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Chain Markov
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Is $E(G)$ recursively presented for finitely presented $G$?

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 that mean $E(G)$ 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.

Chain Markov
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