Quotient of the Baumslag-Solitar group $BS(1,m)=\langle a,b| bab^{-1}=a^m\rangle$.

Question

I would like to prove that in any non-trivial quotient of the Baumslag-Solitar group $BS(1,m)$ defined by, $$BS(1,m)=\langle a,b| bab^{-1}=a^m\rangle$$ the image of one of the generators $a$ or $b$ (by the canonical projection) have finite order. I only know that $BS(1,m)$ is solvable but I don't know if it will help me to prove the statement. By advance thank you. Edgar.

Answer 1

First note that the result is false when $m = 1$, because $G$ is just free abelian of rank $2$ in that case, and the subgroup $K=\langle ab \rangle$ is normal but neither $a$ nor $b$ has finite order $G/K$. So assume that $m \ne 1$. Also, if $m=0$ then $a=1$, so we can assume $m \ne 0$.

For $k \in {\mathbb Z}$, let $a_k = b^kab^{-k}$. So $a=a_0$ and, for all $k \in {\mathbb Z}$, $a_{k+1} = b^{k}bab^{-1}b^{-k} = b^ka^mb^{-k} = a_k^m$.

So, if we let $N = \langle a_k \mid k \in {\mathbb Z} \rangle$, then since $a_l$ is a power of $a_k$ whenever $l \ge k$, any pair of generators of $N$ commute, and hence $N$ is abelian. (It is not hard to see that the map $a_k \mapsto m^k$ induces an isomorphism from $N$ to the additive group of rationals of the form $jm^k$ for $j,k \in {\mathbb Z}$.)

Also, $N$ is centralized by $a=a_0$ and normalized by $b$, so it is normal in $G$ and is in fact the normal closure $\langle a^G \rangle$ of $a$ in $G$.

By putting $a=1$ in the presentation, we see that $G/N$ is infinite cyclic and generated by the image of $b$, so $G$ is the semidirect product $N \rtimes \langle b \rangle$, where $b$ acts on $N$ by mapping $a_k$ to $a_{k+1}$.

Now, since $a_k^m=a_{k+1}$, we can write each element of $N$ as a power of $a_k$ for some fixed $k$. Since $b^{-k}a_kb^k = a_0=a$, we see that every element of $N$ is conjugate in $G$ to a power of $a$.

Now let $K$ be any nontrivial normal subgroup of $G$. If $K \cap N \ne 1$ then, as we just saw, $N \cap K$ contains a nontrivial power of $a$, and so the image of $a$ in $G/K$ has finite order.

Otherwise, since $G = N\langle b \rangle$, $N$ contains an element $nb^j$ for some $n \in N$ with $j > 0$. If $n=1$, then the image of $b$ has finite order in $G/K$. Otherwise, since we are assuming $m \ne 1$, $b$ does not centralize $n$, so the commutator $[b,nb^j]$ is a nontrivial element of $N \cap K$, and we are back in the previous case.