# set $S$ in bijection with the set $G/G_s$

Let $G$ act transitively on a (non-empty) set $S$, and fix $s ∈ S$. Then $S$ is in bijection with the set $G/G_s$ of cosets $gG_s$ of the isotropy group $G_s$ of $s$ in $G$, by $gs \longleftrightarrow gG_s$ Thus, $\text{card} S = [G : G_s]$ Proof: If $hG_s = gG_s$, then there is $x ∈ G_s$ such that $h = gx$, and $hs = gxs = gs$. On the other hand, if $hs = gs$, then $g^{−1}hs = s$, so $g^{−1}h ∈ Gs$, and then $h ∈ gG_s$.

I am not sure how this leads to the conclusion that $S$ is in bijection with the set $G/G_s$. How does $h$ being in $gG_s$ and possibility that there is $x$ in $G_s$ that $hs=gs$ lead to bijection conclusion?

Is the proof saying that all of $gG_s$ is in bijection with $h$?

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The bijection is between $S$ and the set of cosets. So $g G_s$ is being regarded as an element of a set here, not as a set in its own right. – Zhen Lin Aug 23 '12 at 3:38

If $hG_s = gG_s$, then $hs = gs$. Hence there exists a map $\psi\colon G/G_s \rightarrow S$ such that $\psi(gG_s) = gs$. Suppose $\psi(gG_s) = \psi(hG_s)$. Then $gs = hs$. Hence $gG_s = hG_s$ Hence $\psi$ is injective.
Let $t \in S$. Since $G$ acts transitively on $S$, there exists $g \in G$ such that $gs = t$. Hence $\psi(gG_s) = t$. Hence $\psi$ is surjective.
Therefore $\psi$ is bijective.