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 $K_x$.