The alternating group is a normal subgroup of the symmetric group

Question

For an exercise, I need to prove that the alternating group $A_n$ is a normal subgroup of the symmetric group $S_n$.

As clue they say we can use a group homomorphism $\operatorname{sgn} : S_n \to \{-1,1\}$. I really don't see how i can use this.... can somebody help?

Answer 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 $[S_n:A_n]=2$

Answer 2

Add another method to prove in addition to Arpit's answer: conjugation preserves cycle type; so $s a s^{-1} \in A_n$