Normalizer of Subgroup of Symmetric Group that Fixes n

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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