G-congruence equivalence relation
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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
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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
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 |
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up vote
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down vote
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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
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
edited Nov 20 at 11:20
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 $X$.
I see, thank you so much!
– amator2357
Nov 20 at 11:38
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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 $X$.
I see, thank you so much!
– amator2357
Nov 20 at 11:38
add a comment |
up vote
1
down vote
accepted
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 $X$.
I see, thank you so much!
– amator2357
Nov 20 at 11:38
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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 $X$.
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
10.8k1527
10.8k1527
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