TITLE: Generators of a cyclic group
QUESTION [10 upvotes]: In a paper there is a lemma:

Let $G= \langle a,b \rangle$ be a finite cyclic group. Then $G=\langle ab^n \rangle$ for some integer $n$.

The proof is omitted because it's "straightforward" but I'm not able to proof it. How does this work?

REPLY [6 votes]: Yuval's solution, sans Dirichlet: let $n$ be the product of all the primes that divide the order of $G$ but don't divide $x$. Then $\gcd(x+ny,|G|)=1$. 
Proof: Let $p$ be a prime dividing the order of $G$. If $p$ divides $x$, then it doesn't divide $y$ (since $\gcd(x,y,|G|)=1$), and it doesn't divide $n$ (by construction), so it doesn't divide $ny$, so it doesn't divide $x+ny$. 
If $p$ doesn't divide $x$, then it divides $n$ (by construction), so it divides $ny$, so it doesn't divide $x+ny$. So no prime dividing the order of $G$ divides $x+ny$.