[SOLVED] g\dot a=ag^{-1}

Let G be a group and let A = G. Show that the maps defined by $g\cdot a=ag^{-1}, \ \forall g,a\in G$ do not satisfy the axioms of a left group action of G on itself.

I found the opposite. Here is what I did.

Let $g_1,g_2\in G$ .
$g_1\cdot (g_2\cdot a)=g_1\cdot (ag_2^{-1})=ag_2^{-1}g_1^{-1}=a(g_1g_2)^{-1}=(g_1g_2)\cdot a$

$e\cdot a=ae^{-1}=a$

What is wrong?

Correction: It says do satisfy the axioms.