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If a group satisfies $x^3=1$ for all $x$, is it necessarily abelian?

I want to show that group $G$ is abelian (i.e. $ab=ba$) if $a^{3}=e, \forall a\in G.$ I am trying so much but i cant get this so please help me out!

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    Actully, $a^{3}=a$ $\Rightarrow a^{2}a=e.a$ $\Rightarrow a^{2}=e.$2012-07-05

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You are going to have a hard time proving it, since the result is false.

Let $\mathbb{F}_3$ be the field with three elements, $0$, $1$, and $-1$ (with $1+1=-1$). Let $G$ be the group of all matrices of the form $\left(\begin{array}{ccc} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1 \end{array}\right),$ with $a,b,c\in\mathbb{F}_3$.

Denote that element as $(a,b,c)$. Then $(a,b,c)(\alpha,\beta,\gamma) = (a+\alpha, b+\beta, c+\gamma+\beta a).$

Show the group is not abelian, but $x^3 = e$ for all $x$. This is known as the Heisenberg group corresponding to $p=3$.

More generally, for every prime $p\gt 2$, the similar group satisfies $x^p=e$ for all $p$, but is not abelian. A presentation for the group is $\Bigl\langle a,b,c\,\Bigm|\, a^p = b^p = c^p = 1, ac=ca, bc=cb, ba=abc\Bigr\rangle.$