An easy example of a non-quasiconvex subgroup

Question

Let $G$ be a finitely generated group, and consider the surjection $\mu:F(A)\to G$ induced by the set of generators $A$, where $F(A)$ is the free group on $A$. A word $w$ is said to be ($\mu$-)geodesic if is it of minimal length in $\mu^{-1}\mu(w)$. A subgroup $H<G$ is called quasi-convex if there exist a constant $k$ such that for any geodesic word $w=w_1\dots w_n$ such that $\mu(w)\in H$, and for any $0<i<n$, there exist a word $v_i$ of length at most $k$ such that $\mu(w_1\dots w_iv_i)\in H$.

What is the easiest example of a non-quasi-convex subgroup? Here easy may mean with an explicit presentation, easy to prove that is not quasiconvex, or with a very quick and elegant description, depending on the taste of who is answering.

Answer 1

One of the simplest examples is the group called $BS(2,1)$ which has presentation $$ <a, b| aba^{-1}= b^2>. $$ The cyclic subgroup generated by $b$ will be exponentially distorted (hence, non-quasiconvex in your terminology).

In the realm of 3-manifold groups, consider a compact hyperbolic 3-manifold fibered over the circle. Then the fundamental group of the fiber will be exponentially distorted.

Answer 2

Another of the simplest examples is the (discrete) Heisenberg (aka the free nilpotent group of class 2) defined by the presentation $$\langle x,y,z \mid [x,y]=z,[x,z]=[y,z]=1\rangle.$$ The subgroup $\langle z \rangle$ is not quasiconvex, because there is a word $[x^n,y^n]$ of length $4n$ in the generators of $G$ for the element $z^{n^2}$, which has length $n^2$ in the generators of $H$.

But note that the distortion in this example is polynomial, whereas it is is exponential in the Baumslag-Solitar groups.

Answer 3

Here is an example of a non-quasiconvex subgroup of a hyperbolic group. (Quasiconvexity is an important property in the theory of hyperbolic groups, where quasiconvex subgroups are, for example, also hyperbolic.)

In $$G^*=\langle a, b, t\mid a^t=ab^3a, b^t=ba^3b\rangle,$$ the subgroup $H=\langle a, b\rangle$ is not quasiconvex.

This group is an HNN-extension of a free group, but where the base group "grows" under conjugation by $t$.

This is really a single example of a more general construction due to Ilya Kapovich, where the base group can be any non-elementary hyperbolic group $G$ and it embeds non-quasiconvexly into the HNN-extension $G^*$ [1].

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A similar class of examples, with similarly "easy" presentations are the Hydra groups of Dison and Riley [2]:

In $$G_k = \langle a_1,\ldots , a_k, t \mid a_1^t=a_1, a_i^t = a_ia_{i−1} (\forall i > 1) \rangle,$$ the subgroup $H_k = \langle a_1t,\ldots , a_kt\rangle$ has Ackermannian distortion.

(The Ackermann function is a standard example of a recursive function which is not primitive recursive, so this is awesomely mind-bending distortion.) These Hydra groups are not hyperbolic, although Brady+Dison+Riley constructed hyperbolic analogues [3].

[1] see Theorem 4.1 of his paper "A non-quasiconvexity embedding theorem for hyperbolic groups", Math Proc. Cambridge Phil. Soc. 127 (1999), no. 3, pp. 461-486, (arXiv)

[2] W. Dison and T. Riley, "Hydra groups", Commentarii Mathematici Helvetici, 88 (3), (2013), 507-540, (arXiv)

[3] N. Brady, W. Dison, and T. Riley. "Hyperbolic hydra", Groups, Geometry, and Dynamics 7.4 (2013): 961-976. (arXiv)

Answer 4

Another huge class of examples: infinitely generated subgroups of finitely generated groups are not quasi-convex.

[In other words, quasi-convex subgroups of f.g. groups are f.g.].

A simple example is the derived subgroup in a free group on $n\ge 2$ generators.

All examples provided so far are finitely generated subgroups and are thus, in a sense, more refined examples of non-quasi-convex subgroups than the "obvious" case of infinitely generated subgroups (which I view as the "most" possible distorted subgroups in a f.g. group).