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 $G$:



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> If $G$ is finite then $E(G)$ is also finite.
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*Proof*: $|E(G)| \leq |G^G| = |G|^{|G|}$



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> If $G$ is finitely generated then $E(G)$ is also finitely generated.
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*Proof*: If $A$ is a generating set of $G$, then $\{c\_g | g \in A\} \cup \{id\}$ is a generating set of $E(G)$.



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> If $G$ is residually finite then $E(G)$ is also residually finite.
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*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 residual finiteness of $G$.


However, there is also a fourth “finiteness” property I am interested in but do not know how to deal with:



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> If $G$ is finitely presented, does that mean that $E(G)$ is also finitely presented?
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I suspect, it should be, but have no idea how to prove it.


[This question on MSE](https://math.stackexchange.com/q/4330034/407165)



Does this group construction preserve finite presentability?
It seems the answer is no, $E(G)$ can fail to be finitely presented even if $G$ is finitely presented.


I claim that a counterexample is given by $G=F\_2\times F\_2$.


First, as @SeanEberhard explains in the comments, $E(F\_2\times F\_2)$ is isomorphic to the subgroup $H$ of $F\_3\times F\_3$ generated by $\{(x,1),(y,1),(1,x),(1,y),(z,z)\}$, where $\{x,y,z\}$ is a generating set for $F\_3$. Now I claim that $H$ is not finitely presented.


First I claim that $H$ contains the commutator subgroup of $F\_3\times F\_3$. Any simple commutator in $F\_3\times F\_3$ is an ordered pair of simple commutators in $F\_3$, so it suffices to show that any $(g,1)$ for $g$ a simple commutator in $F\_3$ lies in $H$ (and by a parallel argument also $(1,g)$). This is easy to see because given any word in $(x,1),(y,1),(z,1)$ (and their inverses) representing an element of $[F\_3,F\_3]\times\{1\}$, we can replace each instance of $(z,1)$ with $(z,z)$ (and $(z^{-1},1)$ with $(z^{-1},z^{-1})$) and get a word representing the same element of $[F\_3,F\_3]\times\{1\}$ but now lying in $H$.


Now that we know $H$ contains the commutator subgroup, we can use the Bieri-Neumann-Strebel-Renz invariant $\Sigma^2(F\_3\times F\_3)$ to conclude that $H$ is not finitely presented. Indeed, $H$ lies in the kernel of the homomorphism $\chi\colon F\_3\times F\_3 \to \mathbb{Z}$ sending $(z,1)$ to $1$, $(1,z)$ to $-1$, and all of $(x,1),(y,1),(1,x),(1,y)$ to $0$ (I guess it actually equals this kernel), and the computation of $\Sigma^2(F\_3\times F\_3)$ (see, e.g., [Bux-Gonzalez](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/S0024610799007991)) reveals that $[\chi]$ is not in $\Sigma^2(F\_3\times F\_3)$, since the "dead" edge from $(x,1)$ to $(1,x)$ in the defining flag complex has empty "living link". (Actually, I think $\Sigma^2(F\_3\times F\_3)$ is empty, so probably we didn't even need to isolate a particular $\chi$, and in fact every infinite index subgroup of $F\_3\times F\_3$ containing the commutator subgroup fails to be finitely presented, but I'll just leave this analysis as-is.)




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Edit: By request and for the sake of being self-contained, here is the proof that $E(F\_2\times F\_2)\cong H$, compiled from @SeanEberhard's comments. Here $\{x,y,z\}$ is a free basis of $F\_3$ and $H$ is the subgroup of $F\_3\times F\_3$ generated by $\{(x,1),(y,1),(1,x),(1,y),(z,z)\}$.


Let $\{a,b\}$ be a free basis of $F\_2$. We have an epimorphism $\phi\colon F\_3\to E(F\_2)$ given by sending $x$ to $c\_a$, $y$ to $c\_b$, and $z$ to $id\_{F\_2}$, and we claim it is injective (hence an isomorphism). Indeed, given any non-empty reduced word in $c\_a$, $c\_b$, and $id\_{F\_2}$ (and their inverses) representing an element $f$ of $E(F\_2)$, since $F\_2$ is mixed identity-free, we can evaluate $f$ at some element of $F\_2$ to get a non-trivial element of $F\_2$, which means $f$ is a non-trivial element of $E(F\_2)$. Hence $F\_3 \cong E(F\_2)$. This also shows $F\_3\times F\_3 \cong E(F\_2)\times E(F\_2)$. Let us identify $H$ with its isomorphic image in $E(F\_2)\times E(F\_2)$, generated by $\{(c\_a,c\_1),(c\_b,c\_1),(c\_1,c\_a),(c\_1,c\_b),(id\_{F\_2},id\_{F\_2})\}$.


Now consider the monomorphism $\psi \colon F\_2^{F\_2} \times F\_2^{F\_2} \to (F\_2\times F\_2)^{F\_2\times F\_2}$ given by sending $(f,g)$ to the function $f\times g$ defined by $(f\times g)(w,v):=(f(w),g(v))$. The restriction of $\psi$ to the subgroup $H$ is an isomorphism onto its image, which is generated by $c\_{(a,1)}$, $c\_{(b,1)}$, $c\_{(1,a)}$, $c\_{(1,b)}$, and $id\_{F\_2\times F\_2}$, hence is $E(F\_2\times F\_2)$ as desired.


