MODERN ALGEBRA binary operations?

Question

How do I show that every group with identity e and the fact that x*x=e for all x in G is abelian

Answer 1

For any x and y in G, start with:

xyxy = (xy)^2 = e

Multiply both sides of the equation on the right by y:

xyxy^2 = ey = y

But y^2 = e, so
xyx = y

Multiply by x on the right:

xyx^2 = yx

But x^2 = e, so this means that

xy = yx

Answer 2

If x² = e for all x in G, then x = x^-1 for all x in G.
Now let a,b be any 2 elements in G.
Then ab*ab = e, so
ab = (ab)^-1 = b^-1 a^-1= ba,
since every element is its own inverse.
Thus ab = ba and G is abelian.