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Let G denote a group and let g ∈ G of finite order. Let k ∈ Z. Prove that there exists a unique integer r with 0 ≤ r < |g| such that $g^k =g^r$

I know since G is of finite order we know that the order of $g^n$=e for some positive n

I'm a little stuck on how to begin or where to go from here, any direction would be great.

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  • $\begingroup$ Hint: express k with n $\endgroup$ – friedvir Oct 9 at 17:39
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Hint: Divide $k$ by $\lvert g\vert$. That is, write $k$ as $m\times\lvert g\vert+r$, with $r\in\{0,1,\ldots,\lvert g\rvert-1\}$.

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