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: $\S 48 \beta$