$A_n$ normal subgroup of $S_n$

Question

If I want to prove that $A_n$ is a normal subgroup of $\:S_n$
I know how to prove that $A_n$ is a subgroup, but if I want to prove that it is a normal subgroup.
Can I prove like that:
Let $f_1\in S_n$ then we need to provef that: $f_1A_n=A_nf_1$. $\DeclareMathOperator{\sign}{sign}$ So we know that: $\forall h\in A_n:\:\sign\left(f_1\:\circ \:h\right)=\sign\left(f_1\right)\cdot \sign\left(h\right)=\sign\left(f_1\right)$
On another hand: $\forall h\in A_n:\:\sign\left(h\:\circ f_1\right)=\sign\left(h\right)\cdot \sign\left(f_1\right)=\sign\left(f_1\right)$
Because we know that $\sign(h)=1$
So we can say that: $f_1A_n=A_nf1$
Is this correct?

It is sometimes useful to know that a subgroup of index $2$ is normal - every left coset is a right coset, because either you get the subgroup or you get its complement, and there are no other options. — Mark Bennet, Feb 12 '21 at 13:09
There are many different proofs available here. For example, $A_n$ is also the commutator subgroup of $S_n$, and hence normal, see this duplicate. There are several duplicates here, which you can read. Of course, your idea is also there. — Dietrich Burde, Feb 12 '21 at 13:13

score 0 · Accepted Answer · answered Feb 12 '21 at 12:36

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This is correct. However, you could use a more general result. $\mathfrak A_n$ is the kernel of the signature and the kernel of a group homomorphism is always a normal subgroup.

answered Feb 12 '21 at 12:36

mathcounterexamples.net

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Thanks! But what I wrote is enough correct? – Feb 12 '21 at 12:42
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Yes it is correct. – mathcounterexamples.net Feb 12 '21 at 12:42

$A_n$ normal subgroup of $S_n$

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