Return to Answer

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 $\alpha<2.578\dots$. It would be interesting to have better estimates for this constant $\alpha$.

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 $5/\log_2(9)\approx 1.577324$.

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 $$\alpha=\limsup_{m>1,\ m\to\infty}\frac{f(m)}{\log_2\log_2m}.$$ Then the above shows that $\frac{1}{2} < \alpha < 3$, and in fact the paper of Glasby shows that $\alpha<2.578\dots$. It would be interesting to have better estimates for this constant $\alpha$.

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 $\alpha<2.578\dots$. It would be interesting to have better estimates for this constant $\alpha$.

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 $|G|>1$).

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 $$\alpha=\limsup_{m>1,\ m\to\infty}\frac{f(m)}{\log_2\log_2m}.$$ Then the above shows that $\frac{1}{2} < \alpha < 3$, and in fact the paper of Glasby shows that $\alpha<2.578\dots$. It would be interesting to have better estimates for this constant $\alpha$.