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? $\endgroup$ Mar 30, 2017 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. $\endgroup$ Mar 30, 2017 at 16:46
  • $\begingroup$ correct (and you don't need the distorsion for this) $\endgroup$
    – fritz
    Mar 30, 2017 at 16:59

1 Answer 1


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.

  • $\begingroup$ I agree with you and sisto. $\endgroup$
    – fritz
    Mar 30, 2017 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, 2017 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, 2017 at 9:39

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