By the end of this section we will have proven Theorem 3.6, that the countable virtually free groups are demonstrable for V. However the main content of the section will be taken up with proving a smaller result.
Lemma 3.12. The modular group Γ∼=C2∗C3 is contained in the class DV,C2.
As the class of demonstrable groups is closed under passing to subgroups, the lemma implies that any countable free group admits a demonstrative embedding into V. In [3] the authors show that the class of demonstrable groups forV is closed under passage to finite index overgroups, which will then give us the main result.
Free products
We first formally define the free product of groups A∗B, basing our definition on the one given in Chapter 2 of [10]. Suppose A and B are groups such that A∩B = {1}. We define a reduced word over the set A∪B, to be a string g1g2g3. . . gn, where
gi∈(A∪B)\ {1} such that for all 1≤i≤n−1 the elements gi and gi+1 do not lie in
the same group. The reduced word of the identity element is represented by the empty string ε.
For the reduced wordsx=g1. . . gn and y=h1. . . hm we define the productx·y by
x·y=
g1. . . gnh1. . . hm ifgn andh1 are in opposite groups
g1. . . gn−1zh2. . . hm ifgn andh1 are both in the same group
and gnh1 =z6= 1
(g1. . . gn−1)·(h2. . . hm) ifgn andh1 are both in the same group
and gnh1 = 1.
Under the multiplication above the set of reduced words form a group. The inverse of the reduced wordx=g1g2. . . gn is the reduced wordx−1 =g−1n . . . g
−1 2 g
−1
1 . We call this
group the free product ofA and B and denote it by A∗B.
We introduce a function len:A∗B →N0 that takes a reduced word and outputs the number of elements in the string, that is, len(g1. . . gn) = n. Informally we call n the
length of the reduced word.
There also exists an alternative definition for the free product using group presentations.
A∗B has presentation
A∗B ∼=hXtY|RtSi.
3.3.1 Embedding Γ into V
The modular group Γ is the group of M¨obius transformations of the upper half complex plane which have the form
z7→ az+b cz+d
such that ad−bc = 1. For our purposes there are only two properties of Γ that are important. Firstly, that it is isomorphic to the free product of the cyclic group of order two and the cyclic group of order three, namely Γ∼=C2∗C3. Secondly, that it contains
the free group on two generators as a subgroup.
The cyclic group of order m is given by the presentation hx|xmi. Therefore the group
C2∗C3 has presentation hα, β|α2, β3i where subgroups hαi and hβi are isomorphic to
C2 and C3 respectively. The purpose of this section is to find a subgroup of V that is
isomorphic to C2∗C3. In the section following this we will show that this subgroup is
demonstrative.
We begin with two elements from Thompson’s group V, which we callaand b, defined by the two tree pairs below.
1 2 3 4 6 5 a 6 5 4 3 1 2
1 2 5 6 3 4 b 5 6 3 4 1 2
Figure 3.2: The tree pair representative (Db, σb,Rb) b∈V
We denote the tree pairs in Figure 3.1 and Figure 3.2 by (Da, σa,Ra) and (Db, σb,Rb)
respectively. Notice that for each tree pair the domain tree is identical to the range tree, that is Da=Ra and Db =Rb. Thus botha and b are of finite order, determined
by their respective permutations. The permutation on the leaves of (Da, σa,Ra) is the
product of transpositions σa = (1,6)(2,5)(3,4), thusσ2a = 1 and the order ofa is two.
The permutation on the leaves of (Db, σb,Rb) is the product of two disjoint three cycles
σb = (1,3,5)(2,4,6) and thus the order of bis three. Let G=ha, bi be the subgroup of
V generated by the elementsaand b.
Lemma 3.14. The subgroupG=ha, bi ≤V factors as hai ∗ hbi.
To prove Lemma 3.14 we use Fricke and Klein’s well known criterion, the Ping-Pong Lemma. The version we give here is based on the one found in [15].
Lemma 3.15(Ping-Pong Lemma). LetGbe a group acting on a setX and letAandB
be two subgroups ofGsuch that|A| ≥2 and|B| ≥3. Suppose there exist two non-empty
subsets XA and XB such that the following three conditions hold
1. XA6⊂XB
2. for all non-trivial a∈A, (XB)a⊂XA
3. for all non-trivial b∈B, (XA)b⊂XB.
Then hA, Bi ∼=A∗B.
Proof of Lemma 3.15. Let G be a group acting on a set X and let A and B be two
subgroups ofG that satisfy the properties described in the lemma. To prove the result we need to show that any reduced word in {A, B}∗ is non-trivial in hA, Bi. Consider the reduced word w1 =b1a1b2a2. . . bn in {A, B}∗ where ai ∈ A and bi ∈B. Then w1
acts on the setXAand we have the following sequence of containments;
As XA 6⊂ XB the element represented by w1 must have acted on XA in a non-trivial
way and thus cannot be trivial.
Letw2=a1b1a2b2. . . anbe a reduced word and considerb−1w2bwhereb∈B\{1}. Then
b−1w2bis a word in the same form asw1 and hence is non-trivial by the same argument.
Therefore w2 must also be non-trivial. Let w3 = a1b1a2b2. . . anbn be a reduced word
and considerc−1w3cwherec∈B\ {1, b−1n }. Thenc−1w3creduces to a word in the same
form asw1 and thus by the same argument as beforew3 must be non-trivial. Finally let
w4 =b1a1b2a2. . . bnan be a reduced word and consider d−1w4dwhere d ∈ B\ {1, b1}.
Thend−1w4dreduces to a word in the same form asw1 and thus by the same argument
as beforew4 must also be non-trivial. As every non-trivial reduced word in{A, B}∗ has
the same form as eitherw1, w2, w3 orw4 we conclude that they all represent non-trivial
elements inhA, Bi and hencehA, Bi ∼=A∗B.
Proof of Lemma 3.14. We use Lemma 3.15 to prove the result. First we identify the
subgroups hai and hbi with the groups A and B from the lemma. Define the subsets
XA=b111c and XB=b10c.
Immediately we see that XA 6⊂ XB and the first condition of the lemma is satisfied.
Now observe the action of aon the setXB;
(XB)a= (b10c)a=b11110c ⊂ b111c=XA.
This confirms the third requirement. In hbi there are two non-trivial elements, b and
b−1. Observe (XA)b = (b111c)b =b100c ⊂ b10c = XB, and (XB)b−1 = (b111c)b−1 =
b1011c ⊂ b10c =XB. Thus the last of the three conditions in Lemma 3.15 is met and
thusha, bi factors as hai ∗ hbi.
By Lemma 3.14 there exists a copy of the groupC2∗C3 ∼=hα, β|α2, β3i in V given by
the embedding α 7→ a and β 7→ b. We now go on to show that this subgroup inV is demonstrative.
3.3.2 Proving that C2∗C3 is demonstrable for V
In what follows we will prove that the subgroup G is demonstrative in V with demon- stration set b0c. AsG is the free product hai ∗ hbi, each elementg∈G can be written by a unique reduced word. The proof thatGis a demonstrative subgroup forV, begins
Proof. The proof will proceed on the length of reduced words in G.
Suppose g is a reduced word such that len(g) = 1. There are three options, namely a,
b and b−1. Supposeg=a, then b0cg=b11111c. Further, ifg =b, then b0cg =b1010c. Finally suppose g =b−1, then b0cg =b110c. Thus for all reduced words of length one inGthe lemma holds.
We now consider all elements inG with reduced length greater than one. Let P(n) be the statement
b0cg⊆
b111c, ifg ends with generatora
b10c, ifg ends with either of the generatorsb orb−1
for all reduced words g over{a} ∪ {b, b−1}, such that len(g) =n≥2.
We will proceed to prove by induction that P(n) holds for all n≥ 2. Suppose g ∈ G
such that len(g) = 2. There are four options, namely, ab,ab−1,ba and b−1a. Suppose
g=ab, then b0cg=b10011c ⊂ b10c. Suppose g =ab−1, then b0cg=b101111c ⊂ b10c. Suppose g = ba, then b0cg = b1111010c ⊂ b111c. Finally suppose g = b−1a, then
b0cg=b1110c ⊂ b111c. ThereforeP(2) holds.
SupposeP(k) is true for all reduced wordsg withlen(g) =ksuch that 2≤k≤n. Now suppose h is a reduced word such thatlen(h) =n+ 1. Leth0 be the prefix of length n
inh.
Suppose h0 ends with a, then as h is a reduced word there are two possibilities, either
h = h0b or h = h0b−1. As len(h0) = n, by our inductive assumption b0ch0 ⊆ b111c. Thus if h = h0b, then b0ch = b0ch0b ⊆ b111cb =b100c and P(n+ 1) is true. Suppose
h=h0b−1, thenb0ch=b0ch0b−1 ⊆ b111cb−1 =b1011c and againP(n+ 1) is true. Suppose instead thath0 ends with eitherb orb−1. Then there is only one possibility for
h, namely h = h0a. As len(h0) = n, by our inductive assumption b0ch0 ⊆ b10c. Thus
b0ch=b0ch0a⊆ b10ca=b11110c and P(n+ 1) is true.
Therefore for all reduced words h such that len(h) =n+ 1, the statement P(n+ 1) is true, and therefore by induction P(n) must be true for all reduced wordsg such that
len(g)≥2. Therefore, the lemma holds for all elements ofG. Therefore, by Lemma 3.11, we have the following corollary.
Corollary 3.17. The subgroup Gis a demonstrative subgroup of V
Therefore this proves Lemma 3.12, that the modular group Γ∼=C2∗C3is a demonstrable
3.3.3 Proof of Theorem 3.6
Bleak and Salazar-D´ıaz observe the following in Lemma 3.2 [9].
Observation 3.18 (Bleak, Salazar-D´ıaz). Suppose that G is a demonstrative group
with m serving as a demonstration node. Then given any subgroup H ≤G, H is also
demonstrative with demonstration node m.
This observation, together with the fact that the free group on two generators F2 is
isomorphic to the subgrouph[a, b],[a, b−1]i ≤G (where the bracket [x, y] represents the commutatorx−1y−1xy), implies the following corollary.
Corollary 3.19. The free group on two generators, F2, is in the class D(V,C2)
Virtually free groups are groups that contain a free group as a finite index subgroup. While it is known (see [26, 9]) that if a group G embeds in V, then any finite index over-group ofGalso embeds intoV, the paper of Berns-Zieve et al [3], extends this with Theorem 3.3, which we paraphrase below.
Theorem 3.20 (Theorem 3.3 in Berns-Zieve et al [3]). Suppose G is a group which
embeds in R. Thompson’s group V. If G ≤ H where [H : G] = m, for some m ∈ N
and G embeds as a demonstrative subgroup inV, then H also embeds as demonstrative
subgroup of V.
The theorem tells us that D(V,C2) is closed under taking finite index overgroups. As all countable free groups embed into F2 and since virtually free groups are, by definition,
finite index overgroups of free groups, by Corollary 3.19 countable virtually free groups are contained withinD(V,C2). This proves Theorem 3.6.