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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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    It looks correct indeed.2012-12-31
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    @DonAntonio So I don't need to use Lagrange theorem? I just prove that the cosets have a 1-1 correspondence by showing a bijection between group and the orbit?2012-12-31
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    But you still need Lagrange theorem to show that $|G/G_x|=\frac{|G|}{|G_x|}$2012-12-31
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    @Amr Oh ok, thanks! Just to double check the notation, is $G/G_x$ the quotient group? And so the number of elements in the quotient group $|G/G_x|$ can be worked out by doing $\frac{|G|}{|G_x|}$, and by definition of a quotient group $G_X \subset G$ and so thats why I use Lagrange theorem?2012-12-31
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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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