Normal closure in Baumslag-Solitar group

Question

Let $G:=BS(m,n)= \langle a,t \mid ta^mt^{-1}=a^n\rangle$ be a Baumslag-Solitar group (here $m,n\in \Bbb Z\setminus\{\pm 1\})$.

I was trying to determine the normal closure of the element $a\in G$; here, by normal closure I mean $\langle a\rangle^G:=\langle\{gag^{-1}\mid g\in G\}\rangle$.

At first I thought that $\langle a\rangle^G\cong F_\infty$ as the set $\{t^{r}at^{-r}\mid r\in \Bbb Z\}$ is contained in $\langle a\rangle^G$.

But this set is not a free generating set since $a^n=(tat^{-1})^m$.

So I want to know, what is the subgroup $\langle a\rangle^G$ precisely?

Thank you so much in advance!

Answer 1

"determine the normal closure" is not very clear. It seems you want to identify the isomorphism type of the normal closure.

Namely, this kernel $\mathrm{BS}_0(m,n)$ is an infinitely iterated amalgam $$\cdots\ast_\mathbf{Z}\mathbf{Z}\ast_\mathbf{Z}\mathbf{Z}\ast_\mathbf{Z}\mathbf{Z}\ast_\mathbf{Z}\cdots$$ where each left embedding is $k\mapsto nk$ and each right embedding is $k\mapsto mk$. This is a general fact about HNN extensions (description of the isomorphism type of the kernel of the canonical homomorphism from a given HNN-extension onto $\mathbf{Z}$), which can be found in Serre's book. In particular, this kernel $\mathrm{BS}_0(m,n)$ is not free as soon as $\max(|m|,|n|)\ge 2$. Indeed,

• if $\min(|m|,|n|)=1$ (say $m=\pm 1$ and $|n|\ge 2$), this kernel $\mathrm{BS}_0(\pm 1,n)$ is isomorphic to the infinitely generated abelian group $\mathbf{Z}[1/n]$;
• if $\min(|m|,|n|)\ge 2$, this kernel $\mathrm{BS}_0(m,n)$ contains a copy of $\mathbf{Z}^2$ (inside the amalgam $\mathbf{Z}\ast_\mathbf{Z}\mathbf{Z}=\langle a,b\mid a^m=b^n\rangle$, namely $\langle a^m,ab\rangle\simeq\mathbf{Z}^2$), hence is not free.
• (If $|m|=|n|=1$ this kernel $\mathrm{BS}_0(m,n)$ is infinite cyclic.)

On the other hand, the kernel $\mathrm{BS}_{00}(m,n)$ of the canonical homomorphism onto $\mathbf{Z}[m/n,n/m]\rtimes_{m/n}\mathbf{Z}$ is free, since it acts freely on the Bass-Serre tree.

Note: $\mathrm{BS}_0(m,n)$ is locally residually finite. Also $\mathrm{BS}_0(m,n)$ is not residually finite if $|m|,|n|$ are coprime and $\ge 2$: this is due to R. Campbell (1990 Proc AMS). I don't know if it's also non-residually-finite whenever $2\le |m|<|n|$, for instance, $(m,n)=(2,4)$.

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Edit: the free subgroup $\mathrm{BS}_{00}(m,n)$ is infinitely generated as soon as $2\le |m|<|n|$. Since $2\le \min(|m|,|n|)$, it is not trivial, hence contains a loxodromic element for the action on the Bass-Serre tree. Hence it has a unique minimal nonempty subtree (the convex hull of the union of axes of loxodromic elements) for the action on the Bass-Serre tree $T$ of the HNN extension defining $\mathrm{BS}(m,n)$; hence this subtree is invariant under the whole group, and hence by vertex-transitivity, this is the whole tree.

The tree $T$ naturally carries a Busemann-function (defined up to addition with an integral constant), so that every vertex $v$ has $n+m$ adjacent vertices: $m$ vertices $w$ with $b(w)=b(v)-1$ and $n$ vertices $w$ with $b(w)=b(v)+1$. Let $G$ is the group of automorphisms of $T$ preserving $b+\mathbf{Z}$, so the locally compact group $G$ acts cocompactly on $T$ and $\mathrm{BS}(m,n)$ acts through $G$.

Now assume by contradiction that $\mathrm{BS}_{00}(m,n)$ is finitely generated: then since it acts freely and minimally on the tree $T$, the action is cocompact. Hence $\mathrm{BS}_{00}(m,n)$ is a cocompact lattice in $G$. But $G$ is not unimodular, by an easy argument (using $|m|\neq |n|$). This is a contradiction.