Baumslag-Solitar group ($\mathrm{BS}(m,n)$)

The Baumslag-Solitar group is

Represent $y$ by translation $t\mapsto t+1$ and $x$ by $t\mapsto t/3$; the relator holds and $y$ has infinite order. Base-three digits $d_j\in \{0,1,2\}$ give words recursively by $W_0=1$, $W_{j+1}=x^{-1}{W}_{j}x{y}^{d_j}$. These have length at most $4k$ and represent all $y^m$, $0\le m<3^k$. Thus the growth function of a finitely generated group obeys $\beta (4k)\ge 3^k$. Counting freely reduced words in four letters gives $\beta (n)\le 2\cdot 3^n-1$. The growth is therefore exponential up to coarse equivalence.

For $G=⟨a,t\mid t{a}^2{t}^{-1}=a^3⟩$, the surjective endomorphism $\theta (a)=a^2$, $\theta (t)=t$ has nontrivial kernel by Britton's lemma. The commutator subgroup is generated by $a_i=t^{i}a{t}^{-i}$ for $i\in \mathbb{Z}$. For $j>i$, its defining relations give $a_j^{2^{j-i}}=a_i^{3^{j-i}}$. Hence ${\theta }^k([a_i,a_j])=1$ whenever $k\ge j-i$. The union of the increasing kernels is normal, and the images of all $a_i$ commute in its quotient, proving the first inclusion. The second follows from finite residual contains surjective endomorphism kernels applied to every iterate.

In the finite group presentation $⟨a,t\mid ta{t}^{-1}=a^2⟩$, let $a$ act by $x\mapsto x+1$ and $t$ by $x\mapsto 2x$. The conjugates $a_i=t^{-i}a{t}^i$ satisfy $a_{i+1}^2=a_i$, so their increasing cyclic subgroups give the additive group $\mathbb{Z}[1/2]$. Every element has the form $a_i^m{t}^k$, and the affine action proves that this form can be the identity only if $k=m=0$. Thus the displayed semidirect product identification holds. This is a finitely presented group with a series having torsion-free abelian factors, but it is not a poly-infinite cyclic group, since $\mathbb{Z}[1/2]$ is not finitely generated.