[SOLVED] g\dot a=ag

Let G be a group and let A = G. Show that if G is non-abelian then the maps defined by $g\cdot a=ag, \ \forall g,a\in G$ do not satisfy the axioms of a left group action of G on itself.

Since G is non-abelian, there exist $g_1,g_2\in G \ \text{s.t} \ g_1g_2\neq g_2g_1$ .

$eg_1g_2=a^{-1}ag_1g_2=a^{-1}((g_1g_2)\cdot a)=a^{-1}(g_1(g_2\cdot a))=a^{-1}(g_1\cdot (ag_2))=a^{-1}(ag_2g_1)=g_2g_1$

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.