arXiv:1611.08264v2 [] 28 Nov 2016

INVARIABLE GENERATION OF THOMPSON GROUPS
TSACHIK GELANDER, GILI GOLAN, AND KATE JUSCHENKO
Abstract. A subset S of a group G invariably generates G if G = hsg(s) |s ∈ Si for every
choice of g(s) ∈ G, s ∈ S. We say that a group G is invariably generated if such S exists, or
equivalently if S = G invariably generates G. In this paper, we study invariable generation
of Thompson groups. We show that Thompson group F is invariable generated by a finite
set, whereas Thompson groups T and V are not invariable generated.

1. Introduction
Recall that a subset S of a group G invariably generates G if G = hsg(s) |s ∈ Si for every
choice of g(s) ∈ G, s ∈ S. One says that a group G is invariably generated, or shortly IG,
if such S exists, or equivalently if S = G invariably generates G. The term “invariable
generation” was coined by Dixon [Dix92] in his study of generation of Galois groups, where
elements are given only up to conjugacy. Invariably generated groups were studied before
by Wiegold [W76, We77] under different terminology: A group G is invariably generated
if and only if no proper subgroup of G meets every conjugacy class. This is equivalent to
saying that every transitive permutation representation of G on a non-singleton set admits
a fixed-point-free element. Following [KLS15], we say that G is finitely invariably generated,
or shortly FIG, if there is a finite subset S ⊂ G which invariably generates G.
A simple counting argument shows that every finite group is IG. Obviously, abelian
groups are IG. More generally, J. Wiegold [W76] showed that the class of IG groups is
closed under extensions, hence contains all virtually solvable groups. Clearly, this class is
also closed to quotients. On the other hand, the class of IG groups is not closed under
direct unions; for instance the (locally finite) group of finitely supported permutations of N
is clearly not IG, since every element fixes some point in N. Moreover, Wiegold [We77] gave
an example of an IG group whose commutator subgroup is not IG, proving in particular
that the class IG is not subgroup closed.
In [W76], Wiegold proved that the free group F = ha, bi is not IG by producing a list L
of conjugacy class representatives which are jointly independent (i.e., that freely generate
a free group of infinite rank). To recall his construction, let {wn } be conjugacy class
n
representatives which start and end with a non-zero power of b, then take L = {wna :
n ∈ N}. T. Gelander proved in [Gel15] that convergence groups, and in particular Gromov
Key words and phrases. Invariable generation, Thompson groups.
The work of the first author was partially supported by ISF-Moked grant 2095/15. The work of the
second author was partially supported by a Fulbright grant and a postdoctoral scholarship from Bar Ilan
University. The work of the third author was partially supported by NSF CAREER grant DMS 1352173.
1

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TSACHIK GELANDER, GILI GOLAN, AND KATE JUSCHENKO

hyperbolic groups and relatively hyperbolic groups, are not IG, confirming a conjecture from
[KLS15]. Gelander and Meiri [GM16] established various examples of arithmetic groups
possessing the Congruence Subgroup Property which are not IG, providing a negative
answer to a question from [KLS15].
The notion of finite invariant generation is more subtle. For example it is still unknown weather every finitely generated solvable group is FIG. Kantor, Lubotzky and Shalev
[KLS15] proved that a finitely generated linear group is finitely invariably generated if and
only if it is virtually solvable.
The main result of the paper is summarized in the following theorem:
Theorem 1. Thompson’s group F is finitely invariably generated. The Thompson groups
T and V are not invariably generated.
2. Thompson group F
2.1. F as a group of homeomorphisms. Recall that Thompson group F is the group
of all piecewise linear homeomorphisms of the interval [0, 1] with finitely many breakpoints
where all breakpoints are finite dyadic and all slopes are integer powers of 2. The group F
is generated by two functions x0 and x1 defined as follows [CFP96].


t
if 0 ≤ t ≤ 12

1



2t
if
0
≤
t
≤

2t − 1 if 1 ≤ t ≤ 5
4
2
2
8
x0 (t) = t + 41
x1 (t) =
if 14 ≤ t ≤ 12
1
5


t+ 8
if 8 ≤ t ≤ 43
 t + 1 if 1 ≤ t ≤ 1


t 1
2
2
2
+2
if 34 ≤ t ≤ 1
2
The composition in F is from left to right.
Every element of F is completely determined by how it acts on the set Z[ 21 ]. Every number
in (0, 1) can be described as .s where s is an infinite word in {0, 1}. For each element g ∈ F
there exists a finite collection of pairs of (finite) words (ui , vi ) in the alphabet {0, 1} such
that every infinite word in {0, 1} starts with exactly one of the ui ’s. The action of F on a
number .s is the following: if s starts with ui , we replace ui by vi . For example, x0 and x1
are the following functions:


.0α
if t = .0α




if t = .00α
.0α
.10α
if t = .100α
x0 (t) = .10α if t = .01α
x1 (t) =


.110α if t = .101α
.11α if t = .1α



.111α if t = .11α
where α is any infinite binary word.

2.2. Elements of F as pairs of binary trees. Often, it is more convenient to describe
elements of F using pairs of finite binary trees (see [CFP96] for a detailed exposition).
The considered binary trees are rooted full binary trees; that is, every inner vertex (i.e.,

INVARIABLE GENERATION OF THOMPSON GROUPS

3

non-leaf vertex) has two outgoing edges: a left edge and a right edge. A branch in a binary
tree is a simple path from the root to a leaf. If every left edge in the tree is labeled “0” and
every right edge is labeled “1”, then a branch in T has a natural binary label. We rarely
distinguish between a branch and its label.
Let (T+ , T− ) be a pair of finite binary trees with the same number of leaves. (T+ , T− )
is called a tree-diagram. Let u1 , . . . , un be the (labels of) branches in T+ , listed from left
to right. Let v1 , . . . , vn be the (labels of) branches in T− , listed from left to right. We
say that the tree-diagram (T+ , T− ) has the pair of branches ui → vi for i = 1, . . . , n. The
tree-diagram (T+ , T− ) represents the function g ∈ F which takes binary fraction .ui α to
.vi α for every i and every infinite binary word α. We also say that the element g takes the
branch ui to the branch vi . For a finite binary word u, we denote by [u] (resp. (u]) the
dyadic interval [.u, .u1N ] (resp. (.u, .u1N ]). If u → v is a pair of branches of (T+ , T− ), then
g maps the interval [u] linearly onto [v].
A caret is a binary tree composed of a root with two children. If (T+ , T− ) is a tree-diagram
and one attaches a caret to the ith leaf of T+ and the ith leaf of T− then the resulting tree
diagram is equivalent to (T+ , T− ) and represents the same function in F . When we say
that a function f has a pair of branches ui → vi , the meaning is that some tree-diagram
representing f has this pair of branches. Clearly, if u → v is a pair of branches of f , then
for any finite binary word w, uw → vw is also a pair of branches of f . Similarly, if f has
the pair of branches u → v and g has the pair of branches v → w then f g has the pair of
branches u → w.
2.3. Generating sets of F. Let H ≤ F . Following [GS15, Go16], we define the closure
of H, denote Cl(H), to be the subgroup of F of all piecewise-H functions. In other words,
Cl(H) is the topological full group (see, for example, [EM13]) of the group H acting on the
set of finite dyadic fractions in the unit interval [0, 1] with the natural topology. In [Go16],
the first author proved that the generation problem in F is decidable. That is, there is an
algorithm that decides given a finite subset X of F whether it generates the whole F .
Theorem 2. [Go16, Theorem 7.14] Let H be a subgroup of F . Then H = F if and only if
the following conditions are satisfied.
(1) Cl(H) = F .
(2) H[F, F ] = F .
(3) There is an element h ∈ H which fixes a finite dyadic fraction α ∈ (0, 1) such that
h′ (α− ) = 1 and h′ (α+ ) = 2.
Below we would apply Theorem 2 to prove that a given subset of F is a generating set
of F . To verify that Condition (1) in the theorem holds, we will make use of the following
lemma. The lemma is an immediate result of Remark 7.2 and Lemma 10.6 in [Go16].
Lemma 3. Let H be a subgroup of F . If for each of the following pairs of branches there
is an element in H which has the given pair of branches, then Cl(H) = F .
(1) 00 → 0

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TSACHIK GELANDER, GILI GOLAN, AND KATE JUSCHENKO

(2)
(3)
(4)
(5)

1 → 11
01 → 10
010 → 10.
011 → 10.

2.4. Thompson groups T and V. Consider the circle S 1 as the unit interval [0, 1] with
endpoints 0 and 1 identified. In particular, we can refer to dyadic fractions and subintervals of S 1 (including subintervals of the form [a, b] for a > b which contain the point
0 = 1). Thompson group T is the group of all piecewise-linear orientation preserving
homeomorphisms of S 1 which preserve the set of finite dyadic fractions, have finitely many
breakpoints, all of which are at finite dyadic fractions and where all slopes are integer
powers of 2. Thompson group F can be viewed as the subgroup of T which fixes 0.
Thompson group V is the group of left-continuous bijections of S 1 which map finite
dyadic fractions to finite dyadic fractions, that are differentiable except at finitely many
finite dyadic fractions and such that on each maximal interval where the functions are
differentiable, they are linear with slope an integer power of 2. Thompson group T is
clearly a subgroup of V .
Elements of T and V can be represented by tree-diagrams. Let T+ , T− be a pair of trees
with n leaves each. Let σ ∈ Sn be a permutation of {1, . . . , n}. The triple (T+ , σ, T− )
represents an element of V as follows. Let u1 , . . . , un (resp. v1 , . . . , vn ) be the branches
of T+ (resp., T− ), listed from left to right. The function g ∈ V represented by (T+ , σ, T− )
maps .ui α to .vσ(i) α for every i and every infinite binary word α 6≡ 0N . In other words, g
maps the interval (ui ] = (.ui , .ui 1N ] linearly onto the interval (vσ(i) ] = (.vσ(i) , .vσ(i) 1N ]. The
triple (T+ , σ, T− ) is called a tree-diagram of g. As in the case of tree diagrams of elements
of F , one can define equivalent tree-diagrams (which represent the same element of V ) and
are obtained by insertion (or reduction) of common carets. A tree-diagram (T+ , σ, T− ) with
σ ∈ Sn represents an element of Thompson group T if and only if σ is a power of the cycle
(1 2 . . . n) (that is, if and only if σ preserves the cyclic order of {1, . . . , n}).
Let (T+ , σ, T− ) be a tree-diagram of an element g in V . We can consider T+ and T− as
rooted subtrees of the complete infinite binary tree T∞ . If T+ and T− coincide; i.e., have
the same set of branches {u1 , . . . , un }, then σ can be viewed as a permutation on the set
{u1 , . . . , un }. In that case, it is easy to see that g is periodic of the same order as σ. If T+
and T− do not coincide, then since T+ and T− have the same number of carets, T+ must
have a branch v such that v is a strict prefix of some branches vw1, . . . , vwm of T− . Clearly,
m ≥ 2. We shall need the following lemma which follows immediately from [Br04, Lemma
10.5].
Lemma 4. Let g ∈ V be a non periodic element. Then g is represented by a tree-diagram
(T+ , σ, T− ) such that the following holds. If v is a branch of T+ and v is a proper prefix of
branches vw1 , . . . , vwm of T− (for m ≥ 2) then there is k ∈ {1, . . . , m} and r ∈ N such that
the intervals (v], g(v], . . . , g r−1(v] are pairwise disjoint and such that g r (v] = (vwk ].

INVARIABLE GENERATION OF THOMPSON GROUPS

5

We will also need the following two remarks. Remark 5 can be proved in a similar way
to [Sav15, Proposition 3.1] (see also [GS16, Corollary 2.5]). Remark 6 follows easily from
[CFP96, Lemma 4.2].
Remark 5. Let g ∈ V and assume that g fixes an irrational number α. Then g fixes
pointwise a small neighborhhod (α − ǫ, α + ǫ) of α.
Remark 6. Let (a, b) and (c, d) be open sub-intervals of S 1 with dyadic endpoints. Then
there is an element g ∈ T (and in particular, g ∈ V ) which maps (a, b) onto (c, d).
3. Thompson group F is invariably generated
Recall that x0 , x1 are the generators of F defined above. In this section we prove that
Thompson group F is invariably generated by {x0 , x1 , x0 x1 }. We would need the following
two lemmas.
Lemma 7. Let g ∈ F . Let H = h(x0 x1 )g i. Then there is an element f ∈ H such that for
some m, n ∈ N the function f has the pairs of branches
(1) 0m 10 → 1n 0.
(2) 0m 11 → 1n+1 0.
Proof. The function g has pairs of branches 0a → 0b and 1c → 1d for some a, b, c, d ∈ N.
We observe that x0 x1 is composed of the following pairs of branches.

00 → 0



010 → 10

011 → 110



1
→ 111

Let h = x0 x1 . We note that the pairs of branches 00 → 0 and 010 → 10 of h imply that
h has the pair of branches 0a 10 → 10. Indeed, if a = 1, this is clear. If a > 1, then h takes
0a 10 to 0a−1 10 and we are done by induction. Similarly, the pairs of branches 00 → 0 and
011 → 110 of h imply that ha has the pair of branches 0a 11 → 110. The pair of branches
1 → 111 of h implies that hc has the pairs of branches 10 → 12c+1 0 and 110 → 12c+2 0.
Hence ha+c has the pairs of branches 0a 10 → 12c+1 0 and 0a 11 → 12c+2 0.
Let f = (ha+c )g . Then f has the pairs of branches
a

0b 10 → 1c+d+1 0 and 0b 11 → 1c+d+2 0.
Indeed, g −1 takes 0b 10 to 0a 10, then ha+c takes 0a 10 to 12c+1 0. Then g takes 1c 1c+1 0 to
1d 1c+1 0. That gives the first pair of branches. The proof for the second one is similar.
Therefore, f ∈ H is an element as described.

Lemma 8. Let g ∈ F . Let H = hx0 , (x0 x1 )g i. Then Cl(H) = F .

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TSACHIK GELANDER, GILI GOLAN, AND KATE JUSCHENKO

Proof. The element x0 is composed of the following pairs of branches.


00 → 0
x0 = 01 → 10

1 → 11

Since x0 ∈ H, by Lemma 3, it suffices to prove that there is an element in Cl(H) with the
pair of branches 010 → 10 and an element in H with the pair of branches 011 → 10.
By Lemma 7, there is an element f ∈ H with pairs of branches 0m 10 → 1n 0 and
0m 11 → 1n+1 0 for some m, n ∈ N. Let
−(m−1)

h1 = x0

−(n−1)

f x0

−(m−1)

and h2 = x0

f x−n
0 .
−(m−1)

Then h1 , h2 ∈ H. The function h1 has the pair of branches 010 → 10. Indeed, x0
has the pair of branches 010 → 0m 10, then f has the pair of branches 0m 10 → 1n 0, then
−(n−1)
x0
has the pair of branches 1n 0 → 10. Similarly, h2 has the pair of branches 011 → 10.
Therefore, Cl(H) = F .

Theorem 9. Thompson group F is invariably generated by {x0 , x1 , x0 x1 }.
Proof. It suffices to prove that for any h, g ∈ F , the set X = {x0 , xh1 , (x0 x1 )g } is a generating
set of F . Let H be the subgroup of F generated by X. Then H[F, F ] = F (indeed, the
image of X in the abelianization of F contains the image of the generating set {x0 , x1 }).
By Lemma 8, Cl(H) = F . In addition, since x1 fixes 21 and has slope 1 at 12 on the left and
slope 2 on the right, the element xh1 fixes the finite dyadic fraction α = h( 21 ) and has slope

1 at α on the left and slope 2 on the right. Hence, by Theorem 2, we have H = F .
4. Thompson groups T and V are not invariably generated
To prove that T and V are not invariably generated we will need the definition of a
γ-wandering set.
Definition 10. Let γ be an element of V . A subset U ⊆ S 1 is γ-wandering if for every
n ∈ Z with γ n 6= e we have γ n (U) ∩ U = ∅. The set U is weakly γ-wandering if for every
n ∈ Z, either γ n fixes U pointwise or γ n (U) is disjoint from U.
Lemma 11. Let γ ∈ V be a non-periodic element. Then there is an open γ-wandering set.
Proof. Let (T+ , σ, T− ) be a tree-diagram of γ which satisfies the conditions of Lemma 4. By
the discussion preceding that lemma, T+ has a branch v which is a proper prefix of some
branches vw1 , . . . , vwm of T− , where m ≥ 2. Then, by Lemma 4, there is k ∈ {1, . . . , m}
and r ∈ N such that the intervals (v], g(v], . . . , g r−1(v] are pairwise disjoint and such that
g r (v] = (vwk ]. Let j ∈ {1, . . . , m} be distinct from k. We claim that (vwj ] is γ-wandering
(and as such, every open interval contained in (vwj ] is also γ-wandering). Indeed, let
n ∈ Z \ {0}. If γ n (U) ∩ U 6= ∅ then U ∩ γ −n (U) 6= ∅. Hence, we can assume that n is
positive. We write n = ar +b where a ≥ 0 and b ∈ {0, . . . , r −1}. Since γ r (v] = (vwk ] ⊆ (v],
we have γ ar (v] ⊆ (vwk ]. Hence γ ar (vwj ] ⊆ (vwk ]. If b = 0 we are done since j 6= k implies

INVARIABLE GENERATION OF THOMPSON GROUPS

7

that (vwj ] ∩ (vwk ] = ∅. If b 6= 0 then γ ar+b (vwj ] ⊆ γ b (v]. Since b ∈ {1, . . . , r − 1}, γ b (v] is
disjoint from (v]. In particular, γ ar+b (vwj ] is disjoint from (vwj ].

Lemma 12. Let γ be a periodic element of V . Then γ admits an open weakly γ-wandering
set. If γ ∈ T then γ admits an open γ-wandering set (in fact, in that case, any weakly
γ-wandering set is γ-wandering).
Proof. Assume that γ ∈ V has period n. We can assume that γ 6= e (equivalently, that
n > 1). Let b be an irrational number in S 1 such that b is not fixed by γ. Let m be the
period of b under the action of hγi. Since γ, . . . , γ m−1 are left-continuous and do not fix b,
for a small enough interval (a, b], the intervals γ(a, b], . . . , γ m−1 (a, b] are disjoint from (a, b].
Since γ m fixes b and b is irrational, by Remark 6, γ m fixes pointwise an open neighborhood
of b. In particular, we can assume that (a, b] is fixed pointwise by γ m . Then the interval
(a, b] (and as such, the interval (a, b)) is weakly γ-wandering.
Now assume that γ ∈ T . Then the period m of b is equal to the order n of γ. Indeed,
γ being periodic means that every x ∈ S 1 has a finite orbit under the action of γ. But
an orientation preserving homeomorphism of S 1 cannot have two finite orbits of different
periods (see, for example, [KH95]). Thus, every orbit of hγi is of period m. That implies
that the order of γ is m. As noted above, γ(a, b], . . . , γ m−1 (a, b] are disjoint from (a, b].
Hence, (a, b] is a γ-wandering set.

Lemmas 11 and 12 imply the following corollaries. The proof of Corollary 14 is almost
identical to the proof of Corollary 13.
Corollary 13. Let γ ∈ V and let A ⊂ S 1 be a closed subset of S 1 such that A 6= S 1 . Then
there exists g ∈ V such that A is weakly γ g -wandering.
Proof. By Lemmas 11 and 12 there is an open weakly γ-wandering set U ⊆ S 1 . We can
assume that U = (a, b) is an open subinterval of S 1 with dyadic endpoints. Let (c, d) be an
open subinterval of S 1 with dyadic endpoints such that (c, d) ⊇ A. It follows from Remark
6 that there is an element g ∈ T such that g(a, b) = (c, d) ⊇ A. Since g(a, b) is weakly
γ g -wandering, the set A is also weakly γ g -wandering.

Corollary 14. Let γ ∈ T and let A ⊂ S 1 be a closed subset of S 1 such that A 6= S 1 . Then
there exists g ∈ V such that A is γ g -wandering.
Now, we are ready to prove that T and V are not invariably generated. We start with
the proof for T and then adapt it to a proof for Thompson group V .
Theorem 15. Thompson group T is not invariably generated.
Proof. Let In be a set of pairwise disjoint open subintervals of S 1 . Let Cn be the non-trivial
conjugacy classes of T . Let γn ∈ Cn be such that the complement Inc is γn -wandering (γn
exists by Corollary 14). Then for any k ∈ Z, if γnk 6= e then γnk (Inc ) ⊆ In . In particular,
for all i 6= n, γnk (Ii ) ⊆ In . Then by the ping-pong lemma (see, for example, [OS04, Lemma
2.1]) hγn : n ∈ Ni = ∗n∈N hγn i.


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TSACHIK GELANDER, GILI GOLAN, AND KATE JUSCHENKO

Theorem 16. Thompson group V is not invariably generated.
Proof. Let In be a set of pairwise disjoint open intervals converging to the point 0 ∈ S 1 .
We can assume that In ⊆ (0, 14 ) for all n. Let Cn be the non-trivial conjugacy classes of V .
Let γn ∈ Cn be such that the complement Inc is weakly γn -wandering (γn exists by Corollary
13). We claim that the orbit of 0 under the action of H = hγn : n ∈ Ni is contained in
[0, 14 ). Since V acts transitively on the set of finite dyadic fractions in S 1 , that would imply
that H 6= V , as required. Thus, the following lemma completes the proof of the theorem.
Lemma 17. Let α 6= 0 be a point in the orbit of 0 in S 1 under the action of H. Let g ∈ H
be an element of minimal word-length over the alphabet {γn : n ∈ N} such that g(0) = α.
Assume that
g = γik11 · · · γikmm
where k1 , . . . , km 6= 0; for each j = 1, . . . , m − 1, ij+1 6= ij ; and such that |k1 | + · · · + |km |
is the word length of g over {γn : n ∈ N}. Then α ∈ Iim .
Proof. We prove the lemma by induction on m. If m = 1 then g = γik11 . The set Iic1 is
weakly γi1 -wandering. Since γik11 (0) = α 6= 0 and 0 ∈ Iic1 , the set γik11 (Iic1 ) is disjoint from
Iic1 ; i.e., g(Iic1 ) = γik11 (Iic1 ) ⊆ Ii1 . Hence α ∈ Ii1 . Assume that the lemma holds for m and let
k

m+1
g = γik11 · · · γikmm γim+1
.

We let g1 = γik11 · · · γikmm and β = g1 (0). By the induction hypothesis, β ∈ Iim . Since
im 6= im+1 , β ∈ Iim ⊆ Iicm+1 . By the assumption of minimality of the word length of g,
km+1
km+1
α = g(0) = γim+1
(β) 6= β. Hence, γim+1
does not fix pointwise the interval Iicm+1 . Since
km+1 c
Iicm+1 is weakly γim+1 -wandering, that implies that γim+1
(Iim+1 ) ⊆ Iim+1 . Hence α ∈ Iim+1
as required.


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Hebrew University
E-mail address: gelander@math.huji.ac.il
Vanderbilt University
E-mail address: gili.golan@vanderbilt.edu
Northwestern University
E-mail address: kate.juschenko@gmail.com

