Finite subgroups of the multiplicative group of a field are cyclic

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

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 $\mathbb{Z}_m.$