Thread: Order of Abelian Group

Any thoughts?

We gave the definition of a group G under a binary operation. We say a group G is abelian if and and only if the binary operation for G is commutative; that is, g*h = h*g for all g and h in G. For an integer r > 0, write g*g*...*g=g^r.


Suppose an abelian group G has n elements. For any g in G, prove that g^n = e, where e is the identity of G. Do not use Lagrange’s Theorem, or any other theorem, from abstract algebra! [Follow the reasoning of the proofs of Fermat’s Theorem and Euler’s Theorem.]

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.