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Let $G$ be a finite abelian group of order $n$ with identity $e$. If for all $a\in G$, $a^3=e$, then by induction on $n$, show that $n=3^k$ for some non-negative integer $k$.

I am competely stuck on this. Please help anybody.

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    Yes indeed, Calvin. I su$p$pose it is a rather widespread "trick"...2012-12-30

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I don't think you need the assumption that $G$ is abelian. Since $a^3 = e$ for all $a \in G$, every element has order $3$ (Edit: Every element except the identity element which has order 1). If some prime $p$ other than $3$ divided the order $G$, Cauchy's theorem implies there is an element of order $p$, so the order of $G$ is $3^k$ for some $k$.

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    But sureshs proof is better...2012-12-30