Find the derived subgroup of $A_4$ 
Find the derived subgroup of $A_4$.

Since it is $A_4$, for a permutation $\sigma$ to be in $A_4$, $\sigma$ must have a cycle structure of $2$ cycles. Therefore, $\sigma=(ab)(cd)$.
The commutator of such elements, would be obviously another permutation of $2$-cycle (of length $2$).
Does it mean that $A'_4=A_4$?
 A: I don't have enough reputations to comment Alan Wang's solution.
But it should be checked that each factor of a 2 disjoint transpositions is actually single (i.e. odd) permutation which is not in $A_4$.
ex.) (1 2)(3 4) $\in A_4$, but (1 2) $\notin A_4$ nor (3 4) $\notin A_4$.
If (a b c)S = (a b)(c d) for some S$\in A_4$ then
S = (b c d) such that, (a b c) (b c d) = (a b)(c d) where
(d a b) (b c d) (d b a) = (a c b).
Thus, V < $A'_4$.
And, V > $A'_4$ since $A'_4/V\simeq Z_3$ is abelian, we have $A'_4 = V$.
A: Consider the Klein $4$-group $V$ in $A_4$, that is $$\{1,(12)(34),(14)(23),(13)(24)\}$$
It can be checked that $V$ is a normal subgroup of $A_4$.
Since $A_4/V\cong \Bbb{Z}_3$ is abelian, we have $A_4'\le V$.
Since $A_4$ is nonabelian, $A_4'\neq 1$.
Hence there exists a $(ab)(cd)\in A'_4$.
Since $A'_4$ is a normal subgroup of $A_4$ and every product of $2$ disjoint transpositions are conjugate in $A_4$, we have $$A'_4=V$$
There are a few results that are used here:
(i) If $G/N$ is abelian, then $G'\le N$.
(ii) If $G$ is abelian, then $G'=1$
(iii) Two permutations are conjugate in $S_n$ iff they have the same cycle structure. 
