Symmetric Group has Non-Normal Subgroup

Theorem

Then $S_n$ contains at least one subgroup which is not normal.

Let $S_n$ act on the set $S$.

Let $e$ be the identity of $S_n$, by definition the identity mapping $I_S$ on $S$.

As $S$ has at least three elements, three can be arbitrary selected and called $a$, $b$ and $c$.

Let $\rho$ be a transposition of $S_n$, transposing elements $a$ and $b$.

$\rho$ can be described in cycle notation as $\paren {a \ b}$.

From Transposition is Self-Inverse it follows that $\set {e, \rho}$ is a subgroup of $S_n$.

Let $\pi$ be the permutation on $S$ described in cycle notation as $\paren {a \ b \ c}$.

By inspection it is found that $\pi^{-1} = \paren {a \ c \ b}$.

Then we have:

$\ds \pi^{-1} \rho \pi$

$=$

$\ds \paren {a \ c \ b} \paren {a \ b} \paren {a \ b \ c}$

$\ds $

$=$

$\ds \paren {a \ c}$

by evaluation

$\ds $

$\notin$

$\ds \set {e, \rho}$

So, by definition, $\set {e, \rho}$ is not a normal subgroup.

$\blacksquare$