# Bijection between Orbit and Stabilizer

1. Oct 16, 2009

### 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/Gx--->gx

Where H=G/Gx

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.

Last edited: Oct 16, 2009
2. Oct 17, 2009

### Fredrik

Staff Emeritus
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?

3. Oct 17, 2009

### 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/Gx.

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.

Last edited: Oct 17, 2009
4. Oct 17, 2009

### Fredrik

Staff Emeritus
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 $f(gx)=gK_x$.

Last edited: Oct 17, 2009
5. Oct 17, 2009

### 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).

6. Oct 17, 2009

### Fredrik

Staff Emeritus
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.

7. Oct 18, 2009

### chaotixmonjuish

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

8. Oct 20, 2009

### Fredrik

Staff Emeritus
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 $K_x$.