Can I find a series where an abelian series of smallest possible length is different from derived series?

Question

https://groupprops.subwiki.org/wiki/Derived_length

Here I have found two definitions of derived length. How to prove the equivalency of these two definitions. I know that derived series slows down rapidly. I am also getting that for derived series

$G=G^{0}\vartriangleright G^{1}\vartriangleright....\vartriangleright G^{d}\vartriangleright G^{d+1}=1$

$G^{(i)}/G^{(i+1)}$ is abelian so this is an abelian series and this series will indeed represent the smallest possible length of abelian series as if I would have an abelian series $G=G_0\vartriangleright G_1\vartriangleright....\vartriangleright G_n\vartriangleright G_{n+1}=1$ of the smallest possible length

then I will have $G_i \vartriangleright G^{(i)}$ and then $G=G^{0}\vartriangleright G^{1}\vartriangleright....\vartriangleright G^{n}\vartriangleright G^{n+1}=1$ will be the corresponding derived series of that smallest possible length which is $n=d$ here.

Now I have a question that can I find a series where an abelian series of smallest possible length is different from derived series? If so what is the example?

Answer 1

If you are OK with tied for the shortest length you can. For example, consider the dihedral group of order 8. Its derived group has order 2 (and the next term in the derived series has order 1), but there is an abelian series that goes from the whole group to its cyclic subgroup of order 4 to the identity. This has the same length as the derived series.

You can never do better than tie for the length of the derived series, though. In any abelian series $G=G_0\vartriangleright G_1\vartriangleright....\vartriangleright G_n\vartriangleright G_{n+1}=1$, you need $G_1\ge G^{\prime}$ lest $G/G_1$ not be abelian. Using induction, we assume $G_i\ge G^{(i)}$ and we see that in order for $G_i/G_{i+1}$ to be abelian, we need its subgroup $G^{(i)}/G_{i+1}$ to be abelian, and this requires $G_{i+1}\ge G^{(i)\prime}=G^{(i+1)}$, so for any abelian series each term contains the corresponding term of the derived series. So a series of length less than the derived length still contains the last non-trivial term of the derived series in its last term.

This last paragraph also implies the equivalence of the two different definition of derived length that you reference.