Normal Subgroup of Symmetric Group on More than 4 Letters is Alternating Group

Theorem

Let $n \in \N$ be a natural number such that $n > 4$.

Let $S_n$ denote the symmetric group on $n$ letters.

Let $A_n$ denote the alternating group on $n$ letters.

$A_n$ is the only proper non-trivial normal subgroup of $S_n$.

Proof

From Alternating Group is Normal Subgroup of Symmetric Group, $A_n$ is seen to be normal in $S_n$.

It remains to be shown that $A_n$ is the only such normal subgroup of $S_n$.

Aiming for a contradiction, suppose $N$ is a proper non-trivial normal subgroup of $S_n$ such that $N$ is a proper subset of $A_n$.

From Intersection with Normal Subgroup is Normal, we have that:

$A_n \cap N$ is a normal subgroup of $A_n$.

As $N \subseteq A_n$, from Intersection with Subset is Subset‎ we have that:

$A_n \cap N = N$

That is:

$N$ is a normal subgroup of $A_n$.

But from Alternating Group is Simple except on 4 Letters, $A_n$ has no proper non-trivial normal subgroup.

So $N$ cannot be a normal subgroup of $A_n$.

Hence by Proof by Contradiction it follows that $N$ cannot be a normal subgroup of $S_n$.

So $S_n$ has no proper non-trivial normal subgroup apart from $A_n$.

$\blacksquare$