Let $G$ be a group. Define inductively the derived series as follows: $$G^{(0)}=G\\\\ G^{(\lambda)}=\bigcap_{\alpha<\lambda}[G^{(\alpha)},G^{(\alpha)}]$$ Let $\text{sol}(G)=\{\min \alpha\in \text{Ord}:G^{(\alpha)}=G^{(\alpha+1)}\}$. Clearly $\text {sol}(G)\le |G|$.
It is claimed on wikipedia (here) that every ordinal number is the solving length of some group, and the reference provided for this is "Generalized nilpotent algebras and their associated groups" by Maltsev a paper which, unfortunately, is in russian.
To be more formal, let $\Omega:=\{\text{solv}(G): G\text{ is a group}\}\subset \text{Ord}$. How would you go about proving $\Omega=\text{Ord}$? Do you have a reference for it?
What I've done so far:
- For every natural number $n$, $\alpha+n\in \Omega\Rightarrow \alpha\in \Omega$ (just take $G^{(n)}$)
- $\Omega$ is closed under $\sup$: given an increasing sequence $\lambda_i$ of elements in $\Omega$, denote by $G_i$ the associated groups and notice that $G:=\prod G_i$ satisfies $\text{sol}(G)=\sup \lambda_i$
Let $F$ be a field and let $U_n$ be the group of upper triangular $(n \times n)$-matrices with entries in $F$ and only $1$'s on the diagonal.
Then $U_n$ has derived length $d$, where $d$ is the integer such that $2^{d-1} < n \leq 2^d$. See for example Suzuki, Group Theory II, Chapter 4.
So this provides examples of groups with derived length $d$, for any $1 \leq d < \infty$.
I don't know how to give an example for infinite ordinals, and I have no access to the paper by Mal'cev. Perhaps the example of $U_n$ could be generalized to infinite-dimensional vector spaces?
EDIT: You might want to take a look at the following paper, which seems to answer your question: "B. Hartley, The order-types of central series. Proc. Cambridge Philos. Soc. 61 (1965), 303-319.".