Let $G$ be group and $M$ be a set. Furthermore let $G_m$ be a the stabilizer of $m\in M$ and $Gm$ be the orbit of $m$ under the action. Now my problem:
Let $\psi: G /G_m \to Gm$, $gG_m \mapsto g(m)$. I wanna show that $\psi$ is injective. Then the argument is given as follows: $$\begin{align} \psi(gG_m)=\psi(hG_m)& \Rightarrow g(m)=h(m)\\ & \Rightarrow (g^{-1}h)(m)=m\\ & \Rightarrow g^{-1}h \in G_m\\ & \Rightarrow gG_m =hG_m. \end{align}$$ I just don't understand the last part, why we get $g^{-1}h \in G_m \Rightarrow gG_m =hG_m$?
Many thanks for your help!
Here is the fact that is being used:
Here is a proof:
Suppose $xH=yH$. Then $y=y{\cdot}1\in yH=xH$ so we can write $y=xh$ for some $h\in H$. Then $x^{-1}y=h\in H$.
On the other hand suppose $x^{-1}y\in H$, say $x^{-1}y=h$. Notice then that $y=xh$ so given an element $yh'\in yH$ we have $yh'=(xh)h'=x(hh')\in xH$, which shows $yH\subseteq xH$. For the other inclusion notice $x=yh^{-1}$ so if $xh'\in xH$ then $xh'=(yh^{-1})h'=y(h^{-1}h')\in yH$.