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TheAbstractionist · #9 July 10th, 2009, 04:48 PM

Quote:

Originally Posted by halbard

Sorry, Bruno J., but how does $x^2=y^2$ imply that $(xy^{-1})^2=1$ without assuming that $x^2y^{-2}=xy^{-1}xy^{-1}$ ? That is, without assuming that $xy^{-1}=y^{-1}x$ , i.e. that $xy=yx$ ?

Good point. As a counterexample, we can take $G=S_3$ and $x=(1\,2),\ y=(1,3).$ Then $x^2=y^2$ but $xy^{-1}$ has order 3. (Of course $S_3$ is not a group of odd order, but Bruno iddn’t make any assumption about group orders in going from $x^2=y^2$ to $(xy^{-1})^2=1.)$