Group Action on Sets with k Elements

Theorem
Let $\left({G, \circ}\right)$ be a finite group whose identity is $e$.

Let $\Bbb{S} = \left\{{S \subseteq G: \left|{S}\right| = k}\right\}$, that is, the set of all of subsets of $G$ which have exactly $k$ elements.

Let $\operatorname{Stab} \left({x}\right)$ be the stabilizer of $x$ by $G$.

Let $G$ act on $\Bbb{S}$ by the rule $\forall S \in \Bbb{S}: g \wedge S = g S = \left\{{x \in G: x = g s: s \in S}\right\}$.

This is a group action, and:
 * $\forall S \in \Bbb{S}: \left|{\operatorname{Stab} \left({S}\right)}\right| \backslash \left|{S}\right|$

Proof

 * First we need to prove that this is a group action. This is straightforward.

The action is the same as the one defined in Group Action on Coset Space, but this time we are limiting ourselves to the subsets of $G$ which have the same number of elements.

From the result in Group Action on Coset Space, we only have to prove that $\left|{g S}\right| = \left|{S}\right|$.

However, this follows directly from Order of Subset Product with Singleton.


 * Now we need to show that $\left|{\operatorname{Stab} \left({S}\right)}\right| \backslash \left|{S}\right|$.

By definition, $\operatorname{Stab} \left({S}\right) = \left\{{g \in G: g S = S}\right\}$.

It follows that $\forall s \in S: gs \in S$.

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

Thus $\left|{\operatorname{Stab} \left({S}\right) s}\right| \le \left|{S}\right|$.

Because the stabilizer contains the identity, i.e. $e \in \operatorname{Stab} \left({S}\right) \le G$, it follows that $s \in S \implies s \in \operatorname{Stab} \left({S}\right) s$.

(Note that this does not mean that $S \subseteq \operatorname{Stab} \left({S}\right) s$. For any $t \in S$, it does not follow that $t \in \operatorname{Stab} \left({S}\right) s$.)

Thus $\displaystyle S = \bigcup_{t \in S} \operatorname{Stab} \left({S}\right) t$

Since distinct right cosets are disjoint, $S$ consists of a union of disjoint right cosets of $\operatorname{Stab} \left({S}\right)$.

Thus $\displaystyle \bigcup_{t \in S} \operatorname{Stab} \left({S}\right) t$ is a partition of $S$.

As $\forall s \in S: \left|{\operatorname{Stab} \left({S}\right) s}\right| = \left|{\operatorname{Stab} \left({S}\right)}\right|$, the result follows:


 * $\left|{\operatorname{Stab} \left({S}\right)}\right| \backslash \left|{S}\right|$

Comment
This result is key in proving the Sylow Theorems.