited Dec 22 '18 at 17:16
Shaun
10.6k113687
asked Dec 22 '18 at 15:27
ValVal
557
$endgroup$
Let $operatorname{Sym}(x)$ be the symetric group and then $operatorname{Sym}(2)={id,(1,2)}$. Let $X={1,2,3}$. (This might be wrong as i think this set must only contain $1$ and $2$ as elements.)
Define the function $f:operatorname{Sym}(2)times Xto X$, where $f(x,y)=xcirc y$ and $circ$ is the composition of maps.
So therefore $f:{(id,1),(id,2),(id,3),((1,2),1),((1,2),2),((1,2),3)}to{1,2,3}$.
The question now it is how can I apply an action on an element of the domain. So for example $f((1,2),2)=(1,2)circ 2=$?
This has been taken from lecture notes as being an example written as: $S_n$ acts on ${1, 2, . . . , n}$ via application of each map.
group-theory group-actions
group-theory group-actions
edited Dec 22 '18 at 17:16
Shaun
10.6k113687
asked Dec 22 '18 at 15:27
ValVal
557
edited Dec 22 '18 at 17:16
Shaun
10.6k113687
asked Dec 22 '18 at 15:27
ValVal
557
edited Dec 22 '18 at 17:16
Shaun
10.6k113687
edited Dec 22 '18 at 17:16
Shaun
10.6k113687
edited Dec 22 '18 at 17:16
Shaun
10.6k113687
10.6k113687
asked Dec 22 '18 at 15:27
ValVal
557
asked Dec 22 '18 at 15:27
ValVal
557
asked Dec 22 '18 at 15:27
ValVal
557
557
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Dec 22 '18 at 17:56
add a comment |
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Dec 22 '18 at 17:56
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Dec 22 '18 at 17:56
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Dec 22 '18 at 17:56
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
If I understood you correctly, $operatorname{Sym}(2)$ acts on $X$ via $f$ by defining $f: operatorname{Sym}(2)times Xto X$ by $(alpha, x)mapsto alphacirc x$; that is, $f(alpha, x)=alphacirc xstackrel{operatorname{Def}}{=}alpha(x)$, where $alpha(x)$ is $alpha$ evaluated at $xin X$.
It follows that here $X$ can be any set with ${1,2}subseteq X$.
Using your example, $f((12), 2)=(12)circ 2stackrel{operatorname{Def}}{=}(12)(2)=1$.
NB: Here $stackrel{operatorname{Def}}{=}$ means "is, by definition, equal to".
answered Dec 22 '18 at 17:08
ShaunShaun
10.6k113687
$endgroup$
$begingroup$
I m sorry but how did you get (12)(2)=1? I m a bit confuse
$endgroup$
– Val
Dec 22 '18 at 17:20
$begingroup$
Is it clear now, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:26
$begingroup$
Not really, sorry. When you do (12) ∘ 2 I don't really understand what you are following at that bit. I mean what sort of operation you do.
$endgroup$
– Val
Dec 22 '18 at 17:47
$begingroup$
The notation $(12)circ 2$ means "$(12)$ of $2$" because, remember, $(12)$ is a bijection on ${1,2}$; thus "$(12)circ 2$" is synonymous with "$(12)$ evaluated at $2$", which is, in turn, synonymous with "$(12)(2)$", i.e., $1$. Is that any better, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:51
$begingroup$
Is it the fact that 2↦1 in (12)?
$endgroup$
– Val
Dec 22 '18 at 17:56
|
show 9 more comments
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1 Answer
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$begingroup$
If I understood you correctly, $operatorname{Sym}(2)$ acts on $X$ via $f$ by defining $f: operatorname{Sym}(2)times Xto X$ by $(alpha, x)mapsto alphacirc x$; that is, $f(alpha, x)=alphacirc xstackrel{operatorname{Def}}{=}alpha(x)$, where $alpha(x)$ is $alpha$ evaluated at $xin X$.
It follows that here $X$ can be any set with ${1,2}subseteq X$.
Using your example, $f((12), 2)=(12)circ 2stackrel{operatorname{Def}}{=}(12)(2)=1$.
NB: Here $stackrel{operatorname{Def}}{=}$ means "is, by definition, equal to".
answered Dec 22 '18 at 17:08
ShaunShaun
10.6k113687
$endgroup$
$begingroup$
I m sorry but how did you get (12)(2)=1? I m a bit confuse
$endgroup$
– Val
Dec 22 '18 at 17:20
$begingroup$
Is it clear now, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:26
$begingroup$
Not really, sorry. When you do (12) ∘ 2 I don't really understand what you are following at that bit. I mean what sort of operation you do.
$endgroup$
– Val
Dec 22 '18 at 17:47
$begingroup$
The notation $(12)circ 2$ means "$(12)$ of $2$" because, remember, $(12)$ is a bijection on ${1,2}$; thus "$(12)circ 2$" is synonymous with "$(12)$ evaluated at $2$", which is, in turn, synonymous with "$(12)(2)$", i.e., $1$. Is that any better, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:51
$begingroup$
Is it the fact that 2↦1 in (12)?
$endgroup$
– Val
Dec 22 '18 at 17:56
|
show 9 more comments
$begingroup$
If I understood you correctly, $operatorname{Sym}(2)$ acts on $X$ via $f$ by defining $f: operatorname{Sym}(2)times Xto X$ by $(alpha, x)mapsto alphacirc x$; that is, $f(alpha, x)=alphacirc xstackrel{operatorname{Def}}{=}alpha(x)$, where $alpha(x)$ is $alpha$ evaluated at $xin X$.
It follows that here $X$ can be any set with ${1,2}subseteq X$.
Using your example, $f((12), 2)=(12)circ 2stackrel{operatorname{Def}}{=}(12)(2)=1$.
NB: Here $stackrel{operatorname{Def}}{=}$ means "is, by definition, equal to".
answered Dec 22 '18 at 17:08
ShaunShaun
10.6k113687
$endgroup$
$begingroup$
I m sorry but how did you get (12)(2)=1? I m a bit confuse
$endgroup$
– Val
Dec 22 '18 at 17:20
$begingroup$
Is it clear now, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:26
$begingroup$
Not really, sorry. When you do (12) ∘ 2 I don't really understand what you are following at that bit. I mean what sort of operation you do.
$endgroup$
– Val
Dec 22 '18 at 17:47
$begingroup$
The notation $(12)circ 2$ means "$(12)$ of $2$" because, remember, $(12)$ is a bijection on ${1,2}$; thus "$(12)circ 2$" is synonymous with "$(12)$ evaluated at $2$", which is, in turn, synonymous with "$(12)(2)$", i.e., $1$. Is that any better, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:51
$begingroup$
Is it the fact that 2↦1 in (12)?
$endgroup$
– Val
Dec 22 '18 at 17:56
|
show 9 more comments
$begingroup$
If I understood you correctly, $operatorname{Sym}(2)$ acts on $X$ via $f$ by defining $f: operatorname{Sym}(2)times Xto X$ by $(alpha, x)mapsto alphacirc x$; that is, $f(alpha, x)=alphacirc xstackrel{operatorname{Def}}{=}alpha(x)$, where $alpha(x)$ is $alpha$ evaluated at $xin X$.
It follows that here $X$ can be any set with ${1,2}subseteq X$.
Using your example, $f((12), 2)=(12)circ 2stackrel{operatorname{Def}}{=}(12)(2)=1$.
NB: Here $stackrel{operatorname{Def}}{=}$ means "is, by definition, equal to".
answered Dec 22 '18 at 17:08
ShaunShaun
10.6k113687
$endgroup$
If I understood you correctly, $operatorname{Sym}(2)$ acts on $X$ via $f$ by defining $f: operatorname{Sym}(2)times Xto X$ by $(alpha, x)mapsto alphacirc x$; that is, $f(alpha, x)=alphacirc xstackrel{operatorname{Def}}{=}alpha(x)$, where $alpha(x)$ is $alpha$ evaluated at $xin X$.
It follows that here $X$ can be any set with ${1,2}subseteq X$.
Using your example, $f((12), 2)=(12)circ 2stackrel{operatorname{Def}}{=}(12)(2)=1$.
NB: Here $stackrel{operatorname{Def}}{=}$ means "is, by definition, equal to".
answered Dec 22 '18 at 17:08
ShaunShaun
10.6k113687
edited Dec 22 '18 at 17:29
answered Dec 22 '18 at 17:08
ShaunShaun
10.6k113687
answered Dec 22 '18 at 17:08
ShaunShaun
10.6k113687
answered Dec 22 '18 at 17:08
ShaunShaun
10.6k113687
10.6k113687
$begingroup$
I m sorry but how did you get (12)(2)=1? I m a bit confuse
$endgroup$
– Val
Dec 22 '18 at 17:20
$begingroup$
Is it clear now, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:26
$begingroup$
Not really, sorry. When you do (12) ∘ 2 I don't really understand what you are following at that bit. I mean what sort of operation you do.
$endgroup$
– Val
Dec 22 '18 at 17:47
$begingroup$
The notation $(12)circ 2$ means "$(12)$ of $2$" because, remember, $(12)$ is a bijection on ${1,2}$; thus "$(12)circ 2$" is synonymous with "$(12)$ evaluated at $2$", which is, in turn, synonymous with "$(12)(2)$", i.e., $1$. Is that any better, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:51
$begingroup$
Is it the fact that 2↦1 in (12)?
$endgroup$
– Val
Dec 22 '18 at 17:56
|
show 9 more comments
$begingroup$
I m sorry but how did you get (12)(2)=1? I m a bit confuse
$endgroup$
– Val
Dec 22 '18 at 17:20
$begingroup$
Is it clear now, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:26
$begingroup$
Not really, sorry. When you do (12) ∘ 2 I don't really understand what you are following at that bit. I mean what sort of operation you do.
$endgroup$
– Val
Dec 22 '18 at 17:47
$begingroup$
The notation $(12)circ 2$ means "$(12)$ of $2$" because, remember, $(12)$ is a bijection on ${1,2}$; thus "$(12)circ 2$" is synonymous with "$(12)$ evaluated at $2$", which is, in turn, synonymous with "$(12)(2)$", i.e., $1$. Is that any better, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:51
$begingroup$
Is it the fact that 2↦1 in (12)?
$endgroup$
– Val
Dec 22 '18 at 17:56
$begingroup$
I m sorry but how did you get (12)(2)=1? I m a bit confuse
$endgroup$
– Val
Dec 22 '18 at 17:20
$begingroup$
I m sorry but how did you get (12)(2)=1? I m a bit confuse
$endgroup$
– Val
Dec 22 '18 at 17:20
$begingroup$
Is it clear now, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:26
$begingroup$
Is it clear now, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:26
$begingroup$
Not really, sorry. When you do (12) ∘ 2 I don't really understand what you are following at that bit. I mean what sort of operation you do.
$endgroup$
– Val
Dec 22 '18 at 17:47
$begingroup$
Not really, sorry. When you do (12) ∘ 2 I don't really understand what you are following at that bit. I mean what sort of operation you do.
$endgroup$
– Val
Dec 22 '18 at 17:47
$begingroup$
The notation $(12)circ 2$ means "$(12)$ of $2$" because, remember, $(12)$ is a bijection on ${1,2}$; thus "$(12)circ 2$" is synonymous with "$(12)$ evaluated at $2$", which is, in turn, synonymous with "$(12)(2)$", i.e., $1$. Is that any better, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:51
$begingroup$
The notation $(12)circ 2$ means "$(12)$ of $2$" because, remember, $(12)$ is a bijection on ${1,2}$; thus "$(12)circ 2$" is synonymous with "$(12)$ evaluated at $2$", which is, in turn, synonymous with "$(12)(2)$", i.e., $1$. Is that any better, @BurLeXyOOnuTz?
$endgroup$
– Shaun
Dec 22 '18 at 17:51
$begingroup$
Is it the fact that 2↦1 in (12)?
$endgroup$
– Val
Dec 22 '18 at 17:56
$begingroup$
Is it the fact that 2↦1 in (12)?
$endgroup$
– Val
Dec 22 '18 at 17:56
|
show 9 more comments
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$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Dec 22 '18 at 17:56