# The alternating group is a normal subgroup of the symmetric group

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?

• Does $\ker(sgn) = A_n$ help you? – Moritz Mar 9 '15 at 19:06
• yeah, it would but why is ker(sgn)=$A_n$? – Robbe Motmans Mar 9 '15 at 19:10
• Because every element of $A_n$ is mapped to $1$ and every other element in $S_n\setminus A_n$ to $-1$. Note that $1$ is group-unity of $(\{\pm 1\},\cdot)$. – Moritz Mar 9 '15 at 19:12
• oke thnx alot guys, i get it know i think. – Robbe Motmans Mar 9 '15 at 19:13
• Final hint: $\ker(\psi)$ for any group-homomorphism $\psi: G \to H$ between groups $G$ and $H$ is always a normal subgroup of $G$. – Moritz Mar 9 '15 at 19:14

$1$.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 $[S_n:A_n]=2$
• Hint:kernal of given map is $A_n$ – Arpit Kansal Mar 9 '15 at 19:10
Add another method to prove in addition to Arpit's answer: conjugation preserves cycle type; so $s a s^{-1} \in A_n$