Second derived subgroup of Baumslag Solitar group BS(2,3)

Question

I am thinking about the group $G=BS(2,3)=\langle a,b\mid b^{-1}a^2b=a^3\rangle$.

Is it true that $G''\cap\langle a\rangle=1$? By $G''$ I am referring to the second derived subgroup of $G$.

If it is true, is there an elementary proof?

Answer 1

Yes it's true and there's an elementary proof, which works more generally for an arbitrary Baumslag-Solitar group $$\mathrm{BS}(m,n)=\langle t,x\mid tx^mt^{-1}=x^n\rangle;\quad m,n\in\mathbf{Z}\smallsetminus\{0\}.$$ Indeed, let $H$ be the group of upper triangular $2\times 2$ matrices over $\mathbf{Q}$. In $H$ define $$\tau=\tau_{m,n}=\begin{pmatrix} n/m & 0 \\ 0 & 1\end{pmatrix},\quad \xi=\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}.$$ Then $\tau\xi^m\tau^{-1}=\xi^n$. Hence there is a unique homomorphism $\eta:\mathrm{BS}(m,n)\to H$ mapping $(t,x)\mapsto (\tau,\xi)$. In particular, this homomorphism maps $x^k$ ($k\in\mathbf{Z}$) to $\xi^k$, which is zero if and only $k=0$. Therefore $x$ has infinite order and $\langle x\rangle\cap\mathrm{Ker}(\eta)=\{1\}$. Since $H$ is metabelian, we have $G''\subset\mathrm{Ker}(\eta)$. Hence $\langle x\rangle\cap G''=\{1\}$.