# Units of twisted group rings and their correlations to classical group rings

Geoffrey Janssens  ,  Eric Jespers  and  Ofir Schnabel  (Geoffrey Janssens)
Department of Mathematics and Data Science, Vrije Universiteit Brussel,
Pleinlaan $2$, 1050 Elsene, Belgium
E-mail address<EMAIL_ADDRESS>(Eric Jespers)
Department of Mathematics and Data Science, Vrije Universiteit Brussel,
Pleinlaan $2$, 1050 Elsene, Belgium
E-mail address<EMAIL_ADDRESS>(Ofir Schnabel)
Department of Mathematics, ORT Braude College, 2161002 Karmiel, Israel
E-mail address<EMAIL_ADDRESS>
###### Abstract.

Given a central extension $\Gamma$ of some normal subgroup $N$ by a group $G$,
we study the group ring $R\Gamma$ over some domain $R$. We obtain a direct sum
decomposition in terms of various twisted group rings of $G$ and concrete
information on the kernel and cokernel of the projections. This allows to
kick-start the investigation of the unit group ${\mathcal{U}}(R\Gamma)$ via
the unit group of twisted group rings. Among others we construct a new generic
family of units therein. As an application hereof we are able to obtain a
complete description of the unit group of the integral group ring of $G\times
C_{2}^{n}$ from data of $G$.

###### Key words and phrases:

Twisted group rings, unit group, projective representations

###### 1991 Mathematics Subject Classification:

16S35, 16H10, 16U60, 20C25

The first author is grateful to Fonds Wetenschappelijk Onderzoek vlaanderen -
FWO (grant 88258), and le Fonds de la Recherche Scientifique - FNRS (grant
1.B.239.22) for financial support.

###### Contents

  1. 1 Introduction
  2. 2 Cohomological preliminaries
  3. 3 On the decomposition of the twisted group algebras of general extensions and transgression
    1. 3.1 Concrete decomposition
    2. 3.2 Transgression and natural morphisms as projections
    3. 3.3 The case of an $\alpha$-representation group of $G$
  4. 4 When is the unit group of a twisted group ring finite?
  5. 5 Correlations between $R[\Gamma]$ and $R^{\alpha}[G]$ \- a unit group point of view
    1. 5.1 On the cokernel of the generalized transgression map
    2. 5.2 On the kernel of the generalized transgression map
  6. 6 Twisting bicyclic units and their contribution
    1. 6.1 Generalized byclic units versus elementary matrices
    2. 6.2 Twisted bicyclic units - general theory
    3. 6.3 On twisted bicyclic units in cokernel of the transgression map
  7. 7 Method to describe $\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times C_{2}^{n}])$
  8. 8 Full description for elementary abelian 2-groups
    1. 8.1 The starting case $C_{2}\times C_{2}$
    2. 8.2 The general case
  9. 9 Description of $\mathcal{U}({\mathbb{Z}}[D_{8}\times C_{2}^{n}])$

##  1\. Introduction

This paper contributes to the study of a finite group $\Gamma$ via its
representation theory over a number field111The most of the results will in
fact hold in the generality where $F$ is any field with ${\rm
char}(F)\nmid|\Gamma|.$ $F$ and its ring of integers $R$. The overarching
question is which group theoretical invariants are determined by the
$R$-algebra $R\Gamma$ or in other words by the regular $R\Gamma$-module. From
the literature one could somehow distil an approach in two steps to that.
Firstly one considers the decomposition of the group algebra $F\Gamma$ given
by Wedderburn-Artin’s theorem:

$F\Gamma\cong\operatorname{M}_{n_{1}}(D_{1})\oplus\cdots\oplus\operatorname{M}_{n_{q}}(D_{q}),$

where $D_{i}$ are finite dimensional division $F$-algebras. If one now chooses
an order $\mathcal{O}_{i}$ in each $D_{i}$, then we obtain two orders in
$F\Gamma$, namely $R\Gamma$ and
$\bigoplus_{i=1}^{q}\operatorname{M}_{n_{i}}(\mathcal{O}_{i})$. Because $F$ is
a number field, in particular $R$ is a ’nice’ Dedekind domain, it is well
known that the two orders share many properties. In particular their unit
groups have a common subgroup of finite index. The aim of the first step is
two obtain as much information as possible on the $F$-character degrees
$n_{i}$ and the form of the division algebras $D_{i}$. Consequently, it is to
be expected that the number theoretical properties of the number field $F$
will play a special role here.

Recall that there is a bijection between the matrix components above, the
absolutely irreducible $F$-characters and the primitive central idempotents of
$F\Gamma$ (a set denoted $\text{PCI}(F\Gamma)$). A setback of making the
switch from $R\Gamma$ to
$\bigoplus_{i=1}^{q}\operatorname{M}_{n_{i}}(\mathcal{O}_{i})$ is that one is
somehow replacing the group $\Gamma$ by the larger group
$\prod_{e\in\text{PCI}(F\Gamma)}\Gamma e$. Due to this one loses information
on ’ties’. The role of the second step consists then to focus on the given
group $\Gamma$. This might be via representation theoretical methods or group
theoretical ones. To clarify the latter, we now set $F={\mathbb{Q}}$ and so
$R={\mathbb{Z}}$. Then it is well-known that the group isomorphism type of the
unit group ${\mathcal{U}}({\mathbb{Z}}\Gamma)$ and the ring isomorphism type
of ${\mathbb{Z}}\Gamma$ contain the same information222Thus
${\mathbb{Z}}\Gamma\cong{\mathbb{Z}}H$ if and only if
${\mathcal{U}}({\mathbb{Z}}\Gamma)\cong{\mathcal{U}}({\mathbb{Z}}H)$. It is
folklore that it holds more generally for any $\Gamma$-adapted coefficient
ring $R$, i.e. where all prime divisors of $|\Gamma|$ are not divisible in
${\mathcal{U}}(R)$. In short, this follows in that case from the linear
independence of finite subgroups of ${\mathcal{U}}(RG)$.. The gain of this is
that ${\mathcal{U}}({\mathbb{Z}}\Gamma)$ is an arithmetic subgroup in some
linear semsimple algebraic group (in particular it is a finitely presented
group), allowing the use of classical but strong methods in algebraic groups
or geometric group theory.

Briefly said, in this article we will contribute to both steps in a novel way
by regrouping simple components in certain twisted group algebras. More
specifically, we will view $\Gamma$ as some extension

(1) $1\rightarrow N\rightarrow\Gamma\overset{\lambda}{\rightarrow}G\rightarrow
1$

and aim at describing $R\Gamma$ and its unit group in terms of certain twisted
group rings of the smaller group $G$. Before going into details, we first
recall the definition of a twisted group ring. For a $2$-cocycle $\alpha\in
Z^{2}(G,R^{*})$, the twisted group ring $R^{\alpha}[G]$ of $G$ over $R$ with
respect to $\alpha$ is the free $R$-module with basis $\\{u_{g}\\}_{g\in G}$
where the multiplication on the basis is defined via

$u_{g}u_{h}=\alpha(g,h)u_{gh}\ \ \text{for all}\ \ g,h\in G$

and any $u_{g}$ commutes with the elements of $R$. Note that the ring
structure of $R^{\alpha}[G]$ depends only on the cohomology class $[\alpha]\in
H^{2}(G,R^{*})$ of $\alpha$ and not on the particular $2$-cocycle. Of
importance is that there is a 1-1 correspondence between $\alpha$-projective
representations of $G$ and $R^{\alpha}[G]$-modules. As such projective
representations, although not explicity used, are recurrent objects behind the
scenes (see [6, §2.1] for detailed explanation).

In order to keep the introduction notationally light, we will restrict
ourselves to abelian extensions (i.e. $\Gamma$ is split into an extension
where $N$ is abelian). However from Section 2 till Section 5 we will work with
general extensions. When $N$ is abelian, the extension (1) corresponds333The
necessary background is briefly introduced in Section 2. to a cohomology class
$[\alpha]\in H^{2}_{\sigma}(G,N)$ where $\sigma$ is the action of $G$ on $N$.
Via $\sigma$ the group $G$ acts on the set $\operatorname{Lin}(N,F)$ of linear
$F$-characters of $N$ and we denote the orbit space by
$\operatorname{Lin}(N,F)/G$. Now, with a $G$-invariant linear character $\chi$
of $N$ over $F$ one can associate via the transgression map a new $2$-cocycle
$[T_{\alpha}(\chi)]\in H^{2}(G,F^{*})$ where
$T_{\alpha}(\chi)(g,h):=\chi(\alpha(g,h)).$

######  Theorem A (Theorem 3.2 & Proposition 3.1).

With notations as above, $N$ abelian and $[\beta]\in H^{2}(\Gamma,F^{*})$
inflated from $G$. We have that

(2)
$F^{\beta}[\Gamma]\cong\bigoplus_{[\chi]\in\operatorname{Lin}(N,F)/G}(F(\chi)E_{\chi})*G$

for some concrete idempotents $E_{\chi}$ and explicit skewing and twisting of
the crossed product $(F(\chi)E_{\chi})*G$. In particular, if
$\chi\in\operatorname{Lin}(N,F)^{G}$ is a $G$-invariant character, then444With
$\beta.T_{\alpha}(\chi)$ is meant the $2$-cocycle of $G$ with values in
$F(\chi)^{*}$ defined pointwise, i.e.
$(\beta.T_{\alpha}(\chi))(g,h)=\beta(g,h).T_{\alpha}(\chi)(g,h)=\beta(g,h).\chi(\alpha(g,h))$.

$(F(\chi)E_{\chi})*G\cong F(\chi)^{\beta.T_{\alpha}(\chi)}[G].$

It is interesting to pause a second on the classical case when $\beta$ is
trivial, i.e. $F^{\beta}[\Gamma]=F\Gamma$ is a group ring. The above theorem
then tells that even if one is solely interested in non-twisted group rings,
one should still study twisted group rings over finite extensions of the
chosen number field $F$. Important for the application later on is that the
decomposition (2) is not simply an abstract one. Among others, the projections
$p_{\chi}$ onto the direct summands have a down-to-earth description as seen
below. For this we need to fix a section $\mu:G\rightarrow\Gamma$ of $\lambda$
in (1).

######  Proposition B (Proposition 3.3).

Let $R$ be any subring of $F$ and let
$\chi\in\operatorname{Lin}(N,F^{*})^{\Gamma/N}$, using the notations as above.
Then, the projection $p_{\chi}$ viewed over $R$ agrees with the ring
epimorphism

$\Psi_{\chi,\beta}:R^{\beta}[\Gamma]\rightarrow
R[\chi]^{\beta.T_{\alpha}(\chi)}[G]:r\,u_{n.\mu(g)}\mapsto r\chi(n)v_{g}.$

At this point one has another point of view on the first step alluded to at
the start of the introduction. Now we continue with the second part that
focuses on a better understanding of the unit group
${\mathcal{U}}(R^{\beta}[\Gamma])$ for some order $R$ in $F$. For still a
general number field $F$ and extension $\Gamma$, this is the content of
Section 5. In that part we investigate the kernel and cokernel of the induced
group morphism $\widetilde{\Psi}_{\chi,\beta}$ between
${\mathcal{U}}(R^{\beta}[\Gamma])$ and
${\mathcal{U}}(R[\chi]^{\beta.T_{\alpha}(\chi)}[G])$.

######  Theorem C (Theorem 5.2 & Theorem 5.4).

Let $\Gamma$ be some extension (1), $[\beta]\in H^{2}(\Gamma,F^{*})$ inflated
from $G$ and $\chi\in\operatorname{Lin}(N,F)^{G}$. Also let $R$ be an order in
the number field $F$. Then555Again, in order to avoid more notations in the
introduction, some parts of the statements are left vague.:

  1. (1)

$\operatorname{Coker}(\widetilde{\Psi}_{\chi,\beta})$ is finite.

  2. (2)

If $N$ is central, then $\\{\text{torsion units in
}\ker(\widetilde{\Psi}_{\chi,\beta})\\}=\\{\chi(a)^{-1}a\mid a\in N\\}$.

Moreover, we obtain necessary and sufficient conditions for
$\ker(\widetilde{\Psi}_{\chi,\beta})$ to be finite and some computational
reduction for determining
$|\operatorname{Coker}(\widetilde{\Psi}_{\chi,\beta})|$.

Along the way we obtain a version for twisted group algebras of certain
classical theorems of Higman and Berman-Higman. More precisely, in Theorem 4.1
we describe when the unit group of an (aribtrary) twisted group ring is finite
and in Theorem 5.5 we show that torsion units must have trace zero.

All the results till now give new methods to compute ${\mathcal{U}}(R\Gamma)$
for some extension $\Gamma$ which is especially explicit and hence useful for
a central extension. As such, one is now reduced to understanding the unit
group of a twisted group ring over the smaller group $G$, which is a topic of
independent interest. In Section 6 we kick start the investigations of generic
constructions of units therein. More precisely, we now restrict to the case
that $F$ is a cyclotomic field ${\mathbb{Q}}(\zeta_{n})$, with $\zeta_{n}$
some primitive root of unity, $R$ its ring of integers and $[\gamma]\in
H^{2}(G,R^{*})$.

Most notably, we construct a new class of units $\mathcal{B}^{tw}$, see
Definition 6.7, in ${\mathcal{U}}(R^{\gamma}[G])$ which truly makes use of the
twisting $[\gamma]$. These elements might be thought as some deformations of
the classical bicyclic units in (non-twisted) group rings. Due to this there
holds a variant of a theorem of Jespers-Leal [11] which tells us that
$\mathcal{B}^{tw}$ contains enough elementary matrices of each simple
component. More concretely, if $R={\mathbb{Z}}$ and the
$\gamma$-representation group $G_{\gamma}$ (see Section 3.3) is fixed-point
free, then for every $e\in\text{PCI}({\mathbb{Q}}^{\gamma}[G])$ such that
${\mathbb{Q}}^{\gamma}[G]e\cong M_{n}(D)$ with $n\geq 2$ there exists a number
$y\in{\mathbb{Z}}$ such that $1-e+E_{n}(y\mathcal{O})\subset\mathcal{B}^{tw}$
(where $\mathcal{O}$ is an order in the finite dimension division
${\mathbb{Q}}$-algebra $D$).

An intriguing feature of the twisted bicyclic units, which was also the
starting motivation for this project, is that the ’smallest’ ones seem to live
in
$\operatorname{Coker}\left(\widetilde{\Psi}_{\chi}:{\mathcal{U}}({\mathbb{Z}}\Gamma)\rightarrow{\mathcal{U}}({\mathbb{Z}}^{T_{\alpha}(\chi)}[G])\right)$
and hence are truly new type of units from the classical perspective:

######  Question D (6.10).

Is the subgroup of $\mathcal{B}^{tw}(G)$ generated by the ’minimal’ twisted
bicyclic units, as defined in (20), a non-trivial subgroup in
$\operatorname{Coker}(\widetilde{\Psi_{\chi}})$ of finite exponent?

Under the restrictions that ${\mathcal{U}}({\mathbb{Z}}G)$ is finite we are
able in Proposition 6.9 to give a positive answer to the previous question.

Finally, in the last sections we apply all the above machinery to study the
case that $\Gamma=G\times C_{2}^{m}$ for some $m$. Note that the number of
simple components of the group algebra increase exponentially with $m$, which
makes this extensions more subtle as it first looks like. In Section 7 we give
a description of ${\mathcal{U}}({\mathbb{Z}}^{\gamma}[G\times C_{2}^{m}])$ for
$[\gamma]\in H^{2}(G\times C_{2}^{m},{\mathbb{Z}}^{*})$ in terms of
${\mathcal{U}}({\mathbb{Z}}^{\rm Res(\gamma)}[G])$, see Theorem 7.3. From this
we are able to deduce the following:

######  Theorem E (Corollary 7.4).

Let $G$ be a finite group and $[\gamma]\in H^{2}(G\times
C_{2}^{m},{\mathbb{Z}}^{*})$ inflated from a cocycle $[\delta]\in
H^{2}(G,{\mathbb{Z}}^{*})$. If $G$ has a (torsion-free) complement in
${\mathcal{U}}({\mathbb{Z}}^{\delta}[G])$ or satisfies one of the Zassenhaus
conjectures, then the same holds for $G\times C_{2}^{m}$ and
${\mathcal{U}}({\mathbb{Z}}^{\gamma}[G\times C_{2}^{m}])$.

To finish, in Section 8 all the protagonists introduced till now are computed
explicitly in the case that $G\cong D_{8}$. More precisely, we consider
$D_{8}$ as some extension $[\alpha]\in H^{2}(C_{2}\times C_{2},C_{2})$ and
look at one of the projections
$\widetilde{\Psi}_{\chi}:{\mathcal{U}}({\mathbb{Z}}[D_{8}\times
C_{2}^{m}])\rightarrow{\mathcal{U}}({\mathbb{Z}}^{T_{\alpha}(\chi)}[C_{2}^{m+2}])$.
For this map we show that the minimal twisted bicyclic units are exactly the
generators of $\operatorname{Coker}(\widetilde{\Psi}_{\chi})$, see Lemma 9.1.
We expect that this happens in much greater generality. Together with the
general description of $\ker(\widetilde{\psi}_{\chi})$ and E we are able to
pullback a precise description of
${\mathcal{U}}({\mathbb{Z}}^{T_{\alpha}(\chi)}[C_{2}^{m+2}])$, obtained in
Proposition 8.3. Altogether, the main achievement here is a unexpected uniform
description of ${\mathcal{U}}({\mathbb{Z}}[D_{8}\times C_{2}^{m}])$ for all
$m$. This was unexpected due to the fact that with $m$ increasing the
congruence subgroups lie always deeper.

Notational conventions:

  1. (1)

All rings, denoted $R$, will be assumed unital and associative.

  2. (2)

If $f:R\rightarrow S$ is a ring homomorphism, then we denote the induced map
on the unit groups by $\widetilde{f}:\mathcal{U}(R)\rightarrow\mathcal{U}(S)$.

  3. (3)

$\mathcal{U}(R)$ denotes the unit group of $R$. If ${\mathcal{U}}(R)$ appears
as the coefficients of a cohomology group, i.e. $H^{j}(G,{\mathcal{U}}(R))$,
we will instead write $R^{*}$ (thus $H^{j}(G,R^{*})$).

  4. (4)

Except stated otherwise, with ’an order $R$’ we will mean a
${\mathbb{Z}}$-order.

  5. (5)

We will use the convention $g^{h}=h^{-1}gh$ and $\text{conj}(h)(g)=g^{h}$ for
conjugation.

Acknowledgment. We are grateful to Yuval Ginosar for valubale discussions and
for sharing with us example 3.6. During the early stages of this project the
first author was in $2017$ a visitor at the Technion - Israel Institute of
Technology and he would like to thank the institution for the pleasant working
conditions. Thereafter the third author visited several times the VUB and he
warmly thanks the host institution and the mathematics department at VUB.

##  2\. Cohomological preliminaries

Given a finite group $G$ with an action
$\varphi:G\rightarrow\operatorname{Aut}(A)$ on an abelian group $A$,
extensions666Thus $A$ is normal and $G\cong\Gamma/A$. In other words (3) is a
short exact sequence.

(3) $1\rightarrow A\rightarrow\Gamma\overset{\lambda}{\rightarrow}G\rightarrow
1.$

of $A$ by $G$ whose induced action is $\varphi$ are parametrised by the second
group cohomology group $H_{\varphi}^{2}(G,A)$. If the action of $G$ on $A$ is
trivial, in which case we write $H^{2}(G,A)$, the cohomology group
parametrizes the central extensions (i.e. $\mathcal{Z}(A)\subseteq\Gamma$).
Given a $2$-cocycle $\alpha\in Z^{2}(G,R^{*})$ and a commutative ring $R$ one
can associate the twisted group ring $R^{\alpha}[G]$ whose basis we will
denote by $\\{u_{g}\mid g\in G\\}$. Up to $R$-algebra isomorphism,
$R^{\alpha}[G]$ does not depend on $\alpha$ but only on the cohomology class
$[\alpha]\in H^{2}(G,R^{*})$. A $2$-cocycle $\alpha$ is called normalized if
$u_{1}$ is the identity element of $R^{\alpha}[G]$, i.e. if
$\alpha(1,g)=1=\alpha(g,1)$ for all $g\in G$. Notice also that then $u_{1}$ is
the identity of $R^{\alpha}[G]$ and the ring $R=Ru_{1}$ is central in the
twisted group ring $R^{\alpha}[G]$.

######  Example 2.1.

A cohomology class $[\alpha]\in H^{2}(G,{\mathbb{Z}}^{*})$ over ${\mathbb{Z}}$
corresponds to a central extension

(4) $[\alpha]:\quad 1\rightarrow C_{2}\rightarrow\Gamma\rightarrow
G\rightarrow 1.$

As a result, for an abelian group $G$ generated by
$\\{g_{1},g_{2},\ldots,g_{r}\\}$, $[\alpha]\in H^{2}(G,{\mathbb{Z}}^{*})$ is
determined by the values in $\\{\pm 1\\}$ of

(5) $u_{g_{i}}^{\circ(g_{i})},\text{ }1\leq i\leq r,\text{ and the commutators
}[u_{g_{i}},u_{g_{j}}],\text{ }1\leq i<j\leq r,$

in the corresponding twisted group ring ${\mathbb{Z}}^{\alpha}[G]$.

Let $Q$ be a normal subgroup of $G$, let $M$ be an abelian group which we
equip with a trivial $G$-action and let $\pi:G\rightarrow G/Q$ be the quotient
map. Then, for any $\delta\in Z^{i}(G/Q,M)\subset\operatorname{Map}(G/Q,M)$ we
can define $\gamma\in Z^{i}(G,M)$ by

$\gamma(x_{1},\ldots,x_{i})=\delta(\pi(x_{1}),\ldots,\pi(x_{i})).$

The map from $Z^{i}(G/Q,M)$ to $Z^{i}(G,M)$ sending $\delta$ to $\gamma$
induces777It also yields a map from $Z^{2}(G/Q,M)$ to $Z^{2}(G,M)$ which we
also denote by $\operatorname{Inf}$. a map

$\operatorname{Inf}:H^{i}(G/Q,M)\rightarrow H^{i}(G,M)$

which is called the inflation map. Sometimes we will want to emphasize $Q$ and
use the notation $\operatorname{Inf}_{Q}$. Notice also, that for any subgroup
$H$ of $G$, there is a natural restriction map from $H^{i}(G,M)$ to888If $H=Q$
is normal, then ${\rm Im}(\rm Res)$ is easily seen to be contained in the
subgroup $H^{i}(Q,M)^{G/Q}$ of $G/Q$-invariant cocycles. $H^{i}(H,M)$. In case
of a trivial action the Lyndon-Hochschild-Serre spectral sequence yields a
very concrete exact sequence between the lower cohomology groups.

######  Lemma 2.2 (Inflation-Restriction exact sequence).

Let $Q$ be a normal subgroup of $G$ and $M$ a trivial $G$-module. Then one has
an exact sequence

$0\rightarrow
H^{1}(G/Q,M)\overset{\operatorname{Inf}}{\rightarrow}H^{1}(G,M)\overset{\rm
Res}{\rightarrow}H^{1}(Q,M)^{G/Q}\overset{d}{\rightarrow}H^{2}(G/Q,M)\overset{\operatorname{Inf}}{\rightarrow}H^{2}(G,M)$

where the first connecting map $d$ is the transgression map and
$H^{1}(\cdot,\cdot)=\operatorname{Hom}(\cdot,\cdot)$.

In case of the abelian extension (3), i.e. taking $\Gamma$ and $A$ (as
respectively $G$ and $Q$) the first transgression map is a morphism from
$\operatorname{Hom}(A,M)^{G}$ to $H^{2}(G,M)$ whose definition we now recall.
For $\chi\in\operatorname{Hom}(A,M)^{G}$ one can define a $2$-cocycle
$\operatorname{Tra}(\chi)\in Z^{2}(G,M)$ via

$\operatorname{Tra}(\chi)(g_{1},g_{2})=\chi(\alpha(g_{1},g_{2}))$

for any $g_{1},g_{2}\in G$. The cohomology class $[\operatorname{Tra}(\chi)]$
does not depend on the choice of $\alpha$.

######  Definition 2.3.

With the above notations, the map

$\operatorname{Tra}_{\alpha}:\operatorname{Hom}(A,M)^{G}\rightarrow
H^{2}(G,M):\chi\mapsto T_{\alpha}(\chi):=[\operatorname{Tra}(\chi)]$

is called the (first) _transgression map_ associated to $\alpha$.

Using notations as in (3). Let $Q$ be a normal subgroup of $\Gamma$ such that
$A\cap Q=1$. Then $Q\cong\lambda(Q)$ and from now on we will implicitly
identify the both and speak about $G/Q$.

######  Remark 2.4.

Notice that $G/\lambda(Q)\cong G/Q$ and $AQ/Q\cong A/A\cap Q=A$. Hence, if
$\alpha=\operatorname{Inf}_{Q}(\gamma)$ then $\Gamma/Q$ is the central
extension

(6) $1\rightarrow
A\rightarrow\Gamma/Q\overset{\lambda}{\rightarrow}G/Q\rightarrow 1,$

corresponding to $[\gamma]$.

In Section 5.2 we will consider various such normal subgroups $Q$ and the
(co)kernel of the transgression map associated to $\Gamma/Q$. Therefore we
also write $\operatorname{Tra}_{Q}$ for the transgression map associated to
the abelian extension $(\ref{CE over N})$. When the extension or $Q$ are clear
from the context we write $\operatorname{Tra}$.

Finally we record a corollary of $\alpha(g,1)=\alpha(1,1)=\alpha(1,g)$ for all
$g\in\Gamma$, entailed by the $2$-cocycle condition, which will repeatedly be
used without further notice.

######  Corollary 2.5.

Let $G$ be a group, $Q$ a normal subgroup and suppose that $[\alpha]\in
H^{2}(G,R^{*})$ is inflated from a cohomology class $[\gamma]\in
H^{2}(G/Q,R^{*})$. Then, for $g\in G$ and $x\in Q$, $u_{g}$ and $u_{x}$
commute in $R^{\alpha}[G]$ whenever $g$ and $x$ commute in $G$. In particular,
$u_{a}$ is central in $R^{\alpha}[G]$ for any $a\in\mathcal{Z}(Q)$.

Standing assumptions: All cohomology groups will be with respect to a trivial
action, except stated otherwise. Also, given $[\alpha]\in H^{2}(G,A)$, we
always assume a chosen representative and we assume $\alpha\in Z^{2}(G,A)$ to
be normalized.

##  3\. On the decomposition of the twisted group algebras of general
extensions and transgression

In this section, given a central extension as in (3), we will recover the
decomposition [18, Theorem 5.3.] of the twisted group algebra
$F^{\beta}[\Gamma]$ for any $[\beta]\in H^{2}(\Gamma,F^{*})$ inflated from $G$
and $F$ a field with ${\rm char}(F)\nmid|\Gamma|$. Our aim is to use other
methods which allow to work with general extensions (i.e. with $A$ not
necessarily abelian) and to interpret in Section 3.2 the projection maps as a
kind of transgression morphisms. The most general structural result will be
Proposition 3.1, but Theorem 3.2 will be more precise.

###  3.1. Concrete decomposition

Consider an arbitrary short exact sequence

(7) $1\rightarrow N\rightarrow\Gamma\overset{\lambda}{\rightarrow}G\rightarrow
1.$

Fix also a section $\mu$ of $\lambda$. In particular
$\alpha(g,h):=\mu(g)\mu(h)\mu(gh)^{-1}\in N$ for all $g,h\in G$ and
conjugation with $\mu(g)$ gives an outer automorphism of $N$:

(8)
$\sigma:G\rightarrow\operatorname{Out}(N):=\operatorname{Aut}(N)/\operatorname{Inn}(N):g\mapsto\text{conj}(\mu(g)^{-1})$

It is well-known that the associativity of $\Gamma$ means that
$\alpha(g,h)\alpha(gh,k)=\sigma_{g}(\alpha(h,k))\alpha(g,hk).$

Let $F$ be any field with ${\rm char}(F)\nmid|\Gamma|$ and
$[\beta]\in\operatorname{im}(\operatorname{Inf}:H^{2}(G,F^{*})\rightarrow
H^{2}(\Gamma,F^{*}))$. Note that

(9) $F^{\beta}[\Gamma]\cong(FN)*G=\sum_{g\in G}(FN)v_{g}$

is a crossed product of $G$ over $FN$ with the following operations999In
$F^{\beta}[\Gamma]$ an arbitrary basis element has the form $u_{a}u_{\mu(g)}$
for some $a\in N,g\in G$. We are identifying it with $av_{g}$ where $a$ is
considered as a coefficient in $F[N]$. Implicitly we are extending the action
$\sigma$ of $G$ on $N$ to $FN$.:

$v_{g}v_{h}=\beta(g,h)\alpha(g,h)v_{gh}\qquad\text{(twisting)}$

and, for $x\in FN$,

$v_{g}x=\sigma_{g}(x)v_{g}.\qquad\text{(skewing)}$

To give a more precise description we need to consider

$\operatorname{Irr}(N,F):=\\{\text{absolutely irreducible }F\text{-characters
of }N\\}$

and

$\operatorname{Lin}(N,F):=\\{\chi\in\operatorname{Irr}(N,F)\mid\chi(1)=1\\}.$

Recall that, [9, Theorem 3.3.1], $\operatorname{Irr}(N,F)$ can be computed
from $\operatorname{Irr}(N,{\mathbb{C}})$ via Galois descent and that it is in
bijective correspondence with $\text{PCI}(FN)$, the primitive central
idempotents of $FN$. Denote by $e_{\chi}$ be the idempotent corresponding to
$\chi\in\operatorname{Irr}(N,F)$. Note that $G$, interpreted as $\Gamma/N$,
acts on $\operatorname{Irr}(N,F)$ via $\chi^{g}(n):=\chi(\sigma_{g}(n))$ and
this also induces an action on $\text{PCI}(FN)$ (by
$e_{\chi}^{g}:=e_{\chi^{g}}$). The orbit space will be denoted
$\operatorname{Irr}(N,F)/G$.

######  Proposition 3.1.

With notation as above, we have the following:

$F^{\beta}[\Gamma]\cong\bigoplus_{[\chi]\in\operatorname{Irr}(N,F)/G}\big{(}(FNE_{\chi})*G\big{)}$

where $E_{\chi}:=\sum_{\chi^{g}\in[\chi]}e_{\chi}^{g}$ and the skewing of
$(FNE_{\chi})*G$ is given by $\sigma$ and the twisting by101010The notation
needs clarification: we implicitly consider $\beta$ and $\alpha$ both as
having image in $FN$ and hence
$(\beta\cdot\alpha)(g,h):=\beta(g,h).\alpha(g,h)$. Subsequently,
$E_{\chi}.(\beta\cdot\alpha)$ also multiplies the result with the central
idempotent $E_{\chi}$. $E_{\chi}.(\beta\cdot\alpha)$. Moreover, if111111We
denote by $\operatorname{Lin}(N,F)^{\Gamma/N}$ the set of $\Gamma/N$-invariant
linear characters. $\chi\in\operatorname{Lin}(N,F)^{\Gamma/N}$

(10) $(FNE_{\chi})*G\cong F(\chi)^{\beta.T_{\alpha}(\chi)}[G]$

where $F(\chi)$ is the smallest field containing $F$ and the values of $\chi$.

###### Proof.

As mentioned earlier, $F^{\rm
Res(\beta)}[N]=FN=\bigoplus_{\chi\in\operatorname{Irr}(N,F)}FNe_{\chi}.$ In
general the orthogonal idempotents $e_{\chi}$ are not central in
$F^{\beta}[\Gamma]$, however the element $E_{\chi}$ (being the sum of the
orbit under conjugation by $G$) is. These elements are again two-by-two
orthogonal idempotents, hence (9) can be rewritten as
$F^{\beta}[\Gamma]=\bigoplus_{[\chi]\in\operatorname{Irr}(N,F)/G}(FNE_{\chi})*G$.
Since $E_{\chi}$ is central we see that indeed only the twisting changes and
in the way asserted.

If $\chi\in\operatorname{Irr}(N,F)^{\Gamma/N}$ is an invariant character then
its orbit is a singleton and hence $E_{\chi}=e_{\chi}$. Now suppose
$\chi\in\operatorname{Lin}(N,F)^{\Gamma/N}$ is linear. Then multiplying with
$E_{\chi}$ is projecting to the $1$-dimensional component of $FN$ associated
to $\chi$. In other words, for $n\in N$,

(11) $nE_{\chi}=ne_{\chi}=\chi(n)e_{\chi}\quad\text{ and }\quad
FNE_{\chi}\cong F(\chi).$

Also, from the earlier reformulation of the associativity of $\Gamma$, it
readily follows that $\alpha_{\chi}:=\operatorname{Tra}_{\alpha}(\chi)$
satisfies the $2$-cocycle condititon

$\alpha_{\chi}(g,h)\alpha_{\chi}(gh,k)=\alpha_{\chi}(h,k)\alpha_{\chi}(g,hk).$

In particular, $(FNE_{\chi})*G$ is a twisted group algebra where, by the
above, the twisting is indeed
$E_{\chi}.\beta\cdot\alpha=\beta\cdot\operatorname{Tra}_{\alpha}(\chi)$. ∎

In order to obtain a more insightful decomposition we now assume that $N$ is
abelian (and denote it $A$). In this case the extension (7) corresponds to the
cohomology class121212Thus we might have a non-trivial action on the
coefficients given by $\sigma$.  $[\alpha]\in H^{2}_{\sigma}(G,A)$ and
$\operatorname{Irr}(A,F)\subseteq\operatorname{Hom}(A,\overline{F}^{*})$.
Moreover the theorem of Perlis-Walker [9, Theorem 3.3.6.] tells us that

(12) $FA=\bigoplus_{d\mid|A|}F(\zeta_{d})^{\oplus a_{d}},$

where
$a_{d}=k_{d}\frac{[{\mathbb{Q}}[\xi_{d}]:{\mathbb{Q}}]}{[F[\xi_{d}]:F]}$, with
$k_{d}$ is the number of cyclic subgroups of $A$ of order $d$ and $\zeta_{d}$
a primitive $d$-th root of unity. Proposition 3.1 now readily translates into
the following:

######  Theorem 3.2.

With notations as above we have that:

  1. (1)

If $N$ is abelian, then

$F^{\beta}[\Gamma]\cong\bigoplus_{[\chi]\in\operatorname{Lin}(N,F)/G}\big{(}\bigoplus_{\chi\prime\in[\chi]}F(\chi\prime)e_{\chi\prime}*G\big{)}=\bigoplus_{[\chi]\in\operatorname{Lin}(N,F)/G}(F(\chi)E_{\chi})*G$

with the skewing of $(F(\chi)E_{\chi})*G$ given by $\sigma$ and the twisting
by $E_{\chi}.(\beta\cdot\alpha)$.

  2. (2)

If $N$ is central131313As mentioned earlier, this case was already obtained in
[18, Theorem 5.3.], then

$F^{\beta}[\Gamma]\cong\bigoplus_{\chi\in\operatorname{Lin}(N,F)}F(\chi)^{\beta.T_{\alpha}(\chi)}[G].$

###### Proof.

The first part is a direct application of Proposition 3.1 and remarking that
now $\operatorname{Lin}(N,F)=\operatorname{Irr}(N,F)$ and
$F(\chi)=F(\chi\prime)$ for two conjugated characters $\chi$ and $\chi\prime$.
When $N$ is central then
$\operatorname{Irr}(A,F)=\operatorname{Lin}(A,F)^{\Gamma/A}$ and
$e_{\chi}^{g}=e_{\chi}$. Therefore the second part follows from the first and
the last assertion in Proposition 3.1. ∎

###  3.2. Transgression and natural morphisms as projections

Consider again the general extension (7). The projection in Proposition 3.1 of
$F^{\beta}[\Gamma]$ on $FNE_{\chi}*G$, denoted $p_{\chi}$, is given by
multiplying with the central idempotent $E_{\chi}$. If
$\chi\in\operatorname{Lin}(N,F^{*})^{\Gamma/N}$ then this idempotent
$E_{\chi}$ and, in particular, the isomorphism (10) can be made explicit and
moreover we recover some classical constructions (see Example 3.4). More
generally, let $R$ be a domain and $F$ its field of fractions. For
$\chi\in\operatorname{Lin}(N,F^{*})^{\Gamma/N}$ we need its $R$-linear
extension

$\chi_{R}:RN\rightarrow R[\chi]:\sum_{a\in N}r_{a}a\mapsto\sum_{a\in
N}r_{a}\chi(a)$

which is an $R$-algebra map whose kernel we denote by
$I_{\chi}:=\ker(\chi_{R})$.

######  Proposition 3.3.

Let $\Gamma,N,G,F$ be as in Section 3.1,
$[\beta]\in\operatorname{im}(\operatorname{Inf}:H^{2}(G,R^{*})\rightarrow
H^{2}(\Gamma,R^{*}))$ and $\chi\in\operatorname{Lin}(N,F^{*})^{\Gamma/N}$.
Then, the projection $p_{\chi}$ viewed over $R$ agrees with the map defined by
(for $n\in N$ and $g\in G$)

(13) $\Psi_{\chi,\beta}:R^{\beta}[\Gamma]\rightarrow
R[\chi]^{\beta.T(\chi)}[G]:r\,u_{n.\mu(g)}\mapsto r\chi(n)v_{g}.$

which is a ring epimorphism with
$\ker(\Psi_{\chi,\beta})=R^{\beta}[\Gamma]I_{\chi}=R^{\beta}[\Gamma](1-e_{\chi})\cap
R^{\beta}[\Gamma]$. In particular,

$R^{\beta}[\Gamma]/R^{\beta}[\Gamma]I_{\chi}\cong R^{\beta\cdot T(\chi)}[G].$

Moreover, if $\operatorname{im}(\chi)\subseteq R^{*}$, then $I_{\chi}$ is a
free $R$-module with $R$-basis $\\{u_{a}-\chi(a)u_{1}\mid a\in A\\}$.

When $\beta$ is trivial we will simply write $\Psi_{\chi}$. If $R$ is a field
and $N$ central the most of the previous result was known141414If $\chi$ is
the trivial character and $N$ arbitrary, then the statement has to be compared
with [14, Lemma 3.2.12]. (e.g. if $\beta$ is trivial in [14, Theorem 3.2.8.]
and for $\beta\in\operatorname{im}(\operatorname{Inf}_{N})$ in [18,
Proposition 5.2.]).

######  Example 3.4.

  * •

If $\chi$ is the trivial character, i.e. $\chi(n)=1$ for all $n\in N$, then
$\chi_{R}$ is the augmentation map of $RN$. In that case, writing
$[\beta]=\operatorname{Inf}([\gamma])$,

$\Psi_{\chi,\beta}:R^{\beta}[\Gamma]\rightarrow
R^{\gamma}[G]:r\,u_{n.\mu(g)}\mapsto r\,v_{g}$

is the so-called natural homomorphism with respect to $N$ (i.e. induced from
the canoncical map from $\Gamma$ to $G$, denoted $\omega_{N}$. If $\beta$ is
trivial it also is called the relative augmentation with respect to $N$). We
will denote this specific case by $\omega_{\beta,N}$.

  * •

If $\Gamma$ is a central extension, i.e. $N\subseteq\mathcal{Z}(\Gamma)$, then
$\operatorname{Lin}(N,R^{*})^{\Gamma/N}=\operatorname{Hom}(N,R^{*})$ and
$\Psi_{\chi,\beta}$ is the ring morphism associated to the transgression map
$\operatorname{Tra}$ as in [18, Section 5]. Therefore we call the morphism
(13) the generalized transgression morphism.

###### Proof of Proposition 3.3..

A direct calculation verifies that $\Psi_{\chi,\beta}$ is multiplicative when
$\chi$ is an invariant character. Concretely,
$(u_{n_{1}}u_{\mu(g_{1})}).(u_{n_{2}}u_{\mu(g_{2})})=\beta(g_{1},g_{2})u_{n_{1}}u_{\sigma_{g_{1}}(n_{2})}u_{\mu(g_{1}g_{2})}$
which is mapped by $\Psi_{\chi,\beta}$ to
$\beta(g_{1},g_{2})\chi(n_{1}\sigma_{g_{1}}(n_{2}))v_{g_{1}g_{2}}=\beta(g_{1},g_{2})\chi(n_{1})\chi(n_{2})v_{g_{1}g_{2}}$.
The latter is
$\Psi_{\chi,\beta}(u_{n_{1}}u_{\mu(g_{1})}).\Psi_{\chi,\beta}(u_{n_{2}}u_{\mu(g_{2})})$,
as needed. Now it is clear that it is an epimorphism.

Next, because $\chi\in\operatorname{Lin}(N,F^{*})^{\Gamma/N}$, multiplying
with the idempotent $E_{\chi}$, i.e. the value of the projection, was already
explicitly mentioned in (11). By comparing we see that $\Psi_{\chi,\beta}$
indeed agrees with $p_{\chi}$.

Concerning the kernel, it follows from Proposition 3.1 that
$\ker(\Psi_{\chi,\beta})=R^{\beta}[\Gamma](1-e_{\chi})\cap R^{\beta}[\Gamma].$
For the other description, since
${\left.\kern-1.2pt\Psi_{\chi,\beta}\right|_{RN}}=\chi_{R}$, we already have
that $R^{\beta}[\Gamma]I_{\chi}\subseteq\ker(\Psi_{\chi,\beta})$. Also, an
element $y$ of $R^{\beta}[\Gamma]$ can be uniquely written as $y=\sum_{g\in
G}y_{g}u_{\mu(g)}$, with each $y_{g}\in RN$. Since the elements $v_{g}$ are
$R[\chi]$-linearly independent, the concrete form of $\Psi_{\chi,\beta}$
implies that $y\in\ker(\Psi_{\chi,\beta})$ if and only if each
$y_{g}\in\ker(\Psi_{\chi,\beta})$, or equivalently each $y_{g}\in I_{\chi}$.
Consequently, $\ker(\Psi_{\chi,\beta})=\sum_{g\in
G}I_{\chi}u_{\mu(g)}\subseteq R^{\beta}[\Gamma]I_{\chi}$, as asserted.

Finally, suppose $\operatorname{im}(\chi)$ is contained in $R^{*}$ and denote
the free $R$-module generated by the set $\\{u_{a}-\chi(a)u_{1}\mid a\in N\\}$
by $M$. Clearly $M$ is contained in the kernel $I_{\chi}$ of
$\chi_{R}:RN\rightarrow R$. Conversely, if $x=\sum_{a\in
N}r_{a}u_{a}\in\ker(\chi_{R})$, then $\sum_{a\in N}r_{a}\chi(a)u_{1}=0$ and
hence $x=x-\sum_{a\in N}r_{a}\chi(a)u_{1}\in M$. It follows that $I_{\chi}=M$.
∎

In Section 5.1 we will be in the setting of Remark 2.4 where $\chi$ is a
(linear) character of the abelian subgroup $A$, but where once we will be
working with $\Gamma$ and once with some $\Gamma/Q$ for $A\cap Q=1$. In order
to distinguish, we will sometimes write

(14) $\Psi_{\chi,Q}:R^{\beta}[\Gamma/Q]\rightarrow R^{\beta.T(\chi)}[G/Q]$

and hence assume $[\beta]$ is understood from the context.

###  3.3. The case of an $\alpha$-representation group of $G$

Often in this paper we will be concentrating on a fixed $2$-cocycle $\alpha\in
Z^{2}(G,F^{*})$ of $G$. In that case it is useful to consider the following
group which will be recurrent in this paper:

(15) $G_{\alpha}:=\\{\alpha(a,b)u_{c}\mid a,b,c\in G\\}=\langle u_{g}\mid g\in
G\rangle\leq{\mathcal{U}}(F^{\alpha}[G]).$

If $\alpha$ is of finite order151515Since $G$ is finite we know that the
cohomology class $[\alpha]$ has finite order dividing $|G|$, however there has
not to be a representative of finite order. For example the cohomology class
of $C_{2}=\langle x\rangle$ over ${\mathbb{Q}}$ defined by $u_{x}^{2}=2$ has
order $2$ but any $2$-cocycle representative has infinite order. Such a
representative however exists when the values are in a $|G|$-divisible group.,
then every element in $G_{\alpha}$ is of the form $\zeta^{i}u_{g}$ with
$\zeta$ a $o(\alpha)$-primitive root of unity. Hence it is a central extension
of $\langle\zeta\rangle$ by $G$ with $\lambda:G_{\alpha}\rightarrow
G:\zeta^{i}u_{g}\mapsto g.$ Considering the canonical section $\mu(g)=u_{g}$
one has that $\alpha(g,h)=\mu(g)\mu(h)\mu(gh)^{-1}$. Thus
$\operatorname{Hom}(\langle\zeta\rangle,F^{*})=\langle\chi\rangle$ with
$\chi(\zeta)=\zeta$ and $\alpha=\operatorname{Tra}(\chi)$. Therefore we may
now apply Theorem 3.2 to recover [14, Proposition 3.3.8.].

######  Corollary 3.5.

Let $G$ be a finite group, $F$ a number field and $\alpha\in Z^{2}(G,F^{*})$ a
cocycle of finite order. Then

$F[G_{\alpha}]\cong\bigoplus_{i=0}^{o(\alpha)-1}F^{\alpha^{i}}[G].$

Note that the isomorphism class of $G_{\alpha}$ depends on the chosen cocycle.
It turns out that even for cohomologous cocycles of the same finite order the
associated groups might be non isomorphic. The following example was
communicated to us by Y. Ginosar.

######  Example 3.6.

Denote

$G=C_{2}\times C_{2}=\langle x\rangle\times\langle y\rangle$

and let $F=\mathbb{Q}(i)$. Next, we wish to define two coycles
$\alpha,\alpha^{\shortmid}\in Z^{2}(G,F^{*})$. We will take cocycles which are
normalized, that is for any $g\in G$

$\alpha(g,e_{G})=1=\alpha(e_{G},g),\quad\alpha^{\shortmid}(g,e_{G})=1=\alpha^{\shortmid}(e_{G},g).$

Further, define

$\alpha(x,x)=\alpha(x,y)=\alpha(x,xy)=1$
$\alpha(y,y)=1,\,\alpha(y,x)=\alpha(y,xy)=-1$
$\alpha(xy,x)=1,\,\alpha(xy,y)=\alpha(xy,xy)=-1.$

and also

$\alpha^{\shortmid}(x,x)=-1,\quad\alpha^{\shortmid}(x,y)=1,\quad\alpha^{\shortmid}(x,xy)=-1$
$\alpha^{\shortmid}(y,y)=-1,\quad\alpha^{\shortmid}(y,x)=-1,\quad\alpha^{\shortmid}(y,xy)=1$
$\alpha^{\shortmid}(xy,x)=1,\quad\alpha^{\shortmid}(xy,y)=-1,\quad\alpha^{\shortmid}(xy,xy)=-1.$

Now choose a basis $\\{u_{g}\\}_{g\in G}$ for $F^{\alpha}G$ and a basis
$\\{v_{g}\\}_{g\in G}$ for $F^{\alpha^{\shortmid}}G$. Notice that both
cocycles $\alpha,\alpha^{\shortmid}\in Z^{2}(G,F^{*})$ are of order $2$ and
also notice that the cohomology classes which corresponds to these cocycles
are

$[\alpha]:\quad u_{x}^{2}=1,\quad u_{y}^{2}=1,\quad[u_{x},u_{y}]=-1$

and

$[\alpha^{\shortmid}]:\quad v_{x}^{2}=-1,\quad
v_{y}^{2}=-1,\quad[v_{x},v_{y}]=-1.$

It follows that

$G_{\alpha}=\langle u_{g}\mid g\in G\rangle\cong D_{8}\quad\text{ and }\quad
G_{\alpha^{\shortmid}}=\langle v_{g}\mid g\in G\rangle\cong Q_{8}$

and therefore $G_{\alpha}^{*}\not\cong G_{\alpha^{\shortmid}}^{*}$. The
crucial point is that these cocycles are cohomologous over161616The coboundary
here is a map $f:G\rightarrow\mathbb{Q}(i)$ determined by $f(x)=i=f(y)$,
$f(xy)=-1$ and $f(1)=1$. Indeed
$\alpha^{\shortmid}(g_{1},g_{2})=f(g_{1})f(g_{2})f(g_{1}g_{2})^{-1}\alpha(g_{1},g_{2})$
for all $g_{1},g_{2}\in G$. $F=\mathbb{Q}(i)$ but not over $\mathbb{Q}$.

##  4\. When is the unit group of a twisted group ring finite?

Let $F$ be a number field, $R$ an order in $F$ and $\alpha\in Z^{2}(G,R^{*})$
a fixed (normalized) $2$-cocycle. In this section we determine when the unit
group ${\mathcal{U}}(R^{\alpha}[G])$ is finite.

Generally speaking, [9, Corollary 5.5.8.], if $\mathcal{O}$ a is
${\mathbb{Z}}$-order in a finite dimensional semisimple ${\mathbb{Q}}$-algebra
$A$, then ${\mathcal{U}}(\mathcal{O})$ is finite if and only if every simple
component of $A$ is either ${\mathbb{Q}}$, a quadratic imaginary field
extension of ${\mathbb{Q}}$ or a totally definite quaternion algebra over
${\mathbb{Q}}$. As a by-product, if $x\in{\mathcal{U}}(R^{\alpha}[G])$ is of
finite order, say $n$, then $n$ must171717The $F$-subalgebra $F[x]$ of
$F^{\alpha}G$ is a commutative semisimple subalgebra and hence a direct sum of
cyclotomic extensions of ${\mathbb{Q}}(\xi_{m})$, with $\xi_{m}$ a root of
unity of order $m$, a divisor of $n$. Moreover, by the above,
$[{\mathbb{Q}}(\xi_{m}):{\mathbb{Q}}]\leq 2$ and one of the $m$ must be equal
to $n$. Now the Dirichlet unit theorem yields the claim as a
${\mathbb{Z}}$-order in ${\mathbb{Q}}(\xi_{n})$ is finite exactly when $n$
divides $4$ and $6$. divide $4$ or $6$. Consequently, the exponent of
$G_{\alpha}$ is a divisor of $4$ or $6$. Moreover, also ${\mathcal{U}}(R)$
would need to be finite and hence by the above $F={\mathbb{Q}}\text{ or
}{\mathbb{Q}}(\sqrt{-d})$ with $d>0$. In the former case $R={\mathbb{Z}}$ is
the only order in $F$. We obtain the following characterisation, generalizing
a classical result of Higman [9, Theorem 1.5.6] for untwisted group rings.

######  Theorem 4.1.

Let $G$ be a finite group, $F$ be a number field, $R$ a ${\mathbb{Z}}$-order
in $F$ and $\alpha\in Z^{2}(G,R^{*})$ non-trivial (normalized) cocycle. Then
the following are equivalent:

  1. (1)

${\mathcal{U}}(R^{\alpha}[G])$ is finite,

  2. (2)

${\mathcal{U}}(R[G_{\alpha}])$ is finite,

  3. (3)

${\mathcal{U}}(R^{\alpha^{i}}[G])$ is finite for all $i$.

One of the above holds if and only if one of the following conditions is
satisfied:

  1. (i)

$G_{\alpha}$ is an abelian group of exponent $4$ and $F={\mathbb{Q}}$ or
${\mathbb{Q}}(\sqrt{-1})$,

  2. (ii)

$G_{\alpha}$ is an abelian group of exponent $3$ and
$F={\mathbb{Q}}(\sqrt{-3})$,

  3. (iii)

$G_{\alpha}$ is an abelian group of exponent $6$ and $F={\mathbb{Q}}$ or
${\mathbb{Q}}(\sqrt{-3})$ ,

  4. (iv)

$G_{\alpha}$ is a non-abelian Hamiltonian181818A group is called Hamiltonian
if every subgroup is normal. 2-group and $F={\mathbb{Q}}$.

Interestingly, the list of possibilities above which is for a non-trivial
cocycle is more restricted than for a trivial cocycle (it is to emphasize this
that we excluded the trivial case in Theorem 4.1). The group $G_{\alpha}$ is
abelian exactly when the cocycle $\alpha$ is symmetric (i.e.
$\alpha(g,h)=\alpha(h,g)$ for all $h,g\in G$) and $G$ is abelian. In Corollary
4.3 below we will give a concrete interpretation in terms of $G$ and $\alpha$
for case (iv).

###### Proof.

Clearly in any of the cases (1)-(3) $\langle\operatorname{im}(\alpha)\rangle$
is a finitely generated torsion subgroup of $R^{*}\subset F^{*}$ and hence it
is a finite cyclic group. In particular $o(\alpha)<\infty$ which allows to
apply Corollary 3.5:

$F[G_{\alpha}]\cong\bigoplus_{i=0}^{o(\alpha)-1}F^{\alpha^{i}}[G].$

Hence the equivalence between (2) and (3) is a consequence thereof and
commensurability of ${\mathbb{Z}}$-orders in $F[G_{\alpha}]$ (see [9, Lemma
4.6.9]). Due to this also any order in $F^{\alpha^{i}}[G]$ is a direct summand
of an order in $F[G_{\alpha}]$ and hence (2) implies (1). The main bulk of the
proof goes into proving that (1) implies (2). More precisely we will show that
$G_{\alpha}$ and $F$ are of the form (i)-(iv). In those cases one can directly
see that ${\mathcal{U}}(R[G_{\alpha}])$ is finite (e.g. this can be deduced
from [9, Theorem 1.5.6.], handling the case $R={\mathbb{Z}}$, together with
Dirichlet unit theorem [9, Theorem 5.2.4.]).

Suppose ${\mathcal{U}}(R^{\alpha}[G])$ is finite and hence, as noticed
earlier, $\exp(G_{\alpha})\mid 4\text{ or }6$ and $F^{\alpha}[G]$ is a direct
sum of (certain) division $F$-algebras. In particular, it contains no non-zero
nilpotent elements.

Claim 1:  If $u_{k}\in G_{\alpha}$, with $k\in G$ such that $o(u_{k})=o(k)$
then $\langle u_{k}\rangle$ is normal in $G_{\alpha}$. Moreover, this
condition holds when $o(u_{k})$ is square-free and if $o(u_{k})=2$ then
$u_{k}\in\mathcal{Z}(G_{\alpha})$.

For any $t\in G$ the element $(1-u_{k})u_{t}\widetilde{u_{k}}$ with
$\widetilde{u_{k}}:=\sum_{i=0}^{o(u_{k})-1}u_{k}^{i}$ is nilpotent and
$\widetilde{u_{k}}\neq 0$ when $o(u_{k})=o(k)$. By an earlier remark the
nilpotent element is zero and thus
$u_{t}^{-1}u_{k}u_{t}\widetilde{u_{k}}=\widetilde{u_{k}}$. Consider now
$F^{\alpha}[G]$ with its canonical $G$-grading. Then $\text{deg}(u_{k})=k$ and
$\text{deg}(u_{t}^{-1}u_{k}u_{t})=t^{-1}kt.$ Since $o(u_{k})=o(k)$ all
summands of $\widetilde{u_{k}}$ have different degrees and thus a degree
argument shows that $u_{t}^{-1}u_{k}u_{t}=u_{k}^{j}$ for some $j$. In
conclusion $\langle u_{k}\rangle\triangleleft G_{\alpha}$, as desired. Now
suppose $o(u_{k})$ is square-free. It is well-known191919More generally, if
$g_{1}$ and $g_{2}$ commute, then
$[u_{g_{1}},u_{g_{2}}]:=u_{g_{1}}^{-1}u_{g_{2}}^{-1}u_{g_{1}}u_{g_{2}}=\alpha(g_{1},g_{2})$.
Since $u_{k}^{o(k)}$ is central in $R^{\alpha}[G]$, it follows that
$1=[u_{g_{1}}^{o(g_{1})},u_{g_{2}}]=\alpha(g_{1},g_{2})^{o(g_{1})}$. Now
considering $u_{k}^{o(k)}=\alpha(k^{o(k)-1},k)u_{1}$ we obtain the claim. that
if $o(\alpha)$ is finite then $u_{k}^{o(k)}$ is a root of unity in $R^{*}$
dividing $o(k)$. Hence if $o(u_{k})$ is square-free, it is easily seen that
$u_{k}^{o(k)}=1$. The last part follows from the rest.

Claim 2:  Either (I) $G_{\alpha}$ is abelian of exponent dividing 4 or 6, (II)
$G_{\alpha}\cong Q_{8}\times E$ with $E$ an elementary abelian $2$-group.

Note that by Baer-Dedekind’s classification theorem [20, Theorem 1.8.5.] these
cases are exactly those Hamiltonian groups with exponent dividing $4$ or $6$.
Due to claim $1$ it remains to prove that the generators of order $4$ are also
normal. For this recall that [22, Theorem 11.5.12] the unit group of a maximal
order in a quaternion algebra $\left(\frac{-a,-b}{{\mathbb{Q}}}\right)$, with
$a,b\in{\mathbb{N}}$, is cyclic except for the maximal order in
$\left(\frac{-1,-1}{{\mathbb{Q}}}\right)$ and
$\left(\frac{-1,-3}{{\mathbb{Q}}}\right)$. In those cases the unit group is
$SL(2,3)=Q_{8}\rtimes C_{3}$, resp. $Dic_{3}:=C_{3}\rtimes C_{4}$. Now
consider $G_{\alpha}\leq\prod_{e\in\text{PCI}(F^{\alpha}[G])}G_{\alpha}e$.
Since also $\exp(G_{\alpha}e)\mid 4,6$ we see that $G_{\alpha}e$ is either
cyclic or a subgroup of $Q_{8}$ or $D_{6}$. Thus one can check that any
element of order $4$ in the direct product generates a normal therein and in
particular the same holds for $G_{\alpha}$. This finishes the proof of the
second claim.

We will now study both cases in more detail. To start, as already used, we
know that ${\mathcal{U}}(R)$ is finite and hence that $F={\mathbb{Q}}$ or
${\mathbb{Q}}(\sqrt{-d})$ with $d>0$. It remains to restrict the possible
values of $d$ and exclude exponent $2$.

Case (I): Suppose $G_{\alpha}$ is abelian. To start remark
$\exp(G_{\alpha})=2$ is not possible. Indeed, otherwise
$1=(u_{g}u_{h})^{2}=\alpha(g,h)u_{g}^{2}u_{h}^{2}=\alpha(g,h)$ for all $g,h\in
G$, in contradiction with the assumption that $\alpha$ is non-trivial. By
computations of the same type, $\exp(G_{\alpha})=3$ and $F={\mathbb{Q}}$ is
also impossible for non-trivial $\alpha$. The restriction on $F$ for the other
cases will follow from the fact that all simple components have degree at most
$2$ over ${\mathbb{Q}}$. Indeed, suppose $\exp(G_{\alpha})=4$ and write
$F={\mathbb{Q}}(\sqrt{-d})$, with $d$ a square free non-negative integer. Then
$F^{\alpha}[G]$ has a simple component
${\mathbb{Q}}(\sqrt{-d},\sqrt{-1})={\mathbb{Q}}(\sqrt{d},\sqrt{-1})$ which is
of degree at most $2$ if and only if $d=0,$ or $1$. If the exponent is $3$ or
$6$, then there is a simple component ${\mathbb{Q}}(\sqrt{-d},\zeta_{3})$
which would have degree larger than $2$ if $d\neq 0,3$.

Case (II): Finally consider the case that $G_{\alpha}\cong Q_{8}\times E$ is
an Hamiltonian 2-group. Since $F^{\alpha}[G]$ has a simple component
$\left(\frac{-1,-1}{F}\right)$ which needs to be totally definite (in
particular $F$ is totally real), we indeed get that $F={\mathbb{Q}}$. ∎

######  Remark 4.2.

The condition on the coefficient ring $R$ can be generalized further. Namely,
let $F$ be a global field and $S$ be a finite set of places of $F$ containing
the archimedian ones. Denote by $\mathcal{O}_{S}$ the ring of $S$-integers of
$F$, which is well-known to be a Dedekind domain with finite quotients.
Therefore, any $\mathcal{O}_{S}$-order $R$ is commensurable [9, Lemma 4.6.9]
with $\mathcal{O}_{S}$. Since $|\mathcal{U}(\mathcal{O}_{S})|$ is finite only
for $F={\mathbb{Q}}$ or ${\mathbb{Q}}(\sqrt{-d})$, with $d>0$, and
$S=\\{\infty\\}$ (e.g. see [5, Theorem 3.24.]) we see that the conclusion of
Theorem 4.1 also holds for such $R$.

Using Theorem 4.1 one can give an especially precise characterisation when
${\mathcal{U}}(R^{\alpha}[G])$ is a finite non-abelian group in terms of $G$
and $\alpha$. For this recall that any non-abelian Hamiltonian $2$-group $G$
is isomorphic to $Q_{8}\times C_{2}^{n}$ for some $n$. In other words, it can
be written as a stem extension

$1\rightarrow C_{2}\rightarrow G\rightarrow C_{2}^{n+2}\rightarrow 1.$

On the other hand, any non-trivial cohomology class $[\alpha]\in
H^{2}(G,{\mathbb{Z}}^{*})$ corresponds to a central extension of $G$ by
$C_{2}$. Therefore, from Theorem 4.1 (iv) we deduce

######  Corollary 4.3.

Let $G$ be a finite group, $F$ be a number field, $R$ an order in $F$ and
$[\alpha]\in H^{2}(G,R^{*})$ a non-trivial cohomology class. Then
${\mathcal{U}}(R^{\alpha}[G])$ is a finite non-abelian group if and only if
the following conditions are satisfied

  1. (1)

$R=\mathbb{Z}$,

  2. (2)

$G$ is an elementary abelian $2$-group of rank at least $2$,

  3. (3)

$[\alpha]$ is inflated from a cohomology class $[\gamma]\in H^{2}(C_{2}\times
C_{2},C_{2})$ which is determined by $u_{x}^{2}=u_{y}^{2}=[u_{x},u_{y}]=-1$
where $x,y$ are generators of $C_{2}\times C_{2}$ and $-1$ is the generator of
$C_{2}$.

Notice that the cohomology class $[\gamma]\in H^{2}(C_{2}\times C_{2},C_{2})$
above corresponds to $Q_{8}$.

To finish this section, we would like to come back on the proof of Theorem 4.1
which unfortunately is quit indirect. Indeed the implication from (1) to (2)
goes by classifying all the possibilities for $G_{\alpha}$ and $F$ and then
noticing that in all these cases ${\mathcal{U}}(R[G_{\alpha}])$ is finite. A
more natural approach would have been to construct for all $j$ an isogeny
between ${\mathcal{U}}(R^{\alpha}[G])$ and ${\mathcal{U}}(R^{\alpha^{j}}[G])$.
Such a map can however only come from ring (epi)morphism when
$\text{gcd}(j,n)=1$. The statement answers a question of Margolis-
Schnabel202020As is apparent from the proof, the main tool is however their
result over the complex numbers[18, Theorem 3.1.]. [18, remark 3.2.] in case
$[\alpha]$ has a cocycle representative of finite order.

######  Proposition 4.4.

Let $G$ be a finite group, $F$ a field of characteristic zero and $\alpha\in
Z^{2}(G,F^{*})$. If $F^{\alpha}[G]\cong F^{\alpha^{j}}[G]$, then
$\text{gcd}(j,o([\alpha]))=1$. If $\alpha$ has finite order and $F$ is a
number field then also the converse holds.

###### Proof.

First suppose that $F^{\alpha}[G]\cong F^{\alpha^{j}}[G]$. Note that
$\langle\operatorname{im}(\alpha)\rangle$ is a finitely generated abelian
subgroup of $F^{*}$. In particular it lies in some countable subfield, say
$L\subseteq F$. Moreover, $F^{\alpha}[G]\cong F\otimes_{L}L^{\alpha}[G]$ and
so the given isomorphism restricts to an $L$-algebra isomorphism
$L^{\alpha}[G]\cong L^{\alpha^{j}}[G]$. Now, since $L$ is countable it can be
embedded in ${\mathbb{C}}$ and we can view $\alpha$ as having values in
${\mathbb{C}}^{*}$. Consequently, by tensoring with ${\mathbb{C}}\otimes_{L}-$
we see that ${\mathbb{C}}^{\alpha}[G]\cong{\mathbb{C}}^{\alpha^{j}}[G]$.
Therefore we can apply [18, Theorem 3.1.] saying that
$\text{gcd}(j,o([\alpha]))=1$.

Conversely, let $n=o([\alpha])$. When $\text{gcd}(j,n)=1$ we have a
${\mathbb{Q}}$-algebra isomorphism
$\sigma_{j}:{\mathbb{Q}}(\zeta_{o(\alpha)})\rightarrow{\mathbb{Q}}(\zeta_{o(\alpha)}^{j})$
mapping $\zeta_{o(\alpha)}$ to its $j$-th power. With this at hand, define
$\psi:{\mathbb{Q}}(\zeta_{o(\alpha)})^{\alpha}[G]\rightarrow{\mathbb{Q}}(\zeta_{o(\alpha)})^{\alpha^{j}}[G]$
by $\psi(\sum a_{g}u_{g})=\sigma_{j}(a_{g})v_{g}$. Note that for all $g,h\in
G$:

$\psi(u_{g}u_{h})=\psi(\alpha(g,h)u_{gh})=\alpha(g,h)^{j}v_{gh}=v_{g}v_{h}=\psi(u_{g})\psi(u_{h}).$

So now it easily is seen that $\psi$ us a ring epimorphism and hence
isomorphism. An extension of scalars with the larger field $F$ now finishes
the proof. ∎

##  5\. Correlations between $R[\Gamma]$ and $R^{\alpha}[G]$ \- a unit group
point of view

Throughout this section we fix an extension

(16) $1\rightarrow
N\rightarrow\Gamma\overset{\lambda}{\rightarrow}G\rightarrow 1.$

as in (7), fix also a section $\mu$ of $\lambda$ and define

$\alpha(g,h)=\mu(g)\mu(h)\mu(gh)^{-1}.$

In particular when $N$ is abelian we will always choose this $\alpha$ as the
normalized representative of the cohomology class $[\alpha]$ corresponding
(16). We will always assume that the underlying field $F$ is such that ${\rm
char}(F)\nmid|\Gamma|$ and $R$ is some order in $F$

We wish to compare $\mathcal{U}(R[\Gamma])$ and $\mathcal{U}(R^{\alpha}[G])$
with the aim to pullback results from the smaller group $G$, but in the
twisted context, to the larger group $\Gamma$. For this we study the kernel
and cokernel of the generalized transgression $\Psi_{\chi}$ from Proposition
3.3. Also in this section we will work more generally with $R^{\beta}[\Gamma]$
for some $[\beta]\in\operatorname{im}(\operatorname{Inf}_{N})$.

Notation:  Recall that if $f:R\rightarrow S$ is a ring homomorphism, then we
denote the induced map on the unit groups by
$\widetilde{f}:\mathcal{U}(R)\rightarrow\mathcal{U}(S)$.

###  5.1. On the cokernel of the generalized transgression map

In the sequel finiteness of cokernels of group morphisms will always follow
from the following somehow folklore lemma.

######  Lemma 5.1.

Let212121The prototypical example of such a ring is the ring of integers in a
number field. The condition is the minimal needed to use classical methods
with $R$-orders, e.g. that two $R$-orders have commensurable unit groups. $R$
be a Dedekind domain with field of fractions $F$ such that $R/I$ is finite for
all $0\neq I\triangleleft R$ and $A$ and $B$ be semimsimple $F$-algebras.
Furthermore consider an $R$-order $\mathcal{O}_{A}$ in $A$ and
$\mathcal{O}_{B}$ in $B$. If there exists a $R$-algebra epimorphism
$\pi:\mathcal{O}_{A}\rightarrow\mathcal{O}_{B}$, then
$\operatorname{Coker}(\widetilde{\pi})$ is finite.

###### Proof.

By definition $\mathcal{O}_{A}$ contains an $F$-basis of $A$ and similarly for
$\mathcal{O}_{B}$. Therefore we can extend $F$-linearly $\pi$ to an
$F$-algebra epimorphism $\widehat{\pi}:A\rightarrow B$. Formally
$\widehat{\pi}=id_{F}\otimes_{R}\pi$ and we identify $A=F\mathcal{O}_{A}$ with
$F\otimes_{R}\mathcal{O}_{A}$. Due to the semisimplicity of $A$, there exists
a central idempotent $e$ in $A$ such that $Ae\cong B$ via $\widehat{\pi}$. In
other words, when decomposing $A=Ae\oplus A(1-e)$ we can consider
$\widehat{\pi}$ as projection on the first component. To obtain the desired
statement consider the $R$-orders
$\mathcal{O}_{A}\subseteq\mathcal{O}_{A}e\oplus\mathcal{O}_{A}(1-e)$ in $A$.
Due to the conditions on $R$ there exists a positive integer $n$ such that
$n(\mathcal{O}_{A}e\oplus\mathcal{O}_{A}(1-e))\subseteq\mathcal{O}_{A}$ and
see [9, Lemma 4.6.9.],

$[{\mathcal{U}}(\mathcal{O}_{A}e)\times{\mathcal{U}}(\mathcal{O}_{A}(1-e)):{\mathcal{U}}(\mathcal{O}_{A})]\leq[\mathcal{O}_{A}:n\mathcal{O}_{A}]<\infty.$

Consequently also its epimorphic image ${\rm
Im}(\widetilde{\pi})\cong{\mathcal{U}}(\mathcal{O}_{A})e$ is of finite index
in ${\mathcal{U}}(\mathcal{O}_{A}e)\cong{\mathcal{U}}(\mathcal{O}_{B})$ (with
upper bound the above number), as desired. ∎

Note that the proof in fact gives a method to obtain an upper bound on
$|\operatorname{Coker}(\widetilde{\pi})|$. It would be interesting to have a
more explicit one.

Using Proposition 3.3 a first useful incarnation of Lemma 5.1 is with
$\widetilde{\Psi}_{\chi}$ for some
$\chi\in\operatorname{Lin}(\Gamma,R^{*})^{\Gamma/N}$. However, in contrast to
the kernel and despite it to be finite, in general a concrete description (or
even generators) of the cokernel is out of reach. Instead, we will focus on
comparing the cokernel of $\widetilde{\Psi}_{\chi,Q}$, defined in (14), for
certain types of ’nice’ normal subgroups $Q$ of $\Gamma$. The restrictions on
$Q$ will be as in Remark 2.4 and be such that we have the following222222In
the right column one should be careful with the notation $T(\chi)$. More
precisely in the right upper corner $T(\chi)=[\chi\circ\alpha]\in
H^{2}(G,R^{*})$. In the right lower corner $T(\chi)=[\chi\circ\gamma]\in
H^{2}(G/Q,R^{*})$ with $\alpha=\operatorname{Inf}(\gamma)$ as in Equation 6.
In particular
$\operatorname{Tra}_{\alpha}(\chi)=\operatorname{Inf}_{N}(\operatorname{Tra}_{\gamma,N}(\chi))$
and so the going down arrows exist. commutative diagram:

(17)
$\textstyle{\mathcal{U}(R^{\beta}[\Gamma])\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widetilde{\Psi}_{\chi}}$$\scriptstyle{\widetilde{\omega}_{Q,\beta}}$$\textstyle{\mathcal{U}(R^{\beta.T(\chi)}[G])\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widetilde{\omega}_{Q,\beta.T(\chi)}}$$\textstyle{\mathcal{U}(R^{\beta}[\Gamma/Q])\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widetilde{\Psi}_{\chi,Q}}$$\textstyle{\mathcal{U}(R^{\beta.T(\chi)}(\chi)[G/Q])}$

Note that (co)restricting $\widetilde{\Psi}_{\chi}$ yields a morphism

$\widetilde{\Psi}_{ker}:{\mathcal{U}}\left(1+R^{\beta}[\Gamma].I_{Q}\right)\longrightarrow{\mathcal{U}}\left(1+R^{\beta.T(\chi)}[G].I_{Q}\right):1+x\mapsto
1+\Psi_{\chi}(x)$

between $\ker(\widetilde{\omega}_{N})$ and
$\ker(\widetilde{\omega}_{N,\alpha})$ (cf. Proposition 3.3).

######  Theorem 5.2.

Let $\Gamma$ be the extension (16),
$[\beta]\in\operatorname{im}(\operatorname{Inf}_{N}),\,\chi\in\operatorname{Lin}(N,F^{*})^{\Gamma/N}$
and $R$ a Dedekind domain such that $R/I$ is finite for all $0\neq
I\triangleleft R$ and ${\rm char}(R)\nmid\Gamma$. Then,

  1. (1)

$\operatorname{Coker}(\widetilde{\Psi}_{\chi,\beta})$ is finite,

  2. (2)

for any normal subgroup $Q$ of $\Gamma$ such that $Q\cap N=1,\,\alpha(x,y)=1$
for all $x,y\in Q,$ $\beta\in\operatorname{im}(\operatorname{Inf}_{Q})$ and
$\widetilde{\omega}_{Q}$ and $\widetilde{\omega}_{Q,\beta.T(\chi)}$ are
surjective, then

$|\operatorname{Coker}(\widetilde{\Psi}_{ker})|\,.\,|\operatorname{Coker}(\widetilde{\Psi}_{\chi,Q})|=|\operatorname{Coker}(\widetilde{\Psi}_{\chi})|.$

If $Q$ has a complement in $\Gamma$ (which will be our setting in the later
sections), then the maps $\widetilde{\omega}_{Q}$ and
$\widetilde{\omega}_{Q,\beta.T(\chi)}$ are surjective and the inflation
condition on $\alpha$ will also be satisfied. The surjectivity of the
augmentation is also the case when $R$ is an Artinian ring [4, Lemma 3.4.].

###### Proof of Theorem 5.2..

The first statement follows by combining Proposition 3.3 and Lemma 5.1. The
second statment will follow from a diagram chasing like argument starting from
diagram (17). For notation simplicity we will write $\widetilde{\omega}$ for
$\widetilde{\omega}_{Q,\beta}$ and $\widetilde{\omega}_{\chi}$ for
$\widetilde{\omega}_{Q,\beta.T(\chi)}$.

We complete the rows to an exact sequence by adding their kernels and
cokernels with the canonical embedding and projection denoted by
$i_{1},\pi_{1}$ (resp. $i_{Q},\pi_{Q}$). Due to the commutativity of (17),
$\pi_{Q}\circ\widetilde{\omega}_{Q\chi}$ induces a map $F_{Cok}$ between the
cokernels. Altogether we obtain the following diagram where all squares
commute:

$\textstyle{\ker(\widetilde{\Psi}_{\chi})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i_{1}}$$\textstyle{\mathcal{U}(R^{\beta}[\Gamma])\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widetilde{\Psi}_{\chi}}$$\scriptstyle{\pi\circ\widetilde{\omega}}$$\textstyle{\mathcal{U}(R^{\beta.T(\chi)}[G])\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widetilde{\omega}_{\chi}}$$\scriptstyle{\pi_{1}}$$\textstyle{\operatorname{Coker}(\widetilde{\Psi}_{\chi})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{F_{Cok}}$$\textstyle{\frac{\mathcal{U}(R^{\beta}[\Gamma/Q])}{\ker(\widetilde{\Psi}_{\chi,Q})}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widetilde{\Psi}_{\chi,Q}}$$\textstyle{\mathcal{U}(R^{\beta.T(\chi)}[G/Q])\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{Q}}$$\textstyle{\operatorname{Coker}(\widetilde{\Psi}_{\chi,Q})}$

where $\pi$ is the canonical epimorphism from
$\mathcal{U}(R^{\beta}[\Gamma/Q])$ to
$\frac{\mathcal{U}(R^{\beta}[\Gamma/Q])}{\ker(\widetilde{\Psi}_{\chi,N})}$.
The Snake lemma applied to the diagram above now yields an exact sequence:

$\textstyle{\ker(\pi\circ\widetilde{\omega})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widetilde{\Psi}_{ker}}$$\textstyle{\ker(\widetilde{\omega}_{\chi})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\ker(F_{Cok})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d}$$\textstyle{\operatorname{Coker}(\pi\circ\widetilde{\omega})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Coker}(\widetilde{\omega}_{\chi})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Coker}(F_{Cok})}$

with all the morphisms being the straightforward ones and $d$ the connecting
morphism.

Now assume that $\widetilde{\omega}$ and $\widetilde{\omega}_{\chi}$ are
surjective. Then $\operatorname{Coker}(\pi\circ\widetilde{\omega})$ is trivial
and hence, going through the exact sequence above via isomorphism theorems,
$\ker(F_{Cok})\cong\operatorname{Coker}(\widetilde{\Psi}_{ker})$. Also,
$\operatorname{Coker}(F_{Cok})$ is trivial thus altogether we obtain the
desired statement. ∎

###  5.2. On the kernel of the generalized transgression map

Unlike $\operatorname{Coker}(\widetilde{\Psi}_{\chi})$, the kernel of
$\widetilde{\Psi}_{\chi}$ will usually be infinite as already seen with
augmentation maps.

######  Example 5.3.

When $N=\Gamma$ and $\chi$ the trivial character, then
$\widetilde{\Psi}_{\chi}$ is simply the usual augmentation map from
${\mathcal{U}}({\mathbb{Z}}[\Gamma])$ to ${\mathcal{U}}({\mathbb{Z}})=\\{\pm
1\\}$. We now see that
$[{\mathcal{U}}({\mathbb{Z}}[\Gamma]):\ker(\widetilde{\Psi}_{\chi})]=2$ and
therefore the kernel is infinite when ${\mathcal{U}}({\mathbb{Z}}[\Gamma])$ is
infinite which, by a theorem of Higman [9, Th. 1.5.6.], exactly happens when
$\Gamma$ is not an Hamiltonian $2$-group or an abelian group of exponent
dividing $4$ or $6$.

The advantage of the kernel is that thanks to Proposition 3.3 and Theorem 3.2
one has much information on its elements. Combined with an (intensive use of)
Theorem 4.1 we can describe finiteness of $\ker(\widetilde{\Psi}_{\chi})$, in
case of a central extension.

######  Theorem 5.4.

Suppose that the extension (16) is central (i.e.
$N\subseteq\mathcal{Z}(\Gamma)$),
$[\beta]\in\operatorname{im}(\operatorname{Inf}_{N}),\,\chi\in\operatorname{Hom}(N,R^{*})$
and $R$ the ring of integers in a number field232323This restriction on the
coefficient ring is only due to the restriction in Theorem 5.5. With more
work, as in [21, Theorem III.1], one could probably take any domain of
characteristic $0$ such that no prime divisor of $|G|$ is divisible in
$R^{*}$. $F$. Then,

  1. (1)

$\\{\text{torsion units in
}\ker(\widetilde{\Psi}_{\chi,\beta})\\}=\\{\chi(a)^{-1}a\mid a\in N\\}$,

  2. (2)

If $\ker(\widetilde{\Psi}_{\chi})$ is finite if and only if
${\mathcal{U}}(R^{\beta}[\Gamma])$ is finite or $\chi\neq\omega_{N}$ and one
of the following hold:

     * •

$A\cong C_{p}$ for some prime and ${\mathcal{U}}(RG)$ is finite

     * •

$G$ is abelian and $\operatorname{lcm}(\exp(G),\exp(A))$ is a divisor of $4$
or $6$.

To proof the second part we will need the next theorem which is a
generalisation of the classical result of Berman and Higman on torsion units
(see [15, Theorem 2.3.] or [21, Theorem III.1] for the untwisted case).

######  Theorem 5.5.

Let $G$ be a finite group, $R$ the ring of integers in a number field $F$ and
$\beta\in H^{2}(G,R^{*})$. If $x=\sum_{g\in G}a_{g}u_{g}$ is a torsion unit in
$R^{\beta}[G]$ such that $a_{1}\neq 0$, then $x\in{\mathcal{U}}(R)$.

###### Proof.

The proof is along the same lines as [15, Theorem 2.3.]. Let $u=\sum_{g\in
G}r_{g}u_{g}$ be a torsion unit in $R^{\beta}[G]$, say of order $m$. Assume
$r_{1}\neq 0$. Let $n=|G|$ and consider the left regular representation
$\rho:F^{\beta}[G]\rightarrow M_{n}(F)$. Then $\rho(u)$ is of finite order $n$
(note that $\rho$ is injective) and thus it is diagonalisable over the
algebraic closure ${\mathbb{C}}$ of $F$ (recall that $F$ is a number field).
Its eigenvalues $\xi_{1},\cdots,\xi_{n}$ are roots of unity, each of order a
divisor of $m$. Let $\epsilon$ be a primitive $m$-th root of unity. We get
that $\zeta_{1}+\cdots+\zeta_{n}=\sum_{g\in
G}r_{g}\mbox{tr}(\rho(u_{g}))=nr_{1}$. Clearly, $nr_{1}$ then belongs to the
ring of integers of ${\mathbb{Q}}(\epsilon)\subseteq{\mathbb{C}}$ and
$|nr_{1}|=|\zeta_{1}+\cdots\zeta_{n}|\leq n$. Moreover, we get $|nr_{1}|=n$ if
and only if $\zeta_{1}=\zeta_{2}=\ldots=\zeta_{n}$. If this holds then the
diagonalisation of $\rho(u)$ is a multiple of the identity matrix, hence
$\rho(u)$ is a central, i.e. $\rho(u)\in F$. Because $\rho$ is injective this
yields that $u\in F\cap R^{\beta}[G]=R$, as desired.

So, it remains to show that $|nr_{1}|=n$. Assume, the contrary, i.e. suppose
$|nr_{1}|<n$. We claim that, for every
$\sigma\in\mbox{Gal}({\mathbb{Q}}(\epsilon)/{\mathbb{Q}})$, we have
$|\sigma(nr_{1})|<n$. Indeed, note that complex conjugation and elements of
$\mbox{Gal}({\mathbb{Q}}(\epsilon)/{\mathbb{Q}})$ commute and thus if
$n=|\sigma(nr_{1})|=\sigma(nr_{1})\overline{\sigma(nr_{1})}=\sigma(nr_{1})\sigma(\overline{nr_{1}})$
then also $n=\sigma^{-1}(\sigma(nr_{1})\sigma(\overline{nr_{1}}))=|nr_{1}|<1$.
This proves the claim.

Note that $r_{1}\in{\mathbb{Q}}(\epsilon)$. Since also $r_{1}\in R$ and thus
$r_{1}$ is an algebraic integer, it follows that $r_{1}$ is in the ring of
integers of ${\mathbb{Q}}(\epsilon)$. The assumption says that $|r_{1}|<1$ and
from the claim we get that $|\sigma(r_{1})|<1$ for all
$\sigma\in\mbox{Gal}({\mathbb{Q}}((\epsilon)/{\mathbb{Q}})$. It follows that
the norm
$0\neq|N_{\mbox{Gal}({\mathbb{Q}}((\epsilon)/{\mathbb{Q}})}(r_{1})|=|\prod_{\sigma\in\mbox{Gal}({\mathbb{Q}}(\epsilon)/{\mathbb{Q}})}\sigma(r_{1})|=\prod_{\sigma\in\mbox{Gal}({\mathbb{Q}}((\epsilon)/{\mathbb{Q}})}|\sigma(r_{1})|<1$.
However, this yields a contradiction as
$N_{\mbox{Gal}({\mathbb{Q}}((\epsilon)/{\mathbb{Q}})}(r_{1})$ is an algebraic
integer (see for example Proposition 4.1.8 in [9]) and belongs to
${\mathbb{Z}}$; so it can not be strictly between $0$ and $1$. Thus, Berman’s
result remains valid for torsion units in $R^{\beta}[G]$. ∎

We now proceed to the proof of the main theorem

###### Proof of Theorem 5.4..

First note that $\\{\chi(a)^{-1}a\mid a\in N\\}$ is indeed contained in
$\ker(\widetilde{\Psi}_{\chi,\beta})$. Conversely, take a torsion unit
$u\in\ker(\widetilde{\Psi}_{\chi,\beta})$. According to Proposition 3.3
$\ker(\widetilde{\Psi}_{\chi,\beta})=\mathcal{U}(1+R^{\beta}[\Gamma]I_{\chi})$,
allowing to rewrite $u$ as $u=1+\sum_{g\in G}y_{g}u_{\mu(g)}$ with $y_{g}\in
I_{\chi}=\operatorname{span}_{R}\\{u_{a}-\chi(a)u_{1}\mid a\in N\\}.$ In
particular if there would be no $1\neq a\in N$ in $\rm supp(u)$, then $u(1)=1$
and hence by Theorem 5.5 would $u\in{\mathcal{U}}(R)$ and thus $u=1$ (as $u$
is in the kernel).

Now take a non-trivial $a\in N\cap\rm supp(u)$ and consider the element
$z:=u.u_{a^{-1}}\chi(a)$. Note that due tot the centrality of $N$ the element
$z$ is again torsion unit in $\ker(\widetilde{\Psi}_{\chi,\beta})$. However
now $z(1)\neq 0$ and therefore by applying Theorem 5.5 $z=1$. In other words,
$u=\chi(a)^{\textemdash 1}a$ as needed.

For the second part, we will assume that we are not in the trivial case that
${\mathcal{U}}(R^{\beta}[\Gamma])$ is finite. Then using Theorem 3.2 we see
that $\ker(\widetilde{\Psi}_{\chi})$ is finite if and only if
${\mathcal{U}}(R(\chi\prime)^{\beta.T(\chi\prime)}[G])$ for all
$\chi\neq\chi\prime\in\operatorname{Lin}(N,F)$ but, by assumption,
${\mathcal{U}}(R(\chi)^{\beta.T(\chi)}[G])$ not. Remark that the latter can’t
happen if $\chi=\omega_{N}$ is the augmentation.

For the interpretation of those conditions we use Theorem 4.1. Among others,
${\mathcal{U}}(R(\chi\prime))$ must be finite for all $\chi\neq\chi\prime$. In
other words if we decompose $FN\subseteq\mathcal{Z}(F^{\beta}[\Gamma])$ as in
(12), i.e. using the theorem of Perlis-Walker:
$FN=\bigoplus_{d\mid|N|}F(\zeta_{d})^{\oplus a_{d}}$, then there has to be at
most one $\zeta_{d}$ for which the ring of integers in $F(\zeta_{d})$, denoted
$R_{d}$ is infinite. Moreover, for this value we need $a_{d}=1$. Note that
this property is inhered by subgroups of $N$. Also remark that if
${\mathcal{U}}(R_{d})$ is infinite, then also ${\mathcal{U}}(R_{d.m})$ for any
$m\in{\mathbb{N}}$. Therefore, using Dirichlet unit theorem, we see that
either $\exp(N)\mid 4,6$ or $N\cong C_{p}$ for some prime $p$. If $N\cong
C_{p}$ and since $\chi\neq\omega_{N}$, there is only the component
${\mathcal{U}}(RG)$ remaining. This yields the first case mentioned in the
statement.

Now suppose $\exp(A)\mid 4,6$ and $\chi\neq\omega_{N}$. We then already know
that ${\mathcal{U}}(RG)$ is finite. Thus either $G$ is abelian with
$\exp(G)\mid 4,6$ or $G\cong Q_{8}\times E$ with $E$ an elementary abelian
$2$-group. It follows from Corollary 4.3 that the latter case can’t occur.
Finally, if $G$ is abelian one can check that if
$\operatorname{lcm}(\exp(G),\exp(A))$ is not a divisor of $4$ or $6$. Then one
would have to many large cyclotomic fields. ∎

##  6\. Twisting bicyclic units and their contribution

In this section we will introduce a generic construction of units in twisted
group rings (of finite groups) which generalizes the bicyclic units. They
defer by a factor determined by the twisting, hence the suggestive name of
twisted bicyclic units. We will show in Section 6.2 that these units play a
role analogous to the one of elementary matrices inside the special linear
group, building on a generalisation of Jespers-Leal theorem obtained in
Theorem 6.3. Thereafter we return to the context of the previous section and
investigate their contribution to $\operatorname{Coker}(\Psi_{\chi})$.

###  6.1. Generalized byclic units versus elementary matrices

Before specifying to twisted group rings242424The reader only interested
herein, can jump to the next subsection., we wish to write down a general
version of an important result of Jespers-Leal on bicylic units which was only
informally known to some experts. For this section we work as general as
possible : let $B$ be a finite dimensional semisimple ${\mathbb{Q}}$-algebra,
$S$ a finite set of places of $\mathbb{Q}$ such that $\\{\infty\\}\subseteq
S$. Further let $A$ be a finite dimensional semisimple $B$-algebra, $R$ an
$S$-order in $B$.

######  Example 6.1.

Let $S=\\{\infty,(p_{1}),\ldots,(p_{l})\\}$ with $p_{i}$ a prime number and
$\zeta$ some root of unity. Then
$R={\mathbb{Z}}[\frac{1}{p_{1}},\ldots,\frac{1}{p_{l}},\zeta]$ is the ring of
$S$-integers of ${\mathbb{Q}}(\zeta)$. So in this case $B={\mathbb{Q}}(\zeta)$
and $A={\mathbb{Q}}(\zeta)^{\alpha}[G]$ for some $[\alpha]\in
H^{2}(G,{\mathbb{Q}}(\zeta)^{*})$. Another frequent example is $A=M_{n}(D)$
with $D$ a finite dimensional division ${\mathbb{Q}}$-algebra.

Let $\mathcal{A}=\\{g_{1},\ldots,g_{n}\\}$ be a $B$-basis of $A$ and
$\\{x_{1},\ldots,x_{q}\\}$ a ${\mathbb{Z}}_{S}$-basis of $R$. For every
idempotent $f\in A$, there exists a minimal integer $n_{f}\in{\mathbb{Z}}$
such that $n_{f}f\in\operatorname{span}_{R}(\mathcal{A})$. For these consider
now the elements

$b(m_{i}x_{i}g,f)=1+n_{f}^{2}(1-f)m_{i}x_{i}gf\text{ and
}b(f,m_{i}x_{i}g)=1+n_{f}^{2}fm_{i}x_{i}g(1-f).$

with $m_{i}\in{\mathbb{Z}}_{S}$. As usual, these are units because
$(1-f)m_{i}x_{i}gf$ has square zero.

######  Definition 6.2.

Let $\mathcal{F}$ be a set of idempotents in $A$. Then the elements in

$\text{GBic}^{\mathcal{F}}(A,R)=\langle b(m_{i}x_{i}g,f),b(f,m_{i}x_{i}g)\mid
f\in\mathcal{F},1\leq i\leq q,g\in\mathcal{A},m_{i}\in{\mathbb{Z}}_{S}\rangle$

are called generalized bicyclic units corresponding to $\mathcal{F}$.

Classically only the generators with $m_{i}=1$ are called generalized bicyclic
units. The values $m_{i}$ are added in order to still obtain all elements of
$A$ in the middle between $(1-f)$ and $f$ (when taking the group generated
by). An important theorem of Jespers-Leal252525In their version the case
$S=\\{\infty\\}$, i.e. of ${\mathbb{Z}}$-orders was done. [12] says that
generalized bicyclic units should be considered as the analogue of elementary
matrices.

######  Theorem 6.3.

Let $e\in\text{PCI}(A)$ such that $Ae\cong\operatorname{M}_{n}(D)$ with $n\geq
2$. Let $\mathcal{O}$ be a $S$-order in $D$. If $f$ is an idempotent in $A$
such that $fe\notin\mathcal{Z}(Ae)$, then there exists a
$y\in{\mathbb{Z}}_{S}$ such that

$1-e+E_{n}(y\mathcal{O})\subseteq\text{GBic}^{\\{f\\}}(A,R)$

###### Proof.

Multiplying generalized bicyclic units one sees that

$\\{1+n_{f}^{2}f\alpha(1-f),1+n_{f}^{2}(1-f)\alpha f\mid\alpha\in
A\\}\subseteq\text{GBic}^{\\{f\\}}(A,R).$

Writing now $f=E_{11}+\cdots+E_{ll}$, for $0<l<n$ and $\\{E_{i,j}\mid 1\leq
i,j\leq n\\}$ a set of matrix units (which is possible by [9, Lemma 11.2.4]),
it quickly follows that

$f\operatorname{M}_{n}(\mathcal{O})E_{i,j}(1-f)e=\operatorname{M}_{n}(\mathcal{O})E_{i,j}\text{
and
}(1-f)\operatorname{M}_{n}(\mathcal{O})E_{i,j}fe=\operatorname{M}_{n}(\mathcal{O})E_{i,j}$

for all $1\leq i\leq l$ and $l+1\leq j\leq n$. Hence for these $i$ and $j$ we
find that

$1+x\operatorname{M}_{n}(\mathcal{O})E_{i,j}\text{ and
}1+x\operatorname{M}_{n}(\mathcal{O})E_{j,i}$

are contained in $\text{GBic}^{\\{f\\}}(A,R)$ for some $x\in{\mathbb{Z}}_{S}$.
The other indices can now be reached by taking the appropriate commutators and
by doing so one will obtain that $y=x^{2}$ is the multiple you are looking
for. ∎

###  6.2. Twisted bicyclic units - general theory

From now on we consider the setting where $S=\\{\infty\\}$, i.e.
${\mathbb{Z}}$-orders, and $A={\mathbb{Q}}(\zeta_{n})^{\alpha}[G]$ where
$\zeta_{n}$ is some primitive $n$-th root and $[\alpha]\in
H^{2}(G,{\mathbb{Z}}[\zeta_{n}]^{*})\subseteq
H^{2}(G,{\mathbb{Q}}(\zeta_{n})^{*})$ arbitrary. The importance of this case
for the study of ${\mathbb{Q}}[\Gamma]$ for $\Gamma$ some central extension is
highlighted by Theorem 3.2 and Proposition 3.3.

#### Idempotents from trivial units

Suppose there is a $2$-cocycle $\alpha\in
Z^{2}(G,{\mathbb{Z}}[\zeta_{n}]^{*})$ of finite order with $o(\alpha)\mid n$.
Then we can partition $G$ into the following sets:

(18) $\mathcal{G}_{j}^{\alpha}=\\{g\in G\mid u_{g}^{o(g)}=\zeta_{n}^{j}\\}.$

The observation is now that if $g\in\mathcal{G}_{0}^{\alpha}=\\{g\in G\mid
o(u_{g})=o(g)\\}$, then $\widehat{u_{g}}=\frac{1}{o(g)}\widetilde{u_{g}}$ with

$\widetilde{u_{g}}:=1+u_{g}+\ldots+u_{g}^{o(u_{g})-1}$

is again a non-trivial idempotent in ${\mathbb{Q}}(\zeta_{n})^{\alpha}[G]$.
Consequently, $\widetilde{u_{g}}(o(g)-\widetilde{u_{g}})=0$. If
$g\in\mathcal{G}_{j}^{\alpha}$ for a non-zero $j$, then262626Going in the sum
till the usual upper bound $o(g)-1$ will not yield an idempotent.
$\widetilde{u_{g}}=0$. Instead one could make an idempotent out of
$u_{g}^{n/gcd(j,n)}$ but in practice this construction will not be useful.
Hence:

######  Question 6.4.

Does there a generic way to produce a non-trivial idempotent from
$g\in\mathcal{G}_{j}^{\alpha}$ for a non-zero $j$?

######  Remark 6.5.

It is well known that for a cyclic group $C_{n}=\langle g\rangle$ the second
cohomology group over a commutative ring $R$ is isomorphic to
$R^{*}/(R^{*})^{n}$. Due to this the value of $u_{g}^{o(g)}$ is not uniquely
determined by the cohomology class $[\alpha]$. For example, the class
$[\alpha]\in H^{2}(C_{n},{\mathbb{C}}^{*})$ determined by the value
$\lambda=u_{g}^{o(g)}$ is trivial for any $0\neq\lambda\in{\mathbb{C}}$.
Therefore the sets $\mathcal{G}_{j}^{\alpha}$ are unfortunately dependant of
the chosen cocycle. However one could define in an independent way the sets
$\mathcal{G}_{0}$ and $\mathcal{G}_{\neq 0}$ where the latter would be all
$g\in G$ such that $u_{g}^{o(g)}$ is not in the class of $1$ in
$R^{*}/(R^{*})^{n}$. In particular for $R={\mathbb{Z}}$ the situation
simplifies. Indeed due to the previous remark the value $u_{g}^{o(g)}\in\\{\pm
1\\}$ is uniquely determined by the class $[\alpha]$.

For an integral twisted group ring $\mathbb{Z}^{\alpha}G$, for any element of
$g\in G$ of odd order, the restriction of $[\alpha]$ to $\langle g\rangle$ is
trivial. Therefore we may and will always from now on choose a normalized
representative $\alpha$ such that $u_{g}^{o(g)}=1$ in $\mathbb{Z}^{\alpha}G$
for all $g$ of odd order.

#### Twisting a bicyclic unit : construction

We will use the $G$-module structure of the twisted group ring $R^{\alpha}[G]$
described in [6, §3.2] given by conjugation, that is:

$g(u_{x})=u_{g}u_{x}u_{g}^{-1}=\alpha(g,x)\alpha(x,g)^{-1}u_{gxg^{-1}}.$

For every $x\in G$, the restriction of the conjugation representation from $G$
to the centralizer $C_{G}(x)$ admits a $1$-dimensional invariant subspace
spanned by $u_{x}$. By the above this ordinary $1$-dimensional representation
$\chi_{x}:C_{G}(x)\rightarrow R^{*}$ is given by

$\chi_{x}(g)=[u_{g},u_{x}]=\alpha(g,x)\alpha(x,g)^{-1}.$

For $R=\mathbb{Z}$, for any $x\in G$ we can partition $C_{G}(x)$ into two
disjoint sets

$C_{x}^{+}=\\{g\in C_{G}(x)|\chi_{x}(g)=1\\},\quad C_{x}^{-}=\\{g\in
C_{G}(x)|\chi_{x}(g)=-1\\}.$

We remark that if $C_{x}^{-}=\emptyset$ then $x$ (and it’s conjugacy class) is
called $\alpha$-regular and the set of all $\alpha$-regular conjugacy classes
form a basis for the center of $\mathbb{Z}^{\alpha}G$. An extreme case for
that is when $\alpha$ is trivial. In this case $C_{x}^{-}=\emptyset$ for any
$x\in G$.

######  Example 6.6.

A straightforward calculation yields that in $R^{\alpha}G$, if
$[u_{g},u_{h}]=\zeta\in R^{*}$ a root of unity of order $k$, then $k$ is a
divisor of the greatest common divisor of the orders of $g$ and $h$. Hence, in
$\mathbb{Z}^{\alpha}G$ all elements of $G$ of odd order are $\alpha$-regular.

Using this we now introduce a twisted version of the classical bicyclic units.
In order to do so we will use some details from §6.2.

######  Definition 6.7.

The subgroup

$\mathcal{B}^{tw}(G)=\langle
1+(o(g)-\widetilde{u_{g}})(\alpha+\frac{\beta}{o(g)})\widetilde{u_{g}}\mid
g\in\mathcal{G}_{1}^{\alpha},\alpha,\beta\in{\mathbb{Z}}^{\alpha}[G],\,\rm
supp(\beta)\subseteq C_{g}^{-}\rangle$

of ${\mathcal{U}}({\mathbb{Z}}^{\alpha}[G])$ is called the group of twisted
bicyclic units.

For any square zero element $\gamma$, such as
$(o(g)-\widetilde{u_{g}})(\alpha+\frac{\beta}{o(g)})\widetilde{u_{g}}$, one
has a unipotent unit $1+\gamma$ with inverse $1-\gamma$. However at first it
might seem surprising that the above elements are units in
${\mathbb{Z}}^{\alpha}[G]$ and not only in ${\mathbb{Q}}^{\alpha}[G]$. This
follows from the fact that
$u_{g}^{j}u_{h}\widetilde{u_{h}}=(-1)^{j}u_{h}\widetilde{u_{g}}$ for $t\in
C_{g}^{-}$ and hence when $o(g)$ is even272727If $o(g)$ is odd, the last sum
equals $o(g)-1$ and hence in general the element will not be an integral
combination. However, as remarked in Section 6.2, elements of odd order are
$\alpha$-regular (i.e. $C_{g}^{-}=\emptyset$) and thus one can only take
$\beta=0$. In particular we don’t need to specify the order of the elements
from which we built the twisted bicyclic units in order to be a subset of
${\mathbb{Z}}^{\alpha}[G]$.

(19)
$\begin{array}[]{rcl}(o(g)-\widetilde{u_{g}})u_{h}\widetilde{u_{g}}&=&u_{h}\left(o(g)-\sum_{j=0}^{o(g)-1}(-1)^{j}u_{g}^{j}\right)\widetilde{u_{g}}\\\
&=&u_{h}\left(o(g)-\sum_{j=0}^{o(g)-1}(-1)^{j}\right)\widetilde{u_{g}}\\\
&=&o(g)u_{h}\widetilde{u_{g}}.\end{array}$

Furthermore, taking $\mathcal{F}=\\{\widehat{u_{g}}\mid
g\in\mathcal{G}_{1}^{\alpha}\\}$ then we see that
$\text{GBic}^{\mathcal{F}}({\mathbb{Q}}^{\alpha}[G],{\mathbb{Z}})\subseteq\mathcal{B}^{tw}(G).$
Due to this, we can now apply Theorem 6.3 (using the same proof282828The proof
consists in showing that $\widehat{u_{g}}e$ is non-central under the
assumptions written. as in [9, Theorem 11.3.2]) to show that twisted bicyclic
units contain sufficiently many elementary matrices:

######  Corollary 6.8.

Let $G$ be a finite group and $\alpha\in Z^{2}(G,{\mathbb{Z}}^{*})$ such that
$G_{\alpha}$ has no fixed-point free images292929A finite group is called
fixed point free if it has an irreducible complex representation $\rho$ such
that $1$ is not an eigenvalue of $\rho(g)$ for all $1\neq g\in G$. Such groups
are exactly the Frobenius complements [9, Proposition 11.4.6.] and hence those
that are subgroups of $D^{*}$ for some finite dimensional division algebra
$D$.. Then for any $e\in\text{PCI}({\mathbb{Q}}^{\alpha}[G])$ such that
${\mathbb{Q}}^{\alpha}[G]e\cong M_{n}(D)$ with $n\geq 2$ we have that

$1-e+E_{n}(y\mathcal{O})\subseteq\mathcal{B}^{tw}(G)$

with $\mathcal{O}$ an order in $D$ and some $y\in{\mathbb{Z}}$.

######  Remark.

If $\alpha$ is a trivial cocycle, then
$\mathcal{G}_{-1}^{\alpha}=\emptyset=C_{g}^{-}$ for any $g\in G$ and
$\mathcal{G}_{1}^{\alpha}=G$. In particular, in that case
${\mathbb{Q}}^{\alpha}[G]={\mathbb{Q}}[G]$ and
$\text{GBic}^{\mathcal{F}}({\mathbb{Q}}[G],{\mathbb{Z}})=\mathcal{B}^{tw}(G)=\mathcal{B}(G)$,
i.e. we recover exactly the classical bicyclic units.

###  6.3. On twisted bicyclic units in cokernel of the transgression map

We now return to the setting of Section 3.2. More precisely we consider the
central extension (3) corresponding to $[\alpha]\in H^{2}(G,A)$. We also fix a
normalized representant $\alpha\in Z^{2}(G,A)$ of the form
$\alpha(g,h)=\mu(g)\mu(h)\mu(gh)^{-1}$ for a section $\mu$ of
$\lambda:\Gamma\twoheadrightarrow G$.

The twisted bicyclic units from the previous section now allow to construct
elements in $\operatorname{Coker}(\widetilde{\Psi_{\chi}})$ .

######  Proposition 6.9.

Let $\chi\in\operatorname{Hom}(A,{\mathbb{Z}}^{*})$ and $Q$ a central subgroup
of $\Gamma$ such that $A\cap Q=1$. If ${\mathcal{U}}({\mathbb{Z}}G)$ is finite
then, with notations as above, we have that

$\mathcal{B}_{min}:=\\{1+(o(g)-\widetilde{u_{g}})z\frac{u_{h}}{o(g)}\widetilde{u_{g}}\in{\mathbb{Z}}^{\operatorname{Tra}_{\alpha}(\chi)}[G]\mid
g\in\mathcal{G}_{1}^{\operatorname{Tra}_{\alpha}(\chi)},h\in
C_{g}^{-},z\in\\{1-q\,:\,q\in Q\cup 0\\}\\}$

is not in the image of
${\left.\kern-1.2pt\widetilde{\Psi_{\chi}}\right|_{{\mathcal{U}}({\mathbb{Z}}[\Gamma])}}$.

Note that the existence of such an element $h$ indirectly assumes that $g$ is
$\operatorname{Tra}_{\alpha}(\chi)$-irregular. For this
$\operatorname{Tra}_{\alpha}(\chi)$ must be a non-trivial cocycle and hence
$\chi\neq\omega_{A}$.

###### Proof.

To start, using (19), we rewrite the element as
$1+(o(g)-\widetilde{u_{g}})z\frac{u_{h}}{o(g)}\widetilde{u_{g}}=1+zu_{h}\widetilde{u_{g}}.$
Thus an element in
$\Psi_{\chi}^{-1}(1+(o(g)-\widetilde{u_{g}})z\frac{u_{h}}{o(g)}\widetilde{u_{g}})$
is of the form $\tau:=1+z\mu(h)\widetilde{\mu(g)}+x$ for some
$x\in\ker(\Psi_{\chi})$. We need to prove that
$\tau\notin{\mathcal{U}}({\mathbb{Z}}\Gamma)$ for any $x$. Assume otherwise
and start by decomposing $x$ according to Proposition 3.3:

$x=\sum_{a\in\ker(\chi)}x_{a}(a-1)+\sum_{a\in A\setminus\ker(\chi)}x_{a}(a+1)$

with $x_{a}=z_{a}+\sum_{1\neq w\in\Gamma}z_{w}\mu(w)\in{\mathbb{Z}}[\Gamma]$.

Therefore,

$\omega_{A}(\tau)=1+zh\widetilde{g}+2\sum_{a\in A\setminus\ker(\chi)}x_{a}.$

However, by assumption $\omega_{A}(\tau)$ is a torsion unit and hence by
Berman-Higman either $\omega_{A}(\tau)=\pm 1$ or the coefficient of $e_{G}$ in
the above is zero. Reduction modulo $2$ shows that the former is not the case.
For the latter, note that the coefficient of $e_{G}$ is $1+2\sum_{a\in
A\setminus\ker(\chi)}z_{a}$ which is non-zero. This yields the desired
contradiction and finishes the proof. ∎

We expect that the ’smallest’ twisted bicylic units as in Proposition 6.9
contribute to $\operatorname{Coker}(\Psi_{\chi})$ in full generality. Recall
that, Example 3.4, when $[\beta]=\operatorname{Inf}_{Q}([\gamma])$ with
$[\gamma]\in H^{2}(G/Q,R^{*})$ and $\chi$ was the trivial character then the
generalized transgression map was denoted
$\omega_{\beta,Q}:R^{\beta}[G]\rightarrow R^{\gamma}[G/Q].$

######  Question 6.10.

With notations as above and as in Section 5.1, do the following hold? :

  1. (1)

The image of the subgroup

(20) $\langle
1+(o(l)-\widetilde{u_{l}})(x+\frac{y}{o(l)})\widetilde{u_{l}}\mid
l\in\mathcal{G}_{1},0\neq y,x\in{\mathbb{Z}}^{\alpha}G,\,\rm supp(y)\subseteq
C_{l}^{-}\rangle$

of $\mathcal{B}^{tw}(G)$ is a non-trivial subgroup of
$\operatorname{Coker}(\widetilde{\Psi_{\chi}})$ of exponent
$\text{lcm}\\{o(l)\mid l\in\mathcal{G}_{1}\\}$.

  2. (2)

The image of the subgroup

$\langle 1+(o(l)-\widetilde{u_{l}})(z.\frac{y}{o(l)})\widetilde{u_{l}}\mid
l\in\mathcal{G}_{1},0\neq
y\in{\mathbb{Z}}^{\alpha}G,z\in\ker(\omega_{\beta,Q}),\,\rm supp(y)\subseteq
C_{l}^{-}\rangle$

of $\mathcal{B}^{tw}(G)$, where $\beta=\operatorname{Tra}_{\alpha}(\chi)$, is
a non-trivial subgroup of $\ker(F_{Cok})$

Most likely, in general other type of elements can be contained in
$\operatorname{Coker}(\widetilde{\Psi_{\chi}})$. However in Section 8 we will
see that for $\Gamma$ an extension of $C_{2}$ with an elementary abelian
$2$-group $G=C_{2}^{n}$ that these elements will generate the full cokernel.
Moreover in that case $\ker(F_{Cok})\cong C_{2}^{n-2}$. It would be
interesting to know whether this is a general phenomena.

##  7\. Method to describe $\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times
C_{2}^{n}])$

Let $G$ be a finite group and

$C_{2}^{n}=\langle x_{1}\rangle\times\cdots\times\langle x_{n}\rangle$

an elementary abelian $2$-group. In this section we will consider a,
potentially trivial, $2$-cohomology class $[\alpha]$ in the image of the
inflation map

(21) $\operatorname{Inf}:H^{2}(G,{\mathbb{Z}}^{*})\rightarrow H^{2}(G\times
C_{2}^{n},{\mathbb{Z}}^{*}).$

In particular, by Corollary 2.5, the subgroup $C_{2}^{n}$ is central in
${\mathcal{U}}({\mathbb{Z}}^{\alpha}[G\times C_{2}^{n}])$. For such cocycles
we will now present a method to describe
$\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times C_{2}^{n}])$ whenever
$\mathcal{U}({\mathbb{Z}}^{\alpha}[G])$ is known.

Convention: $C_{2}^{i}$ denotes the concrete subgroup $\langle
x_{1},\ldots,x_{i}\rangle$. In particular when we write
${\mathbb{Z}}^{\alpha}[G\times C_{2}^{i}]$ we truly mean the subring of
${\mathbb{Z}}^{\alpha}[G\times C_{2}^{n}]$ generated by $G\times C_{2}^{i}$.

To start, consider for every $1\leq i\leq n$ the projection

$\psi_{i}:{\mathbb{Z}}^{\alpha}[G\times
C_{2}^{i}]\rightarrow{\mathbb{Z}}^{\alpha}[G\times C_{2}^{i-1}]$

induced by mapping $x_{i}$ to $1$, which is a ring morphism303030This is a
ring-morphism due to the centrality of $x_{i}$. In particular for arbitrary
$2$-cocycles this method does not work.. Note that, due to the centrality of
$x_{i}$ that ${\mathbb{Z}}^{\alpha}[G\times
C_{2}^{i}]=\left({\mathbb{Z}}^{\alpha}[G\times C_{2}^{i-1}]\right)[\langle
x_{i}\rangle]$ and therefore $\psi_{i}$ is globally defined by
$\psi(u+vx)=u+v$ for $u,v\in{\mathbb{Z}}^{\alpha}[G\times C_{2}^{i-1}]$. By
definition $\psi_{i}$ induces an epimorphism

$\widetilde{\psi}_{i}:\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times
C_{2}^{i}])\rightarrow\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times
C_{2}^{i-1}]).$

Accordingly, we obtain the following splitting.

######  Lemma 7.1.

Let $K_{i}=\ker(\widetilde{\psi}_{i})$. Then,
$\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times
C_{2}^{i}])=K_{i}\rtimes\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times
C_{2}^{i-1}]).$ Consequently,

$\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times C_{2}^{n}])\cong
K_{n}\rtimes\Big{(}K_{n-1}\rtimes\big{(}\cdots
K_{2}\rtimes(K_{1}\rtimes\mathcal{U}({\mathbb{Z}}^{\alpha}G))\big{)}\Big{)}.$

###### Proof.

Since $\widetilde{\psi}_{i}$ is surjective we obtain an extension

$1\rightarrow K_{i}\hookrightarrow\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times
C_{2}^{i}])\xrightarrow{\widetilde{\psi}_{i}}\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times
C_{2}^{i-1}])\rightarrow 1.$

Clearly, this extension is split by the identity map. The second part now
follows by iteration. ∎

In view of Lemma 7.1 we will now concentrate on describing all the kernels
$K_{i}$. To this end we define the groups

$U_{i}=\\{1+2^{i}u\mid
u\in{\mathbb{Z}}^{\alpha}G\\}\cap\mathcal{U}({\mathbb{Z}}^{\alpha}G).$

Our goal is to show that $K_{i}$ can be built from isomorphic copies of the
groups $U_{1},\ldots,U_{i}$ and the unit group
$\mathcal{U}({\mathbb{Z}}^{\alpha}G)$. Note that, by Theorem 5.5, $U_{i}$ is
torsion-free when $i\geq 2$ and the only torsion in $U_{1}$ is $\\{\pm 1\\}$.

As a tool we need to consider more generally the groups

$U_{k,j}=\\{1+2^{k}u\mid u\in{\mathbb{Z}}^{\alpha}[G\times
C_{2}^{j}]\\}\cap\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times C_{2}^{j}]).$

Note that $U_{k,0}=U_{k}$ and $U_{k_{1},j_{1}}\leq U_{k_{2},j_{2}}$ for
$k_{2}\leq k_{1}$ and $j_{1}\leq j_{2}$. Next to these, the projections

$\varphi_{i}:{\mathbb{Z}}^{\alpha}[G\times
C_{2}^{i}]\rightarrow{\mathbb{Z}}^{\alpha}[G\times
C_{2}^{i-1}]:x_{i}\mapsto-1$

will also be instrumental. The induced epimorphism on the unit groups is
denoted by $\widetilde{\varphi}_{i}$.

######  Lemma 7.2.

For all $1\leq k,j$, with notations as above, we have that

  * •

$K_{j}\cong\widetilde{\varphi}_{j}(K_{j})=U_{1,j-1},$

  * •

$U_{k,j}\cong U_{k+1,j-1}\rtimes U_{k,j-1}$.

###### Proof.

By definition,

$K_{j}=\\{1+u(1-x_{j})\mid u\in{\mathbb{Z}}^{\alpha}[G\times
C_{2}^{j-1}]\\}\cap\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times C_{2}^{j}]).$

Using the centrality of $x_{j}$, hence ${\mathbb{Z}}^{\alpha}[G\times
C_{2}^{j}]=\left({\mathbb{Z}}^{\alpha}[G\times C_{2}^{j-1}]\right)\langle
x_{j}\rangle$, one immediately obtains that $\widetilde{\varphi}_{j}$ is
injective on $K_{j}$. Consequently, $K_{j}\cong\widetilde{\varphi}_{j}(K_{j})$
and more explicitly

$\widetilde{\varphi}_{j}(K_{j})=\\{1+2u\mid u\in{\mathbb{Z}}^{\alpha}[G\times
C_{2}^{j-1}]\\}\cap\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times
C_{2}^{j-1}])=U_{1,j-1}.$

Note that $\widetilde{\varphi}_{j}(K_{j})$ contains all the units above
because by ${\mathbb{Z}}^{\alpha}[G\times C_{2}^{j-1}]$ is truly meant the
subring of ${\mathbb{Z}}^{\alpha}[G\times C_{2}^{j}]$.

For the second statement, remark that $\widetilde{\psi}_{j}$ is the identity
map on $U_{k,i}$ for all $0\leq i<j$ and all $k$. Therefore,
$U_{k,j}=\ker({\left.\kern-1.2pt\widetilde{\psi}_{j}\right|_{U_{k,j}}})\rtimes\widetilde{\psi}_{j}(U_{k,j})$
with $\widetilde{\psi}_{j}(U_{k,j})=U_{k,j-1}$. Next, since
$\ker({\left.\kern-1.2pt\widetilde{\psi}_{j}\right|_{U_{k,j}}})=\\{1+2^{k}u(1-x_{j})\mid
u\in{\mathbb{Z}}^{\alpha}[G\times C_{2}^{j-1}]\\}\cap U_{k,j}$ the map
$\widetilde{\varphi}_{j}$ is injective on it and so
$\widetilde{\varphi}_{j}\left(\ker({\left.\kern-1.2pt\widetilde{\psi}_{j}\right|_{U_{k,j}}})\right)=U_{k+1,j-1}$,
finishing the proof. ∎

An iterative process of the previous lemma decreases the second index of
$U_{k,j}$, with the cost of producing extra complements. Hence it now readily
follows that we can reduce to the groups $U_{i}$.

######  Theorem 7.3.

Let $G$ be a finite group, $[\alpha]\in{\rm Im}(\operatorname{Inf})$ as in
(21). Then,

$\begin{array}[]{rl}\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times
C_{2}^{n}])&=K_{n}\rtimes\Big{(}K_{n-1}\rtimes\big{(}\cdots
K_{2}\rtimes(K_{1}\rtimes\mathcal{U}({\mathbb{Z}}^{\alpha}G))\big{)}\Big{)}\\\
&=\Big{(}\big{(}(K_{n}\rtimes K_{n-1})\rtimes\cdots\big{)}\rtimes
K_{1}\Big{)}\rtimes\mathcal{U}({\mathbb{Z}}^{\alpha}G)\end{array}$

with

$K_{i}\cong\Big{(}\big{(}(U_{i}\rtimes U_{i-1})\rtimes(U_{i-1}\rtimes
U_{i-2})\big{)}\rtimes\cdots\Big{)}\rtimes K_{i-1}$

for all $2\leq i$ and $K_{1}\cong U_{1}$. Moreover,

$\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times
C_{2}^{n}])\cong\Big{(}\big{(}(N_{n}\rtimes
N_{n-1})\rtimes\cdots\big{)}\rtimes
N_{1}\Big{)}\rtimes\big{(}\mathcal{U}({\mathbb{Z}}^{\alpha}[G])\times\langle
x_{1},\ldots,x_{n}\rangle\big{)}$

with $N_{i}$ a torsion-free normal subgroup of $K_{i}$.

###### Proof.

The first part was proven in Lemma 7.1 (that we can rearrange the brackets is
easy to see). The description of $K_{i}$ follows from an iterative use of
Lemma 7.2.

For the last part, due to the centrality of the subgroup $C_{2}^{n}$ and the
obtained form of $K_{i}$, it is sufficient to prove that for every $i$ the
subgroup $K_{i}$ is isomorphic to $N_{i}\times\langle x_{i}\rangle$ for some
(torsion-free) subgroup $N_{i}$ normal in $K_{i-1}$. To this end, remark that
$K_{i}$ contains a copy of $K_{1}$, namely $K_{(i),1}=\\{1+(1-x_{i})u\mid
u\in{\mathbb{Z}}^{\alpha}[G]\\}\cap\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times\langle
x_{i}\rangle])$. Further recall that $\widetilde{\varphi}_{i}$ is injective on
$K_{i}$ and due to Corollary 2.5 ${\mathbb{Z}}^{\alpha}[G\times
C_{2}^{i}]\cong{\mathbb{Z}}^{\alpha}[G]\otimes_{{\mathbb{Z}}}{\mathbb{Z}}[C_{2}^{i}]$.
Consider now the map $Id\otimes\omega_{i}$ where $\omega_{i}$ denotes the
augmentation map of ${\mathbb{Z}}[C_{2}^{i}]$. Then we see that $K_{(i),1}$ is
mapped onto $U_{1}$ with $\langle x_{i}\rangle$ mapped on $\\{\pm 1\\}$.

Now suppose that $\mathcal{U}({\mathbb{Z}}^{\alpha}[G])\cong N\rtimes\pm G$.
Then, $U_{1}=(U_{1}\cap N)\times\\{\pm 1\\}$ and $U_{i}=U_{i}\cap N$ for
$i\geq 2$. Consequently, $K_{(i),1}=\widetilde{\varphi}_{i}^{-1}(U_{1}\cap
N)\times\langle x_{i}\rangle$ which in light of the centrality of $x_{i}$ and
the above also yields the existence of a subgroup $N_{i}$ as we were looking
for. Furthermore if $N$ is torsion-free, then also $U_{1}=U_{1}\cap N$ and
thus the group $N_{i}$ obtained as above would also be torsion-free (since
$U_{i}$ is torsion-free for $i\geq 2$).

Note that we in fact proved that

$\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times
C_{2}^{n}])\cong\Big{(}\big{(}(N_{n}\rtimes
N_{n-1})\rtimes\cdots\big{)}\rtimes
N_{1}\Big{)}\rtimes\big{(}\mathcal{U}({\mathbb{Z}}^{\alpha}[G])\times\langle
x_{1},\ldots,x_{n}\rangle\big{)}$

with $N_{i}$ torsion-free. ∎

Notice that we have only defined recursively a decomposition of $K_{i}$. More
precisely, $K_{i}\cong\mathcal{U}_{i}\rtimes K_{i-1}$ with
$\mathcal{U}_{2}:=U_{2}$ and
$\mathcal{U}_{i}:=\mathcal{U}_{i-1}[1]\rtimes\mathcal{U}_{i-1}$ where
$\mathcal{U}_{i-1}[1]$ looks like $\mathcal{U}_{i-1}$ but with all indices
increased by one. For example, $K_{2}\cong U_{2}\rtimes
K_{1},K_{3}\cong(U_{3}\rtimes U_{2})\rtimes
K_{2},K_{4}\cong\left((U_{4}\rtimes U_{3})\rtimes(U_{3}\rtimes
U_{2})\right)\rtimes K_{3}$ and

$K_{5}\cong(\Big{(}(U_{5}\rtimes U_{4})\rtimes(U_{4}\rtimes
U_{3})\Big{)}\rtimes\Big{(}(U_{4}\rtimes U_{3})\rtimes(U_{3}\rtimes
U_{2})\Big{)})\rtimes K_{4}.$

The decomposition obtained allows to transfer properties of
$\mathcal{U}({\mathbb{Z}}^{\alpha}G)$ to
$\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times C_{2}^{n}])$. The preservation of
(ZC1) was obtained previously by Höfert-Kimmerle [8] in the untwisted case.
Other interesting results in that line can be found in [3, 7].

######  Corollary 7.4.

Let $G$ be a finite group, $[\alpha]\in{\rm Im}(\operatorname{Inf})$ as in
(21). If $\mathcal{U}({\mathbb{Z}}^{\alpha}G)$ satisfy one of the following
properties:

  1. (1)

$\pm G$ has a (torsion-free) complement in
$\mathcal{U}({\mathbb{Z}}^{\alpha}[G])$

  2. (2)

it is commensurable with a direct product of free-by-free groups

  3. (3)

has a non-trivial amalgam decomposition

  4. (4)

$G$ satisfies313131The classical Zassenhaus conjectures are stated for group
rings, however also literally makes sense for twisted group rings. For example
(ZC3) would be that all finite $H\leq V({\mathbb{Z}}^{\alpha}[G])$ are
conjugated over ${\mathbb{Q}}^{\alpha}[G]$ to a subgroup of $G$. (ZC1), (ZC2)
or (ZC3)

then the same holds for $G\times C_{2}^{n}$ and
$\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times C_{2}^{n}])$.

###### Proof.

The first part directly follows from the last decomposition in Theorem 7.3,
since $\big{(}(N_{n}\rtimes N_{n-1})\rtimes\cdots\big{)}\rtimes N_{1}$ is
torsion-free. This decomposition also implies that every torsion subgroup of
$\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times C_{2}^{n}])$ is conjugated to a
torsion subgroup in $\mathcal{U}({\mathbb{Z}}^{\alpha}[G])\times\langle
x_{1},\ldots,x_{n}\rangle$. The fourth statement hence follows.

The second statement is a direct consequence of [13, Theorem 2.1., Lemma 3.1.,
Proposition 3.3.] (or [10]) which altogether323232The fact that a free-by-free
group has cohomological dimension at most $2$ is also needed. asserts that: if
$A$ is a finite dimensional semisimple ${\mathbb{Q}}$-algebra, $\mathcal{O}$
an order in $A$ then $\mathcal{U}(\mathcal{O})$ is commensurable to a direct
product of free-by-free groups if and only the simple quotients of $A$ are
either a field, totally definite quaternion algebra or
$\operatorname{M}_{2}(D)$ with $D$ in a list of certain quadratic imaginary
extensions of ${\mathbb{Q}}$ or certain quaternion algebras over totally real
number fields. In particular the property "commensurable with a direct product
of free-by-free groups" is fully determined by the form of the simple
components. However if $[\alpha]$ is in the image of (21), then

${\mathbb{Q}}^{\alpha}[G\times
C_{2}^{n}]\cong{\mathbb{Q}}^{\alpha}[G]\otimes_{{\mathbb{Q}}}{\mathbb{Q}}[C_{2}^{n}]\cong\underbrace{{\mathbb{Q}}^{\alpha}[G]\oplus\cdots\oplus{\mathbb{Q}}^{\alpha}[G]}_{n-times}.$

Thus the set of isomorphism types of simple quotients of
${\mathbb{Q}}^{\alpha}[G\times C_{2}^{n}]$ is the same as for
${\mathbb{Q}}^{\alpha}[G]$.

Concerning the third statement, note that due to Theorem 7.3
$\mathcal{U}({\mathbb{Z}}^{\alpha}[G])$ is an epimorphic image of
$\mathcal{U}({\mathbb{Z}}^{\alpha}[G\times C_{2}^{n}])$ and hence its amalgam
decomposition can be lifted along the epimorphism. ∎

######  Remark 7.5.

The proof of (2) in Corollary 7.4 in fact shows that any group theoretical
property $\mathcal{P}$ that can be read off the Wedderburn-Artin components
(i.e. the simple quotients) is inherited. Other examples of such properties
are: Kazhdan’s property $(T)$, property FAb and HFA (see [2, 1]). In general,
good candidates for such properties are the ones that are constant on
commensurability classes.

Following Corollary 7.4 the property to have a (torsion-free) normal
complement is preserved, however a concrete description thereof seems
difficult. For example, in [17] it was proven that no normal complement of the
trivial units in ${\mathcal{U}}({\mathbb{Z}}[D_{8}\times C_{2}\times C_{2}])$
is generated by bicyclic units, although a normal complement in
${\mathcal{U}}({\mathbb{Z}}[D_{8}\times C_{2}])$ is
$\operatorname{Bic}(D_{8}\times C_{2})$.

##  8\. Full description for elementary abelian 2-groups

Let $G$ be an elementary abelian $2$-group of rank $n+2$,

(22) $G=C_{2}\times C_{2}\times\ldots\times C_{2}=\langle
g\rangle\times\langle h\rangle\times\langle x_{1}\rangle\times\ldots\langle
x_{n}\rangle,$

and consider the cohomology class $[\alpha]\in H^{2}(G,C_{2})$ determined by
the values

(23) $[u_{g},u_{h}]=-1,\quad
u_{g}^{2}=u_{h}^{2}=u_{x_{i}}^{2}=[u_{g},u_{x_{i}}]=[u_{h},u_{x_{i}}]=[u_{x_{i}},u_{x_{j}}]=1,$

for any $1\leq i\neq j\leq n$. By abuse of notation we think of this
$[\alpha]$ as the cohomology class determining the group $\Gamma\cong
D_{8}\times C_{2}^{n}$ and also as the cohomology class which determines the
twisted group ring ${\mathbb{Z}}^{\alpha}G$. Recall our standing convention of
choosing a normalized $2$-cocycle representant of $[\alpha]$.

To start we handle the case that $n=0$, i.e. we describe
$\mathcal{U}({\mathbb{Z}}^{\alpha}[\langle g,h\rangle])$.

###  8.1. The starting case $C_{2}\times C_{2}$

Denote by $\widetilde{D}$ the subalgebra of $M_{2}(\mathbb{Z})$ consisting of
the elements

$\left\\{\left(\begin{array}[]{cc}a&b\\\ c&d\end{array}\right)\mid
a\equiv_{2}d,\,b\equiv_{2}c\right\\}.$

It can be easily verified that

(24) $\widetilde{D}=\left\\{\left(\begin{array}[]{cc}m+n&k+r\\\
k-r&m-n\end{array}\right)\mid m,n,k,r\in\mathbb{Z}\right\\}.$

Consider now the map $\phi:\mathbb{Z}^{\alpha}[C_{2}\times
C_{2}]\rightarrow\widetilde{D}$ defined by

$\displaystyle u_{1}\mapsto\left(\begin{array}[]{cc}1&0\\\
0&1\end{array}\right),\quad u_{g}\mapsto\left(\begin{array}[]{cc}1&0\\\
0&-1\end{array}\right),\quad u_{h}\mapsto\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right),\quad u_{gh}\mapsto\left(\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right)$

which can easily be seen to be a ring morphism.

######  Proposition 8.1.

With notations as above, ${\mathbb{Z}}^{\alpha}[C_{2}\times
C_{2}]\cong\widetilde{D}$ as rings.

###### Proof.

The map $\varphi$ is surjective as can be seen via

$A=\phi\left(mu_{1}+nu_{x}+ku_{y}+ru_{xy}\right),$

where $A=\left(\begin{array}[]{cc}m+n&k+r\\\ k-r&m-n\end{array}\right)$ is an
arbitrary element of $\widetilde{D}$ (see (24)). Injectivity follows from a
linear Independence argument. ∎

Next we will obtain a presentation of the unit group
$\mathcal{U}(\mathbb{Z}^{\alpha}[C_{2}\times C_{2}])$. For this we need the
following set of twisted bicyclic units (cf. Definition 6.7 and (19) ):

(25) $\mathcal{B}=\\{u_{1}+\left(\pm u_{lt}\pm u_{l}\right)\mid
u_{l}u_{t}=-u_{t}u_{l},\,\,u_{t}^{2}=1\\}.$

######  Proposition 8.2.

Let $\alpha$ be the $2$-cocycle determined by the values $(1,1,-1)$ on
$(u_{g}^{2},u_{h}^{2},[u_{g},u_{h}])$. Then

$\mathcal{U}({\mathbb{Z}}^{\alpha}[C_{2}\times C_{2}])=F_{2}\rtimes D_{8}$

where $F_{2}=\langle v,w\rangle$ is a free group of rank $2$ with
$v=u_{1}+u_{h}-u_{gh}$ and $w=u_{1}+u_{h}+u_{gh}$.

###### Proof.

As explained earlier, the elements $v$ and $w$ are indeed units. Moreover
their image under $\phi$ are equal to

$\phi(v)=\phi(1+u_{h}-u_{gh})=\left(\begin{array}[]{cc}1&0\\\
0&1\end{array}\right)+\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right)-\left(\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right)=\left(\begin{array}[]{cc}1&0\\\ 2&1\end{array}\right)$

and

$\phi(w)=\phi(1+u_{h}+u_{gh})=\left(\begin{array}[]{cc}1&0\\\
0&1\end{array}\right)+\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right)+\left(\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right)=\left(\begin{array}[]{cc}1&2\\\ 0&1\end{array}\right).$

Thus the images $\\{\phi(v),\phi(w)\\}$ generate a well-known free
group333333Sometimes this group is called the Sanov subgroup. of rank $2$
which is of index $24$ in $\operatorname{GL}_{2}({\mathbb{Z}})$ (see e.g. [16,
Example 2]). Therefore, $\\{v,w\\}$ generates a free group of rank $2$ in
$\mathcal{U}\left({\mathbb{Z}}^{\alpha}[C_{2}\times C_{2}]\right)$, say of
index $r$. Denote this subgroup by $H$ and denote by
$T\left(\widetilde{D}\right)=\pm\\{u_{1},u_{g},u_{h},u_{gh}\\}$ the trivial
units of $\tilde{D}$. It is easy to verify that
$T\left(\widetilde{D}\right)\cong D_{8}$ the dihedral group of $8$ elements
and moreover for any $x\neq y\in\phi^{-1}(T\left(\tilde{D}\right))$, the
cosets $xH\neq yH$. Consequently, $[U\left(\mathbb{Z}^{\alpha}[C_{2}\times
C_{2}]\right):H]\geq 8$. As can be seen with the matrices $\phi(u_{h})$ and
$1+E_{12}$, the subgroup
$\phi\left(\mathcal{U}(\mathbb{Z}^{\alpha}[C_{2}\times
C_{2}])\right)=\mathcal{U}(\widetilde{D})$ is not normal in
$GL_{2}(\mathbb{Z})$ and therefore it’s index is at least $3$. Since

$24=[GL_{2}(\mathbb{Z}):H]=[GL_{2}(\mathbb{Z}):\mathcal{U}(\widetilde{D})].[\phi\big{(}\mathcal{U}\left(\mathbb{Z}^{\alpha}[C_{2}\times
C_{2}]\right)\big{)}:H]\geq 3\cdot 8,$

we conclude that $[\mathcal{U}\left(\mathbb{Z}^{\alpha}[C_{2}\times
C_{2}]\right):H]=8$. A simple calculation shows that

$u_{h}vu_{h}^{-1}=w,\quad u_{g}vu_{g}^{-1}=v^{-1},\quad
u_{gh}vu_{gh}^{-1}=w^{-1}.$

In other words, $H$ is normal in
$\mathcal{U}\left(\mathbb{Z}^{\alpha}[C_{2}\times C_{2}]\right)$, and
therefore

$\mathcal{U}\left(\mathbb{Z}^{\alpha}[C_{2}\times
C_{2}]\right)=H\rtimes\phi^{-1}(T\left(\tilde{D}\right))\cong F_{2}\rtimes
D_{8}.$

∎

###  8.2. The general case

Now we consider the general case of $G=\langle g,h\rangle\times C_{2}^{n}$.
More precisely, we will follow the method outlined in Section 7 and therefore
describe the isomorphism type of the groups $U_{1},\ldots,U_{n}$. We use
freely the objects and notations introduced there.

######  Proposition 8.3.

The groups $U_{i}$ satisfy the following:

  * •

$[U_{i}:U_{i+1}]=8$

  * •

$U_{1}\cong F_{3}\times C_{2}$

  * •

If $i\geq 2$, then $U_{i}$ is a free group of rank $n_{i}=1+2.8^{i-1}$.

###### Proof.

By definition $U_{i}=\\{1+2^{i}u\mid u\in{\mathbb{Z}}^{\alpha}[\langle
g,h\rangle]\\}\cap\mathcal{U}({\mathbb{Z}}^{\alpha}[\langle g,h\rangle])$.
Consequently, by Proposition 8.1,

$U_{i}\cong\phi(U_{i})=\\{\left(\begin{array}[]{cc}1+2^{i}a_{11}&2^{i}a_{12}\\\
2^{i}a_{21}&1+2^{i}a_{22}\end{array}\right)\mid
a_{11}\equiv_{2}a_{22},\,a_{12}\equiv_{2}a_{21}\\}\cap\operatorname{GL}_{2}({\mathbb{Z}}).$

Consider now the congruence subgroup of level $2^{i}$ in
$\operatorname{GL}_{2}({\mathbb{Z}})$:

$\Gamma(2^{i})=\ker(\pi_{2^{i}}):\operatorname{GL}_{2}({\mathbb{Z}})\rightarrow\operatorname{GL}_{2}({\mathbb{Z}}/2^{i}{\mathbb{Z}}).$

More concretely,
$\Gamma(2^{i})=\big{(}1+2^{i}\operatorname{M}_{2}({\mathbb{Z}})\big{)}\cap\operatorname{GL}_{2}({\mathbb{Z}})$.
Note that $\Gamma(2^{i+1})\leq\phi(U_{i})\leq\Gamma(2^{i})$. The first
statement is a consequence of the following:

$\textit{{\it Claim: }}[\Gamma(2^{i}):\phi(U_{i})]=2\text{ for all }i\geq
1\textit{ and
}[\phi(U_{i}):\Gamma(2^{i+1})]=\left\\{\begin{array}[]{ll}4&\textit{ if }i\geq
2\\\ 8&\textit{ if }i=1\end{array}\right.$

To start, recall that (see [19, p 146])

(26)
$[\operatorname{SL}_{2}({\mathbb{Z}}):\ker\left(\operatorname{SL}_{2}({\mathbb{Z}})\rightarrow\operatorname{PSL}_{2}({\mathbb{Z}}/n{\mathbb{Z}})\right)]=\left\\{\begin{array}[]{ll}\frac{n^{3}}{2}\prod\limits_{p\,\mid\,n}(1-\frac{1}{p^{2}})&\textit{
if }n>2\\\ 6&\textit{ if }n=2\end{array}\right.$

Hence we need to reinterpret this formula to
$\operatorname{GL}_{2}({\mathbb{Z}})$ and the congruence subgroups
$\Gamma(2^{i})$. For this remark that
$[\operatorname{GL}_{2}({\mathbb{Z}}):\operatorname{SL}_{2}({\mathbb{Z}})]=2$
with coset representatives $1$ and $\left(\begin{array}[]{cc}1&0\\\
0&-1\end{array}\right).$ The same statement hold for the subgroup
$\ker\left(\operatorname{SL}_{2}({\mathbb{Z}})\rightarrow\operatorname{SL}_{2}({\mathbb{Z}}_{2})\right)$
of $\Gamma(2)$. However for higher powers of $2$ that matrix is the only
difference, hence yielding an index of $2$. Also note that, if $i\geq 2$, the
only difference between the kernels when mapping to
$\operatorname{PSL}_{2}({\mathbb{Z}}/2^{i}{\mathbb{Z}})$ or
$\operatorname{SL}_{2}({\mathbb{Z}}/2^{i}{\mathbb{Z}})$ is the central matrix
$-1$. Altogether (26) now yields that

$[\operatorname{GL}_{2}({\mathbb{Z}}):\Gamma(2^{i})]=\left\\{\begin{array}[]{ll}4\,.\,2^{3i}(1-\frac{1}{4})=3.2^{3i}&\textit{
if }i\geq 2\\\ 6&\textit{ if }i=1\end{array}\right.$

Consequently,
$[\Gamma(2^{i}):\Gamma(2^{n+1})]=\frac{[\operatorname{GL}_{2}({\mathbb{Z}}):\Gamma(2^{i+1})]}{[\operatorname{GL}_{2}({\mathbb{Z}}):\Gamma(2^{i})]}=8$
if $i\geq 2$ and $[\Gamma(2):\Gamma(4)]=16$. Further note that
$\left(\begin{array}[]{cc}1&2^{i}\\\
0&1\end{array}\right)\in\Gamma(2^{i})\setminus\phi(U_{i})$ for all $i$.
Therefore the claim would follow if we prove that $A.B\in\phi(U_{i})$ for any
pair of matrices $A=\left(\begin{array}[]{cc}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{array}\right),B=\left(\begin{array}[]{cc}b_{11}&b_{12}\\\
b_{21}&b_{22}\end{array}\right)\in\Gamma(2^{i})\setminus\phi(U_{i})$. A direct
computation shows that $AB\in\phi(U_{i})$ if and only if
$a_{11}+a_{22}\equiv_{2}b_{11}+b_{22}$ and
$a_{12}+a_{21}\equiv_{2}b_{12}+b_{21}$. However it is also easy to prove that
$A$ having determinant $1$ already implies that $a_{11}\equiv_{2}a_{22}$
(analogously for $B$). Hence if $A\in\Gamma(2^{i})\setminus\phi(U_{i})$ then
$a_{12}+a_{21}$ is odd and the same for $b_{12}+b_{21}$. Altogether the
conditions for $AB\in\phi(U_{i})$ is always satisfied which entails that
$[\Gamma(2^{i}):\phi(U_{i})]=2$. The claim343434A concrete matrix in
$\phi(U_{i})$ but not in $\Gamma(2^{i+1})$ is
$\left(\begin{array}[]{cc}1+2^{i}&2^{i}\\\ -2^{i}&1-2^{i}\end{array}\right)$.
now follows from

$[U_{i}:U_{i+1}]=[U_{i}:\Gamma(2^{i+1})]\,.\,[\Gamma(2^{i+1}):U_{i+1}]=[\Gamma(2^{i}):\Gamma(2^{i+1})]=\left\\{\begin{array}[]{ll}8&\textit{
if }i\geq 2\\\ 16&\textit{ if }i=1\end{array}\right.$

For the second statement, notice that by definition $U_{1}$ is a subgroup of
$\mathcal{U}({\mathbb{Z}}^{\alpha}[\langle g,h\rangle])$, however, elements of
$U_{1}$ are of the form $1+2u$ for $u\in{\mathbb{Z}}^{\alpha}[\langle
g,h\rangle]$ and hence the only trivial units of
${\mathbb{Z}}^{\alpha}[\langle g,h\rangle]$ contained in $U_{1}$ are $\pm 1$.
From here, with the notations of the proof of Proposition 8.2 it easily
follows that $U_{1}$ is a subgroup of $\langle
v,w\rangle\times\langle-1\rangle\cong F_{2}\times C_{2}$. Moreover, $U_{1}$ is
a proper subgroup of $\langle v,w\rangle\times\langle-1\rangle$ since
$v=1+u_{h}-u_{gh}$ (and also $w=1+u_{h}+u_{gh}$) cannot be written as $1+2u$.
We will prove that the index of $U_{1}$ in $\langle
v,w\rangle\times\langle-1\rangle$ is $2$. A simple calculation yields

$v^{2}=1+2u_{h}-2u_{gh}=1+2(u_{h}-u_{gh}),\quad
w^{2}=1+2u_{h}+2u_{gh}=1+2(u_{h}+u_{gh}).$

Additionally, $v^{-1}=1-u_{h}+u_{gh}$ and hence

$wv^{-1}=-1-2u_{g}+2u_{gh}=1+2(-1-u_{g}+u_{gh}).$

We conclude that $w^{2},v^{2},wv^{-1}\in U_{1}$. These elements generate a
subgroup of index $2$ in $\langle v,w\rangle$, and therefore these elements
are generators of a free group of rank $3$. Consequently, $U_{1}$ admits a
subgroup $\langle v^{2},w^{2},wv^{-1}\rangle\times\langle-1\rangle\cong
F_{3}\times C_{2}$ which is of index $2$ in $\langle
v,w\rangle\times\langle-1\rangle$ and therefore

$U_{1}=\langle v^{2},w^{2},wv^{-1}\rangle\times\langle-1\rangle\cong
F_{3}\times C_{2}.$

Finally we prove the third statement. By the first part $U_{2}$ is a subgroup
of index $8$ in $U_{1}$. Also it follows from [9, page 17] that $U_{2}$ is a
torsion-free subgroup of $\langle v,w\rangle$. In other words $U_{2}$ is
subgroup of index $8$ in $U_{1}\cap\langle v,w\rangle=\langle
v^{2},w^{2},wv^{-1}\rangle\cong F_{3}.$ Recall now the Nielsen-Schreier
formula which says that a subgroup of index $t$ in a free group of rank $e$ is
a free group of rank $1+t(e-1)$. Thus $U_{2}\cong F_{9}.$ Since $U_{i}\leq
U_{i-1}$ also $U_{i}$ is a free group whose rank is computed with a recursive
use of Nielsen-Schreier’s formula. ∎

The previous proposition combined with Theorem 7.3 yields a full description
of $\mathcal{U}({\mathbb{Z}}^{\alpha}[\langle g,h\rangle\times C_{2}^{n}]).$

##  9\. Description of $\mathcal{U}({\mathbb{Z}}[D_{8}\times C_{2}^{n}])$

Let $G$ be the group $C_{2}^{n+2}=\langle g,h,x_{1},\ldots,x_{n}\rangle$ as in
(22) and

$\Gamma\cong D_{8}\times C_{2}\times\ldots\times C_{2}=\langle
a\rangle\times\langle b\rangle\times\langle y_{1}\rangle\times\ldots\langle
y_{n}\rangle.$

where $D_{8}=\langle a,b\mid a^{4},b^{2},a^{b}=a^{-1}\rangle$. We consider the
epimorphism $\lambda:D_{8}\times C_{2}^{n}\rightarrow G$ determined by
$\lambda(a)=gh,\lambda(b)=g,\lambda(y_{i})=x_{i}$. With the associated
canonical choice of section $\mu:G\rightarrow D_{8}\times C_{2}^{n}$, one
directly checks that the associated $2$-cocycle
$\alpha(s,t)=\mu(s)\mu(t)\mu(st)^{-1}$ is exactly the one determined by (23).
In other words, the central extension $\Gamma$ associated with the
$[\alpha]\in H^{2}(G,C_{2})$ from Section 8 is isomorphic to $D_{8}\times
C_{2}^{n}$.

Taking now $\chi\in\operatorname{Hom}(\langle a^{2}\rangle,{\mathbb{Z}}^{*})$
with $\chi(a^{2})=-1$ we see that
$\Psi_{\chi}:{\mathbb{Z}}[\Gamma]\rightarrow{\mathbb{Z}}^{\operatorname{Tra}(\chi)}[G]$
is nothing else than the ring epimorphism.

(27) $\psi:{\mathbb{Z}}[D_{8}\times
C_{2}^{n}]\rightarrow{\mathbb{Z}}^{\alpha}[G]$

defined by

$a\mapsto u_{gh},\quad b\mapsto u_{g},\quad x_{i}\mapsto u_{y_{i}}.$

In this section we will pullback the description of
$\mathcal{U}({\mathbb{Z}}^{\alpha}[G])$, obtained in section Section 8, by
using the methods from Section 5.2 in order to obtain a description of
$\mathcal{U}({\mathbb{Z}}[D_{8}\times C_{2}^{n}])$. First we invoke Theorem
5.4 and Theorem 5.2 to give a precise description of the (co)kernel of the
induced morphism
$\widetilde{\psi}:\mathcal{U}({\mathbb{Z}}[\Gamma])\rightarrow\mathcal{U}({\mathbb{Z}}^{\alpha}[G])$.

######  Lemma 9.1.

With the notations as above we have that

  * (i)

$\ker(\widetilde{\psi})=\langle-a^{2}\rangle\cong C_{2},$

  * (ii)

$|\operatorname{Coker}(\widetilde{\psi})|=[\mathcal{U}({\mathbb{Z}}^{\alpha}[G]):{\rm
Im}(\widetilde{\psi})]=2^{n+1}.$

###### Proof.

The description of the kernel follows from Theorem 5.4 which says that the
kernel is finite and $\ker(\widetilde{\psi})=\\{1,-a^{2}\\}$, as desired.
Next, as remarked directly after Theorem 5.2, the assumptions for the second
part of that theorem are satisfied and hence

$|\operatorname{Coker}(\widetilde{\psi})|=|\operatorname{Coker}(\widetilde{\psi}_{ker})|.|\operatorname{Coker}(\widetilde{\psi}_{res})|$

where
$\widetilde{\psi}_{res}:\mathcal{U}({\mathbb{Z}}[D_{8}])\rightarrow\mathcal{U}({\mathbb{Z}}^{\rm
Res(\alpha)}[\langle g,h\rangle]):a\mapsto u_{gh},b\mapsto u_{g}$ is a
(co)restriction of $\widetilde{\psi}$.

We start by describing $\operatorname{Coker}(\widetilde{\psi}_{res})$. To this
end, recall that by Proposition 8.2

$\mathcal{U}({\mathbb{Z}}^{\rm Res(\alpha)}[C_{2}\times C_{2}])=\langle
v,w\rangle\rtimes D_{8}$

with $v=u_{1}+\frac{1}{2}(u_{1}-u_{g})u_{h}(u_{1}+u_{g})$ and
$w=u_{1}+\frac{1}{2}(1+u_{g})u_{h}(1-u_{g}).$

Claim: the product of any two generators of $\mathcal{U}({\mathbb{Z}}^{\rm
Res(\alpha)}[\langle g,h\rangle])$ belongs to ${\rm
Im}(\widetilde{\psi}_{res})$.

This claim would imply that the cokernel has size $2$. To start consider the
bicyclic units $b_{1}=1+(1-b)ab(1+b)$ and $b_{2}=1+(1+b)ab(1-b)$. Clearly
$\psi(b_{1})=v^{2}$ and $\psi(b_{2})=w^{2}$. Next a direct computation gives
that

$wv^{-1}=u_{1}-2(u_{g}+u_{hg})=u_{1}-(u_{1}+u_{h})u_{g}(u_{1}-u_{h}).$

Therefore if we consider the bicyclic unit $b_{3}=1-(1+ab)b(1-ab)$ we also
find $\psi(b_{3})=wv^{-1}$. Since the trivial units are clearly also attained,
the claim and therefore the second statement now follows from Proposition 6.9
(which asserts $v\in\operatorname{Coker}(\widetilde{\psi}_{res})$, hence non-
surjectivity).

It remains to compute $|\operatorname{Coker}(\widetilde{\psi}_{ker})|$. For
this we need to determine element in $\operatorname{Coker}(\widetilde{\psi})$
which are also in the kernel of the relative augementation with respect to
$Q=\langle u_{y_{1}},\ldots,u_{y_{n}}\rangle$. From Proposition 6.9 we know
that

$\langle
u_{1}+\frac{1}{2}(u_{1}-u_{g})(u_{1}-u_{y_{i})}u_{h}(u_{1}+u_{g}):1\leq i\leq
n\rangle\cong C_{2}^{n}$

is a subgroup of $\operatorname{Coker}(\widetilde{\psi}_{ker})$. By an
argument as above we can conclude that these are the only elements, hence
altogether yielding part (ii). ∎

Using proposition 8.3 and theorem 7.3 we now immediately obtain the
description we were looking for.

######  Proposition 9.2.

The following hold:

  1. (1)

$\mathcal{U}({\mathbb{Z}}[D_{8}])\cong F_{3}\rtimes(\pm D_{8})$ with $F_{3}$
generated by bicyclic units

  2. (2)

$U_{1}\cong F_{17}\times C_{2}$

  3. (3)

$U_{i}\cong F_{n_{i}}$ with $n_{i}=1+2.8^{i}$ for $i\geq 2$.

  4. (4)

$\pm D_{8}\times C_{2}^{n}$ has a torsion-free normal complement in
$\mathcal{U}({\mathbb{Z}}[D_{8}\times C_{2}^{n}])$

###### Proof.

To start note that $\psi(\pm D_{8})=\langle g,h\rangle\cong D_{8}$. Also it is
not hard to see that the elements $b_{1},b_{2},b_{3}$ from the proof of Lemma
9.1 are in fact generators for $\langle\text{Bic}(D_{8})\rangle$, the group
generated by the bicyclic units of $D_{8}$ (e.g. see [9, part (3) of proof
Example 1.5.4 ]). In particular all elements in
$\langle\text{Bic}(D_{8})\rangle$ are of the form $1+(1-a^{2})u$ with
$u\in{\mathbb{Z}}[D_{8}]$. Hence we already know that

$\langle\text{Bic}(D_{8}),a^{2}\rangle\cong\langle\text{Bic}(D_{8})\rangle\times\langle
a^{2}\rangle\cong\pm 1.\widetilde{\psi}\left(\langle
v^{2},w^{2},wv^{-1}\rangle\right)=U_{1}(\mathbb{Z}^{\alpha}C_{2}^{2})$

with $U_{i}(\mathbb{Z}^{\alpha}C_{2}^{2}):=\\{1+2^{i}u\mid
u\in{\mathbb{Z}}^{\alpha}[\langle
g,h\rangle]\\}\cap\mathcal{U}({\mathbb{Z}}^{\alpha}[\langle g,h\rangle])$. The
latter is isomorphic to $F_{3}\times C_{2}$ by Proposition 8.3. The first part
now follows.

Next, due to the form of the bicyclic units and the Reidemaster-Schreir method
we see that the above even imply that

$U_{i}({\mathbb{Z}}D_{8})=\\{1+2^{i}(1-a^{2})u\mid
u\in{\mathbb{Z}}[D_{8}]\\}\cap\mathcal{U}({\mathbb{Z}}[D_{8}])\cong
U_{i+1}({\mathbb{Z}}^{\alpha}C_{2}^{2})$

for $i\geq 2$ and for $i=1$ the group $U_{1}({\mathbb{Z}}D_{8})$ also contain
the trivial units $\\{\pm 1\\}$. Therefore the second and third statement now
follows from Proposition 8.3. ∎

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