## Multiplicativity of the Phi Function.

It is a classic and fundamental result that the $\phi$function is multiplicative. Meaning,
$\phi (nm)=\phi (n)\phi (m)$
whenver, $\gcd (n,m)=1$.
The proof in my book is not very elegant. Thus, I attempted to find my own proof. I need help with completing it however.
Proof (Incomplete): For any $n\in Z^+$ define a set $G_n=\{x|\gcd(x,n)=1,1\geq x>n\}$. Thus $G_n$ is the set of all integers relatively prime to $n$ but not equal to $n$. Now define a binary operation on this set as multiplication modulo $n$. It can be show that $G_n$ forms a group. (Just a note this is the group we use to show the proof of Euler's Generalized Theorem of Fermat). Now this is the final step, let $\gcd (m,n)=1$ for some $m\in Z^+$. Thus, $G_n\times G_m$ is also a group. This is the incomplete step show that $G_n\times G_m \cong G_{nm}$. The proof is complete because since both $G_n$ and $G_m$ are finite the cardinality of $G_n \times G_m$ is $\phi (n)\phi (m)$. But $G_{nm}$ has cardinality $\phi (nm)$ we finally have that $\phi (n)\phi(m)=\phi (nm)$

This is my 1th Post