# Are these groups solvable?

I am thinking of Baumslag-Solitar groups of type $BS(1,m)=\langle a,b \mid bab^{-1} = a^m\rangle$ as a prototype.

We can think of them as follows: Start with an infinite cyclic group $\langle a\rangle$, choose an injective endomorphism $a\mapsto a^m$, and add a generator $b$ which acts on $\langle a\rangle$ by this endomorphism.

More generally, we can start with the trivial group, and finitely many times add a new generator which acts on the previous group by an injective endomorphism.

Let's call this "the generalized construction".

Examples of groups formed by the generalized construction: $BS(1,m)$ or $\langle a,b,c \mid bab^{-1}=a^m, cac^{-1}=a^n, cbc^{-1}=ba^k\rangle$

Does the generalized construction always yield solvable groups?

My guess: I guess the answer is "yes". My feeling is that those groups are repeated semidirect products of subgroups of $\mathbb{Q}$, and so they should be solvable.

• @user1729: $BS(1,m)=\mathbb{Z}[1/m]\rtimes\mathbb{Z}$, isn't it? – Ben Mar 24 '16 at 14:11
• Yes, sorry, I had interpreted your statement (completely incorrectly!) as "...a semidirect product of $\mathbb{Z}$". – user1729 Mar 24 '16 at 14:14
• The answer to your question is yes, because ${\mathbb Z}$ and ${\mathbb Z}[1/m]$ are both abelian. – Derek Holt Mar 24 '16 at 14:16
• @DerekHolt: Thank you. However, I was asking about the generalized construction, not about $BS(1,m)$. I will edit to make it more clear. – Ben Mar 24 '16 at 14:17
• I think the answer is yes. Assuming that after $d$ iterations the group is solvable of derived length at most $d$, on the next iteration the group $G$will be a semidirect product of a group $K$ by ${\mathbb Z}$ where $A$ is an ascending union of solvable groups of derieved length at most $d$. But then all $d$-fold commutators are trivial in $A$, so $A$ is solvable of derived length at most $d$, and so $G$ is solvable of derived length at most $d+1$. – Derek Holt Mar 24 '16 at 14:36

A useful keyword is "ascending HNN-extension". An ascending extension of a $k$-step solvable group is $(k+1)$-step solvable. The result follows.