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...