Let $G$ be a finite group whose order is divisible by $p^\alpha$ for some prime $p$. Let $X$ be the set of all subsets (not necessarily subgroups) of $G$ whose order is $p^\alpha$. Let $H\in X$, let $1\in H$, and let $G$ act on $X$ by left multiplication. I want to show that $|\text{Stab}_G(H)|$ divides $|H|$.
As far as I understand, for $h\in H$, we have that $\text{Stab}_G(h)$ is the set of all $s\in G$ which fix $h$; i.e. $s\cdot h=h$. Then, $\text{Stab}_G(H)$, intuitively (to me), is the set of all $s\in G$ that fix all of $H$; i.e. $s\cdot h=h$ for all $h\in H$.
(I will note that according to an answer to this post, elements of $\text{Stab}_G(H)$ need only permute the elements of $H$? Is this true? Why? In any case, the question I'm trying to solve did not come with a specification.)
I have shown, at the very least, that $\text{Stab}_G(H)\subseteq H$, as $1\in H$ implies that $s\cdot 1 \ (= s)\in H$.
However, if, say, $|G|=p^\alpha n$ for some natural $n$, how can I show that there does not exist a prime $q\neq p$ such that $q\mid n$ and $q\mid |\text{Stab}_G(H)|$?
I've tried using the Orbit-Stabilizer theorem to no avail. I know that the orbits of $H$ partition $H$, but since I don't know how to find out how many orbits of $H$ there are, I don't see how the Orbit-Stabilizer theorem helps.
What should I do?
Best Answer
The pointwise stabilizer of $\{g\in G\mid gh=h\textrm{ for all }h\in H\}$ of $H$ is the trivial subgroup. This is because groups are cancellative; $gh=h$ for even a single $h$ implies $g=e$. Indeed, the action $G\curvearrowright X$ by left-multiplication defines $gH:=\{gh\mid h\in H\}$, so with respect to this action we must instead be talking about the setwise stabilizer $S=\mathrm{Stab}_G(H)=\{g\in G\mid gH=H\}$.
When a group acts on a set, that set may be partitioned into orbits, and the orbit-stabilizer theorem tells us how to calculate the size of an element's orbit from the size of that element's stabilizer. However, different orbits can have different sizes, and this prevents us from saying much about the group's size or the orbit's size having any sort of divisibility relationship with the set's size.
Things can work out, though. For example, when a group acts on itself by multiplication (the regular action), the stabilizers are trivial and the group is its own orbit. Or, if $K<G$ is a proper subgroup of $G$, then restricting the regular action $G\curvearrowright G$ to $K\curvearrowright G$ means the whole of $G$ is no longer a single orbit, so there are multiple orbits, but stabilizers are still trivial, so all orbits have the same size. Indeed, the orbits are none other than the right cosets of $K$, i.e. the "orbit space" (collection of orbits) is the right coset space $K\backslash G$.
In this case, we have $S\curvearrowright H$. Even though $H$ is not a group, so $S$ is not a subgroup of it (merely a subset), this is still a free action (stabilizers are trivial, i.e. if $h\in H$ and $s\in\mathrm{Stab}_G(H)$ and $sh=h$, then $s=e$), so all orbits are the same size, and $H$ is a disjoint union of these orbits. Thus, $|H|$ is $|S|$ times the number of orbits.