Can x^2 Equal y in a Non-Abelian Group?

gottfried
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Homework Statement


Let (G,.) be an non-abelian group. Choose distinct x and y such that xy≠yx.

Show that if x2≠1 then x2\notin{e,x,y,xy,yx}

The Attempt at a Solution



If x2=x would imply x.x.x-1=x.x-1 and x=e which cannot be.
If x2= xy or x2=yx would imply x=y which also cannot be.

How does one show that x2≠y?
 
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gottfried said:

Homework Statement


Let (G,.) be an non-abelian group. Choose distinct x and y such that xy≠yx.

Show that if x2≠1 then x2\notin{e,x,y,xy,yx}

The Attempt at a Solution



If x2=x would imply x.x.x-1=x.x-1 and x=e which cannot be.
If x2= xy or x2=yx would imply x=y which also cannot be.

How does one show that x2≠y?

x doesn't commute with y. Does x commute with x^2?
 
Yes I believe it does so x2.x=x.x2 which would imply y.x=x.y if x2=y which is a contradiction.

Thanks for the help. You gave me just enough help to move forward with the problem but left some of the satisfaction of figuring it out to me which is cool, thanks.
 
x^2=y→x^3=xy and x^3=yx
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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