Groups whose derived length is logarithmic in the order?

Question

Is there a class of solvable groups $G$ having a derived length $O(\log\lvert G\rvert)$?

See Wikipedia for the definition of Big-Oh ($O$) and the definition of derived series of a group.

Any help would be appreciated. Thank you in advance!

Edit (YCor, after comments below): the intended meaning of big-O is probably not the one linked at, but "$u_n=O(v_n)$" if $c_1v_n<u_n<c_2v_n$ for positive constants $c_1,c_2$ and $n\gg 1$, usually denoted $u_n=\Theta(v_n)$. [The main mathematical use of $u_n=O(v_n)$ being only $u_n<c_2v_n$, which makes the question not interesting.]

Answer 1

A paper of Glasby, "The composition and derived lengths of a soluble group", shows that if a soluble group $G$ has composition length $n$ then its derived length $d$ satisfies $d < 3 \log_2(n) + 9$. Since $n \leqslant \log_2|G|$, this makes $d<3\log_2\log_2|G|+9$ (assuming $|G|>1$).

Edit: As Will Sawin points out, the group $G$ of $s\times s$ upper triangular matrices over $\mathbb{F}_2$ with ones on the diagonal has $|G|=2^{\binom{s}{2}}$, so $\log_2|G|=\binom{s}{2}<s^2$ and $\log_2\log_2|G|< 2\log_2 s$. The derived length is $d=\lceil \log_2 s\rceil$ and so $d> \frac{1}{2}\log_2 \log_2|G|$.

Let $f(m)$ be the maximum derived length of a soluble group of order $m$, and let $$\alpha=\limsup_{m\geqslant 3,\ m\to\infty}\frac{f(m)}{\log_2\log_2m}.$$ Then the above shows that $\frac{1}{2} < \alpha < 3$, and in fact the paper of Glasby shows that $\alpha<2.578$. It would be interesting to have better estimates for this constant $\alpha$.

Edit 2: Using David Speyer's observation and the example of $3^2\!:\!2S_4\leqslant S_9$ of derived length five, we now have $$\frac{5}{\log_2(9)}\ \leqslant\ \alpha\ \leqslant\ 1+\frac{5}{\log_2(9)},$$ where $5/\log_2(9)\approx 1.577324$.

Answer 2

$\def\ZZ{\mathbb{Z}}\def\Id{\text{Id}}$I can improve Will Sawin's bound by a factor of $2$. Let $G$ be the group of $3 \times 3$ matrices with entries in $\ZZ/2^{\ell+1} \ZZ$ that are $\Id_3 \bmod 2$. This group has cardinality $N=2^{9\ell}$ and composition length $n = 9\ell$. I'll show that it has derived length $$\geq \lfloor \log_2 \ell \rfloor= \log_2 n + O(1) = \log_2 \log_2 N + O(1).$$

Let $U_{ij}(a)$ be the matrix $\Id_3 + 2^a e_{ij}$, where $e_{ij}$ is the matrix with a $1$ in position $(i,j)$ and a $0$ everywhere else. A quick computation checks that $$(U_{ij}(a), U_{jk}(b)) = U_{ik}(a+b)$$ for $i$, $j$, $k$ distinct. Thus, by induction, $U_{ij}(2^r)$ is in the $(r-1)$-st derived subgroup for $i \neq j$ and, in particular, the derived length is at least $\lfloor \log_2 \ell \rfloor$.

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Looking at the end of Glasby's paper, he gives a better example.

For a group $A$, let $A \wr S_k$ denote $A^k \rtimes S_k$, where $S_k$ acts on the product $A^k$ by permuting the factors; this is a group of order $|A|^{k} \cdot k!$. Glasby considers the $(r-1)$-fold wreath product $( \cdots ((S_4 \wr S_4) \wr S_4) \cdots \wr S_4) \wr S_4$. It has order $N:=\prod_{j=0}^{r-1} 24^{4^j} = 24^{(4^r-1)/3}$. According to Glasby, $n = (4/3) (4^r-1)= 4 \log_{24} N = c \log_2 N$ and $d = 3r$. (I have partially checked these numbers; I confirm $n = 4 \log_{24} N$ and I find $d=3r$ very plausible, but haven't fully checked it. My earlier comment claiming that Glasby's $n$ looked wrong was mistaken.) So $$d = \tfrac{3}{2} \log_2 n + O(1) =\tfrac{3}{2} \log_2 \log_2 N+O(1).$$