# Finitely generated group that is not finitely presented

The following is Exercise 2.E.27 in Clara Löh's introduction to Geometric Group Theory:

We consider the group $$G := \left\langle s,t \mid \left[t^n s t^{-n} , t^m s t^{-m} \right], \quad m,n \in \mathbb Z \right\rangle.$$ The goal of this exercise is to prove that $$G$$ is not finitely presentable.

The exercise consists of three steps. I get stuck in part 3, but for completeness I include my full solution.

1. Show that $$\displaystyle{G \cong \left\langle s,t \mid \left[ s , t^n s t^{-n} \right], \quad n \in \mathbb N_{>0} \right\rangle}$$.

This I found relatively easy. If the set of relations $$\left[ s , t^n s t^{-n} \right] = e$$ for $$n \in \mathbb N_{>0}$$, then the bigger set of relations $$\left[t^n s t^{-n} , t^m s t^{-m} \right]$$ for $$m,n \in \mathbb Z$$ also hold. Indeed, using that $$\left[ s , t^k s t^{-k} \right] = e$$ for all $$k \in \mathbb N_{>0}$$ and given $$m,n \in \mathbb Z$$, we may assume WLOG that $$m > n$$ and then

\begin{align} \left[t^n s t^{-n} , t^m s t^{-m} \right] &= t^n s \left( t^{m-n} s t^{n-m} \right) s^{-1} t^{m-n} s^{-1} t^{-m} \\ &= t^n \left( t^{m-n} s t^{n-m} \right) s s^{-1} t^{m-n} s^{-1} t^{-m} \\ &= t^{m} s t^{n-m} t^{m-n} s^{-1} t^{-m} \\ &= t^{m} s s^{-1} t^{-m} \\ &= t^{m} t^{-m} \\ &= e. \end{align}

Hence $$G \cong G'$$ and so it suffices to show $$G'$$ is not finitely presentable.

1. For $$N \in \mathbb N_{>0}$$, let $$G_N := \langle s, t \mid \{[s, t^n s t^{−n}] \mid n \in \{1, . . . , N \}\}\rangle.$$ Show that the homomorphism $$π_N\colon G_N \longrightarrow G_{N +1}$$ given by the identity on $$\{s, t\}$$ is surjective but not injective.

Hints. Use the universal property of generators and relations and try to map $$s$$ to the transposition $$(1 \quad 2) \in S_{2\cdot N +3}$$ and $$t$$ to the permutation $$(1 \mapsto 3, 2 \mapsto 4, 3 \mapsto 5, \cdots ) \in S_{2\cdot N +3}$$.

Surjectivity of the natural morphisms $$\pi_N$$ is again relatively easy to see (the image contains the two generators). If $$\pi_N$$ was also injective for some $$N \in \mathbb N_{>0}$$, then $$\pi_N\colon G_{N} \longrightarrow G_{N+1}$$ is an isomorphism of groups and so we have $$[s,t^{N+1} s t^{-N-1}] = e$$ in $$G_N$$. I think we can use the hint to obtain a contradiction:

Define $$f\colon \{ s,t \} \longrightarrow S_{2\cdot N +3}$$ by mapping the generators $$s$$ and $$t$$ as in the hint. Then $$f(s)$$ and $$f(t)$$ satisfy the relations of $$G_N$$, so you can extend $$f$$ to a morphism $$\varphi\colon \{s,t \} \longrightarrow S_{2\cdot N +3}$$. But $$\varphi([s,t^{N+1} s t^{-N-1}]) \neq 1$$, which contradicts the assumption $$[s,t^{N+1} s t^{-N-1}] = e$$ in $$G_N$$. (I did not verify this part in detail but I believe this is the main idea behind the solution.)

1. Use the second part to conclude that $$G$$ is not finitely presentable.

We have found a chain of quotient homomorphisms $$G_1 \longrightarrow G_2 \longrightarrow G_3 \longrightarrow \cdots$$ According to the answer https://math.stackexchange.com/a/547144/909787, $$G'$$ is finitely presented if and only if this sequence eventually stabilizes. I understand that we've proved the latter statement in part 2, but why is $$G'$$ finitely presented (if and) only if the sequence eventually stabilizes?

I get the if-implication: if the sequence eventually stabilizes, then there's some $$N \in \mathbb N_{>0}$$ for which the relations $$[s,t^n s t^{-n}] = e$$ hold in $$G_N$$ for all $$n > N$$, hence $$G_N \cong G'$$.

But why couldn't the sequence not eventually stabilize, while $$G'$$ still be finitely presentable, for instance using some other generators and relations? If the sequence never stabilizes, I get that we can't restrict the set of relations on $$G'$$ to some finite subset, as otherwise we could find some $$N \in \mathbb N_{>0}$$ for which the natural surjective homomorphism $$G_N \longrightarrow G$$ has an inverse. But I don't really understand why never stabilizing is sufficient for $$G$$ to not be finitely presentable at all.

We prove first that, if a general group $$G$$ has two finite generating sets $$X$$ and $$Y$$ and there is a finite presentation on $$Y$$, then there is a finite presentation on $$X$$. This is a standard result in the use of Tietze transformations. Write $$X(Y)$$ and $$Y(X)$$ for sets of words over $$Y$$ for the generators in $$X$$, and vice versa. Then if $$G = \langle\, Y \mid S\, \rangle$$ with $$S$$ a finite set of relators, we have $$G = \langle\, Y \mid S\, \rangle \cong \langle\, Y \cup X \mid S(Y) \cup (X = X(Y))\, \rangle \cong$$ $$\langle\, Y \cup X \mid S(Y) \cup (X = X(Y)) \cup (Y = Y(X))\, \rangle \cong \langle\, X \mid S(Y(X)) \cup (X = X(Y(X)))\, \rangle,$$ which is a finite presentation on the generating set $$X$$.
So if your group $$G$$ is finitely presentable, then there is a finite presentation $$G = \langle s,t \mid T \rangle$$ on the generators $$s$$ and $$t$$. We also know that $$G = \langle s,t \mid R \rangle$$, where $$R$$ is the infinite set $$\{[ts,t^nst^{-n}] : n \ge 0 \}$$. So every element of $$T$$ is a (finite) product of conjugates of elements of $$R$$ in the free group on $$s,t$$. Since $$T$$ is finite, only finitely many of the elements of $$R$$ are involved in all of these products, so $$G = \langle s,t \mid R' \rangle$$, where $$R'$$ is this finite set of elements of $$R$$. This contradicts what you proved in Part 2, so $$G$$ is not finitely presentable.