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I've seen similar proofs which rely on Lagrange's theorem which I'm not allowed to use. Other proofs just showed that there exists some exponent that satisfies the equation.

In my case I want to show (1) that for all $a \in G:a^n=e$, where $n=|G|$ and $e$ is the neutral element of the finite abelian group $G$ and (2) there exists a natural number $k$ so that $a^k=e$ (this k should be the order of the element $a$ noted as $k=ord_G(a)$.

Unfortunately I also count't find any online material. Any help would be appreciated!

mhighwood
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  • It is possible to prove this without Lagrange, see the duplicate, or this post, but it is not very reasonable to avoid Lagrange. Indeed, Lagrange is as basic in group theory as prime numbers in number theory. – Dietrich Burde Jun 27 '23 at 19:09

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