TITLE: Is the commutator subgroup of a commutator subgroup itself?
QUESTION [5 upvotes]: Let $G$ be a multiplicative group. The commutator of $a,b\in G$ is the element $aba^{-1}b^{-1}$. (Now $a,b\in G$ commute iff their commutator is the identity.) The commutator subgroup of $G$, denoted $[G,G]$, is the subgroup generated by all commutators of all elements.
Example. If $G$ is abelian, then every commutator is the identity, so the commutator subgroup $[G,G]$ is the trivial group.
(The commutator subgroup is used to construct the abelianization of $G$ as the quotient $G/[G,G]$, which I've encountered in learning homology.)
Now, an arbitrary element of $[G,G]$ need not be itself a commutator. So, is the commutator subgroup of $[G,G]$ again all of $[G,G]$? If not, is there an easy characterizations of groups $G$ for which the commutator subgroup of $[G,G]$ is again $[G,G]$?

REPLY [7 votes]: A group for which $G=G'$ is called perfect. You might want to read this. All non-abelian simple groups belong to this category.