I'm studying for personal fun and culture group theory, specifically the orbit stabilizer theorem. I've then found this interesting problem:
Let a group $G$ acting on $\Omega$ and take $\alpha,\beta\in\Omega$, $x\in G$ and $\alpha^x=\beta$ (meaning $x$ acts on $\alpha$ to get $\beta$). Show that if $y\in G$ and $\alpha^y=\beta$ $\implies \exists g\in \operatorname{Stab}(\alpha)$ s.t. $y=gx$, i.e. $y\in \operatorname{Stab}(\alpha)x$.
In words, this is tantamount to claim all group elements moving $\alpha$ to $\beta$ belong to the group set given by the stabilizer elements of $\alpha$ followed by a guess element moving $\alpha$ to $\beta$.
To get a better visual mental grasp on it, I've used the rotational symmetries of a cube acting on the set of cube faces. In this case, given a rotation moving a face onto another one, we can get the full rotations which realise the same thing using the stabilizer subgroup of each face. This is the set of rotations around an axis perp to the face mid point, and is isomorphic to a cyclic group of order 4. So, taking each stabilizer rotation followed by the initial rotation from one face to the target one, gives you back the full list of moves.
That said, I've tried to show that such $g$ element exists (problem 3.3 of Groups and Characters book) but I'm not fully sure this demo is complete. It goes like this:
Let's take the element $g\in G$ s.t. $y=gx$, i.e. $g=yx^{-1}$. We can also claim $\alpha^y=\alpha^x$ so, replacing the y: $$\alpha^{gx}=\alpha^x \implies (\alpha^g)^x = \alpha^x$$ But this can be true only iff:$$\alpha^g=\alpha \implies g\in \operatorname{Stab}(\alpha) \implies y\in \operatorname{Stab}(\alpha)x$$
I was wondering if this is enough, or if I'm missing something. I mean: is this enough to proof that the elements $\in \operatorname{Stab}(\alpha)x$ are the only ones bringing $\alpha$ to $\beta$?
I was also wondering if there could be a reductio ad absurdum possible version of this, i.e. "Let's suppose such $g \notin \operatorname{Stab}(\alpha)$ . . . " and deriving from it a contradiction.
Thanks for your precious support in advance.
The proof is enough. Some further clarification:
Claim: $\alpha^x = \alpha^y \implies y \in Stab(\alpha)x$
Proof: As you noted, there exists $g \in G$ such that $y = gx$. Thus,
$$ (\alpha^g)^x = \alpha^x$$
It suffices to show that $\alpha ^ g = \alpha$, as this would imply that $y \in Stab(\alpha)x = \{ gx | \alpha^g = \alpha\}$.
But this follows readily from the existence of an inverse $x^{-1}$ of $x$:
$$ ((\alpha^g)^x)^{x^{-1}} = (\alpha^x)^{x^{-1}}$$
so you would get
$$ \alpha^g = \alpha$$
as desired.