For g, h in G, we consider the element of G given by

gghghh

On one hand, it can be evaluated as

g^2 hg h^2 = ehge = hg

and on the other hand,

g (gh)^2 h = geh = gh.

We conclude hg = gh, and since g,h were arbitrary elements of G, G is abelian.

Let 'x', 'y' are elements of G, and

xy=a. [1]

We need to prove that yx=a. [2]

Starting with [1], multiply both sides with (yx):

(yx)(xy) = (yx)a. {Left multiplication used}

Then, using associativity,

y(xx)y = (yx)a

But gg=e for any 'g' in G, so for g=x and g=y:

yey = (yx)a =>

yy = (yx)a =>

e = (yx)a =>

multiplying with 'a', {Right multiplication}

ea = ((yx)a)a = (yx)(aa) = (yx)e = yx.

Then yx = a = xy,

and this means that G is Abelian.

ab=c=>aab=ac=>eb=ac=>b=ac=>bc=acc=>bc=ae=>bc=a

=>ba=bbc=>ba=ec=>ba=c

so ab=bc and G is Abelian.