Timeline for Groups whose derived length is logarithmic in the order?
Current License: CC BY-SA 4.0
11 events
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| May 1 at 3:51 | comment | added | David E Speyer | Here is an idea on how to improve Glasby. Glasby reduces to the case that $G$ has a unique nontrivial $p$-core, $P$, with $|P/\text{Frat}(P)| = p^b$ and $|\text{Frat}(P)| = p^c$. He then bounds $d(G) \leq d(G/P) + d(P)$, $d(G/P) \leq 5 \log_9 b + O(1)$ and $d(P) \leq \log_2 c + O(1)$. But we have $d(P) \leq \log_2 \text{nil}(P)+O(1)$ (see groupprops.subwiki.org/wiki/… ) and, in examples, it looks like $\text{nil}(P) \approx c/b$, not $c$. Can you make this work? @WillSawin | |
| Apr 30 at 17:43 | comment | added | Will Sawin | Actually the $5/\log_2(9)$ example of a solvable subgroup of $S_n$ with large ratio of derived length to log of $n$ is sharp. This follows from work of Dixon discussed in Newman, the Soluble length of soluble linear groups. Newman also shows his upper bound in the linear case is close to sharp, by just taking this permutation group and linearizing. So if the lower bound is not sharp it must come from a construction other than wreath products and if the upper bound is not sharp it must come from the steps in Glasby's paper. | |
| Apr 30 at 17:14 | comment | added | Will Sawin | Glasby's proof uses two ingredients: This upper bound in the $p$-group case and a bound of Newman that a solvable subgroup of $GL_n(\mathbb F_p)$ has derived length $\leq 5\log_9 (n/8)+14$. It would be great to understand (a) how sharp this bound is and (b) how sharp the argument that combines those two is. The best lower bound I know for the $GL_n$ problem is still the upper-triangular case (or $2\times 2$ block upper triangular for $p=2,3$) which is close to $\log_2 n$, for a gap of $5 \log(2)/2\log(3)=1.577\dots$. | |
| Apr 30 at 16:57 | comment | added | Will Sawin | Your original lower bound with the $3 \times 3$ matrices gives a $p$-group example. For $p$-groups, a better upper bound was known before Glasby's work, due to P. Hall, A contribution to the theory of groups of prime-power order, Proc. London Math. Soc. 36 (1933), 29-95. This upper bound is very close to your lower bound, (also in the natural generalization of matrices over $\mathbb Z/p^{\ell+1}\mathbb Z$) which seems not to have been noticed - Glasby only gives the lower bound from upper triangular matrices. | |
| Apr 30 at 16:37 | comment | added | David E Speyer | Another way to think of Dave Benson's example, which I am finding useful, is that he is considering the group of affine linear transformations of $\mathbb{F}_3^2$, as a subgroup of $S_9$. The derived series can be described as $\text{GL}_2(\mathbb{F}_3) \ltimes \mathbb{F}_3^2$, $\text{SL}_2(\mathbb{F}_3) \ltimes \mathbb{F}_3^2$, $Q_8 \ltimes \mathbb{F}_3^2$, $\{ \pm 1 \} \ltimes \mathbb{F}_3^2$, $\mathbb{F}_3^2$, $\{ 1 \}$. | |
| Apr 30 at 14:24 | comment | added | Dave Benson | The subgroup $3^2:2S_4$ of $S_9$ has derived length $5$, achieving $\alpha=5/\log_2(9)\approx $1.577$, which is slightly better. | |
| Apr 30 at 14:08 | history | edited | David E Speyer | CC BY-SA 4.0 |
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| Apr 30 at 13:59 | history | edited | David E Speyer | CC BY-SA 4.0 |
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| Apr 30 at 13:54 | comment | added | David E Speyer | I'm not sure about this, but it looks like Glasby's computation generalizes to "if $H$ is a solvable subgroup of $S_k$ with derived length $b$, then we can achieve $\alpha = b/\log_2(k)$. Can anyone beat $S_4$ inside itself? | |
| Apr 30 at 13:50 | history | edited | David E Speyer | CC BY-SA 4.0 |
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| Apr 30 at 13:24 | history | answered | David E Speyer | CC BY-SA 4.0 |