Finite Group of Field

Exercise. Show that finite subgroups of the multiplicative group of a field are cyclic.

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To show it, it suffices to show the following lemma. 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.

Lemma

Let $G$ be a finite group with $n$ elements. If for every $d\mid n,\#\left\{x\in G\mid x^d=1\right\}\le d$, then $G$ is cyclic.

\begin{proof} Fix $d\mid n$ and consider the set $G_d$ made up of elements of $G$ with order $d$. Suppose that $G_d\ne \varnothing$, so there exists $y\in G_d$; it is clear that $⟨y⟩\subseteq \left\{x\in G\mid x^d=1\right\}$. But the subgroup $⟨y⟩$ has cardinality $d$, so from the hypothesis we have that $⟨y⟩=\left\{x\in G\mid x^d=1\right\}$. Therefore $G_d$ is the set of generators of the cyclic group $⟨y⟩$ of order $d$, so $\#G_d=\varphi (d)$.

We have proved that $G_d$ is empty or has cardinality $\varphi (d)$, for every $d\mid n$. So we have:

$$\begin{array}{rl}n& =\#G\\ & =\sum _{d\mid n}\#G_d\\ & \le \sum _{d\mid n}\varphi (d)\\ & =n\end{array}$$

Therefore $\#G_d=\varphi (d)$ for every $d\mid n$. In particular $G_n\ne \varnothing$. This proves that $G$ is cyclic. \end{proof}