# Thread: non abelian group problem

1. ## non abelian group problem

Suppose that G = { e, x, x2, y, yx, yx2 } is a non-Abelian group with |x| = 3 and |y|=2. Show xy=yx2. I tried looking at the orders of the elements but i got nowhere doing this, any suggestions?

2. Originally Posted by wutang
Suppose that G = { e, x, x2, y, yx, yx2 } is a non-Abelian group with |x| = 3 and |y|=2. Show xy=yx2. I tried looking at the orders of the elements but i got nowhere doing this, any suggestions?

Hint: show that $xy\neq e,\,x,\,x^2,\,y,\,yx$

Tonio

3. how does this look:
if xy=e, then x=y^-1, so |x|= 2 contradiction.
if xy=e, then y=x^-1, so |y|=3 contradiction.
if xy=x, then y=e contradiction, if x=x(y^-1), then (y^-1)=e contradicition.
suppose xy=x^2, y=x^3=e contradicition,
xy=y, then x=e (by right hand cancellation) contradicition.
I am not sure, what to do if xy=yx, is that a contradicion due to the fact that G is a non-abelian group?

4. Originally Posted by wutang
I am not sure, what to do if xy=yx, is that a contradicion due to the fact that G is a non-abelian group?
Yes. x,y are generators. If they commute, then everything in the subgroup generated by {x,y} (here it's just G itself) commutes with each other. #

5. would I then have to show that xy=y(x^2), or is it enough to show that I have eliminated all other possibilities ( so that xy=y(x^2) by closure?).
Thanks for your help again

6. Originally Posted by wutang
would I then have to show that xy=y(x^2), or is it enough to show that I have eliminated all other possibilities ( so that xy=y(x^2) by closure?).
Thanks for your help again

It is enough since you're given a group and thus $xy$ must be something in that group...

Tonio

### suppose that g =e x x^2 y yx yx^2

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