Timeline for Abelian-by-cyclic subgroups of exponential growth solvable groups
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 6, 2019 at 9:40 | comment | added | YCor | $Z[1/k]$ is not min but $Z[1/k]/Z$ is min. | |
| Dec 6, 2019 at 5:48 | comment | added | ARG | obviously, thanks. So $G = BS(1,k) \simeq \mathbb{Z}[\tfrac{1}{k}] \rtimes \mathbb{Z}$ (where the action of $\mathbb{Z}$ is by multiplication by $k$) is minimax because $\mathbb{Z}$ is Max and $\mathbb{Z}[\tfrac{1}{k}]$ is Min? $\mathbb{Z}[\tfrac{1}{k}]$ cannot be Max (not finitely generated), but I don't see why it's Min (the sequence $\langle k \rangle \geq \langle k^2 \rangle \geq \langle k^3 \rangle \ldots$ does not stabilise). So one needs to split $\mathbb{Z}[\tfrac{1}{k}]$ agin? $G$ cannot be polycyclic since $G^{(1)} / G^{(2)} = \mathbb{Z}[\tfrac{1}{k}]$ is not finitely generated. | |
| Dec 6, 2019 at 1:46 | comment | added | YCor | No, $\langle a\rangle$ is not a normal subgroup in $BS(1,k)=\langle a,b\mid b^{-1}ab=a^k\rangle$ for $k\ge 2$. Indeed $bab^{-1}\notin\langle a\rangle$. | |
| Dec 5, 2019 at 19:00 | comment | added | ARG | About $BS(1,k) = \langle a,b, \mid b^{-1}ab = a^k \rangle$ not being polycyclic: I'm confused. The definition of polycyclic I read is that there is a subnormal serie (of finite length) with cyclic quotients. Here $1 \lhd \langle a \rangle \lhd G$ is such a series. So I'm not using the proper definition of polycyclic? (see here or here ) or I made another silly mistake... or perhaps these definitions are only suited for finite groups? | |
| Dec 4, 2019 at 17:43 | comment | added | YCor | Yes sure the minimax (fg solvable) group are much easier to understand than general f.g. solvable ones; in particular they're virtually nilpotent-by-abelian. | |
| Dec 4, 2019 at 16:16 | comment | added | ARG | so if I replace "subgroup" by "subquotient" in the statement, then it is almost a corollary of Kropholler's result. (I still don't have a good grasp on minimax groups, but I guess it should be simpler to find a polycyclic or abelian-by-cyclic group in their minimax serie) | |
| Dec 3, 2019 at 22:12 | comment | added | YCor | An example is the Baumslag-Solitar group $BS(1,k)$ for $k\ge 2$. | |
| Dec 3, 2019 at 19:02 | comment | added | ARG | apologies for this [well-known] question: what's an easy example of a solvable minimax groups which is not polycyclic? | |
| Dec 3, 2019 at 18:45 | history | edited | ARG | CC BY-SA 4.0 |
added 50 characters in body
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| Dec 3, 2019 at 18:42 | comment | added | ARG | how interesting! is there a relatively recent book on the theory of (infinite) soluble groups? the most recent I could put my hands on was Robinson's "Finiteness Conditions and Generalized Soluble Groups". | |
| Dec 3, 2019 at 16:56 | comment | added | YCor | By a 1984 result of Kropholler, every f.g. solvable group either has a lamplighter $C_p\wr\mathbf{Z}$ subquotient for some prime $p$, or is minimax. Hence, this reduces to proving the result assuming either $G$ minimax (which is quite close to polycyclic in spirit) or $G$ has a such a lamplighter as quotient (or if there is a counterexample, it could be found with either of these properties). | |
| Dec 3, 2019 at 16:13 | history | asked | ARG | CC BY-SA 4.0 |