A short note about diffuse Bieberbach groups

arXiv:1703.04972v3 [] 26 Sep 2017

Rafał Lutowski∗, Andrzej Szczepański∗ and Anna Gąsior†

Abstract
We consider low dimensional diffuse Bieberbach groups. In particular we classify diffuse Bieberbach
groups up to dimension 6. We also answer a question from [7, page 887] about the minimal dimension of
a non-diffuse Bieberbach group which does not contain the three-dimensional Hantzsche-Wendt group.

1

Introduction

The class of diffuse groups was introduced by B. Bowditch in [2]. By definition a group Γ is diffuse, if every
finite non-empty subset A ⊂ Γ has an extremal point, i.e. an element a ∈ A such that for any g ∈ Γ \ {1}
either ga or g −1 a is not in A. Equivalently (see [7]) a group Γ is diffuse if it does not contain a non-empty
finite set without extremal points.
The interest in diffuse groups follows from Bowditch’s observation that they have the unique product
property 1 . Originally unique products were introduced in the study of group rings of discrete, torsion-free
groups. More precisely, it is easily seen that if a group Γ has the unique product property, then it satisfies
Kaplansky’s unit conjecture. In simple terms this means that the units in the group ring C[Γ] are all trivial,
i.e. of the form λg with λ ∈ C∗ and g ∈ Γ. For more information about these objects we refer the reader to
[1], [9, Chapter 10] and [7]. In part 3 of [7] the authors prove that any torsion-free crystallographic group
(Bieberbach group) with trivial center is not diffuse. By definition a crystallographic group is a discrete and
cocompact subgroup of the group O(n) n Rn of isometries of the Euclidean space Rn . From Bieberbach’s
theorem (see [12]) the normal subgroup T of all translations of any crystallographic group Γ is a free abelian
group of finite rank and the quotient group (holonomy group) Γ/T = G is finite.
In [7, Theorem 3.5] it is proved that for a finite group G:
1. If G is not solvable then any Bieberbach group with holonomy group isomorphic to G is not diffuse.
2. If every Sylow subgroup of G is cyclic then any Bieberbach group with holonomy group isomorphic to
G is diffuse.
3. If G is solvable and has a non-cyclic Sylow subgroup then there are examples of Bieberbach groups
with holonomy group isomorphic to G which are and examples which are not diffuse.
Using the above the authors of [7] classify non-diffuse Bieberbach groups in dimensions ≤ 4. One of the
most important non-diffuse groups is the 3-dimensional Hantzsche-Wendt group, denoted in [11] by ∆P . For
the following presentation
∆P = hx, y | x−1 y 2 x = y −2 , y −1 x2 y = x−2 i
the maximal abelian normal subgroup is generated by x2 , y 2 and (xy)2 (see [6, page 154]). At the end of part
3.4 of [7] the authors ask the following question.
∗ Institute

of Mathematics, University of Gdańsk, ul. Wita Stwosza 57, 80-952 Gdańsk, Poland
of Mathematics, Maria Curie-Skłodowska University, Pl. Marii Curie-Skłodowskiej 1, 20-031 Lublin, Poland
E-mail addresses: rafal.lutowski@mat.ug.edu.pl, aszczepa@mat.ug.edu.pl, anna.gasior@poczta.umcs.lublin.pl
2010 Mathematics Subject Classification: Primary 20E40; Secondary 20H15, 20F65.
Key words and phrases: unique product property; diffuse groups, Bieberbach groups.
1 The group Γ is said to have the unique product property if for every two finite non-empty subsets A, B ⊂ Γ there is an
element in the product x ∈ AḂ which can be written uniquely in the form x = ab with a ∈ A and b ∈ B.
† Institute

c 2017. This manuscript version is made available under the CC-BY-NC-ND 4.0 license
http://creativecommons.org/licenses/by-nc-nd/4.0/

Question 1. What is the smallest dimension d0 of a non-diffuse Bieberbach group which does not contain
∆P ?
The answer for the above question was the main motivation for us. In fact we prove, in the next section,
that d0 = 5. Moreover, we extend the results of part 3.4 of [7] and with support of computer, we present the
classification of all Bieberbach groups in dimension d ≤ 6 which are (non)diffuse.

2

(Non)diffuse Bieberbach groups in dimension ≤ 6.

We use the computer system CARAT [10] to list all Bieberbach groups of dimension ≤ 6.
Our main tools are the following observations:
1. The property of being diffuse is inherited by subgroups (see [2, page 815]).
2. If Γ is a torsion-free group, N C Γ such that N and Γ/N are both diffuse then Γ is diffuse (see [2,
Theorem 1.2 (1)]).
Now let Γ be a Bieberbach group of dimension less than or equal to 6. By the first Betti number β1 (Γ)
we mean the rank of the abelianization Γ/[Γ, Γ]. Note that we are only interested in the case when β1 (Γ) > 0
(see [7, Lemma 3.4]). Using a method of E. Calabi [12, Propostions 3.1 and 4.1], we get an epimorphism
f : Γ → Zk , where k = β1 (Γ).

(1)

From the assumptions ker f is a Bieberbach group of dimension < 6. Since Zk is a diffuse group our problem
is reduced to the question about the group ker f.
Remark 1. Up to conjugation in GL(n + 1, R), Γ is a subgroup GL(n, Z) n Zn ⊂ GL(n + 1, Q), i.e. it is a
group of matrices of the form


A a
,
0 1
where A ∈ GL(n, Z), a ∈ Qn . If p : Γ → GL(n, Z) is a homomorphism which takes the linear part of every
element of Γ




A a
A a
p
= A for every
∈ Γ,
0 1
0 1
then there is an isomorphism ρ : G → p(Γ) ⊂ GL(n, Z). It is known that that the rank of the center of a
Bieberbach group equals the first Betti number (see [5, Proposition 1.4]). By [12, Lemma 5.2], the number
of trivial constituents of the representation ρ is equal to k. Hence without lose of generality we can assume
that the matrices in Γ are of the form


A B a
 0 I b
0 0 1
where A ∈ GL(n − k, Z), I is the identity matrix of
a ∈ Qn−k and b ∈ Qk . Then f may be defined by

A
f  0
0

degree k, B is an integral matrix of dimension n − k × k,
B
I
0


a
b  = b
1

and one can easily see that the map F : ker f → GL(n − k + 1, Q) given by




A B a
A a
F  0 I 0 =
0 1
0 0 1
is a monomorphism and hence its image is a Bieberbach group of rank n − k.

2

Now if Γ has rank 4 we know that the only non-diffuse Bieberbach group of dimension less than or equal
to 3 is ∆P . Using the above facts we obtain 17 non-diffuse groups. Note that the list from [7, section 3.4]
consists of 16 groups. The following example presents the one which is not in [7] and illustrates computations
given in the above remark.
Example 1. Let Γ be a crystallographic group denoted by "05/01/06/006" in [3] as a subgroup of GL(5, R).
Its non-lattice generators are as follows




0 1 −1 0 1/2
0 −1 1 0 1/2
0 1 0
−1 0 1 0
0 1/2
0 







0
0 
0 1 0 1/2 and B = −1 1 0
A= 0
.
 0 0 0 −1 0 
0
0 0 −1 1/2
0 0 0
0
1
0
0 0 0
1
Conjugating the above matrices by

1
0

Q=
1
0
0

1
1
0
0
0

0
0
0
1
0

0
0
1
0
0


0
0

0
 ∈ GL(5, Z)
0
1

one gets

1
0

AQ = 
0
0
0

0
−1
0
0
0

0
0
−1
0
0



−1
0 1/2
0
1 0 


Q

0 1/2
 and B =  0
0
1 0 
0
0 1

0 0
1 0
0 −1
0 0
0 0


−1 0
0 1/2

0
0 
.
1
0 
0
1

Now its easy to see that the rank of the center of Γ equals 1 and the kernel of the epimorphism Γ → Z is
isomorphic to a 3-dimensional Bieberbach group Γ0 with the following non-lattice generators:




−1 0 0
0
1 0
0 1/2
 0 1 0 1/2
0 −1 0
0 
0



A0 = 
0 0 −1 1/2 and B =  0 0 −1 0 
0 0 0
1
0 0
0
1
Clearly the center of Γ0 is trivial, hence it is isomorphic to the group ∆P .
Now we formulate our main result.
Theorem 1. The following table summarizes the number of diffuse and non-diffuse Bieberbach groups of
dimension ≤ 6.
Dimension
1
2
3
4
5
6

Total
1
2
10
74
1060
38746

Non-diffuse
0
0
1
17
352
19256

Diffuse
1
2
9
57
708
19490

Proof. If a group has a trivial center then it is not diffuse. In other case we use the Calabi (1) method and
induction. A complete list of groups was obtained using computer algebra system GAP [4] and is available
here [8].
Before we answer Question 1 from the introduction, let us formulate the following lemma:

3

Lemma 1. Let α, β be any generators of the group ∆P . Let γ = αβ, a = α2 , b = β 2 , c = γ 2 . Then the
following relations hold:
[a, b] = 1
aβ = a−1 aγ = a−1
α
−1
[a, c] = 1 b = b
bγ = b−1
(2)
α
−1
β
−1
[b, c] = 1 c = c
c =c
where xy := y −1 xy denotes the conjugation of x by y.
The proof of the above lemma is omitted. Just note that the relations are easily checked if consider the
following representation of ∆P as a matrix group




−1 0 0
0 +
1 0
0 1/2
*
 0 1 0 1/2
0 −1 0 1/2



x=
0 0 −1 0  , y =  0 0 −1 1/2 ⊂ GL(4, Q).
0 0 0
1
0 0
0
1
Proposition 1. There exists an example of a five dimensional non-diffuse Bieberbach group which does not
contain any subgroup isomorphic to ∆P .
Proof. Let Γ be the Bieberbach group enumerated in CARAT
elements γ1 , γ2 , l1 , . . . , l5 where



1
0 1 0 0
0
0
0

1 0 0 0
0
0




0 0 1 0
0 1/2
 and γ2 = 0
γ1 = 
0
0 0 0 −1 0 1/4



0
0 0 0 0 −1 0 
0
0 0 0 0
0
1

as "min.88.1.1.15". It generated by the
0
−1
0
0
0
0

0
0
−1
0
0
0


0 0 1/2
0 0 0 

0 0 0 

−1 0 0 

0 1 1/2
0 0 1

and l1 , . . . , l5 generate the lattice L of Γ:
li :=


I5
0

ei
1



where ei is the i-th column of the identity matrix I5 . Γ fits into the following short exact sequence
π

1 −→ L −→ Γ −→ D8 −→ 1
where π takes the linear part of every element of Γ:


A a
7 A
→
0 1
and the image D8 of π is the dihedral group of order 8.
Now assume that Γ0 is a subgroup of Γ isomorphic to ∆P . Let T be its maximal normal abelian subgroup.
Then T is free abelian group of rank 3 and Γ0 fits into the following short exact sequence
1 −→ T −→ Γ0 −→ C22 −→ 1,
where Cm is a cyclic group of order m. Consider the following commutative diagram
1

T ∩L

T

H = π(T )

1

1

L

π −1 (H)

H

1

We get that H must be an abelian subgroup of D8 = π(Γ) and T ∩ L is a free abelian group of rank 3 which
lies in the center of π −1 (H) ⊂ Γ. Now if H is isomorphic to either to C4 or C22 then the center of π −1 (H) is
4

of rank at most 2. Hence H must be the trivial group or the cyclic group of order 2. Note that as Γ0 ∩ L is
a normal abelian subgroup of Γ0 it must be a subgroup of T :
T ∩ L ⊂ Γ0 ∩ L ⊂ T ∩ L,
hence T ∩ L = Γ0 ∩ L. We get the following commutative diagram with exact rows and columns
1

1

1

1

T ∩L

T

H

1

1

Γ0 ∩ L

Γ0

G

1

1

1

C22

C22

1

1

1

1

where G = π(Γ0 ). Consider two cases:
1. H is trivial. In this case G is one of the two subgroups of D8 isomorphic to C22 . Since the arguments
for both subgroups are similar, we present only one of them. Namely, let
G = hdiag(1, −1, −1, −1, 1), diag(−1, −1, 1, 1, 1).i
In this case Γ0 is generated by the matrices of the form



−1
1 0
0
0 0 x1 − 21
0 −1 0
0

0
0
x
2



0 0 −1 0 0

x3 
 and β =  0
α=
0 0
0

0 −1 0
x4 


1
0 0
0
0
0 1 x5 − 2
0 0
0
0 0
1
0
where xi , yi ∈ Z for i = 1, . . . , 5. If c = (αβ)2 then

0 0 0
0 0 0

0 0 0
cα − c−1 = 
0 0 0

0 0 0
0 0 0

0
−1
0
0
0
0


0 0 0 y1 + 21
0 0 0 y2 − 12 

1 0 0
y3 
,
0 1 0 y4 + 12 

0 0 1
y5 
0 0 0
1

by Lemma 1 cα = c−1 , but

0 0
0

0 0
0


0 0
0
.

0 0
0

0 0 4y5 + 4x5 − 2
0 0
0

Obviously solutions of the equation 4y5 + 4x5 − 2 = 0 are never integral and we get a contradiction.
2. H is of order 2. Then G
γ1 γ2 L and γ2 L, hence

0 −1
1 0

0 0
α=
0 0

0 0
0 0

= D8 and H is the center of G. The generators α, β of Γ0 lie in the cosets
0
0
−1
0
0
0

0 0
0 0
0 0
1 0
0 −1
0 0

x1
x2 −
x3 −
x4 +
x5 +
1




1
1
0
2

1

2  and β = 0
1
0
4

1
0
2
0
5

0
−1
0
0
0
0

0
0
−1
0
0
0

0 0 y1 −
0 0
y2
0 0
y3
−1 0
y4
0 1 y5 −
0 0
1

1
2









1

2

where xi , yi ∈ Z for i = 1, . . . , 5, as before. Setting a = α2 , b = β 2 we get


0 0 0 0 0 2 − 4y1
0 0 0 0 0
0 


0 0 0 0 0
0 


ab − ba = 
0 
0 0 0 0 0

0 0 0 0 0
0 
0 0 0 0 0
0
and again the equation 2 − 4y1 = 0 does not have an integral solution.
The above considerations show that Γ does not have a subgroup which is isomorphic to ∆P .

Acknowledgements
This work was supported by the Polish National Science Center grant 2013/09/B/ST1/04125.

Bibliography
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