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For whatever reason, I am having trouble proving the following claim without using LaGrange's Theorem.

Claim: Let $G$ be a group of order $n < \infty$. Then, $x^{n}=1$, where $1$ is the identity, for all $x \in G$.

If $|x|=m$, then I can use the Division Algorithm (specifically, I can divide $n$ by $m$) to obtain that $x^{n} = x^{r}$, where $r=0,1,2,..., \text{or}\> m-1$. Then, well, I can't see the next step.

dgc1240
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  • Are you sure this is independent of LaGrange? – Alexander Gruber May 18 '14 at 02:19
  • No, I am not sure. In fact, I would love for someone to tell me that it cannot be done without LaGrange's Theorem. For my own edification, I enjoy trying to prove various facts using the most elementary means possible. This claim is easy enough with LaGrange's Theorem; I could not see, however, how to complete the proof without it. – dgc1240 May 18 '14 at 02:24

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