TITLE: Commutator Identity and Commutator Subgroup
QUESTION [22 upvotes]: It is well known that if $G$ is any group in which $[G,G]$ is abelian, then for any $a,b,c\in G$,
$$\bigl[[a,b],c\bigr]\cdot\bigl[[b,c],a\bigr]\cdot\bigl[[c,a],b\bigr]=1$$
Conversely, if this identity holds for all $a,b,c\in G$ then is it necessary that $[G,G]$ is abelian?

REPLY [4 votes]: B. H. Neumann proved that there is a $4$-generator, non-metabelian group, all of whose $3$-generated subgroups are metabelian. So in this group the identity
$$\bigl[[a,b],c\bigr]\cdot\bigl[[b,c],a\bigr]\cdot\bigl[[c,a],b\bigr]=1$$
holds.
Neumann' s example is a finite $2$-group. Compare with Theorem 4 of this paper by Alperin.