Optimal series for solvable groups

In this section we will use potentially infinite series  which may or may not terminate, this forces us to distinguish between ascending and descending series.

Definition Set $G^{(0)} := G$ and for each $i$ define $G^{(i)} = (G^{(i)})' = [G^{i-1},G^{i-1}]$. This gives us the derived (descending) series $$G = G^{(0)} \unrhd G^{(1)} \unrhd G^{(2)} \unrhd G^{(3)} \unrhd \cdots.$$

Prop $G^{(i)} \operatorname{char} G$. proof: unproved

Theorem $G$ is solvable iff the derived series terminates with $G^{(n)}=1$, furthermore it is optimal in the following sense if $$(\star)\,, 1 = G_0 \unlhd G_1 \unlhd G_2 \unlhd \cdots \unlhd G_r = G$$ then we have $G^{(i)} \le G_{r-i}$.
proof: By the commutator characterization of abelian series, each factor of this derived series is abelian. So if $G^{(n)} = 1$ for some $n$ then we have an abelian series and $G$ is solvable. For the converse, suppose $G$ is solvable, let $(\star)$ be an abelian series for it. We will show by induction that $G^{(i)} \le G_{r-i}$. Suppose it is true for $i$, then $G^{(i+1)} \le [G^{(i)},G^{(i)}] \le [G_{r-i},G_{r-i}] = (G_{r-i})' \le G_{r-i-1}$ again by the commutator characterization.

Corollary Every solvable group has an abelian normal series.

Definition If $G$ is solvable the minimal length of series is called it's derived length. show this is the same as the length of the derived series?