Free action of a group and orbits

I'm not sure about how to prove the following about the free action of a group.

Let $G$ be a group acting freely on a set $X$ and $H$ be a subgroup of $G$ of index $n$.Prove that the numbers $s_1$ of orbits by the action of $G$ on $X$ and the number $s_2$ of orbits by the action of H on X, are linked by : $$s_2 = ns_1$$

This is what I did :

We have by the Lagrange's theorem : $|G| = n|H|$. We take $g \in G$ but $g \notin H$, and we take $x$ in relation with $x'$ by : $x=gx'$. We know that $g$ is the only element that put $x$ and $x'$ in relation because otherwise we have $gx = g'x \iff (g^-1g')x' = x'$, and as G acting freely on X, we have $g=g'$. So, $x$ and $x'$ are not in relation by the action of H on X. Then, we have $n$ times more orbits by the action of H than on the G's one on X.

Is it correct ? If it's not, why and how can I prove this relation ?

Sorry for my english, I tried to do my best...

Note that the group $G$ does not have to be finite. Given any $g\in G$ and $x\in X$ we need to show that the orbit $Gx$ is the disjoint union of the $n$ orbits $H_1x,\dots,H_nx,$ where $H_1=H,H_2,\dots,H_n$ are the left cosets of the subgroup $H.$
As $G= H_1\cup\cdots\cup H_n,$ the equality $Gx= H_1x\cup\cdots\cup H_nx$ follows trivially, and the disjointness follows from the fact that the group action is free.
• Thank you for your answer. So if I want to show the disjointness ; let's $y \in H_{i}x \cap H_{j}x = (gH)x \cap (g'H)x$ with $g$, $g' \in G$ and $gH$ and $g'H$ disjoint. So : $y=ghx = g'h'x$, and it implies $((g'h')^{-1}gh)x=x$ and as G acting freely, it implies : $gh=g'h' \in g'H$ : this is not possible as $gH \cap g'H = \emptyset$. Right ? Jun 29 '17 at 19:43