How to find $[A_n,A_n]$ Let $n \in \mathbb{N}$. How could I find 
$$
[A_n,A_n] \quad \cong \quad \langle ghg^{-1}h^{-1} \ : \ g,h \in A_n \rangle
$$ 
My own thoughts
I remembered that any element in $A_n$ can be written as a finite product of cycles of length $3$. Could I use that to write every commutator as some product of commutators of $3$-cycles? In this case, I take two of those cycles $\sigma$ and $\tau$ to see what their commutator looks like.
Whitout loss of generality $\sigma = (123)$. We are interested in a cycle $\tau$ that has one or two numbers in common with $\sigma$. If we'd make other choices, we would get $\sigma \tau \sigma^{-1} \tau^{-1} = e$. This gives us
$$
\tau \ \in \ \{(145),(124),(134) \}
$$
Do you think it would be smart to look at all the commutators I could that thiis way, or would you advise me to do something else.
 A: With $g=(123)$ and $h=(145)$ we have $ghg^{-1}=(245)$. Therefore
$$
ghg^{-1}h^{-1}=(245)h^{-1}=(245)(154)=(124).
$$
Conclusion: If $n\ge5$, then every 3-cycle is a commutator of two 3-cycles. Do you see why? Do you see what it implies?
You need to handle $n=1,2,3,4$ separately, but they are easier cases. You remember that $G$ is abelian, iff $[G,G]=1$. Also if $G/H$ is abelian for some normal subgroup $H\unlhd G$, then $[G,G]\le H$. Also always $[G,G]\unlhd G$. All these points restrict the alternatives for $[A_n,A_n]$, when $n\le4$, to a single subgroup. Provided that you remember a certain exceptional subgroup of $S_4$.
A: In general, if $G$ is a group then $[G, G]$ is a normal subgroup of $G$. This is because $g^{-1}[h, k]g=[g^{-1}hg, g^{-1}kg]$.

Therefore, $[A_n, A_n]\unlhd A_n$.

(In fact, $[G, G]$ is characteristic in $G$, which means that $\phi([G, G])=[G,G]$ for all $\phi\in\operatorname{Aut}(G)$. This is proven using the same argument.)
For $n\geq 5$, what does this allow you to conclude?
