# 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 $e$ 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:

\(\displaystyle \pi^{-1} \rho \pi\) | \(=\) | \(\displaystyle \paren {a \ c \ b} \paren {a \ b} \paren {a \ b \ c}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a \ c}\) | by evaluation | ||||||||||

\(\displaystyle \) | \(\notin\) | \(\displaystyle \set {e, \rho}\) |

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

$\blacksquare$

## Sources

- 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): Chapter $\text{II}$: Groups: Morphisms