Consider Thompson's group (the one commonly referred to as $T$), which is finitely presentable. Consider the Cayley graph, but then forget the coloring and direction on edges. So now we just have an undirected graph. Call this $G$.

Let $B \subset G$ be a ball around some vertex in $G$ of large radius. Since $G$ is vertex-transitive, the choice of vertex is irrelevant. The radius must be large enough that $B$ includes the cycles corresponding to the relations in the presentation.

If $G'$ is a (connected) vertex-transitive graph with a ball isomorphic to $B$, is $G'$ necessarily isomorphic to $G$?

If we make the additional assumption that $G'$ is itself the Cayley graph of some group $H$, then I believe the answer is yes. My reasoning is that since $B$ is large enough to witness both relations, $H$ must be a quotient of $T$, but $T$ is simple. I'm a little uncertain about the claim that $H$ must be a quotient, though.

  • $\begingroup$ For your last question about being isolated within Cayley graphs, I'm not sure about the answer, but within labeled Cayley graphs is yes. Indeed all of $F,T,V$ are isolated as marked groups. For $T,V$, it holds because they are finitely presented simple groups. For $F$, it's because it's finitely presented with any proper quotient abelian; that is was isolated was observed in my J. Algebra paper with Guyot and Pitsch, see arxiv.org/abs/math/0511714 $\endgroup$
    – YCor
    Feb 3, 2015 at 19:26
  • $\begingroup$ I'd guess we can get rid of the labelings in the above fact by showing that if a graphs has all its ball of radius, say, 1000, isomorphic to the balls of Thompson's $F$, then one can recognize the labels of the edges, in the sense that any isometry between balls of radius, say, 500, preserves the labeling. $\endgroup$
    – YCor
    Feb 3, 2015 at 19:31
  • $\begingroup$ Beware that Thompson's group $F$ is not simple. Its abelianization is $\mathbf{Z}^2$. Anyway, as I said, this is not an issue because it's isolated as a marked group. The issue is that the argument only shows isolated among labeled Cayley graphs. $\endgroup$
    – YCor
    Feb 3, 2015 at 19:33
  • $\begingroup$ @Ycor Oops. I guess I'll rephrase to focus on T, although I'm really interested in finitely presented simple groups in general. I should have asked the question that way originally. $\endgroup$ Feb 3, 2015 at 19:35
  • $\begingroup$ Beware that these questions might be sensitive to the choice of Cayley graph. $\endgroup$
    – YCor
    Mar 12, 2018 at 11:25

1 Answer 1


I'll suggest an approach. I'll not focus on $T$ especially: let $H$ be a finitely presentable group, pick a generating set $S$; how can we prove that the unlabeled Cayley graph (denoted $(H,S)$) is isolated among vertex-transitive graphs?

1) a necessary condition for this isolation statement is that $(H,S)$ of $H$ is isolated among labeled Cayley graphs. This condition is simply called, for a group, "isolated", because it does not depend on the generating subset. It is equivalent to be finitely presented and finitely discriminable, where the latter means: having a finite subset $\Omega\subset H\smallsetminus\{1\}$ such that every nontrivial subgroup $N$ of $H$ satisfies $N\cap\Omega\neq\emptyset$. Simple groups, finite groups, are finitely discriminable, but there are many others; for instance Thompson's group $F$.

2) Now here is a condition (X) for $(H,S)$ which implies, if $H$ is isolated, that $(H,S)$ is isolated among vertex-transitive graphs:

(X): there exists $R_0\ge 1$ such that every automorphism of the $R_0$-ball in $(H,S)$ fixing 1 is the identity.

Indeed, assume that $(H,S)$ satisfies (X). Let $Y$ be a (non-empty) vertex-transitive graph whose $R_0$-balls are isomorphic to the balls of $(H,S)$. Let $G$ be the automorphism group of $Y$. We claim that $G$ acts freely on $Y$. Indeed, assume otherwise, hence there exists a vertex $v$ and $g\in G$ fixing $v$ but not fixing some neighbor $w$ of $v$. Since $g$ is an automorphism of the $R_0$-ball around $v$ and by (X), we deduce that $g$ fixes $w$, a contradiction.

Now let us be more precise on the labeling: $(H,S)$ has oriented edges labeled by generators (we can assume wlog $1\notin S$). Now let $v$ be a vertex in $Y$ and $e$ an oriented edge in $Y$ (here I just mean an ordered pair of vertices in $Y$ linked by an edge, since $Y$ is not an oriented graph!), contained in the $R_0$-ball around $v$. Then there exists a unique isomorphism $f_v$ from $B_Y(v,R_0)$ to $B_1((G,S),R_0)$ mapping $v$ to 1, let $s(v,e)\in S$ be the label of the image of $e$ by this isomorphism.

Let us show that $s(v,e)$ does not depend on $v$: it is enough to show that $s(v,e)=s(v',e)$ if $v,v'$ are neighboring vertices such that $e$ is in the $R_0$-ball around both. $f_v$ maps $v'$ to some element $t\in S$. Hence, writing $L_g(h)=gh$ in $H$, we see that $L_t^{-1}f_v$ maps $v'$ to 1, and by uniqueness we get $f_{v'}=L_t^{-1}f_v$. $L_t$ preserves the labeling, we deduce that $s(v,e)=s(v',e)$.

Hence with this labeling, $Y$ has the same $R_0$-balls as $(H,S)$.

On the other hand, the definition of the labeling was canonical and therefore it is invariant by isometries of $Y$. Since, after fixing some vertex 1 in $Y$, $Y$ is the Cayley graph of $G$ with respect to its 1-sphere (call it $S'$), this shows that this is the Cayley labeling of $S'$, modulo the canonical bijection $S'\to S$ mapping $s'$ to the label of $(1,s')$.

If now $H$ is isolated and $R_0$ has been chosen large enough, this implies that $S'\to S$ extends to an isomorphism $G\to H$.

So you should check (X). You first need to check that the isometry group of the Cayley graph is reduced to $H$, I don't know if it holds for $T$, and maybe this requires a good choice of generating set to avoid obvious symmetries.

  • 1
    $\begingroup$ Actually you can replace (X) by (X') the isometry group of $(H,S)$ is reduced to $H$. Maybe you can even replace it by the assumption that the isometry group is discrete. This is a particular case of a result with Romain Tessera, which says that the only way to approximate a (large scale simply connected) graph with (discrete and cocompact isometry group) is from below, ie by coverings. $\endgroup$ Feb 4, 2015 at 8:07

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