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For some group $G$ acting on a set $X$, if we consider the map:

$f: G \rightarrow \mathrm{Orb}(x)$ $ g \mapsto gx $

Then first we can tell that this is a surjection by definition of an orbit. So to prove injection, we can try and see if that for any two elements $g_1, g_2 \in G$, do we get that $g_1 x = g_2x$.

Applying $g_2^{-1}$ we get

$g_2^{-1}g_1 x = g_2^{-1}g_2 x \implies g_2^{-1}g_1 x = x$

Then applying $g_1^{-1}$ we get

$ g_1^{-1} g_2 x = g_1^{-1} g_1 x \implies g_1^{-1} g_2 x= x$

From these two, we see that $g_1 \in g_2 G_x$ and $g_2 \in g_1G_x$ and this happens iff $g_1Gx = g_2Gx$, hence proving an injection and therefore a bijection.

Is this correct?

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    That was a typo, its supposed to say $g_1^{-1}g_2 x = x$, I fixed it now. Ok, instead of saying that shows an injection. Can I say this happens because this tells us that the set of left cosets wrt to $G_x$ now as 1 - 1 relation with the Orbit, i.e $\{aG_x\} \implies Orb(x)$?2012-12-31

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Maybe it is simpler to say that $g_1 x = g_2 x \iff g_2^{-1}g_1x=x$ which means that $g_2^{-1}g_1 = e_G$ (where $e_G$ is the unit of G) so that $g_1 = g_2$, which proves the injectivity.

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    Shouldn't it be mentioned "$\forall x$"?2016-04-17