R. Thompson introduced three groups $F\subset T\subset V$. The question concerning amenability of $F$ is still unanswered and has attracted much attention. I have read that Thompson group $V$ contains a copy of the free group $F_2$ (with two generators), in particular it is not amenable. Does anyone know an explicit embedding or can give me a reference for it? Thank you very much for the help.

$\begingroup$ $T$ itself contains a copy of $F_2$: this is indeed easy to find a pingpong pair starting from 4 disjoint intervals. That $T$ is nonamenable is even easier, because it does not preserve any probability on Borel subsets on the circle (immediate once we check that the only probabilities invariant by $F$ are supported by $\{0,1\}$). $\endgroup$– YCorMar 22, 2016 at 21:20

$\begingroup$ Thank you for the answer. Just one thing, what do you mean by "pingpong pair"? Thank you for the help. $\endgroup$– John N.Mar 22, 2016 at 21:25

$\begingroup$ Given a group with an action on a set, it is a pair satisfying the pingpong lemma. That is, in the language of en.wikipedia.org/wiki/Pingpong_lemma (and the space being the circle), it is a pair $(a_1,a_2)$ as in the pingpong lemma for cyclic subgroups [for future readers: I refer to today's 2016/03/22 version, which can be found in the history in case it doesn't fit]. $\endgroup$– YCorMar 22, 2016 at 21:47

$\begingroup$ Thank you very much. If you copy these comments in an answer I can vote for your answer $\endgroup$– John N.Mar 22, 2016 at 21:50
3 Answers
Thompson's group $T$ of the circle itself contains a copy of $F_2$: this is indeed easy to find a pingpong pair starting from 4 disjoint intervals. That $T$ is nonamenable is even easier, because it does not preserve any probability on Borel subsets on the circle (immediate once we check that the only probabilities on Borel subsets of the interval $[0,1]$ invariant by $F$, are supported by $\{0,1\}$, and hence after identifying $0=1$ the only $F$invariant probability on the circle is the Dirac at $0=1$, which of course is not $T$invariant).
If $\kappa$, $\lambda$, $\mu$ and $\nu$ are generators of Thompson's Group $V$ which satisfy the defining relations given on Page 50 in
Graham Higman, Finitely presented infinite simple groups, Notes on Pure Mathematics, Department of Pure Mathematics, Australian National University, Canberra, 1974. MR0376874 (51 #13049)
an explicit embedding of the free group of rank $2$ is given by $$ \varphi: \ {\rm F}_2 = \langle a, b \rangle \ \rightarrow \ V, \ \ a \mapsto (\kappa \mu)^2, \ b \mapsto \lambda(\kappa \nu)^2 \lambda \kappa. $$ This can be found with GAP as follows:
gap> LoadPackage("rcwa");
gap> k := ClassTransposition(0,2,1,2);; l := ClassTransposition(1,2,2,4);;
gap> m := ClassTransposition(0,2,1,4);; n := ClassTransposition(1,4,2,4);;
gap> V := Group(k,l,m,n); # Thompson's group V with generators as above
<(0(2),1(2)),(1(2),2(4)),(0(2),1(4)),(1(4),2(4))>
gap> F2 := FreeGroup("a","b");;
gap> phi := IsomorphismRcwaGroup(F2);
[ a, b ] > [ <wild rcwa permutation of Z with modulus 8>,
<wild rcwa permutation of Z with modulus 8> ]
gap> IsSubgroup(V,Image(phi)); # we are lucky to have an embedding into V
true
gap> F4 := FreeGroup("k","l","m","n");;
gap> psi := EpimorphismByGenerators(F4,V);
[ k, l, m, n ] > [ ( 0(2), 1(2) ), ( 1(2), 2(4) ), ( 0(2), 1(4) ),
( 1(4), 2(4) ) ]
gap> a := PreImagesRepresentative(psi,Image(phi,F2.1));
(k*m)^2
gap> b := PreImagesRepresentative(psi,Image(phi,F2.2));
l*(k*n)^2*l*k
As already mentioned, a standard strategy to construct (nonabelian) free subgroups in $T$ or $V$ is to play pingpong on the Cantor space. Just for fun, I would like to describe another point of view.
A topological interpretation of $T$ I really like is the following. Let $T_3$ denote the $3$regular tree $T_3$, thought of as drawned on the plane (such that its vertexset is discrete), and let $\mathscr{S}$ denote the ribbon surface obtained by thickening it. For each edge of $T_3$, fix a transverse arc in $\mathscr{S}$ with its endpoints on the boundary of $\mathscr{S}$. This collection of arcs decomposes $\mathscr{S}$ as a union of hexagons. We refer to a homeomorphism of $\mathscr{S}$ which sends all but finitely many hexagons to hexagons as an asymptotically rigid homeomorphism.
Theorem. The asymptotically rigid mapping class group $$\{ \text{orientationpreserving asymptotically rigid homeomorphisms of $\mathscr{S}$} \}/ \text{isotopy}$$ is isomorphic to the PtolemyThompson group $T$.
See for instance Section 1.3 in Kapoudjian and Funar's article The braided PtolemyThompson group is finitely presented (and references therein).
Some isometries of $T_3$ induce asymptotically rigid homeomorphisms of $\mathscr{S}$. For instance, the rotation $\alpha$ of order $3$ around a vertex (say $v$) and the rotation $\beta$ of order $2$ around the middle point of an edge (having $v$ as one of its endpoints). Clearly, the subgroup $\langle \alpha, \beta \rangle$ acts on the subdivision of $T_3$ with a single orbit of edges and with trivial edgestabilisers. It follows that $$\langle \alpha, \beta \rangle = \langle \alpha \rangle \ast \langle \beta \rangle \simeq \mathbb{Z}/2 \mathbb{Z} \ast \mathbb{Z} /3 \mathbb{Z}.$$ Consequently, $T$ contains a nonelementary free product (and a fortiori a nonabelian free subgroup).

$\begingroup$ How is the PtolemyThompson group $T$ related to Thompson's group $T$? or are these the same? $\endgroup$– YCorNov 5, 2020 at 20:11

1$\begingroup$ @YCor The Ptolemy group $T$ and Thompson's group $T$ have different definitions  but happen to be isomorphic. So some authors call $T$ the PtolemyThompson group, but it's all the same $T$. $\endgroup$ Nov 5, 2020 at 22:24