Symmetric Group has Non-Normal Subgroup

Theorem
Let $S_n$ be the (full) symmetric group on $n$ elements, where $n \ge 3$.

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

Proof
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 $\left({a \ b}\right)$.

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

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

By inspection it is found that $\pi^{-1} = \left({a \ c \ b}\right)$.

Then we have:

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