Alternating Group is Normal Subgroup of Symmetric Group

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

For any $\pi \in S_n$, let $\operatorname{sgn} \left({\pi}\right)$ be the sign of $\pi$.

The kernel of the mapping $\operatorname{sgn}: S_n \to C_2$ is called the alternating group on $n$ letters and denoted $A_n$.

It follows that:
 * $A_n$ consists of the set of even permutations of $S_n$.
 * $A_n$ is a normal subgroup of $S_n$ of index $2$.

Proof
We have that $\operatorname{sgn} \left({S_n}\right)$ is onto $C_2$. Thus:


 * From the First Isomorphism Theorem, $A_n$ consists of the set of even permutations of $S_n$


 * A subgroup of index 2 is normal.

Comment
Some authors use $A \left({n}\right)$ for $A_n$.

Note that when $n = 2$, we see immediately that $S_2 \cong C_2$ and the result still holds -- and here $A_n = \left\{{e_{S_2}}\right\}$.