Orbits of Group Action on Sets with Power of Prime Size/Orbit whose Length is Divisible by p

Lemma
Let $G$ be a finite group such that $\order G = k p^n$ where $p \nmid k$.

Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$

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

Then:
 * Each orbit whose length is divisible by $p$ contains no Sylow $p$-subgroups.

Proof
Let $H$ be a Sylow $p$-subgroup of $G$.

By the definition of $\mathbb S$, and because $\order H = p^n$:
 * $H \in \mathbb S$

By Group Action on Sets with k Elements:
 * $g \in G \implies g * H = g H$

which is a left coset of $G$.

However, we know that $g H = H \iff g \in H$.

Now:
 * $g \in \Stab H \iff g * H = H \iff g \in H$

Thus:
 * $\Stab H = H$

From the Orbit-Stabilizer Theorem:
 * $\card {\Orb H} = k$

that is, not divisible by $p$.

As $H \in \Orb H$, the result follows.