I'll write down the definitons I'm using.
Let $G$ be a group and $X$ a set. We say $\bullet$ is a left group action of $G$ on $X$ if the function $$G\times X\to X$$ $$(g,x)\mapsto g \bullet x$$ satisfies
- $e\bullet x=x~\forall x\in X$,
- $g\bullet (h\bullet x)=(gh)\bullet x~\forall x \in X~\forall g,h\in G.$
The orbit of $x\in X$ is the set $G\bullet x=\mathcal O_x=\{g\bullet x:g \in G\}$ (if $\bullet$ is a left group action of $G$ on $X$).
The stabilizer (under the same hypothesis as orbit) of $x$, is the set $$S(x)=\{g\in G: g\bullet x=x\}.$$
The index of a subgroup $H<G$ is the number of left (or right, they have the same cardinality) cosets of $H$. The left cosets of $H$ in $G$ are the sets $gH=\{gh:h\in H\}$ for each $g\in G$.
If $H<G$, then $G/H=\{gH:g\in G\}$ is the quotient of $G$ by $H$, which is a subgroup if $H$ is a normal subgroup of $G$ ($Nx=xN~\forall x\in G$), otherwise it is just a set.
I've alredy proved that the stabilizer is a subgroup of $G$, i.e. $S(x)<G$.
I'm trying to understand the proof for the Orbit-Stabilizer Theorem:
If $G$ is a group and $X$ a set such that $\bullet$ is a left group action of $G$ on $X$, then $$|\mathcal{O}_x|=[G:S(x)]~\forall x \in X.$$
The proof is:
Let $x\in X$. We can define the function $$F:\mathcal{O}_x\to G/S(x)$$ $$g\bullet x\mapsto F(g\bullet x)=gS(x).$$ We need to prove it is well-defined. Let's pick $g,h\in G$ such that $g\bullet x=h\bullet x$, then: $$g\bullet x=h\bullet x\iff (g^{-1}h)\bullet x=x\iff g^{-1}h\in S(x)\color{red}{\iff}gS(x)=hS(x).$$
The red $\color{red}{\iff}$ is what I don't understand. I can see it is true if the subgroup if normal, but why in the case we are just doing the quotient of groups which is a set, that is true? I could think of $g^{-1}h\in S(x)\iff h\in g S(x)$ but not the other way round.
I'd appreciate any help.
This is true for any coset. Namely, let $H\leq G$. Then $gH=H\Leftrightarrow g\in H$. So, $gkH = H\Leftrightarrow gk\in H$. Furthermore, $g^{-1}kH=H \Leftrightarrow g^{-1}k\in H$. But we clearly see that $g^{-1}kH=H\Leftrightarrow kH=gH$ (multiply on the left by $g$).