Symmetric Group has Non-Normal Subgroup

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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 $\paren {a \ b}$.

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.

$\blacksquare$


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