# 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 Subsetâ€Ž 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 $A_n$.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 84 \beta$ - 1989: Ephraim J. Borowski and Jonathan M. Borwein:
*Dictionary of Mathematics*... (previous) ... (next):**alternating group** - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next):**alternating group**