Finite subgroup of a field is cyclic group.

This essay aims to prove that for any field $\mathbb{F}$$\mathbb F$, $(G,\ast )\subseteq \mathbb{F},|G|<\mathrm{\infty }$$(G,*)\subseteq\mathbb F,|G|<\infty$ is a cyclic group.

Let $|G|=n$$|G|=n$, then $\mathrm{\forall }g\in G,\mathrm{ord}(g)|n$$\forall g\in G,\mathrm{ord}(g)|n$. Let $\psi (d)$$\psi(d)$ be the cardinality of $\{g\in G|\mathrm{ord}(g)=d\}$$\{g\in G|\mathrm{ord}(g)=d\}$.

A wrong proof is here:

Now we will use some structure of arithmetic function ring, you can click this blog link here to learn the background knowledge.

Then

$$\begin{array}{}\text{(1)}& \sum _{d|n}\psi (d)=\psi \ast 1(n)=n=\phi \ast 1(n)=\sum _{d|n}\phi (d)\end{array}$$

Using the cancel law of group

We get that

$$\begin{array}{}\text{(2)}& \psi (d)=\phi (d)\end{array}$$

Hence $\psi (n)=\phi (n)\ne 0$$\psi(n)=\varphi(n)\ne 0$.

By definition of $\psi (n)$$\psi(n)$​,

$$\begin{array}{}\text{(3)}& \{g\in G|\mathrm{ord}(g)=n\}\ne \mathrm{\varnothing }\end{array}$$

Hence $G$$G$ is a cyclic group.

So, why this proof is wrong?

Notice that we only get that $\psi \ast 1(n)=\phi \ast 1(n)$$\psi*1(n)=\varphi*1(n)$ for only one $n$$n$, not all the $n\in \mathbb{N}$$n\in\mathbb N$.

Another issue is we only define the function for $G$$G$...

A correct proof is

For fix $d|n$$d|n$, $\psi (d)=0$$\psi(d)=0$ or $\psi (d)\ne 0$$\psi(d)\ne 0$.

For $\psi (d)\ne 0$$\psi(d)\ne 0$, we claim that $\psi (d)=\phi (d)$$\psi(d)=\varphi(d)$. $\psi (d)\ne 0⟹\mathrm{\exists }g,$$\psi(d)\ne 0\implies\exist g,$ generate a group $|⟨g⟩|=d$$|\langle g\rangle|=d$ .

Every element in $⟨g⟩$$\langle g\rangle$ is the root of $x^d=1$$x^d=1$, and in a field, $x^d=1$$x^d=1$ at most have $d$$d$ roots.

Hence $⟨g⟩$$\langle g\rangle$ is all the roots. Then $\psi (d)=\phi (d)$$\psi(d)=\varphi(d)$ by the structure of the finite cyclic group( they are all isomorphic to $\mathbb{Z}/m\mathbb{Z}$$\mathbb Z/m\mathbb Z$) .

Hence we do not have any $d|n,\psi (d)=0$$d|n,\psi(d)=0$ .

Since

$$\begin{array}{}\text{(4)}& \sum _{n|d}\psi (d)=\sum _{d|n}\phi (d)\end{array}$$

Hence $\psi (n)=\phi (n)\ne 0$$\psi(n)=\varphi(n)\ne 0$. That is, $G$$G$ is a cyclic group.

Corollary

$†.{\mathbb{F}}_p^{\ast }$$\dagger.\mathbb F_{p}^*$ is a cyclic group, and ${\mathbb{F}}_p^{\ast }\cong \mathbb{Z}/(p-1)\mathbb{Z}$$\mathbb F_p^*\cong\mathbb Z/{(p-1)}\mathbb Z$.

$†.$$\dagger.$ The number of primitive roots of ${\mathbb{F}}_p$$\mathbb F_p$ is $\phi (p-1)$$\varphi(p-1)$.