Euler-Totient Multiplicative

Question

http://www.oxfordmathcenter.com/drupal7/node/172 By and large, I understand this proof, however I'm struggling to understand how the Chinese remainder theorem implies that there exists some $x \in S_1$ as surely it's not sufficient to show that there is a unique solution to the simultaneous congruencies (as per the Chinese remainder theorem) but one must also that this solution is relatively prime to $mn$ and hence is in $S_1$?

Answer 1

Wait, I think I might have figured it out.

So, assume there is some $x$ that satisfies $x \equiv a (mod\; m)$ and $x \equiv b (mod \; n)$ such that $(x, mn) > 1$ and $(a,m)=1$ and $(b,n)=1$. If $(x,mn)>1$ then they must have at least one prime factor in common, call it $p$, and because $m$ and $n$ are coprime, $p$ must belong to $m$ or $n$, and without loss of generality, we can say it belongs to $m$. Now, from our congruences, we have $x = qm + a$ for some $q$. $\frac{x}{p}$ is an integer, $\frac{qm}{p}$ is an integer, but because $(a,m)=1 \implies (a,p)=1, \frac{a}{p}$ is not an integer (otherwise $(a,m)$ would equal at least $p$), so the left side divided by $p$ is an integer, but the right side isn't, which is a contradiction.

Therefore, our $x$ must be coprime to $mn$.

Is that right?