Orbits of normal subgroup have same size The solution to the following problem can be found here and has been also discussed on SE. However, I was wondering if my solution that does not use transitivity is correct.
"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 came up with the following solution:
Take $a,c \in X$ and let $O_a,O_c$ be their orbits (under the action of H). Assume w.l.o.g. that $|O_a| \geq |O_c|$, where $|O_x|$ gives the size of the orbit.
If $|O_a| = 1$, we're done.
Sup $|O_a| > 1$. Take $b \in O_a$. Then $\exists h \in H$ such that $ha = b$. As H is normal, $a^{-1}ha = a^{-1}b \in H$. Therefore, $a^{-1}bc \in O_c$. Furthermore, this element is uniquely constructed, for if $a^{-1}bc = a^{-1}b'c$, then clearly $b' = b$.
Therefore, for every additional element in $O_a$ we have constructed a unique element in $O_c$, and so the size of these sets must be equal. 
 A: You have $a,b,c\in X$, not $G$, so writing things like $a^{-1}b\in H$ (or $a^{-1}$ at all) is nonsense.
In abstract algebra, a general rule of thumb is to always keep in mind what things look like.
The orbits of $H$ all look like $Hx$ for $x\in X$. Because $G\curvearrowright X$ transitively, if we fix $x\in X$, every other element of $X$ looks like $gx$ for some $g\in G$. Then every orbit looks like $Hgx$, which by normality is the same as $gHx$. Can you think of a bijection $Hx\to gHx$?
Without the transitivity hypothesis, the conclusion may not be true. For instance, consider $G$ acting trivially on the one-point set $\{*\}$ and acting regularly on itself, then let $G\curvearrowright \{*\}\sqcup G$. The $H$ orbit of the point $*$ is just that point, whereas the orbits of $H$ in $G$ all have size $|H|$. The transitivity hypothesis is not strictly necessary though - for instance suppose $G\curvearrowright X$ transitively and then construct $Y$ to be the disjoint union of any number of copies of $X$.
