Consider a Cayley graph of a group $G$ with respect to a symmetric finite generating set $S$. There are some obvious candidates to isometries of this graph - for example, translation by elements of $G$, and group automorphisms which preserve $S$.

In some simple cases, these are everything one has (up to composition) - for example, take $G$ be a the free abelian group $G=\mathbb{Z}^{d}$ with $S$ being $\{\pm e_i\}$, with $\{e_i\}$ being the standard basis. This is true for other generating sets as well. However, it is far from correct, say, for the free group with its standard generators.

I am more inclined toward "$\mathbb{Z}^d$-esque" situations in my research, so I would like to try and know more on the first case example.

For the questions themselves:

(a) Consider $G=\mathbb{Z}^d$ with some generating set. Can all isometries of the Cayley graph be seen as a composition of translations and group Automorphisms (preserving S)?

(b) Is this claim true for "nice" nilpotent groups, e.g. the Heisenberg groups?

(c) Must it be false for hyperbolic groups, considering all possible generating sets?

  • 1
    $\begingroup$ Suppose that we take as connecting set the whole group except for the identity. Then the graph is complete and its automorphism group is the full symmetric group. This is always bigger than the group generated by translations and group automorphisms once |G|>4. In other words, except for a few small exceptions, the answer cannot be yes for all generating sets of a given group. $\endgroup$
    – verret
    Mar 23, 2015 at 22:28
  • $\begingroup$ I Restricted myself to finite, symmetric generating sets in the first sentence exactly for this reason :-). $\endgroup$
    – Miel Sharf
    Mar 24, 2015 at 19:13
  • $\begingroup$ Sorry, I missed the "finite". Please disregard my comment. $\endgroup$
    – verret
    Mar 25, 2015 at 2:12
  • $\begingroup$ (b) has a negative answer if one allows torsion. For instance, the Cayley graph of $\mathbf{Z}\times F$, $F$ a finite group with at least 2 elements, with generating subset $\{-1,0,1\}\times F$, has an uncountable isometry group. Also (c) has a negative answer for a free group with free generating subset. (I'm note that sure of the precise meaning of the question, in case it asks whether there always a generating subset such that there are few isometries, the above does not give a negative answer) $\endgroup$
    – YCor
    Mar 8, 2016 at 3:01
  • 1
    $\begingroup$ For (c), every Cayley graph of a torsion free lattice $\Gamma$ in a simple connected Lie group with finite center and different from $SL(2,\mathbf R)$ has a discrete isometry group by a Theorem of Furman (GAFA 2001). This gives lots of hyperbolic groups, and perhaps the proof can be adapted to show that $\Gamma$ is normal in the isometry group of all its Cayley graphs (which implies that the isometry group of the Cayley is made of group automorphisms). At least for some examples. $\endgroup$ Mar 10, 2016 at 8:56

1 Answer 1


Question (b) (and, therefore, also Question (a)) was answered affirmatively (for all torsion-free nilpotent groups) in 1998 by R.G.Möller and N.Seifter [Europe J. Comb. 19, 597-602] (see Theorem 4.1(1)).

The answer to Question (a) was rediscovered in 2007 by A.A.Ryabchenko [Siberian Math. J. 48 (2007) 919–922] (see the proof of Theorem 2). This short proof is reproduced in Theorem 5.3 of a preprint of J.Morris (arXiv:1502.06114).

A more elementary proof of the answer to Question (b) (for all torsion-free nilpotent groups) is in a preprint by J.Morris, G.Verret, and me (arXiv:1603.01883).

I have no information about Question (c).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .