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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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marked as duplicate by Dylan Moreland, Norbert, Gerry Myerson, Arturo Magidin, lhf Jul 4 '12 at 1:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
I think this has been asked before. One second. –  Dylan Moreland Jul 4 '12 at 1:20
1  
See this. The result is not true. Are you leaving out some hypotheses? –  Dylan Moreland Jul 4 '12 at 1:21
    
Perhaps you mean $a^2=e$, or $a^3=a$? –  Alex Becker Jul 4 '12 at 1:31
    
Actully, $$a^{3}=a$$ $$\Rightarrow a^{2}a=e.a$$ $$\Rightarrow a^{2}=e.$$ –  Kns Jul 5 '12 at 4:10

1 Answer 1

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 $$\Bigl\langle a,b,c\,\Bigm|\, a^p = b^p = c^p = 1, ac=ca, bc=cb, ba=abc\Bigr\rangle.$$

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