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I know that any group satisfying $x^2=1$ for all $x$ is abelian. Is the same true if $x^3=1$? I don't think it is, but I can't find a basic counterexample.

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    In short, no; the result does not hold for *any* $n$ other than $n=1$ and $n=2$.2012-05-21
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    It is however true if there are no elements of order 3!2012-05-21
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    @Steve D: You mean, if $x^3=1$ for all $x$ and $G$ has no elements of order $3$? Well... it's a rather singular situation, don't you think? (-:2012-05-21
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    @ArturoMagidin: hahaha, it's funny that what I wrote is still, somehow, correct. I was thinking of $(xy)^3=x^3y^3$.2012-05-21

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