Isomorphism problem among Thompson's groups 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. 
 A: I would like to add a small comment to Jim Belk's detailed and clear answer.  
There is a preprint Xiaobing Sheng's work describing quasi-isometric embeddings from $T_n$ to $T$ which also provides a partial answer along the lines of Liousse's paper mentioned in J. Belk's answer.  I am not certain of the current publication status of Sheng's paper.  The relevant proposition there is:
Proposition 3.1.2. The order of a torsion element in $T_n$ is a divisor of $l(n − 1) + 1$
for some integer $l \geq 0$.
A: 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, 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.
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:


*

*$X$ is locally compact and Hausdorff, and has no isolated points.

*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.
Bleak and Lanoue use the following technique.  Given a group $G$ of homeomorphisms of a space $X$, let $(g)_p$ denote the germ 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.
