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: $$\forall S \in \mathbb{S}: g \wedge S = g S = \left\{{x \in G: x = g s: s \in S}\right\}$$.

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:

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

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 the definition of the group action, $$g \in G \Longrightarrow g \wedge H = g H$$, which is a left coset of $$G$$.

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

Now $$g \in \operatorname{Stab} \left({H}\right) \iff g \wedge H = H \iff g \in H$$.

Thus we see that $$\operatorname{Stab} \left({H}\right) = H$$.

From our old friend the Orbit-Stabilizer Theorem, it follows that $$\left|{\operatorname{Orb} \left({H}\right)}\right| = k$$, 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 $$g \wedge H = g H$$, 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.