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This problem is extracted from Dummit and Foote, problems 9.5.5 and
9.5.6.

Let $\varphi (n)$ denote the \emph{Euler $\varphi$-function}: for any
positive integer $n$, $\varphi (n)$ is the number of positive integers
less than or equal to $n$ which are relatively prime to $n$.  By
convention, $\varphi (1)=1$.  You may use the following facts about
$\varphi$:

\begin{itemize}
\item If $p$ is prime, then $\varphi (p)=p-1$.
\item If $p^{k}$ is a power of a prime $p$, then $\varphi (p^{k}) =
p^{k-1}(p-1)$.
\item If $a$ and $b$ are relatively prime, then $\varphi (ab) =
\varphi (a) \varphi (b)$.
\item For any positive integer $n$, $\varphi (n)$ is the order of the
group $(\Z/n\Z)^{\times}$.
\end{itemize}

Prove the following:

\begin{enumerate}
\item [(a)] $\sum_{d|n} \varphi (d)=n$.  (The notation here means: for
every divisor $d$ of $n$, add up the numbers $\varphi (d)$.  For
example, if $n=6$, then the sum is $\varphi (1)+\varphi (2)+\varphi
(3)+\varphi (6) = 1 + 1 + 2 + 2$.)
\item [(b)] Let $F$ be a field and let $G$ be a finite subgroup of the
group of units $F^{\times}$.  For any integer $d$, let $\psi (d)$
denote the number of elements of $G$ of order $d$.  Prove that $\psi
(d)=\varphi (d)$ for every divisor $d$ of $|G|=n$.
\item [(c)] Let $d=n$ to conclude that $\psi (n) \geq 1$, so $G$ is
cyclic.  That is, \emph{any finite subgroup of the group of units of a
field is cyclic.}
\end{enumerate}

Hints:
\begin{enumerate}
\item [(a)] Here are two approaches: (i) First prove the formula when
$n$ is a power of a prime.  In general, write $n=p^{m}n'$ for some
prime $p$ and some integer $n'$ not divisible by $p$; show that 
\[
\sum_{d|n} \varphi (d) = \sum_{d''|p^{m}} \varphi (d'') \sum_{d'|n'}
\varphi (d'),
\]
and use induction to finish the proof.  (ii) Let $C_{n}$ be a cyclic
group of order $n$ and show that since $C_{n}$ contains a unique
subgroup of order $d$ for each factor $d$ of $n$, the number of
elements of $C_{n}$ of order $d$ is $\varphi (d)$.  Hence $n=|C_{n}|$
is the sum of $\varphi (d)$ as $d$ ranges over all divisors of $n$.
\item [(b)] For any integer $N$, $x^{N}-1$ has at most $N$ roots in
$F$, and so $\sum_{d|N} \psi (d) \leq N$.  Since $\sum_{d|N} \varphi
(d) = N$, show by induction that $\psi (d) \leq \varphi (d)$ for every
divisor $d$ of $n$.  Since $\sum_{d|n} \psi (d) = n = \sum_{d|n}
\varphi (d)$, conclude that $\psi (d) = \varphi (d)$ for every
divisor $d$ of $n$.
\end{enumerate}

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