Difficult# Derived Series (or Commutator Series)

ABSALG-JAEZ4T

Let $G$ be a *finite* group. Recall that the commutator subgroup $[G,G]=G'=G^{(1)}$ of $G$ is the subgroup generated by the commutators $[g,h]$, $g,h\in G$.

Let $G^{(2)}=[G^{(1)},G^{(1)}]$ be the commutator subgroup of the commutator subgroup, and so on with $G^{(n+1)}=[G^{(n)},G^{(n)}]$ being the commutator subgroup of $G^{(n)}$, $n=1,2,\ldots$.

The chain:

$$G\geq G^{(1)}\geq G^{(2)}\geq\ldots \geq G^{(n)}\geq G^{(n+1)}\ge\ldots$$

...is called the *derived series* of $G$. It is also sometimes called the *commutator series* of $G$.

The derived series is said to *terminate* in the neutral element $e$ of $G$ if $G^{(n)}=\{e\}$ for some finite integer $n\ge1$.

The smallest such $n$, if it exists, is called the *derived length* of $G$ if $G\not=\{e\}$. The derived length of $\{e\}$ is defined to be zero.

Which of the following is true? The groups $G$ are all assumed to be finite.