I'm trying to prove the following. This exercise already appeared multiple times on this site, but I have a specific question about my proof, so please have a look.
Suppose $G$ acts transitively on $X$ and let $N\triangleleft G$. Prove that the orbits of $X$ under $N$ are of equal length, i.e. $|Nx|=|Ny|\quad \forall x,y\in X$.
Proof I will write $N_x$ for the stabilizer of $x$ in $N$. Thus given $x,y\in X$, we know $N_x,N_y\subset N$ are subgroups of $N$. Because $G$ acts transitively on $X$, there is only one orbit, thus $\exists g\in G, g\circ x=y$. We also know $N_{g\circ x}=N_y=gN_xg^{-1}$ and thus $g^{-1}N_yg=N_x$. From the first equality, we conclude $|N_x|\leq |N_y|$, from the second $|N_x|\geq |N_y|$ and thus $|N_x|=|N_y|$.
Then I would use Derek Holt's comment to conclude from $[N:N_y]=[N:gN_yg^{-1}]=[N:N_x]$ that $|Nx|=[N:N_x]=[N:N_y]=|Ny|$, thus the orbits being of equal length indeed. $\square$
Yellow part below is already answered
Now I want to apply the following version of the Orbit-Stabilizer theorem that appears in my syllabus (I quote only the relevant part): $|Gx|=\mathrm{index}[G:G_x]$. From this and $|N_x|=|N_y|$ I want to conclude $|Nx|=\mathrm{index}[N:N_x]=\mathrm{index}[N:N_y]=|Ny|$. Question: can I make this conclusion? I am a bit in doubt about the case where both $N_x$ and $N_y$ are infinite. For example suppose we have the sets $\mathbb{Z},2\mathbb{Z}$ and $3\mathbb{Z}$, then $|2\mathbb{Z}|=|3\mathbb{Z}|$ but $\mathrm{index}[\mathbb{Z}:2\mathbb{Z}]\neq[\mathbb{Z}:3\mathbb{Z}]$. If I can make this conclusion, why? And if not, what would I need to proof the result using my work done this far?
EDIT A classmate pointed out to me that there has to be a mistake in my proof, since I seem to not have used the fact that $N$ was a normal subgroup. However, I can't find a mistake in my proof and I (and two of my classmates) do not see why my proof is not valid. Can anyone take a look?