I am looking for a reference proving the following statement:



> 
> For every $n,m \geq 2$, the groups $T\_n$ and $T\_m$ are isomorphic if and only if $n=m$.
> 
> 
> 


Here, $T\_k$ denotes the variation of Thompson's group $T$ (compared to $F$, $T$ acts on the circle instead of the interval) where dyadic numbers are replaced with $k$-adic numbers. 


I do not focus only on the first appearance of this statement. If several proofs are available in the literature, I am also interested to know them. 



Isomorphism problem among Thompson's groups
As far as I can tell, a solution to this problem has not appeared in the literature. Unless I'm mistaken, the best partial result was obtained by Liousse in [this 2008 paper](https://link.springer.com/article/10.1007/s10711-007-9216-y), where it is proven by examining possible orders of elements that $T\_m$ is not isomorphic to $T\_n$ if there exists a prime that divides $m-1$ but not $n-1$. 


However, as Mark indicates it is possible to use Rubin's work to solve this problem. I will use the theorem of Rubin that appears in the following paper:


[M. Rubin, Locally moving groups and reconstruction problems. *Ordered groups and infinite permutation groups.* Springer, Boston, MA, 1996. 121-157.](https://doi.org/10.1007/978-1-4613-3443-9_5)


The theorem can be stated as follows. Let $G$ be a group acting faithfully by homeomorphisms on a topological space $X$. We say that the action of $G$ on $X$ is a ***Rubin action*** if the following conditions are satified:


1. $X$ is locally compact and Hausdorff, and has no isolated points.
2. For every non-empty open set $U\subseteq X$ and every point $p\in U$, the closure of the set
$$
\{g(p) \mid g\in G\text{ and $g$ is supported on $U$}\}
$$
has nonempty interior.


Rubin refers to condition (2) by saying that the action of $G$ on $X$ is "locally moving". Rubin's theorem is the following.



> 
> **Theorem (Rubin).** Let $G$ and $H$ be groups acting faithfully by homeomorphisms on spaces $X$ and $Y$, respectively, and let $\varphi\colon G\to H$ be an isomorphism. If both actions are Rubin actions, then there exists a homeomorphism $h\colon X\to Y$ such that $\varphi(g)=hgh^{-1}$ for all $g\in G$.
> 
> 
> 


Now, it is easy to check that the action of $T\_m$ on the circle $S^1 = [0,1]/\{0,1\}$ is Rubin. Thus $T\_m$ and $T\_n$ are isomorphic if and only if there exists a homeomorphism $h\colon S^1\to S^1$ such that $T\_m=hT\_nh^{-1}$.


Thus we can distinguish $T\_m$ and $T\_n$ if we can find any property distinguishing them that is invariant under conjugation by a homeomorphism. One useful conjugacy invariant was used by Bleak and Lanoue in the following paper to prove non-isomorphism for Brin's higher-dimensional Thompson groups $nV$.:


[C. Bleak and D. Lanoue, A family of non-isomorphism results. *Geometriae Dedicata* **146.1** (2010): 21-26.](https://link.springer.com/article/10.1007%2Fs10711-009-9423-9)


Bleak and Lanoue use the following technique. Given a group $G$ of homeomorphisms of a space $X$, let $(g)\_p$ denote the [germ](https://en.wikipedia.org/wiki/Germ_(mathematics)) of an element $g\in G$ at a point $p\in X$, and let
$$
(G)\_p = \{(g)\_p \mid g\in G\text{ and }g(p)=p\}.
$$
Note that composition of germs induces a group structure on $(G)\_p$. This is the ***group of germs*** for $G$ at $p$. If $h\colon X\to Y$ is a homeomorphism, it is easy to prove that the mapping
$$
(g)\_p \mapsto (hgh^{-1})\_{h(p)}
$$
is an isomorphism $(G)\_p\to (hGh^{-1})\_{h(p)}$, i.e. the isomorphism type of $(G)\_p$ is invariant under conjugation by a homeomorphism. Bleak and Lanoue prove that $mV$ and $nV$ are non-isomorphic for $m\ne n$ by proving that the isomorphism types of the groups of germs are different.


Now let $T\_m$ act on the circle $S^1 = [0,1]/\{0,1\}$. If $p\in S^1$, it is easy to check that the groups of germs for $T\_m$ at $p$ has the following isomorphism type:
$$
(T\_m)\_p \cong \begin{cases}\mathbb{Z}\times\mathbb{Z} & \text{if }p\text{ is }m\text{-adic} \\ \mathbb{Z} & \text{if }p\text{ is rational but not }m\text{-adic} \\ 1 & \text{if }p\text{ is irrational}.\end{cases}
$$
Since the possible isomorphism types here do not depend on $m$, we cannot use Bleak and Lanoue's method directly. However, it follows from the above considerations that any homeomorphism $h\colon S^1\to S^1$ conjugating $T\_m$ to $T\_n$ must map the set of $m$-adic points in $S^1$ to the set of $n$-adic points in $S^1$.


Thus it suffices to prove that the dynamics of the action of $T\_m$ on the $m$-adic points is somehow different from the dynamics of the action of $T\_n$ on the $n$-adic points. This is actually quite easy: it is well-known that the group $T\_m$ has exactly $m-1$ orbits on the set of ordered pairs $(x\_1,x\_2)$ of distinct $m$-adics in $S^1$. Specifically, if we write
$$
x\_1 = \frac{i}{m^k}\qquad\text{and}\qquad x\_2=\frac{j}{m^k}
$$
for $0\leq i<m^k$ and $i<j< i+m^k$ (so $x\_2$ is represented by a point in the interval $(x\_1,x\_1+1)$), then the orbit of $(x\_1,x\_2)$ is determined by the value of $j-i$ modulo $m-1$. It follows that no homeomorphism can conjugate $T\_m$ to $T\_n$ for $m\ne n$, so by Rubin's theorem these groups are not isomorphic.


