A finitely generated subgroup $H$ of $G$ is said to be undistorted if any word metric on $H$ and any metric on $H$ induced by a word metric of $G$ are roughly equivalent (i.e., they differ by a multiplicative and additive constant; in other words, $H \hookrightarrow G$ is a quasi-isometric embedding). It is said to be distorted otherwise.

Baumslag–Solitar groups $G = \langle a, b \,|\, aba^{-1} = b^n \rangle$ provide easy example of groups $G$ with distorted, infinite cyclic, non-normal subgroups $H$. I was wondering how difficult it is to find examples of groups with distorted, infinite cyclic, normal subgroups.

  • $\begingroup$ You mean Baumslag-Solitar gives examples of distorted infinite cyclic non-normal subgroups, right? $\endgroup$ – Anthony Quas Mar 30 '17 at 15:52
  • $\begingroup$ Just an observation: if there is a distorted cyclic normal subgroup, $\langle a\rangle$, then $gag^{-1}=a^{\pm 1}$ for each $g\in G$ (otherwise you contradict the cyclicity). The centralizer of $a$ is then either all of $G$ or an index 2 subgroup. $\endgroup$ – Anthony Quas Mar 30 '17 at 16:46
  • $\begingroup$ correct (and you don't need the distorsion for this) $\endgroup$ – fritz Mar 30 '17 at 16:59

As explained by A. Sisto here on p.20 and p.21, "The subgroup generated by $z$ in the Heisenberg group $$〈 x,y,z | [ x,y ] = z, [ x,z ] = [ y,z ] = 1 〉$$ is isomorphic to $\mathbf Z$ and distorted." As $\langle z \rangle$ is the center of the Heisenberg group and hence normal, this gives one example.

| cite | improve this answer | |
  • $\begingroup$ I agree with you and sisto. $\endgroup$ – fritz Mar 30 '17 at 17:07
  • 2
    $\begingroup$ Distortion in nilpotent groups is a classical subject. From work of Guivarch (1973) and possibly earlier, if $G$ is finitely generated nilpotent, torsion-free, and $G^i$ is its lower central series, and $x\in G^i-G^{i+1}$ then $|x^n|\simeq n^{1/i}$. In particular, if $G$ is $c$-step nilpotent (that is $G^{c+1}=\{1\}$ and $x\in G^c$, then $x$ is central and hence you get many examples. Also taking a suitable semidirect product of the Heisenberg group and $\mathbf{Z}$, you can get a exponentially distorted central cyclic subgroup. $\endgroup$ – YCor Mar 30 '17 at 20:14
  • $\begingroup$ thanks! Here it is a beautiful illustration of the Cayley graph of the Heisenberg group: conan777.files.wordpress.com/2011/03/c-c-3.jpg The vertical direction is "z", the left-right is x, while the diagonal one is y. The red path is xyx^{-1}y^{-1}. $\endgroup$ – fritz Mar 31 '17 at 9:39

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.