Timeline for If a group $G$ has decidable word problem, must it have a decidable square problem?
Current License: CC BY-SA 4.0
6 events
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| Aug 13, 2018 at 18:17 | comment | added | YCor | There's a quantitative version: for G a f.g. group with word length $|\cdot|$, define, for $g\in G$, $u_2(g)=\min(|h|:h^2=g)$ if $g$ is a square and $u_2(g)=0$ otherwise; define $f_2(n)=\sup(u_2(g):|g|\le n)$. Then (for $G$ with solvable word problem) $f_2$ is bounded above by a recursive function iff $G$ has solvable square problem. For $G$ hyperbolic one would expect $f_2(n)=O(n)$, probably using routine arguments. Many other nice groups should have $f_2(n)=O(n)$. | |
| Aug 13, 2018 at 8:10 | answer | added | Alex Gavrilov | timeline score: 3 | |
| Aug 12, 2018 at 7:28 | comment | added | user6976 | The question is interesting for groups of homeomorphisms such as various Thompson groups because the problem of extracting a root is classical in dynamics. It is also interesting for groups of matrices. | |
| Aug 12, 2018 at 6:17 | comment | added | user6976 | An intuition: It should not have anything to do with the Dehn function; the square root problem should be undecidable even in groups with decidable word problem; we do not know enough about automatic groups to find out one way or another; for relatively hyperbolic groups with good enough parabolic subgroups, the problem should be decidable. | |
| Aug 12, 2018 at 5:58 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
typo in the title
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| Aug 12, 2018 at 5:30 | history | asked | Steven Stadnicki | CC BY-SA 4.0 |