Solvable Nonnilpotent Baumslag-Solitar groups

Question

The Baumslag-Solitar groups are defined by $$BS(m,n)=\langle a,b\mid ba^{m}b^{-1}=a^{n}\rangle,$$

where $m,n$ are non-zero integers.

It is known that for $m=1$ or $n=1$ these groups are solvable (see, for example, the Encyclopedia of Mathematics).

My question is when these groups are solvable and nonnilpotent?

Answer 1

The only nilpotent Baumslag-Solitar group is $\mathbb{Z}^2=\langle a,b\mid [a,b]\rangle$. More generally:

Theorem. A one-relator group is nilpotent if and only if it is abelian, if and only if it is a cyclic group or $\mathbb{Z}^2=\langle a,b\mid [a,b]\rangle$.

Proof. Let $G$ be a one-relator group which is not cyclic nor isomorphic to $\mathbb{Z}^2$, and which has non-trivial centre. Then $G$ is two generated and its centre is infinite cyclic [1]. Moreover, Pietrowski [2] proved that such a group $G$ admits a presentation $$\mathcal{P}:=\langle x_1, \ldots, x_n\mid x_1^{p_1}=x_2^{q_1}, \ldots, x_{n-1}^{p_{n-1}}=x_n^{q_{n-1}}\rangle.$$ Indeed, in this form the centre is the subgroup $\cap\langle x_i\rangle$. (Given a presentation of this form it is not easy to verify if it defines a one-relator group, although an algorithm exists [3], but a simple example is $\langle a,b; a^m=b^n\rangle$.)

Now, abelianise this presentation $\mathcal{P}$. Every one-relator group surjects onto $\mathbb{Z}$, and so this abelianisation also maps onto $\mathbb{Z}$. Therefore, some generator $x_i$ has infinite order in the abelianisation. As the centre is wholly contained in $\langle x_i\rangle$, the centre intersects the derived subgroup trivially. This means that $G$ is not nilpotent, as required, because non-abelian nilpotent groups have a centre which intersects the derived subgroup non-trivially (see, for example, here). QED

In fact, we have a stronger result:

Corollary. A one-relator presentation $\langle \mathbf{x}\mid R\rangle$ defines a nilpotent group if and only if one of the following holds

1. $|\mathbf{x}|=1$ (i.e. the group is obviously cyclic)
2. $|\mathbf{x}|=2$, so $\mathbf{x}=\{a, b\}$ say, and $R$ is conjugate to $[a, b]$ or $[a, b]^{-1}$ in $F(a, b)$.
3. $|\mathbf{x}|=2$ and $R$ is a "primitive element" of $F(\mathbf{x})$, i.e. there exists some word $S\in F(\mathbf{x})$ such that $\langle R, S\rangle=F(\mathbf{x})$.

I won't detail a proof, but the point is that in each case the group is "obviously" nilpotent (1. and 3. correspond to cyclic groups, 2. is $\mathbb{Z}^2$), while certain well-known results in combinatorial group theory give us that a one-relator presentation of a cyclic group or $\mathbb{Z}^2$ must have one of these forms (for example, the main result for (2) is Theorem 3.9 of the book Combinatorial Group Theory by Magnus, Karrass and Solitar).

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[1] Murasugi, K. "The center of a group with a single defining relation." Mathematische Annalen 155.3 (1964): 246-251.

[2] Pietrowski, A. "The isomorphism problem for one-relator groups with non-trivial centre." Mathematische Zeitschrift 136.2 (1974): 95-106.

[3] Metaftsis, V. "An algorithm for stem products and one-relator groups." Proceedings of the Edinburgh Mathematical Society 42.1 (1999): 37-42.

Answer 2

The Baumslag-Solitar solvable group $BS(1,n)$ is always solvable. It is not nilpotent iff $n\ne 1$. Indeed if $n=1$ then the group is abelian, hence nilpotent. If $n\ne 1$ then the derived subgroup is isomorphic to the additive group $\mathbb{Z}[1/n]$. The abelianization is cyclic and the cyclic group acts on the derived subgroup by multiplication by $n$. Hence it is easy to prove that the center is trivial and the group is not nilpotent.

That argument works for every integer $\ne 0, -1$. The case 0 is not possible. If $n=-1$, the group has a quotient $\langle a, b \mid b^2=baba=1\rangle$ which is the infinite dihedral group, so the Baumslag-Solitar group is not nilpotent.