basically it doesn't work because RIGHT multiplication is not a LEFT action:
(gh).a = agh
g.(h.a) = g.(ah) = ahg
if G is not abelian, then hg will be different than gh for at least one pair g,h.
Let G be a group and let A = G. Show that if G is non-abelian then the maps defined by do not satisfy the axioms of a left group action of G on itself.
Since G is non-abelian, there exist .
Next it says this is a contradiction. Didn't the just show it worked? Is it a contradiction since this shows all elements in g commute? I am confused.