number of orbits of a proper subgroup

Question

Let $G$ be a permutation group that acts on (say) $X=\{1,2,..,n\}$, and $H$ be a proper subgroup of $G$. Can one say anything precise about when the number of orbits of $H$ on $X$ will be equal to number of orbits of $G$ on $X$? In other words, is there a nice characterization of this situation, which perhaps gives some certain relation between $H$ and $G$? For instance, if $H$ acts transitively on $X$, so $G$ will also do, then what I say above holds, since both $H$ and $G$ will have single orbit. But, I don't know what one can say in general. I tried to use orbit-counting theorem, but couldn't find a way to get anything that way.

Answer 1

I will write my actions on the left; for right actions, you would get $\mathrm{Stab}(x)H = G$ instead.

Let $G$ be a group acting on a set $X$, let $x\in X$, and let $H$ be a subgroup of $G$. Then $Gx = Hx$ if and only if $H\mathrm{Stab}(x) = G$, where $\mathrm{Stab}(x) = \{g\in G\mid gx=x\}$ is the stabilizer of $x$ in $G$.

Proof. Assume first that $Gx=Hx$. If $g\in G$, then there exists $h\in H$ such that $gx = hx$; hence $h^{-1}gx=x$, so $h^{-1}g\in\mathrm{Stab}(x)$. Hence $g\in h\mathrm{Stab}(x) \subseteq H\mathrm{Stab}(x)$, proving that $G$ is contained in $H\mathrm{Stab}(x)$, hence they are equal. Conversely, if $H\mathrm{Stab}(x) = G$ and $y\in Gx$, then there exists $g\in G$ such that $gx=y$. Since $G=H\mathrm{Stab}(x)$, $g=hu$ for some $h\in H$ and $u\in\mathrm{Stab}(x)$, so $y= gx = (hu)x = hx\in Hx$. This shows $Gx\subseteq Hx$, and the converse inclusion is obvious. $\Box$

Let $G$ act on the finite set $X$, and let $H$ be a subgroup of $G$. The action of $H$ on $X$ by restriction has the same number of orbits as the action of $G$ if and only if for every $x\in X$, $H\mathrm{Stab}(x) = G$.

Note that the action of $H$ partitions the orbits of $G$. Since $X$ is finite, the number of orbits is finite, so $H$ has the same number of orbits on $X$ as $G$ if and only if it has the same orbits as $G$. i.e., if and only if for every $x\in X$, $Gx = Hx$, which yields the characterization. $\Box$

Of course, you really only need to check $H\mathrm{Stab}(x)=G$ for a single element of each orbit; that is, you can restrict the quantifier to a full set of orbit representatives, instead of all elements of $X$.