Subgroup of Symmetric Group that Fixes n

Theorem
Let $S_n$ denote the symmetric group on $n$ letters.

Let $H$ denote the subgroup of $S_n$ which consists of all $\pi \in S_n$ such that:
 * $\map \pi n = n$

Then:
 * $H = S_{n - 1}$

and the index of $H$ in $S_n$ is given by:


 * $\index {S_n} H = n$

Proof
We have that $S_{n - 1}$ is the symmetric group on $n - 1$ letters.

Let $\pi \in S_{n - 1}$.

Then $\pi$ is a permutation on $n - 1$ letters.

Hence $\pi$ is also a permutation on $n$ letters which fixes $n$.

So $S_{n - 1} \subseteq H$.

Now let $\pi \in H$.

Then $\pi$ is a permutation on $n$ letters which fixes $n$.

That is, $\pi$ is a permutation on $n - 1$ letters.

Thus $\pi \in S_{n - 1}$.

So we have that $H = S_{n - 1}$.

We also have that $S_{n - 1}$ is a group, and:
 * $\forall \rho \in S_{n - 1}: \rho \in S_n$

So $S_{n - 1}$ is a subset of $S_n$ which is a group.

Hence $S_{n - 1}$ is a subgroup of $S_n$ by definition.

Then we have: