# Group Action on Sets with k Elements

## Theorem

Let $\struct {G, \circ}$ be a finite group whose identity is $e$.

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

Let $G$ act on $\Bbb S$ by the rule:

$\forall S \in \Bbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$

This is a group action, and:

$\forall S \in \Bbb S: \order {\Stab S} \divides \card S$

where $\Stab S$ denotes the stabilizer of $S$ by $G$.

## 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:

$\card {g S} = \card S$

This follows directly from Order of Subset Product with Singleton.

It remains to be shown that:

$\order {\Stab S} \divides \card S$

By definition of stabilizer:

$\Stab S = \set {g \in G: g S = S}$

It follows that:

$\forall s \in S: g s \in S$

Then:

$\Stab S s \subseteq S$

Thus:

$\card {\Stab S s} \le \card S$

We have from corollary $1$ to Stabilizer is Subgroup that the stabilizer contains the identity:

$e \in \Stab S \le G$

It therefore follows that:

$s \in S \implies s \in \Stab S s$

Thus:

$\displaystyle S = \bigcup_{t \mathop \in S} \Stab S t$

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

Thus $S$ consists of a union of disjoint right cosets of $\Stab S$.

Thus $\displaystyle \bigcup_{t \mathop \in S} \Stab S t$ is a partition of $S$.

Therefore:

$\forall s \in S: \card {\Stab S s} \divides \card S$

As we have:

$\forall s \in S: \card {\Stab S s} = \order {\Stab S}$

the result follows:

$\card {\Stab S} \divides \card S$

$\blacksquare$