Normalizer of 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:
 * $\map \pi n = n$

The normalizer of $H$ is given by:


 * $\map {N_{S_n} } H = \map {N_{S_n} } {S_{n - 1} } = S_{n - 1}$

Proof
We have from Subgroup of Symmetric Group that Fixes n that $N = S_{n - 1}$.

By definition of normalizer:


 * $\map {N_{S_n} } {S_{n - 1} } := \set {\rho \in S_n: \rho S_{n - 1} \rho^{-1} = S_{n - 1} }$

We have from Group is Normal in Itself that:
 * $\forall \rho \in S_{n - 1}: \rho S_{n - 1} \rho^{-1} \in S_{n - 1}$

and so:
 * $S_{n - 1} \subseteq \map {N_{S_n} } {S_{n - 1} }$

It remains to be shown that $\map {N_{S_n} } {S_{n - 1} } \subseteq S_{n - 1}$.

This will be done by demonstrating that:
 * $S_n \setminus S_{n - 1} \cap \map {N_{S_n} } {S_{n - 1} } = \O$

where $\setminus$ denotes set difference.

Let $\rho \in S_n$ such that $\rho \notin S_{n - 1}$.

Thus:
 * $\map \rho n \ne n$

and so:
 * $\exists a \in S_n, a \ne n: \map \rho a = n$

for some $a \in S_{n - 1}$.

Then:
 * $\map {\rho^{-1} } n = a$

Let $\pi \in S_{n - 1}$ such that:
 * $\map \pi a = b$

for some $b \ne a$.

As $\rho$ is a permutation, $\rho$ is by definition a bijection.

Hence:
 * $\map \rho b \ne n$

We have:

Thus $\rho \pi \rho^{-1}$ does not fix $n$.

That is:
 * $\rho \pi \rho^{-1} \notin S_{n - 1}$

Since this is the case for all arbitrary $\rho \in S_n \setminus S_{n - 1}$, it follows that:
 * $S_n \setminus S_{n - 1} \cap \map {N_{S_n} } {S_{n - 1} } = \O$

So from Intersection with Complement is Empty iff Subset:
 * $\map {N_{S_n} } {S_{n - 1} } \subseteq S_{n - 1}$

and so by definition of set equality:
 * $\map {N_{S_n} } {S_{n - 1} } = S_{n - 1}$