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$G$ is a group which acts transitively on a set $S$. $G_s$ is the stabilizer of $s$. How to see the map from $G/G_s$ to $S$ given by $g*G_s \to g(s)$ is one to one?

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Let $g,h\in G$. If $g(s)=h(s)$, then $$g^{-1}(g(s))=g^{-1}(h(s)),$$ which becomes $$(g^{-1}g)(s)=e(s)=s=(g^{-1}h)(s).$$ Thus $g^{-1}h\in G_s$ because $g^{-1}h$ stabilizes $s$. Thus $g^{-1}hG_s=G_s$, and thus $hG_s=gG_s$.

Therefore, given any $gG_s$ and $hG_s$, if $g(s)=h(s)$ then $gG_s=hG_s$. Thus the map is injective (which is one possible meaning of one-to-one).

Some people instead use "one-to-one" to mean bijective. Under this meaning, the map $gG_s\mapsto g(s)$ is still one-to-one. We have already shown the map is injective above; so all that is left is to prove it is surjective, and then we will have shown it is bijective.

Given any $t\in S$, there is a $g\in G$ such that $g(s)=t$ (this is the definition of transitive). Thus given any $t\in S$, there is an element of $G/G_s$ that maps to $t$, namely $gG_s$ where $g$ is any element that sends $g(s)=t$ (and we know at least one must exist).

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$g(s)=h(s)\iff h^{-1}g\in G_s\iff gG_s=hG_s$.

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