The R∞ property for Houghton's groups
We study twisted conjugacy classes of a family of groups which are called Houghton's groups Hn (n∈N), the group of translations of n rays of discrete points at infinity. We prove that the Houghton's groups Hn have the R∞ property for all n∈N.
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Цитувати: | The R∞ property for Houghton's groups / Jang Hyun Jo, Jong Bum Lee, Sang Rae Lee // Algebra and Discrete Mathematics. — 2017. — Vol. 23, № 2. — С. 249–262. — Бібліогр.: 24 назв. — англ. |
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irk-123456789-1560162019-09-02T19:05:03Z The R∞ property for Houghton's groups Jang Hyun Jo Jong Bum Lee Sang Rae Lee We study twisted conjugacy classes of a family of groups which are called Houghton's groups Hn (n∈N), the group of translations of n rays of discrete points at infinity. We prove that the Houghton's groups Hn have the R∞ property for all n∈N. 2017 Article The R∞ property for Houghton's groups / Jang Hyun Jo, Jong Bum Lee, Sang Rae Lee // Algebra and Discrete Mathematics. — 2017. — Vol. 23, № 2. — С. 249–262. — Бібліогр.: 24 назв. — англ. 1726-3255 2010 MSC:20E45, 20E36, 55M20. http://dspace.nbuv.gov.ua/handle/123456789/156016 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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We study twisted conjugacy classes of a family of groups which are called Houghton's groups Hn (n∈N), the group of translations of n rays of discrete points at infinity. We prove that the Houghton's groups Hn have the R∞ property for all n∈N. |
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Jang Hyun Jo Jong Bum Lee Sang Rae Lee |
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Jang Hyun Jo Jong Bum Lee Sang Rae Lee The R∞ property for Houghton's groups Algebra and Discrete Mathematics |
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Jang Hyun Jo Jong Bum Lee Sang Rae Lee |
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Jang Hyun Jo |
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The R∞ property for Houghton's groups |
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The R∞ property for Houghton's groups |
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The R∞ property for Houghton's groups |
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The R∞ property for Houghton's groups |
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The R∞ property for Houghton's groups |
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r∞ property for houghton's groups |
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Інститут прикладної математики і механіки НАН України |
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2017 |
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http://dspace.nbuv.gov.ua/handle/123456789/156016 |
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The R∞ property for Houghton's groups / Jang Hyun Jo, Jong Bum Lee, Sang Rae Lee // Algebra and Discrete Mathematics. — 2017. — Vol. 23, № 2. — С. 249–262. — Бібліогр.: 24 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT janghyunjo therpropertyforhoughtonsgroups AT jongbumlee therpropertyforhoughtonsgroups AT sangraelee therpropertyforhoughtonsgroups AT janghyunjo rpropertyforhoughtonsgroups AT jongbumlee rpropertyforhoughtonsgroups AT sangraelee rpropertyforhoughtonsgroups |
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2025-07-14T08:16:34Z |
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2025-07-14T08:16:34Z |
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1837610075461517312 |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 23 (2017). Number 2, pp. 249–262
c© Journal “Algebra and Discrete Mathematics”
The R∞ property for Houghton’s groups
Jang Hyun Jo, Jong Bum Lee∗ and Sang Rae Lee
Communicated by R. I. Grigorchuk
Abstract. We study twisted conjugacy classes of a family of
groups which are called Houghton’s groups Hn (n ∈ N), the group
of translations of n rays of discrete points at infinity. We prove that
the Houghton’s groups Hn have the R∞ property for all n ∈ N.
Introduction
Let G be a group and ϕ : G → G be a group endomorphism. We
define an equivalence relation ∼ on G, called the Reidemeister action
by ϕ, by
a ∼ b⇔ b = haϕ(h)−1 for some h ∈ G.
The equivalence classes are called twisted conjugacy classes or Reidemeis-
ter classes and R[ϕ] denotes the set of twisted conjugacy classes. The
Reidemeister number R(ϕ) of ϕ is defined to be the cardinality of R[ϕ].
We say that G has the R∞ property if R(ϕ) =∞ for every automorphism
ϕ : G→ G.
In 1994, Fel’shtyn and Hill [10] conjectured that any injective endo-
morphism ϕ of a finitely generated group G with exponential growth
would satisfy that R(ϕ) =∞. Levitt and Lustig ([23]), and Fel’shtyn ([8])
showed that the conjecture holds for automorphisms when G is Gromov
∗The second author was supported by Basic Science Research Program through the
National Research Foundation of Korea(NRF) funded by the Ministry of Education
(No. 2013R1A1A2058693).
2010 MSC: 20E45, 20E36, 55M20.
Key words and phrases: Houghton’s group, R∞ property, Reidemeister number.
250 The R∞ property for Houghton’s groups
hyperbolic. However, in 2003, the conjecture was answered negatively by
Gonçalves and Wong [15] who gave examples of finitely generated groups
with exponential growth which do not have the R∞ property. Since then,
groups with the R∞ property have been known including Baumslag-Solitar
groups, lamplighter groups, Thompson’s groups F and T , Grigorchuk
group, mapping class groups, relatively hyperbolic groups, and some linear
groups (see [2, 3, 6, 9, 11–14, 16, 17, 21, 24] and references therein). For a
topological consequence of the R∞ property, see [16,21,24]. In this article
we show the following.
Theorem 1. The Houghton’s groups Hn have the R∞ property for all
n ∈ N.
It is shown that the conjugacy problem([1]) and the twisted conjugacy
problem([5]) of Hn are solvable for n > 2. In 2010, Gonçalves and Ko-
chloukova [11] proved that there is a finite index subgroup H of Aut(Hn)
such that R(ϕ) =∞ for ϕ ∈ H provided n > 2. Recently the structure
of Aut(Hn) is known from [4] (see Theorem 3 below). In [14], Gonçalves
and Sankaran have studied also the R∞ property of Houghton’s groups.
In this paper we use simple but useful observations of the Reidmeister
numbers and the structure of Aut(Hn) to find equivalent conditions for
two elements of Hn to determine the same twisted conjugacy class under
mild assumptions. In Section 1, we will review definition and some facts
about Houghton’s groups Hn which are necessary mainly to the study of
Reidemeister numbers for Hn. In Section 2, we prove our main result for
n > 2. The case of n = 1 is discussed in Section 3.
1. Houghton’s groups Hn
In this paper we use the following notational conventions. All bijections
(or permutations) act on the right unless otherwise specified. Consequently
gh means g followed by h. The conjugation by g is denoted by µ(g), hg =
g−1hg =: µ(g)(h), and the commutator is defined by [g, h] = ghg−1h−1.
Our basic references are [19,22] for Houghton’s groups and [4] for their
automorphism groups. Fix an integer n > 1. For each k with 1 6 k 6 n,
let
Rk =
{
meiθ ∈ C | m ∈ N, θ = π
2
+ (k − 1)2π
n
}
and let Xn =
⋃n
k=1 Rk be the disjoint union of n copies of N, each arranged
along a ray emanating from the origin in the plane. We shall use the
J. H. Jo, J. B. Lee, S. R. Lee 251
notation {1, · · · , n}×N for Xn, letting (k, p) denote the point of Rk with
distance p from the origin.
A bijection g : Xn → Xn is called an eventual translation if the
following holds:
There exist an n-tuple (m1, · · · , mn) ∈ Z
n and a finite set
Kg ⊂ Xn such that
(k, p) · g := (k, p + mk) ∀(k, p) ∈ Xn −Kg.
An eventual translation acts as a translation on each ray outside a finite
set. For each n ∈ N the Houghton’s group Hn is defined to be the group
of all eventual translations of Xn.
Let gi be the translation on the ray of R1∪Ri+1 by 1 for 1 6 i 6 n−1.
Namely,
(j, p) · gi =
(1, p− 1) if j = 1 and p > 2,
(i + 1, 1) if (j, p) = (1, 1),
(i + 1, p + 1) if j = i + 1,
(j, p) otherwise.
Figure 1. Some examples of Hn.
Figure 1 illustrates some examples of elements of Hn, where points
which do not involve arrows are meant to be fixed. Finite sets Kgi
and Kgj
are singleton sets. The commutator [gi, gj ] of two distinct elements gi and
gj is the transposition exchanging (1, 1) and (1, 2). We will denote this
transposition by α. The last element g is rather generic and Kg consists
of eight points. Johnson provided a finite presentation for H3 in [20] and
the third author gave a finite presentation for Hn with n > 3 in [22] as
follows:
252 The R∞ property for Houghton’s groups
Theorem 2 ([22, Theorem C]). For n > 3, Hn is generated by
g1, · · · , gn−1, α
with relations
α2 = 1, (ααg1)3 = 1, [α, αg2
1 ] = 1, α = [gi, gj ], αg−1
i = αg−1
j
for 1 6 i 6= j 6 n− 1.
From the definition of Houghton’s groups, the assignment g ∈ Hn 7→
(m1, · · · , mn) ∈ Z
n defines a homomorphism π = (π1, · · · , πn) : Hn → Z
n.
Then we have:
Lemma 1 ([22, Lemma 2.3]). For n > 3, we have ker π = [Hn,Hn].
Note that π(gi) ∈ Z
n has only two nonzero values −1 and 1,
π(gi) = (−1, 0, · · · , 0, 1, 0, · · · , 0)
where 1 occurs in the (i + 1)st component. Since the image of Hn under
π is generated by those elements, we have that
π(Hn) =
{
(m1, · · · , mn) ∈ Z
n |
n
∑
i=1
mi = 0
}
,
which is isomorphic to the free Abelian group of rank n−1. Consequently,
Hn (n > 3) fits in the following short exact sequence
1 −→ H′
n = [Hn,Hn] −→ Hn
π
−→ Z
n−1 −→ 1.
The above abelianization, first observed by C. H. Houghton in [19], is the
characteristic property of {Hn} for which he introduced those groups in
the same paper. We may regard π as a homomorphism Hn → Z
n → Z
n−1
given by
π : gi 7→ (−1, 0, · · · , 0, 1, 0, · · · , 0) 7→ (0, · · · , 0, 1, 0, · · · , 0).
In particular, π(g1), · · · , π(gn−1) form a set of free generators for Z
n−1.
By definition, H1 is the symmetric group itself on X1 with finite
support, which is not finitely generated. Furthermore, H2 is
H2 = 〈g1, α | α2 = 1, (ααg1)3 = 1, [α, αgk
1 ] = 1 for all |k| > 1〉,
which is finitely generated, but not finitely presented. It is not difficult to
see that H′
2 = FAlt2.
J. H. Jo, J. B. Lee, S. R. Lee 253
From now on we use the following notations:
• Symn is the full symmetric group of Xn;
• FSymn is the symmetric group of Xn with finite support;
• FAltn is the alternating group of Xn with finite support.
For each n the group FAltn can be seen as the kernel of the sign
homomorphism FSymn → {±1}. The following fact is necessary for our
discussion, see [7].
Remark 1. For any σ ∈ Symn, the conjugation by σ induces auto-
morphisms µ(σ) : FSymn → FSymn and µ(σ) : FAltn → FAltn. Then
µ : Symn → Aut(FAltn) and µ : Symn → Aut(FSymn) are isomorphisms.
Every automorphism of Hn restricts to an automorphism of the cha-
racteristic subgroup H′′
n = [FSymn, FSymn] = FAltn, which induces a
homomorphism res : Aut(Hn) → Aut(FAltn). One can show this map
is injective by using the fact that FAltn is generated by 3-cycles. The
embedding
Res : Aut(Hn)
res
−→ Aut(FAltn)
µ−1
−→ Symn
implies that each automorphism of Hn is given by a conjugation of
an element in Symn. Moreover the composition preserves the normality
Hn = Inn(Hn) � Aut(Hn).
Proposition 1 ([4, Proposition 2.1]). For n > 1, the automorphism
group Aut(Hn) is isomorphic to the normalizer of Hn in the group Symn.
We need an explicit description for the normalizer NSymn
(Hn) to study
Aut(Hn). Consider an element σij ∈ Symn for 1 6 i 6= j 6 n defined by
(ℓ, p) · σij =
(j, p) if ℓ = i
(i, p) if ℓ = j
(ℓ, p) otherwise
for all p ∈ N. Each element σij defines a transposition on n rays iso-
metrically. The subgroup of Symn generated by all σij is isomorphic to
the symmetric group Σn on the n rays. Note that Σn acts on Hn by
conjugation. One can show that NSymn
(Hn) coincides with Hn ⋊ Σn by
using the ray structure (end structure) of the underlying set Xn. An
eventual translation g preserves each ray up to a finite set. Let R∗
i denote
the set of all points of Ri but finitely many. It is not difficult to see that
if φ ∈ Symn normalizes Hn then
(R∗
i )φ = R∗
j
254 The R∞ property for Houghton’s groups
for 1 6 i, j 6 n. Thus φ defines an element σ of Σn, and we see that
φσ−1 ∈ Hn since (R∗
i )φσ−1 = (R∗
j )σ−1 = R∗
i for each i. Consequently,
NSymn
(Hn) has the internal semidirect product of Hn by Σn. Therefore
we have:
Theorem 3 ([4, Theorem 2.2]). For n > 2, we have
Aut(Hn) ∼= Hn ⋊ Σn
where Σn is the symmetric group that permutes n rays isometrically.
2. The R∞ property for Hn, n > 2
We consider the Houghton’s groups Hn with n > 2. Let φ be an auto-
morphism on Hn. Remark that, when n > 3, φ induces an automorphism
φ′ on the commutator subgroup H′
n = FSymn and an automorphism φ̄
on Z
n−1 so that the following diagram is commutative:
1 −−−−→ FSymn
i
−−−−→ Hn
π
−−−−→ Z
n−1 −−−−→ 1
y
φ′
y
φ
yφ̄
1 −−−−→ FSymn
i
−−−−→ Hn
π
−−−−→ Z
n−1 −−−−→ 1
But when n = 2, H′
2 = FAlt2 and H2/H′
2 = Z ⊕ Z2. Since FSym2 is a
normal subgroup of H2, we have the following commutative diagram
1 1
x
x
Z
=
−−−−→ Z
x
x
1 −−−−→ FAlt2 −−−−→ H2 −−−−→ Z⊕ Z2 −−−−→ 1
x
=
x
x
1 −−−−→ FAlt2 −−−−→ FSym2 −−−−→ Z2 −−−−→ 1
x
x
1 1
Let φ ∈ Aut(H2). Then φ restricts to an element φ′ of Aut(H′
2) =
Aut(FAlt2) = Aut(FSym2), and hence induces an automorphism φ̄ on Z
J. H. Jo, J. B. Lee, S. R. Lee 255
so that the following diagram is commutative
1 −−−−→ FSym2 −−−−→ H2 −−−−→ Z −−−−→ 1
y
φ′
y
φ
yφ̄
1 −−−−→ FSym2 −−−−→ H2 −−−−→ Z −−−−→ 1
These diagrams induce an exact sequence of Reidemeister sets
R[φ′]
î
−→ R[φ]
π̂
−→ R[φ̄] −→ 1.
Because π̂ is surjective, we have that if R(φ̄) = ∞, then R(φ) = ∞.
Consequently, we have
Lemma 2. Let φ be an automorphism on Hn, (n > 2). If R(φ̄) = ∞,
then R(φ) =∞.
By Theorem 3, φ = µ(γσ) for some γ ∈ Hn and σ ∈ Σn. First, we
will show that when φ = µ(σ) for σ ∈ Σn the Reidemeister number of φ
is infinity. When σ = id, φ and hence φ̄ are identities. It is easy to see
from definition that R(φ̄) = R(id) =∞, and so R(φ) =∞.
One useful observation in calculating R(µ(σ)) is that a product σ =
σ1σ2 induces a bijection
R[µ(σ1)]←→ R[µ(σ)], (1)
which follows from
b = hah̄σ1 ⇔ bσ2 = h(aσ2)h̄σ1σ2
for all a, b, h ∈ Hn. Note that any product for σ induces a bijection
between the twist conjugacy classes of σ and of the first term in the
product. Recall that a cycle decomposition of a permutation σ allows one
to write σ as a product of disjoint cycles. Since disjoint cycles commute
there exists a bijection between R[µ(σ)] and R[µ(σ1)] for any cycle σ1 in
a cycle decomposition of σ. The following observation plays a crucial role
in the sequel.
Remark 2. For a cycle σ1 in a cycle decomposition of σ ∈ Σn, we have
that R(µ(σ1)) =∞ if and only if R(µ(σ)) =∞.
Recall that the cycle type of a permutation τ ∈ FSymn encodes the
data of how many cycles of each length are present in a cycle decomposition
256 The R∞ property for Houghton’s groups
of τ . Note that two permutations τ and τ ′ have the same cycle type if and
only if they are conjugate in FSymn. In particular two cycles determine
the same conjugacy class if and only if they have the same length. We
extend this to establish a criterion for twisted conjugacy classes of cycles
with respect to an automorphism φ = µ(σ) when σ ∈ Σn is a cycle.
Lemma 3. Suppose σ 6= id ∈ Σn is a cycle and n > 2. A pair of cycles
τ and τ ′ on the same ray determine the same twisted conjugacy class of
φ = µ(σ) if and only if they have the equal length. In particular R(φ) =∞.
Proof. Suppose that τ and τ ′ are cycles on the same ray of the equal
length. We first consider the case when σ permutes rays as an ℓ-cycle
(1 2 · · · ℓ) for some 2 6 ℓ 6 n, and τ and τ ′ are disjoint cycles on R1. Two
cycles τ and τ ′ can be written as
τ = (p1 · · · pm) and τ ′ = (q1 · · · qm)
(by suppressing the ray notation) where m > 2. We need to find an
element h ∈ Hn such that τ ′ = hτµ(σ)(h)−1, or equivalently
hσ = τ ′−1hτ. (2)
Let h1 be the 2m-cycle on R1 given by
h1 = (p1q1 p2 q2 · · · pm qm).
It is direct to check that
τ ′−1h1τ = (qm · · · q1)(p1q1 p2 q2 · · · pm qm)(p1 · · · pm) = h1. (3)
Consider h ∈ H′
n defined by
h = hσℓ−1
1 · · ·hσ
1 h1.
Note that h is a product of ℓ disjoint 2m-cycles each of which is an
‘isometric translation’ of h1 to the ray Rℓ, · · · , R2, R1. More precisely
(k + 1, p)hσk
1 = (1, p)h1σk for all (1, p) ∈ supp(h1) and k = 1, · · · , ℓ − 1.
One crucial observation is that
hσ = h.
The above follows from that σ is a ℓ-cycle and that components of h have
pairwise disjoint supports. Moreover, τ ′ commutes with hσℓ−1
1 · · ·hσ
1 , so
we have
hσ = h = hσℓ−1
1 · · ·hσ
1 h1 = hσℓ−1
1 · · ·hσ
1 (τ ′−1h1τ)
= τ ′−1(hσℓ−1
1 · · ·hσ
1 h1)τ = τ ′−1hτ.
Therefore h satisfies the condition (2), and hence [τ ] = [τ ′] in R[µ(σ)].
J. H. Jo, J. B. Lee, S. R. Lee 257
Applying appropriate conjugations one can extend the above obser-
vations to show that [τ ] = [τ ′] in R[µ(σ)] for any cycle σ ∈ Σn and for
any two disjoint cycles τ and τ ′ on the same ray with the equal length.
Therefore, by the transitivity of the class, we can see that two cycles (not
necessarily disjoint) on a ray belong to the same class for φ = µ(σ) as
long as they have the same length. Indeed, if two m-cycles τ and τ ′ are
not disjoint, one takes another m-cycle τ0 which is disjoint with τ and τ ′
to have [τ ] = [τ0] = [τ ′]. Thus we are done with one direction.
For the converse, suppose there exists h ∈ Hn satisfying the condition
(2) for a cycle σ ∈ Σn even when cycles τ and τ ′ on the same ray have
different lengths m and m′ respectively. Assume m′ > m. Let ℓ be the
order of σ. Applying the identity (2) ℓ times, we have
h = hσℓ
= (τ ′−1)σℓ−1
· · · (τ ′−1)στ ′−1hττσ · · · τσℓ−1
. (4)
Let c′ = (τ ′−1)σℓ−1
· · · (τ ′−1)στ ′−1 and c = ττσ · · · τσℓ−1
be the products
of first and last ℓ terms on the RHS of (4). Note that each component of
c′ is an ‘isometric translation’ of τ ′−1 to different ℓ rays (and similarly
for each component of c). To draw a contradiction, we use the fact that
the size of supp(c′) is strictly greater than that of supp(c). For details we
need to examine how h = c′hc acts on supp(c′). Being a disjoint union,
supp(c′) =
⋃
06k6ℓ−1(supp(τ ′))σk, supp(c′) has size ℓ×m′, while supp(c)
has size ℓ×m. For each P ∈ supp(c′), we have
(P )h = (P )c′hc = (P ′)hc or (P )hc−1 = (P ′)h
where P ′ is a point in the same ray of P but distinct from P . We claim that
(P )h belongs to supp(c). Otherwise c−1 fixes (P )h, forcing (P )h = (P ′)h.
Since P ∈ supp(c′) was arbitrary, a bijection h maps supp(c′) to supp(c).
We conclude that there does not exists h ∈ Hn satisfying the condition
(2) for cycles τ and τ ′ on the same ray with different lengths.
Theorem 4. The Houghton’s groups Hn have the R∞ property for all
n > 2.
Proof. Theorem 3 says that an automorphism φ of Hn is determined by
φ = µ(gσ) for some g ∈ Hn and σ ∈ Σn. As we noted earlier, we may
assume that σ 6= 1. Note that
gσ = σ(σ−1gσ) = σg′
with g′ ∈ Hn. The product in RHS yields a bijection between R[µ(gσ)]
and R[µ(σ)] as in (1). Consider a cycle σ1 in a cycle decomposition of σ.
258 The R∞ property for Houghton’s groups
Remark 2 together with Lemma 3 implies R[µ(σ)] = R[µ(σ1)] = ∞.
Therefore we have R[φ] = R[µ(gσ)] = R[µ(σ)] =∞ for all φ ∈ Aut(Hn)
when n > 2.
We remark that Lemma 2 can be used extensively to establish The-
orem 4. As observed in commuting diagrams above an automorphism
φ = µ(gσ) of Hn induces an automorphism φ on the abelianization Z
n−1,
which is freely generated by π(g1), . . . , π(gn−1). Since µ(g) fixes the gene-
rates g1, . . . , gn−1, se see that φ = µ(σ). The Reidemeister number of an
automorphism on Z
n−1 (n > 2) is well understood. By [18, Theorem 6.11],
R(φ) = ∞ if and only if φ has eigenvalue 1. By using induction on n
one can show that φ = µ(σ) has eigenvalue 1 unless σ is an n-cycle on
the rays R1, . . . , Rn. Now Lemma 3 implies that R(µ(σ)) = ∞, and so
R(φ) = R(φ) =∞.
3. The group H1 and its R∞ property
In this section, we will study the R∞ property for the group H1.
We remark that H1 = FSym1 is generated by the transpositions exchan-
ging two consecutive points of R1. Let φ be an automorphism of H1.
Since Aut(H1) = Aut(FSym1) ∼= Sym1, we have that φ = µ(γ) for some
γ ∈ Sym1.
Lemma 4. Let ϕ : G→ G be an endomorphism. Then for any g ∈ G we
have [g] = [ϕ(g)] in R[ϕ].
Proof. The Lemma follows from
ϕ(g) = (g−1)gϕ(g−1)−1.
An infinite cycle γ ∈ Sym1 is given by a bijection γ : Z → R1.
For convenience we use the 1-to-1 correspondence to denote points of
supp(γ) ⊂ R1 by integers, that is, each point of supp(γ) is denoted by its
preimage. With this notation, each infinite cycle can be realized as the
translation on Z by +1. Remark that if h ∈ FSym1 with supp(h) ⊂ supp(γ)
then the conjugation µ(γ) shifts supp(h) to supp(hγ) by +1;
(k)h = k′ ⇔ (k + 1)hγ = k′ + 1 (5)
for all k ∈ supp(h). We say that an infinite cycle γ conjugates a permu-
tation τ ∈ FSym1 to τ ′ if τ ′ can be written as a conjugation of τ by a
power of γ.
J. H. Jo, J. B. Lee, S. R. Lee 259
Lemma 5. For an infinite cycle γ ∈ Sym1, two transpositions τ and
τ ′ with supp(τ) ⊂ supp(γ) and supp(τ ′) ⊂ supp(γ) determine the same
conjugacy class for φ = µ(γ) if and only if γ conjugates τ to τ ′. In
particular R(µ(γ)) =∞.
Proof. Assume that τ ′ = τγm
or τ ′ = φm(τ), for some m. By Lemma 4,
we have [τ ] = [φ(τ)] = · · · = [φm(τ)] = [τ ′].
For the converse, suppose that there exists h ∈ FSym1 satisfying
hγ = τ ′−1hτ = τ ′hτ (6)
for two transpositions τ and τ ′ with the condition on their supports, one
of which γ does not conjugate to the other. By the shift (5), they can be
written as τ = (0 ℓ) and τ ′ = (m m+ℓ′) for some m > 0 and ℓ 6= ℓ′ > 0.
By Lemma 4, which implies [(0 ℓ′)] = [(m m+ℓ′)] for all m ∈ Z, we may
further assume that τ ′ = (0 ℓ′) and ℓ < ℓ′.
We first claim that (−1)h = −1. If −1 ∈ supp(h), the identity (6) says
(−1)hγ = (−1)τ ′hτ = (−1)hτ 6= −1
since τ and τ ′ fix all negative integers. So −1 ∈ supp(hγ). Now the shift
k ∈ supp(h) ⇔ k + 1 ∈ supp(hγ)
implies −2 ∈ supp(h). Observe that the same argument establishes simul-
taneous induction on k for
−k ∈ supp(h) and − k ∈ supp(hγ)
for all positive k with the above base cases when k = 1. This means that
supp(h) must contain all negative integers. It contradicts that h ∈ FSym1.
Therefore h fixes −1, or equivalently hγ fixes 0. One can also show h fixes
ℓ′ + 1 by verifying
ℓ′ + k ∈ supp(h) and ℓ + k ∈ supp(hγ)
for all positive k if we are given the base case ℓ′ + 1 ∈ supp(h) (and
ℓ′ + 1 ∈ supp(hγ), which follows immediately by (6)). So we also have
(ℓ′ + 1)h = (ℓ′ + 1), and hence (ℓ′ + 1)hγ = (ℓ′ + 1) by (6).
From the fixed point 0 = (0)hγ we have
(0)τ ′hτ = 0 ⇔ (ℓ′)h = ℓ.
The shift (5) says ℓ′ + 1 ∈ supp(hγ). However this contradicts that
(ℓ′ + 1)hγ = (ℓ′ + 1). Therefore τ ′ = (0 ℓ′) does not belong to the class of
τ = (0 ℓ) unless ℓ = ℓ′.
260 The R∞ property for Houghton’s groups
Lemma 6. Suppose two permutations τ, τ ′ ∈ FSym1 are disjoint with a
permutation γ ∈ FSym1. Then τ and τ ′ belong to the same class in R[µ(γ)]
if and only if they have the same cycle type. In particular R(µ(γ)) =∞.
Proof. The statement follows from cycle type criterion for usual conjugacy
classes of the symmetric group on the fixed points of γ ∈ FSym1. Any
permutations on R′
1 = R1 \ supp(γ) with finite supports are conjugate if
and only if they have the same cycle type. For two permutations τ and
τ ′ on R′
1 there exists a permutation h ∈ FSym1 on R′
1 such that
τ ′ = hτh−1
if and only if τ and τ ′ have the same cycle type. Since hγ = h one can
replace h−1 by (h−1)γ in the identity to establish τ ′ = hτ(h−1)γ .
Theorem 5. The group H1 has the R∞ property.
Proof. Recall Aut(H1) = Aut(FSym1) ∼= Sym1. Each automorphism φ is
given by φ = µ(γ) for some γ ∈ Sym1. Consider the orbits of supp(γ) to
form a partition of supp(γ). Observe that γ restricts to a cycle on each
orbit. Thus we see that a cycle decomposition of γ is well defined and so
γ can be expressed as a product of commuting cycles. If γ has an infinite
orbit then it contains an infinite cycle γ1 so that γ can be written as
γ = γ1γ2. (7)
We have a bijection R[µ(γ1)] ↔ R[µ(γ)] from Remark 2. By Lemma 5,
we know that R[µ(γ1)] = ∞, and hence R[µ(γ)] = ∞. If all orbits of γ
are finite then we can express γ as a product (7) with a finite cycle γ1.
From Lemma 6, we see that R[µ(γ1)] = ∞, and so R[µ(γ)] = ∞ due
to the same bijection as above. We have proved that R[φ] = ∞ for all
automorphisms of H1.
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262 The R∞ property for Houghton’s groups
Contact information
Jang Hyun Jo,
Jong Bum Lee
Department of Mathematics,
Sogang University,
Seoul 04107, Korea
E-Mail(s): jhjo@sogang.ac.kr,
jlee@sogang.ac.kr
Web-page(s): http://maths.sogang.ac.kr/jhjo/,
http://maths.sogang.ac.kr/jlee/
Sang Rae Lee Department of Mathematics,
Texas A&M University,
College Station, Texas 77843, USA
E-Mail(s): srlee@tamu.math.edu
Web-page(s): http://www.math.tamu.edu/∼srlee/
Received by the editors: 15.05.2017.
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