Timeline for If G is a finitely generated group with vcd(G) finite, is vcd(H) finite for H, where H is an automorphism group of G?
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| Feb 27, 2023 at 15:36 | comment | added | ADL | @Mike The standard example of a Baumslag-Solitar group with non-finitely generated (outer) automorphism group is $BS(2, 4)$, due to Collins and Levine in the 1980s ("Automorphisms and Hopficity of certain Baumslag-Solitar groups" Arch. Math. (1983)); Levitt gave a geometric reason for this non-finite generation in a 2007 G&T paper (link). Theorem 5.2 of this gives a positive answer to your question in the "nice" case of (Generalised) Baumslag-Solitar groups with no $BS(1, n)$ subgroups, $n>1$. | |
| Feb 27, 2023 at 13:54 | comment | added | Mike | @Agenevois Given the Baumslag-Solitar group is a 1-relator group, IIRC by standard theorems it's going to have a torsion free subgroup of finite index and it should have finite vcd( at most 2). Do you have a reference for the automorphism group or do I need to prove it as an exercise for the reader? ; ) | |
| Feb 27, 2023 at 12:30 | comment | added | HJRW | @ADL: I see it now. Nice! I did look at your paper about HNN extensions of triangle groups when trying to figure out if there was a way to provide examples with base groups of type F. | |
| Feb 27, 2023 at 11:52 | comment | added | ADL | @HJRW Sorry, yes, parameterised by $G$. Then each $\phi_x$ for $x\in N$ is trivial, so this gives a map $Q\to\operatorname{Aut}(H)\to\operatorname{Out}(H)$. These maps can then be shown to be injective here (the map to $\operatorname{Aut}$ is pretty clearly injective, by for example Britton's Lemma, while the map to $\operatorname{Out}$ is also injective as none of these $\phi_x$ are inner because, as $G$ has trivial centre and $N\neq G$, no inner automorphism of $H$ fixes every $g\in G$). | |
| Feb 27, 2023 at 10:27 | comment | added | HJRW | @ADL: I don't quite follow you (perhaps there's a typo?). Surely at the very least $\phi_x$ should be parametrised by the elements of $Q$ (or $G$) rather than $N$? Sorry for being slow on the uptake! | |
| Feb 27, 2023 at 8:35 | comment | added | ADL | @HJRW Finitely presentability can be salvaged from Moishe Kohan's answer, but this likely sacrifices finite dimensionality: Take the HNN-extension $H=\langle G, t\mid t^{-1}xt=x,\:x\in N\rangle$. Then the automorphisms $\phi_x: g\mapsto g, t\mapsto x^{-1}tx$ for $x\in N$ gives an embedding of $Q$ into $\operatorname{Out}(H)$. | |
| Feb 27, 2023 at 7:27 | comment | added | AGenevois | @HJRW: I am curious, what about Baumslag-Solitar groups? Some of them have their automorphism groups not finitely generated. Are they sufficiently big to have an infinite vcd? | |
| Feb 26, 2023 at 21:11 | vote | accept | Mike | ||
| Feb 26, 2023 at 20:42 | comment | added | HJRW | The related question of whether a group $G$ of type F can have a $\mathrm{Out}(G)$ (or $\mathrm{Aut}(G)$) with infinite vcd seems harder. The examples given in Moishe Kohan's answer are finite dimensional, but never finitely presented. It may be possible to use fibre products to get finitely presented examples, but higher finiteness properties seem out of reach to these techniques. Maybe there's some other kind of example known? | |
| Feb 26, 2023 at 15:38 | answer | added | Moishe Kohan | timeline score: 4 | |
| Feb 26, 2023 at 11:39 | comment | added | Matt Zaremsky | For the last question, i.e., are there finitely generated groups with finite vcd but non-finitely generated center, the answer is yes. You can take "Abels's groups" Ab_n, which are the groups of n-by-n upper triangular matrices over Z[1/p] (for any prime p) whose first and last diagonal entries are 1. For n>2, Ab_n is finitely generated, but the center is the copy of Z[1/p] "in the top right" so it's not fin. gen. | |
| Feb 26, 2023 at 3:14 | comment | added | Moishe Kohan | OK, later...... | |
| Feb 26, 2023 at 2:56 | comment | added | Mike | No I don't. How would it be relevant and do you have a reference? | |
| Feb 26, 2023 at 2:18 | comment | added | Moishe Kohan | Do you know the Mikhailova's construction of f.g. subgroups of products of two free groups? | |
| Feb 25, 2023 at 23:28 | comment | added | Mike | I just realized that, to simplify things regarding torsion and finite index subgroups, we may as well assume G is torsion free to start off here. | |
| Feb 25, 2023 at 23:16 | history | edited | YCor | CC BY-SA 4.0 |
added tag, formatting
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| S Feb 25, 2023 at 22:53 | review | First questions | |||
| Feb 26, 2023 at 0:26 | |||||
| S Feb 25, 2023 at 22:53 | history | asked | Mike | CC BY-SA 4.0 |