Find the derived subgroup of $A_4$

Solution 1:

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.

Solution 2:

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 $A'_4 = V$.