It is a classic and fundamental result that the \phifunction 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)

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