Euler totient function is multiplicative

This article is about a fact in number theory that has a direct proof using group theory.

Statement

Verbal statement

The Euler totient function is a multiplicative function.

Statement with symbols

For any natural numbers $m,n$ that are coprime to each other, $\varphi (mn)=\varphi (m)\varphi (n)$.

Proof

Given: $m,n$ coprime positive integers

To prove: $\varphi (mn)=\varphi (m)\varphi (n)$

Proof: Consider a finite cyclic group G of order $mn$.

Denote by $A,B,C$ the set of all elements in G that have orders $m,n,mn$ respectively. We will show that $|C|=|A||B|$.
Define a map $\phi :A\times B\rightarrow C$ by $\phi ((g,h))=gh$. Since the order of $g$ and $h$ are coprime, and $G$ is abelian, their product $gh$ must have order $mn$, so this map is well-defined. If $\phi ((g_{1},h_{1}))=\phi ((g_{2},h_{2}))$ then $g_{1}h_{1}=g_{2}h_{2}$, or equivalently, $g_{2}^{-1}g_{1}=h_{2}h_{1}^{-1}$. They are elements of subgroups of coprime orders that must intersect trivially. So, $g_{1}=g_{2}$ and $h_{1}=h_{2}$, the map is injective. Thus, $|A\times B|\leq |C|$.
Define another map $\psi :C\rightarrow A\times B$ by $\psi (k)=(k^{n},k^{m})$. If $\psi (k)=\psi (l)$ then $k^{n}=l^{n}$ and $k^{m}=l^{m}$. Suppose $k=ul$. Then $k^{m}=u^{m}l^{m}=u^{m}k^{m}$. Therefore $u^{m}=e$. Similarly, $u^{n}=e$. So, $m$ and $n$ must have a common divisor which is the order of $u$. Thus, $u=e$. Hence our map is again injective, so $|C|\leq |A\times B|$.
Therefore, $|C|=|A\times B|=|A||B|$, as desired.