Finitely generated group that is not finitely presented

Question

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.

2. 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.)

3. 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.

Thank you in advance!

Answer 1

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.