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Theorem: For a group $G$ and set $X$, let $\alpha:G\times X\to X$ be an action, where $\alpha(g,x):= g \cdot x$. $\forall x\in X$, $f:G/G_{x}\to G\cdot x$, $f(gG_{x})=g\cdot x$ is a well defined bijection.

Is this theorem true for all group actions?

For example, if I define $\sigma:\mathbb{Z} \times X\to X$ as $\sigma(n,x):=\phi^{n}(x)$ ,where $\phi \in sym(X)$ and X is a finite set, to be an action. If $\phi$ is the identity map, then $G/G_{x}=\mathbb{Z}$. $G \cdot x$ is just $\{x\}$. There is no way we can find a bijection from all integers to $\{x\}$.

I feel like there must exist some sort of constraints on this theorem, but I just don't know what it is.

Irene
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  • The stabilisers are $\Bbb Z$ when the action is the identity. So no contradiction. The theorem is true for any action, but can rule out certain actions on certain sets. – calc ll Sep 19 '22 at 18:54
  • @pipe Yes the stabilizers are ℤ, but the orbit is of x is {x} only. So there is no bijection between these two. Also, can you please elaborate on ruling out certain actions on certain sets. – Irene Sep 19 '22 at 19:02
  • Since $G_x=\Bbb Z$, we have $[G:G_x]=[\Bbb Z:\Bbb Z]=1$. – calc ll Sep 19 '22 at 19:47
  • At any rate, you can draw various conclusions. Here's one: https://math.stackexchange.com/a/4510778/1070376 – calc ll Sep 19 '22 at 19:50
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    @pipe Sorry about it but I'm still a bit confused by this... Orbit-stabilizer theorem states that there is a bijection between $G/G_{x}$ and the orbit of x. You are saying that there exists a bijection from $G$ to $G_{x}$, which is not what I'm asking. – Irene Sep 19 '22 at 20:06
  • No, it's OK. I am saying the first thing (it just so happens that in this case the second is also true). Note that when $G=G_x=\Bbb Z$, we have $G/G_x={e}$. – calc ll Sep 19 '22 at 20:19
  • @pipe Hmm...I think I get what you are saying. Thanks. – Irene Sep 19 '22 at 20:57
  • Ok. Yes the conditions for a group action are enough to assure that the orbit-stabilizer theorem is always true. – calc ll Sep 19 '22 at 20:59

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Here's an example: by the orbit stabiliser theorem, when you have a transitive action (one orbit) , the order of $X$ must divide the order of $G$.

So if say the order of $X$ and $G$ are relatively prime, then there's no transitive action.

calc ll
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