
Abelian groups
Let G be an abelian group and let .
If is finite then prove that, for , iff .
Now the obvious cases are if , and trivially. Same for .
The rest all seems rather straightforward, however I'm having trouble with finding a structure for my proof, especially since n is an integer, not just a natural number.
For the case where I have if then which, by definition of the order of an element cannot be equal to zero. I find it hard to prove this for the similar case of .
Am I making this too difficult or what is it that I'm missing?

Hi
Since and have the same order , proving that for any integer is quite easy. (just need to write what means and use the associativity of )
Now let be an integer such that We can assume because iff (so if we can work with and )
Consider the result of its euclidian division by where
We get But Can you conclude?

Got it, thanks for structuring that for me!!

Re: Abelian groups
Let G be a group. If x^(1)=x, for all x E G, then prove that G is abelian

Re: Abelian groups
suppose that x^{1} = x, for all x in G.
then, for any x,y in G:
(xy)^{1} = xy (since xy is likewise in G, by closure)
y^{1}x^{1} = xy (since (xy)^{1} = y^{1}x^{1}, in any group)
yx = xy (since y^{1} = y, and x^{1} = x, since x,y are in G)
thus G is abelian.