# Torsion subgroups of Whitehead groups of graded division algebras

Huynh Viet Khanh  ,  Nguyen Duc Anh Khoa  and  Nguyen Dinh Anh Khoi
Department of Mathematics and Informatics, HCMC University of Education 280 An
Duong Vuong Str., Dist. 5, Ho Chi Minh City, Vietnam<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this paper, we study the torsion subgroup, which is denoted by ${\rm
TK}_{1}(E)$, of the Whitehead group $E^{*}/[E^{*},E^{*}]$ of a graded division
algebra $E$ which is finite dimensional over its center. In particular, we
provide formulas for ${\rm TK}_{1}(E)$ in some special cases such as when $E$
is unramified, totally ramified or semiramified. Using the formulas, we are
able to compute the group ${\rm TK}_{1}(D)$ of a valued division ring $D$,
which is finite dimensional over a henselian center.

###### Key words and phrases:

division ring; graded division ring; valuation theory; reduced K-theory
2020 Mathematics Subject Classification. 16W60; 19B99; 16K20

##  1\. Introduction and statement of results

Let $D$ be a division ring which is finite dimensional over its center $K$,
and ${\rm Nrd}_{D}:D\to K$ be the reduced norm map. The Whitehead group ${\rm
K}_{1}(D)$ and the reduced Whitehead group ${\rm SK}_{1}(D)$ of $D$ are
respectively defined to be the quotient groups

${\rm K}_{1}(D)=D^{*}/[D^{*},D^{*}]\;\;\;\text{ and }\;\;\;{\rm
SK}_{1}(D)=D^{(1)}/[D^{*},D^{*}],$

where $D^{(1)}$ is the set of elements of $D$ with the reduced norm $1$; that
is,

$D^{(1)}=\\{a\in D^{*}:{\rm Nrd}_{D}(a)=1\\}.$

According to Draxl (see [2, p.125]), the history of reduced ${\rm
K}_{1}$-theory begin with the paper [13] established in 1943 by Y. Matsushima
and T. Nakayama who proved that ${\rm SK}_{1}(D)$ is trivial in case $K$ is a
$p$-adic field. Later, in 1950, Wang Shianghaw [19] obtained a similar result
in case $K$ is an algebraic number field. Due to these results, it was widely
believed that ${\rm SK}_{1}(D)$ is trivial in general. The problem of finding
a suitable division algebra with non-trivial reduced Whitehead group remained
unsolved for a long time and was often referred to as the Tannaka-Artin
Problem. In 1975, the first example of a valued division algebra with non-
trivial reduced Whitehead group was given by Platonov in [14] and [15] with
proofs in [16]. Later, Ershow has provided a successful approach to computing
${\rm SK}_{1}$ for valued division algebras over henselian field ([3]), where
he recovered Platonov’s example and more examples as well.

Given a division algebra $D$ with a valuation, ones can construct its
associated graded division algebra ${\rm
gr}(D)=\bigoplus_{\gamma\in\Gamma_{D}}{\rm gr}(D)_{\gamma}$, where
$\Gamma_{D}$ is the value group of $D$ and each ${\rm gr}(D)_{\gamma}$ arises
from the filtration on $D$ induced by the valuation. As it was illustrated,
this graded ring is often easier to work with than $D$ itself, and not much is
lost in passage from $D$ to its corresponded graded division algebra ${\rm
gr}(D)$. Therefore, ones can find many works by well-known authors devoted to
a systematically study of this correspondence. For example, Boulagouaz ([1]),
Hwang, Tignol and Wadsworth ([7], [6], [18]) have all made significant
contributions to this area. Motivated by these, Hazrat and Wadsworth provided
in [5] a systematic study of the functor ${\rm SK}_{1}$ of a graded division
algebra whose graded group is totally ordered abelian. In this study, they
established analogies to Ershov’s results in graded setting, easily deducing
formulas for ${\rm SK}_{1}$ for graded division algebras in some specific
cases. The study also gave them a shortcut to compute the group ${\rm
SK}_{1}({\rm gr}(D))$ of the corresponded graded division algebra ${\rm
gr}(D)$ of a valued division algebra $D$. From this, they obtained a lot of
information about the group ${\rm SK}_{1}(D)$ of the valued division algebra
$D$ ([5, Corollary 4.10]).

Let $E=\bigoplus_{\gamma\in\Gamma}E_{\gamma}$ be a graded division algebra
(with $\Gamma_{E}$ a torsion-free abelian group) having finite dimension
$n^{2}$ over its center $T$, a graded field. (Note that the dimension of a
graded division algebra over its center is a square; please see the Subsection
2.1 for background knowledge about graded division algebras.) Because $E$ is
an Azumaya algebra ([7, Corollary 2.1]), the reduced norm map ${\rm
Nrd}_{E}:E\to T$ exists for it. Now, denote by $E^{*}$ the multiplicative
group of $E$ consisting of non-zero homogenous elements in $E$. Accordingly,
ones can define, analogously to the non-graded cases, the Whitehead group
${\rm K}_{1}(E)$ and reduced Whitehead group ${\rm SK}_{1}(E)$ for $E$ as

${\rm K}_{1}(E)=E^{*}/[E^{*},E^{*}]\;\;\;\text{ and }\;\;\;{\rm
SK}_{1}(E)=E^{(1)}/[E^{*},E^{*}],$

in which the group $E^{(1)}$ is defined as follows

$E^{(1)}=\\{a\in E^{*}:{\rm Nrd}_{E}(a)=1\\}.$

It was proved in [5] that ${\rm SK}_{1}(E)$ is a torsion group of bounded
degree $n$. Denote by ${\rm TK_{1}}(E)$ the torsion subgroup of the Whitehead
group ${\rm K}_{1}(E)$. As ${\rm SK}_{1}(E)$ is $n$-torsion, it follows that
${\rm SK}_{1}(E)$ is contained in ${\rm TK_{1}}(E)$. It can be observed that
the map ${\rm Nrd}_{E}$ then induces a homomorphism, which is again denoted by
${\rm Nrd}_{D}:{\rm TK}_{1}(D)\to\tau(T^{*})$, where $\tau(T^{*})$ is the
torsion group of the mulitplicative group $T^{*}$ of the graded field $T$.
Since the kernel of this homomorphism is ${\rm SK}_{1}(E)$, the following
sequence is exact:

(1.1) $\displaystyle 1\xrightarrow{\;\;\;\;\;\;}{\rm
SK}_{1}(E)\xrightarrow{\;\;\;\;\;\;\;}{\rm TK}_{1}(E)\xrightarrow{\;\;{\rm
Nrd}_{D}\;\;}{\rm Nrd}_{D}(\tau(T^{*}))\xrightarrow{\;\;\;\;\;\;\;}1.$

Because $T$ is an integral domain, it has the quotient field, say $q(T)$. It
follows $\tau(T^{*})$ is embedded in $\tau(q(T)^{*})$, which is the torsion
subgroup of a field, and is well-understood. (In particular, this implies that
$\tau(T^{*})$ is a locally cyclic group). Thus, the short exact sequence
suggests the close relationship between ${\rm TK}_{1}(E)$ and ${\rm
SK}_{1}(E)$. The group ${\rm SK}_{1}(E)$ was studied in depth by Hazrat and
Wadsworth in [5]. Motivated by this, in this paper, we provide a study of the
group ${\rm TK}_{1}(E)$ of a graded division algebra $E$. As it will turn out,
the group ${\rm TK}_{1}(E)$ enjoys the most important functorial properties of
the ${\rm SK}_{1}(E)$. Before stating the main results, we make the following
assumption in order to simplify the exposition.

STANDING ASSUMPTION. The division rings and graded division rings in this
paper are always assumed to be finite-dimensional over their centers.

Let $E=\bigoplus_{\gamma\in\Gamma}E_{\gamma}$ be a graded division algebra,
where $\Gamma$ is a torsion-free abelian group, with center $T$, a graded
field. Then $E_{0}$ is a division ring whose center $Z(E_{0})$ is a field
containing the field $T_{0}$. Let $E^{*}$ and $E_{0}^{*}$ be the
multiplicative groups of $E$ and $E_{0}$ respectively. It will be proved in
Section 3 that the commutator group $[E^{*},E^{*}]$ subgroup is contained in
$E_{0}^{*}$. Also, for an extension field $K/F$, we denote by $N_{K/F}:K\to F$
the norm map of $K$ into $F$. In what follows, we briefly describe our
principal results. Regarding to the group ${\rm TK}_{1}(E)$ of a graded
division algebra $E$, we have the following theorem:

######  Theorem 1.1.

Let $E$ be a graded division ring with center $T$. So, $Z(E_{0})$ is Galois
over $T_{0}$; let $\mathcal{G}={\rm Gal}(Z(E_{0})/T_{0})$. Let
$\widetilde{N}=N_{Z(E_{0})/T_{0}}\circ{\rm Nrd}_{E_{0}}:E_{0}^{*}\to
T_{0}^{*}$. Let $G$ be a subgroup of $E^{*}$ such that ${\rm
TK}_{1}(E)=G/[E^{*},E^{*}]$. Then, $G$ is a subgroup of $E_{0}^{*}$ for which
$\tau(E_{0}^{*}/[E^{*},E^{*}])=G/[E^{*},E^{*}]$, and the row and column of the
following diagram are exact:

${1}$${{\rm
ker}\widetilde{N}/[E_{0}^{*},E^{*}]}$${\Gamma_{E}/\Gamma_{T}\land\Gamma_{E}/\Gamma_{T}}$${G/[E_{0}^{*},E^{*}]}$${{\rm
TK}_{1}(E)}$${1}$${\tau(T_{0}^{*})\cap\widetilde{N}(E_{0}^{*})}$${1}$$\iota$$\alpha$$\bar{N}$$\bar{\varphi}$

Diagram 1

The map $\alpha$ is given by the following rule: For
$\gamma,\delta\in\Gamma_{E}$, take any non-zero $x_{\gamma}$ and
$x_{\delta}\in E_{\delta}$. Then,
$\alpha((\gamma+\Gamma_{T})\land(\delta+\Gamma_{T}))=[x_{\gamma},x_{\delta}]$
mod $[E_{0}^{*},E^{*}]$.

As a consequence of this theorem, as Hazrat and Wadsworth have done for ${\rm
SK}_{1}(E)$ in [5, Corollary 3.6], we have the following corollary which
provides formulas for calculating ${\rm TK}_{1}(E)$ in some certain cases:

######  Corollary 1.2.

Let $E$ be a graded division ring with center $T$. Then, the following
assertions hold:

  1. (i)

If $E$ is unramified, then ${\rm TK}_{1}(E)\cong{\rm TK}_{1}(E_{0})$.

  2. (ii)

If $E$ is totally ramified, then ${\rm
TK}_{1}(E)\cong\tau(T_{0}^{*})/\mu_{e}(T_{0})$, where $e={\rm
exp}(\Gamma_{E}/\Gamma_{T})$ the exponent of the group
$\Gamma_{E}/\Gamma_{T}$.

  3. (iii)

If $E$ is semiramified, then $E_{0}$ is a field which is Galois over $T_{0}$,
and for $\mathcal{G}={\rm Gal}(E_{0}/T_{0})\cong\Gamma_{E}/\Gamma_{T}$, the
following sequence is exact:

$\mathcal{G}\land\mathcal{G}\longrightarrow
G/I_{\mathcal{G}}(E_{0}^{*})\longrightarrow{\rm TK}_{1}(E)\longrightarrow 1,$

where $G$ is a subgroup of $E_{0}^{*}$ such that
$\tau(E_{0}^{*}/[E^{*},E^{*}])=G/[E^{*},E^{*}]$, and
$I_{\mathcal{G}}(E_{0}^{*})=\left\langle c\sigma(c)^{-1}:c\in
E_{0}^{*},\sigma\in\mathcal{G}\right\rangle$.

  4. (iv)

If $E$ has maximal graded subfields $L$ and $K$ which are respectively
unramified and totally ramified over $T$, then $E$ is semiramified and
$\widehat{H}^{-1}(\mathcal{G},E_{0}^{*})$, the $-1$-Tate cohomology group of
$\mathcal{G}$ with respect to $E_{0}^{*}$, is isomorphic to ${\rm SK}_{1}(E)$;
and the following sequence is exact:

$1\longrightarrow\widehat{H}^{-1}(\mathcal{G},E_{0}^{*})\longrightarrow{\rm
TK}_{1}(E)\longrightarrow\tau(T_{0}^{*})\cap
N_{E_{0}/T_{0}}(E_{0}^{*})\longrightarrow 1.$

As a bridge to the ungraded case, we also establish in the next theorem a
criteria for the group ${\rm TK}_{1}$ of a strongly tame valued division
algebra with a henselian field to be isomorphic to ${\rm TK}_{1}$ of its
associated graded division algebra:

######  Theorem 1.3.

Let $F$ be a field with henselian valuation $v$, and let $D$ be a strongly
tame division algebra with center $F$. If ${\rm TK}_{1}(D)$ contains no
element of order ${\rm char}(\overline{F})$, where $\overline{F}$ is the
residue field of $v$ on $F$, then

${\rm TK}_{1}(D)\cong{\rm TK}_{1}({\rm gr}(D)).$

Using Theorems 1.1 and 1.3, we are able to give the ways calculate ${\rm
TK}_{1}({\rm gr}(D))$ of the associated graded division algebra ${\rm gr}(D)$
of a valued division algebra $D$ in some specific cases:

######  Corollary 1.4.

Let $F$ be a field with henselian valuation, and $D$ be a strongly tame
division algebra with center $F$. If ${\rm TK}_{1}(D)$ contains no element of
order ${\rm char}(\overline{F})$, then the following assertions hold:

  1. (i)

If $D$ is unramified, then ${\rm TK}_{1}(D)\cong{\rm TK}_{1}(\overline{D})$.

  2. (ii)

If $D$ is totally ramified, then ${\rm
TK}_{1}(D)\cong\tau(\overline{F}^{*})/\mu_{e}(\overline{F})$, where $e={\rm
exp}(\Gamma_{D}/\Gamma_{F})$.

  3. (iii)

If $D$ is semiramified, then $\overline{D}$ is a field which is Galois over
$\overline{F}$, and for $\mathcal{G}={\rm
Gal}(\overline{D}/\overline{F})\cong\Gamma_{D}/\Gamma_{F}$, the following
sequence is exact:

(1.2) $\mathcal{G}\land\mathcal{G}\longrightarrow
G/I_{\mathcal{G}}(\overline{D}^{*})\longrightarrow{\rm
TK}_{1}(D)\longrightarrow 1,$

where $G$ is the subgroup of $\overline{D}^{*}$ such that
$\tau(\overline{D}^{*}/[D^{*},D^{*}])=G/[D^{*},D^{*}]$ and
$I_{\mathcal{G}}(\overline{D}^{*})=\left\langle
c\sigma(c)^{-1}:c\in\overline{D}^{*},\sigma\in\mathcal{G}\right\rangle$.

  4. (iv)

If $D$ is nicely semiramified, then $\mathcal{G}={\rm
Gal}(\overline{D}/\overline{F})\cong\Gamma_{D}/\Gamma_{F}$, and
$\widehat{H}^{-1}(\mathcal{G},\overline{D}^{*})$, the $-1$-Tate cohomology
group of $\mathcal{G}$ with respect to $\overline{D}^{*}$, is isomorphic to
${\rm SK}_{1}(D)$; and the following sequence is exact:

$1\longrightarrow\widehat{H}^{-1}(\mathcal{G},\overline{D}^{*})\longrightarrow{\rm
TK}_{1}(D)\longrightarrow\tau(\overline{F}^{*})\cap
N_{\overline{D}/\overline{F}}(\overline{D}^{*})\longrightarrow 1.$

The group ${\rm TK}_{1}(D)$ for a valued division algebra $D$ with a henselian
center was also studied by Motiee in [12] (in which Motiee denoted such a
group by $T(D)$). In this paper, Motiee has completely determined all torsion
abelian groups can occur as the torsion group of the Whitehead group of a
division algebra of prime index ([12, Theorem 2]). This result can be
considered as a generalization of May’s Theorem about the torsion group of the
multiplicative group of a field ([10, Theorem 3]). The key idea that was used
to prove [12, Theorem 2] is that for a given torsion locally cyclic group $A$,
Motiee constructed a valued division algebra $D$ of prime index over a
henselian field such that ${\rm TK}_{1}(D)\cong A$, using [12, Theorem 10]
which is a connection between ${\rm TK}_{1}$ of the valued division ring $D$
and that of the residue division ring $\overline{D}$. In the current paper, we
give another way to obtain this result as well as extend it using the study in
graded setting.

The paper is organized as follows. In Section 2, we briefly recall some
background on the theory of graded division algebras indexed by a totally
ordered abelian group as well as the theory of valued division ring. For a
full reference for these topics, we refer the reader to the excellent book by
Tignol and Wadsworth [18]. Section 3 is devoted to a study of the group ${\rm
TK}_{1}$ of a graded division algebra $E$ which is finite dimensional over its
center. The main purpose of this section is to give a proof of Theorem 1.1,
which is an analogous version of [5, Theorem 3.4] by Hazrat and Wadsworth,
yielding formulas for ${\rm TK}_{1}$ of unramified, semiramified, and totally
ramified graded division algebras given in Corollary 1.4. In Section 4, we
provide the proofs of Theorem 1.3 and of Corollary 1.4. Also, in each section,
we present some examples of calculating ${\rm TK}_{1}$ of graded division
algebras and valued division algebras.

##  2\. Preliminaries

This section is devoted to recalling basic concepts from the theory of graded
division algebras indexed by a totally ordered abelian group that will be used
throughout the current paper. For an extensive study of such algebras, we
refer the readers to an excellent book by Tignol and Wadsworth [18].

###  2.1. Graded division algebras

A unital ring $R$ is said to be a graded ring if
$R=\bigoplus_{\gamma\in\Gamma}R_{\gamma}$, where $\Gamma$ is an abelian group,
and each $R_{\gamma}$ is a subgroup of $(R,+)$ such that
$R_{\gamma}R_{\delta}\subseteq R_{\gamma+\delta}$ for all
$\gamma,\delta\in\Gamma$. Put

$\displaystyle\Gamma_{R}=\\{\gamma\in\Gamma:R_{\gamma}\neq 0\\},\text{ the
grade set of $R$};$ $\displaystyle
R^{h}=\bigcup_{\gamma\in\Gamma_{R}}R_{\gamma},\text{ the set of homogeneous
elements of $R$}.$

For a homogeneous elements of degree $\gamma$, that is, an $r\in
R_{\gamma}\backslash\\{0\\}$, we write $\deg(r)=\gamma$. Note that $R_{0}$ is
a subring of $R$, and for each $\gamma\in\Gamma_{R}$, the additive group
$R_{\gamma}$ is a left and right $R_{0}$ module. A subring $S$ of $R$ is a
graded subring if $S=\bigoplus_{\gamma\in\Gamma_{R}}(S\cap R_{\gamma})$. For
example, the center of $R$, denoted by $Z(R)$, is a graded subring of $R$.

Let $R$ be a graded ring. Then, a graded left $R$-module $M$ is a left
$R$-module with a grading $M=\bigoplus_{\gamma\in\Gamma}M_{\gamma}$, where
each $M_{\gamma}$ is an abelian group, such that
$R_{\gamma}M_{\delta}\subseteq M_{\gamma+\delta}$ for all
$\gamma,\delta\in\Gamma$. As an analogy to $\Gamma_{R}$ and $R^{h}$, the set
graded set $\Gamma_{M}$ and the set of homogenous element $M^{h}$ of $M$ are
defined similarly. We say that $M$ is a graded free $R$-module if it has a
base as a free $R$-module consisting of homogenous elements.

A graded ring $E=\bigoplus_{\gamma\in\Gamma}E_{\gamma}$ is called a graded
division ring if $\Gamma$ is a torsion-free abelian group and every non-zero
homogenous element of $E$ has a multiplicative inverse. Recall that the grade
set $\Gamma_{E}$ is then actually a group. Note also that $E_{0}$ is a
division ring, and each $E_{\gamma}$ is a 1-dimensional left and right
$E_{0}$-vector space for every $\gamma\in\Gamma_{E}$. The requirement that
$\Gamma$ be torsion-free is made because in this paper we regularly consider
graded division algebra arising from valued division rings, in which all the
grade groups appearing there are torsion-free. It can be proved (see e.g. [18,
Proposition 2.3, p.35] that $E$ has no zero divisors and that the
multiplicative group of units $E^{*}$ of $E$ coincides with
$E^{h}\backslash\\{0\\}$. Therefore, the degree map

${\rm deg}:E^{*}\to\Gamma_{E}$

is a group epimorphism with kernel $E_{0}^{*}$.
It can be proved that every graded module $M$ over a graded division ring $E$
is graded free, and every two homogenous bases have the same cardinality.
Thus, it can be said that $M$ is a graded vector space over $E$, an so we can
write ${\rm dim}_{E}(M)$ for the rank of $M$ as a graded free $E$-module. Let
$S\subseteq E$ be a graded subring which is a graded division ring. Then, we
can view $E$ as a graded left $S$-vector space, and we write $[E:S]$ for
$\dim_{S}(E)$. Then, we have the “Fundamental Equality”,

$[E:S]=[E_{0}:S_{0}]|\Gamma_{E}:\Gamma_{S}|,$

where $[E_{0}:S_{0}]$ is the dimension of $E_{0}$ as a vector space over
$S_{0}$ and $|\Gamma_{E}:\Gamma_{S}|$ denotes the index in the group
$\Gamma_{E}$ of its subgroup $\Gamma_{S}$ (see [18, Section 2.1.2]).

A graded field $T$ is a commutative graded division ring. Such a $T$ is an
integral domain, so it has a quotient field, which is denoted by $q(T)$. If
$E$ is a graded division algebra, then its center $Z(E)$ is clearly a graded
subfield of $E$. A graded division algebra $E$ with center $T$ is said to be
unramified if $\Gamma_{E}=\Gamma_{T}$. In this case, it follows from above
equality that $[E:S]=[E_{0}:T_{0}]$. On the other hand, $E$ is said to be
totally ramified if $E_{0}=T_{0}$. In a case in the middle, $E$ is said to be
semiramified if $E_{0}$ is a field and
$[E_{0}:T_{0}]=[\Gamma_{E}:\Gamma_{T}]={\rm ind}(E)$, where the index of $E$
is defined by ${\rm ind}(E)^{2}=[E:T]$.

###  2.2. Some group theory and the group ${\rm TK}_{1}$ of graded division
algebras

Before giving the definition of the group ${\rm TK}_{1}$ of a graded division
algebra, it is convenient to recall some basic concepts from group theory. For
a group $G$ and $a,b\in G$, we set $[a,b]=aba^{-1}b^{-1}$, the multiplicative
commutator of $a$ and $b$. If $H$ and $K$ are subgroup of $G$, then we let
$[H,K]$ denotes the subgroup of $G$ generated by all commutators of form
$[a,b]$ where $a\in H$ and $b\in K$; that is,

$[H,K]=\left\langle[h,k]:h\in H,k\in K\right\rangle.$

In the case when $K=H=G$, we have $[G,G]=G^{\prime}$ which is the commutator
subgroup of $G$.

Every group $G$ has a unique maximal torsion normal subgroup, denoted by
$\tau(G)$. The next lemma is made to assure the existence of such a subgroup
$\tau(G)$; and, in particular, determines $\tau(G)$ in the case when $G$ is an
abelian group. Although the proof is elementary, we still include it for
readers’ convenience.

######  Lemma 2.1.

Let $G$ be an arbitrary group. Then the unique maximal torsion normal subgroup
$\tau(G)$ of $G$ always exists, containing all torsion normal subgroup of $G$.
Further, $\tau(G)$ is a characteristic subgroup of $G$. If $G$ is assumed
further to be abelian, then $\tau(G)$ consists of all torsion element of $G$.

###### Proof.

Let $\mathcal{B}$ be a family of subgroups which are torsion and normal in
$G$; that is,

$\mathcal{B}=\left\\{X\leq G\;|\;X\text{ is torsion and }X\unlhd G\right\\}.$

Then $(\mathcal{B},\subseteq)$ is a partially ordered set which is non-empty
as the subgroup $\\{1\\}\in\mathcal{B}$. Let $T$ be a chain of subgroups in
$\mathcal{B}$. Put $B=\cup_{X\in T}X$. Then, it is straightforward to check
that $B$ is a torsion subgroup which is normal in $G$, and so
$B\in\mathcal{B}$. As $X\leq B$ for all $X\in T$, we get that $B$ is an upper
bound for $T$ in $\mathcal{B}$. By Zorn’s Lemma, $\mathcal{B}$ has maximal
element, say $\tau(G)$, which is a maximal torsion normal subgroup of $G$. To
see the uniqueness of $\tau(G)$, take $H$ to be a maximal torsion normal
subgroup of $G$. Our aim is to prove that $H=\tau(G)$. Indeed, it is clear
that $\tau(G)\leq H\tau(G)$. Because both $H$ and $\tau(G)$ are torsion and
$H$ normalizes $\tau(G)$, the group $H\tau(G)$ is torsion too. Moreover, the
normality of $H$ and $\tau(G)$ in $G$ implies that $H\tau(G)\unlhd G$. It
follows that $H\tau(G)$ belongs to $\mathcal{B}$, yielding that
$H\tau(G)=\tau(G)$ due to the maximality of $\tau(G)$ in $\mathcal{B}$. This
implies that $\tau(G)=H$, proving the uniqueness of $\tau(G)$.

Next, we prove that $\tau(G)$ is characteristic in $G$. For this purpose, pick
$\varphi\in{\rm Aut}(\tau(G))$. Then $\varphi(\tau(G))$ is certainly a torsion
normal subgroup of $G$. The uniqueness of $\tau(G)$ implies that
$\varphi(\tau(G))=\tau(G)$, proving that $\tau(G)$ is a characteristic
subgroup of $G$.

For the final conclusion, assume that $G$ is abelian, and denote $T(G)$ the
set of all torsion element of $G$. Then, it is clear that $\tau(G)\subseteq
T(G)$. Moreover, as $G$ is abelian, it is easy to see that $T(G)$ is indeed a
subgroup which is normal in $G$. It follows that $T(G)=\tau(G)$ by the
uniqueness of $\tau(G)$. ∎

In view of Lemma 2.1, every abelian group $G$ has a unique subgroup $\tau(G)$
consisting of all torsion elements of $G$, which is called the torsion
subgroup of $G$. If $\tau(G)=G$, then $G$ is a torsion group; such a group can
be expressed in just one way as a direct product $G=\prod G_{p}$, where $p$
runs over all primes and, $G_{p}$, the $p$-primary component of $G$, consists
of all elements of $p$-power order. If the group $G$ is locally ciyclic, i.e.
all its finitely generated subgroups are cyclic, then its $p$-primary
component $G_{p}$ is either cyclic of order $p^{r}$, where $r$ is a non-
negative integer, or of type $Z_{p^{\infty}}$ (the group of all $p^{r}$th
roots of 1 in $\mathbb{C}$, for all $r$). It is well-known that the torsion
subgroup $\tau(F^{*})$ of the multiplicative group $F^{*}$ of a field $F$ is
alway locally cyclic.

Now, we return our attention to giving the definition of the group ${\rm
TK}_{1}$ of a graded division algebra. Let
$E=\bigoplus_{\gamma\in\Gamma}E_{\gamma}$ be a graded division ring with
center $T$, where $\Gamma$ is a torsion-free abelian group. The group ${\rm
TK}_{1}(E)$ is defined to be torsion subgroup $\tau({\rm K}_{1}(E))$ of the
Whitehead group ${\rm K}_{1}(E)$. Then, because the Whitehead group ${\rm
K}_{1}(E)$ is an abelian group, as a consequence of Lemma 2.1, the group ${\rm
TK}_{1}(E)$ consists of all elements of finite order of ${\rm K}_{1}(E)$.
Moreover, in view of the exact sequence (1.1), the factor group ${\rm
TK}_{1}(E)/{\rm SK}_{1}(E)$ is isomorphic to a subgroup of $\tau(T^{*})$ which
is a locally cyclic group.

###  2.3. Valued division rings and the associated graded division rings

Let $D$ be a division algebra which is finite-dimensional over its center $F$,
with a valuation $v:D^{*}\to\Gamma$. Then, by definition of a valuation, we
have: For all $a,b\in D^{*}$, we have

  1. (1)

$v(ab)=v(a)+v(b)$;

  2. (2)

$v(a+b)\geq{\rm min}\\{v(a),v(b)\\}\;\;(b\neq-a)$.

For the valuation $v$, let

$\displaystyle V_{D}=\\{a\in D^{*}:v(a)\geq 0\\}\cup\\{0\\},\text{ the value
ring of }v;$ $\displaystyle M_{D}=\\{a\in D^{*}:v(a)>0\\}\cup\\{0\\},\text{
the unique maximal ideal of }V_{D};$
$\displaystyle\overline{D}=V_{D}/M_{D},\text{ the residue division ring of
}v\text{ on }D;\text{ and }$ $\displaystyle\Gamma_{D}={\rm im}(v),\text{ the
value group of the valuation}.$ $\displaystyle\text{Set }U_{D}=\\{d\in
D^{*}:v(d)=0\\}\text{ so that }U_{D}=V_{D}^{*}.$

The graded division ring ${\rm gr}(D)$ associates to the valued division
algebra $D$ is defined as follows: For each $\gamma\in\Gamma_{D}$, put

$\displaystyle D^{\geq\gamma}=\\{d\in
D^{*}:v(d)\geq\gamma\\}\cup\\{0\\},\text{ an additive subgroup of }D;$
$\displaystyle D^{>\gamma}=\\{d\in D^{*}:v(d)>\gamma\\}\cup\\{0\\},\text{ an
additive subgroup of }D^{\geq};\text{ and }$ $\displaystyle{\rm
gr}(D)_{\gamma}=D^{\geq\gamma}/D^{>\gamma}.$

Define

${\rm gr}(D)=\bigoplus_{\gamma\in\Gamma_{D}}{\rm gr}(D)_{\gamma}.$

Then, the condition
$D^{>\gamma}D^{\geq\delta}+D^{\geq\gamma}D^{>\delta}\subseteq
D^{>(\gamma+\delta)}$, for all $\gamma,\delta\in\Gamma_{D}$, assures that the
multiplication on ${\rm gr}(D)$ induced by multiplication on $D$ is well-
defined. It can be checked that ${\rm gr}(D)$ is indeed a graded division
ring, called the associated graded division ring of $D$. Also, it is clear
that ${\rm gr}(D)_{0}=\overline{D}$ and $\Gamma_{{\rm gr}(D)}=\Gamma_{D}$. For
each $d\in D^{*}$, we write $\tilde{d}$ for the image $d+D^{>v(d)}$ of $d$ in
${\rm gr}(D)_{v(d)}$. Then, the map

$D^{*}\to{\rm gr}(D)^{*}\;\;\;\text{ given by }\;\;\;d\mapsto\tilde{d}$

is a group epimorphism with kernel $1+M_{D}$. It was shown in [5, Corollary
4.4] that the reduced norm maps for $D$ and ${\rm gr}(D)$ are related by

(2.1) $\displaystyle{\rm Nrd}_{{\rm gr}(D)}(\widetilde{a})=\widetilde{{\rm
Nrd}_{D}(a)},\text{ for all }a\in D^{*}.$

The restriction $v|_{F}$ of the valuation $v$ on $D$ to its center $F$ is a
valuation on $F$, which induces a corresponding graded field ${\rm gr}(F)$. It
can be checked that ${\rm gr}(D)$ is a graded ${\rm gr}(F)$-algebra which
satisfies:

(2.2) $\displaystyle[{\rm gr}(D):{\rm
gr}(F)]=[\overline{D}:\overline{F}]|\Gamma_{D}/\Gamma_{F}|\leq[D:F]<\infty.$

Let $F$ be a field with a henselian valuation $v$. Recall that a finite degree
field extension $L$ of $F$ is defined to be unramified over $F$, with respect
to the unique extension of $v$ to $L$, if $[\overline{L}:\overline{F}]=[L:F]$
and $\overline{L}$ is separable over $\overline{F}$ (hence $L$ is separable
over $F$). Also, a field extension $L$ of $F$ of finite degree $n$ is said to
be tamely ramified or tame over $F$ if $\overline{L}$ is a separable field
extension of $\overline{F}$ and ${\rm char}(\overline{F})\nmid
n/[\overline{L}:\overline{F}]$. Such an $L$ satisfies the equality
$[L:F]=[\overline{L}:\overline{F}]|\Gamma_{L}/\Gamma_{F}|=[{\rm gr}(L):{\rm
gr}(F)]$.

Let $D$ be a division ring with center $F$ and $[D:F]<\infty$, then the
valuation $v$ on $F$ extends uniquely to a valuation on $D$. With respect to
this valuation, we say that $D$ is tamely ramified or tame if
$Z(\overline{D})$ is separable over $\overline{F}$ and ${\rm
char}(\overline{F})\nmid{\rm ind}(D)/({\rm
ind}(\overline{D})[Z(\overline{D}):\overline{F}])$. Recall from [9,
Proposition 1.7] that whenever the field extension
$Z(\overline{D})/\overline{F}$ is separable, it is abelian Galois. Also, it
follows from [7, Propostion 4.3] that $D$ is tame if and only if $[{\rm
gr}(D):{\rm gr}(F)]=[D:F]$ and $Z({\rm gr}(D))={\rm gr}(F)$, if any only if
$D$ is split by a maximal tamely ramified extension of $F$, if and only if
${\rm char}(\overline{F})=0$ or ${\rm char}(\overline{F})=p\neq 0$ and the
$p$-primary component of $D$ is inertially split, i.e., split by a maximal
unramified extension of $F$. We say that $D$ is strongly tame if ${\rm
char}(\overline{F})\nmid{\rm ind}(D)$. It is clear that strong tameness
implies tameness. Note that the term “tameness” used in [12] is exactly the
term “strong tameness” here.

Let $D$ be a division ring with a henselian center $F$. Then the valuation on
$F$ extends uniquely to a valuation on $D$. With respect to this valuation, we
say that $D$ is unramified or innertial over $F$ if
$[\overline{D}:\overline{F}]=[D:F]$ and $Z(\overline{D})$ is separable over
$\overline{F}$. We say that $D$ is semiramified if $\overline{D}$ is a field
and $[\overline{D}:\overline{F}]=|\Gamma_{D}:\Gamma_{F}|={\rm ind}(D)$. At the
other extreme, $D$ is said to be totally ramified over $F$ if
$|\Gamma_{D}:\Gamma_{F}|=[D:F]$, or equivalently, $\overline{D}=\overline{F}$
and $D$ is defectless over $F$. Recall from [9, p.128] that a division algebra
is called nicely semiramified if it has a maximal subfield inertial over $F$
and another maximal subfield totally ramified over $F$.

##  3\. Torsion subgroup of graded division algebras

Now, for further using, let us make some observations about some certain
subgroups of the multiplicative group $E^{*}$ of the graded division ring $E$.
Recall first that $E^{*}$ coincides with the set of all non-zero homogenous
elements of $E$. The next remark follows directly from [5, Proposition 3.2]
and is used regularly throughout the present paper.

######  Remark 1.

Let $E$ be a graded division ring with center $T$, and let $n={\rm ind}(E)$.

  1. (i)

If $K$ is any graded subfield of $E$ containing $T$ and $a\in K$, then

${\rm Nrd}_{E}(a)=N_{K/T}(a)^{n/[K:T]}.$

  2. (ii)

For $\gamma\in\Gamma_{E}$, if $a\in E_{\gamma}$ then ${\rm Nrd}_{E}(a)\in
E_{n\gamma}$.

  3. (iii)

Set $\delta={\rm ind}(E)/({\rm ind}(E_{0})[Z(E_{0}):T_{0}])$. If $a\in E_{0}$,
then

${\rm Nrd}_{E}(a)=N_{Z(E_{0})/T_{0}}{\rm Nrd}_{E_{0}}(a)^{\delta}\in T_{0}.$

Remark 1 helps us to prove the following useful lemma:

######  Lemma 3.1.

Let $E$ be a graded division ring and $a\in E^{*}$. If $a$ is radical over
$E^{(1)}$, then $a\in E^{0}$. In particular, we have
$[E_{0}^{*},E^{*}]\subseteq[E^{*},E^{*}]\subseteq E^{(1)}\subseteq E_{0}^{*}$
and $\tau(E^{*})\subseteq E_{0}^{*}.$

###### Proof.

As we have mentioned before, $E^{*}$ only consists of homogeneous elements of
$E$. Since $a$ is a unit, it follows that $a\in E_{\gamma}$ for some
$\gamma\in\Gamma_{E}$. Let $k\geq 1$ be an integer such that $a^{k}=b\in
E^{(1)}$. We have

$1={\rm Nrd}_{E}(b)={\rm Nrd}_{E}(a^{k})={\rm Nrd}_{E}(a)^{k}.$

Since ${\rm Nrd}_{E}(a)\in E_{n\gamma}$ (Remark 1(ii)), we have $1={\rm
Nrd}_{E}(a)^{k}\in E_{nk\gamma}$. This implies that $\gamma=0$, and so $a\in
E_{0}^{*}$. The remaining assertion is obvious. ∎

######  Corollary 3.2.

Let $E$ be a graded division ring. Then $E_{0}^{*}$ is a normal subgroup of
$E^{*}$.

###### Proof.

The assertion follows from the fact that $[E^{*},E^{*}]\subseteq E_{0}^{*}$. ∎

######  Lemma 3.3.

Let $E$ be a graded division ring. If $G$ and $G_{0}$ are subgroups of $E^{*}$
and $E_{0}^{*}$ such that $\tau(E^{*}/[E^{*},E^{*}])=G/[E^{*},E^{*}]$ and
$\tau(E_{0}^{*}/[E^{*},E^{*}])=G_{0}/[E^{*},E^{*}]$ respectively, then
$G=G_{0}$.

###### Proof.

For any $x\in G$, there exists $k\in\mathbb{N}$ such that $x^{k}=1$. Then, for
$\overline{x}=x\left[{E^{*},E^{*}}\right]$, we have

$\overline{x}^{k}=x^{k}\left[{E^{*},E^{*}}\right]=\left[{E^{*},E^{*}}\right]\Longrightarrow
x^{k}\in\left[{E^{*},E^{*}}\right].$

Thus, $G$ is radical over $[E^{*},E^{*}]\subseteq E^{(1)}$; and so, by Lemma
3.1, we obtain that $G\subseteq E_{0}^{*}$. It follows that $G/[E^{*},E^{*}]$
is a (torsion) subgroup of $E_{0}^{*}/[E^{*},E^{*}]$. As
$G/[E^{*},E^{*}]\unlhd E_{0}^{*}/[E^{*},E^{*}]$, we get that
$G/[E^{*},E^{*}]\subseteq G_{0}/[E^{*},E^{*}]$, which implies that $G\subseteq
G_{0}$. For opposite inclusion, we note first that
$E_{0}^{*}/[E^{*},E^{*}]\unlhd E^{*}/[E^{*},E^{*}]$ as $E_{0}^{*}\unlhd
E^{*}$. Now, because $G_{0}/[E^{*},E^{*}]$ is characteristic in
$E^{0}/[E^{*},E^{*}]$, it follows that $G_{0}/[E^{*},E^{*}]\unlhd
E^{*}/[E^{*},E^{*}]$. Thus, $G_{0}/[E^{*},E^{*}]$ is a torsion subgroup which
is normal in $E^{*}/[E^{*},E^{*}]$. It follows that
$G_{0}/[E^{*},E^{*}]\subseteq G/[E^{*},E^{*}]$, and so $G_{0}\subseteq G$. The
Lemma is proved. ∎

######  Remark 2.

Let $E$ be a graded division ring with center $T$. Then, the degree map
$E^{*}\to\Gamma_{E}$ induces a surjective map $E^{*}\to\Gamma_{E}/\Gamma_{T}$
given by $a\mapsto{\rm deg}(a)+\Gamma_{T}$ with kernel $T^{*}E_{0}^{*}$.
Therefore, we have the following short exact sequence:

(3.1) $\displaystyle
1\xrightarrow{\;\;\;\;\;\;}T^{*}E_{0}^{*}\xrightarrow{\;\;\;\;\;\;\;}E^{*}\xrightarrow{\;\;\;\;\;\;\;}\Gamma_{E}/\Gamma_{T}\xrightarrow{\;\;\;\;\;\;\;}1.$

Next, we need to recall the definition of the cohomology group. Let $G$ be a
finite abelian group and $M$ be a $G$-module, with operation written
multiplicatively; that is the group action $G\times M\to
M:(\sigma,c)\mapsto\sigma(c)$ satisfies
$\sigma(m_{1}m_{2})=\sigma(m_{1})\sigma(m_{2})$ and
$(\sigma_{1}\sigma_{2})(m)=\sigma_{1}(\sigma_{2}(m))$.

######  Lemma 3.4.

Let $G$ be a finite group and $M$ a $G$-module. Let $N_{G}:M\to M$ be the
$G$-norm map given by $c\mapsto\prod_{\sigma\in G}\sigma(c)$. Put
$I_{G}(M)=\left\langle c\sigma(c)^{-1}:c\in M,\sigma\in G\right\rangle$. Then
$I_{G}(M)$ is a subgroup of ${\rm ker}N_{G}$.

###### Proof.

It suffices to show that $I_{G}(M)\subseteq{\rm ker}N_{G}$. For this purpose,
we need to prove $N_{G}(c\sigma(c)^{-1})=1$ for any $c\in M$ and $\sigma\in
G$. Indeed, we have

$\displaystyle N_{G}(c\sigma(c)^{-1})$
$\displaystyle=\prod\limits_{\sigma_{1}\in
G}{\sigma_{1}\left({c\sigma\left(c\right)^{-1}}\right)}$
$\displaystyle=\prod\limits_{\sigma_{1}\in
G}{\sigma_{1}\left(c\right)}\prod\limits_{\sigma_{1}\in
G}{\sigma_{1}\left({\sigma\left(c\right)^{-1}}\right)}$
$\displaystyle=\prod\limits_{\sigma_{1}\in
G}{\sigma_{1}\sigma\left(c\right)}\prod\limits_{\sigma_{1}\in
G}{\sigma_{1}\left({\sigma\left(c\right)^{-1}}\right)}$
$\displaystyle=\prod\limits_{\sigma_{1}\in
G}{\sigma_{1}\left({\sigma\left(c\right)}\right)}\prod\limits_{\sigma_{1}\in
G}{\sigma_{1}\left({\sigma\left(c\right)^{-1}}\right)}$
$\displaystyle=\prod\limits_{\sigma_{1}\in G}{\sigma_{1}\left(1\right)}=1.$

∎

######  Definition 3.5.

The $-1$-Tate cohomology group of $G$ with respect to $M$ is defined to be the
following factor group:

$\widehat{H}^{-1}(G,M)={\rm ker}N_{G}/I_{G}(M).$

Let $K/F$ be a finite Galois extension of fields. Set $\mathcal{G}={\rm
Gal}(K/F)$. Then, for any $c\in K$, it is well-known that the norm of $c$ is
given by

$N_{K/F}(c)=\prod_{\sigma\in\mathcal{G}}\sigma(c).$

######  Lemma 3.6.

Let $K/F$ be a finite Galois extension of fields. Set $\mathcal{G}={\rm
Gal}(K/F)$. Then $|\mathcal{G}|=[K:F]<\infty$. Define an action
$\mathcal{G}\times K^{*}\to K^{*}$ given by $(\sigma,c)=\sigma(c)$. Then
$K^{*}$ becomes a $\mathcal{G}$-module. Moreover, the $K^{*}$-norm map
$N_{\mathcal{G}}:K^{*}\to K^{*}$ coincides with the ordinary norm field
$N_{K/F}$ and $I_{\mathcal{G}}(K^{*})=\left\langle c\sigma(c)^{-1}:c\in
K^{*},\sigma\in\mathcal{G}\right\rangle$. Consequently, we have

$\widehat{H}^{-1}(\mathcal{G},K^{*})={\rm ker}N_{K/F}/I_{\mathcal{G}}(K^{*}).$

###### Proof.

For each $a\in K^{*}$, we have

$N_{\mathcal{G}}(a)=\prod_{\sigma\in\mathcal{G}}\sigma(c)=N_{K/F}(a).$

This implies that $N_{\mathcal{G}}$ coincides with $N_{K/F}$. Also, it is
clear that $I_{\mathcal{G}}(K^{*})=\left\langle c\sigma(c)^{-1}:c\in
K^{*},\sigma\in\mathcal{G}\right\rangle$. The final conclusion follows
immediately. ∎

Before give a proof of Theorem 1.1, let us explain the notation
$\Gamma_{E}/\Gamma_{T}\land\Gamma_{E}/\Gamma_{T}$ appearing in Diagram 1. Such
an explanation can be found in [5, Lemma 3.5]. However, for reader’s
convenience, we also include the construction of the group
$\Gamma_{E}/\Gamma_{T}\land\Gamma_{E}/\Gamma_{T}$ here. Let $P$ be a group,
and $S$ a subgroup of $P$ such that $[P,P]\subseteq S$. Let $Q=P/(SZ(P))$.
Then, we have $[[P,P],[P,P]]\subseteq[S,P]$, from which it follows that
$[S,P]$ is a normal subgroup of $[P,P]$ with abelian factor. Consider the map

$\eta:P\times P\to[P,P]/[S,P]\text{\;\;\;\; given by \;\;\;\;}(a,b)\mapsto
aba^{-1}b^{-1}[S,P].$

For any $a,b,c\in P$, we have $[a,bc]=[a,b][b,[a,c]][a,c]$. As
$[b,[a,c]]\in[H,G]$, we have $\eta(a,bc)=\eta(a,b)\eta(a,c)$. So $\eta$ is
multiplicative in the second variable. Likewise, $\eta$ is multiplicative in
the first variable.

Let $x$ is an element of $[SZ(P),P]$, there exists $a\in S,b\in Z(P),c\in P$
such that $x=abc(ab)^{-1}c^{-1}=abcb^{-1}a^{-1}c^{-1}$. As $b\in Z(P)$, we
have $x=aca^{-1}c^{-1}\in[S,P]$, so $[SZ(P),P]\subset[S,P]$.

Let $\eta^{\prime}:Q\times Q\to[P,P]/[S,P]$ given by
$(aSZ(P),bSZ(P))\mapsto[a,b][S,P]$. Take any $a,b,c,d\in P$ such that
$(aSZ(P),bSZ(P))=(cSZ(P),dSZ(P))$. We have
$[a,b][S,P]=[a,b][a^{-1}c,b][c,b^{-1}d][S,P]=[c,b][c,b^{-1}d][S,P]=[c,d][S,P]$,
which means $\eta^{\prime}$ is well-defined and multiplicative. So
$\eta^{\prime}$ induces a well-defined group homomorphism
$\eta^{\prime\prime}:Q\otimes_{\mathbb{Z}}Q\to[P,P]/[S,P]$ given by
$aSZ(P)\otimes bSZ(P)\mapsto[a,b][S,P]$, which is obviously surjective. As
$\eta^{\prime\prime}(\zeta\otimes\zeta)=[S,P]$ for all $\zeta\in Q$, we have
an induced epimorphism $B\land B\to[P,P]/[S,P]$.

The proof of Theorem 1.1. That $G$ is a subgroup of $E_{0}^{*}$ follows
immediately from Lemma 3.1, and that
$\tau(E_{0}^{*}/[E^{*},E^{*}])=G/[E^{*},E^{*}]$ follows from Lemma 3.3(i).
Now, we consider the column in Diagram 1. For convenience, we present the
sequence here:

$1\xrightarrow{\;\;\;\;\;\;\;\;}{\rm
ker}\widetilde{N}/[E_{0}^{*},E^{*}]\xrightarrow{\;\;\;\;\iota\;\;\;\;}G/[E_{0}^{*},E^{*}]\xrightarrow{\;\;\;\;\bar{N}\;\;\;\;}\tau(T_{0}^{*})\cap\widetilde{N}(E_{0}^{*})\xrightarrow{\;\;\;\;\;\;\;\;\;}1$

We first show how the map $\iota$ is defined. In view of Remark 1(iii), we
obtain that ${\rm ker}\widetilde{N}\subseteq E^{(1)}$. Because ${\rm
SK}_{1}(E)=E^{(1)}/[E,E]$ is a torsion group, we get that $E^{(1)}\subseteq
G$. It follows that ${\rm ker}\widetilde{N}\subseteq G$, which implies that
the embedded $\iota$ in above sequence is well-defined.

Now we construct the surjective map $\bar{N}$ in the column. In view of Lemma
3.3, we conclude that $G\subseteq E_{0}^{*}$. For each $g\in G$, there exists
$m\geq 0$ such that $g^{m}\in[E^{*},E^{*}]$, and so $1={\rm
Nrd}_{E}(g^{m})={\rm Nrd}_{E}(g)^{m}$. Thus, by Remark 1(i), we get that

$1={\rm Nrd}_{E}(g^{m})={\rm Nrd}_{E}(g)^{m}=N_{Z(E_{0})/T_{0}}{\rm
Nrd}_{E_{0}}(g)^{\delta m}=\widetilde{N}(g)^{\delta m}.$

This means that
$\widetilde{N}(G)\subseteq\tau(T_{0}^{*})\cap\widetilde{N}(E_{0}^{*})$.
Moreover, as $[E_{0}^{*},E^{*}]\subseteq{\rm ker}\widetilde{N}$, the map

$\bar{N}:G/[E_{0}^{*},E^{*}]\to\tau(T_{0}^{*})\cap\widetilde{N}(E_{0}^{*})\text{
\;\;\;\; given by \;\;\;\; }a[E_{0}^{*},E^{*}]\mapsto\widetilde{N}(a)$

is well-defined. To prove that this map is surjective, we take $b$ to be an
arbitrary element in $\tau(T_{0}^{*})\cap\widetilde{N}(E_{0}^{*})$. As $b$ is
an element of $\widetilde{N}(E_{0}^{*})$, there exists an element $a$ of
$E_{0}^{*}$ such that $b=\widetilde{N}(a)$. Moreover, because
$b\in\tau(T_{0}^{*})$, there is an integer $k$ for which $b^{k}=1$, yielding
that $\widetilde{N}\left(a^{k}\right)=1$. It follows that
$a^{k}\in\ker\widetilde{N}\subseteq G$, and so we can find an integer $m$ such
that $a^{km}=\left(a^{k}\right)^{m}\in[E^{*},E^{*}]$. Hence,
$a[E^{*},E^{*}]\in\tau(E^{*}_{0}/[E^{*},E^{*}])=G/[E^{*},E^{*}]$, which leads
to $a\in G$. Therefore, we finally obtain that
$\bar{N}(a[E_{0}^{*},E^{*}])=b$, proving that the map $\bar{N}$ is surjective.

Next, we prove that the column is exact at $G/[E_{0}^{*},E^{*}]$. To prove
this, as ${\rm im}\iota={\rm ker}\widetilde{N}/[E_{0}^{*},E^{*}]$, we need to
show that ${\rm ker}\bar{N}={\rm ker}\widetilde{N}/[E_{0}^{*},E^{*}]$. Let
make the following computations:

$\displaystyle{\rm ker}\bar{N}$ $\displaystyle=\\{a[E_{0}^{*},E^{*}]|a\in
G\text{ and }\bar{N}(a[E_{0}^{*},E^{*}])=1\\}$
$\displaystyle=\\{a[E_{0}^{*},E^{*}]|a\in G\text{ and }\widetilde{N}(a)=1\\}$
$\displaystyle=\\{a[E_{0}^{*},E^{*}]|a\in\ker\widetilde{N}\\}$
$\displaystyle={\rm ker}\widetilde{N}/[E_{0}^{*},E^{*}].$

The above computations show that the column is exact. Now, we consider the row
of Diagram 1.

$\Gamma_{E}/\Gamma_{T}\land\Gamma_{E}/\Gamma_{T}\xrightarrow{\;\;\;\alpha\;\;\;}G/[E_{0}^{*},E^{*}]\xrightarrow{\;\;\;\;\bar{\varphi}\;\;\;\;}{\rm
TK}_{1}(E)\xrightarrow{\;\;\;\;\;\;\;\;\;}1$

First, let us explain the construction of the map $\bar{\varphi}$. Let
$\varphi:G\to G/[E,E]={\rm TK}_{1}(E)$ be the natural epimorphism. Because
$[E_{0}^{*},E^{*}]\subseteq[E^{*},E^{*}]$, the map $\varphi$ induces an
epimorphism

$\bar{\varphi}:G/[E_{0}^{*},E^{*}]\to{\rm TK}_{1}(E)\text{\;\;\;\; given by
\;\;\;\;}a[E_{0}^{*},E^{*}]\mapsto a[E^{*},E^{*}].$

Now, we show that the sequence is exact at $G/[E_{0}^{*},E^{*}]$. By the
definition of $\alpha$, a simple calculation shows that ${\rm
im}\alpha=[E^{*},E^{*}]/[E_{0}^{*},E^{*}]$. Moreover, we have

$\displaystyle\ker\bar{\varphi}$
$\displaystyle=\\{a[E_{0}^{*},E^{*}]|a[E^{*},E^{*}]=[E^{*},E^{*}]\\}$
$\displaystyle=\\{a[E_{0}^{*},E^{*}]|a\in[E^{*},E^{*}]\\}$
$\displaystyle=[E^{*},E^{*}]/[E_{0}^{*},E^{*}].$

This proves the exactness of the column. The proof of Theorem 1.1 is complete.

Before giving a proof of Corollary 1.2, let us make a lemma for further use.

######  Lemma 3.7.

Let $E$ be a graded division ring with center $T$. If $E$ is semiramified,
then $E_{0}$ is a field which is Galois over $T_{0}$, and for
$\mathcal{G}={\rm Gal}(E_{0}/T_{0})\cong\Gamma_{E}/\Gamma_{T}$, we have
$E^{(1)}/[E_{0}^{*},E^{*}]=\widehat{H}^{-1}(\mathcal{G},E_{0}^{*})$. In
particular, we have $[E^{*},E_{0}]\cong I_{\mathcal{G}}(E_{0}^{*})$ and
$E^{(1)}={\rm ker}N_{E_{0}/T_{0}}$.

###### Proof.

Since $E$ is semiramified, we have $[E_{0}:T_{0}]=|\Gamma_{E}/\Gamma_{T}|={\rm
ind}(E)$, and $E_{0}$ is a field which is Galois over $T_{0}$. Therefore, the
group $E_{0}^{*}$ becomes a $\mathcal{G}$-module with the action is defined as
those in Lemma 3.6, and the map $N_{\mathcal{G}}$ coincides with
$N_{Z(E_{0})/T_{0}}$. Consider the map $\widetilde{N}$ defined in Proposition
1.1. Because $E_{0}$ is a field, the map ${\rm Nrd}_{E_{0}}$ is the identity
map on $E_{0}$, from which it follows that $\widetilde{N}=N_{E_{0}/T_{0}}$.
Consequently, we have ${\rm ker}N_{\mathcal{G}}={\rm ker}N_{E_{0}/T_{0}}={\rm
ker}\widetilde{N}$. Moreover, it was proved in [5, Corollary 3.6(iii)] that,
if $E$ is semiramified, we have

$\widehat{H}^{-1}(\mathcal{G},E_{0}^{*})={\rm
ker}N_{\mathcal{G}}/I_{\mathcal{G}}(E_{0}^{*})={\rm
ker}\widetilde{N}/[E_{0}^{*},E^{*}]=E^{(1)}/[E_{0}^{*},E^{*}].$

This implies that have $[E^{*},E_{0}]=I_{\mathcal{G}}(E_{0}^{*})$ and
$E^{(1)}={\rm ker}N_{E_{0}/T_{0}}$, and so

$E^{(1)}/[E_{0}^{*},E^{*}]=\widehat{H}^{-1}(\mathcal{G},E_{0}^{*}).$

The proof of the lemma is now completed. ∎

The proof of Corollary 1.2. (i) Since $E$ is unramified, we have
$\Gamma_{E}=\Gamma_{T}$, and so $\Gamma_{E}/\Gamma_{T}=1$. Thus, the row of
Diagram 1 says that $G/[E_{0}^{*},E^{*}]={\rm TK}_{1}(E)$. Moreover, it
follows from (3.1) that $E^{*}=E_{0}^{*}T^{*}$, and so
$[E^{*},E^{*}]=[E_{0}^{*},E^{*}]=[E_{0}^{*},E_{0}^{*}]$. Let $G_{0}$ be the
subgroup of $E_{0}$ such that
$G_{0}/[E_{0}^{*},E^{*}]=\tau(E_{0}^{*}/[E_{0}^{*},E^{*}])$. As
$[E^{*},E^{*}]=[E_{0}^{*},E_{0}^{*}]$, we obtain that
$G_{0}/[E_{0}^{*},E^{*}]={\rm TK}_{1}(E_{0})$. Furthermore, in view of Lemma
3.3(2), we get that $G=G_{0}$, and so ${\rm TK}_{1}(E)={\rm TK}_{1}(E_{0})$.
The part (1) is proved.

(ii) Consider the column in Diagram 1. Since $E_{0}=T_{0}\subseteq T$, we get
that $\widetilde{N}$ is the identity map on $T_{0}$ and
$[E^{*},E_{0}^{*}]=[E^{*},T_{0}^{*}]=1$. Therefore, the exactness of the
column implies that $G\cong\tau(T_{0}^{*})$. Moreover, in view of [7,
Propositon 2.1], we have $[E^{*},E^{*}]\cong\mu_{e}(T_{0})$. Thus, ${\rm
TK}_{1}(E)=G/[E^{*},E^{*}]\cong\tau(T_{0}^{*})/\mu_{e}(T_{0})$.

(iii) This assertion follows from Lemma 3.7 and the row of Diagram 1.

(iv) As $L$ is unramified over $T$, we get that
$[L:T]=[L_{0}:T_{0}]\leq[E_{0}:T_{0}]$. The maximality of $L$ in $E$ implies
that ${\rm ind}(E)=[L:T]$. It follows that ${\rm TK}_{1}(E)\cong
G/I_{\mathcal{G}}(E_{0}^{*})$.

(3.2) ${\rm ind}(E)=[L_{0}:T_{0}]\leq[E_{0}:T_{0}].$

Similarly, because $K$ is totally ramified over $T$, we have
$[K:T]=|\Gamma_{K}:\Gamma_{T}|\leq|\Gamma_{E}:\Gamma_{T}|$. As $K$ is maximal
in $E$, we conclude that ${\rm ind}(E)=[L:T]=[K:T]$, and so

(3.3) ${\rm ind}(E)=|\Gamma_{K}:\Gamma_{T}|\leq|\Gamma_{E}:\Gamma_{T}|.$

Moreover, it follows from the Fundamental Equality that

(3.4) ${\rm ind}(E)^{2}=[E:T]=[E_{0}:T_{0}]|\Gamma_{E}:\Gamma_{T}|.$

It follows from (3.2), (3.3) and (3.4) that $[L_{0}:T_{0}]=[E_{0}:T_{0}]$ and
$|\Gamma_{K}:\Gamma_{T}|=|\Gamma_{E}:\Gamma_{T}|$ which implies that
$E_{0}=L_{0}$ and $\Gamma_{K}=\Gamma_{E}$. This implies that $E$ is
semiramified over $T$, and (iii) applies. In other words, we have ${\rm
ker}\tilde{N}/[E_{0}^{*},E^{*}]\cong\widehat{H}^{-1}(\mathcal{G},E_{0}^{*})$
and $\mathcal{G}\cong\Gamma_{E}/\Gamma_{T}$. Take
$\nu,\eta\in\Gamma_{E}/\Gamma_{T}$, and let $a,b$ be the inverse images of
$\nu,\eta$ in $E^{*}$. The left map of (1.2) sends $\nu\wedge\eta$ to
$(aba^{-1}b^{-1})I_{\mathcal{G}}(E_{0}^{*})$. Because $\Gamma_{E}=\Gamma_{K}$,
the elements $a$ and $b$ can be chosen in $K$ so that they commute. Thus, the
left map of (1.2) is trivial, which implies that and ${\rm TK}_{1}(E)\cong
G/[E_{0}^{*},E^{*}]$. Put these information into the exact column, the
sequence is exact. Finally, the fact that
$\widehat{H}^{-1}(\mathcal{G},E_{0}^{*})\cong{\rm SK}_{1}(E)$ was proved in
[5, Corollary 3.6].

######  Remark 3.

Let $E$ be a graded division ring with center $T$. Then, the finite-
dimensionality assures that $E$ has a quotient division ring $q(E)$ obtain by
central localization; that is, $q(E)=E\otimes_{T}q(T)$, where $q(T)$ is the
quotient field of $T$. Then, we have $Z(q(E))=q(T)$ and ${\rm ind}(E)={\rm
ind}(q(E))$. It was shown in [5] that ${\rm SK}_{1}(q(E))\cong{\rm
SK}_{1}(E)$, which is a very difficult result. This result motivates the
question that whether ${\rm TK}_{1}(q(E))\cong{\rm TK}_{1}(E)$ or not? It is
not hard to see that ${\rm TK}_{1}(E)\subseteq{\rm TK}_{1}(q(E))$. However,
whether the inverse inclusion holds or not is not clear.

Now, let us give some examples to demonstrate how to use Corollary 1.2 to
compute the group ${\rm TK}_{1}(E)$ for some graded division algebras $E$. The
constructions of graded division algebras we use in these examples can be
directly adopted from [17, Examples 11.13 and 11.14, p.544].

######  Example 1.

Let $D$ be a division ring which is finite-dimensional over its center $F$,
and $x$ an indeterminate. Let $E=D[x,x^{-1}]$ be the ring of Laurent
polynomials with coefficients taken from $D$. Then, with the usual
$\mathbb{Z}$ grading, $E$ is a graded division ring which is unramified over
its center $F[x,x^{-1}]$ with $E_{0}=D$. Thus, by Corollary 1.2(i), we have
${\rm TK}_{1}(E)\cong{\rm TK}_{1}(D)$.

In the next example, we use the $\mathcal{T}$ construction which was
investigated in [17, 9.1.3 Example, p.461]. For convenience, we briefly
present this construction here. Let $n_{1},\dots,n_{r}$ be integers with
$n_{i}\geq 2$ for all $i$. Let

$n=n_{1}\dots n_{r}\text{ and }m={\rm lcm}(n_{1},\dots,n_{r}).$

Let $k$ be any field containing a primitive $m$-th root of unity $\omega$, and
let $\omega_{i}=\omega^{m/n_{i}}$ for $i=1,\dots,r$; so $\omega_{i}$ is a
primitive $n_{i}$-th root of unity. Let $x_{1},y_{1},\dots,x_{r},y_{r}$ be
$2r$ independent indeterminates over $k$. Put

$F=k((x_{1}))((y_{1}))\dots((x_{r}))((y_{r})),$

the iterated Laurent series (cf. [17, Example 1.1.3, p.4]; and put

$\mathcal{T}(k;n_{1},\dots,n_{r})=(x_{1},y_{1}/F)_{\omega_{1},n_{1}}\otimes_{F}\cdots\otimes_{F}(x_{r},y_{r}/F)_{\omega_{r},n_{r}},$

where each $(x_{i},y_{i}/F)_{\omega_{i},n_{i}}$ is the graded symbol algebra
of degree $n_{r}$ (see [17, Definition 2.18, p.49]. Then, the field $F$ is
henselian for the $(x_{1},y_{1},\dots,x_{r},y_{r})$-adic valuation, with
residue field $\overline{F}=k$ and $\Gamma_{F}=\mathbb{Z}^{2r}$
lexicographically ordered from right to left.

######  Example 2.

Let $D=\mathcal{T}(k;n_{1},\dots,n_{r})$ be as above. Then, $D$ is a tame
totally ramified division algebra with center $Z(D)=F$. Let $E={\rm gr}(D)$,
which is a graded division algebra totally ramified over its center $T={\rm
gr}(F)$. According to [17, Proposition 9.8], we have ${\rm ind}(E)={\rm
ind}(D)=n$ and ${\rm exp}(\Gamma_{E}/\Gamma_{T})=m$. It follows from Corollary
1.2(ii) that ${\rm TK}_{1}(E)\cong\tau(k^{*})/\mu_{m}(k)$. It was shown in
[17, Example 11.14] that ${\rm SK}_{1}(E)\cong\mu_{n}(k)/\mu_{m}(k)$, where
$\mu_{m}(k)$ is the group of $m$-th roots of unity; and thus ${\rm SK}_{1}(E)$
is a cyclic group of order $|\mu_{n}(k)|/m$. Therefore, the following short
sequence is exact:

$1\xrightarrow{\;\;\;\;\;\;}\mu_{n}(k)/\mu_{m}(k)\xrightarrow{\;\;\;\;\;\;\;}{\rm
TK}_{1}(E)\xrightarrow{\;\;\;\;\;\;}\tau(k^{*})/\mu_{n}(k)\xrightarrow{\;\;\;\;\;\;\;}1.$

Thus, by using $\mathcal{T}$ construction, ones may produce many examples of
totally ramified graded division algebras $E$ for which ${\rm TK}_{1}(E)$ is
an extension of a finite cyclic group by a locally cyclic group.

Before closing this section, let us make the following remark.

######  Remark 4.

Let $E$ be a graded division ring of prime index $p$ with center $T$. In view
of the short exact sequence (1.1), we get that

${\rm TK}_{1}(E)/{\rm SK}_{1}(E)\cong{\rm Nrd}_{D}(\tau(T^{*})),$

which is a locally cyclic group. Let $q(E)$ be the quotient division ring of
$E$. Then, we have ${\rm ind}(E)={\rm ind}(q(E))=p$. It was proved in [5,
Theorem 5.7] that ${\rm SK}_{1}(E)\cong{\rm SK}_{1}(q(E))$. Thus, by Wang’s
Theorem, we have ${\rm SK}_{1}(q(E))=1$, and hence ${\rm SK}_{1}(E)=1$. It
follows that ${\rm TK}_{1}(E)$ is a locally cyclic group. Therefore, the
torsion group of the Whitehead group of a graded division algebra of prime
index is a locally cyclic group. At this point, a question is naturally
raised: Given a locally cyclic group $A$, is there a graded division graded
algebra $E$ of prime index with ${\rm TK}_{1}(E)\cong A$? This may be
considered as a graded version of [12, Theorem 2] by Motiee, where he has
completely determined all torsion abelian groups that can occur as the torsion
subgroup of the Whitehead group of a division algebra of prime index.

##  4\. ${\rm TK}_{1}$ of associated graded division ring of a valued division
algebra

Let $F$ be a field with a valuation $v$. For each $a\in\tau(F^{*})$, there
exists $m\in\mathbb{Z}$ such that $a^{m}=1$. It follows that
$mv(a)=v(a^{m})=v(1)=0$. Since $\Gamma_{F}$ is torsion free, we get that
$v(a)=0$, and so $a\in U_{F}$. Thus, the reduction map $F\to\overline{F}$
induces a map

$U_{F}\to\overline{F}^{*}\text{\;\;\;\; given by \;\;\;\;}a\mapsto a+M_{F}$

with kernel $1+M_{F}$. In other words, we have an short exact sequence:

$\displaystyle
1\xrightarrow{\;\;\;\;\;\;}1+M_{F}\xrightarrow{\;\;\;\;\;\;\;}U_{F}\xrightarrow{\;\;\;\;\;\;\;}\overline{F}^{*}\xrightarrow{\;\;\;\;\;\;\;}1.$

Before proving the main theorems of the current section, we first give some
auxiliary results which focus on some certain subgroups of $\tau(F^{*})$ of a
valued field $F$. Although these results are made as a preparation for further
use, they are interesting in their own rights.

######  Lemma 4.1.

Let $F$ be a valued field. If there exists $0\neq a\in M_{F}$ such that
$(1+a)^{k}=1$ for some $k\in\mathbb{Z}$, then $0\neq{\rm
char}(\overline{F})|k$.

###### Proof.

Put $p={\rm char}(\overline{F})$. Since $a\in M_{F}$, we get $v(a)>0$. Write
$1=(1+a)^{k}=1+ka+ba$ with $b\in M_{F}$, we get that $(k+b)a=0$ and so
$k+b=0$. It follows that $v(k)=v(b)>0$. This implies that
$\overline{k}=\overline{0}$ in $\overline{F}$, and so ${\rm
char}\overline{F}=p>0$ and $p|k$. ∎

######  Corollary 4.2.

Let $F$ be a valued field. If $\tau(1+M_{F})\neq 1$, then $p={\rm
char}(\overline{F})>0$.

###### Proof.

Let $1+a$ be a non-trivial torsion element of $1+M_{F}$. Let $k$ be the order
of $1+a$. The non-triviality implies that $a\neq 0$. It follows from above
lemma that $p|k$. ∎

Recall that if ${\rm char}=p>0$, then ${\rm char}\overline{F}=p$; and, in case
that ${\rm char}F=0$ then ${\rm char}\overline{F}$ is either $0$ or $p$ for
some prime $p$. So, if we write the characteristic of $F$ and $\overline{F}$
as a pair, then we get $({\rm char}F,{\rm
char}\overline{F})\in\\{(0,0),\;(0,p),\;(p,p)\\}$.

######  Corollary 4.3.

Let $F$ be a valued field. If $-1\in\tau(F^{*})_{p}$, the $p$-primary
component of $\tau(F^{*})$ for some odd prime $p$, then ${\rm char}F=2$.

###### Proof.

Since $-1\in\tau(1+M_{F})$, Corollary 4.2 implies that ${\rm
char}\overline{F}=p>0$. Let $k\in\mathbb{Z}$ such that $(-1)^{p^{k}}=1$. Since
$p^{k}$ and $2$ are coprime, there are $m,n\in\mathbb{Z}$ such that $2m+pn=1$.
Then,

$-1=(-1)^{2m+p^{k}n}=(-1)^{2m}(-1)^{p^{k}n}=1.$

It follows that $2=0$ in $F$, and so $p={\rm char}\overline{F}={\rm char}F=2$.
∎

To clarify the situation of Corollary 4.3, we give an example of a henselian
field $F$ with $\tau(1+M_{F})\neq 1$.

######  Example 3.

Let $(\mathbb{Q}_{2},v_{2})$ be the field of $2$-adic numbers, and $v_{2}$ the
$2$-adic valuation on $\mathbb{Q}_{2}$. Then $(\mathbb{Q}_{2},v_{2})$ becomes
a henselian valued field with $\overline{\mathbb{Q}_{2}}=\mathbb{F}_{2}$, the
field of $2$ elements. Then, we have $v_{2}(-2)=v_{2}(2)=1>0$, which implies
that $-2\in M_{\mathbb{Q}_{2}}$. It follows that
$-1=1+(-2)\in\tau(1+M_{\mathbb{Q}_{2}})$, so $\tau(1+M_{\mathbb{Q}_{2}})\neq
1$.

######  Corollary 4.4.

Let $F$ be a field with a henselian valuation $v$, and let $D$ be a strongly
tame $F$-central division algebra with ${\rm ind}(D)=n$. Suppose that $G$ is a
subgroup of $D^{*}$ containing $[D^{*},D^{*}]$ such that $G/[D^{*},D^{*}]$ is
torsion of bounded degree $n$. Then $G\cap(1+M_{F})=1$.

###### Proof.

Assume by contradiction that $G\cap(1+M_{F})\neq 1$. Then, there exists $0\neq
a\in M_{F}$ such that $(1+a)^{n}\in[D,D]$. It follows that

$1={\rm Nrd}_{D}((1+a)^{n})=(1+a)^{n^{2}}.$

It follows from Lemma 4.1 that ${\rm char}(\overline{F})|n$, a contradiction.
∎

######  Corollary 4.5.

Let $F$ be a field with a henselian valuation $v$, and let $D$ be a strongly
tame $F$-central division algebra with ${\rm ind}(D)=n$. Suppose that $G$ is a
subgroup of $D^{*}$ containing $[D^{*},D^{*}]$ such that $G/[D^{*},D^{*}]$ is
torsion. If $G/[D^{*},D^{*}]$ contains no element of order ${\rm
char}(\overline{F})$, then $G\cap(1+M_{F})=1$.

###### Proof.

Assume by contradiction that $G\cap(1+M_{F})\neq 1$. Then, there exists $0\neq
a\in M_{F}$ such that $(1+a)^{m}\in[D^{*},D^{*}]$ for some $m\in\mathbb{Z}$.
It follows that

$1={\rm Nrd}_{D}((1+a)^{m})=(1+a)^{mn}.$

It follows from Lemma 4.1 that ${\rm char}(\overline{F})|{mn}$. Since $D$ is
strongly tame, we conclude that ${\rm char}(\overline{F})|m$, which would
imply that $G/[D^{*},D^{*}]$ contains an element of order ${\rm
char}(\overline{F})$, a contradiction. ∎

######  Lemma 4.6 ([4, Lemma 1]).

Let $F$ be a field and $D$ be a $F$-central division algebra with ${\rm
ind}(D)=n$. If $N$ is a normal subgroup of $D^{*}$, then
$N^{n}\subseteq(F^{*}\cap N)[D^{*},N]$.

Next we present another version of Congruence Theorem, which is considered as
an analogous version of Congruence Theorem ([4, Theorem 2]) for ${\rm
SK}_{1}(D)$ or of [12, Theorem 9]. To give the proof of this theorem, we learn
the techniques which were used by Motiee in the proof of [12, Theorem 9].

######  Theorem 4.7 (Congruence Theorem).

Let $F$ be a field with a henselian valuation $v$, and let $D$ be a strongly
tame $F$-central division algebra with ${\rm ind}(D)=n$. If $G/[D^{*},D^{*}]$
contains no element of order ${\rm char}(\overline{F})$, then $(1+M_{D})\cap
G\subseteq[D^{*},D^{*}]$.

###### Proof.

The proof for this case was given in the proof of [12, Theorem 9]. For
completeness we include a proof here. Since $1+M_{D}$ is a normal subgroup of
$D^{*}$, we may apply Lemma 4.6 to conclude that
$(1+M_{D})^{n}\subseteq(1+M_{F})[D^{*},(1+M_{D})]$. Since ${\rm
char}(\overline{F})\nmid n$, Hensel’s Lemma shows that
$(1+M_{D})^{n}=(1+M_{D})$. It follows that
$(1+M_{D})^{n}=(1+M_{F})[D^{*},(1+M_{D})]$. Moreover, in view of Corollary
4.5, we have $(1+M_{F})\cap G=1$, from which it follows that $(1+M_{D})\cap
G\subseteq[D^{*},D^{*}]$. ∎

Theorem 4.7 has two corollaries which are interesting results proved by Motiee
and Hazrat respectively. Before presenting these corollaries, we need the
Primary Decomposition Theorem which decomposes the group ${\rm TK}_{1}(D)$ of
a central division ring $D$ into its primary components, given by Motiee in
[12, Theorem 5]. Recall that if $D$ is a division ring with center $F$ of
index $n=p_{1}^{r_{1}}p_{2}^{r_{2}}\dots p_{k}^{r_{k}}$, with distinct prime
numbers $p_{j}$’s, then it can be decomposed as
$D=D_{p_{1}}\otimes_{F}D_{p_{2}}\otimes_{F}\dots\otimes_{F}D_{p_{k}}$, where
each $D_{p_{j}}$ is a division ring with center $F$ of index $p^{r_{j}}$.

######  Theorem 4.8 (Primary decomposition).

Let $F$ be a field, and $D$ be an $F$-central division algebra. If
$D=D_{p_{1}}\otimes_{F}D_{p_{2}}\otimes_{F}\dots\otimes_{F}D_{p_{k}}$ is the
primary decomposition of $D$, then

${\rm TK}_{1}(D)=\left(\prod_{i=1}^{k}{\rm
TK}_{1}(D_{p_{j}})_{p_{j}}\right)\times\left(\prod_{p\in\mathfrak{P}}\tau(F^{*})_{p}\right),$

where $\mathfrak{P}$ runs through the all primes in
$\mathbb{N}\backslash\\{p_{1},p_{2},\dots,p_{k}\\}$.

######  Corollary 4.9 ([12, Theorem 9]).

Let $F$ be a field with a henselian valuation $v$, and let $D$ be a strongly
tame $F$-central division algebra. Suppose that $G$ is a subgroup of $D^{*}$
containing $[D^{*},D^{*}]$ such that $G/[D^{*},D^{*}]$ is torsion. If ${\rm
char}(F)={\rm char}(\overline{F})$, then $(1+M_{D})\cap
G\subseteq[D^{*},D^{*}]$.

###### Proof.

We will prove that $G/[D^{*},D^{*}]$ contains no element of order ${\rm
char}(\overline{F})$. In fact, there is nothing to prove if ${\rm
char}(\overline{F})=0$. Thus, we may assume that ${\rm
char}(\overline{F})=p>0$. Assume by contradiction that there exists $g\in
G\backslash[D^{*},D^{*}]$ such that $g^{p}\in[D^{*},D^{*}]$. It follows that
the $p$-component ${\rm TK}_{1}(D)_{p}$ of ${\rm TK}_{1}(D)$ is non-trivial.
By Primary Decomposition Theorem, we get that $\tau(F^{*})_{p}\cong{\rm
TK}_{1}(D)_{p}\neq 1$. This is impossible because it is well-known that if
${\rm char}F=p>0$ then $\tau(F^{*})$ contains no element of order $p$. ∎

######  Corollary 4.10 ([4, Theorem 2]).

Let $F$ be a field with a henselian valuation $v$, and let $D$ be a strongly
tame $F$-central division algebra. Then $(1+M_{D})\cap
D^{(1)}\subseteq[D^{*},D^{*}]$.

###### Proof.

Let $n={\rm ind}(D)$. Because ${\rm SK}_{1}(D)=D^{(1)}/[D^{*},D^{*}]$ is
$n$-torsion and $D$ is strongly tame, we conclude that $D^{(1)}/[D^{*},D^{*}]$
contains no element of order ${\rm char}(\overline{F})$. Therefore the result
follows from Theorem 4.7. ∎

In view of the proof of Corollary 4.9, we conclude that if ${\rm char}(F)={\rm
char}(\overline{F})$, then $G/[D^{*},D^{*}]$ contains no element of order
${\rm char}(\overline{F})$. However, the converse does not hold, as we will
see in Example 5 later. Therefore, the result obtained in Theorem 4.7 properly
covers that of Corollary 4.9.

######  Remark 5.

The statement in Corollary 4.10 is also true if the hypothesis “strongly tame”
is replaced by a weaker hypothesis “tame”. This is a very beautiful result
which was proved by Hazrat and Wadsworth in [5, Theorem B.1] with a very long
and complex argument which is very difficult to follow. The authors of the
current paper still don’t know whether Theorem 4.7 true or not if the words
“strongly tame” are placed by “tame”. It seems that the arguments used in the
proof of [5, Theorem B.1] cannot be applied here.

Before giving a proof of Theorem 1.3, we need record an auxiliary lemma.

######  Lemma 4.11.

Let $F$ be a field with a henselian valuation $v$, and let D be an $F$-central
division algebra. If $D$ is strongly tame, then ${\rm TK}_{1}({\rm gr}(D))$
contains no element of order ${\rm char}(\overline{F})$.

###### Proof.

There is nothing to prove if ${\rm char}(\overline{F})=0$. Thus, assume that
${\rm char}(\overline{F})=q>0$. Let $Q$ be the quotient division ring of ${\rm
gr}(D)$. Then ${\rm ind}(Q)={\rm ind}({\rm gr}(D))$, and so, by (2.2), we have

$\displaystyle{\rm ind}(Q)^{2}=[{\rm gr}(D):{\rm
gr}(F)]=[\overline{D}:\overline{F}]|\Gamma_{D}/\Gamma_{F}|.$

Moreover, according to [11, Theorem 3], we have the “Ostrowski theorem”:

$\displaystyle{\rm
ind}(D)^{2}=[D:F]=q^{k}[\overline{D}:\overline{F}]|\Gamma_{D}/\Gamma_{F}|,$

where $k\in\mathbb{Z}$ with $k\geq 0$. The two equations show that ${\rm
ind}(Q)$ divides ${\rm ind}(D)$. Assume by contradiction that there exists
$1\neq a\in{\rm gr}(D)^{*}$ such that $a^{q}\in[{\rm gr}(D)^{*},{\rm
gr}(D)^{*}]$. Consider $a$ as an element of $Q^{*}$. Because $[{\rm
gr}(D)^{*},{\rm gr}(D)^{*}]\leq[Q^{*},Q^{*}]$, we get that
$a^{q}\in[Q^{*},Q^{*}]$. This implies that the $q$-primary component ${\rm
TK}_{1}(Q)_{q}$ of ${\rm TK}_{1}(Q)$ is non-trivial. At the other extreme, as
$\overline{F}$ is contained in $Q$, we conclude that ${\rm char}(Q)={\rm
char}(\overline{F})$. It follows that $Z(Q)$ is a field with ${\rm
char}(Z(Q))=q$, and so $\tau(Z(Q^{*}))_{q}=1$. Put $n={\rm ind}(Q)$ and write
$n=p_{1}^{r_{1}}p_{2}^{r_{2}}\dots p_{k}^{r_{k}}$. Let

$Q=Q_{p_{1}}\otimes_{Z(Q)}Q_{p_{2}}\otimes_{Z(Q)}\dots\otimes_{Z(Q)}Q_{p_{k}},$

where each $Q_{p_{j}}$ is a division ring with center $Z(Q)$ of index
$p^{r_{j}}$. By Primary Decomposition Theorem applied for $Q$, we get

${\rm TK}_{1}(Q)=\left(\prod_{i=1}^{k}{\rm
TK}_{1}(Q_{p_{j}})_{p_{j}}\right)\times\left(\prod_{p\in\mathfrak{P}}\tau((Z(Q))^{*})_{p}\right),$

where $\mathfrak{P}$ runs through the all primes in
$\mathbb{N}\backslash\\{p_{1},p_{2},\dots,p_{k}\\}$. Because
$\tau(Z(Q^{*}))_{q}=1$ and ${\rm TK}_{1}(Q)_{q}\neq 1$, we conclude that $q$
must be one of the $p_{i}$’s in the decomposition of ${\rm TK}_{1}(Q)$. This
implies that $q$ divides ${\rm ind}(Q)$. Moreover, as ${\rm ind}(Q)$ divides
${\rm ind}(D)$, we obtain that $p$ divides ${\rm ind}(D)$. This violates the
assumption that $D$ is strongly tame over $F$. ∎

With Theorem 4.7 in hand, we are now able give a proof of Theorem 1.3 which
indicates that the ${\rm TK}_{1}$ of a valued division ring coincides with
that of its associated graded division ring under a certain circumstance.

The proof of Theorem 1.3. Let $G$ and $\mathbf{G}$ be subgroup of $D^{*}$ and
${\rm gr}(D)^{*}$ such that ${\rm TK}_{1}(D)=G/[D^{*},D^{*}]$ and ${\rm
TK}_{1}({\rm gr}(D))=\mathbf{G}/[{\rm gr}(D)^{*},{\rm gr}(D)^{*}]$
respectively. Consider the surjective group homomorphism $\rho:D^{*}\to{\rm
gr}(D)^{*}$ given by $a\to\widetilde{a}$ with ${\rm ker}\rho=1+M_{D}$. We
claim that $\rho(G)=\mathbf{G}$. Consider the following diagram:

${1}$${(1+M_{D})\cap D^{\prime}}$${D^{\prime}}$${{\rm
gr}(D)^{\prime}}$${0}$${1}$${(1+M_{D})\cap G}$${G}$${\mathbf{G}}$${0}$ $\iota$
$\rho$ $\iota$$\rho$

where $\iota$ denotes the inclusion map. It is clear that $1+M_{D}\subseteq
D^{*}$. Thus, $\mathrm{im}\iota=(1+M_{D})\cap D^{*}=1+M_{D}=\mathrm{ker}\rho$.
Moreover, $\iota$ is a monomorphism and $\rho$ is an epimorphism so the top
row of the diagram is exact.

Next, we prove that the lower row of the diagram is exact. First, we claim
that $\rho(G)=\mathbf{G}$. First, we prove that $\rho(G)\subseteq\mathbf{G}$.
Indeed, for any $g\in G$, there exist $m\in\mathbb{Z}$ and such that $g^{m}=a$
for some $a\in[D^{*},D^{*}]$. As $\rho([D^{*},D^{*}])\subseteq[{\rm
gr}(D)^{*},{\rm gr}(D)^{*}]$, we get that

$\rho(g)^{m}=\rho(g^{m})=\rho(a)\in[{\rm gr}(D)^{*},{\rm gr}(D)^{*}].$

This implies that $\rho(G)\subseteq\mathbf{G}$. For the converse, we note
first, by Lemma 4.11, that ${\rm TK}_{1}({\rm gr}(D))$ contains no element of
order ${\rm char}(\overline{F})$. Let $b\in\mathbf{G}$. According to Lemma
3.1, we get that $\mathbf{G}\subseteq{\rm gr}(D)_{0}$. Then, there exists
$k\in\mathbb{Z}$ such that $b^{k}\in[{\rm gr}(D),{\rm gr}(D)]$. As ${\rm
gr}(D)_{0}=V_{D}/M_{D}=\overline{D}$, there exists $a\in U_{D}$ such that
$\widetilde{a}=b$. By (2.1), we have

$\overline{{\rm Nrd}_{D}(a^{k})}=\widetilde{{\rm Nrd}_{D}(a^{k})}={\rm
Nrd}_{{\rm gr}(D)}(\widetilde{a}^{k})={\rm Nrd}_{{\rm gr}(D)}(b^{k})=1,$

which implies that ${\rm Nrd}_{D}(a^{k})\in 1+M_{F}={\rm Nrd}_{D}(1+M_{D})$.
As ${\rm TK_{1}}({\rm gr}(D))$ contains no element of order ${\rm
char}(\bar{F})$, we have ${\rm char}(\bar{F})\nmid k$. By Hensel’s Lemma, we
obtain that $\left(1+M_{D}\right)^{k}=1+M_{D}$. So, let $c\in 1+M_{D}$ such
that ${\rm Nrd}_{D}(c)={\rm Nrd}_{D}(a^{k})^{-1}$. Then, there exists $d\in
1+M_{D}$ such that $d^{k}=c$. We have ${\rm Nrd}_{D}((ad)^{k})={\rm
Nrd}_{D}\left(a^{k}\right){\rm Nrd}_{D}\left(c\right)=1$, which implies
$(ad)^{k}\in D^{(1)}\subseteq G$. It follows that $(ad)^{kl}\in[D^{*},D^{*}]$
for some $l\in\mathbb{Z}$. Hence, $ad\in G$ and

$\rho(ad)=\rho(a)\rho(d)=\rho(a)=b,$

which means that $b\in\rho(G)$. Therefore, the lower row is exact at
$\mathbf{G}$. Moreover, we observe that the kernel of the restriction of
$\rho$ to $G$ is $(1+M_{D})\cap G$, which shows that the lower row is also
exact at $G$. Therefore, the lower row of the diagram is exact.

Moreover, by Congruence theorem, we get that the left vertical map of the
diagram is an isomorphism. Thus, by snake lemma, we obtain that ${\rm
TK}_{1}(D)\cong{\rm TK}_{1}({\rm gr}(D))$. The theorem is proved.

The proof of Corollary 1.4. As $D$ is strongly tame, we have ${\rm ind}({\rm
gr}(D))={\rm ind}(D)$ and $Z({\rm gr}(D))={\rm gr}(F)$. Therefore, we may
apply Corollary 1.2 to the graded division algebra ${\rm gr}(D)$ here, and
then all assertions may follows from the fact that ${\rm TK}_{1}(D)\cong{\rm
TK}_{1}({\rm gr}(D))$.

(i) As $D$ is unramified, by definition, we get that
$[\overline{D}:\overline{F}]=[D:F]$, from which it follows that $[{\rm
gr}(D):Z({\rm gr}(D))]=[\overline{D}:\overline{F}]=[{\rm gr}(D)_{0}:Z({\rm
gr}(D))_{0}]$. This means that ${\rm gr}(D)$ is a unramified graded division
algebra, and so (i) follows from Corollary 1.2(i).

(ii) Because $D$ is totally ramified, we have ${\rm
gr}(D)_{0}=\overline{D}=\overline{F}=Z({\rm gr}(D))_{0}$, which implies that
${\rm gr}(D)$ is a totally ramified graded division algebra, and so (ii)
follows from Corollary 1.2(ii).

(iii) Assume that $D$ is semiramifield. Then, by definition, we get that
$\overline{D}$ is a field and
$[\overline{D}:\overline{F}]=|\Gamma_{D}:\Gamma_{F}|={\rm ind}(D)$. As
$\Gamma_{{\rm gr}(D)}=\Gamma_{D}$ and $\Gamma_{{\rm gr}(F)}=\Gamma_{F}$, it
follows that $[{\rm gr}(D)_{0}:Z({\rm
gr}(D))_{0}]=[\overline{D}:\overline{F}]=[\Gamma_{{\rm gr}(D)}:\Gamma_{{\rm
gr}(F)}]={\rm ind}({\rm gr}(D))$. It follows that ${\rm gr}(D)$ is a
semiramified graded division algebra, and so (iii) follows from Corollary
1.2(iii).

(iv) As $D$ is nicely semiramified, it has a maximal subfield $K$ unramified
over $F$ and another maximal subfield $L$ totally ramified over $F$. Then, it
can be checked that ${\rm gr}(K)$ and ${\rm gr}(L)$ are graded maximal
subfields of ${\rm gr}(D)$, with ${\rm gr}(K)$ unramified over ${\rm gr}(F)$
and ${\rm gr}(L)$ totally ramified over ${\rm gr}(F)$. Therefore the assertion
(iv) follows immediately from Corollary 1.2(iv).

Now, we present some examples of strongly tame division algebras over a
henselian center, and compute the torsion subgroup of its Whitehead group
using Corollary 1.4. We start with using $\mathcal{T}$ construction to produce
examples of strongly tame totally ramified division ring.

######  Example 4.

Let $D=\mathcal{T}(k;n_{1},\dots,n_{r})$ be the division ring considered in
Example 2. Then, we have $F:=Z(D)=k((x_{1}))((y_{1}))\dots((x_{r}))((y_{r}))$
which is henselian for the $(x_{1},y_{1},\dots,x_{r},y_{r})$-adic valuation,
with residue field $\overline{F}=k$. Because ${\rm ind}(D)=n_{1}\dots n_{r}$,
by suitable choices of $k$ and the $n_{i}$’s, we get ${\rm char}(k)\nmid{\rm
ind}(D)$; that is, $D$ is strongly tame. Moreover, because ${\rm char}(F)={\rm
char}(k)={\rm char}(\overline{F})$, the proof of Corollary 4.9 shows that
${\rm TK}_{1}(D)$ contains no element of order ${\rm char}(\overline{F})$, and
so Corollary 1.4(ii) is applied. It follows that there is a short exact
sequence:

$1\xrightarrow{\;\;\;\;\;\;}\mu_{n}(k)/\mu_{m}(k)\xrightarrow{\;\;\;\;\;\;\;}{\rm
TK}_{1}(D)\xrightarrow{\;\;\;\;\;\;}\tau(k^{*})/\mu_{n}(k)\xrightarrow{\;\;\;\;\;\;\;}1,$

in which $\mu_{n}(k)/\mu_{m}(k)$ is a finite cyclic group and
$\tau(k^{*})/\mu_{n}(k)$ is a locally cyclic group. Using this, ones may
produce many examples of strongly tame totally ramified division ring $D$ for
which ${\rm TK}_{1}(D)$ is an extension of a finite cyclic group by a locally
cyclic group.

In what follows, we give an examples of a nicely semiramified divsion ring $D$
over an henselian valuded field and compute ${\rm SK}_{1}(D)$ using 1.4(iv).

######  Example 5.

Let $(\mathbb{Q}_{p},v_{p})$ be the field of $p$-adic numbers with $p$ a
prime, and $v_{p}$ the $p$-adic valuation on $\mathbb{Q}_{p}$. Then
$(\mathbb{Q}_{p},v_{p})$ becomes a henselian valued field with
$\overline{\mathbb{Q}_{p}}=\mathbb{F}_{p}$, the field of $p$ elements. Let $L$
be an unramified extension of $\mathbb{Q}_{p}$ of degree $2$. Then, with
respect to the valuation $v$ extending the valuation $v_{p}$ on
$\mathbb{Q}_{p}$, we have $\overline{L}=\mathbb{F}_{p^{2}}$, and $L$ is cyclic
Galois over $\mathbb{Q}_{p}$ as the valuation is henselian and $\overline{L}$
is cyclic Galois over $\overline{\mathbb{Q}_{p}}$. Let ${\rm
Gal}(L/\mathbb{Q}_{p})=\langle\sigma\rangle$. Take $\pi\in\mathbb{Q}_{p}$ such
that $v(\pi)=1$. Let $D$ be the cyclic algebra
$D=(L/\mathbb{Q}_{p},\sigma,\pi)=L\oplus Lx$, where $xc=\sigma(c)x$ for all
$c\in L$ and $x^{2}=\pi$ (see [8, p.218] for the definition of a cyclic
algebra). Then, $v$ extends to a valuation on $D$ given by
$v(a+bx)=\min\\{v(a),v(b)+1/2\\}$. Moreover, if we set $K=\mathbb{Q}_{p}(x)$,
then $K$ is a maximal subfield of $D$ totally ramified over $\mathbb{Q}_{p}$.
It follows from [9, Theorem 4.4] that $D$ is a nicely semiramified valued
division ring with $\overline{D}=\overline{K}=\mathbb{F}_{p^{2}}$, and thus
Corollary 1.4(iv) is applied here. As ${\rm ind}(D)=2$, according to Wang’s
Theorem (see [2, Corollary 4, p.164]), we get that ${\rm SK}_{1}(D)=1$. Thus,
Corollary 1.4(iv) implies that ${\rm
TK}_{1}(D)\cong\tau(\mathbb{F}_{p}^{*})\cap
N_{\mathbb{F}_{p^{2}}/\mathbb{F}_{p}}(\mathbb{F}_{p^{2}}^{*})$ which contains
no element of order $p$.

## References

  * [1] M. Boulagouaz, Le gradúe d’une algèbre à division valuée, Comm. Algebra 23 (1995) 4275-4300.
  * [2] P. Draxl, Skew Fields, London Math. Soc. Lecture Note Ser., 81 Cambridge University Press, Cambridge, 1983.
  * [3] Yu. Ershov, Henselian valuations of division rings and the group ${\rm SK}_{1}$, Mat. Sb. (N.S.) 117 (1982) 60-68 (Russian), Math. USSR Sb. 45 (1983) 63-71 (English translation).
  * [4] R. Hazrat, Wedderburn’s factorization theorem application to reduced K-theory, Proc. Am. Math. Soc. 130(2) (2002) 311-314.
  * [5] R. Hazrat and A. R. Wadsworth, $\rm SK_{1}$ of graded division algebras, Israel J. Math. 183 (2011) 117-163.
  * [6] Y.-S. Hwang and A. R. Wadsworth, Algebraic extensions of graded and valued fields, Comm. Algebra 27 (1999) 821-840.
  * [7] Y.-S. Hwang and A. R. Wadsworth, Correspondences between valued division algebras and graded division algebras, J. Algebra 220 (1999) 73-114.
  * [8] T. Y. Lam, A first course in noncommutative rings, 2nd edn, GTM 131, Springer-Verlag, New York, 2001.
  * [9] B. Jacob and A. R. Wadsworth, Division algebras over henselian fields, J. Algebra 128 (1990) 126-179.
  * [10] W. May, Multiplicative groups of fields, Proc. London Math. Soc 24(3) (1972) 295-306.
  * [11] P. Morandi, The Henselization of a valued division algebra, J. Algebra 122 (1989) 232-243.
  * [12] M. Motiee, On torsion subgroups of Whitehead groups of division algebras, Manuscripta Math. 141(3-4) (2013) 717-726.
  * [13] T. Nakayama and Y. Matsushima, Über die multiplikative Gruppe einer $p$-adischen Divisions algebra, Proc. Imp. Acad. Japan 19 (1943) 622-628.
  * [14] V. P. Platonov, On the Tannaka-Artin problem. Dokl. Akad. Nauk SSSR 221(5) (1975) 1038-1041. English trans.: Soviet Math. Dokl. 16 (1975) 468-473.
  * [15] V. P. Platonov, The Tannaka-Artin problem, and groups of projective conorms. Dokl. Akad. Nauk SSSR 222(6) (1975) 1299-1302. English trans.: Soviet Math. Dokl. 16 (1975) 782-786.
  * [16] V. P. Platonov, Reduced K-theory and approximation in algebraic groups. Trudy Mat. Inst. Steklov. 142 (1976) 198-207. English trans.: Proc. Steklov Inst. Math. 142 (1979) 213-224.
  * [17] J.-P. Tignol and A. R. Wadsworth, Totally ramified valuations infinite-dimensional division algebras. Trans. Am. Math. Soc 302(1)(1987) 223-250.
  * [18] J.-P. Tignol and A. R. Wadsworth, Value functions and associated graded rings for semisimple algebras, Trans. Amer. Math. Soc. 362 (2010) 687-726.
  * [19] Wang Shianghaw, On the commutator group of a simple algebra, Amer. J. Math. 72 (1950) 323-334.