Re: Order of Abelian Group

Hey ncshields.

If you have g*h = h*g for all h and g, then try and prove that h must be the inverse of g for all g.

But by the uniqueness theorem of inverses, this means that only one inverse exists and that only one g exists in this group since we know that the inverse is unique and this applies for all g.

So the group only has two elements which are g and g inverse, but if this is the case then then the only group that has this is the trivial group with an identity.

You can show this by showing that g^(-1) = e by using gh = hg implies

g = hgh^(-1) and

g^(-1) = (hgh^(-1))^-1 = h*g^(-1)*h^(-1) = g

so g^(-1) = g which means that g = e.

You may have to add a little more detail for the lecturer but that's the basic idea.