How to calculate $Card(Stab(\mathbb{N}^*_k) $?

Question

Let a group $G$ operate on a set $X$, we define $\displaystyle{G_x=Stab(x)= \left\{ g \in G \mid g.x=x\right\}}$,and $Orb(x)=\left\{g.x \mid g \in G\right\}$.

moreover we have: $Card(Orb(x))=[G:G_x]. \quad(*)$

Now let $G=\mathcal{S}_n$ , $X=\mathcal{P_{n,k}}=$ set of all $k$-subsets formed from $\mathbb{N}^*_n=\{1,2,3,\cdots,n\}$.

the action is defined by $(\sigma,A)\mapsto \sigma(A)$. $Orb(\mathbb{N}^*_k)=\mathcal{P}_{n,k}\,$ .

if we apply $(*)$, we have $$Card(Orb(\mathbb{N}^*_k)=Card(\mathcal{P}_{n,k}\;)=\dfrac{|\mathcal{S}_n|}{Card(Stab(\mathbb{N}^*_k)}.$$ I know that $Card(Stab(\mathbb{N}^*_k)=k!(n-k)!$ but i can't prove it!

Thanks in advance.

Answer 1

The point is that if $\sigma$ stabilizes $\mathbb N_k^*$ (that's a terrible notation by the way), then it induces a permutation of $\mathbb N_k^*$, and a permutation of $\mathbb N_n^*\setminus\mathbb N_k^* = \{k+1,...,n\}$.

So you get a morphism $Stab(\mathbb N_k^*)\to \mathcal S_n \times \mathcal S_{n-k}$ : to each $\sigma$ you associate the induced permutations.

It's easy to check that this is actually bijective, and so this tells you what $|Stab(\mathbb N_k^*)|$ is.