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\begin{center}
\textbf{Rep\#2a: Finite subgroups of multiplicative groups of fields}
Darij Grinberg
[not completed,
%NOT actually theorems proofread
not proofread]
\bigskip
\end{center}
This note is mostly an auxiliary note for Rep\#2. We are going to prove a fact
which is used rather often in algebra:
\begin{quote}
\textbf{Theorem 1.} Let $A$ be a field, and let $G$ be a finite subgroup of
the multiplicative group $A^{\times}$. Then, $G$ is a cyclic group.
\end{quote}
This theorem generalizes the (well-known) fact that the multiplicative group
of a finite field is cyclic. Most proofs of this fact can actually be used to
prove Theorem 1 in all its generality, so there is not much need to provide
another proof here. Nevertheless, let us sketch a proof of Theorem 1 that
requires only basic number theory. The downside is that it is very ugly.
First, an easy number-theoretical lemma:
\begin{quote}
\textbf{Lemma 2.} Let $i$, $g$ and $a$ be three integers such that $a$ is
positive, such that $g\mid a$, and such that $i$ is coprime to $g$. Then,
there exists an integer $I$ such that $I\equiv i\operatorname{mod}g$ and such
that $I$ is coprime to $a$.
\end{quote}
\textit{Proof of Lemma 2.} For every integer $n$, let us denote by
$\operatorname{PF}n$ the set of all prime divisors of $n$. By the unique
factorization theorem, for any positive integer $n$, the set
$\operatorname{PF}n$ is finite and satisfies $n=\prod\limits_{p\in
\operatorname{PF}n}p^{v_{p}\left( n\right) }$.
Clearly, $a\neq0$ (since $a$ is positive) and $g\neq0$ (since $a\neq0$ and
$g\mid a$). Now, $g\mid a$ yields $\operatorname{PF}g\subseteq
\operatorname{PF}a$. We have%
\[
a=\prod\limits_{p\in\operatorname{PF}a}p^{v_{p}\left( a\right) }%
=\prod\limits_{p\in\operatorname{PF}g}p^{v_{p}\left( a\right) }\cdot
\prod\limits_{p\in\operatorname{PF}a\setminus\operatorname{PF}g}%
p^{v_{p}\left( a\right) }\ \ \ \ \ \ \ \ \ \ \left( \text{since
}\operatorname{PF}g\subseteq\operatorname{PF}a\right) .
\]
In other words, $a=a_{1}a_{2}$, where $a_{1}=\prod\limits_{p\in
\operatorname{PF}g}p^{v_{p}\left( a\right) }$ and $a_{2}=\prod
\limits_{p\in\operatorname{PF}a\setminus\operatorname{PF}g}p^{v_{p}\left(
a\right) }$.
The number $g$ is not divisible by any prime $p\in\operatorname{PF}%
a\setminus\operatorname{PF}g$ (because if $g$ is divisible by a prime $p$,
then $p\in\operatorname{PF}g$, so that $p$ cannot lie in $\operatorname{PF}%
a\setminus\operatorname{PF}g$). Hence, $g$ is coprime to $p^{v_{p}\left(
a\right) }$ for every $p\in\operatorname{PF}a\setminus\operatorname{PF}g$.
Consequently, $g$ is coprime to the product $\prod\limits_{p\in
\operatorname{PF}a\setminus\operatorname{PF}g}p^{v_{p}\left( a\right) }$. In
other words, $g$ is coprime to $a_{2}$ (since $\prod\limits_{p\in
\operatorname{PF}a\setminus\operatorname{PF}g}p^{v_{p}\left( a\right)
}=a_{2}$). Thus, by Bezout's Theorem\footnote{\textbf{Bezout's theorem} states
that if $\lambda_{1}$ and $\lambda_{2}$ are two coprime integers, then there
exist integers $\rho_{1}$ and $\rho_{2}$ such that $\rho_{1}\lambda_{1}%
+\rho_{2}\lambda_{2}=1$.}, there exist integers $\rho_{1}$ and $\rho_{2}$ such
that $\rho_{1}g+\rho_{2}a_{2}=1$. Thus, $1-\rho_{1}g=\rho_{2}a_{2}%
\equiv0\operatorname{mod}a_{2}$. Now, let $I=i-\left( i-1\right) \rho_{1}g$.
Then, $I=i-\left( i-1\right) \rho_{1}g\equiv i\operatorname{mod}g$. Hence,
$I$ is coprime to $g$ (since $i$ is coprime to $g$). Hence, $I$ is not
divisible by any prime $p\in\operatorname{PF}g$. Thus, $I$ is coprime to
$p^{v_{p}\left( a\right) }$ for every $p\in\operatorname{PF}g$.
Consequently, $I$ is coprime to the product $\prod\limits_{p\in
\operatorname{PF}g}p^{v_{p}\left( a\right) }$. In other words, $I$ is
coprime to $a_{1}$ (since $\prod\limits_{p\in\operatorname{PF}g}%
p^{v_{p}\left( a\right) }=a_{1}$). On the other hand, $I$ is coprime to
$a_{2}$ (since%
\[
I=i-\left( i-1\right) \rho_{1}g=i\underbrace{\left( 1-\rho_{1}g\right)
}_{\equiv0\operatorname{mod}a_{2}}+\rho_{1}g\equiv\rho_{1}g\equiv\rho
_{1}g+\rho_{2}a_{2}=1\operatorname{mod}a_{2}%
\]
). Hence, $I$ is coprime to $a_{1}a_{2}$ (since $I$ is coprime to $a_{1}$ and
to $a_{2}$). In other words, $I$ is coprime to $a$ (since $a_{1}a_{2}=a$).
This proves Lemma 2.
\textit{Proof of Theorem 1.} We first notice that the group $G$ is a subgroup
of the abelian group $A^{\times}$ (since $A$ is a field), and thus itself
abelian. We shall thus use the standard properties of abelian groups (such as
the identity $\left( \alpha\beta\right) ^{m}=\alpha^{m}\beta^{m}$ for all
$m\in\mathbb{Z}$ and $\alpha,\beta\in G$) without further mention.
Now, we shall show that%
\begin{align}
& \text{if }\alpha\text{ and }\beta\text{ are two elements of }G\text{, then
there exists }\gamma\in G\text{ such that}\nonumber\\
& \alpha\in\left\langle \gamma\right\rangle \text{ and }\beta\in\left\langle
\gamma\right\rangle . \label{3proof1}%
\end{align}
\textit{Proof of (\ref{3proof1}).} Let $a$ be the order of $\alpha$ in $G$,
and let $b$ be the order of $\beta$ in $G$. Let $g$ be $\gcd\left(
a,b\right) $. Then, $g\mid a$ and $g\mid b$. Thus, $\left( a\diagup
g\right) \mid a$ and $\left( b\diagup g\right) \mid b$.
The order of $\alpha$ in $G$ is $a$. Hence, the order of $\alpha^{a\diagup g}$
in $G$ is $\dfrac{a}{a\diagup g}=g$ (since $\left( a\diagup g\right) \mid
a$). Consequently, the elements $\left( \alpha^{a\diagup g}\right) ^{0},$
$\left( \alpha^{a\diagup g}\right) ^{1},$ $...,$ $\left( \alpha^{a\diagup
g}\right) ^{g-1}$ are pairwise distinct, and we have $\left( \alpha
^{a\diagup g}\right) ^{g}=1$. Now, for every $i\in\left\{
0,1,...,g-1\right\} $, we have $\left( \left( \alpha^{a\diagup g}\right)
^{i}\right) ^{g}=\left( \underbrace{\left( \alpha^{a\diagup g}\right)
^{g}}_{=1}\right) ^{i}=1$, and thus the element $\left( \alpha^{a\diagup
g}\right) ^{i}$ is a root of the polynomial $X^{g}-1\in A\left[ X\right] $.
In other words, the elements $\left( \alpha^{a\diagup g}\right) ^{0},$
$\left( \alpha^{a\diagup g}\right) ^{1},$ $...,$ $\left( \alpha^{a\diagup
g}\right) ^{g-1}$ are roots of the polynomial $X^{g}-1\in A\left[ X\right]
$. Since we know that these elements $\left( \alpha^{a\diagup g}\right)
^{0},$ $\left( \alpha^{a\diagup g}\right) ^{1},$ $...,$ $\left(
\alpha^{a\diagup g}\right) ^{g-1}$ are pairwise distinct, we thus see that
the elements $\left( \alpha^{a\diagup g}\right) ^{0},$ $\left(
\alpha^{a\diagup g}\right) ^{1},$ $...,$ $\left( \alpha^{a\diagup g}\right)
^{g-1}$ are pairwise distinct roots of the polynomial $X^{g}-1\in A\left[
X\right] $. But the polynomial $X^{g}-1\in A\left[ X\right] $ can only have
at most $g$ roots (since any nonzero polynomial of degree $g$ over a field can
only have at most $g$ roots), so these roots $\left( \alpha^{a\diagup
g}\right) ^{0},$ $\left( \alpha^{a\diagup g}\right) ^{1},$ $...,$ $\left(
\alpha^{a\diagup g}\right) ^{g-1}$ must be all the roots of the polynomial
$X^{g}-1\in A\left[ X\right] $. Consequently, the polynomial $X^{g}-1$
equals a constant times $\left( X-\left( \alpha^{a\diagup g}\right)
^{0}\right) \left( X-\left( \alpha^{a\diagup g}\right) ^{1}\right)
...\left( X-\left( \alpha^{a\diagup g}\right) ^{g-1}\right) $. But the
constant just mentioned must be $1$ (since the polynomials $X^{g}-1$ and
\newline$\left( X-\left( \alpha^{a\diagup g}\right) ^{0}\right) \left(
X-\left( \alpha^{a\diagup g}\right) ^{1}\right) ...\left( X-\left(
\alpha^{a\diagup g}\right) ^{g-1}\right) $ have the same leading term);
hence, this becomes
\[
X^{g}-1=\left( X-\left( \alpha^{a\diagup g}\right) ^{0}\right) \left(
X-\left( \alpha^{a\diagup g}\right) ^{1}\right) ...\left( X-\left(
\alpha^{a\diagup g}\right) ^{g-1}\right) .
\]
In other words, $X^{g}-1=\prod\limits_{i=0}^{g-1}\left( X-\left(
\alpha^{a\diagup g}\right) ^{i}\right) $. Applying this identity to
$X=\beta^{b\diagup g}$, we obtain $\left( \beta^{b\diagup g}\right)
^{g}-1=\prod\limits_{i=0}^{g-1}\left( \beta^{b\diagup g}-\left(
\alpha^{a\diagup g}\right) ^{i}\right) $. Since $\left( \beta^{b\diagup
g}\right) ^{g}-1=\beta^{b}-1=0$ (since $b$ is the order of $\beta$, and thus
$\beta^{b}=1$), this becomes $0=\prod\limits_{i=0}^{g-1}\left( \beta
^{b\diagup g}-\left( \alpha^{a\diagup g}\right) ^{i}\right) $. Hence, there
must exist some $i\in\left\{ 0,1,...,g-1\right\} $ such that $\beta
^{b\diagup g}-\left( \alpha^{a\diagup g}\right) ^{i}=0$ (because if a
product of elements of a field is zero, then one of the factors must be zero).
Consequently, this $i\in\left\{ 0,1,...,g-1\right\} $ satisfies
$\beta^{b\diagup g}=\left( \alpha^{a\diagup g}\right) ^{i}$. Similarly,
there exists some $j\in\left\{ 0,1,...,g-1\right\} $ satisfying
$\alpha^{a\diagup g}=\left( \beta^{b\diagup g}\right) ^{j}$. Thus,
$\alpha^{a\diagup g}=\left( \underbrace{\beta^{b\diagup g}}_{=\left(
\alpha^{a\diagup g}\right) ^{i}}\right) ^{j}=\left( \left( \alpha
^{a\diagup g}\right) ^{i}\right) ^{j}=\left( \alpha^{a\diagup g}\right)
^{ij}$, so that $1=\dfrac{\left( \alpha^{a\diagup g}\right) ^{ij}}%
{\alpha^{a\diagup g}}=\left( \alpha^{a\diagup g}\right) ^{ij-1}$. Since the
order of the element $\alpha^{a\diagup g}$ is $g$, this yields $g\mid ij-1$,
so that $ij\equiv1\operatorname{mod}g$. Hence, $ij$ is coprime to $g$, so that
$i$ must also be coprime to $g$. Thus, by Lemma 2, there exists an integer $I$
such that $I\equiv i\operatorname{mod}g$ and such that $I$ is coprime to $a$.
Since $I\equiv i\operatorname{mod}g$, we have $g\mid I-i$, and thus $\left(
\alpha^{a\diagup g}\right) ^{I-i}=1$ (since $g$ is the order of
$\alpha^{a\diagup g}$), so that
\begin{equation}
\left( \alpha^{a\diagup g}\right) ^{I}=\left( \alpha^{a\diagup g}\right)
^{\left( I-i\right) +i}=\underbrace{\left( \alpha^{a\diagup g}\right)
^{I-i}}_{=1}\left( \alpha^{a\diagup g}\right) ^{i}=\left( \alpha^{a\diagup
g}\right) ^{i}=\beta^{b\diagup g}. \label{3proof5}%
\end{equation}
Now, the integers $a\diagup g$ and $b\diagup g$ are coprime (since
$\gcd\left( a\diagup g,b\diagup g\right) =\underbrace{\gcd\left(
a,b\right) }_{=g}\diagup g=g\diagup g=1$); hence, by Bezout's Theorem, there
exist integers $u$ and $v$ such that $u\cdot a\diagup g+v\cdot b\diagup g=1$.
Now, let $\gamma=\alpha^{Iv}\beta^{u}$. Then, $\gamma\in G$ and%
\begin{align*}
\gamma^{b\diagup g} & =\left( \alpha^{Iv}\beta^{u}\right) ^{b\diagup
g}=\underbrace{\left( \alpha^{Iv}\right) ^{b\diagup g}}_{=\alpha^{Iv\cdot
b\diagup g}}\underbrace{\left( \beta^{u}\right) ^{b\diagup g}}_{=\left(
\beta^{b\diagup g}\right) ^{u}}=\alpha^{Iv\cdot b\diagup g}\left(
\underbrace{\beta^{b\diagup g}}_{\substack{=\left( \alpha^{a\diagup
g}\right) ^{I}\\\text{(by (\ref{3proof5}))}}}\right) ^{u}=\alpha^{Iv\cdot
b\diagup g}\underbrace{\left( \left( \alpha^{a\diagup g}\right)
^{I}\right) ^{u}}_{=\left( \alpha^{a\diagup g}\right) ^{Iu}=\alpha^{Iu\cdot
a\diagup g}}\\
& =\alpha^{Iv\cdot b\diagup g}\alpha^{Iu\cdot a\diagup g}=\alpha^{Iv\cdot
b\diagup g+Iu\cdot a\diagup g}=\alpha^{I}%
\end{align*}
(since $Iv\cdot b\diagup g+Iu\cdot a\diagup g=I\underbrace{\left( u\cdot
a\diagup g+v\cdot b\diagup g\right) }_{=1}=I$). Since $I$ is coprime to $a$,
there exist integers $x$ and $y$ such that $xI+ya=1$ (according to Bezout's
theorem). Thus,%
\begin{align*}
\alpha & =\alpha^{1}=\alpha^{Ix+ay}\ \ \ \ \ \ \ \ \ \ \left( \text{since
}1=xI+ya=Ix+ay\right) \\
& =\underbrace{\alpha^{Ix}}_{=\left( \alpha^{I}\right) ^{x}}%
\underbrace{\alpha^{ay}}_{=\left( \alpha^{a}\right) ^{y}}=\left(
\underbrace{\alpha^{I}}_{=\gamma^{b\diagup g}}\right) ^{x}\left(
\underbrace{\alpha^{a}}_{\substack{=1\text{ (since }a\text{ is}\\\text{the
order of }\alpha\text{)}}}\right) ^{y}=\left( \gamma^{b\diagup g}\right)
^{x}1^{y}=\left( \gamma^{b\diagup g}\right) ^{x}\in\left\langle
\gamma\right\rangle .
\end{align*}
On the other hand, since $\gamma=\alpha^{Iv}\beta^{u}$, we have%
\begin{align*}
\gamma^{a\diagup g} & =\left( \alpha^{Iv}\beta^{u}\right) ^{a\diagup
g}=\underbrace{\left( \alpha^{Iv}\right) ^{a\diagup g}}_{\substack{=\alpha
^{Iv\cdot a\diagup g}=\alpha^{\left( a\diagup g\right) \cdot Iv}\\=\left(
\alpha^{a\diagup g}\right) ^{Iv}=\left( \left( \alpha^{a\diagup g}\right)
^{I}\right) ^{v}}}\cdot\underbrace{\left( \beta^{u}\right) ^{a\diagup g}%
}_{=\beta^{u\cdot\left( a\diagup g\right) }}=\left( \underbrace{\left(
\alpha^{a\diagup g}\right) ^{I}}_{\substack{=\beta^{b\diagup g}\\\text{(by
(\ref{3proof5}))}}}\right) ^{v}\cdot\beta^{u\cdot\left( a\diagup g\right)
}\\
& =\underbrace{\left( \beta^{b\diagup g}\right) ^{v}}_{=\beta^{\left(
b\diagup g\right) \cdot v}=\beta^{v\cdot\left( b\diagup g\right) }}%
\cdot\,\beta^{u\cdot\left( a\diagup g\right) }=\beta^{v\cdot\left( b\diagup
g\right) }\cdot\beta^{u\cdot\left( a\diagup g\right) }=\beta^{v\cdot\left(
b\diagup g\right) +u\cdot\left( a\diagup g\right) }\\
& =\beta^{1}\ \ \ \ \ \ \ \ \ \ \left( \text{since }v\cdot\left( b\diagup
g\right) +u\cdot\left( a\diagup g\right) =u\cdot a\diagup g+v\cdot b\diagup
g=1\right) \\
& =\beta,
\end{align*}
and therefore $\beta=\gamma^{a\diagup g}\in\left\langle \gamma\right\rangle $.
Altogether, we have proven that $\gamma\in G$, that $\alpha\in\left\langle
\gamma\right\rangle $ and that $\beta\in\left\langle \gamma\right\rangle $.
This proves (\ref{3proof1}).
Now, let us finally prove Theorem 1: Clearly, there exists a subset $P$ of the
group $G$ such that $G=\left\langle P\right\rangle $ (in fact, the whole group
$G$ is an example of such a subset $P$). Let $U$ be such a subset with the
smallest number of elements.\footnote{Indeed, such a $U$ exists, because the
set of all subsets of the group $G$ is finite (since $G$ itself is finite).}
Then, $U$ is a subset of the group $G$ such that $G=\left\langle
U\right\rangle $, but there is no subset $U^{\prime}$ of $G$ with less
elements than $U$ that satisfies $G=\left\langle U^{\prime}\right\rangle $.
We let $k=\left\vert U\right\vert $, and we write the set $U$ as $U=\left\{
u_{1},u_{2},...,u_{k}\right\} $, where $u_{1}$, $u_{2}$, $...$, $u_{k}$ are
the $k$ (pairwise distinct) elements of $U$. Assume now that $k>1$. Then,
$u_{1}$ and $u_{2}$ are well-defined. Now, there exists an element $\gamma\in
G$ such that $u_{1}\in\left\langle \gamma\right\rangle $ and $u_{2}%
\in\left\langle \gamma\right\rangle $ (by (\ref{3proof1}), applied to
$\alpha=u_{1}$ and $\beta=u_{2}$), and therefore $u_{i}\in\left\langle
\gamma,u_{3},u_{4},...,u_{k}\right\rangle $ for every $i\in\left\{
1,2,...,k\right\} $ \ \ \ \ \footnote{In fact, three cases are possible:
either $i=1$, or $i=2$, or $i\geq3$. If $i=1$, then $u_{i}\in\left\langle
\gamma,u_{3},u_{4},...,u_{k}\right\rangle $ follows from $u_{1}\in\left\langle
\gamma\right\rangle \subseteq\left\langle \gamma,u_{3},u_{4},...,u_{k}%
\right\rangle $. If $i=2$, then $u_{i}\in\left\langle \gamma,u_{3}%
,u_{4},...,u_{k}\right\rangle $ follows from $u_{2}\in\left\langle
\gamma\right\rangle \subseteq\left\langle \gamma,u_{3},u_{4},...,u_{k}%
\right\rangle $. Finally, if $i\geq3$, then $u_{i}\in\left\langle \gamma
,u_{3},u_{4},...,u_{k}\right\rangle $ is trivial. Thus, $u_{i}\in\left\langle
\gamma,u_{3},u_{4},...,u_{k}\right\rangle $ holds in all cases.}. Hence,
$\left\langle u_{1},u_{2},...,u_{k}\right\rangle \subseteq\left\langle
\gamma,u_{3},u_{4},...,u_{k}\right\rangle $, so that%
\[
G=\left\langle U\right\rangle =\left\langle \left\{ u_{1},u_{2}%
,...,u_{k}\right\} \right\rangle =\left\langle u_{1},u_{2},...,u_{k}%
\right\rangle \subseteq\left\langle \gamma,u_{3},u_{4},...,u_{k}\right\rangle
=\left\langle \left\{ \gamma,u_{3},u_{4},...,u_{k}\right\} \right\rangle
=\left\langle U^{\prime}\right\rangle ,
\]
where $U^{\prime}$ denotes the subset $\left\{ \gamma,u_{3},u_{4}%
,...,u_{k}\right\} $ of $G$. But clearly, also $G\supseteq\left\langle
U^{\prime}\right\rangle $. Thus, $G=\left\langle U^{\prime}\right\rangle $.
Besides, the subset $U^{\prime}$ of $G$ has less elements than $U$ (because
$U^{\prime}=\left\{ \gamma,u_{3},u_{4},...,u_{k}\right\} $ has at most $k-1$
elements, while $U$ has $\left\vert U\right\vert =k$ elements). This
contradicts the fact that there is no subset $U^{\prime}$ of $G$ with less
elements than $U$ that satisfies $G=\left\langle U^{\prime}\right\rangle $.
This contradiction shows that our assumption $k>1$ was wrong. Hence, $k\leq1$,
so that $k=1$ or $k=0$. If $k=0$, then $\left\vert U\right\vert =k=0$ and thus
$U=\varnothing$, which leads to $G=\left\langle \varnothing\right\rangle
=\left\{ 1\right\} $, so that $G$ is a cyclic group. If $k=1$, then
$\left\vert U\right\vert =k=1$, so that $U=\left\{ u\right\} $ for some
$u\in G$, and therefore $G=\left\langle U\right\rangle =\left\langle \left\{
u\right\} \right\rangle =\left\langle u\right\rangle $ is a cyclic group.
Hence, in both cases, $G$ is a cyclic group. This proves Theorem 1.
Here is an easy consequence of Theorem 1:
\begin{quote}
\textbf{Lemma 3.} Let $A$ be a field. Let $n$ be a positive integer, and for
every $i\in\left\{ 1,2,...,n\right\} $, let $\xi_{i}$ be a root of unity in
$A$. Then, there exists some root of unity $\zeta$ of $A$ and a sequence
$\left( k_{1},k_{2},...,k_{n}\right) $ of nonnegative integers such that
$\left( \xi_{i}=\zeta^{k_{i}}\text{ for every }i\in\left\{
1,2,...,n\right\} \right) $ and $\gcd\left( k_{1},k_{2},...,k_{n}\right)
=1$.
\end{quote}
\textit{Proof of Lemma 3.} Let $G$ be the subgroup $\left\langle \xi_{1}%
,\xi_{2},...,\xi_{n}\right\rangle $ of the multiplicative group $A^{\times}$.
Then, the map%
\begin{align*}
\Phi:\left\langle \xi_{1}\right\rangle \times\left\langle \xi_{2}\right\rangle
\times...\times\left\langle \xi_{n}\right\rangle & \rightarrow\left\langle
\xi_{1},\xi_{2},...,\xi_{n}\right\rangle \ \ \ \ \ \ \ \ \ \ \text{defined
by}\\
\left( x_{1},x_{2},...,x_{n}\right) & \mapsto x_{1}x_{2}...x_{n}%
\end{align*}
is surjective (because every element of $\left\langle \xi_{1},\xi_{2}%
,...,\xi_{n}\right\rangle $ has the form $\prod\limits_{i=1}^{n}\xi_{i}%
^{f_{i}}$ for some $n$-tuple $\left( f_{1},f_{2},...,f_{n}\right) $ of
integers\footnote{Again, we are using the fact that $G$ is abelian (since $G$
is a subgroup of the abelian group $A^{\times}$).}, and thus is $\Phi\left(
\xi_{1}^{f_{1}},\xi_{2}^{f_{2}},...,\xi_{n}^{f_{n}}\right) $), and the set
$\left\langle \xi_{1}\right\rangle \times\left\langle \xi_{2}\right\rangle
\times...\times\left\langle \xi_{n}\right\rangle $ is finite (since the set
$\left\langle \xi_{i}\right\rangle $ is finite for every $i\in\left\{
1,2,...,n\right\} ,$ because $\xi_{i}$ is a root of unity). Hence, the set
$\left\langle \xi_{1},\xi_{2},...,\xi_{n}\right\rangle $ is finite. Thus,
$G=\left\langle \xi_{1},\xi_{2},...,\xi_{n}\right\rangle $ is a finite
subgroup of $A^{\times}$. Hence, by Theorem 1, this group $G$ is cyclic, so
that there exists some $\tau\in G$ such that $G=\left\langle \tau\right\rangle
$. Now, if $u$ is the order of $\tau$ in the group $G$, then $\left\langle
\tau\right\rangle =\left\{ \tau^{0},\tau^{1},...,\tau^{u-1}\right\} $.
Hence, for every $i\in\left\{ 1,2,...,n\right\} $, there exists some
nonnegative integer $\ell_{i}$ such that $\xi_{i}=\tau^{\ell_{i}}$ (since
$\xi_{i}\in G=\left\langle \tau\right\rangle =\left\{ \tau^{0},\tau
^{1},...,\tau^{u-1}\right\} $). Now, let $\ell=\gcd\left( \ell_{1},\ell
_{2},...,\ell_{n}\right) $. Let $\zeta=\tau^{\ell}$, and let $k_{i}=\ell
_{i}\diagup\ell$ for every $i\in\left\{ 1,2,...,n\right\} $%
\ \ \ \ \footnote{If $\ell=0$, then we take $k_{i}=1$ for all $i\in\left\{
1,2,\ldots,n\right\} $ instead. Note that all of the $\ell_{1},\ell
_{2},\ldots,\ell_{n}$ are $0$ in this case, because of $\gcd\left( \ell
_{1},\ell_{2},...,\ell_{n}\right) =\ell=0$.}. Then, $\ell_{i}=\ell k_{i}$ for
every $i\in\left\{ 1,2,...,n\right\} $.
Now we know that $\zeta$ is a root of unity (since $\zeta\in G$, and thus
Lagrange's theorem yields $\zeta^{\left\vert G\right\vert }=1$), and for every
$i\in\left\{ 1,2,...,n\right\} $ we have $\xi_{i}=\tau^{\ell_{i}}=\tau^{\ell
k_{i}}=\left( \underbrace{\tau^{\ell}}_{=\zeta}\right) ^{k_{i}}=\zeta
^{k_{i}}$. Finally, recall that $k_{i}=\ell_{i}\diagup\ell$ for every
$i\in\left\{ 1,2,...,n\right\} $. Thus, $\gcd\left( k_{1},k_{2}%
,...,k_{n}\right) =\gcd\left( \ell_{1}\diagup\ell,\ell_{2}\diagup
\ell,...,\ell_{n}\diagup\ell\right) =\underbrace{\gcd\left( \ell_{1}%
,\ell_{2},...,\ell_{n}\right) }_{=\ell}\diagup\ell=1$. Thus, Lemma 3 is proven.
\end{document}