G-congruence equivalence relation
Let $G$ act on $X$. An equivalence relation ~ on $X$ is called a $G$-congruence if whenever $x,y in X$ then $x$~$y$ implies $xg$~$yg$ for all $g in G$.
Suppose ~ is a $G$-congruence on $X$. Show that $G$ acts on the set $X/$~ of equivalence classes.
I'm new to group actions (and most of the group theory in fact) and I'm just trying to make sure that I understand things correctly. So basically, we want to show that if we take an equivalence class of some $x in X$, denote it by $[x]$ and act (act on all elements of that class?) on it with $G$ we would get another (or possibly the same?) equivalence class back, is that correct? Since the equivalence relation that we are using to factor $X$ is of $G$-congruence type we know that by acting with G on $[x]$ we would get bunch of elements (as many as in $[x]$) in $X$ which are also equivalent to each other. But how do we know that they'd actually create another equivalence class from $X/$~, which I believe is what we want to show? It is probably something really simple but somehow I am struggling to see it.
group-theory group-actions
asked Nov 20 at 10:44
amator2357
76
add a comment |
Let $G$ act on $X$. An equivalence relation ~ on $X$ is called a $G$-congruence if whenever $x,y in X$ then $x$~$y$ implies $xg$~$yg$ for all $g in G$.
Suppose ~ is a $G$-congruence on $X$. Show that $G$ acts on the set $X/$~ of equivalence classes.
I'm new to group actions (and most of the group theory in fact) and I'm just trying to make sure that I understand things correctly. So basically, we want to show that if we take an equivalence class of some $x in X$, denote it by $[x]$ and act (act on all elements of that class?) on it with $G$ we would get another (or possibly the same?) equivalence class back, is that correct? Since the equivalence relation that we are using to factor $X$ is of $G$-congruence type we know that by acting with G on $[x]$ we would get bunch of elements (as many as in $[x]$) in $X$ which are also equivalent to each other. But how do we know that they'd actually create another equivalence class from $X/$~, which I believe is what we want to show? It is probably something really simple but somehow I am struggling to see it.
group-theory group-actions
asked Nov 20 at 10:44
amator2357
76
To prove that the induced action of $G$ on $X/sim$ is well-defined, you have to prove that, for all $g in G$ and $x,y in X$, we have $x sim y Rightarrow xg sim yg$, which follows immediately from the definition of a $G$-congruence.
– Derek Holt
Nov 20 at 11:29
Thank you Derek! Clearly I didn't quite understand what the question was asking me to do.
– amator2357
Nov 20 at 11:37
add a comment |
Let $G$ act on $X$. An equivalence relation ~ on $X$ is called a $G$-congruence if whenever $x,y in X$ then $x$~$y$ implies $xg$~$yg$ for all $g in G$.
Suppose ~ is a $G$-congruence on $X$. Show that $G$ acts on the set $X/$~ of equivalence classes.
I'm new to group actions (and most of the group theory in fact) and I'm just trying to make sure that I understand things correctly. So basically, we want to show that if we take an equivalence class of some $x in X$, denote it by $[x]$ and act (act on all elements of that class?) on it with $G$ we would get another (or possibly the same?) equivalence class back, is that correct? Since the equivalence relation that we are using to factor $X$ is of $G$-congruence type we know that by acting with G on $[x]$ we would get bunch of elements (as many as in $[x]$) in $X$ which are also equivalent to each other. But how do we know that they'd actually create another equivalence class from $X/$~, which I believe is what we want to show? It is probably something really simple but somehow I am struggling to see it.
group-theory group-actions
asked Nov 20 at 10:44
amator2357
76
Let $G$ act on $X$. An equivalence relation ~ on $X$ is called a $G$-congruence if whenever $x,y in X$ then $x$~$y$ implies $xg$~$yg$ for all $g in G$.
Suppose ~ is a $G$-congruence on $X$. Show that $G$ acts on the set $X/$~ of equivalence classes.
I'm new to group actions (and most of the group theory in fact) and I'm just trying to make sure that I understand things correctly. So basically, we want to show that if we take an equivalence class of some $x in X$, denote it by $[x]$ and act (act on all elements of that class?) on it with $G$ we would get another (or possibly the same?) equivalence class back, is that correct? Since the equivalence relation that we are using to factor $X$ is of $G$-congruence type we know that by acting with G on $[x]$ we would get bunch of elements (as many as in $[x]$) in $X$ which are also equivalent to each other. But how do we know that they'd actually create another equivalence class from $X/$~, which I believe is what we want to show? It is probably something really simple but somehow I am struggling to see it.
group-theory group-actions
group-theory group-actions
asked Nov 20 at 10:44
amator2357
76
asked Nov 20 at 10:44
amator2357
76
edited Nov 20 at 11:20
asked Nov 20 at 10:44
amator2357
76
asked Nov 20 at 10:44
amator2357
76
asked Nov 20 at 10:44
amator2357
76
76
To prove that the induced action of $G$ on $X/sim$ is well-defined, you have to prove that, for all $g in G$ and $x,y in X$, we have $x sim y Rightarrow xg sim yg$, which follows immediately from the definition of a $G$-congruence.
– Derek Holt
Nov 20 at 11:29
Thank you Derek! Clearly I didn't quite understand what the question was asking me to do.
– amator2357
Nov 20 at 11:37
add a comment |
To prove that the induced action of $G$ on $X/sim$ is well-defined, you have to prove that, for all $g in G$ and $x,y in X$, we have $x sim y Rightarrow xg sim yg$, which follows immediately from the definition of a $G$-congruence.
– Derek Holt
Nov 20 at 11:29
Thank you Derek! Clearly I didn't quite understand what the question was asking me to do.
– amator2357
Nov 20 at 11:37
To prove that the induced action of $G$ on $X/sim$ is well-defined, you have to prove that, for all $g in G$ and $x,y in X$, we have $x sim y Rightarrow xg sim yg$, which follows immediately from the definition of a $G$-congruence.
– Derek Holt
Nov 20 at 11:29
To prove that the induced action of $G$ on $X/sim$ is well-defined, you have to prove that, for all $g in G$ and $x,y in X$, we have $x sim y Rightarrow xg sim yg$, which follows immediately from the definition of a $G$-congruence.
– Derek Holt
Nov 20 at 11:29
Thank you Derek! Clearly I didn't quite understand what the question was asking me to do.
– amator2357
Nov 20 at 11:37
Thank you Derek! Clearly I didn't quite understand what the question was asking me to do.
– amator2357
Nov 20 at 11:37
add a comment |
1 Answer
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First of all I'm going to use left action $gx$ instead of right $xg$ because it is more common. And these are (sort of) equivalent anyway.
Show that $G$ acts on the set $X/$~ of equivalence classes.
Well, this statement is a bit trivial. Because if $Y$ is any set then $G$ acts on $Y$ via $gx:=x$. There are many different ways that $G$ can act on $Y$.
But I assume that the author meant something more interesting, i.e. that there's an induced action of $G$ on $X/sim$. Induced from the action of $G$ on $X$.
So basically, we want to show that if we take an equivalence class of some $x in X$, denote it by $[x]$ and act (act on all elements of that class?) on it with $G$ we would get another (or possibly the same?) equivalence class back, is that correct?
Yes, this is correct. I mean except for act on all elements of that class. No, you don't act on elements of the class, you act on the class itself.
What you want to show is that you have some function $Gtimes Ato A$ that satisfies group action properties. In this case $A=X/sim$.
So as I've mentioned earlier there are multiple ways to define such a group action. But I assume that what the author really wanted is the action induced from the action on $X$. In other words this is what we are looking for:
$$Gtimes X/simto X/sim$$
$$(g,[x])mapsto [gx]$$
or for short $g[x]:=[gx]$. So first of all you have to show that this is a well defined function. In other words you have to show that if $[x]=[y]$ then $[gx]=[gy]$. But this follows precisley from the fact that $sim$ is a $G$-congruence.
All that is left is to show that $g[x]$ is a group action, i.e. that $g(h[x])=(gh)[x]$ and $e[x]=[x]$. But this follows from the fact that $gx$ is a group action on $X$.
answered Nov 20 at 11:29
freakish
11.3k1629
I see, thank you so much!
– amator2357
Nov 20 at 11:38
add a comment |
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First of all I'm going to use left action $gx$ instead of right $xg$ because it is more common. And these are (sort of) equivalent anyway.
Show that $G$ acts on the set $X/$~ of equivalence classes.
Well, this statement is a bit trivial. Because if $Y$ is any set then $G$ acts on $Y$ via $gx:=x$. There are many different ways that $G$ can act on $Y$.
But I assume that the author meant something more interesting, i.e. that there's an induced action of $G$ on $X/sim$. Induced from the action of $G$ on $X$.
So basically, we want to show that if we take an equivalence class of some $x in X$, denote it by $[x]$ and act (act on all elements of that class?) on it with $G$ we would get another (or possibly the same?) equivalence class back, is that correct?
Yes, this is correct. I mean except for act on all elements of that class. No, you don't act on elements of the class, you act on the class itself.
What you want to show is that you have some function $Gtimes Ato A$ that satisfies group action properties. In this case $A=X/sim$.
So as I've mentioned earlier there are multiple ways to define such a group action. But I assume that what the author really wanted is the action induced from the action on $X$. In other words this is what we are looking for:
$$Gtimes X/simto X/sim$$
$$(g,[x])mapsto [gx]$$
or for short $g[x]:=[gx]$. So first of all you have to show that this is a well defined function. In other words you have to show that if $[x]=[y]$ then $[gx]=[gy]$. But this follows precisley from the fact that $sim$ is a $G$-congruence.
All that is left is to show that $g[x]$ is a group action, i.e. that $g(h[x])=(gh)[x]$ and $e[x]=[x]$. But this follows from the fact that $gx$ is a group action on $X$.
answered Nov 20 at 11:29
freakish
11.3k1629
I see, thank you so much!
– amator2357
Nov 20 at 11:38
add a comment |
First of all I'm going to use left action $gx$ instead of right $xg$ because it is more common. And these are (sort of) equivalent anyway.
Show that $G$ acts on the set $X/$~ of equivalence classes.
Well, this statement is a bit trivial. Because if $Y$ is any set then $G$ acts on $Y$ via $gx:=x$. There are many different ways that $G$ can act on $Y$.
But I assume that the author meant something more interesting, i.e. that there's an induced action of $G$ on $X/sim$. Induced from the action of $G$ on $X$.
So basically, we want to show that if we take an equivalence class of some $x in X$, denote it by $[x]$ and act (act on all elements of that class?) on it with $G$ we would get another (or possibly the same?) equivalence class back, is that correct?
Yes, this is correct. I mean except for act on all elements of that class. No, you don't act on elements of the class, you act on the class itself.
What you want to show is that you have some function $Gtimes Ato A$ that satisfies group action properties. In this case $A=X/sim$.
So as I've mentioned earlier there are multiple ways to define such a group action. But I assume that what the author really wanted is the action induced from the action on $X$. In other words this is what we are looking for:
$$Gtimes X/simto X/sim$$
$$(g,[x])mapsto [gx]$$
or for short $g[x]:=[gx]$. So first of all you have to show that this is a well defined function. In other words you have to show that if $[x]=[y]$ then $[gx]=[gy]$. But this follows precisley from the fact that $sim$ is a $G$-congruence.
All that is left is to show that $g[x]$ is a group action, i.e. that $g(h[x])=(gh)[x]$ and $e[x]=[x]$. But this follows from the fact that $gx$ is a group action on $X$.
answered Nov 20 at 11:29
freakish
11.3k1629
I see, thank you so much!
– amator2357
Nov 20 at 11:38
add a comment |
First of all I'm going to use left action $gx$ instead of right $xg$ because it is more common. And these are (sort of) equivalent anyway.
Show that $G$ acts on the set $X/$~ of equivalence classes.
Well, this statement is a bit trivial. Because if $Y$ is any set then $G$ acts on $Y$ via $gx:=x$. There are many different ways that $G$ can act on $Y$.
But I assume that the author meant something more interesting, i.e. that there's an induced action of $G$ on $X/sim$. Induced from the action of $G$ on $X$.
So basically, we want to show that if we take an equivalence class of some $x in X$, denote it by $[x]$ and act (act on all elements of that class?) on it with $G$ we would get another (or possibly the same?) equivalence class back, is that correct?
Yes, this is correct. I mean except for act on all elements of that class. No, you don't act on elements of the class, you act on the class itself.
What you want to show is that you have some function $Gtimes Ato A$ that satisfies group action properties. In this case $A=X/sim$.
So as I've mentioned earlier there are multiple ways to define such a group action. But I assume that what the author really wanted is the action induced from the action on $X$. In other words this is what we are looking for:
$$Gtimes X/simto X/sim$$
$$(g,[x])mapsto [gx]$$
or for short $g[x]:=[gx]$. So first of all you have to show that this is a well defined function. In other words you have to show that if $[x]=[y]$ then $[gx]=[gy]$. But this follows precisley from the fact that $sim$ is a $G$-congruence.
All that is left is to show that $g[x]$ is a group action, i.e. that $g(h[x])=(gh)[x]$ and $e[x]=[x]$. But this follows from the fact that $gx$ is a group action on $X$.
answered Nov 20 at 11:29
freakish
11.3k1629
First of all I'm going to use left action $gx$ instead of right $xg$ because it is more common. And these are (sort of) equivalent anyway.
Show that $G$ acts on the set $X/$~ of equivalence classes.
Well, this statement is a bit trivial. Because if $Y$ is any set then $G$ acts on $Y$ via $gx:=x$. There are many different ways that $G$ can act on $Y$.
But I assume that the author meant something more interesting, i.e. that there's an induced action of $G$ on $X/sim$. Induced from the action of $G$ on $X$.
So basically, we want to show that if we take an equivalence class of some $x in X$, denote it by $[x]$ and act (act on all elements of that class?) on it with $G$ we would get another (or possibly the same?) equivalence class back, is that correct?
Yes, this is correct. I mean except for act on all elements of that class. No, you don't act on elements of the class, you act on the class itself.
What you want to show is that you have some function $Gtimes Ato A$ that satisfies group action properties. In this case $A=X/sim$.
So as I've mentioned earlier there are multiple ways to define such a group action. But I assume that what the author really wanted is the action induced from the action on $X$. In other words this is what we are looking for:
$$Gtimes X/simto X/sim$$
$$(g,[x])mapsto [gx]$$
or for short $g[x]:=[gx]$. So first of all you have to show that this is a well defined function. In other words you have to show that if $[x]=[y]$ then $[gx]=[gy]$. But this follows precisley from the fact that $sim$ is a $G$-congruence.
All that is left is to show that $g[x]$ is a group action, i.e. that $g(h[x])=(gh)[x]$ and $e[x]=[x]$. But this follows from the fact that $gx$ is a group action on $X$.
answered Nov 20 at 11:29
freakish
11.3k1629
answered Nov 20 at 11:29
freakish
11.3k1629
answered Nov 20 at 11:29
freakish
11.3k1629
answered Nov 20 at 11:29
freakish
11.3k1629
11.3k1629
I see, thank you so much!
– amator2357
Nov 20 at 11:38
add a comment |
I see, thank you so much!
– amator2357
Nov 20 at 11:38
I see, thank you so much!
– amator2357
Nov 20 at 11:38
I see, thank you so much!
– amator2357
Nov 20 at 11:38
add a comment |
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To prove that the induced action of $G$ on $X/sim$ is well-defined, you have to prove that, for all $g in G$ and $x,y in X$, we have $x sim y Rightarrow xg sim yg$, which follows immediately from the definition of a $G$-congruence.
– Derek Holt
Nov 20 at 11:29
Thank you Derek! Clearly I didn't quite understand what the question was asking me to do.
– amator2357
Nov 20 at 11:37