# Does this group construction preserve finite presentability?

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 residually finite then $$E(G)$$ is also residually finite.

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:

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.

This question on MSE

• Just as sort of a smell test: If $G=\langle x,y,z,w\mid [x,z]=[y,z]=[x,w]=[y,w]=1\rangle$ (so $G\cong F_2\times F_2$), then $E(G)$ satisfies $[[id,c_g],c_z]=1$ for every $g\in G$ (these square brackets are all commutators). Can these infinitely many relations really be boiled down to finitely many?.... Dec 20, 2021 at 0:33
• @MattZaremsky I don't understand. If you take $g = z$ and evaluate $[[\mathrm{id}, c_g], c_z]$ at $w$ you get $[[w,z],z]$, which is nontrivial. Maybe you mean for every $g \in \langle x, y\rangle$? Dec 20, 2021 at 10:56
• @MattZaremsky $\def\i{\mathrm{id}}$Since $c_z$ already commutes with $c_x$ and $c_y$, your relations are equivalent to $c_z$ commuting with $g^\i$ for $g\in\langle c_x,c_y\rangle$ (where $g^h=h^{-1}gh$). Since $-^\i$ is an automorphism, these things belong to $\langle c_x^\i,c_y^\i\rangle$; thus, since $C(c_z)$ is a subgroup, it is enough that $c_z$ commutes with $c_x^\i$ and $c_y^\i$. Dec 20, 2021 at 12:02
• @MattZaremsky Oh right, good point. I should have said that if $F_3 = F\{x,y,z\}$, $E(G)$ isomorphic to the subgroup of $F_3 \times F_3$ generated by $(x,1), (y,1), (1, x), (1,y), (z,z)$, but that's not quite as useful Dec 20, 2021 at 12:11
• @MattZaremsky I'm almost certain. Consider the map $F_3 \to E(F_2)$ defined by sending $x, y$ to the constants $c_x, c_y$ and sending $z$ to $\mathrm{id}$. I'm pretty sure this map is an isomorphism (equivalently, there is no identity $w(x, y, v) = 1$ holding for all $v \in F_2$). Assuming this, we have an isomorphism $F_3 \times F_3 \to E(F_2) \times E(F_2)$. The image of $(z, z)$ is $(\mathrm{id}, \mathrm{id}) = \mathrm{id}$, so the image of $\langle (x, 1), (y, 1), (1, x), (1, y), (z, z)\rangle$ is $E(F_2 \times F_2)$. Dec 20, 2021 at 15:22

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) 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.)

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.

• There’s a theorem of Baumslag and Roseblade that classifies the finitely presented subdirect products of free groups: IIRC they’re all finite index, which would confirm your final parenthetical remark. There’s also some relevant work of Bridson—Howie—Miller—Short, which relates the finiteness properties of subdirect products of free groups to the lower central series.
– HJRW
Dec 20, 2021 at 17:11
• Sorry but for completeness it would be better to include an accurate proof that $E(F_2\times F_2)$ is isomorphic to $H$. I do buy the argument by @SeanEberhard but still it is sort of dispersed through several comments, so... Jan 10, 2022 at 6:02
• @მამუკაჯიბლაძე Good point, I added that. Jan 10, 2022 at 19:38