The group \(A_3\) is cyclic of order three, so already abelian.
In \(A_4\), let \(V=\{1,(12)(34),(13)(24),(14)(23)\}\).
Products and inverses stay in this four-element set, and conjugation
permutes its three nonidentity elements. Hence \(V\) is normal.
The quotient has order three, so it is cyclic and
\([A_4,A_4]\subseteq V\). Direct multiplication gives
\([(123),(124)]=(12)(34)\). Conjugating this element by the
three-cycle \((123)\) gives the other two double transpositions.
The commutator subgroup is normal, so it contains all of \(V\);
equality follows.
For \(d\geq 5\), put \(g=(12)(34)\) and \(h=(12)(35)\) in \(A_d\).
Multiplication gives \([g,h]=(345)\). Every three-cycle is conjugate
to this one within \(A_d\): first choose a conjugating permutation
in \(S_d\); if it is odd, compose on the right with a transposition
of two symbols outside \(\{3,4,5\}\). This centralizes \((345)\)
and changes the parity. Thus the normal subgroup \([A_d,A_d]\)
contains every three-cycle.
Three-cycles generate all even permutations. Write an even
permutation as a product of an even number of transpositions and
pair them. Equal transpositions cancel; two sharing one symbol give
a three-cycle; a disjoint pair satisfies
\((ab)(cd)=(acb)(acd)\). Hence every pair is a product of three-cycles,
proving \([A_d,A_d]=A_d\).