arXiv:1607.03790v1 [] 13 Jul 2016

ON THE FINITENESS OF THE CLASSIFYING SPACE FOR THE FAMILY OF
VIRTUALLY CYCLIC SUBGROUPS
TIMM VON PUTTKAMER AND XIAOLEI WU

Abstract. Given a group G, we consider its classifying space for the family of virtually
cyclic subgroups. We show for many groups, including for example, one-relator groups,
acylindrically hyperbolic groups, 3-manifold groups and CAT(0) cube groups, that they do
not admit a finite model for this classifying space unless they are virtually cyclic. This
settles a conjecture due to Juan-Pineda and Leary for these classes of groups.

Introduction
Given a group G, we denote by EG = EVCY (G) a G-CW-model for the classifying space
for the family of virtually cyclic subgroups. The space EG is characterized by the property
that the fixed point set EG H is non-empty and contractible for all virtually cyclic subgroups
H of G and empty otherwise. In the formulation of the Farrell–Jones Conjecture EG plays
an important role (see for example [7, 19] for more information). Due to this, there has
been a growing interest in studying EG, see for example [6, 13, 15, 17, 18]. Recall that a
G-CW-complex X is said to be finite if it has finitely many orbits of cells. Similarly, X is
said to be of finite type if it has finitely many orbits of cells of dimension n for any n. In
[13, Conjecture 1], Juan-Pineda and Leary formulated the following conjecture:
Conjecture A. [13, Juan-Pineda and Leary] Let G be a group admitting a finite model for
EG. Then G is virtually cyclic.
The conjecture may be surprising at the beginning as there are many examples of groups
with a finite model for the classifying space for the family consisting only of the trivial
subgroup or for the family of finite subgroups. Juan-Pineda and Leary demonstrated the
validity of the conjecture for hyperbolic groups [13, Corollary 12]. Later, Kochloukova,
Martı́nez-Pérez and Nucinkis verified the conjecture for elementary amenable groups [15].
Groves and Wilson gave a simplified proof for elementary amenable groups [10]. So far,
almost all the proofs boil down to analyzing whether EG has a finite 0-skeleton. It turns
out that having a finite 0-skeleton for EG is equivalent to the following purely algebraic
condition (see 1.3)
Date: July, 2016.
2010 Mathematics Subject Classification. 55R35, 20B07.
Key words and phrases. Classifying space, virtually cyclic subgroup, acylindrically hyperbolic groups, 3manifold groups, HNN-extension, one-relator groups, CAT(0) cube groups.
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TIMM VON PUTTKAMER AND XIAOLEI WU

(BVC) G has a finite set of virtually cyclic subgroups {V1 , V2 , . . . , Vn } such that every
virtually cyclic subgroup of G is conjugate to a subgroup of some Vi .
Following Groves and Wilson, we shall call this property BVC and the finite set {V1 , V2 ,
. . . , Vn } a witness to BVC for G. In this paper, we give a systematic study of the property
BVC. Our main theorem can be stated as follows
Theorem. The following classes of groups satisfy Conjecture A:
(a) HNN extensions of free groups of finite rank (2.11),
(b) one-relator groups (2.12),
(c) acylindrically hyperbolic groups (3.2),
(d) 3-manifold groups (3.6),
(e) CAT(0) groups that contain a rank-one isometry or Z2 (4.13), in particular CAT(0)
cube groups (4.16).
In fact, we show that any finitely generated group in these classes has BVC if and only
if it is virtually cyclic. Our result also suggests that the following conjecture could be true.
Conjecture B. Let G be a finitely presented group which has BVC. Then G is virtually
cyclic.
Remark.

(a) The assumption of having a finitely presented group is necessary here

since Osin [25] has constructed finitely generated torsion-free groups with exactly
two conjugacy classes. In particular these groups have BVC.
(b) We do not know whether Conjecture B holds for all elementary amenable groups.
Groves and Wilson showed that solvable groups satisfy it [10].
(c) If we know the Flat Closing Conjecture, then it would follow that any CAT(0) group
satisfies Conjecture B. See 4.14 for more information.
Our paper is organized as follows. In Section 1, we first summarize what we already
know about groups admitting a finite model for EG, then we study basic properties of
groups with BVC and deduce that many groups cannot have this property. In section 2,
we study HNN extension of groups and show for example that any HNN extension of a
finitely generated free group does not have BVC. Using this, we prove that any non-cyclic
one-relator group does not have BVC. In section 3, we show that acylindrically hyperbolic
groups and finitely generated non virtually cyclic 3-manifold groups do not have BVC. In
the last section, we study groups acting on CAT(0) spaces. We show for example, that
CAT(0) cube groups do not have BVC unless they are virtually cyclic.
Results of this paper will also appear as a part of the first author’s thesis.
Acknowledgements. The first author was supported by an IMPRS scholarship of the
Max Planck Society. The second author would like to thank the Max Planck Institute for
Mathematics at Bonn for its support.

ON THE FINITENESS OF THE CLASSIFYING SPACE FOR THE FAMILY OF VIRTUALLY CYCLIC SUBGROUPS 3

1. Properties of groups admitting a finite model for EG
In this section we first review properties of groups admitting a finite model for EG.
We then proceed to prove many useful lemmas for groups with BVC. We also use these
lemmas to show many groups cannot have BVC.
We denote by EG, resp. EG the G-CW classifying space for the family of finite subgroups resp. for the family consisting only of the trivial subgroup. We summarize the
properties of groups admitting a finite model for EG as follows
Proposition 1.1. Let G be a group admitting a finite model for EG, then
(a) G has BVC.
(b) G admits a finite model for EG.
(c) For every finite subgroup of H ⊂ G, the Weyl group WG H is finitely presented and
of type FP∞ . Here WG H := NG (H)/H, where NG (H) is the normalizer of H in G.
(d) G admits a finite type model for EG. In particular, G is finitely presented.
(e) G has finitely many conjugacy classes of finite subgroups. In particular, the order
of finite subgroups of G is bounded.
Proof Note that if G has a finite or finite type model for EG, then G is finitely presented
and has finitely many conjugacy classes of finite subgroups [16, 4.2]. Now (a), (b), (c) can
be found for example in [15, Section 2]. Part (d) can be deduced from (c) by taking H to
be the trivial group.
Remark 1.2. If one replaces finite by finite type in the assumptions of the above proposition, then the conclusions still hold except one has to replace finite by finite type in (b).
The following lemma is well-known to experts.
Lemma 1.3. Let G be a group. Then there is a model for EG with finite 0-skeleton if and
only if G has BVC.
Proof Suppose X is a G-CW model for EG with a finite 0-skeleton. Let σ1 , σ2 , . . . , σn
be a set of representatives for each orbit of 0-cell. Let V1 , V2 , . . . , Vn be the corresponding
virtually cyclic stabilizers. Since for every virtually cyclic subgroup V, the set of fixed
points X V is non-empty, there exists some vertex of X that is fixed by V. Since this vertex
stabilizer is a conjugate to some Vi , the subgroup V is subconjugate to Vi . Conversely,
suppose G has BVC, and let V1 , V2 , . . . , Vn be witnesses. We can construct a model for
`
EG with finite 0-skeleton as follows. Consider the G-set S := ni=1 G/Vi . The complete

graph ∆(S ) spanned by S , i.e., the simplicial graph that contains an edge for every two elements in ∆(S ), carries a cocompact simplicial G-action. The first barycentric subdivision
¯ ) of ∆(S ) is a G-CW-complex. Note that ∆(S
¯ ) is again cocompact. Moreover, ∆(S
¯ )H
∆(S

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TIMM VON PUTTKAMER AND XIAOLEI WU

is nonempty when H is virtually cyclic and empty otherwise. Now we can adding equi¯ ) and make ∆(S
¯ )H contractible for all virtually cyclic
variant cells of dimension ≥ 1 to ∆(S
subgroup by induction. Thus we get a model for EG at the end with finite 0-skeleton.



The following structure theorem about virtually cyclic groups is well known, see for
example [13, Proposition 4] for a proof.
Lemma 1.4. Let G be a virtually cyclic group. Then G contains a unique maximal normal
finite subgroup F such that one of the following holds
(a) the finite case, G = F;
(b) the orientable case, G/F is the infinite cyclic group;
(c) the nonorientable case, G/F is the infinite dihedral group.
Note that the above lemma implies that a torsion-free virtually cyclic group is either
trivial or isomorphic to Z. Thus we have the following
Corollary 1.5. Let G be a torsion-free group, then G has BVC if and only if there exist
elements g1 , g2 , . . . gn in G such that every element in G is conjugate to a power of some gi .
The following lemma is key to many of our proofs.
Lemma 1.6. Let V be a virtually cyclic group and let g, h ∈ V be two elements of infinite
order, then there exist p, q ∈ Z such that g p = hq . Furthermore, there exists v0 ∈ V such
p

that for any v ∈ V of infinite order there exist nonzero p0 , p such that v0 0 = v p with

p0
p

∈ Z.

Proof By 1.4, there exists an finite normal subgroup of F such that V/F is isomorphic
to Z or Z ⋊ Z/2. We denote the quotient map by q. Then if g is of infinite order and
V/F  Z ⋊ Z/2, then q(g) ∈ Z ⋊ {0}. Thus we can always assume V/F  Z. In this case
V  F ⋊ f Z, where Z acts on F via the automorphism f . Now for any g = (x, m) ∈ F ⋊ f Z
we can choose k = |F|| f | where |F| is the order of the finite group F and | f | is the order of
the automorphism f , then gk = (0, km). Let h = (y, n), then hk = (0, kn). Now we see that
k
gkn = hkm . If we choose v0 = (0, 1) ∈ F ⋊ f Z then for any v = (x, m) ∈ V, vmk
0 =v .



Note that in any virtually cyclic group there are only finitely many distinct finite subgroups up to conjugacy (using 1.4 or the fact that virtually cyclic groups are CAT(0)).
Using this fact one immediately obtains
Lemma 1.7. If a group G has BVC, then G has finitely many conjugacy classes of finite
subgroups. In particular, the order of finite subgroups in G is bounded.
In a group G, we call an element g primitive if it cannot be written as a proper power.
Then 1.5 implies the following
Lemma 1.8. Let G be a torsion-free group. If G has infinitely many conjugacy classes of
primitive elements, then G does not have BVC.

ON THE FINITENESS OF THE CLASSIFYING SPACE FOR THE FAMILY OF VIRTUALLY CYCLIC SUBGROUPS 5

Note that without the assumption of G being torsion-free, the previous lemma does not
hold. In fact, even a virtually cyclic group could contain infinitely many conjugacy classes
of primitive elements.
Example 1.9. Let S n be the symmetric group of order n with n ≥ 3, then S n × Z has
infinitely many primitive conjugacy classes. In fact, let (xi , 2i ) ∈ S n × Z with xi an odd
element in S n . Then (xi , 2i ) is primitive for any i ≥ 1. Since if (xi , 2i ) = (x, y)k for some
k > 1, then ky = 2i . In particular, k is even and thus xk cannot equal the odd element
xi . The elements (xi , 2i ) cannot be conjugate to each other since the second coordinate is
different.
Lemma 1.10. Let G = A ∗ B be a free product with A and B nontrivial, then G has BVC if
and only if G is virtually cyclic.
Proof If A and B are finite groups, then A ∗ B is a virtually free group and hence hyperbolic. So the lemma holds in this case. Now we assume that A is not finite, then A ∗ B is
not virtually cyclic. Let a1 , a2 . . . , an , . . . , be a sequence of different elements in A and let
b ∈ B be a non-trivial element. Then {ai b | i ≥ 1} in G form infinitely many conjugacy
classes of primitive elements in G. Moreover, when i , j, ai b and any conjugates of a j b
cannot lie in a virtually cyclic group. In fact, if this were the case, by 1.6 we would have
that (ai b)m is conjugate to (a j b)n in G for some m, n , 0. But this is impossible by the

choices of ai and b and [20, IV.2.8]. Hence G does not have BVC in this case.
Lemma 1.11. [15, 5.6] If a group G has BVC, then any finite index subgroup also has
BVC.
Combining this with the main result of [10], we have
Proposition 1.12. Virtually solvable groups have BVC if and only if they are virtually
cyclic.
Lemma 1.13. [10, 2.2] Let G be a group with property BVC. Then the following assertions
hold.
(a) The group G satisfies the ascending chain condition for normal subgroups.
(b) If L and M are normal subgroups of G with M < L and L/M a torsion group, then
there are only finitely many normal subgroups K of G such that M ≤ K ≤ L.
(c) Let
1 = Gn ≤ Gn−1 ≤ · · · ≤ G1 ≤ G0 = G
be a series of normal subgroups of G. Then the number of factors Gi /Gi−1 that are
not torsion groups is bounded by the number of infinite groups in a witness to BVC
for G.
The following lemma allows us to show that many groups cannot have BVC.

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Lemma 1.14. Let π : G → Q be a surjective group homomorphism. If Q is a torsion-free
group that does not have BVC, then G does not have BVC.
Proof Suppose G has BVC and let V1 , V2 . . . , Vn be a witness for BVC of G. Note that
any quotient of a virtually cyclic group is again virtually cyclic. Hence π(Vi ) are virtually
cyclic subgroups in Q. Since Q is torsion-free and does not have BVC, we can find a
nontrivial element q ∈ Q such that q cannot be conjugated to π(Vi ) for any i. Now take
g ∈ G such that π(g) = q, then we can find c ∈ G such that cgc−1 ∈ Vi for some i. But then
we would have π(c)π(g)π(c−1) = π(c)qπ(c−1) ∈ π(Vi ) which is a contradiction.

Corollary 1.15. If G is a group having BVC, then the abelianization H1 (G, Z) is finitely
generated of rank at most one.
Proof Let A be the abelianization of G and let T be the torsion subgroup of A. By 1.14
the torsion-free abelian group A/T has to be trivial or infinite cyclic. Hence A = T or
A  T × Z. Now 1.13(b) implies that the torsion group T has to be finite.



Example 1.16. The Thompson groups are a family of three finitely presented groups, F ⊂
T ⊂ V. Thompson’s group F can be defined by the presentation hA, B | [AB−1 , A−1 BA] =
[AB−1, A−2 BA2] = 1i. Since H1 (F)  Z2 , it follows that F does not have BVC. Since the
order of finite subgroups in T and V is unbounded, we see that T and V also do not have
BVC. See [3] for more information about Thompson groups.
Recall that a group satisfies the strong Tits alternative if any finitely generated subgroup has a finite index subgroup which is either solvable, or has a non-abelian free quotient. Since virtually solvable groups and free groups have BVC if and only if they are
virtually cyclic, we have the following by 1.11 and 1.14,
Lemma 1.17. If a group G satisfies the strong Tits alternative, then a finitely generated
subgroup of G has BVC if and only if it is virtually cyclic.

Since Coxeter group and right angled Artin groups are known to satisfy the strong Tits
alternative by [24] and [1, 1.6], we have the following
Corollary 1.18. Let G be a Coxeter group or a right angled Artin group, then a finitely
generated subgroup of G satisfies BVC if and only if it is virtually cyclic.
Note that 1.15 reduces the study of finitely generated groups with BVC to the case
where H1 (G, Z) is of rank one or zero. If H1 (G, Z) is of rank one, then up to replacing G
by a finite index subgroup, we can assume H1 (G, Z)  Z. Then G becomes a semidirect
product of the form H ⋊ Z. We proceed to study groups of this type.
Given an automorphism φ of G, we say that two elements g, h in G are φ-conjugate if
g = xhφ(x−1 ) for some x ∈ G. This is an equivalence relation whose equivalence classes

ON THE FINITENESS OF THE CLASSIFYING SPACE FOR THE FAMILY OF VIRTUALLY CYCLIC SUBGROUPS 7

are called φ-twisted conjugacy classes. For φ = idG one recovers the usual notion of
conjugacy.
Lemma 1.19. Let φ be an automorphism of H such that H has infinitely many φ-twisted
conjugacy classes, then the semidirect product H ⋊φ Z does not have BVC.
Proof Note that in H ⋊φ Z, the elements (g, 1) and (h, 1) are primitive and they are in
the same conjugacy class if and only if g and h are in the same φ-twisted conjugacy class
in H. In fact, (g, 1) is conjugate to (h, 1) in H ⋊φ Z if and only if we can find (x, k) such
that (x, k)(g, 1)(x, k)−1 = (xφk (g)φ(x−1 ), 1) = (h, 1). This is equivalent to saying that g is φconjugate to φk (g) in H. But g and φ(g) are φ-conjugate in H since φ(g) = g−1 gφ(g). Hence
g and h are in the same φ-twisted conjugacy class in H if and only if (g, 1) is conjugate to
(h, 1) in H ⋊φ Z.
Since H has infinitely many φ-twisted conjugacy classes, we have infinitely many primitive elements of the form (g, 1) ∈ H ⋊φ Z that are not conjugate to each other. If H ⋊φ Z has
BVC, then we can choose infinitely many elements (g1 , 1), (g2, 1), . . . , (gn , 1), . . . that are
not conjugate to each other, but they lie in the same virtually cyclic subgroup. In particular
the group V generated by (g1 , 1), (g2, 1), . . . , (gn , 1) . . . in G is virtually cyclic. But this is
impossible. Consider the quotient map q : H ⋊ Zφ → Z, which is onto when restricted to
V. Hence the kernel must be finite. This means V  F ⋊ Z for some finite group F and the
image of (gi , 1) is of the form ( fi , 1). This leads to a contradiction since there are infinitely
many (gi , 1) but only finitely many ( fi , 1).

Recall that a group is said to have property R∞ if it has infinitely many φ-twisted conjugacy classes for any automorphism φ.
Corollary 1.20. Let G be a group with the property R∞ , then any semidirect product H⋊φ Z
does not have BVC.
Note that there are many groups with property R∞ , for example hyperbolic groups
that are not virtually cyclic, relatively hyperbolic groups and most generalzied BaumslagSolitar groups. For more information about groups with property R∞ and further examples,
see [8].
2. HNN extension of groups and one-relator groups
In this section, we study HNN extension of groups. We first give a quick review of
basic results concerning the structure of these types of groups. Then we prove that any
non-ascending HNN extension of a group and any HNN extension of a finitely generated
free group do not have BVC. Using this, we are able to show that one-relator groups have
BVC if and only if they are virtually cyclic.
Recall that given a group H and an isomorphsim θ : A → B between two subgroups A
and B of H, we can define a new group H∗θ , called the HNN extension of H along θ, by

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the presentation hH, t | txt−1 = θ(x), x ∈ Ai. In the study of HNN extensions of groups,
Britton’s Lemma and Collin’s Lemma play an important rule. We give a quick review of
the two lemmas and refer to [20, IV.2] for more details.
Definition 2.1. A sequence g0 , tǫ1 , g1 , . . . , tǫn , gn of elements with gi ∈ H and ǫi ∈ {−1, +1}
is said to be reduced if there is no consecutive sequence t, gi , t−1 with gi ∈ A or t−1 , g j , t
with g j ∈ B.
Lemma 2.2 (Britton’s Lemma). If the sequence g0 , tǫ1 , g1 , . . . , tǫn , gn is reduced and n ≥ 1,
then g0 tǫ1 g1 · · · tǫn gn , 1 in H∗θ .
In the following we will not distinguish between a sequence of words as above and the
element it defines in the HNN extension H∗θ .
Give any g ∈ H∗θ , we can write g in a reduced form. Let w = g0 tǫ1 g1 · · · tǫn gn , 1
be any reduced word in H∗θ which represents g. The we define the length of g, written as |g|, to be the number n of occurences of t± in w. Moreover we call an element
w = g0 tǫ1 g1 · · · tǫn gn , 1 cyclically reduced if all cyclic permutations of the sequence
g0 , tǫ1 , g1 , . . . , tǫn , gn are reduced. Every element of H∗θ is conjugate to a cyclically reduced
element.
Lemma 2.3 (Collin’s Lemma). Let G = hH, t | txt−1 = θ(x), x ∈ Ai be an HNN extension.
Let u = g0 tǫ1 g1 · · · tǫn gn and v be cyclically reduced elements of G that are conjugate, n ≥ 1.
Then |u| = |v|, and u can be obtained from v by taking a suitable cyclic permutation v∗ of v,
which ends in tǫn , and then conjugating by an element z, where z ∈ A if ǫn = 1, and z ∈ B if
ǫn = −1.
We are ready to prove the following:
Lemma 2.4. Let H be a group and let θ : A → B be an isomorphism between two subgroups of H. If [H : A], [H : B] ≥ 2, then the corresponding HNN extension G = H∗θ does
not have BVC
Proof Choose α ∈ H \ B and β ∈ H \ A, define
wn = t−1 αtn+1 β
for n ≥ 1. Note that the elements wn are of infinite order and cyclically reduced. By
Collin’s Lemma, they are not conjugate to each other. If G had BVC, there would exist
a virtually cyclic subgroup V ⊆ G such that there is an infinite subsequence {wni } with
each wni conjugate to an infinite order element of V. But this cannot happen as wnpn is not
conjugate to wmpm for any pn , pm , 0 when n , m. In fact, first note that wnpn is cyclically
p

p

reduced for any n. So if wn n is conjugate to wmm , their lengths must coincide by Collin’s
Lemma. Hence we have an equation |pn |(n + 2) = |pm |(m + 2). On the other hand, there is a
p

p

p

p

canonical quotient map q : G → hti  Z. If wn n is conjugate to wmm , then q(wn n ) = q(wmm ).

ON THE FINITENESS OF THE CLASSIFYING SPACE FOR THE FAMILY OF VIRTUALLY CYCLIC SUBGROUPS 9

This means pn n = pm m. But the two equations can never hold at the same time when
n, m ≥ 1 unless n = m. Hence we have a contradiction by 1.6.



Now when H = A or H = B, we would have an ascending HNN extension. It seems
to us that this case cannot be handled as easily as before. In the following we will analyze
the case of an ascending HNN extension of a free group F of finite rank in detail. We will
first deal with the case that θ : F → F is injective with its image lying in the commutator
subgroup of F. Given a group G, we will write G′ for the commutator subgroup [G, G]
and we will denote the r-th term in the lower central series by Γr (G) = [Γr−1 (G), G] where
Γ1 (G) = G. Let us first recall the following facts:
Lemma 2.5. The lower central series has the following properties:
(a) [Γr (G), Γ s(G)] ⊆ Γr+s (G) for any group G.
T
(b) r≥1 Γr (F) = {1} for any free group F.
(c) Γr (F)/Γr+1 (F) is a free abelian group for any r and any free group F of finite rank.
Proof (a) can be found in [11, Corollary 10.3.5]. (b) and (c) can be found in [11, 11]. 
Corollary 2.6. Let θ : F → F be an injective map of the finitely generated free group F
with the image of θ lying in the commutator subgroup of F. If x ∈ Γr (F), then θ(x) ∈ Γ2r (F).
Proof If x ∈ Γ1 (F) = F, then by assumption on θ we have θ(x) ∈ [F, F] = Γ2 (F). Let
r ≥ 2 and suppose that for any s < r the claim holds. If x ∈ Γr (F) = [Γr−1 (F), F], then by
induction and 2.5 we get θ(x) ∈ [Γ2(r−1) (F), Γ2 (F)] ≤ Γ2r (F).

Lemma 2.7. Let G = hH, t | txt−1 = θ(x), x ∈ Hi be an ascending HNN extension of a
group H. Then any element of G can be written in the form t−p htq with p, q ≥ 0 and h ∈ H.
S
Moreover the normalizer NG (H) of H in G is given by NG (H) = i≥0 t−i Hti .

Proof The claim about the form elements of G take follows since for any h ∈ H, th =
θ(h)t and similarly ht−1 = t−1 θ(h) in G. For the second part, notice that t−i Hti ⊂ NG (H)
for any i. Since G/NG (H)  hti, we have that if g = t−p htq ∈ NG (H), then p = q. Thus
S
NG (H) = i≥0 t−i Hti .

Lemma 2.8. Let G = hF, t | txt−1 = θ(x), x ∈ Fi be any ascending HNN extension of a free
group F of finite rank with im(θ) ≤ [F, F]. Suppose that x, y ∈ F are non-primitive in G

and generate a free subgroup of rank 2. Then xy is primitive in G.
Proof Suppose x, y and xy are all non-primitive. Let x = um , y = vn , xy = wl for some
u, v, w ∈ G and m, n, l ≥ 2. Let q be the canonical quotient map from G to hti  Z mapping
F to 0. Then u, v, w lie in the kernel since x and y lie in the kernel. Note that the kernel is
just the normalizer of F in G. By 2.7, there exist some p ≥ 0 such that u, v, w lie in the free
subgroup t−p Ft p . But by [21], the equation um vn = wl has a solution in a free group only if
u, v, w generate a free subgroup of rank 1. This contradicts our hypothesis on x and y.



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We need the following lemma.
Lemma 2.9. Let f : A → A be an automorphism of a free abelian group A. If f (ka) = la
for some a , 0 and positive integers k, l, then k = l.
Proof We can assume without loss of generality that A has infinite rank, so A 

L

i∈I

Z

for some infinite index set I. We call a non-trivial element a ∈ A prime if the common
divisor of its finitely many non-zero coordinates is trivial. Note that any non-trivial a ∈ A
can be written as a = d · x with x being prime. Since f is an automorphism, it will preserve
prime elements. So now suppose that f (ka) = la with k, l ∈ N and a , 0. We write a = dx
as above with x prime. Then k f (x) = lx and by cancelling common factors we might as
well assume that k and l are coprime. Since k divides all coordinates of the prime element
x it has to equal to one and the same holds for l since f (x) is prime.



Proposition 2.10. Let G = hF, t | txt−1 = θ(x), x ∈ Fi be an ascending HNN extension
of a free group F of finite rank, where the image of θ lies in the commutator subgroup of
F. If x, y ∈ F \ [F, F] generate a free subgroup of rank 2 in F and x is primitive, then
the elements {xk yxk y−1 | k ≥ 2} form pairwise distinct primitive conjugacy classes. In
particular, G does not have BVC.
Proof Note first that xk yxk y−1 does not lie in [F, F] and is primitive when k ≥ 2 by 2.8.
Note that every element in G can be written in the form t−p wtq for some p ≥ 0, q ≥
0 and w ∈ Fn by 2.7. Now if xk yxk y−1 is conjugate to xl yxl y−1 for some k , l, then
xk yxk y−1 = t−p wtq xl yxl y−1 t−q w−1 t p for some p, q ≥ 0 and w ∈ F. Hence θ p (xk yxk y−1 ) =
wθq (xl yxl y−1 )w−1
If p , q, the equation never holds. In fact, assume p > q. We can further assume
θq (x) ∈ Γr (Fn ) \ Γr+1 (F) for some r ≥ 2 by 2.5 (b). Then θ p (x) ∈ Γr+1 (F) by 2.6 and thus
θ p (xk yxk y−1 ) ∈ Γr+1 (F). On the other hand, θq (xl ) ∈ Γr (F) \ Γr+1 (F) for any l > 0 since
Γr (F)/Γr+1 F is a free abelian group by 2.5 (c). Now θq (xl yxl y−1 ) = θq (x2l )[θq (x−l ), θq (y)]
and [θq (x−l ), θq (y)] ∈ Γr+1 (F) by 2.6, we have θq (xl yxl y−1 ) ∈ Γr (F) \ Γr+1 (F). But the right
hand side wθq (xl yxl y−1 )w−1 ∈ Γr (F) \ Γr+1 (F), hence the equation cannot hold.
If p = q, then the equation again cannot hold unless k = l. In fact, assume θ p (x) ∈
Γr (F)\Γr+1 (F), then both sides lie in Γr (F)\Γr+1 (F) by the same argument above. By taking
the quotient by Γr+1 (F) we obtain an equation in the free abelian group Γr (F)/Γr+1 (F).
Then we would have k([θ p (x)] + [θ p (yxy−1 )]) = l([wθ p (x)w−1 ] + [wθ p (yxy−1 )w−1 ]). Note
that Γr (F)/Γr+1 (F) is a free abelian group of infinite rank by 2.5 (c) and the action of w on
Γr (F)/Γr+1 (F) induced by conjugation is an isomorphism. Thus the equation k([θ p (x)] +

[θ p (yxy−1 )]) = l([wθ p (x)w−1 ] + [wθ p (yxy−1 )w−1 ]) can never hold unless k = l by 2.9.
We are ready to prove the following.
Theorem 2.11. Let G be an HNN extension of a free group of finite rank, then G does not
have BVC.

ON THE FINITENESS OF THE CLASSIFYING SPACE FOR THE FAMILY OF VIRTUALLY CYCLIC SUBGROUPS11

Proof By 2.4, we can assume G = hFn , t | txt−1 = θ(x), x ∈ Fn i, where θ : Fn → Fn is
injective, Fn is a free group of rank n. For n = 1 the group G is solvable but not virtually
cyclic. Thus G does not have BVC by 1.12. So in the following we assume that Fn is a
free group of rank bigger than 1.
Note first that we have an induced map θ̄ : Fn /[Fn , Fn ] → Fn /[Fn , Fn ]  Zn . Since
the rank of the abelian group is finite, there exists some k ≥ 1 such that rank(ker(θ̄k+1 )) =
rank(ker(θ̄k )). But since Zn is free abelian, it follows that ker(θ̄k+1 ) = ker(θ̄k ), and we will
denote this group by K. This implies that θ̄k induces an injective endomorphism of Zn /K.
If K is a proper subgroup of Zn , we consider the induced quotient map Fn ∗θk → (Zn /K)∗θ̄k .
Note that the quotient is a torsion-free metabelian group which is not virtually cyclic.
Hence Fn ∗θk does not have BVC by 1.12 and 1.14. As Fn ∗θk is a finite index subgroup
of Fn ∗θ (see for example [14, 2.2]) we conclude that the latter group does not have BVC
by 1.11.
If K = Zn , we are in the situation that the image of θk lies in the commutator subgroup
of Fn . By 2.10 the group Fn ∗θk does not have BVC. Again by 1.11 it follows that Fn ∗θ
does not have BVC.

We now want to apply the previous results to verify Conjecture B for the class of onerelator groups. Recall that a one-relator group is a group G which has a presentation with
a single relation, so G = hx1 , . . . , xn | ri where r is a word in the free group F on the letters
x1 , . . . , xn . The group G is torsion-free precisely when r, as an element of the free group
F, is not a proper power. If r = sn for some maximal n ≥ 2 and s ∈ F, then s, considered
as an element in G, is of order n. In all cases there exists a finite G-CW model for EG, see
for example [17, 4.12].
For one-relator groups G with torsion, Newman’s Spelling Theorem [23] implies that
G is a hyperbolic group. In particular, the one-relator groups containing torsion satisfy
Conjecture B. However, our proof of the following theorem does not depend on this fact.
Theorem 2.12. A one-relator group has BVC if and only if it is cyclic.
Proof Let G be a one-relator group.
If the one-relator presentation of G contained three or more generators then G would
surject to Z2 , in particular G would not have BVC by 1.15. Thus we can restrict to the case
that G has two generators, so
G = ha, b | R(a, b) = 1i
for some word R(a, b) in the free group on the two generators a, b. By [20, Lemma V.11.8]
we can moreover assume that the exponent sum of one of the generators in the single
relator equals to zero, say for the generator a. The following rewriting procedure, which
we outline for the reader’s convenience, is standard. The proofs of the mentioned facts
can be found in [20, IV.5]. We let bi = ai ba−i for all i ∈ Z. Then R can be rewritten as

12

TIMM VON PUTTKAMER AND XIAOLEI WU

a cyclically reduced word R′ in terms of these, so R′ = R′ (bm , . . . , b M ) for some m ≤ M,
such that the elements bm , b M occur in R′ . If m = M, then R(a, b) = bm for some m ∈ Z
and thus G  Z or G  Z ∗ Z/|m| where |m| ≥ 2. Note that by 1.10 the latter group does not
have BVC. So in the following we can assume that m < M. We let
H = hbm , . . . , b M | R′ (bm , . . . , b M ) = 1i.
Moreover we define A to be the subgroup of H generated by bm , . . . b M−1 and we let B to
be the subgroup of H generated by bm+1 , . . . b M . Then A and B are free subgroups of the
one-relator group H and G is isomorphic to the HNN extension H∗θ where θ : A → B is
the isomorphism defined by θ(bi ) = bi+1 for m ≤ i < M.
If [H : A] ≥ 2 and [H : B] ≥ 2, then G does not have BVC by 2.4. Otherwise G is
an ascending HNN extension, say with H = A. Since A was free, G is an ascending HNN

extension of a finitely generated free group. The claim now follows from 2.11.
3. Groups with some hyperbolicity
In this section, we first show that acylindrically hyperbolic groups do not have BVC.
Using this we show that any finitely generated 3-manifold group does not have BVC.
We first give a quick definition of acylindrically hyperbolic group and refer to [12] and
[26] for more information. Recall the action of a group G on a metric space S is called
acylindrical if for every ε > 0 there exist R, N > 0 such that for every two points x, y with
d(x, y) ≥ R, there are at most N elements g ∈ G satisfying d(x, gx) ≤ ε and d(y, gy) ≤ ε.
Given a hyperbolic space S , we use ∂S to denote its Gromov boundary.
Definition 3.1. A group G is called acylindrically hyperbolic if there exists a generating
set X of G such that the corresponding Cayley graph Γ(G, X) is hyperbolic, |∂Γ(G, X)| > 2,
and the natural action of G on Γ(G, X) is acylindrical.
Proposition 3.2. An acylindrically hyperbolic group does not have BVC.
Proof Let G be an acylindrically hyperbolic group. By [26, Theorem 1.2] this is equivalent to saying that G contains a proper infinite hyperbolically embedded subgroup. By [5,
Lemma 6.14], we further have a subgroup K = F2 × K(G) inside G which is hyperbolically
embedded, where K(G) is some maximal normal finite subgroup of G and F2 is the free
group of rank 2. The following two statements are copied from the proof of [12, VI.1.1]
(they can be easily decduced from [5, Proposition 4.33]): (1) every element of infinite
order which is primitive in K is also primitive in G; (2), if two elements of K of infinite
order are conjugate in G, they are conjugate in K. Now since K contains F2 as a direct
factor, there exists infinitely many primitive elements of infinite order g1 , g2 , . . . , gn , . . . in
F2 ⊂ K such that gi is not conjugate to g j or g−1
j when i , j. But any two such elements
or conjugates of them cannot lie in the same virtually cyclic subgroup. In fact, let gi , g j
be two such elements and suppose gi , xg j x−1 lie in the same virtually cyclic subgroup in

ON THE FINITENESS OF THE CLASSIFYING SPACE FOR THE FAMILY OF VIRTUALLY CYCLIC SUBGROUPS13
n
n −1
G, where x ∈ G. Then we have gm
for some m and n by 1.6. Thus gm
i and g j
i = xg j x

are conjugate in G. Hence they are also conjugate in F2 , which is a contradiction. Thus
these primitive elements of infinite order cannot lie in finitely many conjugacy classes of
virtually cyclic subgroups. Hence G does not have BVC.

Corollary 3.3. The following classes of groups does not have BVC:
(a) Hyperbolic groups that are not virtually cyclic;
(b) Non-elementary groups that are hyperbolic relative to proper subgroups;
(c) The mapping class group MCG(Σg , p) of a closed surface of genus g with p punctures except for g = 0 and p ≤ 3 (in these exceptional cases, MCG(Σg , p) is finite);
(d) Out(Fn ) where n ≥ 2;
(e) Groups which act properly on proper CAT(0) spaces and contain rank-one elements;
Proof These groups are all acylindrically hyperbolic, we refer to [12, I.1,page 4] and [26,
Section 8] for the detailed reference.

Corollary 3.4. Let φ : A → G be a surjective group homomorphism and suppose that G is
an acylindrically hyperbolic group. Then A does not have BVC.
Proof As in the proof of 3.2, there exists infinitely many primitive conjugacy classes of
elements {gi }i≥1 inside a certain free group F2 ⊂ G such that gi pi is not conjugate to any
g j p j for i , j and pi , p j , 0. Now we take any preimage g′i of gi in A. Then the g′i are
′

′

primitive and g′i pi is not conjugate to any g′j p j for i , j and p′i , p′j , 0. By 1.6, they cannot
lie in finitely many conjugacy classes of virtually cyclic subgroups in A. Hence A does not
have BVC.



By a 3-manifold group we mean a group that can be realized as the fundamental group
of a 3-manifold, which may be open or have boundary. Note that, by Scott’s theorem [29],
every finitely generated 3-manifold group is itself the fundamental group of a compact
3-manifold. Minsanyan and Osin prove in [22, 2.8] the following:
Lemma 3.5. Let G be a subgroup of the fundamental group of a compact 3-manifold, then
exactly one of the following holds:
(a) G is acylindrically hyperbolic.
(b) G contains a normal infinite cyclic subgroup N such that G/N is acylindrically
hyperbolic.
(c) G is virtually polycyclic.
Now combining this with 1.12, 3.2 and 3.4, we have the following

14

TIMM VON PUTTKAMER AND XIAOLEI WU

Proposition 3.6. Let G be the subgroup of the fundamental group of a compact 3-manifold,
then G has BVC if and only if G is virtually cyclic. In particular, if G is a finitely generated
3-manifold group, then G has BVC if and only if G is virtually cyclic.
4. Groups acting on CAT(0) spaces
In this section, we study groups acting on CAT(0) spaces and show that many of them
do not have BVC. We first give a quick review of properties of CAT(0) spaces and groups
that we may need and refer to [2] for more details.
Definitions 4.1. [2, II.6.1] Let X be a metric space and let g be an isometry of X. The
displacement function of g is the function dg : X → R+ = {r ≥ 0 | r ∈ R} defined by
dg (x) = d(gx, x). The translation length of g is the number |g| := inf{dg (x) | x ∈ X}. The
set of points where dg attains this infimum will be denoted by Min(g). More generally,
T
if G is a group acting by isometries on X, then Min(G) := g∈G Min(g). An isometry g
is called semi-simple if Min(g) is non-empty. An action of a group by isometries of X is
called semi-simple if all of its elements are semi-simple.
The following theorem is known as the Flat Torus Theorem [2, II.7.1].
Theorem 4.2. Let A be a free abelian group of rank n acting properly by semi-simple
isometries on a CAT(0) space X. Then:
T
(a) Min(A) = α∈A Min(α) is non-empty and splits as a product Y × En , here En
denotes Rn equipped with the standard Euclidean metric.
(b) Every element α ∈ A leaves Min(A) invariant and respects the product decomposition; α acts as the identity on the first factor Y and as a translation on the second
factor En .
(c) The quotient of each n-flat Y × En by the action of A is an n-torus.
It is clear that the translation length is invariant under conjugation, i.e. |hgh−1| = |g| for
any g, h ∈ G. Moreover, for g semi-simple, we have that |gn | = |n| · |g| for any n ∈ Z, e.g.
by the Flat Torus Theorem. It turns out that the translation length can also be defined by
the following limit for g a semi-simple isometry
1
d(x, gn x),
n
where x is an arbitrary point of the metric space X [2, II.6.6].
Note that if a group acts properly and cocompactly on a CAT(0) space via isometries,
|g| = lim

n→∞

we call the group a CAT(0) group. In this case, the action is semi-simple.
Proposition 4.3. If a group G acts properly and cocompactly by isometries on a CAT(0)
space X, then
(a) G is finitely presented.
(b) G has only finitely many conjugacy classes of finite subgroups.

ON THE FINITENESS OF THE CLASSIFYING SPACE FOR THE FAMILY OF VIRTUALLY CYCLIC SUBGROUPS15

(c) Every solvable subgroup of G is virtually abelian.
(d) Virtually abelian subgroups of G satisfy the ascending chain condition.
(e) G has a finite model for EG.
(f) There is a finite-dimensional model for EG.
Proof (a) - (c) can be found in [2, III.Γ.1.1], (d) can be found in [2, II.7.5]. Since G acts
on X properly and cocompactly via isometry, X is proper by [2, I.8.4(1)]. With this (e) is

implied by [27, Proposition B]. The last statement was proven in [18].
Lemma 4.4. Let V be an infinite virtually cyclic group which acts on a CAT(0) space via
semi-simple isometries. Then there exist an element v0 ∈ V such that for any element v ∈ V,
the translation length |v| of v is an integer multiple of the translation length of v0 .
Proof When v is of finite order, then |v| = 0. So let us assume in the following that v is
′

of infinite order, in this case the translation length is strictly positive. If vk = wk for some
′

w ∈ V and k, k′ ∈ Z, then |k||v| = |vk | = |wk | = |k′ ||w|. Now by 1.6, there exist v0 ∈ V such
that for any v ∈ V, there exist nonzero p, p0 such that v0p0 = v p with pp0 ∈ Z. This implies
that |v| is a multiple of |v0 |.

The lemma leads us to define the following terminology.
Definition 4.5. We define a subset A of the real numbers to be finitely divisor dominated
if there are finitely many real numbers x1 , x2 , . . . , xn such that every a ∈ A can be written
in the form kxi for some k ∈ Z, or equivalently
n
[
A⊂
Z · xi .
i=1

In this case we say that A is finitely divisor dominated by x1 , x2 , . . . , xn .
Note that for a CAT(0) group the set of translation lengths { |g| | g ∈ G} is a discrete
subset of R [2, II.6.10 (3)]. We obtain first the key property of the set of translation lengths
for a group acting on a CAT(0) space with BVC.
Lemma 4.6. Let G be a group acting properly on a CAT(0) space via semi-simple isometries. Let L = { |g| | g ∈ G} ⊂ R≥0 be the set of translation lengths of G. If G has BVC, then
L is finitely divisor dominated.
Proof Note first that if g and h are conjugate, then they have the same translation length.
Now assume G has BVC, and let V1 , V2 , . . . , Vn be witnesses. We only need to consider
those Vi that are infinite since a finite order element has vanishing translation length.
By 4.4, we can choose for each Vi an element vi such that the translation length of any
infinite order element of Vi is a multiple of the translation length |vi |. Now L is finitely
divisor dominated by |v1 |, |v2 |, . . . , |vn |.



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TIMM VON PUTTKAMER AND XIAOLEI WU

Remark 4.7. For a hyperbolic group acting on its Cayley graph with respect to some
fixed generating set S , we define the algebraic translation length using the limit |g| =
limn→∞ 1n dS (1, gn) [2, III.Γ.3.13], where dS denotes the word metric with respect to S .
Gromov [2, III.Γ.3.17] showed that the set of algebraic translation lengths in this case is a
discrete subset of the rational numbers and the denominators are bounded. In particular,
the set of algebraic translation lengths of a hyperbolic group is finitely divisor dominated.
We need do some preparation before we prove our main result in this section.
Lemma 4.8. Let x > 0, y ≥ 0 be two rational numbers and let
(
)
q
2
A = λn | λn = x + (y + n) , n ∈ N .
Then A is not finitely divisor dominated.
Proof Since x and y are rational numbers, we can choose d to be the smallest positive
integer such that 2yd and (y2 + x)d are integers. Then we can consider the quadratic polynomial
f (n) = d(x + (y + n)2 ) = dn2 + 2ydn + (y2 + x)d
which has coprime integer coefficients. Note that f (n) is irreducible over R since d, x > 0.
Now by an old result of Ricci [28], there exists infinitely many positive integers n such that
the integer f (n) is square-free.
Now if A was finitely divisor dominated, there would exist finitely many positive real
numbers x1 , x2 , . . . , xm such
A⊂

n
[

Z · xi

i=1

In particular, there exists some i0 and an infinite sequence n1 , n2 , . . . , n j , . . . of natural
numbers such that λn j = k j xi0 , with k j ∈ Z and f (n j ) square-free. This implies there are
infinitely many k j > k1 such that
λ2n j
λ2n1

=

k2j
x + (y + n j )2
=
x + (y + n1 )2 k12

This further implies
n2j + 2yn j + y2 + x =

k2j
k12

(n21 + 2yn1 + y2 + x)

Multiplying both sides by d, we obtain
dn2j + 2ydn j + (y2 + x)d =

k2j
k12

(dn21 + 2ydn1 + (y2 + x)d)

Now since f (n) = d(x + (y + n)2 ) = dn2 + 2ydn + (y2 + x)d is a polynomial in n with positive
integer coefficients that have no common divisor, the left hand side of the above equation
must be a positive integer and f (n1 ) = dn21 + 2ydn1 + (y2 + x)d is also a positive integer.

ON THE FINITENESS OF THE CLASSIFYING SPACE FOR THE FAMILY OF VIRTUALLY CYCLIC SUBGROUPS17
k2

But since k j > k1 are positve integers, the value of f (n j ) = k2j (dn21 + 2ydn1 + (y2 + x)d) is
1
not square-free. This leads to a contradiction as we have chosen the n j such that f (n j ) is
square-free.



Lemma 4.9. Let n0 , q0 ∈ N and x > 0, y ≥ 0 be two real numbers, such that there are
infinitely many intergers m > n with

x2 +(y+m)2
x2 +(y+n0 )2

=

p2
q20

for some p ∈ N. Then y ∈ Q, x2 ∈ Q.

Proof Let m1 , m2 , . . . , mi , . . ., be an infinite sequence of positive integers greater than
2
2
p2
i)
n0 such that xx2+(y+m
= q2i . Let us write x2 + (y + n0 )2 = q20 t for some t > 0, then
+(y+n0 )2
0

x2 + (y + mi )2 = p2i t. Subtracting this by the previous equation, we get
(mi − n0 )(2y + mi + n0 ) = (p2i − q20 )t.
Now comparing this with the same equality for m j , we obtain
(4.1)

p2 − q20
(mi − n0 )(2y + mi + n0 )
= 2i
(m j − n0 )(2y + m j + n0 ) p j − q20

Since mi , m j , n0 , pi , p j are all integers, we have y is rational unless
p2i − q20
p2j − q20

=

m2i − n20
m2j − n20

But this cannot happen as long as mi , m j . In fact, note first that we can assume
without loss of generality that n0 = 0 (via shifting y by some integer). Now let r =
(p2i − q20 )/(p2j − q20 ). Then equation 4.1 above leads to
2y(mi − m j r) = rm2j − m2i
We cannot solve for y if mi = m j r. But if this happens r =

m2i
m2j

=

mi
mj ,

hence mi = m j .

This also immediately implies x2 ∈ Q using the equation x2 + (y + n0 )2 = q20 t.



Combining 4.8 and 4.9, we have the following
Corollary 4.10. Let x > 0, y ≥ 0 be two real numbers and let A = {λn | λn =
N}. Then A is not finitely divisor dominated.

p

x2 + (y + n)2 , n ∈

Proposition 4.11. Let G be a group acting properly on a CAT(0) space X via semi-simple
isometries. If G contains Z2 as a subgroup, it does not have BVC.
Proof Assume G has BVC, then the set of translation lengths L = { |g| | g ∈ G} is finitely
divisor dominated by 4.6. On the other hand, by the Flat Torus Theorem, we have that Z2
acts on a flat plane E2 inside X and the translation length of any g = (z, w) ∈ Z2 is just
d(g, gx0) for some base point x0 ∈ E2 . Let a be the translation vector of (1, 0) ∈ Z2 and
b for (0, 1) ∈ Z2 . Let (a1 , a2 ), (b1 , b2 ) be the coordinate of a and b in the Euclidean plane
E2 . Without loss of generality, we can assume a1 > 0, a2 ≥ 0 and (b1 , b2 ) = (1, 0). Then

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TIMM VON PUTTKAMER AND XIAOLEI WU

the translaton length of (1, k) ∈ Z2 is the length of the vector a + kb = (a1 , a2 ) + k(0, 1) =
(a1 , a2 + k), which is
λ(1,k) =

q
a21 + (a2 + k)2 .

Now if a set is finitely divisor dominated,
then any subset
 of it is also finitely divisor
q
2
2
a1 + (a2 + k) | k ∈ N is finitely divisor dominated
dominated. In particular, the subset
for some a1 > 0, a2 ≥ 0. But this contradicts 4.10.

Recall that a semi-simple isometry is called hyperbolic if it has positive translation
length. Now if g acts properly on a CAT(0) space X via a hyperbolic isometry, by the Flat
Torus Theorem, we have an axis E1 on which g acts via translation by the length |g|.
Definition 4.12. Supppose g is a hyperbolic isometry of a CAT(0) space X. Then g is
called rank one if no axis of g bounds a flat half plane in X.
Note that if a group acts on a CAT(0) space X properly and cocompactly via isometries,
then X is proper [2, I.8.4(1)]. Combining this with 3.3 (e) and 4.11, we have the following
Theorem 4.13. Let G be a subgroup of a CAT(0) group which contains a subgroup ismorphic to Z2 or a rank-one isometry, then G does not have BVC.
Remark 4.14. The Flat Closing Conjecture [9, 6.B3] predicts that X contains a d-dimensional
flat if and only if G contains a copy of Zd as a subgroup. In particular, it implies that a
CAT(0) group is hyperbolic if and only if it does not contain a subgroup isomorphic to Z2 .
Thus the Flat Closing Conjecture together with 4.13 would also imply that a CAT(0) group
has BVC if and only if it is virtually cyclic.
Recall that a CAT(0) cube group is a group which acts properly and cocompactly on a
CAT(0) cube complex via isometries.
Lemma 4.15. Let G be a group which acts on a CAT(0) cube complex X properly and cocompactly via isometries and suppose that G is not virtually cyclic. Then either G contains
a rank one isometry or G contains a free abelian subgroup of rank 2.
Proof This is essentially due to Caprace and Sageev [4]. Note first that X is locally finite,
see for example [2, I.8.4(1)]. Note also that G acts on X without fixed points and essentially
(see [4, 1.1] for the terminology). Now by [4, Theorem A] and remarks below it, we have
that either X is a product of two unbounded CAT(0) cube subcomplexes or G contains an
element acting on X as a rank one isometry. Note that since G acts on X cocompactly, if X
is a product of two CAT(0) cube complexes, by [4, Corollary D], it follows that X contains
a free abelian subgroup of rank 2.


ON THE FINITENESS OF THE CLASSIFYING SPACE FOR THE FAMILY OF VIRTUALLY CYCLIC SUBGROUPS19

Corollary 4.16. Let G be a CAT(0) cube group. Then G has BVC if and only if G is
virtually cyclic.

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University of Bonn, Mathematical Institute, Endenicher Allee 60, 53115 Bonn, Germany
E-mail address: tvp@math.uni-bonn.de
Max Planck Institut für Mathematik, Vivatsgasse 7, 53111 Bonn, Germany
E-mail address: hsiaolei.wu@mpim-bonn.mpg.de

