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?

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 09 '15 at 19:12
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 09 '15 at 19:14

score 7 · Accepted Answer · edited Apr 13 '17 at 12:20

7

$1$.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 $[S_n:A_n]=2$

edited Apr 13 '17 at 12:20

Community

1

answered Mar 09 '15 at 19:06

Arpit Kansal

10,489

oke thnx, then i can prove it! – Robbe Motmans Mar 09 '15 at 19:08
do you also know how i can use the given group homomorphism to prove the same? – Robbe Motmans Mar 09 '15 at 19:09
Hint:kernal of given map is $A_n$ – Arpit Kansal Mar 09 '15 at 19:10

score 2 · Answer 2 · answered Oct 26 '17 at 18:48

2

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

answered Oct 26 '17 at 18:48

Daniel Li

3,340
15
31

The alternating group is a normal subgroup of the symmetric group

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