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This is very simple, but as far as I can tell it has not been asked yet.

Let the group $G$ act on the set $S$ and define an equivalence relation by $x \sim x'$ if there exists a $g \in G$ for which $gx=x'$.

Proving reflexivity and transitivity is easy, so let's look at the symmetric property: Say $x \sim x'$ with $gx=x'$. Then we have $x=x'g^{-1}$. So $x$ is equal to $x'$ multiplied by an element of $G$, but does this work since we are now using right multiplication? Can we do something 'clever' like $ex=x'g^{-1} \Rightarrow xe=g^{-1}x' \Rightarrow x=g^{-1}x' \Rightarrow x' \sim x$? Something about that last bit seems foul to me.

The group theory tag isn't really appropriate here. I would create a 'group actions' tag if I were able.

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Why do you have $x = x'g^{-1}$? If the action is on the left, you should just have $x = g^{-1}gx = g^{-1}x'$ –  user29743 Apr 28 '12 at 2:31
    
There need not be any such thing as "right multiplication". An action tells you what $gs$ is for any $g\in G$ and $s\in S$, i.e. it is a function $f:G\times S\to S$ (satisfying a few axioms). –  Zev Chonoles Apr 28 '12 at 2:35
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up vote 6 down vote accepted

You're idea is right, but all your actions should be on the same side. Here's the correct version of the argument you were trying to construct:

If $gx=x^\prime$, then $x=ex=(g^{-1}g)x=g^{-1}(gx)=g^{-1}x^\prime$

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Makes sense. I should really stop trying to do math when I'm tired! Thanks. –  Alex Petzke Apr 28 '12 at 3:08
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