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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?

Kaish
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    Your last equality states that an element of $G$ is equal to an element of $X$, which make no sense. Also $f$ is clearly not always injective, though you seem to state that explicitly (consider the trivial action of a non-trivial group $G \neq {0}$ on some set). – Nils Matthes Dec 31 '12 at 14:01
  • 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 ${aG_x} \implies Orb(x)$? – Kaish Dec 31 '12 at 14:31
  • Kaish you're supposed to tag using "@". @NilsMatthes, Kaish replied...I think the issue is still not quite resolved though. I think "Then first we can tell that this is a surjection by definition of an orbit. So to prove injection" is confusing because the reader would think the asker is about the prove the injectivity of $f$, where $f$ is not given injective. –  Oct 16 '19 at 05:34

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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.

Alan Simonin
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