Group Action on Prime Power Order Subset

Lemma
Let $G$ be a finite group.

Let $\mathbb{S} = \left\{{S \subseteq G: \left|{S}\right| = p^n}\right\}$ where $p$ is prime (that is, the set of all subsets of $G$ whose cardinality is the power of a prime number).

Let $G$ act on $\mathbb{S}$ by the group action defined in Group Action on Sets with k Elements: $\forall S \in \mathbb{S}: g \wedge S = g S = \left\{{x \in G: x = g s: s \in S}\right\}$.

Then:
 * 1) $\operatorname{Stab} \left({S}\right)$ is a $p$-subgroup of $G$;
 * 2) If $p^n$ is the maximal power of $p$ dividing $\left|{G}\right|$, and if $p \nmid \left|{\operatorname{Orb} \left({S}\right)}\right|$, then $\forall s \in S: \operatorname{Stab} \left({S}\right) s = S$.

Proof

 * First we show that $\operatorname{Stab} \left({S}\right)$ is a $p$-subgroup of $G$:

From Group Action on Sets with k Elements, $\forall S \in \mathbb{S}: \left|{\operatorname{Stab} \left({S}\right)}\right| \backslash \left|{S}\right|$.

So $\left|{\operatorname{Stab} \left({S}\right)}\right| \backslash p^\alpha$ and thus $\operatorname{Stab} \left({S}\right)$ is a $p$-group, and thus by Stabilizer is Subgroup, a $p$-subgroup of $G$.


 * Now, suppose $p^n$ is the maximal power of $p$ dividing $\left|{G}\right|$.

Let $p \nmid \left|{\operatorname{Orb} \left({S}\right)}\right|$.

We know from Group Action on Coset Space that $\forall s \in S: \operatorname{Stab} \left({S}\right) s \subseteq S$.

From the Orbit-Stabilizer Theorem, we have $\left|{G}\right| = \left|{\operatorname{Orb} \left({S}\right)}\right| \times \left|{\operatorname{Stab} \left({S}\right)}\right|$.

As $p \nmid \left|{\operatorname{Orb} \left({S}\right)}\right|$, it must be the case that $p^n \backslash \left|{\operatorname{Stab} \left({S}\right)}\right|$.

Thus, $\left|{\operatorname{Stab} \left({S}\right)}\right| \ge p^n$.

Now we note that $\left|{\operatorname{Stab} \left({S}\right)}\right| = \left|{\operatorname{Stab} \left({S}\right) s}\right|$ from Cosets are Equivalent.

Thus we also have that $\left|{\operatorname{Stab} \left({S}\right) s}\right| \ge p^n$.

But since $\left|{S}\right| = p^n$ and $\operatorname{Stab} \left({S}\right) s \subseteq S$, it must follow that $\left|{\operatorname{Stab} \left({S}\right) s}\right| = p^n$ and $\operatorname{Stab} \left({S}\right) s = S$.