Timeline for When is a finitely generated group finitely presented?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 19, 2022 at 0:49 | comment | added | biringer | Ok, more commentary, just for posterity: Use the presentation $L=<e,t | e^2, [t^net^{-n},e]>$. To show no finite subset of relators suffices, you need to say that it's impossible to derive the relation $[t^net^{-n},e]=1$ from the relation $e^2=1$ and the relations $[t^iet^{-i},e]=1$, where $|i|\leq n$. The easiest way to do this is to map $e,t$ to the permutations $(12)$ and $(12\cdots (2n+1))^2$ in the symmetric group $S_{2n+1}$. In that group, the latter relations hold, but $[t^net^{-n},e]=1$ doesn't. | |
| Oct 19, 2022 at 0:39 | comment | added | biringer | IIRC, technically you can only say that if a group $G$ is f.p. and $G=<X|R>$ where $X$ is finite, then there's a finite subset R' of $R$ where $G=<X|R'>$. There's a formal proof of this in Kapovich-Drutu, for instance. I can't remember a counterexample when $X$ is infinite at the moment, but I think they exist. Anyway, in the lamplighter example, you should then start with an infinite presentation with finitely many generators if you want to show it's not f.p. (You only need $a_0,t$ to generate.) | |
| Sep 23, 2021 at 1:42 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
fixed arxiv front-end link, added doi link for paper
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| Jun 3, 2019 at 10:26 | history | made wiki | Post Made Community Wiki by Todd Trimble | ||
| Jun 2, 2019 at 17:50 | history | edited | user6976 | CC BY-SA 4.0 |
Added a result of BierI and Strebel
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| Feb 10, 2011 at 14:27 | comment | added | user6976 | ... of course, you need to use the fact that if $G$ has finite presentation with one generating set, then for any other finite generating set, it also has finite presentation (just rewrite all relations in the new generating set). | |
| Feb 10, 2011 at 13:36 | comment | added | user6976 | If a finitely generated group $G$ is finitely presented with finite set of relators $Q$, then every infinite presentation $R$ has a finite subset that is also a presentation. Indeed, consider any proof of $Q$ using $R$. It involves only finite subset $R'$ of $R$. The relations from $R'$ define the group $G$. Indeed, let $G'$ be the group defined by $R'$. Then the identity map on the generating set induces a hom. $\phi: G'\to G$. All relations of $Q$ hold in $G'$, so the kernel of $\phi$ is trivial, and $G'=G$. | |
| Feb 10, 2011 at 13:26 | comment | added | aaron | Can this type of argument be used to show that the group has no finite presentation? (Or does it only show what it seems to show, namely that the group has no finite sub-presentation of a given infinite presentation.) | |
| Feb 10, 2011 at 3:39 | history | edited | user6976 | CC BY-SA 2.5 |
added 94 characters in body
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| Feb 10, 2011 at 2:36 | history | answered | user6976 | CC BY-SA 2.5 |