Prove that $A_n$ is a Normal subgroup of $S_n$?

Question

I know how to prove that $A_n$ is a subgroup of $S_n$ but i do not know how to prove this using the index.

What is the additional properties one needs to show a subgroup is normal? — Prince M, Nov 03 '16 at 10:33
HINT: $A_n$ is by definition the kernel of some homomorphism (the homomorphism mapping every permutation to its sign), hence it is obviously normal. — Crostul, Nov 03 '16 at 10:34

score 5 · Answer 1 · answered Nov 03 '16 at 10:37

5

The index is $2$ so automatically left cosets will coincide with right cosets.

If $\sigma\in A_n$ then $\sigma A_n=A_n=A_n\sigma$.

If $\sigma\notin A_n$ then $\sigma A_n=S_n\setminus A_n=A_n\sigma$.

answered Nov 03 '16 at 10:37

drhab

1 Answers1