A nice Property of Euler function, prove via multi set.

We know that Euler function

$$\begin{array}{}\text{(2)}& \phi (n)=n\prod _{p|n}((1-\frac{1}{p})\end{array}$$

Proposition.

Let $gcd(s,t)=g$$\gcd(s,t)=g$

$$\begin{array}{}\text{(3)}& \phi (g)\phi (st)=g\phi (s)\phi (t)\end{array}$$

Proof.

The left hand side is

$$\begin{array}{}\text{(4)}& gst\prod _{p|g}(1-\frac{1}{p})\prod _{p^{\prime }|st}(1-\frac{1}{p^{\prime }})\end{array}$$

and the right hand side is

$$\begin{array}{}\text{(5)}& gst\prod _{p|s}(1-\frac{1}{p})\prod _{p^{\prime }|t^{\prime }}(1-\frac{1}{p^{\prime }})\end{array}$$

So we only need to prove that

$$\begin{array}{}\text{(6)}& t\downarrow \coprod s\downarrow =st\downarrow \coprod gcd(s,t)\end{array}$$

Which is extremely obvious. You doubt count the gcd part in the left hand side.

The notation is from coproduct, but if it is coproduct, we only can talk about isomorphism. So what I mean here is union of multi set.

 

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)$.