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 $$\sgn \left({\pi}\right)$$ be the sign of $\pi$.

The kernel of the mapping $$\sgn: S_n \to C_2$$ 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 $$\sgn \left({S_n}\right)$$ 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 $$A_n = \left\{{e_{S_2}}\right\}$$.