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 $h$?