Bijection between Orbit and Stabilizer

[chaotixmonjuish]

So I know this is the orbit-stabilizer theorem. I saw it in Hungerford's Algebra (but without that name).

So we want to form a bijection between the right cosets of the stabilizers and the orbit. Could I define the bijection as this:

f: gG/G_x--->^gx

Where H=G/G_x

f(hx)=gx h in H

^ Is that what the function is suppose to look like? I'm really stuck on understanding the proof since it doesn't show me this function but it guarantees that it is a bijection. I'm not sure if there is a better way to word this question except, perhaps, what is the function between the right cosets and the orbit.

 

[Fredrik]

Have a look at theorem 3 on page 5 of this article. Is the function that the author defines in the proof what you're looking for?

 

[chaotixmonjuish]

Sort of, but I'm getting kind of thrown off by his notation.

So the function that is defined is sending f: g*H---->Orb(x) where H=G/G_x.

Is this just saying that f(gh)=gx or f(g)=gx because H is everything that doesn't move x so its not really worth mentioning.

 

[Fredrik]

He's saying that if y\in\mathcal O_x, then there exists g\in G such that y=gx. Then he defines f(y)=gK_x, where K_x is the stabilizer subgroup for x. It's not obvious that this f is well-defined, because there could exist a g'\neq g such that y=gx=g'x. To prove that the above actually defines a function, we must prove that gx=g'x implies gK_x=g'K_x.

Here's the proof: Suppose that z\in gK_x. Then there's a k\in K_x such that z=gk=g'g'^{-1}gk. This is a member of g'K_x if g'^{-1}gk\in K_x, and it's not hard to see that it is:

g'^{-1}gkx=g'^{-1}gx=g'^{-1}g'x=x

This proves that gK_x\subset g'K_x, and we can of course repeat the argument with g and g' swapped.

So f is a well-defined function from the orbit \mathcal O_x into the set of right cosets G/K_x. The equation that defines f can also be written as f(gx)=gK_x.

 

[chaotixmonjuish]

The very last function you defined is from the orbit to the stabilizer (I know its a bijection but I just want to make sure).

 

[Fredrik]

No, gK_x=\{gk|k\in K_x\} is a right coset, so

f:\mathcal O_x\rightarrow G/K_x​

if we use the notation G/K_x for the set of right cosets.

 

[chaotixmonjuish]

Great, it all makes sense. However, how does this bijection prove that O*G_x:K=G. To me it just proves they are equal, but that can't always be the case.

 

[Fredrik]

The theorem says that the action restricted to \matcal O_x is equivalent (in a technical sense) to the canonical action on G/K_x. This splits the problem of understanding a group action into two pieces: 1. Find the orbits. 2. Study the action on G/K_x, for each different K_x.