Finite subgroups of the multiplicative group of a field are cyclic In Grove's book Algebra, Proposition 3.7 at page 94 is the following

If $G$ is a finite subgroup of the multiplicative group $F^*$ of a field $F$,
  then $G$ is cyclic.

He starts the proof by saying "Since $G$ is the direct product of its Sylow subgroups ...". But this is only true if the Sylow subgroups of $G$ are all normal. How do we know this? 
 A: We know that if $G$ is a finite abelian group, $G$ is isomorphic to a direct product $\mathbb{Z}_{(p_1)^{n_1}} \times \mathbb{Z}_{(p_2)^{n_2}} \times \cdots \times  \mathbb{Z}_{(p_r)^{n_r}}$ where $p_i$'s are prime not necessarily distinct.
Consider each of the $\mathbb{Z}_{(p_i)^{n_i}}$ as a cyclic group of order $p_i^{n_i}$ in multiplicative notation. Let $m$ be the $lcm$ of all the $p_i^{n_i}$ for $i=1,2,\ldots,r.$ Clearly $m\leq {p_1}^{n_1}{p_2}^{n_2}\cdots{p_r}^{n_r}.$ If $a_i \in \mathbb{Z}_{(p_i)^{n_i}}$ then $(a_i)^{({p_i}^{n_i})}=1$ and hence $a_i^m=1.$ Therefore for all $\alpha \in G,$ we have $\alpha^m=1;$ that is, every element of $G$ is a root of $x^m=1.$
However, $G$ has ${p_1}^{n_1}{p_2}^{n_2}\cdots{p_r}^{n_r}$ elements, while the polynomial $x^m-1$ can have at most $m$ roots in $F.$ So, we deduce that $m={p_1}^{n_1}{p_2}^{n_2}\cdots{p_r}^{n_r}.$ Therefore $p_i$'s are distinct primes, and the group $G$ is isomorphic to the cyclic group $\mathbb{Z}_m.$
A: Note that this result is not true if $F$ is a skew field (division ring), as is illustrated by the quaternion group $Q_8$ inside the quaternions. So one must use commutativity somewhere, and this usually happens implicitly by using that the polynomial $X^d-1$ can have at most $d$ roots in $F$; this is for instance the case in the answer by Andrea, where the proof of the lemma does not use commutativity. Here is a somewhat different approach that exploits commutativity a second time.
Lemma. The set of orders of elements in a finite Abelian group is closed under taking least common multiples.
(Edit: This happens to be the subject of another math.SE question. It may seem quite hard, unless one realises that in Abelian torsion groups, different prime factors can be considered independently due to a canonical direct sum decomposition, after which the question becomes trivial. Here I'll leave my original proof below, which follows another answer to that question.)
Proof. The set of orders (in any group) is certainly closed under taking divisors: if $x$ has order $n$ and $d\mid n$ then $x^{n/d}$ has order $d$. Now if $a,b$ are orders of elements in an Abelian group and $\def\lcm{\operatorname{lcm}}m=\lcm(a,b)$, then there are relatively prime $a',b'$ with $a'\mid a$, $b'\mid b$, and $a'b'=m$: it suffices to retain in $a'$ those and only those prime factors of $a$ whose multiplicity in $a$ is at least as great as in $b$, and to retain in $b'$ all other prime factors of $b$ (those whose multiplicity exceeds those in $a$). Now if $x$ has order $a'$ and $y$ has order $b'$, then these orders are relatively prime, whence $\langle x\rangle\cap\langle y\rangle=\{e\}$, and their product is$~m$ so that
$$
 x^iy^i =e\iff x^i=e=y^i\iff (\lcm(a',b')=a'b'=)\; m\mid i,
$$
and therefore $xy$ has order $m$. QED
Now to prove the proposition, let $n=\#G$, and let $m$ be the least common multiple of all the orders of elements of $G$. By Lagrange's theorem the order of every element divides$~n$, whence $m\mid n$ by the property of least common multiples. But one also has $n\leq m$ since all $n$ elements of $G$ are roots of the polynomial $X^m-1$ in the field$~F$. Therefore $n=m$, and by the lemma (using that $G$ is commutative since $F$ is so) $G$ has an element $g$ of order $m=n=\#G$, so that $G=\langle g\rangle$ is cyclic.
A: There's a simple proof which doesn't use Sylow's theory.
Lemma. Let $G$ a finite group with $n$ elements. If for every $d \mid n$,  $\# \{x \in G \mid x^d = 1 \} \leq d$, then $G$ is cyclic.
If $G$ is a finite subgroup of the multiplicative group of a field, then $G$ satisfies the hypothesis because the polynomial $x^d - 1$ has $d$ roots at most.
Proof.  Fix $d \mid n$ and consider the set $G_d$ made up of elements of $G$ with order $d$. Suppose that $G_d \neq \varnothing$, so there exists $y \in G_d$; it is clear that $\langle y \rangle \subseteq \{ x \in G \mid x^d = 1 \}$. But the subgroup $\langle y \rangle$ has cardinality $d$, so from the hypothesis we have that $\langle y \rangle = \{ x \in G \mid x^d = 1 \}$. Therefore $G_d$ is the set of generators of the cyclic group $\langle y \rangle$ of order $d$, so $\# G_d  = \phi(d)$.
We have proved that $G_d$ is empty or has cardinality $\phi(d)$, for every $d \mid n$. So we have:
$$\begin{align}
n &= \# G\\
& = \sum_{d \mid n} \# G_d \\
&\leq \sum_{d \mid n} \phi(d) \\
&= n.
\end{align}$$
Therefore $\# G_d = \phi(d)$ for every $d \vert n$. In particular $G_n \neq \varnothing$. This proves that $G$ is cyclic. QED
A: Let's have a number-theoretic proof .
Let $o(G) = n$ .
Then , $d|n$ $\Rightarrow$ $x^{d}-1|x^{n}-1$ . 
Or, $x^{n}-1=g(x).(x^{d}-1)$ , where , $g(x)$ is a polynomial of degree , $n-d$ . 
Now, if $x^{d}-1$ has less than d distinct roots , F being a field,$g(x)$ can't have more than $n-d$ roots resulting less than $n$ distinct roots for $x^{n}-1$ , which is absurd as $o(G)=n$ .
Thus , $G'=(\alpha\in G : \alpha^{d}=1)$ , is a subgroup of $G$ .
Let, $\psi(d)$ be the number of elements of order d in $G'$ , then , $\sum_{c|d} \psi(d)= d$ .
By , Möbius Inversion formula ,
$\psi(d)=\sum_{c|d}\mu(c).\frac{d}{c}$.....(1) , where $\mu$ is Möbius function.
We know , $\sum_{c|d}\phi(c)=d$ . Applying , Möbius Inversion formula
to it we observe , $\sum_{c|d}\mu(c).\frac{d}{c}$....(2).
Using (1) and (2) , we may conclude, $\psi(d)=\phi(d)\geq1$ .
Putting , n in place of d , the theorem is proven.
A: Let $G$ be a finite group with $n$ elements of a field with the field's multiplication operation. Let $d \: | \: n$ and consider the set $G_d$ containing elements of $G$ with order $d$. Suppose $G_d \neq \varnothing$, so there is $y \in G_d$.
Let's generate a cyclic group $\langle y \rangle$. From group theory, generators of $\langle y \rangle$ are elements of $\langle y \rangle$ that are relatively prime to $|\langle y \rangle| = d$. And because any other elements having orders different than $d$ could not generate $\langle y \rangle$, $G_d$ contains all elements that generate $\langle y \rangle$, so $\#G_d$ = $\phi(d)$.
($\#G_d$ is number of elements of $G_d$; $\phi$ is Euler's totient function.)
Let's partition $G$ into sets of elements with the same order. For example if $G = Z_{10}$, then it would be partitioned into sets of elements: $\{0\}$, $\{5\}$, $\{2, 4, 6, 8\}$ and $\{1, 3, 7, 9\}$ of order $1$, $2$, $5$, and $10$ respectively. By Lagrange's theorem, all elements of a group must have their orders divide the order of the group, so by partitioning G into all $d \: | \: n$, we would have the whole $G$, and having a formula as following:
$$n = |G| = \sum_{d|n} \#G_d = \sum_{d|n} \phi(d)$$
Borrowing Gauss' divisor sum we also have the following formula:
$$\sum_{d|n} \phi(d) = n$$
To recapitulate, we have shown that $\#G_d = 0$ or $\#G_d = \phi(d)$, but if $\#G_d$ could be equal to $0$, then we would have:
$$n = |G| = \sum_{d|n} \#G_d \leq \sum_{d|n} \phi(d) = n$$
which is false because we would have $|G|$ less than $n$. So $\#G_d = \phi(d)$, could not be $0$, and in particular for $d = n$ and $y_n \in G_n \neq \varnothing$, we are guaranteed to have $y_n$, so is $\langle y_n \rangle$ as well.
Then note that $\langle y_n \rangle = \{x \in G \: | \: x^n = 1\}$, and as $|\langle y_n \rangle| = n$, that is all elements of $G$ (because $G$ is already defined as a finite group with $n$ elements). This proves that $G \cong \langle y_n \rangle$ hence cyclic.
