Timeline for Infinitely presented group where every finite sub-presentation is virtually free
Current License: CC BY-SA 4.0
21 events
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| Mar 25, 2021 at 12:19 | comment | added | YCor | @HJRW Indeed most existing uses of "finitely presented" are meant to be "finitely presentable". | |
| Mar 25, 2021 at 12:15 | comment | added | HJRW | @YCor: when talking about presentations, we sometimes distinguish between "finitely presented" and "finitely presentable": a group is finitely presentable if it has a finite presentation, while it is finitely presented if the presentation we are given is finite. Unfortunately, we don't distinguish between "finitely generated" and "finitely generable"! | |
| Feb 21, 2019 at 12:53 | comment | added | YCor | @BenjaminSteinberg Thanks. This shows that for every nonempty finite subset $I\subset\mathbf{N}_{>0}$ there exists $k\notin I$ such that $G_{I\cup\{k\}}$ is not virtually free (and even that for each such $I$ and every $k_0$, there exists $k\ge k_0$ such that $G_{I\cup\{k\}}$ is not virtually free). | |
| Feb 21, 2019 at 12:11 | comment | added | Benjamin Steinberg | @YCor, A RA Coxeter group is virtually free iff the graph is chordal. This is not the case for a 4-gon or higher. | |
| Feb 21, 2019 at 11:50 | comment | added | YCor | About your last additional question: it might be that $G_{\{1,k\}}$ is not virtually free for any $k\ge 3$. This would be so if it's known that the RA Coxeter group whose commuting graph on generators is a $(k+1)$-gon is not virtually free, and I tend to believe that it's not. But I don't know enough about RA Coxeter groups to answer this right away. | |
| Feb 21, 2019 at 11:41 | history | edited | DavidHume | CC BY-SA 4.0 |
More specific question asked
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| Feb 21, 2019 at 11:41 | comment | added | Florian Lehner | @DavidHume I think if you keep the relators $R_i$ and $R_j$, but not $R_{i+j}$ and $R_{i-j}$ (where $R_k = [a,t^kat^{-k}]$), then the subgroup generated by $\{a_0,a_i,a_j,a_{i+j}\}$ is $(\mathbb Z_2*\mathbb Z_2) \times(\mathbb Z_2*\mathbb Z_2)$. This in particular includes $G_{\{1,k\}}$ when $k \neq 2$. | |
| Feb 21, 2019 at 11:23 | comment | added | YCor | @IanAgol sure, but this (removing $a^2$) is too obvious. Actually I think that the question should be amended by restricting $I$ to range over all finite subsets containing some given finite subset $I_0$ of the set of relators. | |
| Feb 21, 2019 at 9:03 | comment | added | DavidHume | Thanks for all the comments. Can the statement about $G_{1,3}$ not being virtually free be extended to show that $G_{\{1,k\}}$ is virtually free only if $k=1,2$? Or more generally that the only free finite sub-presentations are the ones coming from segments? | |
| Feb 21, 2019 at 8:55 | history | rollback | DavidHume |
Rollback to Revision 1
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| Feb 21, 2019 at 8:52 | history | edited | DavidHume | CC BY-SA 4.0 |
Question clarified to avoid good counterexamples provided in the comments.
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| Feb 21, 2019 at 6:14 | answer | added | Ian Agol | timeline score: 3 | |
| Feb 21, 2019 at 4:49 | comment | added | Ian Agol | @YCor: I suppose even more easily, when $a^2$ is not a relator, then the kernel of the map to $\mathbb{Z}$ will be a right-angled Artin group, hence not free when any other relator is present. | |
| Feb 21, 2019 at 0:09 | comment | added | YCor | @IanAgol yes the isomorphism with the RA Coxeter is true (I checked it in a 2006 Geom. Dedicata paper). So you're right, $\Gamma_{\{1,3\}}$ is not virtually free. Actually, a similar argument implies that $\Gamma_{\{2,3\}}$ is not virtually free, namely considering the subgroup generated by $\{a_0,a_2,a_3,a_5\}$. | |
| Feb 20, 2019 at 23:58 | comment | added | Ian Agol | @YCor: I think the group $G_{\{1,3\}}$ is not virtually free. The kernel of the map to $\mathbb{Z}$ given by $a\mapsto 0, t\mapsto 1$ is a right-angled Coxeter group with vertices labelled by $a_k, k\in\mathbb{Z}$ ($a_k=t^kat^{-k}$ in $G_I$). Then I think $\langle a_0, a_1, a_2, a_3\rangle \cong (\mathbb{Z}_2 \ast \mathbb{Z}_2) \times (\mathbb{Z}_2\ast \mathbb{Z}_2)$, so is not virtually free. | |
| Feb 20, 2019 at 23:31 | comment | added | YCor | At this moment I'm unable to determine whether $\Gamma_{\{2,3\}}$ is virtually free. [[In my previous comment, ignore "is a segment": I mean that $\Gamma_{\{1,\dots,n\}}$ is virtually free for every $n$, as a HNN-extension of a finite group, namely of $C_2^{n+1}$ over the isomorphism $C_2^n\times\{0\}\to \{0\}\times C_2^n$ given by $(x_1,\dots,x_n,0)\mapsto (0,x_1,\dots,x_n)$.]] | |
| Feb 20, 2019 at 23:29 | comment | added | YCor | @DerekHolt I have to confess that I already had to consider infinite generating subsets of finitely generated groups (including for which the diameter of the group remains infinite) | |
| Feb 20, 2019 at 19:24 | comment | added | YCor | The truncated presentations $\Gamma_J=\langle a,t\mid a^2,[a,t^kat^{-k}]:k\in J\rangle$ are HNN extensions of finite groups, hence virtually free, when $J$ is a segment (i.e. equal to $\{1,\dots,n\}$ -- we can ignore those relators for $k\le 0$ since they are redundant). | |
| Feb 20, 2019 at 19:21 | comment | added | Derek Holt | @YCor I have to confess that I have been guilty of writing "Let $G = \langle X \rangle$ be a finitely generated group'' without explicitly saying that $X$ is assumed to be finite. | |
| Feb 20, 2019 at 19:18 | comment | added | YCor | If you assume $S$ to be finite, you should say it. | |
| Feb 20, 2019 at 17:20 | history | asked | DavidHume | CC BY-SA 4.0 |