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 $$\operatorname{Stab} \left({S}\right) s = S$$.