Timeline for Groups whose derived length is logarithmic in the order?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 1 at 0:50 | comment | added | David E Speyer | Ah, Corollary 1 in "The soluble length of soluble linear groups", Newman, 1972, cited by Will Sawin: If $A \subset S_m$ and $T \subset S_k$ are solvable transitive permutations groups, then $(A \wr T) \subset S_{mk}$ is a solvable transitive permutation group and $d(A \wr T) = d(A) + d(T)$. Used inductively, $d(( \cdots ((H \wr H) \wr H) \cdots \wr H) \wr H) = r d(H)$, where there are $r$ $H$'s in the repeated wreath product. | |
| Apr 30 at 14:56 | comment | added | David E Speyer | Now that we have a possible value beating $1.5$, we should actually fill in the details in computing the derived length of $(\cdots ((H \wr H) \wr H) \cdots \wr H) \wr H$. It looks to me like every group in the derived sequence looks something like $L \rtimes R$ where $L$ is the kernel of a map from $(\cdots ((H \wr H) \wr H) \cdots \wr H)$ to an abelian group and $R$ is in the derived sequence of $H$, with the result that each wreath product contributes a factor of $d(H)$ to the derived length of the wreath product. Can you (or someone else) verify this? | |
| Apr 30 at 14:47 | history | edited | Dave Benson | CC BY-SA 4.0 |
added 236 characters in body
|
| Apr 30 at 11:24 | history | edited | Dave Benson | CC BY-SA 4.0 |
Changed $m>1$ to $m\geqslant 3$ to avoid dividing by zero.
|
| Apr 30 at 7:51 | history | edited | Dave Benson | CC BY-SA 4.0 |
added 668 characters in body
|
| Apr 30 at 1:06 | comment | added | Will Sawin | The $n\times n$ upper-triangular matrices have derived length $(1/2) \log_2 \log_2 |G| +O(1)$ since $|G|= p^{n^2}$ so $\log_2 |G|= n^2 \log p$ and the derived length is $\log_2 n$. So the bound is sharp to within a factor of $6$. | |
| Apr 29 at 21:49 | comment | added | Dave Benson | I should add that this bound is not particularly sharp, but seems about the right order of magnitude. It would be interesting to know what the actual asymptotics are, in the sense of limb soup. | |
| Apr 29 at 20:35 | comment | added | LSpice | @YCor, re, I take the intended question to be one with what I would call $\Theta$ (or just $\asymp$) in place of $O$, per what Wikipedia says is the use in computer science. | |
| Apr 29 at 20:32 | comment | added | YCor | Could you be explicit on what you call "the intended question"? I can't guess it. So far, abelian groups do the job. | |
| Apr 29 at 20:03 | history | edited | LSpice | CC BY-SA 4.0 |
Link to paper
|
| Apr 29 at 19:55 | comment | added | Dave Benson | I've tried to answer the intended question, rather than take big-O literally. | |
| Apr 29 at 19:44 | history | edited | Dave Benson | CC BY-SA 4.0 |
deleted 28 characters in body
|
| Apr 29 at 19:18 | history | edited | Dave Benson | CC BY-SA 4.0 |
added 19 characters in body
|
| Apr 29 at 19:09 | history | answered | Dave Benson | CC BY-SA 4.0 |