
g\cdot a=ag
Let G be a group and let A = G. Show that if G is nonabelian then the maps defined by $\displaystyle g\cdot a=ag, \ \forall g,a\in G$ do not satisfy the axioms of a left group action of G on itself.
Since G is nonabelian, there exist $\displaystyle g_1,g_2\in G \ \text{s.t} \ g_1g_2\neq g_2g_1$.
$\displaystyle eg_1g_2=a^{1}ag_1g_2=a^{1}((g_1g_2)\cdot a)=a^{1}(g_1(g_2\cdot a))=a^{1}(g_1\cdot (ag_2))=a^{1}(ag_2g_1)=g_2g_1$
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.

Re: g\cdot a=ag
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.