I'm told that gcd(a,n)=1 and gcd(b,n)=1, a,b,n are natural.
I need to prove that gcd(ab,n)=1.
I don't succeed :-\ Tried millions of things!
Say $\displaystyle \gcd(ab,n)=d > 1$. Then $\displaystyle d|ab$ and $\displaystyle d|n$. Note $\displaystyle \gcd(d,a)=1$ since $\displaystyle d|n$ and that would imply $\displaystyle \gcd(a,n) \geq d > 1$ a contradiction, so $\displaystyle \gcd(d,a)=1$. But then $\displaystyle d|ab \implies d|b$. And so $\displaystyle \gcd(b,n)\geq d > 1$ a contradiction. Thus, $\displaystyle d=1$.
I agree. that's one of the reasons i like it. it is not so obvious at the moment you see the problem. NonCommAlg's way is something you would do on impulse to see how it works out, which is the charm of that method. good for tests when you can't think clearly or be fancy, but are racing against the clock and just need to get the job done