The alternating group is a normal subgroup of the symmetric group
Solution 1:
$1$.Note that kernal of sign homomorphism is precisely $A_n$ and kernal of a homomorphism is a normal subgroup.
$2$. Recall that every Subgroup of index 2 is Normal and note that $[S_n:A_n]=2$
Solution 2:
Add another method to prove in addition to Arpit's answer: conjugation preserves cycle type; so $s a s^{-1} \in A_n$