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” properties of $G$:
If $G$ is finite then $E(G)$ is also finite.
Proof: $|E(G)| \leq |G^G| = |G|^{|G|}$
If $G$ is finitely generated then $E(G)$ is also finitely generated.
Proof: If $A$ is a generating set of $G$, then $\{c_g | g \in A\} \cup \{id\}$ is a generating set of $E(G)$.
If $G$ is finitely approximated then $E(G)$ is also finitely approximated.
Proof: Consider the following class of maps $\pi_g: E(G) \to G, f \mapsto f(g)$ for all $g \in G$. All $\pi_g$ are homomorphisms and each non-trivial element of $E(G)$, maps to a non-zero element of $G$ under some of $\pi_g$. The rest follows from finite approximated ness of $G$.
However, there is also a fourth “finiteness” property I am interested in but do not know how to deal with:
If $G$ is finitely presented, does that mean that $E(G)$ is also finitely presented?
I suspect, it should be, but have no idea how to prove it.