In Dummit and Foote's Abstract algebra (p.45 exercise 15), there's a problem asking
Let $G$ a group and $A=G$, prove that the maps defined by $g\cdot a =ag^{-1}$ for all $g, a\in G$ satisfy the axioms of a left group action.
So I know I have to prove that $g_1\cdot(g_2\cdot a) = (g_1g_2)\cdot a$. But when you attempt that, you get $$g_1(g_2a) = g_1(ag_2^{-1}) = a(g_2g_1)^{-1} \stackrel{?}{=} a(g_1g_2)^{-1} = (g_1g_2)a.$$ But because $G$ need not be abelian, this need not be true. What is it I'm missing here?
You made an edit within the five minutes, so it doesn't appear. But note that the group action does indeed satisfy: $$g_1\cdot(g_2\cdot a)=g_1\cdot(ag_2^{-1})=ag_2^{-1}g_1^{-1}=a(g_1g_2)^{-1}=(g_1g_2)\cdot a.$$
Note that being abelian doesn't matter here, and you are best off leaving the '$\cdot$' in your calculations, just so you don't get tempted to use your group multiplication (as opposed to using the given group action).
Note that $(ab)^{-1}=b^{-1}a^{-1}$. This doesn't rely on commutativity. You wish to find the inverse of $ab$, then naturally if you want to invert this from the right, you will first invert by $b$ and then by $a$, $abb^{-1}a^{-1}=e$. (I.e. you've made an error in the edit, on the third equality.)