# If $G$ is a finite abelian group of order $n$ where $a^3=e$ for all $a\in G$, then $n=3^k$ for some non-negative integer $k$

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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Could you elaborate a bit on what results you have already seen? There are a lot of ways to show this, depending on how advanced results you have seen. – Tobias Kildetoft Dec 30 '12 at 3:55
This sounds way too much like the group theory homework that Neptune has been posting in the past few minutes. – Calvin Lin Dec 30 '12 at 3:56
Yes indeed, Calvin. I suppose it is a rather widespread "trick"... – DonAntonio Dec 30 '12 at 4:30

## 1 Answer

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 how should i able to solve this problem using induction only.? – TUMO Dec 30 '12 at 4:13
@TUMO An inductive proof would be easy in the abelian case. Every subgroup is normal, so the quotient by $\langle a \rangle$ for any $a\in G$ has order $3^k$, whence...? – Alexander Gruber Dec 30 '12 at 6:09
But sureshs proof is better... – Alexander Gruber Dec 30 '12 at 6:10