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:

$\map \pi n = n$


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:

\(\ds \index {S_n} {S_{n - 1} }\) \(=\) \(\ds \dfrac {\order {S_n} } {\order {S_{n - 1} } }\) Definition of Index of Subgroup
\(\ds \) \(=\) \(\ds \dfrac {n!} {\paren {n - 1}!}\) Order of Symmetric Group
\(\ds \) \(=\) \(\ds \dfrac {n \paren {n - 1}!} {\paren {n - 1}!}\) Definition of Factorial
\(\ds \) \(=\) \(\ds n\)

$\blacksquare$


Sources

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