# Orbit of g

• May 6th 2009, 09:52 AM
Roland25
Orbit of g
Hi,

I know that if G acts on set X then the orbit is
Attachment 11301

But say G=X.

Does it mean that I put two g's into the above?
• May 6th 2009, 10:55 AM
NonCommAlg
Quote:

Originally Posted by Roland25
Hi,

I know that if G acts on set X then the orbit of x is $\color{red}Gx=\{g*x : \ g \in G \}.$

But say G=X. Does it mean that I put two g's into the above?

if $G=X,$ and $x \in X,$ then $\text{orbit}_G(x)=Gx=\{g*x: \ g \in G \}.$ clearly if $*$ is the same as multiplication operation of G, then $Gx=G,$ for all $x \in X=G.$
• May 6th 2009, 11:44 AM
Roland25
Sorry, I meant G(x) in the above.
What I mean is that if Gx={g*x : g in G}
what is Gg if g is in G.
Would it just be Gg={G} ?
• May 6th 2009, 12:04 PM
NonCommAlg
Quote:

Originally Posted by Roland25
Sorry, I meant G(x) in the above.
What I mean is that if Gx={g*x : g in G}
what is Gg if g is in G.
Would it just be Gg={G} ?

it depends how you define the group action $*$. if you define that for all $x,g \in G: \ g*x=gx,$ then $Gx=G.$ but if you define $g*x=gxg^{-1},$ for all $g,x \in G,$ then $Gx=\{gxg^{-1}: \ g \in G \},$

which is the conjugacy class of $x.$
• May 6th 2009, 12:23 PM
Roland25
thanks for the help so far.

I think * was multiplication in my case and so Gx=G kinda seems more appropriate.

What I still don't get however is what the actual different between Gx and Gg would be if X=G?

Say X doesn't equal G, would Gx and Gg differ then?

This is where g is in G
• May 6th 2009, 12:55 PM
NonCommAlg
Quote:

Originally Posted by Roland25
thanks for the help so far.

I think * was multiplication in my case and so Gx=G kinda seems more appropriate.

What I still don't get however is what the actual different between Gx and Gg would be if X=G?

Say X doesn't equal G, would Gx and Gg differ then?

This is where g is in G

$x \in X$ is fixed and $g$ is any element of $G.$ this should be clear from the definition: $Gx= \{g*x: \ g \in G \}.$