# $H\le G$. Number of orbits of the induced $H$-action of a transitive $G$-action: why assuming $H\unlhd G$

group-actionsgroup-theory

This post claims:

$$G$$ is a transitive group action. Normal subgroup of transitive group $$G$$ has at most $$|G:N|$$ orbits, and if $$|G:N|$$ is finite, then the number of orbits of $$N$$ divides $$|G:N|$$.

My question deals with the part in italics, as I seemingly proved it without using the normality of $$N$$. So I wonder whether I inadvertently used it somewhere, or is my proof just wrong (or, thirdly, the result is more general and holds for every subgroup $$H\le G$$, whereas perhaps the normality is used to prove the second part of the claim). Hereafter what I did:

Let $$H\le G$$. Every $$g\in G$$ lays in some right coset of $$H$$ in $$G$$. So, denoted with $$R\subseteq G$$ a complete set of coset representatives, for $$x\in X$$ we get:
\begin{alignat}{1} \operatorname{Orb}_G(x) &= \{g\cdot x, g\in G\} \\ &= \{(hg_i)\cdot x, g_i\in R \text{ and } h\in H\} \\ &= \{h\cdot(g_i\cdot x), g_i\in R \text{ and } h\in H\} \\ &= \{h\cdot y_i, y_i\in Y(x) \text{ and } h\in H\} \\ &= \bigcup_{y_i\in Y(x)}\{h\cdot y_i, h\in H\} \\ &= \bigcup_{y_i\in Y(x)}\operatorname{Orb}_H(y_i) \\ \end{alignat}
where $$Y(x):=\{g_i\cdot x, g_i\in R\}$$. Now, $$|Y(x)|\le |R|=[G:H]$$, so if the $$G$$-action is transitive, then the $$H$$-action has at most $$[G:H]$$ orbits.

Indeed, if $$G\curvearrowright\Omega$$ transitively then the (general version of the) orbit-stabilizer theorem says $$\Omega\cong G/S$$ as a $$G$$-set, where $$S$$ is the stabilizer of any element in $$\Omega$$ (these stabilizers are conjugate so the corresponding coset spaces are equivalent). Which in turn means if $$H\le G$$ then the orbit space $$H\backslash \Omega$$ is in bijection with the double coset space $$H\backslash G/S$$, and there is in turn a surjection to the double coset space from the right coset space $$H\backslash G$$, which makes $$[G:H]$$ an upper bound. This is basically a different perspective on your own proof.
So you're right the first part of the theorem doesn't require the hypothesis that $$H$$ is normal. But the second part of the theorem does require this hypothesis. If $$N$$ is normal, then the quotient group $$G/N$$ acts on the orbit space $$N\backslash\Omega$$, transitively because $$G\curvearrowright\Omega$$ was transitive, which by orbit stabilizer implies the size of the orbit space divides the size of $$G/N$$. But if $$H$$ is not normal, the $$H$$-orbits can have different sizes, and there is no guaranteed well-defined projection from $$G/H$$ to $$H\backslash\Omega$$ with equal-size fibers. And indeed, we can cook up a counterexample by looking at small groups presumably.