7
$\begingroup$

Suppose $G$ is a group. Consider the set $G^G$ of all functions $G \to G$, which forms a group under elementwise multiplication. Now, for all $g \in G$ let’s define $c_g \in G^G$ as the constant function $c_g(x) \equiv g$, and $id \in G^G$, as the identity map $id(x) = x$. Now, consider the subgroup $E(G) = \langle \{c_g | g \in G\} \cup \{id\} \rangle$.

$E(G)$ preserves several finiteness conditions, like finiteness, finite generatedness and residual finiteness. However, it does not preserve finite presentability: a counterexample $E(F_2 \times F_2)$ was constructed by @MattZaremsky

However, that counterexample inspired me for a new question about a weaker condition:

If $G$ is finitely presented, does this imply that $E(G)$ is recursively presented?

The counterexample to the previous question is recursively presented by Higman embedding theorem, as it can be embedded in a finitely presented group $F_3 \times F_3$. However, maybe there is some other counterexample, that isn’t.

Improve this question
$\endgroup$
3
  • 8
    $\begingroup$ 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. $\endgroup$
    – Benjamin Steinberg
    Commented Dec 21, 2021 at 18:25
  • $\begingroup$ 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 $V_G$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jan 8, 2022 at 17:17
  • $\begingroup$ Since I answered that other question, I figured I should just pop in here to say, I have no idea! $\endgroup$
    – Matt Zaremsky
    Commented Jan 10, 2022 at 19:40

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .