It is a classic and fundamental result that the $\displaystyle \phi$function is multiplicative. Meaning,

$\displaystyle \phi (nm)=\phi (n)\phi (m)$

whenver, $\displaystyle \gcd (n,m)=1$.

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 $\displaystyle n\in Z^+$ define a set $\displaystyle G_n=\{x|\gcd(x,n)=1,1\geq x>n\}$. Thus $\displaystyle G_n$ is the set of all integers relatively prime to $\displaystyle n$ but not equal to $\displaystyle n$. Now define a binary operation on this set as multiplication modulo $\displaystyle n$. It can be show that $\displaystyle G_n$ 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 $\displaystyle \gcd (m,n)=1$ for some $\displaystyle m\in Z^+$. Thus, $\displaystyle G_n\times G_m$ is also a group. This is the incomplete step show that $\displaystyle G_n\times G_m \cong G_{nm}$. The proof is complete because since both $\displaystyle G_n$ and $\displaystyle G_m$ are finite the cardinality of $\displaystyle G_n \times G_m$ is $\displaystyle \phi (n)\phi (m)$. But $\displaystyle G_{nm}$ has cardinality $\displaystyle \phi (nm)$ we finally have that $\displaystyle \phi (n)\phi(m)=\phi (nm)$

This is my 1th Post