Prove Group Commutativity: (G,*) w/ x*x=eG

  • Thread starter Thread starter tasha10
  • Start date Start date
tasha10
Messages
7
Reaction score
0
(A) Let (G,*) be a group such that x*x=eG for all x in G. Prove G is commutative.
(B) Give a specific example of an infinite group (G,*) such that x*x=eG for all x in G.

I have not gotten very far, just to let two variable x,y be in G and I know that (x*y)*(x*y) = eG .. I'm not sure where to go from here..
 
Physics news on Phys.org
Well, for a, you know ...
<br /> (xy)^{-1} = y^{-1}x^{-1}<br />

Try multiplying that to both sides of the equality you presented, and see what you get.
 
hmm can i ask, what's "eG" means?? identity?
 
so, will i just get eG on both sides? does this prove that it is commutative? I'm confused.
 
yes, it is the identity
 
You won't get eG on both sides. I'm saying, multiply what I showed you to both sides of

<br /> xyxy = e_{G}<br />
 
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...
Back
Top