Euler-phi function is multiplicative if coprime

Verbal statement
The Euler-phi function is multiplicative on coprime natural numbers.

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 $$g_1h_1=g_2h_2$$, or equivalently, $$g_2^{-1}g_1=h_2h_1^{-1}$$. 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, $$|A \times B| \le |C|$$. 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 $$|C| \le |A \times B|$$. Therefore, $$|C|=|A \times B|=|A||B|$$, as desired.