Bijective mapping from a subgroup H of G to an orbit of the element $x\in G$

Question

We let $H\leq G$, where $G$ is a finite group. We consider that $H$ acts on $G$ by left multiplication. I was trying to prove that the map $\phi:H\rightarrow \mathcal{O_x}$ (where $x\in G$) is bijective.

When I am proving that it is surjective, I encounter the fact that for $g\in \mathcal{O_x}$, $\exists\; h\in H:g=hx\Rightarrow x=h^{-1}g$.

We may show that $\phi(h^{-1}g)=g$ but how do I show that $h^{-1}g$ is a member of $H$?

Answer 1

Since every element of $\mathcal{O_x}$ is of the form $hx$ for some $h \in H$ this shows that the map $\phi$ is surjective (because $\exists h \in H s.t. \phi(h) = hx $).

To prove injectivity suppose $\phi(h_1) = \phi(h_2)$ then $h_1x = h_2x \Rightarrow h_1xx^{-1} = h_2x x^{-1} \Rightarrow h_1 = h_2$.