# 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:

\(\ds \map {\rho \pi \rho^{-1} } n\) | \(=\) | \(\ds \map {\rho \pi} a\) | as $\map {\rho^{-1} } n = a$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map \rho b\) | as $\map \pi a = b$ | |||||||||||

\(\ds \) | \(\ne\) | \(\ds n\) | as $\map \rho b \ne n$ |

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:

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

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 48 \beta$