Group Action on Sets with k Elements

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

Let $\Bbb S = \left\{{S \subseteq G: \left\vert{S}\right\vert = k}\right\}$, 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:
 * $\forall S \in \Bbb S: \left\vert{\operatorname{Stab} \left({S}\right)}\right\vert \mathrel \backslash \left\vert{S}\right\vert$

Proof
First it is necessary to prove that this is a group action.

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:
 * $\left\vert{g S}\right\vert = \left\vert{S}\right\vert$

This follows directly from Order of Subset Product with Singleton.

It remains to be shown that:
 * $\left\vert{\operatorname{Stab} \left({S}\right)}\right\vert \mathrel \backslash \left\vert{S}\right\vert$

By definition of stabilizer:
 * $\operatorname{Stab} \left({S}\right) = \left\{{g \in G: g S = S}\right\}$

It follows that:
 * $\forall s \in S: g s \in S$.

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

Thus:
 * $\left\vert{\operatorname{Stab} \left({S}\right) s}\right\vert \le \left\vert{S}\right\vert$

We have that the stabilizer contains the identity.

That is:
 * $e \in \operatorname{Stab} \left({S}\right) \le G$

It therefore follows that:
 * $s \in S \implies s \in \operatorname{Stab} \left({S}\right) s$

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

By Congruence Class Modulo Subgroup is Coset, distinct right cosets are disjoint.

Thus $S$ consists of a union of disjoint right cosets of $\operatorname{Stab} \left({S}\right)$.

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

Therefore:
 * $\forall s \in S: \left\vert{\operatorname{Stab} \left({S}\right)s}\right\vert \mathrel \backslash \left\vert{S}\right\vert$

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


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

Also see

 * Sylow Theorems: this result is key.