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Proving the stabilizer is a subgroup of the group to prove the Orbit-Stabiliser theorem

OS theorem is: for some group $G$ actincg on a set $X$, it's stabiliser $G_x$ and orbit, $\mathrm{Orb}(x)$, we get

$|G| = |\mathrm{Orb}(x)| \cdot |G_x|$

To prove it, what I thought about doing was saying that we can re-arrange the this to be

$|\mathrm{Orb}(x)| = \frac{|G|}{|G_x|}$

And we can first prove the RHS by Lagrange theorem and the prove the whole equality by showing the Orbit has 1 - 1 correspondence with the left cosets of $G$. But then I though, if I prove the coset thing, then as the orbit has 1 - 1 correspondence with the left cosets of $G$, then multiplying the orbit by another number will mean that it is still a multiple of $G$ and so I wont need to prove that we can divide them.

Is this correct?

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    Simply exhibit a bijection $G_x\times Gx\to G$. Actually, you can just as well consider Lagrange a special case of the orbit stabilzer theorem.2012-12-31

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