The Baumslag-Solitar groups $BS(m,n) = \langle b,s\mid s^{-1}b^ms = b^n\rangle$, with $mn\neq 0$, are important examples (more often, counter-examples) in group theory. They are residually finite if, and only if, either $m$ and $n$ are equal in absolute value, or one of $m$ and $n$ has absolute value equal to $1$. In that case, they are Hopfian, and also when $m$ and $n$ have the same prime divisors. Otherwise, they are non-Hopfian. For $m=n$, we get examples of one-relator groups with non-trivial center. The group $BS(2,2)$ is an example in which the Howson property fails.

I can only recall having seen these groups defined by means of a presentation. I would like to know whether these groups (apart from the obvious special cases such as the metabelian ones) can be realized by some other fairly elementary and concrete mechanism. My question is:

Does $BS(m,n)$ occur "in nature"?

(For example, the Sanov matrices $\left(\begin{matrix} 1&2 \\ 0&1 \end{matrix}\right)$ and $\left(\begin{matrix} 1&0 \\ 2&1\end{matrix}\right)$ generate a free subgroup of $SL(2,\mathbb{Z})$, so I would say that free groups occur "in nature".)

Obviously, since the Baumslag-Solitar groups are often non-Hopfian, they cannot be constructed as groups of matrices. But, perhaps there is some other concrete realization of these groups.

If there isn't a general construction, it would still be useful to get a concrete realization for the non-Hopfian group $BS(2,3)$.

  • 3
    $\begingroup$ It is a theorem of Wise that for each finitely presented group $Q$ there is a finitely generated residually finite group $N$ such that $Q\leq \operatorname{Out}(N)$. So, in some ways, every finitely presented group occurs in nature...(See "A residually finite version of Rips's construction", Corollary 3.3). $\endgroup$
    – ADL
    Aug 31, 2011 at 13:50

1 Answer 1


For $m,n \gt 0, m\ne n$, $BS(m,n)$ acts on $\mathbb H^2$ by isometries with a common ideal fixed point. In particular, you can represent it by the action on the upper half plane of

$S = \bigg(\begin{matrix}\sqrt{m/n} & 0 \\\ 0 &\sqrt{n/m}\end{matrix}\bigg),$ $B = \bigg(\begin{matrix}1 & \alpha \\\ 0 &1\end{matrix}\bigg)$

where $\alpha$ is arbitrary.

There are elements of $BS(m,n)$ which act trivially, but you can lift this to a free action on a topological $T\times \mathbb R$ where $T$ is a $n+m$-regular directed tree with outdegree $m$ and indegree $n$ so that for any path $P \subset T$, $P\times \mathbb R$ is a copy of $\mathbb H^2$ so that the sets $\{p\}\times \mathbb R$ are concentric horocycles (like horizontal lines in the upper half plane).

This is related to the tilings of the hyperbolic plane by horobricks, e.g., there are tiles with arbitrarily small diameter which tile $\mathbb H^2$ which are really fundamental domains of $BS(m,n)$, and analogously there are polyhedra of arbitrary Dehn invariant which tile $\mathbb H^3$.


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