4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ @user1729: $BS(1,m)=\mathbb{Z}[1/m]\rtimes\mathbb{Z}$, isn't it? $\endgroup$ – Ben Mar 24 '16 at 14:11
  • $\begingroup$ Yes, sorry, I had interpreted your statement (completely incorrectly!) as "...a semidirect product of $\mathbb{Z}$". $\endgroup$ – user1729 Mar 24 '16 at 14:14
  • $\begingroup$ The answer to your question is yes, because ${\mathbb Z}$ and ${\mathbb Z}[1/m]$ are both abelian. $\endgroup$ – Derek Holt Mar 24 '16 at 14:16
  • $\begingroup$ @DerekHolt: Thank you. However, I was asking about the generalized construction, not about $BS(1,m)$. I will edit to make it more clear. $\endgroup$ – Ben Mar 24 '16 at 14:17
  • 1
    $\begingroup$ 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 $d+1$. $\endgroup$ – Derek Holt Mar 24 '16 at 14:36
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.