Timeline for Is $E(G)$ recursively presented for finitely presented $G$?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| S Jan 16, 2022 at 18:02 | history | bounty ended | CommunityBot | ||
| S Jan 16, 2022 at 18:02 | history | notice removed | CommunityBot | ||
| Jan 10, 2022 at 19:40 | comment | added | Matt Zaremsky | Since I answered that other question, I figured I should just pop in here to say, I have no idea! | |
| Jan 8, 2022 at 17:17 | comment | added | მამუკა ჯიბლაძე | No idea if this can be used in any way but $E(G)$ can be realized as a free algebra in a variety of universal algebras. Extend the variety $\mathbf{Gr}$ of groups to the variety $G{\downarrow}\mathbf{Gr}$ by adding constants from $G$ in a standard way. That is, add constants $c_g$, one for each $g\in G$, and identities $c_xc_y=c_{xy}$ for all $x,y\in G$. Then, $G$ itself is an algebra in this variety, with $c_g=g$. Next, let $V_G$ be the subvariety of $G{\downarrow}\mathbf{Gr}$ generated by $G$. Then $E(G)$ is the free algebra on a single generator in $V_G$. | |
| S Jan 8, 2022 at 16:47 | history | bounty started | Chain Markov | ||
| S Jan 8, 2022 at 16:47 | history | notice added | Chain Markov | Draw attention | |
| Dec 21, 2021 at 18:25 | comment | added | Benjamin Steinberg | Notice of you use generators the generators of G for the c_g and x for the identity map, then you can present E(G) by the relations for G plus all words w involving x for which the universally quantified statement for all x\in G w=1 is true. So if the universal theory of G is decidable then E(G) is recursively presented. There are finitely generated nilpotent groups with undecidable universal theory but I don't know if you can use such simple sentences. | |
| Dec 21, 2021 at 18:07 | history | edited | Chain Markov | CC BY-SA 4.0 |
added 6 characters in body
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| Dec 21, 2021 at 12:49 | history | asked | Chain Markov | CC BY-SA 4.0 |