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?
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