# Is the commutator subgroup of a commutator subgroup itself?

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 $[G,G]$?

Let $G=S_3$. Then $[G,G]=A_3$ (see answers to this question) which is Abelian so $[A_3,A_3]=\langle e\rangle\ne[G,G]$.
A group for which $G=G'$ is called perfect. You might want to read this. All non-abelian simple groups belong to this category.
• In particular, this answer led to an answer to my question (from the Wikipedia page on perfect groups): a group $G$ satisfying $G' = G''$ is called quasi-perfect, ie. its commutator subgroup $G'$ is perfect, ie. the commutator subgroup of the commutator subgroup is the commutator subgroup. May 30 '17 at 20:24
The commutator subgroup is often called the derived subgroup and denoted $G'$. This makes it easy to write its derived subgroup as $G''$. An example of a group with $G''\ne G'$ is $G=A_4$.
If instead we consider groups of upper triangular matrices over a finite field, we can get sequences $G\supset G'\supset G''\supset\cdots$ which finally stabilise after any given finite number of steps.