TITLE: Lifting units from modulus n to modulus mn.
QUESTION [9 upvotes]: Background
In his beautifully short answer to a previous question of mine, Robin Chapman asserted the following.

Let $m,n,r$ be natural numbers with $r$ coprime to $n$.  Then there is $r' \equiv r \mod n$ which is coprime to 
  $mn$.

Letting $C_n$ denote the cyclic group of order $n$, the above statement is equivalent to this:

Every automorphism of $C_n$ lifts to an automorphism of $C_{nm}$ for all $m$.

Since that is the context of the question I asked, I thought that this fact ought to have an elementary group-theoretical derivation, but alas I have been unable to find one.  I asked a number theorist colleague of mine and he gave me this "sledgehammer proof" (his words):
Since $r$ is coprime to $n$, the arithmetic progression $r + kn$ for $k=1,2,\ldots$ contains an infinite number of primes (by a theorem of Dirichlet's).  Since only a finite number of those primes can divide $m$, there is some $k$ for which $r'= r+kn$ is a prime which does not divide $m$, and hence neither does it divide $nm$.
Question
Is there an elementary (and preferably group-theoretical) proof of this result?

REPLY [5 votes]: There is an explicit formula for $r'$, and no need to invoke the Chinese Remainder Theorem; $$r'=r+kn{\rm\ where\ }k=\prod_{p\mid m,p\nmid r}p.$$ We need to show that $r'$ is relatively prime to $mn$. It suffices to show that $r'$ is relatively prime to $m$. Let $p$ be a prime dividing $m$. If $p$ divides $r$, then it doesn't divide $n$ (since $r$ and $n$ are relatively prime), and it doesn't divide $k$ (by construction of $k$), so it doesn't divide $kn$, so it doesn't divide $r'$. If $p$ doesn't divide $r$, then, by construction, it divides $k$, so it divides $kn$, so it doesn't divide $r'$. 
Schinzel showed me this construction 35 years ago.