Orbits of Group Action on Sets with Power of Prime Size

Lemma
Let $G$ be a finite group such that $\left|{G}\right| = k p^n$ where $p \nmid k$.

Let $\mathbb S = \left\{{S \subseteq G: \left|{S}\right| = p^n}\right\}$

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

Then:
 * 1) The length of every orbit of this action is divisible by $k$
 * 2) Each orbit whose length is not divisible by $p$ contains exactly one Sylow $p$-subgroup
 * 3) Each orbit whose length is divisible by $p$ contains no Sylow $p$-subgroups.

Proof

 * First we show that the length of every orbit of this action is divisible by $k$:

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

From Group Action on Prime Power Order Subset, $\operatorname{Stab} \left({S}\right)$ is a $p$-subgroup of $G$.

Therefore $k \nmid \left|{\operatorname{Stab} \left({S}\right)}\right|$ and therefore $k \backslash \left|{\operatorname{Orb} \left({S}\right)}\right|$.


 * Next, each orbit whose length is not divisible by $p$ contains at least one Sylow $p$-subgroup:

From the reasoning in the First Sylow Theorem, we know that at least one $S \in \mathbb S$ is such that $p \nmid \operatorname{Orb} \left({S}\right)$, and that for such a set, $\operatorname{Stab} \left({S}\right)$ is a Sylow $p$-subgroup of $G$.

From Group Action on Prime Power Order Subset, we have $\forall s \in S: \operatorname{Stab} \left({S}\right) s = S$.

Therefore $\operatorname{Stab} \left({S}\right) = S s^{-1}$.

Thus for any $s \in S$, $S s^{-1}$ is a Sylow $p$-subgroup of $G$.

It also follows that $s^{-1} \operatorname{Stab} \left({S}\right) s$ is also a subgroup of $G$ with the same number of elements as $\operatorname{Stab} \left({S}\right)$.

Thus $s^{-1} \operatorname{Stab} \left({S}\right) s = s^{-1} \left({S s^{-1}}\right) s = s^{-1} S$ is a Sylow $p$-subgroup of $G$ in the orbit of $S$.

From the Orbit-Stabilizer Theorem:
 * $\left|{\operatorname{Orb} \left({S}\right)}\right| = k$

as $\left|{\operatorname{Stab} \left({S}\right)}\right| = p^n$.

So if an orbit has length not divisible by $p$, then this orbit contains a Sylow $p$-subgroup of $G$ (that is, $\operatorname{Stab} \left({S}\right)$ and also $s^{-1} \operatorname{Stab} \left({S}\right) s$), and the length of the orbit is $k$.


 * Next we show that every Sylow $p$-subgroup of $G$ has to be in one of the orbits whose size is not divisible by $p$.

Let $H$ be a Sylow $p$-subgroup of $G$. By the definition of $\mathbb S$, and because $\left|{H}\right| = p^n$, $H \in \mathbb S$.

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:
 * $g \in \operatorname{Stab} \left({H}\right) \iff g * H = H \iff g \in H$

Thus:
 * $\operatorname{Stab} \left({H}\right) = H$

From the Orbit-Stabilizer Theorem:
 * $\left|{\operatorname{Orb} \left({H}\right)}\right| = k$

that is, not divisible by $p$, and of course $H \in \operatorname{Orb} \left({H}\right)$.


 * Now we show that each orbit whose size is not divisible by $p$ contains no more than one Sylow $p$-subgroup:

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

Now $\operatorname{Orb} \left({H}\right)$ consists of all $g * H = g H$, that is, all the right cosets of $H$.

But the only right coset of $H$ which is a subgroup of $G$ is in fact $H$ itself.

None of the rest of the elements of $\mathbb S$ can actually be a subgroup of $G$.

Thus, no orbit can contain more than one Sylow $p$-subgroup.