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