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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?

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    If $G$ is cyclic then by definition it is generated by one element, say $G = \langle x \rangle$. Then $a=x^\alpha$ and $b=x^{\beta}$ for some $\alpha, \beta$. Then $ab^n=x^{\alpha+n\beta}$; you need to show that, given that $a,b$ cenerate $G$, an $n$ exists such that $x^{\alpha+n\beta}=x$.2012-08-06
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    This doesn't always work, consider for example the case $\alpha=2$, $\beta=3$, $|G|=6$. Then $2 + 3n \not{\equiv} 1 \pmod{6}$.2012-08-06
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    @YuvalFilmus On the other hand, $2+3=-1$, which is also a generator of $\mathbb{Z}/(6)$.2012-08-06
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    @KReiser: absolutely, the theorem still holds in this case; but @Yuval’s point I think was that @Clive’s outlined solution doesn’t always work, with (2,3,6) giving a counterexample.2012-08-07

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