If $G$ is a finite group which acts transitively on $X$, and if $H$ is a normal subgroup of $G$, show that the orbits of the induced action of $H$ on $X$ all have the same size.
I don't know how to prove this problem. I hope to get your help. Thank you.
If $Hx$ and $Hy$ are two orbits, then by transitivity there exists $g\in G$ with $gx=y$. Then $z\mapsto gz$ is a map $Hx\to Hy$ because for $hx\in Hx$, we have $ghx=hgx=hy\in Hy$. Its obvious inverse $z\mapsto g^{-1}z$ makes it a bijection.